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

[electrodacus]

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).

I'm not talking about the flywheel effect.  I talking about the pressure differential energy storage and that is fairly large.
If you agree with the fact that there are more air particles on one side of the propeller than the other side while propeller rotates then you agree that there is a pressure differential and that contains stored energy.
You can only argue about the amount of energy stored there but I already showed a few post back that is more than enough to accelerate blackbird to 13m/s (the speed record).

You are not offering facts just strong opinions.

[PlainName]

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 ?

It would move the sail and thus the vehicle along a little bit, no?

[electrodacus]

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 ?

It would move the sail and thus the vehicle along a little bit, no?

Yes but relative to ground vehicle and sail will move in the same direction the ball moved.
He specifies that sail moves in opposite direction compared to vehicle in both cases relative to ground. That is only the case for a direct upwind vehicle tho he tries to explain a direct downwind one.

[PlainName]

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 ?

It would move the sail and thus the vehicle along a little bit, no?

Yes but relative to ground vehicle and sail will move in the same direction the ball moved.
He specifies that sail moves in opposite direction compared to vehicle in both cases relative to ground. That is only the case for a direct upwind vehicle tho he tries to explain a direct downwind one.

Not at all. If the ball hits the sail what is there to stop the vehicle heading in the opposite direction? There are no balls hitting that, so the only force is on the sail.

But also doesn't it depend on the gearing? That is, the sub-question is how far the vehicle moves compared to the sail.

[electrodacus]

Not at all. If the ball hits the sail what is there to stop the vehicle heading in the opposite direction? There are no balls hitting that, so the only force is on the sail.

But also doesn't it depend on the gearing? That is, the sub-question is how far the vehicle moves compared to the sail.

If the vehicle moves in the other direction relative to ground compared to the ball that will be a direct upwind so a different type of vehicle that requires a different discussion.

Before any discussion can start we need to decide if we are talking about direct downwind or direct upwind version of the vehicle as you can not use the same vehicle without modifications to do both things.

[PlainName]

If the vehicle moves in the other direction relative to ground compared to the ball that will be a direct upwind so a different type of vehicle that requires a different discussion.

Not at all. The problem with it is it's stationary so the sails aren't moving because they aren't being driven. It's effectively a turbine analogue rather than a propeller analogue.

[Naej]

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.

Nah I said wind speed IF the car speed is equal to twice the wind speed.

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.

Yes, and yes there is a chance.

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 ?

It's called mechanics. Forces mean there's an acceleration.

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.

Let's say efficiency is defined as  force*car speed/mechanical power on input so that your argument works.
Now: what if efficiency is 200%?
Of course it's impossible if there's no wind, but there is wind.
If you want to compute the efficiency of a propeller I gave you the formulae.

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)

Assuming the two velocities are in the same direction, you have:
P_w approximately equal to 0.5 * air density * area * (wind speed)² * |wind speed - vehicle speed|
And the approximation is good when the relative speed is large.
So where do you disagree then? Either it's the amount of kinetic energy per kg, or it's the mass flow.

[electrodacus]

If the vehicle moves in the other direction relative to ground compared to the ball that will be a direct upwind so a different type of vehicle that requires a different discussion.

Not at all. The problem with it is it's stationary so the sails aren't moving because they aren't being driven. It's effectively a turbine analogue rather than a propeller analogue.

Not sure I get what you want to say.

Is a direct upwind a different vehicle than direct downwind ?
If yes then it is important to specify what we are discussing else we will talk about completely different things.

It is also important to use the same reference like vehicle and sail relative to ground not vehicle relative to ground and then sail relative to vehicle.
The sail/sails are part of the vehicle so they will move around the vehicle but together with the vehicle.

