Now, airplane taking off from a treadmill still takes an effort to figure out.https://www.youtube.com/watch?v=Y64ZdSaDdoo (https://www.youtube.com/watch?v=Y64ZdSaDdoo)
A wind powered craft going faster than A TAIL wind speed.
It's for real...
Anyone here thinks it's impossible?
If so, watch the video again...
Against the wind already would be possible with a craft like this, as long as the drag force of the blades is higher than the craft, it would be working like a windmill,
I only needed 1 important fact in advance to know that it was possible. That sail-boats who tack at an angle achieve a relative forward velocity greater than the wind speed forward velocity because of the air-foil effect on curve bulge of the sail. Then I knew it could be done.
I'm not convinced the inverter analogy made by 'IanB' is correct as you even loose power/wattage on the higher voltage side due to efficiency, but I guess I can go with it.
I've always been intrigued by a similar idea myself: Could a wind powered craft go AGAINST the wind? Basically you're still harnessing the wind's energy so it's not really overunity but at same time the idea of going against the wind by using the wind feels like it breaks laws of thermodynamics. I think it would be equivalant to using a hydro electric generator to pump water back up the dam.Sure, put a wind turbine on a car and use the power generated to turn the wheels moving the car forwards into the wind. I would guess that only losses would limit how fast it can go.
No need to go to that sort of complexity at all. You can already tack into the wind on a normal sailboat. https://en.wikipedia.org/wiki/Tacking_(sailing) (https://en.wikipedia.org/wiki/Tacking_(sailing)) Edit: unless you want to go directly against the wind. Some thing similar to the craft in the video would probably work too but as a windmill rather than a propeller. Edit2: which rerouter has already pointed out.I've always been intrigued by a similar idea myself: Could a wind powered craft go AGAINST the wind? Basically you're still harnessing the wind's energy so it's not really overunity but at same time the idea of going against the wind by using the wind feels like it breaks laws of thermodynamics. I think it would be equivalant to using a hydro electric generator to pump water back up the dam.Sure, put a wind turbine on a car and use the power generated to turn the wheels moving the car forwards into the wind. I would guess that only losses would limit how fast it can go.
Now, airplane taking off from a treadmill still takes an effort to figure out.
I've always been intrigued by a similar idea myself: Could a wind powered craft go AGAINST the wind? Basically you're still harnessing the wind's energy so it's not really overunity but at same time the idea of going against the wind by using the wind feels like it breaks laws of thermodynamics. I think it would be equivalant to using a hydro electric generator to pump water back up the dam.They've been doing this since sails were invented. You can't go DIRECTLY into the wind, you have to "tack" to one side or the other.
I've always been intrigued by a similar idea myself: Could a wind powered craft go AGAINST the wind? Basically you're still harnessing the wind's energy so it's not really overunity but at same time the idea of going against the wind by using the wind feels like it breaks laws of thermodynamics. I think it would be equivalant to using a hydro electric generator to pump water back up the dam.They've been doing this since sails were invented. You can't go DIRECTLY into the wind, you have to "tack" to one side or the other.
The problem of moving faster than the wind is more tricky to under stand: At the moment the vehicle drives as fast as the wind, a sail would not see any apparent wind speed, so no force to drive the boat and also no wind to drive a kind of wind turbine. The trick is to take power from the forward movement of the wheels and from that drive the propeller to drive the vehicle forward. So the propeller is diving the vehicle and the wheels are opposing the movement, slowing the vehicle down. The propeller sees a low wind speed (essentially 0 if at the speed of the wind), while the wheels see the full speed relative to ground. So with the same power the propeller can have more force than the wheels brake.
I took it as the fan not exactly working as a fan. Instead they are using it to have a large high drag surface appearing to move towards the wind. Increasing its apparent relative velocity higher than the craft
The blade appears to be moving towards the wind at a given point. This apparent velocity allows the wind to push it a little faster than the actual wind speed. But not to get it twisted, the wind is pushing on the fan blades. Not the other way around if it makes sense. So the energy is still coming from the wind. And the theoretical max gets lost in energy to spin the blades.
If this assumption of mine is correct. The delta % between wind speed and actual speed should remain pretty constant until the crafts drag begins to dominate
I know where you are coming from, but I have the perfect example to explain, why we shouldn't just rely on the energy conservation law.I only needed 1 important fact in advance to know that it was possible. That sail-boats who tack at an angle achieve a relative forward velocity greater than the wind speed forward velocity because of the air-foil effect on curve bulge of the sail. Then I knew it could be done.
I'm not convinced the inverter analogy made by 'IanB' is correct as you even loose power/wattage on the higher voltage side due to efficiency, but I guess I can go with it.
Like, what? You think sailboats achieve 100% efficiency?
The analogy is straightforward: A large mass of wind moving at slow speed can lead to a smaller object moving at a higher speed. As long as the kinetic energy (and momentum) of the craft is lower than the kinetic energy (and momentum) of the wind that it harnessed, then there is no violation of the conservation laws.
In electricity, a larger current at a lower voltage can lead to a smaller current at a higher voltage, minus some losses.
By analogy:
Speed of wind/object == Voltage
Mass of wind/object == Current
The physics of this is all about levers and force multipliers.
I don't get all the apparent resistance from the people once the principle is explained. I somehow have no issues with it and don't really need a full scale replica to believe it.
Now, airplane taking off from a treadmill still takes an effort to figure out.
Who will win the bet?
https://www.youtube.com/watch?v=yCsgoLc_fzI (https://www.youtube.com/watch?v=yCsgoLc_fzI)
Who will win the bet?
Derek said at the end of the video that he won the bet, and the physics prof paid up the $10k.
Settled?
All you need to do to prove it is make a toy with a propeller where when you blow into it, say with a fan, it travels toward the wind. Demonstrate this, moving/accelerating directly into the wind, then it is obvious you can gear the device in the other direction to accelerate faster than the wind directly behind you.Or just make a toy that outruns the wind.
It's called a teaser...Who will win the bet?Derek said at the end of the video that he won the bet, and the physics prof paid up the $10k.
Settled?
That video is great in explaining the KEY factor in why this DOES work - the fact that the propeller is driving the wheels (a point that was made in the Veritassium video).
The ratios involved simply provide the mathematical constraints - which also allows critical assessment of alternative models that others might offer as disproving the claim.
But, then again, it only takes ONE example to prove that it is possible!
That video is great in explaining the KEY factor in why this DOES work - the fact that the propeller is driving the wheels (a point that was made in the Veritassium video).
:palm: He means it was his own click bait for the video.It's called a teaser...Who will win the bet?Derek said at the end of the video that he won the bet, and the physics prof paid up the $10k.
Settled?
Teaser for what?
He flat out said the professor had conceeded the bet and transfer the money to his account, and that he's now giving the money away.
What's left?
I only needed 1 important fact in advance to know that it was possible. That sail-boats who tack at an angle achieve a relative forward velocity greater than the wind speed forward velocity because of the air-foil effect on curve bulge of the sail. Then I knew it could be done.
I'm not convinced the inverter analogy made by 'IanB' is correct as you even loose power/wattage on the higher voltage side due to efficiency, but I guess I can go with it.
Like, what? You think sailboats achieve 100% efficiency?
So much for the "experts" and their big "opinions". I often notice that people seen as "celebrities" who are also held in great esteem, as if their analyses are flawless ALL the time, and they can do no wrong, are the ones who rest on their laurels, and see some things as "beneath them" (not necessarily in this case, but who knows!). It's almost as if they are SO confident in their assertions, that they are beyond reproach (Bill Nye and NG Tyson, the former I have seen being proven wrong MULTIPLE times, the latter I know [nor care] little about).
Now, airplane taking off from a treadmill still takes an effort to figure out.Airplane needs only air and an engine with a propeller to fly. If it will start from the treadmill, or not - it doesn't matter (if it has ideal weels with an ideal bearing).
Now, airplane taking off from a treadmill still takes an effort to figure out.Airplane needs only air and an engine with a propeller to fly. If it will start from the treadmill, or not - it doesn't matter (if it has ideal weels with an ideal bearing).
To be fair, I don't think either of them catagorically said that Derek was wrong, they just erred on the side he might be wrong. Just like any one of us that hasn't done a full deep dive into the maths and details to prove something can have an educated opinion, a hunch, a spidey sense that somethings right or wrong, or somewhere inbetween. Most technical people just don't care enough to actually go ino the details enough. This is why there are so few people who debunk things, it can be a lot of work.
And the science "celebrities" are often asked their opinion on something, and they feel a need to give their best educated opinion on something, the details of which could very well prove to be wrong.
I'm asked all the time to give my opinion on this and that technical thing, is it bullshit or not, is it viable or not, etc.
The explanation Derek made will violate the law of conservation of energy.
Of course this vehicle is not working as explained it uses stored energy to temporarily exceed wind speed. I explained that both to the creator of that vehicle as well as had a long conversation with Derek and they still do not get it.
Nobody that understand conservation of energy should have ever claimed that a vehicle can travel faster than wind in any direction forward or backwards in same exact direction at speed above wind speed for an unlimited amount of time since a vehicle traveling at wind speed is already a 100% efficient vehicle so only possible in theory and above 100% efficient is not allowed in this universe.
Flawed tests and flawed calculations are what results in wrong conclusions.
Nobody that understand conservation of energy should have ever claimed that a vehicle can travel faster than wind in any direction forward or backwards in same exact direction at speed above wind speed for an unlimited amount of time since a vehicle traveling at wind speed is already a 100% efficient vehicle so only possible in theory and above 100% efficient is not allowed in this universe.
It still needs to be proven under proper controlled conditions continuously.
But the professor was convinced enopugh that he paid up.
Applying oversimplified principles incorrectly leads to wrong conclusions. Why do you conclude (or simply state without demonstrating) that a vehicle travelling at wind speed is "100% efficient"? What does efficiency mean in that case?
You can read my attempt at an explanation of the mechanics of this a few pages back. The only remaining issue is understanding the operation of and limitations of propellers. I think the device has been sufficiently demonstrated that there isn't any serious question that it works and is not any sort of over-unity efficiency scam or trick.
Once you understand the idea of using the propeller to slow down the wnd even if the vehicle is moving faster than the wind, it is not so hard to understand the principle. The calculations are sound and the experiments, especially the one on the treadmill are convincing.
If you add the addtional requirement to have just simple passive wheels the system does not work, but this is not claimed or required.
There is no violation in energy conservation: the vehicle still gets enough energy from the wheels to drive to propeller. The trick is that the wheels relative to the ground are realatively fast, while the propeller relative to the wind is relatively slow. The power is force times speed. So from the forward force produced by the propeller the faster moving wheels can produce enough power to drive the propeller. Just a little faster than the wind, the propeller sees a very low speed - so it works even with a pretty low propeller efficiency, but it gets increasingly more difficult at higher speeds. The propeller effiency will limit the maximum obtainable speed.
all available power is used meaning 100% efficient travel.
If the propeller based vehicle was ruining under water (non compressible material) so a water propeller then vehicle will have no energy storage and exceeding water speed can not be demonstrated even for a sort amount of time as it is the case with air.
That reasoning is all wet! It might be impractical to made an underwater version of this, but the device does not depend on a compressible medium. Please, I implore you, go back and find my previous post on this subject since it explains all this using incompressible mediums (belts and friction instead of air and propellers).
Just look at my first diagram.
There is no energy storage involved (other than the normal kineric energy). The small treadmill modes are fully open, no hidden parts. The large one drove for quite some distance faster than the wind, not just a few seconds.
The suggestion with the day with no wind - shows that you don't understand the principle. The final speed of the vehicle would be something like 1.2 times the speed of the wind (because of the limited prop efficiency I doubt it could get much beyound 2 times).So 1.2 times zero wind speed would be zero. There is no more acceleration beyond a speed limit set by the gear ratio, so there is a limit to the speed, but no limit to how long this can be maintained.
One difficulty in the build is to find the right gear ratio / prop angle: the faster the design speed the more breaking force from the wheels and the larger the chance it would not work at all. The speed dependent prop efficiency and additional friction (e.g. from the wheels) the speed range is limited. So it may not work well at very low or very high wind speed.
One could imagine a purely mechanical analog, and there are even toy like this: consider a vehicle that uses the wheels to wind up a string and that pull on that string. The vehicle will come to you faster than you pull. Nothing really magic, but essentially the same pinciple, just pulling and not pushing.
I just did and I don't see your point. If you simply couple the two wheels (assume what you have drawn is a drive belt between the two so they go in the same direction at a 1:1 ratio), the whole thing will go from left to right at double your track speed, assuming there is no slippage anywhere.
If an ice yacht can tack downwind with a velocity made good (VMG) of over double the wind speed, which is very well documented*, its obvious that a land vehicle with a variable angle of attack wind turbine geared to wheels can also beat the wind downwind if its got sufficient blade area and great care is taken to minimize friction.
The problem is getting past the stagnation point in a steady wind where the velocity is equal to the wind which is going to demand some sort of mechanism so stored energy in the rotor can briefly accelerate the vehicle past the wind speed, then flipping the angle of attack will let the rotor continue to draw power from the relative wind.
I *think* the vehicle in the video BrianHG linked at the start of this topic gets around the apparent need for variable gearing by utilizing the gustiness of the wind. At the trailing edge of a gust, the vehicle can be already travelling faster than the wind, so its just a matter of picking the moment to flip the blade pitch control through the dead spot, where the blades are effectively flat to the plane of rotation, to continue accelerating.
Any sort of conveyor belt or wind tunnel modelling of such a vehicle, or anything even vaguely similar, that does not have a variable rotor to wheels gear ratio and doesn't model the gustiness of real wind, is doomed to failure . . . :horse:
* e.g. http://www.nalsa.org/Articles/Cetus/Iceboat%20Sailing%20Performance-Cetus.pdf (http://www.nalsa.org/Articles/Cetus/Iceboat%20Sailing%20Performance-Cetus.pdf)
If you use a 1:1 gear ratio and assume ideal case no friction then vehicle will not move in any direction. In a real case where there is some friction the vehicle will move from right to left.
Not quite sure how you came with the 2x track speed with a 1:1 gear ratio. I guess you imagined a powered off treadmill dragged under the vehicle as that is what people thing is equivalent when it is not at all.
I'm of above average intelligence
I'm assuming no slippage, not no friction. And I did miscalculate, 1:1 won't actually work in practice as the vehicle would try for infinite velocity and something would have to slip or break. However, with an appropriate gearing between the two wheels, you can get the vehicle to move in either direction at an arbitrary speed.
Let's say V1 is the belt speed, where going from right to left as your arrow is drawn is the positive direction. Let's say V2 is the speed of the outer circumference of the right wheel (G) and V3 is the speed of the left wheel (M), with a positive value reflecting a clockwise rotation. The gear ratio R of your drive mechanism is V3/V2, so a value of 0.5 means that M is spinning in the same direction as G but half as fast, and so on. V4 is the speed of the vehicle from left to right relative to the stationary block on the right and the stationary part of the system containing the moving belt, all presumably attached to some reference like the ground. What you want is to determine V4 given R and V1.
Now assuming no slippage, the velocity of the vehicle V4 has to be the same as V3, which equals R*V2. At the same time you should see that V2 has to equal the sum the belt speed and the vehicles speed, so V2 = V1 + V4.
Well you are wrong about the fact that vehicle can travel in any direction based on a gearbox ratio. That is not true and the vehicle I show can only travel from right to left it can not advance on the treadmill.
What you are thinking of is maybe something like this
I'm not thinking about random videos, I'm applying basic principles that I thoroughly understand to your example. Now even though I have that understanding, I still could make a mistake, so check my work. I explained where I think your main error lies, and that is in assuming that conservation of energy is somehow relevant here. It isn't. If you assume a belt moving at a given speed, as you have, then it would be possible to extract any arbitrary amount of energy from that subject to actual physical limitations in practice. Stop thinking in terms of energy when you can solve the whole thing with elementary mechanics.
So please, at least solve those basic equations and show me if there is any part of the solution that involves any error like infinite force, division by zero or the like. Even if you can't solve the equations, you should be able to see that for R=1/2, the vehicle will move from left to right at V1.
Conservation of energy is relevant everywhere. Nobody has proven that conservation of energy law can be broken tho many have claimed that including here with blackbird.
Nobody has shown an example where the vehicle in my diagram works as it is impossible. Not quite sure why that is not obvious.
Say you have two identical vehicles (say they are electric vehicles but with a gearbox) and so they have exactly same power motors but one is in first gear and the other in 4th gear and they are back to back connected trough a rope and so question is witch one of the two vehicle will win?
But what that represents is a vehicle traveling against the wind direction at much lower speed than the wind. The wind in that case is the treadmill and the road is the fixed body of the treadmill. So it demonstrates that a vehicle can travel down wind at lower speed than the wind and nobody will disagree with that. But people including that guy interprets this wrongly and adds the speed of the wind to the speed of the vehicle and consider that to be the speed of the vehicle.
A similar vehicle would go faster than the band, if the gear ratio is negative, so invert the direction of rotation.
The sytem with the moving ground on one side or the wind blowing has plenty of energy to go around. So it is not a good idea to argue with conservation of energy. Even a standing vehicle would not conserve the enery and standing in the wind is definitely possible. It is more about arguing with a balance / imbalance of forces.
What is wrong in my simple explanation where all I say is the following.
-In my diagram ideal case vehicle can not move from left to right because any breaking power at wheel G will need to be fully available to wheel M just so that vehicle remains stationary.
A perfect example of where your thinking is going wrong. Power is completely irrelevant. Torque--which does not mean power--will determine which way the vehicles go, subject to the limitations of the strength of the rope and the friction of the tires with the ground. If the motors have the same torque, then the one with the greater torque multiplication--the one in first gear--will be the one pulling harder.
What is wrong is that force is not power. It is possible (and completely ordinary) to continuously exert a force without expending any power. Braking force is what is important, if the components are stationary, then the braking power is zero. If they are moving, braking power is negative, but that's another issue altogether.
with no energy source
When changing reference frames
Force is contained in power. Conservation of energy will not mention force as energy is power over time so force in isolation is irrelevant.
Braking power is what is important. A breaking force alone will say nothing (provide no useful information).
That's unparsable gibberish to me....
You will not be able to answer that question.
But if I say something like I apply a breaking force of 10Ws for 1 second you will be able to say that vehicle has now just 490Ws of kinetic energy so lower speed since weight is unlikely to have changed.
Also in keeping with this example if you want to get back to the same speed you where at before the 10Ws for one second breaking you will need to apply the same in propulsion so if you stored those 10Ws in a battery or capacitor instead of wasting as heat then you will need all that energy to get back to where you where and there is no way to use those 10Ws to increase the vehicle speed (meaning increase kinetic energy).
Knowing the mass "m" (ie. 1000kg) of the vehicle you can, imho.
OK let me give you an example. You have an ideal vehicle driving at constant speed and say has 500Ws of stored kinetic energy.
Then I say to you that I apply a 10N breaking force for 1 second so then I ask. What is the vehicle current kinetic energy?
You will not be able to answer that question..
Aside from the fact that 10W is not a 'force', OK. Say a force such that 10W is dissipated. So what?
What does all that have to do with the example we are talking about?
One thing about physics is that often different sets of rules apply to the same problem and the solutions have to work in all of them, otherwise you have made an error. You can try to solve an energy equation here if you like, but you should also be able to come up with a solution in simple mechanics, Newton's laws and so forth, and if the methods don't yield the same result, then one of them hasn't been done right or doesn't apply. The simplest trick exam question involves firing a bullet into a pendulum and then determining the resulting motion of the pendulum afterwards. In that case, the common mistake is to use conservation of energy instead of conservation of momentum to try and solve the problem. That doesn't work because (coherent) kinetic energy is not conserved in that case. You're making a similar mistake here--this is not a closed system and you cannot use conservation of energy laws to solve it.
Knowing the mass "m" (ie. 1000kg) of the vehicle you can, imho.
KE=p^2/(2m)
KE=10Ns*10Ns/(2m)=100/2000=0.05Ws
500-0.05=499.95Ws
The point is that taking energy from vehicle wheel and then transferring all that to another wheel or to a propeller will not allow you to accelerate the vehicle.
Not in the example where vehicle is only powered by wind and it is at wind speed or higher and vehicle direction is exact same as wind direction.
OK. |O Would you like to bet $10,000? :-DD
Perhaps they see an easy mark who if they believe their pet theory strongly enough may take a sucker bet! The physics is validated by the existence of several types of wind powered vehicles that have been demonstrated to sustainably exceed the wind speed down wind when the vehicle's velocity is resolved in the true downwind direction. Smells like a potentially profitable bet to me . . . .
OK. |O Would you like to bet $10,000? :-DD
What is up with people and obsession with gambling.
Perhaps they see an easy mark who if they believe their pet theory strongly enough may take a sucker bet! The physics is validated by the existence of several types of wind powered vehicles that have been demonstrated to sustainably exceed the wind speed down wind when the vehicle's velocity is resolved in the true downwind direction. Smells like a potentially profitable bet to me . . . .
Problem is that is not indefinitely it is just for at most a few minutes depending on amount of stored energy and amount of vehicle friction.
Perhaps they see an easy mark who if they believe their pet theory strongly enough may take a sucker bet! The physics is validated by the existence of several types of wind powered vehicles that have been demonstrated to sustainably exceed the wind speed down wind when the vehicle's velocity is resolved in the true downwind direction. Smells like a potentially profitable bet to me . . . .
I had not claimed blackbird and similar vehicle can not exceed wind speed in the same direction as the wind. Problem is that is not indefinitely it is just for at most a few minutes depending on amount of stored energy and amount of vehicle friction.
Vehicle will get to peak speed and then slowly decelerate until it is below wind speed.
If an ice yacht can tack downwind with a velocity made good (VMG) of over double the wind speed, which is very well documented*, its obvious that a land vehicle with a variable angle of attack wind turbine geared to wheels can also beat the wind downwind if its got sufficient blade area and great care is taken to minimize friction.
The problem is getting past the stagnation point in a steady wind where the velocity is equal to the wind which is going to demand some sort of mechanism so stored energy in the rotor can briefly accelerate the vehicle past the wind speed, then flipping the angle of attack will let the rotor continue to draw power from the relative wind.
Except that you are not able to demonstrate or prove this to the satisfaction of others.
It has been shown both by experiment and mathematics that a vehicle with a trailing wind can travel faster than the wind speed indefinitely, if the vehicle is constructed appropriately.
The vehicle does of course need to have the correct design. A simple sailboat cannot do this.
One basic point to make clear is that the power is force times speed. So at a lower speed one has a higher force at the same power level. Confusing power and force is a mistake sometimes made by beginners.
The basic idea behind the vehicle is to take up power from the wheels and use this power to drive the prop. The prop than produces force relative to the wind.
So it looks a bit like the fake over unit devices, but there is a twist to it, as the two parts have different systems to act on:
With the vehicle moving in the direction of the wind, the speed of the wheels to the ground is relatively high.
With the vehicle moving with about the speed of the wind, the relative speed between the prop and wind is relatively low and with a speed lower than the wind speed the wind is even driving the prop. So even with a not so good efficiency of the prop it is possible to create more forward force ( = power / speed) than the backward force from the wheels. So the net force can be forward.
The energy comes from slowing down the wind. This is seen as the backwards force from the wheels. With the fan blowing in reverse this even works when moving faster than the wind or also when moving at the speed of the wind. So there is no dead point or special case when the speed is the same.
The difficulty in actually making this work comes from getting good enough an efficiency from the prop and gears. No need for over unity, just not too small (e.g. < 30%) to not overcome the friction forces and parasitic drag.
One can test the easy purely mechanical (no aero dynamic) analog. The nice point here is that it does not depend on speed so much and can be done hands on on the desk: Have a spool of thread with the outer rims larger than the diameter where the thread is. To improve the gearing ratio maybe add some rubber rings to the outer rims. With the spool on the desk slowly pull the thread horzontally. The thread is the analog to the wind.
With the thread on the top side the spool moves slower the thread is pulled, so nothing funny happening.
With the thread on the bootom side the spool moves faster than the thread is pulled. On a slippery surface it may slip.
You can not claim that if you use forces and speed in your calculation the result will be different than if you use power as if applied correctly the result will need to be exactly the same.This is true, but so far the calculation based on power alone is wrong.
In my wheels only example there is a G (generator) wheel and a M (motor) wheel and say there is a 2:1 gear ratio that means the generator wheel will make 2 full rotations while the motor wheel will do just one.Up to this point the argument is correct. Just accept it not just think it must be wrong, because is contradics intuition and the misconception that the systen should not work. The net force is real and it will move the vehehicle - that's it.
So say treadmill speed is 5m/s and say you apply 10W of breaking power that will mean a breaking force of 2N now if you look at the motor wheel you will see that spins at half the speed 2.5m/s but twice the force 4N assuming ideal vehicle where all 10W from generator are available at the motor wheel.
Now if you look just at the forces you will think that since there are 4N at the motor wheel and only 2N breaking force at generator the vehicle will advance forward ...
Say wheels are mechanically connected with a chain then those forces will be transferred to the chain not to the vehicle body and so since power is equal on G and M wheels the vehicle will actually not move in any direction.
:blah: :blah: :blah: :blah:
I think the trick comes from the propeller construction itself. It has nothing to do with wheels or gears..
The propeller's blades are built in form of a wing.
This is true, but so far the calculation based on power alone is wrong.What do you mean by that ? I made a mistake in my calculations, or are you saying this vehicle is a special case and this clear rule will not apply to it.
Up to this point the argument is correct. Just accept it not just think it must be wrong, because is contradics intuition and the misconception that the systen should not work. The net force is real and it will move the vehehicle - that's it.No it will not and if you do not believe that I encourage you to test it. Forces at the wheel act on the transmission not directly on the vehicle body.
A chain drive as a gear will just transfer power. The force of the chain is balanced by the structure between the axels. So there is no magic force to stop the vehicle. Otherwise a simple chain drive could move a space craft.
Having the same power does not mean it would not move - power is not moving things, it is force. Power is only needed to generate a force in the direction of the movement. Head-lights are incresddibly inefficient breaks.
This is embarrassing. Please stop.
You have clearly misunderstood the mechanism and continue to argue "physics" when you are not using a model which describes the system that has actually demonstrated the effect.
Yes, there are parameter constraints for the propeller and gearing for this to work, but within those constraints, it does work as claimed.
... and - as other have stated - no energy storage is required and as long as the wind is consistent and there is room to drive, it will continue, unabated.
That is why I asked the question about two vehicle with same exact motor power and different gear ratio and nobody was brave enough to answer. The answer to that is also that none of the vehicle will win and it will be a tie.
:palm: - there still is the misconception about force and power. :horse:
Under normal condition (e.g. no ice) one gets a higher maximum force in the lower gear and thus the lower gear car wins.
This is why you don't drive up a steep hill in the higherst gear.
Please do a little more reading on the basics, before wirting even more stupid and wrong claims.
If your answer to my question is yes then you do not understand physics and since my words will not convince you the only thing you can to is test it in real life.
I will have tested this for you but I do not own a treadmill and also if I will do the test you will think I cheat in some way.
And I hope I do not sound to condescending but I do not have great social skills and on top of that I see this sort of wrong answers from people for the past few weeks. Even people that should understand this like university level physics professors and science communicators like Derek.
You don't need a treadmill, you can just substitute a sliding block as you won't have to move it very far to prove yourself wrong. If you make the gear ratio 2:1, so the M turns one revolution for every two of G, in the same direction, you will find that as you slide the right hand block towards the fixed block, the car will move to the right with equal speed. Or you could prove your skills at math and physics by solving the very basic equations I gave you. Either experimentally or mathematically these should be trivially easy for someone claiming to be knowledgeable about physics.
I don't think the problem is a lack of understanding of physics by forum members--there are fairly accomplished people here, some with actual degrees in the subject. I also no longer think the actual issue here is your lack of understanding of physics, rather it is a fixation on one particular aspect of this that you think you understand intuitively, but are wrong about. The amazing part is that you yourself have provided a blindingly obvious way to demonstrate a correct solution and understanding of the issue--which boils down to the fact that power equals force x speed--and yet you not only recognize it, but won't even respond directly when it is explicitly pointed out. Several times I thought I had clearly explained this in the past--both to you and previously--but apparently my explanations aren't clear enough or something.
I think you are very close to understand what I'm saying you just need to put a bit of effort in reading what I write.
Like I said, your purported understanding of basic principles--as represented here--is highly flawed, but that's not the real problem. At this point, from my perspective you are failing the Turing test....
you will also know vehicle can not be powered by wind when vehicle speed is higher than wind speed in the same direction
You keep writing lots and lots of words trying to explain that this is not possible. The difficulty you face is that the universe does not agree with you. It is provably possible to construct a vehicle that does what you claim is impossible. Your challenge, should you accept it, is to explain how actual experiments, in the real world, do this "impossible" thing, since they actually do work.
Theory is good up to a point, but all of physics is backed by experiment. Every theory, every formula, every equation, is consistent with experimental evidence. Here, the experimental evidence is clear--a suitably constructed vehicle can run forever at a speed faster than the wind when the wind is directly behind it. If you want to disagree with the experiments, you need to show what is wrong with the experimental evidence. You can't do that with theory. You have to do it with practice. What part of the experiment has the wrong method, and why?
You just do not understand that with vehicle above wind speed in the exact same direction or opposite direction to wind there is no power available from the wind.
There is just no mechanism to allow a vehicle be powered by wind when above wind speed in same direction as the wind.
My problem was setup in ideal conditions meaning no wheel slip allowed on any of the vehicle's. The vehicle perfectly identical except for the gear ratio.
I guess you can not solve a problem as you need to see an experimental result to know what will happen in a system.
And keep in mind that your example of a vehicle climbing a hill has no relevance here at all.
That vehicle has the power source on board it is not taking power from one wheel and transferring it to another so as long as it has traction it can always move forward.
Look again at my diagram and answer this simple question.
Can the vehicle as shown in the diagram using any gear ration you want between G wheel and M wheel be able to move from left to right with no external power source?
(http://electrodacus.com/temp/Winds.png)
If your answer to my question is yes then you do not understand physics and since my words will not convince you the only thing you can to is test it in real life.
I will have tested this for you but I do not own a treadmill and also if I will do the test you will think I cheat in some way.
Any "examples" that do not include the propeller are fundamentally wrong and discussion on such is completely irrelevant
Oh, but there IS.
The fact you can't see this simply underlines your understanding of the system is lacking. Any "examples" that do not include the propeller are fundamentally wrong and discussion on such is completely irrelevant.
The answer is yes, yet surprisingly I *do* understand physics. This is a simple 1st year mechanics problem, and again the answer is 100% yes, you can make the cart move in either direction by changing the gear ratio. This is actually exactly equivalent to the demo from the veritasium video (https://youtu.be/yCsgoLc_fzI?t=808).
In your example, I define positive velocity to the right and positive angular frequency to be clockwise. Both wheels are assumed to have radius r, although you could let that radius vary instead of the gear ratio. Then I can write three equations:
$$v = \omega_G \cdot r \\
v = \omega_G \cdot r + v_t\\
\omega_G = X\omega_R
$$
The first two are the center of mass velocity of the two wheels (which both must be equal to v). $v_t$ is the velocity of the treadmill. The third is the gear ratio constraint where X is the gear ratio.
The solution to these three equations is:
$$v = v_t / (1 - X)$$
If $v_t$ is negative but you want positive velocity X must be >1. If X < 1, the cart will move in the same direction as the treadmill. If X equals 1, then you get divide by zero but that just means there is no solution: the cart will just slip. Close to the 1:1 gear ratio you can theoretically generate arbitrarily high speed in either direction but the forces involved will be quite high and it will likely slip. If you want to build this yourself, a 1:2 or 2:1 ratio should work well, although as I said the video above already does exactly this demonstration.
The point of the controversy is that the vehicle demonstated that it is possible to extract energy from the wind when moving at the speed and direction of the wind. It is not easy to understand, but it works. It is proven by experiment and also the calcultions show that is should work, if the prob is reasonable efficient.
One can not understand a system when one starts with the proposition that the system does not work.
The idea of looking at the case with 2 platforms is a way to get away from the complications of aerodynamics. So it is a good way to understand the principle. Even electrodacus did the math correct on this :-+ . For some reason he just refused to accept the result and than argued that the force is not real.
If you replace the 2nd platform with a string / thread, one can easily do the experiment hands one. A nearly empty spool of thread on a not so slippery desk is all it takes to try. The gear ratio is replaced with different diamaters.
When I first saw this effect it confused me for a short time - like 5 minutes for 5 year old.
Oh, but there IS.
Please explain what IS ?
I wasn't going to go quite that far - I simply meant to say that the propeller is a key part of how this mechanism works.Quote from: BrumbyAny "examples" that do not include the propeller are fundamentally wrong and discussion on such is completely irrelevant
That is not actually true, or at least you don't have to include any fluid dynamics.
.... but the blackbird is fundamentally a simple gear ratio problem and doesn't need any aerodynamics to understand the basic operation. If you stood behind the blackbird and poked the propeller blade with a stick, it would move forward faster than your stick, because the propeller would turn "into" the stick as you pushed. Of course you couldn't go very far before the stick would run off the edge of the blade so the wind is a lot more convenient.I quite like this explanation. :-+
There is no experiment showing vehicle runs forever above wind speed. The most I think was about one minute for the blackbird when they did the 28mph record.
no wind energy is available above wind speed (not sure why that is not obvious)
and also as mentioned wind is not necessary at all
If there is a velocity difference between the air and the ground, then a vehicle in contact with both air and ground can extract energy from the system. This can happen even if the vehicle is moving faster than the differential velocity.This was the concept that made it "click" for me, originally stated a few pages back.
In both the Blackbird and treadmill experiments the vehicle accelerated. Blackbird accelerated so vigorously they had to put the brakes on to slow it down. If an unpowered vehicle accelerates it is gaining energy from the surroundings. If it accelerates under its own steam, it will run forever, by definition.
If there is a velocity difference between the air and the ground, then a vehicle in contact with both air and ground can extract energy from the system. This can happen even if the vehicle is moving faster than the differential velocity.
If course wind is necessary! If there is no wind, there is no motive power for the vehicle, and it will coast to a standstill and stop moving.
This was the concept that made it "click" for me, originally stated a few pages back.
...Really? If that was the case, then an airplane flying exactly downwind faster than the wind speed would drop out of the sky!
Should (it is for me) be easy to visualize that no air molecules moving at wind speed in same direction as vehicle can hit a vehicle that moves faster than the wind. ...
As far as vehicle is concerned when you are at exact same speed as wind speed (like in the treadmill example) air molecules do not move relative to the vehicle meaning air molecules can not transfer any energy to the vehicle.
Really? If that was the case, then an airplane flying exactly downwind faster than the wind speed would drop out of the sky!
As far as vehicle is concerned when you are at exact same speed as wind speed (like in the treadmill example) air molecules do not move relative to the vehicle meaning air molecules can not transfer any energy to the vehicle.
You make so many mistakes in your analysis that it is hard to keep track of them all, let alone the glaring inconsistencies in your reasoning. But here is at least one lucid example of where you have gone wrong. The air molecules movement relative to the vehicle is not the point, it is the movement of the air molecules relative to the parts of the propeller with which they are interacting. That is the whole reason the propeller is there. Now you may say that the propeller as a whole is moving at the same speed as the vehicle, which is true, but from the perspective of an air molecule, the surface of the propeller will still appear to be coming at it as the propeller rotates. That is how propellers work.
And as to previous comments about your "intuitive understanding" of physics, give it up--that's a delusion. The rule in physics and similar sciences is that if you can't do the math, you don't really understand it. Period. ejeffrey has laid out clearly the equations that I wanted you to find and solve on your own--which you didn't even attempt. Look at those and if you don't understand them, your 'intuition' is worthless. One of your other misconceptions has to do with your own example where you claim that analyzing the power at M and G proves your point. You've simply miscalculated because you just don't grasp the implications of the fact that power is force x speed, or in another form, torque x rpm (or ω for actual science types). In your own diagram it is obvious that G will be spinning faster than M, thus at equal force (any steady state where the car is not accelerating) G will always produce more power than is consumed by driving M.
My math was done quite a few times here and that is power at the propeller is smaller than power taken from the wheel thus vehicle willslow down without any form of energy storage. accelerate.
You've done no math, but you have just accidentally actually stated the reason that the vehicle in question works as advertised. The horizontal force from the propeller is greater than the horizontal force the other way at the wheels, yet the power generated at the wheels is greater than what is used by the propeller.
You've done no math, but you have just accidentally actually stated the reason that the vehicle in question works as advertised. The horizontal force from the propeller is greater than the horizontal force the other way at the wheels, yet the power generated at the wheels is greater than what is used by the propeller.
I'm fairly certain your expertise is not in electrical engineering else you will not make such a statement.
At starting point the vehicle has zero kinetic energy since it is stationary and in order to increase the kinetic energy (also vehicle speed) you will need have an imbalance of power not forces.
Since the only imbalance of power you can have is lower power at M motor wheel and higher power at G generator wheel that means kinetic energy will increase but not in the direction you will like meaning vehicle can only move from right to left and not the other way around.
I'm fairly certain your expertise is not in electrical engineering else you will not make such a statement.
At starting point the vehicle has zero kinetic energy since it is stationary and in order to increase the kinetic energy (also vehicle speed) you will need have an imbalance of power not forces.
Since the only imbalance of power you can have is lower power at M motor wheel and higher power at G generator wheel that means kinetic energy will increase but not in the direction you will like meaning vehicle can only move from right to left and not the other way around.
as you do not know how to apply the formula in that case
you can not get a different result by using forces and speeds if you apply the formula correctly.
That's gibberish. Really. But if I take a wild guess at what you are saying, what is the basis for your notion that the car will somehow only travel in the direction of the wheel with 'lower power'?
But the power doesn't have any directionality. That is one of your fundamental misunderstandings. The backwards force on the wheels is less than the forward force of the prop. Net force: forward. The power input at the wheels is less than the power consumed by the prop. Net power: positive. The fact that the force at the wheels is backwards is *irrelevant*
The power input at the wheels is less than the power consumed by the prop.Your sentence makes no sense. Wheels powers the propeller so in an ideal system the two can be at best equal and in a real system power at the propeller will be lower than power generated at the wheels as wheels provide power to propeller and nothing else.
Kinetic energy is not a vector and has no direction. No equation in units of energy or power tells you anything about direction of motion. You need to look at force or velocity. I worked out your treadmill problem above. You haven't addressed it. Can you find a *mathematical* flaw in it? Or you can actually do the experiment. You obviously haven't done it.
Indeed you should get the same result if you use two different methods to determine the same thing, but the trick is in figuring out which one is most appropriate given the problem. I cannot fathom why you don't understand Newton's laws of motion, or at least are ignoring them. Do you actually not believe that the car is going to go in whatever direction the net force is?
I'm beginning to think we are being trolled.
In this particular case I think using power is the simplest and simple means less chance to mess up the calculation.
.... it seems you just need to run the experiment as you can not properly calculate the result.
Well let's try power again and let's not use any tricky delays, storage or anything else, just the generator and the motor as you originally suggested. Lets see what we can agree on so we can find out where we diverge?
So for the starting position, lets say the car is not moving, the belt is moving from right to left at a given speed, I think you mentioned 1m/s, but it doesn't matter. We load the generator wheel to produce power, which creates a force that acts on the wheel and is transmitted to the belt, which provides the usual equal but opposite reaction and since the wheel is held to the body of the car by an axle, that force is transferred to the car. Simultaneously we use M to hold the car in place by preventing M from turning. The force from G through the body of the car is transmitted to M and then to the fixed platform, which provides and equal but opposite reaction. You don't need to worry about following or balancing the forces since you want to jump straight to energy, but I'm stating it now so there's no confusion later. So now we are generating energy, but not using it anywhere and the car is not moving. Any problems so far?
if vehicle is ideal then vehicle will just stay in place the power from G is all transferred to M so since one opposes the other result will be zero
if vehicle is ideal then vehicle will just stay in place the power from G is all transferred to M so since one opposes the other result will be zero
OK, lets just focus on that so I understand your position. If the car is staying in place and G is generating the power, how does M use the power in an ideal machine?
.... it seems you just need to run the experiment as you can not properly calculate the result.Sage advice. You should try following it.
But the power doesn't have any directionality. That is one of your fundamental misunderstandings. The backwards force on the wheels is less than the forward force of the prop. Net force: forward. The power input at the wheels is less than the power consumed by the prop. Net power: positive. The fact that the force at the wheels is backwards is *irrelevant*
Power has directionality if that is what I chose.
in order to know what direction the vehicle has moved you will want to have a sign (your choice of convention).
The power input at the wheels is less than the power consumed by the prop.Your sentence makes no sense. Wheels powers the propeller so in an ideal system the two can be at best equal and in a real system power at the propeller will be lower than power generated at the wheels as wheels provide power to propeller and nothing else.
As mentioned direction can be added to both power and kinetic energy as you start with a stationary vehicle co zero kinetic energy and if you want to know the direction you can decide to use a sign to indicate the direction of travel.
As I offered to others here if you can prove a vehicle as shown in my diagram moves from left to right I will pay for your expenses related to the experiment.
Your explanation involves larger force at wheel M but that is irrelevant as long as power is lower than on wheel G.
Your explanation involves larger force at wheel M but that is irrelevant as long as power is lower than on wheel G.
I'm not explaining anything. I'm asking you to clarify yours by explaining what happens in the simple, specific situation where M is held fixed so that the car cannot move and G is generating power. Where does the power go?
I don't understand your conditions which seem arbitrary and capricious since you think that it matters if the applied motion is from the top (as in the demo from veritasium) or is a board pushed by hand rather than an actual treadmill. So I don't think you would accept any demo you saw. As far as I am concerned the demo Derek did with the little cart was identical to your proposal, just with a board pushed from the top side instead instead of a treadmill and a horizontal linkage connecting the wheels. So I really don't know how to do a demo you would accept. Your only argument about the existing demo is "can't you see it is different" I can't, so I can't reasonably do a demo.
Therefore I will make this super concrete so that nobody has to do a demo.
The treadmill is moving at 10 cm/second to the left (that would be negative in the sign convention I suggested). The wheels both have a radius of 5 cm. The gear ratio is such that for every one rotation of G, M rotates 0.5 turn in the same direction. Please calculate the steady-state velocity of the cart as well as the rotation rate of each wheel (radians per second, or RPM if you prefer). Numerical answers, and explicitly state the direction of motion for each (left/right for the cart, clockwise/anticlockwise for the wheel).
Can you solve that simple problem? Feel free to use whatever formulas you like, but again: give the answer in the steady state when the cart is no longer accelerating. I will check to make sure that your answers aren't slipping.
Can you or can you not solve this problem using your equations? I will post mine here after you try.
But you don't. You think you do - but you don't..... it seems you just need to run the experiment as you can not properly calculate the result.Sage advice. You should try following it.
Why will I do that ? I know what the result will be
both by brain simulationVery scientific ... NOT
and simple power equation.Simply wrong equations ... or at the very least, incomplete.
It will be a waste of time for me to confirm what is super clearExcept, you are still wrong - and seem unable to acknowledge that possibility. The mechanism is NOT "super clear" and the fact that you have made this statement condemns your claims as being made from a position of preconception - and not open to thinking beyond your expectation.
and there is no evidence proposed to the contrary (no experiment showing that or equation predicting anything other than vehicle moving from right to left).There is an experiment - and it's freely available to view, review and consider. The information is there - IF you care to expand your thinking past "the obvious" ... which you consistently refuse to do.
My conditions are not capricious.
What Derek from Veritasium shows is fairly different form what I have in my diagram and also different from what he claims it represents.
In his case the floor is the input source for the vehicle driving the small wheels (generator wheels) and those in turn will drive the large wheel (M wheel) that drives from left to right on the lumber. The floor that you can consider the wind as it is what powers the vehicle moves in the opposite direction about 2.8x time faster than how fast the large wheel travels on the lumber.
How is M held fixed ? That is not the case in real world If you fix the wheel M to the vehicle then you will no longer call that a wheel.
A rotating propeller creates a forward lift independent of the surrounding wind direction and wind speed (under typical conditions).
At the beginning the vehicle is pushed forward by the force of the wind where the propeller acts as simple sail (no blade's wing effect, no additional propeller lift, the max achievable vehicle speed is equal the tail wind speed).
The moving mass of the vehicle incl. inertia of the propeller and mechanically coupled wheels creates a storage of the energy compensating fluctuations of the wind speed/direction and by the propeller induced lift (ie. like a filter capacitor in our electronics).
At a specific vehicle/blades speed the propeller blades start to act as the wings (as they are designed that way) and start to create the lift.
The by the propeller created lift/speed adds up to the speed of the wind.
With already rotating propeller (the blades create lift):
with the wind coming from front: v_vehicle = j*v_vehicle + k*prop_lift - v_wind
with the wind coming from back : v_vehicle = i*v_vehicle + k*prop_lift + v_wind
where j and i are some params related to vehicle losses, k is related to propeller lift efficiency..
With a variable-pitch propeller you would be able to regulate the lift at specific configuration - thus to maximize the vehicle speed.
The max speed of the vehicle would be determined by the losses (friction in gear, wheels, aerodynamics, etc.) and the lift efficiency of the propeller.
For example with 50% "propeller's lift creating efficiency" and 40% mechanical vehicle losses the speed of the vehicle will be higher than the speed of the tail wind.
Simply wrong equations ... or at the very least, incomplete.
And as mentioned the most convincing experiment for you (all) will be to push the vehicle to designed speed and see how vehicle accelerates faster than pushed speed while there is no wind (this only works in air not in water).
Please show how my equation is wrong or incomplete.
If you can demonstrate that then you can convince me that I'm wrong.
The equation is both correct and complete for what it is needed to know.
To remind you power at the generator will always be higher than power at the motor (equal only in ideal theoretical case) and since Motor is powered by generator only motor power will be lower resulting in vehicle moving from right to left.
If you can not disprove that simple (common sense to me) equation then you can not say using a different method of calculating will provide a different result since that will mean your method is wrong.
I say the board is the wind, the floor is the ground which is stationary, and the cart moves to the right faster than the board. This is exactly as in the blackbird. Why is that not ok? To be clear you can label things differently and change your reference frame if you want, but why is my choice and Derek's unacceptable or different than the blackbird?
The vehicle won't accelerate faster than pushed speed when there is no wind, because there is no wind! It won't accelerate at all, it will slow down and stop.
So we can do this experiment of pushing the vehicle when there is no wind, and the vehicle will coast to a standstill. And we will not be surprised, because this is the expected outcome. There is nothing for us to be convinced about, because we are already convinced about this.
This is the whole point we've been trying to hammer into your head for the past umpteen posts. You are completely ignoring the basic principle of force multiplication. Your so-called knowledge of physics doesn't even encompass the basic principles of Archimedes, let alone Newton. A less-powerful motor can easily overcome a more powerful one with force multiplication via a gearset or some other mechanism. This is how the propeller-powered vehicle can move upwind, which you your self seem to have acknowledged as a workable concept. Similarly whatever force is applied to get your G wheel to generate power, that can be overcome with an arbitrarily lesser-powered motor given sufficient gear reduction. The initial case I'm trying to get you to consider just starts with zero speed and thus zero power. But you deflect this point with the objection that if we keep it from turning it then is not a wheel. Really? And so what? And even if you persist with that, then instead let motor M be a very small, low-powered motor with a very high gear reduction ratio. Each force will push the car in opposite directions, but the one with greater force will win, period.
What you are thinking about is a tractor with low power motor pulling a vehicle with higher power motor but the reason that happen's is very different.
The tractor will have better traction either because of weight or amount of surface contact with the ground or a combination of both so the lower traction vehicle will be dragged no mater how powerful the motor is or what gear ratio he is using since he just has lower traction so it can not use the available power.
It will accelerate faster because you create a pressure differential when pushing the vehicle. The wheel drives the propeller so while there is no wind there will be a high pressure (high air density) behind the propeller and low air density in front and this pressure differential is a form of energy storage so when you let go after pushing it to sufficient speed (sufficient speed means this pressure differential power on vehicle needs to be higher than vehicle friction) the vehicle will continue to accelerate.
No, I'm not thinking of a situation where one vehicle loses traction. Suppose you have two very heavy vehicles that are otherwise identical except one of them has a small motor with great deal of gear reduction and the second a more powerful motor without gear reduction. In fact, just to make calculations easy, lets make the second one have a motor that consists of 4 of the small motors in the first vehicle all connected axially so that they function identically except that in combination they have 4 times the power and torque. OTOH, the first vehicle will have 8:1 mechanical gearing, the second will be 1:1. Further assume that neither vehicle's motor is powerful enough to make the wheels slip. Which one wins?
Without wind, pushing the vehicle to design speed is easy, as the design speed is something like 1.2 times the wind speed and without wind this is zero :-DD. Startuing a zero speed and no wind is pretty boring. But maybe this experiment is enough convince electrodacus. :-DD
For the model system with constant velocities one is free to choose a suitable identification of the refrence frames. The gear ratios work in all cases and it does not depend which system you choose as ground and which system is moving. This is the whole idea behind using the treadmill and similar analog model.
@electrodacus: If you have problems with math / equations, please do a simple hand on experiment:
Have an axle with 2 wheels at the ends and a string wound up around the axle. This could be something like a relatively empty spool of thread, or just use 2 round pieces of cardboard (for the wheels) and a pencil. The string kind of represents the wind - this time pullung, because we are not good at pushing the string.
Than pull horizontal on the string and watch it move. The interresting case is with the string at the underside of the axle.
It is not as mind-blowing as the experiment with the prop driven car, but enough to surprize.
No, this will not happen the way you think. Storing energy in compressed gas (air) only works in enclosed vessels like gas cylinders. The space behind the propeller is not a closed space, it is open to the surroundings, so it is not possible to store any compression energy there.
You can think about this as a very large gas cylinder orders of magnitude larger than what you are thinking of with also orders of magnitude lower pressure and a very large opening in the tank.
You know that the pressure downstream of a propeller is actually lower than atmospheric pressure, right? The propeller does not compress the air, it creates a vacuum. You can see that demonstrated here:
https://youtu.be/f2QfVJe7yEg?t=198
If none of the vehicle can lose traction the one with 4 small motors will win fairly easy and just damage the single small motor on the other vehicle.
It is probably hard for you to think about power.....what you are thinking of is not 4x higher power.
You need to understand that wind can power the vehicle as you mention as long as wind speed is below vehicle speed and as soon as that is not the case zero power will be available from the wind (and yes we are just talking about the special case where wind direction is the exact same as vehicle direction). For any other wind direction wind can still power the vehicle no matter how fast the vehicle is moving. This was a well known fact but unfortunately someone decided to build a useless vehicle and confuse a lot of people even smart well educated people that only know physics and do not understand it (my definition of what understanding means).
For a simple sail it is impossible to extract power from the wind when the vehicle is moving at the speed of the wind or faster with the wind from behind. Surprisingly this is no longer true for the vehicle with the propeller and wheels to interact with the ground. This is the slightly confusing effect the whole discussion is about. For the discussion you can not start with the assumption that this is not possible - it is not a well known fact, but a more or less widespread misconception.
It is known that one can extract wind energy it the wind comes from a different direction. The wings of the prop move and the apparent wind is no longer coming from the front. So in principle they can extract wind power. So you see that with moving winds it is no longer clear one can not extract power. I would not explain the effect this way (the vehicle used a different principle), but it shows that a clever construction can do things a fixed sail can not do.
An ideal propeller is kind of working like a srew in a solid, though real world the efficiency is lower.
With such an ideallized prop the problem get transformes to something like 2 platforms / referenc surfaces moving relativ to each other. With a simple mechanical systen, like with wheels and pullies, it is relatively easy to understand that the difference in speed can be used to power a vehicle at any sensible speed and direction. This part may confuse a child, but is not really surprising.
It only take basic mechanics to understand. For understanding it may be easier to just calculate speeds and use the wheels, links and gears as conditions to link different movements. So all without, without looking at power or forces, just at position and coordinates. This usually leads to linear equations that are not so hard so solve. If such a mechanical system allows a movement (solution for the equations) it would do so with only friction forces when forced to by the boundary conditions (e.g. moving plantforms).
The arguments with motor power to decide which side is stronger tricky as motor power sometimes means maximum power and real motors are more limited in torque and not actual physical power (in W). This is especially true at low speed or starting from 0.
The point how good a prop can rally work is hard to calculate - I fully understant if one does not follow the math behind the limits there - that is hard core aero dynamics and may not have closed solutions. It is sensible that a suitable size prob can have enough efficiency to get at least a speed slightly higher than the wind speed. Working with a relatively low relative speed it should be even easier for a prop to work in water.
For a simple sail it is impossible to extract power from the wind when the vehicle is moving at the speed of the wind or faster with the wind from behind. Surprisingly this is no longer true for the vehicle with the propeller and wheels to interact with the ground. This is the slightly confusing effect the whole discussion is about. For the discussion you can not start with the assumption that this is not possible - it is not a well known fact, but a more or less widespread misconception.
It is known that one can extract wind energy it the wind comes from a different direction. The wings of the prop move and the apparent wind is no longer coming from the front. So in principle they can extract wind power. So you see that with moving winds it is no longer clear one can not extract power. I would not explain the effect this way (the vehicle used a different principle), but it shows that a clever construction can do things a fixed sail can not do.
An ideal propeller is kind of working like a srew in a solid, though real world the efficiency is lower.
With such an ideallized prop the problem get transformes to something like 2 platforms / referenc surfaces moving relativ to each other. With a simple mechanical systen, like with wheels and pullies, it is relatively easy to understand that the difference in speed can be used to power a vehicle at any sensible speed and direction. This part may confuse a child, but is not really surprising.
It only take basic mechanics to understand. For understanding it may be easier to just calculate speeds and use the wheels, links and gears as conditions to link different movements. So all without, without looking at power or forces, just at position and coordinates. This usually leads to linear equations that are not so hard so solve. If such a mechanical system allows a movement (solution for the equations) it would do so with only friction forces when forced to by the boundary conditions (e.g. moving plantforms).
The arguments with motor power to decide which side is stronger tricky as motor power sometimes means maximum power and real motors are more limited in torque and not actual physical power (in W). This is especially true at low speed or starting from 0.
The point how good a prop can rally work is hard to calculate - I fully understant if one does not follow the math behind the limits there - that is hard core aero dynamics and may not have closed solutions. It is sensible that a suitable size prob can have enough efficiency to get at least a speed slightly higher than the wind speed. Working with a relatively low relative speed it should be even easier for a prop to work in water.
I looked at all explanations available in details but the current explanation is wrong.
Propeller can not magically have access to wind energy when wind speed is below vehicle speed and both are in the exact same direction.
A sail is the most efficient way to use wind energy and ideal case a sail that is at the same speed as the wind speed has nothing more to extract from the wind and you will call that a 100% efficient sail powered vehicle (not in real world just ideal case).
No matter what other device you use to replace the sail you will not be able to extract energy from the wind when vehicle and that device (propeller or anything else) drives above wind speed.
Think about it this way. Say you have a wind turbine on wheels and when stationary you extract the most you can from the wind (real wind turbine is just at most around 40% efficient but that is irrelevant). Now you start to move the wind turbine in the same direction as wind direction so when you get at half wind speed the wind turbine will experience just half of the wind speed so just 12.5% of the power level compared to stationery wind turbine.
When your turbine move at the same speed as wind it will experience zero wind speed so it will not be able to produce anything and above that things will not change as apparent wind will now be from the other side and if you try to access that by rotating the wind turbine 180 degree you will slow the vehicle down way more than you produce from the wind turbine.
The same applies here with a propeller used as a fan and the only difference is that pressure differential energy storage witch allows vehicle as you see in tests to exceed momentarily the wind speed may be a few minutes depending on design and amount of stored energy.
Yes a propeller works as a screw in a solid now imagine a piece of wood or butter :) for lower friction where a screw stile propeller will advance and below this wood the vehicle moves on the ground. As long as wood moves above ground (floating) the vehicle can be powered by this wood but as soon as vehicle speed same as wood speed there is no longer any way to power the vehicle as taking power from vehicle wheel and transferring that to the screw will result in vehicle slowing down as not all power from the wheel can be transferred to the screw.
What you are confusing is the apparent wind speed direction relative to vehicle. Below wind speed vehicle will see a positive power available above wind speed the power will be negative meaning counter productive so vehicle will be slowed down rather than accelerated.
I know the limitations of a real motor but we are talking about power here so it is assumed the motor at whatever speed is capable of that power. The same is true about a generator so you need to remember that a generator powers the motor and while this is an electrical analogy using sprockets and a chain to transfer power from one wheel to the other will be constrained by the same rules.
Yes fluid dynamics is not easy (I had done both mechanics and fluid dynamic curses at university) but you do not need to know fluid dynamics (you will usually have a computer to simulate something like this) all you need to know is that if your only power source is the generator wheel that will not be able to provide the motor with more power than the breaking power.
Breaking (generating power) will be from the vehicle kinetic energy and if that kinetic energy is not put all back the vehicle will slow down.
I won't claim to understand this well enough to make a definitive statement. But if you look at the original video's explanation all of the reference is to the similarity to tacking. Your argument that a sail cannot go faster than the wind is only true when the sail is going directly downwind. Light speedy sailboats can easily have a water speed much greater than the wind speed when tacking cross wind.
I believe that the key here is that the windspeed relative to the vehicle is not the pertinent answer. It is the wind direction in the coordinate frame of the propeller blade. I still haven't wrapped my head around all of this to be able to explain it more simply than others, but that line of thinking gives me a way to believe it is possible, and I choose to believe that the vehicle is not a scam, with hidden batteries and electric motors. With that assumption facts speak louder than any theory. Just like conventional aerodynamics had no explanation for how bees could fly for decades. Nevertheless they did and do. And finally in the 1980s as I recall it was finally figured out how it worked.
I have recommended and will recommend again a book written by Arthur C. Clarke, "Profiles of the Future". The book attempts to identify traps in predicting the future, methods for predicting it, and attempts to define the real limits on what is possible. In chapter one of the book he describes what he calls failure of nerve and gives numerous examples of highly trained people proving something was impractical or even impossible. Using well established facts and theorems and impeccable math. But applied in ways that turned out to be incorrect. Some of the examples: proof that heavier than air flight was impossible, then once that had happened, proof that airplanes could never carry multiple passengers or travel hundreds of miles per hour, proof that no man-made object could be sent to the moon and proof that electric lighting would never work for private houses. He calls it failure of nerve because in several cases the individuals providing these proofs had access to descriptions of how the problems they were analyzing could be avoided, but just couldn't bring themselves to believe that they really worked because they contradicted their sense of how the world worked. While the book was written in 1961-1962 it has stood the test of time well. It is well worth finding and reading. It is both cautionary and humbling for those who are assessing new ideas, and particularly worth paying attention if such assessments are part of your profession.
A more recent example of such failures of nerve are the many proofs that it is impossible to intercept a ballistic missile (hit a bullet with a bullet was the popular way of describing it). During the 1980s many prominent scientists outlined why it was "impossible". While it is still debatable whether such systems can be effective enough, cost effective, or politically wise, impossible is an adjective that definitely does not apply.
The propeller does not compress the air, it creates a vacuum. You can see that demonstrated here:
What you see there is a stream passing over a tube sucking the contents of the tube out. That stream might be very compressed indeed but it will still cause a vacuum in that tube so long as it's whizzing past the end.
When students arrive to study engineering, one of the first things that happens in the "Fluid Mechanics 101" class is to dispel all sorts of wrong intuition about the way flowing fluids behave. It takes various experiments to show what really happens, vs. what students think is going to happen.
What happens on both sides of the propeller is described in this diagram that I posted earlier
When students arrive to study engineering, one of the first things that happens in the "Fluid Mechanics 101" class is to dispel all sorts of wrong intuition about the way flowing fluids behave. It takes various experiments to show what really happens, vs. what students think is going to happen.
Unfortunately professors are not immune to making mistakes.
Unfortunately professors are not immune to making mistakes.
Where does that diagram come from?
But electrodacus is immune, right? ;)
No one tell this guy about heat pumps.:-DD
No one tell this guy about heat pumps.
Comes from a quick google search and I selected the one that looked to contains the info I wanted to show.
Seems photo is from a Wikipedia article on axial fans https://en.wikipedia.org/wiki/Axial_fan_design (https://en.wikipedia.org/wiki/Axial_fan_design)
You see blackbird and treadmill model exceed wind speed and you immediately think is wind that powers the vehicle (in a way it is but is stored wind energy so you will not stay above wind speed indefinitely).
OK, but I don't see how that refutes the experiment IanB showed. Pressure in a moving stream of air is anisotropic so the fact that one pressure vector is higher as shown doesn't mean that the orthogonal one won't be lower. And of course you have to consider how that operates on the propeller, which is another matter. (hint--it's pointing away!)
You keep saying this but a propeller doesn't work by building up a huge reservoir of pressure behind it and then having that pressure push it forward, unless perhaps you are taking off out of a small cave or some ridiculous contrived scenario. Propellers do something entirely different. The idea that you can store enough energy in a pressure field behind the vehicle in the open atmosphere with nothing confining it to keep the vehicle moving above wind speed for 'several minutes' seems absolutely ludicrous to me and you have presented no math, citations or cogent explanation to support that quite wild theory. If you suggested that energy was stored in the angular momentum of the propeller, perhaps I'd be willing to look at the math before dismissing that, but air pressure seems pretty far out there.
I mentioned my math proof many times before is about the power on motor not possible to be higher than on the generator since motor is supplied from the generator.
