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

[PlainName]

Blackbird doesn't need accelerating. When the prop is 'turned on' Blackbird is already at speed. Don't you take in anything I tell you, even several times?

Further, even if acceleration had to be taken into account, the longer it takes the more power you acquire from the wind. It's just going to be slower, not impossible.

[electrodacus]

Blackbird doesn't need accelerating. When the prop is 'turned on' Blackbird is already at speed. Don't you take in anything I tell you, even several times?

Further, even if acceleration had to be taken into account, the longer it takes the more power you acquire from the wind. It's just going to be slower, not impossible.

Black bird can self start it is just slower (takes longer) than pushing it and in that case the people that push help charge the energy storage faster (create that pressure differential).
Not quite sure you even know what acceleration means.
In any case I spent 5 minutes or so to give you a proper reply including the math and you 10 second response has no value (pun intended as you provided no equations or numbers).  But you have a false strong intuition that wind somehow can power a vehicle driving in same direction at higher than wind speed. 

[Naej]

Yes the propeller is powered by the wheels. Yes it is silly. Many things in physics are silly yet correct.

Yes air applies a force on the car, which pushes it. Amazing, no?

By the way, you didn't tell what was the engine providing 3MW to a wind turbine so that it can stay at the same place. 

Wheels are not a power source (they are not radioactive and not being burned to extract energy from them). Wheels are an intermediary but power to any part of the (wind powered vehicle) can only be provided by air particles hitting the vehicle body (any part of the body including the propeller blades).
When air particles hit the direct downwind vehicle from the back (wind speed higher than vehicle speed) the wind power is available to vehicle to accelerate in the desired direction and or store that.

When vehicle speed equal wind speed there is zero available wind power to vehicle and when vehicle exceeds wind speed the vehicle will now hit the air particles so wind power available to vehicle will be negative meaning will slow down the vehicle unless as it is the case here vehicle also has access to earlier stored pressure differential that will be used to cover vehicle friction losses including air drag. 

When vehicle speed equal wind speed there is plenty available wind power.
You just need to power your propeller with your wheels, and get more force out of it than you took from the wheels so that your wheels turn faster.
The air is accelerated by the propeller. So in the Earth reference frame, wind decreased behind the car: the kinetic energy of the wind is now in the car.

That's how it works and if you can't understand you should read about Newton laws and stop thinking in terms of energy until you understand them.

The amount of power available to a wind powered vehicle (any design) is 1/2*mass flow*wind speed^2. If the vehicle seed is twice the wind speed, you get mass flow=density*area*wind speed.

Your computations with the wind turbine are quite funny. What if there is wind not for 1ms, but for 1 week?
What if I replace the wind turbine by a water one? There you go: take a dam (1 million ton of concrete, say), and it produces 1 GW for 10 years. What is the speed of the dam?

[PlainName]

Not my problem if you spent 5 mins thinking up rubbish.

Watch the video again. Blackbird accelerates up to near wind speed and then the prop is unfeathered. And, again, none of the demos started at zero - all of them start at speed.

And, as I just said, it may take 7Wh or whatever to accelerate however many kg to some arbitrary speed, but all the time the wind is blowing it is applying power. If it takes longer to get to speed that just means the wind power has been applied over more time, so 7Wh or 100Wh is meaningless unless you specify the time limit for allow the wind to blow. It could give you 1W over 7 hours, for instance. Or 2W for 7 hours while you nick 1W for something else.

[electrodacus]

When vehicle speed equal wind speed there is plenty available wind power.
You just need to power your propeller with your wheels, and get more force out of it than you took from the wheels so that your wheels turn faster.
The air is accelerated by the propeller. So in the Earth reference frame, wind decreased behind the car: the kinetic energy of the wind is now in the car.

That's how it works and if you can't understand you should read about Newton laws and stop thinking in terms of energy until you understand them.

The amount of power available to a wind powered vehicle (any design) is 1/2*mass flow*wind speed^2. If the vehicle seed is twice the wind speed, you get mass flow=density*area*wind speed.

Your computations with the wind turbine are quite funny. What if there is wind not for 1ms, but for 1 week?
What if I replace the wind turbine by a water one? There you go: take a dam (1 million ton of concrete, say), and it produces 1 GW for 10 years. What is the speed of the dam?

How much wind power do you think is available when vehicle speed equal wind speed and what is the equation describing that ?
What you say sounds exactly like "generator supplies the motor and the motor supplies the generator" not only that but there is also plenty of extra energy.