So for a direct down wind the 1.2kg ball is hitting the sail transferring the kinetic energy to the vehicle.
When vehicle gets at exactly wind speed (balls speed) there is no way for the ball to heat any part of the vehicle including the moving sails as sails move around the vehicle.
When vehicle moves faster than wind the sails (part of the vehicle) will heat the balls so bals get kinetic energy from the vehicle slowing the vehicle down.
The balls are not a gas so you do not have the advantage of pressure differential.
The example with bals shows why vehicle can not exceed wind speed unless you have stored energy.

[PlainName]

When vehicle gets at exactly wind speed (balls speed) there is no way for the ball to heat any part of the vehicle including the moving sails as sails move around the vehicle.

You've got yourself in a mess there. The vehicle is moving at ball speed, right? So the balls are stationary with respect to the vehicle (or vice versa).

Now, the sails are moving relative to the vehicle, so they must be moving relative to the balls as well. How about that, eh!

Nevertheless, somehow you think the balls ain't gonna hit them sails. Well, you're right because it's the sails hitting the balls, but that's just a reference preference. I wonder how you can explain that one away.

[electrodacus]

Let's say efficiency is defined as  force*car speed/mechanical power on input so that your argument works.
Now: what if efficiency is 200%?
Of course it's impossible if there's no wind, but there is wind.
If you want to compute the efficiency of a propeller I gave you the formulae.

   wow
How will efficiency ever be above 100%
Wind is providing all the input power to vehicle (that if we ignore the pressure differential energy storage as you seems to want to).

So if at any point in time available wind power is say 1000W (just a round number)
You have the choice to
a) use all of it to accelerate the vehicle
b) as it is the case with direct downwind blackbird split it into multiple parts with total still 1000W
An example of splitting the 1000W available will be use 600W to accelerate and take 400W at the wheel to deliver to a 70% efficient propeller so 280W propeller output.
So if the medium is not a compressible fluid the 280W will end up accelerating the vehicle so add to the 600W total  880W worth of acceleration.
880W output/ 1000W input = 88% efficient.

Then there is the real case where those 280W will not all accelerate the vehicle but only small part say 80W with the difference of 200W being used to increase the pressure differential thus stored for later use.
Then you have the same 88% efficiency but only 680W are used for acceleration (vehicle will accelerate slower) but there is that 200W put in to storage than can be used later as it accumulates to exceed wind speed for some limited amount of time.

Assuming the two velocities are in the same direction, you have:
P_w approximately equal to 0.5 * air density * area * (wind speed)² * |wind speed - vehicle speed|
And the approximation is good when the relative speed is large.
So where do you disagree then? Either it's the amount of kinetic energy per kg, or it's the mass flow.

Another modification but that is fine at least you provided an equation that can be tested.


Will use 10m/s as wind speed
And equivalent area of 1m²

We have vehicle at:
0m/s
Available wind power according to your equation
0.5 * 1.2 * 1* 10² * |10-0| = 600W

According to the equation I claim to be correct
0.5 * 1.2 * 1 * (10-0)³ = 600W

5m/s
your equation
0.5 * 1.2 * 1* 10² * |10-5| = 300W

the one I claim to be correct:
0.5 * 1.2 * 1 * (10-5)³ = 75W

9m/s
your equation
0.5 * 1.2 * 1* 10² * |10-9| = 60W

the one I claim to be correct:
0.5 * 1.2 * 1 * (10-9)³ = 0.6W

10m/s
your equation
0.5 * 1.2 * 1* 10² * |10-10| = 0W

the one I claim to be correct:
0.5 * 1.2 * 1 * (10-10)³ = 0W

15m/s
your equation
0.5 * 1.2 * 1* 10² * |10-15| = 300W

the one I claim to be correct:
0.5 * 1.2 * 1 * (10-15)³ = -75W

20m/s
your equation
0.5 * 1.2 * 1* 10² * |10-20| = 600W

the one I claim to be correct:
0.5 * 1.2 * 1 * (10-20)³ = -600W

30m/s
your equation
0.5 * 1.2 * 1* 10² * |10-30| = 1200W

the one I claim to be correct:
0.5 * 1.2 * 1 * (10-30)³ = -4800W

50m/s
your equation
0.5 * 1.2 * 1* 10² * |10-50| = 2400W

the one I claim to be correct:
0.5 * 1.2 * 1 * (10-50)³ = -38400W


Can you see a problem with results provided by your equation ?
The only two points where the two equations provide the same result is for vehicle speed of 0m/s and for vehicle speed equal with wind speed.

