Electroboom: How Right IS Veritasium?! Don't Electrons Push Each Other??

[electrodacus]

Completely irrelevant diversion. Settle on the equation and then you can think about values. You don't change things to suit the result you've already decided upon.

So you think that just providing a random (made up) equation was what I was looking for ?
I only asked for the equation that provides available wind power to vehicle in ideal case.

By adding the P_p you subtracted the power needed to supply the propeller and that is fine but is now less than available wind power.
To correct for this you decided that "wash speed" will reflect the output from the propeller but for conservation of energy to hold there will be a relation between P_p and wash speed so that P_w can never be higher than zero at wind speed.

It is you thinking that 100W put in to propeller can provide you with 101W in propulsion which is not possible.

P_p can not be higher than 0.5 * air density * area * (wind speed - vehicle speed)³
And similarly "wash speed" can not be larger than wind speed.
And the relation between the P_p and "wash speed" is so that wind power can never be higher at any vehicle speed than the original equation provides that means zero wind power when vehicle speed equals wind speed.

Keep in mind the above is true for the case where energy can not be stored (vehicle with no energy storage). Because I can already read you writing that Blackbird exceeds wind speed and that is true only because there is an energy storage device there called pressure differential and it is thanks to combination of propeller and compressible fluid.

[PlainName]

Completely irrelevant diversion. Settle on the equation and then you can think about values. You don't change things to suit the result you've already decided upon.

So you think that just providing a random (made up) equation was what I was looking for ?

Not at all. I gave you what you asked for. What, exactly, is wrong with it?

By adding the P_p you subtracted the power needed to supply the propeller and that is fine but is now less than available wind power.

What? The available wind power is the same - it hasn't disappeared.

To correct for this you decided that "wash speed" will reflect the output from the propeller but for conservation of energy to hold there will be a relation between P_p and wash speed so that P_w can never be higher than zero at wind speed.

Mate, wind speed of what? The thrust is acting as a sail to the wind, so when the thrust is static against the wind (zero relative wind speed) the vehicle will be traveling faster. I know that's probably blown your mind, but that's how it could be. You need to work out the values to discover if it actually occurs, and without doing that you can't say.

It is you thinking that 100W put in to propeller can provide you with 101W in propulsion which is not possible.

You're making things up: I have never suggested that. In fact I am saying right now you wouldn't get even that 100W out.

Further, you're mistakingly assuming that the thrust is providing ALL the motive force. It is not - the thrust could be really very teeny tiny.  A mere breath.

I've highlighted that so that when you skim this you might at least pause to read that bit and digest it.

P_p can not be higher than 0.5 * air density * area * (wind speed - vehicle speed)³
And similarly "wash speed" can not be larger than wind speed.

Mate, you're the one that's saying that. Nowhere has there been a suggestion that the wash speed is large than the wind speed. In fact, as above, it is very VERY much less than the wind speed

And the relation between the P_p and "wash speed" is so that wind power can never be higher at any vehicle speed than the original equation provides that means zero wind power when vehicle speed equals wind speed.

Oh, for crying out loud! Tell you what, why don't YOU come up with an euqation that covers the propeller turning and sucking power out of the wheels? It is there - you can SEE the damn thing turning in the video. You  KNOW the wheels turn it. Even if you think it works the other way (prop drive the wheels) it is happening, and yet your equation ignores it completely. So, write one that accounts for the prop.

I bet you can't, because what you're good at is cut'n'pasting stuff off t'web. That equation you posted was the best you could find to give power, but it is meaningless except for the exact situation where it ends up as zero. It tells you nothing else and doesn't apply to this situation.

So, write one that does apply, that does show the props effect. Even if it you think it works the wrong way (or not works) you can write the equation to show that, so do it. Shit or get off the loo.

[electrodacus]

So, write one that does apply, that does show the props effect. Even if it you think it works the wrong way (or not works) you can write the equation to show that, so do it. Shit or get off the loo.

I see that you get frustrated but important question is what is the available wind power to vehicle.
For ideal case (so best case scenario) that equation provides the wind power available to any wind powered vehicle driving directly downwind not just a simple sail based one.

What you do with that available wind power is your choice but if the vehicle does not have an energy storage device then it can not exceed wind speed
And if it has an energy storage device as it is the case with Blackbird then it can exceed wind speed even significantly but only for a limited amount of time proportional with the amount of stored energy.