Power and energy is all you need to use the forces and speed (voltage and current) are not relevant when you try to do a energy balance calculation.

Do you agree that a typical propeller is just around 70% efficient  for propulsion in air while a wheel is well over 90 to 95% efficient.
So when you are at wind speed what is the difference between using the energy generate at the back wheels in to the front wheel/wheels instead of putting it in a less efficient propeller ?


If you take 1000W from the wheels for 1ms that will be 1Ws when vehicle is at wind speed vehicle kinetic energy will be reduced by this exact amount 1Ws as there is no wind power available.
Then when you take this 1Ws and put in to the 70% efficient propeller the kinetic energy of the vehicle increases by 0.7Ws so vehicle ends up with a deficit of 0.3Ws meaning vehicle has lower speed after you do this.

The amount of power available to a wind powered vehicle (any design) is 1/2*mass flow*wind speed^2. If the vehicle seed is twice the wind speed, you get mass flow=density*area*wind speed.

We are making progress as now you say that wind power available to vehicle is the same as I always say but just with the equation presented a bit different

You wrote:
0.5 * (air density * area * wind speed) * wind speed²

That is correct and the same with
0.5 * air density * area * wind speed³

All you need to clarify now is that wind speed in this equation is wind speed relative to vehicle and not to the ground.
The wind speed relative to vehicle = (wind speed relative to ground - vehicle speed relative to ground) and now you will have the equation I provided you with many times.

Now when vehicle is at 2x the wind speed you have

0.5 * air density * area * (wind speed - (2 * wind speed))³  = 0.5 * air density * area * (-wind speed)³

Notice the negative sign in front of wind speed meaning negative power relative to vehicle meaning vehicle will be slowed down / decelerated at that rate and not accelerated.
The vehicle hits the air not the air hits the vehicle when vehicle speed is 2x wind speed.

If you replace air (compressible fluid) with water (incompressible fluid) you can no longer store energy in pressure differential thus you can no longer demonstrate 2x or 3x fluid flow speed for that vehicle.

[Naej]

The amount of power available to a wind powered vehicle (any design) is 1/2*mass flow*wind speed^2. If the vehicle seed is twice the wind speed, you get mass flow=density*area*wind speed.

We are making progress as now you say that wind power available to vehicle is the same as I always say but just with the equation presented a bit different

You wrote:
0.5 * (air density * area * wind speed) * wind speed²

That is correct and the same with
0.5 * air density * area * wind speed³

All you need to clarify now is that wind speed in this equation is wind speed relative to vehicle and not to the ground.
The wind speed relative to vehicle = (wind speed relative to ground - vehicle speed relative to ground) and now you will have the equation I provided you with many times.

I didn't write this. You did.
I'm glad you agree with my correct formula (yours is not).
I can clarify: wind speed is relative to ground.

Now when vehicle is at 2x the wind speed you have

0.5 * air density * area * (wind speed - (2 * wind speed))³  = 0.5 * air density * area * (-wind speed)³

Notice the negative sign in front of wind speed meaning negative power relative to vehicle meaning vehicle will be slowed down / decelerated at that rate and not accelerated.
The vehicle hits the air not the air hits the vehicle when vehicle speed is 2x wind speed.

If you replace air (compressible fluid) with water (incompressible fluid) you can no longer store energy in pressure differential thus you can no longer demonstrate 2x or 3x fluid flow speed for that vehicle.

Notice how there's no minus sign in my formula (the correct one).
Also you didn't compute the dam speed.

[electrodacus]

I didn't write this. You did.
I'm glad you agree with my correct formula (yours is not).
I can clarify: wind speed is relative to ground.

You did wrote exactly that.
Wind speed in your equation is not relative to the ground.

Just think about a 1.2kg ball traveling at say 10m/s relative to ground hitting a stationary vehicle (vehicle zero speed relative to ground)
Then imagine same ball with same 10m/s relative to ground hitting a vehicle that drives at 9m/s relative to ground in the same direction.
The speed of the ball relative to vehicle is just 1m/s (10m/s-9m/s) and so it will transfer way less energy to vehicle.
Then imagine the vehicle at say 15m/s relative to ground in same direction as the ball that travels at just 10m/s and in this case what happens is that vehicle will hit the ball at 5m/s so vehicle will be slowed down by this interaction.