You forced your equation to provide only positive numbers even tho when vehicle speed is above wind speed the direction of air molecules relative to vehicle changes.
I guess now but you will also claim this equation only applies to blackbird tho the equation contains absolutely nothing specific to blackbird.
You will not be able to claim the equation applies to a sail vehicle, to a wind turbine or to a vehicle when we are talking about drag power that is the equation I provided that applies to all that.
Your equation still predicts zero wind power when vehicle speed and wind speed are equal. So how will the vehicle exceed wind speed even the ideal one with no friction ?
What about a real one with say a total of 60W of friction? as according to your equation it can not exceed 9m/s since from there and up to 11m/s wind power available is less than 60W so less than friction losses.

Your equation is not describing what is seen in real world test not the blackbird and not the treadmill model.

[electrodacus]

You've got yourself in a mess there. The vehicle is moving at ball speed, right? So the balls are stationary with respect to the vehicle (or vice versa).

Now, the sails are moving relative to the vehicle, so they must be moving relative to the balls as well. How about that, eh!

Nevertheless, somehow you think the balls ain't gonna them sails. Well, you're right because it's the sails hitting the balls, but that's just a reference preference. I wonder how you can explain that one away.

The balls are 1m apart so if vehicle is longer than 1m there may be one or two balls stuck between the sails but they do nothing as they have the same speed as the sails or maybe fall down when says go under due to gravity.
Sails will require energy from somewhere in order to be able to hit the balls. It will be either from some energy storage device if the vehicle is to accelerate forward or from vehicle kinetic energy but that will slow the vehicle down so such a vehicle can never exceed wind speed.

[Naej]

Let's say efficiency is defined as  force*car speed/mechanical power on input so that your argument works.
Now: what if efficiency is 200%?
Of course it's impossible if there's no wind, but there is wind.
If you want to compute the efficiency of a propeller I gave you the formulae.

   wow
How will efficiency ever be above 100%
Wind is providing all the input power to vehicle (that if we ignore the pressure differential energy storage as you seems to want to).

So if at any point in time available wind power is say 1000W (just a round number)

Did you read the first part?
It says: Let's say efficiency is defined as  force*car speed/mechanical power on input so that your argument works.

So it's not efficiency with respect to wind power.
If you have 1000W in wind energy you can for example leave 700W in the air, use 100W for the drag of the vehicle, 100W to compensate tire friction, 100W for accelerating.
So the propeller gets 300W out of 100W in mechanical input, or 300% efficiency.

Note that I used a definition of efficiency to match your use of the word.

Assuming the two velocities are in the same direction, you have:
P_w approximately equal to 0.5 * air density * area * (wind speed)² * |wind speed - vehicle speed|
And the approximation is good when the relative speed is large.
So where do you disagree then? Either it's the amount of kinetic energy per kg, or it's the mass flow.

Another modification

Wrong it's the same.
Relative wind speed is |wind speed - vehicle speed|.

You forced your equation to provide only positive numbers even tho when vehicle speed is above wind speed the direction of air molecules relative to vehicle changes.

Yes kinetic energy is always positive.

I guess now but you will also claim this equation only applies to blackbird tho the equation contains absolutely nothing specific to blackbird.

Wrong again.

You will not be able to claim the equation applies to a sail vehicle, to a wind turbine or to a vehicle when we are talking about drag power that is the equation I provided that applies to all that.

Of course it's not about drag power. It's the kinetic energy in the wind.

