EEVblog Electronics Community Forum
General => General Technical Chat => Topic started by: electrodacus on November 17, 2022, 11:16:10 pm
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I have the diagram below.
The vehicle has only two points of contact.
Front wheel the one of right sits on a treadmill witch can apply a force F1 to that wheel
Back wheel the one of the left is on the ground (red box is connected to ground) same as treadmill body witch is also connected to ground.
The question:
a) What will happen in a theoretical case? where there is no wheel slip and no components can deform in any way elastic or plastic including the belt.
b) What happens in a real setup? where both slip and deformation exist and can not be get rid of.
(http://electrodacus.com/temp/Windup.png)
Edit:
Seems that is a bit of a boring subject or not sure why there are no comments but to help things out I will add a video with what happens in a real setup
https://odysee.com/@dacustemp:8/wheel-cart-energy-storage-slow:8 (https://odysee.com/@dacustemp:8/wheel-cart-energy-storage-slow:8)
What you see in the video is a slowed down video of a toy vehicle with the same design as in the diagram.
The front wheels are on a moving paper simulating the treadmill and the back wheels are on the ground.
What happens in the video is as follow (my interpretation so if you disagree please explain).
When treadmill starts to move the vehicle remains stationary while the front wheel rotates charging the build in energy storage (stretching the rubber belt).
The force F1=F2 increases until force is high enough for the front wheel to slip and at that exact moment the energy stored in the belt is used to rotate the back wheel and move the vehicle from left to right so in the opposite direction from treadmill surface.
Once the stored energy is used up the cycle repeats and this happens many times per second so much so that in most cases is not possible to realize what happens without watching a slowed down video.
So the reason why this locked gearbox vehicle can move opposite to the treadmill direction (the only thing powering the system) is a combination of energy storage and stick slip histeresis.
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To properly model this situation, you need to account for momentum (both linear and angular), and therefore the masses of the components also. Having a springy belt makes it quite difficult to model; a thin tiny-mass thread or wire would be better –– modelled in real life by making the wheel masses much larger, and using e.g. a toothed belt.
One could model the springy belt using a torsion spring conforming to Hooke's law (https://en.wikipedia.org/wiki/Hooke%27s_law), F=-kϕ, where ϕ is the angular difference between the two wheels, but it is not at all obvious it matches how a springy elastic belt behaves.
Stiction (static friction) and dynamic friction are key here, and depend on the materials used, and definitely affect exactly what will happen. Oscillating motion at the static-dynamic friction boundary is annoyingly difficult to model, because tiny changes drastically alter the behaviour of the system; in real world, you get chaotic effects, in the simulation, rounding errors and such can cause unphysical oscillations or oscillation dampening.
While I can create a simulation model for you, it's a lot of work; and I don't really see the point here.
If the stiction on the two surfaces is sufficiently great –– say, toothed wheels and matching rack-like surface on both –– then the device will move right (opposite direction to the top surface of the treadmill).
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A better model would be to have two parallel racks, with a chain drive between, with top surfaces level. Create a three-wheeled "car", running on top, that has a center sprocket as the front wheel, and two pinion wheels at the back, with front and back connected using a gearbox so pinions rotate faster than the sprocket. (You only need one pinion wheel, but with two it will stay upright and not try to skew.)
You can do this easily if you have suitable Technics Lego sets (specifically, some gears, racks, and chain). I might... :-[
The "car" will always move opposite to the top surface of the chain. But what does this prove? Do you need a video to prove this?
You can also use Lego Digital Designer (Windows 7) or Bricklink Studio (Windows, Mac) to design the models digitally first, but AFAIK they do not model technics in operation, only statically.
So the reason why this locked gearbox vehicle can move opposite to the treadmill direction (the only thing powering the system) is a combination of energy storage and stick slip histeresis.
No, I don't think so.
Yes, the vehicle can definitely move opposite to the treadmill direction, but you do not actually need energy storage to achieve this; all you need is sufficient friction between the wheels and the treadmill and the static surface.
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Yes, the vehicle can definitely move opposite to the treadmill direction, but you do not actually need energy storage to achieve this; all you need is sufficient friction between the wheels and the treadmill and the static surface.
Thanks for taking the time to replay. It seems this is not an exciting subject.
If you take a theoretical case where the mass of the vehicle is zero and you exclude any energy storage (no elastic deformation or gravitational energy storage as it will happen with a louse chain).
What then will be the mechanism allowing the vehicle to move opposite to the treadmill direction?
The only reason I see for F2 to exist is as the pair of F1 thus equal and opposite.
I can show the same behaviour with gears as you can just not get rid of elastic deformation or gravitational energy storage (if some parts lift up) in real world but what I can do is reduce the friction at the back wheels so that the back wheels slip before the front and then this is what happens https://odysee.com/@dacustemp:8/stick-slip-removed-from-front-wheels:0 (https://odysee.com/@dacustemp:8/stick-slip-removed-from-front-wheels:0)
As I expected eliminating the slip at front wheel will result in vehicle being dragged (as it is a locked gearbox) in the direction that force is applied by the treadmill.
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My reading, with no elasticity, the 'car' should advance at 1/2 the speed of the treadmill - due to the gearing ratio. Otherwise, it will stutter as energy pumps and dumps in the band. As Nominal Animal states, energy transfer is chaotic.
Another way to think of your problem is, what would happen if you replaced a metal bicycle chain with a band made from bungie rope? How hard would you have to pedal to overcome the elasticity in the bungie rope before the back wheel turns - and would the pedals feel like they were made from jelly? This would make an impossible bike to ride :)
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Yikes! You're back!
As explained clearly to you earlier, this is not a Newtonian physics problem, it is basic Archimedes-era problem. Your knowledge here is not hundreds of years behind, but rather a few thousand. You simply need to understand the force multiplication possible with levers and then go one step further and use that to understand torque multiplication by pulleys and gears. A big hint--gears and pulleys in the static case are the same as levers.
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Yes, the vehicle can definitely move opposite to the treadmill direction, but you do not actually need energy storage to achieve this; all you need is sufficient friction between the wheels and the treadmill and the static surface.
I can show the same behaviour with gears as you can just not get rid of elastic deformation or gravitational energy storage (if some parts lift up) in real world but what I can do is reduce the friction at the back wheels so that the back wheels slip before the front and then this is what happens https://odysee.com/@dacustemp:8/stick-slip-removed-from-front-wheels:0 (https://odysee.com/@dacustemp:8/stick-slip-removed-from-front-wheels:0)
You may need some kind of energy storage to be able to accelerate, but I do not think it is necessary for steady-state operation.
(I'd need to model the entire system, including friction and momentum and losses, to be able to examine the system phase space, and determine whether acceleration is possible without energy storage or not.)
Also: aw crappola, I topsy-turvied myself with the gearing ratio. :-[ I thought inverse to what I wrote. It is a typical error for me, too. Dammit.
Teaches me right for not testing it with a model first. Apologies.
If we ignore losses, it makes no sense to look at the forces involved. The mechanics of the situation suffice.
If we use \$v_t\$ for the speed of the surface of the treadmill (going in the opposite direction), \$v_d\$ for the speed for (the surface of) the driven wheel on the treadmill, and \$v_c\$ for the speed of the car itself (and therefore also (the surface of) the driving wheels), the machine requires
$$\left\lbrace \begin{aligned}
v_d &= v_t + v_c \\
v_c &= \lambda v_d \\
\end{aligned} \right.$$
where \$\lambda\$ is the gearing ratio, i.e. number of turns of the driving wheels per each turn of the driven wheel, assuming their diameters is the same.
The lower equation describes the locked gearbox.
The upper equation describes that the driven wheel must have a surface speed that is the sum of the treadmill surface and "car" speeds, or something has to give (slip).
Combining the two, and solving for \$v_c\$, we have, in the steady state ignoring any losses,
$$v_c = v_t \frac{\lambda}{1 - \lambda}$$
so there is a steady state for \$0 \le \lambda \lt 1\$ only.
In your example, \$\lambda \gt 1\$, and there is no steady state.
However, do check what happens if you reverse the gearing. For example, with \$\lambda = 0.6\$ (driving wheel surface speed 3/5 of that of the driven wheel, or if their diameters are the same, the driving wheel turning 3 times in the time the driven wheel turns 5 times), the solution is \$v_c = 1.5 v_t\$, i.e. the car runs at 1.5 times the treadmill speed.
When \$\lambda = 0.5\$, \$v_c = v_t\$. When \$\lambda = 0.2\$, \$v_c = 0.25 v_t\$.
I do believe that losses can be adequately modeled by reducing the gearing ratio. If the actual ratio is \$\lambda\$, then the model \$\lambda^\prime = (1 - \epsilon)\lambda\$, where \$\epsilon\$ represent the overall losses in the system, should match physical results.
A real world vehicle does probably need energy storage for acceleration and to avoid tiny losses causing a catastrophic failure, and an adjustable gearing \$0 \lt \lambda \lt 1\$, so it can optimize the gear ratio to the current car velocity \$v_c\$. Achieving any ratio \$\lambda \gt 0.5\$ is sufficient to travel faster than the treadmill.
In case anyone is wondering where the necessary energy is coming from, the answer is obvious: from the treadmill. When the "car" is placed on the treadmill, the larger \$\lambda\$ is, the more energy is needed to keep the treadmill moving at a given velocity \$v_t\$.
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Seems that is a bit of a boring subject
More likely it's already been done to death in previous threads and no-one has the energy (ha ha) to go through all that again.
or not sure why there are no comments
Probably because we know full well that whatever explanations, theories, laws of physics, etc. are posted, nothing will drag you from your insistence that it is all about energy storage (and, as a corollary, the movement against the power source cannot happen).
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A simplified example is a pair of wheels with almost as large a pulley in between, with a thread rotated on the pulley. What happens when you pull the thread?
If you start examination at the zero-movement state, you'll find that the angle at which the thread exits the spool, with respect to gravity down, largely determines whether it starts moving or not. If you pull the thread exactly level, the forces do not leverage the wheel to turn so that it would travel opposite to the direction the thread is being pulled to. If you pull the thread slightly upwards or downwards from level, the forces are no longer balanced, resulting a small torque on the wheel, so it'll start traveling opposite to the direction the thread is being pulled to. If you pull the thread nearly vertical, the torque is maximized, and it is easiest to get the wheel to travel (away from where you are pulling the thread).
(The same applies to acceleration: it will only accelerate as you pull harder, if the thread is not exactly level. The minimum angle for a constant velocity depends on the losses, and the minimum angle for acceleration depends on losses and rotational inertia.)
This is why omitting any forces, and only looking at some, will often lead to unphysical results. Would you have considered the effect of the thread angle? It is not obvious until you examine all of the forces acting here, not to me at least.
When the pulley is exactly the size of the wheel, you cannot get it to move (in the opposite direction of where you pull the thread).
When the pulley is larger than the wheel, the angle of the thread no longer matters, and it is easy to get the wheel to move (in the opposite direction of where you pull the thread).
Here, the treadmill is a continuous thread, and the pulley/wheel size ratio represents the gearing between the driven and driving wheels of the "car".
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My reading, with no elasticity, the 'car' should advance at 1/2 the speed of the treadmill - due to the gearing ratio. Otherwise, it will stutter as energy pumps and dumps in the band. As Nominal Animal states, energy transfer is chaotic.
Another way to think of your problem is, what would happen if you replaced a metal bicycle chain with a band made from bungie rope? How hard would you have to pedal to overcome the elasticity in the bungie rope before the back wheel turns - and would the pedals feel like they were made from jelly? This would make an impossible bike to ride :)
On a bicycle the rider provides the force and against ground and it is connected to bicycle frame so there is no analogy that can be made with this vehicle that is powered from outside by the treadmill and the vehicle frame is floating.
The elasticity of the belt only exaggerate what will happen with any mechanism including a bicycle chain.
This is a locked gearbox since F1 = F2
Please show how F2 can be larger than F1. As F1 can not exist without the equal and opposite F2 in this particular example.
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Yikes! You're back!
As explained clearly to you earlier, this is not a Newtonian physics problem, it is basic Archimedes-era problem. Your knowledge here is not hundreds of years behind, but rather a few thousand. You simply need to understand the force multiplication possible with levers and then go one step further and use that to understand torque multiplication by pulleys and gears. A big hint--gears and pulleys in the static case are the same as levers.
There can not be a force multiplication in this particular problem as the vehicle has only two points of contact.
Think about this way:
A rope tied to a tree and you pull the rope. Is there any chance that rope can get shorter and pull you towards the tree using your own power ?
Cut the rope in two and add anything you like that has no internal energy source and that can pull you towards the tree while you pull away from it.
It is the same problem here not sure why so many people fail to see this even with the clear video evidence showing exactly what I'm describing.
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A real world vehicle does probably need energy storage for acceleration and to avoid tiny losses causing a catastrophic failure, and an adjustable gearing \$0 \lt \lambda \lt 1\$, so it can optimize the gear ratio to the current car velocity \$v_c\$. Achieving any ratio \$\lambda \gt 0.5\$ is sufficient to travel faster than the treadmill.
In case anyone is wondering where the necessary energy is coming from, the answer is obvious: from the treadmill. When the "car" is placed on the treadmill, the larger \$\lambda\$ is, the more energy is needed to keep the treadmill moving at a given velocity \$v_t\$.
The claim I make and it is proved by the video's I shwed is that no vehicle including this one can move from left to right without both energy storage and stick slip systeresis.
I showed what happens when you have both energy storage and stick slip hysteresis https://odysee.com/@dacustemp:8/wheel-cart-energy-storage-slow:8
And also showed what happens if you remove the ability of the front wheel to slip https://odysee.com/@dacustemp:8/stick-slip-removed-from-front-wheels:0
Against all evidence and theoretical explanation of how it works people think is not true.
The friction of the back wheel slip in second video is significantly higher than the internal friction of the rotating parts of the vehicle. And is clear vehicle is dragged as that type of gearbox for this particular application is locked.
I think I will need to make a drawing showing the lever type equivalent of this vehicle as that may be more easy to understand.
You need to twist the belt (infinite symbol shape) in order for this mechanism to not be locked so changing the gear ratio will not help with anything.
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There can not be a force multiplication in this particular problem as the vehicle has only two points of contact.
That's just silly. But perhaps think about motion amplification or reduction instead of forces if that helps. If you want to use a basic lever analysis, your 'third point of contact' is the net force against or motion of the body of the vehicle itself. And that might require Newtonian mechanics if you let it progress beyond the static case, but you have to understand Archimedes first.
As far as the video, it clearly shows just the opposite of what you say it does. Your interpretation of the video appears to be an optical delusion.
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Probably because we know full well that whatever explanations, theories, laws of physics, etc. are posted, nothing will drag you from your insistence that it is all about energy storage (and, as a corollary, the movement against the power source cannot happen).
I gave this example earlier but do you think you can add any mechanism (not powered) that can pull you towards the tree while you pull away from the tree ?
Or to be more equivalent pusing a solid pipe against a huge wall and the pipe using your energy to push you away from the wall?
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That's just silly. But perhaps think about motion amplification or reduction instead of forces if that helps. If you want to use a basic lever analysis, your 'third point of contact' is the net force against or motion of the body of the vehicle itself. And that might require Newtonian mechanics if you let it progress beyond the static case, but you have to understand Archimedes first.
As far as the video, it clearly shows just the opposite of what you say it does. Your interpretation of the video appears to be an optical delusion.
Motion amplification ? What will that even mean ?
Are you saying power output of the back wheel can be higher than power input on the front wheel ? Since it sounds like that.
What video and please explain what you see if you do not agree with my explanation of what happens.
I clearly see a vehicle that is not moving relative to the ground and a front wheel that moves while belt is stretched so force increases and so is power that it is stored (stretched belt is stored energy).
Then when force is large enough the front wheel slips allowing the back wheel to push the vehicle using the energy stored earlier in the belt and when that belt is discharged the cycle repeats.
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Wasn't this already shown in one of the threads on people being completely wrong about electricity?
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A simplified example is a pair of wheels with almost as large a pulley in between, with a thread rotated on the pulley. What happens when you pull the thread?
If you start examination at the zero-movement state, you'll find that the angle at which the thread exits the spool, with respect to gravity down, largely determines whether it starts moving or not. If you pull the thread exactly level, the forces do not leverage the wheel to turn so that it would travel opposite to the direction the thread is being pulled to. If you pull the thread slightly upwards or downwards from level, the forces are no longer balanced, resulting a small torque on the wheel, so it'll start traveling opposite to the direction the thread is being pulled to. If you pull the thread nearly vertical, the torque is maximized, and it is easiest to get the wheel to travel (away from where you are pulling the thread).
(The same applies to acceleration: it will only accelerate as you pull harder, if the thread is not exactly level. The minimum angle for a constant velocity depends on the losses, and the minimum angle for acceleration depends on losses and rotational inertia.)
This is why omitting any forces, and only looking at some, will often lead to unphysical results. Would you have considered the effect of the thread angle? It is not obvious until you examine all of the forces acting here, not to me at least.
When the pulley is exactly the size of the wheel, you cannot get it to move (in the opposite direction of where you pull the thread).
When the pulley is larger than the wheel, the angle of the thread no longer matters, and it is easy to get the wheel to move (in the opposite direction of where you pull the thread).
Here, the treadmill is a continuous thread, and the pulley/wheel size ratio represents the gearing between the driven and driving wheels of the "car".
I know the tread problem is the same and same stick slip hysteresis and energy storage is involved.
But I think the belt vehicle is an easier example to understand and see the effects. Especially with a long stretchy belt and slowed down video.
The charge discharge frequency can be quite high for most mechanisms and will be impossible to see without slow motion video.
But I should make a drawing where all forces will be placed (currently the forces acting on vehicle body are not drawn) and I will make it just as a lever system to maybe be even clearer that F2 can not be different from F1 as F1 is the reason F2 exists and vice versa.
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Motion amplification ? What will that even mean ?
If you have a lever with a fulcrum and the ends of the lever have different distances to the fulcrum, then a small motion on the short end becomes a large motion on the long end and vice versa. Like I said, Archimedes.
Are you saying power output of the back wheel can be higher than power input on the front wheel ? Since it sounds like that.
I didn't say that. You appear to be 2000+ years short of even understanding the term 'power'.
What video and please explain what you see if you do not agree with my explanation of what happens.
I clearly see a vehicle that is not moving relative to the ground and a front wheel that moves while belt is stretched so force increases and so is power that it is stored (stretched belt is stored energy).
Then when force is large enough the front wheel slips allowing the back wheel to push the vehicle using the energy stored earlier in the belt and when that belt is discharged the cycle repeats.
Your video. I don't see any wheel slippage at all. I see a bit of jerkiness and stretching of the rubber band, all accentuated by the stop-action of the slow frame rate, but none of what you describe. Even if the band were entirely inflexible or the equivalent function was implemented with gears, the result would be the same.
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Wasn't this already shown in one of the threads on people being completely wrong about electricity?
Do not change the subject. This is a problem where everything can be seen and you are still getting things wrong.
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Reciprocal: a connecting rod between a piston and a wheel is replaced with a shock absorber. Does the wheel rotate if the shock absorber is 'storing' energy?
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The claim I make and it is proved by the video's I shwed is that no vehicle including this one can move from left to right without both energy storage and stick slip systeresis.
Sure it can. Just swap the gearing ratio, and replace the rubber band with a non-springy belt.
You need to twist the belt (infinite symbol shape) in order for this mechanism to not be locked so changing the gear ratio will not help with anything.
No, you don't. Just make the floor-wheel gear or pulley larger than the treadmill gear or pulley, and be amazed. The bigger the ratio (rear size to front size), the easier it moves.
Please, try it. I'm not tricking you. I showed you math that *proves* it works. The only reason for you to not try it, would be to avoid being proven wrong. Keep your mind open, friend, and try it.
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If you have a lever with a fulcrum and the ends of the lever have different distances to the fulcrum, then a small motion on the short end becomes a large motion on the long end and vice versa. Like I said, Archimedes.
Let me know if you see the image and if you agree with the lever option as analog to that vehicle.
[attachimg=1]
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Sure it can. Just swap the gearing ratio, and replace the rubber band with a non-springy belt.
All materials have elastic and plastic deformation in real world so you can not get rid of that.
You can reduce the size of energy storage but that is irrelevant the frequency of charge discharge will increase and it will still work the same way.
No, you don't. Just make the floor-wheel gear or pulley larger than the treadmill gear or pulley, and be amazed. The bigger the ratio (rear size to front size), the easier it moves.
Please, try it. I'm not tricking you. I showed you math that *proves* it works. The only reason for you to not try it, would be to avoid being proven wrong. Keep your mind open, friend, and try it.
See attachment and let me know if you agree with that analogy.
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See attachment and let me know if you agree with that analogy.
I won't tell you, before you actually retry the experiment with swapped gear ratio, and describe your observations here.
It is a very small, simple thing for you to do, after all. You've already posted videos of two variants above, so how much more effort would a third one be?
Or is it that you did it, and observed exactly what I said you would, and are now annoyed that your model is broken?
I'm not doing this –– responding here –– because I want to break your model, or break anything at all.
I am responding here only because I want you to have a *working model*.
Please understand that.
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See attachment and let me know if you agree with that analogy.
I won't tell you, before you actually retry the experiment with swapped gear ratio, and describe your observations here.
It is a very small, simple thing for you to do, after all. You've already posted videos of two variants above, so how much more effort would a third one be?
Or is it that you did it, and observed exactly what I said you would, and are now annoyed that your model is broken?
I'm not doing this –– responding here –– because I want to break your model, or break anything at all.
I am responding here only because I want you to have a *working model*.
Please understand that.
I tested all variants and there is no difference the vehicle works the same way with charging and stick slip hysteresis as there is no other way for the vehicle to work.
Please see that attach diagram and let me know how is not that a locked system that will not be able to move to the right without both energy storage and stick slip hysteresis and to the left without slip.
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I have the diagram below.
The vehicle has only two points of contact.
Front wheel the one of right sits on a treadmill witch can apply a force F1 to that wheel
Back wheel the one of the left is on the ground (red box is connected to ground) same as treadmill body witch is also connected to ground.
The question:
a) What will happen in a theoretical case? where there is no wheel slip and no components can deform in any way elastic or plastic including the belt.
The vehicle will move to the right, obviously.
Until you can see that, your reasoning is defective, and many things you try to deduce with your defective reasoning will be wrong.
The reason nobody will debate with you is not because it is boring, but because it is pointless. Since you cannot, or will not, understand, there is no point trying to explain anything to you.
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The vehicle will move to the right, obviously.
Until you can see that, your reasoning is defective, and many things you try to deduce with your defective reasoning will be wrong.
The reason nobody will debate with you is not because it is boring, but because it is pointless. Since you cannot, or will not, understand, there is no point trying to explain anything to you.
Have you looked at the lever version? See attachment if not.
Do you still think it will move to the right ? If so explain how F2 can be different from F1.
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The vehicle will move to the right, obviously.
Until you can see that, your reasoning is defective, and many things you try to deduce with your defective reasoning will be wrong.
The reason nobody will debate with you is not because it is boring, but because it is pointless. Since you cannot, or will not, understand, there is no point trying to explain anything to you.
Have you looked at the lever version? See attachment if not.
Do you still think it will move to the right ? If so explain how F2 can be different from F1.
The green bar will move to the right, obviously (again).
I do not care about the value of F2 (or indeed F1). They are irrelevant to the outcome.
See the attached animation for the wheeled cart example. The cart moves to the left when the belts under the wheels move to the right. This is a pure mathematical model. There is no slip, no energy storage, no hysteresis, no anything of that. The observed outcome is a simple mathematical fact arising from the configuration of belts, wheels and axles.
(https://s4.gifyu.com/images/cart-animated.gif) (https://gifyu.com/image/ShH1I)
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The green bar will move to the right, obviously (again).
I do not care about the value of F2 (or indeed F1). They are irrelevant to the outcome.
See the attached animation for the wheeled cart example. The cart moves to the left when the belts under the wheels move to the right. This is a pure mathematical model. There is no slip, no energy storage, no hysteresis, no anything of that. The observed outcome is a simple mathematical fact arising from the configuration of belts, wheels and axles.
(Well, it seems the forum strips the animation. I'll have to try another way.)
Yes there is no animation but I think I saw that some time ago if I'm not mistaken.
You need a physical not mathematical model. Even the lever can move if you apply an external force other than F1 but there is nothing other than F1 in this setup.
It seems you ignore the lever model. You do care what F1 is in relation to F2 if you want to understand how this vehicle can move. Because is clear it can not move if you remove the stick slip and or energy storage.
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the vehicle works the same way with charging and stick slip hysteresis as there is no other way for the vehicle to work
That, there, illustrates why you will never get this. You start out 'knowing' it won't work, so therefore nothing anyone can do or suggest will show it working because, well, you know it can't.
Just for once, try NA's suggestion without a preconceived idea and see what happens. Once you have uncontaminated experimental data you can argue about how it came about and what it means, and refine the experiment for more data if necessary.
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It seems you ignore the lever model. You do care what F1 is in relation to F2 if you want to understand how this vehicle can move. Because is clear it can not move if you remove the stick slip and or energy storage.
And yet the green bar moves to the right with no stick slip or energy storage needed. How do you explain that?
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And yet the green bar moves to the right with no stick slip or energy storage needed. How do you explain that?
The green bar will not move to the right without both stick slip and and energy storage.
I also saw your animation, and you need to keep one of two surfaces fixed (ground reference). Then see if your math still works.
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I also saw your animation, and you need to keep one of two surfaces fixed (ground reference).
Why? You think it is less likely for the cart to move with one belt stationary than with both belts moving?
If you want the left belt to be stationary, just make that the frame of reference and follow it along with your eye. It is exactly the same equations.
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That, there, illustrates why you will never get this. You start out 'knowing' it won't work, so therefore nothing anyone can do or suggest will show it working because, well, you know it can't.
Just for once, try NA's suggestion without a preconceived idea and see what happens. Once you have uncontaminated experimental data you can argue about how it came about and what it means, and refine the experiment for more data if necessary.
I made the experiment's and all show exactly what I'm saying.
Not sure you saw the video where I remove the slip at the front wheel and vehicle as expected was dragged in the direction of the treadmill as it is as I say a locked mechanism.
It should be easy to see that the mechanism is locked if you look at the lever version of the same mechanism.
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It should be easy to see that the mechanism is locked if you look at the lever version of the same mechanism.
Is the orange line a rope? And are the yellow circles hinges? And if so, how can it be locked with so many movable elements present?
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Why? You think it is less likely for the cart to move with one belt stationary than with both belts moving?
If you want the left belt to be stationary, just make that the frame of reference and follow it along with your eye. It is exactly the same equations.
I guess you can not do what you are asking me to do.
Fortunately I can imagine what happens if you have the left belt stationary. I guess you need to do the math to know that.
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Is the orange line a rope? And are the yellow circles hinges? And if so, how can it be locked with so many movable elements present?
Yes and yes.
There are movable elements but treadmill can apply F1 to the left so the mechanism as it is, is no different from a solid rock.
And I could add the other rope below the green beam but it makes no sense for the forces in this direction as that plays no role here.
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Fortunately I can imagine what happens if you have the left belt stationary.
No need to imagine, we know that the cart moves to the left.
I guess you need to do the math to know that.
Yes, it is the math that gives us that result.
There are movable elements but treadmill can apply F1 to the left so the mechanism as it is, is no different from a solid rock.
And I could add the other rope below the green beam but it makes no sense for the forces in this direction as that plays no role here.
Why are you bringing forces into the analysis? This is just a misdirection to send the discussion in the wrong direction.
This is a simple problem of mechanics. The only important variables are the (x, y) coordinates of the pivot points in the mechanism and the connections between them. If you write down the linking equations you will see that the mechanism is not at all fixed and is quite free to move.
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No need to imagine, we know that the cart moves to the left.
Yes, it is the math that gives us that result.
So you have the fixed ground on the left and the treadmill that moves to the right 3 squares.
How many squares does the vehicle move to the left ?
I think you need to make that animation to see the problem.
Why are you bringing forces into the analysis? This is just a misdirection to send the discussion in the wrong direction.
This is a simple problem of mechanics. The only important variables are the (x, y) coordinates of the pivot points in the mechanism and the connections between them. If you write down the linking equations you will see that the mechanism is not at all fixed and is quite free to move.
So asking to prove by showing how F2 can be larger than F1 is a misdirection ?
"free to move" how ? the only two places you can interact with that are at F1 and F2
Since you can not have F1 without F2 and F2 is opposite and equal it means it can not move.
Just build that mechanism have the leg with F1 in one in right hand and the leg with F2 in the left hand.
Keep the left hand stationary and push with the right hand towards the left hand and see if anything moves.
Whatever force F1 you apply with the right hand will be equal and opposite to the force you will feel against the left hand.
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So you have the fixed ground on the left and the treadmill that moves to the right 3 squares.
How many squares does the vehicle move to the left ?
I think you need to make that animation to see the problem.
The animation is already made, it is posted above.
With the fixed ground on the left the treadmill moves to the right one square. When this happens the vehicle moves two squares to the left.
Just build that mechanism have the leg with F1 in one in right hand and the leg with F2 in the left hand.
Keep the left hand stationary and push with the right hand towards the left hand and see if anything moves.
Whatever force F1 you apply with the right hand will be equal and opposite to the force you will feel against the left hand.
You have the question, you build it. This one is simple to make, you can construct it out of cardboard, paper fasteners (https://www.staples.com/OIC-Brass-Plated-Round-Head-Fasteners-1-2-Head-1-1-2-Capacity/product_615716), sticky tape and string.
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The animation is already made, it is posted above.
With the fixed ground on the left the treadmill moves to the right one square. When this happens the vehicle moves two squares to the left.
I guess you can not see the problem as you concentrate only on the math with zero connection to physics.
It is OK for you that an invisible hand pushes the vehicle and is not actually powered by the treadmill.
Also do you observe that in reality ? The vehicle with this gear ratio moves at 2x the speed relative to ground vs the treadmill 1x relative to ground ?
You have the question, you build it. This one is simple to make, you can construct it out of cardboard, paper fasteners (https://www.staples.com/OIC-Brass-Plated-Round-Head-Fasteners-1-2-Head-1-1-2-Capacity/product_615716), sticky tape and string.
Why will I need to build that ? I already had the vehicle which is the same thing.
You just seems to not understand what the problem is.
F1 is the only force acting on the vehicle there is no other magical force or energy source.
The direction of F1 is towards the left in my diagram so the vehicle no matter what it will move to the left if at all without involving energy storage.
I drawn all forces in that mechanism 6 of them in total each group of two equal according to Newton's 3'rd law.
F1=F2 F3=F4 and F5=F6
You need to show how F1 different from F2 if you want to claim the vehicle can move.
And I already showed how F2 becomes larger for short bursts due to energy storage released by slip at the front wheel.
