Newton's third law problem.

[electrodacus]

We covered this in another thread, and it appears that a simple square sail is not very efficient at all. A sail self-evidently has zero efficiency when stationary, and zero efficiency when moving at the same speed as the wind. It seems the maximum efficiency of a simple square sail occurs somewhere in between, and is not much more than 15% at best. See this post for an explanation:

https://www.eevblog.com/forum/chat/mess-with-your-minds-a-wind-powered-craft-going-faster-than-a-tail-wind-speed/msg3888830/#msg3888830

Yacht sails are clearly much more efficient, but that is because they use angles and aerodynamics to do a better job.

What you call a stationary sail is not stationary it moves at the same speed as the earth move as it is anchored to earth.
When you release the brakes what is the acceleration power that the vehicle experiences assuming no rolling resistance ?

Measure the distance the vehicle moved in one second after the brakes were released and calculate based on that the amount of kinetic energy vehicle gained then work backwards to calculate the amount of average power the wind applied in order to gain that amount of kinetic energy.

Also calculate the amount of brake power needed (energy to be wasted as heat) to maintain vehicle speed at 1m/s while wind speed is 11m/s
See what you think about the efficiency of a sail after that.

[Kleinstein]

That is not good enough.
If you want to say that equation is wrong you need to provide the correct one so we can compare the prediction each of them makes.

No, to proof that the euqation is wrong it is enough to show an example where the equation is wrong. That example is given (both with the math and as real world example in the videos in several examples).

The true burden is to you to show that the equation is correct to start with. Just a link to a dubious web site that seems to do the same mistake or a remotely similar equation for the stationary vehicle is not a proof. Given that it is pooven wrong there no hope to find a proof.

[electrodacus]

No, to proof that the euqation is wrong it is enough to show an example where the equation is wrong. That example is given (both with the math and as real world example in the videos in several examples).

The true burden is to you to show that the equation is correct to start with. Just a link to a dubious web site that seems to do the same mistake or a remotely similar equation for the stationary vehicle is not a proof. Given that it is pooven wrong there no hope to find a proof.

You (all) are talking nonsense.

It is time for you to make a real world prediction and we will find a way to setup the experiment.

So say a small maybe toy like vehicle with a simple shape say a cube on wheels.

Say a cube with 20x20x20cm should be fine but we can change the dimensions if it makes it easier to do the experiment.

Drag coefficient for a cube is 1.05
 
Say cube will need to move upwind at a slow speed of 1m/s and wind speed is 10m/s

Based on all this I say the electric motor driving the cube requires this amount of power to overcome drag

Pd = 0.5 * 1.2 * (0.2*0.2*1.05) * (10+1)³ = 33.5W

Time for your prediction and please show how it is calculated and we will find a way to do the experiment so that we all agree it is done correctly.

[Kleinstein]

The (..)³ equation is wrong by having the wind speed inside for all 3 factors. One factor is only the vehicle speed. So the toy would only need 1/11 the power.

The difference gets even larger, when we assume a smaller speed, like 1 mm/s and than 11 m/s of wind. In this relatively extreme slow speed example it gets obvious that your prediction is wrong:
The mechanical power is speed times force and the force thus power divided by speed. So the froce would be 33.5 W /0.001 m/s = 33.5 kN. That would be way too much force to be produced by the wheels. So your equation would predict that a vehicle can not go slow against a heat wind. With the cazy prediction of needing more force the slower it wants to move and thus no way to start from a standstill as this naturally innitially is very slow.

[electrodacus]

The (..)³ equation is wrong by having the wind speed inside for all 3 factors. One factor is only the vehicle speed. So the toy would only need 1/11 the power.

The difference gets even larger, when we assume a smaller speed, like 1 mm/s and than 11 m/s of wind. In this relatively extreme slow speed example it gets obvious that your prediction is wrong:
The mechanical power is speed times force and the force thus power divided by speed. So the froce would be 33.5 W /0.001 m/s = 33.5 kN. That would be way too much force to be produced by the wheels. So your equation would predict that a vehicle can not go slow against a heat wind. With the cazy prediction of needing more force the slower it wants to move and thus no way to start from a standstill as this naturally innitially is very slow.