Your equation still predicts zero wind power when vehicle speed and wind speed are equal. So how will the vehicle exceed wind speed even the ideal one with no friction ?
What about a real one with say a total of 60W of friction? as according to your equation it can not exceed 9m/s since from there and up to 11m/s wind power available is less than 60W so less than friction losses.

Your equation is not describing what is seen in real world test not the blackbird and not the treadmill model.

Maybe you should read what I wrote then??? "And the approximation is good when the relative speed is large."
As I explained before - but you won't listen - in still air, you still have some air flowing through a propeller when it is powered, but it is due to power input.
Now I could find a better approximation but since you don't really read what I write it makes little sense.

So where do you disagree then? Either it's the amount of kinetic energy per kg, or it's the mass flow.

[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.

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

We don't need to consider any windmill since that would not apply to a car with sails.  You do know a windmill is not a sail, right?  This model uses sails, as does your equation that you insist on applying to everything.  This is very simple to understand and analyze.  That's why it is proposed.  So, of course you are going to try to change it entirely. 

Forget about air and wind

Nope.  I've proposed a simple, clear model of a car in the wind traveling down wind that will clearly be capable of going down wind.  Let's stick to that.

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.

No, you don't understand.  Relative to the car, the sail moves toward the wind.  With 1:1 gearing relative to the ground, the sail doesn't actually move at all, with 4:3 gearing it moves relative to the ground at 1/4 the speed of the car. 

Stop trying to muddy the waters.  This car will clearly move down wind, at speeds greater than the wind, only limited by the frictional losses.  You insistence in applying an erroneous calculation to the Blackbird car is fixed by this model where your equation does apply, since the wind is blowing into sails, not the air movement produced by the propeller. 

[electrodacus]

Did you read the first part?
It says: Let's say efficiency is defined as  force*car speed/mechanical power on input so that your argument works.

So it's not efficiency with respect to wind power.
If you have 1000W in wind energy you can for example leave 700W in the air, use 100W for the drag of the vehicle, 100W to compensate tire friction, 100W for accelerating.
So the propeller gets 300W out of 100W in mechanical input, or 300% efficiency.

Note that I used a definition of efficiency to match your use of the word.

Please define exactly the "force*car speed/mechanical power on input"
What applies that force ?
car speed relative to ground or wind?
and what is exactly mechanical power on input?
 
There is no such thing as 1000W of wind energy  (Watt is unit for power not energy)
If mechanical input to a propeller is 100W the output will be in ideal case 100W and in real world more like 70W for an efficient propeller in air.

You have very significant problems in understanding so I was quite wrong about you.
It makes no sense to replay to the rest of your comments as long as you think efficiency above 100% is possible and you confuse power with energy.

[gnuarm]

Not at all. If the ball hits the sail what is there to stop the vehicle heading in the opposite direction? There are no balls hitting that, so the only force is on the sail.

But also doesn't it depend on the gearing? That is, the sub-question is how far the vehicle moves compared to the sail.

The point of this vehicle, is to arrange a sail that does not move as fast as the car, so that the power equation he keeps insisting is the right one, really is the right one.  So  now, the sail moves to the wind end of the car as the car moves in the direction of the wind.  As the sail reaches the back of the car (the windward end) it folds up and no longer catches the wind, but moves back to the front with the belt.  Other sails continue to unfold at the front of the car (the leeward end) where they start catching the wind again and move to the back of the car.

To keep people from freaking out about a sail that doesn't move at all, consider the gearing to be 4:3, so as the car moves forward (with the wind) by 4 units, the sail moves backwards (toward the wind) on the car by 3 units, which means it moves relative to the ground forward (with the wind) by 1 unit. 

The power equation now provides significant power to the sails when the car is moving the same speed as the wind, since the sail is moving with the wind, at only 1/4 the speed of the wind.  When the sail is moving with the wind, at half the speed of the wind, it still receives significant power, and the car is now moving at twice the speed of the wind.  QED!