There is no alternative explanation and I showed in the video that is exactly what happens. What more can I do ?
Is the same reason a vehicle driving at 30m/s with no wind experiences the same drag as if it drives at 10m/s in a 20m/s head wind again something that to many people get wrong.
As far as I can see is from not understanding that every action has an equal and opposite reaction. I can only imagine what Newton had to deal with.
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Notice how there is no-one else responding to your posts?
It is because they are all tired of how you keep advancing nonsensical arguments that are not supported by mathematics, physics or logic, and then you keep asking people to do work to disprove your fallacies.
We don't care that you demonstrate an improper understanding of science, engineering or mathematics. It is not our problem, and it doesn't affect us that you are wrong.
Please stop dragging up the same tired old rubbish, go away, and keep it to yourself.
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Notice how there is no-one else responding to your posts?
It is because they are all tired of how you keep advancing nonsensical arguments that are not supported by mathematics, physics or logic, and then you keep asking people to do work to disprove your fallacies.
We don't care that you demonstrate an improper understanding of science, engineering or mathematics. It is not our problem, and it doesn't affect us that you are wrong.
Please stop dragging up the same tired old rubbish, go away, and keep it to yourself.
Maybe they starting to question their understanding about the subject.
As mentioned earlier your math doesn't add up. According to you 1x treadmill speed relative to ground equals 2x vehicle speed also relative to ground in the opposite direction.
That is not what is observed in reality.
In fact with just math you can show a 1:1 gear ratio where vehicle moves 2 squares while treadmill is not moving at all. Will that represent reality ? A non moving treadmill powers a 1:1 geared vehicle ? How will that work.
If your theory was adding up and was able to predict what happens I will happy to change my mind and admit I was wrong and also apologize for all the wasted time.
But as of now what I think happens perfectly describes what is observed and is according to accepted physics like Newton's laws and conservation of energy.
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Maybe they starting to question their understanding about the subject.
Absolutely not.
IanB is spot on with this analysis:
Notice how there is no-one else responding to your posts?
It is because they are all tired of how you keep advancing nonsensical arguments that are not supported by mathematics, physics or logic, and then you keep asking people to do work to disprove your fallacies.
We don't care that you demonstrate an improper understanding of science, engineering or mathematics. It is not our problem, and it doesn't affect us that you are wrong.
Please stop dragging up the same tired old rubbish, go away, and keep it to yourself.
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The reason why you have such a significant problem is because you are stubborn, close-minded and have preconceived ideas which prevents you from being open to learning.
Here is a perfect example:
..... as there is no other way for the vehicle to work.
* sigh *
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The reason why you have such a significant problem is because you are stubborn, close-minded and have preconceived ideas which prevents you from being open to learning.
Here is a perfect example:
..... as there is no other way for the vehicle to work.
* sigh *
Maybe I do not express myself correctly and that should read "there is no other way I can see or was presented that could work"
And main point IanB makes was debunked.
Hes mathematical model can output anything and while consistent with math is not consistent with physics.
As I mentioned earlier his own model can predict the vehicle will move say a random 2 squares to the left with a vehicle that has a 1:1 gear ratio while treadmill will not move at all.
I hope you agree that it will be absurd to claim that vehicle that moved (work was done) was powered by a treadmill that did not move at all.
So yes vehicle can move if you push it but it can not be powered by the treadmill without involving energy storage and stick slip hysteresis.
It is not only that energy storage and stick slip histeresis perfectly explains what happens but it can also be seen happening in the video I provided.
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Electrodacus, I have proven your assertion wrong, and proven a vehicle constructed as I described, works.
I constructed a simple vehicle with a thread spool and a worm gear on a vertical axis, and a pinion gear and driving wheels on a horizontal axis. I also created a LeoCAD (https://www.leocad.org/) model of the vehicle, so that anyone with Windows, Linux, or Mac, can examine the vehicle in 3D, and build one or a similar one with whatever parts they may have handy. Just save the attached trike.txt as trike.ldr, and it will open in LeoCAD. It only uses parts that are available in LeoCAD by default, I believe.
When the spool turns clockwise, the single wheel is the rear wheel.
When the spool turns counterclockwise, the single wheel is the front wheel.
Using strong, non-springy polyester thread (that I normally use for sewing buttons back on), configured either way, pulling the thread off from the spool makes the vehicle move; even when you are pulling the thread in the opposite direction.
Specifically, if the two wheels point left and one wheel right, and the thread comes off the spool counterclockwise and through a hole in the black holder, pulling the thread right makes the vehicle travel left.
Conversely, if the thread comes off the spool clockwise and through a hole in the gray holder, pulling the thread left makes the vehicle travel right. In other words, this works –– so that pulling the thread makes the vehicle travel in the other direction –– in both directions and configurations; all you need to do is swap the direction the spool turns when you pull the thread off it.
Because this is a worm gear drive, there is absolutely no energy storage. When you stop pulling the thread, the vehicle stops, because the wheels cannot turn the spool (as a pinion cannot drive a worm, only the worm can drive the pinion). The spool axis has minimal inertia, too; when you stop pulling the thread, the spool does not unwind on its own.
I can take video of it tomorrow if you insist –– I need to make some kind of a holder for my phone and get better lighting, because my hand-held video was horrible ––, but I'd prefer you yourself build and test a similar vehicle. It does not need to be exact same, just make the gearing ratio small enough so that there is sufficient stiction for the wheels to drive the vehicle forward. Note that I had to use a worm gear because I didn't have pinions of different sizes in this set, and because the driving wheels are so much larger than the spool, I'd have needed a large reduction anyway to get the surface speed ratio below 1:1.
I seem to have misplaced my Lego tubs, but I did have one unopened set, Lego Technic 9395 "Pick-up tow truck", that I got as a present from a friend. This limited the types of gears I had available. As the set has no chain elements, the best I could do was a thread spool and a single worm-pinion reduction. I would have used spur gear reduction gearbox if I had suitable gears; alas, this set does not have sufficient gears to do that. But it does have the worm and pinion.
To turn this into the original treadmill model I described, I would need a Lego chain, two sprockets for it, and replace the spool with a sprocket at the bottom of the trike. Two idlers for the chain on the trike, so that the chain makes an Ω-shaped loop around the sprocket, would help ensure the chain does not slip. If you don't want to run the chain vertically, you need to replace the worm and pinion with spur gear reduction instead. Remember, the reduction has to get *surface speed ratio* below 1:1 for stable running to be possible, but that does *not* limit the ratio of how fast the vehicle travels compared to the thread or treadmill speed (see \$v_c\$ and \$v_t\$ in my earlier post, and how they relate via \$\lambda\$, the surface speed ratio.)
The smaller the ratio, the easier it is to get movement without losing traction. Remember, *any* movement of the driving wheels moves the vehicle away from the direction you're pulling the thread or the treadmill surface travels, so something like a 1:10 reduction just makes sure you're not tripping on insufficient traction/stiction and such issues.
I do not think anyone should trust my word for it if they truly doubt this is possible (and nothing out of the ordinary, just a trivial mechanism).
This is why I show the CAD model of the exact vehicle I built and verified works as I described –– and Electrodacus claims is impossible ––: so that you can examine it and build it or your own version of it, and prove it yourself. There is absolutely no trickery here.
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Nominal Animal, you have fallen into the trap laid by Electrodacus. He is not interested in truth, he is only interested in persuading other people to spend (waste) their time making things or constructing things in an attempt to prove him wrong. If you spend your time on such fruitless endeavors, you are simply encouraging him. He is a sociopath, whose only interest is in manipulating others. He feels a sense of power and control when he gets other people to do what he wants.
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Nominal Animal, you have fallen into the trap laid by Electrodacus. He is not interested in truth, he is only interested in persuading other people to spend (waste) their time making things or constructing things in an attempt to prove him wrong. If you spend your time on such fruitless endeavors, you are simply encouraging him. He is a sociopath, whose only interest is in manipulating others. He feels a sense of power and control when he gets other people to do what he wants.
You have not answered me to the last few comments.
I asked if your model can show a vehicle with a 1:1 gear ratio moving 2 squares to the left while the treadmill is not moving at all.
If you agree with that let me know how a non moving treadmill can power the vehicle ?
If you disagree let me know how ?
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Trap or not, we have physical proof now.
Besides, the subject is something that others may wonder about.
It is one thing to argue with math and diagrams and even animations, but a completely different thing to show a model one can build and examine, and build the proof with their own hands.
I hate the idea of others stubling on this thread, and not having any way to determine the truth. I want to provide a way they can verify for themselves.
Plus, I like to show that when I provide some math, I can also provide physical examples to back it up. You don't need to trust me, but ignore me at your own peril.
As of right now, it seems that the cheapest way to obtain sufficient parts would be one 42133 set, and one or two 42132 sets. The first one has the worm and structural parts, the second has drivable wheels (but alas only one of each size) and even a chain (but alas a short one), so two sets of the second are needed for a robust trike model similar to mine, but using a chain. The sets cost 10€ apiece here right now, so 30€ total.
I won't be spending more effort on this, unless someone wants me to build the treadmill/chain model with build instructions using one 42133 and two 42132 sets (LeoCAD can be used to make Lego build instructions), and is willing to cover the price of the three sets for me, though. It might be an interesting hands-on device and experiment on why intuition might fail here (thinking that it would be impossible for the vehicle to travel in the opposite direction of the treadmill surface or chain or pulled thread), so might have some educational value.
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It is one thing to argue with math and diagrams and even animations, but a completely different thing to show a model one can build and examine, and build the proof with their own hands.
Someone built a model before, but Electrodacus simply rejected it and said it couldn't possibly work without including energy storage in the explanation.
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This is why I show the CAD model of the exact vehicle I built and verified works as I described –– and Electrodacus claims is impossible ––: so that you can examine it and build it or your own version of it, and prove it yourself. There is absolutely no trickery here.
I want to first mention that I appreciate your long description and I read it at least twice.
I do not need to build the vehicle you mentioned to know it will work and never claimed your device or mine works as shown in many videos.
The question is not if it works or not but how it works.
Both my simpler belt model and your use the same principle is just simple to explain how it works on the model that is easier to visualize.
It is not possible to get rid of energy storage so saying that the string you used is not elastic will not be true as all stings will have elasticity it will just be less elastic than the rubber belt I used.
It is also very likely that the string is not even the energy storage used for your vehicle or not the main one.
For example on the geared vehicle I used (gears instead of the belt) the actual frame of the vehicle was the part that stores most of the energy and you could see how the vehicle body deforms with a regular patterns as energy was charged and discharged.
So the question is the same and in the simplest form I can think of it looks like this
(http://electrodacus.com/temp/WindupL.png)
I did not added the force pairs related to gravity as they are not relevant.
In this case the applied force is F1 and all the resulting forces are added in the diagram.
Now if you prefer you can reverse the direction of the treadmill so F1 faces away from F2 to be more similar to your vehicle setup but nothing else will actually change.
So question is in what circumstances the F2 can be larger than F1 to allow the vehicle to be accelerated in the opposite direction the F1 is applied.
The only solution I can see is energy storage so it can be the orange string that is stretched or it can be the green body that is compressed or even a combination of both.
This storing of energy alone will not be sufficient as the energy should be discharged in order to move the vehicle and for that to happen something will need to slip and it is also important that there is a delta between the force needed to slip and the one needed to stick so that energy can be charged and discharged repeatedly.
I actually showed video evidence of what I'm describing but the critique was that vehicle is just not build great and that observed behavior is just some sort of coincidence and not needed.
One thing that can be done is to measure the force needed to move the vehicle and show that is exactly the force needed for the front wheel to slip.
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While this is an interesting topic (originally "downwind faster than the wind"), I am skipping this discussion because we have already discussed it to death with electrodacus here: https://www.eevblog.com/forum/chat/mess-with-your-minds-a-wind-powered-craft-going-faster-than-a-tail-wind-speed/ (https://www.eevblog.com/forum/chat/mess-with-your-minds-a-wind-powered-craft-going-faster-than-a-tail-wind-speed/) and here: https://www.eevblog.com/forum/chat/veritasium-wrong-about-faster-than-wind-direct-down-wind/ (https://www.eevblog.com/forum/chat/veritasium-wrong-about-faster-than-wind-direct-down-wind/). Don't bother trying to explain it to him, unless you enjoy repeatedly banging your head against a brick wall.
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It is one thing to argue with math and diagrams and even animations, but a completely different thing to show a model one can build and examine, and build the proof with their own hands.
Someone built a model before, but Electrodacus simply rejected it and said it couldn't possibly work without including energy storage in the explanation.
Oh, I missed that.
This one, with a worm and a pinion, eliminates the energy storage argument. The worm can turn the pinion, but the pinion cannot turn the worm at all: it immediately locks up. Any energy storage would therefore have to be in the spool axis. Because the spool stops turning immediately when you stop pulling on the thread (which also means the trike immediately stops), there is evidence of no energy storage.
Attached are couple of real-life photos of it. (I removed the front black thread guide block for these, so you can see the structure better, but otherwise it matches the LeoCAD model.)
Both my simpler belt model and your use the same principle
No, they absolutely do not.
Your model uses a gearing ratio which means no stable motion is possible.
This model uses a gearing ratio where stable motion is possible, and indeed, if you pull the thread at a constant velocity, the vehicle travels at a constant velocity as well. If you relax the thread, the motion stops immediately. There is no energy storage. Even the forward momentum of the vehicle itself does not and cannot act as a energy storage, because of the worm-pinion gearing: the forward momentum and wheels cannot feed back energy to the worm. Lego worm gear is steep, and immediately locks up if you try to back-drive the worm with the pinion gear.
Your model requires energy storage because of the inverse gearing ratio. Mine works completely without any kind of energy storage, for the reasons I described earlier.
It is not possible to get rid of energy storage
Yes, it is. I described how, and implemented it in real life. If you think about how a worm-pinion gearbox works, and demonstration shows the worm side (spool side) does not retain any energy, because as soon as you relax the thread, the spool stops turning, and that immediately stops the trike.
In particular, the vehicle does not move if you just repeatedly tug on the thread. You need to pull consistently, without any oscillation, to make the vehicle move.
Because the spool, worm, and pinion are in a very rigid structure above the driving wheel axis, any flex in the rest of the structure is irrelevant.
You cannot consider the moment of inertia in the wheels as energy storage, because it cannot be back-fed over the worm-pinion. It is just rotational inertia in the wheels. To prove this, switch to smaller, more lightweight wheels (thus less rotational inertia), and the vehicle will move even easier!
(The rotational inertia in the driving wheels means you must pull consistently, as dropping the pull rate even slightly will lead to lockup, because the wheels will try to drive the worm via the pinion, which leads to the worm-pinion locking up. With less rotational inertia, you can drop the pull rate slightly without lockup, because the wheels won't freewheel as much.)
we have already discussed it to death with electrodacus here: https://www.eevblog.com/forum/chat/mess-with-your-minds-a-wind-powered-craft-going-faster-than-a-tail-wind-speed/ (https://www.eevblog.com/forum/chat/mess-with-your-minds-a-wind-powered-craft-going-faster-than-a-tail-wind-speed/) and here: https://www.eevblog.com/forum/chat/veritasium-wrong-about-faster-than-wind-direct-down-wind/ (https://www.eevblog.com/forum/chat/veritasium-wrong-about-faster-than-wind-direct-down-wind/). Don't bother trying to explain it to him, unless you enjoy repeatedly banging your head against a brick wall.
Oh. Thanks. I avoided those threads, because I knew a physics view would be lost in the noise in such a thread. And the physics view is that whenever there is wind with respect to the vehicle, energy can be extracted; and no Earthly wind is stable enough for the switchover (when vehicle goes exactly at downwind speed) to last for any appreciable time, so it is just a matter of efficiency and reducing losses.
It seems I was indeed trapped in a fool's errand, because I did not realize this thread is just a continuation or restatement of those. :'(
Having played with Lego Technic extensively as a child, I discovered many non-intuitive powering mechanisms on my own, so I know how unintuitive many of them are; I spent many tens of hours just examining the mechanisms. Here, I just didn't associate the treadmill or "providing linear mechanical power opposite to the desired movement" with the faster than tail wind stuff at all. :palm:
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And the physics view is that whenever there is wind with respect to the vehicle, energy can be extracted
Indeed that is what is apparent at first glance, but it isn't actually true, at least not if you mean with respect to the body of the vehicle. What is actually needed is simply any reference (the ground in the actual vehicle) and a wind relative to that reference. The point at which the wind speed equals the vehicle body speed isn't really that special at all because the action of the wind on the body is not the main event.
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No, they absolutely do not.
Your model uses a gearing ratio which means no stable motion is possible.
Yes, it is. I described how, and implemented it in real life. If you think about how a worm-pinion gearbox works, and demonstration shows the worm side (spool side) does not retain any energy, because as soon as you relax the thread, the spool stops turning, and that immediately stops the trike.
In particular, the vehicle does not move if you just repeatedly tug on the thread. You need to pull consistently, without any oscillation, to make the vehicle move.
Your vehicle is in no way different in functioning principle.
The gear ratio while larger on your machine it is not different.
On my vehicle the wheels with the smaller pulley is the generator wheel (input) and in your case the input is the spool.
If you can measure the tension in the thread (a small strain gauge load cell connected to an oscilloscope) you will be able to see the charge discharge cycles.
It will obviously look smooth the same way the vehicle I showed looks smooth in real life no slow down video and a bit higher speed when the vehicle kinetic energy storage smooths things same way as adding a capacitor in an electrical circuit will do.
Look at this zoomed in gear and slowed down video 120FPS using my soldering microscope https://odysee.com/@dacustemp:8/120fps24:9
See at around second 9 in the video the gear changing direction for a few video frames imperceptible with the naked eye. That is the point the energy was discharged and that will repeat at regular intervals. The video is upside down but not relevant.
The point is that you are unable to explain how it works without energy storage and that is my main question. There are plenty of variants of this same vehicle and all of them work the same. For some strange reason you think that your vehicle is different than mine when it is not and no the gear ratio is not reversed as there is no such thing (you just rotate my vehicle 180 degree and it what you are asking witch is the same thing).
Also if I move the paper that is under the back wheels instead of the front the vehicle works exactly the same as it is irrelevant what surface moves relative to the other or in what direction.
In the case of your vehicle if you spool the thread in one direction you get the same vehicle I have and if you spool in the other direction you get the version I will have if I were to twist the belt like an infinity sign so no longer a locked gear and vehicle will move in the direction of the applied force.
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And the physics view is that whenever there is wind with respect to the vehicle, energy can be extracted
Yes energy can be extracted but you are confusing energy with power.
Power needed to drive against wind direction (upwind) is equal with the power wind already exert against the vehicle in the opposite direction.
So if based on vehicle area and wind speed 100W act against the vehicle then vehicle needs 100W + friction losses to drive at any speed against the wind direction.
So the only way to apply more than 100W is to store energy and then release it.
Driving at say 30m/s with no wind will have the same power required to overcome drag as driving at 10m/s in a 20m/s head wind.
So you agree with this above statement ? Because is not that is the the main disagreement that we have and this while not easy can be tested.
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A simplified example is a pair of wheels with almost as large a pulley in between, with a thread rotated on the pulley. What happens when you pull the thread?
If you start examination at the zero-movement state, you'll find that the angle at which the thread exits the spool, with respect to gravity down, largely determines whether it starts moving or not. If you pull the thread exactly level, the forces do not leverage the wheel to turn so that it would travel opposite to the direction the thread is being pulled to. If you pull the thread slightly upwards or downwards from level, the forces are no longer balanced, resulting a small torque on the wheel, so it'll start traveling opposite to the direction the thread is being pulled to. If you pull the thread nearly vertical, the torque is maximized, and it is easiest to get the wheel to travel (away from where you are pulling the thread)
Underline added by me.
I tried this experiment a while back with a roll of solder that had several layers used. The unused solder was about 5mm down from the cheeks of the plastic spool. With the solder extending parallel to the table top and coming from the underside of the roll you could pull the solder and it would roll up in a curious sort of way. It almost looked like an optical illusion. Something like you could pull the solder 20mm and the roll would move toward your hand 150mm, winding the solder back onto the roll as it went. Best results were when the extended solder was halfway between the two cheeks so the spool wouldn't pull to one side.
A roll of solder is a good choice because it is reasonably heavy so the edges of the spool can have a grip on the table top without slipping, and everybody who's anybody has a roll of solder to try it with.
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Something like you could pull the solder 20mm and the roll would move toward your hand 150mm, winding the solder back onto the roll as it went. Best results were when the extended solder was halfway between the two cheeks so the spool wouldn't pull to one side.
We are discussing about the opposite of your experiment. You pull the sting and the roll will go away from your hand.
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And the physics view is that whenever there is wind with respect to the vehicle, energy can be extracted
Indeed that is what is apparent at first glance, but it isn't actually true, at least not if you mean with respect to the body of the vehicle. What is actually needed is simply any reference (the ground in the actual vehicle) and a wind relative to that reference. The point at which the wind speed equals the vehicle body speed isn't really that special at all because the action of the wind on the body is not the main event.
Sure. If you don't want to limit to any particular mechanism, then "wind with respect to ground" is better and more accurate.
(I was thinking of devices without any energy storage, which are much more limited. My one-track mind and all. I now wish I had written "respect to the ground" instead.)
Even a simple mechanism, say a VAWT (vertical axis wind turbine), which extracts energy from wind in any direction perpendicular to its axis, can be made much more interesting by using say three of them rotating around a common vertical axis. As long as the common axis rotates, all three experience a different relative wind speed and direction (relative to its own axis, that is) –– unless the wind is actually a vortex around that common vertical axis.
(In this case the rotation around the common vertical axis can be considered energy storage, but it is a consequence of the mechanism.)
Such mechanisms are very interesting to model and build, but I agree, it is not useful to discuss their theoretical properties in a forum like this: the signal to noise ratio is too low, with noise generated by well-meaning people basing their argument on intuition and experience with dissimilar devices, instead of physics.
As to the simplest possible example of the pull-thread-thing-moves-the-other-way, just put a spool on an axis, and make two straight rails for the axis. If you pull the thread horizontally below the rails, the pulley will go in the opposite direction. Just make sure the axle and rails have sufficient friction, so that the axis rotates instead of sliding.
With a heavy spool, the friction of the spool to the ground is so good that instead of sliding, pulling on the thread (or solder wire) so that it comes out from below, will actually cause the spool to rotate towards you, even when that causes the spool to rewind itself; and the axis of the spool will travel faster than you are pulling on the thread/solder. This is what Circlotron observed.
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And here is the simplest, most intuitive treadmill example I can think of, using Lego Technic.
If you rename the attached treadmill.txt to treadmill.ldr, it too can be opened in LeoCAD. I omitted two rack pieces, and cut the long holey bar short, so you can see what is happening better.
If you turn the blue handle clockwise, the upper surface of the treadmill chain goes right, but the gearwheel rotates counterclockwise and moves left.
If you turn the blue handle counterclockwise, the upper surface of the treadmill chain goes left, and the gearwheel moves right, rotating clockwise.
This is not a "locked gearbox", because the geared wheel can move horizontally.
The racks and chain are not necessary for correct operation, they just provide lots of friction. The same works even with thread and smooth wheels, as long as you have sufficient friction that nothing slips.
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This is not a "locked gearbox", because the geared wheel can move horizontally.
The racks and chain are not necessary for correct operation, they just provide lots of friction. The same works even with thread and smooth wheels, as long as you have sufficient friction that nothing slips.
It is a locked gear and I already did the experiment see video https://odysee.com/@dacustemp:8/gear-slow30p2:9
And here is the slow motion and zoomed bit unfortunately is upside down but around second 9 you can see the energy stored is released https://odysee.com/@dacustemp:8/120fps24:9
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No, electrodacus. You're letting your preconceptions override your search for the facts.
This is the complete model of the situation:
(https://www.nominal-animal.net/answers/spool.svg)
Let t be the surface velocity of the treadmill, positive right.
Let c be the velocity of the center of the gears, positive left.
Let R be the radius of the treadmill gear (blueish), and r the ground gear (pale yellow-orange).
Finally, let φ be the angular velocity of the gears.
For the ground gear to not slip, we need φ=c/r.
For the treadmill gear to not slip, we need φ=(c+t)/R.
Combining the two rules, we have an equation that holds whenever the gears do not slip: c/r=(c+t)/R. Solving this for c yields c=t*r/(R-r).
The only impossible situation is when the two gears are the exact same size. This is the locked-up gearbox case, where non-slip motion is not possible.
In the case where r<R, c will be positive, and the center of the gears will move in the opposite direction compared to the surface of the treadmill.
Circlotron observed the situation when R<r, about R≃0.867r. Then, c=t*(-7.5). Because c is negative, it travels in the same direction the surface of the treadmill travels (or equivalently the wire is pulled). Because the magnitude of the ratio is so large, the forces are such that you need spool to have excellent traction (stiction, static friction) to ground: a heavy soldering wire spool is an excellent test case.
Then, when you pull the wire towards yourself, the spool will also rotate towards yourself but much faster. When you pull the wire 20mm, the spool will travel 20mm×7.5 ≃ 150mm, just as Circlotron described. (I wonder if circlotron agrees that the diameter of amount of solder in their spool was about 0.867 of the outer diameter of the spool? In any case, if you happen to have a heavy spool yourself, you can easily check the math here.)
Although I called t and c and φ velocities, the math stands exactly the same if you consider them displacements instead. That is, when the treadmill surface moves right by t (left if negative), the angle of the axis of the gears changes by φ, and the spool/axis of the gears moves left by c (right if negative).
There is no energy storage needed. You can start from a standstill, move the treadmill surface by a fixed amount, and measure how far the spool/axis of the gears moved. If the spool/gears were heavy enough with enough friction/traction/stiction, so that there was no slippage, the above formulae will hold.
There is no strangeness related to energy conservation either. If you do the heavy almost-full spool test, you'll find that it is quite hard to pull the wire. In other words, it is the treadmill that provides all the energy here, at every instant in time. It will all be spent in the friction/traction/stiction, if you do the test from standstill to standstill. All pure mechanics, no slapstick, no aether, no fancy theories. Plain ol' classical mechanics here.
Sizes and ratios do matter in practice, though. For example, if your heavy spool is 99% full (meaning, the diameter or radius of the wire in it is 99% of the diameter or radius of its outer edges), the ratio is 1/(0.99-1) = -100. This means that every millimeter you manage to pull the wire, the spool will travel 100 mm. It is unlikely that there is enough friction to see this happen; instead, the spool will slip. So, to see the phenomena better, use a spool with somewhat less wire.
I so wish BigClive would try this. He's got good cameras, nice bench setup, and suitable spools at the top of his shelves. Or maybe Dave would?
Me and cameras don't mix too well. It does look funky, and is a perfect example of how our intuition can lead us astray, which is the reason I answered to this thread in the first place.
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I so wish BigClive would try this. He's got good cameras, nice bench setup, and suitable spools at the top of his shelves. Or maybe Dave would?
There's really no point.
You know that thing children do, when they are playing a game and about to lose, they find some way of cheating so they can try to avoid the outcome? Electrodacus is playing that game here. There is no evidence you can provide, no experiment anyone can perform that will persuade him, because he will just come up with some kind of nonsensical word salad to dispute the result.
Notice how he never does any analysis himself, never shows any equations, but always tries to make other people do the work? It's a game for him, trying to make people jump to his command, and then getting satisfaction from the "power" that gives him. As I said above, this has all the signs of sociopathic behavior. It is really best not to enable it.
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No, electrodacus. You're letting your preconceptions override your search for the facts.
This is the complete model of the situation:
Let t be the surface velocity of the treadmill, positive right.
Let c be the velocity of the center of the gears, positive left.
Let R be the radius of the treadmill gear (blueish), and r the ground gear (pale yellow-orange).
Finally, let φ be the angular velocity of the gears.
I did not quote everything just to keep it readable but I read all your repay in details.
I mentioned before but while this mechanism looks simpler is more complex than the one with the belt.
When R=r the gear is locked as you already mentioned.
When R>r the gear is still locked as nothing was changed just the size of the gears.
You can have a working gear box with a gear ratio of 1:1 so the gear ratio is not what makes a gearbox locked or unlocked.
Set the speed of the player at 0.25x and pay close attention to what happens https://odysee.com/@dacustemp:8/gear-slow30p2:9
It is irrelevant witch of the two surfaces move the system will work the same way.
The reason it works has to do with the shape of the tooths on most gear as they allow the gear assembly to lift up when you apply a force.
So energy storage in this case is gravitational the wheel is lifted when charging then falls back when discharged.
If you make custom gear's with the shape of the tooth so that it will not allow this lifting of the gear you can eliminate the energy storage and you will no longer be able to move it.
With typical gears as the ones in my video the horizontal applied forces allow the gear to lift up and thus store potential gravitational energy.
If you push a triangle against an upside down triangle horizontally the triangles will slip against each other and lift up. So charge the shape of the tooth's to eliminate this and you will see that it no longer works.
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I so wish BigClive would try this. He's got good cameras, nice bench setup, and suitable spools at the top of his shelves. Or maybe Dave would?
There's really no point.
You misunderstand. I meant that as a short video, it would be interesting, exactly because with sufficient weight and thus friction, the spool does behave unintuitively. Moreso because we can't "see" how hard one must tug on the wire to get it to move; it looks unintuitive.
Like the trick with rails that become wider as they go up, and how a suitable double cone seems to roll uphill along such rails. Or how you can accidentally power an IC via an I/O pin.
Electrodacus is playing that game here. There is no evidence you can provide, no experiment anyone can perform that will persuade him, because he will just come up with some kind of nonsensical word salad to dispute the result.
True; I've lost all hope of being able to help electrodacus here.
Yet, there is still a possibility that someone else reading this thread –– say, arriving here via a web search –– might start thinking about whether they just assume things because they're intuitive, and learn to question their intuition, and maybe even how to find out for themselves. If that happens, all my effort has not been in vain.. My last couple of posts have been trying to round the topic up in case that does happen.
If I were still believing I might be able to help electrodacus see, I would have added another figure that contains the force vectors. The key vectors would obviously be the two torques around the gear axis, because once one realizes their importance on how this system works, everything else including friction/stiction/traction becomes obvious and straightforward. But no, I haven't drawn such an image, and will not. Anyone truly interested in the subject can use any (classical) mechanics 101 book, and draw this themselves, and work it all out. Or indeed grab some Technic Lego, and build working models, and compare their behaviour to what I described.
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There's really no point.
You know that thing children do, when they are playing a game and about to lose, they find some way of cheating so they can try to avoid the outcome? Electrodacus is playing that game here. There is no evidence you can provide, no experiment anyone can perform that will persuade him, because he will just come up with some kind of nonsensical word salad to dispute the result.
Notice how he never does any analysis himself, never shows any equations, but always tries to make other people do the work? It's a game for him, trying to make people jump to his command, and then getting satisfaction from the "power" that gives him. As I said above, this has all the signs of sociopathic behavior. It is really best not to enable it.
