Author Topic: Mess with your minds: A wind powered craft going faster than a tail wind speed.  (Read 147327 times)

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Offline Microdoser

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It did take some head scratching before I got it, but I can totally see how it works.

It does require the surface it rides on to be moving towards the rear of the vehicle faster than the air though, which is where it gets the energy to move upwind. You don't get free energy.

This is why it works on a treadmill as well as with a tailwind. I imagine it would need a push start too because stationary props don't provide any thrust.

You do need a really really low friction gearing setup too.
 

Offline electrodacus

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Lol

I didn't read the whole thread.  How long has electrodacus been insisting the car that goes down wind faster than the wind doesn't?  How long has he insisted energy involves time and power doesn't?

I find his arguments mildly entertaining.  The diagram posted seems to me to be a lot of fluff.  Especially amusing is his stored energy argument to explain "impossible actions" when it is just poor construction at work.  Then he says a better constructed model has "micro storage" of energy or something. 

If the wheels are geared 2:1, applying a force to the 1 wheel by moving the belt it is riding on will exert say 1 unit of force in the direction of the belt.  The gearing will result in twice the movement in the 2 wheel and half the force.  The two forces are opposite, so the vehicle has a half unit of force pushing it in the direction of the moving belt with the movement relative to the ground being twice the velocity of the movement of the belt.

I'm willing to say I'm wrong when someone shows that to me.  Happy to, in fact.  But someone who doesn't understand that "rate of change" in regards to values like power and energy fundamentally involves time can't possibly construct a rational argument.  While rate of change can refer to some other variable, in the case of power, it is the rate of change of energy wrt time.  Power involves time.  Energy does not. 

This is such an insane argument!  I won't argue with the guy because his arguments are so specious, yet I want so badly to help him understand.  But I'm not going to chase impossible dreams.  They guy clearly does not want to understand, so no one can help him.

I have never claimed blackbird is not going faster than the wind.
What I was saying is that it can only do so using energy storage and in case of blackbird that energy storage is the pressure differential created by the propeller.
A gear can not amplify power thus output power will always be lower than the input.

Offline electrodacus

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It did take some head scratching before I got it, but I can totally see how it works.

It does require the surface it rides on to be moving towards the rear of the vehicle faster than the air though, which is where it gets the energy to move upwind. You don't get free energy.

This is why it works on a treadmill as well as with a tailwind. I imagine it would need a push start too because stationary props don't provide any thrust.

You do need a really really low friction gearing setup too.

There is no need for push start.  The way it works is that it stores energy while below wind speed and uses that stored energy to exceed wind speed for a limited amount of time.
For blackbird including the small treadmill model the energy is stored in the pressure differential created by the propeller and only works because air is a compressible fluid.
If you where to replace air with say water (non compressible fluid) then it will no longer work.
I can build a vehicle that uses a sail add to it a energy storage device say a super capacitor and a generator/motor and have it perform the same (actually better) than blackbird or the small treadmill model.

Offline electrodacus

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I need to bring back the discussion to what is relay important.

What is the available wind power to direct down wind vehicle.

I say that available wind power is defined by this equation  0.5 * air density * area * (wind speed - vehicle speed)^3

If anyone disagree with that please provide the correct equation so we can compare the predictions.

Offline Kleinstein

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One may call the pressure that drives the prop forward stored energy, if one really wants too. However this a energy in a dynamic balance: loosing energy to drive the vehicle and to the less than perfect aerodynamics and getting new energy from the rear wheels.

The point is that the driven prop can get you more drag than a sail. It can do it a zero speed (e.g. no wind and not moving and thus zero drag for the sail) - this case is the abvious one I think.  Similar it can do it when the vehicle is driving at the speed of the wind and thus zero relativ speed between the vehicle and the wind.

When starting at low speed the slow moving prop would still have some drag, not much, but it should be enough to get it starting. The prop would need not extra power and at low speed the prop would actually generate extra forward power and act as a windmill. It would need some minimum wind speed to get it going.