You are right about one thing and that is 1m/s is still way to fast for such a small vehicle especially if we will test this in a small DIY wind tunnel.
Will still use 10m/s for wind as I think that should be possible even in DIY wind tunnel and use 0.02m/s (2cm per second) this way a 30cm run will take about 15 seconds enough to see the power measurement.

With this new data we have

Pd = 0.5 * 1.2 * (0.2*0.2*1.05) * (10+0.02)³ = 25.3W

I is time for your prediction as it makes no sense to discuss since you have a wrong idea of how air particles and the object/vehicle interact.

[electrodacus]

The mechanical power is speed times force and the force thus power divided by speed. So the froce would be 33.5 W /0.001 m/s = 33.5 kN. That would be way too much force to be produced by the wheels. So your equation would predict that a vehicle can not go slow against a heat wind. With the cazy prediction of needing more force the slower it wants to move and thus no way to start from a standstill as this naturally innitially is very slow.

I see no prediction from you. Do you need more time ?

In the meantime I can show how that what you think as the correct equation violates the energy conservation law.
Say you had a 100% efficient wind generator (obviously a wind turbine will be only around 40% efficient but the discussion is ideal case).
For easy calculation say the wind generator swept area is 1m² and there is nothing more. It is on wheels but we consider those super narrow so they do not add to wind drag. Also wind speed is 10m/s
A stationary vehicle with this characteristics can generate 0.5 * 1.2 * 1 * (10)³ = 600W again ideal case 100% efficiency.
What happens if vehicle moves very slowly upwind say 1m/s upwind ?
The wind turbine will see a 11m/s wind speed so it can now generate 798W that is 198W more than when stationary.

According to your equation only 0.5 * 1.2 * 1 * (11)²  * 1m/s = 72.6W are needed.

So inputting 72.6W and getting out 198W will violate energy conservation. Keep in mind we already discuss about an ideal 100% efficient system.

While the equation I consider to be the correct one will predict that you need 798W for the motor to drive at 1m/s against a 10m/s headwind (just drag power) so there is no energy conservation conflict.

Also if you want to travel down wind powered by the wind at 1m/s you need to apply a brake at the wheel with the power equal with 0.5 * 1.2 * 1 (10-1)³ = 437W so you can do what you want with those 437W say light some incandescent lamps and be able to maintain a constant 1m/s downwind powered by the wind.

So a stationary vehicle in 10m/s wind with 1m² capture area has 600W available to accelerate downwind and if it wants to move upwind it needs to add extra on top of that 600W to start accelerating the mass in that direction. 

[AVGresponding]

I do enjoy a good circular argument   

[Kleinstein]

The mechanical power is speed times force and the force thus power divided by speed. So the froce would be 33.5 W /0.001 m/s = 33.5 kN. That would be way too much force to be produced by the wheels. So your equation would predict that a vehicle can not go slow against a heat wind. With the cazy prediction of needing more force the slower it wants to move and thus no way to start from a standstill as this naturally innitially is very slow.

I see no prediction from you. Do you need more time ?

In the meantime I can show how that what you think as the correct equation violates the energy conservation law.
Say you had a 100% efficient wind generator (obviously a wind turbine will be only around 40% efficient but the discussion is ideal case).
For easy calculation say the wind generator swept area is 1m² and there is nothing more. It is on wheels but we consider those super narrow so they do not add to wind drag. Also wind speed is 10m/s
A stationary vehicle with this characteristics can generate 0.5 * 1.2 * 1 * (10)³ = 600W again ideal case 100% efficiency.
What happens if vehicle moves very slowly upwind say 1m/s upwind ?
The wind turbine will see a 11m/s wind speed so it can now generate 798W that is 198W more than when stationary.

According to your equation only 0.5 * 1.2 * 1 * (11)²  * 1m/s = 72.6W are needed.

So inputting 72.6W and getting out 198W will violate energy conservation. Keep in mind we already discuss about an ideal 100% efficient system.

This is a good, but tricky point. By moving against the wind the volume of wind that is used is increased.  So there is extra air to use. The exact calculation is tricky as the wind interacts with the slowed down air.  The limited efficiency of a wind turbine is not just for technical reasons, the Betz limit is there for prociple reasons and a 100% efficient wind turbine thus already in in conflict with theroy and by itself cause contradictions.

While the equation I consider to be the correct one will predict that you need 798W for the motor to drive at 1m/s against a 10m/s headwind (just drag power) so there is no energy conservation conflict.