I know I'm bad at explaining things so maybe I can find someone that can do a better job explaining.
And round and round we go! I very carefully explained why the motor power does not need to be higher than the generator power and your limited answers demonstrated that you somehow don't believe in simple mechanics.
And there's a big clue for you. I saw a quote recently that seems appropriate--"whatever cannot be said clearly is probably not being thought clearly either" You may be entirely convinced of whatever it is you are convinced of for whatever reason, but your inability to demonstrate this to multiple people, all well capable of understanding the matter should serve as an indicator that perhaps you have made an error somewhere.
No one tell this guy about heat pumps.Heat pumps are not more than 100% efficient (not even close to that). The term used there is coefficient of performance not efficiency.
It is a pump so pumps heat from outside to inside or inside to outside depending on application heating or cooling.
In some sense the situation with the vehicle with the prop driven by the wheels similar: At first galne it may look like over unity efficiency, and thus may be mind-blowing to some. Hower it is not: the prob produces more forward force than the wheels need backwrard force to drive the prob. But this is just the force. Over unity gain for the force is nothing new, known from the lever and pully.
For the power / energy side it is not over unity: the wheels see a relatively high speed to the ground and thus generate high power from a given force, while the prop works relative to the wind. So at a speed of the vehicle a little faster than the wind the prop gives out litte power (energy), because the speed (relative to the wind) is low. So it is a bit like with a pully: high speed and low fore on one side and low speed and high force at the other side.
wind has now changed direction so it is counterproductive.
I'll leave the rest of your statement alone, but this is another thing you insist on that is simply wrong. The wind has not changed direction relative to the parts of the propeller that it is interacting with. That's the whole point of the propeller.
This power available from wind will continue to drop to zero as vehicle gets to wind speed and at this point there is no power from wind available to vehicle but vehicle still have a lot of stored energy (that pressure differential I was always talk about).
You overlook the fact that when the vehicle is moving at wind speed there is no direct impulse force available from the wind,
You overlook the fact that when the vehicle is moving at wind speed there is no direct impulse force available from the wind, but at this point the ground is moving very fast beneath the vehicle's wheels. There may be no force available from the wind, but there is significant power available from the ground as it turns the wheels of the vehicle. By capturing this wheel power and turning it into backward thrust the vehicle is able to continue moving ahead faster than the wind.
No comparison can be done between a heat pump and what happens here.
This is the main reason I asked you to test the simple wheel only vehicle in my diagram the one with generator wheel on a fast moving treadmill and the motor wheel on a fixed box on the ground (box is there so that vehicle is level with the treadmill). Then you can see the vehicle can only move backwards from right to left no matter what sort of gear ration you want to setup between G wheel and M wheel.
No comparison can be done between a heat pump and what happens here.
There can if I chose to.
I think we have all looked at that diagram, and I think we can all see that the vehicle can easily be made to move forwards from left to right.
The reason for this is, as has already been explained to you, power does not have a direction. It is a scalar quantity. Therefore, if the vehicle can be made to move at all by picking up power from the treadmill wheel, it can be made to move in whatever direction we like.
You (all) should prove that to me either with a test or with math. You will not be able to do that since it is not possible.
Let me ask you a question:
Can I make a wheeled vehicle that when facing head on into the breeze from a fan, will use the wind energy to drive directly towards the fan?
Yes a vehicle can drive against wind as it has access to wind power
Yes a vehicle can drive against wind as it has access to wind power
In that case your treadmill cart can drive against the treadmill as it has access to the treadmill power.
(motor wheels in from of the generator wheels while generator wheels are still on the treadmill)
Quote(motor wheels in from of the generator wheels while generator wheels are still on the treadmill)
Just to be clear, and remind me of what this is about...
In your diagram the treadmill is pushing the G wheel, yes? Absent the drive belt the G wheel will thus rotate and the vehicle will remain static (more or less - we are ignoring friction effects for the moment). With the treadmill turning the G wheel, and still absent the drive belt, we should be able to place a hand or other attachment on the vehicle and move it left or right without apparent bias or effort, right?
At this point I think you need to either agree with the above or state explicitly where it is wrong.
Now, the M wheel has been free to turn, the vehicle has been free to move and the G wheel is rotating clockwise. If we add the drive belt, the difference is that now the M wheel is rotating clockwise. Since it has a grip on the platform - just like the treadmill has a grip on the G wheel - the vehicle will move right. There is nothing to stop it moving right since we agreed above that the G wheel being rotated by the treadmill is not imparting motion, merely causing the wheel to turn. So the vehicle moves right and the G wheel rotates faster, which imparts faster rotation of the M wheel which causes the vehicle to move faster to the right.
Of course, it won't be quite like that in practice because there will be friction resisting the turning of the G wheel, which will impart a small movement of the vehicle to the left. However, I think that should be ignored since otherwise you could add X amount of friction sufficient to stop the wheel turning at all and prove nothing except the experiment is testing the wrong thing.
Which bit did I get wrong?
Now if you connect the wheel G and M with an ideal belt and gear ratio is 1:1 then vehicle will still be standing in the same place (no movement in any direction).
Propeller is connected to vehicle so they will travel together.
Now if you connect the wheel G and M with an ideal belt and gear ratio is 1:1 then vehicle will still be standing in the same place (no movement in any direction).
Let's observe that if nothing slips, then this situation is impossible.
If wheel G is in contact with the belt and the belt is moving, then wheel G is turning clockwise. If there is a 1:1 belt between wheel G and wheel M, then M is turning clockwise at the same speed as wheel G. If wheel M is turning clockwise, and wheel M is in contact with a fixed platform, then wheel M is moving from left to right. If wheel M is moving from left to right, then the whole cart is moving from left to right.
With 1:1 gear ratio and no slip is allowed the treadmill will be locked so there will be no movement.Indeed. But if there is a generator to motor connection instead of a 1:1 belt, then there can be torque conversion and differential wheel speeds. In this case the system will not be locked.
The treadmill will try to push the vehicle back while the back wheel will try to move the vehicle forward so there will be equal but opposite power acting on the belt. In real world the M wheel will slip unless friction between wheel and red box is higher than treadmill power so assuming treadmill is not overpowered the vehicle will move backwards (right to left).Which wheel slips depends on the coefficient of friction and the weight above the wheel. If there is a heavier weight on M wheel, then maybe the G wheel will slip and the M wheel will grip. Or maybe both wheels will slip a bit. In any given case, you cannot state what will happen without defining further details.
Now if you connect the wheel G and M with an ideal belt and gear ratio is 1:1
the treadmill will be locked
Quotethe treadmill will be locked
What is locking it? It would only be locked if the vehicle is prevented from moving.
We've agreed that without the belt the G wheel is free to rotate and hence doesn't move the vehicle, so with the belt, the only thing working against the G wheel is the M wheel.
The only thing preventing the M wheel from turning is the vehicle, and since the vehicle is free to move the M wheel will move the vehicle.
the G wheel is breaking
Quotethe G wheel is breaking
OK, to be explicit the G wheel is being braked by the mass of the vehicle acting through the M wheel, yes? In other words, if the M wheel had no grip then the G wheel would again be free to rotate and the vehicle wouldn't move either way?
You can compare forces in this particular case as gear ratio is 1:1
assuming the M wheel is the one that slips
QuoteYou can compare forces in this particular case as gear ratio is 1:1
Why does it have to be 1:1? Suppose it is 2:1 (or 1:2) - doesn't that make a difference?Quoteassuming the M wheel is the one that slips
Why do we assume that? Let's instead assume the M wheel is grippier than the G wheel.
Is this all a Monty Python skit? :-DD
It can be any gear ratio the result will be the same as far as vehicle direction of travel.
2:1 will mean G wheel turns 2x in the same time M will turn only once but since also the force on G is half of that on M (again ideal case) Power at M and G wheel will be the same just opposite direction so vehicle can not move as with no delta in power you can not change the vehicle kinetic energy.
0.5:1 (1:2) means M wheel can spin 2x faster than G wheel but then force at M is half that at G thus again Power applied by both wheels is equal and so no change in vehicle kinetic energy thus vehicle id not moved.
All of the above is based on ideal case in real case power at wheel M will always be slightly lower than at wheel G due to friction and so vehicle will move from right to left if M wheel slips else is G wheel slips then vehicle will stay where it is and energy balance will all be wasted as heat at the interface between the treadmill surface and the wheel.
So if wheel M has perfect grip vehicle remains stationary while energy delta will end up as heat at the G wheel interface with treadmill moving surface.
Just make a test.
"Braking", not "breaking". Please. I beg you. Make it stop.
.... and for Pete's sake, can we please abandon these absurd, asinine, daft, foolish, futile, harebrained, idiotic, illogical, imbecilic, laughable, meaningless, mindless, pointless, puerile, ridiculous, senseless, silly and worthless discussions about two wheels and motor/generator function on the vehicle. There is absolutely NO relevance between these discussions and the mechanism which operates between a wheel and a propeller.
The propeller is essential to the mechanism and ANY discussion which excludes it is simply a waste of time.
Sheesh :palm:
That grates on me also ... but that is the least of the transgressions.
The power is the same (actually opposite sign) on both sides only in the ideal stationary case. In the ideal case there is no extra power needed to change the kinetic energy as the speeds are constant. In the non ideal case the power on both side does not have to be the same. If needed to overcome friction or to accelerate the vehicle the power on both sides can be different (e.g. more power going in to accelecate, or more power coming out to slow down). The system has plenty of power (force exchanged through the vehicles times paltform speed) - so there is no real arguing about not enough power to increase the kinetic energy. The vehicle could move and still produce extra power, e.g. for lights.
---------------------------
To calculate the stationary speed we don't even need to look at forces or power, just look at the velocities. The calculation was shown in a similar way before and is still correct:
The gear ratio (k) connects the speeds of the two wheels and thus the speed of the vehicle relative to the 2 platforms (V1 and V2). So as an equation V2 = k * V1. The 2 plattforms move relative to each other at a speed V0. So V2 = V1 + V0.
Do the math with the 2 equations, and one gets V1 = V0 / (k-1). This gives a valid solution for all gear ratios except k = 1. So the 1:1 gear ratio does not work as we have seen before.
With a suitable value for the gear ratio you can have V1 at nearly any speed you want: both directions ( sign of V1) and also faster or slower than the relative morement of the platforms. In a reall life situation very high speed ratios may not work well due to friction, but gear rations of -1, 0.5, 1.5 or 3 are not a problem. This would give you half and twice the relative speed of the platforms in both directions.
The equations show the the vehicle could move at the calculated speed to avoid slip or tearing the vehicle appart. The platforms will provide the necessary forces and power to make the vehicle move this way, as this is the path of least restance (only friction).
There is absolutely NO relevance between these discussions and the mechanism which operates between a wheel and a propeller.
The propeller is essential to the mechanism and ANY discussion which excludes it is simply a waste of time.
Sheesh :palm:
.
.
As mentioned many time before. If you are not convinced by my correct calculations you just need to test. And I promise if you (any of you) can demonstrate vehicle moving from left to right I will pay you back for the experiment.
OK, so it is a Monty Python skit....We should be so lucky.
I think we should all get together and make a video.Veritasium already did a vid... ..... ..... Oh - you mean that sort of video. Hmmmm...
Can't say I agree with that.
The propeller, IMO, is a red herring.
Although it is a possible source for misunderstanding the machine,I would change "possible" to "certain"
the basic misconceptions here go way beyond propeller-related issues. Simplifying the the problem to take out some of the variables is usually good way to work past such errors.I quite agree!
See my reply #20. I would have hoped that once it was understood that a vehicle driven by an overhead belt could be made to go either direction at an arbitrary speed, subbing the propeller and wind back in for the overhead wheel and belt would have made it all easier to understand. Unfortunately this does not appear to have worked all that well.Personally, I found your variation more confusing - but that may be because you moved to a somewhat different framework. Kudos for making the effort and being creative.
But what can you do with someone who claims an advancedAgree completely on this question - and I am struggling to come up with any suggestions to answer it.knowledgeintuition about physics but fails to understand Archimedes, Newton, levers, gears and propellers and tries to applies conservation of energy principles to a system that is not closed?
Yes propeller is essential for energy storage in air.
And say breaking power is 100W say that is 10N at 10m/s
Yes propeller is essential for energy storage in air.
Serious question: How do you store energy in air which is not contained in any way?
If you are not convinced by my correct calculations
The pole you are getting up on is getting very high, remember the distance to fall is also getting longer all the time.....
Serious question: How do you store energy in air which is not contained in any way?
Your previous hand-waving and "intuition" has not been informative, let alone convincing.
Can you cite any examples
All experiments you seen before has the vehicle flipped and that changes everything
Air compresses, so although the storage is leaky (and then some) isn't there some brief storage?
Musical wind instrument?
You keep saying that. Why does it change everything? And in what way?
I posted before the diagram from Wikipedia that shows the pressure differential for an axial fan. You an see P1 and P2 is much higher than PA (ambient pressure)
And I linked you to a video of a real experiment which shows the pressure in front of a fan is lower than the ambient pressure. Which means that diagram does not show what you think it shows.
Why do you keep asking us to do experiments, but when I show you an example of an experiment you refuse to believe it?
You have no idea how to interpret that experiment and neither the person that made the video.
It does not need to be interpreted to observe the simple fact that it demonstrates, which is that the pressure in front of the fan is lower than the ambient pressure.
The fact is that fans and propellers do not compress air. What they do is accelerate the air passing through them. This acceleration generates a reaction force that drives the propeller forwards.
Unfortunately, if we show you the experiment where the cart in your diagram moves from left to right, you will dispute that too, and will claim that experiment also is being misinterpreted. So it's not worth our time showing you results that you will refuse to believe.
It does not need to be interpreted to observe the simple fact that it demonstrates, which is that the pressure in front of the fan is lower than the ambient pressure.
I will like to show the below diagram but hopefully that will not create even more confusion for you.
Oh, absolutely it is showing what I think it is.QuoteIt does not need to be interpreted to observe the simple fact that it demonstrates, which is that the pressure in front of the fan is lower than the ambient pressure.
Once again, I don't think it is showing what you (and the chap in the video) think it is. Blow across a pipe and it will suck the air out.
Last time I raised this issue you didn't catch on for whatever reason, so here's a video I just ran off to demonstrate:I did catch on, and I gave the explanation.
That's an ally pipe with a plastic bag cellotaped on the end. The vacuum cleaner nozzle is actually the output of my vacuum, so blowing instead of sucking.That's excellent that you did the experiment. Far better than drawing pictures and just imagining what will happen.
(FYI it's a DeWalt DC500 so not particularly strong, but I think we can agree that it is higher pressure than ambient and absolutely not a vacuum.)No, I can't agree there, as you will see. It is actually a vacuum.
As you can see, blowing air across the pipe sucks air out of the bag, giving the exact same effect as the chap in your video.Yes, I can see what you see, and here is the explanation.
How does he, and you, know that he is not recreating this situation?
But also, the low pressure air inside the canister is the same air the the blower is ejecting. This is partly why the stream of air coming out is at lower pressure than the surrounding air.
Yes, most likely. The air stream can only increase in pressure as it slows down and mixes with the surrounding air, so the pressure inside the nozzle will be at its lowest point. In fluid mechanics, the place where the flow is fastest and the stream is narrowest is called the vena contracta. This is also the point where the pressure is lowest.But also, the low pressure air inside the canister is the same air the the blower is ejecting. This is partly why the stream of air coming out is at lower pressure than the surrounding air.
Are you saying the air within the nozzle, before it gets out, is below ambient atmospheric pressure?
And does it follow that a vacuum cleaner is incapable of generating a pressure from it's nozzle when you connect to the out port?No, it does not follow. A flowing fluid obeys the energy conservation law, and each element of fluid has two components to its energy: its pressure energy and its kinetic energy. The fast flowing fluid has higher kinetic energy, which is why the pressure decreases as it goes faster (the total energy must remain the same). However, if you capture that fast flowing fluid inside a container, some of the kinetic energy can be converted back to pressure energy, and this will cause the pressure to rise. Since the vacuum blower has given a lot of kinetic energy to the fast flowing air, if you slow down that air you can get back a higher pressure than it started with. This is the principle of operation of centrifugal compressors.
(the total energy must remain the same).
This is the principle of operation of centrifugal compressors.
Well it has! What's the blue wheel, the green thing, etc? Is the green thing touching the blue wheel? What do the arrows mean? It would be less confusing if you told us what the things represent instead of leaving us to guess (and then tell us we're wrong about our conclusions based on those quesses) :-//
It does not need to be interpreted to observe the simple fact that it demonstrates, which is that the pressure in front of the fan is lower than the ambient pressure.
The fact is that fans and propellers do not compress air. What they do is accelerate the air passing through them. This acceleration generates a reaction force that drives the propeller forwards.
Unfortunately, if we show you the experiment where the cart in your diagram moves from left to right, you will dispute that too, and will claim that experiment also is being misinterpreted. So it's not worth our time showing you results that you will refuse to believe.
I almost feel like I'm talking with flat earthers and so nothing I can say will convince you. The best I can suggest is you do the experiment and see that what happens is exactly what I predict.
So have a vehicle as shown in my diagram (not flipped) and observe that whatever you do in therms of gear ratio the vehicle will move from right to left.
The flipped vehicle is completely different due to the way the forces act on the vehicle are different and that vehicle can travel in both directions based on gear ratio. That flipped vehicle represents the analog of the propeller vehicle where propeller is the generator (not a fan) and so it drives against the wind direction.
What you can observe on that one if you want is that it can not exceed the wind speed meaning vehicle relative to stationary ground can not travel faster than the speed of the treadmill.
So the vehicle in my diagram is very different from the vehicle that was demonstrated where vehicle was flipped.
I will like to show the below diagram but hopefully that will not create even more confusion for you. That shows what happens when you flip a vehicle and top of the wheel will touch the surface instead of the bottom of the wheel.
This diagram was to explain Derek what his wheel vehicle with the floor and lumber represents.
(http://electrodacus.com/temp/Wind3.png)
In any case just build the vehicle in my original diagram and experience the fact that vehicle will always move from right to left. This will be analog to getting a flat earther in to space to see for himself that earth is an approximately spherical object.
| (https://www.eevblog.com/forum/chat/mess-with-your-minds-a-wind-powered-craft-going-faster-than-a-tail-wind-speed/?action=dlattach;attach=1258408;image) | :palm: |
Here you go. The setup is a little janky, but clearly demonstrates the effect. The right wheel rotates ~2.5 times for every rotation of the left wheel. I don't have a treadmill so I used a piece of paper which I slide back and forth under the right wheel to push on it. I can push with the paper "treadmill" in either direction and the cart moves in the opposite direction: towards the application of force. This is due to the gear ratio, and the formula which I posted above that you ignored and claimed had to be wrong without describing why. No slippage is occurring on the wheels or the belt. It works at very low speeds or moderately fast speeds, if I go faster it obviously starts to slip.
Apparently the eevblog forum won't allow video embeds regardless of size, so sorry about the imgur link:
https://imgur.com/a/lTqAFg6
We are still talking about "motor wheels" and "generator wheels"....??
Nomenclature is not the problem. It's the use of two 'sets' of wheels - it is completely invalid for the mechanism being debated.
Unfortunately, if we show you the experiment where the cart in your diagram moves from left to right, you will dispute that too, and will claim that experiment also is being misinterpreted. So it's not worth our time showing you results that you will refuse to believe.
Here you go.
Sorry a piece of paper is not a treadmill but you can build a simple treadmill made of paper.
Sorry a piece of paper is not a treadmill but you can build a simple treadmill made of paper. The paper will need to be glued in a loop and then use two cylinders with axis fixed to the table.
The reason the piece of paper is not a treadmill is because the paper moves relative to the ground/table where a treadmill will not do that. It is like powering OFF a treadmill and push it under the vehicle.
The paper will not push against the generator wheels it will just slide under
Your wheels argument does not reflect the real world environment governing the blackbird AT ALL.
The vehicle doesn't care care what is going on with the rest of the treadmill except at the point of contact. The paper is moving exactly the same way as the top surface of a treadmill would. Whether the end returns to the beginning in a loop or whether the paper is supported by another object that doesn't move is completely irrelevant.
Simplifying the the problem to take out some of the variables is usually good way to work past such errors.
But two 'sets' of wheels is not an appropriate analogue.
Excluding the propeller and it's interaction with the air is as relevant to this discussion as me saying I want to test a space suit by jumping into a pool with it. Whatever else it might simulate, such an exercise does not reflect the real world environment.
Your wheels argument does not reflect the real world environment governing the blackbird AT ALL.
Continuing to do so simple underlines your ignorance, stubbornness and, dare I say (what is increasingly difficult to dismiss) your delusion.
Your wheels argument does not reflect the real world environment governing the blackbird AT ALL.
Wait a second! The wheeled models are demonstrating a principle, they don't have to cover all aspects of the real world operation of the original vehicle in question. The fact that we can't agree on the basic principle of how forces can be made to operate given two mediums/objects/surfaces/whatever moving relative to one another makes it pointless to continue on further to discuss the full vehicle. Sort of like if we can't agree on how levers work, there's no point in discussing gears.
OK.
Since you claim to have all the answers with the physics knowledge and maths to support your conclusions, I put this challenge to you....
Take all reasonable references from the blackbird example and generate the maths to describe its operation. Once you have done that, the run time limitation of such an experiment - which you claim will occur - should naturally fall out of those equations. Please tell us what that time limitation will be, so that it can be validated experimentally.
The onus on doing this falls on YOU - not everybody else here - as
(1) You are the one claiming what was demonstrated in the Veritasium video is incomplete
and
(2) Those of us who understand the principle as it was presented agree that the demonstration proves the point.
You need to prove your point.
Or, you can pull a number of of thin air - just make sure the error bar is included.
For example if you say "5 minutes" and the experiment runs for 10 minutes at higher than wind speed, then don't turn around and say - "Oh, it must be longer".
Pick a number and stick to it.
To disprove the current explanation all I need is a single phrase.
A vehicle power only by the wind driving in the same direction as the wind will be 100% efficient when it is at same speed as the wind (possible only in theory) and from this since no claim can be made that above 100% efficient device was ever proven higher than wind speed is impossible if powered only the the wind and no energy storage.
To disprove the current explanation all I need is a single phrase.This is not true, as going at 100% the wind speed with a sail is not 100% efficient, as least not as energy efficency. You actually use no wind-engy at all. So at best you could call it a zero divide by zero to get 100%, but math does not approve this.
A vehicle power only by the wind driving in the same direction as the wind will be 100% efficient when it is at same speed as the wind (possible only in theory) and from this since no claim can be made that above 100% efficient device was ever proven higher than wind speed is impossible if powered only the the wind and no energy storage.
Seems super intuitive for me as no air particle can get to a vehicle that travels at or above wind speed in same direction so is clear wind can not power this.The effect is a bit complicated, so one should be care ful with the intuition - this is why the thread title is "Mess with your minds: ..", as this is an effect that may conflict with simple intuition.
The explanation to get around this was that vehicle can extract energy from the difference between the two mediums that is ground and air but that is disprove by my simple wheels only diagram as there treadmill has a different speed than ground and vehicle can not use that energy to move forward from left to right.
Seems super intuitive for meIn other words, you've already made up your mind.
as no air particle can get to a vehicle that travels at or above wind speed in same direction so is clear wind can not power this.It's not just wind. It's wind plus something else.
The explanation to get around this was that vehicle can extract energy from the difference between the two mediums that is ground and airCareful ... you're almost getting it!
but that is disprove by my simple wheels only diagramUtter rubbish. Your "wheels only diagram" is not relevant - never has been - but as you continue to insist it is, you are condemned to never understand correctly.
The interesting thing about the treadmill experiment is that they missed an opportunity to show something more. They let the model vehicle fly forwards and drive off the end of the treadmill. But if they had simply anchored it in place with some thread, they could have shown it pulling against the thread indefinitely while the treadmill was running.Now THAT is a brilliant idea. Put a tension measurement device on that thread and you could generate data of tension vs speed with alternative gearing and propeller parameters.
Quote
As you can see, blowing air across the pipe sucks air out of the bag, giving the exact same effect as the chap in your video.
How does he, and you, know that he is not recreating this situation?
Yes, I can see what you see, and here is the explanation.
Firstly, there is no such thing as "suction". Therefore, the pipe cannot suck air out of the bag. When we talk about suction this is just loose and imprecise terminology.
Gases (like air) consist of particles that are all pushing each other apart and trying to spread out. In the absence of overriding gravitational or accelerating fields, gases will spread out to evenly fill any container. The gas pushing against the walls of the container manifests as pressure, which can be measured. There is no such thing as "suction" because gases are pushing apart and trying to spread out, not pulling together. To have "suction" requires a "pull", which gases do not have.
The fact you may see a huge air stream flowing into the intake part of the propeller
The interesting thing about the treadmill experiment is that they missed an opportunity to show something more. They let the model vehicle fly forwards and drive off the end of the treadmill. But if they had simply anchored it in place with some thread, they could have shown it pulling against the thread indefinitely while the treadmill was running.Now THAT is a brilliant idea. Put a tension measurement device on that thread and you could generate data of tension vs speed with alternative gearing and propeller parameters.
You could chart the results and get a very good visual understanding of the behaviour.
Now imagine if Blackbird plus driver weighs 200 kg, and is travelling at 11 m/s, with a 10 m/s tailwind, both with respect to the ground, so Blackbird's speed in the frame of reference of the air is 1 m/s. With respect to the ground, Blackbird has a kinetic energy of 12,100 joules (mv2/2). Suppose that the wheels now extract 1,000 joules of this energy, reducing the kinetic energy (with respect to the ground) to 11,100 joules and therefore the ground speed to 10.54 m/s. With respect to the air, Blackbird now has a speed of 0.54 m/s and a therefore a kinetic energy of 29 joules. Suppose that the propeller, after losses, manages to add 700 joules to this kinetic energy, increasing it to 729 joules. The speed with respect to the air will now be 2.7 m/s, so the speed with respect to the ground will be 12.7 m/s. This is greater than the starting figure of 11 m/s. So Blackbird can indeed accelerate beyond wind speed without breaking any laws of physics or requiring energy storage.
Quoteas no air particle can get to a vehicle that travels at or above wind speed in same direction so is clear wind can not power this.It's not just wind. It's wind plus something else.
Quotebut that is disprove by my simple wheels only diagramUtter rubbish. Your "wheels only diagram" is not relevant - never has been - but as you continue to insist it is, you are condemned to never understand correctly.
So your mistake is that you decided the entire vehicle moved in to air same as the entire vehicle will have moved on to that other lane.
This is not true, as going at 100% the wind speed with a sail is not 100% efficient, as least not as energy efficency. You actually use no wind-engy at all. So at best you could call it a zero divide by zero to get 100%, but math does not approve this.
And your mistake, among many many others, is that you don't see that this distinction doesn't matter. The vehicle is simultaneously rolling on the road and 'in the air'. You made a similar error when you claim that the surface of a treadmill is somehow different than the surface of a moving piece of paper. It's ridiculous and I don't think there's much point in discussing it further when your intuition misguides you so badly and you refuse to actually use math in a meaningful way.
Yes please test a treadmill to see the difference from moving piece of paper. The difference is huge you just can not see that without doing the experiment.
Yes please test a treadmill to see the difference from moving piece of paper. The difference is huge you just can not see that without doing the experiment.
What is your theory or explanation as to why they are different?
The treadmill belt moves relative to ground, what on Earth you smoking? BC bud?
The part of the treadmill that stays fixed to the ground is equivalent to the guy's feet in the paper sheet video. :-//
I probably mentioned this before so if you can not see the difference between them you will just need to make a test. The treadmill can be just a simple paper loop rolling on two cylinders.
Is like having a powered treadmill vs one that is not powered but instead pushed towards the vehicle.
I know it may look non intuitive to you but please think more about this or do a test.
I know it may look non intuitive to you but please think more about this or do a test.
It's not about intuition, it's about logic and math. In each case the two surfaces are moving at the same rate in the same direction. What other characteristic could that surface have that would change how it interacts with the car?
The paper moves relative to the table while the treadmill is not moving relative to the table.
I am but a humble amputated brain ape, but what on Earth do you mean by "The paper moves relative to the table while the treadmill is not moving relative to the table"?
Obviously the paper is the belt in the treadmill? Would you prefer there to be another sheet of paper between the table and the first sheet, moving in the opposite direction?
Now, I accept that the terms being used are incorrect, in that 'suction' is perhaps confusing in a literal sense. However, from my point of view there is high pressure across the top of that pipe - if a sheet of something was placed there, at right angles to the flow, it would be pushed away from the hose outlet. Now, you might say that this is merely air flowing, but if it is flowing then it is being pushed against something. Either the sheet of whatever or just ambient air - there is a restriction against which the flowing air is being pushed, and that must surely be increasing the pressure. A fan is taking air from one side and pushing it to the other despite there already being air there, so the pressure must increase.There is a directional component when you measure the pressure of a flowing air stream. If you measure the pressure head-on, you will see a higher pressure than if you measure the pressure at right angles. The flowing air has mass and momentum, so when you measure the pressure head-on it "pushes" against the measurement device and increases the effective pressure. The following Wikipedia article talks about this:
The way I understood there being an area of low pressure, which the air in the pipe will flow to fill, is because the air moving across the end of the pipe drags molecules of air at the pipe entrance along with the flow.This is a real effect, but it is relevant only in special kinds of vacuum pump. It is not applicable here.
So I am suggesting that the chap is measuring an artifact caused by the act of measuring. Instead of the pipe, what if he put, say, a flap of paper in the airstream. Would that bend toward the blades? I doubt it. But if he placed the paper edge-on, would it be pulled into the stream? It likely would (if it could retain it's shape) but only because the surface of the paper is causing the low pressure that pulls it in.The paper experiment is a good one to do. Get two sheets of paper and hold them in a V shape around the air jet from the vacuum nozzle (blow the air into the V shape between the parallel sheets of paper). You will see the two sheets of paper get squeezed tightly together. The harder you blow, the more tightly the paper will get squeezed. This is another way of demonstrating there is a vacuum in the space between the two sheets of paper. The dragging of air molecules cannot apply here, as the paper does not have any free molecules to get dragged.
I'm sorry I do not have the ability to properly explain this but that dragged paper will not have the same effect on vehicle as a treadmill.
How does the wheel know the difference in your diagram that clearly shows the arrow on the belt? How is it different if it's just the paper that moves and the person pulling the paper is the "mill"?
Seems like a simple question. I don't care if you don't have the ability to "properly" explain it, just try improperly explaining.
Was this wheel instead of treadmill more helpful ?
Was this wheel instead of treadmill more helpful ?
Not really. Please address my 5-step question. You don't have explain why, just at what stage in those 5 steps it goes from working as we say it does to working the way you say.
For the treadmill version in order for vehicle to advance from left to right the G wheel will need to spin clockwise faster than the treadmill belt but that is impossible since the G wheel is not powered by anything other than the treadmill belt and in order to get power from the G wheel you will need to break the G wheel so vehicle will be moved backwards more than M wheel can push back.The red parts are wrong, get the 2 points right, and you should be allright.
The red parts are wrong, get the 2 points right, and you should be allright.
1) When the vehicle moves left to right, the wheel will spin faster (unless it slips). One does not have to power the wheel, it is enough to move the vehicle. After all this is the G wheel and not the motor. So the generator does not have to more the vehicle, it just follows where the motor says is should go.
2) the M wheel can more the vehicle to the right, if it can provide enough force to do so. The force needed is the breaking force of the motor. This force not power power that needs to be overcome. The lower the speed of the vehicle relative to the fixed frame, the more force the motor can provide from a given power.
The final point is, that for the case that the motor drives to the right and the generator brakes, the power available to the generator is higher than the power needed by the motor. If you don't believe: take the simple example of the motor in stop (no power needed) and the generator is still driven by the treadmill speed. So there is plenty of power to spare and slowly move to the right.
Do the math and you see that it even works at higher speed if the efficiency is good (e.g. > 50%).
The vehicle will only see the upper side of the belt of the treadmill and it can not tell appart if the is a belt, or a piece of paper or rubber pulled by a human.That is your wrong intuition that they will be the same but they are not even close.
It is just stupid to insist on a difference between the paper and belt of the treadmill. For the sake of the discussion they are the same: a surface that is moved to the left.
for your response of the 2 simple points:
1) The vehicle moving to the right is the assumption you introduce in the sentence. So for the point 1) you assume it moves to the right - no matter what.
If the vehicle moves to the right, the wheel has to rotate faster - that is the geometriy and condition for the wheel no to slip. That is kind of the definition of a wheel. The treadmil and the motor combined drive the wheel - some of the power comes from the treadmill, that in the simple picture has plenty of power. Using the conservation of energy is a tricky argument in this comtext, as the treadmill constantly provides power. The conservation of energy does not prohibit the movement to the right - that is kind of handled in the 2.nd point.
2) The motor does not have the same power as the generator to more the vehicle. It only has to have the same (or more if there is friction) force.
The force is power divided by speed. So if you keep the speed low enough there is always enough force.
(knowing the difference between power and force is really important here)
Gear ratio is another way to express the speed of the motor. The gear ratio of 1:1 is the critical case that does not work - so don't use it. But a gear ration so that the motor side is rotating slower (e.g. half the speed) does, and it provides enough force to do this against the generator pulling in the other direction.
If the final point works for the dragged paper, it also work for the treadmil, as the 2 cases are exactly the same. How should the vehicle tell the difference ?
Are you telling me a person standing on a stopped treadmill is going to have to run in another direction when the treadmill starts, than someone who has the rug under them pulled in the same direction?
A person on a treadmill is not between two separate mediums at the same time unless it has a foot on treadmill and one on the ground.
A person on a treadmill is not between two separate mediums at the same time unless it has a foot on treadmill and one on the ground.
That's not what I asked. Which way will you fall or run if a treadmill starts, or a carpet is pulled under you? Then we can add your separate mediums and being between dimensions or whatever.
You're in Time Cube territory now. You're not even making sense in your own problem space.
Which way will you run when a treadmill starts? Which way will you run if someone pulls a rug from under you in the same direction?
Stop avoiding the question, stop making up weird scenarios, stop prevaricating.
Stick to the question as presented.
The rug is pulled at constant speed and I'm siting on it and I can just relax. On a treadmill at constant speed I still need to burn plenty of calories.Here is someone standing on a treadmill, feeling very relaxed, and not burning any calories:
There are no weird scenarios is just youThere, fixed that for you.(most of you)(all of us) not able to understand there is a big difference between a treadmill and that rug.
The rug is pulled at constant speed and I'm siting on it and I can just relax. On a treadmill at constant speed I still need to burn plenty of calories.Here is someone standing on a treadmill, feeling very relaxed, and not burning any calories:
https://youtu.be/GvfF4TeXz7UQuoteThere are no weird scenarios is just youThere, fixed that for you.(most of you)(all of us) not able to understand there is a big difference between a treadmill and that rug.
The way peoples brain work is very different likely way more different than you imagine. You should search about Aphantaisia that affects maybe 1 to 3% of the population the others have a spectrum of capabilities in creating a mental image. And you should also look-up "Internal monologue" where an estimated 30% of the population has no internal monologue. You will be blown away.
I can create a mental image (sort of average) and I sure have an internal monologue so I think more abstract using language.
I can simulate fairly accurately in my head how a simple cart like those in my diagram will work including what wheel is the generator and witch is the motor and I can know the outcome.
I can also imagine air as a multitude of small particles moving at a relatively constant speed above the ground and that have elastic forces keeping them apart like repelling magnets and then imagine a vehicle driving in the same direction at higher speed than this air molecules and can see that air molecules can no longer help accelerate the vehicle (impossible) but I can also imagine this now higher pressure behind the vehicle meaning higher density (more air particle in the same volume now with higher forces keeping them apart same as force increases when you try to bring two opposing magnets closer).
This is the so called stored energy and as vehicle continues to accelerate this pressure will reduce meaning the density of air molecules drops up to the point that there is not enough pressure to cover the vehicle losses and so vehicle will start to decelerate until it will get below wind speed.
If this description created for you a mental image (visualizing this moving animation) then you will also know vehicle can not be powered by wind when vehicle speed is higher than wind speed in the same direction.
With this mental model of air, what you have been saying about energy storage, and other characteristics of gases, would make some sense.
Unfortunately this is a very poor model for air, or any other gas. For example, the particles in your model appear to only be moving at the speed of the wind, so it is very easy for a vehicle to outrun them. In reality, the molecules of nitrogen, oxygen and other gases that make up the air, are moving randomly and at speeds comparable to and often higher than the speed of sound. There are also no significant forces between the molecules keeping them apart, except briefly when they collide with one another or with other objects.
A far better model is that used in the kinetic theory of gases. The Wikipedia page on this subject includes an animation that may help you to form a much more useful mental image. https://en.wikipedia.org/wiki/Kinetic_theory_of_gases Once you have moved over to using that model, a lot of the things that others on this thread have said should begin to make sense.
I think that your other sources of disagreement with the rest of us may be due to poorly defined frames of reference. In particular, the wrong choice of reference frame can obscure the simple relationships between force, distance and energy or work done. I tried to explore this earlier with my thought experiment with a passenger walking down the aisle of an aircraft.
I have aphantasia myself, so I do have some sympathy with you. I can't form any kind of internal visual images, but I can form what I can best describe as invisible wire-frame images, that I can freely manipulate and rotate. I can't see them in any visual sense, but I can sense them.
No when there is wind the particles move in the same chaotic way but on top of that there is a net movement in a particular direction and that will be called wind.That is right, but the wind speed in all the circumstances that we have been discussing is a tiny fraction of the average individual speeds of the air molecules.
When I think in my head I imagine the vehicle still moving and the road is what is stationary so I do not modify the frame of reference.That is fine when considering the interaction between the wheels and the stationary platform, but when considering the interaction between the other wheels and the treadmill, the stationary frame of reference is not the best choice.
That is right, but the wind speed in all the circumstances that we have been discussing is a tiny fraction of the average individual speeds of the air molecules.
That is fine when considering the interaction between the wheels and the stationary platform, but when considering the interaction between the other wheels and the treadmill, the stationary frame of reference is not the best choice.
However you can still do it, as long as you remember to include the work done by the treadmill on the vehicle, as well as the work done by the vehicle on the treadmill. What you seem to keep doing is forgetting the work done by the treadmill, and then concluding that conservation of energy prevents the vehicle from moving to the right.
Going back to my aircraft thought experiment. The aircraft is travelling at 200 m/s, I get out my seat and walk forwards at 1 m/s, I weigh 70 kg. How much energy did I have to expend to accelerate from 200 m/s to 201 m/s in the frame of reference of the ground? From the ground frame of reference my kinetic energy has increased from 1,400,000 joules to 1,414,035 joules. A difference of 14,035 joules.
Any experiment where my diagram is respected will show that vehicle can not move from left to right. There is not even one experiment showing that.
What do you mean by aircraft ? For me an aircraft will be something that travels in air not on ground so something like an airplane.
I can not imagine you getting out of an aircraft and then walking.
But say you are referring to an airplane driving very close to ground and then you get out of that (it will be very painful) as that 200m/s speed of aircraft is relative to stationary ground so if you get out of the airplane as soon as you touch the ground all you kinetic energy will be reduced close to zero in seconds so you will likely not survive let alone be able to walk.
Have you ever got down from a moving vehicle like maybe a train ? It will need to move very slowly so that you are able to absorb all that change in kinetic energy else you will get hurt.
There are at least two, and you've posted one yourself. I doubt if there will be another because it's obvious you will find some issue that allegedly negates the experiment.
How about you show the experiment that proves your hypothesis?
My apologies, but I thought it was obvious. I am a passenger in a normal aeroplane flying at normal cruising altitude at 200 m/s. I stand up from my seat and walk down the aisle of the aeroplane towards the front of the aeroplane at 1 m/s, relative to the aeroplane. I weigh 70 kg.
To an observer on the ground (using the ground frame of reference), by getting up and walking along the aisle, I have accelerated from 200 m/s to 201 m/s. To this observer on the ground, I have increased my kinetic energy by 14,035 joules. Where did this energy come from?
That is because you do not understand the difference between my diagram and a flipped vehicle.
even if I show it is now working you will think I did something to trick you
My apologies, but I thought it was obvious. I am a passenger in a normal aeroplane flying at normal cruising altitude at 200 m/s. I stand up from my seat and walk down the aisle of the aeroplane towards the front of the aeroplane at 1 m/s, relative to the aeroplane. I weigh 70 kg.
To an observer on the ground (using the ground frame of reference), by getting up and walking along the aisle, I have accelerated from 200 m/s to 201 m/s. To this observer on the ground, I have increased my kinetic energy by 14,035 joules. Where did this energy come from?
I did not understand because your proposal demonstrates nothing.
You are not between two mediums you are just on one medium that moves relative to the ground. So you walking inside the plane requires energy that you will provide by burning some extra calories while in motion at 1m/s and as soon as you stop walking you will be back at 200m/s.
I am not talking about operating between two mediums. I'm taking a step back and considering a simpler case, which will turn out to be very relevant to the original more complex case.
I ask you again, to the observer on the ground (in my thought experiment above), I have increased my kinetic energy by 14,035 joules. Where did this energy come from?
I may not have been very clear. In your tough experiment there are two sources of energy both contributing to your increase in kinetic energy.
Out of that 14035J you contributed 35J and the plane contributed with 14000J. You used some of your stored energy from food while plane used kerosene or whatever that plane was using for fuel.
This will not apply to vehicle having a single energy source (wind energy).
That is correct. So relative to the ground, I have accelerated to a speed slightly in excess of that of of the aeroplane, and I was able to do that by taking the majority of the required energy from the aeroplane. You have repeatedly claimed that it is impossible to take energy from a moving medium when moving faster than that medium, whether that medium is a treadmill, a sliding piece of paper, the air, or the floor of the aisle of an aeroplane. In your calculation above you have just demonstrated that this is indeed possible.
Yes it does apply. It shows that the generator and motor of the vehicle only have to provide a portion of the kinetic energy required to increase the speed of the vehicle in excess of the speed of the moving medium. The remainder of the kinetic energy is provided directly by the moving medium. So while the motor will itself return a little less kinetic energy to the vehicle than was taken by the generator, the additional direct contribution of the medium will more than compensate for this. So the vehicle can accelerate beyond the speed of the moving medium.
In my example above my internal motor (my muscles) only had to supply 35 joules to gain 14,035 joules (relative to the ground). So if my muscles were powered by some small hypothetical wind turbine, held in my hand and in contact with the air outside of the aeroplane, that generator would only have had to extract 35 joules out of my 1,400,000 joules of kinetic energy (travelling at 200 m/s). So I would lose 35 joules of kinetic energy through the drag of the hypothetical generator, but gain about 14,035 joules of kinetic energy through the propulsion of my internal motor.
Excluding energy storage in the experiments is not easy - at least I sone see a very simple way. It is clear that there is no battery or spring storage - they are just not there and a bayttery would also need a motor. Kinetic energy (like flywheel) goes up when the vehicle gets faster, so it can not provide energy. It is only theoretical air pressure in the free space before / behind the fan. Without a container the energy/ pressure field would not stay there very long - more like lenghtscale divided by speed of sound - so in the low ms range. Compared to this the experiment with the free running vehicle on the tradmill was allready very long. A few seconds may not look long, but it is sufficiently long. Even the vehicle size divided by wind speed would be short compared.
Similar for the larger backbrid vehicle, with a pressure field that may last a few 1/100 of a second and time traveled at speed higher than the wind of more like a minute or more.
There is another point that shows that the vehicle did harmess the power of the wind, even if at the wind speed: When held at position at the treadmill there was a forward force, as when left free the vehicle accelerated forward. This forward force means the the wheels could use even more force and generate extra power. So the vehicle would be able to generate some power even if at the speed of the wind. The time the vehicle was hold in position pretty long and the time is definitely sufficient to esteblisch a steady state air pressure around the prop. There is also absolutely no reason to assume one could not hold the vehicle longer before letting it run forward. So this is steady state and thus has nothing to do with energy storage. It may be against some peoples intuition, but the experiment still shows it.
The though experiment with the man walking in the plane is good, as it shows that the use of just power / energy to avoid looking at forces can get pretty complicated when looking at different refrence frames. Similar complications also apply to the system on the treadmill: you get different energies / power when you use different refrence systems. So looking at the power is tricky and prone to error. It is much easier to look at force and speed seprate. The forces don't change when one changes the reference system, and the speeds simply add.
Assuming without checking or agreeing that your math is correct...
6Wh = 21600Ws
4800Ws = 1.5Wh.
and so on. Not keeping your units tidy can make things look different--and wrong.
and then you have to ask by what mechanism is the propeller compressing this air if the wind is pushing the vehicle and not the other way around?
The wind is pushing the vehicle and the blackbird takes a big part of that energy and put it back in the propeller generating an "artificial wind" in the opposite direction. When vehicle is way below half the wind speed the so called artificial wind will be stronger than the actual wind and this is where energy is stored as pressure differential.
This is Wind Cube territory, with apologies to real companies called Wind Cube.
The wind is pushing the vehicle and the blackbird takes a big part of that energy and put it back in the propeller generating an "artificial wind" in the opposite direction. When vehicle is way below half the wind speed the so called artificial wind will be stronger than the actual wind and this is where energy is stored as pressure differential.
How can there be an 'artificial wind' coexisting with and opposite to the actual wind? Can you demonstrate that somehow with a fan or something?
And setting that aside, if the resultant artificial wind is stronger than the real wind, where would the energy for that come from? Wouldn't that violate your conservation of energy principles?
Think about this way. You have a large wind turbine generator in low wind speed supplying a much smaller fan that creates even hurricane level winds but since the diameter of the fan is much smaller much less wind is actually moved so is like a gear box.
So say you will want to use the fan to supply the wind turbine then that will not work as power output is much smaller from the fan than what the wind turbine will need to supply the fan.
The propeller is like a diode it can leave air molecules travel from upwind to down wind but not the other way around so it acts as a one way sail while spinning.
But here you only have one fan, the propeller, and it obviously the same size as itself and is pushing directly back at the same wind that is powering it. So first, I don't see how your 'gearbox' analogy works and second, you didn't respond to the question of how the wind and the 'artificial wind', which you claim can be stronger, can coexist in the same space.
It will be 4800Ws. But there are quite a few seconds during the acceleration phase when the propeller at maybe 70% can compress the air behind in a huge volume of tens of thousand of liters (20m^2 propeller area times the length behind the propeller where pressure drops in a gradient to ambient pressure).
Except for the huge problem that propellers do not compress air. As has been shown by more than one experiment in this thread, the air behind a propeller is actually a vacuum.
The facts are simple: no walls, no containment, no compression, no pressure differential.
As I mentioned before Wikipedia is not always a source of reliable information but it is sure way better for this than the youtube video performed by someone with no knowledge about the subject.
But instead of using all wind energy to accelerate fast you divert large part of the energy to power the propeller that creates a sort of artificial wind eventually at multiple times the wind speed.
You have a problem here in that experiments show what happens in real life. The observations from an experiment are a hard, solid, fact, independent of the knowledge of the person performing the experiment. You could have a two year old perform an experiment, and without them knowing anything, the experiment would still display the factual behavior of the universe.
On the other hand, a written document is simply what someone wrote. It may or may not be accurate, and conveys no factual weight in and of itself.
In this case, there is an obvious flaw. The diagram shows an abrupt change in pressure between P1 and P2, but it shows no simultaneous change in velocity. Since the mass flow between 1 and 2 must be equal by continuity, and since a change in pressure requires a change in density, it follows that the velocity must change abruptly if the pressure changes abruptly. Therefore, the diagram cannot be accepted at face value and must be rejected as an accurate description.
So lets focus on just that for now. How can there be two winds in opposite directions in the same space? After all, the original wind has to continue to reach the prop to power it, right?
I do understand air and molecules, thanks. But this misses the point. In air, the molecules are always moving about and something like wind represents a net movement of all of them as an average. What you seem to be proposing is that a wind that hits a propeller or fan will somehow generate a counterwind as or more powerful than the original wind, yet still get power from the original wind. If so, what wind would an observer standing behind the vehicle see?
Is clear you do not know how a fan works or/and what air is.
There is a fairly sudden change in pressure, density and speed of air molecules from one side to the other of the blades.
Is clear you do not know how a fan works or/and what air is.
There is a fairly sudden change in pressure, density and speed of air molecules from one side to the other of the blades.
In the diagram, the curve of "C" shows the velocity (speed) of the air flow. Since you say there should be a sudden change in speed, and yet the diagram shows no such change in speed, you agree that the diagram is bogus?
So that is a good test that should convince anyone about pressure differential energy storage.
I'm not persuaded one bit by any of that....and you didn't answer the question about what an observer behind the vehicle would feel.
Here's one last thought. If your theory about pressure accumulation behind the vehicle is true, then that pressure would dissipate equally in all directions, right? If so, the least area that the pressure could be confined to would be a cylinder the diameter of the propeller and as long as the distance travelled during the storage phase. Since the pressure is dissipating equally in all directions and only the part that represents the vehicles propeller can actually get any benefit from it, it seems that this would be very, very inefficient. You'd have to calculate how far the vehicle would travel during the time it takes your pressure balloon to accumulate the needed energy (reduced by this inefficiency) but it seems to me you would only recover a tiny fraction of the total and the rest would just go off into space.
No it will not dissipate equally in all directions as it can not do that where the propeller sweep area is that large 20m^2 disc created by the moving propeller and so air molecules hitting that will just give their energy to the vehicle.
OK, please describe the area or whatever that stores the energy--the physical dimensions, the method of energy storage (pressure?), etc. I'm not getting that part.
If working at all, energy storage in the pressure field around the vehicle is only short time. This is a bit like tryoing to store energy in a rather leaky capacitor. The pressure will not only push the vehicle, but just dissipare as sound to all other directions. The time constant for the "discharge" of the hypotetical energy storrage is on the order of dimensions divided by the speed of sound. So for the balckbird this are some 5 m divided by 330 m/s, some thing like 2/100 seconds. So any energy stored would vanish fast, even if the order of magnitude estimate is off by a factor of 2 or Pi it is just a faktor of 1000 to short to keep significant energy.
The argument is about the time scale, so no need to calculate the energy - after some 100 time constants there is essentially nothing left, no matter how much was there to start with.
But not sure how we got to discus energy storage when the treadmill vs paper discussion is not solved.It is solved for everyone except you.
I think explaining pressure differential energy storage is more complex than the wheel only vehicle diagram solution and nobody confirmed that they agree with my claim that vehicle can not move from left to right.Everybody, without exception, disagrees with your claim.
That when understood will demonstrate that exceeding wind speed without energy storage is not possible and then we can discus about more seriously about the pressure differential energy storage.You are doing this backwards. Once understood that the vehicle can move from left to right, it then becomes possible to understand that exceeding wind speed without energy storage is also possible.
You are doing this backwards. Once understood that the vehicle can move from left to right, it then becomes possible to understand that exceeding wind speed without energy storage is also possible.
If you start by assuming something is impossible, rather than trying to understand how it works, you become stuck in a hole you cannot get out of.
The point is not capacitor vs inductor, the point is that it there is any energy stored in the pressure field is dissipates quite fast to the environment, not to the vehicle. The time scale is so short, that even 1/10 of a second is a long time. So the experiments shown are well long enough to be in essentially steady state.
Aerodynamics is a bit difficult, so I avoided going in the details there. The intuition says that if all the energy would rise when the vehicle would go faster, so the ernergy storrage would be more like adding a small part to the energy when moving and thus would work in the other direction. So the energy in the air would be available only when the vehicle slows down and can thus not increase the speed of the vehicle. So I doubt the air pressure could "help" (explain the higher speed) even at the bery short time scale.
The problem with the mis-understanding of paper vs treadmill could be solved if you tell us why the two should be different or how the vehicle could notice the difference. Most of us have a problem spotting the difference.
My formula is super simple power at motor wheel is smaller always than power power at generator wheel since generator wheel is what powers the motor wheel thus easy conclusion that vehicle can not advance from left to right.That simple "fomula" (it is more like an agumant in words) is simple, but simply wrong.
You did not took energy from the aeroplane.
I may not have been very clear. In your tough experiment there are two sources of energy both contributing to your increase in kinetic energy.
Out of that 14035J you contributed 35J and the plane contributed with 14000J.
If the aeroplane did not had a constant speed controller say maybe even engines turned off then you will have slowed down the aeroplane while you where traveling at 1m/s.
... power at motor wheel ... ... power at generator wheel ...
Put a stationary bowling ball on a OFF treadmill belt. Put another stationary bowling ball on a stationary rug.
With me so far? Do you need a diagram?
As I mentioned the big misconception for you is to think that a treadmill is the same thing with a dragged piece of paper and that is not the case.
The top of the treadmill is a flat surface moving from right to left under the front wheels of the cart, relative to a stationary surface on which the back wheels rest.
The piece of paper is a flat surface moving from right to left under the front wheels of the cart, relative to a stationary surface on which the back wheels rest.
Explain where is the difference between these two cases?
Can you not see the problem when you have the treadmill running at 2m/s and it is also pushed at 2m/s that case b)
Wait... you think the entire treadmill moves too? Maybe English isn't your native language, please be more precise.
What is being "also" pushed here?
Do you understand that the treadmill doesn't move? Like in a gym? It's only the belt? The belt is the black thing that people stand on.
The top of the treadmill is a flat surface moving from right to left under the front wheels of the cart, relative to a stationary surface on which the back wheels rest.
The piece of paper is a flat surface moving from right to left under the front wheels of the cart, relative to a stationary surface on which the back wheels rest.
Explain where is the difference between these two cases?
Have you even read my entire post ? That post already explains what happens.
No. Do not go off on a tangent. Do not start introducing other topics.
Compare my two sentences. Explain what is the difference between those two scenarios?
Hi , Really interesting and cool,
Yes this is totally possible ,and should work .
as the subject says Mess with your mind . there are 4 forces all working together here . Plus the fan is not just a propeller .
This experiment just would not work on a tread mill as in real life they ran it on a very large flat open surface without opticals .
The bit were the fan started wobbling but as the wind increased it stop .. this is caused by fan tip speeds and centrifugal force which in turn stabilizes
the craft and the tip air flow will be faster than wind speed . So when all the forces balance it will travel faster .
These guys really went to the next level .
:popcorn:
Ok you introduce the case with the powerd of an pushed treadmil as a 3rd scenario. If it help you to make clear were the diffence is Ok.No. Do not go off on a tangent. Do not start introducing other topics.
Compare my two sentences. Explain what is the difference between those two scenarios?
Please the the case b) intermediary state to understand that there is a big difference between a treadmill that is not moving relative to the ground and a powered off treadmill moved relative to the ground (powered off treadmill is your piece of paper or moving flat surface).
I think case b) is a good way to explain/visualize the difference between a) treadmill and c) piece of paper
The Fan is not just a Propeller as it has several functions in this type of setup . I was sorry that they did not use a smoker to show how the air was moving.Hi , Really interesting and cool,
Yes this is totally possible ,and should work .
as the subject says Mess with your mind . there are 4 forces all working together here . Plus the fan is not just a propeller .
This experiment just would not work on a tread mill as in real life they ran it on a very large flat open surface without opticals .
The bit were the fan started wobbling but as the wind increased it stop .. this is caused by fan tip speeds and centrifugal force which in turn stabilizes
the craft and the tip air flow will be faster than wind speed . So when all the forces balance it will travel faster .
These guys really went to the next level .
:popcorn:
Yes it seems it messes with many peoples mind.
What do you mean by the fan is not just a propeller ?
The experiment worked perfectly fine on a treadmill see the propeller based cart showing as moving forward on a treadmill.
The experiment worked perfectly fine on a treadmill see the propeller based cart showing as moving forward on a treadmill.Yes I saw that experiment with a small model and a good wind source but it needs the distance to get all forces balanced very hard .
Ok you introduce the case with the powerd of an pushed treadmil as a 3rd scenario. If it help you to make clear were the diffence is Ok.
However I still don't see, if the powered off and manually pushed treadmill is more like the running treadmill, or more like the piece of paper pulled across the floor.
The next question is, how does the vehicle know which of the 2 (or 3) cases is present and which way to go ? So there should be a difference to the vehicle, not just to your mind.
The Fan is not just a Propeller as it has several functions in this type of setup . I was sorry that they did not use a smoker to show how the air was moving.
the wind is turning the fan . that is connected to the drive wheels . that energy move it forward . the fan/Propeller starts to spin faster . The tips of the propeller
are now creating a rotating vortex . Screw effect . so the air to the side of the vortex will move outwards > So the when equilibrium is reached the front
should have a lower air pressure . less resistance (vacuum) the propeller is now blowing the craft . Yes there is ground resistance but it gets faster the
propeller will try to tilt back and will stay . More like a gyroscope would Ballance.
I see how this works but find it hard to explain exactly why .
Please the the case b) intermediary state to understand that there is a big difference between a treadmill that is not moving relative to the ground and a powered off treadmill moved relative to the ground (powered off treadmill is your piece of paper or moving flat surface).
I think case b) is a good way to explain/visualize the difference between a) treadmill and c) piece of paper
The Fan is not just a Propeller as it has several functions in this type of setup . I was sorry that they did not use a smoker to show how the air was moving.Hi , Really interesting and cool,
Yes this is totally possible ,and should work .
as the subject says Mess with your mind . there are 4 forces all working together here . Plus the fan is not just a propeller .
This experiment just would not work on a tread mill as in real life they ran it on a very large flat open surface without opticals .
The bit were the fan started wobbling but as the wind increased it stop .. this is caused by fan tip speeds and centrifugal force which in turn stabilizes
the craft and the tip air flow will be faster than wind speed . So when all the forces balance it will travel faster .
These guys really went to the next level .
:popcorn:
Yes it seems it messes with many peoples mind.
What do you mean by the fan is not just a propeller ?
The experiment worked perfectly fine on a treadmill see the propeller based cart showing as moving forward on a treadmill.
the wind is turning the fan . that is connected to the drive wheels . that energy move it forward . the fan/Propeller starts to spin faster . The tips of the propeller
are now creating a rotating vortex . Screw effect . so the air to the side of the vortex will move outwards > So the when equilibrium is reached the front
should have a lower air pressure . less resistance (vacuum) the propeller is now blowing the craft . Yes there is ground resistance but it gets faster the
propeller will try to tilt back and will stay . More like a gyroscope would Ballance.
I see how this works but find it hard to explain exactly why .QuoteThe experiment worked perfectly fine on a treadmill see the propeller based cart showing as moving forward on a treadmill.Yes I saw that experiment with a small model and a good wind source but it needs the distance to get all forces balanced very hard .
Update whats with the treadmill . this is thinking back 100years when people thought if a Car went faster than 7mph you would get sucked out
....
What do you think will power the vehicle after the wind speed is exceeded ? As wind can no longer power the vehicle that travels in the exact same direction?
As said I find it hard to explain exactly how this happens.
The Fan is not just a Propeller as it has several functions in this type of setup . I was sorry that they did not use a smoker to show how the air was moving.
the wind is turning the fan . that is connected to the drive wheels . that energy move it forward . the fan/Propeller starts to spin faster . The tips of the propeller
are now creating a rotating vortex . Screw effect . so the air to the side of the vortex will move outwards > So the when equilibrium is reached the front
should have a lower air pressure . less resistance (vacuum) the propeller is now blowing the craft . Yes there is ground resistance but it gets faster the
propeller will try to tilt back and will stay . More like a gyroscope would Ballance.
I see how this works but find it hard to explain exactly why .
You describe what happens before vehicle exceeds wind speed.
What do you think will power the vehicle after the wind speed is exceeded ? As wind can no longer power the vehicle that travels in the exact same direction?
a treadmill that is not moving relative to the ground
- is a flat surface moving from right to left under the front wheels of the cart, relative to a stationary surface on which the back wheels rest.
a powered off treadmill moved relative to the ground
- is a flat surface moving from right to left under the front wheels of the cart, relative to a stationary surface on which the back wheels rest.
There is no difference between these two cases.
As said I find it hard to explain exactly how this happens.
Can I give you another example . A torpedo travelling though water (sea) with currents in any direction .
As travels it is spinning and creates a tubercular vortex around it so the out casing is no longer in contact with the water .
The new Torpedo's can have almost zero surface resistance losses
Just like when you pull the plug out the bath you get that lovely vortex whirlpool that if your are careful one put ones finger in the middle without getting wet .
The propeller also blows
The same thing happens with air it just harder to see
Did that help !!
Totally correct It would help others if maybe some could do a 3D model :-+ :popcorn:
Imaging a small zeppelin (an airship) with the same propeller as it is in the Blackbird configuration. The zeppelin will fly, say, 5m high above the ground.
Will the zeppelin fly faster then the tail wind speed?
a treadmill that is not moving relative to the ground
- is a flat surface moving from right to left under the front wheels of the cart, relative to a stationary surface on which the back wheels rest.
a powered off treadmill moved relative to the ground
- is a flat surface moving from right to left under the front wheels of the cart, relative to a stationary surface on which the back wheels rest.
This is really simple. If you think there is something wrong with one of these two statements, change some of the green words to make it correct.
..What powers the propeller on your zeppelin ?With the zeppelin case I've been trying to disconnect the "propeller" and "wheels" concepts here.
..What powers the propeller on your zeppelin ?With the zeppelin case I've been trying to disconnect the "propeller" and "wheels" concepts here.
I can also ask you the same: "What powers the propeller on you Blackbird?"
Imaging a small zeppelin (an airship) with the same propeller as it is in the Blackbird configuration. The zeppelin will fly, say, 5m high above the ground.