Notice how there's no minus sign in my formula (the correct one).
Also you didn't compute the dam speed.

Look closer. I considered your example where vehicle speed is 2 * wind speed and it is correctly applied in the equation as wind speed relative to vehicle is minus wind speed since vehicle will see a head wind.
Maybe the ball example above is more clear.

[Naej]

I didn't write this. You did.
I'm glad you agree with my correct formula (yours is not).
I can clarify: wind speed is relative to ground.

You did wrote exactly that.
Wind speed in your equation is not relative to the ground.

No. Don't tell me what I meant.
The amount of power available to a wind powered vehicle (any design) is 1/2*mass flow*wind speed². If the vehicle seed is twice the wind speed, you get mass flow=density*area*wind speed.

Just think about a 1.2kg ball traveling at say 10m/s relative to ground hitting a stationary vehicle (vehicle zero speed relative to ground)
Then imagine same ball with same 10m/s relative to ground hitting a vehicle that drives at 9m/s relative to ground in the same direction.
The speed of the ball relative to vehicle is just 1m/s (10m/s-9m/s) and so it will transfer way less energy to vehicle.
Then imagine the vehicle at say 15m/s relative to ground in same direction as the ball that travels at just 10m/s and in this case what happens is that vehicle will hit the ball at 5m/s so vehicle will be slowed down by this interaction.

No. Let's say a guy in the vehicle slaps the ball at 10 m/s. Then the ball lost its kinetic energy, and the car is propelled.

Notice how there's no minus sign in my formula (the correct one).
Also you didn't compute the dam speed.

Look closer. I considered your example where vehicle speed is 2 * wind speed and it is correctly applied in the equation as wind speed relative to vehicle is minus wind speed since vehicle will see a head wind.
Maybe the ball example above is more clear.

Look closer, there is no minus sign: The amount of power available to a wind powered vehicle (any design) is 1/2*mass flow*wind speed². If the vehicle seed is twice the wind speed, you get mass flow=density*area*wind speed.

[electrodacus]

No. Don't tell me what I meant.
The amount of power available to a wind powered vehicle (any design) is 1/2*mass flow*wind speed². If the vehicle seed is twice the wind speed, you get mass flow=density*area*wind speed.

Look closer, there is no minus sign: The amount of power available to a wind powered vehicle (any design) is 1/2*mass flow*wind speed². If the vehicle seed is twice the wind speed, you get mass flow=density*area*wind speed.

So what you are saying is that vehicle speed is nowhere in your equation meaning non important ?
If you are not saying that then show where the vehicle speed fits.

No. Let's say a guy in the vehicle slaps the ball at 10 m/s. Then the ball lost its kinetic energy, and the car is propelled.

Yes but what powers the guy ? And is the ball stationary relative to the guy ?
If ball moves relative to the guy then amount of kinetic energy it will gain will be very different depending on both speed and direction.

[Naej]

No. Don't tell me what I meant.
The amount of power available to a wind powered vehicle (any design) is 1/2*mass flow*wind speed². If the vehicle seed is twice the wind speed, you get mass flow=density*area*wind speed.

Look closer, there is no minus sign: The amount of power available to a wind powered vehicle (any design) is 1/2*mass flow*wind speed². If the vehicle seed is twice the wind speed, you get mass flow=density*area*wind speed.

So what you are saying is that vehicle speed is nowhere in your equation meaning non important ?
If you are not saying that then show where the vehicle speed fits.

I'm not.
Mass flow is close to the density*area*relative wind speed.
(Not equal because when you have a relative wind speed equal to 0, you can still get some mass flow on your propeller)

No. Let's say a guy in the vehicle slaps the ball at 10 m/s. Then the ball lost its kinetic energy, and the car is propelled.

Yes but what powers the guy ? And is the ball stationary relative to the guy ?
If ball moves relative to the guy then amount of kinetic energy it will gain will be very different depending on both speed and direction.

The wheel. Perhaps the guy is some gears connected to the wheel.
And the ball stops moving relative to the ground, it loses its kinetic energy.

[electrodacus]

I'm not.
Mass flow is close to the density*area*relative wind speed.
(Not equal because when you have a relative wind speed equal to 0, you can still get some mass flow on your propeller)

The wheel. Perhaps the guy is some gears connected to the wheel.
And the ball stops moving relative to the ground, it loses its kinetic energy.