Also if you want to travel down wind powered by the wind at 1m/s you need to apply a brake at the wheel with the power equal with 0.5 * 1.2 * 1 (10-1)³ = 437W so you can do what you want with those 437W say light some incandescent lamps and be able to maintain a constant 1m/s downwind powered by the wind.

Your prediction is also causing a problem with energy conservation. You predict the 437 W from the vehicle moving down wind, but the wind turbine on the vehicle still sees 9 m/s of wind and could use that wind to creat additional power. Chances are it could gain more than 170 W from this - likely even with an efficiency within the Betz limit. When moving with the wind, there is less (at least not more) of the wind actually used - so that point does not work here either.  So it is your prediction that violates the conservation of energy.

[electrodacus]

This is a good, but tricky point. By moving against the wind the volume of wind that is used is increased.  So there is extra air to use. The exact calculation is tricky as the wind interacts with the slowed down air.  The limited efficiency of a wind turbine is not just for technical reasons, the Betz limit is there for prociple reasons and a 100% efficient wind turbine thus already in in conflict with theroy and by itself cause contradictions.

Sorry but Benz limit only refers to wind turbines. And even if we used the 59% Benz limit the problem is not changed. You add a small amount of energy to an a system and you get more in return (significantly more but even just 1% extra will violate the energy conservation law).
The example is there just to prove that equation you and others proposed is not correct.
 

Your prediction is also causing a problem with energy conservation. You predict the 437 W from the vehicle moving down wind, but the wind turbine on the vehicle still sees 9 m/s of wind and could use that wind to create additional power. Chances are it could gain more than 170 W from this - likely even with an efficiency within the Betz limit. When moving with the wind, there is less (at least not more) of the wind actually used - so that point does not work here either.  So it is your prediction that violates the conservation of energy.

There is no extra power. Those 437W are all that it is available and you are already extracting everything that is it available to be extracted with the generator at the wheel.
At 9m/s which is the wind speed relative to vehicle (all that is important) you have 0.5 * 1.2 * 1 * (10-1)³ = 437.4W  There is nothing more.
A wind turbine as you already mentioned can extra way less than this if it was to be used.
So whatever a wind turbine where to extract it will be subtracted from those 437.4W.

You can say that a sail vehicle on wheels is much more efficient than a propeller but the vehicle is moving and so you will need a flexible cable to connect the vehicle and at some point you need to stop and come back and you will wasting time not generating and use energy to return to a start point.
So overall efficiency if it was to be used as a wind turbine will not be that great and way more expensive. Plus it will be close to the ground where wind speed is much lower.
That is the reason I say a sail is as close to 100% efficiency as possible in using wind power to accelerate a vehicle.
All that calculated drag power is available to accelerate the vehicle.
So staying with this example when stationary the vehicle has available 600W of wind power so as soon as you remove the brakes say for 1ms the kinetic energy of the vehicle will increase by  600W * 0.001s = 0.6Ws (0.6J) so the speed of the vehicle after 1ms will depend on vehicle weight.
I know I use unusual methods of calculating all this compared to how this things are teached in schools but I think talking only about forces in schools while correct is the wrong approach in properly understanding what happens and human intuition about this things is wrong thus the wrong equation circulating around.
I wish people still did experiments in the classrooms then it will be easy to spot this incorrect equations.

I also think this has way more implications than in this highschool level physics as you can not work in any field of physics or engineering and not be affected by wrongly understanding this subject.
I really want for this misinformation to be corrected if possible. But no idea how to do this. The only thing that will convince most people will be the real world experiment.
The simplest I can think of is a vehicle driving slowly in a head wind and measuring the motor consumption. 
 

[Kleinstein]

Your prediction is also causing a problem with energy conservation. You predict the 437 W from the vehicle moving down wind, but the wind turbine on the vehicle still sees 9 m/s of wind and could use that wind to create additional power. Chances are it could gain more than 170 W from this - likely even with an efficiency within the Betz limit. When moving with the wind, there is less (at least not more) of the wind actually used - so that point does not work here either.  So it is your prediction that violates the conservation of energy.

There is no extra power. Those 437W are all that it is available and you are already extracting everything that is it available to be extracted with the generator at the wheel.
At 9m/s which is the wind speed relative to vehicle (all that is important) you have 0.5 * 1.2 * 1 * (10-1)³ = 437.4W  There is nothing more.
A wind turbine as you already mentioned can extra way less than this if it was to be used.
So whatever a wind turbine where to extract it will be subtracted from those 437.4W.