I don't think they needed the adjustable pitch during driven. This is likely just a thing of finding the best setting to get the highest speed. Once set right it should be OK to keep it there. The prop is not very heavy and not going very fast, as it is optimized to work at a relatively low relativ air speed. Remember it only sees the difference, so normally not the full wind speed. So the kinetic energy in the prop is low compared to the kinetic energy in the main vehicle. The model on the treadmil works with a fixed prop.
 

Offline bdunham7

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I have never claimed blackbird is not going faster than the wind.

But you are claiming that it cannot sustain the faster-than-wind travel.  Or are  you beginning to realize that you are wrong?

Quote
I say that available wind power is defined by this equation  0.5 * air density * area * (wind speed - vehicle speed)^3

As I said earlier, the 'available wind power' does not depend on the vehicle speed where the vehicles mechanisms can use the ground as a fixed reference, only the windspeed relative to the ground matters.  So the prediction is that a properly configured vehicle can travel both upwind (at some rate) and downwind (faster than the wind) indefinitely.  And, it does exactly that experimentally.
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Offline electrodacus

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One may call the pressure that drives the prop forward stored energy, if one really wants too. However this a energy in a dynamic balance: loosing energy to drive the vehicle and to the less than perfect aerodynamics and getting new energy from the rear wheels.

I do not see how else you can call the pressure differential created by the propeller other than a form of energy storage.
There is loop but it is as follow (proportions are just an example).
Wind power available to vehicle will be split say in two equal parts.  One half accelerates the vehicle directly thus ends up as kinetic energy the other half will be taken from wheels and sent to propeller witch then increases the pressure differential (you can look at this as increasing the apparent wind speed relative to vehicle).
So for a sail only vehicle all available wind power is converted to kinetic energy thus it will accelerate much faster than blackbird.
Since there is no other energy storage device other than kinetic energy all potential wind energy is converted to kinetic energy in case of sail cart and so that can never exceed wind speed unless it is not driving directly down wind.
For blackbird since it takes in this example half of the available power and stores that as pressure differential basically increasing the available potential energy it allows the blackbird to exceed wind speed for some limited amount of time as when above wind speed direct downwind there is no longer any wind power available to vehicle and it is starting to use the energy stored as pressure differential but it will continue to use just half of the power provided by the pressure differential to accelerate (increase kinetic energy) and then the other half it will put back in to increasing the pressure differential.
Obviously since only half is put back the overall pressure differential will drop and vehicle acceleration rate will continue to drop until there is not enough and it will need to start to slow down.


The point is that the driven prop can get you more drag than a sail. It can do it a zero speed (e.g. no wind and not moving and thus zero drag for the sail) - this case is the abvious one I think.  Similar it can do it when the vehicle is driving at the speed of the wind and thus zero relativ speed between the vehicle and the wind.

When starting at low speed the slow moving prop would still have some drag, not much, but it should be enough to get it starting. The prop would need not extra power and at low speed the prop would actually generate extra forward power and act as a windmill. It would need some minimum wind speed to get it going.

I don't think they needed the adjustable pitch during driven. This is likely just a thing of finding the best setting to get the highest speed. Once set right it should be OK to keep it there. The prop is not very heavy and not going very fast, as it is optimized to work at a relatively low relativ air speed. Remember it only sees the difference, so normally not the full wind speed. So the kinetic energy in the prop is low compared to the kinetic energy in the main vehicle. The model on the treadmil works with a fixed prop.

Yes as shown by the treadmill model the variable pitch is not needed for this to work.

Offline electrodacus

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But you are claiming that it cannot sustain the faster-than-wind travel.  Or are  you beginning to realize that you are wrong?

Yes that is correct the acceleration rate as seen in all tests drops and at some point acceleration rate will be zero and the deceleration phase will start.