So you changed now to relative wind speed but you need to do the same on the other part of the equation.

You specifically mentioned "guy in the vehicle slaps the ball at 10m/s"
If vehicle is at wind speed all balls will look stationary so hitting the ball at 10m/s means the guy will have needed energy.
There is the vehicle kinetic energy that the guy could have used but that means reduction in vehicle kinetic energy which also means reduction in speed.

Wind power is only available to vehicle when vehicle speed is lower than wind speed while both vehicle and air particles travel in the same direction.

I guess the confusion comes from the fact that you think that using power taken at the wheels (breaks the vehicle) can result in more power output at the propeller (propulsion power) witch is obviously not allowed by energy conservation law.


You have potential energy that will be based on wind speed relative to vehicle and that can be converted into kinetic energy.

No direct downwind vehicle powered by wind can exceed wind speed as the max kinetic energy of that wind only powered vehicle can be ideal case equal with potential wind energy.
In non ideal case you have friction so max speed will be significantly lower than wind speed for a direct downwind vehicle but if you add an energy storage device then you can get to almost any speed as vehicle kinetic energy can get as high as the energy storage capacity of the device minus the friction losses.


a) Will you agree that I can build a sail vehicle (no propeller) where I add a super capacitor or similar energy storage device that I charge while vehicle is below wind speed and the exceed wind speed for some limited amount of time with that vehicle using the stored energy?

b) Will you agree that pressure differential created by the propeller is an energy storage device that can easily contain way more energy than needed to accelerate the blackbird to that record 13m/s ?

But if you still insist your equation is correct and different than the one I provided
P_w = 0.5 * air density * area * (wind speed - vehicle speed)³

Then just write it in the simplest form and we will put that to the test (see what prediction it makes in a few different cases and see if that matches reality).
To my understanding and your latest modifications this is what you say all speeds relative to ground.
P_w = 0.5 * air density * area * (wind speed)² * (wind speed - vehicle speed)

[gnuarm]

I should have known.  You are ^LITERALLY _incapable of understanding anything anyone tells you.

The design of the sails that move backward on the car, means the wind vs. sail relative speed is maintained as the rest of the car speed increases.  So the available energy never changes by your equation. 

[gnuarm]

Let's try this one more time. 

Car has upright sails mounted on a track, like a treadmill.  The track runs backwards when the car moves forward, as the track is geared to the wheels. 

When the sails reach the rear of the car and would otherwise go under the car with the track, they fold up so as to no longer interact with the wind.  When the sails reach the front of the car, they rise upward with the track and unfurl. 

The rest of the car is adequately aerodynamic so that any wind resistance is minute compared to the forces on the sails.

As the car moves forward, the sails move backward at the same speed, relative to the car, so the sails while unfurled, are motionless with respect to the ground.  Now the vehicle speed in YOUR equation is the speed of the sails, not the car.  This speed is always zero.

It should be apparent, that the speed of the car is irrelevant to this thought experiment.  The speed that does matter, is the speed of the sails, which is always zero.  Relative to the wind, the sail speed is always the wind speed alone, never changing, so the full power of the wind is always available. 

The maximum speed of the car will only depend on the losses in this imaginary sail mechanism and any real wind resistance of the car along with rolling resistance.  The wind will continue to push against the sails transferring the full power of the wind (calculated by any formula you want to think up) accelerating the car until the losses in the car equal that wind power.

[electrodacus]

I should have known.  You are ^LITERALLY _incapable of understanding anything anyone tells you.

The design of the sails that move backward on the car, means the wind vs. sail relative speed is maintained as the rest of the car speed increases.  So the available energy never changes by your equation.

Are you talking about a vehicle driving perpendicular to wind direction ? As that is a very different case that direct downwind.
If you are talking about a direct downwind vehicle no matter the design it can not exceed wind speed without energy storage. And obviously since stored energy is used it will only exceed wind speed for a limited amount of time and it will need to drop well below wind speed in order to recharge and repeat the cycle.

The one incapable of understanding may be you.
Do you notice that you provide no equations to support your claim?

All is needed is an equation describing the amount of wind power available to vehicle since the one I provided shows no wind power available when vehicle speed is at and above wind speed for a direct down wind vehicle.
I did not invent that equation it is used by many people in multiple scenarios.
It is simple high school level physics and just classical mechanics.  Not being able to understand this simpler case of energy conservation will basically make it impossible to understand the problem discussed in this main thread that also has to do with energy conservation and energy storage.
For you (and apparently many others) tho air seems to be as invisible and magic as the electrons.