The moving wind turbine would still see the 9 m/s wind can it can thus still extract power from that wind.  It gets even more obvious when you consider moving the turbine only at snails pace. That would chance essentially nothing with the wind. So there would still be nearly the full wind and with correct formular also only very little power on the wheels. Your formular has just way too much power for the wheels and than the problem that there is overall too much power to be gained.

Too much power from the wheels comes with another problem: the force gets too large: 437 W with a speed of 1 m/s would mean a force of 437 N , which is too high by about a factor of 10. The errir gets even more rediculous if slower. Essentially constant power even at slow speed just does not work. 

[electrodacus]

Your prediction is also causing a problem with energy conservation. You predict the 437 W from the vehicle moving down wind, but the wind turbine on the vehicle still sees 9 m/s of wind and could use that wind to create additional power. Chances are it could gain more than 170 W from this - likely even with an efficiency within the Betz limit. When moving with the wind, there is less (at least not more) of the wind actually used - so that point does not work here either.  So it is your prediction that violates the conservation of energy.

There is no extra power. Those 437W are all that it is available and you are already extracting everything that is it available to be extracted with the generator at the wheel.
At 9m/s which is the wind speed relative to vehicle (all that is important) you have 0.5 * 1.2 * 1 * (10-1)³ = 437.4W  There is nothing more.
A wind turbine as you already mentioned can extra way less than this if it was to be used.
So whatever a wind turbine where to extract it will be subtracted from those 437.4W.

The moving wind turbine would still see the 9 m/s wind can it can thus still extract power from that wind.  It gets even more obvious when you consider moving the turbine only at snails pace. That would chance essentially nothing with the wind. So there would still be nearly the full wind and with correct formular also only very little power on the wheels. Your formular has just way too much power for the wheels and than the problem that there is overall too much power to be gained.

Too much power from the wheels comes with another problem: the force gets too large: 437 W with a speed of 1 m/s would mean a force of 437 N , which is too high by about a factor of 10. The errir gets even more rediculous if slower. Essentially constant power even at slow speed just does not work. 

You will need to be more precise in describing what you are thinking off.

As it is in this example the vehicle is a cube with 1m sides so there is a 1m² surface interacting with air and for simplicity I considered the coefficient of drag to be 1
Now when you say a wind turbine is added to this is that on top of this 1m² or are you replacing the cube with a wind turbine that has the same 1m² swept area ?

You need to understand that interaction between air and vehicle is based on elastic collisions between air particles and vehicle boddy.
While vehicle speed is 1m/s the 1m²  side of the cube is hit by a cubic meter of air about 1.2kg 9 times per second and that kinetic energy is transferred to vehicle. Unless you get rid of that by say taking that energy at the wheel the vehicle will continue to accelerate so not be able to maintain that 1m/s

So you have this as air kinetic energy of one meter cube of air

KE = 0.5 * 1.2kg * 9m/s² = 48.6Ws * 9 of this hitting the vehicle each second  = 437.4Ws so same result.

[electrodacus]

We got stuck at the equation for wind power available to vehicle. But I can prove the same thing without using that equation at all and just using equations you all agree with.


We all agree with the Drag force equation:
for upwind vehicle it will look like:
Fd = 0.5 * air density * coefficient for drag * area * (wind speed + vehicle speed)²

Another equation needed for my proof is equation for work done
W= Fd * distance

As we will use an example that ignores the rolling resistance for simplicity the work done will all end up as vehicle kinetic energy.

Vehicle 10000kg with
1m² frontal area
coefficient of drag of 1
wind speed 10m/s
air density 1.2kg/m³

Fd= 60N
Work = 60N * 1m = 60 Joules (I prefer to use Ws)
Work = delta KE and since at start KE was zero at the start of the experiment the current vehicle KE is now 60Ws

So how much energy do you need to move the vehicle back to the starting point?

[Kleinstein]

The drag force is calculated with still 10 m/s. So the calculation assume a very low speed (which is OK for the very high weight).
The Force needed to push the vehicle back would be the same 60 N. One would first need to stop brake instantly - in the phisical sense this needs no energy, but one could even get the energy back.
So in the ideal case one would need the same 60 Joules.