Just put him on your ignore list, one less source of noise.
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If I were still believing I might be able to help electrodacus see, I would have added another figure that contains the force vectors. The key vectors would obviously be the two torques around the gear axis, because once one realizes their importance on how this system works, everything else including friction/stiction/traction becomes obvious and straightforward. But no, I haven't drawn such an image, and will not. Anyone truly interested in the subject can use any (classical) mechanics 101 book, and draw this themselves, and work it all out. Or indeed grab some Technic Lego, and build working models, and compare their behaviour to what I described.
The interesting and important thing about the analysis of kinematic structures, is that forces or force vectors are not required in the analysis, in fact trying to use forces just makes the system harder to understand.
You have a system of rigid elements connected at various points by pivots or hinges, or in the case of wheels or gears the contact points. Since all the elements are rigid, there are equations involving only (x, y) coordinates that relate all the points in the system. All you have to do is to move one point, and all the other points will move in unison as governed by the connections. It is a pure problem of spatial geometry, with no force, energy, power or momentum involved in the analysis. (For example, consider a pantograph.)
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It is a pure problem of spatial geometry, with no force, energy, power or momentum involved in the analysis.
That is your problem. It is not just about geometry.
With that even 1:1 gear will work fine.
A 1:1 will not work even in theory and close to 1:1 will not work in practice due to losses. The higher gear ratios work due to energy storage but you can not know that if all you look is geometry.
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Why is this started up again, electrodacus? No-one can change your mind, and you can't change the laws of physics or the opinions of anyone not you. That's been amply demonstrated in more than one lengthy thread previously.
So... why did you start this again? The only reason I could think of if someone else had done so is because they still couldn't solve the problem and wanted some help to clear up some points. But you don't - you know what you know and nothing is going to change that. Even if there was something you were unsure of, you wouldn't accept what anyone explained or demonstrated anyway unless it fit your already decided solution.
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If I were still believing I might be able to help electrodacus see, I would have added another figure that contains the force vectors. The key vectors would obviously be the two torques around the gear axis, because once one realizes their importance on how this system works, everything else including friction/stiction/traction becomes obvious and straightforward. But no, I haven't drawn such an image, and will not. Anyone truly interested in the subject can use any (classical) mechanics 101 book, and draw this themselves, and work it all out. Or indeed grab some Technic Lego, and build working models, and compare their behaviour to what I described.
The interesting and important thing about the analysis of kinematic structures, is that forces or force vectors are not required in the analysis, in fact trying to use forces just makes the system harder to understand.
I know and agree. If you look at my posts in this thread, I've said so myself, and instead described the system behaviour in terms of surface velocities or displacements in the math snippets.
If we had managed to progress to the "okay, I now see how this behaves so; but when and how does the slippage I saw occur? And why didn't my original approach describe the system?" stage, then we could have moved on to the forces. Not to describe the behaviour of the system, but to describe why any simplified force-based description fails. It is a pedagogical pattern that has often worked well: you look at the situation from a completely different approach, and only if/when you finally grasp the behaviour, you close the loop by examining how and why the original approach failed.
Here, you would only care about the forces if you wanted to find out the limits where slippage occurs; the coefficients of friction and system weight that are required for no-slip operation at a given gear ratio. Starting from that is silly.
Where the insistence of energy storage being involved here stems from, I can only imagine. There is no need for one. The same math is valid both at standstill and at any velocity, continuously. Even if one were to examine the forces, one would see they balance perfectly at each instance; with the only "energy storage" being the linear and angular momentum in the "car" or "spool", both perfectly aligned in the direction of travel. No vertical motion at all, unless slippage occurs; and the systems I've described are explicitly in the no-slip cases. You can make all sorts of things "work" if they slip a bit, but these ones do not need slippage or energy storage to behave as I've described.
:-//
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Why is this started up again, electrodacus? No-one can change your mind, and you can't change the laws of physics or the opinions of anyone not you. That's been amply demonstrated in more than one lengthy thread previously.
So... why did you start this again? The only reason I could think of if someone else had done so is because they still couldn't solve the problem and wanted some help to clear up some points. But you don't - you know what you know and nothing is going to change that. Even if there was something you were unsure of, you wouldn't accept what anyone explained or demonstrated anyway unless it fit your already decided solution.
I need to share the world with all of you. So it is in my interest that people understand how the world works as I depend on that.
What you (all) are claiming violates energy conservation so since that is not possible the explanation is wrong.
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Where the insistence of energy storage being involved here stems from, I can only imagine. There is no need for one. The same math is valid both at standstill and at any velocity, continuously. Even if one were to examine the forces, one would see they balance perfectly at each instance; with the only "energy storage" being the linear and angular momentum in the "car" or "spool", both perfectly aligned in the direction of travel. No vertical motion at all, unless slippage occurs; and the systems I've described are explicitly in the no-slip cases. You can make all sorts of things "work" if they slip a bit, but these ones do not need slippage or energy storage to behave as I've described.
:-//
I guess you did not read my reply to your latest comment.
There is slip with gears also due to the gear shapes you are pushing an upside-down triangle on to another triangle so the two triangle will slip past each other lifting the gear that in this particular case is representing the vehicle.
So there is both slip and energy storage in the gear vehicle example.
And if you want to know about why energy conservation is relevant here then let me explain.
Any gearbox (witch is what this vehicles are) have an input and an output. The output power of the gearbox can not be higher than the input power and this is what you are all claiming without realizing you do. The energy storage and stick slip hysteresis is necessary for this to work while not violating the energy conservation law.
Fact:
Power needed to overcome drag for a vehicle driving at 30m/s with no wind is exactly the same as for the same vehicle driving at 10m/s in to a 20m/s head wind.
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Fact:
Power needed to overcome drag for a vehicle driving at 30m/s with no wind is exactly the same as for the same vehicle driving at 10m/s in to a 20m/s head wind.
And 40m/s with a 10m/s tailwind, right?
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Fact:
Power needed to overcome drag for a vehicle driving at 30m/s with no wind is exactly the same as for the same vehicle driving at 10m/s in to a 20m/s head wind.
And 40m/s with a 10m/s tailwind, right?
Yes.
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Where the insistence of energy storage being involved here stems from, I can only imagine.
:-//
My (admittedly cynical) guess is a desire to jump on the novel energy storage solution/harvesting bandwagon...
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Where the insistence of energy storage being involved here stems from, I can only imagine.
Here's the genesis:
* Because of electro's unwillingness to properly consider frames of reference, he believes that it is impossible for a wind-powered vehicle travel directly downwind FTTW (Faster Than The Wind).
* So, he presumes that energy used for this physically demonstrated FTTW speed has been stored as pressure behind the vehicle while it was travelling slower TTW.
* When presented with simple physical or conceptual models that demonstrate analogous behavior he clings to the "energy storage" model, invoking things like "slip-stick hysteresis". (this does exist in some cases, but is irrelevant here).
So it all goes back to the original propeller/wheel vehicle that violates electro's misunderstood laws of physics.
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Here's the genesis:
* Because of electro's unwillingness to properly consider frames of reference, he believes that it is impossible for a wind-powered vehicle travel directly downwind FTTW (Faster Than The Wind).
* So, he presumes that energy used for this physically demonstrated FTTW speed has been stored as pressure behind the vehicle while it was travelling slower TTW.
* When presented with simple physical or conceptual models that demonstrate analogous behavior he clings to the "energy storage" model, invoking things like "slip-stick hysteresis". (this does exist in some cases, but is irrelevant here).
So it all goes back to the original propeller/wheel vehicle that violates electro's misunderstood laws of physics.
You're confusing things.
There are two vehicles
a) the direct upwind discussed in this thread as it should be simpler to explain and has analog version using just wheels and this one uses small capacity energy storage and stick slip hysteresis to trigger when energy is being charged and discharged.
b) the direct downwind faster than wind not discussed here that uses a large capacity energy storage called pressure differential and created by the propeller.
The only way wind interacts with any vehicle is through collisions between air particles and vehicle and depending on direction vehicle gains or loses kinetic energy.
That is why the equation for wind power has no variable related to the vehicle design so all wind powered vehicles have access to the same amount of wind power.
WindPower = 0.5 * air density * (area * coefficient of drag) * (wind speed - vehicle speed)3
This above equation is universal for any wind powered vehicle so for direct downwind equation shows there is zero wind power available when vehicle speed equals wind speed.
And for direct upwind discussed in this thread the equation shows the power that winds acts against the vehicle as well as the theoretical max power available to vehicle and so since this two are ideal case equal the vehicle can not move unless it first store energy then uses that stored energy to move a little bit against the wind direction and then repeat this cycle as many times as you want.
So that equation is the only prove I need to show that any vehicle powered only by wind requires energy storage in order to drive faster than wind directly downwind or at any speed directly upwind.
If you want to contradict what I say you will need to provide a wind power equation that matches what is observed and measured in reality.
And my best example of what happens and it is measured in reality was the one about the power needed to overcome drag (same equation) is the same for a vehicle driving at 30m/s with no wind as it is for the same vehicle driving at 10m/s with a 20m/s head wind.
The basic physics law that you try to violate is Newton's third law the subject of this thread and by extension the energy conservation.
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Never been sailing? You can tack into the wind, effectively using wind power to propel you in the direction the wind is coming from.
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Never been sailing? You can tack into the wind, effectively using wind power to propel you in the direction the wind is coming from.
I have never been on a sailboat but I can explain how it works.
A sailboat can not sail directly upwind but it can sail at an angle to gain kinetic energy then rotate to direct upwind and use that kinetic energy to sail for a small amount of time upwind then it will need to change direction again to recharge the kinetic energy.
So it can take advantage of kinetic energy storage if it is allowed to change direction but if that is not allowed it can not sail directly upwind.
A vehicle perpendicular to wind direction will always have access to max wind power no matter the vehicle speed as wind speed relative to vehicle is always the same.
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Never been sailing? You can tack into the wind, effectively using wind power to propel you in the direction the wind is coming from.
And if you're using a hydrofoil, you can do it faster than the wind is blowing at you. Wrong but true. Google America's Cup. On the subject of friction reduction, is anyone using the downward force of gravity g in their equations? Does this thing still work in microgravity?
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And if you're using a hydrofoil, you can do it faster than the wind is blowing at you. Wrong but true. Google America's Cup. On the subject of friction reduction, is anyone using the downward force of gravity g in their equations? Does this thing still work in microgravity?
Yes you can because you are allowed to change direction and thus take advantage of the vehicle kinetic energy.
An ideal setup where you sail perpendicular to wind direction can get the vehicle at any speed say 3x the wind speed then all you need to do is change direction to either direct upwind or direct downwind and since there is no friction loss in an ideal set up maintain that 3x speed forever.
In real world you can do the same but you need to repeat the change in direction as the stored kinetic energy will be used up to counter the losses.
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And my best example of what happens and it is measured in reality was the one about the power needed to overcome drag (same equation) is the same for a vehicle driving at 30m/s with no wind as it is for the same vehicle driving at 10m/s with a 20m/s head wind.
And by extension a vehicle traveling at 0 m/s with a 30 m/s head wind...
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And by extension a vehicle traveling at 0 m/s with a 30 m/s head wind...
Exactly right.
There is potential power available just not used by vehicle if it is anchored to earth but if it is not then that power will be accelerating that vehicle in the same direction as the wind.
So after 1ms of this the vehicle speed will no longer be zero but some value that can be calculated based on vehicle weight and friction losses as you calculate the gained kinetic energy and from that you can find out the new vehicle speed.
You will need to integrate as the wind speed relative to vehicle changes as the vehicle speed increases.
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So it all goes back to the original propeller/wheel vehicle that violates electro's misunderstood laws of physics.
I see. Thanks for letting me know!
This reminds me of the simple thought experiment on conservation of momentum and kinetic energy in elastic collisions.
Let's say you have a spaceship of mass M traveling at velocity V. There is a projectile of mass m and velocity v on the same trajectory (v > V, both in the same exact direction). They impact, but elastically, so that neither deforms, they just bounce without any losses. What are the resulting velocities V' and v'?
Conservation of momentum says that MV + mv = MV' + mv'. In a perfectly elastic collision, kinetic energy is also conserved, MV^2/2 + mv^2/2 = MV'^2/2 + mv'^2/2. Solving the system of two equations for V' and v' yields two answers: one is V'=V, v'=v, i.e. no change. The other is V' = (2mv+MV-mV)/(m+M), v' = (2MV+mv-Mv)/(m+M). (Feel free to check, e.g. here (https://en.wikipedia.org/wiki/Conservation_of_momentum#Application_to_collisions).)
What happens when the projectile goes twice as fast as the ship, and weighs twice as much as the ship, i.e. m=2M, v=2V?
You work out the math, and out comes the unintuitive but physically correct and easily verifiable (using e.g. an air track) V'=7/3V≃2.333V, v'=4/3V≃1.333V.
In other words, the projectile loses one third of its velocity, and the ship gains four thirds; and the ship will end up traveling faster than the incoming projectile originally was.
If the velocities are a significant fraction of light speed, then one needs to switch to generalized momentum (https://en.wikipedia.org/wiki/Conservation_of_momentum#Lorentz_invariance), (Newtonian momentum multiplied by the Lorentz factor γ, p = γmv) and relativistic kinetic energy (https://en.wikipedia.org/wiki/Kinetic_energy#Relativistic_kinetic_energy) (E=(1-γ)mc²) that are conserved, but at v<<c, the two yield the same answer to within rounding error.
This also answers the question, "Can you accelerate a spaceship to a velocity higher than at which you can lob boulders at it?", with "Yes. Just use boulders with more mass than the ship has."
This is also the reason why one wants solar sails to be reflective, and not absorb the photons. If the solar sail absorbs the photon, the craft gains the momentum of the photon. However, if the solar sail reflects the photon, the craft gains up to two times the momentum of the original photon, depending on the angle of reflection, with maximum achieved when the photon is reflected back the way it came from.
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This reminds me of the simple thought experiment on conservation of momentum and kinetic energy in elastic collisions.
Elastic collisions is what you have between the air particles and the vehicle and that is how kinetic energy is transferred from the air molecules to the vehicle.
All the kinetic energy of the air molecule will end up transferred to the vehicle because it will in average bounce back and forth between the air molecule in the back and the vehicle body.
That is how you get to the wind power equation that I always mention as best case wind power available to any wind powered vehicle.
In that equation you have the air density about 1.2kg/m^2 the aerodynamic drag and projected surface area of the vehicle and the vehicle and wind speed.
Nothing else is needed to know what is available in ideal case to accelerate the vehicle.
So wind powered vehicle can be wind powered only in the stationary to wind speed region in the same direction.
You have the max available wind power when wind speed relative to vehicle is highest so stationary or traveling perpendicular to wind direction.
And zero wind power is available for a vehicle with any design traveling at the same speed as wind since air molecules can no longer collide with the vehicle to provide the increase in kinetic energy.
Vehicle can take energy from the wheel but that will result in a proportional reduction in vehicle kinetic energy so speed and appling that energy extracted at the wheel to another wheel or propeller or any other form of propulsion can not give a net gain even in ideal case.
Thus the only option to exceed wind speed directly down wind or drive directly upwind is to use energy storage.
Exceeding wind speed at an angle to wind direction is possible because there is still wind relative to vehicle and the only limitation when perpendicular to wind direction are the frictional losses.
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Elastic collisions is what you have between the air particles and the vehicle and that is how kinetic energy is transferred from the air molecules to the vehicle.
Only if the vehicle is basically a stiff box, and has no mechanism at all.
Consider autogyro (https://en.wikipedia.org/wiki/Autogyro). If what you claimed was correct, it could not fly. Yet, it does. Lift is provided by freely rotating rotor blades, due to a pressure differential between different sides of the blade. The rotation of the blades is caused by perpendicular airflow below the rotor, generated by a fore or aft propeller engine.
Things like airfoils (https://en.wikipedia.org/wiki/Airfoil) (aeroplane wings, wings in vertical wind turbines, etc.) are not elastic collisions, as the pressure varies around the airflow. Air is a compressible fluid, and has properties which can be exploited in a mechanism, for example via by conversion between different forms of energy (which means it behaves non-elastically). Linear momentum is not conserved, but kinetic energy is, if and only if we include temperature (thermal energy) as part of the kinetic energy.
A particular example of conversion between different forms of energy in a compressible fluid (air) is de Laval nozzle (https://en.wikipedia.org/wiki/De_Laval_nozzle) in supersonic engines, converting thermal energy to kinetic energy.
The maximum speed a vehicle can attain depends on the losses. To simplify, the power (energy transferred or converted per unit time) used matches the power lost, when the vehicle travels at its maximum velocity. However, the losses can be made arbitrarily small. (As an example, if you dimple your car's surface like a golfball, you'll reduce aerodynamic drag significantly, and increase fuel efficiency, as demonstrated famously by MythBusters.) So, it is just a question of how to harvest energy from the wind, using a platform traveling downwind at velocities below and above the windspeed. It is just an engineering problem, really, hiding inside a scenario that seems counterintuitive to most people.
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The maximum speed a vehicle can attain depends on the losses.
Only true for a vehicle traveling perpendicular to wind direction.
For particular cases of interest here:
Direct downwind limit is wind speed.
Direct upwind limit is zero.
Both direct downwind faster than wind and direct upwind is possible using energy storage.
For direct downwind as soon as you are above wind speed there is no longer any wind power available so you can drive above that speed only for as long as the stored energy permits.
For direct upwind wind the same is true is just that due to fast cycles of charge and discharge effect is not visible in most cases without slow motion video.
To simplify, the power (energy transferred or converted per unit time) used matches the power lost, when the vehicle travels at its maximum velocity. However, the losses can be made arbitrarily small. (As an example, if you dimple your car's surface like a golfball, you'll reduce aerodynamic drag significantly, and increase fuel efficiency, as demonstrated famously by MythBusters.) So, it is just a question of how to harvest energy from the wind, using a platform traveling downwind at velocities below and above the windspeed. It is just an engineering problem, really, hiding inside a scenario that seems counterintuitive to most people.
The only way the vehicle is powered by wind is by air particles colliding and providing their kinetic energy to vehicle.
So to get a real world example we can take a Tesla model 3
First google search say Cd = 0.23 and frontal area 2.22m2
So if vehicle is stationary with the front facing the wind (so upwind) and wind speed is say a constant 30m/s (maybe a wind tunnel or something that can provide this perfectly constant air flow).
Then vehicle will experience a wind power trying to push vehicle backwards of
Pw = 0.5 * 1.2 * 0.23 * 2.22 * 303 = 8272W
So this is both the power that wind wants to accelerate the vehicle backwards (as front faces the wind) so the max theoretical available wind power and the power the vehicle will need if it wants to stay in place without the brakes engaged.
So if the vehicle wants to drive at 1m/s upwind it will need the following amount of power
P = 0.5 * 1.2 * 0.23 * 2.22 * (30+1)3 = 9127W
Hope you see that it is impossible for a vehicle even ideal one with no friction losses to drive upwind at any speed as power out can not be higher than power in.
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[...]
Direct upwind limit is zero.
Both direct downwind faster than wind and direct upwind is possible using energy storage.
For direct downwind as soon as you are above wind speed there is no longer any wind power available so you can drive above that speed only for as long as the stored energy permits.
For direct upwind wind the same is true is just that due to fast cycles of charge and discharge effect is not visible in most cases without slow motion video.
Hope you see that it is impossible for a vehicle even ideal one with no friction losses to drive upwind at any speed as power out can not be higher than power in.
And with this you have completely lost the plot. Direct upwind is possible, has been demonstrated in the real world, and requires no energy storage, no "charge and discharge cycle". You have not learned a thing from these tedious discussions.
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And with this you have completely lost the plot. Direct upwind is possible, has been demonstrated in the real world, and requires no energy storage, no "charge and discharge cycle". You have not learned a thing from these tedious discussions.
Of course direct upwind is possible using energy storage.
The explanation of how it works is wrong nobody is denying the experiments.
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Of course direct upwind is possible using energy storage.
The explanation of how it works is wrong nobody is denying the experiments.
And if you average the upwind movement over thousands or millions of cycles, then at the limit it becomes continuous upwind movement, which can last forever, without stopping.
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And if you average the upwind movement over thousands or millions of cycles, then at the limit it becomes continuous upwind movement, which can last forever, without stopping.
It is never a continues upwind movement even if it will look like that the same way as a youtube video looks continues to you even if is just 30 or 60 separate frames.
The point is that power that wind pushes against the vehicle even if you capture all of that at 100% efficiency is not enough to advance against wind direction so you need to wait while you charge a small energy storage device then use that stored energy to advance a tiny amount against the wind direction.
When you sail upwind this cycles are much larger so that you can actually see them as you actually drive at an angle where you can using directly wind power then when you picked up enough speed (so increased the vehicle kinetic energy) you can rotate even direct upwind and you will advance in that direction while the speed and thus kinetic energy drops.
Direct down wind will also be visible if you did the test for long enough to see how acceleration rate drops as stored energy is used up then see how vehicle slows down all the way below wind speed where you can recharge and start the cycle again.
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And if you average the upwind movement over thousands or millions of cycles, then at the limit it becomes continuous upwind movement, which can last forever, without stopping.
It is never a continues upwind movement even if it will look like that the same way as a youtube video looks continues to you even if is just 30 or 60 separate frames.
The point is that power that wind pushes against the vehicle even if you capture all of that at 100% efficiency is not enough to advance against wind direction so you need to wait while you charge a small energy storage device then use that stored energy to advance a tiny amount against the wind direction.
And yet, the net motion is still upwind. How can this possibly be happening, given your "stored energy" theory??? (hint: You are wrong.)
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You have not learned a thing from these tedious discussions.
Apparently some of us haven't either.... :-DD
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And yet, the net motion is still upwind. How can this possibly be happening, given your "stored energy" theory??? (hint: You are wrong.)
If you are not moving wait for 1 minute to charge your energy storage then move at any low speed upwind for even a second will you not have an average upwind speed higher than zero ?
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You have not learned a thing from these tedious discussions.
Apparently some of us haven't either.... :-DD
You're right. Although I can quit anytime I like... (he said)
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You guys are arguing with either a troll or an idiot, possibly both.
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The only way the vehicle is powered by wind is by air particles colliding and providing their kinetic energy to vehicle.
Wrong. You'll never get how it works whilst you persist on thinking this.
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You guys are arguing with either a troll or an idiot, possibly both.
Harsh dude 8)
Maybe these wikis will help OP:
https://en.m.wikipedia.org/wiki/Fluid_dynamics
https://en.m.wikipedia.org/wiki/Radiation_pressure
https://en.m.wikipedia.org/wiki/Yarkovsky_effect
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You guys are arguing with either a troll or an idiot, possibly both.
I wish I was a troll then I will have enjoyed this.
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The only way the vehicle is powered by wind is by air particles colliding and providing their kinetic energy to vehicle.
Wrong. You'll never get how it works whilst you persist on thinking this.
Please enlighten me.
How do air particle transfer energy to vehicle if not through collision ?
And as you claim to know how it works please also provide the equation describing the amount of wind power available to vehicle.
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Harsh dude 8)
Maybe these wikis will help OP:
https://en.m.wikipedia.org/wiki/Fluid_dynamics
https://en.m.wikipedia.org/wiki/Radiation_pressure
https://en.m.wikipedia.org/wiki/Yarkovsky_effect
Quote from your first link:
"Fluids are composed of molecules that collide with one another and solid objects."
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The only way the vehicle is powered by wind is by air particles colliding and providing their kinetic energy to vehicle.
Wrong. You'll never get how it works whilst you persist on thinking this.
Please enlighten me.
How do air particle transfer energy to vehicle if not through collision ?
And as you claim to know how it works please also provide the equation describing the amount of wind power available to vehicle.
In the previous threads I tried stepping you though it very simply (and it was tough keeping you ontrack and not racing ahead down distraction rabbit holes), but once you saw where it was going you disappeared. Why would this time be any different?
Nevertheless, your problem is starting with the wind power first and then trying to apply that to obvious stuff. Forget where the power comes from for the moment and look at how it's actually working, then figure how to power it.
First, you have a propeller which is pushing the vehicle along. We know that works because jets take off - if the prop is turning then it is pushing the vehicle even if ever so slightly. It's pushing against the prevalent wind, and you even agreed earlier in this thread that this kind of thing is additive. If the prop can push a 5m/s and the wind is 10m/s then the vehicle will go at 15m/s.
That's uncontestable physics and fact, but feel free to disagree if you dare.
So, what we're stuck for is the power that drives the prop. We know it is coming from the wheels (that is, the wheels drive the prop and not vice versa), and the wheels get it from the vehicle being pushed by the wind. But... the vehicle is going faster than the wind so what does the wind push on? It is the backwash from the propeller - remember the 5m/s plus whatever the prevalent wind is, which is this case is maybe -2m/s, so a small forward movement for a larger backwards jet.
Hang on, isn't this free energy or something? No, because of drag, which will require more power to overcome, and loss of power as more is extracted. Just think about it for a moment: the wind is pushing the vehicle (yes, due to air bouncing off it) at almost wind speed and there is no drag. All that's required is for the prop to turn however, slowly, and produce a minute amount of thrust which will increase the vehicle speed a tiny amount and it's game on, but go a bit faster and you need more power to overcome the drag and similar losses. But if you have drag you are going faster than the wind. Can the wheels extra that tiny amount of power without slowing the vehicle? That's where leverage, size of prop and angle of blades come in.
Equations? Dunno. I'm not clever enough to work one out that you would reject or ignore anyway, and I doubt there is one on t'web to copy like you do. But I can point to video of the actual vehicle and many models working continuously and without any energy storage facilities.
Edit: forgot to point out that the wind power increases as the vehicle slows. There is far more power available than is consumed to push the vehicle forward - it's just that there is nothing to suck it up normally so the vehicle accelerates. At wind speed, there is 0 power pushing the car forward, but all the power of the wind is still available if you can work out how to access it. That's where the wheels are getting apparently free power from.
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How do air particle transfer energy to vehicle if not through collision ?
Modeling the macro-scale properties of compressible fluids as colliding solid particles is just not tenable. I know, because my proper field is molecular dynamics simulations.
In particular, a molecule does not have a well-determined surface at all; there is just a "shroud" of delocalized outermost electrons, and we assign an arbitrary electron density as the "surface".
Molecule interactions can be split into four categories: Pauli exclusion principle, electrostatic interactions between permanent charges (ions and multipoles), induction/polarization ("Debye force"), and London dispersion. "Collision" does not even begin to describe all these. At the molecular scale, even defining "a collision" is difficult, as there is no strict distance at which interactions start or stop: they interact, exchanging energy. Not just "kinetic" energy –– as in the movement of the center of mass of the molecule ––, but also energy related to internal excitation states (often part of the "thermal" energy of that molecule).
When we are dealing with air, the key physics field to consider is aerodynamics (https://en.wikipedia.org/wiki/Aerodynamics). In particular, most useful effects are related to changes in pressure, not "collisions". However, the case where the vehicle is in connection to ground, differs a lot from aeroplanes, because the key velocity is not airspeed, i.e vehicle velocity with respect to surrounding air, but groundspeed, because the vehicle uses the contact with ground for acceleration.
And as you claim to know how it works please also provide the equation describing the amount of wind power available to vehicle.
It depends on the vehicle. For a ground vehicle, what matters is that it depends not on the vehicle airspeed (speed of vehicle relative to the surrounding air), but on the velocity and pressure of air with respect to ground. This is because the vehicle spends its energy in an effort to accelerate with respect to ground, not with respect to wind; to compensate for the drag and friction (i.e., losses) the vehicle experiences.
Yes, aeroplanes and ground vehicles differ significantly in this.
An easy to understand vehicle that can extract energy from air when their relative speeds are the same, is a cart with a vertical axis airfoil wind turbine. The airfoils do not "catch" wind, they act like aeroplane wings, generating torque from the pressure differential on different sides of the blade. During any motion of the vehicle, the VAAWT is rotating. When the vehicle travels downwind at the same speed as the wind, the blades of the VAAWT can still extract energy from the wind in all phases except when the blade is in its extremum position with respect to the travel direction. It is obvious, since the blade is moving wirth respect to air surrounding it. Because of the pressure differential principle, the direction of the wind is easily compensated by changing the blade orientation/attack angle. The only limiting factor is how efficient you can make your vehicle, how small you can make the losses in its mechanisms.
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Thanks for taking the time to replay.
I also need to mention for others that below discussion is about direct downwind since this thread was about the upwind mainly.
In the previous threads I tried stepping you though it very simply (and it was tough keeping you ontrack and not racing ahead down distraction rabbit holes), but once you saw where it was going you disappeared. Why would this time be any different?
Nevertheless, your problem is starting with the wind power first and then trying to apply that to obvious stuff. Forget where the power comes from for the moment and look at how it's actually working, then figure how to power it.
I think that is the logical place to start from figuring out the amount of wind power available then decide what you can do with that.
If you forget about that you will think that you can do anything.
First, you have a propeller which is pushing the vehicle along. We know that works because jets take off - if the prop is turning then it is pushing the vehicle even if ever so slightly. It's pushing against the prevalent wind, and you even agreed earlier in this thread that this kind of thing is additive. If the prop can push a 5m/s and the wind is 10m/s then the vehicle will go at 15m/s.
That's uncontestable physics and fact, but feel free to disagree if you dare.
If by pushing you mean accelerating then you need to apply power to the propeller.
So if the starting point is that vehicle is at 15m/s driving in the same direction as wind witch is at just 10m/s and there is zero friction so ideal case the vehicle will be able to maintain that speed but will be unable to accelerate if there is no stored energy other than the vehicle kinetic energy.
So, what we're stuck for is the power that drives the prop. We know it is coming from the wheels (that is, the wheels drive the prop and not vice versa), and the wheels get it from the vehicle being pushed by the wind. But... the vehicle is going faster than the wind so what does the wind push on? It is the backwash from the propeller - remember the 5m/s plus whatever the prevalent wind is, which is this case is maybe -2m/s, so a small forward movement for a larger backwards jet.
Getting 100W as an example from the wheels means you are decelerating the vehicle by that exact amount and so if you provide that to an 100% efficient propeller you will get back 100W worth of acceleration thus you end up with nothing.
In a real vehicle doing such a thing will result in deceleration as for 100W taken at the wheel you will be able at most to say provide 80W worth of thrust thus you are decelerating at a 20W rate (slowing down).
Hang on, isn't this free energy or something? No, because of drag, which will require more power to overcome, and loss of power as more is extracted. Just think about it for a moment: the wind is pushing the vehicle (yes, due to air bouncing off it) at almost wind speed and there is no drag. All that's required is for the prop to turn however, slowly, and produce a minute amount of thrust which will increase the vehicle speed a tiny amount and it's game on, but go a bit faster and you need more power to overcome the drag and similar losses. But if you have drag you are going faster than the wind. Can the wheels extra that tiny amount of power without slowing the vehicle? That's where leverage, size of prop and angle of blades come in.