Will the zeppelin fly faster then the tail wind speed?
What powers the propeller on your zeppelin ?
For this zeppelin to be equivalent you will need to have a wheel ruining on the ground connected trough a cord that may be also the electric cable and then energy generated from that will power the propeller. In this case that is the same as blackbird the zeppelin can not exceed wind speed other than for a few moments based on pressure differential energy storage. The drag from the generator wheel will be higher than the trust you can get from the propeller powered only the the power coming from the wheel on the ground.
I think your confusion comes from the shape of the treadmill...
There is a very big difference between the two cases and key in understanding what will happen in my diagram and why vehicle will not be able to move from left to right if you use a treadmill.
Nope - forget the vehicle, no vehicle here, no cables, no wheelsImaging a small zeppelin (an airship) with the same propeller as it is in the Blackbird configuration. The zeppelin will fly, say, 5m high above the ground.
Will the zeppelin fly faster then the tail wind speed?
What powers the propeller on your zeppelin ?
For this zeppelin to be equivalent you will need to have a wheel ruining on the ground connected trough a cord that may be also the electric cable and then energy generated from that will power the propeller. In this case that is the same as blackbird the zeppelin can not exceed wind speed other than for a few moments based on pressure differential energy storage. The drag from the generator wheel will be higher than the trust you can get from the propeller powered only the the power coming from the wheel on the ground.
The idea with the zeplin and towed genrator is good. One just has to be carefull in how to calculate the power:
Ignoring friction the towed gerator can gerate a power of tow-force times velocity relative to ground.
The prop on the zeplin needs some power to generate a given forward thrust. This power is independen of the wind speed, as the zeplin does not see the ground, except for the cable to transfer the thrust to the ground vehicle.
It depends on the speed of the wind, how much drag is needed to create a given power. Ignoring friction the drag is power divided by speed relative to ground. Given enough speed (wind plus the little movement of the zeplin) the drag can be smaller than the thrust of the zeplin.
Yes it is fairly simple but you are the one that is not seeing the difference.
Nope - forget the vehicle, no vehicle here, no cables, no wheels
Only a zeppelin with the propeller mounted on the zeppelin the same way as it was on the Blackbird..
The one that is confused here is you.
Frankly, I do not see a difference between the two cases - Zeppelin vs. Blackbird..Nope - forget the vehicle, no vehicle here, no cables, no wheels
Only a zeppelin with the propeller mounted on the zeppelin the same way as it was on the Blackbird..
What will be the role of the propeller on a zeppelin if the propeller is not powered ?
Zeppelin will not be able to go faster than the wind speed.
What do we say about such behavior?
Frankly, I do not see a difference between the two cases - Zeppelin vs. Blackbird..Nope - forget the vehicle, no vehicle here, no cables, no wheels
Only a zeppelin with the propeller mounted on the zeppelin the same way as it was on the Blackbird..
What will be the role of the propeller on a zeppelin if the propeller is not powered ?
Zeppelin will not be able to go faster than the wind speed.
Both are powered by the tail wind..
Blackbird drives on the ground and Zeppelin drives in the air so no contact with ground.
Blackbird takes power from the wheel to spin the propeller and since your Zeppelin has no contact with ground it can not power the propeller.
How could the Blackbird's propeller take the power from the wheels? Where the "power of the wheels" comes from?Frankly, I do not see a difference between the two cases - Zeppelin vs. Blackbird..Nope - forget the vehicle, no vehicle here, no cables, no wheels
Only a zeppelin with the propeller mounted on the zeppelin the same way as it was on the Blackbird..
What will be the role of the propeller on a zeppelin if the propeller is not powered ?
Zeppelin will not be able to go faster than the wind speed.
Both are powered by the tail wind..
Blackbird drives on the ground and Zeppelin drives in the air so no contact with ground.
Blackbird takes power from the wheel to spin the propeller and since your Zeppelin has no contact with ground it can not power the propeller.
On the video with the bet for 10K with the professor.. Derek wins the Bet and get the 10K .. So Why cant every one understand .
This works its been proven to work . and the concept is quite simple . You just have to think out of the BOX .
The same as the aeroplane cant take off from a rolling road .
and a Zeppelin Can not fly faster than tail wind because it flies in 1 (ONE medium air) Need two . But can burn faster with a good tail wind :-DD
How could the Blackbird's propeller take the power from the wheels? Where the "power of the wheels" comes from?
For both Zeppelin and Blackbird the only source of energy I can see is the tail wind..
First as far as I am concerned Professors are just people who think up a theory that is impossible to solve . so some one will sponsor them to solveOn the video with the bet for 10K with the professor.. Derek wins the Bet and get the 10K .. So Why cant every one understand .
This works its been proven to work . and the concept is quite simple . You just have to think out of the BOX .
The same as the aeroplane cant take off from a rolling road .
and a Zeppelin Can not fly faster than tail wind because it flies in 1 (ONE medium air) Need two . But can burn faster with a good tail wind :-DD
Neither the professor or Derek understood how vehicle works.
Bet was that vehicle can or can not exceed wind speed and Derek proved to professor with the treadmill prototype that vehicle can exceed wind speed.
Since neither of them seen the energy storage device that I call pressure differential they where not able to understand how the vehicle works.
I do not claim that Blackbird vehicle can not exceed wind speed as based on my theory it works exactly the same as shown in tests. What I claim is that stored energy is used for that and so if test will have been long enough the slow down will have been seen.
You seem to be the most intelligent here so I'm sure you will be able to get what I'm saying.
Here is my wheel only analogy first without energy storage so vehicle can not move from left to right
(http://electrodacus.com/temp/Winds.png)
And second with energy storage the spiral spring that emulates the air pressure differential. The spring connects the motor to the wheel
(http://electrodacus.com/temp/Windsw.png)
This one with the spiral spring (same as you get on a measure tape) will work exactly like the propeller based treadmill model in Derek's video.
How could the Blackbird's propeller take the power from the wheels? Where the "power of the wheels" comes from?
For both Zeppelin and Blackbird the only source of energy I can see is the tail wind..
Yes exactly the right question.
Below wind speed power at the wheel comes from wind and above wind speed power comes from stored energy in pressure differential so it will be for a limited amount of time than blackbird will be above wind speed not indefinitely as claimed.
Blackbird or the Zeppelin can not use any energy from the tail wind when vehicle speed is above that tail wind speed.
The red part is repeated over and over again as an argument, but it is plain wrong. Not so obvious, but still wrong.How could the Blackbird's propeller take the power from the wheels? Where the "power of the wheels" comes from?
For both Zeppelin and Blackbird the only source of energy I can see is the tail wind..
Yes exactly the right question.
Below wind speed power at the wheel comes from wind and above wind speed power comes from stored energy in pressure differential so it will be for a limited amount of time than blackbird will be above wind speed not indefinitely as claimed.
Blackbird or the Zeppelin can not use any energy from the tail wind when vehicle speed is above that tail wind speed.
Just like the convener belt . It has more questions than answers .
Bit like the riddle .. Have a 2 meter round pond with a Frog dead in the middle, How far does the frog have to jump to leave the pond area ???
.
.
Well . no prize for getting it right
@ yes I understand or what you are Getting at .
Oh and a spring is kinetic energy a propeller is not .. Forget the the conveyor you are confusing yourself and thinking on the wrong lines .
nice drawing .. But does not prove any thing sorry you have missed the real concept as i did try to explain how it works but .. maybe in another life time
The red part is repeated over and over again as an argument, but it is plain wrong. Not so obvious, but still wrong.
Blackbird or the Zeppelin can not use any energy from the tail wind when vehicle speed is above that tail wind speed.
1. Both the Zeppelin and the Blackbird are "vehicles" with the same propeller design.
2. Both the Zeppelin and the Blackbird are powered by the energy of the tail wind only.
3. The Blackbird can go faster than the speed of the tail wind (based on the experiments we have seen).
4. Based on the above the Zeppelin should fly faster than the speed of the tail wind..
:phew:
like that dragged paper instead of treadmill and claim they are equivalent when they are not even close to be equivalent
Quotelike that dragged paper instead of treadmill and claim they are equivalent when they are not even close to be equivalent
You keep saying that but don't say what the difference (an apparently massive difference) is. So... I don't have a treadmill and I can't believe that your magic only works with a treadmill. What other means could be used to prove, or not, your intuition? I think you will need to be quite explicit so no-one can be accused of not doing it quite right. If things need to be a certain way up you need to say that up front.
Treadmill is not moving relative to ground the pushed treadmill is clearly moving relative to ground.
But yes you can make changes to my diagram like that dragged paper instead of treadmill and claim they are equivalent when they are not even close to be equivalent.
Check post #351 and I explained why it is very different yesterday.
The intermediary case b) should make it clear why there is a difference between a) and c)
Treadmill is not moving relative to ground the pushed treadmill is clearly moving relative to ground.
In both cases the top surface of the treadmill is moving relative to ground.
Say we go back to propeller based vehicle.
I will make this problem as a reply to all of you and even do the calculation for you just let me know if I'm wrong.
Start conditions:
Wind speed 0m/s
Vehicle speed 0m/s
Vehicle mass 1kg
Available energy 10Ws
Wheel efficiency 90%
Propeller efficiency 70%
So vehicle will end up with 9Ws of Kinetic energy if energy is used to power the wheel and 7Ws if propeller is used.
Thus you probably agree that using wheel for propulsion is best option thus 9Ws is kinetic energy and vehicle is at 4.24m/s
Now at this point with vehicle at 4.24m/s will there be any logic in taking energy from the wheel and putting in to propeller ?
I hope you agree it will make no sense.
Now let say same conditions with vehicle at 4.24m/s but now there is a wind speed of 2m/s in same direction as the vehicle.
Will this change anything ? Will it make sense to take energy from the wheel and power the propeller ?
If your answer is yes please provide the calculation showing that vehicle can increase the current 9Ws kinetic energy using that 2m/s wind.
I will make this problem as a reply to all of you and even do the calculation for you just let me know if I'm wrong.
Start conditions:
Wind speed 0m/s
Vehicle speed 0m/s
Vehicle mass 1kg
Available energy 10Ws
Wheel efficiency 90%
Propeller efficiency 70%
So vehicle will end up with 9Ws of Kinetic energy if energy is used to power the wheel and 7Ws if propeller is used.
Thus you probably agree that using wheel for propulsion is best option thus 9Ws is kinetic energy and vehicle is at 4.24m/s
Now at this point with vehicle at 4.24m/s will there be any logic in taking energy from the wheel and putting in to propeller ?
I hope you agree it will make no sense.
Now let say same conditions with vehicle at 4.24m/s but now there is a wind speed of 2m/s in same direction as the vehicle.
Will this change anything ? Will it make sense to take energy from the wheel and power the propeller ?
If your answer is yes please provide the calculation showing that vehicle can increase the current 9Ws kinetic energy using that 2m/s wind.
OK, that is easy enough. Assuming your wheel generator has the same 90% efficiency and the propeller 70%, load the generator so that there is a 1N force (backwards) on the car. I can't be more detailed because I don't know the radius of the wheel, but that won't matter. ( 1N * 4.24m/s * 0.9 efficiency ) = 3.816W. Now if you take the 3.816W and use the propeller to generate a force, you will get a larger force because...power is force * speed and you don't have to generate that force at 4.24m/s, but rather2.422.24m/s, the airspeed that the propeller sees. So the force generated will be (3.816W * 0.7 efficiency /2.422.24m/s) =1.1041.1925N. The net force (propeller force minus wheel force) will then be 0.11925N and the vehicle will accelerate at 0.11925m/s2. Thus the vehicle will continue to accelerate even though it is already travelling at over twice the wind speed. In this example, the propeller is providing 2.6712W of power, which is obviously less than is taken out at the wheels, but the wind provides (1.1925 * 2m/s) =2.0282.385W of additional power for a total of4.69925.0562W, which is more than is being taken out at the wheels, so a net gain of (4.69925.0562- 4.24) = 0.8162W. And then you can check the math to make sure that the net power I just gave you results in the acceleration stated--which is why I had to go back and fix my numerical transpositions.But I think I've made an additional error, so wait a bit.
Edit: Should be all good now. If not, someone point out the errors.
Good effort but you forgot something super important.
Forces have no role here
OK, in a universe where forces have no role in basic mechanics, then I have nothing to offer. My entire education was a complete waste.
You missed / cut out the more important point: for the dicussion the red sentence in question is not an agument, but part in question.The red part is repeated over and over again as an argument, but it is plain wrong. Not so obvious, but still wrong.
Blackbird or the Zeppelin can not use any energy from the tail wind when vehicle speed is above that tail wind speed.
The red part is not wrong and there is no trick to make that right as if it was to be right it will violate the conservation of energy.
You always mention force when force has nothing to do with conservation of energy.
You only care about power and since propeller is powered by the wheel generator the power you take from the wheel is the most in ideal case you put in the propeller. In real world propeller output may be just 70% of what is available at the wheel so without stored energy the vehicle could never exceed wind speed.
Because of this misconception you also think that the wheel only vehicle in my diagram can move from left to right when that is not possible and you will never be able to prove something impossible.
The result may be confusing, but the math is simple::horse:
The power needed by the prop to generate a fixed amount of thrust is independent of the wind speed.
The power the generator can produce is force times speed. The speed is wind-speed plus a little from the movement in the air.
Plot the two in a graph (X= speed relative to ground, Y = power) and the result is clear:
Given enough wind speed the generator can produce more power than the prop needs to pull it.
I will make this problem as a reply to all of you and even do the calculation for you just let me know if I'm wrong.
Start conditions:
Wind speed 0m/s
Vehicle speed 0m/s
Vehicle mass 1kg
Available energy 10Ws
Wheel efficiency 90%
Propeller efficiency 70%
So vehicle will end up with 9Ws of Kinetic energy if energy is used to power the wheel and 7Ws if propeller is used.
Thus you probably agree that using wheel for propulsion is best option thus 9Ws is kinetic energy and vehicle is at 4.24m/s
Now at this point with vehicle at 4.24m/s will there be any logic in taking energy from the wheel and putting in to propeller ?
I hope you agree it will make no sense.
Now let say same conditions with vehicle at 4.24m/s but now there is a wind speed of 2m/s in same direction as the vehicle.
Will this change anything ? Will it make sense to take energy from the wheel and power the propeller ?
If your answer is yes please provide the calculation showing that vehicle can increase the current 9Ws kinetic energy using that 2m/s wind.
OK, that is easy enough. Assuming your wheel generator has the same 90% efficiency and the propeller 70%, load the generator so that there is a 1N force (backwards) on the car. I can't be more detailed because I don't know the radius of the wheel, but that won't matter. ( 1N * 4.24m/s * 0.9 efficiency ) = 3.816W. Now if you take the 3.816W and use the propeller to generate a force, you will get a larger force because...power is force * speed and you don't have to generate that force at 4.24m/s, but rather2.422.24m/s, the airspeed that the propeller sees. So the force generated will be (3.816W * 0.7 efficiency /2.422.24m/s) =1.1041.1925N. The net force (propeller force minus wheel force) will then be 0.11925N and the vehicle will accelerate at 0.11925m/s2. Thus the vehicle will continue to accelerate even though it is already travelling at over twice the wind speed. In this example, the propeller is providing 2.6712W of power, which is obviously less than is taken out at the wheels, but the wind provides (1.1925 * 2m/s) =2.0282.385W of additional power for a total of4.69925.0562W, which is more than is being taken out at the wheels, so a net gain of (4.69925.0562- 4.24) = 0.8162W. And then you can check the math to make sure that the net power I just gave you results in the acceleration stated--which is why I had to go back and fix my numerical transpositions.But I think I've made an additional error, so wait a bit.
Edit: Should be all good now. If not, someone point out the errors.
Good effort but you forgot something super important. What is the kinetic energy of the vehicle before and after you did the break ?
Say you maintain that 3.816W break for one second now the vehicle kinetic energy is reduce by 3.816Ws
Now you apply 3.816W * 0.7 = 2.671W for one second and you can see you end up with lower kinetic energy than you started with.
And is irrelevant if you do that for 1 full second or a 1ms or even 1us the end result is that vehicle speed will be reduced.
The wind can not provide anything unless you refer to stored energy. As long as wind speed is lower than vehicle speed there is nothing to gain from that.
Forces have no role here since all you care is vehicle kinetic energy change. If you end up with lower kinetic energy then vehicle has slowed down.
Do you see a difference in kinetic energy gain between a vehicle traveling at 2.24m/s with no wind and one traveling at 4.24m/s and 2m/s wind when power is applied to the propeller ?
If you want to know if vehicle will increase or decrease in speed you do not need to be concerned with forces just with power. Breaking power and propulsion power. As long as breaking power is more than propulsion power vehicle will slow down unless there is something else that can offer propulsion.
To be honest education of most people seems to have been a waste of time. Only knowing the formula and not knowing how to apply or what result you should expect is sort of pointless.
If you think you are frustrated try to be in my place where almost nobody understand how basic physics works including people teaching others.
If you want to know if vehicle will increase or decrease in speed you do not need to be concerned with forces just with power.
What if wind speed was 6.24m/s so higher wind speed than vehicle speed ? Based on how you applied the formula this will be a fairly bad thing.
100 years ago we all thought that the Plus & Minus on a battery that electricity flowed . From Plus to Minus
Every one was happy with this .. But it was found to be WRONG it goes from Minus to Plus .. :wtf: :blah:
Most people now except this as this is what you are told in school / Uni or what ever.
weather right or wrong the world maybe Flat . Depends on your perspective thinking . What you see is not always what is .
Mother Nature does not have to play by the rules of mea Humans . So what when you see something that you Don't understand your Brain
will try to fit some thing in to stop it blowing a fuse ..
A computer if it does not understand will return ERROR . No Data . because the Human did not enter it .
Somethings are as is . Appling maths formulators to do quick fix may solve your problem but not be Right solution
Murphy's Law is always true .. Humans will Always Get it round the wrong way.
:-DD I'm sure that they don't "understand" it the way you do. You seem to have adopted 'conservation of energy' as a religion of some sort without comprehending what it means nor understanding where or how to apply it.I understand perfectly how to apply it. If you have 10W for propulsion and your propulsion is 70% efficient you will get 7W worth of thrust no matter if your vehicle drives at 2m/s or 4m/s or any other speed if you apply this 7W worth of thrust for one second the vehicle kinetic energy will increase by 7Ws (7 Joules if you prefer).
Ok, so imagine your 1kg vehicle is travelling at 4.24m/s and you drive it into an infinitely strong and rigid concrete wall. Calculate the power that the wall provides to stop the vehicle.Wrong way of asking the question as there is not enough data to be able to answer that. The question that can easily be answered with the amount of data available is calculate the energy the vehicle acted against the wall and that is fairly simple since we know the vehicle speed and mass we can calculate the kinetic energy of the vehicle 9Ws and so since before hitting the wall vehicle had that and after the wall that will be zero (assuming non elastic collision) then it will be 9Ws.
A bad thing? So plug in the numbers to whatever formula you think I used and post the numbers. How bad can they be? :palm:
Wrong way of asking the question as there is not enough data to be able to answer that. The question that can easily be answered with the amount of data available is calculate the energy the vehicle acted against the wall and that is fairly simple since we know the vehicle speed and mass we can calculate the kinetic energy of the vehicle 9Ws and so since before hitting the wall vehicle had that and after the wall that will be zero (assuming non elastic collision) then it will be 9Ws
Wrong way of asking the question as there is not enough data to be able to answer that. The question that can easily be answered with the amount of data available is calculate the energy the vehicle acted against the wall and that is fairly simple since we know the vehicle speed and mass we can calculate the kinetic energy of the vehicle 9Ws and so since before hitting the wall vehicle had that and after the wall that will be zero (assuming non elastic collision) then it will be 9Ws
So I'm unclear here. Are you saying the the wall provides 9Ws of energy to stop the car or are you saying that it is not possible to calculate the energy provided by the wall?
You asked about power not energy and that was the wrong thing to ask for as power is not the same with energy.
To answer the power question you will need to know much more about the vehicle shape and construction and you will also need to know at what exact moment you needed to know the power.
Question: Do you agree with the red arrows I added to your diagram?
(And JFC if someone could explain to me how the #%$#%$!! you get attachments to show up properly!!)
(Attachment Link)
You asked about power not energy and that was the wrong thing to ask for as power is not the same with energy.
To answer the power question you will need to know much more about the vehicle shape and construction and you will also need to know at what exact moment you needed to know the power.
OK, a point for you. I meant energy, and my followup is clear enough--are you saying the wall provides 9Ws of energy to stop the car or did you mean something else?
Excellent. How can the G wheel "see" or be influenced by the force/movement/energy/power indicated by the red arrows?
You say since there is higher force it means vehicle will accelerate in a certain direction when that is wrong as the force you refer to is force acting against the propeller or against the wheel not against the body of the vehicle.
Force at the wheel is not the same with force on the body when the wheels are acting against two isolated mediums.
I did not meant something else wall needed to be able to absorb the 9Ws else it will no longer be called a wall after the impact.
But what about my other replays where I say power in and out of the system is a better/simpler way to predict if vehicle will slow down or accelerate. Looking at the balance of forces is not useful as those are forces at the wheel / propeller not the forces acting against the vehicle since vehicle is between two isolated mediums that have different speeds.
Are there any of Newton's laws that you agree with?
So you are saying that the rate of change of momentum of a body over time is not just proportional to the net external force. It also depends on what that force or those forces are acting against.
Are there any of Newton's laws that you agree with?
The Power equation is hard to mess up
Again, your arguments and suppositions easily reduce to absurdities, but for some reason you don't see it. An inflexible wall cannot absorb or provide any energy. It can provide a reaction force sufficient to stop the car, but no energy is exchanged between the wall and the car. Now if you try to bob and weave by saying the wall is not infinitely rigid, then you need to do more calculations. If you do, you will find that any actual energy exchange is miniscule.
Edit: And reflecting, I think the reason my mind allowed me to type 'power' when I meant 'energy' is that there actually is enough data to calculate the power provided by the wall--zero, the same as the energy. So I'll take back my point.
That's just nonsense. If the wheel and propeller are fixed to the vehicle, as they are at least on the x-axis, then forces acting on them along that axis will act on the vehicle through reaction forces. Perhaps you don't understand what a 'force' is?
The Power equation is hard to mess up
Apparently it is really easy to mess up, as you keep doing so with your treadmill example and the nonsense conclusion that the cart cannot move from left to right when powered by the treadmill.
It is clear that if the G wheel can pick up power then the M wheel can receive power. If the M wheel can receive power, then it can equally well be made to turn clockwise or anti-clockwise (it just depends on the gearing). If the M wheel turns clockwise the cart will move from left to right.
There is nothing at all preventing the M wheel from turning clockwise. If it receives power, it can turn.
Are there any of Newton's laws that you agree with?
So now you can look at speeds and say wheel speed is 5m/s it means breaking force was 2N
And if say the other medium you pushed against only had 2.5m/s you will have needed 4N to be able to push with 10W for 1 second and get the vehicle kinetic energy back to where it was.
You see the 4N is higher than 2N and you think vehicle will gain speed but that is not the case and in ideal case the vehicle will just be able to maintain speed.
The problem is you are confusing the force at the wheel with forces acting against the vehicle.
Thanks, I'll take that as a no.
Wall is fixed to the ground so that energy is transferred to earth. When vehicle accelerated it has pushed against earth accelerating the earth in opposite direction then when vehicle hit the wall all that energy was put back in to earth in opposite direction to initial acceleration.
Vehicle had 9Ws of kinetic energy before hitting the wall and 0Ws after. So where do you think that 9Ws disappeared ?
You seems to be the one not understanding that force at the wheel (or propeller) is not the same with force against the body of the vehicle.
Think about a gearbox only that is 2:1 ideal so no friction.
I'm on the input side putting 1N and you are at the output side and will need to put 2N so that gearbox is in equilibrium no movement.
have you got the equilibrium part ? There is a huge difference in forces but no movement in any direction.
You did not read my answer if that is your conclusion. Newton laws are correct you just do not know how to apply them when vehicle acts against two separate mediums.
Wall is fixed to the ground so that energy is transferred to earth. When vehicle accelerated it has pushed against earth accelerating the earth in opposite direction then when vehicle hit the wall all that energy was put back in to earth in opposite direction to initial acceleration.
Vehicle had 9Ws of kinetic energy before hitting the wall and 0Ws after. So where do you think that 9Ws disappeared ?
Nope! If you do the math properly, you'll find that the energy transferred to that much larger system is absolutely miniscule. This is a variant on a standard high-school level physics question that is often initially answered wrongly because the wrong law (conservation of energy) is applied when the correct result is obtained by using the law (conservation of momentum). So yes, the 9Ws 'disappears', why don't you figure out where it goes. Or, reflect on the fact that you apparently don't know where it goes. It's a very basic question.QuoteYou seems to be the one not understanding that force at the wheel (or propeller) is not the same with force against the body of the vehicle.
Think about a gearbox only that is 2:1 ideal so no friction.
I'm on the input side putting 1N and you are at the output side and will need to put 2N so that gearbox is in equilibrium no movement.
have you got the equilibrium part ? There is a huge difference in forces but no movement in any direction.
Since you said gearbox, I'll assume you mean newton-meter (torque). Or you could use a lever example if you like. In any case, the only reason you imagine that works is because you do not understand the laws of motion.
For every action there is an equal and opposite reaction.
Now for your gearbox example, indeed the lever-action of the gears will result in no movement, provided the gearbox body is held in place. Whatever is holding it in place will have to provide 1N/m of torque as a reaction, otherwise the body of the gearbox will just rotate. If you see an example of unequal forces but no movement, it is because you have not properly accounted for all of the forces.
First part with the wall you probably think all of that ends as heat and that is not the case. I get that earth is so large that 9Ws means nothing but you can say that vehicle moved or the wall (entire earth moved).
Think of a very heavy wall but on wheels then you can understand that while wall moved very little because is so much heavier than the vehicle it did moved.
Think about those swing-balance you find on playground but imagine one that has an arm longer than the other. To keep that from moving you will need to apply different forces at the end.
The same happens between two wheels connected trough a chain with different gear ration than 1:1 Forces will be different but no motion.
.
If you think you are frustrated try to be in my place where almost nobody understand how basic physics works including people teaching others.
.
Please provide the correct power equation so I can see where I did the mistake.
To get any energy from G wheel the vehicle will need to move backwards (right to left) and putting that generated energy back in to M wheel will be enough in ideal case to bring the vehicle back to original position thus not possible for vehicle to move from left to right.
Excellent. How can the G wheel "see" or be influenced by the force/movement/energy/power indicated by the red arrows?
Not quite sure you know what you are asking. Do you notice the two larger black triangles ? Those fix the treadmill to the ground and there will be forces acting against those when vehicle is pushed down from the treadmill. I guess you can make an analogy with how pulleys work.
The important part is that treadmill is not moving relative to the ground. If you turn off the treadmill rotation and push the treadmill then it will move relative to the ground and then you can reverse that movement and say that ground moves relative to the treadmill. So it is not the same thing.
Well, it could be heat or some other change in the potential energy of the wreckage of the car. Anyhow, lets say the wall is in fact on frictionless rollers and is very heavy (say 100,000kg). And presume a completely non-elastic collision. Are you saying that after the collision the 9Ws energy will be found as kinetic energy of the now very slowly moving wall? |O
And the fulcrum will supply a reaction force equal to the sum of both forces. Same for the wheels. You really are refuting Newton's laws! :-DD
And that, my dear, is exactly why you should start to question yourself if you in fact are such a supreme person that knows everything MUCH better than ALL OTHER, or, maybe, just maybe, you could be wrong ?
I will have questioned myself if you (any of you) provided me with an explanation that worked and did not violate the energy conservation law.'
The G wheel is a generator. It only needs to turn on its axle to generate power, it doesn't need to move. Since the belt is moving it can turn the G wheel to generate power without the cart needing to move left or right. The power generated by the G wheel can be transferred to the M wheel, and as noted before, the M wheel can turn clockwise or anti-clockwise at any speed according to gearing. If we gear it to turn clockwise it can move the cart from left to right. With low gearing it can move the cart slowly, and this slow movement will require less power than was generated in the G wheel.
I believe this was asked before and I don't think you answered (just link it if you did).
What is the energy conservation law you are referring to? Can you state the law fully?
You can not create or destroy energy you can just convert it from one form to another.
OK, so in what way have any of the examples or calculations I have proposed violated this statement of the law of conservation of energy?
I do not have that much time to make a list but check all my old reply's.
If you are in between this two extreme cases so extraction energy from the wheel and putting part of it in to motor wheel to push the vehicle back will result in a slower movement from right to left as that is the only direction this vehicle can go if the only source of energy is the treadmill.
If you are in between this two extreme cases so extraction energy from the wheel and putting part of it in to motor wheel to push the vehicle back will result in a slower movement from right to left as that is the only direction this vehicle can go if the only source of energy is the treadmill.
It cannot move from right to left as the powered motor wheel is turning clockwise and moving it from left to right.
You are ignoring the fact that power at the motor is lower in real life than power at generator and this powers are opposite.
You are ignoring the fact that power at the motor is lower in real life than power at generator and this powers are opposite.
No, I'm not ignoring this. Only very little power is needed by the motor for the M wheel to turn slowly.
You can understand this if we brake the M wheel so it cannot turn. In this case the cart will not move, but the brake also requires zero power while the generator produces lots of power. Now we can take just a little power from the generator to make the M wheel start turning.
The ones left here start to become boring.You mean threatening of your beliefs. (It's not science, because you refuse to look at the physics objectively. You have your preconceptions and you won't even consider exploring any possibility that disagrees with those.)
So I will not be back here for some time.Yeah, I'd run away now if I were you.
Maybe I will be back some time latter to see if you learned something new.Maybe if you come back we will see if you have learned something new.
Hi I am Back..
I was racking my brains how to explain how this is Possible .
First clear your minds and think of a Bumblebee .. Ok you say this is off topic . well not really.
According to all known laws of aviation, there is no way that a bee should be able to fly. Its wings are too small to get its fat little body off the ground. The bee, of course, flies anyway. Because bees don't care what humans think is impossible. :-+ Yes you ..
That this machine can go faster than the tail wind for only very short time . because .
The wind first pushes it upto the tail wind speed .
Now comes the tricky part . First the wind MUST NOT CHANGE SPEED at the point of acceleration . HARD as wind is variable .
The propeller pitch is changed to the right angle ( those who missed that part in the video watch again) .
now at this moment there is no longer wind pushing from the back . Only from the front . when travelling at the same wind speed .
an air pocket is forming at the front of the machine . Which is at a LOWER air pressure vacuum bubble . That is rotating round the front in a
kind of whirlpool the centre is now acting like a vacuum suction . The air resistance will drop so there is a state where there is no longer any
front air drag slowing it down . The Propeller still has air flow & ground power . but the friction lose is smaller .
This is when their Red marker flag moved backwards .
This is the point it enters the Air vortex tunnel . and starts to get sucked as well as the tip velocity of the propeller is at its highest .
and is Now Blowing . Houston we have Go for acceleration .
This will only last while the tail wind remains constant maybe minute or 2 or just seconds .
When the tail wind speed changes the front air pocked is broken .
The point is it is possible maybe only for short times maybe a few minutes on a good day.
World records have been beaten in micro seconds . So I reckon a minute is a :-+
Couldn't this easily be proven or disproven using a miniature wind tunnel and small version?Properly yes as long as the model scaling was exact to the original .
I guess the length of the tunnel required would be a bit long.
The experiment with the treadmill is the controles environment "wind tunnel" replacement, just with a different system of reference.No .. Not the same .. you just contradicted your self from a previous post .
That was my point, not everyone agrees that a treadmill is a valid analog.. There seems to be a fixation on treadmills etc.
So a wind tunnel seems like the only way to convince those people.
. There seems to be a fixation on treadmills etc.That's because a treadmill is a convenient mechanism to demonstrate the principle. This is because of the validity of the concept of "Frames of reference".
I will give a just a few reasons why they are not used other than exercising the Dog .This will be entertaining....
First they have a very high friction as the belt runs over the rollers the surface gets warm to hot .So? How does that affect things?
the bottom of the belt that hangs down at the bottom has to be loose . So it flaps .So? How does that affect things?
So? As a factor that is somewhat minimal, how does that affect things?
Having to moving surfaces running in parallel in opposite directions causes side air distortion and turbulence. .
Even at airports they did away with this due to belts getting hot . Now they use just rolling beds .. Belt-less.So? The temperature of belts is not going to affect the experiment unless they've burst into flames or gone so tacky as to be unable to allow rolling. This is a non-argument.
In other places the belts were replaced with slated belts that were self cooling ..
The aerospace industry , Lockheed , NASA , and all the others .Don't see this being significant for the scale of test we are talking about.
Don't use this in a wind tunnels or for replacing the use of a wind tunnel due to there significance and disturbance in testing.
The test item in a wind tunnel is usably suspended . in the car test rollers are used and the wheels are placed between 2 controlledOK. Fine. The rollers are a perfect analogy for travelling over the ground (Just as a treadmill belt is, BTW) - but the only thing is that they constrain the movement of the wheels. In such an experiment, the rollers would need to be programmed to increase speed to match the performance of the vehicle, while the wind speed remains constant.
rollers . 2 at the front and 2 at the back . and are computer controlled . The rollers are covered by a plate so no other side effects
mess with the final tests .
Treadmills etc work great on paper and for demonstrating science projects . And not to for get the GYM .OK - a dismissive wave of the hand and it all goes away...? That is classic denial.
or moving lots of odd shaped stuff from containers etc .
So Please get over the treadmill . .. hamsters likes them .
.... not everyone agrees that a treadmill is a valid analog.
I am sorry and not being rude but you have not got any idea in real world industrial physics. There seems to be a fixation on treadmills etc.That's because a treadmill is a convenient mechanism to demonstrate the principle. This is because of the validity of the concept of "Frames of reference".QuoteI will give a just a few reasons why they are not used other than exercising the Dog .This will be entertaining....
Answer You have No idea of what the difference between a treadmill that you have in your house and a wind tunnelQuoteFirst they have a very high friction as the belt runs over the rollers the surface gets warm to hot .So? How does that affect things?
Answer The Added Friction will give bad results ..Quotethe bottom of the belt that hangs down at the bottom has to be loose . So it flaps .So? How does that affect things?
Answer You are really Funny that you don't know why, Must have slept thought entire physics lessonQuoteSo? As a factor that is somewhat minimal, how does that affect things?
Having to moving surfaces running in parallel in opposite directions causes side air distortion and turbulence. .
Answer : OMG somewhat minimal :palm: If this was your oscilloscope giving these error would you except itQuoteEven at airports they did away with this due to belts getting hot . Now they use just rolling beds .. Belt-less.So? The temperature of belts is not going to affect the experiment unless they've burst into flames or gone so tacky as to be unable to allow rolling. This is a non-argument.
In other places the belts were replaced with slated belts that were self cooling ..
Answer it proves that the industry has excepted they are a waste of energy and uselessQuoteThe aerospace industry , Lockheed , NASA , and all the others .Don't see this being significant for the scale of test we are talking about.
Don't use this in a wind tunnels or for replacing the use of a wind tunnel due to there significance and disturbance in testing.
Answer the smaller the item being tested the greater the accuracy has to beQuoteThe test item in a wind tunnel is usably suspended . in the car test rollers are used and the wheels are placed between 2 controlledOK. Fine. The rollers are a perfect analogy for travelling over the ground (Just as a treadmill belt is, BTW) - but the only thing is that they constrain the movement of the wheels. In such an experiment, the rollers would need to be programmed to increase speed to match the performance of the vehicle, while the wind speed remains constant.
rollers . 2 at the front and 2 at the back . and are computer controlled . The rollers are covered by a plate so no other side effects
mess with the final tests .
Do that and you'll find the Blackbird does work. Oh now you agree with me ..
what's this .. I said " computer controlled " your remark "would need to be programmed " has to be spiting hairsQuoteTreadmills etc work great on paper and for demonstrating science projects . And not to for get the GYM .OK - a dismissive wave of the hand and it all goes away...? That is classic denial. answer :-DD By whom
or moving lots of odd shaped stuff from containers etc .
So Please get over the treadmill . .. hamsters likes them .
But the zero relative windspeed case is absolutely the point of most interest for this particular problem.
When such arguments come from those who cannot understand fundamental principles, their process is bound to be flawed and their conclusions will be of no value to anyone but their delusion.
Tally me on the skeptic side.
Not only do I not believe this vehicle’s speed can exceed the wind speed when aligned downwind, I don’t believe it can equal the wind speed due to frictional losses. Believers will have to explain to me what source of energy can accelerate this vehicle above the wind speed once the speed is equivalent, at which point every molecule of air surrounding and in contact with it has 0 relative velocity, and overcoming frictional losses at the same time!
Set this up with anemometers along the path to confirm the constant wind velocity, evidence of perfect alignment with the wind, a long enough course to eliminate the possible effects of wind gusts, and an unedited, continuous video, and you’ll have my attention. Until then, to me this myth is busted.
I think there is not much point demonstrating things with experiments, because if people don't like what the experiment shows they will find reasons to dismiss what they see.
I am sorry and not being rude but you have not got any idea in real world industrial physicsPassive-aggressive much? :-DD
If this had been your own personal project Then you would look at it differently and in that case I would not have cared .
This a wind tunnel from a car manufacture .. If they could do this on a Treadmill, That are relatively cheap .
.
NO.. They go too all that trouble in building a wind tunnel that costs $ Millions .
:scared: They could have just got a Treadmill from BangUgood for $200 .
Enjoy the video .
<< Link to totally irrelevant video >>
I think there is not much point demonstrating things with experiments, because if people don't like what the experiment shows they will find reasons to dismiss what they see.
Sure, some have definitely made up their mind and wont accept any evidence. but i don't think that describes everyone.
But that seems to be a subtlety beyond your comprehension.
.. The sailboat sail is not traveling along the same vector as the wind. The propellor on this contraption is. At ground speed equivalent to wind velocity, there is no longer any wind force being applied to the propellor airfoil, unlike a tacking boat sail. ..
He sounds like the kind of guy who will argue that no one should use a dead carbon-zinc battery when learning Ohm's Law ... and will describe fifteen different kinds of batteries.
Then after six weeks of exhaustive and exhausting explanation, will reluctantly agree that the battery is just an illustration...
Then just as everyone is ready for lunch, he'll say "yeah but no one uses incandescent lights anymore!"
This perfectly illustrates the fact that you have no clue about the actual processes involved. You should be embarrassed - but I doubt you will.For sure not . I'm jealous. That I could No way match your supreme gift in Scepticism.
https://youtu.be/ZqUvbYb-x2Y?t=70This defiantly messed with the mind .
but...
Still is hard for me to understand why such a large majority thinks traveling above wind speed in same direction indefinitely can be possibleMaybe it's because you are wrong.
(since is clearly not the case as it is not allowed by the conservation of energy).But there IS energy available. You are just limiting your vision to one aspect and, thus, are blinding yourself to the possibility. Until you get past that blindness, you will never understand.
While vehicle is above wind speed there is no longer wind energy available to the vehicle.
The wheels again....? :palm:Still is hard for me to understand why such a large majority thinks traveling above wind speed in same direction indefinitely can be possibleMaybe it's because you are wrong.
Quote(since is clearly not the case as it is not allowed by the conservation of energy).But there IS energy available. You are just limiting your vision to one aspect and, thus, are blinding yourself to the possibility. Until you get past that blindness, you will never understand.
While vehicle is above wind speed there is no longer wind energy available to the vehicle.
..... and Lord knows there have been many attempts to help you see this.
Do you not agree with that fact that all air molecules move in the opposite direction of travel when vehicle is above wind speed traveling in the wind direction?No, we do not agree.
If you agree with that then how will the vehicle have any usable energy from the wind to be able to accelerate ?We do not agree.
If you explain the same way as all others with taking energy from the wheels (reducing vehicle kinetic energy) and then use that to power the propeller then you same as many others are wrong as that will be called an over-unity device and nobody has ever proved that.The energy comes from the wind. It is not over-unity because the vehicle is powered by the wind.
You need more energy put in to propeller than you took from the wheels just to maintain speed let alone accelerate.Yes, the energy comes from the wind.
The only reason the blackbird continues to accelerate is the pressure differential stored energy while vehicle was below wind speed and was able to access wind energy.No, this is wrong. Experiments clearly show such vehicles being able to run for as long as the wind holds up.
I'm not wrong. If I was wrong one of you will have been able to explain where I'm wrong.
If I was wrong one of you will have been able to explain where I'm wrong.We have tried. Several dozen times. What prevents us from being successful is your steadfast refusal to consider, let alone accept the arguments that have been put before you - and your continued reliance on your own, flawed beliefs. Even there, we have tried to explain the flaws, but you seem incapable of accepting that possibility.
The real life video footage had so many easy tests that could have been implementedInstead of a smoke trail, they used a single ribbon. That's all that was needed for this proof of concept demonstration.
Smoke trail , camera angles , High speed cameras 360 degree . wind speed constant monitoring ground level . was it really flat? .
Yes he Won the Bet . But there seems to be a lot of missing data . Weight of the blackbird . Forward and back thrust being produced by the
propeller real time measurements . There are really too many incomplete variables that are not shown .
Do you not agree with that fact that all air molecules move in the opposite direction of travel when vehicle is above wind speed traveling in the wind direction?No, we do not agree.
QuoteI'm not wrong. If I was wrong one of you will have been able to explain where I'm wrong.
Conversely, if you were right one of us would have been able to understand your explanation.
You're using a non-sequitur as proof, and not just in this example.
I promise I will never refuse a sound argument.If I was wrong one of you will have been able to explain where I'm wrong.We have tried. Several dozen times. What prevents us from being successful is your steadfast refusal to consider, let alone accept the arguments that have been put before you - and your continued reliance on your own, flawed beliefs. Even there, we have tried to explain the flaws, but you seem incapable of accepting that possibility.
You think you're right - and that's as far as your thinking goes.
How can we explain where you're wrong when you won't even listen?
The typical molecule velocity from the thermal movement is about at the speed of sound.
The propeller in the prop driven vehicle makes the air around the propeller to more slower than the wind. So the wind can push against that moving air.
Relative to the vehicle that air mores in the opposite direction of the travel, but relative to ground the air pushed by the prop is still moving in the direction of the wind and vehicle.
since the wheels are connected (by an elastic rubber band)
Quotesince the wheels are connected (by an elastic rubber band)
Don't use an elastic band - it is introducing something that's not necessary and just confuses things (such as, allowing you to claim energy storage). Ideally use a toothed belt, but failing that a tight piece of string would be better.
still significant amount of energy storage in the vehicle body flex that can be seen in the video I just posted.
So the movement is not smooth it is start/stop
Quotestill significant amount of energy storage in the vehicle body flex that can be seen in the video I just posted.
Well done for posting the experiment.
The lack of wheel grip was expected. I wouldn't have called it energy storage, but I suppose you could get away with that, despite the amount being trivial.
QuoteSo the movement is not smooth it is start/stop
Can you point to any part of this discussion that demands the progress be actually smooth rather than having the appearance of same? AFAIA, the requirement was simply that the vehicle moves left to right, something you insisted was impossible.
More importantly, whether or not it does that in a hop, skip or jump, it will continue to do so for as long as the paper lasts (or forever if on a treadmill). Your supposition, as I understood it, was that such effects would be temporary and last only as long as the initial energy storage could give a boost.
I say blackbird will stay above wind speed only as long as there is still stored energy so depending on vehicle construction (amount of energy that can be stored and vehicle friction losses) from a few seconds to a few minutes.
Is fairly clear that vehicle stores energy and when that is large enough the vehicle will use it to advance forward. Motion seems smoother the smaller the storage capacity is since charge discharge cycles happen faster than our brain can perceive.
Without any amount of energy storage the vehicle will not be able to move forward left to right.
I say blackbird will stay above wind speed only as long as there is still stored energy so depending on vehicle construction (amount of energy that can be stored and vehicle friction losses) from a few seconds to a few minutes.
I know I'm jumping in here, but how do you reconcile your claim with the fact that Blackbird (and demonstrations with similar craft) show the craft maintains a significant multiple of the true wind speed while continuing to accelerate? The typical, and measured wind gusts come nowhere close to exceeding the craft speed as evidenced by the wind-indicator ribbons. And even if there were significant gusts, I see no energy storage system (inertia, flex, etc), especially in the tiny treadmill devices, that could maintain speed for any appreciable time.
One factor I haven't deeply studied (and may have already been answered in this thread), is how the craft gets started from an initial zero ground speed? If the propeller were symmetrical in push/pull then wouldn't the craft be driven upwind? I see two possibilities:
1) The propeller has more drag than lift when spinning "backwards", or
2) There is an initial"push" given to the craft that gets the propeller spinning fast enough to generate lift. This is certainly the case in the treadmill tests, but I don't see it in the Blackbird tests.
But even if #2 is the case, it doesn't invalidate the general principal being demonstrated.
Is fairly clear that vehicle stores energy and when that is large enough the vehicle will use it to advance forward. Motion seems smoother the smaller the storage capacity is since charge discharge cycles happen faster than our brain can perceive.
Without any amount of energy storage the vehicle will not be able to move forward left to right.
I think you're just seeing stiction (https://en.wikipedia.org/wiki/Stiction (https://en.wikipedia.org/wiki/Stiction)). Eliminate stiction (but some amount of friction is still OK), and there will be no jumps, only steady motion.
If I where to compete with a similar vehicle just using a round 5.3m diameter sail instead of the propeller I will win the race any time against the blackbird.For such a race is depends on the lenght of the race. Initially the backbrid needs additional energy for the kinetik energy of the propeller - that is the main extra energy storrage. So in the acceleration phase to maybe 90% of the wind speed the sail drive may be more efficient, but later the propeller extra pushing action can give it an edge and finally reaching the higher sustained speed (e.g. 120% the wind speed), while the sail may only reach 95%.
What will happen is that I will be the first to accelerate while blackbird will waste a lot of time initially to store energy and while blackbird will have higher top speed the average speed will be lower due propeller being less efficient than a sail.
So if the race is about top speed then blackbird can win but if it is about getting first to finish line then sail vehicle will win.
Where is the energy storage in a lever? Where is it in a gear ratio? Energy storage is not required here.
If I where to compete with a similar vehicle just using a round 5.3m diameter sail instead of the propeller I will win the race any time against the blackbird.For such a race is depends on the lenght of the race. Initially the backbrid needs additional energy for the kinetik energy of the propeller - that is the main extra energy storrage. So in the acceleration phase to maybe 90% of the wind speed the sail drive may be more efficient, but later the propeller extra pushing action can give it an edge and finally reaching the higher sustained speed (e.g. 120% the wind speed), while the sail may only reach 95%.
What will happen is that I will be the first to accelerate while blackbird will waste a lot of time initially to store energy and while blackbird will have higher top speed the average speed will be lower due propeller being less efficient than a sail.
So if the race is about top speed then blackbird can win but if it is about getting first to finish line then sail vehicle will win.
If there is extra energy storred this extra would go up with higher speed. So that extra energy could not be used to accelerate - this would be the wrong way round, like using the gain in potential energy to go up faster up-hill, which obviously does not work. There just is not significant energy storrage that would help to accelerate the vehicle (only to slow down the slow down).
The energy storage is just there to confuse and maybe find an explaination for seeing the experiment but not believing. The idea with energy storrage just does not work.
The example with wheels can be calculated without, with very basic math - not even physics. The movement to the right is the only one compatible with the gear ratio and no slip at the wheeels. Trying to convince us of the opposite as nearly as crazzy as claiming that 3x3 is 7. :horse:
I am now halfway through this entire thread, and I apologize to all for beating a thoroughly dead horse.
Have you watched my slow motion video https://odysee.com/@dacustemp:8/wheel-cart-energy-storage-slow:8I watched the video, and when taking the paper as the reference and the ground as the driving system, the average velocitiy is larger - the vehicle moved to the right in the picture after all, so more to the right than the ground has moved.
The energy storage is fairly visible there and you can see vehicle peak speed even above paper speed but average speed is much lower. Same will happen to blackbird if you waited long enough to see multiple cycles then it will look the same as my toy cart.
There can not be a sustained speed above wind speed and that I what I try to make people understand.However this is only a claim and not a valid argument. This claim is also wrong - any most of us here have recognized this and try to explain how it is possible. The videos show it - just believe what you see and don't thing it is the red hering making this happen. It is much simpler.
Have you watched my slow motion video https://odysee.com/@dacustemp:8/wheel-cart-energy-storage-slow:8I watched the video, and when taking the paper as the reference and the ground as the driving system, the average velocitiy is larger - the vehicle moved to the right in the picture after all, so more to the right than the ground has moved.
The energy storage is fairly visible there and you can see vehicle peak speed even above paper speed but average speed is much lower. Same will happen to blackbird if you waited long enough to see multiple cycles then it will look the same as my toy cart.
The movement is a bit jerky, because the paper is not moved at a constant speed. Move the paper at a constant speed and vehicle would also more at a constant speed.There can not be a sustained speed above wind speed and that I what I try to make people understand.However this is only a claim and not a valid argument. This claim is also wrong - any most of us here have recognized this and try to explain how it is possible. The videos show it - just believe what you see and don't thing it is the red hering making this happen. It is much simpler.
Do you not agree with that fact that all air molecules move in the opposite direction of travel when vehicle is above wind speed traveling in the wind direction?No, we do not agree.
Please provide details if you do not agree.
My statement is not only very simple but also correct so I wait for you explain how my statement is incorrect
We do not agree that there is an apparent violation of conservation of energy -- because there isn't: the vehicle extracts energy from the velocity difference between the wind and the ground and uses that to make the vehicle go faster.
We don't agree that kinetic energy is a vector. Because it is not, as taught in every physics class since Newton. Momentum is a vector, kinetic energy is a scalar. This is your most fundamental, most wrong misunderstanding of basic physics. See: Kinetic energy is a scalar (https://www.khanacademy.org/science/physics/work-and-energy/work-and-energy-tutorial/a/what-is-kinetic-energy#:~:text=Kinetic%20energy%20is%20not%20a%20vector.)
We don't agree that "the resultant velocity must be in the direction of the net work" because that is a meaningless statement based on the above two points.
We don't agree that energy storage beyond the kinetic energy of the vehicle is relevant in any of the demonstrations because they are not, and you have not been able to quantify or explain them
We solved the free body diagram to find the net torque and direction of motion of the mechanical analog. You don't agree that we did math correctly, but you didn't explain why other than your misguided "net work" argument.
And now that you finally agree that a moving piece of paper is equivalent to a treadmill, you have invented a new energy storage theory that also makes no sense. I did the experiment you said would prove you right or wrong, it proved you wrong, and you don't believe it. I was sitting right there and I can tell you that the belt was stretching a negligible amount and not storing energy.
Multiple experiments demonstrate that you are incorrect, so yes we disagree.
Front wheel powers the back wheels and so if power at the back wheel can only be smaller than power on the front wheel then vehicle can not advance from left to right.
Front wheel powers the back wheels and so if power at the back wheel can only be smaller than power on the front wheel then vehicle can not advance from left to right.
Why not?
Power does not have a direction, so whether left to right, or right to left is not relevant. If the vehicle can move, it can move in any direction we want: left, right, or sideways. It only depends on the gearing. There are no physical constraints that prevent this.
The aerospace industry , Lockheed , NASA , and all the others .
Don't use this in a wind tunnels or for replacing the use of a wind tunnel due to there significance and disturbance in testing.
So Please get over the treadmill . .. hamsters likes them .
2 W on the motor is more than enough to move the vehicle from left to right. So if you have 10 W from the generator, you have 2 W to turn the motor and 8 W spare to overcome friction and other losses. Since 2 W is much less than 10 W, there is no problem with conservation of energy.
Remember that it takes no power at all to keep the vehicle stationary. All you have to do to achieve that is to lock the motor axle so the wheels cannot turn. If it takes zero power to keep the vehicle stationary, it follows that any power greater than zero can move it to the right.
People seems not to get the generator part as I hear many times in my business people saying things like while not charge my RV battery while I drive as there is power available at the alternator that is free to use since I still need to drive anyway.Those figures are a bit dated, a modern Atkinson cycle engine can reach 40%. And what makes more sense is to charge the batteries using regenerative braking.
They have a hard time understanding that taking 1kWh from the alternator and putting it in the battery will cost them an extra liter of gasoline so about $1/kWh or about 50x more expensive than getting that 1kWh from a solar panel.
So there is no free energy available at alternator since as soon as you take power out of alternator say 1kW then you add an extra 2kW load to the engine (alternators are typically just 50% efficient so the other 1kW is just heat in the alternator).
A liter of gasoline contains about 9kWh of energy if you burn it but since engine is typically just around 20% efficient you can see how 1 liter is needed for 1kWh took out from alternator.
Those figures are a bit dated, a modern Atkinson cycle engine can reach 40%. And what makes more sense is to charge the batteries using regenerative braking.
Of course there is a direction as the generator wheels in order to produce the power will need to oppose the moving paper direction so whatever you extract from there you can call that breaking power.
So if you have 10W of breaking power you will need a motor on the other wheels (back wheels) that also receives 10W just for vehicle to remain stationary(ideal vehicle in real world you will need more power on the back wheels to also cover the friction losses).
Of course there is a direction as the generator wheels in order to produce the power will need to oppose the moving paper direction so whatever you extract from there you can call that breaking power.
So if you have 10W of breaking power you will need a motor on the other wheels (back wheels) that also receives 10W just for vehicle to remain stationary(ideal vehicle in real world you will need more power on the back wheels to also cover the friction losses).
We are actually maybe getting somewhere here. The paper does positive work on the cart. Work (energy) is force * displacement while power = force * velocity. If the force and velocity are in the same direction then the power input is positive and can increase the kinetic energy of the car. If they are in opposite directions it the work is negative and will decrease kinetic energy. While the paper is moving backwards and the cart is moving forwards, the bottom surface of the wheel is moving backwards so the work done is actually positive. This power input is what allows the cart to accelerate, or in the steady state with no acceleration provides energy to overcome frictional losses. Due to the gear ratios the acceleration of the cart is in the opposite direction of the force applied to the wheel, but that is no big deal because energy doesn't have a direction.
The table provides a forward force but since it is stationary and so is the bottom of the wheel it does no net work. The table is still important because we still need a forward force to balance F=ma. But the table can provide that balancing force without doing any work, positive or negative.
The change of reference frame is helpfull. If one understands how to do it, it is a realtively simple step - though it can confuse others.
If for some reason you want to look at this system from a modified frame of reference and switch the vehicle characteristics with ground and then continue the experiment from there then you need to look at now moving ground kinetic energy to know what happens with vehicle speed.
I do not get why people will want to change reference frame and deal with all the complexities of that (seems people wrongly think changing reference frame has no implications and then get to wrong conclusions).
You can see the effect of how blackbird works in my slow down video version.is completely invalid. You presume your presentation models the Blackbird mechanism - which it clearly does not.
"Flywheel storage" - that is a concept which works.
Why the rotating propeller and rotating wheels cannot be considered (in the Blackbird's case) an "energy storage"?
https://en.wikipedia.org/wiki/Flywheel_energy_storage
Jesus Christ! STILL can't spell BRAKING correctly? |ONot quite sure what that fictional character has to do with me not spelling braking correctly but I get your point and I try to improve. I did not made the English grammar rules where two words with completely different meaning have the same pronunciation but different spelling. And the limited spell checker is not able to help me here.
Look, the spinning propeller creates a virtual sail behind the vehicle. It pushes against the wind, if you sum the velocities you end up with 0, but since it's blowing backwards, it adds to the ground velocity of the vehicle.
You can view this virtual sail as storing the initial energy of the vehicle being blown from rest to its cruising velocity.
(Attachment Link)
is completely invalid. You presume your presentation models the Blackbird mechanism - which it clearly does not.
You also need to consider what the example represents.
Moving paper is the energy source for the vehicle and is the analog of win in a wind powered vehicle. So in my video you see the equivalent of a vehicle moving in the opposite direction to wind direction at around half the wind speed.
If you want to see things in modified reference frame
The flyheel effect of the propeller can not be used to accelerate the vehicle, unless they have a variable gear box that they would switch. The speed is linkt to the wheels and thus the speed to ground. So like heavy wheels the flywheel part acts similar to additional mass of the vehicle, making it harder to accelerate. From that argument there is also no real way to osciallate energy between the pressure field around the vehicle and the propeller RPM.
Keep in mind that the time scale for the pressure field is the speed of sound divided by the dimensions - so this is less than 1/10 s. Without a container, compressed air is not really effective storrage. So if at all you may get a very short time effect, hard to see on a normal video. So the air pressure around the vehicle can not be the magical energy storrage to drive the vehicle.
It is the stationary pressure field that builds up that drives the vehicle. The wind and the power from the wheels transferred to the prop keep this pressure field around the vehicle alife. Usually one uses a simpler pricture of the working the prop and calls it the prop driving the vehilcle.
Sure, it was your demo proposal, not mine. You claimed this was a demo that would satisfy you and that it was impossible for the cart to go the opposite direction from the applied force. It clearly is possible as I have demonstrated and as I have explained both in terms of forces and torques as well as power transfer.
You can alter it to go the other way if you want, or faster than the paper in either direction with a suitable choice of gear ratio. The traction requirement becomes larger so you will probable need to add some weight to the cart and pull carefully. I can sort of get that to work with the same cart just putting the paper under the other wheel but it's hard to do without slip.
QuoteIf you want to see things in modified reference frame
I would caution you against trying to think in multiple reference frames at once,. It's true that you can solve the problem in any reference frame you want and get the right answer but in practice it is an easy way to get led astray. This is because it is easy to come up with "invariants" that are not actually true in other frames.
One example is that the change of frame itself does not conserve energy. In fact in any reference frame other than the lab frame the kinetic energy of the ground is "infinite" so you basically can't use global conservation of energy. Also if you solve part of the problem in one frame and switch you will get inconsistent results. Another problem people have is that they accidentally use an accelerating reference frame such as a vehicle that isn't moving at constant speed.
All this is why I stick to the lab frame in all of these discussions.
Yes it is impossible without energy storage to move the vehicle from left to right. The force at wheel difference is a bad explanation as is about power at the wheel not force.:bullshit: :bullshit: :horse:
Vehicle can not travel faster than paper unless it uses energy storage.
Yes it is impossible without energy storage to move the vehicle from left to right.
Using force is the easier way and the normal way one does the calculation if one understand what force is.
Using power works, but has to take the different velocities into account and than gets the same result as using power.
QuoteYes it is impossible without energy storage to move the vehicle from left to right.
I see two situations here:
1. There is no energy storage and you are wrong.
2. There is energy storage but it doesn't matter because the vehicle moves from left to right as long as the paper is moving.
That's the crucial thing, is it? That the movement is not a temporary 'until the energy is gone' thing. Even if the alleged energy store was storing and releasing, storing and releasing, the thing is moving left to right without at any point moving right to left. And it will keep doing that for as long as the paper holds out. So how it does that in movement terms, in what way the energy is tranferred, is just a sideshow and irrelevant.
You entire supposition of the energy store is that it cannot last indefinitely, so the Blackbird can only slow down to below wind speed once it has used the store to get above wind speed. But here your store (that is, your energy store, wherever you happen to think it is) is able to keep supplying the left-to-right motion forever.
The problem with the motor example is that a motor with a given fixed efficency is kind of including an automatic gear box to adjust the torq to the speed. So it does not make sense to consider such an ideallized motor with a gear reduction. So the example just does not make sense. With only one vehichel on one groud the wheels see the same speed and thus in this special case with ideallized motors power is directly proportional to force. So in this special case there is no consequence in confusing power and force - except that you can't transfer this to other cases.
So the example is a bad example.
Things get different if the 2 wheels are on different grounds moving relative to each other.
Vehicle is powered by the paper (hope you agree with that)
so it can move forever
but speed of the paper will always be higher than average speed of the vehicle
Irrelevant and meaningless. Does the vehicle move left to right? Yes! Does it move right to left? No! That's it - how, when, with what, at what speed, anything else is not relevant and just adding to the noise.
It moves from left to right because there is energy storage
for a limited amount of time proportional with the amount of energy storage capacity available
This represents a vehicle with a wind turbine driving against the wind direction at about 0.5x
Here's a demonstration I like (most have probably already seen it), and it's pretty similar to the moving paper test. In this one the upper board represents the wind, and the floor represents the ground, so there's no reference frame issue. And there's no energy storage, and it can continue, faster than the "wind", forever.
1. Sorry but how can you say there is no energy storage when is so clear seen in video and is clear to see the stretching of the rubber belt.
I don't think it really matters if the wood is the wind or the floor is the wind.
The important factor is that two forces together can move the object faster than either of them are moving.
It's the best demonstration of how this can be possible that i have seen so far.
1. Sorry but how can you say there is no energy storage when is so clear seen in video and is clear to see the stretching of the rubber belt.
Here is a classic example of YOU not understanding physics.
The jerkiness of motion as seen in your video is the direct cause of friction in your system. You are taking a practical experiment - with all its flaws - and creating a theoretical model based on a completely wrong interpretation.
Remove the friction and it will run smoothly. Use an inelastic chain and it will run smoothly, too. In the ideal scenario, there is NO energy storage, just energy transfer. Your demonstration is nowhere near ideal.
You are only compounding the lack of understanding you have in physics. You are making it very clear that you are either a troll or the epitome of Dunning-Kruger.
It maters what is the input and what is the output.
Is like looking at the gear box input and output power and saying if you look from this side efficiency is 90% but if you look from this other side is over 100%
Do you understand what I'm trying to say ?