But as I mentioned before the ones that know the least are the most confident. It is not that different from people that ^LITERALLY and confidently think feries and other similar creatures are real or think earth is flat despite all the evidence to the contrary.

[electrodacus]

Let's try this one more time. 

Car has upright sails mounted on a track, like a treadmill.  The track runs backwards when the car moves forward, as the track is geared to the wheels. 

When the sails reach the rear of the car and would otherwise go under the car with the track, they fold up so as to no longer interact with the wind.  When the sails reach the front of the car, they rise upward with the track and unfurl. 

The rest of the car is adequately aerodynamic so that any wind resistance is minute compared to the forces on the sails.

As the car moves forward, the sails move backward at the same speed, relative to the car, so the sails while unfurled, are motionless with respect to the ground.  Now the vehicle speed in YOUR equation is the speed of the sails, not the car.  This speed is always zero.

It should be apparent, that the speed of the car is irrelevant to this thought experiment.  The speed that does matter, is the speed of the sails, which is always zero.  Relative to the wind, the sail speed is always the wind speed alone, never changing, so the full power of the wind is always available. 

The maximum speed of the car will only depend on the losses in this imaginary sail mechanism and any real wind resistance of the car along with rolling resistance.  The wind will continue to push against the sails transferring the full power of the wind (calculated by any formula you want to think up) accelerating the car until the losses in the car equal that wind power.

I think you will need to make a drawing as your explanation of how the vehicle is constructed and works makes little sense.
It sort of seems like you are talking about a direct upwind rather than a direct down wind vehicle.  Is that right ?
A vehicle driving directly upwind has always access to wind energy unlike a vehicle driving directly downwind when it exceeds wind speed and no wind power is available to accelerate the vehicle.

As for the direct upwind vehicle that you seem to describe it will also not work without energy storage (a much smaller capacity charged and discharged multiple times per second but still necessary else it will not work).
What prevents a direct upwind vehicle to work without energy storage is the Newton's third law the one about equal and opposite reaction.
Fortunately energy storage is always present in basically any real system especially when we are talking about a super small storage capacity like a rubber belt or even a loose chain or gears.

[Alex Eisenhut]

[Naej]

I'm not.
Mass flow is close to the density*area*relative wind speed.
(Not equal because when you have a relative wind speed equal to 0, you can still get some mass flow on your propeller)

The wheel. Perhaps the guy is some gears connected to the wheel.
And the ball stops moving relative to the ground, it loses its kinetic energy.

So you changed now to relative wind speed but you need to do the same on the other part of the equation.

I didn't change anything and I don't need to change anything.

There is the vehicle kinetic energy that the guy could have used but that means reduction in vehicle kinetic energy which also means reduction in speed.

Yes. But the recoil means you accelerate the car and it is larger.

I guess the confusion comes from the fact that you think that using power taken at the wheels (breaks the vehicle) can result in more power output at the propeller (propulsion power) witch is obviously not allowed by energy conservation law.

No. What I think is that you get more force, but you don't see the difference between force and energy so 

Then just write it in the simplest form and we will put that to the test (see what prediction it makes in a few different cases and see if that matches reality).
To my understanding and your latest modifications this is what you say all speeds relative to ground.
P_w = 0.5 * air density * area * (wind speed)² * (wind speed - vehicle speed)

First, it wasn't changed. Second, the formula about kinetic energy was proven a long time ago.
So where do you disagree then? Either it's the amount of kinetic energy per kg, or it's the mass flow.

[electrodacus]

I didn't change anything and I don't need to change anything.

If you say so. The relative wind speed was just wind speed before. But not relevant anyway, I just need a final equation you think describes what happens in reality.

Yes. But the recoil means you accelerate the car and it is larger.

So will it be able to "recoil" without energy storage. There is no chance to put more power in to propeller than you take at the wheel.