Quote
I say that available wind power is defined by this equation  0.5 * air density * area * (wind speed - vehicle speed)^3

As I said earlier, the 'available wind power' does not depend on the vehicle speed where the vehicles mechanisms can use the ground as a fixed reference, only the windspeed relative to the ground matters.  So the prediction is that a properly configured vehicle can travel both upwind (at some rate) and downwind (faster than the wind) indefinitely.  And, it does exactly that experimentally.

Please provide the equation describing what you are saying. If you can not do that you can not claim you understand how this vehicle works.
It is super obvious that vehicle speed while driving directly down wind affects the amount of available wind power since the wind speed relative to vehicle drops (wind speed - vehicle speed).

Offline fourfathom

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But you are claiming that it cannot sustain the faster-than-wind travel.  Or are  you beginning to realize that you are wrong?

Yes that is correct the acceleration rate as seen in all tests drops and at some point acceleration rate will be zero and the deceleration phase will start.

Quote
I say that available wind power is defined by this equation  0.5 * air density * area * (wind speed - vehicle speed)^3

As I said earlier, the 'available wind power' does not depend on the vehicle speed where the vehicles mechanisms can use the ground as a fixed reference, only the windspeed relative to the ground matters.  So the prediction is that a properly configured vehicle can travel both upwind (at some rate) and downwind (faster than the wind) indefinitely.  And, it does exactly that experimentally.

Please provide the equation describing what you are saying. If you can not do that you can not claim you understand how this vehicle works.
It is super obvious that vehicle speed while driving directly down wind affects the amount of available wind power since the wind speed relative to vehicle drops (wind speed - vehicle speed).

The applicable equation is some_factor * air density * area * (wind speed - ground speed)^3 

Since ground speed = 0, the equation becomes:
some_factor * air density * area * (wind speed)^3

The difference is that your equation assumes that the vehicle has no contact with the ground, and behaves something like a balloon.  This is certainly not the case with the DDWFTTW vehicles we ae discussing.
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Offline bdunham7

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Please provide the equation describing what you are saying. If you can not do that you can not claim you understand how this vehicle works.

I have and I just did.  Simply remove '-vehicle speed' from your equation.  And I can claim anything I like, so I'm going to claim that power has very little to do with how it works.

Quote
It is super obvious that vehicle speed while driving directly down wind affects the amount of available wind power since the wind speed relative to vehicle drops (wind speed - vehicle speed).

Yes, at first glance it seems 'super obvious', but that's the whole point that makes it interesting--the 'super obvious' answer is wrong.  And that has been demonstrated and explained in detail--and understood clearly by everyone but you, as far as I can tell.  Do  you think that YouTube and EEVBlog, or the entire world for that matter, is gaslighting you? 
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Online IanB

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I do not see how else you can call the pressure differential created by the propeller other than a form of energy storage.

A propeller does not create a pressure differential. A propeller moves ("propels") air. It creates thrust by the law of conservation of momentum, by accelerating the air that goes through it. The air leaving a propeller has a lower pressure because it is moving. You can see this described here:

https://www.quora.com/Why-is-the-static-pressure-before-a-fan-higher-than-after-the-fan

And illustrated here by experiment:

https://youtu.be/f2QfVJe7yEg?t=198

Given that there is no "pressure bubble" behind the propeller, there is no energy storage there to push back on the propeller. Once the air has gone through the propeller it has done its work, there is no more work to give.
 

Offline electrodacus

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The applicable equation is some_factor * air density * area * (wind speed - ground speed)^3 

Since ground speed = 0, the equation becomes:
some_factor * air density * area * (wind speed)^3

The difference is that your equation assumes that the vehicle has no contact with the ground, and behaves something like a balloon.  This is certainly not the case with the DDWFTTW vehicles we ae discussing.