Too much nonsense to be able to provide a constructive replay.
Equations? Dunno. I'm not clever enough to work one out that you would reject or ignore anyway, and I doubt there is one on t'web to copy like you do. But I can point to video of the actual vehicle and many models working continuously and without any energy storage facilities.
Edit: forgot to point out that the wind power increases as the vehicle slows. There is far more power available than is consumed to push the vehicle forward - it's just that there is nothing to suck it up normally so the vehicle accelerates. At wind speed, there is 0 power pushing the car forward, but all the power of the wind is still available if you can work out how to access it. That's where the wheels are getting apparently free power from.
There is no problem copying an equation from either the web or your mind if it is the right one for the subject.
There is no video showing a direct downwind vehicle driving continuously. As far as I know large scale model is just one the Blackbird and all the treadmill models have a very limited space before they fall out on the other side.
But what you can do is take a video from the side on one of the treadmill models and you will see the rate of acceleration decreasing meaning it at some point get to zero and start decelerating.
"Free power" :) There is no such thing.
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Direct upwind is trivial to demonstrate: just put an efficient horizontal turbine on a cart, and use a very low gearing to the wheels (i.e. turbine turns many tens of times per one rotation of the wheels). Make it a VAWT and the cart will move regardless of the wind direction.
This matches my Lego trike I demonstrated earlier, and whose exact behaviour I described mathematically (in terms of velocities or displacements, and gear ratio); instead of wind, it used a spool of thread, or a Lego chain/track.
If one were to build a model where the ground wheel is actually a pulley with a wire looped around it and fixed to ground at both ends, it shows that no slip is required at all for it to function as described. Indeed, if the wire is looped many times around the pulley, and fixed to the pulley at the midpoint, then no slip is even possible if one uses e.g. thin steel cable.
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Fixed the topic for you.
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Modeling the macro-scale properties of compressible fluids as colliding solid particles is just not tenable. I know, because my proper field is molecular dynamics simulations.
That is because of our limited compute capability. That will not change the fact that air molecules collide with the vehicle body and that is how energy is transferred.
The equation we have and I provided many times is doing a great approximation of the wind power available without calculating energy provided by each air molecule. We just select an air density typical 1.2kg/m3 and the area and coefficient of drag of the solid surface it interacts with.
It depends on the vehicle. For a ground vehicle, what matters is that it depends not on the vehicle airspeed (speed of vehicle relative to the surrounding air), but on the velocity and pressure of air with respect to ground. This is because the vehicle spends its energy in an effort to accelerate with respect to ground, not with respect to wind; to compensate for the drag and friction (i.e., losses) the vehicle experiences.
Yes, aeroplanes and ground vehicles differ significantly in this.
An easy to understand vehicle that can extract energy from air when their relative speeds are the same, is a cart with a vertical axis airfoil wind turbine. The airfoils do not "catch" wind, they act like aeroplane wings, generating torque from the pressure differential on different sides of the blade. During any motion of the vehicle, the VAAWT is rotating. When the vehicle travels downwind at the same speed as the wind, the blades of the VAAWT can still extract energy from the wind in all phases except when the blade is in its extremum position with respect to the travel direction. It is obvious, since the blade is moving wirth respect to air surrounding it. Because of the pressure differential principle, the direction of the wind is easily compensated by changing the blade orientation/attack angle. The only limiting factor is how efficient you can make your vehicle, how small you can make the losses in its mechanisms.
A lot of words and no equations.
I demonstrated that I can predict exactly what happens using very few simple and well known equations.
While all I get from you is "It depends on the vehicle".
Facts:
- Yes a vehicle with no energy storage like a simple sail vehicle can drive only using available wind power thus up to wind speed in wind direction and no speed directly upwind.
- A vehicle that uses some form of energy storage can drive both upwind and for a limited amount of time proportional with amount of stored energy directly downwind.
Understanding the difference between power and energy may be critical in understanding the above two facts.
So all you need to now are this:
a) P = 0.5 * air density * area * coefficient of drag * (wind speed - vehicle speed)3 to get ideal so best case scenario for wind power available.
b) KE = 0.5 * mass * (vehicle speed)2
c) details about the vehicle energy storage if it has any.
Knowing the above 3 things you can predict exactly what happens and so understand how all this vehicle work.
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Direct upwind is trivial to demonstrate: just put an efficient horizontal turbine on a cart, and use a very low gearing to the wheels (i.e. turbine turns many tens of times per one rotation of the wheels). Make it a VAWT and the cart will move regardless of the wind direction.
This matches my Lego trike I demonstrated earlier, and whose exact behaviour I described mathematically (in terms of velocities or displacements, and gear ratio); instead of wind, it used a spool of thread, or a Lego chain/track.
If one were to build a model where the ground wheel is actually a pulley with a wire looped around it and fixed to ground at both ends, it shows that no slip is required at all for it to function as described. Indeed, if the wire is looped many times around the pulley, and fixed to the pulley at the midpoint, then no slip is even possible if one uses e.g. thin steel cable.
I will assume you have basic electricity knowledge.
Vehicle has a wind turbine connected to an electric generator.
The electric generator is connected to an electric motor that drives the wheels.
Based on wind speed and wind turbine swept area 1000W are available from the wind.
That means wind pushes against the vehicle with 1000W so if all 1000W output from the wind turbine are applied to an ideal 100% efficient electric motor the best that the vehicle can do is stay in the same place it is.
So the problem of you not understanding can come from
a) Not understanding that wind wants to accelerate the vehicle in the direction of the wind and so if your wind turbine output 1000W electrical of mechanical power you need 1001W at the wheels in order to begin to accelerate the vehicle mass against wind direction with 1W
b) Not understanding what power is and that if you include time you deal with energy
So you can wait for 2 seconds and store (in some energy storage device) 2000Ws worth of energy then you can use that stored energy to accelerate for say 1 second at 2000W or for 200ms at 10kW.
No energy storage means you can not accelerate the vehicle directly upwind.
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Fixed the topic for you.
What ?
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A lot of words and no equations.
Well, it is better than unphysical nonsense and equations unrelated to the physical phenomena at hand. You seem to believe that if an equation applies in some specific situation, it must apply always, and that just isn't so.
Like I explained, if a vehicle uses the fixed ground instead of surrounding air in its propulsion mechanism, the amount of energy it can obtain from wind is not relative to the vehicle speed, it is relative to the ground. So no subtracting vehicle speed from the wind speed relative to ground.
Extracting the energy is only an engineering problem.
Because of the Venturi effect (https://en.wikipedia.org/wiki/Venturi_effect), the exact wind speed just isn't an issue wrt. harvesting energy from it.
You insist on using static surfaces, ignoring all pressure-related effects, as if the vehicle is and has to be a simple rectangular box, with no internal mechanisms and only interacting with its environment via elastic collisions. It is silly, and quite annoying.
You need to move past your preconceptions. For example, you could start gently, and consider a vehicle with a very aerodynamic shape, say saucer-shape with a sharp edge, with upper and lower curvatures different (to balance out any vertical forces, including lift, due to the pressure differential), and with a Gorlov helical turbine (https://en.wikipedia.org/wiki/Gorlov_helical_turbine) poking up on top. It has minimal drag, and the turbine efficiency is about 35%. It typically operates at tip speed ratio over 1, which means that the induced flow rate due to rotation of the turbine is greater than the flow rate with respect to the axis. Go read Gorlov's 2001 paper (http://www.math.le.ac.uk/people/ag153/homepage/Gorlov2001.pdf) on it, but do recall that it only considers the properties of a stationary turbine.
And before you assert it, no, a moving turbine is not the same thing as stationary turbine in zero-wind situation. It only matches if we assume the turbine is not moving in either case, and that's not going to happen with a moving turbine. Because of the tip speed ratio being over 1, even when the wind speed approaches zero relative to the turbine axis, the turbine isn't going to stop (unless it has bad bearings or other significant losses). Because it can keep accelerating the vehicle all the way, it will be less efficient at that particular point compared to speeds immediately above and below, but it isn't zero even there.
What makes that seemingly simple construction interesting, is the fact that airflow speed varies as a distance from static surfaces like the ground, below the boundary layer (https://en.wikipedia.org/wiki/Boundary_layer). This means that along the vertical turbine axis, there is always a range of wind speeds relative to the axis, and because the wind direction does not affect the turbine, it can extract energy from the wind even when the vehicle is traveling downwind at the nominal wind speed. This applies even to wind tunnels (and in those, it is most noticeable, because they have laminar flow patterns by design; in nature, wind tends to be a bit turbulent). That sort of a vehicle does not even need wheels, it can even use a propeller for the propulsion, as long as it keeps close to a static surface (ground), and still achieve faster than nominal wind speed downwind or upwind.
When you wrap your mind around that, go look at aerodynamics, propulsion, and so on, and expand your understanding from there. Reconsider the models and practical examples I've shown earlier in this thread. Don't get stuck in your preconceived notions just because you're emotionally heavily invested in them. Learn to entertain ideas and theories and models that you do not believe are true, and learn how to rationally examine why your beliefs and the idea/theory/model do not agree.
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If by pushing you mean accelerating then you need to apply power to the propeller.
I stopped reading there since it's obvious you're already predetermined to go off on a distractions again. Recall:
"Forget where the power comes from for the moment and look at how it's actually working"
You even quoted that, yet the first second thing you say is "you need to apply power to the propeller".
And, to round it off, if you're pushing along at 5m/s or 10m/s or <any>m/s you are NOT accelerating. If it were accelerating I would say "it's accelerating".
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Well, it is better than unphysical nonsense and equations unrelated to the physical phenomena at hand. You seem to believe that if an equation applies in some specific situation, it must apply always, and that just isn't so.
Like I explained, if a vehicle uses the fixed ground instead of surrounding air in its propulsion mechanism, the amount of energy it can obtain from wind is not relative to the vehicle speed, it is relative to the ground. So no subtracting vehicle speed from the wind speed relative to ground.
Extracting the energy is only an engineering problem.
There is no such thing as ground energy. There is only wind energy.
But more importantly we are only talking about Power
You can get as much energy from the wind as long as direct downwind vehicle is below wind speed.
You can do anything you want with that energy like:
Wasting it as heat (maybe apply brakes).
Increase vehicle kinetic energy and thus speed.
Storing it for later use
Or a combination of the above.
You insist on using static surfaces, ignoring all pressure-related effects, as if the vehicle is and has to be a simple rectangular box, with no internal mechanisms and only interacting with its environment via elastic collisions. It is silly, and quite annoying.
Surfaces can be static or dynamic like a propeller they will act as a sail and vehicle can gain kinetic energy if air particle hit the static or moving part of the vehicle and will lose kinetic energy if say propeller hits a air particle that has the same speed as the vehicle or it moves in the opposite direction before being hit by the propeller blade.
You need to move past your preconceptions. For example, you could start gently, and consider a vehicle with a very aerodynamic shape, say saucer-shape with a sharp edge, with upper and lower curvatures different (to balance out any vertical forces, including lift, due to the pressure differential), and with a Gorlov helical turbine (https://en.wikipedia.org/wiki/Gorlov_helical_turbine) poking up on top. It has minimal drag, and the turbine efficiency is about 35%. It typically operates at tip speed ratio over 1, which means that the induced flow rate due to rotation of the turbine is greater than the flow rate with respect to the axis. Go read Gorlov's 2001 paper (http://www.math.le.ac.uk/people/ag153/homepage/Gorlov2001.pdf) on it, but do recall that it only considers the properties of a stationary turbine.
And before you assert it, no, a moving turbine is not the same thing as stationary turbine in zero-wind situation. It only matches if we assume the turbine is not moving in either case, and that's not going to happen with a moving turbine. Because of the tip speed ratio being over 1, even when the wind speed approaches zero relative to the turbine axis, the turbine isn't going to stop (unless it has bad bearings or other significant losses). Because it can keep accelerating the vehicle all the way, it will be less efficient at that particular point compared to speeds immediately above and below, but it isn't zero even there.
What makes that seemingly simple construction interesting, is the fact that airflow speed varies as a distance from static surfaces like the ground, below the boundary layer (https://en.wikipedia.org/wiki/Boundary_layer). This means that along the vertical turbine axis, there is always a range of wind speeds relative to the axis, and because the wind direction does not affect the turbine, it can extract energy from the wind even when the vehicle is traveling downwind at the nominal wind speed. This applies even to wind tunnels (and in those, it is most noticeable, because they have laminar flow patterns by design; in nature, wind tends to be a bit turbulent). That sort of a vehicle does not even need wheels, it can even use a propeller for the propulsion, as long as it keeps close to a static surface (ground), and still achieve faster than nominal wind speed downwind or upwind.
When you wrap your mind around that, go look at aerodynamics, propulsion, and so on, and expand your understanding from there. Reconsider the models and practical examples I've shown earlier in this thread. Don't get stuck in your preconceived notions just because you're emotionally heavily invested in them. Learn to entertain ideas and theories and models that you do not believe are true, and learn how to rationally examine why your beliefs and the idea/theory/model do not agree.
OK I will give you a concrete example with numbers that you can verify in a real world test.
35% efficient wind turbine installed on top of a vehicle. Wind turbine swept area say for simplicity 1m2
a) vehicle drives at +10m/s in a day with no wind so wind speed 0m/s
b) vehicle is stationary 0m/s and head wind speed is -10m/s
c) vehicle speed is +10m/s and head wind speed is -20m/s
d) vehicle speed is +10m/s and tail wind is +10m/s
e) vehicle speed is +10m/s and tail wind is +6m/s
For all this cases you use this equation to find what the wind turbine power output will be.
Pw = 0.5 * 1.2 * 1 * 0.35 * (wind speed - vehicle speed)3
If you can prove that this equation will not provide the correct result for any of the a) to e) cases or any other case you can think of then you can say you proved me wrong.
The mistake you make is to think there is not always an equal and opposite reaction to any action basically the title of this thread.
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And, to round it off, if you're pushing along at 5m/s or 10m/s or <any>m/s you are NOT accelerating. If it were accelerating I would say "it's accelerating".
You can not get there (above wind speed directly downwind) unless you push the vehicle or stored energy pushes vehicle.
So from that point you mentioned 50% above wind speed without any stored energy your vehicle can only decelerate due to frictional losses and there is no way to accelerate as wind power available there is zero and there is no such thing as ground power.
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Nominal, what did I tell you???
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Nominal, what did I tell you???
I'm an easily socially manipulated sucker, I know. :-[
But the demand to "prove them wrong" just goes over the top. As if the physical Lego model that behaves as I described and contrary to their description was not proof enough.
Time to use the Ignore list, I guess.
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:palm: :horse:
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I'm an easily socially manipulated sucker, I know. :-[
But the demand to "prove them wrong" just goes over the top. As if the physical Lego model that behaves as I described and contrary to their description was not proof enough.
Time to use the Ignore list, I guess.
You do not have the explanation of how the Lego model works.
I do and the equations exactly predict what the Lego model shows in real world.
a) The input power to the Lego is the force and speed you apply to the string.
b) The output power can not be higher so vehicle could not move away from the direction the string is pulled without including energy storage.
So while you pull the string if you do that slow enough you will notice that you apply both a force and there is a speed for the string but the vehicle is not moving meaning the input energy was stored unless you can see that radiated as heat to the outside world (part of it will be radiated but a small percentage).
Then when enough energy was stored to allow the vehicle to move against the direction the sting moves it will do so but then it will need to again stop accelerating to charge and then accelerate again.
This pause to charge then accelerate cycles are fast enough that your brain will not be able to detect same as you can not detect that a video is made out of individual still frames so for a few ms the image is unchanged but your brain (and mine and anyone else's) will think that motion is continues.
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And, to round it off, if you're pushing along at 5m/s or 10m/s or <any>m/s you are NOT accelerating. If it were accelerating I would say "it's accelerating".
You can not get there (above wind speed directly downwind) unless you push the vehicle or stored energy pushes vehicle.
So from that point you mentioned 50% above wind speed without any stored energy your vehicle can only decelerate due to frictional losses and there is no way to accelerate as wind power available there is zero and there is no such thing as ground power.
"Forget where the power comes from for the moment and look at how it's actually working"
See, I told you it was utterly pointless and useless to try explaining to you. You simply cannot get your mind out of the groove it's worn, and you will drop in any distraction going in order to wear that groove deeper.
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"Forget where the power comes from for the moment and look at how it's actually working"
See, I told you it was utterly pointless and useless to try explaining to you. You simply cannot get your mind out of the groove it's worn, and you will drop in any distraction going in order to wear that groove deeper.
So are you saying that is irrelevant to know the input power available for this wind only powered vehicle ?
Without knowing the answer to that question you will not be able to predict anything about the motion of that vehicle.
So either post the equation describing the wind power available to vehicle or you are just wasting my time and others that read this.
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So are you saying that is irrelevant to know the input power available for this wind only powered vehicle ?
At this point, yes. Later we will figure out what power we need and where it comes from, but right now we don't even know what use it would be! Well, obviously, we do but you don't.
Edit: rhetorical. I have no intention of spending another 2 months battling your mental illness or whatever it is that causes you to stick in your rut.
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So are you saying that is irrelevant to know the input power available for this wind only powered vehicle ?
At this point, yes. Later we will figure out what power we need and where it comes from, but right now we don't even know what use it would be! Well, obviously, we do but you don't.
Edit: rhetorical. I have no intention of spending another 2 months battling your mental illness or whatever it is that causes you to stick in your rut.
The first thing you need to predict anything is the input wind power available to vehicle.
If there is no input wind power available then vehicle will not be able to move so there is nothing you can say or predict about.
After you know that you can look at design to figure if it has or not an energy storage device.
If it exceeds wind speed directly downwind or it drives at any speed directly upwind you know for sure energy storage is involved as else it can not work based on currently agreed laws of physics.
Then if you do the test correctly and look close enough at the data you will see that yes energy storage is involved in all of this cases.
I showed both the equations and the experimental data (for upwind variant) and both the equations and the experimental data fully agree.
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The first thing you need to predict anything is the input wind power available to vehicle.
This is not true. We don't need to know how much power is available, it is not relevant. If the vehicle has any power available at all it is able to move in any chosen direction.
If there is no input wind power available then vehicle will not be able to move so there is nothing you can say or predict about.
Yes, of course. See above.
After you know that you can look at design to figure if it has or not an energy storage device.
There is no reason to do this, since it is not relevant.
If it exceeds wind speed directly downwind or it drives at any speed directly upwind you know for sure energy storage is involved as else it can not work based on currently agreed laws of physics.
This is not true.
Then if you do the test correctly and look close enough at the data you will see that yes energy storage is involved in all of this cases.
This is not true. If the mathematical models do not include energy storage, then there cannot be any energy storage.
I showed both the equations and the experimental data (for upwind variant) and both the equations and the experimental data fully agree.
You have never shown any such thing.
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This is not true.
This is not true.
This is not true.
That is all that you said "This is not true" . No equation or proof for any of those not true claims.
This is a physics not a math question.
As mentioned with math you can also say a locked 1:1 gear ratio can drive upwind.
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As mentioned with math you can also say a locked 1:1 gear ratio can drive upwind.
This is not true either. If you think it is true, demonstrate it.
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This is not true either. If you think it is true, demonstrate it.
You did that yourself with your gif animation.
You had a 3:2 gear ratio
Showed a wheel traveling 3 squares and the other 2 squares while vehicle also moved one square.
I mentioned to fix one of the moving surfaces to represent the earth and your answer was
"With the fixed ground on the left the treadmill moves to the right one square. When this happens the vehicle moves two squares to the left."
I asked if you observe such a thing in reality meaning treadmill moves one square relative to ground while vehicle moves two squares also relative to ground.
I obviously got no answer from you.
In the same way you can claim that with a 1:1 gear ratio the vehicle moves 2 squares while treadmill does not move at all.
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You did that yourself with your gif animation.
You had a 3:2 gear ratio
Showed a wheel traveling 3 squares and the other 2 squares while vehicle also moved one square.
I mentioned to fix one of the moving surfaces to represent the earth and your answer was
"With the fixed ground on the left the treadmill moves to the right one square. When this happens the vehicle moves two squares to the left."
I asked if you observe such a thing in reality meaning treadmill moves one square relative to ground while vehicle moves two squares also relative to ground.
I obviously got no answer from you.
It didn't deserve an answer. Of course it would do the same thing in reality. It is a model of reality.
In the same way you can claim that with a 1:1 gear ratio the vehicle moves 2 squares while treadmill does not move at all.
Equally, of course you cannot claim this. Because no accurate model will predict such a thing.
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If it exceeds wind speed directly downwind or it drives at any speed directly upwind you know for sure energy storage is involved as else it can not work based on currently agreed laws of physics.
There are no "currently agreed" laws of physics that would claim this. You are the only person who says this. Every textbook, every physics professor, everyone in this forum, every qualified engineer in the world would disagree with you.
You keep making incorrect statements, such as suggesting that power has a direction like "left" or "right", "forwards" or "backwards", when everyone here has repeatedly told you that power is a scalar quantity, it has no direction.
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Equally, of course you cannot claim this. Because no accurate model will predict such a thing.
Your mathematical model will can predict almost anything with not many constraints.
And yes you can show your vehicle moving two squares relative to the grid and of course 2 squares on the fixed treadmill representing ground and also two squares on the non moving treadmill.
But if you insist in treadmill moving you can move it 1 square and vehicle can move any random value relative to ground.
The thing that your model is not considering is Newton's 3'rd law.
The gear is locked and treadmill applying a force to the generator wheel will not create a larger force at the motor wheel because that is physically possible without some other external force or force provided by stored energy.
Since there are no other external sources it is clear is energy storage. (Also demonstrated in my slow motion video).
I'm still not sure if is the wrong understanding of the difference between power and energy or the fact that any action has an equal and opposite reaction.
Power at the motor wheel will always be lower than power at the generator wheel meaning vehicle can not move against the direction of the treadmill powering the generator wheel.
Power out (at the motor wheel) will need to be larger than power extracted at the generator wheel in order for the vehicle to move against the treadmill direction and the only way to do that is to store energy then use that stored energy to power the motor wheel at any power you want (higher power for lower duration or lower power for longer duration but always more than generated at that time in order to be able to move in the other direction).
Why do none of you use the electric model ? It should be harder to make mistakes for someone with electrical knowledge (at least I will think so).
But no matter if electrical or mechanical system the output power will always be lower unless you add energy storage and then for small portion of time you can output higher power by using stored energy.
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If it exceeds wind speed directly downwind or it drives at any speed directly upwind you know for sure energy storage is involved as else it can not work based on currently agreed laws of physics.
There are no "currently agreed" laws of physics that would claim this. You are the only person who says this. Every textbook, every physics professor, everyone in this forum, every qualified engineer in the world would disagree with you.
You keep making incorrect statements, such as suggesting that power has a direction like "left" or "right", "forwards" or "backwards", when everyone here has repeatedly told you that power is a scalar quantity, it has no direction.
Actually there are and law of conservation of energy is a law for a good reason.
Your output (propulsion) power can not be larger than input generated power.
I showed the wind power equation used by any engineer in the world that needs it and it shows that wind power available is highest when the wind speed is highest relative to vehicle and zero when there is no wind speed relative to vehicle.
So a vehicle can move against the wind direction only when output (propulsion power) is higher than input wind power.
Since that can not happen due to energy conservation law you need to use energy storage to move a vehicle upwind.
If you ever swimmed in a river against the current without being anchored to the ground you will have needed a minimum power just to maintain your position relative to the ground.
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Equally, of course you cannot claim this. Because no accurate model will predict such a thing.
Your mathematical model will can predict almost anything with not many constraints.
But the model does have constraints. It is constrained by an accurate representation of the system. What the model does, the real system does. And vice versa.
And yes you can show your vehicle moving two squares relative to the grid and of course 2 squares on the fixed treadmill representing ground and also two squares on the non moving treadmill.
But if you insist in treadmill moving you can move it 1 square and vehicle can move any random value relative to ground.
And what's wrong with that?
The thing that your model is not considering is Newton's 3'rd law.
Because it is not relevant.
The gear is locked and treadmill applying a force to the generator wheel will not create a larger force at the motor wheel because that is physically possible without some other external force or force provided by stored energy.
This is demonstrably not true. If I have a 10:1 lever, I can apply a force of 1 N on the longer end and get a force of 10 N at the shorter end. Gears are no different from levers. Even if the wind exerts a force of 1 N on the vehicle, we can use gearing to make the motor exert a force of 10 N at the wheels in the opposite direction.
Since there are no other external sources it is clear is energy storage. (Also demonstrated in my slow motion video).
Clearly not true, as above.
I'm still not sure if is the wrong understanding of the difference between power and energy or the fact that any action has an equal and opposite reaction.
Everyone has been asking you to look inside yourself for the wrong understanding, but you won't do it.
Power at the motor wheel will always be lower than power at the generator wheel
True
meaning vehicle can not move against the direction of the treadmill powering the generator wheel.
Not true. Any small amount of power at the motor wheel can move the vehicle with appropriate gearing.
Power out (at the motor wheel) will need to be larger than power extracted at the generator wheel in order for the vehicle to move against the treadmill direction and the only way to do that is to store energy then use that stored energy to power the motor wheel at any power you want (higher power for lower duration or lower power for longer duration but always more than generated at that time in order to be able to move in the other direction).
Not true. See above.
Why do none of you use the electric model ? It should be harder to make mistakes for someone with electrical knowledge (at least I will think so).
But no matter if electrical or mechanical system the output power will always be lower unless you add energy storage and then for small portion of time you can output higher power by using stored energy.
The electrical model is even easier. As long as there is some small power at the electric motor the vehicle can move. The stronger the wind, the more power, so the easier it is for the vehicle to move.
If it exceeds wind speed directly downwind or it drives at any speed directly upwind you know for sure energy storage is involved as else it can not work based on currently agreed laws of physics.
There are no "currently agreed" laws of physics that would claim this. You are the only person who says this. Every textbook, every physics professor, everyone in this forum, every qualified engineer in the world would disagree with you.
You keep making incorrect statements, such as suggesting that power has a direction like "left" or "right", "forwards" or "backwards", when everyone here has repeatedly told you that power is a scalar quantity, it has no direction.
Actually there are and law of conservation of energy is a law for a good reason.
Conservation of energy is not a useful analysis here.
Your output (propulsion) power can not be larger than input generated power.
True, but irrelevant.
I showed the wind power equation used by any engineer in the world that needs it and it shows that wind power available is highest when the wind speed is highest relative to vehicle and zero when there is no wind speed relative to vehicle.
As has previously been explained by many people, it is the wind speed relative to the ground that decides how much power can be extracted by the vehicle, not the speed relative to the vehicle. If you take an equation and apply it incorrectly, you will draw incorrect conclusions.
So a vehicle can move against the wind direction only when output (propulsion power) is higher than input wind power.
As repeated many times, not true.
Since that can not happen due to energy conservation law you need to use energy storage to move a vehicle upwind.
Nonsense.
If you ever swimmed in a river against the current without being anchored to the ground you will have needed a minimum power just to maintain your position relative to the ground.
Yes, but the vehicle is anchored to the ground. It has wheels with friction in contact with the ground. So the swimming analogy does not apply. The swimming analogy would only apply to a free floating airship above the ground (a body which would be "swimming" in the air).
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But the model does have constraints. It is constrained by an accurate representation of the system. What the model does, the real system does. And vice versa.
This is demonstrably not true. If I have a 10:1 lever, I can apply a force of 1 N on the longer end and get a force of 10 N at the shorter end. Gears are no different from levers. Even if the wind exerts a force of 1 N on the vehicle, we can use gearing to make the motor exert a force of 10 N at the wheels in the opposite direction.
This must be a record. I was not going to replay but there are parts that I want to help you with.
Your constraints are just geometrical. Not enough for physical simulation.
You can only get 10N out with 1N in if the body of the gearbox is not floating as it is for this example vehicle.
In the case where gearbox body is floating (not attached to anything) your output force can not be anything other than equal with input force.
Take a pair of scissors have it open it half way and glue some adhesive tape between the tip of the cutting blade on one side and at half the distance between the tip and the pivot point on the other side.
Now try and pull apart the two handles and see if you can have two different forces at the two handles
If you can prove that force is different at one of the handles I will be happy to admit I was wrong.
But I hope you do not even need to do the test to know you can not get a different force at one handle than on the other one.
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All usable gearboxes have 3 points of contact.
1) Input
2) Output
3) Body
If the body of the gearbox is not connected to anything as it is the case for all this vehicles then output force can not be different from input force.
If you have a vehicle you can imagine the gearbox body being disconnected from the vehicle body and if say it is also locked the force at the input will be exactly the same as on the output.
You can not have a usable gearbox with only input and output connected with the body floating. You always need 3 points for the gearbox to work.
So this is not a usable gearbox with just two points of contact in total.
(http://electrodacus.com/temp/Windup.png)
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All usable gearboxes have 3 points of contact.
1) Input : Ground
2) Output: Propeller / Air
3) Body: Chassis
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This must be a record. I was not going to replay but there are parts that I want to help you with.
Then stop trying to help. You cannot be a successful student if you keep trying to teach instead of learning.
Your constraints are just geometrical. Not enough for physical simulation.
I gave you the example of a pantograph. Look it up. It has only geometrical constraints.
You can only get 10N out with 1N in if the body of the gearbox is not floating as it is for this example vehicle.
In the very previous post we had the point where it is not floating. It is sitting on the ground.
In the case where gearbox body is floating (not attached to anything) your output force can not be anything other than equal with input force.
Then you need to enlarge your understanding of how gears work.
Take a pair of scissors have it open it half way and glue some adhesive tape between the tip of the cutting blade on one side and at half the distance between the tip and the pivot point on the other side.
Now try and pull apart the two handles and see if you can have two different forces at the two handles
If you can prove that force is different at one of the handles I will be happy to admit I was wrong.
But I hope you do not even need to do the test to know you can not get a different force at one handle than on the other one.
Yes, this is exactly the example to think about. Suppose the force between the handles is 10 N. What is the force (tension) on the tape?
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So this is not a usable gearbox with just two points of contact in total.
(http://electrodacus.com/temp/Windup.png)
Having F1 equal to F2 is not relevant to how the vehicle will move. Forces are useful in static analysis problems. They do not help with kinematic problems like this one. The vehicle will move in accordance with how all the parts are connected. It is pure geometry. Unless you think the real world does not obey geometric constraints?
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Thanks for taking the time to replay.
Sorry, but:
(http://[attachimg=1])
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Having F1 equal to F2 is not relevant to how the vehicle will move. Forces are useful in static analysis problems. They do not help with kinematic problems like this one. The vehicle will move in accordance with how all the parts are connected. It is pure geometry. Unless you think the real world does not obey geometric constraints?
:) OK so you understood the floating gearbox problem.
Static analysis is the important part here. Vehicle is stationary to start so you need a higher F2 if you want to accelerate the vehicle to the right.