To get back to a real world example the direct down wind powered vehicle with 1m² equivalent area facing the wind the wind speed of 6m/s and vehicle speed of 3m/s total wind power ideal case is 16.2W

If you take that 16.2W from the wheels and say use it to supply an incandescent lamp the vehicle will no longer accelerate as all wind power will be used to supply the lamp.
So vehicle will no longer accelerate.
You can disconnect the lamp and vehicle will continue to accelerate using this 16.2W at that point dropping as vehicle speed increases. Now say vehicle speed is maybe 4m/s and you again connect that 16.2W lamp.
What will happen is that at 4m/s wind power available to vehicle is just 4.8W so the delta at that moment of 16.2 - 4.8W will be provided by the vehicle stored kinetic energy so vehicle will decelerate until will get again at the equilibrium point that 3m/s where wind power available to vehicle is equal exactly the 16.2W the lamp requires.

Now you can do whatever you want with the wind power available to vehicle but you will never be able to accelerate faster than a sail vehicle or exceed the wind speed unless you add an energy storage device.

So if all you have are those 16.2W you can apply that to a wheel (most efficient) or to a propeller in a non compressible gas all that you will get ideal case is 16.2W.
Applying that to a propeller in air will result in less thrust as good part of the energy will be stored in pressure differential resulting in lower acceleration rate but the advantage of using the stored energy latter.

[PlainName]

So, write one that does apply, that does show the props effect. Even if it you think it works the wrong way (or not works) you can write the equation to show that, so do it. Shit or get off the loo.

I see that you get frustrated

Yes, sorry for the intemperate language. This is very trying, you know.

but important question is what is the available wind power to vehicle.

That's partly where you're going wrong. Your euqation shows the wind speed at which power is zero. It also shows the power available at any particular speed. But... so what?

All it's showing is the possible power, but how much does the vehicle need to move? When you look up on t'web how much power something needs to do, say, 15mph it's not relevant to this because most of that power will be overcoming drag. There isn't any drag here. Actually, it is negative drag, which is what's pushing the thing along. So you have all this power available but where is it going? It's accelerating the vehicle.

Now, if there were a little less power available, perhaps because something is sucking it off, would that make the vehicle go backwards? Of course not! It will just accelerate a little less. It is literally a brake on the wheels, but there is still enough to power available from the wind to drive the vehicle forwards.

The vehicle, without the prop, would never get to wind speed. It would be very very close but there is friction, so it's not quite there. Your formula will tell you how much power you have at that 'very close to wind speed' speed, and I'll bet you it's a surprisingly small amount because there is no drag to overcome. So just think how much extra power would be available if the vehicle was a little bit slower than that.

So, that formula tells us how much power we have and when it would theoretically run out. It doesn't tell us how fast we go, how much we can divert for other usage, how quickly we can accelerate, nothing.

And if it has an energy storage device as it is the case with Blackbird

Give it up. There is no energy storage device there. If you're convinced there is, prove it with your new modified equation that shows how Blackbird can (temporarily if you wish) exceed wind speed.

[electrodacus]

That's partly where you're going wrong. Your euqation shows the wind speed at which power is zero. It also shows the power available at any particular speed. But... so what?

All it's showing is the possible power, but how much does the vehicle need to move? When you look up on t'web how much power something needs to do, say, 15mph it's not relevant to this because most of that power will be overcoming drag. There isn't any drag here. Actually, it is negative drag, which is what's pushing the thing along. So you have all this power available but where is it going? It's accelerating the vehicle.

Now, if there were a little less power available, perhaps because something is sucking it off, would that make the vehicle go backwards? Of course not! It will just accelerate a little less. It is literally a brake on the wheels, but there is still enough to power available from the wind to drive the vehicle forwards.

The vehicle, without the prop, would never get to wind speed. It would be very very close but there is friction, so it's not quite there. Your formula will tell you how much power you have at that 'very close to wind speed' speed, and I'll bet you it's a surprisingly small amount because there is no drag to overcome. So just think how much extra power would be available if the vehicle was a little bit slower than that.

So, that formula tells us how much power we have and when it would theoretically run out. It doesn't tell us how fast we go, how much we can divert for other usage, how quickly we can accelerate, nothing.

Yes available wind power is accelerating the vehicle plus it covers frictional losses (but we can ignore those if you talk about ideal case where there is no friction just to get the best case scenario).

If the power you take at the wheels is less than available wind power the vehicle acceleration rate will just decrease.
If you take more power than available from wind then vehicle will decelerate as in the example I gave before when 4.8W of wind power was available with vehicle speed at 4m/s and and 16.2W of braking power was applied to wheels in order to power a 16.2W incandescent lamp. In that case the vehicle continued to decelerate until vehicle speed dropped to 3m/s and at that point wind power was 16.2W exactly covering the power taken at the wheels for the lamp.