That will be the correct equation for something that is not moving relative to ground like a wind turbine. For a wind turbine that is exactly the equation just that you add the turbine efficiency.
If you install the wind turbine on a vehicle and then drive direct down wind then equation of the power output for that wind turbine will again contain (wind speed - vehicle speed) since the wind speed relative to that moving wind turbine will be lower the faster the vehicle moves.

Offline electrodacus

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A propeller does not create a pressure differential. A propeller moves ("propels") air. It creates thrust by the law of conservation of momentum, by accelerating the air that goes through it. The air leaving a propeller has a lower pressure because it is moving. You can see this described here:

https://www.quora.com/Why-is-the-static-pressure-before-a-fan-higher-than-after-the-fan

And illustrated here by experiment:

https://youtu.be/f2QfVJe7yEg?t=198

Given that there is no "pressure bubble" behind the propeller, there is no energy storage there to push back on the propeller. Once the air has gone through the propeller it has done its work, there is no more work to give.

Sorry but you are just ignoring the fact that air is a compressible fluid.
Are you saying this graph below is completely wrong ?
https://en.wikipedia.org/wiki/Axial_fan_design

Online IanB

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I say that available wind power is defined by this equation  0.5 * air density * area * (wind speed - vehicle speed)^3

The equation  0.5 * (air density) * (area) * (wind speed)^3  gives the total flow of kinetic energy from wind moving through a particular cross section of area, and is applicable to wind turbine design as the air is moving through the turbine. A turbine could never extract more power than this from the wind (it will always be less than this). Note that this equation only works if the wind moves through the area of interest, like the disk area of a turbine.

Since a sail is a solid barrier, the wind does not, and cannot, move through the sail. Therefore, this equation cannot be applied to sails.

Another way of looking at this is to observe that the equation is (kinetic energy of wind per unit mass) x (mass flow of wind moving through control area). Since the mass flow of wind through the sail is zero, the energy flow must necessarily be zero.

When working with sails, you have to use force vectors and force balances.
 

Offline Kleinstein

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I say that available wind power is defined by this equation  0.5 * air density * area * (wind speed - vehicle speed)^3
I say this is wrong, and the PDF on wind power you linked also says something different:
They get the wind power from the change in kinetic energy for the air passing though the "turbine" area. The same way of calculations,  when slowing down the air to the vehicle speed this would be 0.5 * air density * area * (wind speed^2 - vehicle speed^2)*(wind speed - vehicle speed)
 = 0.5 * air density * area * (wind speed - vehicle speed)^2*(wind speed + vehicle speed).
With a simple sail the power would be from drag force time vehicle speed and thus within the approximations in these equations:
P_sail =  0.5 * air density * area * (wind speed - vehicle speed)^2* vehicle speed

With the active driven prop there can be even more power available, as the air may slow down even more. This is especially true for a speed close to or higher than the speed of the wind. This is because this way the wind can be slowed down to a speed below the vehicle speed.
 

Offline electrodacus

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I say that available wind power is defined by this equation  0.5 * air density * area * (wind speed - vehicle speed)^3

The equation  0.5 * (air density) * (area) * (wind speed)^3  gives the total flow of kinetic energy from wind moving through a particular cross section of area, and is applicable to wind turbine design as the air is moving through the turbine. A turbine could never extract more power than this from the wind (it will always be less than this). Note that this equation only works if the wind moves through the area of interest, like the disk area of a turbine.

Since a sail is a solid barrier, the wind does not, and cannot, move through the sail. Therefore, this equation cannot be applied to sails.

Another way of looking at this is to observe that the equation is (kinetic energy of wind per unit mass) x (mass flow of wind moving through control area). Since the mass flow of wind through the sail is zero, the energy flow must necessarily be zero.

When working with sails, you have to use force vectors and force balances.