I'm not sure, but I think ED is playing dumb so he doesn't have to deal with the reality of this car which clearly can move down wind faster than the wind.  He will insist on changing the case and turn it into something where he can obfuscate and confuse the facts. 

[electrodacus]

No, you don't understand.  Relative to the car, the sail moves toward the wind.  With 1:1 gearing relative to the ground, the sail doesn't actually move at all, with 4:3 gearing it moves relative to the ground at 1/4 the speed of the car. 

Stop trying to muddy the waters.  This car will clearly move down wind, at speeds greater than the wind, only limited by the frictional losses.  You insistence in applying an erroneous calculation to the Blackbird car is fixed by this model where your equation does apply, since the wind is blowing into sails, not the air movement produced by the propeller.

OK so both the sail and the car moves in the same direction as the wind just at different speeds.
Getting to 1.2kg balls traveling at 10m/s
When vehicle speed relative to ground is below 10m/s balls will hit the sail and transfer kinetic energy to vehicle.
When vehicle speed is the same as wind speed no balls (no air particles) will be able to hit the moving sail's
The sails move around the vehicle. You may need to see an animation if you can not visualize that.
The 1.2kg ball has the exact same speed as the vehicle so it will follow the vehicle but never be able to hit any sail mounted to the vehicle even if the sails move around the vehicle they move half the time in the direction the vehicle moves faster than the vehicle (folded under the vehicle) and half the time they move in the opposite direction above the vehicle but the average will be the speed of the vehicle as they are part of the vehicle. 

[electrodacus]

I'm not sure, but I think ED is playing dumb so he doesn't have to deal with the reality of this car which clearly can move down wind faster than the wind.  He will insist on changing the case and turn it into something where he can obfuscate and confuse the facts.

The vehicle you propose can move faster than wind if you provide that vehicle with an energy source (battery and an electric motor for example).
If wind power alone needs to power the vehicle then it will also work but only at a speed below wind speed if no energy storage of any type is available like pressure differential that you prefer to ignore it exist (air is a compressible fluid if you understand the meaning of that).
Exchange the air for 1.2kg balls or water where there is no pressure differential possible as water is a non compressible fluid and your vehicle can not exceed wind speed.
 
With air it will exceed wind speed thanks to pressure differential energy storage but it will be above wind speed only for as long as that stored energy can accelerate the vehicle. When that stored energy is used up the vehicle will start to decelerate (due to friction losses) until it drops below wind speed.
You have seen incomplete tests as both blackbird stopped before the vehicle used up all stored energy and treadmill model has to little space before it runs out of track.
Especially with treadmill where there are less variable and more controlled environment you can see how acceleration rate drops as stored energy is used up and you can calculate exactly when it will stop accelerating based on the rate at which the acceleration rate drops.

[electrodacus]

Maybe as this is an electrical forum people here have a bit better intuition about electricity (possibly just wishful thinking on my part).

Input is a CV-CC lab power supply rated at 200W 20V and 10A max

Say you want 30V.  Is it possible to add a circuit between the output of the 200W lab power supply that has no energy storage device and get 30V on the other side?
What about 15A ?  What about more than 200W for some limited amount of time?

Is any of that possible without having a energy storage device in that circuit ?

[gnuarm]

No, you don't understand.  Relative to the car, the sail moves toward the wind.  With 1:1 gearing relative to the ground, the sail doesn't actually move at all, with 4:3 gearing it moves relative to the ground at 1/4 the speed of the car. 

Stop trying to muddy the waters.  This car will clearly move down wind, at speeds greater than the wind, only limited by the frictional losses.  You insistence in applying an erroneous calculation to the Blackbird car is fixed by this model where your equation does apply, since the wind is blowing into sails, not the air movement produced by the propeller.