It maters what is the input and what is the output.
Is like looking at the gear box input and output power and saying if you look from this side efficiency is 90% but if you look from this other side is over 100%
Do you understand what I'm trying to say ?
yes, I understand what you mean now. It does matter which is wind and which is ground.
But I still think the demonstration shows how blackbird can move downwind faster than the wind.
It seems that most peoples arguments against the results of blackbird are that the wind was going faster than stated.
Here's a demonstration I like (most have probably already seen it), and it's pretty similar to the moving paper test. In this one the upper board represents the wind, and the floor represents the ground, so there's no reference frame issue. And there's no energy storage, and it can continue, faster than the "wind", forever.
That demonstration is wrongly interpreted.
The floor is the wind and the vehicle travels on the moving lumber.
Small wheels are the generator wheels and the large wheel is the motor wheel. Please look closer to understand how that works.
So what is shown there is a vehicle traveling at about 2.8x lower speed on the lumber (road) powered by the floor (wind) in the opposite direction.
A good demo might be
- Toroidal pipe, maybe 4inch diameter and 1m toroid radius.
- Angled air in/out tubes to get known air flow moving around the tube.
- Small cart with propeller and ball bearing wheels touching 3 or 4 sides of the tube. (so it can move freely around the toroid)
- Could test with water or air flowing around the tube, water might make coupling/gearing easier to show the effect.
- The relationship of air flow into tube and speed of cart around toroid should be very different if the effect happens vs does not happen.
It seems that most peoples arguments against the results of blackbird are that the wind was going faster than stated.
So this would allow windspeed to be known/measurable
And I say you are the one misinterpreting the demonstration. The floor is the ground and the board is the wind.
What the difference between a vehicle going from point A to point B faster than the wind on water vs on land.
Whenever I look at attempts to disprove that blackbird can travel continuously downwind faster than the wind. I always come back to the fact that, if a sailboat can do it then how can it suddenly become impossible on land.
Regardless of the design or implementation of blackbird, if it's possible on water, it should be possible on land. And if it is possible on land, how would it be done?
I think the core of the issue is going back to first principles. What the difference between a vehicle going from point A to point B faster than the wind on water vs on land.
Because it's accepted fact that you can do it on water without breaking any physics
I can absolutely understand that one would have trouble with the explaination base on the sail going around the shaft. I think this is even wrong.
However the case with driving the prop from the wheels is relatively easy - just get the difference between forces and power. This way one can avoid al the complication from aerodynamics and wing profiles.
When vehicle speed equal wind speed there is zero wind power available and the only way to exceed that speed while only moving directly down wind is by storing energy before that point and using that stored energy to do so.We are running in circles here. The video showed that it works (2 experiments) and give an explaination how, it just is not so easy to understand, and thus the title.
Edit: Sorry - removed a post after considering I might be being sucked into something I know little about.
When vehicle speed equal wind speed there is zero wind power available and the only way to exceed that speed while only moving directly down wind is by storing energy before that point and using that stored energy to do so.We are running in circles here. The video showed that it works (2 experiments) and give an explaination how, it just is not so easy to understand, and thus the title.
The point is, that there is power from the wind available, even when moving with the wind at exactly the speed of the wind or somewhat faster. This does not work with a passive sail, but it does work with the fan.
The energy storage idea to explain the experiment is just not working: both the large vehicle driving outside and the small one on the treadmill have quite some friction. There is nowhere enough stored energy to drive the vehicle for the time they where going faster that the wind or against the treadmill. The kinetic energy does not count here as it is not available to accelerate or keep the speed.
... and everybody agrees it can not exceed wind speed directly down wind and can also not travel at any speed directly upwind.
Quote... and everybody agrees it can not exceed wind speed directly down wind and can also not travel at any speed directly upwind.
Yes, but we can't agree why. There is plenty of energy available from the wind, it's just that we can't reap it because there is no relative movement between the vehicle and the wind. It's not like the faster the vehicle goes the more energy it needs to overcome air friction, because there is less of that when we travel with the wind.
The problem can thus be reduced to merely finding some means of making the wind move relative to us, and that's what the prop does.
You can stay as long as you want below wind speed and have as much energy as you want
QuoteYou can stay as long as you want below wind speed and have as much energy as you want
Well, this is it, isn't it? It's not the vehicle that is extracting the energy from the wind, and is therefore dependent on relative wind speed, but the prop. Apply your formula to the prop instead of the vehicle and things are a bit different, no?
Of course, the vehicle is being pushed by the wind, but when it is at wind speed everything is balanced - there is enough push to keep it moving but nothing extra to let it go faster. But the prop is not moving at wind speed and can therefore extract power from the wind. Add that to the balance and not the vehicle can go faster.
And the idea that energy storage is necessary (the underpinning of the electrodacus theory) is just plain wrong. Yes, air is compressible, but there is no storage in the system that is adequate to accelerate the craft beyond windspeed for more than a very brief period. The various craft have demonstrated an ability to move well beyond windspeed, effectively continuously, no energy storage required.
And I still claim that that demonstration with the wheeled fixture, rolling on the floor, with the board above it being pushed, is a good analogy. The floor represents the ground, the board represents the wind, and the large coupled wheel represent the propeller. The fixture moves "downwind" faster than the "wind". electrodacus disagrees. But regardless, the "energy storage" notion is wrong, and this nullifies the electrodacus theory.
What craft has demonstrated direct down wind faster than wind continuously ?
The wheel analogy represents a direct upwind and it also uses energy storage but in a very different way and it is not pressure differential.
Here is the formula for wind power
0.5 * air density * area * (w-v)^3
air density is usually around 1.2kg/m^3
area is the sail area or equivalent (the area facing the wind direction) so direct down wind this is also a constant for a sail
and then is that w (wind speed) minus v (vehicle speed) so there there is max power when vehicle just starts moving super low vehicle speed and decreases significantly as vehicle approaches wind speed and equal with zero when at wind speed.
Here is the formula for wind power
0.5 * air density * area * (w-v)^3
air density is usually around 1.2kg/m^3
area is the sail area or equivalent (the area facing the wind direction) so direct down wind this is also a constant for a sail
and then is that w (wind speed) minus v (vehicle speed) so there there is max power when vehicle just starts moving super low vehicle speed and decreases significantly as vehicle approaches wind speed and equal with zero when at wind speed.
This formula is quite wrong. It is not w (wind speed) minus v (vehicle speed), it is w (wind speed) - g (ground speed), where g is usually zero (unless on a treadmill).
This means the available power from the wind does not depend on the speed of the vehicle, and so it does not matter if the vehicle is travelling at the speed of the wind or faster than it. As long as the vehicle has contact with the ground (wheels with friction), and contact with the air (some kind of sail, fan or propeller), then the vehicle can extract power from the wind/ground system and can go faster then either.
Wheel friction is important here. If the vehicle was skating on ice, then it couldn't go faster than the wind speed, since it would have no grip on the surface between it.
Similarly, it is easy to make a wheeled vehicle run directly into a headwind, and any number of experiments can be built to demonstrate this.
Wind is powering the propeller not the wheels .
The wind tread mill is not correct .
Sorry you can't emerlate wind with a tread mill
When going at the speed of the wind or faster than the wind, the wheels are driving the propeller, not the other way around.
It is a nice graphics from electrodacus, but for the formulars one has too look at what they actually calculate, and what the symbols stand for:
In the pictures the vehicle is standing still with respect to the picture and the belt like platforms should show the relative movements.
The right side is easy, that is the power generated by generator. Positive power giving power given off.
For the left side it is a bit more complicated: The "belt" velocity is (w-v). The power of the motor is F_m*(w-v). Here a negative sign means the motor takes up power and a positive sign means the motor works as a generator and produces power.
Together with the power from the genrator side this gives the net power. A postive values means excess power available.
So the lable P_out is a bit misleading here. P_net would be more suitable.
The center formula gives the power from the motor, as the generator power is again subtracted.
So the labels there are somewhat mixed up ! And thus the wrong conclusions.
In the upper case both side produce power and thus the high excess power.
In the middle case the motor stands still, and thus no power and thus the 10 W power from the generator.
In the lower case there is still a 6 W of excess power.
So all three cases show net power excess to overcome friction. So all 3 cases are possible.
Also there was never a sail based vehicle demonstrated to exceed wind speed directly down wind (not even close not even in ideal case).True. I don't think that a sailboat can beat a balloon to a goal directly downwind -- even if the boat jibes.
Unless you have an energy storage device you can not exceed wind speed directly down wind.Using a sail. But a propeller is not a sail, and a boat on the water is not a wheeled vehicle on the land.
I have no problem understanding a propeller . I was flying twin engine Doves out of Hurn airport UK in the 60s . ( Not the white birds that flap around going tweet tweet )Wind is powering the propeller not the wheels .
The wind tread mill is not correct .
Sorry you can't emerlate wind with a tread mill
The propeller was replaced by the M wheel here not to have any confusion about what a propeller is (just a wheel for traveling trough materials instead of on the material surface).
People think propellers have magical proprieties so I prefer to use wheels as people have less misunderstandings about how those work. :wtf: :-DD
I have no problem understanding a propeller . I was flying twin engine Doves out of Hurn airport UK in the 60s . ( Not the white birds that flap around going tweet tweet )
WE used the wheels to keep the fuselage from scraping along the ground . They also came in handy for Landing ..
A wheel is a wheel .. Propeller is a propeller that's it .
The design of a propeller is an important factor the one they use was a variable pitch . ie the blades can have their pitch altered
this is not in your formulae.. There for you are wrong . Angle , degree etc etc etc Please add these in .
Then the Bumble Bee will fly :-DD
The wind power the vehicle and thus the wheels. The power from the wind is given as thrust from the propeller times speed of the wind.When going at the speed of the wind or faster than the wind, the wheels are driving the propeller, not the other way around.
What powers the wheels ?
For the math it does not matter if one has a motor or generator. Esepcially wheels and gears work in both directions. In the examples generator power was fixed positive and there is no porblem ignoring the possible extra power in case 1, it works without it too.
I wanted to make sure is clear Pout includes Pin that is why a Pnet was need to show what the state of the vehicle will be.
Derek used the wrong formula with (v-w) instead of correct (w-v) and only looked at the case C so vehicle speed above wind speed.
The G wheel is a generator only and M wheel is a motor only same as with blackbird where wheel is only a generator and propeller is only for propulsion (they even have a freewheel device installed to make sure power from propeller can not be transferred to wheel).
In case B that is at the limit so in real world you can not get exactly zero speed but all power from generator is needed at the motor else vehicle will move backwards (decelerate if any power is generated at G wheel and not all of it is put in the M wheel).
If someone suggested you take energy from back wheel and power the front wheel it will have been considered stupid (by most not all) but suggesting taking power and supplying a propeller is even worse if you exclude that energy storage in pressure differential that nobody seems to understand.
I have no problem understanding a propeller . I was flying twin engine Doves out of Hurn airport UK in the 60s . ( Not the white birds that flap around going tweet tweet )
WE used the wheels to keep the fuselage from scraping along the ground . They also came in handy for Landing ..
A wheel is a wheel .. Propeller is a propeller that's it .
The design of a propeller is an important factor the one they use was a variable pitch . ie the blades can have their pitch altered
this is not in your formulae.. There for you are wrong . Angle , degree etc etc etc Please add these in .
Then the Bumble Bee will fly :-DD
I hope not to sound rude but unis something is different from understanding how it works.
Variable pitch is the equivalent of a gearbox before the wheel.
A propeller has no "magic" proprieties and it is the exact analog of a wheel just that a wheel is much more efficient than a propeller.
So if you wanted to be more efficient you will have just took the power from the back wheel on blackbird and put that in the front wheel. But doing so provides you with no energy storage as road is not compressible like air.
If someone suggested you take energy from back wheel and power the front wheel it will have been considered stupid (by most not all) but suggesting taking power and supplying a propeller is even worse if you exclude that energy storage in pressure differential that nobody seems to understand.
If someone suggested you take energy from back wheel and power the front wheel it will have been considered stupid (by most not all)Yes, that would be stupid.
but suggesting taking power and supplying a propeller is even worse if you exclude that energy storage in pressure differential that nobody seems to understand.Well, somebody doesn't understand, and it's you. There is no speed difference between the road and the road. There is a speed difference between the wind and the road.
The wind power the vehicle and thus the wheels. The power from the wind is given as thrust from the propeller times speed of the wind.
For the math it does not matter if one has a motor or generator. Esepcially wheels and gears work in both directions. In the examples generator power was fixed positive and there is no porblem ignoring the possible extra power in case 1, it works without it too.
Whether one uses (v-w) or (w-v) is only a question of the direction of the force, so it depends which side of the balance to look at.
The case B is very easy: one needs no power to stop the motor. So here zero power for the motor is obvious.
Well, somebody doesn't understand, and it's you. There is no speed difference between the road and the road. There is a speed difference between the wind and the road.
The 3 examples really show a lack of understanding.Well, somebody doesn't understand, and it's you. There is no speed difference between the road and the road. There is a speed difference between the wind and the road.
Well yes that is what I showed in my 3 examples.
That difference at case C is not helping you so it will be even worse that putting that energy in the front wheel.
The 3 examples really show a lack of understanding.
The 3 examples really show a lack of understanding.
Are you saying that wrong formula was applied ?
If so please provide yours for all 3 cases.
The motor power is given by force times velocity seen by the motor. The velocity seen by the motor is w-v, as the indicated speed in the graphics.
So this gives: P_motor = F_M * (W-V). For the steady state with no extra friction F_M = F_G.
A positive value would be power generated by the motor and one may want to set this to 0 instead. A negative power is actual power send to the motor.
The net power or excess power available to overcome friction is then P_net = P_in + P_motor.
So for the 3 example cases this would be 30 W (or 10 W if the motor is not allowed to generate power), 10 W and 6 W of net power.
You are wasting your time arguing with formulas, when actual, real-world experiments prove the case. All of us know we can build a craft that will run faster than the wind indefinitely, and all of us know we can build a craft that will run directly into the wind.
Since this is a fact that is provable by experiment, it doesn't really matter that you keep trying to say that it is "impossible". It is like trying to argue that it is impossible for airplanes to fly, when airplanes are flying over your head.
We don't need to prove anything to you, because we know already what the facts are. You are the one who needs to figure out for yourself how these things work. It is not our problem to convince you.
The motor power is given by force times velocity seen by the motor. The velocity seen by the motor is w-v, as the indicated speed in the graphics.
So this gives: P_motor = F_M * (W-V). For the steady state with no extra friction F_M = F_G.
A positive value would be power generated by the motor and one may want to set this to 0 instead. A negative power is actual power send to the motor.
The net power or excess power available to overcome friction is then P_net = P_in + P_motor.
So for the 3 example cases this would be 30 W (or 10 W if the motor is not allowed to generate power), 10 W and 6 W of net power.
Motor is what is says a motor not a generator. The back wheel is always powered by the front wheel that is the generator.
Same vehicle is used in all 3 case the vehicle construction can not be changed and will be the same for all 3 cases.
If I understand right your answer is net power acting against vehicle is always positive and this are the values.
A) 10W
B) 10W
C) 6W
Is that correct ?
It seems that you agree in steady state (referenced to vehicle) FG = FM due to Newton third law.
Without any connection between generator and motor both generator and motor will free spin with no friction (all ideal case).
You can decide what power you want to take out of generator and send to motor in my example I used 10W
Then if 100% of that generated power is sent to the motor Pout will always equal Pin plus or minus some extra depending on speed and direction of the wind treadmill.
[...] small capacity energy storage and stick slip hysteresis with charge discharge cycles of many times per second. [...]
I don't agree with the red part: the net power is not acting against the vehicle - power has no geometric direction. A positive net power means there is power to spare, e.g. to turn on some lights or overcome friction. So not all power needs to be send to the motor in the steady state for the 3 cases shown. If all the power is send to the motor it would accelerate the vehicle and thus no longer the steady state.
The orange part is some-how self-contradicting, or I don't understand what it should mean: if all the power is send to the motor, Pout=P_motor= Pin and thus no net power.
[...] small capacity energy storage and stick slip hysteresis with charge discharge cycles of many times per second. [...]
This is all so wrong... No wonder you can't figure it out.
To make it easier to replay I will post again the photo with the 3 cases
Pout is the power provided by the wind (wind treadmill in this example) FM * (w-v) plus what you supply the motor with so Pin
Case A: The vehicle can move to the right with zero power from the generator and zero power at the motor.
Case B: The vehicle can move to the right with any power greater than zero at the motor. Less than 1 W would be enough.
Case C: The vehicle can move to the right with a power greater than about 4 W at the motor. Since this is less than 10 W, the generator can supply it.
In each case, the pictures clearly show the cart being able to move to the right.
Case A: If there is no power at the motor the motor will free spin there is no friction. So in all cases if motor is not connected to generator vehicle will not move in any direction.
Case B: Vehicle can only move to the right if there is an energy storage device and slip stick hysteresis. In this theoretical analysis there is no specified energy storage device so no movement is possible as net power will always be zero.
You can only have 1W at the motor if you get that from the generator and the two are opposite so will cancel Pnet = Pout - Pin (conservation of energy)
Case C: There is no way for that vehicle to move to the right. Not quite sure how your intuition works but you have a vehicle with no on board energy source and is on two treadmills both moving from right to left.
For direct upwind you can travel forever using those mini charge discharge cycles and stick slip hysteresis as you always have access to wind energy
QuoteFor direct upwind you can travel forever using those mini charge discharge cycles and stick slip hysteresis as you always have access to wind energy
Whoah! Have I read that correctly? You actually admit that going faster than the wind, with the wind directly behind, is doable indefinitely? I don't care if you have to butter it with this stick slip stuff, that simple acknowledgement makes this entire thread superfluous :-+
Case A: If there is no power at the motor the motor will free spin there is no friction. So in all cases if motor is not connected to generator vehicle will not move in any direction.
Incorrect, the wheel can be locked so it does not spin freely (a worm drive will do that). A wheel that is not turning requires zero power, since power = speed of rotation x torque.QuoteCase B: Vehicle can only move to the right if there is an energy storage device and slip stick hysteresis. In this theoretical analysis there is no specified energy storage device so no movement is possible as net power will always be zero.
You can only have 1W at the motor if you get that from the generator and the two are opposite so will cancel Pnet = Pout - Pin (conservation of energy)
Again, incorrect. If the motor wheel is stationary then it consumes no power, the vehicle is stationary, and 10 W is available to be used from the generator. Take any small amount of power from the generator and supply it to the motor, and the wheel can turn clockwise, moving the vehicle to the right. We have a perfect system with no slipping anywhere.QuoteCase C: There is no way for that vehicle to move to the right. Not quite sure how your intuition works but you have a vehicle with no on board energy source and is on two treadmills both moving from right to left.
This is the point. I don't use intuition, since intuition can give wrong answers. I use engineering analysis instead.
The cart can move to the right if the motor wheel turns faster than 4 m/s. There is ample power available from the generator to power the motor, since the generator wheel is turning at 10 m/s.
If you want a different way of looking at this, the "wind" and the "road" are moving at different speeds (4 m/s and 10 m/s), and the vehicle is in contact with both of them. The vehicle is therefore able to use the difference in speeds (10 − 4 = 6 m/s) as a power source to tap into, and it can use this power source to move in any direction it wishes by turning the motor appropriately.
I seen many of your comments on this forum and you look more like a troll thus not sure if is worth wasting my time.This is classical troll behavior, switching to ad hominem attacks when you can't debate the issue at hand.
If you think I'm wrong please provide the "correct" equation.You are wrong. I provided the correct analysis above.
The one Derek provided in his video is clearly wrong as it will make bad predictions for both case A and case C sine his equation include (v-w) instead of the correct (w-v) w-wind speed v-vehicle speed.Kindly don't change the subject. We are looking at your pictures, not Derek's equations. This is classic troll behavior--when you don't like the answer, switching to a different topic.
Also there was never a sail based vehicle demonstrated to exceed wind speed directly down wind (not even close not even in ideal case).True. I don't think that a sailboat can beat a balloon to a goal directly downwind -- even if the boat jibes.
That doesn't sound right from what I have read, granted there is lots of misinformation out there and i have not looked into this much. But I was under the impression it was accepted scientific fact that a sailboat can beat a floating balloon in a race from point A to Point B by traveling much faster than the wind (at an angle) and periodically changing directions to enable navigating to point B indirectly. (verses the balloon that travels directly at wind speed).
Is that actually not true?
I changed my strategy on a different forum so I will do the same here.
The claim of blackbird is that it can travel indefinite well above wind speed and even continue to accelerate.
Now imagine you are in your car traveling on a long strait highway directly downwind at 2x the wind speed and now you pop up a large propeller and connect that to the back wheels.
According to current theory you can stop the engine (no need to waste gasoline) and likely you have extra to even heat or cool the vehicle.
Does this seem reasonable to anyone ?
Again, you are trying to change the subject by making false comparisons.
The claim is that the Blackbird can do what it does.
Nobody is claiming you can stick a propeller on a big heavy car on the highway and achieve the same result.
So your question is irrelevant.
The correcrt equations / explainartions were given multiple time - just read them and understand or show, were you think they are wrong.
If you think I'm wrong please provide the "correct" equation.
The one Derek provided in his video is clearly wrong as it will make bad predictions for both case A and case C sine his equation include (v-w) instead of the correct (w-v) w-wind speed v-vehicle speed.
Blackbird plus driver has about 300kg. That is not that far of from a lightweight car.The balckbird example showed that it works, with an optimized vehicle (low friction) and relatively large prop, though not at twice the speed of the wind, but still faster than the wind.
Add a larger propeller I do not care the question is still valid. Cost is no issue.
I've not the time to waste any more on this silliness.
The correcrt equations / explainartions were given multiple time - just read them and understand or show, were you think they are wrong.
If you think I'm wrong please provide the "correct" equation.
The one Derek provided in his video is clearly wrong as it will make bad predictions for both case A and case C sine his equation include (v-w) instead of the correct (w-v) w-wind speed v-vehicle speed.
Are we back into this brick wall headbutting exercise again...? :palm:
The simple fact is that the Blackbird can do what is claimed. It has been proven experimentally - except someone doesn't want to admit something than runs counter to their intuition.
Many members have pointed out errors in their argument, yet they refuse to concede any of them. Stubborn and ignorant ... it's a fool's errand to try and correct them. I've not the time to waste any more on this silliness.
Dunning–Kruger is strong in this one.
You care about the wind power available to vehicle as that is what allows you or not to accelerate and the correct formula for that in case of a vehicle traveling directly downwind is
0.5 * air density * area * (wind speed - vehicle speed)^3
The conclusion for this correct formula is that there will be zero wind power available to any wind powered vehicle traveling directly downwind.
So no wind powered vehicle can ever exceed wind speed unless it has an additional energy source or as is the case here an energy storage device.
And as I keep telling you, that is not the correct formula. The correct formula is:
available power from wind = 0.5 * air density * area * (wind speed − ground speed)^3
If you keep putting vehicle speed in there you will get the wrong conclusion.
ground speed relative to what ?Speed of the ground, relative to the ground. In other words, zero.
the vehicle is in contact with the ground using wheels that do not slip.
ground speed relative to what ?Speed of the ground, relative to the ground. In other words, zero.
the vehicle is in contact with the ground using wheels that do not slip.
And even if the wheels did slip, as long as there's some traction (friction) the vehicle can still draw energy. Of course the system efficiency will be less with slipping wheels.
But for the sake of analysis, and for all practical purposes, the wheels do not slip.
you do not have an extension cord to a stationary location
you do not have an extension cord to a stationary location
You do have an extension cord to a stationary location. It is the wheels of the vehicle, which are sitting on the ground, which is stationary.
electrodacus, perhaps you should ponder the demonstrations with the small vehicles on the treadmill. I recall that you dismissed these demonstrations, but do reconsider. In these there is no appreciable energy storage, and the vehicle is propelled faster than the treadmill speed for extended periods of time. In these, the frame of reference has shifted, with still air and moving ground. Perhaps this will help.
If the vehicles work, but your models and equations don't, you should suspect that your models and equations are wrong.
The treadmill model showed in Derek's video works exactly according to my explanation.
While you set the vehicle on the treadmill you need to keep it there until all energy storage sources are fully charged up then you can release.
During that period you charge all flywheels (vehicle wheels and propeller so all spinning mass) and the important part you are also charging the pressure differential as tis is what will power the vehicle against the treadmill travel direction. The thing is that this stored energy is large enough that the treadmill is way to short to demonstrate how vehicle will start to slow down.
If you try to interact with the vehicle by keeping it moving forward you are just charging the pressure differential back to the same initial level.
So keeping the vehicle forced in place will result in you having an inefficient treadmill powered fan.
Now that propeller may be 50% efficient so you will end up with 25W of thrust.
25W of thrust
The difference between force and power is important here.
When you are holding the vehicle in place on the moving treadmill (relative airspeed zero) the vehicle is pulling on the string. Using your equation there should be zero force, but there obviously is. Remember, if you swapped the reference plane, this is identical to the vehicle rolling across the ground at windspeed. In your equation, where is this stored energy coming from when (vehicle speed = windspeed)?
Hint: There is no appreciable stored energy. There is potential energy that can be extracted.
25W of thrust
I don't want to rehash our much earlier discussion, but you saying this just another iteration of the fundamental misunderstanding that I was unable to get you to see then. This misunderstanding--and your statement here--is exactly where you are repeatedly going wrong. It's not a typo, lack of clarity or your self stated inability to explain things clearly. It is the fundamental error that is preventing you from seeing reality here.
The lack of understanding is coming from you.
The lack of understanding is coming from you.
Really? Do you not see the inherent fallacy in the term "25W of thrust"?
The main point of the propeller in this design is to store energy so that it can be used to accelerate the vehicle above wind speed.
There is absolutely nothing wrong in the way I presented thrust and you will understand that if you knew what power is.
The main point of the propeller in this design is to store energy so that it can be used to accelerate the vehicle above wind speed.
No, the main point of the propeller in this design is the change the effective wind velocity at the vehicle. If the vehicle was travelling at wind speed with a sail, the difference in speed between the wind and the sail would be zero. No force available. But if the vehicle is travelling at wind speed with a rotating propeller, the difference in speed between the wind and the moving surfaces of the propeller can be greater than zero. Hence, with a propeller, the wind can still provide force.
There is absolutely nothing wrong in the way I presented thrust and you will understand that if you knew what power is.
Thrust has dimensions of force and is measured in Newtons.
Power has dimensions of rate of doing work, and is measured in Newton-meters per second, or Watts.
Same thing can be had with a sail type vehicle just add an electric generator/motor and a battery and you can charge the battery while below wind speed then use that stored energy to accelerate above wind speed.
QuoteSame thing can be had with a sail type vehicle just add an electric generator/motor and a battery and you can charge the battery while below wind speed then use that stored energy to accelerate above wind speed.
Contrary to your presumed intuition, having the propeller feathered until reaching wind speed would be better since it would present less drag. So no storage of anything before then to release after.
In fact, I think that's demonstrated with the treadmill models, which go from nothing to tread speed almost instantly.
If air was not a compressible fluid then having the propeller will have been very useless and vehicle will not have been able to exceed wind speed.
Do you have a sense of how much energy is 0.25Ws ?
You can have Force without having power means vehicle speed and kinetic energy will not change.
Power is a more than Force especially in this context.
The treadmill model is being held not dropped on the treadmill. During those few seconds the treadmill charges all energy storage devices available on the vehicle the multiple flywheel devices and the pressure differential.
Air being a compressible fluid is not a means of energy storage, as the moving air in the propeller stream is less dense (lower pressure) than the still air in the surroundings. This can be learned in every physics or aerodynamics textbook as Bernoulli's principle (https://en.wikipedia.org/wiki/Bernoulli%27s_principle).
There is no way that the lower pressure air can "push" from behind.
No, while the vehicle is being held on the treadmill it is reaching equilibrium. If after release it was burning off stored energy it would instantly begin to slow down. Instead, it accelerates.
Correct answer is this below but if you disagree please feel free to offer the correct equation.
0.5 * air density * area * (wind speed - vehicle speed)^3
Correct answer is this below but if you disagree please feel free to offer the correct equation.
0.5 * air density * area * (wind speed - vehicle speed)^3
0.5 * air density * area * (wind speed- vehicle speed)^3
The equation for total power available does not have vehicle speed in it. :)
The equation is correct as I stated for a direct down wind vehicle.
Wind speed relative to a directly downwind vehicle is wind speed - vehicle speed.
True but irrelevant. The power theoretically available to the vehicle, if you want to try and use that as the basis for some calculations, depends on the wind speed relative to the ground.
You have a 100% efficient wind turbine so output will be 0.5 * air density * swept area * w^3
Now you put that wind turbine on top of a vehicle and drive directly downwind at say at a quarter wind speed
Then your wind turbine will output this 0.5 * air density * swept area * (w -(w/4))^3
Anyone that has ever calculate the wind power available to a sail vehicle going directly downwind will know the equation
0.5 * air density * area * (wind speed - vehicle speed)^3
That is the reason a sail vehicle can never exceed wind speed directly downwind.
A propeller can create a pressure differential in air and that is where energy is stored.
See below graph https://en.wikipedia.org/wiki/Axial_fan_design (https://en.wikipedia.org/wiki/Axial_fan_design)
(https://upload.wikimedia.org/wikipedia/commons/thumb/f/f1/Axial_fan_slipstream_theory.svg/600px-Axial_fan_slipstream_theory.svg.png)
A propeller can create a pressure differential in air and that is where energy is stored.
See below graph https://en.wikipedia.org/wiki/Axial_fan_design (https://en.wikipedia.org/wiki/Axial_fan_design)
(https://upload.wikimedia.org/wikipedia/commons/thumb/f/f1/Axial_fan_slipstream_theory.svg/600px-Axial_fan_slipstream_theory.svg.png)
You misunderstand that graph. It does not show what you think it shows. In fluid flow theory, there is static pressure and total pressure. Total pressure is the sum of the static pressure and the velocity component of the fluid. The total pressure may be higher downstream of the fan, but it is pointing in a direction away from the fan, so there is no way it can act as a store of energy. The static pressure points in all directions, and could potentially be a store of energy behind the fan. Except the static pressure is lower than the surroundings, and therefore it is more like a vacuum dragging the fan backwards. To store energy in a compressible fluid you need to contain it within walls, as in a storage tank. There is no tank here, there are no walls, there is no containment, therefore no store of pressure energy. Once the air leaves the fan, its energy quickly dissipates in all directions away from the fan.
I showed you a video where someone measured the pressure in the air stream leaving a fan to verify that it is lower than the surroundings. Unless you can show how that video was somehow faked or manipulated, you have to accept experimental evidence as fact.
Are you able to read a graph? or are you saying that the graph is incorrect ?
Is P2 higher than ambient pressure PA and much higher than P1 ? Thus a pressure differential potential energy.
If it has stored energy say that 0.07mWh I calculated
You have a 100% efficient wind turbine so output will be 0.5 * air density * swept area * w^3This calculation neglegts the force on the turbine. This gives an additional energy as drag force times vehichle speed.
Now you put that wind turbine on top of a vehicle and drive directly downwind at say at a quarter wind speed
Then your wind turbine will output this 0.5 * air density * swept area * (w -(w/4))^3
Anyone that has ever calculate the wind power available to a sail vehicle going directly downwind will know the equation
0.5 * air density * area * (wind speed - vehicle speed)^3
That is the reason a sail vehicle can never exceed wind speed directly downwind.
QuoteIf it has stored energy say that 0.07mWh I calculated
Stored where?
I was wrong -- a fast sailboat *can* beat a balloon directly downwind.
Here are the polars for one of the 2013 America's Cup races (you may recall that these were extremely fast catamarans):
(http://www.cupinfo.com/images/ac-polars-sog-r-5.png)
Look at the polar diagram for 20 kts true windspeed at a true wind angle of 135 degrees (TWS and TWA are ground-referenced values shown here, not the boat-referenced AWS and AWA). You will see that with a 20 kt wind the boat can sail 45 kts at 135 degrees (180 degrees is directly downwind). Jibing +/- 45 degrees, to beat the balloon the boat would only have to exceed (20 / 0.707) or 28.28 kts. I haven't tried to figure out the optimum downwind jibing angle, but at 45 degrees off DDW there is plenty of margin. By the way, these conditions would give you (on each 180 +/- 45 degree jibe) an AWS of 33.9 kts, AWA 34 degrees.
So this is a jibing boat, similar in many ways to a spinning propeller.
This calculation neglegts the force on the turbine. This gives an additional energy as drag force times vehichle speed.
Even for just a passive sail the energy is vehichle speed times force and thus
0.5 * air density * area * (wind speed - vehicle speed)^2 * vehicle speed
The passive sail is however not the best that can be done. So the theoretical available energy can be larger than that. The blackbird vehicle showed one way to do that. A sail boat going at a suitable angle is another way - though it complicates the area part. Anyway it still gets power when the speed component parallel to the wind is faster than the wind speed so there is no inherent zero at that speed. Yes it is sideways, but for a though expoeriment one can consider the sideways area as still part of the vehicle size, so the movement would be only inside (like the fan going in circles).
One could consider the fan blades as sails going zig-zag. So this kind of circumvents the going straight in the direction of the wind.
The turbine is supplied by wind power so you can not add that again.
Everything on a wind only powered vehicle is supplied by the wind power.
You can search research papers or similar and see that any direct down wind sail vehicle powered only by the wind will use this wind power formula
0.5 * air density * area * (wind speed - vehicle speed)^3
The passive sail is actually the most efficient and an ideal one will be 100% efficient.No : the sail is not most efficient way to harness the wind, at least not in all cases. It is obviously not at low speed, aspecially 0 speed, as it is 0 efficiency there. It is also not the most efficient way when the vehicle speed approaches the wind speed, as there it is zero efficiency too. The vehicle in question here is shown to be more efficient, as it can still get power from the wind, even at the speed of the wind and a little faster. You don't have to believe it up front, but you also can't exclude it up front from intuition. Doing this is trying a circular argument so it is not a valid argument.
Sail boat going at an angle is a geometric problem as the sail will still have access to wind speed since it will not be driving faster than wind relative to wind direction.That claim is not true: - by going zig zag with the wind a sail boat and especially an ice sail can go faster than a ballon, just not a straight line. Just look at the graph in the answer before yours. They get some 1.6 times the speed of the wind / ballon when doing a zig zag with a 45 degree angle and even more with a smaller angle like 20 degree.
The formular would be OK for the power a turbine can generate on the moving vehincle, but this is not relevant here.
The energy added to a vehicle with sail (usually in the form of kinetic energy or energy available to generate from a genrator at it's wheels) it is the formula I gave. The power is simply force times speed. Wind force is proportional to the square of relative velocity and the speed is the speed of the vehicle. You need the vehicle speed to get zero power from the sail when not moving at all, which is the trivial case: no speed means no power from a sail. You don't want the house to gain kinetic energy from the wind !
1m^2 swept area wind turbine with 40% efficiency in a 10m/s wind speed vs 1m^2 sail
If thery speed relative to the ground is zero then of course sail will do nothing and there will just be a static force as on any wall where the wind turbine output power will be 0.5 * 1.2 * 1 * 10^3 * 0.4 = 240W
So stationary wind turbine wins 240W vs 0W
Now install the wind turbine and the sail on a vehicle driving directly downwind at 0.1m/s, 5m/s and 9m/s
0.1m/s
wind turbine 0.5 * 1.2 * 1 * (10-0.1)^3 * 0.4 = 232.87W
sail (can install an electric generator at the wheel say that is 90% efficient) 0.5 * 1.2 * 1 * (10-0.1)^3 * 0.9 = 523.96W
5m/s
wind turbine 0.5 * 1.2 * 1 * (10-5)^3 * 0.4 = 30W
sail 0.5 * 1.2 * 1 * (10-5)^3 * 0.9 = 67.5W
9m/s
Wind turbine 0.5 * 1.2 * 1 * (10-9)^3 * 0.4 = 0.24W
sail 0.5 * 1.2 * 1 * (10-9)^3 * 0.4 = 0.54W
1m^2 swept area wind turbine with 40% efficiency in a 10m/s wind speed vs 1m^2 sail
If thery speed relative to the ground is zero then of course sail will do nothing and there will just be a static force as on any wall where the wind turbine output power will be 0.5 * 1.2 * 1 * 10^3 * 0.4 = 240W
So stationary wind turbine wins 240W vs 0W
Now install the wind turbine and the sail on a vehicle driving directly downwind at 0.1m/s, 5m/s and 9m/s
0.1m/s
wind turbine 0.5 * 1.2 * 1 * (10-0.1)^3 * 0.4 = 232.87W
sail (can install an electric generator at the wheel say that is 90% efficient) 0.5 * 1.2 * 1 * (10-0.1)^3 * 0.9 = 523.96W
5m/s
wind turbine 0.5 * 1.2 * 1 * (10-5)^3 * 0.4 = 30W
sail 0.5 * 1.2 * 1 * (10-5)^3 * 0.9 = 67.5W
9m/s
Wind turbine 0.5 * 1.2 * 1 * (10-9)^3 * 0.4 = 0.24W
sail 0.5 * 1.2 * 1 * (10-9)^3 * 0.4 = 0.54W
So if, by your own admission, the sail power is zero when the vehicle is stationary at 0 m/s, how can the sail power suddenly be 524 W when the vehicle is traveling at 0.1 m/s? (Which is nearly stationary.)
Would the sail power be higher or lower when the vehicle is traveling at 0.01 m/s? How about 0.001 m/s?
Can you make a graph of sail power vs vehicle speed? What does it look like?
You can calculate so for 0.001m/s is 0.5 * 1.2 * 1 * (10-0.001)^3 * 0.9 = 539.8WWith some 500W at only 0.001 m/s would require a force of 500 kN from the sail.
With some 500W at only 0.001 m/s would require a force of 500 kN from the sail.
This value is obviously wrong (way to high) and shows that the formular used above must be wrong, not just at zero speed.
With some 500W at only 0.001 m/s would require a force of 500 kN from the sail.
This value is obviously wrong (way to high) and shows that the formular used above must be wrong, not just at zero speed.
The formula used above is correct you used the wrong speed to calculate the force on the sail. You need to use 10-0.001 = 9.999m/s
An ideal 1m^2 sail in 10m/s speed will see a 60N force. I considered the generator 90% efficient in that formula that is why is less than 600W
OK, so 60N of force x 0.001m/s is how much power?
That is not how you will make the calculation. It will be 60N * (10m/s-0.001m/s).
This sort of things is what also make people think Blackbird can exceed wind speed powered only the the wind with no energy storage.
That is not how you will make the calculation. It will be 60N * (10m/s-0.001m/s).
This sort of things is what also make people think Blackbird can exceed wind speed powered only the the wind with no energy storage.
OK, if that is the case, then what is the power at 0 m/s?
No motion of the vehicle means zero power can be generated at the wheel.
No motion of the vehicle means zero power can be generated at the wheel.
Let's stick with that for a moment.
So you say that for any non-zero speed, the power is (whatever your formula is) or about 60N * (wind speed - vehicle speed), but right at zero the power goes to zero?
So if I'm sitting there and the vehicle is stopped with brakes, there's no energy being dissipated by the brake, but if I allow the vehicle to creep forward at 1mm/s, now the brakes are dissipating almost 600W?
Yes in theory if you want to keep the vehicle at 1mm/s brakes will need to dissipate 600W
Going to 0.5 m/s in only 2 ms sounds like a lot of acceleration - actually 250 m/s² and thus about 25 G, which for a 10 kg vehicle would need 2500 N of force. Something in this calculation does not sound right ! :-//
Yes in theory if you want to keep the vehicle at 1mm/s brakes will need to dissipate 600W
One of us has a fundamental misunderstanding of a very basic principle of physics. I think the brakes will dissipate approximately 60mW under those conditions.
If you apply only 60W of breaking power
If you apply only 60W of breaking power will get in about 1 second to around half the wind speed so 5m/s maybe just slightly more I will need to calculate exactly.
We can test this and see that a real test will exactly be predicted by my theory (is not my theory but the one I presented to you).
If you apply only 60W of breaking power
You need just under 60N of braking force, not 'braking power'. Then the power would be 60N X 0.001m/s = 60mW. Power is force x speed.
Not sure if you are familiar with power (it is an electronics forum) but do you find that a 10m/s (22mph) wind pushing on 1m^2 (10.7sqft) sail can be slowed down to 0.001m/s by 60mW ?
If the vehicle speed is nearly zero the power is nearly zero because Power = Force x (very small number).
Not sure if you are familiar with power (it is an electronics forum) but do you find that a 10m/s (22mph) wind pushing on 1m^2 (10.7sqft) sail can be slowed down to 0.001m/s by 60mW ?
Yes!
Not sure if you are familiar with power (it is an electronics forum) but do you find that a 10m/s (22mph) wind pushing on 1m^2 (10.7sqft) sail can be slowed down to 0.001m/s by 60mW ?
The mistake you make is using the 0.001m/s instead of (10m/s-0.001m/s)
Not sure if you are familiar with power (it is an electronics forum) but do you find that a 10m/s (22mph) wind pushing on 1m^2 (10.7sqft) sail can be slowed down to 0.001m/s by 60mW ?
The mistake you make is using the 0.001m/s instead of (10m/s-0.001m/s)
I'm not sure if you can say you are 'slowing the wind down' to the vehicle speed, but since you said it, lets use that.
If I can slow the wind down from 10m/s to ZEROusingby dissipating 0 power, why is it so difficult to believe that I can slow it down to 0.001m/swithdissipating 60mW?
You need to use brakes and sufficient large contact with ground to transfer that static force to ground so then ground deals with that.
Let's take this in reverse and say you have an onboard battery and electric motor and want to drive upwind at 0.001m/s in to a 10m/s wind with a 1m^2 frontal area and COD (coefficient of drag) of 1.
You think you can manage that with 60mW ? (ideal vehicle no friction).
Let's take this in reverse and say you have an onboard battery and electric motor and want to drive upwind at 0.001m/s in to a 10m/s wind with a 1m^2 frontal area and COD (coefficient of drag) of 1.
You think you can manage that with 60mW ? (ideal vehicle no friction).
You need to use brakes and sufficient large contact with ground to transfer that static force to ground so then ground deals with that.
Let's take this in reverse and say you have an onboard battery and electric motor and want to drive upwind at 0.001m/s in to a 10m/s wind with a 1m^2 frontal area and COD (coefficient of drag) of 1.
You think you can manage that with 60mW ? (ideal vehicle no friction).
I think perhaps you don't realize that "the ground deals with that" in all of the situations, not just the zero speed one?
Yes, if the sail is producing about 60N force, then 60mW will get me upwind at 1mm/s, provided there are no losses.
Let's take this in reverse and say you have an onboard battery and electric motor and want to drive upwind at 0.001m/s in to a 10m/s wind with a 1m^2 frontal area and COD (coefficient of drag) of 1.
You think you can manage that with 60mW ? (ideal vehicle no friction).
Yes, of course.
The force acting on the sail is 60 N. The vehicle speed required is 0.001 m/s. Therefore the power required from the motor is 60 N x 0.001 m/s = 60 mW.
Where is the 10m/s headwind in all this ?
Where is the 10m/s headwind in all this ?
It produces a force on the sail of 60 N.
No motion of the vehicle means zero power can be generated at the wheel.
Let's stick with that for a moment.
So you say that for any non-zero speed, the power is (whatever your formula is) or about 60N * (wind speed - vehicle speed), but right at zero the power goes to zero?
So if I'm sitting there and the vehicle is stopped with brakes, there's no energy being dissipated by the brake, but if I allow the vehicle to creep forward at 1mm/s, now the brakes are dissipating almost 600W?
Having said all of that, it has been regularly asserted that a sailboat can tack downwind faster than a balloon blown down wind. If this is indeed true the advantage must be small and dependent on perfect execution and perhaps on boat configurations which perform poorly in other conditions. I say this because racing yachts consistently set spinnakers and run nearly directly downwind rather than tacking to get there faster. These people spend millions of dollars to win, they wouldn't ignore any consistent advantage.
Electrodacus is in love with his model. A common problem with simulation people. Simulations are wonderful things. They can provide insight that is hard to obtain with actual physical tests, allow measurements that are literally impossible with physical tests and often save tremendous amounts of time and money.
But simulations have two fundamental flaws, that are often difficult to recognize. First, they are all approximations and do not provide comprehensive information on when the omissions are important. Second, usually a small problem but huge here, is that simulations no matter how large and wonderful do not inherently match the problem being simulated. The math is all correct but doesn't represent the physics of the situation.
In Electrodacus case there is a pretty obvious problem which he is overlooking. He is not intrinsically wrong with his (10m/s -0.001m/s) formulation. That is one way of presenting the problem. But he is overlooking the fact that 0.001 m/s second of incremental velocity requires only a trivial amount of power. Regardless of source. In his own thought process there is zero drag in the condition of interest since the vehicle is moving at wind speed. So the only consumer of power is the increase of momentum due to making the vehicle move faster. The flawed thinking is ignoring the power required to maintain the initial 10 m/s velocity.
Having said all of that, it has been regularly asserted that a sailboat can tack downwind faster than a balloon blown down wind. If this is indeed true the advantage must be small and dependent on perfect execution and perhaps on boat configurations which perform poorly in other conditions. I say this because racing yachts consistently set spinnakers and run nearly directly downwind rather than tacking to get there faster. These people spend millions of dollars to win, they wouldn't ignore any consistent advantage.
Having said all of that, it has been regularly asserted that a sailboat can tack downwind faster than a balloon blown down wind. If this is indeed true the advantage must be small and dependent on perfect execution and perhaps on boat configurations which perform poorly in other conditions. I say this because racing yachts consistently set spinnakers and run nearly directly downwind rather than tacking to get there faster. These people spend millions of dollars to win, they wouldn't ignore any consistent advantage.
It wasn't until recently that sailboats have been able to reduce drag sufficiently to be able to beat a balloon DDW. Iceboats can easily do this, but traditional "displacement" sailboats (such as mine) just can't overcome the drag to be able to do it. When I fly a spinnaker in heavy wind my best angle is DDW, not jibing, because I am limited by my "hull speed". In lighter wind I will jibe back and forth, using the spinnaker, and make better progress than I would by sailing DDW. The more modern "ultralight" boats have reduced drag to where they can beat the balloon by jibing. Foiling boats (such as the Americas Cup catamarans that I posted the polars for) have even less drag and can very comfortably beat that balloon, and not by a small margin. These boats don't even carry spinnakers as, even downwind jibing, the AWA is so far forward that the spinnaker would not be an efficient sail shape.
(AWA = Apparent Wind Angle, DDW = Dead Down Wind)
Recongnizing that the equation should conver different cases without changing the equation is a good point: this also applies to looking at the vehicle moving a snails speed aroung zero. The correct equation incudes this case.
Derek's proof was done using (vehicle speed - wind speed) and that is just demonstrably wrong as correct equation will have (wind speed - vehicle speed).
Same equation will need to apply to all conditions meaning when vehicle below wind speed and when vehicle above wind speed as you can not just change the equation when vehicle is at a different speed.
I say this because racing yachts consistently set spinnakers and run nearly directly downwind rather than tacking to get there faster. These people spend millions of dollars to win, they wouldn't ignore any consistent advantage.
Recongnizing that the equation should conver different cases without changing the equation is a good point: this also applies to looking at the vehicle moving a snails speed aroung zero. The correct equation incudes this case.
The slow speed case does not directly apply to the vehicle going faster that the wind. However it clearly shows the the equation proposed by electrodacus is flawed and leads to obvious problems there. Looking at this point may also help him correcting his understanding of force and power. Doing the calculations only with power can be tricky if there are processes with less than 100% efficiency (e.g. like the sail).
Having an equation with (w-v) or (v-w) is not such a big difference, it is just the sign and this may be different depending on how one defines the direction of force or an axis. One has to look not just at the one line with the formula, but also the explaination fo the symbols.
I don't think Derek's proof has a major mistake in the equations, as it leads to the right conclusion - just getting a result that is not obvious to everybody is no proof that there is an error. Usually math is way more reliable than intuition.
Does anyone have access to a treadmill and propeller model? ISTM the way to settles this is to get that model doing its thing, but have it tethered by some string so it can't move forward. Either the string is kept taught and energy storage is rubbish, or it goes loose and some store has clearly run out of puff.
If your vehicle has no mechanical brakes and you want to keep the vehicle from moving in 10m/s wind with a 1m^2 sail then you will need to apply 600W to that motor (all ideal case so ideal motor and no friction losses).
If your vehicle has no mechanical brakes and you want to keep the vehicle from moving in 10m/s wind with a 1m^2 sail then you will need to apply 600W to that motor (all ideal case so ideal motor and no friction losses).
That's the point of contention. I'm saying that this statement is not true. What is your theory/proof/argument/evidence that supports it?
So what do you think is needed if not 600W to prevent the vehicle from being pushed by the wind ?
We can do this test relatively easily.
The equation showing that is simple and it is this one you can find everywhere for ideal case 0.5 * air density * area * (wind speed)^3
So 0.5 * 1.2 * 1 * 10^3 = 600W
An arbitrarily small amount of power is needed to hold it in place because power is force x speed. Because the speed is zero, you can use as much gear reduction as you like.
You try pedaling at 1km/s direct upwind with 35km/h wind and let me know how much power you need to do that.
1 km/s is going to be spectacular, with or without the headwind.
Seriously, what sort of argument is that? If your motor setup is stupidly inefficient, it could take any amount of input power--even much more than 600W if you like--but at zero speed the output power is zero, so if you improve the design you can make it as efficient as you like, thus my 'arbitrarily small' statement. For the .001m/s upwind example, you can make it arbitrarily close to 60mW because that is the output power.
Just take a bicycle if you have one and try to drive direct upwind at as slow of a speed as you need to maintain balance (maybe 1km/h is a bit slow) and you will see you do not need mW but hundreds of watts with 10m/s (36km/h) as clearly seen in that graph I posted.
Having said all of that, it has been regularly asserted that a sailboat can tack downwind faster than a balloon blown down wind. If this is indeed true the advantage must be small and dependent on perfect execution and perhaps on boat configurations which perform poorly in other conditions. I say this because racing yachts consistently set spinnakers and run nearly directly downwind rather than tacking to get there faster. These people spend millions of dollars to win, they wouldn't ignore any consistent advantage.
It wasn't until recently that sailboats have been able to reduce drag sufficiently to be able to beat a balloon DDW. Iceboats can easily do this, but traditional "displacement" sailboats (such as mine) just can't overcome the drag to be able to do it. When I fly a spinnaker in heavy wind my best angle is DDW, not jibing, because I am limited by my "hull speed". In lighter wind I will jibe back and forth, using the spinnaker, and make better progress than I would by sailing DDW. The more modern "ultralight" boats have reduced drag to where they can beat the balloon by jibing. Foiling boats (such as the Americas Cup catamarans that I posted the polars for) have even less drag and can very comfortably beat that balloon, and not by a small margin. These boats don't even carry spinnakers as, even downwind jibing, the AWA is so far forward that the spinnaker would not be an efficient sail shape.
(AWA = Apparent Wind Angle, DDW = Dead Down Wind)
But if the wind pushes against a vehicle and a motor used to keep the wheels from rolling, there's definitely force, but the motor will also burn Watts to keep the wheels from rolling. Watts is power.
And this actually has nothing to do with the reality of the demonstrated sustained DDWFTTW vehicles.
A (nearly) stationary bicycle with a wind is a different problem with different results.
Going with 1 km/h against a heat wind of 35 km/h would need the force of going 36 km/h with no wind (irgnoring wheel friction and similar other power needs) with only 1/km/h instead of 36 km/h this would be 1/36 of the power needed for 36 km/h with zero wind. So taking the graph below to get a number, this for be bit below 300 W for 36 km/h and some 0.8 W for 1 km/h.
Other than for a rough number the chart is not really helping.
You can push against a brick wall with lots of force, but no work will be done, so there's no power.
If the wind pushes against a vehicle with a the wheels locked and not rolling, no work will be done, so there's no power.
But if the wind pushes against a vehicle and a motor used to keep the wheels from rolling, there's definitely force, but the motor will also burn Watts to keep the wheels from rolling. Watts is power.
And this actually has nothing to do with the reality of the demonstrated sustained DDWFTTW vehicles.
You need 300W to cycle at 36km/h with no wind and also 300W to cycle at 1km/h with a 35km/h head wind.That is absoulutely nonsense. It may be bit windy to drive a bike, 1 km/h is more like very slow walking and walking against the wind is still relatively easy.
That is where you are wrong. The problem is basically the same. Why do you think 300W are needed constantly to maintain that speed ? Is the air drag and you have that if you are cycling at 36km/h with no wind or at 1km/h with a 35km/h head wind. Same drag sosame amount of power required.
All of you seems to have some misconception about this that is why I insist.
You need 300W to cycle at 36km/h with no wind and also 300W to cycle at 1km/h with a 35km/h head wind.
Not according to this: https://www.omnicalculator.com/sports/cycling-wattage (https://www.omnicalculator.com/sports/cycling-wattage)
Its ~10W at 1km/h vs ~300W at 36km/h
The relative wind speed produces, as you say, the same force. That force, multiplied by the speed of the bicycle, is the power required. The power required is not the same in each case.
That's a nice calculator. A screen grab is attached below.
That calculator does not even have the input data it will need to calculate the correct result as there is no wind direction and no frontal area.
tops 0.408, hoods 0.324, drops 0.307, aerobars 0.2914
Most of that excess power difference is due to rubber tires on asphalt which is really affected by tire pressure.You need 300W to cycle at 36km/h with no wind and also 300W to cycle at 1km/h with a 35km/h head wind.
Not according to this: https://www.omnicalculator.com/sports/cycling-wattage (https://www.omnicalculator.com/sports/cycling-wattage)
Its ~10W at 1km/h vs ~300W at 36km/h
That calculator does not even have the input data it will need to calculate the correct result as there is no wind direction and no frontal area.
Wind direction is toward the rider (head wind, it says if you hover over the box).
Frontal area (Cd*A) is determined by Position of the rider, roughly. Its described on the right hand page.Quotetops 0.408, hoods 0.324, drops 0.307, aerobars 0.2914
Sorry to say but that is complete nonsense. Not quite sure where do you get this sort of intuition.
Have you ever used a bicycle or pedaled upwind ? It is the same principle as driving on an incline.
It's not exactly the same, but let's go with it. On a uniform incline of θ from horizontal, there will be a constant force of sin(θ) * g * m (mass) downhill. How much energy does it take to climb the incline a) slowly and b) faster? Do you agree or not that climbing the incline very slowly can be done with a very small amount of energy?
We are talking about two very different things here. Power and energy are not the same thing.
How much energy will depend on the distance you need to travel uphill and the weight of the vehicle.
not a perfect analogy as you will have gravity as you start to move if it is to compare with wind drag.
In any case all those bicycle calculators are wrong so this is a very widespread problem.
Most of that excess power difference is due to rubber tires on asphalt which is really affected by tire pressure.
I know that over-inflated 125psi high pressure narrow racing tires take half the effort compared having the same tires inflated to 80psi. Even worse compared to fat underinflated mountain-bike tires.
Like a train, if your bike had metal wheels driving on a flat steel beam, then wind preasure would be more dominating.
Also, the wattage in that calculator, it that the human muscle exertion's calorie burn, or torque at the pedals?
Yes, the wind is still a major factor as it also pushes on every moving surface.
OK :) I needed to have fun so apparently a human has no problem pedaling upwind at 1km/h with headwind of 230km/h
(Attachment Link)
OK :) I needed to have fun so apparently a human has no problem pedaling upwind at 1km/h with headwind of 230km/h
(Attachment Link)
Keep in mind 300W is not "no problem" for most people, its a lot of effort.
Do you know what 230km/h sustain speed means ?
That person even if tied to the ground will have the meat removed from the bones.
Do you know what 230km/h sustain speed means ?
That person even if tied to the ground will have the meat removed from the bones.
So how do people go that fast on motorcycles without being defleshed? It's a lot of wind and the turbulence might be tough on a bicycle, but if you were on a heavy, stable pedal-driven machine with approximately the frontal area of an upright rider you could indeed pedal into a 230km/h wind.
Really? 'Everyone' is wrong? How far are you willing to go with that? You never doubt yourself even a little? ::)
:-XQuoteIn any case all those bicycle calculators are wrong so this is a very widespread problem.
Really? 'Everyone' is wrong? How far are you willing to go with that? You never doubt yourself even a little? ::)
It's just about impossible to consider any other alternative. Even the Dunning-Kruger explanation is at a reach.Really? 'Everyone' is wrong? How far are you willing to go with that? You never doubt yourself even a little? ::)
You guys do realize we are dealing with a troll here, right?
Have you observed that our friend disagrees with, or contradicts, every statement made in this thread, however basic, and finds a reason to reject every experimental result and every external authority?Reason? Most of the time it seems to be a simple statement of "that is wrong" - with little, if any, valid argument.
Nobody could disagree with 100% of all points made, without doing it deliberately. Every normal person would have at least some points of agreement.Looking at it from the reverse angle - that's the magic of lying/misdirecting properly. Have just a taste of accurate bits mixed up in a soup of ignorance. Then you will be greeted with indignation about the "accurate bits" when you were actually challenging the crap.
Nobody could disagree with 100% of all points made, without doing it deliberately. Every normal person would have at least some points of agreement.
No vehicle without an external energy source or energy storage device can exceed wind speed direct down wind.
The resistance drop on under inflated tires tends towards an exponential curve, not linear and it also depends on the weight of the driver. I used to be 250lb, an 80psi tire was mushed close to the rip and believe me, I was adding a good additional 50-75 watts just to maintain 25km/h compared to 120psi tires at that weight.Most of that excess power difference is due to rubber tires on asphalt which is really affected by tire pressure.
I know that over-inflated 125psi high pressure narrow racing tires take half the effort compared having the same tires inflated to 80psi. Even worse compared to fat underinflated mountain-bike tires.
Like a train, if your bike had metal wheels driving on a flat steel beam, then wind preasure would be more dominating.
Definitely not half the effort for slightly higher inflated tires.
The difference in rolling resistance for 100 vs 120psi on GP5000 tires at 29km/h is ~0.7W: https://www.bicyclerollingresistance.com/road-bike-reviews/continental-grand-prix-5000-2018 (https://www.bicyclerollingresistance.com/road-bike-reviews/continental-grand-prix-5000-2018)
So 80 vs 125psi would be ~3W at most (10-13W total).
Sure fat mountainbike tires will be worse, but, calculator is not using them.
I showed the correct equation for wind power that relates to such a vehicle 0.5 * air density * area * (wind speed - vehicle speed)^3That is the power theoretical available to a wind turbine on the vehicle, but not the power needed to more the vehicle. So a "correct" formular. but for a different problem.
0.5 * 1.2 * 0.5 * (64)^3 = 78.6kW of drag with just 0.3kW on input power.I am quite shocked by this result - not by the number, but by the units: the drag is a force and not a power ! :horse:
How much more extreme you need the values to be in order to seems shocking.
How many here would like to see that happen?
No vehicle without an external energy source or energy storage device can exceed wind speed direct down wind.
Then your argument has nothing to do with blackbird, you're trying to argue against the accepted scientific facts about sailing boats.
The fact that the boat achieves sailing directly downwind faster than the wind from point A to B through a series of indirect zig-zags is irrelevant, since the length of each zig/zag has no effect on mechanism causing faster than wind travel.
If the sail boat could change direction in an instant with no time needed to stop, rotate and move the sails, then it could move downwind faster than the wind through a series of zig-zags that were so tiny that they were imperceptible and it would, in effect, be a straight line downwind faster than the wind.
That is the power theoretical available to a wind turbine on the vehicle, but not the power needed to more the vehicle. So a "correct" formular. but for a different problem.We are discussing the best case here and ignoring frictional losses. And so even in this ideal case no more wind power is available to vehicle than that is calculated with that formula.
0.5 * 1.2 * 0.5 * (64)^3 = 78.6kW of drag with just 0.3kW on input power.I am quite shocked by this result - not by the number, but by the units: the drag is a force and not a power ! :horse:
How much more extreme you need the values to be in order to seems shocking.
Ok, give me a vote to go ahead and flip 'electrodacus' head this weekend making him convince himself that a wind powered craft going faster than a tail wind speed with his own logic...
How many here would like to see that happen?
Are you building something? Just don't use a rubber-band instead of gears or chain-drive. We wouldn't want any of that "stick-slip hysteresis energy storage" to cloud the discussion.
Feel free to use force if that is your favorite but that is how mistakes are made. I personally prefer to use power as it already includes the speed also and there can be no confusion.
We have to put this in terms with no room for misunderstanding.
If the wind is blowing from east to west at 20 mph, then a boat can sail 20 miles due west in less than one hour.
It is the air resistance and if you could move that you can get to almost any speed on a bicycle.The air resistance is a force. So the force is the same, however that does not mean that the power is the same.
There is no difference between you moving through air or heaving a similar speed headwind.
Feel free to use force if that is your favorite but that is how mistakes are made. I personally prefer to use power as it already includes the speed also and there can be no confusion.
That's like saying you personally prefer to use watts rather than volts because watts include amps....
The most basic physics instruction will teach you that you need to use the correct units to get the correct answer. Using the complete form of the units all the way through the equation is how you determine whether 'the units come out'. Having the units come out to a nonsensical expression indicates that you made an error somewhere, either small typo-like mistake or a fundamental misunderstanding. Try it--use the complete form of all the units in your equations and at the end you will have some nonsensical result like square watts per meter-second or something like that.