The fun part when you calculate the time is needs to reach a m distance.: this is t² = 2 * 1m * Mass / force  and thus t = 18.26s. For the 60 Joules of energy this gives an average power of a little under 4 W.

[electrodacus]

The drag force is calculated with still 10 m/s. So the calculation assume a very low speed (which is OK for the very high weight).
The Force needed to push the vehicle back would be the same 60 N. One would first need to stop brake instantly - in the phisical sense this needs no energy, but one could even get the energy back.
So in the ideal case one would need the same 60 Joules.

The fun part when you calculate the time is needs to reach a m distance.: this is t² = 2 * 1m * Mass / force  and thus t = 18.26s. For the 60 Joules of energy this gives an average power of a little under 4 W.

Maybe my example made things even more confusing than they were before.

It takes just a bit over 0.1 seconds to accelerate the vehicle to just over 0.1m/s
The work is done on the air so the air mass has delivered the vehicle 60 Joules that ended up as kinetic energy and so if after you did that you hide the sail (sail area equal zero) the vehicle can move forever at that 0.1m/s with no extra energy needed as there is no rolling resistance in this example.

So it was the air that moved 1 meter not the vehicle. The air mass while moving 1m lost 60 Joules to the vehicle that barely moved in that same period of time but gained those 60 Joules and can not continue to move forever at just over 0.1m/s about 0.11m/s

Again sorry for the confusing example.  Is not the vehicle that moves 1m as vehicle can move forever any distance after it started to move.

[Kleinstein]

The drag force is calculated with still 10 m/s. So the calculation assume a very low speed (which is OK for the very high weight).
The Force needed to push the vehicle back would be the same 60 N. One would first need to stop brake instantly - in the phisical sense this needs no energy, but one could even get the energy back.
So in the ideal case one would need the same 60 Joules.

The fun part when you calculate the time is needs to reach a m distance.: this is t² = 2 * 1m * Mass / force  and thus t = 18.26s. For the 60 Joules of energy this gives an average power of a little under 4 W.

Maybe my example made things even more confusing than they were before.

It takes just a bit over 0.1 seconds to accelerate the vehicle to just over 0.1m/s
The work is done on the air so the air mass has delivered the vehicle 60 Joules that ended up as kinetic energy and so if after you did that you hide the sail (sail area equal zero) the vehicle can move forever at that 0.1m/s with no extra energy needed as there is no rolling resistance in this example.

So it was the air that moved 1 meter not the vehicle. The air mass while moving 1m lost 60 Joules to the vehicle that barely moved in that same period of time but gained those 60 Joules and can not continue to move forever at just over 0.1m/s about 0.11m/s

Again sorry for the confusing example.  Is not the vehicle that moves 1m as vehicle can move forever any distance after it started to move.

No : to accerate the 10 ton vehicle to 0.1 m/s  in 0.1 seconds it needs way more force: the accelration is  dV/dt and F = m *A  = 10000 kg * 0.1m/s / 0.1 s = 10000 N. That is well more than 60 N you calculated for the drag force.

The assumption of 100% efficiency for the wind is wrong.

[electrodacus]

No : to accerate the 10 ton vehicle to 0.1 m/s  in 0.1 seconds it needs way more force: the accelration is  dV/dt and F = m *A  = 10000 kg * 0.1m/s / 0.1 s = 10000 N. That is well more than 60 N you calculated for the drag force.

The assumption of 100% efficiency for the wind is wrong.

Keep in mind this is a theoretical problem so there is no rolling resistance.
The vehicle barely moved in that 0.1s and the 0.11m/s is not much
 
KEvehicle = 0.5 * 10000kg * (0.11m/s)² = 60 Joules

Yes the assumption of 100% efficiency for a theoretical sail is correct. And even in real world that will still be around the same.
What happened in this setup is that about 1m³ of air collides with the 1m² sail and transfers all that kinetic energy to vehicle.

KEair = 0.5 * 1.2kg * (10m/s)² = 60 Joules  In that 0.1s just 1 cubic meter of air can collide with the vehicle body.

Do not ignore the conservation of energy when solving any problem. Here the mistake is to not consider air made out of particles that have elastic collisions with vehicle body.  You do that for calculating drag force but then you want to ignore that when calculating power needed to overcome drag.