OK so both the sail and the car moves in the same direction as the wind just at different speeds.
Getting to 1.2kg balls traveling at 10m/s
When vehicle speed relative to ground is below 10m/s balls will hit the sail and transfer kinetic energy to vehicle.
When vehicle speed is the same as wind speed no balls (no air particles) will be able to hit the moving sail's

You aren't understanding.  The sails move at a quarter the rate of the vehicle.  So when the vehicle is moving at twice the speed of the wind, the wind is still hitting the sails at one half the wind speed relative to the ground. 

The sails move around the vehicle. You may need to see an animation if you can not visualize that.

Uh, no, the sails move from the front of the vehicle to the rear in a straight line along a belt.  The sails collapse on reaching the rear, so they can follow the belt under the vehicle and be pulled back up on reaching the front of the car.  I created this model.  If you don't understand, I suggest you create an animation to help you visualize it.

The 1.2kg ball has the exact same speed as the vehicle so it will follow the vehicle but never be able to hit any sail mounted to the vehicle even if the sails move around the vehicle they move half the time in the direction the vehicle moves faster than the vehicle (folded under the vehicle) and half the time they move in the opposite direction above the vehicle but the average will be the speed of the vehicle as they are part of the vehicle.

You literally can't understand this can you?  The only speed of the sail that matters, is the speed when it is catching the wind.  At that time it is moving wrt to the ground slower than the wind, so receives power from the wind, while the car is moving forward, faster than the wind. 

Please create the animation.  But if you think the velocity of the sail when it is folded up has anything to do with the problem, there is nothing anyone can say to you to help you understand, or.... you do understand, but you are not capable of acknowledging you have been wrong all this time.  Even money, either way.

[gnuarm]

I'm not sure, but I think ED is playing dumb so he doesn't have to deal with the reality of this car which clearly can move down wind faster than the wind.  He will insist on changing the case and turn it into something where he can obfuscate and confuse the facts.

The vehicle you propose can move faster than wind if you provide that vehicle with an energy source (battery and an electric motor for example).
If wind power alone needs to power the vehicle then it will also work but only at a speed below wind speed if no energy storage of any type is available like pressure differential that you prefer to ignore it exist (air is a compressible fluid if you understand the meaning of that).
Exchange the air for 1.2kg balls or water where there is no pressure differential possible as water is a non compressible fluid and your vehicle can not exceed wind speed.

It doesn't matter how many times you say this, it still won't be true.

This is a simple case as you can get.  You provided the equation of the wind interacting with the sail which shows this will work.  That equation does not depend on any storage of energy or any other nonsense you come up with.  It is as fundamental as the downwind sailboat example it was created for.  Now you are just kidding yourself.

With air it will exceed wind speed thanks to pressure differential energy storage

Blah, blah, blah.  The sail does not exceed the speed of the wind, so the equation applies perfectly. 

but it will be above wind speed only for as long as that stored energy can accelerate the vehicle.

Blah, blah, blah...  you can't squirm your way out of this with your nonsense.  None of that applies to a sail boat.  This is just a sail boat geared to the car.  Wind on sail, moving the car faster than the wind, while the sail moves slower than the wind.  There is no energy storage bunkum like you want to apply to the Blackbird.

When that stored energy is used up the vehicle will start to decelerate (due to friction losses) until it drops below wind speed.
You have seen incomplete tests as both blackbird stopped before the vehicle used up all stored energy and treadmill model has to little space before it runs out of track.
Especially with treadmill where there are less variable and more controlled environment you can see how acceleration rate drops as stored energy is used up and you can calculate exactly when it will stop accelerating based on the rate at which the acceleration rate drops.

You are so full of it. 

[Naej]

Maybe as this is an electrical forum people here have a bit better intuition about electricity (possibly just wishful thinking on my part).

Input is a CV-CC lab power supply rated at 200W 20V and 10A max

Say you want 30V.  Is it possible to add a circuit between the output of the 200W lab power supply that has no energy storage device and get 30V on the other side?
What about 15A ?  What about more than 200W for some limited amount of time?

Is any of that possible without having a energy storage device in that circuit ?