The force of the wind on a sail is expressed in Newtons or the equivalent in another system. Period. It doesn't matter whether it is a big sail in a small wind or vice versa. Now you may protest that a system will behave differently in the two cases as the sail moves, which is true, but that is because the force, expressed only in Newtons, will change differently as the sail and wind speeds change. You simply cannot express force in energy or power terms as it isn't what the word means--and the units won't come out.
So this boat with a wind turbine and an energy storage device can finish the race much faster than 1h
The air resistance is a force. So the force is the same, however that does not mean that the power is the same.
Is there any problem understanding the very basic equation for mechanical power being force time velocity ?
Oh my! What a load of.....????
What was said was that a sailboat without any energy storage can sail the 20 miles in less than an hour. In fact, give enough distance and a steady 20mph wind from the east, it could sail west indefinitely at more than 20mph.
And yes I fully agree that you should know what to expect as a result."Knowing" (or better some intuition) what to expect as the result, however does not mean to put you intuition for the result over the calculation. A carefully done calculation is way more reliable than the intuition. There are some effects in physics that confuse a simple mind and some effects even surprise an expert (though less often). The vehicle in the video is such a case that may cause irritation, though this still a relatively simple one.
There is no problem in using force instead of power but in that case you need to know what is the correct speed you need to multiply with to get the correct result.
The power actually transfered to the vehincle or needed to more the vehicle is not the same as the power theoretical available from the wind. For the power transfered to the vehicle or needed to move it against the wind, the relevant speed is the speed of the vehicle relative to the ground (the reference system you transfer the force to or calculate the kinetic energy in).
That force will multiply with air speed to get the power available. If you transfer all that force to ground through a bake then vehicle is part of the earth.
As soon as you want the vehicle to move relative to earth vehicle is on his own and it needs that specific power to be able to move against the wind direction.
I do not get your reaction. I proved to you that a vehicle that has an energy storage device can get there in less than an hour.
Blackbird has an energy storage device (pressure differential created by the propeller).
So using the full wind speed for the power transferred to the vehicle is obviously wrong.
That's irrelevant. A sailboat, with no turbine, and no energy storage, can sail 20 miles downwind in less than an hour with a 20 mph following wind. It could do this indefinitely.
If you are referring to zigzag (not direct down wind travel like blackbird) then you can use the kinetic energy as an energy storage device.
If you are referring to zigzag (not direct down wind travel like blackbird) then you can use the kinetic energy as an energy storage device.
How? The boat could go 20 miles downwind and 5 to the side without changing direction, or it could make one turn halfway through the course. In that turn, which only takes a few seconds out of that hour, its speed before the turn and after the turn are the same, so what kinetic energy?
At a vehicle speed much smaller than the air speed it does not matter if one subtracts the vehicle speed. At zero vehicle speed your (wrong) formular gives the full power theoretical power in the wind, and this is not the power actually captured. A sail is not at all 100% efficient in converting the energy ! Going against the wind actually needs extra power, while the could provide power if used in a different way. At zero speed the efficency is zero.So using the full wind speed for the power transferred to the vehicle is obviously wrong.I do not use the full wind power since I use (wind speed - vehicle speed) for a direct downwind vehicle unlike the wrong (vehicle speed - wind speed) Derek used in his video.
That will not be directly downwind it will be at an angle and it will get to another location compared to someone that went direct down wind.
At a vehicle speed much smaller than the air speed it does not matter if one subtracts the vehicle speed. At zero vehicle speed your (wrong) formular gives the full power theoretical power in the wind, and this is not the power actually captured. A sail is not at all 100% efficient in converting the energy ! Going against the wind actually needs extra power, while the could provide power if used in a different way. At zero speed the efficency is zero.
When the vehicle moves at the speed of the wind it does not matter if w-v or v-w is used, both would be zero. So even if the sign is wrong it would not make a difference at that point. When using the correct formular for the power, there is (w-v)² * v and in the square the sign makes no difference.
That is potential power if you want so it will not do any work but as soon as vehicle moves it will do work.Some ( a small fraction when at low speed) of the power would be used if going with the wind, none of this power would be used when moving against the wind. It is wrong to assume that whole available power would be transferred to the vehicle when moving with the wind - only a small fration ( vehicle speed / wind speed) is.
Obviously at vehicle speed zero w-v or v-w will make no difference other than the sign of potential power showing direction in witch the power can be used if you start moving.How do you define a direction for the power ? power is a scalar value, not a vector so it has no direction. The sign shows wether a system takes up power or gives up power, but this not a direction of geometric sense. A direction in space is a feature of the force.
The sign is important as it will show if vehicle accelerates or decelerates and w-v is correct as it will show vehicle can accelerate when vehicle speed is below wind speed directly downwind while using v-w will mean vehicle decelerates while vehicle speed is below wind speed and that will not match any real experiment.The importance of the sign is correct, but with the (w-v)³ formular this gets you the wrong result (allways positve) near zero. This is because the fomular is for the theoretical possible power, but not the actual transfered power or power needed to push the vehicle. The formula for the actual power under the condition ( for W>V) , goes with (w-v)²*v gives the right change in sign at aroung v=0 and thus include the transition between against and with the wind. This formular was never meant to also include the case v > w. To also incluse the reversal of the wind direction at the vehicle it needs an extra factor for the direction. So the form for the full range is more like (w-v)2 * v * (w-v) / |w-v| for the direction of the drag force.
How come nobody answered the time it will take such a vehicle to get to half the wind speed so 5m/s ?
I'm not an expert in all area of physics but I have large amounts of experience in energy storage of any type and renewable energy generation.
People on a motorcycle are shielded by the motorcycle body there is no 0.5m^2 of human exposed to those speeds.
The resistance drop on under inflated tires tends towards an exponential curve, not linear and it also depends on the weight of the driver. I used to be 250lb, an 80psi tire was mushed close to the rip and believe me, I was adding a good additional 50-75 watts just to maintain 25km/h compared to 120psi tires at that weight.
Some ( a small fraction when at low speed) of the power would be used if going with the wind, none of this power would be used when moving against the wind. It is wrong to assume that whole available power would be transferred to the vehicle when moving with the wind - only a small fration ( vehicle speed / wind speed) is.
That 437.4W of wind power will be available to vehicle as long as wind matains 10m/s
If you understand this example and explanation you should be able to understand my entire explanation about how blackbird works.
I wanted to answer the other questions but I feel we need to concentrate on this and once this part is understood the rest should be simple.
Try doing your math with the units!
Can you explain what you mean ? I think I used units everywhere in my example.
If I miss one somewhere please point that out.
Can you explain what you mean ? I think I used units everywhere in my example.
If I miss one somewhere please point that out.
Take the expression " 0.5 * 1.2 * 1 * (10-1)^3 = 437.4W" and put the actual units in everywhere the number refers to units.
So the 0.5 refers to area of the sail, so 0.5m2, I don't know offhand what the 1.2 or 1 refer to, the (10-1)3 refers to speed, so (9m/s)3 becomes 729m3s-3 and so on, then your result will be some monstrosity of units until you simplify it. If it doesn't simplify down to a usable unit, then your formula or math are wrong.
Sorry OK I see what you mean. Here is the equation with units.
0.5 * 1.2kg/m^3 * 1m^2 * (10m/s-1m/s)^3 = 437.4W
Sorry OK I see what you mean. Here is the equation with units.
0.5 * 1.2kg/m^3 * 1m^2 * (10m/s-1m/s)^3 = 437.4W
OK, now if you could just explain what each of the terms is, actually just the 1.2kg/m3 term--I see the 1m2 is area and the others are the speed of the wind and the sail. Then instead of magically arriving at 'Watts' at the end, do all the operations with the units and see if what you get simplifies to watts or not.
The 0.5 is a constant has no unit
1.2kg/m^3 is the air density
1m^2 is the equivalent are of the sail
then there is the wind speed and vehicle speed in m/s
I use all international system of unit so I do not need to be as careful as if using all sort of strange unrelated units.
And yes that result is in Watt's
If all the power or only a small farction is actually used makes a big difference. The problem with the assumption all the theoretically possible power would be added to the kinetic energy is that it is rather hard (essentially impossible) to do this.Some ( a small fraction when at low speed) of the power would be used if going with the wind, none of this power would be used when moving against the wind. It is wrong to assume that whole available power would be transferred to the vehicle when moving with the wind - only a small fraction ( vehicle speed / wind speed) is.
When you are start moving slowly in to wind direction you need to apply brakes so some small part of the power will be used to increase the vehicle kinetic energy while acceleration and the difference will be burned by the friction brakes (assuming that is what you use to slow down the vehicle).
So if you use no brakes then vehicle will use all that wind power to accelerate the vehicle (increase kinetic energy).
Thus as soon as the vehicle moves the power is either all used to increase vehicle kinetic energy thus super fast get to a large fraction of the wind speed or if you want to maintain that low speed constantly then you need to do something with all that wind power available either convert to heat using friction brakes or generate electricity and store that in some sort of energy storage.
I think this is important to understand so I will try to insist on this point.
Except from this error I have not seen a problem with the units.
If all the power or only a small farction is actually used makes a big difference. The problem with the assumption all the theoretically possible power would be added to the kinetic energy is that it is rather hard (essentially impossible) to do this.
With a 100 kg vehicle to add the first 0.5 Ws to the kinetic energy it would need a speed of 0.1 m/s. If the sail would provide a full 500 W to add to the kineatic energy, this would be only 1 ms to reach 0.1 m/s. It is only a low speed, but 0.1 m/s / 1 ms is still 100 m/s² and thus a bit more than 10 times gravity for the acceleration needed (on averge). Things get even more carzy when looking at lower speed / shorter time. So there must be an error in the calculation.
The error in this example is in the idea that a "sail" would be 100 % energy efficient in converting to kinetic energy. Energy is still conserved, but most is converted to heat and not to the kinetic energy of the vehicle.
I somehow have the feeling the concept of force is not really understood. Using a power for the drag is an indication. Drag is a force and not a power.
Except from this error I have not seen a problem with the units. The desire to do the calculations with examples instead of the letters is often found with beginners who try to use intuation instead of math. However even beginners are usually way better in realizing that they may be wrong.
Just a quick summary on the calculations and units:
The wind force in the simple case of the wind directly hitting a flat surface is given by: (force) = (wind pressure) x (area) x (drag coefficient)
We have been neglecting the drag coefficient in the discussions so far.
The transmitted power where force and speed are in the same direction is then given by: (power) = (force) x (speed)
The wind pressure is given by: (wind pressure) = (one half) x (air density) x (apparent wind velocity)^2
The drag coefficient varies a lot, but for a flat, perpendicular surface it could be about 1.1 or so.
The apparent wind velocity is the velocity experienced by the sail.
The density of air is about 1.2 kg/m3, so for a 10 m/s wind hitting a 1 m2 flat surface moving downwind at 1 m/2, the wind pressure would be 0.5 x 1.2 x (10-1)^2 x 1.1 ~= 53 N
The transmitted power would then be 53 N x 1 m/s = 53 W.
Then 46.8N * 9m/s = 437.4W so same exact result I got.
Then 46.8N * 9m/s = 437.4W so same exact result I got.
Just a quick summary on the calculations and units:
The wind force in the simple case of the wind directly hitting a flat surface is given by: (force) = (wind pressure) x (area) x (drag coefficient)
Then 46.8N * 9m/s = 437.4W so same exact result I got.
But the 46.8 N is not moving at 9 m/s. If you make the sail move at 9 m/s, then the apparent wind velocity is going to be 1 m/s.
In that case: Force on sail = 0.5 * 1.2 * (10-9)^2 * 1 = 0.6 N
Power = 0.6 N * 9 m/s = 5.4 W
You cannot have a high wind pressure on the sail, and also make the sail move at the speed of the wind. That is double accounting, and it does not work.
Just a quick summary on the calculations and units:
The wind force in the simple case of the wind directly hitting a flat surface is given by: (force) = (wind pressure) x (area) x (drag coefficient)
That omits the laminar drag portion, but I'm guessing perhaps that is negligible in this case? (I don't actually know)
And that does appear to be the equation, more or less, that he is using. I didn't recognize it because of the way he was expressing it, but it works out--as I showed above!
Just think about such a vehicle and the fact that you need to apply brakes to maintain the speed at this 1m/s while wind speed is 10m/s and relative to vehicle 9m/s
Your brakes will need to deal with 437.4W else vehicle speed will increase. With brakes you just heat the brake pads and air but you can replace those with an electric generator and find some better use for it.
The equation works in general for any shape of object. The area is calculated in some standard way (usually the cross section facing the wind), and the drag coefficient accounts for the shape of the object. Basically, all the complex geometry, skin friction, turbulent effects and everything else is wrapped up in the Cd value. It can vary a lot, from as little as 0.01 to as much as 2.
Just think about such a vehicle and the fact that you need to apply brakes to maintain the speed at this 1m/s while wind speed is 10m/s and relative to vehicle 9m/s
Your brakes will need to deal with 437.4W else vehicle speed will increase. With brakes you just heat the brake pads and air but you can replace those with an electric generator and find some better use for it.
If you want to get 437 W from a vehicle moving at 1 m/s then you can use the formula: power = force x velocity
You have:
437 W = force x 1 m/s
So the force on the sail would have to be 437 N.
From the sail equation:
Force = 0.5 x (air density) x (apparent wind velocity)^2 x (area of sail)
437 N = 0.5 x 1.2 kg/m3 x va^2 x 1
va^2 = 437 / (0.5 x 1.2) = 728
va = 27 m/s
This tells that the wind speed would have to be 27 + 1 = 28 m/s.
If you apply no brakes at all then those 437W will all be available to accelerate the vehicle.
What I meant is that there is another term in the full equation, the laminar drag. This is the low-speed drag component dependent on speed, the viscosity of the medium and constant based on the shape and size of the object. the viscosity of the medium.
No, it won't. You don't accelerate a vehicle with power, you do it with force. At 1m/s with a 10m/s wind, you have your 46.8 Newtons available to accelerate the vehicle, which at 1m/s represents 46.8 watts of power. The remainder of the wind's power will be dissipated. Of course that number changes as the speed change. And b/t/w, from an earlier post, 46.8 * 9 = 421.2, not 437.4. Not a big issue, but it was making something not look right.
.... that a 100% efficient wind powered vehicle can not exceed wind speed directly downwind and thus any vehicle no matter how it is build can only exceed wind speed directly downwind if it has some sort of energy storage device or a external energy source.Taking the above literally - which is what you have been doing all along - and you will never be able to see the full mechanism. The above is the "obvious" bit - but it's not the full story.
.... that a 100% efficient wind powered vehicle can not exceed wind speed directly downwind and thus any vehicle no matter how it is build can only exceed wind speed directly downwind if it has some sort of energy storage device or an additional energy source.
Here is the key - there IS an additional energy source, but it is not as obvious as wind on a sail (which is why "intuition" is a really bad influence - really bad.). It is, however, fully contained within the system consisting of only the Blackbird, the wind and the ground. It is not external to that system and it is not an energy storage mechanism.
If you cannot OPENLY consider such a possibility and work through the relevant calculations, you will never understand.
Such closed-mindedness would have kept the study of physics from progressing any further than Newton.
If you look back at earlier example I stated that it takes around 100ms to get to 1m/s and the vehicle kinetic energy at 1m/s is 50Ws
So according to your theory it will take more like one full second for this vehicle to get to 1m/s instead of just 100ms.
If you look back at earlier example I stated that it takes around 100ms to get to 1m/s and the vehicle kinetic energy at 1m/s is 50Ws
So according to your theory it will take more like one full second for this vehicle to get to 1m/s instead of just 100ms.
I don't know what numbers you are using for mass, etc, but:
Where v = speed (m/s), a = acceleration (m/s2), m = mass (kg), t = time (s), F = force (N or kg * m/s2))
F = ma => a = F/m
v = at = t*F/m
t = v/a = mv/F
Pro tip for physics beginners when confused: solve for t where possible
So provide the time it takes for this vehicle to get to 1m/s and 5m/s
I say 103ms and 8.3s
So provide the time it takes for this vehicle to get to 1m/s and 5m/s
I say 103ms and 8.3s
I don't know and I don't know the mass you are using, but in any event how does this help resolve anything? What would knowing the initial acceleration do as far as understanding faster-then-wind travel? Or is there another point?
I say 103ms and 8.3s
In any case this vehicle is 100kg wind speed is 10m/s, sail area is 1m^2 and air density 1.2kg/m^3
Blackbird will accelerate much slower than an equivalent area sail but Blackbird is using that extra time below wind speed to store energy so that it can then exceed wind speed for some limited amount of time.
Accelerating a vehicle from a stop to 1m/s in 100ms is an acceleration of 10m/s2 (slightly over 1g) and would take 1000 newtons of force. What were the forces calculated earlier, 53N at zero speed and 46.8 at 1m/s? So maybe an average of about 50N? So I'd say that is going to take more like 2 seconds.
That isn't actually all that crazy, but in order for that to be a workable theory you would have to demonstrate that there is a reasonable place to store the energy and that, in fact, the Blackbird was using that energy during its above-windspeed run. The only place where any energy can be stored is in the propeller. Energy storage and on-demand recovery in uncontained air is something you're going to have to affirmatively demonstrate experimentally before anyone believes that--and I don't think you'll succeed. Anyhow, as for the propeller, it does indeed store some energy but now you have to calculate how much, and more importantly, show that the Blackbird is actually using that energy by slowing the propeller down. You can easily put an upper bound on the energy in the propeller by estimating its mass and the tangential velocity of the tips, then positing that all of its mass is at the tips. So throw some numbers at that and see what you get. Does the video have a high enough framerate to determine the rotational speed of the propeller?
Even if all of that worked out and you can demonstrate that the Blackbird stores some amount of energy in the propeller, that doesn't mean that the stated principle (stated by the rest of us) isn't valid or that the Blackbird will slow down as this stored energy is exhausted.
It will be 60N at zero vehicle speed
Vehicle at 1m/s has a kinetic energy of 0.5 * 100kg * 1m/s^2 = 50Ws (same as 50 Joules)
So since average was around 500W available over that period about 100ms (0.1 seconds) are needed to transfer that amount of energy from wind to vehicle kinetic energy.
I know we take very different routes in calculating the same thing but results needs to be the same
so a cell phone battery could do that test over 200000 times with power at the wheel and 100000 times with a 50% efficient propeller
Nope, just words...How many here would like to see that happen?
I'd like a ticket for this, please.
Are you building something? Just don't use a rubber-band instead of gears or chain-drive. We wouldn't want any of that "stick-slip hysteresis energy storage" to cloud the discussion.
Just so we are clear. This calculator is fairly accurate.Not according to this: https://www.omnicalculator.com/sports/cycling-wattage (https://www.omnicalculator.com/sports/cycling-wattage)
Its ~10W at 1km/h vs ~300W at 36km/h
That's a nice calculator. A screen grab is attached below.
(https://www.eevblog.com/forum/chat/mess-with-your-minds-a-wind-powered-craft-going-faster-than-a-tail-wind-speed/?action=dlattach;attach=1350590)
That isn't actually all that crazy, but in order for that to be a workable theory you would have to demonstrate that there is a reasonable place to store the energy and that, in fact, the Blackbird was using that energy during its above-windspeed run. The only place where any energy can be stored is in the propeller. Energy storage and on-demand recovery in uncontained air is something you're going to have to affirmatively demonstrate experimentally before anyone believes that--and I don't think you'll succeed. Anyhow, as for the propeller, it does indeed store some energy but now you have to calculate how much, and more importantly, show that the Blackbird is actually using that energy by slowing the propeller down. You can easily put an upper bound on the energy in the propeller by estimating its mass and the tangential velocity of the tips, then positing that all of its mass is at the tips. So throw some numbers at that and see what you get. Does the video have a high enough framerate to determine the rotational speed of the propeller?
Even if all of that worked out and you can demonstrate that the Blackbird stores some amount of energy in the propeller, that doesn't mean that the stated principle (stated by the rest of us) isn't valid or that the Blackbird will slow down as this stored energy is exhausted.
I will see things the other way around. I demonstrated that using correct equations an ideal 100% efficient wind powered vehicle can not exceed wind speed without energy storage.
But of course I did calculated how much energy needed to be stored for blackbird to get to 28mph with 10mph average wind speed (real wind speed average was larger but I was generous).
Blackbird is around 300kg including the driver the wheels are bicycle wheels so super low rolling resistance.
Kinetic energy of the vehicle at 28mph (12.5m/s) will be 0.5 * 300kg * 12.5^2 = 23437Ws = 6.5Wh
To get there even with no wind to start with is not that much of a problem even 3 small 3000F 2.7V super capacitors can store that about 3Wh per capacitor.
Even a cell phone battery can deliver this energy twice it just can not deliver that fast but supercapacitors can charge and discharge in seconds.
Now the propeller is just massive at 5.3m diameter 20m^2 swept area is larger than the floor area in my livingroom and not much pressure differential is needed to store this sort of energy needed to accelerate to 28 or 30mph even without the initial help from the wind.
So I did looked at all the numbers to make sure they all fit correctly.
If I finish my other work by Sunday, it will be time to tackle 'electrodacus' on the wind powered craft.
The green statement is true, the red one is not.
As you've stated, the maximum force is 60N and the mass is 100kg. So at what point in the following do you disagree?
a) F=ma => a = F/m.
b) v = at => t = v/a
c) t = mv/F
d) substituting in 100kg for m, 1m/s for v and 60N for F (60N is the maximum force, thus the result will not be perfectly accurate, but an upper bound for acceleration and thus a lower bound for time)
we get t = 100kg * 1m/s / 60N.
e) Solving we get: t = 100/60 kg m s-1 N-1 or 5/3s (1.66666666s). Integrating F over the the interval would give us a somewhat lower average F and thus a longer t, but our differences here are large enough not to worry about this yet.
To resolve this, you need to point out exactly what is wrong with my assertions, not repeat your own. In the case of your statement (in red) I would specifically point to the error being your belief that the power transmitted to the vehicle is 500W or so, when in fact this more accurately represents the power not transmitted to the car (dissipated). In fact at the very beginning when the speed is zero, the power transmitted to the car is ~60N * vehicle speed, or nil at zero, a small amount in the beginning stages and then more as the speed picks up--to a point.
Fair points so I took a look on what pressure will you have on a flat 1m^2 wall in 10m/s wind.
So I will say this is a good explanation of what happens as there will be super high air density just next to the wall but I think my way of calculating what happens is way easier and accurate.
Maybe I'm just biased but I'm confident my results are correct as they match any real world experiment.
Fair points so I took a look on what pressure will you have on a flat 1m^2 wall in 10m/s wind.
So I will say this is a good explanation of what happens as there will be super high air density just next to the wall but I think my way of calculating what happens is way easier and accurate.
Maybe I'm just biased but I'm confident my results are correct as they match any real world experiment.
You appear to have realized that your energy transfer model would require the initial acceleration to be very, very high--diverging to infinity as the speed is zero. And so it won't work.
Keeping in mind that the Blackbird is not a wall or a sail, you are going to have to stipulate what the equations or formula or calculator or whatever-have-you are going to produce for the force on the sail, if we're using sails to demonstrate something. I thought we had that all worked out with formulas that you and IanB agreed on and the result was 60N at zero vehicle speed and 53N at 1m/s vehicle speed. If you want to change that, fine--but justify it somehow.
And perhaps this wouldn't be so hard to verify experimentally, and perhaps someone else already has and such results can be googled.
The calculation with only energy and avoiding forces at all costs gets quite complicated and error prone if there are elements (like a sail) that don't have 100% efficiency. It may look easier to calculate with energy only, but I am afraid this is only an easy way to the wrong result. A sail is not 100% energy efficient - with low speed it is actually rather low efficiency.
To me all seems to line up and make sense both calculations if done properly will have the same result I provided. For me the one involving power and energy works better (simpler).
The calculation with only energy and avoiding forces at all costs gets quite complicated and error prone if there are elements (like a sail) that don't have 100% efficiency. It may look easier to calculate with energy only, but I am afraid this is only an easy way to the wrong result. A sail is not 100% energy efficient - with low speed it is actually rather low efficiency.
To me all seems to line up and make sense both calculations if done properly will have the same result I provided. For me the one involving power and energy works better (simpler).
Even if you think you got the energy based calculation done, please also have a look at the way with forces, without taking the result from the other calculation for granted. The force based calculation is not that complicated and the way it is usually done.
All this calculations are done for ideal case so to get the best case and show that even with best case (ideal) no vehicle can exceed wind speed direct downwind without energy storage.We have spend quite some time with the low speed case to show you that the formula provided is wrong. They produce a diverging force / acceleration (e.g. accelaration to 0.1 m/s in less than 1 ms) near zero speed and are thus obviously wrong. :horse:
But in terms of efficiency (converting wind power to kinetic energy) the sail is by far the most efficient.
.....
All this calculations are done for ideal case so to get the best case and show that even with best case (ideal) no vehicle can exceed wind speed direct downwind without energy storage.We have spend quite some time with the low speed case to show you that the formula provided is wrong. They produce a diverging force / acceleration (e.g. accelaration to 0.1 m/s in less than 1 ms) near zero speed and are thus obviously wrong. :horse:
But in terms of efficiency (converting wind power to kinetic energy) the sail is by far the most efficient.
.....
A sail is not the most efficient way to harness the wind. It is actually very low effciency at low speed. Assuming 100% efficiency for the sail is one cause of getting the rediclulous fast acceleration. At zero vehicle speed the sail has zero energy efficiency - that is a very simple fact. A Wind turbine has a efficiency better than zero and is thus higher effciency than the sail (proven here for the case when the windmill is not moving).
With just a sail in a 1 D world (straight downwind) one can not reach a speed higher than the wind.
However the sail is not the only option and with a more intelligent way (e.g. the prop drive like in the Blackbird) it is possible to get more energy from the wind than the sail and move faster than the wind.
is in violation of energy conservation.
is in violation of energy conservation.
What's that?
That energy storage device will be charged during the initial acceleration phase
Energy cannot be created or destroyed just converted from one form to another.
The way pressure varies at the back of a sail / wall is exponential and that is perfectly matching that decreasing rate of acceleration shown using power and energy. Not to mention they perfectly match any experimental test.
You just think sail is low efficiency because you use incorrect formula.
QuoteThat energy storage device will be charged during the initial acceleration phase
Where is the acceleration phase on the treadmill?
How would that be violated by a sail dissipating all or most of the wind energy and only transferring a small portion of that energy to the vehicle?
The formula used earlier that seems to be widely accepted as correct (the drag equation) is also exponential and doesn't diverge to infinity as the vehicle speed approaches zero. In fact, it doesn't matter whether the vehicle is stationary or moving beyond how that changes the relative speed of the wind and vehicle.
So now you are positing some new drag or sail formula that would give you an extremely high pressure and thus an extremely high initial acceleration for a 10m/s wind and a stationary vehicle. So what would the pressure and acceleration be if you had an 11m/s wind and a vehicle moving at 1m/s? Would that be different?
What experimental tests are you referring to?
QuoteThat energy storage device will be charged during the initial acceleration phase
Where is the acceleration phase on the treadmill?
I think I explained that before here but it is when you put the vehicle on the treadmill.
When you put the vehicle on the treadmill and hold it in place, this is equivalent to the vehicle traveling downwind exactly at windspeed [ (wind speed - vehicle speed) = 0 in your equation ] So according to you, there's no energy to be stored -- and you're right, at least about that. But when you let go, the vehicle instantly rolls forward, faster than the wind.
When you put the vehicle on the treadmill and hold it in place, this is equivalent to the vehicle traveling downwind exactly at windspeed [ (wind speed - vehicle speed) = 0 in your equation ] So according to you, there's no energy to be stored -- and you're right, at least about that. But when you let go, the vehicle instantly rolls forward, faster than the wind.
There is maybe a second or two when you touch the vehicle to treadmill until all energy storage devices are fully charged up that means all wheels get to same speed as the treadmill the propeller gets to nominal speed and the pressure differential is created.
For that small treadmill model we are talking about 0.25 to maybe 1Ws (1J) of total stored energy.
The kinetic energy needed for that vehicle to accelerate from zero (when you release) up to 1m/s (never got that fast in the video) is 0.5 * 0.5kg * 1m/s^2 = 0.25Ws (0.25J) basically nothing is needed to be stored in pressure differential to be able to accelerate that small vehicle to 1m/s even to 2m/s will not take much and vehicle was only demonstrated for way lower speed. The treadmill is just to short to show the vehicle accelerating to whatever max speed it is designed for and then show also how it will slow down (that will take again as much as to get there).
What stored energy is there when the vehicle is stationary? You claim that there is zero available energy when the relative windspeed is zero, so there's nothing to store. By your thinking, all the storage occurs when the vehicle is traveling slower than windspeed.
So, how long would we need to hold that vehicle stationary for all that "stored energy" to dissipate? And why do we care about max speed? Isn't any speed beyond wind speed proof that it works?
Your model, and your understanding, is flawed.
What do you mean by stationary vehicle ? Vehicle before it touches the treadmill ? if yes then of course there is no stored energy.
What do you mean by stationary vehicle ? Vehicle before it touches the treadmill ? if yes then of course there is no stored energy.
No. By "stationary" I mean the vehicle is being held stationary on the treadmill. The wheels are in contact with the treadmill, turning, and the propeller is spinning. The vehicle speed and the windspeed are equal. This is exactly equivalent to a vehicle traveling directly downwind at the speed of the wind. And no matter how long you hold the vehicle stationary, when released the vehicle will accelerate.
You claim there is no available energy when the vehicle is stationary, since "windspeed - vehicle speed = 0". Even if we ignore your equation, and at windspeed there is stored energy in a pressure differential (which I dispute -- I claim there is no stored energy adequate for even a brief burst of acceleration), how long do you think this stored energy will persist before dispersing into the surrounding environment?
There is measurable stored kinetic energy in the spinning propeller, and rotating wheels, and the mass of the vehicle itself, but unless you have some sort of gearshift or prop-feathering (which the treadmill vehicle does not) none of these will cause the vehicle to accelerate, they will only slow the rate of friction-caused deceleration. And yet, the vehicle accelerates.
Also my point is that no vehicle can exceed wind speed directly downwind unless it has an energy storage device.And that is wrong and there is no sense in repeating this all over !. :horse: :horse: :horse: :horse:
I'm very confident that online calculator that I liked is accurate in describing pressure on either a wall or a sail directly downwind.Very confident is has to be seen with a grain of salt: The calculator is for a different problem: the force a wall has to withstand, when close to an edge (kind of clif if you want). The Zones are not the place where the pressure in the air is measured, but the distance of the wall from the edge. Close to the edge there is some concentration of the wind - so the real value to compare is the one far away from the edge.
The crazy explanation that vehicle when above wind speed directly down wind will take power from the wheel to supply the propeller is exactly like saying a motor and generator connected together with a belt are a free energy generator.No, it's not. Your motor and generator are operating at the same speed - but the the two sides of the Blackbird mechanism are not.
That is because when you take energy from the wheel that is directly taken out of the vehicle kinetic energy Ws or Joules and since propeller is maybe at best 70% efficient can only put back 70% of that back in to kinetic energy and so vehicle will have lower speed.Yes, there will be a reduced efficiency in the power extracted from the wheels through to the propeller - but the propeller is pushing against the wind. With the blackbird travelling at wind speed, the wind will appear as a stationary wall of air - and THIS is what the propeller is pushing against. <== IMPORTANT POINT!
What you can see if you take a video from the side is the decrease acceleration rateIt doesn't matter even if the acceleration rate drops to zero. If the resulting "terminal velocity" is higher than wind speed, then we have found the equilibrium point which supports Blackbird's achievement.
and that will not correspond to Derke's or anyone else's prediction.You say "will not". That sounds like an intuitive belief rather than a mathematical proof.
In short, energy is extracted from a 20km/h input and is applied to an output working against a 0 km/h reference.
And that is wrong and there is no sense in repeating this all over !. :horse: :horse: :horse: :horse:
Very confident is has to be seen with a grain of salt: The calculator is for a different problem: the force a wall has to withstand, when close to an edge (kind of clif if you want). The Zones are not the place where the pressure in the air is measured, but the distance of the wall from the edge. Close to the edge there is some concentration of the wind - so the real value to compare is the one far away from the edge.
The calculated force is not that impressive to support the claimed high power from a sail.
So I am glad you are confident in a calculator that contradicts your claims.
Better have more confidence in the calculator for the bicycle power that showed that it is possible to drive with relatively low power against an head wind.
@electrodacus
A Bumblebee can't fly .!!
Same as electrodacus can't believe a simple
Solution. That if everyone says the 🌎 is round .
He would argue its flat because my spirit level says so .. :palm:
So if everyone here has proven it does work .
Plus there has been a video showing it works
. There for Bumblebee can fly also faster the tail wind .
Your ideas of what is true and false , was is, and what can .
Maybe beyond your comprehension .
It wouldn't matter if Albert Einstein could tell you that your wrong .
You are living in some sort of self denial.
Get over it and except that it can and does work.
No, it's not. Your motor and generator are operating at the same speed - but the the two sides of the Blackbird mechanism are not.
Yes, there will be a reduced efficiency in the power extracted from the wheels through to the propeller - but the propeller is pushing against the wind. With the blackbird travelling at wind speed, the wind will appear as a stationary wall of air - and THIS is what the propeller is pushing against. <== IMPORTANT POINT!
Warning: The following uses terms like power and energy in a conversational style - NOT an engineering one! Change the words as you see fit - but follow the intent.
- Let's say wind speed is 20 km/h
- Power is extracted from the wheels which are rotating at a tangential speed of 20km/h
- This power is transferred through a gearbox to rotate a propeller. (The ratio of the gearing is important here - which includes propeller characteristics.)
- The propeller then pushes against a wall of air that appears to be stationary ( 0 km/h ) from the perspective of the vehicle.
In short, energy is extracted from a 20km/h input and is applied to an output working against a 0 km/h reference.
The extracted energy will cause a drop in speed, but the additional thrust will cause an increase. Finding the equilibrium point is the trick. Is is below wind speed? Is it above? Is it equal to wind speed?
Clearly shows you do not understand power and energy. 20km/h ~ 5.5m/s
If you take 5.5W form the wheel (1N * 5.5m/s) and put all that in to propeller (ideal case) the most propeller can output is 5.5W thus no acceleration possible even in ideal case.
The gear ratio is irrelevant since you can still not output more power than you input.
This is exactly what others including Derek will try to say it is so silly I do not even know what to say. Luckily there are people that understand all this they just seems to be unwilling to waste time with you (all).
If you think with a 5.5W input the propeller can output more then I have a free energy generator for sale.
The gear ratio is very relevant, as it determines how much thrust / force the propeller will produce. With a good gear ratio propeller can create more than 1 N of thrust from the given 5.5 W. This is possible as the prop sees essentially standing air. 1 N at zero relative speed is zero ouput power from the prop.
What is a "free" energy generator good for that can create 0 W when the wind is blowing. There is quite some energy taken from the wind.
The gear ratio is very relevant, as it determines how much thrust / force the propeller will produce. With a good gear ratio propeller can create more than 1 N of thrust from the given 5.5 W. This is possible as the prop sees essentially standing air. 1 N at zero relative speed is zero ouput power from the prop.
What is a "free" energy generator good for that can create 0 W when the wind is blowing. There is quite some energy taken from the wind.
When you take 5.5W say for 1 second then that 5.5Ws (5.5J) energy will be subtracted from the vehicle kinetic energy and since kinetic energy is proportional to speed speed will also be reduced.
Now all you have is this 5.5Ws (5.5J) and no matter what gear ratio you have you can not deliver more than the same 5.5J to the vehicle kinetic energy assuming ideal case.
So yes you can have a gearbox with say 2:1 ratio so input is 5.5m/s 1N and output can be 2.75m/s at 2N it will still only put back 5.5Ws (5.5J) in the vehicle kinetic energy since gear box is not magic and can not output higher power than it has available at the input.
So you just get confused as you do not use the correct speeds in your equations like Derek subtracting wind speed from vehicle speed (ridiculous).
The gear box does not increase the power, but the prop or wheels on the treadmill work relative to a different speed. Because of the lower speed the gear box can reduce the speed so much to generate more than 1 N ( e.g. 2 N). With 2 N of force to drive the vehicle forward and 1 N of force to work against it, there would be still 1 N going forward and thus an increase in speed. It is the force that decides which way the vehicle moves, not the power.
The gain in power is from slowing down the wind with those 2 N. 2 N pushing against the 5.5 m/s wind is a power source of 11 W and thus plenty of power. It is not free energy, but energy taken from the wind.
You will never (I mean never) get 11W with an input of 5.5W. Thinking like this is that resulted in this wild explanations.
Sure you will, and the Blackbird does just that. There's no big mystery how, either. You simply don't understand the energy conservation laws that you continually cite.
The only way you get that 11W on the output is if you add another power supply in parallel or series with the first one.
I like that analogy, but the DCDC converter would be more like a step down one to get more current, though at a lower voltage.The only way you get that 11W on the output is if you add another power supply in parallel or series with the first one.
Yes, that's the principle. The propeller and the wind are 'in series', so to speak. If you like bad analogies, you can compare the various 'speed' of the players with voltage. If the wind and the vehicle are at the same speed you can compare that to a power source and a sink--say a charger and battery--that are both at the same voltage so that no power flows. But if you add an isolated DC-DC converter across the source and then put its output in series between the source and sink, now power will flow.
So the wheel-to-propeller link can be considered as a sort of boost converter that allows power to go from an original source that is the same or even lower voltage than the sink.
The only way you get that 11W on the output is if you add another power supply in parallel or series with the first one.
Yes, that's the principle. The propeller and the wind are 'in series', so to speak. If you like bad analogies, you can compare the various 'speed' of the players with voltage. If the wind and the vehicle are at the same speed you can compare that to a power source and a sink--say a charger and battery--that are both at the same voltage so that no power flows. But if you add an isolated DC-DC converter across the source and then put its output in series between the source and sink, now power will flow.
So the wheel-to-propeller link can be considered as a sort of boost converter that allows power to go from an original source that is the same or even lower voltage than the sink.
And to be clear a boost DC-DC converter outputs will be lower power than input and also any boost converter will have energy storage devices either an inductor a capacitor or both.
And to be clear a boost DC-DC converter outputs will be lower power than input and also any boost converter will have energy storage devices either an inductor a capacitor or both.
What about a DC-DC converter with a solar panel providing 1 watt in sunlight added to the output, and used outside?
What then? How much power do you measure at the input vs the output?
Suppose a 5V to 12V boost converter you can buy on eBay, it's 90% efficient. Power to the input from a USB power pack at 5V with enough current capacity. I added a solar panel and circuitry to add the power to the output. I used it outside in sunlight no clouds with the solar panel pointed at the sun.
I measure 12V at 1A through a resistor. Do you agree this is 12 watts of output power?
Then a 90% efficient supply would need 13.3W, so 2.7A at 5V.
But 1W is coming from the solar panel.
How much current will the supply actually need at 5V?
2.5A? Why? 12W at 90% efficiency = 13.3W needed, MINUS 1W from the panel, 12.3W needed from the 5V supply.
Clear? The *SUN* has added power. Happy?
Here are the correct equations for wind power available to a vehicle traveling directly down wind
0.5 * air density * area * (wind speed - vehicle speed)^3 yes this is the correct equation. Notice how you have highest power at low vehicle speed and power drops to zero by the time vehicle speed equals wind speed.
You Quoted ::@electrodacus
A Bumblebee can't fly .!!
Same as electrodacus can't believe a simple
Solution. That if everyone says the 🌎 is round .
He would argue its flat because my spirit level says so .. :palm:
So if everyone here has proven it does work .
Plus there has been a video showing it works
. There for Bumblebee can fly also faster the tail wind .
Your ideas of what is true and false , was is, and what can .
Maybe beyond your comprehension .
It wouldn't matter if Albert Einstein could tell you that your wrong .
You are living in some sort of self denial.
Get over it and except that it can and does work.
Solution is not simple and I never believe anything I need to understand.
All works as shown in tests and is fully described by my theory.
On the other hand the equations you proposed do not predict what happens in real tests.
We are drifting in all sort of analogies but to try and answer what happens imagine that vehicle with the solar panel having the solar panels rotate away from the sun as the speed increases. That is what happens to a wind powered vehicle that drives directly downwind.
Here are the correct equations for wind power available to a vehicle traveling directly down wind
0.5 * air density * area * (wind speed - vehicle speed)^3 yes this is the correct equation. Notice how you have highest power at low vehicle speed and power drops to zero by the time vehicle speed equals wind speed.
QuoteHere are the correct equations for wind power available to a vehicle traveling directly down wind
0.5 * air density * area * (wind speed - vehicle speed)^3 yes this is the correct equation. Notice how you have highest power at low vehicle speed and power drops to zero by the time vehicle speed equals wind speed.
So surely you agree that the total available wind power at a certain point (no vehicle or sail involved) is 0.5*airdensity * windspeed3, correct?
So where does the rest of that wind power go if you can only get the fraction of it that your formula specifies? That power is still there somewhere, isn't it?
Please read again my formula as you forgot a very important part and that is the area that wind pushes against.
0.5 * air density * area * windspeed3
This is the formula anyone designing a wind turbine will use except that is is ideal 100% of available wind power and to this they need to add the turbine efficiency usually around 40% and the generator efficiency that can be over 90%
For a vehicle driving directly down wind ideal case all you need is to subtract vehicle speed from wind speed since wind speed relative to vehicle is what can power the vehicle.
This formula
0.5 * air density * area * (wind speed - vehicle speed)3 is what will apply to any wind only powered vehicle driving directly downwind.
This is also the ideal case so absolutely max that is available to the vehicle meaning that if you do not add an energy storage device to this vehicle or some external energy source like gasoline and an engine then your vehicle can not exceed wind speed.
Clearly shows you do not understand power and energy. 20km/h ~ 5.5m/sUsing this, you have 5.5W spinning a propeller. This propeller creates a thrust force against an air mass which is stationary with respect to the vehicle.
If you take 5.5W form the wheel (1N * 5.5m/s) and put all that in to propeller (ideal case) the most propeller can output is 5.5W
thus no acceleration possible even in ideal case.How can a non-zero force NOT create a change in velocity?
Mea culpa, I forgot to put in 'area'. But you've dodged the actual question. Since all of the energy in the first question represents the total wind energy available for that particular area and the second equation represents only part of that energy, where does the rest of it go?
It's really necessary to work with power, rather than energy.
For a vehicle to travel at any steady speed, v, the power it requires is the power required to overcome the frictional resistance--the rolling resistance on the ground and the drag from the surrounding air.
Let's suppose that a vehicle wants to travel at a speed, v, that is faster than the wind, then it will require some motive power, Pv.
If a vehicle can obtain at least that much power from the wind, using a turbine or other mechanism, then it can travel at that speed. It becomes an engineering problem of how to design that mechanism, that will necessarily include such items as turbines, fans, gears, shafts, even possibly a motor-generator combination.
There is no violation of conservation of energy implied, since clearly the wind has unlimited power available (if you want more power, just use more area for your fan/turbine).
To be clear, sailors know this, since they routinely sail downwind faster than the wind, as noted by my example earlier in the thread. If the wind speed is 20 mph due west, then racing yachts can travel 20 miles due west at an average speed greater than 20 mph. And they can do this indefinitely. Energy storage cannot be claimed for a boat that is travelling for an hour or more at these speeds.
(Specifically, if point B is 20 miles due west of point A, and the wind is blowing at 20 mph due west, then a boat can travel from A to B in less than 1 hour.)
Using this, you have 5.5W spinning a propeller. This propeller creates a thrust force against an air mass which is stationary with respect to the vehicle.
How can a non-zero force NOT create a change in velocity?
...as boat is still slower in the direction of the wind.
The equation is for wind power available to vehicle not wind energy. Wind power will be stored over time as kinetic energy and pressure differential in the case of direct downwind blackbird.
If there was just kinetic energy like it is the case with a sail vehicle then that vehicle could not exceed wind speed unless it was driving at an angle and not directly down wind.
You all seems to be confused on what power and energy is so most of your questions do not make sense.
I will rather talk with GPT-3 :)
...as boat is still slower in the direction of the wind.
No, the boat is faster than the wind in the direction of the wind. That is the whole point of how the boat can travel 20 mph downwind in less time than the wind takes to cover the same distance.
(If you disagree with this, you will have to tell the sailors that what they are doing is impossible. They will be very interested to hear that.)
OK, you're right. It's turtles all the way down.
I know exactly what you are saying here ... and it is what you would naturally expect from applying intuitive thinking.
Using this, you have 5.5W spinning a propeller. This propeller creates a thrust force against an air mass which is stationary with respect to the vehicle.
How can a non-zero force NOT create a change in velocity?
Yes but those 5.5W will not create more than 5.5W at the propeller output. So the kinetic energy / speed reduction to get those 5.5W will require the same 5.5W just to be where you started from in ideal case.
...as boat is still slower in the direction of the wind.
No, the boat is faster than the wind in the direction of the wind. That is the whole point of how the boat can travel 20 mph downwind in less time than the wind takes to cover the same distance.
(If you disagree with this, you will have to tell the sailors that what they are doing is impossible. They will be very interested to hear that.)
Let me try this again. In ideal case a boat can get to 2x even more wind speed at an angle and then since there is no friction it can just turn directly down wind and travel forever at 2x. In real world there is friction so sailor will get the boat to speed means increased kinetic energy (stored kinetic energy) and then it can turn straight and for a few minutes stay above wind speed as kinetic energy and boat speed reduces then he can turn again to an angle to get to speed and repeat. If friction is sufficient low average may be higher than wind speed but even if it is not above wind speed it is still faster as most sail boats direct downwind can do just 0.5x wind speed.
So in this case kinetic energy is used as energy storage. In case of blackbird driving at an angle is not allowed so kinetic energy can not be used as energy storage in the same way it is used in the example above.
QuoteYes but those 5.5W will not create more than 5.5W at the propeller output. So the kinetic energy / speed reduction to get those 5.5W will require the same 5.5W just to be where you started from in ideal case.I know exactly what you are saying here ... and it is what you would naturally expect from applying intuitive thinking.
However, the mechanism utilises a mechanical advantage (if you will) with the power of the wind augmenting the effectiveness of that 5.5W.
This is how Blackbird can do the same thing:
But keep in mind the main discussion here is about blackbird that is not allowed to travel at any angle other than direct down wind.
There is no mechanical system or even non mechanical that can output any more than the input in therms of power.It's not outputting any more. It's extracting power from a wheel system that is travelling at one speed (frame of reference is the ground) and then applying that power to a propeller system travelling at a different speed (frame of reference is the wind).
... when the obvious solution was to look for an energy storage device.No. The obvious solution would be to see if you could find out a way it COULD work. This is something you continually refuse to do.
It is exactly like those motor generators setups where a motors drives a generator usually trough a gear ratio change and generator supplies the motor.It's nothing like that - but you seemed hung up on standing your ground with this concept and your totally flawed "modelling" of that.
There people use tricks like adding a large flywheel (mechanical energy storage device) or even hide a battery to trick people.Just because you keep saying it, doesn't make it true.
Here there was no intent in deceiving (at least I do not think it was) but just did not understood that there is an energy storage device and that is what powers the vehicle above wind speed.
The sad part is that people with a PhD in physics get tricked by this.No. The sad part is the people who think they know better - and and either won't listen or cannot comprehend an objective process to describe what is going on.
That university professor that lost the bet (his fault for gambling) said nothing wrong in Derek's video but since it was unable to explain how the treadmill prototype worked it was forced to pay the money not to be seen as not honoring his obligations. He should have asked for help and try to understand why it works then make that information public.Poor, dumb professor.
Unfortunately school is mostly for those that can memorize facts and equations an understanding is completely irrelevant.Sounds like someone we all know here.... ::)
But keep in mind the main discussion here is about blackbird that is not allowed to travel at any angle other than direct down wind.
Hmmm. Is that true for the propeller blades? Are they going straight downwind?
It's not outputting any more. It's extracting power from a wheel system that is travelling at one speed (frame of reference is the ground) and then applying that power to a propeller system travelling at a different speed (frame of reference is the wind).
This is the concept you continually dismiss, bypass or ignore. It is THE heart of the Blackbird doing what it can - and has - done. By refusing to even consider this concept as a possibility you are condemned to remain ignorant.
QuoteThat university professor that lost the bet (his fault for gambling) said nothing wrong in Derek's video but since it was unable to explain how the treadmill prototype worked it was forced to pay the money not to be seen as not honoring his obligations. He should have asked for help and try to understand why it works then make that information public.Poor, dumb professor.
I do suggest you spend more time thinking about this (valid for all people here)."There Are None So Blind As Those Who Will Not See."
I spent enough time trying to help and I feel I repeated my self more than I needed to so it will not be helpful for me to help if you do not want to put the brain to work.
Try to learn what power and energy is as that should help.
Do you understand that power you take from the wheel will brake the vehicle ?I certainly do - and the fact you want to make a point of this seems to be a desperate move to discredit me.
Please read again my formula as you forgot a very important part and that is the area that wind pushes against.
0.5 * air density * area * windspeed3
This is the formula anyone designing a wind turbine will use except that is is ideal 100% of available wind power and to this they need to add the turbine efficiency usually around 40% and the generator efficiency that can be over 90%
For a vehicle driving directly down wind ideal case all you need is to subtract vehicle speed from wind speed since wind speed relative to vehicle is what can power the vehicle.
This formula
0.5 * air density * area * (wind speed - vehicle speed)3 is what will apply to any wind only powered vehicle driving directly downwind.
This is also the ideal case so absolutely max that is available to the vehicle meaning that if you do not add an energy storage device to this vehicle or some external energy source like gasoline and an engine then your vehicle can not exceed wind speed.
For direct downwind blackbird is clear to me that energy is stored in pressure differential generated by the propeller and that is what allows it to exceed wind speed even if it may be for just a few minutes depending on the design, how fast it will use that stored energy.
I certainly do - and the fact you want to make a point of this seems to be a desperate move to discredit me.
Do you realise that while that extraction of power will slow the vehicle, that the loss of velocity will result in the wind directly applying more force to the vehicle as it will be travelling below wind speed? This results in an increase in total power applied to the vehicle. Pull 5.5W from the wheels and, as the vehicle slows, the wind adds more power. Even if that additional power is just 2W or 1W or even 1mW, the overall power delivered to the vehicle by the wind has increased.
You have never mentioned this increased power, which indicates your understanding is less than comprehensive and your analysis certainly incomplete.
And we haven't even looked at the harvested power component. Apply that harvested power into driving the propeller and now the backward thrust will be trying to slow the wind, but the wind (having a practically infinite energy store) will just push against this - applying even more power into the system.
The extra power required for the Blackbird to do its thing, comes precisely from the wind as the wind pushes against those forces which come from the elements that are trying to slow the vehicle down down.
At this point, even your intuition must be curious as to where this total system will settle - and what parameters will affect that point.
I would really encourage you to do a proper job of investigating this - rather than do your typical handwaving dismissal. The total power has increased, so you cannot just say the additional power is of no significance without providing real numbers.
power is instantaneous contains no time dimension
Here are the correct equations for wind power available to a vehicle traveling directly down wind
0.5 * air density * area * (wind speed - vehicle speed)^3 yes this is the correct equation. Notice how you have highest power at low vehicle speed and power drops to zero by the time vehicle speed equals wind speed.
The point is not in reading the euqations, the point is understand what they are actually calculating. The first one is the power theortically available from the wind as the kinetic energy in the moving air.
The second equation is giving the energy available at the reduces wind speed, e.g. for a wind-turbine on the moving vehicle. This is not the maximum energy available to the vehicle.
The wind turbine also creates some drag force, comparable to a sail and this drag force also creates power to a moving vehicle. This power contribution is force times vehicle speed. For a sail this is the only way it creates power.
The wind turbine (=porp) on the vehicle could create some extra power directly. For the sail the drag force is proportional to (wind_speed-vehicle_speed)². For the prop the drag force (called thrust when caused by the driven prop) also depends on the speed the prop rotates. depending on the direction and speed this can be less, but also more than just a sail. With the driven prop there can be drag/thrust even with zero wind speed.
But why is this the correct equation?
Please derive this equation step by step from first principles, stating any assumptions made at each step.
Then you will understand where and how this equation can be applied, and what limitations exist in its usage.
One part will be stored as kinetic energy the other part will be stored as both rotational kinetic energy in the rotating propeller and pressure differential by increasing the difference in pressure between upwind side of the propeller and down wind side.
(https://en.wikipedia.org/wiki/File:Axial_fan_slipstream_theory.svg)
Then this stored pressure differential energy storage is what will allow blackbird to exceed wind speed for some limited amount of time.
There are multiple resources here is one
http://web.mit.edu/wepa/WindPowerFundamentals.A.Kalmikov.2017.pdf (http://web.mit.edu/wepa/WindPowerFundamentals.A.Kalmikov.2017.pdf)
The equation can be applied anywhere starting with wind turbine design, power available to any wind powered vehicle, drag power needed for vehicle design .....
Unfortunately school is mostly for those that can memorize facts and equations an understanding is completely irrelevant.
One part will be stored as kinetic energy the other part will be stored as both rotational kinetic energy in the rotating propeller and pressure differential by increasing the difference in pressure between upwind side of the propeller and down wind side.
(https://en.wikipedia.org/wiki/File:Axial_fan_slipstream_theory.svg)
Then this stored pressure differential energy storage is what will allow blackbird to exceed wind speed for some limited amount of time.
But the pressure after the propeller is lower than the pressure in front of the propeller: https://www.quora.com/Why-is-the-static-pressure-before-a-fan-higher-than-after-the-fan (https://www.quora.com/Why-is-the-static-pressure-before-a-fan-higher-than-after-the-fan)
I have shown you a video illustrating this. Do you think that video was manipulated or faked in some way? Every engineer knows that video shows the truth. You can do the same experiment yourself. Just use any household fan. You will find the pressure in the moving airstream is lower than the surroundings.
You should stop looking at that Wikipedia article. It does not say what you think it says.
No, I want you to derive it yourself, and give your assumptions in the derivation.
If you cannot do that, you are merely quoting facts and equations from others, which in your own words shows a lack of understanding.Unfortunately school is mostly for those that can memorize facts and equations an understanding is completely irrelevant.
Not sure what you mean by after and front. Here is the graph for pressure difference.
Never claimed that anything was faked. It works exactly as shown from the equations and theory I presented.
(https://upload.wikimedia.org/wikipedia/commons/thumb/f/f1/Axial_fan_slipstream_theory.svg/400px-Axial_fan_slipstream_theory.svg.png)
Never claimed that anything was faked.
The difference is that sail vehicle (an ideal one) takes all this available power and stores it as kinetic energy thus not able to exceed wind speed.So why is there no pressure differential energy stored behind the sail?
Blackbird will split this available wind power in to two. One part will be stored as kinetic energy the other part will be stored as both rotational kinetic energy in the rotating propeller and pressure differential by increasing the difference in pressure between upwind side of the propeller and down wind side.Stored rotational kinetic energy can not cause a fixed-geometry propeller to increase rotational speed. It can at best reduce the rate that the propeller rotation will slow down.
Then this stored pressure differential energy storage is what will allow blackbird to exceed wind speed for some limited amount of time.
Do not be afraid just open the pdf and look at page 4.
So why is there no pressure differential energy stored behind the sail?
Stored rotational kinetic energy can not cause a fixed-geometry propeller to increase rotational speed. It can at best reduce the rate that the propeller rotation will slow down.
In your models and equations, stored pressure differential energy will not cause the vehicle to accelerate to above windspeed, since any pressure differential will diminish to zero at windspeed, and go negative when the vehicle is above windspeed.
Your errors in analysis have been explained many times.
Do not be afraid just open the pdf and look at page 4.
Do you understand that power you take from the wheel will brake the vehicle ?
If you talk about power and taking a power from the wheel then there is no speed change as power is instantaneous contains no time dimension.
as power is instantaneous contains no time dimension.Now that's just funny ... in a sad, sad way.
Do you understand thatI presume you mean energy.poweryou take from the wheel will brake the vehicle ?
Let's revisit:Do you understand thatI presume you mean energy.poweryou take from the wheel will brake the vehicle ?
I certainly do - and the fact you want to make a point of this seems to be a desperate move to discredit me.
Do you realise that while that extraction of energy will slow the vehicle, that the loss of velocity will result in the wind directly applying more force to the vehicle as it will be travelling below wind speed? This results in an increase in total power applied to the vehicle. Pull 5.5W from the wheels and, as the vehicle slows, the wind adds more power and, thus, more energy. Even if that additional power is just 2W or 1W or even 1mW, the overall energy delivered to the vehicle by the wind has increased.
The above is perfectly reasonable and reads pretty clearly, even before throwing in the word "energy". There is no conflict in using the terms power or energy. Each describes a particular quantity and those terms can be mixed quite happily. It is for the reader to understand what information each quantity provides. The only difference is time and the context gives enough structure for the appropriate consideration of time.
Below are the correct equations
Below are the correct equations
If you would stop making assertions like this, and instead ask questions like, "Are these the correct equations?" or "What is the correct way to analyze this system?", then you would become far more enlightened.
As it is, you are unable to make any progress in your understanding.
If you disagree post what you think are the correct ones. I do not make a video without checking multiple times and making sure what I say is correct. I can of course still make mistakes as anyone else.
Looking forward to your results for that problem.
The case with the wheels is a bit different from the prop case, but I think it contains the same difficulty in understanding. As an advantage the basics of mechanics are a bit simpler and less of an approximation. So it may be OK to look again at this simpler picture for the start.
To judge if the equations are correct or not the first thing is to make clear what is meant with P_out and P_net mean and the condition for the vehicle.
For the condition of the vehicle the situation to look at, the most useful case would be the vehicle standing still in the picture and with F_G = F_M to have a stationary case with no acceleration.
So what do P_out and P_net stand for ?
What does output of the motor wheel mean ? This is not really clear to me.
It that the same as the power needed to supply the motor = the ouput power of the motor ?
For now it is just about the equation, so we don' t need to dicuss the 3 cases. We need to understand the equation first.
Do you understand that power you take from the wheel will brake the vehicle ?If you talk about power and taking a power from the wheel then there is no speed change as power is instantaneous contains no time dimension.
Which is it? You can't have it both ways.
power is instantaneous contains no time dimension.
I'm just pointing the fact that you do not know what power is and also do not know the difference between power and energy.You are so hung up about that ... and you couldn't be more wrong.
I can of course still make mistakes as anyone else.But you can never admit to them when they go against your beliefs and your responses are the worst kind of ignorance. You introduce contradictions and deny their existence.
You are either trolling, getting people to spend their time posting, to see how long you can make it continue. Or you lack sufficient understanding to support any kind of dialog, and lack any capacity to learn.Options are:
Either way, you are wasting people's time.Agreed.
power is instantaneous contains no time dimension.
It doesn't? Power = Energy/Time. Joules per second. Ergs per eon. Electron-volts per attosecond.
Speaking an an electronics nerd who skipped Physics, I think of power in Watts. Watts = Volts * Amps, and there is no time dimension unless you consider that Amps = Coloumbs per second. But that is a rate, not a duration. Even Watt-hours is rate, or a quantity, with no particular duration (unless specified). I look forward to some clarification.
You seemed to have "overlooked" this:Do you understand that power you take from the wheel will brake the vehicle ?If you talk about power and taking a power from the wheel then there is no speed change as power is instantaneous contains no time dimension.
Which is it? You can't have it both ways.
Care to answer?
I'm fairly bad at explainingNo kidding. You are even worse at understanding.
But honestly, trying to fine-tune and correct these equations is fairly pointless when they do not reflect the actual characteristics ofthese DDWFTTW vehiclesany system.
power is instantaneous contains no time dimension.
It doesn't? Power = Energy/Time. Joules per second. Ergs per eon. Electron-volts per attosecond.
Power contains no time component just the energy contains time.
Power is measured in Watt while energy is measured in Joules same as Watt second
So yes Power = Energy/Time but the part containing time is Energy not Power that is why you divide by time to get the power.
That's some extraordinarily convoluted babbling, even for you. Power literally has no definition other than energy per unit time, or rate of energy transfer. An amount of energy can be stored (as potential energy, like a compressed spring) for an arbitrary amount of time.
There is nothing wrong with that definition if you understand what it means.
There seems to be a lot of difficulty working through the maths for the situation when the Blackbird is travelling at wind speed - so let's back away from that for a moment.
How about we put together the equations that describe the system for the Blackbird when it is travelling at, say, half wind speed?
This will require consideration of all the components including wind input, propeller output, wheel motion and whatever power train is in place between wheels and propeller. ALL things must be considered and nothing can be dismissed out of hand as they are all, very clearly, involved.
If done correctly, then we follow this mathematical model and see where it ends up.
If you can't be bothered to look at this, properly, then you've revealed yourself as a troll.
There is nothing wrong with that definition if you understand what it means.
What definition? ::) Where have you 'defined' power other than to say that it has no time dimension?
Power is just the rate at which energy is converted to another form so it is not containing time.
Power is just the rate at which energy is converted to another form so it is not containing time.
|O
Alright, can you define the term 'rate' then?
The word rate should be self explanatory not sure if I can define that better for you but it has nothing to do with time.
So, power is energy measured per unit of......??
No really, I am not sure what is going on here,but in general, this shape somehow slows down other air around the craft and thus converts this energy into a greater speed for the craft. So conservation is retained as the other air is slowed down. This could have to do with a strong negative energy source against the positive one, caused by the positive one.
There is nothing wrong with that definition if you understand what it means.
What definition? ::) Where have you 'defined' power other than to say that it has no time dimension?