Even to travel 1m the vehicle will require less than 10 seconds assuming you remove the sail after the first 100ms and there will be no extra energy loss or gain so still the same 60 Joules of vehicle stored kinetic energy.

[Kleinstein]

Do not ignore the conservation of energy when solving any problem. Here the mistake is to not consider air made out of particles that have elastic collisions with vehicle body.  You do that for calculating drag force but then you want to ignore that when calculating power needed to overcome drag.

The conservation of energy is OK, but than you have to also include the friction in the air, converting kinetic energy of the wind to heat. In this case it just not very practical to use conservation of energy as the starting point. It is much easier to use the drag force and the basic newton's laws.

To drive against the wind, the vehicle has to overcome the drag force. So it does not matter if the force is from a large sail in low wind or small sail in high wind. In a though experiment the wheel / motor part does not even know if the force is from wind or electromagnetic or friction to the ground.

The ideallized sail may theoretical approach 100% efficiency - but only when moving at nearly the speed of the wind. Going at 10% the speed of the wind it is more like 10% efficient.
With wind resistance there is not 100% transfer of kinetic energy. Generally much of the ernergy is converted to heat. The collisions are not 100% elastic - that is only an approximation sometimes used in some simplified calculations. Fast flying planes do get hot from the air resistance.

[electrodacus]

The conservation of energy is OK, but than you have to also include the friction in the air, converting kinetic energy of the wind to heat. In this case it just not very practical to use conservation of energy as the starting point. It is much easier to use the drag force and the basic newton's laws.

To drive against the wind, the vehicle has to overcome the drag force. So it does not matter if the force is from a large sail in low wind or small sail in high wind. In a though experiment the wheel / motor part does not even know if the force is from wind or electromagnetic or friction to the ground.

The ideallized sail may theoretical approach 100% efficiency - but only when moving at nearly the speed of the wind. Going at 10% the speed of the wind it is more like 10% efficient.
With wind resistance there is not 100% transfer of kinetic energy. Generally much of the ernergy is converted to heat. The collisions are not 100% elastic - that is only an approximation sometimes used in some simplified calculations. Fast flying planes do get hot from the air resistance.

Air is not an ideal gas but close enough in relation to how much of his kinetic energy is transferred to the vehicle.
There is really no difference between vehicle colliding with stationary air particles (no wind) and moving air particles colliding with a stationary vehicle.
If you change the reference frame the result will need to be exactly the same else mistakes were made.

The sail size vs wind speed is relevant.  Say you want the same 60N at just 5m/s instead of 10m/s wind.
Then your sail size will need to be 4x larger so 4m² instead of just 1m² needed with 10m/s wind.
But then the power available is 2x lower than before even if force is the same.

Fd = 0.5 * 1.2 * 1 * (10)² = 60N
Fd = 0.5 * 1.2 * 4 * (5)² =60N

Pd = Fd * 10 = 600W
Pd = Fd * 5 = 300W

If you want the same power at half the wind speed you need 8x the sail area.  The wind speed is relevant not just the force.
You will not make this assumptions if you considered the vehicle powered by ideal collisions with large balls that you can see.
It is the same thing just that air is made out of super small (low weight) particles even different size that collide almost perfectly elastic with the vehicle body.

[Kleinstein]

The collisions for the molecules with a solid are not really elastic. One sees this as heat can be exchanges from a gas and solid. Even of only small fraction of the energy is lost to heat, the argumet with energy conservation breaks down and should be avoided. At best it is a weak argument - so one of the first points to question.

There is no physical law of conservation of kinetic energy. There is hower conservation of momentum and thus a balance of force.  The assumption of conservation of kinetic energy is a rather weak one - it can work for explite eleastic collisions or frictionless ideallized mechnical  systems. However it does not work for systems with friction - air resistance is a type of friction.

The picture of moving particles is not really practical to understand air resistance.
It is not a transfer of kinetic energy to kinetic energy (most of the energy in the gas is thermal energy and the wind is more like a minor shift in the distribution).
The interaction is more like a transfer of momentum and force. Much of the air particels never actuall hit the obstacle but hit other air that diverts it to the sides and can create turbulance that than dissipates the energy away from the obstacle.

[electrodacus]

The collisions for the molecules with a solid are not really elastic. One sees this as heat can be exchanges from a gas and solid. Even of only small fraction of the energy is lost to heat, the argumet with energy conservation breaks down and should be avoided. At best it is a weak argument - so one of the first points to question.