No. What I think is that you get more force, but you don't see the difference between force and energy so 

What has force to do with anything ? Looking at force alone is a bad idea and that is what likely got Derek in trouble.
The one that needs to see the difference between force and power in not me.
Take the case of vehicle at 2x the wind speed (say wind speed is 10m/s and vehicle speed 20m/s)
Apply a 10N force at the wheel for 1ms (with the generator) and the power you take out ideal case will be 10N * 20m/s = 200W and say you do so for 1ms just so that we can also talk about energy and you have 200W * 0.001s = 0.2 Joules = 0.2Ws
Now take this energy 0.2Ws as that is all you have and apply it for 1ms to the propeller again ideal propeller 100% efficient at propulsion
What you get in therms of propulsion is exactly what you got out by slowing down the vehicle.
So 200W applied at propeller for 1ms will be 20N * (10m/s-20m/s) = -200W so -0.2Ws back into increasing vehicle kinetic energy the negative sign is your choice you can have that for generated power or for propulsion power is non-relevant is just to show direction is different putting in or taking out from the vehicle stored kinetic energy.
So looking at force it is 2x as large but due to speed being half the amount of thrust power and so the amount of kinetic energy you put back in vehicle is the same as the amount you took out in ideal case. In real case it will of course be way less.
 

First, it wasn't changed. Second, the formula about kinetic energy was proven a long time ago.
So where do you disagree then? Either it's the amount of kinetic energy per kg, or it's the mass flow.

Please just confirm that the one I wrote is what you mean. If is not write the one you think is correct. Just one equation do not split it in two parts.
P_w = 0.5 * air density * area * (wind speed)² * (wind speed - vehicle speed)

[PlainName]

Please just confirm that the one I wrote is what you mean. If is not write the one you think is correct. Just one equation do not split it in two parts.
Pw = 0.5 * air density * area * (wind speed)2 * (wind speed - vehicle speed)

I already did that way back, and you've yet to say precisely what's wrong with it (except you 'think it's wrong'), nor have you produced an equation that takes into account all aspects of the device. The one you keep quoting is the wrong one for this situation. As long as you keep fooling yourself about this, you will never get out of your rut.

[electrodacus]

Please just confirm that the one I wrote is what you mean. If is not write the one you think is correct. Just one equation do not split it in two parts.
Pw = 0.5 * air density * area * (wind speed)2 * (wind speed - vehicle speed)

I already did that way back, and you've yet to say precisely what's wrong with it (except you 'think it's wrong'), nor have you produced an equation that takes into account all aspects of the device. The one you keep quoting is the wrong one for this situation. As long as you keep fooling yourself about this, you will never get out of your rut.

I provide the one that is correct many times. You can have it in the same format:
P_w= 0.5 * air density * area * (wind speed - vehicle speed)² * (wind speed - vehicle speed)

Or the cleaner one that I have posted maybe hundreds of times
P_w = 0.5 * air density * area * (wind speed - vehicle speed)³

Once Naej confirm it is what he means we can test the equation at different vehicle speeds and see what it predicts and if the prediction is what we observe in reality. Spoiler alert it will not.
But maybe my last replay where I explained that higher force does not mean higher power and so no increase in speed even for ideal vehicle will convince him.

[PlainName]

I provide the one that is correct many times.

No, you haven't. You keep posting the wrong one. Where in that does it account for the propeller or drive to it? Are you saying they don't exist? They are the thing that makes this different from a sailboat, and yet your equation ignores them completely.

It is the wrong one. Really.

[electrodacus]

No, you haven't. You keep posting the wrong one. Where in that does it account for the propeller or drive to it? Are you saying they don't exist? They are the thing that makes this different from a sailboat, and yet your equation ignores them completely.

It is the wrong one. Really.

What has the design of the vehicle to do with available wind power ? Other the equivalent area that interacts with air there is no aspect of the vehicle that will influence the amount of wind power available to vehicle.
If you have had the back wheels generating energy for the front wheel. Will you have considered that as important in any way or make any sense ?
The main difference between the propeller being used as propulsion instead of the front wheel is the fact that propeller is also a large capacity energy storage device due to pressure differential that it creates.
That stored energy is a separate matter that is calculated separately as it has nothing to do with the question I posted and that is the available wind power relative to vehicle speed.

I know you have a feeling that propeller is important but it is because it is an energy storage device not his role as a propulsion device.
For example if you replace air with a non compressible fluid like water then the propeller is exactly like the front wheel no significant energy storage just good for propulsion and in that case the vehicle could not exceed wind speed directly downwind as there will not have been an energy storage device to help with that.
Same could be done if you add an energy storage device between the generator wheels and say a motor wheel in the front like say adding a capacitor or battery in between the two or even a flywheel as this things will be the equivalent of the compressible fluid.