What helps you have a higher F2 is both energy storage in the rubber belt and stick slip hysteresis as shown clearly in the slow motion video.
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Sorry, but:
(http:// (Attachment Link) )
Thanks for the correction. I'm sure you can find way more than that if you search.
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In the very previous post we had the point where it is not floating. It is sitting on the ground.
Can you be more specific ? Where there 3 points of contact for the gearbox ?
Then you need to enlarge your understanding of how gears work.
Like I mentioned a gearbox with floating body is not actually a gearbox.
Yes, this is exactly the example to think about. Suppose the force between the handles is 10 N. What is the force (tension) on the tape?
Force at the tape is irrelevant as that represents the belt and the only points making contacts are the two handles one is the input the other the output with the floating body and non relevant internal mechanism because the body is floating.
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All usable gearboxes have 3 points of contact.
1) Input : Ground
2) Output: Propeller / Air
3) Body: Chassis
What vehicle are you referring here ?
I'm going to guess the direct downwind version.
Yes input wheel is in contact with ground and the propeller is the output but the body of the gearbox is floating even if it just happens that the body of the gearbox can also be called the body.
The problem is that input is not between the body and the wheels. There is no internal engine or motor witch will be connected with the stator to body and rotor to gearbox input.
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Static analysis is the important part here. Vehicle is stationary to start so you need a higher F2 if you want to accelerate the vehicle to the right.
What helps you have a higher F2 is both energy storage in the rubber belt and stick slip hysteresis as shown clearly in the slow motion video.
Firstly, there is no rubber belt and no stick slip hysteresis because this is a CAD model. It does not contain those elements.
Secondly, no such invention is needed. If you move the treadmill, then by necessity the wheel on the treadmill will turn, and the belt will move, and the other wheel will turn, and the cart will move.
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Can you be more specific ? Where there 3 points of contact for the gearbox ?
The axles (or bearings), the input shaft and the output shaft.
Like I mentioned a gearbox with floating body is not actually a gearbox.
But it is not actually floating. When the input shaft rotates the gearbox is not free to rotate with it, it is fixed to the body of the cart. It would only be a floating gearbox if it was free to rotate with the input shaft.
Force at the tape is irrelevant as that represents the belt and the only points making contacts are the two handles one is the input the other the output with the floating body and non relevant internal mechanism because the body is floating.
Not at all irrelevant, because forces come in pairs (Newton's 3rd law: every force has an equal and opposite reaction). In the same way that there is no such thing as a voltage at a single point (a voltage is always a difference between two points), there is no such thing as a force at a single point. There is always a reacting force paired with it. So the force pair on the handle of the scissors is like the input voltage on the primary of a transformer, and the force pair on the blades of the scissors is like the output voltage on the secondary of the transformer. Bear in mind that a transformer can often be floating; it does not have to be grounded at all.
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Firstly, there is no rubber belt and no stick slip hysteresis because this is a CAD model. It does not contain those elements.
Secondly, no such invention is needed. If you move the treadmill, then by necessity the wheel on the treadmill will turn, and the belt will move, and the other wheel will turn, and the cart will move.
Exactly. Since the diagram makes no mention about belt or any other part being elastic and since slip is also not mentioned then this diagram as a problem will represent a system that can not move as F2 can not be larger than F1 no matter how large the F1 is the system will not move.
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Exactly. Since the diagram makes no mention about belt or any other part being elastic and since slip is also not mentioned then this diagram as a problem will represent a system that can not move as F2 can not be larger than F1 no matter how large the F1 is the system will not move.
This is faulty reasoning. The movement of the cart has nothing to do with the size of F2 and F1. They don't affect the analysis.
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The axles (or bearings), the input shaft and the output shaft.
The input shaft is connected to treadmill and the output shaft to ground.
That is all the rest represent the gearbox body that is not connected to anything.
But it is not actually floating. When the input shaft rotates the gearbox is not free to rotate with it, it is fixed to the body of the cart. It would only be a floating gearbox if it was free to rotate with the input shaft.
The gearbox body is not connected to anything so it is free to rotate with the input shaft if there is enough friction in the front wheel bearing.
Not at all irrelevant, because forces come in pairs (Newton's 3rd law: every force has an equal and opposite reaction). In the same way that there is no such thing as a voltage at a single point (a voltage is always a difference between two points), there is no such thing as a force at a single point. There is always a reacting force paired with it. So the force pair on the handle of the scissors is like the input voltage on the primary of a transformer, and the force pair on the blades of the scissors is like the output voltage on the secondary of the transformer. Bear in mind that a transformer can often be floating; it does not have to be grounded at all.
Not quite sure the analogy works.
But an electronic component with just two connections will have the same current at input and output.
So F1 is the equal and opposite force in this setup and so they will always be equal unless energy storage and stick slip hysteresis or some similar trigger for charge discharge exist.
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This is faulty reasoning. The movement of the cart has nothing to do with the size of F2 and F1. They don't affect the analysis.
That is your mathematical approach only. The cart can not move if all forces acting on it are equal and opposite.
In the math model you do not care about physics just geometry so you select an arbitrary distance that you want to move the vehicle and then plot the movement of all parts but in real world there is nothing that can move the vehicle if there is no energy storage and stick slip hysteresis.
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The gearbox body is not connected to anything so it is free to rotate with the input shaft if there is enough friction in the front wheel bearing.
Don't be absurd. The gearbox is fixed to the cart, so it can't rotate unless the cart rotates, which clearly the cart doesn't.
So F1 is the equal and opposite force in this setup and so they will always be equal unless energy storage and stick slip hysteresis or some similar trigger for charge discharge exist.
F1 acting on the wheel has an equal and opposite force F1' acting on the belt the wheel is in contact with. F2 can be anything different from F1, depending on the design of the system, and F2 definitely doesn't have to be equal and opposite to F1.
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Don't be absurd. The gearbox is fixed to the cart, so it can't rotate unless the cart rotates, which clearly the cart doesn't.
F1 acting on the wheel has an equal and opposite force F1' acting on the belt the wheel is in contact with. F2 can be anything different from F1, depending on the design of the system, and F2 definitely doesn't have to be equal and opposite to F1.
The cart body just happens to be the same with the gearbox body.
There will be no F2 without F1 and if slip is not allowed and there is no energy storage then F2 will be equal and opposite of F1.
This device can be called a cart or a gearbox it just depends on the definition you want to give but there is nothing more than just a gearbox.
It has the input connected to treadmill the output to ground and the body not connected to anything.
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The cart body just happens to be the same with the gearbox body.
And since the cart doesn't fly through the air doing somersaults, we can agree that the cart/gearbox doesn't rotate, right?
There will be no F2 without F1 and if slip is not allowed and there is no energy storage then F2 will be equal and opposite of F1.
There is no law of physics that says this.
This device can be called a cart or a gearbox it just depends on the definition you want to give but there is nothing more than just a gearbox.
It has the input connected to treadmill the output to ground and the body not connected to anything.
Just like car's gearbox then? It has the input connected to the engine, the output to the road, and the body not connected to anything? And yet, somehow, the gearbox in a car works.
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And since the cart doesn't fly through the air doing somersaults, we can agree that the cart/gearbox doesn't rotate, right?
It could fly if the friction is high enough or the acceleration rate of the treadmill is large enough.
The point is the body is floating.
There is no law of physics that says this.
It is for this particular gearbox configuration as it is locked basically no different from a rock
Just like car's gearbox then? It has the input connected to the engine, the output to the road, and the body not connected to anything? And yet, somehow, the gearbox in a car works.
The car gearbox body is connected to the motor stator or engine body.
It is like the treadmill body was connected to this vehicle body instead of the ground.
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It could fly if the friction is high enough or the acceleration rate of the treadmill is large enough.
Don't be absurd.
The point is the body is floating.
No, the point is that it is fixed on a horizontal plane and is not free to rotate.
There is no law of physics that says this.
It is for this particular gearbox configuration as it is locked basically no different from a rock
Saying that it is locked is a conclusion reached before doing the analysis. It is clearly not locked, since it has all sorts of axles and freely rotating wheels and belts. It would only be locked if it had brakes, and there are no brakes in the model. The wheels are free to turn.
Just like car's gearbox then? It has the input connected to the engine, the output to the road, and the body not connected to anything? And yet, somehow, the gearbox in a car works.
The car gearbox body is connected to the motor stator or engine body.
It is like the treadmill body was connected to this vehicle body instead of the ground.
This is meaningless. No clue what you are trying to say.
Anyway, good night. We will see tomorrow if anybody else feels like playing with the troll.
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Don't be absurd.
No, the point is that it is fixed on a horizontal plane and is not free to rotate.
Saying that it is locked is a conclusion reached before doing the analysis. It is clearly not locked, since it has all sorts of axles and freely rotating wheels and belts. It would only be locked if it had brakes, and there are no brakes in the model. The wheels are free to turn.
This is meaningless. No clue what you are trying to say.
Anyway, good night. We will see tomorrow if anybody else feels like playing with the troll.
You mentioned forces come in pairs.
The only force pair is F1 relative to ground not relative to vehicle body.
So vehicle/gearbox has input connected to F1 and output to ground and nothing else.
I should have insisted more on the floating gearbox as it seems it was not obvious.
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Anyway, good night. We will see tomorrow if anybody else feels like playing with the troll
I admire your perseverance and composure in the face of insurmountable odds :-+
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Don't be absurd.
Here a link to a proper vehicle physics calculator https://www.electromotive.eu/?page_id=12&lang=en (https://www.electromotive.eu/?page_id=12&lang=en)
Just change the head wind to 100km/h leave all the rest as default (it is for an ebike) see what power is needed for the biker pedaling at 20km/h in a 100km/h headwind and is almost 20kW as they use the proper equation.
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Here a link to a proper vehicle physics calculator https://www.electromotive.eu/?page_id=12&lang=en (https://www.electromotive.eu/?page_id=12&lang=en)
That calculator is next to useless. It has no facility to include coefficient of drag so it doesn't know the difference between a teardrop shape and a box shape.
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Here a link to a proper vehicle physics calculator https://www.electromotive.eu/?page_id=12&lang=en (https://www.electromotive.eu/?page_id=12&lang=en)
That calculator is next to useless. It has no facility to include coefficient of drag so it doesn't know the difference between a teardrop shape and a box shape.
It has.
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That calulator is useless an using the wrong equations: compare the case of 100 km/h headwind and 20 km/h with 110 and 10 and the calculator gives out the same result. The actual power needed would be half.
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And that calculator is irrelevant, or at least applicable to only a small part of the DDWFTTW discussion and that part can be ignored when studying the general principles.
electro' still insists that the treadmill models are "transmission-locked"
electro' still insists that his "wind turbine generator powering a motor that turns the wheels" is a valid analogy to the demonstrated DDWFTTW vehicles.
Neither of these assertions is remotely true.
Pedaling bicycles upwind is a distraction that confuses rather than clarifies.
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That calulator is useless an using the wrong equations: compare the case of 100 km/h headwind and 20 km/h with 110 and 10 and the calculator gives out the same result. The actual power needed would be half.
The calculator actually uses the correct equation and the result will be the same for vehicle at 20km/h with 100km/h head wind as it will be for vehicle at 110km/h and just 10km/h head wind.
Not quite sure where the wrong equation many of you are using originated. Can someone point to a source for that equation and I'm not referring to Wikipedia.
That wrong equation seems to have significant implications and I will like to get rid of it. Anyone that will do an actual experiment will realize the correct equation is the one I mentioned hundreds of times and also the same used in that calculator.
Here is another website with the correct equation and definition https://scienceworld.wolfram.com/physics/DragPower.html
it reads correctly as v being the speed of the fluid relative to the body (vehicle in this case).
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And that calculator is irrelevant, or at least applicable to only a small part of the DDWFTTW discussion and that part can be ignored when studying the general principles.
electro' still insists that the treadmill models are "transmission-locked"
electro' still insists that his "wind turbine generator powering a motor that turns the wheels" is a valid analogy to the demonstrated DDWFTTW vehicles.
Neither of these assertions is remotely true.
Pedaling bicycles upwind is a distraction that confuses rather than clarifies.
The equation (correct one) used in that calculator is key to understand upwind version and also key to understand direct downwind version tho that will actually require even more understanding of fluid mechanics.
But I think once we can establish that is the correct equation (and there are relatively simple tests that can prove that is the case) there will be zero way to claim that direct upwind can happen without energy storage.
As that equation states that to move upwind you need more power than what is available to vehicle in ideal case from the wind. Ideal case vehicle will be in equilibrium.
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there will be zero way to claim that direct upwind can happen without energy storage.
As that equation states that to move upwind you need more power than what is available to vehicle in ideal case from the wind. Ideal case vehicle will be in equilibrium.
"And yet it moves."
Energy storage? Think about it a bit. If you can move upwind or downwind with energy storage, you can move upwind or downwind with sufficiently low gearing -- no storage required. Storage is irrelevant and unnecessary.
And now I have hit my daily limit for this thread.
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"And yet it moves."
Energy storage? Think about it a bit. If you can move upwind or downwind with energy storage, you can move upwind or downwind with sufficiently low gearing -- no storage required. Storage is irrelevant and unnecessary.
And now I have hit my daily limit for this thread.
A gearbox can not amplify power.
If to drive upwind requires more power than available form the wind. How do you get the extra power if not from stored energy?
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A gearbox can not amplify power.
If to drive upwind requires more power than available form the wind. How do you get the extra power if not from stored energy?
There is not need to have more power than the wind, it is only about having more force available than the wind.
With the wind much faster than the movement to ground the force from the wind is approximatedly constant. The power needed is just force times speed. So zero power at zero speed and little power needed at low speed.
The gearbox can not amplify power, but it can amplifiy the torque / force.
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There is not need to have more power than the wind, it is only about having more force available than the wind.
With the wind much faster than the movement to ground the force from the wind is approximatedly constant. The power needed is just force times speed. So zero power at zero speed and little power needed at low speed.
The gearbox can not amplify power, but it can amplifiy the torque / force.
Vehicle will not need zero power to keep the vehicle at equilibrium point and zero speed.
You are thinking about having the vehicle anchored to ground by applying brakes but that just makes the entire earth part of the vehicle so no longer separated.
Please imagine a vehicle that has no brakes then think about the amount of power needed to keep the vehicle stationary based on wind speed and vehicle effective projected area facing the wind.
If you apply no power the vehicle will be accelerated in the direction of the wind and you need to match that power in ideal case to stay at zero speed then you need extra in order to move upwind.
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Not quite sure where the wrong equation many of you are using originated. Can someone point to a source for that equation and I'm not referring to Wikipedia.
And yet it's OK for you to copy random equations from t'web to prove ... something. Why not actually write yours1 from scratch instead of copying someone elses thing meant for some different scenario? You might learn something.
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[1] You don't seem to have made one yet. You've copypasta'd many.
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You might learn something.
Learning something is not the goal of trolling. He's obviously just yanking your chain, no real person is that stupid or dense.
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And yet it's OK for you to copy random equations from t'web to prove ... something. Why not actually write yours from scratch instead of copying someone elses thing meant for some different scenario? You might learn something.
Not quite sure what you are talking about. The equation is universal and is just one applicable to all of this cases.
You may make your own equations if you come up with new physics but there is nothing new in relation with what we discuss here.
The correct and universal equation for Drag Power is this that I mentioned for months here and all of you where disagreeing with.
Pd = 0.5 * air density * (area * coefficient of drag) * (wind speed - vehicle speed)3
The wrong equation that many of you are claiming to be correct is the one below,
Pd = 0.5 * air density * (area * coefficient of drag) * (wind speed - vehicle speed)2 * vehicle speed
Both this link https://scienceworld.wolfram.com/physics/DragPower.html (https://scienceworld.wolfram.com/physics/DragPower.html) and this calculator https://www.electromotive.eu/?page_id=12&lang=en (https://www.electromotive.eu/?page_id=12&lang=en) use the correct equations so they agree with what I'm saying.
I provided those links to show that I'm not inventing new physics but that all this is known for a very long time.
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The correct and universal equation for Drag Power is this that I mentioned for months here and all of you where disagreeing with.
Pd = 0.5 * air density * (area * coefficient of drag) * (wind speed - vehicle speed)3
a) How did you derive those terms? You could post anything and say that's correct, or copy anything and say it's correct "because someone else uses it too", but unless you know how the terms were derived it's meaningless to you.
b) Which term(s) cover the propeller and wheel operation? The propeller, and connection to the wheels, is kind of super-important since it's the thing that makes it work, and yet it doesn't exist in your copied equation. How can that equation possibly account for it, then?
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The formular for the drag power is the frist one, but this is the power theoretical possible to take from the wind and that is not the power needed for the vehicle to drive. The power for the vehicle is more like the 2nd formular.
It is obvious that the first formula can not be corrent for the power of the vehicle: if the vehicle changes direction from going against the wind to going with the wind it changes from needing power to drive to gaining power from the wind. So the sign in the power has to change at around 0 vehicle speed.
Another way is to look at the force need: mechnical power is force times speed, more or less by definition. The force from the wind is limited and approaching zero speed the power this also has the approach zero.
The relevant question for the vehicle is if the motor can provide sufficient force. The power only determines how fast the vehicle can go. With low speed very little power is sufficient.
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Both this link https://scienceworld.wolfram.com/physics/DragPower.html (https://scienceworld.wolfram.com/physics/DragPower.html) and this calculator https://www.electromotive.eu/?page_id=12&lang=en (https://www.electromotive.eu/?page_id=12&lang=en) use the correct equations so they agree with what I'm saying.
I provided those links to show that I'm not inventing new physics but that all this is known for a very long time.
You've scoured the internet to find 'examples' where someone has made an error somewhere that supports your argument, but those errors are painfully obvious with just a little bit of a fair-minded consideration.
In the Wolfram example, they clearly have only considered the case where the vehicle speed is the same as the speed of the fluid relative to the body, as in driving in still air. That is obvious simply looking at the equations.
In your other example, if you bother to read their explanations elsewhere on their website under "Vehicle Physics" they say:
Now, the propulsion power can be calculated as a product of driving resistance and vehicle speed:
But from the operation of the malfunctioning calculator app, you can see that they have not incorporated this into their algorithm.
In neither case do they use the equation you claim they do.
The Wolfram example has at least one part right before they go and screw it up. P = FD * Vvehicle
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a) How did you derive those terms? You could post anything and say that's correct, or copy anything and say it's correct "because someone else uses it too", but unless you know how the terms were derived it's meaningless to you.
b) Which term(s) cover the propeller and wheel operation? The propeller, and connection to the wheels, is kind of super-important since it's the thing that makes it work, and yet it doesn't exist in your copied equation. How can that equation possibly account for it, then?
a) not only I know how that is derived but also know it is correct as I build things not just play with numbers.
b) The equation for available wind power to any wind power vehicle is independent of the design other than shape and area interacting with air particles.
The wheel propeller connection is a separate issue and using that data you can calculate how much of the available wind power is used to increase vehicle kinetic energy and how much is diverted to propeller witch will use it to store it mainly in the form of pressure differential (this is the explanation for why direct downwind version can exceed wind speed for a limited amount of time).
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The formular for the drag power is the frist one, but this is the power theoretical possible to take from the wind and that is not the power needed for the vehicle to drive. The power for the vehicle is more like the 2nd formular.
It is obvious that the first formula can not be corrent for the power of the vehicle: if the vehicle changes direction from going against the wind to going with the wind it changes from needing power to drive to gaining power from the wind. So the sign in the power has to change at around 0 vehicle speed.
Another way is to look at the force need: mechnical power is force times speed, more or less by definition. The force from the wind is limited and approaching zero speed the power this also has the approach zero.
The relevant question for the vehicle is if the motor can provide sufficient force. The power only determines how fast the vehicle can go. With low speed very little power is sufficient.
I do not think that second formula describes anything. I'm fairly certain one or more people that do not understand what power is came up with that as they were likely thinking the correct one outputs a value larger than they expected or wishing.
Yes as you see the sign changes correctly when vehicle speed is zero.
Equation contains this (wind speed-vehicle speed) so max wind power to accelerate the vehicle is proportional with (wind speed - 0)3
If vehicle drives downwind it can be powered by wind at (wind speed - vehicle speed)3
If vehicle wants to drive upwind it requires a power proportional with (wind speed - (-vehicle speed))3 same as saying (wind speed + vehicle speed)3
So is clear from this that no vehicle can exceed wind speed powered only by the wind and can not drive at any speed upwind.
And yes all this is correct. Any wind powered vehicle that exceeds wind speed will do so with either stored energy or another energy source other than the wind and same is true for a vehicle driving upwind at any speed.
When you think at a zero speed vehicle you are wrongly thinking about a vehicle anchored to the ground. But that anchoring to ground just means vehicle is now part of earth.
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You've scoured the internet to find 'examples' where someone has made an error somewhere that supports your argument, but those errors are painfully obvious with just a little bit of a fair-minded consideration.
In the Wolfram example, they clearly have only considered the case where the vehicle speed is the same as the speed of the fluid relative to the body, as in driving in still air. That is obvious simply looking at the equations.
In your other example, if you bother to read their explanations elsewhere on their website under "Vehicle Physics" they say:
Now, the propulsion power can be calculated as a product of driving resistance and vehicle speed:
But from the operation of the malfunctioning calculator app, you can see that they have not incorporated this into their algorithm.
In neither case do they use the equation you claim they do.
The Wolfram example has at least one part right before they go and screw it up. P = FD * Vvehicle
It took me all of 3 or 4 minutes to find those two links.
You should read again to understand what that (the thing you highlighted) actually means.
A vehicle on frictionless wheels will be no different from a floating balloon.
Yes when friction is added that is subtracted from the wind power available to vehicle.
They properly incorporated the algorithm and that calculator outputs the correct results.
Look at what is defined as v in that equation as it is not the vehicle speed.
Quote from the text under that equation
"v is speed of the fluid relative to the body"
Which translates for this case in to v = (wind speed - vehicle speed)
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You should read again to understand what that (the thing you highlighted) actually means.
A vehicle on frictionless wheels will be no different from a floating balloon.
Both of those links purport to calculate the power needed to drive a vehicle against a fluid drag, not the other way around.
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Both of those links purport to calculate the power needed to drive a vehicle against a fluid drag, not the other way around.
One link shows the equation the other is a calculator using that equation and that equation will provide you with any question you may have about power needed to overcome drag when driving upwind and it can also calculate the max wind power available to vehicle when driving downwind.
You can just use negative sign in that calculator to calculate wind power available to accelerate so tailwind.
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One link shows the equation the other is a calculator using that equation and that equation will provide you with any question you may have about power needed to overcome drag when driving upwind and it can also calculate the max wind power available to vehicle when driving downwind.
Nope, every word of that is entirely wrong.
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Nope, every word of that is entirely wrong.
That is your unqualified opinion. You will need to measure and prove it is not true assuming you even understand what you will need to measure and 3.5 digit will be enough :)
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a) How did you derive those terms? You could post anything and say that's correct, or copy anything and say it's correct "because someone else uses it too", but unless you know how the terms were derived it's meaningless to you.
b) Which term(s) cover the propeller and wheel operation? The propeller, and connection to the wheels, is kind of super-important since it's the thing that makes it work, and yet it doesn't exist in your copied equation. How can that equation possibly account for it, then?
a) not only I know how that is derived but also know it is correct as I build things not just play with numbers.
So tell us then. Show us your workings so we can go "Doh! Of course, now we can see it."
b) The equation for available wind power to any wind power vehicle is independent of the design other than shape and area interacting with air particles.
The wheel propeller connection is a separate issue ...
It isn't a separate issue. Consider two scenarios:
1. The propeller blades are perpendicular to the wind. In this case the propeller might as well be a flat surface with area the same as the blades, and the power sucked from the wheel (if that's what you think happens) will be near enough zero.
2. The propeller blades are parallel to the wind. In this case the propeller is more or less non-existent so far as the wind pressure is concerned, but it is sucking maximum power from the wheels.
How can you say it's irrelevant when it can affect the system so much?
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b) The equation for available wind power to any wind power vehicle is independent of the design other than shape and area interacting with air particles.
The wheel propeller connection is a separate issue ...
It isn't a separate issue. Consider two scenarios:
1. The propeller blades are perpendicular to the wind. In this case the propeller might as well be a flat surface with area the same as the blades, and the power sucked from the wheel (if that's what you think happens) will be near enough zero.
2. The propeller blades are parallel to the wind. In this case the propeller is more or less non-existent so far as the wind pressure is concerned, but it is sucking maximum power from the wheels.
How can you say it's irrelevant when it can affect the system so much?
1 and 2. Do you even read what I write ?
"independent of the design other than shape and area interacting with air particles"
I see your interest is in the direct downwind faster than wind vehicle and blackbird to be specific as the same thing can be done without any propeller and in that case the propeller is
a) a sail with increasingly large effective area as the rotational speed increases.
b) propulsion device
c) part of the energy storage device.
The discussion here is much more basic and asks the amount of wind power available to any wind powered vehicle. All you need to know about vehicle design is shape (so drag coefficient) and vehicle area facing the wind.
You seems to disagree with that based on your false impression that above wind speed direct downwind blackbird is powered by wind (not the case).
While blackbird is above wind speed it is powered by stored energy only. And yes that means acceleration rate will drop if you can not control the rate at witch you use that stored energy. With the blackbird you can control that as pilot can change the propeller pitch. The treadmill model has no pitch control so on that it will be very easy to observe decrease in acceleration rate as stored energy is used up.
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The treadmill model has no pitch control so on that it will be very easy to observe decrease in acceleration rate as stored energy is used up.
And yet the videos showed them accelerating. Must've been accumulating energy for storage, eh.
https://www.youtube.com/watch?v=RGUzFs7gT0s (https://www.youtube.com/watch?v=RGUzFs7gT0s)
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And yet the videos showed them accelerating. Must've been accumulating energy for storage, eh.
Do you understand what decrease acceleration rate means ?
That video is poorly done and does not show actually anything.
Every time the vehicle body is connected to ground either through a device or the hand of the experimenter the energy storage is being charged.
So there are only very brief moments where vehicle is shown discharging the energy storage.
Once you understand all elements involved you will understand why this is a poorly done experiment.
The one done by Derek (Veritasium) was much better and they could have measured the decrease in acceleration rate by taking a video from the moment of release until vehicle left the treadmill and then calculate from the video frames what the acceleration rate was for the first half of the run and the second run showing clearly that acceleration rate drops as stored energy is being used up.
Based on that they could have calculated how far the vehicle will have traveled before it stopped accelerating and then how long it will have took for the vehicle speed to drop to zero.
So no need for a very long treadmill to make measurements and show what happens.
Another way to deal with a limited length treadmill was to increase the vehicle mass as then max speed will have been much lower and so the distance traveled.
But nobody even looked for this as they all assumed vehicle will forever drive in the opposite direction of the force applied (ridiculous to think if you understand the basics of classical mechanics).
The professor that lost the bet understood that this is not possible thus his high confidence. Unfortunately he did not know how the vehicle actually works and that energy storage was involved.
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But nobody even looked for this as they all assumed vehicle will forever drive in the opposite direction of the force applied (ridiculous to think if you understand the basics of classical mechanics).
If you sling a rope over a pulley and pull downwards, the weight on the other end will get pulled upwards (the opposite direction of the force applied). This is ridiculous to think if you understand the basics of classical mechanics, so we have to assume that pulleys can only work by stored energy and stick-slip hysteresis.
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If you sling a rope over a pulley and pull downwards, the weight on the other end will get pulled upwards (the opposite direction of the force applied). This is ridiculous to think if you understand the basics of classical mechanics, so we have to assume that pulleys can only work by stored energy and stick-slip hysteresis.
Pulley in order to work will have 3 points of contact the input the output and the center of the pulley all have a connection to something.
The vehicle used as example for upwind has only input and output connected the body is floating.
It is like you connecting the positive input of a DC-DC converter to battery+ the negative of the battery is connected to earth the positive output of the DC-DC converter to earth and the negative input and output of the DC-DC converter connected to each other but nothing else and you expecting that DC-DC converter to do anything useful.
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It is like you connecting the positive input of a DC-DC converter to battery+ the negative of the battery is connected to earth the positive output of the DC-DC converter to earth and the negative input and output of the DC-DC converter connected to each other but nothing else and you expecting that DC-DC converter to do anything useful.
That configuration is a little unusual, but can still work if the DC/DC converter has a lower output than input voltage and can thus provide more current at the output than it needs at the input. It is not a very praktical (efficient) solution but it would still generate a negative voltage. With real world parts it may need so help to start up, depending how well the DCDC converter handles load to the output if there is no input drive.
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It is like you connecting the positive input of a DC-DC converter to battery+ the negative of the battery is connected to earth the positive output of the DC-DC converter to earth and the negative input and output of the DC-DC converter connected to each other but nothing else and you expecting that DC-DC converter to do anything useful.
That configuration is a little unusual, but can still work if the DC/DC converter has a lower output than input voltage and can thus provide more current at the output than it needs at the input. It is not a very praktical (efficient) solution but it would still generate a negative voltage. With real world parts it may need so help to start up, depending how well the DCDC converter handles load to the output if there is no input drive.
I just tried it with an old Traco TEF2011 module that I found in my junk box, 10-30V in 5V out. It needed an additional 5V power supply connected to the output to get it started, but once started the additional supply could be disconnected. As expected it worked just fine, and even continued to run with the input power supply reduced to 2V, giving 7V across the converter input terminals.
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If you apply no power the vehicle will be accelerated in the direction of the wind and you need to match that power in ideal case to stay at zero speed then you need extra in order to move upwind.
If you are staying at zero speed then no work is being expended. A force maybe, but if there is no movement then no power is being used.
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It is like you connecting the positive input of a DC-DC converter to battery+ the negative of the battery is connected to earth the positive output of the DC-DC converter to earth and the negative input and output of the DC-DC converter connected to each other but nothing else and you expecting that DC-DC converter to do anything useful.
That configuration is a little unusual, but can still work if the DC/DC converter has a lower output than input voltage and can thus provide more current at the output than it needs at the input. It is not a very praktical (efficient) solution but it would still generate a negative voltage. With real world parts it may need so help to start up, depending how well the DCDC converter handles load to the output if there is no input drive.
Did you understood the setup ?
There are only to points connected on the DC-DC converter. The positive input and the positive output and nothing else.
Can you explain how you will get higher current at output than at the input ?
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I just tried it with an old Traco TEF2011 module that I found in my junk box, 10-30V in 5V out. It needed an additional 5V power supply connected to the output to get it started, but once started the additional supply could be disconnected. As expected it worked just fine, and even continued to run with the input power supply reduced to 2V, giving 7V across the converter input terminals.
Sorry but can you be more exact in describing your setup ?
You can of course have a voltage drop on the converter if you have a current flowing through the converter depending on the type of converter.
What you can not have is a higher voltage at output than input, a negative voltage or different current at output than at the input.