I need to say that since you think it has a time dimension. I was refereeing to the definition you provided.
Power is just the rate at which energy is converted to another form so it is not containing time. The thing that contains time is energy.
So when I say 10W of braking power and I do not mention any time interval for witch this power is applied then time is not involved at all and no work it is being done.
You may know speed as in case A 2m/s and you can calculate force 5N but that is about all you can know. You can not say about any changes to vehicle speed or kinetic energy as time is not mentioned anywhere.
As soon as I say the 10W is applied for 1ms then you have the time and also the energy 0.01Ws and thus you can calculate what happens to vehicle over that time period including change in kinetic energy and speed.
To me it just seems you do not see power for what it is and it just seems you confuse that with Energy.
Power is in units of work per time, J/s. The 10W already includes the time element as it is joules per unit time. Joules is not defined with any regard to time. It can be 10 W for one second or 1 W for ten seconds. 10 W is 10 J/s. So that could be 10 joules in 1 second, or 100 joules in 10 seconds, but it is 10 joules per second.
Interesting that neither work nor power are considered SI base units, but that's not important. I don't know if this matters either, but NIST shows work (energy) as derived from force and distance. They show power as defined by work per unit time (J/s).
NIST Diagram (https://www.nist.gov/sites/default/files/styles/2800_x_2800_limit/public/images/2021/08/23/NIST.SP_.1247.png?itok=CAFgVO0D)
Does that help?
What brought it home for me was when the video talked about the difference in ground speed and air speed. The wheels are working at ground speed while the propeller is dealing with the wind speed. The higher ground speed means the "gearing" (including the propeller pitch, et. al.) allows the faster ground speed to create less resistance than the force generated by the prop.
As others have pointed out the propeller blows air backwards from the car reducing the wind speed and lowering the wind's energy content conserving energy.
That should cover it, no?
Here's a thought experiment. Instead of wind and a propeller, what if the wind were a conveyor belt and the prop were another wheel on that conveyor belt?
Now the wheels can be geared so the ground movement applies a force to the conveyor belt. Can the car move forward faster than the conveyor belt?
Yes, as the conveyor ramps up speed the car would be propelled faster than the conveyor from the start. Give the wheels a 2:1 gear ratio and the car will move at twice the conveyor speed. Like with the propeller, the ground wheels turn faster, so the conveyor wheels can exert more torque resulting in a net forward force.
This is stirring a memory of some toy I had years and years ago. It also is making me wonder if that could be used for something practical.
Power is in units of work per time, J/s. The 10W already includes the time element as it is joules per unit time. Joules is not defined with any regard to time. It can be 10 W for one second or 1 W for ten seconds. 10 W is 10 J/s. So that could be 10 joules in 1 second, or 100 joules in 10 seconds, but it is 10 joules per second.
Interesting that neither work nor power are considered SI base units, but that's not important. I don't know if this matters either, but NIST shows work (energy) as derived from force and distance. They show power as defined by work per unit time (J/s).
NIST Diagram (https://www.nist.gov/sites/default/files/styles/2800_x_2800_limit/public/images/2021/08/23/NIST.SP_.1247.png?itok=CAFgVO0D)
Does that help?
Sorry not sure where you have this wrong definition of what power and energy is.
Power is measured in Watt
Energy measured in Joules
Applying a power of 10W for one second will result in 10J or what I prefer 10Ws.
So the 10W includes not time is just a rate of work nothing to do with time while Energy includes time thus why I prefer to use Ws instead of Joules.
One Joule is power of 1W applied for 1 second but can also be 10W applied for 100ms or 1kW applied for 1ms. You get the idea that power has nothing to do with time it is the rate at witch work is done.
Not sure why Power and Energy are so misunderstood and the way they are defined by NIST is juts super bad and understandably confusing.
Why will you ever define Power using Energy ?
No wonder why so many are confused about how blackbird works.
There is nothing wrong with anything I've said. All the examples you give agree with my post entirely. The error is in your thinking that somehow time is associated with energy. You can pervert the units as much as you want, but a joule is a unit of energy or work that has no time factor in it.
I see you talk about "rate" of work. That is the same thing as J/s. In this context rate implies time. So work is independent of time as shown by your examples above. The same work results from different powers with different times, but the products must always be the same. Why? Because power is a rate, a function of time and must be multiplied by time to get work... to eliminate the time factor in work.
Do you think the same way about speed and distance? Speed is analogous to power and distance is analogous to work. Speed times time is distance, power times time is work. Do you consider distance to be a function of time? The only difference is we usually don't have a particular name for speed, rather it's distance per time like miles per hour, although they do use knots. So is distance Knot·hours?
I suppose none of this is going to matter. You seem to be pretty entrenched in your thinking. That's fine. We can agree to disagree.
I don't think I'm willing to watch a 25 minute video that is hard to understand when I find an error in the first four minutes. You define the power from the wind as some stuff * (w-v). You claim this is the reason why a propeller vehicle can't exceed the wind speed. This ignores the rotation of the propeller which is the whole point!
I also skipped ahead to 19 minutes where you claim the geared wheels and stick are not being analyzed correctly. I don't see how you can say anything of the sort. The analogy is not perfect between the propeller car and the gear car because the gear ratio of the gear car makes it move instantly while the propeller car has to accelerate to beat the wind speed.
Maybe I'll watch the rest of this video someday, but for now it's just too much work.
So how far would they need to drive the propeller car for you to say it's not running on stored power?
Ok, if you are going to invoke relativity to discuss distance, then I think we will never find a common ground to discuss.
Enjoy
The naming is a bit confusing, but it is OK to calculate this way, if the stick to the vehicle drawing frame as the reference and also use it for the rest of the calulation. In this case however we must have a closer look at the center part, especially the question, what is the meaning of P_net ?What does output of the motor wheel mean ? This is not really clear to me.
It that the same as the power needed to supply the motor = the ouput power of the motor ?
For now it is just about the equation, so we don' t need to dicuss the 3 cases. We need to understand the equation first.
Output at the motor wheel is the input power in the motor from generator wheel plus the power that wind treadmill provide to that same wheel or subtract from the wheel as it is the case at C.
There are just two sources of energy available to vehicle one is the road treadmill and one is the wind treadmill. This is because we are looking in the frame of reference of the vehicle so vehicle is the reference thus is not moving.
You can salve the same problem in multiple way this is just the one I selected mostly based on the fact that people seems to prefer looking at this problem form the vehicle frame of reference and that is how Derek also solved this same problem.
You can look from the ground reference frame where ground / Road treadmill is not moving at 2m/s but vehicle moves forward at 2m/s left to right and in that case the wind treadmill will have moved at 6m/s the wind speed relative to ground/road but it will still be just 4m/s relative to vehicle.
We can agree to disagree.
What happens is that energy is stored as pressure differential (propeller uses some of the wind power to increase pressure differential).
This stored energy while vehicle is below wind speed is then used to accelerate for some limited amount of time above wind speed.
QuoteWe can agree to disagree.
What!??!!! After 958 comments? NOOOOOooooooo
A question, if I may....
Let's say the Blackbird has gone through all this "energy storage" exercise and has been running along the ground for long enough that the "stored energy" is all gone. We are looking for the steady state where the speed is constant - no acceleration nor deceleration. (I don't care how long this will take as we can imagine a test track as long as we want and a nice, consistent wind.)
electrodacus - what do you say it's speed will be?
1. Greater than wind speed
2. Equal to wind speed
3. Less than wind speed
The naming is a bit confusing, but it is OK to calculate this way, if the stick to the vehicle drawing frame as the reference and also use it for the rest of the calulation. In this case however we must have a closer look at the center part, especially the question, what is the meaning of P_net ?
edit:
Looking at it again, I realized that the formular is calculating the power in a difference reference frame. So I don't see a good way for calulating this way. It kind of lead nowhere. You get a number but no good way use to it, at least I don't see one.
The logical way is to calculate the power that is needed to power the motor, as that is what we would have to compare to the power from the generator.
To analyze the system correctly, a different approach is required.
I didn't read the whole thread.
How long has electrodacus been insisting the car that goes down wind faster than the wind doesn't? How long has he insisted energy involves time and power doesn't?
A question, if I may....
Let's say the Blackbird has gone through all this "energy storage" exercise and has been running along the ground for long enough that the "stored energy" is all gone. We are looking for the steady state where the speed is constant - no acceleration nor deceleration. (I don't care how long this will take as we can imagine a test track as long as we want and a nice, consistent wind.)
electrodacus - what do you say it's speed will be?
1. Greater than wind speed
2. Equal to wind speed
3. Less than wind speed
Lol
I didn't read the whole thread. How long has electrodacus been insisting the car that goes down wind faster than the wind doesn't? How long has he insisted energy involves time and power doesn't?
I find his arguments mildly entertaining. The diagram posted seems to me to be a lot of fluff. Especially amusing is his stored energy argument to explain "impossible actions" when it is just poor construction at work. Then he says a better constructed model has "micro storage" of energy or something.
If the wheels are geared 2:1, applying a force to the 1 wheel by moving the belt it is riding on will exert say 1 unit of force in the direction of the belt. The gearing will result in twice the movement in the 2 wheel and half the force. The two forces are opposite, so the vehicle has a half unit of force pushing it in the direction of the moving belt with the movement relative to the ground being twice the velocity of the movement of the belt.
I'm willing to say I'm wrong when someone shows that to me. Happy to, in fact. But someone who doesn't understand that "rate of change" in regards to values like power and energy fundamentally involves time can't possibly construct a rational argument. While rate of change can refer to some other variable, in the case of power, it is the rate of change of energy wrt time. Power involves time. Energy does not.
This is such an insane argument! I won't argue with the guy because his arguments are so specious, yet I want so badly to help him understand. But I'm not going to chase impossible dreams. They guy clearly does not want to understand, so no one can help him.
It did take some head scratching before I got it, but I can totally see how it works.
It does require the surface it rides on to be moving towards the rear of the vehicle faster than the air though, which is where it gets the energy to move upwind. You don't get free energy.
This is why it works on a treadmill as well as with a tailwind. I imagine it would need a push start too because stationary props don't provide any thrust.
You do need a really really low friction gearing setup too.
I have never claimed blackbird is not going faster than the wind.
I say that available wind power is defined by this equation 0.5 * air density * area * (wind speed - vehicle speed)^3
One may call the pressure that drives the prop forward stored energy, if one really wants too. However this a energy in a dynamic balance: loosing energy to drive the vehicle and to the less than perfect aerodynamics and getting new energy from the rear wheels.
The point is that the driven prop can get you more drag than a sail. It can do it a zero speed (e.g. no wind and not moving and thus zero drag for the sail) - this case is the abvious one I think. Similar it can do it when the vehicle is driving at the speed of the wind and thus zero relativ speed between the vehicle and the wind.
When starting at low speed the slow moving prop would still have some drag, not much, but it should be enough to get it starting. The prop would need not extra power and at low speed the prop would actually generate extra forward power and act as a windmill. It would need some minimum wind speed to get it going.
I don't think they needed the adjustable pitch during driven. This is likely just a thing of finding the best setting to get the highest speed. Once set right it should be OK to keep it there. The prop is not very heavy and not going very fast, as it is optimized to work at a relatively low relativ air speed. Remember it only sees the difference, so normally not the full wind speed. So the kinetic energy in the prop is low compared to the kinetic energy in the main vehicle. The model on the treadmil works with a fixed prop.
But you are claiming that it cannot sustain the faster-than-wind travel. Or are you beginning to realize that you are wrong?
QuoteI say that available wind power is defined by this equation 0.5 * air density * area * (wind speed - vehicle speed)^3
As I said earlier, the 'available wind power' does not depend on the vehicle speed where the vehicles mechanisms can use the ground as a fixed reference, only the windspeed relative to the ground matters. So the prediction is that a properly configured vehicle can travel both upwind (at some rate) and downwind (faster than the wind) indefinitely. And, it does exactly that experimentally.
But you are claiming that it cannot sustain the faster-than-wind travel. Or are you beginning to realize that you are wrong?
Yes that is correct the acceleration rate as seen in all tests drops and at some point acceleration rate will be zero and the deceleration phase will start.QuoteI say that available wind power is defined by this equation 0.5 * air density * area * (wind speed - vehicle speed)^3
As I said earlier, the 'available wind power' does not depend on the vehicle speed where the vehicles mechanisms can use the ground as a fixed reference, only the windspeed relative to the ground matters. So the prediction is that a properly configured vehicle can travel both upwind (at some rate) and downwind (faster than the wind) indefinitely. And, it does exactly that experimentally.
Please provide the equation describing what you are saying. If you can not do that you can not claim you understand how this vehicle works.
It is super obvious that vehicle speed while driving directly down wind affects the amount of available wind power since the wind speed relative to vehicle drops (wind speed - vehicle speed).
Please provide the equation describing what you are saying. If you can not do that you can not claim you understand how this vehicle works.
It is super obvious that vehicle speed while driving directly down wind affects the amount of available wind power since the wind speed relative to vehicle drops (wind speed - vehicle speed).
I do not see how else you can call the pressure differential created by the propeller other than a form of energy storage.
The applicable equation is some_factor * air density * area * (wind speed - ground speed)^3
Since ground speed = 0, the equation becomes:
some_factor * air density * area * (wind speed)^3
The difference is that your equation assumes that the vehicle has no contact with the ground, and behaves something like a balloon. This is certainly not the case with the DDWFTTW vehicles we ae discussing.
A propeller does not create a pressure differential. A propeller moves ("propels") air. It creates thrust by the law of conservation of momentum, by accelerating the air that goes through it. The air leaving a propeller has a lower pressure because it is moving. You can see this described here:
https://www.quora.com/Why-is-the-static-pressure-before-a-fan-higher-than-after-the-fan (https://www.quora.com/Why-is-the-static-pressure-before-a-fan-higher-than-after-the-fan)
And illustrated here by experiment:
https://youtu.be/f2QfVJe7yEg?t=198 (https://youtu.be/f2QfVJe7yEg?t=198)
Given that there is no "pressure bubble" behind the propeller, there is no energy storage there to push back on the propeller. Once the air has gone through the propeller it has done its work, there is no more work to give.
I say that available wind power is defined by this equation 0.5 * air density * area * (wind speed - vehicle speed)^3
I say that available wind power is defined by this equation 0.5 * air density * area * (wind speed - vehicle speed)^3I say this is wrong, and the PDF on wind power you linked also says something different:
I say that available wind power is defined by this equation 0.5 * air density * area * (wind speed - vehicle speed)^3
The equation 0.5 * (air density) * (area) * (wind speed)^3 gives the total flow of kinetic energy from wind moving through a particular cross section of area, and is applicable to wind turbine design as the air is moving through the turbine. A turbine could never extract more power than this from the wind (it will always be less than this). Note that this equation only works if the wind moves through the area of interest, like the disk area of a turbine.
Since a sail is a solid barrier, the wind does not, and cannot, move through the sail. Therefore, this equation cannot be applied to sails.
Another way of looking at this is to observe that the equation is (kinetic energy of wind per unit mass) x (mass flow of wind moving through control area). Since the mass flow of wind through the sail is zero, the energy flow must necessarily be zero.
When working with sails, you have to use force vectors and force balances.
Sorry but you are just ignoring the fact that air is a compressible fluid.
Are you saying this graph below is completely wrong ?
https://en.wikipedia.org/wiki/Axial_fan_design (https://en.wikipedia.org/wiki/Axial_fan_design)
(https://upload.wikimedia.org/wikipedia/commons/thumb/f/f1/Axial_fan_slipstream_theory.svg/400px-Axial_fan_slipstream_theory.svg.png)
This equation perfectly works for sails.
If you install the wind turbine on a vehicle and then drive direct down wind then equation of the power output for that wind turbine will again contain (wind speed - vehicle speed) since the wind speed relative to that moving wind turbine will be lower the faster the vehicle moves.
P_sail = 0.5 * air density * area * (wind speed - vehicle speed)^2* vehicle speed
With the active driven prop there can be even more power available, as the air may slow down even more. This is especially true for a speed close to or higher than the speed of the wind. This is because this way the wind can be slowed down to a speed below the vehicle speed.
If you install the wind turbine on a vehicle and then drive direct down wind then equation of the power output for that wind turbine will again contain (wind speed - vehicle speed) since the wind speed relative to that moving wind turbine will be lower the faster the vehicle moves.
This is exactly correct, at least up to (vehicle speed = wind speed), for a turbine/generator bolted onto a moving vehicle. However, that is not what the rest of us are discussing; the Blackbird and related DDWFTTW vehicles.
In the comparison you forgot one important point - the point with vehicle speed zero.
P_sail = 0.5 * air density * area * (wind speed - vehicle speed)^2* vehicle speed
With the active driven prop there can be even more power available, as the air may slow down even more. This is especially true for a speed close to or higher than the speed of the wind. This is because this way the wind can be slowed down to a speed below the vehicle speed.
Say wind speed is 10m/s and vehicle speed is half that 5m/s directly down wind
What sail will see is a 10 - 5 = 5m/s wind relative to it and vehicle thus correct equation for a 1m^2 sail will be
0.5 * 1.2 * 1 * (10-5)^3 = 75W
Using your equation 0.5 * 1.2 * 1 * ((10-5)^2 * 5) = 75W
Seems to provide the same value for half wind speed but
Vehicle speed of 2m/s
0.5 * 1.2 * 1 * (10-2)^3 = 307.2W
0.5 * 1.2 * 1 * ((10-2)^2 * 2) = 76.8W no longer correct
As I pointed out much earlier, starting with reply #20 to this thread, you can pretty much explain it without much more than the principles of Archimedes.
In the comparison you forgot one important point - the point with vehicle speed zero.
In this case the (w-v)³ type gives 0.5 * 1.2 * 1 * (10)^3 = 600W.
For sail driven vehicle at zero speed there is only force and not power transfered to the vehicle. So this equation is obviously wrong at that point. :horse:
It gets even worse when you drive agains the wind :-DD
If you remove the brakes those 600W will be available to vehicle instead of earth.
As I pointed out much earlier, starting with reply #20 to this thread, you can pretty much explain it without much more than the principles of Archimedes.
Indeed, and for reference, here is an analysis of the cart/wheel/belt system.
Refer to the diagram attached below.
(https://www.eevblog.com/forum/chat/mess-with-your-minds-a-wind-powered-craft-going-faster-than-a-tail-wind-speed/?action=dlattach;attach=1355879;image)
Assume that it is an ideal system, no friction losses, and both wheels are able to spin freely on their bearings.
There are five variables: \$v_W\$, \$v_R\$, \$v_M\$, \$v_G\$ and \$v\$
\$v_M\$ and \$v_G\$ are the rim speeds of the wheels, in m/s.
\$v_W\$ and \$v_R\$ are given, leaving three degrees of freedom.
Two equations can be written relating the belt velocities, the rim speeds of the wheels, and the cart speed:
$$v=v_M-v_W$$ $$v=v_G-v_R$$
These two equations take up two degrees of freedom, leaving one degree of freedom remaining.
With no additional constraints this system has an infinite number of solutions. The cart velocity, \$v\$ can have any value we choose.
This makes sense, intuitively, since the wheels are free running, and without friction losses any initial cart velocity will remain the same forever.
To remove the remaining degree of freedom in the system, the wheels can be linked with a connecting gear. If we let the gear ratio be \$\alpha\$, then we can write a linking equation:
$$v_G=\alpha\,v_M$$
For each revolution of wheel M, wheel G makes \$\alpha\$ revolutions.
(Note that the gearing could be mechanical, or it could be electrical. Mechanical is simpler.)
By making a suitable choice of \$\alpha\$, we can arrange for a desired cart speed, \$v\$.
For instance, suppose we wish the cart to move at 1 m/s to the right.
We set \$v\$ = 1 m/s, which gives \$v_M=v+v_W\$ = 5 m/s, and \$v_G=v+v_R\$ = 11 m/s.
From this, we have: \$\alpha = v_G / v_M\$ = 11 / 5 = 2.2
Therefore, with a gear ratio of 2.2, the cart will move to right at a speed of 1 m/s.
[...]
Therefore, with a gear ratio of 2.2, the cart will move to right at a speed of 1 m/s.
If you remove the brakes those 600W will be available to vehicle instead of earth.
And if you release the brakes, but not all the way, so that the vehicle creeps, isn't most of your 600W still 'available to the earth'? The vehicle is still pushing on the earth...
I appreciate the effort you put in to this but no vehicle without energy storage can move in this example from left to right.
I appreciate the effort you put in to this but no vehicle without energy storage can move in this example from left to right.
Look again. Actually think about it.
He just looked at what gear ratio he will need if he wanted to push the vehicle with 1m/s (by push I mean apply external power to the system as that is the only way that vehicle can move from left to right).
No need to build a model, just have think: The mechanical power from a sail is transferered as force times vehicle speed. So the force would be power divided by speed. Does it really make sense to get an even increasing force from the sail, the slower you make the vehicle. Thas is obviously not going to happen. The wind power of some 600 W is theoretically (that is with an ideallized wind turbine, ignoring the Beetz limit and similar details) available at zero speed, but a simple sail can not harvest it. A sail is not 100% efficient. So doing the math with a sail you can not exclude anything for a vehichle used other means than a simple sail. Using the wrong equations comes one top of this.If you remove the brakes those 600W will be available to vehicle instead of earth.
And if you release the brakes, but not all the way, so that the vehicle creeps, isn't most of your 600W still 'available to the earth'? The vehicle is still pushing on the earth...
As soon as you start moving those now less than 600W will be split between vehicle acceleration and brake friction (lost as heat).
Say you want to maintain a 1m/s then brakes will need to deal with 0.5 * 1.2 * 1 * (10-1)^3 = 437.4W
If your brakes where not designed to be able to get rid of those 437.4W then they will overheat and get damaged.
We can setup a scale down model of this if you do not believe that will be the case.
No need to build a model, just have think: The mechanical power from a sail is transferered as force times vehicle speed. So the force would be power divided by speed. Does it really make sense to get an even increasing force from the sail, the slower you make the vehicle. Thas is obviously not going to happen. The wind power of some 600 W is theoretically (that is with an ideallized wind turbine, ignoring the Beetz limit and similar details) available at zero speed, but a simple sail can not harvest it. A sail is not 100% efficient. So doing the math with a sail you can not exclude anything for a vehichle used other means than a simple sail. Using the wrong equations comes one top of this.
As soon as you start moving those now less than 600W will be split between vehicle acceleration and brake friction (lost as heat).
He just looked at what gear ratio he will need if he wanted to push the vehicle with 1m/s (by push I mean apply external power to the system as that is the only way that vehicle can move from left to right).
The system does, of course, have external power. The two belts are not moving by themselves, so we must assume there is some kind of motor driving them, and if we make any attempt to slow down the belts then the motors must supply more power to maintain the belt speed. So the cart can pick up any power it needs from the difference in speed of the belts.
But leaving that aside, power is only required to overcome friction or to accelerate the cart. In an ideal system, in steady conditions, with no friction, then the power requirement is zero. So no external power is actually needed for the cart to be moving at any steady velocity.
As soon as you start moving those now less than 600W will be split between vehicle acceleration and brake friction (lost as heat).
Is the car (and the wind through the car) not still pushing on the earth?
When I mentioned external power I was excluding the two belts / treadmills.OK, no belts or treadmills. They are excluded.
You need to be able to demonstrate that using the energy from road treadmill (generator wheel) and applying that to Motor wheel you are able to advance forward (meaning accelerate from left to right).It seems that won't be possible, as the treadmills are excluded. If the road treadmill is excluded it cannot supply energy.
You will not be able to do that without adding energy storage.Energy storage from where? The treadmills are excluded, so they cannot supply any energy to store. There is no other source of energy in the system.
I do not think I can continue to argue here as your level of understanding is way below what is required. Not sure if this is a failing of the education system or a limitation of most humans brain (or maybe a combination of both).It's certainly true that your brain is different from most humans. You are exceptional. Congratulations!
Even if I setup an experiment and show conclusively that my theory is correct all you will do is defend the new theory without understanding what you are defending.
The mechanical power from a sail is transferered as force times vehicle speed.
I do not see how else you can call the pressure differential created by the propeller other than a form of energy storage.I would not call it energy storage - but I see why you are saying that.
There is loop but it is as follow (proportions are just an example).I'm not sure I agree with the proportions but, as you say, they are just to illustrate the example. The important part here is they represent two non-zero values.
Wind power available to vehicle will be split say in two equal parts. One half accelerates the vehicle directly thus ends up as kinetic energy the other half will be taken from wheels and sent to propeller witch then increases the pressure differential
(you can look at this as increasing the apparent wind speed relative to vehicle).You have just given the answer as to why the Blackbird could be capable of exceeding wind speed.
... ... ...Again, I'm not enthusiastic about some of the terminology and I'd be cautious about the proportions used - but as a qualitative description, that's not too bad.
For blackbird since it takes in this example half of the available power and stores that as pressure differential basically increasing the available potential energy it allows the blackbird to exceed wind speed for some limited amount of time as when above wind speed direct downwind there is no longer any wind power available to vehicle and it is starting to use the energy stored as pressure differential but it will continue to use just half of the power provided by the pressure differential to accelerate (increase kinetic energy) and then the other half it will put back in to increasing the pressure differential.
Obviously since only half is put back the overall pressure differential will drop and vehicle acceleration rate will continue to drop(Again, proportions) - But this is not unexpected. Acceleration will tend to zero.
until there is not enoughAs in not enough "pressure differential" (as you call it) to provide acceleration? I can agree with that. There will be a point where everything balances out. This will be the "steady state" situation I was asking about earlier.
and it will need to start to slow down.Now this statement is where (as I see it) you have made your error.
OK, no belts or treadmills. They are excluded.
QuoteYou need to be able to demonstrate that using the energy from road treadmill (generator wheel) and applying that to Motor wheel you are able to advance forward (meaning accelerate from left to right).It seems that won't be possible, as the treadmills are excluded. If the road treadmill is excluded it cannot supply energy.QuoteYou will not be able to do that without adding energy storage.Energy storage from where? The treadmills are excluded, so they cannot supply any energy to store. There is no other source of energy in the system.QuoteI do not think I can continue to argue here as your level of understanding is way below what is required. Not sure if this is a failing of the education system or a limitation of most humans brain (or maybe a combination of both).It's certainly true that your brain is different from most humans. You are exceptional. Congratulations!
Even if I setup an experiment and show conclusively that my theory is correct all you will do is defend the new theory without understanding what you are defending.
Now this statement is where (as I see it) you have made your error.
When acceleration drops to zero, it does not mean the Blackbird slows down. It means the Blackbird's velocity does not change. As long as the wind blows the same, it will continue at that same speed.
As long as the Blackbird is moving, it is continually producing (not storing) this "pressure differential" as you call it and the "apparent wind speed" will be greater than the actual wind speed.
How this translates into the Blackbird's final velocity is very dependent on the proportions you mentioned. These proportions and the mechanisms that determine them need to be properly worked out.
I must be very bad at explaining myself. The two treadmills are the only thing interacting with vehicle. So they need to power the vehicle not some external source if you do not want to accept that there is a energy storage device then you need to explain how it will move from left to right using just the two treadmills.
I appreciate the effort you put in to this but no vehicle without energy storage can move in this example from left to right.When you make this statement, you are starting with a conclusion, "No vehicle can...", and working backwards to a justification. But that is not how reasoning works. You need to start with some definitions and a statement of the problem, and work forward from there using logical reasoning to reach a conclusion.
You just looked at speed here no forces ? How will that even make sense for this particular problem ?There are no forces because this is a steady state analysis of an ideal system with no friction. There are no forces, and there is no power involved. This is a pure problem in mechanics.
I wish I knew how to help you from here but you are just way off thus I'm sorry but I cant help.
After vehicle is above wind speed wind can not offer anything to the vehicle so vehicle is on his ownAt wind speed, the wind itself is not directly driving the vehicle. It's like a wall of air that never makes contact with the vehicle - that part is true. However, we still have a rotating propeller which is producing thrust. It is this thrust which is pushing against that wall of air, giving the "apparent wind speed" phenomenon YOU mentioned.
I need to bring back the discussion to what is relay important.
What is the available wind power to direct down wind vehicle.
I say that available wind power is defined by this equation 0.5 * air density * area * (wind speed - vehicle speed)^3
If anyone disagree with that please provide the correct equation so we can compare the predictions.
+ speed of propeller thrust)^3
+ speed of propeller thrust)^3
Don't go there! Thrust is force and force doesn't have a speed, although I can understand what you are getting at. I think what you want is the effective pitch of the propeller blades multiplied by its angular velocity. So if the pitch, which can be stated as distance per revolution, is 0.1m/rev and the propeller is spinning at 100rpm, then it's pitch speed (probably the wrong term) is 10m/s.
Besides, I use the word "thrust" here simply as a descriptive term. If you want something less controversial, then I could have said:
0.5 * air density * area * (wind speed - vehicle speed + speed of the air being blown backwards by the propeller)^3
There is nothing you need to do to that equation. The correct one is the one I presented.Wrong.
The extra bit that is missing the the stored energy. The propeller was powered by wind power when well below wind speed then as the wind power decreased the stored energy started to provide most of the power to vehicle.Wow. Just, wow. That is just so wrong. There is no energy storage as you maintain. There can't be.
The way that stored energy is calculated is a bit complicated but it can be done as I already did in the spreadsheet calculator.Again, you are aware of the phenomenon, but you continually misrepresent it! The "pressure differential" is thrust. This thrust is not stored energy - it is continuously created moment by moment. Disconnect the drive shaft and it will cease as soon as the rotational momentum of the propeller is spent.
Say 60% of wind power is sent to propeller (from the wheel) then the other 40% gets to accelerate the vehicle increasing the vehicle kinetic energy and obviously speed.
The 60% that gets to propeller will be contributing to increased pressure differential (of course propeller may be 50% efficient so half of that 60% ends up as stored energy) You can see this increase in pressure differential about the same way as if natural wind speed has increased.
In the calculator all that energy put in the propeller increases the potential wind energy and with such a mechanism 2x even 3x wind speed is not a problem but since this is stored energy as soon as it is all used up vehicle will slow down below wind speed.You need to get over this "stored energy" thing. It's stopping you from seeing the correct picture.
All that it will need to be done is to run the experiment fully not stop the experiment just before vehicle will start to slow down.Now this is a problem I have with you. This is an assertion on your part. Aside from your own, self-serving claims, there is (as far as I know) NO evidence that this will happen. You can't say "I've proven it" because you are biased. Your opinion does not count. What is needed is an INDEPENDENT, authoritative source.
It is a shame that this problem is wrongly presented in some schools and I wish I could do something about it. I was much more optimistic about being able to explain this but it seems so far I'm failing to do so.Oh - you are doing an EXCELLENT job in helping people understand what is going on. By presenting an absolutely flawed explanation - one that has been disproven experimentally - you have engaged curious minds in a process of examination of the problem. As a result, they are much more aware of how the mechanism actually works.
Besides, I use the word "thrust" here simply as a descriptive term. If you want something less controversial, then I could have said:The trust is a force and not power ( please don't make the same error as electrodacus). So the is only a square. If you want the power, than one has to multiply by a speed, that may depend on the reference system used and this often is not the same as the one used for the thrust.
0.5 * air density * area * (wind speed - vehicle speed + speed of the air being blown backwards by the propeller)^3
Besides, I use the word "thrust" here simply as a descriptive term. If you want something less controversial, then I could have said:The trust is a force and not power ( please don't make the same error as electrodacus). So the is only a square.
0.5 * air density * area * (wind speed - vehicle speed + speed of the air being blown backwards by the propeller)^3
Even in the verbose form it is still not allways right to use a simple speed difference to the 3rd power. There is a 2nd power for the force and than an often different speed for the speed to make is a power.Besides, I use the word "thrust" here simply as a descriptive term. If you want something less controversial, then I could have said:The trust is a force and not power ( please don't make the same error as electrodacus). So the is only a square.
0.5 * air density * area * (wind speed - vehicle speed + speed of the air being blown backwards by the propeller)^3
Again - I used the term "thrust" in a conversational sense, not an engineering one, which is why I offered the verbose form.
Again, you are aware of the phenomenon, but you continually misrepresent it! The "pressure differential" is thrust. This thrust is not stored energy - it is continuously created moment by moment. Disconnect the drive shaft and it will cease as soon as the rotational momentum of the propeller is spent.
Then you are saying there is a fundamental problem with our friend's original formula?
Then you are saying there is a fundamental problem with our friend's original formula?
An irrelevant question that has no bearing on the problem:
Is the "sail" in that equation just a planar surface? Actual sailboat sails, even those used for DDW sailing, are actually cut to create a three-dimensional airfoil, and are trimmed (tensioned with ropes) to optimize airflow and create lift. So the efficiency numbers may be different than the Bernoulli numbers.
With a failure to recognize even such a simple point, I think I will give up on convincing electrodacus. Not sure if he does not know better or is acting as a troll just to keep the discussion running in circles, repeating his wrong claims over and over again.
Wow. Just, wow. That is just so wrong. There is no energy storage as you maintain. There can't be.
The problem is in what the formula stands for: The (w-v)³ type formular is for the power available to a moving wind turbine (ignoring the power needed of vailable form the movement of the turbine). It is the the same as the normal ergy availabel to a wind turbine, just with a different wind speed.
This is different from it is totally different from the power from a sail on the vehicle: that power is force times vehichle velocity as also shown in the calculation of IanB. So the 2 parts together may make more sense.
Anyway the equation for the power a sail vehicle could use is not even relevant for the calculation that shows that the Backbird vehicle can work.
With only using a formula for a sail based setup on can not calculate the prop driven vehincle, as this is something different. When you do the calculation for the simple (e.g. 1 D world directly downwind) sail driven vehicle the result is that it can not go faster than the wind. This still does not say anything about a sail driven vehicle going zig-zag. This is known to be able to go faster.
I only took a more detailed look with the case of zero and low speed, to show that the 0.5 * air density * area * (wind speed - vehicle speed)^3 form is obviously wrong for a sail based vehicle.
For some reason electrodacus has difficulties in accepting that he can be wrong with his equation. It is not the only point he is wrong, but I had hope he would recognize the error in a equation that is not really related (at least I don't see a good way to use it one way or the other) to the main question.
With a failure to recognize even such a simple point, I think I will give up on convincing electrodacus. Not sure if he does not know better or is acting as a troll just to keep the discussion running in circles, repeating his wrong claims over and over again.
The sail in the calculation is just the planar surface with wind perpendicular (down wind). So not sophisticated areodynamics included.
Correct formula for a direct down wind vehicle in therms of available wind power is this: 0.5 * air density * area * (wind speed - vehicle speed)^3
Not only the formula is correct it also shows that without energy storage no direct down wind vehicle can exceed wind speed.
Correct formula for a direct down wind vehicle in therms of available wind power is this: 0.5 * air density * area * (wind speed - vehicle speed)^3
Not only the formula is correct it also shows that without energy storage no direct down wind vehicle can exceed wind speed.
Were do you get that magic formula from ? It is not supported by the wind power PDF you linked and I have not seen it in any books. The standard formular is just the 3rd power of the wind speed, without the vehicle moving. With the idealization of the Betz limit this is OK. The moving wind turbine is a rather exotic problem and thus usually no read made tabulated solution for this problem and one has to derive it from the general principles.
This problem is impossible to solve without the correct information .
Firstly the Prop . being variable pitch it also has centre clutch assembly [...]
I suppose this makes sense to electrodacus, but I've sailed thousands of miles downwind at less than windspeed. There is no significant energy storage, you use it or lose it.Yes your correct .
So say this vehicle needs to drive on highway at 120km/h about 33m/s and there is no wind.
Then vehicle will need 0.5 * 1.2 * 0.75 * (33)^3 = 16.2kW just to deal with the drag. A bit more will of course be needed to deal with other friction losses internal and rolling resistance but this air drag will be the most significant.
So no vehicle with those characteristics 0.75m^2 effective frontal area can ever claim it needs less than 16.2kW/120km/h = 135Wh/km
QuoteSo say this vehicle needs to drive on highway at 120km/h about 33m/s and there is no wind.
Then vehicle will need 0.5 * 1.2 * 0.75 * (33)^3 = 16.2kW just to deal with the drag. A bit more will of course be needed to deal with other friction losses internal and rolling resistance but this air drag will be the most significant.
So no vehicle with those characteristics 0.75m^2 effective frontal area can ever claim it needs less than 16.2kW/120km/h = 135Wh/km
When the vehicle is at wind speed there is no air drag. Nothing. 0. At faster than wind speed there is a small amount (assuming just slightly faster), increasing as the vehicle exceeds wind speed more. The increased drag prevents infinite acceleration, but at just over wind speed there is no drag to speak of, hence the power needed to combat drag could be as small as you can arrange to measure.
And also this formula is used if you want to know how what is the minimum power your vehicle engine or motor requires in order to overcome the drag due to wind.
Most small passenger vehicles have around 2.5m^2 of frontal area and if they are aerodynamic the Coefficient of Drag may be around 0.3 thus effective frontal area 2.5m^2 * 0.3 = 0.75m^2
So say this vehicle needs to drive on highway at 120km/h about 33m/s and there is no wind.
Then vehicle will need 0.5 * 1.2 * 0.75 * (33)^3 = 16.2kW just to deal with the drag. A bit more will of course be needed to deal with other friction losses internal and rolling resistance but this air drag will be the most significant.
So no vehicle with those characteristics 0.75m^2 effective frontal area can ever claim it needs less than 16.2kW/120km/h = 135Wh/km
To be fair, while I initially was thinking "but DDW at windspeed there's no drag at all", in further reflection I read this as an interesting digression from the DDWFTTW arguments, and an illustration of how drag affects automobiles at speed. I don't think that electrodacus is claiming that a vehicle driving 120 km/h downwind in a 120 km/h wind will see 16.2 kW of drag.
In the electrodacus theory, this pressure differential is stored and accumulated while the vehicle is below wind speed (sort of like inflating a balloon behind the vehicle). As the vehicle reaches wind speed, the balloon begins to deflate, blowing the vehicle above windspeed for a while. Once the balloon is deflated the vehicle slows down until it is again under windspeed.
I don't buy it. There is no stored pressure differential.
Wow. Just, wow. That is just so wrong. There is no energy storage as you maintain. There can't be.
Is air a compressible fluid ? If so using a propeller/fan will create a pressure differential.
...
Having a pressure delta is similar to having a compressed spring so energy can be stored that way.
Quote...
Having a pressure delta is similar to having a compressed spring so energy can be stored that way.
A compressed spring contains energy within that spring - but only while the compression is held. That is, the energy is contained. There is no containment in an unbounded air mass, therefore energy cannot be stored. A balloon can store energy, as the air under compression is bounded by the balloon, but once ruptured, that energy is dissipated within milliseconds.
Not only the formula is correct it also shows that without energy storage no direct down wind vehicle can exceed wind speed.
Well, obviously you can compress a spring by pressing on only one end of it, while leaving the other end free... :-//
Actually, if you look at the car/wheel/belt system carefully, and make the following assumptions:
1. The system is in steady state (no acceleration)
2. The system is ideal (no friction losses from bearings or air resistance)
Then from these assumptions the power everywhere is zero, as power is only required to overcome losses or produce acceleration.
Since the power is everywhere zero, it is not possible to analyze this system in terms of power flows, and no conclusions can be drawn using that approach.
To analyze the system correctly, a different approach is required.
I have never claimed blackbird is not going faster than the wind.
What I was saying is that it can only do so using energy storage and in case of blackbird that energy storage is the pressure differential created by the propeller.
A gear can not amplify power thus output power will always be lower than the input.
Yes, don't assume zero losses. Assume some losses from the air resistance and rolling resistance. It doesn't need to be complex. Just consider that you can't move through the air with zero loss.
Not only the formula is correct it also shows that without energy storage no direct down wind vehicle can exceed wind speed.
Since this "energy storage" mechanism is so critical to your theory, I would ask that we take a moment to examine it.
You seem keen to find ways to help us understand, so I would invite you to put together a diagram (or series of diagrams) showing how this energy is captured, stored and released. Include any necessary description and formulae that would be required for it to stand up to examination.
If vehicle was to drive at just 20km/h but with a headwind of 100km/h then it will need the same amount of power to maintain that speed.
If vehicle was to drive at just 20km/h but with a headwind of 100km/h then it will need the same amount of power to maintain that speed.
Nope. It will need 1/6 of the power.
That is exactly what it seems meaning there is no wind at all but since vehicle travel trough air at 120km/h the power that vehicle will need to overcome drag is the one I calculated using the same formula.Why do you keep going back to old things that we have proved to you are not true?
If vehicle was to drive at just 20km/h but with a headwind of 100km/h then it will need the same amount of power to maintain that speed.
Is air a compressible fluid ? If so using a propeller/fan will create a pressure differential.
I will no longer post the graph but you can check that here https://en.wikipedia.org/wiki/Axial_fan_design
Having a pressure delta is similar to having a compressed spring so energy can be stored that way.
I will make an attempt to explain pressure differential.This is pretty straightforward. You didn't need to explain this as we are all well aware of this.
Think about a stationary fan (stationary as in not moving but rotation blades) there will be a lower than ambient pressure on one side and higher than ambient pressure on the the side as shown in that diagram on wikipedia that I posted here quite a few times.
Now start to move this fan but at the same time the higher the fan moves the higher the rotational speed of the blades. While pressure differential will drop it will not be sudden as it is in part compensated by increased blade rotational speed.
Blackbird propeller swept area is a massive 20m^2 and all the energy Blackbird needs to get to that record 28mph speed is just around 6Wh easily stored with not much pressure on such a massive swept area.Are you saying that energy is stored across the entire swept area of the propeller?
The analogy to the inductor is good and it includes 2 not so convenient effects:
If any of you has good electrical knowledge the analogy I like to make is with an air inductor storing energy in the magnetic field it creates.
Many that do not understand fully an inductor may not consider inductor an energy storage device same way it will not think propeller can store energy in the pressure differential.
Yes, don't assume zero losses. Assume some losses from the air resistance and rolling resistance. It doesn't need to be complex. Just consider that you can't move through the air with zero loss.
I suggest it is better to not complicate the key discussion with real world minutiae. It is far more practical to work out the basic principles using ideal elements. That is, to work out if the desired result is possible.
Once that is established, then add in the real world losses to see what sort of experimental result you can expect.
If it can't be found in the ideal situation, then there's no point in even considering losses.
Actually, the analysis does not change if you introduce friction losses. All that happens is that any losses due to friction are overcome by the power of the motors driving the belts. You just have to assume that the motors are powerful enough to achieve the stated belt speeds, and that the wheels of the cart do not slip on the belts.
Not only the formula is correct it also shows that without energy storage no direct down wind vehicle can exceed wind speed.
Since this "energy storage" mechanism is so critical to your theory, I would ask that we take a moment to examine it.
You seem keen to find ways to help us understand, so I would invite you to put together a diagram (or series of diagrams) showing how this energy is captured, stored and released. Include any necessary description and formulae that would be required for it to stand up to examination.
I will make an attempt to explain pressure differential.
Think about a stationary fan (stationary as in not moving but rotation blades) there will be a lower than ambient pressure on one side and higher than ambient pressure on the the side as shown in that diagram on wikipedia that I posted here quite a few times.
Now start to move this fan but at the same time the higher the fan moves the higher the rotational speed of the blades. While pressure differential will drop it will not be sudden as it is in part compensated by increased blade rotational speed.
Blackbird propeller swept area is a massive 20m^2 and all the energy Blackbird needs to get to that record 28mph speed is just around 6Wh easily stored with not much pressure on such a massive swept area.
If any of you has good electrical knowledge the analogy I like to make is with an air inductor storing energy in the magnetic field it creates.
Many that do not understand fully an inductor may not consider inductor an energy storage device same way it will not think propeller can store energy in the pressure differential.
All this is possible because air is a compressible fluid and it is much heavier than you actually imagine.
This is also the reason I insist that none of the wheels only vehicles can represent direct down wind faster than wind and all of them represent direct upwind faster than wind as there pressure differential energy storage is not used.
I saw some comments about my example with vehicle drag.
That is exactly what it seems meaning there is no wind at all but since vehicle travel trough air at 120km/h the power that vehicle will need to overcome drag is the one I calculated using the same formula.
If vehicle was to drive at just 20km/h but with a headwind of 100km/h then it will need the same amount of power to maintain that speed.
I was making a point of the fact that the equation I insist on is one of the most important equations in anything that has to deal with air from wind turbines to any type of wind powered vehicles and to any aerodynamic drag calculation.
And yes using power to define drag is absolutely the right type of unit. When you design a vehicle you need to know the drag power as that is the most significant part affecting vehicle consumption.
I see people comparing Tesla model 3 a very small super aerodynamic vehicle with a large SUV in therms of consumption at highway driving speeds and they think that the SUV was just badly designed with a very inefficient drive-train where the reality is that the larger air drag is by far the largest factor in an EV consumption.
That is exactly what it seems meaning there is no wind at all but since vehicle travel trough air at 120km/h the power that vehicle will need to overcome drag is the one I calculated using the same formula.Why do you keep going back to old things that we have proved to you are not true?
If vehicle was to drive at just 20km/h but with a headwind of 100km/h then it will need the same amount of power to maintain that speed.
A normal person cannot ride a bicycle at 60 km/h. However, a normal person can comfortably ride a bicycle at 10 km/h against a headwind of 50 km/h. This is because the correct formula for power is (drag force) x (vehicle speed). If the vehicle speed is lower, the power required is less. This is common sense. It takes more power to go faster.
You can see this for yourself at the bicycle calculator we previously showed you: https://www.omnicalculator.com/sports/cycling-wattage (https://www.omnicalculator.com/sports/cycling-wattage)
I believe you said you could not analyze the power flow because it was non-existent in the ideal case. So I'm saying don't assume a perfectly ideal case and you can analyze the power flow.
I think he is talking about only the wind effects, so he is right.
A normal person cannot ride a bicycle at 60 km/h. However, a normal person can comfortably ride a bicycle at 10 km/h against a headwind of 50 km/h. This is because the correct formula for power is (drag force) x (vehicle speed). If the vehicle speed is lower, the power required is less. This is common sense. It takes more power to go faster.
You can see this for yourself at the bicycle calculator we previously showed you: https://www.omnicalculator.com/sports/cycling-wattage (https://www.omnicalculator.com/sports/cycling-wattage)
If you keep going back and re-stating wrong things that have been corrected earlier, then this thread is going round in circles and cannot make progress. That is why we think you are trolling.
Insisting on something doesn't make you right, no matter how often you repeat it. In order to demonstrate the correctness of what you are saying, you have to be able to prove it with appropriate equations and logic, which is something you consistently fail to do. Moreover, you keep ignoring inconvenient facts when they go against your preconceptions.
No. He said the power required to overcome wind resistance is the same when cycling at 60 km/h as it is when cycling at 10 km/h against a 50 km/h headwind. That is wrong. It takes much less power to go at 10 km/h.
(This has nothing to do with rolling resistance. This is only about wind effects.)
Why, when there is no rolling resistance isn't the apparent wind the determining factor? Apparent wind speed is (the vector sum of) wind speed + ground speed.
Why, when there is no rolling resistance isn't the apparent wind the determining factor? Apparent wind speed is (the vector sum of) wind speed + ground speed.
So, if you are moving at 60 km/h in still air, the speed is 60 km/h, and the drag force corresponds to a wind velocity of 60 km/h.
On the other hand, if you are moving at 10 km/h against a 50 km/h headwind, the speed is 10 km/h, and the drag force corresponds to an apparent wind velocity of 10 + 50 = 60 km/h (same).
The drag force is the same in both cases, but in the second case the speed is 1/6 of the first case, so the power required is also 1/6.
Full disclosure, I skipped quite a few pages of this discussion, for obvious reasons.
In any case, how I interpret the downwind faster than wind is just as a simple lever. The wind has arbitrarily large amount of energy (size of the propeller is not specified), it's just a matter of devising a mechanism to push against the huge energy source that is the wind.
A simple demonstration is a yo-yo (yes, the spiny toy on a string).
(Attachment Link)
Put the yo-yo on the table with the string coming from the underside of the bobbin part and pull. The yo-yo will catch up with you.
If you don't have a yo-yo at hand, substitute with a spool of solder wire.
Full disclosure, I skipped quite a few pages of this discussion, for obvious reasons.
In any case, how I interpret the downwind faster than wind is just as a simple lever. The wind has arbitrarily large amount of energy (size of the propeller is not specified), it's just a matter of devising a mechanism to push against the huge energy source that is the wind.
A simple demonstration is a yo-yo (yes, the spiny toy on a string).
(Attachment Link)
Put the yo-yo on the table with the string coming from the underside of the bobbin part and pull. The yo-yo will catch up with you.
If you don't have a yo-yo at hand, substitute with a spool of solder wire.
Can you not see that direction of travel is the same as pulling direction ? This has nothing to do with either direct down wind when above wind speed or direct upwind version of Blackbird.
Also all those yo-yo games are based on energy storage.
The yo-yo toy is based on stored energy, but here it is only about the same toy, used in a different way.
The yo-yo is moving in the same direction as the string is pulled, but faster than the string is pulled.
This is just like blackbird going the direchtion of the wind, and faster than the wind.
The thing you seems to ignore is that vehicle being faster than wind in same direction can not be pushed by the wind.
There is no such problem in the yo-yo example.
The thing you seems to ignore is that vehicle being faster than wind in same direction can not be pushed by the wind.
There is no such problem in the yo-yo example.
The yo-yo going faster than the string in the same direction as the string cannot be pulled by the string.
Yesterday I tried that yo-yo experiment, using a spool of rope (I have spools of rope I use on my boat). I wanted to check that the difference in circumference between the spool and the level of the rope gave the same result as the "geared wheels on two conveyor belt". It was as close as I could measure. I would have been astounded if the result were otherwise, but since this whole subject can be non-intuitive I figured it was a test worth doing. I could shoot a video, but that's already been done.
If the string is being pulled along at, say, 1 cm/s, then the yo-yo cannot go faster than 1 cm/s in the same direction, because then it would be going faster than the string that is pulling it, and it could no longer receive any energy from the string ;)
Don't you see how obvious this is?
QuoteBlackbird propeller swept area is a massive 20m^2 and all the energy Blackbird needs to get to that record 28mph speed is just around 6Wh easily stored with not much pressure on such a massive swept area.Are you saying that energy is stored across the entire swept area of the propeller?
No, no, no, you don't understand. If the string is being pulled along at, say, 1 cm/s, then the yo-yo cannot go faster than 1 cm/s in the same direction, because then it would be going faster than the string that is pulling it, and it could no longer receive any energy from the string ;)
Don't you see how obvious this is?
I draw attention to my earlier question:QuoteAre you saying that energy is stored across the entire swept area of the propeller?
The yo-yo toy is based on stored energy, but here it is only about the same toy, used in a different way.
The yo-yo is moving in the same direction as the string is pulled, but faster than the string is pulled.
This is just like blackbird going the direchtion of the wind, and faster than the wind.
The thing you seems to ignore is that vehicle being faster than wind in same direction can not be pushed by the wind.
Good thing the propeller is creating its own wind going the other way, therefore adding to the total speed.
The Earth's wind is pushing against the propeller wind, which are both pushing against the propeller.
Good thing the propeller is creating its own wind going the other way, therefore adding to the total speed.
The Earth's wind is pushing against the propeller wind, which are both pushing against the propeller.
Propeller is powered by what ?
This is a wind powered only vehicle. No wind power then no power. Unless you can accept that there is energy stored in pressure differential that pushes the vehicle even when above wind speed. Of course as any stored energy it will be finite thus vehicle will slow down below wind speed after that is used up.
Yes that bicycle calculator is incorrect (you are not the only one to get this wrong so that includes those people that did the calculator)
That calculator will say you can drive at 1km/h in 230km/h headwind with just 300W (easy for a cyclist)
This is maybe 80km/h no where near 230km/h and here is what happens
youtube.com/watch?v=IoX-JUPvrwo
By stored energy, ie the kinetic energy of the moving vehicle as it moves up to wind speed. It never experiences a relative zero wind speed as it approaches wind speed, as the propeller is always blowing backwards. It will reach some sort of equilibrium state but wind is not that steady.
The kinetic energy of the vehicle is being drained by the propeller contraption, luckily, the wind keeps blowing, much like this thread.
Here's the main question; you surely consider yourself deeply scientific: so, for a theory to be scientific it must be falsifiable. What evidence would you accept to show your theory is false?
https://en.wikipedia.org/wiki/Falsifiability
Yes that bicycle calculator is incorrect (you are not the only one to get this wrong so that includes those people that did the calculator)
That calculator will say you can drive at 1km/h in 230km/h headwind with just 300W (easy for a cyclist)
This is maybe 80km/h no where near 230km/h and here is what happens
youtube.com/watch?v=IoX-JUPvrwo
That is not "maybe 80km/h", it says in the article "over 100km/h". How would you possibly stay upright?
and how would you stay upright traveling at 1km/h?
But all the theoretical analysis like Derek's (vehicle speed - wind speed) will predict increased rate of acceleration and that is just never observed or measured in any real test.
So there is not even the need for any other test than the ones already performed to show that the equations Derek and others claim to express how this vehicle works are not matching reality.
The geared-wheel and the pulled-yo-yo models don't predict continuous acceleration. They show downwind travel at some fixed multiple of windspeed. Once they accelerate to that speed they will remain at that speed.
Again (I repeated this many times) [...]
There are none so blind as those who will not see.
I draw attention to my earlier question:QuoteAre you saying that energy is stored across the entire swept area of the propeller?
Yes the entire swept area of the propeller acts as a barrier keeping the low and high pressure volumes separate.
What if the propeller is spinning slower than the moving air it is cutting into?I draw attention to my earlier question:QuoteAre you saying that energy is stored across the entire swept area of the propeller?
Yes the entire swept area of the propeller acts as a barrier keeping the low and high pressure volumes separate.
How does that work? (I think this needs a diagram.)
I think the big problem for electrodacus is that at wind speed (and faster) there seems to be no relative wind to power anything. The only way to go faster than what's pushing is via some stored energy, ergo there must be some stored energy somewhere. Where? The only possible place, given that the prop is quite important, is the prop wash.
Which seems reasonable if you haven't seen this thing work, and all the explanations so far seem to run up against the problem that at wind speed there is no push (or even pull), so there is no obvious way this could be working.
So... how about an alternative explanation without the maths that no-one can agree on...
ISTM the most important thing to bear in mind is the prop is rotating and pushing air back. For a given rotational speed, the air will flow backwards at some derived speed - the faster the prop goes the faster the air flows back. You could use this to push the vehicle at a given speed by altering the speed of the prop, so we could say that the prop turning at xrpm is equivalent to airflow of ym/s. It works in reverse, of course - push the vehicle and the airflow will cause the prop to turn proportionally to the vehicle speed through the air.
So, back to wind speed and no push. But there is a push - the prop wash is effectively a wall (albeit quite flimsy!) moving at a negative speed relative to the vehicle and tied to the vehicle. If the vehicle is moving at wind speed, the prop wash is moving backwards at some speed relative to the vehicle and thus the wind. The wind can push against the prop wash to continue pushing the vehicle a little faster. In effect, the prop wash is an infinite length of the string we see with that yo-yo.
By what powers the prop? It is still the wind because it is able to be pushing the vehicle still. With the prop wash moving backwards and effectively part of the vehicle, the vehicle is not at maximum speed relative to the wind, so the wind can keep pushing it.
Here it is. I posted this several times but people try to ignore it or say it is incorrect.I draw attention to my earlier question:QuoteAre you saying that energy is stored across the entire swept area of the propeller?
Yes the entire swept area of the propeller acts as a barrier keeping the low and high pressure volumes separate.
How does that work? (I think this needs a diagram.)
Do you agree air is a compressible gas ?
QuoteDo you agree air is a compressible gas ?
Yes, but I don't think it's relevant. Without air there would be no puzzle since there would be no wind to move faster than :)
I think the crux of it is that you're missing the effect of the wall of air. That wall could be a moving sail, being pushed back by the vehicle, except that there is a finite distance to which it could be pushed. With air, there is no limit because there is always air in front of you to push back. There is no infinite mechanism to do the same for a sail.
But let's assume for the moment that it is a sail, and the vehicle is infinitely long. The sail starts at the front and moves backwards at 1ms, say. There is a wind pushing the vehicle at 2ms. But that's pushing against the sail, so the wheel speed across the ground is actually 3ms - the 2ms wind plus 1ms sail.
That's how it works. The vehicle speed across ground is the wind speed plus the prop wash speed. Simple as that.
Ah, but the prop is powered by the wheels. Which is fine because the real vehicle speed will never reach the theoretical maximum. As has already been discussed, as the max is approached the wind power drops off and the speed stops increasing. So, in our example above, the vehicle may be doing 2.5m/s and can't go faster because of the lack of power as the vehicle-prop velocity approaches wind speed. But, as also noted, as the vehicle speed drops the apparent wind power will increase. Given the available wind power (acting against the diameter of the prop wash) is greater than the frictional losses of pushing the wheels to turn the prop, the realised speed can be higher than the winds ground speed (which is what all this is about).
Do you agree air is a compressible gas?
Do you agree air is a compressible gas?
No, I don't agree. All of the wind turbine power equations you have posted treat air as an incompressible fluid. All the equations you have posted have a constant density for air, which you state as 1.2 kg/m3.
If air were to be considered a compressible gas then the density would vary as a function of pressure, and this would appear in the equations. Since none of your equations have a variable density in them, you are stating that air is modeled as an incompressible fluid.
So if you try to say air is a compressible gas you are contradicting yourself.
What you call that wall of air is the wind (air particles moving in the same direction in average)
That equation I posted shows the available wind power the stored energy is a separate equation that makes no sense to talk about until you understand that vehicle has no wind power available once wind speed is exceeded.
Wise words : now I understand why some of the equations given by electrodacus make no sense.
The disagreement is about whether it is possible to get wind energy when already faster than the wind or not. So by definition we do no agree on this and can not use this in our arguments, as it would be a circular argument.
What you call that wall of air is the wind (air particles moving in the same direction in average) and once your vehicle exceeds that the vehicle will face the opposite as not vehicle is no longer pushed by wind
OK, so if we ignore the prop and have an infinitely long vehicle with a sail moving from the front to back, don't you agree that the SAIL could be moved at wind speed and the vehicle would then move faster? Ignore for the moment the power needed to move the sail - we are just interested in if we are together at this point.
If you don't agree, please say exactly why not.
Not quite sure I understand your infinitely long vehicle
OK, so if we ignore the prop and have an infinitely long vehicle with a sail moving from the front to back, don't you agree that the SAIL could be moved at wind speed and the vehicle would then move faster? Ignore for the moment the power needed to move the sail - we are just interested in if we are together at this point.
If you don't agree, please say exactly why not.
Not quite sure I understand your infinitely long vehicle but you seems to say the sail will move directly down wind relative to vehicle and then that movement will be used to power the vehicle.
If that is correct then no the vehicle can not move faster than wind since at the point vehicle speed and wind speed are equal the sail will no longer move relative to vehicle body and no movement means no power.
But also when vehicle is above wind speed the sail will move in the opposite direction relative to vehicle thus will slow the vehicle down.
OK, just very long then. Let's say it is 30m long. The sail is mounted on the vehicle just like Domagoj T's affair, but a single vertical sail. The sail starts at the front and stays there.
Now:
1. the body of the vehicle is streamlined so the only power transferred to it by the wind is via the sail. With the sail fixed in place and the wind blowing at 2m/s, are we agreed that the vehicle will move at 2m/s (or close enough)?
If not, why not?
2. There is no wind. A motor turns the pulley moving the sail to the back at 1m/s. The power for that motor comes from a battery - we are not concerned with that at the moment, but it has nothing to do with the wind, the road, anything. Are we agreed that the vehicle moves forward at around 1m/s?
If not, why not?
3. There is a 2m/s wind, and the sail is moving backwards at 1m/2. Why cannot the vehicle be moving at ~3m/s when we know the sail will be pushed at 2m/s and the vehicle is pushed by the sail at 1m/s?
I will assume you now removed that battery from the vehicle. So what moves the sail backwards ?
No, the battery is still driving the motor. It is simply a combination of 1 and 2: wind blowing in the sail, sail being moved along the vehicle.
Do you agree the vehicle ground speed would be close to 3m/s?
Here it is. I posted this several times but people try to ignore it or say it is incorrect.Yes the entire swept area of the propeller acts as a barrier keeping the low and high pressure volumes separate.
How does that work? (I think this needs a diagram.)
No, the battery is still driving the motor. It is simply a combination of 1 and 2: wind blowing in the sail, sail being moved along the vehicle.
Do you agree the vehicle ground speed would be close to 3m/s?
Yes but this just proves my point that an energy storage device is needed to exceed wind speed.
The diagram might be correct for what it's describing - but it is does not match what is happening with the Blackbird.
The first thing I see is that there is no confinement surrounding the Blackbird propeller - and that is clearly an important part of the diagram you presented. Does this not compromise what you are saying?
(https://www.eevblog.com/forum/chat/mess-with-your-minds-a-wind-powered-craft-going-faster-than-a-tail-wind-speed/?action=dlattach;attach=1358987;image)
No, the battery is still driving the motor. It is simply a combination of 1 and 2: wind blowing in the sail, sail being moved along the vehicle.
Do you agree the vehicle ground speed would be close to 3m/s?
Yes but this just proves my point that an energy storage device is needed to exceed wind speed.
No, the battery is still driving the motor. It is simply a combination of 1 and 2: wind blowing in the sail, sail being moved along the vehicle.