So question is just for available wind power vs vehicle speed and that is the same for any vehicle design.  You just noticed this particular vehicle design exceeds wind speed and concluded that it is magic instead of knowing magic is not real and you need to look for an energy storage device as that is the only thing allowing a wind only powered vehicle to exceed wind speed directly downwind.

[gnuarm]

The car is going down wind.  The wheels on the car are connected to a belt so the top side moves backwards, opposite the direction of travel of the car. 

There are sails attached to the belt.  When the car moves forward, the belt moves backwards and the sails with it.  Since the gearing is 1:1, the sails actually stand still as the car moves under them. 

I only want to consider the sails on the top side, so when not on the top side, a mechanism collapses the sails to no size, so they do not interact with the wind. 

When the wind pushes on the sails, it will push on the car as a whole.  The car will move forward.  The belt is geared to the wheels, so moves backwards wrt the car, but stationary wrt the ground, just like the sail.  The wind speed blowing on the sail is always the same, no matter how fast the car moves.  So the available power from the wind is always the same according to YOUR formula. 

The result is the car will be propelled down wind, at a speed determined only by the various power losses vs. the wind power, otherwise having nothing to do with the wind speed.  In other words, the car can move faster than the wind, with no other theoretical limitations than the power in the wind vs. the frictional losses.

If you don't like the gearing that leaves the belt stationary, change it to 4:3 gearing so the belt moves backwards at 3/4 speed of the car moving forward.  Then the wind relative speed of the sail does not drop to zero until the car is moving at 4 times the speed of the wind.  So at some speed, there will be equilibrium between the wind power and the losses in the car.  For that speed to be greater than the wind speed, only requires practical design considerations in limiting the losses.

[PlainName]

I know you have a feeling that propeller is important but it is because it is an energy storage device not his role as a propulsion device.

It is no more an energy storage device than a set of tyres are.

Your closed mind prevents you from seeing how it works. I think we are all resigned to that now, but you keep spouting all this wrong stuff and that needs to be corrected or it gets perpetuated as fact (on the basis that no-one disputes it).

[electrodacus]

The car is going down wind.  The wheels on the car are connected to a belt so the top side moves backwards, opposite the direction of travel of the car. 

There are sails attached to the belt.  When the car moves forward, the belt moves backwards and the sails with it.  Since the gearing is 1:1, the sails actually stand still as the car moves under them. 

I only want to consider the sails on the top side, so when not on the top side, a mechanism collapses the sails to no size, so they do not interact with the wind. 

When the wind pushes on the sails, it will push on the car as a whole.  The car will move forward.  The belt is geared to the wheels, so moves backwards wrt the car, but stationary wrt the ground, just like the sail.  The wind speed blowing on the sail is always the same, no matter how fast the car moves.  So the available power from the wind is always the same according to YOUR formula. 

The result is the car will be propelled down wind, at a speed determined only by the various power losses vs. the wind power, otherwise having nothing to do with the wind speed.  In other words, the car can move faster than the wind, with no other theoretical limitations than the power in the wind vs. the frictional losses.

If you don't like the gearing that leaves the belt stationary, change it to 4:3 gearing so the belt moves backwards at 3/4 speed of the car moving forward.  Then the wind relative speed of the sail does not drop to zero until the car is moving at 4 times the speed of the wind.  So at some speed, there will be equilibrium between the wind power and the losses in the car.  For that speed to be greater than the wind speed, only requires practical design considerations in limiting the losses.

I do not not think you fully understand how the vehicle you describe will work.
Seems like what you just did was replace the axial propeller with a savonius type design

Forget about air and wind and imagine there is a automatic gun shutting 1.2kg balls leaving exactly 1m distance between them at say 6m/s relative to ground so that means each second your vehicle can be hit by six 1.2kg balls traveling at 6m/s

Now one of this balls hits the sail that you mentioned and so vehicle can gain that kinetic energy.

Say your vehicle was stationary first ball hits one of those sails. What happens to the vehicle and the sail relative to the ground ? What direction do they move relative to ground ?
Since you say vehicle is direct down wind the vehicle will move in the same direction the ball was moving and so will the sail even if you you seome gear ratio so that sail moves more than the vehicle or the other way around both sail and vehicle will need to move in the same direction relative to ground.

Only if you are talking about a direct upwind vehicle you can say the sail will move in the opposite direction to vehicle direction.