And yes the slower the vehicle speed the more wind power is available but if you can not store energy then that power will just accelerate the vehicle witch is a form of energy storage (kinetic energy) but not the type that will allow you to exceed wind speed.
     
I think I mentioned before but I can (anyone can) build a vehicle that uses no propeller and exceeds the wind speed exactly the same way Blackbird is doing and actually be more efficient.
All that is needed is a sail (collapsible will be best to get rid of the drag when above wind speed) the propeller is doing this in a natural way).
Then 3 or 4 super capacitors 3000F 2.7V as each can store about 3Wh so in the same range as the large Blackbird.
The capacitors will be fully charged well before the vehicle gets to half the wind speed and from there the wind power and sail are no longer needed as the stored energy allows a 300kg (same weight as Blackbird) to be accelerated to about 3x the wind speed maybe even 4x will low enough friction losses.

To accelerate a 300kg vehicle from say 3m/s (half the wind speed of 6m/s) to 13m/s (a bit higher speed than Blackbird record of around 12.4m/s).
Vehicle kinetic energy at 13m/s is 0.5 * 300 * 13² = 25350Ws = 7Wh to this some frictional losses will need to be added but it will not be much.
So if I charge 4x 3Wh = 12Wh super capacitors (way smaller and less dangerous than a huge 20m² swept area propeller) I can easily exceed blackbird speed record.
I don't want a 20m₂ sail as that will be to crazy but with say a very manageable 2m²  sail 10x smaller than blackbird swept propeller area I will need to spend:
Say charging when vehicle is at 1m/s that is 6-1 = 5m/s wind speed relative to vehicle 0.5 * 1.2 * 2 * 5³ = 150W so I will say generator is just 80% efficient still 120W available to charge the supercapacitors.
I need 12Wh to fully charge that is 43200Ws so I need 360 seconds (a bit slow if nobody pushes the vehicle like it was the case with Blackbird) and if there are no wind gusts above 6m/s to help still 6 minute charge time is reasonable then just a few minutes to accelerate to top speed of 13m/s maybe even more dempensing on friction losses.
Or I can increase the sail size to 20m² and then I can charge the capacitors in just over half a minute.

[Naej]

I guess you will agree that wind power available to this vehicle (ideal case no friction involved) is 0.5 * air density * area * (wind speed - vehicle speed)³

No.
This is the wind power available to a wind turbine not moving and getting a wind of the speed you specified.

I may be repeating myself a bit, but this is NOT A WIND TURBINE.
This is A PROPELLER.
Don't use a WIND TURBINE formula on a PROPELLER even if they look alike.

The mechanical power given to the car is equal to the speed of the car multiplied by the force on the air.
You must, of course, subtract the power used by the propeller, which is the friction force with the road multiplied by the speed of the car (with efficiency 1 etc.).


Perhaps you really like the frame of reference of the car. In this case, the kinetic energy of the car is 0 (it's not moving with respect to itself), so you need 0 power. 

[iMo]

You have to change the subject of this thread..
"Don't Electrons Push the Blackbird?"
PS: I think it could be so, as the electrons could be somehow involved in the processes inside the Sun.. So let you continue..
 

[PlainName]

And yes the slower the vehicle speed the more wind power is available but if you can not store energy then that power will just accelerate the vehicle witch is a form of energy storage (kinetic energy) but not the type that will allow you to exceed wind speed.

OK, we are getting close. As noted previously, the power needed to maintain speed at wind speed is very low (no drag, only friction). How much? Well, it's barely anything since we can reach full speed as near as dammit.

Now, the prop provides thrust. How much thrust could it provide with, say, 5% of the available power? Not a lot, but we're not looking for a lot. Let's say it gives us 3% of the wind speed.

So, the question is: at a reduced 97% of wind speed, is the additional power available more or less than 5% of the total?

Obviously, actual values and balances will vary, but the big thing to remember is that at close to wind speed there is no drag. If the prop did manage to push the vehicle over wind speed then there will be drag and thus a limit to how much faster it could go. Be that's not important - just 0.01m/s over would be enough, and I think that's doable. (Actually, we all know it is because we've seen it many times now.)

[cbutlera]

For ideal case (so best case scenario) that equation provides the wind power available to any wind powered vehicle driving directly downwind not just a simple sail based one.

The equation you gave is more or less appropriate for a stationary wind turbine (Betz's law gives a better theoretical upper limit), but this vehicle is not stationary and is propeller driven.