This equation perfectly works for sails.
So for a sail cart that drives directly down wind this is the equation 0.5 * (air density) * (area) * (wind speed - vehicle speed)^3
And you can have a much more efficient wind turbine this way just have a sail on a cart and connect the electric generator to the wheels of the cart.
This can get you 2x the efficiency of typical propeller type wind turbine.
The problem is that you need a flexible cable connected to the cart and then every minute or so you need to bring the vehicle back to the start of the road.
It will not make much sense to do that even if for that part of the time efficiency is much higher.
Sail is the most efficient device to convert wind power in to kinetic energy.

Online IanB

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Sorry but you are just ignoring the fact that air is a compressible fluid.
Are you saying this graph below is completely wrong ?
https://en.wikipedia.org/wiki/Axial_fan_design


Yes. The given equations are no doubt fine, but the picture is wrong or misleading. That can happen with Wikipedia, since the articles do not get a rigorous peer review.

What are you going to trust more: an actual experiment, on video, where you can see the real world outcome, or a picture that someone posted on the internet?
 

Online IanB

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This equation perfectly works for sails.

So the equation is (kinetic energy of wind per unit mass) x (mass flow of wind moving through the sail).

What is the mass flow of wind moving through the sail?

Why do you say it works perfectly in this case?
 

Offline fourfathom

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If you install the wind turbine on a vehicle and then drive direct down wind then equation of the power output for that wind turbine will again contain (wind speed - vehicle speed) since the wind speed relative to that moving wind turbine will be lower the faster the vehicle moves.

This is exactly correct, at least up to (vehicle speed = wind speed), for a turbine/generator bolted onto a moving vehicle.  However, that is not what the rest of us are discussing; the Blackbird and related DDWFTTW vehicles.
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Offline electrodacus

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P_sail =  0.5 * air density * area * (wind speed - vehicle speed)^2* vehicle speed

With the active driven prop there can be even more power available, as the air may slow down even more. This is especially true for a speed close to or higher than the speed of the wind. This is because this way the wind can be slowed down to a speed below the vehicle speed.


Say wind speed is 10m/s and vehicle speed is half that 5m/s directly down wind
What sail will see is a 10 - 5 = 5m/s wind relative to it and vehicle thus correct equation for a 1m^2 sail will be
 0.5 * 1.2 * 1 * (10-5)^3 = 75W

Using your equation 0.5 * 1.2 * 1 * ((10-5)^2 * 5) = 75W

Seems to provide the same value for half wind speed but
Vehicle speed of 2m/s

0.5 * 1.2 * 1 * (10-2)^3 = 307.2W

0.5 * 1.2 * 1 * ((10-2)^2 * 2) = 76.8W  no longer correct


Vehicle speed or 8m/s

0.5 * 1.2 * 1 * (10-8)^3 = 4.8W

0.5 * 1.2 * 1 * ((10-8)^2 * 8.0) = 19.2W again incorrect.


Still even your incorrect formula will predict no wind power available when vehicle speed equal wind speed and above.

 


Offline electrodacus

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If you install the wind turbine on a vehicle and then drive direct down wind then equation of the power output for that wind turbine will again contain (wind speed - vehicle speed) since the wind speed relative to that moving wind turbine will be lower the faster the vehicle moves.

This is exactly correct, at least up to (vehicle speed = wind speed), for a turbine/generator bolted onto a moving vehicle.  However, that is not what the rest of us are discussing; the Blackbird and related DDWFTTW vehicles.

Wind power available for any direct down wind vehicle no matter how it is designed is always 0.5 * air density * area * (wind speed - vehicle speed)^3 and yes this includes DDWFTTW vehicles.
This if you understand is sufficient proof that without energy storage or an external energy source no vehicle can exceed wind speed.

Offline Kleinstein

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P_sail =  0.5 * air density * area * (wind speed - vehicle speed)^2* vehicle speed

With the active driven prop there can be even more power available, as the air may slow down even more. This is especially true for a speed close to or higher than the speed of the wind. This is because this way the wind can be slowed down to a speed below the vehicle speed.