Do you agree the vehicle ground speed would be close to 3m/s?
Yes but this just proves my point that an energy storage device is needed to exceed wind speed.
OK, so now we go back to scenario 2 and replace the moving sail with a static propeller. Recall that:
* There is no wind
* The propeller is powered by a battery
If the speed of the propeller wash is 1m/s, do you agree that the vehicle will move forward at around 1m/s?
* There is no windReplacing the moving sail with a propeller will not be the same thing if the propeller is not powered by something.
* The propeller is powered by a battery
If the speed of the propeller wash is 1m/s, do you agree that the vehicle will move forward at around 1m/s?
So propeller will still rotate but slower and slower same thing that actually happens after the vehicle used all the pressure differential and vehicle starts to slow down.You keep forgetting 1 thing . This type of propeller has an integral clutch . Which means it can spin Free
You keep forgetting 1 thing . This type of propeller has an integral clutch . Which means it can spin Free
with just wind force . only . so it can be used as a sail .(kind of ) When the clutch engages it can ether give
power to the wheels or from . Depending on the setting of the pitch angle of the Blades .
I believe we can analyze, or at least demonstrate this with a fixed-pitch, fixed-gearing propeller. That's how the treadmill model is built. I don't know if this model can start from a full-stop, but it does accelerate once past windspeed.
I am not sure how this can be done as both these propellers have totally different properties .Your know about boats . It would be like saying a life raft is the same as a boat or worse still a Ship . They both travel or go in water . (spot the difference)You keep forgetting 1 thing . This type of propeller has an integral clutch . Which means it can spin Free
with just wind force . only . so it can be used as a sail .(kind of ) When the clutch engages it can ether give
power to the wheels or from . Depending on the setting of the pitch angle of the Blades .
I believe we can analyze, or at least demonstrate this with a fixed-pitch, fixed-gearing propeller. That's how the treadmill model is built. I don't know if this model can start from a full-stop, but it does accelerate once past windspeed.
I believe we can analyze, or at least demonstrate this with a fixed-pitch, fixed-gearing propeller. That's how the treadmill model is built. I don't know if this model can start from a full-stop, but it does accelerate once past windspeed.
That model will be able to start from a full stop assuming high enough wind speed so that power based on vehicle equivalent back area can overcome vehicle friction.
The thing you (all) seems to be ignoring is that air speed relative to vehicle is what can power the vehicle and so when air speed / wind is equal with vehicle speed in exact same direction then there is no way for the air to push the vehicle.
And somehow you imagine that propeller can push air backwards but that will require an energy source and wind is no longer there to help when vehicle is at or above wind speed.
You just think that you can take energy from the wheel but the problem with that is that when above wind speed that wheel no longer takes energy from the wind but will need to take energy from the vehicle kinetic energy and so vehicle will just slow down if you ignore a energy storage device like the pressure differential witch continues to push the vehicle basically taking the role of the wind for a few seconds or minutes depending on the design.
Yes the propeller to push back against the air and thus the wind does require some power and an energy source. So this sounds like you agree that with a given power (from a battery or whatever source) the prop can drive the vehicle a little faster than the wind speed or at exactly windspeed (this is the case that is a bit easier to calculate).
Now comes the slightly tricky question: how much power is needed for the prop to create a given thrust (force) to overcome friction ?
The thing you (all) seems to be ignoring is that air speed relative to vehicle is what can power the vehicle and so when air speed / wind is equal with vehicle speed in exact same direction then there is no way for the air to push the vehicle.
[... fixed-pitch propellers ...]
I am not sure how this can be done as both these propellers have totally different properties .Your know about boats . It would be like saying a life raft is the same as a boat or worse still a Ship . They both travel or go in water . (spot the difference)
Do you understand 2+2? Then why don't you understand 2 - (-2)?
No, the battery is still driving the motor. It is simply a combination of 1 and 2: wind blowing in the sail, sail being moved along the vehicle.
Do you agree the vehicle ground speed would be close to 3m/s?
Yes but this just proves my point that an energy storage device is needed to exceed wind speed.
OK, so now we go back to scenario 2 and replace the moving sail with a static propeller. Recall that:
* There is no wind
* The propeller is powered by a battery
If the speed of the propeller wash is 1m/s, do you agree that the vehicle will move forward at around 1m/s?
Replacing the moving sail with a propeller will not be the same thing if the propeller is not powered by something.
And since you do not have a battery and you do not accept there is an energy storage device (pressure differential) then there is nothing that can power the propeller other than the kinetic energy of the vehicle but taking energy from that will result in reduced vehicle speed. So propeller will still rotate but slower and slower same thing that actually happens after the vehicle used all the pressure differential and vehicle starts to slow down.
OK, so now we go back to scenario 2 and replace the moving sail with a static propeller. Recall that:
* There is no wind
* The propeller is powered by a battery
Against the wind already would be possible with a craft like this, as long as the drag force of the blades is higher than the craft, it would be working like a windmill,
The thing you seem to be ignoring is that the propeller is part of the vehicle and is blowing its own wind backwards. The effective wind speed is both.
Do you understand 2+2? Then why don't you understand 2 - (-2)?
Mate, why did you create a strawman? Well, obviously, it was to knock him down but, really, it's so blatant. You even quoted me where I said:QuoteOK, so now we go back to scenario 2 and replace the moving sail with a static propeller. Recall that:
* There is no wind
* The propeller is powered by a battery
Can you grasp that? The propeller is using B.A.T.T.E.R.Y. power to turn, so the rest of your post knocking down your strawman is completely irrelevant.
So, once again and PLEASE don't jump ahead, make things up or change the subject:
With a propeller fixed in position, powered solely by battery and with no wind, if the prop wash is thrust back at 1m/s, do you NOW agree that the vehicle will move forward at around 1m/s?
Also, please bear in mind that this actual experiment is carried out across the world at all times of day an night, so if you still disagree I will be expecting an extraordinary explanation as to why.
Why include a battery in the equation
For the vehicle to maintain 3m/s with a 2m/s tail wind the battery will need to provide enough power to cover all frictional losses.
Trust me, it is a fucking HUGE battery. But if it pleases you we can make it nuclear-powered instead, just in case you're worried about the thing running out of charge.
So, are we at this point agreed that, ignoring how the propeller is powered, the vehicle with a 1m/s prop and a 2m/s tail wind will move forward at close to 3m/s?
Trust me, it is a fucking HUGE battery. But if it pleases you we can make it nuclear-powered instead, just in case you're worried about the thing running out of charge.
So, are we at this point agreed that, ignoring how the propeller is powered, the vehicle with a 1m/s prop and a 2m/s tail wind will move forward at close to 3m/s?
It is obvious I will agree. You can drive at almost any speed you want if you have a battery.
That's not the point. The question is *how* *fast* will the vehicle move under the conditions above. 3 m/s, right? We are trying to find agreement on forces and speeds.
What powers the propeller when vehicle is above wind speed ? It is a simple question for with you have no answer.
That's not the point. The question is *how* *fast* will the vehicle move under the conditions above. 3 m/s, right? We are trying to find agreement on forces and speeds.
We just agreed on an arbitrary number of 3m/s and I said yes it will be possible. What else will you want me to answer ?
Battery is an energy source and since there is no electrochemical battery on Blackbird I look forward to see what that will be replaced with.
Trust me, it is a fucking HUGE battery. But if it pleases you we can make it nuclear-powered instead, just in case you're worried about the thing running out of charge.
So, are we at this point agreed that, ignoring how the propeller is powered, the vehicle with a 1m/s prop and a 2m/s tail wind will move forward at close to 3m/s?
It is obvious I will agree. You can drive at almost any speed you want if you have a battery.
So what is next ?
The energy which is extracted from slowing down the wind.
On blackbird, the propeller is not a windmill, it's a literal propeller exactly like on an airplane that pushes the air backwards. Effective airspeed speed at the propeller is higher than windspeed, causing the airmass to slow down, which results in excess energy which is used to push the vehicle forwards.
You agreed that the yoyo toy can move faster than the string. Earlier I replaced a hand pulling the string with a sail system that is basically infinite. Sails always move slower than the wind, but the vehicle itself is faster. That model should be perfectly clear.
The next step is slightly more tricky: how much power is needed to drive the prop to create the 1 m/s air speed ?
The power would obvious depend on the size of the prop. In ideal world this would be comparable (maybe smaller by some factor somewhat smaller than 1, as neither the wind turbine nor prop are 100% efficient) to the power a wind turbine could produce.
Yoyo has always access to energy trough the string as long as the string is under tension and it can only do so as long as there is enough string so it has a limited range of motion.The design I proposed in Reply #1102 on page 45 of this thread has string in a loop, so it has infinite range.
For my own interest, I made a geometrical model to show how a cart with appropriate gearing can travel in the opposite direction to the moving belts it is sitting on.
When the belts are moving to the right the cart moves to the left, and vice versa.
The works because one belt is moving twice as fast as the other, and the two wheels are the cart are connected by a geared pulley, so that both wheels turn together. The difference in belt speeds is used to make the wheels of the cart turn in the opposite direction of the belts.
(https://i.imgur.com/rAWZKcX.gif)
Your model is missing a physics engine.
Nothing has any mass so there are no forces involved at all.
Your model describes the steady state for an ideal vehicle and an ideal vehicle since there is no friction will continue to keep whatever steady state is in without requiring any energy. If in real world anyone will be able to demonstrate something like this it will be called a perpetuum mobile.
Your model is missing a physics engine.
Nothing has any mass so there are no forces involved at all.
Your model describes the steady state for an ideal vehicle and an ideal vehicle since there is no friction will continue to keep whatever steady state is in without requiring any energy. If in real world anyone will be able to demonstrate something like this it will be called a perpetuum mobile.
Forces and mass play no part in the analysis, it is pure geometry. If you were to build this model using exactly the same geometry, it would behave in exactly the same way.
If you say the real world model needs power to work, that is true, but the power comes from the belts it sits on. You move the belts under the wheels and the cart follows as shown. It is just the same as a gear chain. If you turn one gear, all the others follow according to the gear ratios.
Air is a fluid trough with your vehicle moves and while driving directly down wind with vehicle speed lower than fluid speed the fluid can push your vehicle but as soon as fluid speed and vehicle speed are the same the fluid can no longer provide any power to the vehicle and if by some chance your vehicle is above fluid speed the fluid will oppose your vehicle motion thus not only there is no way for your vehicle to accelerate but your vehicle will be slowed down.This is true for a simple sail-driven vehicle, but is not necessarily true for a vehicle with a more complex mechanism like a propeller.
And mathematically tat will be seen as the direction of the fluid motion will change relative to vehicle.Relative to the vehicle, but not necessarily relative to the pitch angle of the propeller.
All that moves is the fluid relative to ground and vehicle also relative to ground and as soon as fluid and vehicle have the same speed there is no way for the vehicle to be powered by the fluid.This is an assertion without proof. Nor is it something you can prove, since once you use words like "there is no way", you are required to prove that no possibility exists out of all the infinite ways it might be done. You cannot feasibly analyze an infinity of designs and eliminate all of them.
It will behave the same way if you push the cart with your hand. If cart needs to be powered by the two treadmill then it will not behave anywhere close to this it will just move backwards.If you say the cart will behave differently than the diagram when powered by the treadmill, then you need to explain what part of the geometric model will behave differently in the physical model. What will be different in the real world?
There is a lot of confusion as you seen those toy cars that move against the paper moving direction but that will also not work that way without energy storage and stick slipIt's a geometric diagram. Where is the energy storage and stick slip in a drawing on the screen? It cannot have any of those things.
This is true for a simple sail-driven vehicle, but is not necessarily true for a vehicle with a more complex mechanism like a propeller.
Relative to the vehicle, but not necessarily relative to the pitch angle of the propeller.
This is an assertion without proof. Nor is it something you can prove, since once you use words like "there is no way", you are required to prove that no possibility exists out of all the infinite ways it might be done. You cannot feasibly analyze an infinity of designs and eliminate all of them.
The energy which is extracted from slowing down the wind.
Yes that is a perfectly fine explanation for when vehicle is below wind speed. But how you do that when you are above wind speed and apparent wind speed changes direction?
The energy which is extracted from slowing down the wind.
Yes that is a perfectly fine explanation for when vehicle is below wind speed. But how you do that when you are above wind speed and apparent wind speed changes direction?
In the blackbird type craft the propeller blows the car forward in the same direction as the wind. In that case the propeller blows against the wind slowing it. Even when the car is traveling down wind faster than the wind, the exhaust from the prop leaves the prop faster than the wind recedes and so is slower (relative to the ground) than the wind.
The gearing between the prop and the wheels provides the power to turn the prop. The wind is pushing against the prop wash and the combination of forces between the prop and the wheels provides the energy, i.e. the power to make this work. The fact that the wheels move with the full velocity of the vehicle while the wind speed is a part and the prop wash is the other part means the wheels get power from the force of the wind and the prop, but only have to power the prop. Net power input is from the wind.
The energy which is extracted from slowing down the wind.
Yes that is a perfectly fine explanation for when vehicle is below wind speed. But how you do that when you are above wind speed and apparent wind speed changes direction?
In the blackbird type craft the propeller blows the car forward in the same direction as the wind. In that case the propeller blows against the wind slowing it. Even when the car is traveling down wind faster than the wind, the exhaust from the prop leaves the prop faster than the wind recedes and so is slower (relative to the ground) than the wind.
The gearing between the prop and the wheels provides the power to turn the prop. The wind is pushing against the prop wash and the combination of forces between the prop and the wheels provides the energy, i.e. the power to make this work. The fact that the wheels move with the full velocity of the vehicle while the wind speed is a part and the prop wash is the other part means the wheels get power from the force of the wind and the prop, but only have to power the prop. Net power input is from the wind.
Just noticed your signature so I will assume you may be owning an EV.
Imagine your EV has an empty battery (completely empty) and you have a strong wind from the back of the vehicle and you are on a perfectly flat road.
Now the wind will push your vehicle and your vehicle max speed while pushed by the wind can not exceed wind speed. (I hope you will agree with that).
But now you can use your regenerative brake to start charging battery while your vehicle wind speed will be way lower than wind speed since now there is a lot of resistance as you take wind energy and store it in to the battery and at some point you decide you got enough maybe 1% SOC and even with that you can exceed wind speed for a few minutes.
That is exactly how blackbird works other than propulsion is delivered by a propeller fan (instead of the more efficient wheel) and the energy is not stored in a lithium battery but in the air as pressure differential so as you charge you also start using some of the energy.
And to make it even more clear say your Tesla has a flat rear end with a total surface area of 2m^2 (not aerodynamic at all so CoD of 1).
We will also make the assumption that vehicle drive train is perfectly efficient no friction or even rolling resistance.
Also let say wind speed is a constant 25m/s = 90km/h = 55.9mph
Then direct down wind Tesla will have this amount of wind power available 0.5 * 1.2kg/m^3 * 2m^2 * (25m/s-vehicle speed)^3
If vehicle speed is zero (friction brakes ON then there is of course no wind power available to charge the battery).
Then as soon as you remove the friction brakes wind power will be available.
Say vehicle barely moves at 2m/s then if you want to maintain this low speed all you need to do is charge the battery at this rate
0.5 * 1.2 * 2 * (25-2)^3 = 14.6kW (it may be just half of this in a real vehicle as there is a wind gradient due to interaction with road and vehicle has quite a bit of friction loss).
If you where to drive at 10m/s while you where charging then less wind power will be available 0.5* 1.2 * 2 * (25-10)^3 = 4.05kW
And at 20m/s vehicle speed it is only 0.5* 1.2 *2 * (25-20)^3 = 150W
You can see where this is going.
There is no wind power available to a direct down wind vehicle even if it is ideal case no friction loss and if there is no energy storage device then it is not possible to exceed wind speed.
because I can understand conservation of energy.
Except my car has no battery and neither does the blackbird. Also, the blackbird has both a wheel and a propeller while my car ONLY has a wheel.
Ok, now you are going into pointless details because this is not how the blackbird works.
Nothing, literally NOTHING you have talked about here has to do with the blackbird, so just stop the insanity!
It is clear that you will never understand anything anyone tells you that doesn't agree with your idea because you reject it out of hand without understanding it.
I keep coming to the conclusion that you can't be taught anything, but I get suckered in when I see a point that is so crystal clear to anyone else that it must be possible to show it to you. However, as others have pointed out, instead of understanding what they tell you, you simply duck the issue and start talking about something different.
So no more replying to your nonsense.
No, you really don't and that's where you are failing hard. You do not understand or correctly apply the basic tenets of junior high-school level physics, namely:
1. The principles of Archimedes
2. Newton's laws of motion
3. The principle of conservation of energy--the one you keep bringing up but clearly have no clue as to how to apply it.
equation for ideal 100% power available to any direct down wind powered vehicle is 0.5 * air density * area * (wind speed - vehicle speed)^3 (Fact 2)
equation for ideal 100% power available to any direct down wind powered vehicle is 0.5 * air density * area * (wind speed - vehicle speed)^3 (Fact 2)
Except:
a) You cannot derive this equation from first principles when asked to do so,
b) You have not pointed to any authoritative source that gives this equation (online or textbook).
Therefore you cannot take this as fact, and if you rely on it all your arguments are flawed.
OK ... Trying to get a feel as to where this energy storage is taking place.
Is this what you mean?
(https://www.eevblog.com/forum/chat/mess-with-your-minds-a-wind-powered-craft-going-faster-than-a-tail-wind-speed/?action=dlattach;attach=1361096;image)
If not, then where?
Thus this sort of wrong understanding is fairly common as it ended up in Wikipedia.
No, they got it exactly right here and we've all been telling you the exact same thing....
OK then why that second formula that should represent the same thing provide a different value ?
Because it doesn't represent the same thing. You are 'intuiting' (assuming wrongly) what the result should be and then assuming the formula is wrong when it doesn't match your preconceived notion. The result of the formula, which is correct, is that it takes more power to go 20km/h in still air than it does to go 10km/h against a 10km/h headwind.
This is not a question of general knowledge is a question of logic alone. So you did not need to learn this in school to understand the two cases are the same as far as drag is concerned.
This is not a question of general knowledge is a question of logic alone. So you did not need to learn this in school to understand the two cases are the same as far as drag is concerned.
I don't know what you should and shouldn't learn in school, but I do agree that the question is extremely simple. And you have simply gotten it wrong. The drag is the same, the power is not. Drag is a force and is the same in each case. Thus when I pedal the bicycle, I will have to apply a certain amount of force to the pedals to counter that drag. If I'm going 20km/h in still air however, I have to pedal twice as fast with the same force as I do pedaling an identical bicycle at 10km/h into a 10km/h wind. Same force, twice as fast. Twice the power. Extremely simple. No calculus. Everybody understands it except you.
Thus this sort of wrong understanding is fairly common as it ended up in Wikipedia.
This is not a question of general knowledge is a question of logic alone. So you did not need to learn this in school to understand the two cases are the same as far as drag is concerned.
Also if you double the speed drag power increases 8x not 2x or even 4x
The problem is not with the 2nd formular, but with the 1st.OK then why that second formula that should represent the same thing provide a different value ?
No, they got it exactly right here and we've all been telling you the exact same thing....
Except my car has no battery and neither does the blackbird. Also, the blackbird has both a wheel and a propeller while my car ONLY has a wheel.
It was my assumption that you own a Tesla due to referral program mentioned in your signature.
Why do you think a propeller that is less efficient than a wheel can help ? The propeller itself has no advantages over wheels other than it is to travel trough air and it can store energy as pressure differential.
Ok, now you are going into pointless details because this is not how the blackbird works.
Nothing, literally NOTHING you have talked about here has to do with the blackbird, so just stop the insanity!
It is clear that you will never understand anything anyone tells you that doesn't agree with your idea because you reject it out of hand without understanding it.
I keep coming to the conclusion that you can't be taught anything, but I get suckered in when I see a point that is so crystal clear to anyone else that it must be possible to show it to you. However, as others have pointed out, instead of understanding what they tell you, you simply duck the issue and start talking about something different.
So no more replying to your nonsense.
No, you really don't and that's where you are failing hard. You do not understand or correctly apply the basic tenets of junior high-school level physics, namely:
1. The principles of Archimedes
2. Newton's laws of motion
3. The principle of conservation of energy--the one you keep bringing up but clearly have no clue as to how to apply it.
OK answer this.
Max wind power available to a direct down wind powered vehicle is when vehicle just starts moving so low speed (Fact 1).
equation for ideal 100% power available to any direct down wind powered vehicle is 0.5 * air density * area * (wind speed - vehicle speed)^3 (Fact 2)
What powers the vehicle when vehicle speed is above wind speed ?
Clearly not wind power (at least not directly but stored wind power).
I'm starting to get very tiered by this level of stupidity.
I asked for the equation showing the available wind power for Blackbird and nobody presented one that will match the observed results in both blackbird and treadmill model.
equation for ideal 100% power available to any direct down wind powered vehicle is 0.5 * air density * area * (wind speed - vehicle speed)^3 (Fact 2)
Except:
a) You cannot derive this equation from first principles when asked to do so,
b) You have not pointed to any authoritative source that gives this equation (online or textbook).
Therefore you cannot take this as fact, and if you rely on it all your arguments are flawed.
It is a fact. Please provide an equation that you think is correct.
This equation is used everywhere from wind turbine design to vehicle drag so it is one of the most used equations and you can find it literally everywhere.
Your problem seems to be not understanding power and working with forces only not understanding what speed will correspond to get the correct result.
OK then why that second formula that should represent the same thing provide a different value ?
Because it doesn't represent the same thing. You are 'intuiting' (assuming wrongly) what the result should be and then assuming the formula is wrong when it doesn't match your preconceived notion. The result of the formula, which is correct, is that it takes more power to go 20km/h in still air than it does to go 10km/h against a 10km/h headwind.
This is not a question of general knowledge is a question of logic alone. So you did not need to learn this in school to understand the two cases are the same as far as drag is concerned.
I don't know what you should and shouldn't learn in school, but I do agree that the question is extremely simple. And you have simply gotten it wrong. The drag is the same, the power is not. Drag is a force and is the same in each case. Thus when I pedal the bicycle, I will have to apply a certain amount of force to the pedals to counter that drag. If I'm going 20km/h in still air however, I have to pedal twice as fast with the same force as I do pedaling an identical bicycle at 10km/h into a 10km/h wind. Same force, twice as fast. Twice the power. Extremely simple. No calculus. Everybody understands it except you.
I don't want to go through the entire history of this. Why would that be true? Are they taking into account the other losses like tire drag? The air drag would only be a matter of bike vs. air, no?
Thus this sort of wrong understanding is fairly common as it ended up in Wikipedia.
But every physicist in the world, every engineer in the world, every textbook in the world, agrees with that formula in Wikipedia. So are you going to tell the whole world they have been getting it wrong for the past 200 years and re-write all the physics textbooks?This is not a question of general knowledge is a question of logic alone. So you did not need to learn this in school to understand the two cases are the same as far as drag is concerned.
This is interesting. That formula that appears in Wikipedia is not derived from logic. It is derived from experiment. Many experiments. Over the years, scientists and experimenters measured the amount of power required in different situations, and they found that all of their experiments match the formula given by Wikipedia. What is more, you can repeat those experiments yourself, and if you do so, you will also obtain results matching that formula.
Also if you double the speed drag power increases 8x not 2x or even 4x
It is darg force, not power.The problem is not with the 2nd formular, but with the 1st.OK then why that second formula that should represent the same thing provide a different value ?
No, they got it exactly right here and we've all been telling you the exact same thing....
If to claculations don't agree one (at least one) has to be wrong (or using different approximations). If you get one of the formulas from text books and many sources chances are that that's the good one and the other is the bad one.
For the second formular there is a simle derivation: The drag force is proportional to relative wind speed squared and the mechanical power is force times relative speed (in this case bicycle relative to ground).
Why is anyone concerned about drag? I'm being serious. On the blackbird the speeds are low enough that the drag has to be pretty minimal. The issue is more one of showing the forces to create acceleration once you reach wind speed. No? The forces from the propeller are going to be much greater than wind drag at 15 mph.
But every physicist in the world, every engineer in the world, every textbook in the world, agrees with that formula in Wikipedia. So are you going to tell the whole world they have been getting it wrong for the past 200 years and re-write all the physics textbooks?That is your opinion and not a fact. Anyone that designs any sort of vehicle will know that second formula is not correct.
This is interesting. That formula that appears in Wikipedia is not derived from logic. It is derived from experiment. Many experiments. Over the years, scientists and experimenters measured the amount of power required in different situations, and they found that all of their experiments match the formula given by Wikipedia. What is more, you can repeat those experiments yourself, and if you do so, you will also obtain results matching that formula.
Anyone that designs any sort of vehicle will know that second formula is not correct.
therms of drag
Anyone that designs any sort of vehicle will know that second formula is not correct.
Can you cite any actual person that 'designs any kind of vehicle? Is not the designer of the Blackbird such a person? Do you think he agrees with you?Quotetherms of drag
|O :-DD
Why not 'pints of drag'? Or perhaps 'acres of drag'.
There is absolute need to have only the vehicle speed in there as one factor:
Consider not driving the bicycle directly, but use a rope or pully to pull it. No doubt that with some extra gearing from the pully the force is lower and thus less power needed to pull the rope with a constant speed. So half the speed needs half the power.
If you have the (V_w-v_v) factor instead the power needed would be essentially constant, and the pully would magically need to increase the power. So this is obvious nonsense and thus the (w-v)³ form is wrong for calculating the power needed to move the vehicle against or with the wind.
The (w-v)³ type formula is "correct" (an approximation ignoring the Betz limit) for a wind turbine on the vehicle to produce power there (e.g. to power the light). This is a different thing from the power needed to drive against a head wind or a sail uses to drive a vehicle with head wind. It is more like the 2 parts together make up the wind power. The sail uses force times speed and the wind turbine used the (w-v)³ part. It would still need a good explaination if the 2 parts together are actually the maximum avialable power.
So both formulas can be correct, but are made for different problems. For resistors you also know formulas for 2 parallel and 2 series resistors - different formulas for different problems.
There is absolute need to have only the vehicle speed in there as one factor:
Consider not driving the bicycle directly, but use a rope or pully to pull it. No doubt that with some extra gearing from the pully the force is lower and thus less power needed to pull the rope with a constant speed. So half the speed needs half the power.
If you have the (V_w-v_v) factor instead the power needed would be essentially constant, and the pully would magically need to increase the power. So this is obvious nonsense and thus the (w-v)³ form is wrong for calculating the power needed to move the vehicle against or with the wind.
The (w-v)³ type formula is "correct" (an approximation ignoring the Betz limit) for a wind turbine on the vehicle to produce power there (e.g. to power the light). This is a different thing from the power needed to drive against a head wind or a sail uses to drive a vehicle with head wind. It is more like the 2 parts together make up the wind power. The sail uses force times speed and the wind turbine used the (w-v)³ part. It would still need a good explaination if the 2 parts together are actually the maximum avialable power.
So both formulas can be correct, but are made for different problems. For resistors you also know formulas for 2 parallel and 2 series resistors - different formulas for different problems.
Both a wind turbine and a sail will have available the same wind power but the wind turbine is limited by Betz limit around 59% while a sail has no such limitation as wind speed behind a sail can be zero thus a sail is much more efficient in theory 100%
So while stationary a vehicle with a wind turbine can at most have 59% of that ideal Wind power equation while a sail can have 100% in ideal case.
Here again the wrong formula https://en.wikipedia.org/wiki/Drag_(physics) (https://en.wikipedia.org/wiki/Drag_(physics))
(https://wikimedia.org/api/rest_v1/media/math/render/svg/5e127a8e4fb6ade15a4d76e5e2f26514d8b6c2ca)
They add vehicle speed and wind speed for force part of the equation but not for the power part ?
How will that make any sense unless people just understand force but have no clue what power is.
That last therm also need to be vo+vw then result will be correct.
This can be tested relatively easy why is this not done at universities ? If they did this test they will realize their formula is just wrong and use the correct one.
Check this https://www.omnicalculator.com/sports/cycling-wattage (https://www.omnicalculator.com/sports/cycling-wattage) use 50kg cyclist weight 20kg bike weight 1km/h (0.277m/s) bike speed and 230km/h (63.88m/s) headwind.
The Cd*A for tops is 0.408
So 0.5 * 0.408 * 1.225 * (63.88 + 0.277)2 * 0.277 = 284.9 less than they get 300W with no elevation but they add the other less significant parts as riling resistance
So yes you think 300W is enough to deal with 230km/h head wind ? If so you never experienced strong winds (I'm sure nobody here experienced 230km/h).
The most I have ever experienced is 110km/h gusts (not even continues wind speed) and this is 8x less damaging than 230km/h
Correct answer for that is 0.5 * 0.408 * 1.225 * (63.88 + 0.277)2*(63.88 + 0.277) = 66kW
Unless you are a super hero with super human strength you will not going to pedal against a 230km/h wind at any speed not even 1km/h
The direct downwind sail analysis is here:
https://www.eevblog.com/forum/chat/mess-with-your-minds-a-wind-powered-craft-going-faster-than-a-tail-wind-speed/msg3888830/#msg3888830 (https://www.eevblog.com/forum/chat/mess-with-your-minds-a-wind-powered-craft-going-faster-than-a-tail-wind-speed/msg3888830/#msg3888830)
Such a sail can be at most 15% efficient.
If a sail is traveling directly downwind at the speed of the wind, then (w-v) is zero, (w-v)^2 is zero, (w-v)^3 is zero. Any way you look at it, if the sail is moving at the same speed as the wind, then the power will be zero.
Consider the worst case where the speed of the vehicle has dropped infinitesimally toward zero. In that case vo has approximated zero and so the power required to move the car has approximated zero.
How much power does that formula say is required for a zero bike speed? The correct answer is zero. I think anyone who understand work and power understand there is no work if there is no motion, so no power also. Otherwise there are a lot of buildings doing a lot of work which we should be able to tap into. We could make windmills without any movement! That would be a big plus!!!
Consider the worst case where the speed of the vehicle has dropped infinitesimally toward zero. In that case vo has approximated zero and so the power required to move the car has approximated zero.
Consider the worst case where the speed of the vehicle has dropped infinitesimally toward zero. In that case vo has approximated zero and so the power required to move the car has approximated zero.
How much power does that formula say is required for a zero bike speed? The correct answer is zero. I think anyone who understand work and power understand there is no work if there is no motion, so no power also. Otherwise there are a lot of buildings doing a lot of work which we should be able to tap into. We could make windmills without any movement! That would be a big plus!!!
That is not the correct answer. You assume there is brake. Remove the brake and think again what will be the power needed to maintain around zero speed while you have a headwind.
You will need to remove the brakes in order to move so at very low speed you have no brakes to anchor you to the ground.
I didn't say anything about a brake. I said zero motion. Work is force over a distance. You can push as hard as you want against an object and if it does not move you have done no work on it. You may have sweated up a storm, but that's internal inefficiencies. Actually, I can lean into a wall without any real effort on my part. I'm just using my weight and gravity to provide the force.
I would construct some examples to show you there is no work being done, but it really gets tiresome that you just can't understand the basic concepts.
So you think that holding an object in a fixed position against a force requires performing work on the object? Do you understand what is meant by "performing work" in the physics sense? If the wind is blowing, and something is pushing against the object, but not moving it against the wind, is the something doing work?
My earlier example is perfectly correct and for a bicycle to move at 1km/h against a 230km/h head wind it will require a minimum of 66kW and gears can not help with anything as a cyclist may be able to provide 300W for minutes and maybe peak around 1kW it will never be able to provide 66kW and so never be able to move against a 230km/h head wind let a lone the possibility that 300W is enough to do that as the wrong formula and that calculator will imply.
I don't hink the high power makes sense: We all know mechanical power is force times speed.
With a constant power at a low speed this would need an awful lot of force - more force the slower you go. With this logic it would be impossible the slowly walk against an even weak head-wind, as it would need near inifite force at a very slow speed.
I didn't say anything about a brake. I said zero motion. Work is force over a distance. You can push as hard as you want against an object and if it does not move you have done no work on it. You may have sweated up a storm, but that's internal inefficiencies. Actually, I can lean into a wall without any real effort on my part. I'm just using my weight and gravity to provide the force.
I would construct some examples to show you there is no work being done, but it really gets tiresome that you just can't understand the basic concepts.
So you think that holding an object in a fixed position against a force requires performing work on the object? Do you understand what is meant by "performing work" in the physics sense? If the wind is blowing, and something is pushing against the object, but not moving it against the wind, is the something doing work?
I remove the brakes from a bicycle you can sit on that and then in a 230km/h head wind and see if you can have a low speed close to zero relative to ground. In fact you can keep the brakes and you will still have no change against 230km/h wind.
A wall is anchored to ground.
I see what bdunham7 meant...
Ok, how about a 20 km/h wind? I don't even need to work. I just have to keep pressure on one pedal which I can do easily without doing work on the bicycle. No motion, no work. Do you agree with that?
I debated with myself over the stationary vs. infinitesimal speed issue realizing that either way you would find some inane point to argue about. 230 km/h head wind is pretty much reductio ad absurdum.
No vehicle can move directly upwind powered only by wind without energy storage (Fact).
I see what bdunham7 meant...
Ok, how about a 20 km/h wind? I don't even need to work. I just have to keep pressure on one pedal which I can do easily without doing work on the bicycle. No motion, no work. Do you agree with that?
I debated with myself over the stationary vs. infinitesimal speed issue realizing that either way you would find some inane point to argue about. 230 km/h head wind is pretty much reductio ad absurdum.
Yes 230km/h is absurd that is why I selected that speed. I set 1km/h for the bike so it is low speed but possible while keeping your balance then increased the head wind in the calculator until power required was equal with 300W as 300W can be done by a fit person.
Obviously power needed is not 300W else you will see electric bikes with 300W motor and 230km/h top speed.
20km/h (5.55m/s) head wind is no challenge as power need to slyly move forward will be
0.5 * 1.225 * 0.408 * (5.55)^3 = 42W so basically a breeze.
And yes you can just stay still by leaving your body weight on the pedal at just 20km/h head wind.
What would you say if someone showed you a wind powered vehicle that moves directly upwind without energy storage?
I see what bdunham7 meant...
Ok, how about a 20 km/h wind? I don't even need to work. I just have to keep pressure on one pedal which I can do easily without doing work on the bicycle. No motion, no work. Do you agree with that?
I debated with myself over the stationary vs. infinitesimal speed issue realizing that either way you would find some inane point to argue about. 230 km/h head wind is pretty much reductio ad absurdum.
Yes 230km/h is absurd that is why I selected that speed. I set 1km/h for the bike so it is low speed but possible while keeping your balance then increased the head wind in the calculator until power required was equal with 300W as 300W can be done by a fit person.
Obviously power needed is not 300W else you will see electric bikes with 300W motor and 230km/h top speed.
20km/h (5.55m/s) head wind is no challenge as power need to slyly move forward will be
0.5 * 1.225 * 0.408 * (5.55)^3 = 42W so basically a breeze.
And yes you can just stay still by leaving your body weight on the pedal at just 20km/h head wind.
Ok, so if the user is stationary in a 20 km/h head wind, how much power is required to maintain this position? If you come up with any answer other than zero, you can't explain how a brake works. The brake can maintain this position into the 20 km/h wind while dissipating no power. A rider can do the same thing by simply standing on the pedal preventing it from rising up. One foot on the ground for balance, one foot on the pedal to maintain position. No power transfer. ZERO
You seem to have already agreed that this is correct. In that case you must agree there is no power transfer at the wheels for a stationary bike/car/blackbird and that the equation that predicts power at the wheels based solely on the wind relative speed must be wrong.
A very high wind speed will give very high forces - there is nothing ridiculous about that. The problem is with the wrong equation it gets ridiculous as it predicts too high a power at low speeds, like 20 km/h wind and 0.001 m/s movement relative to ground. 42 W of power don't looks so bad, but at the snails pace that would be still 42 kN if you calculate the power as force times speed.
I know the snails speed may not be so common, but things always start slow.
You claimed the power would be the power needed to drive against the wind - I (and most others here) were in doubt of that. That is your theory that gives the wrong result ! I just enteres a very small speed and used the very basic force = power / speed formula for a mechanical movement.
The 42W are not for that 0.001m/s but for the effect of drag due to a fluid traveling in the opposite direction at 20km/h.
The 42kN are valid just in theoretical world in reality you will not be able to drive at 0.001m/s as the wheels will slip due to this huge force so speed will become higher very fast so fast that your brain (or mine) will not be able to see the transition.
That is like saying no vehicle can ever leave from standing still as the force needed to move that vehicle will be to high.
Wheel will have a force at with it will start to slip and that will be the max force you will be able to provide at the wheel.
You can see the effect when you try to accelerate a vehicle to fast and wheel will spin.
What would you say if someone showed you a wind powered vehicle that moves directly upwind without energy storage?
That is not possible so I will show him where energy storage is. Maybe for this particular case I will ask him to take a video with a high speed camera to notice how the movement is not constant but it fluctuates as energy storage is charged and discharged.
You claimed the power would be the power needed to drive against the wind - I (and most others here) were in doubt of that. That is your theory that gives the wrong result ! I just enteres a very small speed and used the very basic force = power / speed formula for a mechanical movement.
Just imagine this.
You are stationary anchored to the ground. You can chose to "lift the anchor" and your vehicle will be accelerated direct down wind or down stream (boat on a river). This power to accelerate is provided to you by the wind or stream of water but if you decide to drive swim upstream then you need way more power as you are going against the stream.
That is the wrong answer. Maybe you would like to try again?
What would you say if someone showed you a wind powered vehicle that moves directly upwind without energy storage?
That is not possible so I will show him where energy storage is. Maybe for this particular case I will ask him to take a video with a high speed camera to notice how the movement is not constant but it fluctuates as energy storage is charged and discharged.
That is the wrong answer. Maybe you would like to try again?
I will ask you the same thing as it is a decently good analogy.
Have you ever had the chance to swim in a river ?
How is the energy storage relevant? If the vehicle continues against the wind indefinitely, it is moving against the wind. Who cares if there are tiny variations in the speed? This sort of storage of energy is not relevant to the issue. At no time does the velocity fall below the wind speed.
That is the wrong answer. Maybe you would like to try again?
I will ask you the same thing as it is a decently good analogy.
Have you ever had the chance to swim in a river ?
https://youtu.be/Qf03U04rqGQ?t=134
That is the wrong answer. Maybe you would like to try again?
I will ask you the same thing as it is a decently good analogy.
Have you ever had the chance to swim in a river ?
https://youtu.be/Qf03U04rqGQ?t=134
How is that video that you already posted relevant to my question about swimming in a river ?
Do you think air is not a fluid ?
Ok, so if the user is stationary in a 20 km/h head wind, how much power is required to maintain this position? If you come up with any answer other than zero, you can't explain how a brake works. The brake can maintain this position into the 20 km/h wind while dissipating no power. A rider can do the same thing by simply standing on the pedal preventing it from rising up. One foot on the ground for balance, one foot on the pedal to maintain position. No power transfer. ZERO
You seem to have already agreed that this is correct. In that case you must agree there is no power transfer at the wheels for a stationary bike/car/blackbird and that the equation that predicts power at the wheels based solely on the wind relative speed must be wrong.
If there are no sort of brakes it will require around 42W. A brake will anchor the vehicle to the ground thus no work is done on the vehicle.
A rider with sufficient weight standing on the pedal will still be a form of brake. It is a gravitational based one but still a brake.
There is potential wind power that can not be used because of the brake but there are 42W available at 20km/h wind speed and that 0.408m^2 equivalent area.
The equilibrium state will be the bike being pushed at 20km/h relative to ground so that there is no more force on the bike and no potential energy relative to air but there is now a potential energy relative to ground based on vehicle weight and speed relative to ground.
Without any energy storage the bike can be between zero speed relative to ground if anchored to ground and wind speed if there is no friction loss so not anchored to ground.
The vehicle can be anywhere between this two speed directly down wind at wind speed relative to ground and zero speed relative to ground and at this ends it will have potential energy storage relative to ground or to air.
This kinetic energy while it is a form of energy storage can not help the vehicle get outside this speed limits if it travels directly down wind at all times.
This kinetic energy can be used if vehicle travels at an angle to the wind direction that is how a sail vehicle can exceed wind speed.
In case of blackbird direct downwind version pressure differential is used to exceed wind speed for a limited amount of time.
In case of blackbird direct upwind version elastic and or gravitational energy storage in combination with stick slip hysteresis is what is used to drive at any speed (limited by frictional losses) for any amount of time.
....
My earlier example is perfectly correct and for a bicycle to move at 1km/h against a 230km/h head wind it will require a minimum of 66kW and gears can not help with anything as a cyclist may be able to provide 300W for minutes and maybe peak around 1kW it will never be able to provide 66kW and so never be able to move against a 230km/h head wind let a lone the possibility that 300W is enough to do that as the wrong formula and that calculator will imply.
Swiming in the rive is something for a red herring: it removes the 2nd moving plane and thus the whole point.You claimed the power would be the power needed to drive against the wind - I (and most others here) were in doubt of that. That is your theory that gives the wrong result ! I just enteres a very small speed and used the very basic force = power / speed formula for a mechanical movement.
It is not my claim as in I did not discovered anything new.
Just imagine this.
You are stationary anchored to the ground. You can chose to "lift the anchor" and your vehicle will be accelerated direct down wind or down stream (boat on a river). This power to accelerate is provided to you by the wind or stream of water but if you decide to drive swim upstream then you need way more power as you are going against the stream.
So while your vehicle in the wind is stationary or boat on a river is stationary relative to ground you are likely anchored.
Have you ever had the chance to swim in a river ? What amount of power do you need to swim upstream compared to down stream ?
Your reply is a red herring. Swimming in a river has nothing to do with my question.
Let me ask you again: what would you say if someone showed you a demonstration of a vehicle moving directly against the wind, powered by the wind, with no energy storage?
It's a very simple question. It doesn't involve rivers, or swimming, or any bullshit like that.
Swiming in the rive is something for a red herring: it removes the 2nd moving plane and thus the whole point.
Your reply is a red herring. Swimming in a river has nothing to do with my question.
Let me ask you again: what would you say if someone showed you a demonstration of a vehicle moving directly against the wind, powered by the wind, with no energy storage?
It's a very simple question. It doesn't involve rivers, or swimming, or any bullshit like that.
My question is super relevant but you likely do not understand what air is.
And as already answer no vehicle powered only by wind can go directly upwind without energy storage.
Your reply is a red herring. Swimming in a river has nothing to do with my question.
Let me ask you again: what would you say if someone showed you a demonstration of a vehicle moving directly against the wind, powered by the wind, with no energy storage?
It's a very simple question. It doesn't involve rivers, or swimming, or any bullshit like that.
My question is super relevant but you likely do not understand what air is.
And as already answer no vehicle powered only by wind can go directly upwind without energy storage.
That's still the wrong answer.
If you cannot answer a simple question like this one, which almost anyone else on the planet would know how to answer, then continuing the debate in this thread is rather pointless, wouldn't you agree?
Your reply is a red herring. Swimming in a river has nothing to do with my question.
Let me ask you again: what would you say if someone showed you a demonstration of a vehicle moving directly against the wind, powered by the wind, with no energy storage?
It's a very simple question. It doesn't involve rivers, or swimming, or any bullshit like that.
My question is super relevant but you likely do not understand what air is.
And as already answer no vehicle powered only by wind can go directly upwind without energy storage.
Can't believe this. Are you afraid that he'll actually produce the goods and show you? Why else wouldn't jump at this and say "Alright, show me then"?
Gosh, he hasn't even asked you to renounce your stated opinion on stuff, just asking what you'd say. Of course, after 1000 posts we know jolly well that if he did manage to demonstrate this you would still deny it in some way, probably by some distraction technique like changing the subject or introducing a strawman or similar :(
You think it is the wrong answer but that is not the case. And not sure why you say that I did not answered ?
Can't believe this. Are you afraid that he'll actually produce the goods and show you? Why else wouldn't jump at this and say "Alright, show me then"?
Gosh, he hasn't even asked you to renounce your stated opinion on stuff, just asking what you'd say. Of course, after 1000 posts we know jolly well that if he did manage to demonstrate this you would still deny it in some way, probably by some distraction technique like changing the subject or introducing a strawman or similar :(
Why will I be afraid ? There are many examples he can show and claim they do not use energy storage when they are but he can not understand that.
Is the same like saying the direct upwind version of blackbird is not using energy storage when I know it is using energy storage. What good will a link to the upwind version of blackbird do.
This guy reminds me of Andrea Rossi, the cold fusion guy who claims to have a device that will generate more heat than the amount of energy added electrically. He is never direct with anyone, only giving out the information he wants to give out and never letting anyone actually have access to the device.
ED never answers a question directly, only answering the parts he wants and trying to steer the discussion away from the points he can't successfully discuss.
I guess there really is no point in trying to reason with someone who doesn't properly understand that energy is a simple scalar quantity like distance and power is a rate wrt time like speed.
Or is it that you think swimming in a river against the current is different than a vehicle driving with a headwind ?
QuoteYou think it is the wrong answer but that is not the case. And not sure why you say that I did not answered ?
You stopped answering my questions just when it was leading to something interesting. You're afraid that you might be wrong and wiggle out of things if they look like showing that.
Oh, no, we don't think that. We know that swimming in a river against the current is different than a vehicle driving with a headwind.
The question is, do you?
This guy reminds me of Andrea Rossi, the cold fusion guy who claims to have a device that will generate more heat than the amount of energy added electrically. He is never direct with anyone, only giving out the information he wants to give out and never letting anyone actually have access to the device.
ED never answers a question directly, only answering the parts he wants and trying to steer the discussion away from the points he can't successfully discuss.
I guess there really is no point in trying to reason with someone who doesn't properly understand that energy is a simple scalar quantity like distance and power is a rate wrt time like speed.
This is the sort of claims you make (overunity) is just that you do not realize you are doing this. The use of energy storage brakes no laws and is how blackbird works.
Why none of you answered the swimming in a river question ? Is that because none of you had the chance to swim in a river ? Or is it that you think swimming in a river against the current is different than a vehicle driving with a headwind ?
Oh, no, we don't think that. We know that swimming in a river against the current is different than a vehicle driving with a headwind.
The question is, do you?
It is not different. Are you saying air is not a fluid same as water ?
I saw a better example but here is one. Notice the non constant vehicle movement ?
Yes, of course. He is never going to understand the physics because he is too invested in his position. There is literally nothing anyone can do to show him he doesn't understand the problem. Actually, that might not be true. It is very possible that he knows he is wrong, but wants to save face.
I know I've been in that position a few times and had to admit I was wrong. Heck, I think I started out saying (or at least thinking) the downwind vehicle could not exceed the wind speed until I realized the distinction between the wheels on the pavement only at the vehicle speed and the force on the propeller was from the power generated by the wheels (vehicle speed) AND the wind speed together.
Then I had to rethink again on the vehicle going into the wind. It was when I saw the blades nearly feathered that I realized again, it is a matter of gearing/prop pitch to get the right forces.
I had to change my argument in both cases when I saw the light. This guy wants to deny the light and live in the dark.
It is pretty impressive that he can generate these elaborate constructions to obfuscate the real physics. Yes, very creative.
Swimming in a river is like an airplane in the air, not a car into the wind.
I saw a better example but here is one. Notice the non constant vehicle movement ?
No. Every frame, the vehicle moves forward by the same amount.
I saw a better example but here is one. Notice the non constant vehicle movement ?
https://www.youtube.com/watch?v=J7XYzE2T-MA (https://www.youtube.com/watch?v=J7XYzE2T-MA)
and another one a bit better but still not the one I was searching for.
https://www.youtube.com/watch?v=ecwl-wuqSQU (https://www.youtube.com/watch?v=ecwl-wuqSQU)
The first video doesn't seem to show much in the way of speed variation. The second video shows the gear driving the track slipping from time to time and I'm not sure, but I think the fan is being turned on and off so it doesn't blow so hard. If it blows too hard the gear slips and the vehicle goes nowhere.
Either way, the storage has nothing to do with nothing. You just provided two examples of vehicles moving INTO the wind being powered by the wind. The first one even goes up a 30 or 40 degree ramp!
When are you going to respond to my prior post about the power to hold a vehicle stationary? You've already said the power was zero if the bike rider simply puts his weight on the pedal into a 20 km/s wind. So doesn't the rest follow, the equations are as I've shown them?
Quote from: electrodacus on Today at 07:33:23 pm
A rider with sufficient weight standing on the pedal will still be a form of brake. It is a gravitational based one but still a brake.
Let's just deal with one detail at a time. So do you acknowledge that the correct equation for the power at the wheels to move the vehicle into the wind must contain a factor which is just the velocity of the vehicle relative to the ground the wheel is pushing against?
The force from the wind is
Fd = 1/2 · Cd · A · p · (vw + vo)^2
The power required at the vehicle wheels to maintain a speed into the wind is
Pv = 1/2 · Cd · A · p · (vw + vo)^2 · vo
Correct?
I saw a better example but here is one. Notice the non constant vehicle movement ?
No. Every frame, the vehicle moves forward by the same amount.
Can you not see the vehicle stopping while the propeller still rotates ? Meaning power stored and then released.
If vehicle was to be directly powered by the propeller/wind turbine then there will be a smooth constant movement forward.
The stopping is irrelevant. It is moving INTO the wind. You were trying to argue that when moving downwind it was on average moving slower than the wind because of some speed variation that no one else thought was relevant. Moving into the wind the only requirement is that the speed must be greater than zero.
The first video clearly shows the vehicle moving with a positive velocity at all times. So clearly any energy storage is incidental. The second video clearly shows the gear slipping on the belt rather than energy storage. Even so, why does that matter? Both vehicles move INTO the wind. Energy storage is not relevant. They do what we are talking about - moving into the wind while powered by the wind.
If you want to use a river example, how about we use a sailboat on the water sailing into the wind? This happens all the time... literally!
The first video doesn't seem to show much in the way of speed variation. The second video shows the gear driving the track slipping from time to time and I'm not sure, but I think the fan is being turned on and off so it doesn't blow so hard. If it blows too hard the gear slips and the vehicle goes nowhere.
Either way, the storage has nothing to do with nothing. You just provided two examples of vehicles moving INTO the wind being powered by the wind. The first one even goes up a 30 or 40 degree ramp!
When are you going to respond to my prior post about the power to hold a vehicle stationary? You've already said the power was zero if the bike rider simply puts his weight on the pedal into a 20 km/s wind. So doesn't the rest follow, the equations are as I've shown them?
Quote from: electrodacus on Today at 07:33:23 pm
A rider with sufficient weight standing on the pedal will still be a form of brake. It is a gravitational based one but still a brake.
Let's just deal with one detail at a time. So do you acknowledge that the correct equation for the power at the wheels to move the vehicle into the wind must contain a factor which is just the velocity of the vehicle relative to the ground the wheel is pushing against?
The force from the wind is
Fd = 1/2 · Cd · A · p · (vw + vo)^2
The power required at the vehicle wheels to maintain a speed into the wind is
Pv = 1/2 · Cd · A · p · (vw + vo)^2 · vo
Correct?
I think is fairly clear to see the stop start movement in both videos while second video is better quality and easier to see. You just try to find some other non existent reasons but the reason it moves that way is due to energy storage charge and discharge.
Like I mentioned even presenting clear evidence you will try to find some other reasons to exclude the energy storage witch is essential.
Evacuation for force is correct for power it is not and correct one is below.
Pv = 1/2 · Cd · A · p · (vw + vo)^2 · (vo+vw)
The stopping is irrelevant. It is moving INTO the wind. You were trying to argue that when moving downwind it was on average moving slower than the wind because of some speed variation that no one else thought was relevant. Moving into the wind the only requirement is that the speed must be greater than zero.
The first video clearly shows the vehicle moving with a positive velocity at all times. So clearly any energy storage is incidental. The second video clearly shows the gear slipping on the belt rather than energy storage. Even so, why does that matter? Both vehicles move INTO the wind. Energy storage is not relevant. They do what we are talking about - moving into the wind while powered by the wind.
If you want to use a river example, how about we use a sailboat on the water sailing into the wind? This happens all the time... literally!
The stopping is relevant as it confirms my energy storage theory. The video shows a upwind version not sure why you include the downwind in discussion as that is competently different.
For upwind as shown in the video there is a small capacity energy storage that charges and discharges even a few times a second and that is what allows the vehicle to move against the wind direction for unlimited amount of time.
For the down wind version energy storage is the pressure differential much, much larger energy storage capacity and so the vehicle will move above wind speed for a few minutes before staring to slow down below wind speed.
Without energy storage neither upwind at any speed or down wind at higher than wind speed will be possible.
The stopping is relevant as it confirms my energy storage theory. The video shows a upwind version not sure why you include the downwind in discussion as that is competently different.
For upwind as shown in the video there is a small capacity energy storage that charges and discharges even a few times a second and that is what allows the vehicle to move against the wind direction for unlimited amount of time.
For the down wind version energy storage is the pressure differential much, much larger energy storage capacity and so the vehicle will move above wind speed for a few minutes before staring to slow down below wind speed.
Without energy storage neither upwind at any speed or down wind at higher than wind speed will be possible.
The energy storage explanation will not violate the conservation of energy.
Ok, tons of energy storage is happening. Nuclear bombs of energy is being stored and released. Doesn't matter all the energy is coming from the wind, so the wind is propelling the vehicle upwind, just like the wind propels the vehicle downwind faster than the wind in that case.
All your energy storage BS is just that, BS and adds nothing to the discussion.
[another mic drop]
Also to store that energy the vehicle will accelerate much slower when below wind speed compared to a sail vehicle that will not store anything other than kinetic energy.
Also to store that energy the vehicle will accelerate much slower when below wind speed compared to a sail vehicle that will not store anything other than kinetic energy.
I've asked you this before, but why does a sail not store energy when below windspeed, but a propeller does?
I hope you keep your microphone on the floor and I'm getting tired to explain the same thing multiple times to someone that will not understand power, energy and conservation of energy.
Also to store that energy the vehicle will accelerate much slower when below wind speed compared to a sail vehicle that will not store anything other than kinetic energy.
I've asked you this before, but why does a sail not store energy when below windspeed, but a propeller does?
The blades move through the air pushing air behind creating a higher pressure behind and a low pressure in front. This pressure gradient is what ED claims is storing energy. It does, but not very much. It dissipates very quickly, almost instantaneously, when the blades are stopped.
The blades move through the air pushing air behind creating a higher pressure behind and a low pressure in front. This pressure gradient is what ED claims is storing energy. It does, but not very much. It dissipates very quickly, almost instantaneously, when the blades are stopped.
But the blades of the Blackbird never stop. True?
They rotate as long as the vehicle moves as they are connected to wheels. But the vehicle will get to a peak speed / kinetic energy when all stored energy (pressure differential) is used up then it will start to slow down well below wind speed.
They rotate as long as the vehicle moves as they are connected to wheels. But the vehicle will get to a peak speed / kinetic energy when all stored energy (pressure differential) is used up then it will start to slow down well below wind speed.
This kind of response, overshooting and then coming back down again, can only be observed in a system with second order dynamics, and a second order system requires some kind of inertia or momentum term. A compressed gas is a first order system, so it cannot explain this kind of system behavior. Where is the second order inertial term found in Blackbird?
They rotate as long as the vehicle moves as they are connected to wheels. But the vehicle will get to a peak speed / kinetic energy when all stored energy (pressure differential) is used up then it will start to slow down well below wind speed.But won't the propeller still be pushing back and causing another pressure differential?
What are you even talking about.
Of course energy stored in pressure differential will exactly result in this sort behavior where as the pressure differential drops the rate of acceleration drops until there is no longer any acceleration and deceleration phase will start.
But won't the propeller still be pushing back and causing another pressure differential?
But where does the stored energy in the pressure differential come from? There is no external power source to charge it up. The vehicle starts from a standstill and gets blown along by the wind like a sail, until it speeds up and goes faster than the wind. Until this point, all the wind energy is being used to make Blackbird go faster. Where does the extra energy come from to charge up a pressure bubble?
But where does the stored energy in the pressure differential come from? There is no external power source to charge it up. The vehicle starts from a standstill and gets blown along by the wind like a sail, until it speeds up and goes faster than the wind. Until this point, all the wind energy is being used to make Blackbird go faster. Where does the extra energy come from to charge up a pressure bubble?
It comes from the wind while vehicle is well below wind speed. As soon as the vehicle starts to move pushed by the wind the propeller will start to rotate and not only push air back but increase the equivalent sail area. So in the initial phase the sail size increases from the vehicle back are plus the propeller blades area up to full propeller swept area.
Only small part of the wind energy is used to accelerate the Blackbird that ratio is given by the gear ratio between wheel and propeller and the propeller pitch.
So wind pushes against vehicle body and later propeller swept area and that power is split say 60% to propeller and 40% to acceleration so even more time spend charging as the acceleration is slowed then from those 60% say about 70% (assuming 70% efficient propeller) is stored but that stored energy will increase a bit the pressure differential so part of what you just stored a moment ago also starts to power the vehicle and at some point (below half wind speed according to my calculations) the pressure differential becomes more significant than the wind power.
So most of the charged energy happens well before vehicle gets to half the wind speed.
I still don't get why the pressure differential is more efficient in pushing the vehicle forward when the vehicle is above wind speed than opposed to below the wind speed. So what slows down the conversion of energy so much that is needs quite some time from reaching the speed of the wind to reaching the maximum speed ?
But where does the stored energy in the pressure differential come from? There is no external power source to charge it up. The vehicle starts from a standstill and gets blown along by the wind like a sail, until it speeds up and goes faster than the wind. Until this point, all the wind energy is being used to make Blackbird go faster. Where does the extra energy come from to charge up a pressure bubble?
It comes from the wind while vehicle is well below wind speed. As soon as the vehicle starts to move pushed by the wind the propeller will start to rotate and not only push air back but increase the equivalent sail area. So in the initial phase the sail size increases from the vehicle back are plus the propeller blades area up to full propeller swept area.
Only small part of the wind energy is used to accelerate the Blackbird that ratio is given by the gear ratio between wheel and propeller and the propeller pitch.
So wind pushes against vehicle body and later propeller swept area and that power is split say 60% to propeller and 40% to acceleration so even more time spend charging as the acceleration is slowed then from those 60% say about 70% (assuming 70% efficient propeller) is stored but that stored energy will increase a bit the pressure differential so part of what you just stored a moment ago also starts to power the vehicle and at some point (below half wind speed according to my calculations) the pressure differential becomes more significant than the wind power.
So most of the charged energy happens well before vehicle gets to half the wind speed.
Well it is very relevant as the direct downwind vehicle will not travel indefinitely at 2x or 3x the wind speed but only for a limited amount of time proportional with the amount of stored energy.
Also to store that energy the vehicle will accelerate much slower when below wind speed compared to a sail vehicle that will not store anything other than kinetic energy. So Blackbird direct down wind version spends more time below wind speed to charge the energy storage (pressure differential) then uses that stored energy to accelerate above wind speed and stay a bit above that speed before returning to below wind speed.
I still don't get why the pressure differential is more efficient in pushing the vehicle forward when the vehicle is above wind speed than opposed to below the wind speed. So what slows down the conversion of energy so much that is needs quite some time from reaching the speed of the wind to reaching the maximum speed ?
I'm still waiting for you to respond to my wind powered toy design from page 45.
It has no storage, operates in steady state continuously and has infinite range. At no point do extended sails travel faster than the wind, but the vehicle as a whole does.
I'm waiting.
The never ending diversion from the real issues. Just like Andrea Rossi. This guy should try the cold fusion business, he seems a natural.
Is hard to know what the real issue is but currently my best guess is the wrong equation for wind power.
The real issue is your habit of diverting when it looks like something is going places. As I pointed out earlier, you're afraid that you might be shown to be wrong, so you do or say whatever is necessary to ensure we never get there.
I might remind you of the outstanding questions I asked which were designed to be small steps along such a route, but you appear to have figured there might be something in it and dropped it pretty quickly.
All the wheels only toys where air is not involved are the equivalent of that direct upwind version of blackbird using small capacity energy storage device maybe charging in a few ms and then discharging triggered by stick slip hysteresis either internally in the mechanism or externally at the wheels.
The wrong equation for power either drag or generation is what made the math possible without energy storage. If the correct equation for power is used then is clear from the math that neither direct downwind faster than wind nor direct up wind at any speed is not possible without energy storage.
There may have been other before, but the first to bring up the mistake here was electrodacus.
People sees to like to use force for witch in most case they use the correct equation that includes (wind speed - vehicle speed)2 but then when it gets the time to calculate power they only multiply with vehicle speed instead of multiplying with (wind speed - vehicle speed).
Basically the wind speed is considered for force but ignored for power.
Not sure witch person did this mistake first but it seems it got everywhere.
The real issue is your habit of diverting when it looks like something is going places. As I pointed out earlier, you're afraid that you might be shown to be wrong, so you do or say whatever is necessary to ensure we never get there.
I might remind you of the outstanding questions I asked which were designed to be small steps along such a route, but you appear to have figured there might be something in it and dropped it pretty quickly.
If there was some question you asked and I did not answered was likely because it was useless for me to do so and probably answered before.
Most important equation for any wind powered vehicle is the one showing the available wind power to the vehicle and of course if that is wrong you will get to wrong conclusions.
The wheel based versions can be interpreted both ways. It is just a question on which plane is identified as the ground or "wind". It is somewimes a bit trikcky to look at the same thing with different referene frames, but this a major point of doing though experiments.