Consider the case where the vehicle is travelling at much greater than wind speed, with a driven propeller.  The propeller is adjusted so that the vehicle leaves in its wake a trail of stationary air, in other words, the change in velocity of the air as it passes through the propeller disc is equal to the wind speed.  So the vehicle is effectively harvesting all of the kinetic energy from the air that it encounters, and the faster it travels through the air the greater the rate at which it can harvest that kinetic energy.

The available kinetic energy per unit volume of the air is 0.5 * air_density * wind_speed²

So if the vehicle is travelling at a speed of (vehicle_speed - wind_speed) with respect to the air, it can harvest wind energy at the rate of

P_w = 0.5 * air_density * wind_speed² * area * (vehicle_speed - wind_speed)

Note that this equation is appropriate for a vehicle speed that is significantly greater than wind speed.  Close to wind speed, a more careful analysis is required, along the same lines as the derivation of Betz's law given on the Wikipedia page https://en.wikipedia.org/wiki/Betz%27s_law

Assuming a hypothetical vehicle with no losses, the wheels of the vehicle can recover all of the energy expended in turning the propeller and recycle it back to the propeller through the transmission.  The additional energy harvested from the wind goes into further accelerating the vehicle.

As the hypothetical vehicle has no losses, conservation of energy not only allows the vehicle to accelerate while travelling faster than the wind, it demands it.  Otherwise where does the energy from the harvested from the wind go?  The real vehicle of course does have losses, but the harvested energy can still be sufficient to overcome these and propel it beyond wind speed.

[electrodacus]

No.
This is the wind power available to a wind turbine not moving and getting a wind of the speed you specified.

I may be repeating myself a bit, but this is NOT A WIND TURBINE.
This is A PROPELLER.
Don't use a WIND TURBINE formula on a PROPELLER even if they look alike.

The mechanical power given to the car is equal to the speed of the car multiplied by the force on the air.
You must, of course, subtract the power used by the propeller, which is the friction force with the road multiplied by the speed of the car (with efficiency 1 etc.).


Perhaps you really like the frame of reference of the car. In this case, the kinetic energy of the car is 0 (it's not moving with respect to itself), so you need 0 power. 

This formula is not specific to a wind turbine it is generic used in many case.
If you want to find the frictional losses needed for a vehicle you will use this same formula.

You need to ignore the propeller as propeller will not change in any way the available wind power other than propeller acting as a sail.

I can build a non conventional energy generating device.
I take a simple sail vehicle on wheels add a generator at the wheel and take so much power from the generator so that vehicle always move at a constant speed of 1m/s
Say wind speed is 10m/s and sail area 1m²
This vehicle generator will be way more efficient than any propeller type wind turbine even a theoretical one working at Benz limit that is about 59%

Generator connected at wheels can produce 0.5 * 1.2 * 1 * (10-1)³ = 437.4W
No stationary wind turbine can get anywhere close to this.
The problem is that the generator is on wheels and moves at 1m/s so you need to connect it to your house (say that is what you want to power) with a flexible cable and after some distance that the cable can reach you will need to stop maybe fold the sail and return to start producing again so not very convenient.
But if what you want to power with wind is a vehicle that needs to move in the same direction as the wind anyway then a sail is extremely efficient and the equation that describe wind power available to vehicle is the one I provided.

Now in this example when vehicle is at 1m/s in a wind of 10m/s driving directly downwind you can take all this 437.4W available wind power and you can apply that to a wheel of the vehicle or to a propeller but the output from those will never be higher than 437.4W as that is not allowed by the conservation of energy law.
So you are better just disconnecting the generator and let the wind push the vehicle as wind at that moment in time provides 437.4W worth of acceleration. 

[electrodacus]

OK, we are getting close. As noted previously, the power needed to maintain speed at wind speed is very low (no drag, only friction). How much? Well, it's barely anything since we can reach full speed as near as dammit.

Now, the prop provides thrust. How much thrust could it provide with, say, 5% of the available power? Not a lot, but we're not looking for a lot. Let's say it gives us 3% of the wind speed.

So, the question is: at a reduced 97% of wind speed, is the additional power available more or less than 5% of the total?

Obviously, actual values and balances will vary, but the big thing to remember is that at close to wind speed there is no drag. If the prop did manage to push the vehicle over wind speed then there will be drag and thus a limit to how much faster it could go. Be that's not important - just 0.01m/s over would be enough, and I think that's doable. (Actually, we all know it is because we've seen it many times now.)