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If you are staying at zero speed then no work is being expended. A force maybe, but if there is no movement then no power is being used.
Have you ever had an electric motor providing a force / torque without using energy ? Or even simpler an electromagnet providing a force without using energy.
Even you using your arm to provide a force will be using energy even tho there is no movement and yes that energy will be dissipated as heat.
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It is like you connecting the positive input of a DC-DC converter to battery+ the negative of the battery is connected to earth the positive output of the DC-DC converter to earth and the negative input and output of the DC-DC converter connected to each other but nothing else and you expecting that DC-DC converter to do anything useful.
That configuration is a little unusual, but can still work if the DC/DC converter has a lower output than input voltage and can thus provide more current at the output than it needs at the input. It is not a very praktical (efficient) solution but it would still generate a negative voltage. With real world parts it may need so help to start up, depending how well the DCDC converter handles load to the output if there is no input drive.
Did you understood the setup ?
There are only to points connected on the DC-DC converter. The positive input and the positive output and nothing else.
Can you explain how you will get higher current at output than at the input ?
The effective input voltage see by the DC/DC converter is the sum of input voltage + output voltage (not considering a sign). So the available output current / load current between the DCDC converters outputs is the difference of the input current minus the output current. So if the DCDC converter is something like a 24 V input to 6 V output type with 75% efficiency one would have 3 times the ouput courrent at the DCDC converter an 2 times the output current from the input current. So one could get something like 18V 1A in and -6 V 2 A out.
There is nothing magic with the circuit - depending one the DCDC is may just need some help on start up.
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The effective input voltage see by the DC/DC converter is the sum of input voltage + output voltage (not considering a sign). So the available output current / load current between the DCDC converters outputs is the difference of the input current minus the output current. So if the DCDC converter is something like a 24 V input to 6 V output type with 75% efficiency one would have 3 times the ouput courrent at the DCDC converter an 2 times the output current from the input current. So one could get something like 18V 1A in and -6 V 2 A out.
There is nothing magic with the circuit - depending one the DCDC is may just need some help on start up.
Maybe I need to be more clear as I do not think you understood the setup.
Get a plastic box and put in it whatever you want as long as it is not an energy source like a battery. So you can put a DC-DC converter inside.
The box will have just two wires one on each end and nothing else.
Now can you have a difference in current between the two wires ? Have you ever seen a two pin IC that could have a current input at one of the pins and a different current on the other pin.
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I just tried it with an old Traco TEF2011 module that I found in my junk box, 10-30V in 5V out. It needed an additional 5V power supply connected to the output to get it started, but once started the additional supply could be disconnected. As expected it worked just fine, and even continued to run with the input power supply reduced to 2V, giving 7V across the converter input terminals.
Sorry but can you be more exact in describing your setup ?
The setup was exactly and precisely the one that you described, other than the temporary connection of a second power supply across the output terminals of the DC/DC converter to get the thing started. Once started the second power supply was disconnected.
You can of course have a voltage drop on the converter if you have a current flowing through the converter depending on the type of converter.
What you can not have is a higher voltage at output than input, a negative voltage or different current at output than at the input.
The output voltage was not higher than the input. The output of the DC/DC converter was 5V. The external power supply, set to 2V, was connected between the input and output of the DC/DC converter just as you described, which boosted this voltage to 7V. So there was 7V at the input to the DC/DC converter.
The input current and output current are necessarily equal. With no load, the circuit naturally settles at an operating current such that the 2V boost to the output voltage exactly compensates for the losses in the DC/DC converter.
This experiment took less than 10 minutes to perform. Rather than arguing about it, please just try it for yourself.
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This experiment took less than 10 minutes to perform. Rather than arguing about it, please just try it for yourself.
You expect a troll to try an experiment? He just got you to waste your time setting one up, and he's going to continue to argue that it can't work despite the fact that it does.
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The setup was exactly and precisely the one that you described, other than the temporary connection of a second power supply across the output terminals of the DC/DC converter to get the thing started. Once started the second power supply was disconnected.
The output voltage was not higher than the input. The output of the DC/DC converter was 5V. The external power supply, set to 2V, was connected between the input and output of the DC/DC converter just as you described, which boosted this voltage to 7V. So there was 7V at the input to the DC/DC converter.
The input current and output current are necessarily equal. With no load, the circuit naturally settles at an operating current such that the 2V boost to the output voltage exactly compensates for the losses in the DC/DC converter.
This experiment took less than 10 minutes to perform. Rather than arguing about it, please just try it for yourself.
The setup was surely not as I asked.
But let me simplify things and make it even simpler to understand.
You get a plastic box and put whatever you want in the box as long as there is no energy source like a battery.
And out of that box you have two wires and nothing else.
Now what you say is that you will put a DC-DC converter in that box and get the positive input and positive output out of that box.
My understanding will be that you connected the positive input to your lab power supply positive output and the positive output of the DC-DC converter to the negative of your lab power supply.
But then it will not make sense to say you measured 7V if the lab power supply was set to 5V. Where will you measure 7V ?
I may guess you charged some large capacitor on the DC-DC converter and then you measured that in series with your lab power supply voltage.
So the plastic box with just two wires coming out seems a better simplification and you can get inside whatever you want. You will not be able to have a different current through one of the two wires.
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I think you guys should get back to wheels, gears, and surfaces. Misunderstood switching regulator topology when the model doesn't even represent the thing being analyzed is getting you nowhere.
Or, DDWFTTW: The cart wheels drive a generator, the generator powers a fan blowing backwards. To start, the wind pressure gives the idle fan a little push. The cart starts rolling downwind, spinning the generator which powers the fan. Take it from there. But remember the wheels rolling on the ground. For starters, assume no slip, no energy storage, no friction, no loss.
And while interesting, I realize that this discussion is a lost cause. Fortunately for me, I enjoy some lost causes.
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Classic troll behavior. When losing the debate on the original problem, come up with some entirely different problem as a distraction, and try to argue that. And when that argument doesn't go well, keep trying to move the goalposts in order to get the conclusion desired. And than try to say that any conclusion drawn from the entirely different problem must also apply to the original problem.
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Or, DDWFTTW: The cart wheels drive a generator, the generator powers a fan blowing backwards. To start, the wind pressure gives the idle fan a little push. The cart starts rolling downwind, spinning the generator which powers the fan. Take it from there. But remember the wheels rolling on the ground. For starters, assume no slip, no energy storage, no friction, no loss.
You do realize you try to describe an overunity device ? If vehicle is direct down wind aboe wind speed.
Wind pushes the vehicle so instead of leaving it at that you brake the vehicle to extract energy at the wheel than then you put in to the propeller to get at best about 70% back in to propulsion.
What happens is as you describe when vehicle starts and that is the vehicle is pushed by wind same as any sail vehicle then instead of using all that to accelerate the vehicle you take a big part of the energy at the wheel and use the propeller to create a pressure differential (basically store that energy).
So if you have 1000W available as wind power you use say only 200W to accelerate the vehicle while taking the other 800W at the wheel and due to efficiency maybe put around 500W in storage so after 1ms of this you get to store about 0.5Ws
As vehicle accelerates the amount of wind power available from wind drops but the amount provided by the stored energy increases and this allows vehicle to get to a speed well above wind speed but then after stored energy is used up it will slow down below wind speed.
I think a simple vehicle to understand that will do exactly the same thing will be:
A simple sail vehicle with an electrical generator at the wheel so when vehicle is well below wind speed you can charge a supercapacitor or a battery but a supercapacitor will be sufficient then after some seconds when you consider enough energy was stored you can accelerate direct downwind for as long as you have stored energy. So you can do a fast acceleration to say 2x the wind speed maybe in 30 seconds or you can do a very slow acceleration say over 5 minutes to same 2x wind speed but in both cases you only have enough energy to accelerate the vehicle to that peak speed as energy stored in supercapacitor is converted in vehicle kinetic energy.
And from that peak speed say 2x the wind speed it will again take a few minutes to slow down to wind speed depending on the amount of friction losses vs the amount of gained kinetic energy.
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Classic troll behavior. When losing the debate on the original problem, come up with some entirely different problem as a distraction, and try to argue that. And when that argument doesn't go well, keep trying to move the goalposts in order to get the conclusion desired. And than try to say that any conclusion drawn from the entirely different problem must also apply to the original problem.
Sorry I must have missed the part where you demonstrated that the equation for available wind power was wrong ?
If you want to say I lost the argument then you need to show proof the equation everyone is using is wrong.
So show the output from this free online calculator is not corresponding with real world experiments https://www.electromotive.eu/?page_id=12&lang=en (https://www.electromotive.eu/?page_id=12&lang=en)
If you can do that I will admit to be wrong and apologize to everyone.
If not that equation alone is enough to prove that this sort of vehicle's do not work as you claim and most certainly work the way I explain and guaranteed they use energy storage.
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Or, DDWFTTW: The cart wheels drive a generator, the generator powers a fan blowing backwards. To start, the wind pressure gives the idle fan a little push. The cart starts rolling downwind, spinning the generator which powers the fan. Take it from there. But remember the wheels rolling on the ground. For starters, assume no slip, no energy storage, no friction, no loss.
You do realize you try to describe an overunity device ?
Of course it isn't - there are limits to how much it works. The many practical examples should have shown that it does work, it's just that with your closed and single-track mind you'll never grasp it. Perhaps you're afraid of doing so after all the nay saying you've staked your reputation on (what little of it remains).
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Of course it isn't - there are limits to how much it works. The many practical examples should have shown that it does work, it's just that with your closed and single-track mind you'll never grasp it. Perhaps you're afraid of doing so after all the nay saying you've staked your reputation on (what little of it remains).
There is a simple question.
What is the wind power available to a wind powered vehicle ?
My answer (and not just mine) is this https://scienceworld.wolfram.com/physics/DragPower.html
That equation alone is what proves "your" explanation of how this vehicle's work wrong.
All you need to do to prove me wrong is do an experiment that shows that equation is incorrect.
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The equation / expression alone does not prove anything: one has to use the formulas in the right context. Just because a formula applies to one context does not mean it is also correct in another context.
An important point constantly wrong by electrodacus is assumung a balance of power to decide if a vehicle can move. The seems to be his starting point leeding to lots of stupid mistakes and claims.
The more relevant point is the balance of forces.
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You do realize you try to describe an overunity device ? If vehicle is direct down wind aboe wind speed.
But I didn't describe an over-unity device. If I had, the device would accelerate with zero (ground-referenced) wind speed. Nobody is claiming this, and no experiments have shown this, because this *ISN'T* over-unity.
Your cherished equation for "wind power available to a wind powered vehicle" applies to an isolated wind-generator mounted on a moving vehicle, or mounted on the ground, assuming the equation is correct, which may be the case. Yes, of course, at zero relative windspeed that wind-generator will deliver no power. But that's not the situation, and your equation does not apply.
Stop thinking of the propeller as a wind-generator. That's not how it's operating in the DDWFFTW situation. The propeller provides the propulsion -- it's a "pusher prop".
Please go back to the wheels, gears, and two surfaces model. Try to understand how the gears are not locked. Otherwise you are doomed.
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The equation / expression alone does not prove anything: one has to use the formulas in the right context. Just because a formula applies to one context does not mean it is also correct in another context.
An important point constantly wrong by electrodacus is assumung a balance of power to decide if a vehicle can move. The seems to be his starting point leeding to lots of stupid mistakes and claims.
The more relevant point is the balance of forces.
Yes you are correct in the fact that formula needs to be applied in the correct context also.
So here is the simplest test that shows what I'm claiming.
Have an electric vehicle (simpler that way) even an ebike drive directly upwind within a known wind speed and see what power is needed to start moving.
All of you seems to be saying this below is power needed by the motor (correct me if you think something else) when vehicle wants to move upwind:
P = Fd * vehicle speed
I (and others) say:
P = Fd * (wind speed - (-vehicle speed)) = Fd * (wind speed + vehicle speed).
The difference is so large that it will be easy to measure.
This calculator can be used for the equation I agree with https://www.electromotive.eu/?page_id=12&lang=de (https://www.electromotive.eu/?page_id=12&lang=de)
A day with 30km/h winds should be easy to find (right now at my location is 50km/h with 70km/h gust)
Anyone that biked with a 30km/h wind gust will know what a struggle that is so with my prediction power needed to be provided to drive at 5km/h is
We use 0.827 as effective projected area as default in that calculator
Fd = 0.5 * 1.2 * 0.827 * (35/3.6)2 = 46.9N (we all agree with this)
Then power needed for propulsion I say
P = 46.9 * (35/3.6) = 456W a bit less than what that calculator will show 554W as that takes rolling resistance and drivetrain efficiency in to account.
But you say
P = 46.9 * (5/3.6) = 65W and that is such a large difference that is very easy to verify.
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If you are staying at zero speed then no work is being expended. A force maybe, but if there is no movement then no power is being used.
Have you ever had an electric motor providing a force / torque without using energy ? Or even simpler an electromagnet providing a force without using energy.
Even you using your arm to provide a force will be using energy even tho there is no movement and yes that energy will be dissipated as heat.
A superconducting motor or electromagnet would continue to supply a non-moving force without any ongoing power being consumed once initially energised. For that matter, so would a permanent magnet. I have seen that “supplying continuous power without movement” argument used by the free energy from magnets crowd.
As far as your arm needing energy to produce a non moving force, that’s true, but you could also get that force by for example driving a wedge into a narrow space. Once you have stopped pressing it in you are no longer putting energy into it but it continues to exert a force against an object. Are we to assume it continues to dissipate energy into this object indefinitely? A force, certainly, but there is no movement, so where is the energy?
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Driving 5 km/h against a 30 km/h head wind is relatively easy, if you have suitable low gears (e.g. montain bike). The difficult point is more to keep the balance at slow speed, if gusty.
5 km/h is about walking speed and when walking 30 km/h head wind is not hard either.
Providing 460 W is in contrary only possible for a short time for a top athlet.
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A superconducting motor or electromagnet would continue to supply a non-moving force without any ongoing power being consumed once initially energised. For that matter, so would a permanent magnet. I have seen that “supplying continuous power without movement” argument used by the free energy from magnets crowd.
As far as your arm needing energy to produce a non moving force, that’s true, but you could also get that force by for example driving a wedge into a narrow space. Once you have stopped pressing it in you are no longer putting energy into it but it continues to exert a force against an object. Are we to assume it continues to dissipate energy into this object indefinitely? A force, certainly, but there is no movement, so where is the energy?
Opposing magnets and springs can store energy. So there is a potential stored energy. So that wedge you mention takes advantage of materials elastic properties.
You do miss the point that none of your examples will work without brakes.
You have a vehicle on wheels with no brakes that needs to keeps his relative position on the ground using an electric motor.
The main question is the one currently discussed with Kleinstein. That is how much power is needed to advance at any low speed against the wind.
There is an equation for that and we disagree on which one is the correct one.
If the correct one is the one I claim it is then that alone is enough proof that no vehicle can drive upwind without energy storage.
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Driving 5 km/h against a 30 km/h head wind is relatively easy, if you have suitable low gears (e.g. montain bike). The difficult point is more to keep the balance at slow speed, if gusty.
5 km/h is about walking speed and when walking 30 km/h head wind is not hard either.
Providing 460 W is in contrary only possible for a short time for a top athlet.
I know all this about walking speed and the 300 to 500W sustained by a human is possible for at least a few minutes that is why I selected the values for speed.
The difference is just huge 460W to 560W when frictional losses are added is very different from 65 to 150W you may be claiming.
As an example that calculator will say 85W for 5km/h with no wind so your claim is that wind makes almost no difference when that is not the case.
Also your theory is not consistent.
For example if you release the brakes and let the wind pushing you what will the power provided by the wind be with wind at 30km/h ?
Changing gears will not help with amount of power needed.
If you change in a lower gear yes the force you need to apply at the pedal is lower but the speed will be higher so exactly the same amount of power will be needed.
You will say that since bicycle is at 0km/h the wind power available to accelerate the bicycle will be zero.
While I say is almost 300W witch again correspond to what you will observe in reality.
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Here's some of the actual physics related to this phenomena, for anyone interested:
Let \$v_v\$ be the vehicle speed with respect to ground, and \$v_w\$ be the wind speed with respect to ground.
For a vehicle that is in continuous contact with ground (through e.g. wheels), the amount of power available from the wind is
$$P_\text{in} = \frac{\kappa}{2} \rho A_\text{in} v_w^3 \tag{1}\label{1}$$
where \$\kappa\$ is the extraction efficiency, up to about 0.593 in theory (Betz's coefficient), and up to about 0.475 in existing devices; \$\rho\$ is the density of air; and \$A_\text{in}\$ is the cross-sectional area of the extraction device.
In addition to the ground friction losses, the amount of power lost via drag is
$$P_\text{out} = \frac{C_D}{2} \rho A_\text{out} \lvert v_w - v_v \rvert^3 \tag{2}\label{2}$$
where \$C_D\$ is the drag coefficient depending on the shape of the vehicle (varies between approx. \$0.02\$ for airfoils to \$0.5\$ for golfballs and similar), and can be different for \$v_w \gt v_v\$ than for \$v_w \lt v_v\$ (i.e. front-to-back drag coefficient can be and often is different than back-to-front drag coefficient); and \$A_\text{out}\$ is the cross-sectional area of the vehicle.
The only situation where using \$P_\text{out}\$ to model the amount of energy available from airflow to a vehicle makes any sense at all, is when that vehicle is in flight and only uses airflow propulsion, and \$C_D\$ is used to denote the energy extraction efficiency \$\kappa\$, noting that there is absolutely no physical reason or mathematical relation tying \$\kappa\$ to \$D_C\$.
For example, existing airfoil-type wind turbines, ignoring the tower, have \$A_\text{in} = A_\text{out}\$, \$v_v = 0\$ (stationary), and \$\kappa \gt C_D\$. In current commercial wind turbines, \$C_D \lt 0.1\$, with \$\kappa \gt 0.4\$, i.e. \$\kappa \approx 4 C_D\$. If they had \$P_\text{out} = P_\text{in}\$, they could not produce any power at all. It is the difference between \$\kappa\$ and \$C_D\$ that allows a stationary object to extract power from wind.
The reason there is no \$v_v\$ term (typo fixed) in \$\eqref{1}\$ is that the fact that the contact to ground, even if rolling, allows the mechanism to balance static forces with forces against the ground. A similar effect with the keel is extensively used in sailing; the keel having very little drag in the direction of travel, but enormous drag in the perpendicular direction, allowing the keel to be used to balance the forces exerted by the wind to the vehicle. In recent decades, hydrofoils are being used more and more, lifting the vessel upwards, reducing drag (\$C_D\$) due to smaller cross-sectional area \$A_\text{out}\$, without increasing the total drag much at all.
In very rough terms, a sufficiently clever land vehicle can balance the forces from the wind with static forces through the contact with the ground. Such mechanisms will have some small losses, but at least in theory, they can be reduced without any known lower limit.
While we currently believe the upper maximum limit for \$\kappa\$ is indeed about 0.593 as described by Betz's limit (at essentially \$C_D \approx 0\$), there are no known physical limitations for the ratio of \$\kappa\$ to \$C_D\$: that is just an engineering problem.
Indeed, the Mythbusters experiment by dimpling a Ford Taurus with over a thousand golfball-like dimples, increasing the fuel efficiency from 26 to 29 MPG, shows that many widely used vehicles have an unreasonably large \$C_D\$. This affects what humans perceive as normal, and results in quite normal but efficient mechanisms obeying Newtonian physics to defy human intuition. Those interested in such quirks, may find the misconceptions related to Bumblebee flight (https://en.wikipedia.org/wiki/Bumblebee#Misconception_about_flight) informative.
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Mechanical power is force times (inner product if using vectors) speed. So at a low speed there is little power needed. How much depends on the motor and gearing. With a suitable low gear the power can be very low and the motor internal friction may be enough. To provide some braking (intentionally avoid the word power here). Common languish english seems to use the word "power" also for a few different things than the physical power.
One can use this also to convert physical power back to the force. So the force is power divided by the speed.
I think it should get obvious that the idea of an essentially fixed power even at low speed leads to a contradiction, as this means the force about doubles when you half the speed.
So there must be some wrong point leading to deverging large forces: The formula or power = force times speed is universally accepted, sometimes even used as definition. So the wrong point is assuing the essentially (vehicle much slower than the wind) constant power from the kinetic energy of the wind as the power needed to move the vehicle.
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You will say that since bicycle is at 0km/h the wind power available to accelerate the bicycle will be zero.
While I say is almost 300W witch again correspond to what you will observe in reality.
The mechanical power is force times speed. So how on earth can you get 300 W with 0 speed as one factor.
A limited force and zero speed naturally gives zero power - where is the problem ?
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Here's some of the actual physics related to this phenomena, for anyone interested:
Let \$v_v\$ be the vehicle speed with respect to ground, and \$v_w\$ be the wind speed with respect to ground.
For a vehicle that is in continuous contact with ground (through e.g. wheels), the amount of power available from the wind is
$$P_\text{in} = \frac{\kappa}{2} \rho A_\text{in} v_w^3 \tag{1}\label{1}$$
where \$\kappa\$ is the extraction efficiency, up to about 0.593 in theory (Betz's coefficient), and up to about 0.475 in existing devices; \$\rho\$ is the density of air; and \$A_\text{in}\$ is the cross-sectional area of the extraction device.
In addition to the ground friction losses, the amount of power lost via drag is
$$P_\text{out} = \frac{C_D}{2} \rho A_\text{out} \lvert v_w - v_v \rvert^3 \tag{2}\label{2}$$
OK so you agree with the correct equation for power available to any wind powered vehicle. Keep in mind that first equation is only for a stationary vehicle.
All you did is add the efficiency while I took the ideal case with 100% efficiency to be generous.
Glad you find the equations somewhere and agree with what I'm saying.
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The mechanical power is force times speed. So how on earth can you get 300 W with 0 speed as one factor.
A limited force and zero speed naturally gives zero power - where is the problem ?
Wind speed is not zero is 30km/h
Vehicle speed is zero.
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So the wrong point is assuing the essentially (vehicle much slower than the wind) constant power from the kinetic energy of the wind as the power needed to move the vehicle.
That is exactly the case when vehicle drives upwind. Once you understand that you will understand all that I was saying.
To give you maybe a more intuitive example.
Imagine the vehicle is colliding with balls say 8 balls of 1.2kg each per second. That amount of lost kinetic energy will need to be replaced by the motor in order to maintain speed.
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Common languish english seems to use the word "power" also for a few different things than the physical power.
I'm not sure how it translates in German, but in English technical language, thermodynamics has the two concepts of work and heat (first law: ΔU = Q - W). Mechanical work is force times distance, while heat relates to the molecular energy of materials. It gets confusing with power, because power applies to the rate of doing work, and also to the rate of transfer of heat (there are no different words for mechanical power and thermal power). Thus, you can have a 10 kW gas furnace that does no work at all.
Because of this, careful thermodynamic arguments tend to use Work and Heat as the terms of reference in order to be precise.
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There is a simple question.
What is the wind power available to a wind powered vehicle ?
My answer (and not just mine) is this https://scienceworld.wolfram.com/physics/DragPower.html
That equation alone is what proves "your" explanation of how this vehicle's work wrong.
No it doesn't. Let's just assume that the equation is correct (there is some doubt) - you're not applying it appropriately. Specifically, you're ignoring the thrust from the propeller. There is thrust (it's a turning propeller, after all) so where does that figure in your equation?
All you need to do to prove me wrong is do an experiment that shows that equation is incorrect.
Pointless. There are now a large number of such experiments showing that the thing works - pick any of those and there is your proof. All you're saying there is "I want you to waste your time and then I'll find some dubious reason why I'll still ignore the evidence." If the equation says to you that those experiments that work can't actually work, then either the equation is wrong or you're using the wrong one.
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Keep in mind that first equation is only for a stationary vehicle.
No, don't keep that in mind. The formula is also for moving vehicles. Stop this ridiculous behavior of changing what people say and then pretending to agree with them.
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No, don't keep that in mind. The formula is also for moving vehicles. Stop this ridiculous behavior of changing what people say and then pretending to agree with them.
Where in that equation do you see the vehicle speed ?
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Where in that equation do you see the vehicle speed ?
Also, stop this ridiculous behavior of asking snarky and patronizing questions.
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No it doesn't. Let's just assume that the equation is correct (there is some doubt) - you're not applying it appropriately. Specifically, you're ignoring the thrust from the propeller. There is thrust (it's a turning propeller, after all) so where does that figure in your equation?
The equation is correct.
We are discussing the simpler upwind version.
The propeller trust in the direct downwind version can only be lower than the braking at the wheel witch powers that propeller.
Pointless. There are now a large number of such experiments showing that the thing works - pick any of those and there is your proof. All you're saying there is "I want you to waste your time and then I'll find some dubious reason why I'll still ignore the evidence." If the equation says to you that those experiments that work can't actually work, then either the equation is wrong or you're using the wrong one.
"The thing works" When did I claim it does not ?
It just does not work the way you think it does.
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Also, stop this ridiculous behavior of asking snarky and patronizing questions.
I think that is a perfectly normal question to ask. If you think my questions are patronizing then that is your problem not mine.
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No, don't keep that in mind. The formula is also for moving vehicles. Stop this ridiculous behavior of changing what people say and then pretending to agree with them.
Where in that equation do you see the vehicle speed ?
If you had bothered to read Nominal's post for content, you would have seen this:
The reason there is no vw term in (1) is that the fact that the contact to ground, even if rolling, allows the mechanism to balance static forces with forces against the ground.
He should have written "no vv term", (there is a vW term in equation 1, but no vv), but from context the meaning should have been obvious.
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OK so you agree with the correct equation for power available to any wind powered vehicle.
Finally! Yes, thank you. Have you finally realized your error?
Keep in mind that first equation is only for a stationary vehicle.
Nope, you need to learn to read.
All you did is add the efficiency while I took the ideal case with 100% efficiency to be generous.
No, I wrote down the correct equations.
What you do, is take the amount of drag a vehicle experiences, and somehow equate that with the available power.
That is beyond stupid, replacing the entire field of aerodynamics with a singular concept, "drag", that somehow explains all phenomena.
You must be a troll, or someone on the autistic spectrum with severe learning and understanding difficulties. Nothing else can explain this denseness.
Any wagers on which one it is, anyone?
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If you had bothered to read Nominal's post for content, you would have seen this:
I had read his entire post. I know what I was asking.
He is getting confused by the two equations and how they apply.
If you look at the original post here the main discussion is about the direct downwind verizon as that is simpler to debunk.
There is only one equation that describes wind power available to a vehicle for any type of wind powered vehicle but applies also to a stationary wind turbine and also to power needed to overcome drag.
They are all one and the same thing as is all about elastic collisions between an object and air particles.
So a vehicle driving at 120km/h in a day without wind will need the same amount of power to overcome drag as a vehicle driving at 20km/h with a 100km/h headwind.
It is always about the wind speed relative to vehicle and if vehicle is stationary you will remain only with the wind speed in that equation.
It is irrelevant if air, vehicle or both move is all about the relative speed between the two.
So if a heavy car at 120km/h hits a stationary lightweight motorcycle the energy of the impact will be the same as if motorcycle drove at 100km/h and collided head on with the car driving at 20km/h.
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No, I wrote down the correct equations.
What you do, is take the amount of drag a vehicle experiences, and somehow equate that with the available power.
That is beyond stupid, replacing the entire field of aerodynamics with a singular concept, "drag", that somehow explains all phenomena.
You must be a troll, or someone on the autistic spectrum with severe learning and understanding difficulties. Nothing else can explain this denseness.
Any wagers on which one it is, anyone?
That is correct the equation is universal it can calculate the power required to overcome drag and that is also the same with ideal case available wind energy.
That is not stupid it is the reality.
If you have a stationary vehicle and release the brakes wind will accelerate that with the amount of power provided by that single equation.
Also if you want to know what is the amount of power to overcome drag so same vehicle wants to move upwind the equation will provide minimum power required to start moving assuming drag is the only loss.
I guess you need to do the experiment as you do not understand the concept that from any reference frame the result will be the same so it is irrelevant if air particles collide with vehicle or vehicle collides with air particles.
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He is getting confused by the two equations and how they apply.
Nope. They are perfectly in agreement with the experiments that you claim are misconstructed or exhibit some special slip-stick hysteresis or magical energy storage pixies.
They are also perfectly in alignment with my old University Physics textbooks on Classical Mechanics. I have done the work and passed those courses without problems. (Modern/quantum mechanics too, but that's not relevant here.)
The only thing there that can be discussed – among physicists – is how balancing static forces against the ground means the relative wind speed is irrelevant. To laymen, the analog to keels and hydrofoils in sailing vessels provides the concept and the practical example.
Nothing can affect the misconceptions your kind of people have. You refuse to acknowledge or even perceive anything that is contrary to your preconceptions, and instead disregard them as being "wrong" somehow. That really reminds me of the people who insist that despite hundreds of millions of murders, communism is still a viable political system; it's just that because they themselves have not been in the lead, nobody has implemented it truly correctly yet. Their argument, too, is that until you can prove it would not work when they are in charge, it is proven to work and everybody else is wrong. Invalid logic, irrational thinking, and evasive random argumentation, just to prop up your ego and misplaced beliefs of your own 'understanding'.
It is horrible, and yet interesting, in the pathological sense. There are those who interview monsters, so why not engage with delusional lunatics pushing irrational concepts?
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He is getting confused by the two equations and how they apply.
Nope. They are perfectly in agreement with the experiments that you claim are misconstructed or exhibit some special slip-stick hysteresis or magical energy storage pixies.
They are also perfectly in alignment with my old University Physics textbooks on Classical Mechanics. I have done the work and passed those courses without problems. (Modern/quantum mechanics too, but that's not relevant here.)
The only thing there that can be discussed – among physicists – is how balancing static forces against the ground means the relative wind speed is irrelevant. To laymen, the analog to keels and hydrofoils in sailing vessels provides the concept and the practical example.
Nothing can affect the misconceptions your kind of people have. You refuse to acknowledge or even perceive anything that is contrary to your preconceptions, and instead disregard them as being "wrong" somehow. That really reminds me of the people who insist that despite hundreds of millions of murders, communism is still a viable political system; it's just that because they themselves have not been in the lead, nobody has implemented it truly correctly yet. Their argument, too, is that until you can prove it would not work when they are in charge, it is proven to work and everybody else is wrong. Invalid logic, irrational thinking, and evasive random argumentation, just to prop up your ego and misplaced beliefs of your own 'understanding'.
It is horrible, and yet interesting, in the pathological sense. There are those who interview monsters, so why not engage with delusional lunatics pushing irrational concepts?
You know why engineers exist ? It is so that physicist have someone to look-up to :)
Is sad but you will be only convinced by doing the test and seeing the measurement results. They will match what this calculator provides https://www.electromotive.eu/?page_id=12&lang=en (https://www.electromotive.eu/?page_id=12&lang=en)
You do not have a good understanding of what power and energy is and always use just forces that clearly are not properly understood also.