Yes. Excellent example. The energy storage you're searching for is: the wheels, the gears, the propeller blades and the vehicle.

[Naej]

Did you read the first part?
It says: Let's say efficiency is defined as  force*car speed/mechanical power on input so that your argument works.

So it's not efficiency with respect to wind power.
If you have 1000W in wind energy you can for example leave 700W in the air, use 100W for the drag of the vehicle, 100W to compensate tire friction, 100W for accelerating.
So the propeller gets 300W out of 100W in mechanical input, or 300% efficiency.

Note that I used a definition of efficiency to match your use of the word.

Please define exactly the "force*car speed/mechanical power on input"
What applies that force ?
car speed relative to ground or wind?
and what is exactly mechanical power on input?

Again, I defined the word YOU used so you should know.
The propeller applies the force on the car, car speed relative to ground, and mechanical power on input is what comes from the wheels (by gears).

If mechanical input to a propeller is 100W the output will be in ideal case 100W and in real world more like 70W for an efficient propeller in air.

What you said is correct in still air, and completely incorrect when there's a wind.
Of course you didn't prove what you said so you shouldn't be surprised it's completely incorrect.

If you prefer, you take my example above where the propeller takes 1000W of wind power, and puts 300W in the car so it's "30% efficient" (for a different definition of efficiency, obviously).

[iMo]

The world land speed record for a wind-powered vehicle was broken on 26 March 2009 by Briton Richard Jenkins in his yacht Greenbird with a speed of 126.1 mph (202.9 km/h). Wind speeds were fluctuating between 30–50 mph (48–80 km/h) at that time.

https://www.wikiwand.com/en/Land_sailing

[PlainName]

I'm not sure, but I think ED is playing dumb so he doesn't have to deal with the reality of this car which clearly can move down wind faster than the wind.  He will insist on changing the case and turn it into something where he can obfuscate and confuse the facts.

Exactly that. It would be the prefect definition of trolling technique were it not that he believes what he's posting.

[electrodacus]

You aren't understanding.  The sails move at a quarter the rate of the vehicle.  So when the vehicle is moving at twice the speed of the wind, the wind is still hitting the sails at one half the wind speed relative to the ground. 

Uh, no, the sails move from the front of the vehicle to the rear in a straight line along a belt.  The sails collapse on reaching the rear, so they can follow the belt under the vehicle and be pulled back up on reaching the front of the car.  I created this model.  If you don't understand, I suggest you create an animation to help you visualize it.

You literally can't understand this can you?  The only speed of the sail that matters, is the speed when it is catching the wind.  At that time it is moving wrt to the ground slower than the wind, so receives power from the wind, while the car is moving forward, faster than the wind. 

Please create the animation.  But if you think the velocity of the sail when it is folded up has anything to do with the problem, there is nothing anyone can say to you to help you understand, or.... you do understand, but you are not capable of acknowledging you have been wrong all this time.  Even money, either way.

I can create an animation in my head and since I know the physics involved it will be consistent with reality.
The sail move at a quarter of the vehicle speed when it is above the vehicle and at one and a quarter the vehicle speed when bellow the vehicle with average sail speed being exactly the speed of the vehicle.
That is what it means to move around the vehicle.

Since no balls can hit any part of the vehicle including the sail when their average speed is equal the vehicle kinetic energy can not increase thus speed can not increase.

Blah, blah, blah.  The sail does not exceed the speed of the wind, so the equation applies perfectly.

On blackbird the sail exceeds wind speed when vehicle exceeds the wind speed as they do have the exactly same speed.

Blah, blah, blah...  you can't squirm your way out of this with your nonsense.  None of that applies to a sail boat.  This is just a sail boat geared to the car.  Wind on sail, moving the car faster than the wind, while the sail moves slower than the wind.  There is no energy storage bunkum like you want to apply to the Blackbird.

No sail boat or sail vehicle can exceed wind speed directly downwind.