If all frictional losses of the vehicle at wind speed are 1W then vehicle needs a battery or any other energy storage device to provide that 1W for the vehicle to maintain wind speed as it will not get anything from wind when vehicle has the same speed as the wind.

If vehicle just find itself at wind speed maybe because there was a wind gust with higher speed to bring the vehicle there then vehicle will start to slow down if it has no energy storage device on board to provide the 1W for friction losses.

You can calculate exactly how long it will take the vehicle to slow down to final speed relative to wind speed if you know the wind speed and weight of the vehicle and what powers the vehicle is the vehicle kinetic energy and as that drops the vehicle speed drops and it will get after some time to a speed low enough so that wind power available to vehicle will be 1W needed to cover the friction and that will be the vehicle constant speed if wind speed remains constant.

What you have seen many times is a vehicle exceeding wind speed powered from the pressure differential stored energy. What you have not seen because the experiment ended to early is how vehicle will slow down as stored energy is all used up.   

[electrodacus]

The equation you gave is more or less appropriate for a stationary wind turbine (Betz's law gives a better theoretical upper limit), but this vehicle is not stationary and is propeller driven.

Consider the case where the vehicle is travelling at much greater than wind speed, with a driven propeller.  The propeller is adjusted so that the vehicle leaves in its wake a trail of stationary air, in other words, the change in velocity of the air as it passes through the propeller disc is equal to the wind speed.  So the vehicle is effectively harvesting all of the kinetic energy from the air that it encounters, and the faster it travels through the air the greater the rate at which it can harvest that kinetic energy.

The available kinetic energy per unit volume of the air is 0.5 * air_density * wind_speed²

So if the vehicle is travelling at a speed of (vehicle_speed - wind_speed) with respect to the air, it can harvest wind energy at the rate of

P_w = 0.5 * air_density * wind_speed² * area * (vehicle_speed - wind_speed)

Note that this equation is appropriate for a vehicle speed that is significantly greater than wind speed.  Close to wind speed, a more careful analysis is required, along the same lines as the derivation of Betz's law given on the Wikipedia page https://en.wikipedia.org/wiki/Betz%27s_law

Assuming a hypothetical vehicle with no losses, the wheels of the vehicle can recover all of the energy expended in turning the propeller and recycle it back to the propeller through the transmission.  The additional energy harvested from the wind goes into further accelerating the vehicle.

As the hypothetical vehicle has no losses, conservation of energy not only allows the vehicle to accelerate while travelling faster than the wind, it demands it.  Otherwise where does the energy from the harvested from the wind go?  The real vehicle of course does have losses, but the harvested energy can still be sufficient to overcome these and propel it beyond wind speed.

Wind turbine is only one application of that equation. I can use the same equation to calculate how much power a vehicle will require at a certain vehicle speed with or without any wind speed.

The rate at which a vehicle can harvest wind energy is this

Pw = 0.5 * air_density * (wind_speed-vehicle speed)² * area * (wind_speed-vehicle speed)

A vehicle traveling at faster than wind speed will not be able to harvest any wind energy to allow increasing vehicle kinetic energy relative to ground.
 
What you want to say is that you have an ideal wind turbine (not affected by Benz limit) installed in front of the vehicle.
Say wind speed is 10m/s and vehicle speed is 50m/s directly downwind and it is irrelevant how vehicle got to that moment.
Vehicle experience a 40m/s headwind witch it can use to generate power with say a 1m² ideal generator area

P_G = 0.5 * 1.2 * 1 * (50-10)³ = 38400W

This is also exactly the amount of braking power experienced by the vehicle if you take this generate power and use it to say supply to a heating element.
You will need to take all this 38400W and supply to an ideal electric motor driving the wheels in order for the vehicle to just maintain speed (no acceleration and no deceleration as it is the ideal case no friction losses anywhere).

You are just talking about the motor generator (free energy generator scam) without realizing.

[gnuarm]

Ok, that makes everything perfectly clear now.  I'm sure everyone else understands it now as well.  Thank you. 

No need to continue with this off-topic conversation. 

[electrodacus]

Ok, that makes everything perfectly clear now.  I'm sure everyone else understands it now as well.  Thank you. 

No need to continue with this off-topic conversation.

I'm fairly certain you are being sarcastic.

As for being offtopic it is not as offtopic as you think.
Both direct downwind faster than wind and how electrical energy is transferred from source are incorrectly presented by Derek Veritasium because his lack of understanding of energy storage.