Forces come in pairs always equal and opposite unless they accelerate an object.
Your first equation is a general case for a stationary object relative to ground like a wind turbine fixed to ground.
Second equation is the same equation but for the case where objects moves relative to ground so the speed is the relative speed between wind and the object and it is the same equation I linked quite a few times https://scienceworld.wolfram.com/physics/DragPower.html (https://scienceworld.wolfram.com/physics/DragPower.html) and all of you disagreed with.
The above linked calculator implements this equation correctly.
I need to remind you that all of you not even just on this forum used this equation that is completely wrong meaning it has no applications.
Same speed will be used for calculating the Force to overcome drag as it to calculate power to overcome drag.
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Second equation is the same equation but for the case where objects moves relative to ground so the speed is the relative speed between wind and the object and it is the same equation I linked quite a few times https://scienceworld.wolfram.com/physics/DragPower.html (https://scienceworld.wolfram.com/physics/DragPower.html) and all of you disagreed with.
That equation describes the power needed to overcome drag. It has nothing to do with harvesting power from the wind, because harvesting power from the wind is not based on drag; the most efficient ones are based on aerodynamic lift (https://en.wikipedia.org/wiki/Lift_(force)). Lift and drag are not collinear or opposite, they are typically perpendicular forces, and their magnitudes are only very loosely coupled together, based on the shape of the airflow, its orientation with respect to the airflow, and the relative speed of the airflow. In most useful airfoils, lift is much, much greater than drag.
Scoop-type turbines, like the Savonius vertical-axis wind turbine, are based on drag, but they have a very low efficiency. There are other "drag" type airfoils that rely on the difference in drag between the two airfoil surfaces. All those are less efficient than "lift" type airfoils, and their tip velocity – the radial velocity at the blades – is slower than the relative wind speed. For "lift" type airfoils, the radial velocity is much higher than the relative wind speed; they rotate faster than the wind.
Thus, the drag power equation has no relevance at all to limits of harvesting power from the wind. None. It only describes aerodynamic losses, and certain very poor turbine models, nothing more.
As an example, consider the Gorlov helical turbine I described some time back. It has three airfoils, narrow wing-like elements, that "coil" around the common axis. In wind tunnel tests, it has reached \$\kappa = 0.35\$, with drag coefficient on the order of \$C_D \approx 0.05\$). More interestingly, even the drag force vectors are not parallel to the wind, so a vehicle using one could possibly use the small drag as kind of a keel. Moreover, it typically rotates much faster than the wind speed relative to its axis.
(In water, according to Bachant and Wosnik's experiments in 2011 (https://tethys.pnnl.gov/sites/default/files/publications/Bachant-and-Wosnik-2011.pdf), Gorlov helical turbine lift to drag ratio can reach 70, i.e. \$\kappa \approx 70 C_D\$.)
I don't know how more clearly that can be stated. You claim I don't know this or that, only because you live in your own private little world with your private little custom physics. I do not know those, that's for sure; nobody can.
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Second equation is the same equation but for the case where objects moves relative to ground so the speed is the relative speed between wind and the object and it is the same equation I linked quite a few times https://scienceworld.wolfram.com/physics/DragPower.html (https://scienceworld.wolfram.com/physics/DragPower.html) and all of you disagreed with.
That equation describes the power needed to overcome drag. It has nothing to do with harvesting power from the wind, because harvesting power from the wind is not based on drag; the most efficient ones are based on aerodynamic lift (https://en.wikipedia.org/wiki/Lift_(force)). Lift and drag are not collinear or opposite, they are typically perpendicular forces, and their magnitudes are only very loosely coupled together, based on the shape of the airflow, its orientation with respect to the airflow, and the relative speed of the airflow. In most useful airfoils, lift is much, much greater than drag.
Scoop-type turbines, like the Savonius vertical-axis wind turbine, are based on drag, but they have a very low efficiency. There are other "drag" type airfoils that rely on the difference in drag between the two airfoil surfaces. All those are less efficient than "lift" type airfoils, and their tip velocity – the radial velocity at the blades – is slower than the relative wind speed. For "lift" type airfoils, the radial velocity is much higher than the relative wind speed; they rotate faster than the wind.
Thus, the drag power equation has no relevance at all to limits of harvesting power from the wind. None. It only describes aerodynamic losses, and certain very poor turbine models, nothing more.
As an example, consider the Gorlov helical turbine I described some time back. It has three airfoils, narrow wing-like elements, that "coil" around the common axis. In wind tunnel tests, it has reached \$\kappa = 0.35\$, with drag coefficient on the order of \$C_D \approx 0.05\$). More interestingly, even the drag force vectors are not parallel to the wind, so a vehicle using one could possibly use the small drag as kind of a keel. Moreover, it typically rotates much faster than the wind speed relative to its axis.
(In water, according to Bachant and Wosnik's experiments in 2011 (https://tethys.pnnl.gov/sites/default/files/publications/Bachant-and-Wosnik-2011.pdf), Gorlov helical turbine lift to drag ratio can reach 70, i.e. \$\kappa \approx 70 C_D\$.)
I don't know how more clearly that can be stated. You claim I don't know this or that, only because you live in your own private little world with your private little custom physics. I do not know those, that's for sure; nobody can.
There is only one way that air interacts with an object and that is trough elastic collisions between object and air molecules.
So the same equation applies to any type of wind turbine as well as to any wind powered vehicle and also to any other vehicle when you want to know the power needed to overcome drag.
This will be that equation
P = 0.5 * air density * (area * drag coefficient) * (wind speed - vehicle speed)3
So if it is a propeller type wind turbine you use the swept area of the propeller thus there is no drag coefficient involved or you can consider that as 1 and the vehicle speed is zero if the wind turbine is mounted on the ground.
So for wind turbine that same equation will look like (still the same equation)
P = 0.5 * air density * swept area * (wind speed)3 * efficiency (basically the first equation you mentioned in an earlier comment).
If the wind turbine is installed on a vehicle that drives say upwind same equation will look like this
P = 0.5 * air density * swept area * (wind speed-(-vehicle speed))3 * efficiency
If it is a vehicle driving direct downwind same equation will look like
P = 0.5 * air density * area * coefficient of drag * (wind speed-vehicle speed)3 this is the wind power available to vehicle ideal case
If vehicle drives upwind the same equation shows the power needed to overcome drag
P = 0.5 * air density * area * coefficient of drag * (wind speed-(-vehicle speed))3
So there is no other equation it just looks a bit different depending on what it applies to but is the exact same one in all cases.
Look at energy balance for example.
You agree with drag force and I hope you agree with the Kinetic energy loss or gain (depending on what powers what) is drag force times distance so basically the amount of work done.
KE = Fd * d
Getting back to that 5km/h bicycle example with a 30km/h headwind
distance traveled in one second will be 5km/h / 3.6 = 1.39m/s so 1.39m
KEvehicle = (0.5*1.2*0.827*(35/3.6)2) * 1.39 = 65.2Ws
Now you need to calculate the energy needed to deal with the air drag
That is almost the same just you need to add the distance air traveled over that same 1 second period
KEair = (0.5*1.2*0.827*(35/3.6)2) * 8.33 = 391Ws
Then you will add this two 391 + 65.2 = 456Ws
So total energy that should be provided by vehicle propulsion for one second is 456Ws so basically the same as what average power needs to be.
Also say you want to be powered by the wind so you do not use the motor just let the wind push the bike your acceleration power will be 391W and will drop as the bicycle speed increases so you will need to integrate.
So say over 1ms that 391W will not change then bicycle gained 0.391Ws of kinetic energy and from that and mass of the bike + rider you get the speed.
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The interaction of the wind with the vehicle is no way close to 100% efficient - with a standing obstacle it is more like 0%. Even a good wind turbine is not close to 100% efficient. So it does not make sense to base the calculation on the power that the wind could provide with 100% efficiency.
So it is not about an energy or power balance, but about the forces.
You agree with drag force and I hope you agree with the Kinetic energy loss or gain (depending on what powers what) is drag force times distance so basically the amount of work done.
KE = Fd * d
Getting back to that 5km/h bicycle example with a 30km/h headwind
distance traveled in one second will be 5km/h / 3.6 = 1.39m/s so 1.39m
KEvehicle = (0.5*1.2*0.827*(35/3.6)2) * 1.39 = 65.2Ws
Now you need to calculate the energy needed to deal with the air drag
That is almost the same just you need to add the distance air traveled over that same 1 second period
KEair = (0.5*1.2*0.827*(35/3.6)2) * 8.33 = 391Ws
The power you calculate for the air drag is coming from the wind and not form vehicle to drive. It gets obvious for a stationary obstacle - not power needed by the obstacle to keep standing. There is no sudden change from stationaly objects and very slow moving objects. The air drag power is taken from wind and converted to heat. A wind turbine could use part of that power for other purposes, like driving the vehicle.
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If it is a vehicle driving direct downwind same equation will look like
P = 0.5 * air density * area * coefficient of drag * (wind speed-vehicle speed)3 this is the wind power available to vehicle ideal case
This is where you are going wrong. Let's try and fix it...
Once again I will say that previously I would have given you one step and let you agree on it before moving to the next, because otherwise you just jump to the chase and insist it can't be done, skipping everything about how we get there. But life's too short and it's your loss.
So, what we want to know is how much power is available to make the vehicle move (because, ultimately, we need power to make it move faster than the wind is blowing). Your Someone's copied equation can be made simpler thus:
Terms:
Pu - Useful power. This is what we want to use to move the vehicle
Vw - Wind velocity. Importantly, this is relative to the ground and in the same direction as the vehicle moves.
Vv - Vehicle velocity. Again, relative to the ground and in the same direction as the wind
The details (right now) of how Pu is derived from Vw and/or Vv are unimportant - it is sufficient to say that Pu is proportional to Vw - Vv. That is:
Pu ∝ (Vw - Vv)
Your premise is that when the vehicle is at wind speed there is no power available. That is, Pu = 0. And that's correct for this simple case. What we're interested in is a case where Vw - Vv = 0 and Pu > 0.
Moving on, there's the small problem of the propeller, which is driven by the wheels. The propeller is sucking power, via the wheels, so let's add another term:
Pv - Vehicle power. This is the power the wheels soak up when driving the propeller.
That changes things thus:
(Pu + Pv) ∝ (Vw - Vv)
or
Pu ∝ (Vw - Vv) - Pv
Two things to note here: first that now when Pu = 0, Vw - Vv > 0. Second, Pw is proportional to the vehicle speed since it depends on how fast the wheels turn:
Pv ∝ Vv
Next, another term is needed for the effect of the propeller:
Vt = Thrust velocity. This is the speed of the air being pushed backwards relative to the vehicle.
We know from seeing jets take off that the vehicle speed is wind speed plus thrust, and you even agreed earlier in this thread.
Vv = Vw + Vt
What makes this complicated is that Vt is also proportional to the vehicle speed since the power to turn the prop is derived from the speed of the wheels:
Vt ∝ Vv
OK, so to fill in the blanks, the usable power is proportional to the wind speed less the vehicle speed less the thrust velocity, minus the power to produce the thrust:
Pu + Pv ∝ Vw - (Vv - Vt)
or
Pu ∝ (Vw - (Vv - Vt)) - Pv
Essentially, this means that Vw = Vv - Vt can't occur because of the power Pv used up by the prop. In words, there isn't enough power from the wind to have the vehicle speed less thrust get up to wind speed. But, by the same token, if the thrust Vt is high enough then the vehicle speed Vv can be higher than wind speed whilst the same constraint exists. Pv, thus Vt, is proportional to wheel speed, so the faster the wind the higher these are and at some value (left for someone clever to work out if they can be arsed) we will find that Vt > Vw when Pu != 0.
Edit: noted that Vt is relative to the vehicle whereas Vw and Vv are reltive to the ground.
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The interaction of the wind with the vehicle is no way close to 100% efficient - with a standing obstacle it is more like 0%. Even a good wind turbine is not close to 100% efficient. So it does not make sense to base the calculation on the power that the wind could provide with 100% efficiency.
So it is not about an energy or power balance, but about the forces.
You are correct that a wind turbine is no more than 40 to 45% efficient but a sail is way more efficient approaching ideal.
In any case the reason for using ideal numbers is so that there can be no discussion about improving efficiency as even with ideal case no vehicle can drive upwind without using energy storage and a trigger mechanism.
No sail based vehicle can drive directly upwind not because the wind power is zero at that point is actually maximum is because that power can only be used to accelerate the vehicle direct downwind not direct upwind.
So if a stationary vehicle has 300W of wind power available when stationary it can accelerate down wind using all that but it can not move upwind at all unless it has >300W
Changing the sail for anything else like a less efficient wind turbine will not help things as you still need more than wind turbine power in order to accelerate upwind so what happens is that a very small energy storage is involved and that will be charged then used to move upwind even for a super small distance then energy storage can be recharged and things will repeat many times a second but human brain will translate that as continues motion so you will in most cases need a slow motion video to observe what happens.
And I showed that video many times and it was dismissed.
The power you calculate for the air drag is coming from the wind and not form vehicle to drive. It gets obvious for a stationary obstacle - not power needed by the obstacle to keep standing. There is no sudden change from stationary objects and very slow moving objects. The air drag power is taken from wind and converted to heat. A wind turbine could use part of that power for other purposes, like driving the vehicle.
Actually the stationary obstacle is moving is just locked to earth so it is moving together with earth. The earth is accelerated by that wind hitting the obstacle but since on the planet wind is moving in multiple direction it all basically cancels out not to mention the earth mass vs air density.
So you will need to ignore the case where brakes are engaged.
Also thinking in terms of forces gets you much higher chances of making mistakes but in both cases you will get the same result.
A sail vehicle can also drive upwind but in order to do so it will need first to charge the only type of energy storage it has (kinetic energy storage) and so drive at an angle maybe even perpendicular to wind direction and when it has enough stored energy it can just change direction to directly upwind and yes it will slow down as stored energy is used up but it will move significant distance upwind.
This things happen much slower so anyone can observe but the propeller based moves directly upwind at all times is just that it accelerates and stops accelerating multiple times per second as the very small energy storage is charged and discharged at many Hz so human brain perceive/interpret that as smooth motion.
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Pu ∝ (Vw - Vv)
Your premise is that when the vehicle is at wind speed there is no power available. That is, Pu = 0. And that's correct for this simple case. What we're interested in is a case where Vw - Vv = 0 and Pu > 0.
That is the same thing with what I say is just that is correct for any case unless it involves energy storage. You may call that stored wind power.
Moving on, there's the small problem of the propeller, which is driven by the wheels. The propeller is sucking power, via the wheels, so let's add another term:
Pv - Vehicle power. This is the power the wheels soak up when driving the propeller.
That changes things thus:
(Pu + Pv) ∝ (Vw - Vv)
All this and what followed is incorrect.
I will try to explain so please read:
This Pu is all that it is so when you take what you call Pv from the wheel you are actually subtracting that from Pu
If that Pu is 1000W that is what is available to accelerate the vehicle in the wind direction and so if you take 700W at the wheel you are left with only 300W to accelerate the vehicle so the vehicle acceleration is reduced.
Now you can take those 700W and send it to propeller that will be used for propulsion and so if this was 100% efficient and air was not a compressible fluid then all this 700W will end up accelerating the vehicle adding to the existing 300W so 1000W in total thus you did nothing compared to original where all 1000W where directly provided by wind.
The reality is that air is a compressible fluid so very little of those 700W you put in propeller will result as net thrust as it pushes against a compressible fluid the equivalent of a spring so most of that is stored so pressure differential is increased.
The more energy you put there the more pressure differential will increase and sometime around when vehicle speed is about 0.3x wind speed direct down wind the power provided by the stored energy matches the power available from wind directly and from that point the stored energy is the main force pushing the vehicle and when vehicle is at wind speed all power comes from this created pressure differential that is also maintained moved along with the vehicle but is more complex to visualize.
This is also what pushes the treadmill vehicle that is indoor with zero wind speed. The treadmill provide vehicle with power as long as human keeps the vehicle constrained so all energy is used to create the pressure differential then when it is full it will be maintained and so when human releases the vehicle the force pushing the vehicle comes from this stored pressure differential and there is no longer any power coming from the treadmill.
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This Pu is all that it is so when you take what you call Pv from the wheel you are actually subtracting that from Pu
If that Pu is 1000W that is what is available to accelerate the vehicle in the wind direction and so if you take 700W at the wheel you are left with only 300W to accelerate the vehicle so the vehicle acceleration is reduced.
You are missing it still.
Yes, Pv is subtracted from Pu (that's what the equation says after all!) and, no doubt to your vast delight, when Vw-(Vv-Vt) is 0, Pu is 0 so there is nothing to extract from, hence no power to turn the prop. That's why Vv-Vt can never reach wind speed.
So, Vv-Vt must be slower than wind speed because around the point Pu is 0, Pv is still being sucked out leading to an actual negative Pu. The whole thing will slow (although, in reality accelerate1) until Pu = Pv.
If that Pu is 1000W
Stop with the putting in values until you have a proper equation to insert them into, and currently you don't. All you're doing is confusing things by having some arbitrary number you need to distribute somewhere, but you don't really know where.
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[1] Kind of. As we see from the experiments, it needs a helping had (some models, literally) to get up to speed whereupon the model will take over on its own. This is really just saying that it would slow down if for some reason (wind gust, kick up the pants) it did go overspeed.
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Stop with the putting in values until you have a proper equation to insert them into, and currently you don't. All you're doing is confusing things by having some arbitrary number you need to distribute somewhere, but you don't really know where.
I use values as people do not see a limit.
There is only wind power as input. There is absolutely nothing else.
There is a single equation describing the amount of power available to any wind powered vehicle and as you started to realize it is proportional with wind speed minus vehicle speed.
This equation alone is proof that no vehicle can exceed wind speed directly downwind as equation shows wind power available is zero.
And also same equation shows that no vehicle can drive directly upwind because in order to move upwind you need more than wind power.
So there is nothing to discuss other than if this equation is the correct one or not. Once you agree with that equation the conclusion is clear.
Now if you want to know how those vehicle work because they work then that is a separate discussion and involves energy storage.
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This equation alone is proof that no vehicle can exceed wind speed directly downwind as equation shows wind power available is zero.
You are applying it by ignoring the propeller output and power sink, so the equation is useless for this vehicle. I've shown you how the missing things fit in and you can't even say why that is wrong, other than "it's all incorrect".
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This equation alone is proof that no vehicle can exceed wind speed directly downwind as equation shows wind power available is zero.
You are applying it by ignoring the propeller output and power sink, so the equation is useless for this vehicle. I've shown you how the missing things fit in and you can't even say why that is wrong, other than "it's all incorrect".
That's because he's a troll, he's succeeding in wasting your time and frustrating you which is exactly what trolls seek to do. We already know that he's wrong because multiple people have built vehicles that do exactly what he says is not possible, there's no point in arguing.
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You're right, of course. I thought it useful to me to crystalise my thinking in that post, so not completely wasted.
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You are applying it by ignoring the propeller output and power sink, so the equation is useless for this vehicle. I've shown you how the missing things fit in and you can't even say why that is wrong, other than "it's all incorrect".
Propeller is powered by wind so it will not add up.
All you have as input is wind power.
Equation for available wind power cares only about the vehicle shape (coefficient of drag) and projected frontal area.
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That's because he's a troll, he's succeeding in wasting your time and frustrating you which is exactly what trolls seek to do. We already know that he's wrong because multiple people have built vehicles that do exactly what he says is not possible, there's no point in arguing.
Where I ever claimed any of those vehicles do things that are not possible?
They all work as shown but not for the reasons you think they do.
They will also not work as you will predict they will based on flawed theory.
For example the direct downwind faster than wind vehicles both the large blackbird and the small treadmill model will accelerate to a max speed (can be 2x or even 3x the wind speed) and then they will slow down as there is no longer any stored energy to cover the friction losses.
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Stop with the putting in values until you have a proper equation to insert them into, and currently you don't. All you're doing is confusing things by having some arbitrary number you need to distribute somewhere, but you don't really know where.
I use values as people do not see a limit.
Doing much of the math with numbers can be confusing, but it can sometimes also help. Getting obviously wrong resucts (e.g. reduculous numbers that contradict experiance) is one way to show that a claimed equation is obviously wrong. However this still does not work with electrodacus: he refuses to see how rediculous his claims are (e.g. walking against a 30 km/h head wind would need some 400 W of power and thus be a kind of heavy sports).
There is only wind power as input. There is absolutely nothing else.
There is a single equation describing the amount of power available to any wind powered vehicle and as you started to realize it is proportional with wind speed minus vehicle speed.
This equation alone is proof that no vehicle can exceed wind speed directly downwind as equation shows wind power available is zero.
And also same equation shows that no vehicle can drive directly upwind because in order to move upwind you need more than wind power.
So there is nothing to discuss other than if this equation is the correct one or not. Once you agree with that equation the conclusion is clear.
Now if you want to know how those vehicle work because they work then that is a separate discussion and involves energy storage.
The problem is that AFAIK there is no such simple, easy to accept equation for the maximum available power. At least I don't see such an equation that I would take for granted. So it would still need a proof or at least good plausibile explaination.
Aerodynamics can be be quite tricky and I would consider getting such an euqation and proving it rather hard. So the way via the available power is a hard one.
The equation claimed by electrodacus is known to be wrong. AFAIR his reasoning kind of works for a simple sail, but it does not apply for the fan/ porp case. There are cases when a prop can be more efficient than a sail. The calculations for the prop drivne vehicle show that one can get more power and one can go faster than the wind. So the equation claimed by electrodacus is proven wrong - it is just that he refuses to accept (and maybe understand - though it is relatively easy) the proof.
Beside the calculation there are also videos to show that it is possible to go faster than the wind / or the equivalent case of moving against the moving treadmil, driven by a prop. With the plausible forms of ernergy storage the times shown are well long enough to consider them to show a stationary case - so the energy storrage argument / read herring is dead. :horse:
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For example the direct downwind faster than wind vehicles both the large blackbird and the small treadmill model will accelerate to a max speed (can be 2x or even 3x the wind speed) and then they will slow down as there is no longer any stored energy to cover the friction losses.
Would you mind providing more information on this energy stored in the pressure differential behind the propeller? The amount, and duration of this storage would be helpful. There are equations for the dissipation of a pressurized area -- the one I've seen apply to bomb blast overpressure, and the rate of dissipation seems quite high, the pressure dissipates much too quickly to provide power beyond a few milliseconds. Of course there is also kinetic energy stored in the spinning propeller blades, and in the momentum of the vehicle. Do you have numbers for those as well?
You have obviously studied this in some detail, and I'm sure we would all be interested in what you have learned.
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Doing much of the math with numbers can be confusing, but it can sometimes also help. Getting obviously wrong resucts (e.g. reduculous numbers that contradict experiance) is one way to show that a claimed equation is obviously wrong. However this still does not work with electrodacus: he refuses to see how rediculous his claims are (e.g. walking against a 30 km/h head wind would need some 400 W of power and thus be a kind of heavy sports).
You may not get strong winds at your location so your intuition may be wrong. Also keep in mind that wind speed close to ground is lower and this is a theoretical calculation that asume same wind speed on the entire surface.
The problem is that AFAIK there is no such simple, easy to accept equation for the maximum available power. At least I don't see such an equation that I would take for granted. So it would still need a proof or at least good plausibile explaination.
Aerodynamics can be be quite tricky and I would consider getting such an euqation and proving it rather hard. So the way via the available power is a hard one.
The equation claimed by electrodacus is known to be wrong. AFAIR his reasoning kind of works for a simple sail, but it does not apply for the fan/ porp case. There are cases when a prop can be more efficient than a sail. The calculations for the prop drivne vehicle show that one can get more power and one can go faster than the wind. So the equation claimed by electrodacus is proven wrong - it is just that he refuses to accept (and maybe understand - though it is relatively easy) the proof.
Beside the calculation there are also videos to show that it is possible to go faster than the wind / or the equivalent case of moving against the moving treadmil, driven by a prop. With the plausible forms of ernergy storage the times shown are well long enough to consider them to show a stationary case - so the energy storrage argument / read herring is dead. :horse:
It will be your job to prove the equation is wrong as that equation is used by almost all engineers to build things that work in real world as predicted by it.
Equation is universal and the fan/prop is powered by the wind power provided by that equation.
I can build a vehicle with sail that exceeds wind speed and will not involve any propeller.
All you do is take wind energy at the wheel and store it in a battery. The same thing is done by the propeller version but instead of a battery it uses pressure differential to store energy.
If you could prove that vehicle (direct downwind blackbird or the analog treadmill model) can maintain above wind speed indefinitely then you will be right but nobody has proved that because they can not.
All those examples are incomplete and people rushed to get the wrong conclusion about how they work. Going as far as to modify correct equations and posible one of the reasons all those wrong equations are everywhere.
The large heavy blackbird needed about 6Wh less than half the energy in a cell phone battery to achieve that speed record that with pressure differential on each side of a 20m2 swept area propeller.
The small lightweight treadmill model will need orders of magnitude less energy storage.
I showed that calculating using that equation and alternative kinetic energy equation will provide the same result.
You will need to prove the same that calculating in this two different ways you get the same result.
The only alternative I seen to my equation is that incorrect one where for drag force you have (w-v) correctly but for power you want to use just vehicle speed.
All anyone needs to prove me wrong is show a vehicle powered by an electric motor upwind at very low speed and show that power needed for vehicle is lower than the equation I claim to be correct predicts.
And why do you not contact this guys https://www.electromotive.eu/?page_id=12&lang=en (https://www.electromotive.eu/?page_id=12&lang=en) to let them know they have a defective calculator and explanation on their website ?
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For example the direct downwind faster than wind vehicles both the large blackbird and the small treadmill model will accelerate to a max speed (can be 2x or even 3x the wind speed) and then they will slow down as there is no longer any stored energy to cover the friction losses.
Would you mind providing more information on this energy stored in the pressure differential behind the propeller? The amount, and duration of this storage would be helpful. There are equations for the dissipation of a pressurized area -- the one I've seen apply to bomb blast overpressure, and the rate of dissipation seems quite high, the pressure dissipates much too quickly to provide power beyond a few milliseconds. Of course there is also kinetic energy stored in the spinning propeller blades, and in the momentum of the vehicle. Do you have numbers for those as well?
You have obviously studied this in some detail, and I'm sure we would all be interested in what you have learned.
Yes you can watch my video starting at 10:30 https://www.youtube.com/watch?v=4Hol57vTIkE&t=630s (https://www.youtube.com/watch?v=4Hol57vTIkE&t=630s) calculations there do not include friction and that is important if you want to know how long it will take until vehicle will slow down below wind speed.
And starting at 15:35 you can see the details about pressure differential
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All anyone needs to prove me wrong is show a vehicle powered by an electric motor upwind at very low speed
The previously posted video shows the electric motor version. It DOES NOT WORK AT LOW SPEED as we keep pointing out - it has to get up to speed somehow and then it works fine.
I also stressed that in the explanation of why your love child of an equation is inappropriate.
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All anyone needs to prove me wrong is show a vehicle powered by an electric motor upwind at very low speed
The previously posted video shows the electric motor version. It DOES NOT WORK AT LOW SPEED as we keep pointing out - it has to get up to speed somehow and then it works fine.
I also stressed that in the explanation of why your love child of an equation is inappropriate.
I remember some video with an electric wind turbine probably battery and motor. Is that what you refer to or something else ?
The minimum wind speed is related to wind turbine. Wind speed needs to be above a minimum threshold 3 to 4m/s in order to be enough wind power to cover the turbine friction losses. That is valid for any wind turbine no matter if it is connected to an electric generator or directly to wheels.
The equation is both correct and appropriate as it predicts exactly what it is observed. You did not provide an alternative equation that works (can do correct predictions).
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It will be your job to prove the equation is wrong as that equation is used by almost all engineers to build things that work in real world as predicted by it.
This will be that equation
P = 0.5 * air density * (area * drag coefficient) * (wind speed - vehicle speed)3
Luckyly the equation is mainly used by fools and not be engineers, because it is flawed.
The part sometimes used is the special case the vehicle speed = 0 and with a slightly different coeffcient - that it get right. Adding the vehicle speed is a point making a flawed.
It would actually your part to prove the equation, not my part to proof it is false.
I can still give you the proof by showing how the backbird vehichle works and gets power out from the wind, when driving at the speed of the wind. So giving an example that poofs the equation wrong.
A prop running in still air needs a given power P to provide a thrust force of F. We actually don't care very much how good the prop is, it just has to work in the ideal case with no extra friction loss.
The force from the prob can than be used to have equal force on the wheels to counter-act the movement. This way the wheels provide a power of vehichel speed times force-. This power can than be used to drive the Fan and excess power could be used for what ever other task. That excess power would be generated from the wind, as the wind is the only power soruce to the system.
To make this there needs to be excess power available. So one need F*V_vehicle > P_prop or V_vehicle > P_prop / F. So given enough speed for the vehicle = wind this condition can be met.
For only somewhat realistic prop quality it does not even need a high speed, nowhere near the speed of sound where you can expect the equations to break down.
So this gives a proof that one can get power at the speed of the wind and thus that the equation with (wind speed - vehicle speed)3 is wrong.
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The minimum wind speed is related to wind turbine. Wind speed needs to be above a minimum threshold 3 to 4m/s in order to be enough wind power to cover the turbine friction losses.
It is not a wind turbine, it's a propeller. And the effect it more pronounced the fast the wind is because the propller thrust is proportional to the vehicle speed (not the wind speed). Thus the faster the vehicle goes over the ground, the greater the power sucked up by the wheels and the greater the thrust.
Think about it (I know, you need someone else to do if for you, but ask them nicely) - at zero ground speed there is no propeller power regardless of how fast the wind is blowing, or how many megawatts your favoured equation says there is.
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It will be your job to prove the equation is wrong as that equation is used by almost all engineers to build things that work in real world as predicted by it.
This will be that equation
P = 0.5 * air density * (area * drag coefficient) * (wind speed - vehicle speed)3
Luckyly the equation is mainly used by fools and not be engineers, because it is flawed.
That is not good enough.
If you want to say that equation is wrong you need to provide the correct one so we can compare the prediction each of them makes.
The other equation that floats around is wrong and is so easy to show it is as it will predict zero power available for a stationary vehicle also the values for available power will be way off what is measured in real world.
That calculator comes from "electromotive engineering & consulting" they use the same equation and they must have at least one engineer based on the company name so equation is used by engineers.
You can not move a vehicle with the brakes engaged so when you want to start moving you need to disengage the brakes thus you need to deal with the power applied by the headwind if you want to move in that direction.
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The problem is that AFAIK there is no such simple, easy to accept equation for the maximum available power. At least I don't see such an equation that I would take for granted. So it would still need a proof or at least good plausible explanation.