Say wind speed is 10m/s and vehicle speed is half that 5m/s directly down wind
What sail will see is a 10 - 5 = 5m/s wind relative to it and vehicle thus correct equation for a 1m^2 sail will be
 0.5 * 1.2 * 1 * (10-5)^3 = 75W

Using your equation 0.5 * 1.2 * 1 * ((10-5)^2 * 5) = 75W

Seems to provide the same value for half wind speed but
Vehicle speed of 2m/s

0.5 * 1.2 * 1 * (10-2)^3 = 307.2W

0.5 * 1.2 * 1 * ((10-2)^2 * 2) = 76.8W  no longer correct

In the comparison you forgot one important point - the point with vehicle speed zero.
In this case the  (w-v)³ type gives 0.5 * 1.2 * 1 * (10)^3 = 600W.
For sail driven vehicle at zero speed there is only force and not power transfered to the vehicle. So this equation is obviously wrong at that point.  :horse:
It gets even worse when you drive agains the wind  :-DD

 

Online IanB

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As I pointed out much earlier, starting with reply #20 to this thread, you can pretty much explain it without much more than the principles of Archimedes.

Indeed, and for reference, here is an analysis of the cart/wheel/belt system.

Refer to the diagram attached below.



Assume that it is an ideal system, no friction losses, and both wheels are able to spin freely on their bearings. No slip occurs between the wheels and the belts.

There are five variables: \$v_W\$, \$v_R\$, \$v_M\$, \$v_G\$ and \$v\$

\$v_M\$ and \$v_G\$ are the rim speeds of the wheels, in m/s.

\$v_W\$ and \$v_R\$ are given, leaving three degrees of freedom.

Two equations can be written relating the belt velocities, the rim speeds of the wheels, and the cart speed:
$$v=v_M-v_W$$ $$v=v_G-v_R$$
These two equations take up two degrees of freedom, leaving one degree of freedom remaining.

With no additional constraints this system has an infinite number of solutions. The cart velocity, \$v\$ can have any value we choose.

This makes sense, intuitively, since the wheels are free running, and without friction losses any initial cart velocity will remain the same forever.

To remove the remaining degree of freedom in the system, the wheels can be linked with a connecting gear. If we let the gear ratio be \$\alpha\$, then we can write a linking equation:
$$v_G=\alpha\,v_M$$
For each revolution of wheel M, wheel G makes \$\alpha\$ revolutions.

(Note that the gearing could be mechanical, or it could be electrical. Mechanical is simpler.)

By making a suitable choice of \$\alpha\$, we can arrange for a desired cart speed, \$v\$.

For instance, suppose we wish the cart to move at 1 m/s to the right.

We set \$v\$ = 1 m/s, which gives \$v_M=v+v_W\$ = 5 m/s, and \$v_G=v+v_R\$ = 11 m/s.

From this, we have: \$\alpha = v_G / v_M\$ = 11 / 5 = 2.2

Therefore, with a gear ratio of 2.2, the cart will move to right at a speed of 1 m/s.
« Last Edit: December 22, 2021, 01:18:22 am by IanB »
 
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Offline electrodacus

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In the comparison you forgot one important point - the point with vehicle speed zero.
In this case the  (w-v)³ type gives 0.5 * 1.2 * 1 * (10)^3 = 600W.
For sail driven vehicle at zero speed there is only force and not power transfered to the vehicle. So this equation is obviously wrong at that point.  :horse:
It gets even worse when you drive agains the wind  :-DD

Yes that calculation for vehicle speed zero is correct. There is a potential 600W available but with no movement there is no power available to vehicle. The vehicle is stuck to earth so those 600W push the earth trough the vehicle that is attached/anchored with a brake to earth.
If you remove the brakes those 600W will be available to vehicle instead of earth.
 

Offline bdunham7

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If you remove the brakes those 600W will be available to vehicle instead of earth.

And if you release the brakes, but not all the way, so that the vehicle creeps, isn't most of your 600W still 'available to the earth'?  The vehicle is still pushing on the earth...
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