Power out can not be larger than power in without energy storage being involved and when that is the case (almost always since energy storage is as important to understand as friction).
It is also important to know which one is power in and which one is power out as Derek and many others confuse input with output and decare overunity is possible.

Electrical energy as I mentioned before means electrical power integrated over time.
Electrical power is the product of electrical potential and electrical current.
If electrical current is zero then electrical power and by extension electrical energy is zero.
Electrical current is defined as a stream of charged particles and since you only have moving charged particles through the wire and not through air electrical energy is transferred through wires and not through air.
And as at some point someone mentioned vacuum diode as an example where energy travels through vacuum that is a special case not applying to Derek's experiment where electrons (charged particles) have enough energy to travel through vacuum.
In that special case you can say that energy travels outside the wire trough that vacuum gap between wires.

[PlainName]

Power out can not be larger than power in without energy storage being involved

It is only you that suggests there is more power out than in, and you're no doubt only doing that so you can wheezle in your favourite energy storage rubbish. At least a stuck record would eventually wear out...

[electrodacus]

Power out can not be larger than power in without energy storage being involved

It is only you that suggests there is more power out than in, and you're no doubt only doing that so you can wheezle in your favourite energy storage rubbish. At least a stuck record would eventually wear out...

When you say that power taken at the wheel make sense as you out that power in to propeller.
You slow down the vehicle when you extract power at the wheel and if you where to put the exact same amount of power at the propeller there is no gain so it will not make sense to do that.

Wind can provide power to vehicle only when vehicle speed is lower than wind speed and if you say get 100W at some point from wind and you steal 40W at the wheel to put ideal case 40W in to propeller for propulsion you still have net 100W so you do nothing.

a) Vehicle can be accelerated by all 100W from wind. 
b) Vehicle can be accelerated by 60W from wind + 40W you take at the wheels and put all in propeller still 100W ideal case.

So please let me know if you disagree with b) then let me know how it works without involving energy storage and without violating the energy conservation law.

[PlainName]

You slow down the vehicle when you extract power at the wheel and if you where to put the exact same amount of power at the propeller there is no gain so it will not make sense to do that.

How do you know? Have you done the sums? I know you haven't because a) they're not simple and b) you reject the only equation that makes sense (for no good reason - what, exactly, is wrong with it? You won't say, other than something something too big something).

[PlainName]

b) Vehicle can be accelerated by 60W from wind + 40W you take at the wheels and put all in propeller still 100W ideal case.

Wrong, sorry.

It takes almost nothing to maintain speed. It doesn't take 60W to accelerate - it takes whatever is available over the practically nothing that maintains speed. There is no drag the normal speed vs power curves you might want to consult are not relevant.

And you don't take 40W or whatever from the wheel - just a bit more than the practically nothing that maintains speed would be fine. So we are looking at maybe 5% being taken via the wheel, and whatever slower speed relates to 5% less power from the direct wind. If you want 100% to be 100W I'm sure you can do the maths. (Actually, it could be anything from 0.001% to 99.9% - solving the equation would tell you what value, if there is a working value, would do the biz.)

And that is it. The question is whether that 5% of power can produce thrust that at least equals the lost speed due to 5% less direct wind power. But recall that most of the power is used in accelerating up to speed, so the reduction may not be very much at all.

Forget your 100/60/40 splits - that might be appropriate when the thing is stationary but not when it is in full flight.

[SiliconWizard]

So. Have we revolutionized EE yet?

[electrodacus]

b) Vehicle can be accelerated by 60W from wind + 40W you take at the wheels and put all in propeller still 100W ideal case.

Wrong, sorry.

It takes almost nothing to maintain speed. It doesn't take 60W to accelerate - it takes whatever is available over the practically nothing that maintains speed. There is no drag the normal speed vs power curves you might want to consult are not relevant.

And you don't take 40W or whatever from the wheel - just a bit more than the practically nothing that maintains speed would be fine. So we are looking at maybe 5% being taken via the wheel, and whatever slower speed relates to 5% less power from the direct wind. If you want 100% to be 100W I'm sure you can do the maths. (Actually, it could be anything from 0.001% to 99.9% - solving the equation would tell you what value, if there is a working value, would do the biz.)

And that is it. The question is whether that 5% of power can produce thrust that at least equals the lost speed due to 5% less direct wind power. But recall that most of the power is used in accelerating up to speed, so the reduction may not be very much at all.

Forget your 100/60/40 splits - that might be appropriate when the thing is stationary but not when it is in full flight.

Thanks for answering.
Ideal case you need no energy to maintain speed.

But to accelerate the vehicle you need energy.