Aerodynamics can be be quite tricky and I would consider getting such an euqation and proving it rather hard. So the way via the available power is a hard one.
There are some ways of making estimates, however.
It doesn't help Electrodacus that he makes assertions that are demonstrably untrue and then relies on them to derive conclusions. For example, saying that a sail is nearly 100% efficient at extracting wind energy:
You are correct that a wind turbine is no more than 40 to 45% efficient but a sail is way more efficient approaching ideal.
We covered this in another thread, and it appears that a simple square sail is not very efficient at all. A sail self-evidently has zero efficiency when stationary, and zero efficiency when moving at the same speed as the wind. It seems the maximum efficiency of a simple square sail occurs somewhere in between, and is not much more than 15% at best. See this post for an explanation:
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)
Yacht sails are clearly much more efficient, but that is because they use angles and aerodynamics to do a better job.
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It is not a wind turbine, it's a propeller. And the effect it more pronounced the fast the wind is because the propller thrust is proportional to the vehicle speed (not the wind speed). Thus the faster the vehicle goes over the ground, the greater the power sucked up by the wheels and the greater the thrust.
Think about it (I know, you need someone else to do if for you, but ask them nicely) - at zero ground speed there is no propeller power regardless of how fast the wind is blowing, or how many megawatts your favoured equation says there is.
This is getting old. How many times I need to mention that there are two very different vehicle being discussed.
Direct downwind uses a propeller for propulsion
Direct upwind uses a propeller as a generator so wind turbine.
There is power available for a vehicle at 0 speed while wind is blowing and in fact that is when the max wind power is available for a direct down wind vehicle.
By applying the brakes you transfer all this wind power directly to earth.
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We covered this in another thread, and it appears that a simple square sail is not very efficient at all. A sail self-evidently has zero efficiency when stationary, and zero efficiency when moving at the same speed as the wind. It seems the maximum efficiency of a simple square sail occurs somewhere in between, and is not much more than 15% at best. See this post for an explanation:
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)
Yacht sails are clearly much more efficient, but that is because they use angles and aerodynamics to do a better job.
What you call a stationary sail is not stationary it moves at the same speed as the earth move as it is anchored to earth.
When you release the brakes what is the acceleration power that the vehicle experiences assuming no rolling resistance ?
Measure the distance the vehicle moved in one second after the brakes were released and calculate based on that the amount of kinetic energy vehicle gained then work backwards to calculate the amount of average power the wind applied in order to gain that amount of kinetic energy.
Also calculate the amount of brake power needed (energy to be wasted as heat) to maintain vehicle speed at 1m/s while wind speed is 11m/s
See what you think about the efficiency of a sail after that.
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That is not good enough.
If you want to say that equation is wrong you need to provide the correct one so we can compare the prediction each of them makes.
No, to proof that the euqation is wrong it is enough to show an example where the equation is wrong. That example is given (both with the math and as real world example in the videos in several examples).
The true burden is to you to show that the equation is correct to start with. Just a link to a dubious web site that seems to do the same mistake or a remotely similar equation for the stationary vehicle is not a proof. Given that it is pooven wrong there no hope to find a proof.
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No, to proof that the euqation is wrong it is enough to show an example where the equation is wrong. That example is given (both with the math and as real world example in the videos in several examples).
The true burden is to you to show that the equation is correct to start with. Just a link to a dubious web site that seems to do the same mistake or a remotely similar equation for the stationary vehicle is not a proof. Given that it is pooven wrong there no hope to find a proof.
You (all) are talking nonsense.
It is time for you to make a real world prediction and we will find a way to setup the experiment.
So say a small maybe toy like vehicle with a simple shape say a cube on wheels.
Say a cube with 20x20x20cm should be fine but we can change the dimensions if it makes it easier to do the experiment.
Drag coefficient for a cube is 1.05
Say cube will need to move upwind at a slow speed of 1m/s and wind speed is 10m/s
Based on all this I say the electric motor driving the cube requires this amount of power to overcome drag
Pd = 0.5 * 1.2 * (0.2*0.2*1.05) * (10+1)3 = 33.5W
Time for your prediction and please show how it is calculated and we will find a way to do the experiment so that we all agree it is done correctly.
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The (..)³ equation is wrong by having the wind speed inside for all 3 factors. One factor is only the vehicle speed. So the toy would only need 1/11 the power.
The difference gets even larger, when we assume a smaller speed, like 1 mm/s and than 11 m/s of wind. In this relatively extreme slow speed example it gets obvious that your prediction is wrong:
The mechanical power is speed times force and the force thus power divided by speed. So the froce would be 33.5 W /0.001 m/s = 33.5 kN. That would be way too much force to be produced by the wheels. So your equation would predict that a vehicle can not go slow against a heat wind. With the cazy prediction of needing more force the slower it wants to move and thus no way to start from a standstill as this naturally innitially is very slow.
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The (..)³ equation is wrong by having the wind speed inside for all 3 factors. One factor is only the vehicle speed. So the toy would only need 1/11 the power.
The difference gets even larger, when we assume a smaller speed, like 1 mm/s and than 11 m/s of wind. In this relatively extreme slow speed example it gets obvious that your prediction is wrong:
The mechanical power is speed times force and the force thus power divided by speed. So the froce would be 33.5 W /0.001 m/s = 33.5 kN. That would be way too much force to be produced by the wheels. So your equation would predict that a vehicle can not go slow against a heat wind. With the cazy prediction of needing more force the slower it wants to move and thus no way to start from a standstill as this naturally innitially is very slow.
You are right about one thing and that is 1m/s is still way to fast for such a small vehicle especially if we will test this in a small DIY wind tunnel.
Will still use 10m/s for wind as I think that should be possible even in DIY wind tunnel and use 0.02m/s (2cm per second) this way a 30cm run will take about 15 seconds enough to see the power measurement.
With this new data we have
Pd = 0.5 * 1.2 * (0.2*0.2*1.05) * (10+0.02)3 = 25.3W
I is time for your prediction as it makes no sense to discuss since you have a wrong idea of how air particles and the object/vehicle interact.
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The mechanical power is speed times force and the force thus power divided by speed. So the froce would be 33.5 W /0.001 m/s = 33.5 kN. That would be way too much force to be produced by the wheels. So your equation would predict that a vehicle can not go slow against a heat wind. With the cazy prediction of needing more force the slower it wants to move and thus no way to start from a standstill as this naturally innitially is very slow.
I see no prediction from you. Do you need more time ?
In the meantime I can show how that what you think as the correct equation violates the energy conservation law.
Say you had a 100% efficient wind generator (obviously a wind turbine will be only around 40% efficient but the discussion is ideal case).
For easy calculation say the wind generator swept area is 1m2 and there is nothing more. It is on wheels but we consider those super narrow so they do not add to wind drag. Also wind speed is 10m/s
A stationary vehicle with this characteristics can generate 0.5 * 1.2 * 1 * (10)3 = 600W again ideal case 100% efficiency.
What happens if vehicle moves very slowly upwind say 1m/s upwind ?
The wind turbine will see a 11m/s wind speed so it can now generate 798W that is 198W more than when stationary.
According to your equation only 0.5 * 1.2 * 1 * (11)2 * 1m/s = 72.6W are needed.
So inputting 72.6W and getting out 198W will violate energy conservation. Keep in mind we already discuss about an ideal 100% efficient system.
While the equation I consider to be the correct one will predict that you need 798W for the motor to drive at 1m/s against a 10m/s headwind (just drag power) so there is no energy conservation conflict.
Also if you want to travel down wind powered by the wind at 1m/s you need to apply a brake at the wheel with the power equal with 0.5 * 1.2 * 1 (10-1)3 = 437W so you can do what you want with those 437W say light some incandescent lamps and be able to maintain a constant 1m/s downwind powered by the wind.
So a stationary vehicle in 10m/s wind with 1m2 capture area has 600W available to accelerate downwind and if it wants to move upwind it needs to add extra on top of that 600W to start accelerating the mass in that direction.
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I do enjoy a good circular argument :popcorn:
https://www.youtube.com/watch?v=3oM7hX3UUEU (https://www.youtube.com/watch?v=3oM7hX3UUEU)
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The mechanical power is speed times force and the force thus power divided by speed. So the froce would be 33.5 W /0.001 m/s = 33.5 kN. That would be way too much force to be produced by the wheels. So your equation would predict that a vehicle can not go slow against a heat wind. With the cazy prediction of needing more force the slower it wants to move and thus no way to start from a standstill as this naturally innitially is very slow.
I see no prediction from you. Do you need more time ?
In the meantime I can show how that what you think as the correct equation violates the energy conservation law.
Say you had a 100% efficient wind generator (obviously a wind turbine will be only around 40% efficient but the discussion is ideal case).
For easy calculation say the wind generator swept area is 1m2 and there is nothing more. It is on wheels but we consider those super narrow so they do not add to wind drag. Also wind speed is 10m/s
A stationary vehicle with this characteristics can generate 0.5 * 1.2 * 1 * (10)3 = 600W again ideal case 100% efficiency.
What happens if vehicle moves very slowly upwind say 1m/s upwind ?
The wind turbine will see a 11m/s wind speed so it can now generate 798W that is 198W more than when stationary.
According to your equation only 0.5 * 1.2 * 1 * (11)2 * 1m/s = 72.6W are needed.
So inputting 72.6W and getting out 198W will violate energy conservation. Keep in mind we already discuss about an ideal 100% efficient system.
This is a good, but tricky point. By moving against the wind the volume of wind that is used is increased. So there is extra air to use. The exact calculation is tricky as the wind interacts with the slowed down air. The limited efficiency of a wind turbine is not just for technical reasons, the Betz limit is there for prociple reasons and a 100% efficient wind turbine thus already in in conflict with theroy and by itself cause contradictions.
While the equation I consider to be the correct one will predict that you need 798W for the motor to drive at 1m/s against a 10m/s headwind (just drag power) so there is no energy conservation conflict.
Also if you want to travel down wind powered by the wind at 1m/s you need to apply a brake at the wheel with the power equal with 0.5 * 1.2 * 1 (10-1)3 = 437W so you can do what you want with those 437W say light some incandescent lamps and be able to maintain a constant 1m/s downwind powered by the wind.
Your prediction is also causing a problem with energy conservation. You predict the 437 W from the vehicle moving down wind, but the wind turbine on the vehicle still sees 9 m/s of wind and could use that wind to creat additional power. Chances are it could gain more than 170 W from this - likely even with an efficiency within the Betz limit. When moving with the wind, there is less (at least not more) of the wind actually used - so that point does not work here either. So it is your prediction that violates the conservation of energy.
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This is a good, but tricky point. By moving against the wind the volume of wind that is used is increased. So there is extra air to use. The exact calculation is tricky as the wind interacts with the slowed down air. The limited efficiency of a wind turbine is not just for technical reasons, the Betz limit is there for prociple reasons and a 100% efficient wind turbine thus already in in conflict with theroy and by itself cause contradictions.
Sorry but Benz limit only refers to wind turbines. And even if we used the 59% Benz limit the problem is not changed. You add a small amount of energy to an a system and you get more in return (significantly more but even just 1% extra will violate the energy conservation law).
The example is there just to prove that equation you and others proposed is not correct.
Your prediction is also causing a problem with energy conservation. You predict the 437 W from the vehicle moving down wind, but the wind turbine on the vehicle still sees 9 m/s of wind and could use that wind to create additional power. Chances are it could gain more than 170 W from this - likely even with an efficiency within the Betz limit. When moving with the wind, there is less (at least not more) of the wind actually used - so that point does not work here either. So it is your prediction that violates the conservation of energy.
There is no extra power. Those 437W are all that it is available and you are already extracting everything that is it available to be extracted with the generator at the wheel.
At 9m/s which is the wind speed relative to vehicle (all that is important) you have 0.5 * 1.2 * 1 * (10-1)3 = 437.4W There is nothing more.
A wind turbine as you already mentioned can extra way less than this if it was to be used.
So whatever a wind turbine where to extract it will be subtracted from those 437.4W.
You can say that a sail vehicle on wheels is much more efficient than a propeller but the vehicle is moving and so you will need a flexible cable to connect the vehicle and at some point you need to stop and come back and you will wasting time not generating and use energy to return to a start point.
So overall efficiency if it was to be used as a wind turbine will not be that great and way more expensive. Plus it will be close to the ground where wind speed is much lower.
That is the reason I say a sail is as close to 100% efficiency as possible in using wind power to accelerate a vehicle.
All that calculated drag power is available to accelerate the vehicle.
So staying with this example when stationary the vehicle has available 600W of wind power so as soon as you remove the brakes say for 1ms the kinetic energy of the vehicle will increase by 600W * 0.001s = 0.6Ws (0.6J) so the speed of the vehicle after 1ms will depend on vehicle weight.
I know I use unusual methods of calculating all this compared to how this things are teached in schools but I think talking only about forces in schools while correct is the wrong approach in properly understanding what happens and human intuition about this things is wrong thus the wrong equation circulating around.
I wish people still did experiments in the classrooms then it will be easy to spot this incorrect equations.
I also think this has way more implications than in this highschool level physics as you can not work in any field of physics or engineering and not be affected by wrongly understanding this subject.
I really want for this misinformation to be corrected if possible. But no idea how to do this. The only thing that will convince most people will be the real world experiment.
The simplest I can think of is a vehicle driving slowly in a head wind and measuring the motor consumption.
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Your prediction is also causing a problem with energy conservation. You predict the 437 W from the vehicle moving down wind, but the wind turbine on the vehicle still sees 9 m/s of wind and could use that wind to create additional power. Chances are it could gain more than 170 W from this - likely even with an efficiency within the Betz limit. When moving with the wind, there is less (at least not more) of the wind actually used - so that point does not work here either. So it is your prediction that violates the conservation of energy.
There is no extra power. Those 437W are all that it is available and you are already extracting everything that is it available to be extracted with the generator at the wheel.
At 9m/s which is the wind speed relative to vehicle (all that is important) you have 0.5 * 1.2 * 1 * (10-1)3 = 437.4W There is nothing more.
A wind turbine as you already mentioned can extra way less than this if it was to be used.
So whatever a wind turbine where to extract it will be subtracted from those 437.4W.
The moving wind turbine would still see the 9 m/s wind can it can thus still extract power from that wind. It gets even more obvious when you consider moving the turbine only at snails pace. That would chance essentially nothing with the wind. So there would still be nearly the full wind and with correct formular also only very little power on the wheels. Your formular has just way too much power for the wheels and than the problem that there is overall too much power to be gained.
Too much power from the wheels comes with another problem: the force gets too large: 437 W with a speed of 1 m/s would mean a force of 437 N , which is too high by about a factor of 10. The errir gets even more rediculous if slower. Essentially constant power even at slow speed just does not work. :horse:
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Your prediction is also causing a problem with energy conservation. You predict the 437 W from the vehicle moving down wind, but the wind turbine on the vehicle still sees 9 m/s of wind and could use that wind to create additional power. Chances are it could gain more than 170 W from this - likely even with an efficiency within the Betz limit. When moving with the wind, there is less (at least not more) of the wind actually used - so that point does not work here either. So it is your prediction that violates the conservation of energy.
There is no extra power. Those 437W are all that it is available and you are already extracting everything that is it available to be extracted with the generator at the wheel.
At 9m/s which is the wind speed relative to vehicle (all that is important) you have 0.5 * 1.2 * 1 * (10-1)3 = 437.4W There is nothing more.
A wind turbine as you already mentioned can extra way less than this if it was to be used.
So whatever a wind turbine where to extract it will be subtracted from those 437.4W.
The moving wind turbine would still see the 9 m/s wind can it can thus still extract power from that wind. It gets even more obvious when you consider moving the turbine only at snails pace. That would chance essentially nothing with the wind. So there would still be nearly the full wind and with correct formular also only very little power on the wheels. Your formular has just way too much power for the wheels and than the problem that there is overall too much power to be gained.
Too much power from the wheels comes with another problem: the force gets too large: 437 W with a speed of 1 m/s would mean a force of 437 N , which is too high by about a factor of 10. The errir gets even more rediculous if slower. Essentially constant power even at slow speed just does not work. :horse:
You will need to be more precise in describing what you are thinking off.
As it is in this example the vehicle is a cube with 1m sides so there is a 1m2 surface interacting with air and for simplicity I considered the coefficient of drag to be 1
Now when you say a wind turbine is added to this is that on top of this 1m2 or are you replacing the cube with a wind turbine that has the same 1m2 swept area ?
You need to understand that interaction between air and vehicle is based on elastic collisions between air particles and vehicle boddy.
While vehicle speed is 1m/s the 1m2 side of the cube is hit by a cubic meter of air about 1.2kg 9 times per second and that kinetic energy is transferred to vehicle. Unless you get rid of that by say taking that energy at the wheel the vehicle will continue to accelerate so not be able to maintain that 1m/s
So you have this as air kinetic energy of one meter cube of air
KE = 0.5 * 1.2kg * 9m/s2 = 48.6Ws * 9 of this hitting the vehicle each second = 437.4Ws so same result.
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We got stuck at the equation for wind power available to vehicle. But I can prove the same thing without using that equation at all and just using equations you all agree with.
We all agree with the Drag force equation:
for upwind vehicle it will look like:
Fd = 0.5 * air density * coefficient for drag * area * (wind speed + vehicle speed)2
Another equation needed for my proof is equation for work done
W= Fd * distance
As we will use an example that ignores the rolling resistance for simplicity the work done will all end up as vehicle kinetic energy.
Vehicle 10000kg with
1m2 frontal area
coefficient of drag of 1
wind speed 10m/s
air density 1.2kg/m3
Fd= 60N
Work = 60N * 1m = 60 Joules (I prefer to use Ws)
Work = delta KE and since at start KE was zero at the start of the experiment the current vehicle KE is now 60Ws
So how much energy do you need to move the vehicle back to the starting point?
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The drag force is calculated with still 10 m/s. So the calculation assume a very low speed (which is OK for the very high weight).
The Force needed to push the vehicle back would be the same 60 N. One would first need to stop brake instantly - in the phisical sense this needs no energy, but one could even get the energy back.
So in the ideal case one would need the same 60 Joules.
The fun part when you calculate the time is needs to reach a m distance.: this is t² = 2 * 1m * Mass / force and thus t = 18.26s. For the 60 Joules of energy this gives an average power of a little under 4 W.
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The drag force is calculated with still 10 m/s. So the calculation assume a very low speed (which is OK for the very high weight).
The Force needed to push the vehicle back would be the same 60 N. One would first need to stop brake instantly - in the phisical sense this needs no energy, but one could even get the energy back.
So in the ideal case one would need the same 60 Joules.
The fun part when you calculate the time is needs to reach a m distance.: this is t² = 2 * 1m * Mass / force and thus t = 18.26s. For the 60 Joules of energy this gives an average power of a little under 4 W.
Maybe my example made things even more confusing than they were before.
It takes just a bit over 0.1 seconds to accelerate the vehicle to just over 0.1m/s
The work is done on the air so the air mass has delivered the vehicle 60 Joules that ended up as kinetic energy and so if after you did that you hide the sail (sail area equal zero) the vehicle can move forever at that 0.1m/s with no extra energy needed as there is no rolling resistance in this example.
So it was the air that moved 1 meter not the vehicle. The air mass while moving 1m lost 60 Joules to the vehicle that barely moved in that same period of time but gained those 60 Joules and can not continue to move forever at just over 0.1m/s about 0.11m/s
Again sorry for the confusing example. Is not the vehicle that moves 1m as vehicle can move forever any distance after it started to move.
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The drag force is calculated with still 10 m/s. So the calculation assume a very low speed (which is OK for the very high weight).
The Force needed to push the vehicle back would be the same 60 N. One would first need to stop brake instantly - in the phisical sense this needs no energy, but one could even get the energy back.
So in the ideal case one would need the same 60 Joules.
The fun part when you calculate the time is needs to reach a m distance.: this is t² = 2 * 1m * Mass / force and thus t = 18.26s. For the 60 Joules of energy this gives an average power of a little under 4 W.
Maybe my example made things even more confusing than they were before.
It takes just a bit over 0.1 seconds to accelerate the vehicle to just over 0.1m/s
The work is done on the air so the air mass has delivered the vehicle 60 Joules that ended up as kinetic energy and so if after you did that you hide the sail (sail area equal zero) the vehicle can move forever at that 0.1m/s with no extra energy needed as there is no rolling resistance in this example.
So it was the air that moved 1 meter not the vehicle. The air mass while moving 1m lost 60 Joules to the vehicle that barely moved in that same period of time but gained those 60 Joules and can not continue to move forever at just over 0.1m/s about 0.11m/s
Again sorry for the confusing example. Is not the vehicle that moves 1m as vehicle can move forever any distance after it started to move.
No : to accerate the 10 ton vehicle to 0.1 m/s in 0.1 seconds it needs way more force: the accelration is dV/dt and F = m *A = 10000 kg * 0.1m/s / 0.1 s = 10000 N. That is well more than 60 N you calculated for the drag force.
The assumption of 100% efficiency for the wind is wrong.
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No : to accerate the 10 ton vehicle to 0.1 m/s in 0.1 seconds it needs way more force: the accelration is dV/dt and F = m *A = 10000 kg * 0.1m/s / 0.1 s = 10000 N. That is well more than 60 N you calculated for the drag force.
The assumption of 100% efficiency for the wind is wrong.
Keep in mind this is a theoretical problem so there is no rolling resistance.
The vehicle barely moved in that 0.1s and the 0.11m/s is not much
KEvehicle = 0.5 * 10000kg * (0.11m/s)2 = 60 Joules
Yes the assumption of 100% efficiency for a theoretical sail is correct. And even in real world that will still be around the same.
What happened in this setup is that about 1m3 of air collides with the 1m2 sail and transfers all that kinetic energy to vehicle.
KEair = 0.5 * 1.2kg * (10m/s)2 = 60 Joules In that 0.1s just 1 cubic meter of air can collide with the vehicle body.
Do not ignore the conservation of energy when solving any problem. Here the mistake is to not consider air made out of particles that have elastic collisions with vehicle body. You do that for calculating drag force but then you want to ignore that when calculating power needed to overcome drag.
Even to travel 1m the vehicle will require less than 10 seconds assuming you remove the sail after the first 100ms and there will be no extra energy loss or gain so still the same 60 Joules of vehicle stored kinetic energy.
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Do not ignore the conservation of energy when solving any problem. Here the mistake is to not consider air made out of particles that have elastic collisions with vehicle body. You do that for calculating drag force but then you want to ignore that when calculating power needed to overcome drag.
The conservation of energy is OK, but than you have to also include the friction in the air, converting kinetic energy of the wind to heat. In this case it just not very practical to use conservation of energy as the starting point. It is much easier to use the drag force and the basic newton's laws.
To drive against the wind, the vehicle has to overcome the drag force. So it does not matter if the force is from a large sail in low wind or small sail in high wind. In a though experiment the wheel / motor part does not even know if the force is from wind or electromagnetic or friction to the ground.
The ideallized sail may theoretical approach 100% efficiency - but only when moving at nearly the speed of the wind. Going at 10% the speed of the wind it is more like 10% efficient.
With wind resistance there is not 100% transfer of kinetic energy. Generally much of the ernergy is converted to heat. The collisions are not 100% elastic - that is only an approximation sometimes used in some simplified calculations. Fast flying planes do get hot from the air resistance.
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The conservation of energy is OK, but than you have to also include the friction in the air, converting kinetic energy of the wind to heat. In this case it just not very practical to use conservation of energy as the starting point. It is much easier to use the drag force and the basic newton's laws.
To drive against the wind, the vehicle has to overcome the drag force. So it does not matter if the force is from a large sail in low wind or small sail in high wind. In a though experiment the wheel / motor part does not even know if the force is from wind or electromagnetic or friction to the ground.
The ideallized sail may theoretical approach 100% efficiency - but only when moving at nearly the speed of the wind. Going at 10% the speed of the wind it is more like 10% efficient.
With wind resistance there is not 100% transfer of kinetic energy. Generally much of the ernergy is converted to heat. The collisions are not 100% elastic - that is only an approximation sometimes used in some simplified calculations. Fast flying planes do get hot from the air resistance.
Air is not an ideal gas but close enough in relation to how much of his kinetic energy is transferred to the vehicle.
There is really no difference between vehicle colliding with stationary air particles (no wind) and moving air particles colliding with a stationary vehicle.
If you change the reference frame the result will need to be exactly the same else mistakes were made.
The sail size vs wind speed is relevant. Say you want the same 60N at just 5m/s instead of 10m/s wind.
Then your sail size will need to be 4x larger so 4m2 instead of just 1m2 needed with 10m/s wind.
But then the power available is 2x lower than before even if force is the same.
Fd = 0.5 * 1.2 * 1 * (10)2 = 60N
Fd = 0.5 * 1.2 * 4 * (5)2 =60N
Pd = Fd * 10 = 600W
Pd = Fd * 5 = 300W
If you want the same power at half the wind speed you need 8x the sail area. The wind speed is relevant not just the force.
You will not make this assumptions if you considered the vehicle powered by ideal collisions with large balls that you can see.
It is the same thing just that air is made out of super small (low weight) particles even different size that collide almost perfectly elastic with the vehicle body.
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The collisions for the molecules with a solid are not really elastic. One sees this as heat can be exchanges from a gas and solid. Even of only small fraction of the energy is lost to heat, the argumet with energy conservation breaks down and should be avoided. At best it is a weak argument - so one of the first points to question.
There is no physical law of conservation of kinetic energy. There is hower conservation of momentum and thus a balance of force. The assumption of conservation of kinetic energy is a rather weak one - it can work for explite eleastic collisions or frictionless ideallized mechnical systems. However it does not work for systems with friction - air resistance is a type of friction.
The picture of moving particles is not really practical to understand air resistance.
It is not a transfer of kinetic energy to kinetic energy (most of the energy in the gas is thermal energy and the wind is more like a minor shift in the distribution).
The interaction is more like a transfer of momentum and force. Much of the air particels never actuall hit the obstacle but hit other air that diverts it to the sides and can create turbulance that than dissipates the energy away from the obstacle.
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The collisions for the molecules with a solid are not really elastic. One sees this as heat can be exchanges from a gas and solid. Even of only small fraction of the energy is lost to heat, the argumet with energy conservation breaks down and should be avoided. At best it is a weak argument - so one of the first points to question.
There is no physical law of conservation of kinetic energy. There is hower conservation of momentum and thus a balance of force. The assumption of conservation of kinetic energy is a rather weak one - it can work for explite eleastic collisions or frictionless ideallized mechnical systems. However it does not work for systems with friction - air resistance is a type of friction.
The picture of moving particles is not really practical to understand air resistance.
It is not a transfer of kinetic energy to kinetic energy (most of the energy in the gas is thermal energy and the wind is more like a minor shift in the distribution).
The interaction is more like a transfer of momentum and force. Much of the air particels never actuall hit the obstacle but hit other air that diverts it to the sides and can create turbulance that than dissipates the energy away from the obstacle.
Understanding what air is and how it interacts with the vehicle is actually key to understanding this problems.
With no wind you seem to agree with the fact that all that is there are elastic collisions but then when air moves you try to find some alternative explanation and that is not the case as all that changed is the frame of reference.
If you take the example of the 1m2 frontal area and coefficient of drag of 1 traveling through air at 10m/s
Then you will say drag force is 60N and power needed to overcome this drag is 600W
If the same vehicle travels at half this speed you again will say drag force is 15N and power deeded to overcome drag is 75W
So all that changed is the vehicle speed trough stationary air and reducing the speed to half reduced the drag force 4x but reduced the power by 8x
If you look at what happens and the fact that each second vehicle collides with 10 cubic meters of air when driving at 10m/s and collides with 5 cubic meters when driving at 5m/s
When you calculate the collision energy you get a perfect transfer of energy like ideal elastic collisions took place.
But then if air moves instead of vehicle you want to claim that is no longer the same thing when nothing has actually charged other than the frame of reference.
You are not using different equations for drag force but you insist on two different equations for drag power. Is likely just based on the wrong intuition of what happens.
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No: the collisions are not elastic. With the usually 2 atom molecules (N2,O2) in air they are likely quite far away from elestic, with strong coupling between the linear and rotational degrees of freedom.
Even if the collision are nearly elestic (e.g. in helium or argon) it would still not matter. The picture of the air mass hitting the vehicle is still wrong. I don't care about the details of gas - that is only going off topic.
I somewhat agree (have not really checked, but sounds reasonable) with the 60 N of drag force, but there is not such thing as drag power of 600 W and no balance of powers (that is politics).
The power needed to drive agains the wind (or any other resistance) is Force time vehicle speed relative to the ground (where the propulsion transfers the force to). So at 1 m/s this would be 60 W and at
1mm/s this would be 6 mW. Power = force times speed is directly dirived from energy = force times distance, just by dividing by the time or taking the derivative.
If you can't agree with this, it gets hopeless.
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No: the collisions are not elastic. With the usually 2 atom molecules (N2,O2) in air they are likely quite far away from elestic, with strong coupling between the linear and rotational degrees of freedom.
Even if the collision are nearly elestic (e.g. in helium or argon) it would still not matter. The picture of the air mass hitting the vehicle is still wrong. I don't care about the details of gas - that is only going off topic.
I somewhat agree (have not really checked, but sounds reasonable) with the 60 N of drag force, but there is not such thing as drag power of 600 W and no balance of powers (that is politics).
The power needed to drive agains the wind (or any other resistance) is Force time vehicle speed relative to the ground (where the propulsion transfers the force to). So at 1 m/s this would be 60 W and at
1mm/s this would be 6 mW. Power = force times speed is directly dirived from energy = force times distance, just by dividing by the time or taking the derivative.
If you can't agree with this, it gets hopeless.
Take it the other way around vehicle driving in stationary air (no wind).
At 1m/s 0.6W that is exactly what perfectly elastic collision with 1m3 of air will result in as 1 cubic meter of air has 1.2kg
At 1mm/s 60nW again the kinetic energy of that volume of air.
At 10m/s 600W exactly matching the change in kinetic energy of that volume of air after collision with the vehicle.
Reversing this and having air particles move and collide with the vehicle can not change the results as you are just changing the reference frame.
In order to justify using a different equation for when vehicle moves in stationary air compared to moving air and stationary vehicle you will need to prove that the difference in energy is lost in some other ways and that will just be insanity.
As I mentioned before the energy of impact if a bus collides heads on with a stationary car will be the exact same as if vehicle collides with a stationary buss assuming exactly the same speed in both cases.
So in order for what you say to make sense you need to chose one equation for both cases. It needs to be the same both for vehicle moving through stationary air and vehicle in moving air also anything in between.
Also the same equation when driving in to headwind and driving with tail wind.
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I had the chance to converse with ChatGPT (language based AI model).
While it has many limitations it is still very impressive.
You can see part of my conversation about wind powered vehicles in the attached screenshots