Quite some wheeled models may show some slip stick like action, but this does not say that this is essentially for them to work, there are some with little visible slip stick.
The equation for the power from a sail vehicle is not used in the calculatoins for the backbird at all, as there is no sail involved, only an active driven prop. So there is no need and no sense in using the equation for a sail dirven vehicle.
Trying to use the equation to show that the blackbird vehickle would not work also makes little sense, as at best this would only show that with passive sails it would not work downwind. This is accepted and one would even get the some conclusion with the wrong (w-v)³ type form and the correct (w-v)²*v type form. However this does not proof that a different type of vehicle could no work - it just does not apply.
To proof that the backbrid vehicle would not work the way would be to calculate the power available from the wheels and the power needed to drive the prop. If the prop needs more power than the wheels can provide, it does not work (at least not without energy storage or other methods not inlcuded in the model). So this already quite close to the calculation in the video. Just need to calculate (e.g. get an upper / lower limit) the power needed to drive to prop, so kind of the other way around from a wind turbine, maybe include the Betz limit or a similar factor for the prop.
There may have been other before, but the first to bring up the mistake here was electrodacus.
The form with (wind speed - vehicle speed)2 only applies to vehicle speed < wind speed or would need to include the sign of (wind speed - vehicle speed), but at least it gets the low velocity range right.
... may be the right equation, but for the wrong purpose
The equation 0.5 * air density * area * (wind speed - vehicle speed)3 may be the right equation, but for the wrong purpose.
In determining the power required to move the vehicle into (or away from) the wind, the above equation is wrong. It uses (wind speed - vehicle speed) as the speed of the applied force which would be talking about the power of the wind. No one is asking about the power from the wind. The question is the power applied to the VEHICLE to maintain a speed relative to the ground, "vehicle speed".
So the full equation for the power applied to the vehicle through the wheels is
0.5 * air density * area * (wind speed - vehicle speed)² * vehicle speed
It's very simple. You need to apply the right equation to the right purpose. This equation tells you how much power is needed to move the vehicle. I don't know for sure what the other equation is telling you. It probably includes the power lost in turbulence and all the hard to calculate effects when trying to calculate turbulent air flow. Many PhDs have been earned in that field.
Listen and learn son... listen and learn.
So the full equation for the power applied to the vehicle through the wheels is
0.5 * air density * area * (wind speed - vehicle speed)² * vehicle speed
This has long since stopped being a technical discussion and has become a psychological interview. I'm very interested in understanding ED's motivation.YES this exactly what he is doing . I mentioned that he was doing this about 10 pages back .
It could just be troll baiting, trying to keep this thread going as long as possible.
This has long since stopped being a technical discussion and has become a psychological interview. I'm very interested in understanding ED's motivation.
It could just be troll baiting, trying to keep this thread going as long as possible.
It could be that he really does think he is the only person to understand this issue. Clearly he is wrong and so either is not capable of understanding the real issues or is just playing with us.
Orrrr... he might know he is wrong, but figures as long as he is not forced to admit it, he isn't technically wrong.
Hmmm.... I wonder which it is.
So the full equation for the power applied to the vehicle through the wheels is
0.5 * air density * area * (wind speed - vehicle speed)² * vehicle speed
Are you sure about that? When wind speed = vehicle speed, that equation still reduces to zero.
I don't care about the wind power. This equation was being tossed about when there seemed to be a discrepancy of whether the power to move a vehicle falls to zero when the vehicle approaches a zero velocity. To determine that, the force is required which is
0.5 * air density * area * (wind speed - vehicle speed)²
This is the force provided by the wind which must then be applied by the wheels to move the vehicle at a speed in the face of the wind. To find the power that must be applied to the vehicle you multiply by the relative speed of the vehicle and the point applying the force, the ground, or "vehicle speed". The equation for the applied power then becomes
0.5 * air density * area * (wind speed - vehicle speed)² * (vehicle speed)
There's no point in shifting the conversation to rivers or moats or spacecraft. We are talking about a ground vehicle in the wind. Your equation is for power in the wind and the wind alone and has nothing to do with the power required to move the vehicle.
Until you understand your mistake here, you have no possibility of understanding the blackbird. You really need to listen to those who are trying to help you understand. This is not a battle, this is your friends who want to see you succeed in learning the correct way to understand this issue.
I have no intention for this to continue in fact my goal is to get this done as soon as possible.
I'm sure not the only person understanding this as many product designs depend on using the correct equation.
I work in the field where understanding power and energy is super important (I designed my own wind turbine) and my business is in the renewable energy storage thus I investigated all energy storage sources including but not limited to eletrochemical energy storage, kinetic energy storage and thermal energy storage.
There will always be people that do not understand some parts of physics and I will not get bothered by this particular problem if it was not such a wide spread misinformation involving science communicators and university professors.
Are you sure about that? When wind speed = vehicle speed, that equation still reduces to zero.
The equation for force is correct and includes wind speed - vehicle speed the equation for power will also need to include both.
Just imagine air as a solid hitting the vehicle. Can you see why there power available and not just force will include the speed of that solid ?
This is just wrong equation0.5 * air density * area * (wind speed - vehicle speed)² * (vehicle speed)the correct one is this for all applications
0.5 * air density * area * (wind speed - vehicle speed)² * (wind speed - vehicle speed)
So the full equation for the power applied to the vehicle through the wheels is
0.5 * air density * area * (wind speed - vehicle speed)² * vehicle speed
Are you sure about that? When wind speed = vehicle speed, that equation still reduces to zero.
Yes, that is correct. When the wind speed and the vehicle speed are the same, there is no wind force on the vehicle and so no force required to maintain the vehicle speed... other than friction which we are not factoring in here. This is just about the wind forces.
Again, not trying to calculate the power in the wind. We are calculating the power to move the object in the wind. That power is the force times the vehicle velocity. You can dance around all you want, but the fact remains you are applying the wrong equation to this.
Imagine the sail is stationary and the wind blows on it. The force on the sail is applied through linkages to the vehicle. Now the vehicle is not in the wind, only the sail. Clearly in this situation the force that must be applied to the vehicle to maintain a constant velocity is the force on the sail from the wind and the ONLY velocity that is relevant to the vehicle is the vehicle speed, so
0.5 * air density * area * (wind speed - vehicle speed)² * (vehicle speed)
I've explained it, I've derived it. You are not capable of understanding it or you are just refusing to acknowledge you understand it in public.
The equation 0.5 * air density * area * (wind speed - vehicle speed)3 may be the right equation, but for the wrong purpose.
OK. I will shut up and wait until friction is re-introduced. I want to see the situation when the vehicle speed = windspeed, and force has to be found to overcome friction. I know this works, but how the math applies to the actual real-world blades, gears, and wheels still isn't clear to me.
The equation 0.5 * air density * area * (wind speed - vehicle speed)3 may be the right equation, but for the wrong purpose.
There is an equation, 0.5 x (air density) x (area) x (wind speed)^3
This equation gives the flow of wind kinetic energy through the swept area of a turbine.
It breaks down into three parts:
1) kinetic energy of wind per unit mass of air
2) mass flow of air per unit area
3) swept area of turbine
Hence:
(flow of kinetic energy) = (energy per unit mass of air) x (mass flow per unit area) x (area)
If any one of those terms is zero, the available energy will be zero.
If you try to apply this to a sail, you find that the mass flow of air though the sail is zero (because the sail is a wind barrier). So you cannot use this equation in this form for sails.
When I brought up this point with ED and asked him to explain it, he quickly changed the subject and avoided giving an answer.
OK. I will shut up and wait until friction is re-introduced. I want to see the situation when the vehicle speed = windspeed, and force has to be found to overcome friction. I know this works, but how the math applies to the actual real-world blades, gears, and wheels still isn't clear to me.
So each time a particle hits that area of the sail it will lose half of the kinetic energy since the other half will be transferred to sail.
Actually a sail on ideal wheels friction less will be 100% efficient in using that available wind power.
For Blackbird, the spinning prop adds some extra thrust, allowing the (force applied to vehicle) to be positive even when exceeding the wind speed. Therefore, Blackbird can reach a steady speed faster than the wind where it is still true that (power supplied) = (power consumed).
So each time a particle hits that area of the sail it will lose half of the kinetic energy since the other half will be transferred to sail.
Say what? Is this another gem that 'university professors' don't understand?
Newton's first law of motion says that an object will not change its motion unless a force acts on it. If the sail boat is moving at the same speed as the wind there is no wind force: (w-v) = 0. If there is no friction, there is no friction force. Therefore no force at all. If there is no friction, the boat requires no power at all to continue moving at the same speed, so it neither needs nor gets any power from the wind. The wind power in this situation is zero, so the efficiency must be 0%.
As for direct down wind the ideal vehicle can get to same speed as wind speed but not above that without having some stored energy or an external energy source other than wind.
Blackbird is not using wind power when above wind speed
https://projects.m-qp-m.us/donkeypuss/wp-content/uploads/2009/06/Bauer-Faster-Than-The-Wind-The-Ancient-Interface.pdf
https://projects.m-qp-m.us/donkeypuss/wp-content/uploads/2009/06/Bauer-Faster-Than-The-Wind-The-Ancient-Interface.pdf
Good find. Thank you!
That will take some time to digest.
So the full equation for the power applied to the vehicle through the wheels is
0.5 * air density * area * (wind speed - vehicle speed)² * vehicle speed
Are you sure about that? When wind speed = vehicle speed, that equation still reduces to zero.
Yes, that is correct. When the wind speed and the vehicle speed are the same, there is no wind force on the vehicle and so no force required to maintain the vehicle speed... other than friction which we are not factoring in here. This is just about the wind forces.
OK. I will shut up and wait until friction is re-introduced. I want to see the situation when the vehicle speed = windspeed, and force has to be found to overcome friction. I know this works, but how the math applies to the actual real-world blades, gears, and wheels still isn't clear to me.
The equation 0.5 * air density * area * (wind speed - vehicle speed)3 may be the right equation, but for the wrong purpose.
There is an equation, 0.5 x (air density) x (area) x (wind speed)^3
This equation gives the flow of wind kinetic energy through the swept area of a turbine.
It breaks down into three parts:
1) kinetic energy of wind per unit mass of air
2) mass flow of air per unit area
3) swept area of turbine
Hence:
(flow of kinetic energy) = (energy per unit mass of air) x (mass flow per unit area) x (area)
If any one of those terms is zero, the available energy will be zero.
If you try to apply this to a sail, you find that the mass flow of air though the sail is zero (because the sail is a wind barrier). So you cannot use this equation in this form for sails.
When I brought up this point with ED and asked him to explain it, he quickly changed the subject and avoided giving an answer.
OK lets compare the output of the two equations
air density 1.2kg/m3
area 1m2
Wind speed 20m/s
Vehicle speed a) 0m/s, b) 10m/s, c) 20m/s
0.5 * air density * area * (wind speed - vehicle speed)2 * (wind speed - vehicle speed)
a) 4800W
b) 600W
c) 0W
0.5 * air density * area * (wind speed - vehicle speed)2 * (vehicle speed)
a) 0W (how will your wind powered vehicle ever start)
b) 600W
c) 0W
Any comments ?
Seems even with your wrong equation there is no wind power available to any wind powered vehicle when at and above wind speed.
https://projects.m-qp-m.us/donkeypuss/wp-content/uploads/2009/06/Bauer-Faster-Than-The-Wind-The-Ancient-Interface.pdf
Good find. Thank you!
That will take some time to digest.
Do not waste your time it is the same rubbish from another person that has no understanding of power and energy.
Is that clear?
OK lets compare the output of the two equations
air density 1.2kg/m3
area 1m2
Wind speed 20m/s
Vehicle speed a) 0m/s, b) 10m/s, c) 20m/s
0.5 * air density * area * (wind speed - vehicle speed)2 * (wind speed - vehicle speed)
a) 4800W
b) 600W
c) 0W
0.5 * air density * area * (wind speed - vehicle speed)2 * (vehicle speed)
a) 0W (how will your wind powered vehicle ever start)
b) 600W
c) 0W
Any comments ?
Seems even with your wrong equation there is no wind power available to any wind powered vehicle when at and above wind speed.
The first equation is the power available in the wind. Not relevant to the discussion of how much power it takes to move the vehicle.
The second equation is the power required to move the vehicle at that speed against the wind. So when the vehicle is not moving, it takes no power since there is no motion.
OK lets compare the output of the two equations
air density 1.2kg/m3
area 1m2
Wind speed 20m/s
Vehicle speed a) 0m/s, b) 10m/s, c) 20m/s
0.5 * air density * area * (wind speed - vehicle speed)2 * (wind speed - vehicle speed)
a) 4800W
b) 600W
c) 0W
0.5 * air density * area * (wind speed - vehicle speed)2 * (vehicle speed)
a) 0W (how will your wind powered vehicle ever start)
b) 600W
c) 0W
Any comments ?
Seems even with your wrong equation there is no wind power available to any wind powered vehicle when at and above wind speed.
The first equation is the power available in the wind. Not relevant to the discussion of how much power it takes to move the vehicle.
The second equation is the power required to move the vehicle at that speed against the wind. So when the vehicle is not moving, it takes no power since there is no motion.
The first equation is the wind power available to any wind powered vehicle. Second equation is not representing anything as it is just the wrong version of the first one.
The first equation (the correct one is valid for all cases).
When vehicle wants to travel direct upwind the vehicle speed will be negative so equation will look like this
0.5 * air density * area * (wind speed - (-vehicle speed))2 * (wind speed - (-vehicle speed)) = 0.5 * air density * area * (wind speed + vehicle speed)3
And yes for any vehicle speed direct down wind you need the wind power available plus some more meaning it is impossible for any wind powered vehicle to drive directly upwind without energy storage or external energy source.
Ok, so what is the equation for the power applied to a vehicle to maintain a constant speed into a constant headwind? Lets simplify the messy bits and assume the force generated by the wind on the vehicle is
Fw = Kv * (wind speed + vehicle speed)2
Where Kv is an accumulation of all the factors that relate the force to the relative wind speed.
So if Fw is the force required to maintain a speed of the vehicle, what is the equation for the power that is exerted on the vehicle to maintain the vehicle speed?
Now, forget wind. Lets say there is a drag on the vehicle defined by
Fdrag = Kdrag * (vehicle speed)2
What is the power required to maintain the speed of this vehicle?
I bet $50 he won't answer this and ducks the question. I suppose there's a chance he comes up with something amazingly convoluted. Probably starts talking about driving the vehicle on a river bottom.
I really can't believe this guy designs anything. He must be pulling our legs.
Ok, so what is the equation for the power applied to a vehicle to maintain a constant speed into a constant headwind? Lets simplify the messy bits and assume the force generated by the wind on the vehicle is
Fw = Kv * (wind speed + vehicle speed)2
Where Kv is an accumulation of all the factors that relate the force to the relative wind speed.
So if Fw is the force required to maintain a speed of the vehicle, what is the equation for the power that is exerted on the vehicle to maintain the vehicle speed?
Now, forget wind. Lets say there is a drag on the vehicle defined by
Fdrag = Kdrag * (vehicle speed)2
What is the power required to maintain the speed of this vehicle?
I bet $50 he won't answer this and ducks the question. I suppose there's a chance he comes up with something amazingly convoluted. Probably starts talking about driving the vehicle on a river bottom.
I really can't believe this guy designs anything. He must be pulling our legs.
It is quite simple. Power needed to counter drag will be
Fw * (wind speed + vehicle speed).
If there is no headwind then
Fdrag * (0 + vehicle speed)
So to give real numbers.
If vehicle speed is 20m/s it will require the same power to maintain speed as if vehicle is at 10m/s with a headwind of 10m/s (ignoring the difference in friction loss inside the vehicle and roiling resistance).
There is only one formula for all cases
0.5 * air density * area * (wind speed - vehicle speed)3
When you go against wind direction the vehicle speed is negative so (wind speed - (-vehicle speed)) thus basically (wind speed + vehicle speed)
You ignored the rest of my post just as I expected you would. if you looked at the calculations you would see that the force is only multiplied by the vehicle speed to get the power required to maintain speed. But you don't want to admit that, so you ignored it. That was my first prediction, that you would duck the question.
You are hopeless. You know you are wrong, but refuse to discuss it in any meaningful way. Silly rabbit. Trix are for kids.
Yeah, I do pity anyone depending on you for sound engineering. I don't believe you even have a technical job. How may quarters???
Unfortunately there are enough people that do not understand what conservation of energy really means and so with that there is a bad understanding of how the world works.
It seems the education system is failing in this regards. I say this seeing university level physics professors getting this wrong.
See this link and scroll down to air drag https://en.wikipedia.org/wiki/Bicycle_performance (https://en.wikipedia.org/wiki/Bicycle_performance)
You will find two equations
(https://wikimedia.org/api/rest_v1/media/math/render/svg/f636021ff7f54d76fa02a87786ea8d91b118597e)
This is the same that I use for max wind power available.
And this
(https://wikimedia.org/api/rest_v1/media/math/render/svg/534a6faf09414b48071ddd66ad96cb82852b3ccf)
That is incorrect (you can not always expect much from wikipedia).
Here again the wrong formula https://en.wikipedia.org/wiki/Drag_(physics) (https://en.wikipedia.org/wiki/Drag_(physics))
(https://wikimedia.org/api/rest_v1/media/math/render/svg/5e127a8e4fb6ade15a4d76e5e2f26514d8b6c2ca)
They add vehicle speed and wind speed for force part of the equation but not for the power part ?
How will that make any sense unless people just understand force but have no clue what power is.
That last therm also need to be vo+vw then result will be correct.
This can be tested relatively easy why is this not done at universities ? If they did this test they will realize their formula is just wrong and use the correct one.
There will always be people that do not understand some parts of physics and I will not get bothered by this particular problem if it was not such a wide spread misinformation involving science communicators and university professors.
You just believe the same absurd thing that those people that designed those online bicycle calculators.
Like 300W is plenty to bike at 1km/h against a 230km/h head wind. People that say that is a possibility are just clueless.
OK, we are making progress here. We have established that university professors are getting it wrong. We have established that Wikipedia is getting it wrong. We have established that the bicycle power calculators are getting it wrong.
What now remains is to figure out where the physics textbooks are getting it wrong, then we can have them recalled and pulped, and replaced with corrected versions. It is astonishing how the whole world has been getting this wrong for so long.
And again you manage to deflect and avoid answering.I'm still waiting for you to respond to my wind powered toy design from page 45.
It has no storage, operates in steady state continuously and has infinite range. At no point do extended sails travel faster than the wind, but the vehicle as a whole does.
I'm waiting.
There are two very different versions of blackbird. There is the direct downwind that works based on pressure differential energy storage and there is the direct upwind version that uses small capacity internal storage and stick slip hysteresis and that version since it always have access to wind power it can work continuously as long as there is wind.
All the wheels only toys where air is not involved are the equivalent of that direct upwind version of blackbird using small capacity energy storage device maybe charging in a few ms and then discharging triggered by stick slip hysteresis either internally in the mechanism or externally at the wheels.
The wrong equation for power either drag or generation is what made the math possible without energy storage. If the correct equation for power is used then is clear from the math that neither direct downwind faster than wind nor direct up wind at any speed is not possible without energy storage.
And again you manage to deflect and avoid answering.
My design is different than Blackbird, so let's stick to mine, please.
I'm not talking about upwind, only downwind.
There is no stick slip hysteresis in my design (substitute string with chain, and wheels and ground with rack and pinion, if you so choose).
There is no energy storage in my design, unless you consider sails themselves to be storage, in which case sure, whatever, doesn't matter, let's not mention it ever again since it's irrelevant.
The steady state of this design is that the vehicle as a whole will continuously travel above wind speed, directly downwind.
and no vehicle can exceed wind speed unless it uses an energy storage device or an external energy source.
...This only applies if you use the power to drive the vehicle with a prop.
It is quite simple. Power needed to counter drag will be
Fw * (wind speed + vehicle speed).
OK lets compare the output of the two equationsThe 2nd expression (it is not even an equation, as there is no "=" sign ) never claimed to calculate the available wind power, but the power available from a sail (which is not the most power efficient way) or the power needed to drive against the wind, like a sail in reverse.
air density 1.2kg/m3
area 1m2
Wind speed 20m/s
Vehicle speed a) 0m/s, b) 10m/s, c) 20m/s
0.5 * air density * area * (wind speed - vehicle speed)2 * (wind speed - vehicle speed)
a) 4800W
b) 600W
c) 0W
0.5 * air density * area * (wind speed - vehicle speed)2 * (vehicle speed)
a) 0W (how will your wind powered vehicle ever start)
b) 600W
c) 0W
Any comments ?
Seems even with your wrong equation there is no wind power available to any wind powered vehicle when at and above wind speed.
Is it possible to get an summary from start to finish about why it is impossible?
Examples and illustrations along to way would be helpful so that I can follow all the steps.
I'm really struggling to follow the thought process here. Guess I'm not smart enough. So please put it as plainly as possible.
That can be proved with the correct equation for wind power available to any wind powered vehicle
Pw = 0.5 * air density * area * (wind speed - vehicle speed)3
It is irrelevant how vehicle is build as long as wind power is the only source then the above is the correct formula for an ideal system (so absolutely best case scenario).
That can be proved with the correct equation for wind power available to any wind powered vehicle
Pw = 0.5 * air density * area * (wind speed - vehicle speed)3
It is irrelevant how vehicle is build as long as wind power is the only source then the above is the correct formula for an ideal system (so absolutely best case scenario).
So do the wheels in contact with the ground make no difference? Your equation might be appropriate for a hot-air balloon.
The calculation this way is absolutely not easy, as there is no accepted formula for the maximum power available to a vehicle. So the difficulty is in deriving (not just propose a solution and claim it must but ture as an act of god). As there are multiple possible ways to harness the wind power this is a really difficult task. To make it a proof it needs a really good explaination that most people would agree with.
The easiest part will be to show that directly down wind faster than wind is not possible without some sort of energy storage device or an external energy source.
Let me know if you agree with the above and then I can continue with how the Blackbird can actually exceed wind speed for a limited amount of time using energy storage in pressure differential created by the propeller with part of the wind power.
The calculation this way is absolutely not easy, as there is no accepted formula for the maximum power available to a vehicle. So the difficulty is in deriving (not just propose a solution and claim it must but ture as an act of god). As there are multiple possible ways to harness the wind power this is a really difficult task. To make it a proof it needs a really good explaination that most people would agree with.
Just presenting an expression that most people think is wrong is far from supporting the claim. It is more like showing poor understanding of logic and science in general.
The V_w * F_w form was never claimed to be the available wind power, but the power needed by the vehicle. Just like for another force to push against the power is force times speed (by definition and not by mistake). Thinking the power would stay constant, essentially independent of the speed is just a rediculous idea, that causes obvious contradictions. One such contraticion would be that the power when going in the same direction as the wind would be the same as the maximum avialable power and thus all wind genrator would be 100% efficient. I don't think that sounds plausible.
The formula presented is not an accepted formula for the available wind power for the moving vehicle - it is not even correct. If it would be show us a reliable source ! If you can't , just give up on repeating that formula.
The calculation this way is absolutely not easy, as there is no accepted formula for the maximum power available to a vehicle. So the difficulty is in deriving (not just propose a solution and claim it must but ture as an act of god). As there are multiple possible ways to harness the wind power this is a really difficult task. To make it a proof it needs a really good explaination that most people would agree with.
There is an accepted formula is the one that I posted and that is the ideal case (best case if you want) of available wind power to any vehicle you can imagine.
You can see in the formula what influences the available wind power and that is wind speed relative to vehicle and area that wind/air has to interact with.
That is all it can not be any simpler than that.Just presenting an expression that most people think is wrong is far from supporting the claim. It is more like showing poor understanding of logic and science in general.
The V_w * F_w form was never claimed to be the available wind power, but the power needed by the vehicle. Just like for another force to push against the power is force times speed (by definition and not by mistake). Thinking the power would stay constant, essentially independent of the speed is just a rediculous idea, that causes obvious contradictions. One such contraticion would be that the power when going in the same direction as the wind would be the same as the maximum avialable power and thus all wind genrator would be 100% efficient. I don't think that sounds plausible.
What are you even talking about ? When have I ever said that wind power available to vehicle is not dependent on speed (it is clearly show in the equation).
I even provided an example yesterday showing highest wind power available is when vehicle is just starting to move super low by the time it gets to half the wind speed and zero when vehicle speed equals wind speed.
The easiest part will be to show that directly down wind faster than wind is not possible without some sort of energy storage device or an external energy source.
That can be proved with the correct equation for wind power available to any wind powered vehicle
Pw = 0.5 * air density * area * (wind speed - vehicle speed)3
It is irrelevant how vehicle is build as long as wind power is the only source then the above is the correct formula for an ideal system (so absolutely best case scenario).
The formula presented is not an accepted formula for the available wind power for the moving vehicle - it is not even correct. If it would be show us a reliable source ! If you can't , just give up on repeating that formula.
The point of essentially independet of velocity was for the low velocity case (e.g. vehicle much slower than the wind). Sorry forgot about hat detail.
The slower one goes the more rediculous your claim gets.
It's a very circular argument:
"Why can no vehicle go downwind faster than the wind speed? Because this correct formula says so."
"Why is this formula correct? Because no vehicle can go downwind faster than the wind speed."
Each conclusion depends on the other.
I don't think there is an generally accepted ready made formula for the maximum available wind power for the moving vehicle. At least I don't know one.It's a very circular argument:
"Why can no vehicle go downwind faster than the wind speed? Because this correct formula says so."
"Why is this formula correct? Because no vehicle can go downwind faster than the wind speed."
Each conclusion depends on the other.
Provide an equation that you think is the correct one for a wind powered vehicle.
The one I provided is valid for any case and it perfectly predicts what happens in reality.
Even the incorrect formula used by many predicts no wind power available above wind speed so that is not an argument either.
I don't think there is an generally accepted ready made formula for the maximum available wind power for the moving vehicle. At least I don't know one.
It is not up to us to provide a formula for a rather tricky problem. Chances are you would not understand or accept it anyway. It is your turn to show a good source, ideally with a good explaination.
The formula with the (w-v)²*v is the power available to a simple sail for velocities lower than the speed of the wind. However this does not have much relevance to a vehicle driven by a prop. It shows that going faster than the wind does not work with a simple (e.g. spinnacker like) sail, when going straight down the wind.
This is not really surprising and know for a long time.
it is driven by wind power that pushes against the equivalent back area of the vehicle
Quoteit is driven by wind power that pushes against the equivalent back area of the vehicle
So close! It's pushing against the rearward-moving airflow from the propeller, not any part of the vehicle (which would be receding forwards).
But while vehicle is above wind speed there is no more wind power available
You're falling into your hole again. When the vehicle is at or above wind speed, the airflow from the propeller is still going backwards at slower than wind speed. Thus the wind is still pushing against that airflow and generating power.
The prop doesn't need much power to do that. Just enough to turn it and move the vehicle through almost static air - any drag is from the vehicle going faster than the wind, and we only need it to be tiny to score.
Is it possible to get an summary from start to finish about why it is impossible?
Examples and illustrations along to way would be helpful so that I can follow all the steps.
I'm really struggling to follow the thought process here. Guess I'm not smart enough. So please put it as plainly as possible.
I have made a video but is maybe a bit long and boring https://www.youtube.com/watch?v=4Hol57vTIkE&t=142s (https://www.youtube.com/watch?v=4Hol57vTIkE&t=142s)
In any case I will try to make a summary.
The easiest part will be to show that directly down wind faster than wind is not possible without some sort of energy storage device or an external energy source.
That can be proved with the correct equation for wind power available to any wind powered vehicle
Pw = 0.5 * air density * area * (wind speed - vehicle speed)3
It is irrelevant how vehicle is build as long as wind power is the only source then the above is the correct formula for an ideal system (so absolutely best case scenario).
The air density can be considered a constant so not very relevant and then there are just two other therms the area of the vehicle that in the particular case of blackbird will increase with speed up to a max of swept area of the propeller but while that increase in area helps make more power available to vehicle it will only be valid as long as vehicle speed is smaller than wind speed since as it can be seen in the equation if vehicle speed equals wind speed the wind power will be zero.
This formula is all that is needed in order to demonstrate that any wind powered only vehicle will need an energy storage device in order to exceed wind speed directly down wind.
It seems that many people do not work with power and prefer to work with force and speed separately and this is how a wrong equation ended everywhere.
The equation for force is correct
Fw = 0.5 * air density * area * (wind speed - vehicle speed)2
But then when they want to calculate wind power they just multiply by vehicle speed and leave the wind speed out of the equation. The most likely reason they do that is because someone made this mistake first and they just copy paste the same wrong equation everywhere without thinking to much or testing to see if it is true.
They just think that if vehicle speed is zero the wind power available to that vehicle is zero and they think that since they imagine zero speed as a vehicle with brakes engaged. That seems as a super silly mistake to make is like saying that a sail boat has zero wind power available when boat speed is zero because the boat is anchored to the ground.
Let me know if you agree with the above and then I can continue with how the Blackbird can actually exceed wind speed for a limited amount of time using energy storage in pressure differential created by the propeller with part of the wind power.
Is it possible to get an summary from start to finish about why it is impossible?
Examples and illustrations along to way would be helpful so that I can follow all the steps.
I'm really struggling to follow the thought process here. Guess I'm not smart enough. So please put it as plainly as possible.
In any case I will try to make a summary.
The easiest part will be to show that directly down wind faster than wind is not possible without some sort of energy storage device or an external energy source.
That can be proved with the correct equation for wind power available to any wind powered vehicle
Pw = 0.5 * air density * area * (wind speed - vehicle speed)3
It is irrelevant how vehicle is build as long as wind power is the only source then the above is the correct formula for an ideal system (so absolutely best case scenario).
The air density can be considered a constant so not very relevant and then there are just two other therms the area of the vehicle that in the particular case of blackbird will increase with speed up to a max of swept area of the propeller but while that increase in area helps make more power available to vehicle it will only be valid as long as vehicle speed is smaller than wind speed since as it can be seen in the equation if vehicle speed equals wind speed the wind power will be zero.
This formula is all that is needed in order to demonstrate that any wind powered only vehicle will need an energy storage device in order to exceed wind speed directly down wind.
It seems that many people do not work with power and prefer to work with force and speed separately and this is how a wrong equation ended everywhere.
The equation for force is correct
Fw = 0.5 * air density * area * (wind speed - vehicle speed)2
But then when they want to calculate wind power they just multiply by vehicle speed and leave the wind speed out of the equation. The most likely reason they do that is because someone made this mistake first and they just copy paste the same wrong equation everywhere without thinking to much or testing to see if it is true.
They just think that if vehicle speed is zero the wind power available to that vehicle is zero and they think that since they imagine zero speed as a vehicle with brakes engaged. That seems as a super silly mistake to make is like saying that a sail boat has zero wind power available when boat speed is zero because the boat is anchored to the ground.
Let me know if you agree with the above and then I can continue with how the Blackbird can actually exceed wind speed for a limited amount of time using energy storage in pressure differential created by the propeller with part of the wind power.
Ah ok. So the only thing relevant to the vehicle speed is the air pushing it (assuming everything else ideal). The part that is confusing me is the spinning propeller. I know from experience that those do push or pull air depending on orientation when spinning. And spinning this does since it is driven by the wheels (which are also spinning), So shouldn't this factor into the problem in some way? And sorry I don't have access to youtube here.
Hey guys, isn't it absurd to think that one could move into 500km/h winds with only 1W of power at a speed of 0.000001km/h? That is clearly WRONG. The equation is WRONG.
Here is a video of 500km/h winds destroying a car to prove my point.
:palm:
Its like you haven't heard of gear reduction or understand that taking equations to their limit makes the numbers seem "wrong" (as pointed out already (https://en.wikipedia.org/wiki/Reductio_ad_absurdum)).
A racing bike shown is not capable of riding at 1km/h at cadence, the gear ratio simply does not allow it.
I think 1km/h on a bike should be possible if you have good enough balance. It is about 30cm per second so 1ft per second.
But yes getting things to extreme is a good way to test if an equation provide correct results.
The correct equation will not provide this insane results for 1km/h bicycle speed.
If all you have is around 300W max power available then my equation correctly predicts that you can drive at 1km/h in head winds of 35km/h
35km/h + 1km/h is 36km/h round number as it is 10m/s
So 0.5 * 1.2 * 0.408 * 103 = 244.8W will say that the rest to 300W is friction and rolling resistance.
So if you have 300W available you can drive at 36km/h or a max headwind of 36km/h so (bicycle speed + wind speed) < 36km/h
Nothing as absurd as 230km/h head wind and can be tested.
It is 100% not possible on a racing bike with wind blowing in gusts.
The highest gear ratio (https://www.bikecalc.com/speed_at_cadence) (53/39 + 25 cassette + 60rpm) puts you at 12km/hr. Try to pedal at 6rpm (10s per rotation) with any power.
If you were crazy enough to build a high enough gear ratio, and have guides preventing the bike from falling over, it should work.
Keep in mind, as pointed out to you already, moving 0 km/h takes ZERO power. So its a matter of interpolating after that point. 0.001km/h might take 1W say. It sounds "wrong" but its not.
I provide you with a bike that has no brakes and put you in a head wind of 36km/h. If you input no power your bike will accelerate in theory up to wind speed 36km/h in practice it will be less than that due to friction.Not necessarily. I could just put my feet on the pedals and stop them from turning. No need for brakes.
So in order for you to keep some speed around 0km/h relative to ground (exactly 0 will not be practical same as balancing something on the edge of a knife)...Why do you introduce absurdly irrelevant things into the discussion? It is a tricycle. It does not fall over when stationary.
...but some arbitrary low speed 1km/h or 2km/h against the head wind will require 300W.This is against all experimental evidence. For this experiment we have an appropriately low gear ratio. One turn of the pedals moves the bike forwards 30 cm. It will be effortless to move the bike against the headwind. Nothing like 300 W required. Not even breaking a sweat. (Note: 30 cm/s = 1 km/h)
In the other direction direct down wind with friction brakes enabled to maintain 1 or 2km/h the friction brakes will need to be capable to dissipate around 300W as heat.Again, no experiment shows this. If it were true, the brakes would get hot and start smoking. But they don't.
"extra thrust" ? You get that vehicle is only powered by wind ? There is no extra thrust and all the thrust as in the case of a sail is provided by the wind.But the propeller is spinning. A spinning propeller pushes air. This is called thrust and it must exist in the local region around the propeller. Surely this thrust has an effect when applied to the surrounding air.
Not necessarily. I could just put my feet on the pedals and stop them from turning. No need for brakes.
Why do you introduce absurdly irrelevant things into the discussion? It is a tricycle. It does not fall over when stationary.I see you did not get my analogy. The analogy was referring to being able to keep the speed at zero like trying to swim in a river against the current and maintain zero speed relative to ground. You may average around zero but it will me more or less that that at any one time.
This is against all experimental evidence. For this experiment we have an appropriately low gear ratio. One turn of the pedals moves the bike forwards 30 cm. It will be effortless to move the bike against the headwind. Nothing like 300 W required. Not even breaking a sweat. (Note: 30 cm/s = 1 km/h)
Again, no experiment shows this. If it were true, the brakes would get hot and start smoking. But they don't.
"extra thrust" ? You get that vehicle is only powered by wind ? There is no extra thrust and all the thrust as in the case of a sail is provided by the wind.But the propeller is spinning. A spinning propeller pushes air. This is called thrust and it must exist in the local region around the propeller. Surely this thrust has an effect when applied to the surrounding air.
.........
The propeller is a wheel for traveling trough a medium instead of traveling on top surface of a medium like a regular wheel.
Propeller is powered by wind and or pressure differential depending on vehicle speed.
Since output power of the propeller is lower than input power from the wheel there could not be any net gain.
The role of the propeller is to compress air creating that pressure differential where energy is stored to allow vehicle to exceed wind speed for a limited amount of time.
...
I draw a picture to help here...
So there are 2 cases:
A) the fan is NOT spinning (but wheel is, clutch?)
B) fan is being spun from the wheel through some gearbox or similar
so in case A the wind is pushing the vehicle, but then we turn the gearbox on and now the fan is spinning... The wind is still pushing the same, and now the fan also adds some to the vehicle. You say the power needed to turn the fan from the wheel and generate thrust is higher than just having the fan locked and not spinning (like a sail)?
...
Do you agree that a propeller is less efficient than a wheel ? Propeller maybe a realistic 70% while a wheel can easily be 95% efficient.
If you agree then why not just take the power from wind pushing the vehicle from the back wheel and outing that in to driving the front wheel ?
Do you see the problem?
The difference is that road is a solid thus not compressible and you can not store any energy.
The only reason the propeller works is because you can store energy by increasing the pressure differential.
If you where to replace the air (compressible fluid) with water (incomprehensible fluid) the it will be the same useless stuff as getting the power from back wheels and putting in to front wheels.
So neither a gear box nor a propeller are magic devices they can only output less power than you put in so the thing that helps here is the air (compressible fluid).
So all you do when you take part of the wind power from the wheel and putting it in to propeller is storing energy.
We lose some power to spin the propeller, but we gain it (some?) back because it is now easier to move forward.
like we decrease drag, so we can go a little faster using same power.
I see it like we move energy around in the system, but the total is still pretty much the same (minus some loss here and here).
We make something less efficient (spinning the thing isn't free) but make something else more efficient (lower pressure infront of us, woho here we come).
so the total energy is the same for the 2 cases, but the speed isn't. How much difference? Now is where all the %'s come in.
But I still see it as "possible" to go faster without "cheating". Or where did my thoughts go wrong?
But I still see it as "possible" to go faster without "cheating". Or where did my thoughts go wrong?
The only source of energy is wind.
Wind can power the vehicle only as long as vehicle speed is lower than wind speed.
To exceed wind speed you need to store energy else is just not possible for the vehicle to exceed wind speed.
It is simple to test just wait until vehicle acceleration speed drops to zero then see how vehicle decelerates.
No test has done that so with the incomplete tests come wrong conclusion about what the test showed.
Repeating something that is wrong over and over and over does not make it right.
I'm extremely impressed with the knowledgeable people here that tries again and again to make you realize that you are soooo wrong. It takes an impressive amount of patience to do what they do.
Now, I could handle it if you were "just wrong", but the "god-like" way you try to convince people about that they are wrong and you are right "because you say so", is really something that turns my stomach around |O
Repeating something that is wrong over and over and over does not make it right.
I'm extremely impressed with the knowledgeable people here that tries again and again to make you realize that you are soooo wrong. It takes an impressive amount of patience to do what they do.
Now, I could handle it if you were "just wrong", but the "god-like" way you try to convince people about that they are wrong and you are right "because you say so", is really something that turns my stomach around |O
I'm sorry if you can not understand complex dynamic interactions. The only way it will convince you is a test.
Tho even with a test you may not understand what happens as I showed a very clear example for the direct upwind version and people still try to find some other alternative explanations of what happens when it seems super clear what it happens there.
"extra thrust" ? You get that vehicle is only powered by wind ? There is no extra thrust and all the thrust as in the case of a sail is provided by the wind.But the propeller is spinning. A spinning propeller pushes air. This is called thrust and it must exist in the local region around the propeller. Surely this thrust has an effect when applied to the surrounding air.
If propeller has no input power or if the input power is taken from the output power (always smaller than input) then vehicle will just slow down and not accelerate.
You skipped over the question.
I asked about the interaction between this thrust and the surrounding air.
Friends, I found this thread by accident. There appears to be no consensus. 54 pages is beyond my strength, but I saw the film. This is incredible! New horizons are before us!
Unfortunately, I still have questions. First of all - the car moves faster than the wind, that is, it moves in the oncoming air flow ?! Tell me, why is this impossible with complete calm? After some initial impulse, of course. No, no, a perpetual motion machine is impossible.
And one more thing, it seems to me that the ribbon (speed indicator) is in the aerodynamic shadow (or inside the vortex) and much lower than the propeller, where the wind is slower.
Finally, regarding the background for the invention. Under certain conditions, the movement of a sailboat at an angle to the wind can have a higher speed than the wind. But not in the direction the wind is blowing!
Finally, regarding the background for the invention. Under certain conditions, the movement of a sailboat at an angle to the wind can have a higher speed than the wind. But not in the direction the wind is blowing!
Friends, I found this thread by accident. There appears to be no consensus. 54 pages is beyond my strength, but I saw the film. This is incredible! New horizons are before us!
It is 100% not possible on a racing bike with wind blowing in gusts.
The highest gear ratio (https://www.bikecalc.com/speed_at_cadence) (53/39 + 25 cassette + 60rpm) puts you at 12km/hr. Try to pedal at 6rpm (10s per rotation) with any power.
If you were crazy enough to build a high enough gear ratio, and have guides preventing the bike from falling over, it should work.
Keep in mind, as pointed out to you already, moving 0 km/h takes ZERO power. So its a matter of interpolating after that point. 0.001km/h might take 1W say. It sounds "wrong" but its not.
Not quite sure you understand.
I provide you with a bike that has no brakes and put you in a head wind of 36km/h. If you input no power your bike will accelerate in theory up to wind speed 36km/h in practice it will be less than that due to friction.
So in order for you to keep some speed around 0km/h relative to ground (exactly 0 will not be practical same as balancing something on the edge of a knife) but some arbitrary low speed 1km/h or 2km/h against the head wind will require 300W.
In the other direction direct down wind with friction brakes enabled to maintain 1 or 2km/h the friction brakes will need to be capable to dissipate around 300W as heat.
A bike that is not moving because is anchored to the ground has nothing to do with this problem as you basically become one with the earth and so that power accelerates the earth rotation witch considering the earth mass is just ridiculously low and nobody will ever care about that and so earth is just considered stationary.
You can be offended if you want but you do not even deserve an answer as it will be a long and involved one from me and you will have no clue of what I just said.
My answer was generic and included all possible vehicle types powered only by wind directly down wind and no vehicle can exceed wind speed unless it uses an energy storage device or an external energy source.So let's focus not on generic, but on this specific example.
Yes, there was.... but let me pick up from your next statement:You skipped over the question.
I asked about the interaction between this thrust and the surrounding air.
There is nothing skipped.
Propeller will create a pressure differential thus energy is being stored.
Friends, I found this thread by accident. There appears to be no consensus. 54 pages is beyond my strength, but I saw the film. This is incredible! New horizons are before us!
Certainly there is consensus. A set of equations describing how to design such a vehicle has been put together using established principles of physics, and accepted as correct by qualified professors of engineering. Both full size and scale models have been constructed and shown to work.
This was first done as long ago as 1969. Reference here: https://projects.m-qp-m.us/donkeypuss/wp-content/uploads/2009/06/Bauer-Faster-Than-The-Wind-The-Ancient-Interface.pdf
When going at a very low speed, that is low RPM on the pedeal on will not be able to get 300 W. The driver has the same problem as the vehicle: it is hard to get the same power, as the forces have to go up.
The problem why going 1 km/h again a 40 km/h head wind with a normal bicyle is not possible is not because the 300 W are needed, but because it is hard to produce even 30 W at such a low speed, as this needs very high force.
The is just not enough force to produce 100 W at a very low speed. So if the magic (..)³ expression would be true one would never be able to start from a stand still aginst any headwind. One allways starts with infinitesimal low speed and thus infintesimal small (essentially zero) mechanical power.
There is no need to repeat the supposed formula for the maximum available wind power for the moving vehicle. Without a good source or maybe an acceptable explaination this is worthless like repeating "The earth is flat.".
The (w-v)³ type equation is not only withput a good source, but in addition also proven wrong (see above).
But more important that that the equation works and produces the correct results as seen in any real test.
Don't worry about my feelings, I'm a big boy, I can handle it. Will I'll have no clue what you would have said because you think I have reduced mental capability or because you just can't provide an argument?
You keep saying that at wind speed a sail has no power, yet when I give you an example of sail traveling below wind speed, you deflect and avoid.
So let's focus not on generic, but on this specific example.
And yes, this vehicle does have an external energy source - the wind.
It is clearly seen form the equation that I constantly post here.
Other people call it thrust - and you call it a pressure differential - but the question is on the energy storage.
Rather than running around with words - where we all seem to get nowhere - it would make life so much easier if you could provide the formula which tells us how much energy is stored.
Unfortunately, regardless of your claims your equation does not apply to the Blackbird-type vehicles, where there is a propeller connected to wheels rolling on the ground.
Other people call it thrust - and you call it a pressure differential
Have you not seen the (w-v)3 in enough places ? It is basically everywhere.
Don't worry about my feelings, I'm a big boy, I can handle it. Will I'll have no clue what you would have said because you think I have reduced mental capability or because you just can't provide an argument?
You keep saying that at wind speed a sail has no power, yet when I give you an example of sail traveling below wind speed, you deflect and avoid.
So let's focus not on generic, but on this specific example.
And yes, this vehicle does have an external energy source - the wind.
I do think people mental capabilities are very different.
There is no wind power available to a sail vehicle traveling at wind speed. There is of course plenty of power available to a sail traveling below wind speed. The slower it travels relative to wind speed the higher the available wind power.
It is clearly seen form the equation that I constantly post here.
Pw = 0.5 * air density * area * (wind speed - vehicle speed)3
I even provided examples. Maybe you got here more recently and did not see them but here is one example just for you
area 1m2
air density 1.2kg/m3
Wind speed 20m/s
Vehicle speed:
a)0m/s
b)5m/s
c)10m/s
d)15m/s
e)20m/s
Wind power available to vehicle
a)4800W
b)2025W
c)600W
d)75W
e)0W
...
The main point I want to make since this is ester to understand (at least is what I was thinking) is that there is no wind power available to a direct down wind traveling at or above wind speed.
I provided the correct equation and anyone that is not agreeing with that equation is welcome to provide the correct one.
Nobody can claim to understand a wind powered only vehicle without being able to provide an equation describing the amount of wind power available to vehicle.
The energy storage is a separate equation and it is not relevant since I claim energy storage is involved
You ignored the rest of my post just as I expected you would. if you looked at the calculations you would see that the force is only multiplied by the vehicle speed to get the power required to maintain speed. But you don't want to admit that, so you ignored it. That was my first prediction, that you would duck the question.
You are hopeless. You know you are wrong, but refuse to discuss it in any meaningful way. Silly rabbit. Trix are for kids.
Yeah, I do pity anyone depending on you for sound engineering. I don't believe you even have a technical job. How may quarters???
:) So I correct your mistake and you are saying that I ignored your post ?
You should do a test and see that your assumption is wrong.
You just believe the same absurd thing that those people that designed those online bicycle calculators.
Like 300W is plenty to bike at 1km/h against a 230km/h head wind. People that say that is a possibility are just clueless.
Now, forget wind. Lets say there is a drag on the vehicle defined by
Fdrag = Kdrag * (vehicle speed)2
What is the power required to maintain the speed of this vehicle?
You can be offended if you want but you do not even deserve an answer as it will be a long and involved one from me and you will have no clue of what I just said.
@electrodacus,
Is it possible to get an summary from start to finish about why it is impossible?
Examples and illustrations along to way would be helpful so that I can follow all the steps.
I'm really struggling to follow the thought process here. Guess I'm not smart enough. So please put it as plainly as possible.
@ electrodacus
Your Forum is also full of stupid answers .. Now you have plagued us with your stupidity of Not seeing what is true life Working .
If it works don't brake it . Your Maths is no better than a 1st grader . Just picking up numbers @ random will never solve this .
Nor will you ever learn . and worst still your not willing to learn .
As hard as it is for you to except you are making a terrible mistake .
Maybe it would have been a wise questions to ask what are our Degree's :popcorn:
"extra thrust" ? You get that vehicle is only powered by wind ? There is no extra thrust and all the thrust as in the case of a sail is provided by the wind.But the propeller is spinning. A spinning propeller pushes air. This is called thrust and it must exist in the local region around the propeller. Surely this thrust has an effect when applied to the surrounding air.
If propeller has no input power or if the input power is taken from the output power (always smaller than input) then vehicle will just slow down and not accelerate.
You skipped over the question.
I asked about the interaction between this thrust and the surrounding air.
Friends, I found this thread by accident. There appears to be no consensus. 54 pages is beyond my strength, but I saw the film. This is incredible! New horizons are before us!
Unfortunately, I still have questions. First of all - the car moves faster than the wind, that is, it moves in the oncoming air flow ?! Tell me, why is this impossible with complete calm? After some initial impulse, of course. No, no, a perpetual motion machine is impossible.
And one more thing, it seems to me that the ribbon (speed indicator) is in the aerodynamic shadow (or inside the vortex) and much lower than the propeller, where the wind is slower.
Finally, regarding the background for the invention. Under certain conditions, the movement of a sailboat at an angle to the wind can have a higher speed than the wind. But not in the direction the wind is blowing!
Friends, I found this thread by accident. There appears to be no consensus. 54 pages is beyond my strength, but I saw the film. This is incredible! New horizons are before us!
Unfortunately, I still have questions. First of all - the car moves faster than the wind, that is, it moves in the oncoming air flow ?! Tell me, why is this impossible with complete calm? After some initial impulse, of course. No, no, a perpetual motion machine is impossible.
And one more thing, it seems to me that the ribbon (speed indicator) is in the aerodynamic shadow (or inside the vortex) and much lower than the propeller, where the wind is slower.
Finally, regarding the background for the invention. Under certain conditions, the movement of a sailboat at an angle to the wind can have a higher speed than the wind. But not in the direction the wind is blowing!
What film have you seen ? The one made by Derek / Veritasium ?
It is not impossible as demonstrated to drive faster than wind direct down wind but the way that happens is by storing energy before exceeding wind speed and then using that stored energy to exceed wind speed for a limited amount of time.
The wind resistance is low compared to the force generated by the propeller driving the wheels because the angle of the blades is nearly parallel with the wind. The lift of the airfoil is what provides the force that rotates the propeller and therefore the wheels.
It is a very simple process once it is properly understood.
When going at a very low speed, that is low RPM on the pedeal on will not be able to get 300 W. The driver has the same problem as the vehicle: it is hard to get the same power, as the forces have to go up.
The problem why going 1 km/h again a 40 km/h head wind with a normal bicyle is not possible is not because the 300 W are needed, but because it is hard to produce even 30 W at such a low speed, as this needs very high force.
The is just not enough force to produce 100 W at a very low speed. So if the magic (..)³ expression would be true one would never be able to start from a stand still aginst any headwind. One allways starts with infinitesimal low speed and thus infintesimal small (essentially zero) mechanical power.
There is no need to repeat the supposed formula for the maximum available wind power for the moving vehicle. Without a good source or maybe an acceptable explaination this is worthless like repeating "The earth is flat.".
The (w-v)³ type equation is not only withput a good source, but in addition also proven wrong (see above).
It is clearly seen form the equation that I constantly post here.
Unfortunately, regardless of your claims your equation does not apply to the Blackbird-type vehicles, where there is a propeller connected to wheels rolling on the ground.
Your assumption that power is required to accelerate the vehicle is not correct. At a stand still the force can be anything, but with velocity zero the power is zero. f = a • m, p = f • v
If the force applied equals the force from the wind, the net acceleration is zero and the velocity remains at zero so no power.
No fancy math required, just a clear understanding of how to apply the science and math.
Other people call it thrust - and you call it a pressure differential - but the question is on the energy storage.
Rather than running around with words - where we all seem to get nowhere - it would make life so much easier if you could provide the formula which tells us how much energy is stored.
If this was a non compressible fluid like water then there will be no pressure differential and no energy storage just thrust.
Simplest analogy that I can think of right now and it is visual enough will be pushing a vehicle with another vehicle having just a solid bar connecting them (non compressible fluid like water) or having a spring between them (compressible fluid like air).
The spring is an energy storage device and it is moving together with the pushed vehicle. As any analogy it has limitations but a compressible fluid can be used to store energy and a propeller can be used to increase the pressure differential thus energy can be stored.
This may be hard for some to understand as most will think that there is no way to maintain this pressure differential and there is not much I can do about this.
The main point I want to make since this is ester to understand (at least is what I was thinking) is that there is no wind power available to a direct down wind traveling at or above wind speed.
I provided the correct equation and anyone that is not agreeing with that equation is welcome to provide the correct one.
Nobody can claim to understand a wind powered only vehicle without being able to provide an equation describing the amount of wind power available to vehicle.
Your assumption that power is required to accelerate the vehicle is not correct. At a stand still the force can be anything, but with velocity zero the power is zero. f = a • m, p = f • v
If the force applied equals the force from the wind, the net acceleration is zero and the velocity remains at zero so no power.
No fancy math required, just a clear understanding of how to apply the science and math.
I really don't think that Kleinstein was saying that huge power is actually required to accelerate the vehicle. To the contrary, Kleinstein was pointing out how unreasonable such an idea would be.
@ gnuarm
You haven't seen his website or his personal forum.
It's worth looking at .. :phew:
I was thinking more on the lines of us collectively, I know that every one here, are more than Qualified if not over qualified .@ gnuarm
You haven't seen his website or his personal forum.
It's worth looking at .. :phew:
Why am I apprehensive............? :scared:
The wind resistance is low compared to the force generated by the propeller driving the wheels because the angle of the blades is nearly parallel with the wind. The lift of the airfoil is what provides the force that rotates the propeller and therefore the wheels.
It is a very simple process once it is properly understood.
It is fairly simple, but you have it backwards. The propeller is driven by the wheels. If you posit the wind driving the propeller somehow, ED's theories would be right since the interaction between the propeller and the wind is dependent on the windspeed relative to the vehicle. The way it works when the vehicle speed is the same as the windspeed is the propeller reacts with apparently still air, neglecting a bit of turbulence, and that allows it to continue to push the vehicle even when it already at or above windspeed.
Stored pressure-difference energy? I was looking up explosive shock-wave overpressure propagation, and even at only 1 PSI, the pressure-wave is moving about 70 MPH (about 700 MPH at much higher pressures). The idea that wind-pressure builds up behind a surface (and a surface that is slowly accelerating towards windspeed), and somehow this accumulated open-space pressure will persist for minutes, or even seconds, is pretty ridiculous.
@ gnuarm
You haven't seen his website or his personal forum.
It's worth looking at .. :phew:
Then he takes close up photos of gears nearly slipping and talking as if the micro level of energy involved in that is providing the necessary energy for faster travel. It's pure BS.
Maybe we are talking past each other. In the downwind case the wheels drive the propeller. In the upwind case the propeller drives the wheels. That's why I had trouble understanding the upwind case, because if the wind pushes on the propeller very hard it will run downwind, with the propeller turning backwards by the wheels. But because the propeller is nearly feathered there is very little resistance to the wind. However the airfoil gives enough lift to spin the propeller in the correct direction which pushes the vehicle upwind.I did mention this about the feathering of this type of propeller a good many pages back . and also the vortex affect
Then he takes close up photos of gears nearly slipping and talking as if the micro level of energy involved in that is providing the necessary energy for faster travel. It's pure BS.
The whole stick-slip hysteresis energy storage explanation is silly. If there is a stick-slip cycle then energy is stored and released during the cycle. If ED's explanation were true, the vehicle would be varying above and below windspeed at the cycle rate. But the downwind speed of the vehicle through many of these nonexistent "cycles" remains faster than windspeed.
Now, replace the brakes with a stalled motor. There is still no movement, just two opposing forces (just as in the previous situation). But the motor requires power (V*A) to remain stalled. This power is being turned into heat. I assume that we just consider that the motor is operating at 0% efficiency? But it is providing torque.
Here's something that had me confused:
Bicycle stopped in the wind. Wheels are not turning. Brakes are locked (or wheels are bolted to the ground -- same thing). There is force, but no movement, so no power.
Now, replace the brakes with a stalled motor. There is still no movement, just two opposing forces (just as in the previous situation). But the motor requires power (V*A) to remain stalled. This power is being turned into heat. I assume that we just consider that the motor is operating at 0% efficiency? But it is providing torque.
I don't follow at all about the front wheels lifting for lower drag causing the car to accelerate. Do you mean the drag of the wheels on the road?I think you will find the blackbird gearbox is on the back wheels as I remember it's a chain and cog drive . So the front wheels are only for supporting the front & steering..
I think you will find the blackbird gearbox is on the back wheels as I remember it's a chain and cog drive . So the front wheels are only for supporting the front & steering..
In thrust the noise will lift . It's an aerodynamics it will rise against thrust due to centrifugal force from the prop and air under the fuselage.
I am Not going into a mile of advanced math in aerodynamics. Take it as is . It does and will.
Remember that gearing can be placed between the motor and the wheels. [...]
No, I'm comfortable with all that, it's just the fundamental discussion of work, energy, force, etc. And I'm actually pretty good with those as well. It's just where does the motor (or solenoid) power go, if not into motion? I think the answer is "heat".
Right, but the key thing is the motor the doesn't need any input power if the vehicle is stationary, because the vehicle is held in place by the gears that can't turn backwards. So from that stationary starting position, any tiny bit of power applied to the motor can make it turn forwards. There is no need for any "holding torque" on the motor to waste power as heat.
you will need more than "any tiny bit of power" to get the motor to turn
Why can you put 300W and go at low speed when climbing a hill ?
Why do you think driving in head wind is any different in that regard?
The bicycle speed will be low but your gear ration can be set so that your legs move very fast thus even with low force needed at the pedals you can still produce 300W
There is a limit based on when the traction wheel will start to slip but 300W even at 1km/h is possible on a bicycle.
Quoteyou will need more than "any tiny bit of power" to get the motor to turn
I think you may be picturing this in your mind, seeing pretty big gears or pulleys, and figuring you need a substantial motor even so. But perhaps those big gears are several orders of magnitude too small, and the stress may break gear teeth in reality but for the purpose of this exercise it's not going to happen.
Yes and no.
My thought experiment posits a motor directly connected to the wheel, via gears if you like, but in a completely linear fashion where the wind can cause an un-powered motor to rotate in either direction.
But even with your one-way gearing, you will need more than "any tiny bit of power" to get the motor to turn. The motor needs to generate sufficient torque to overcome the (head wind) wind-force. Whatever your gear-ratio, there will be some minimum amount of power needed to make the motor spin. Otherwise the motor is stalled and power turns into heat.
Yes and no.
My thought experiment posits a motor directly connected to the wheel, via gears if you like, but in a completely linear fashion where the wind can cause an un-powered motor to rotate in either direction.
But even with your one-way gearing, you will need more than "any tiny bit of power" to get the motor to turn. The motor needs to generate sufficient torque to overcome the (head wind) wind-force. Whatever your gear-ratio, there will be some minimum amount of power needed to make the motor spin. Otherwise the motor is stalled and power turns into heat.
The distinction between torque (rotation force) and power is where this thread was coming unstuck before.
Suppose, for a moment, that we have ideal, frictionless gears. It's not going to happen in the real world, but suppose we have Teflon gears and roller bearings and whatever.
Now, suppose we wish to move our vehicle forward at 0.1 m/s against a 100 N force of headwind. We can calculate the required power as 0.1 x 100 = 10 W. So, if there are no losses in our ideal gear train, then the motor needs to output 10 W to achieve this rate of forward progress.
If, maybe, we only have a 1 W motor, then we cannot go this fast. However, we could go at 0.01 m/s, since 0.01 x 100 = 1 W.
How big the motor is determines how fast we can go, but if we just want to go at any speed at all, then we can introduce ludicrous gear ratios and make the motor as tiny as we like.
This is why I say "any tiny bit of power". In the real world, of course, some power is required to overcome the friction in the gears, and more gears will have more friction, so there is a law of diminishing returns. However, in principle, a wind up clock mechanism could make a vehicle move against a gale force headwind, albeit at a glacial pace.
- There is a force acting on the mechanism. Let it be a head wind. With no brakes and no friction, the mechanism will roll backwards at wind speed.
- An electric motor is used to counteract this force. Just enough power is applied to the motor (volts * amps) to hold the armature in one place. This is a locked rotor.
- Since the mechanism is not moving, there is no mechanical work being done.
- But the motor is still consuming power, turning it into heat. The motor is operating at 0% efficiency, but still performing a useful function.
This -- electrical power being used by the motor while no mechanical work is being done -- may be one of the confusing factors when discussing the system.
Here's something that had me confused:
Bicycle stopped in the wind. Wheels are not turning. Brakes are locked (or wheels are bolted to the ground -- same thing). There is force, but no movement, so no power.
Now, replace the brakes with a stalled motor. There is still no movement, just two opposing forces (just as in the previous situation). But the motor requires power (V*A) to remain stalled. This power is being turned into heat. I assume that we just consider that the motor is operating at 0% efficiency? But it is providing torque.
Or replace the motor with an electromagnetic solenoid that holds the bicycle in place against the wind -- really the same situation as the stalled motor. In both cases power is being consumed, but no work is being done other than heating the air?
It seems to me that some of ED's confusion is related to this. I know I still struggle with the terms, since I always go back to the volts and amps. ED's models using generator and motor also add to the potential (no pun intended) confusion.
Dear friends, thank you for a little adventure. Very funny!
To electrodacus. Give up! opposition is useless! This device really works. Although the principle is not obvious. :)
Yes, I have not read everything here. Perhaps I will repeat someone's words, sorry.
Quite a long time ago, my friend voiced a banal thing - the mathematical apparatus is secondary. Understanding the principle is primary. In this case (electrdoacus) understanding did not happen. Therefore, all formulas are useless.
And now, in essence. Perhaps this will help. The entire mechanism with a transmission and a propeller makes it possible to perceive wind energy at almost any speed. The basic ratio of wheel circumference to propeller pitch is 0.7. In other words, the propeller always "advances" the movement of the cart. Everything is very simple.
Are the scientists wrong, or are they deliberately misleading us for personal gain?Good question. I would say both.