You have some wind power available to vehicle depending on wind speed relative to vehicle (wind speed - vehicle speed) and equivalent surface area of the vehicle on witch wind interacts with.
I just gave an example of 100W available from wind to vehicle.
All this 100W will accelerate the vehicle. If you take anything at the wheel I gave 40W as an example then only 60W are available for vehicle acceleration so acceleration rate will be lower. If you then use the 40W at the propeller ideal case you get back to 100W of acceleration power but nothing more.


I will like to know exactly what part you disagree with but to me it seems that you think taking 40W at the wheel will not reduce the available acceleration by exactly that amount.

[electrodacus]

So. Have we revolutionized EE yet?

There is nothing to revolutionize. Conservation of energy is nothing new is just people seems to not understand it is a law and nobody has ever proved a case where it is not valid.
Wind power is the only input to this vehicle and it is a positive value only when vehicle speed is below wind speed for direct downwind case.

When vehicle finds himself above wind speed for any reason taking energy from the wheel will reduce the vehicle kinetic energy by that exact amount and that means reducing vehicle speed unless there is some energy storage that was charged when vehicle was below wind speed.

Do you have an opinion are are you just a spectator ?

[PlainName]

b) Vehicle can be accelerated by 60W from wind + 40W you take at the wheels and put all in propeller still 100W ideal case.

Wrong, sorry.

It takes almost nothing to maintain speed. It doesn't take 60W to accelerate - it takes whatever is available over the practically nothing that maintains speed. There is no drag the normal speed vs power curves you might want to consult are not relevant.

And you don't take 40W or whatever from the wheel - just a bit more than the practically nothing that maintains speed would be fine. So we are looking at maybe 5% being taken via the wheel, and whatever slower speed relates to 5% less power from the direct wind. If you want 100% to be 100W I'm sure you can do the maths. (Actually, it could be anything from 0.001% to 99.9% - solving the equation would tell you what value, if there is a working value, would do the biz.)

And that is it. The question is whether that 5% of power can produce thrust that at least equals the lost speed due to 5% less direct wind power. But recall that most of the power is used in accelerating up to speed, so the reduction may not be very much at all.

Forget your 100/60/40 splits - that might be appropriate when the thing is stationary but not when it is in full flight.

Thanks for answering.
Ideal case you need no energy to maintain speed.

But to accelerate the vehicle you need energy.

You have some wind power available to vehicle depending on wind speed relative to vehicle (wind speed - vehicle speed) and equivalent surface area of the vehicle on witch wind interacts with.
I just gave an example of 100W available from wind to vehicle.
All this 100W will accelerate the vehicle. If you take anything at the wheel I gave 40W as an example then only 60W are available for vehicle acceleration so acceleration rate will be lower. If you then use the 40W at the propeller ideal case you get back to 100W of acceleration power but nothing more.


I will like to know exactly what part you disagree with but to me it seems that you think taking 40W at the wheel will not reduce the available acceleration by exactly that amount.

I disagree with taking 40% for the wheel, mainly because I think the power required to product thrust increases proportionally to the speed of the thrust - drag is involved there. So low thrust is better, and it isn't going to need anywhere near 40%.

But... let's say it did take 40%. How much slower will the vehicle be? We don't know; we only know that the acceleration is affected so the question is whether it could accelerate to a high enough speed for the thrust to be in with a shout. Maybe it could, maybe it couldn't.

I referred previously to the feathered Blackbird blades. In all the demos I've seen, none of them (except blackbird) started from zero speed. That is, none of them accelerated up to wind speed - all of them started at effective wind speed. Blackbird was the exception but it used feathered blades (which wouldn't suck much power). In the other demos I would just assume that it's difficult to start from zero, but the fact that Blackbird had to have feathered blades may be significant.

[electrodacus]

I disagree with taking 40% for the wheel, mainly because I think the power required to product thrust increases proportionally to the speed of the thrust - drag is involved there. So low thrust is better, and it isn't going to need anywhere near 40%.

But... let's say it did take 40%. How much slower will the vehicle be? We don't know; we only know that the acceleration is affected so the question is whether it could accelerate to a high enough speed for the thrust to be in with a shout. Maybe it could, maybe it couldn't.

I referred previously to the feathered Blackbird blades. In all the demos I've seen, none of them (except blackbird) started from zero speed. That is, none of them accelerated up to wind speed - all of them started at effective wind speed. Blackbird was the exception but it used feathered blades (which wouldn't suck much power). In the other demos I would just assume that it's difficult to start from zero, but the fact that Blackbird had to have feathered blades may be significant.

40% was just a random example.

If you provide a vehicle 100W of power to accelerate for 10ms that means you provide the vehicle with 100W * 0.01s = 1Ws = 1 Joule of energy.
It at that time vehicle was at 3m/s and had a weight of 10kg his kinetic energy initially was 0.5 * 10kg * (3m/s)² = 45Ws
So due to 1Ws of energy applied (100W for 10ms) the vehicle new kinetic energy will be 45Ws + 1Ws = 46Ws
That means vehicle new speed will be sqrt(46/(0.5*10)) = 3.033ms

If you instead apply a brake of 100W the kinetic energy of the moving vehicle will drop meaning the speed will drop.

So in my example the choice was 100W of acceleration power from the wind directly or ideal case 100W with extra complications like taking some power at the wheels and putting in to a propeller or another wheel at 100% efficiency.

If you have 100W of wind power than that is it. And if you are at wind speed and have 0W of available power to accelerate that is it and above wind speed power available is negative meaning deceleration.
The only way for any wind powered vehicle to exceed wind speed directly downwind is to have an energy storage device (other than the vehicle kinetic energy).

It is theoretically impossible for any wind only powered vehicle to exceed wind speed directly downwind or travel at any speed directly upwind.

[rfeecs]

electrodacus, here is a paper that analyzes the problem from an energy point of view:

"Theory and Design of Flow Driven Vehicles Using Rotors for Energy Conversion"
https://backend.orbit.dtu.dk/ws/portalfiles/portal/3748519/2009_28.pdf

They conclude:

By applying basic concepts from mechanics and fluid mechanics we have derived the equations applicable for the performance of vehicles “powered” by a difference in velocity between two media. A myriad of intriguing results emerge from these relations. 

• It is theoretically and practically possible to build a wind driven car that can go directly upwind (using a generator/wind turbine in the air).  ...

• It is theoretically possible to build a wind driven car that can go in the downwind direction faster than the free stream wind speed (using a propeller in the air)

This idea has been around a long time.  It's been demonstrated.  You have been proven wrong.

Despite your hundreds of posts you have failed to convince anyone by mindlessly repeating your flawed theory.

[electrodacus]

electrodacus, here is a paper that analyzes the problem from an energy point of view:

"Theory and Design of Flow Driven Vehicles Using Rotors for Energy Conversion"
https://backend.orbit.dtu.dk/ws/portalfiles/portal/3748519/2009_28.pdf

They conclude:

By applying basic concepts from mechanics and fluid mechanics we have derived the equations applicable for the performance of vehicles “powered” by a difference in velocity between two media. A myriad of intriguing results emerge from these relations. 

• It is theoretically and practically possible to build a wind driven car that can go directly upwind (using a generator/wind turbine in the air).  ...

• It is theoretically possible to build a wind driven car that can go in the downwind direction faster than the free stream wind speed (using a propeller in the air)

This idea has been around a long time.  It's been demonstrated.  You have been proven wrong.

Despite your hundreds of posts you have failed to convince anyone by mindlessly repeating your flawed theory.

The paper seems to concentrate on direct upwind witch uses different type of energy storage and can be debunked even easier.
All you need to do is take a high speed video of any direct upwind device and see how it moves in discrete steps with time to charge then discharge the energy storage device tens of charge discharge cycles per second depending on design.

Here is my video proof for direct upwind is just a 20 second slow motion video https://odysee.com/@dacustemp:8/wheel-cart-energy-storage-slow:8
Input power is provided here by the middle paper the one that moves relative to camera frame and the vehicle advances against the direction that paper is moving only because it first stores energy then it releases the energy to move a bit forward.
Wheels in front the one on moving paper are the generator wheels (same as the wind turbine on a direct upwind vehicle) and the back wheels are the motor wheels powered by the energy stored in the rubber belt. If there was no energy storage then all times Pin = Pout and vehicle will not move just sit in place or be dragged back.
But since the belt is charged (elastic potential energy) and then when the back wheel spins because it slips there is power available at the back wheel from the belt that is now discharging and the process repeats.


Edit: Looking a bit more at that paper and that Figure 1 contradicts what the paper is trying to prove.
Vg is wind speed and direction is left to right Fg and Fp are correct same value and opposite directions
And then Vp is the same direction as wind direction meaning the vehicle is being dragged directly downwind at significantly less than wind speed.
They try to explain how direct upwind will work and the only diagram they have in there correctly predicts that without energy storage the vehicle will either not move or be slowly dragged in the wind direction.