There is no physical law of conservation of kinetic energy. There is hower conservation of momentum and thus a balance of force.  The assumption of conservation of kinetic energy is a rather weak one - it can work for explite eleastic collisions or frictionless ideallized mechnical  systems. However it does not work for systems with friction - air resistance is a type of friction.

The picture of moving particles is not really practical to understand air resistance.
It is not a transfer of kinetic energy to kinetic energy (most of the energy in the gas is thermal energy and the wind is more like a minor shift in the distribution).
The interaction is more like a transfer of momentum and force. Much of the air particels never actuall hit the obstacle but hit other air that diverts it to the sides and can create turbulance that than dissipates the energy away from the obstacle.

Understanding what air is and how it interacts with the vehicle is actually key to understanding this problems.

With no wind you seem to agree with the fact that all that is there are elastic collisions but then when air moves you try to find some alternative explanation and that is not the case as all that changed is the frame of reference.

If you take the example of the 1m² frontal area and coefficient of drag of 1 traveling through air at 10m/s

Then you will say drag force is 60N and power needed to overcome this drag is 600W
If the same vehicle travels at half this speed you again will say drag force is 15N and power deeded to overcome drag is 75W

So all that changed is the vehicle speed trough stationary air and reducing the speed to half reduced the drag force 4x but reduced the power by 8x

If you look at what happens and the fact that each second vehicle collides with 10 cubic meters of air when driving at 10m/s and collides with 5 cubic meters when driving at 5m/s
When you calculate the collision energy you get a perfect transfer of energy like ideal elastic collisions took place.

But then if air moves instead of vehicle you want to claim that is no longer the same thing when nothing has actually charged other than the frame of reference.

You are not using different equations for drag force but you insist on two different equations for drag power. Is likely just based on the wrong intuition of what happens.

[Kleinstein]

No: the collisions are not elastic. With the usually 2 atom molecules (N2,O2) in air they are likely quite far away from elestic, with strong coupling between the linear and rotational degrees of freedom.

Even if the collision are nearly elestic (e.g. in helium or argon) it would still not matter. The picture of the air mass hitting the vehicle is still wrong. I don't care about the details of gas - that is only going off topic.

I somewhat agree (have not really checked, but sounds reasonable) with the 60 N of drag force, but there is not such thing as drag power of 600 W and no balance of powers (that is politics).

The power needed to drive agains the wind (or any other resistance) is Force time vehicle speed relative to the ground (where the propulsion transfers the force to). So at 1 m/s this would be 60 W and at
1mm/s this would be 6 mW. Power = force times speed is  directly dirived from energy = force times distance, just by dividing by the time or taking the derivative.
If you can't agree with this, it gets hopeless.

[electrodacus]

No: the collisions are not elastic. With the usually 2 atom molecules (N2,O2) in air they are likely quite far away from elestic, with strong coupling between the linear and rotational degrees of freedom.

Even if the collision are nearly elestic (e.g. in helium or argon) it would still not matter. The picture of the air mass hitting the vehicle is still wrong. I don't care about the details of gas - that is only going off topic.

I somewhat agree (have not really checked, but sounds reasonable) with the 60 N of drag force, but there is not such thing as drag power of 600 W and no balance of powers (that is politics).

The power needed to drive agains the wind (or any other resistance) is Force time vehicle speed relative to the ground (where the propulsion transfers the force to). So at 1 m/s this would be 60 W and at
1mm/s this would be 6 mW. Power = force times speed is  directly dirived from energy = force times distance, just by dividing by the time or taking the derivative.
If you can't agree with this, it gets hopeless.

Take it the other way around vehicle driving in stationary air (no wind).

At 1m/s 0.6W that is exactly what perfectly elastic collision with 1m³ of air will result in as 1 cubic meter of air has 1.2kg
At 1mm/s 60nW again the kinetic energy of that volume of air.
At 10m/s 600W exactly matching the change in kinetic energy of that volume of air after collision with the vehicle.

Reversing this and having air particles move and collide with the vehicle can not change the results as you are just changing the reference frame.
In order to justify using a different equation for when vehicle moves in stationary air compared to moving air and stationary vehicle you will need to prove that the difference in energy is lost in some other ways and that will just be insanity.

As I mentioned before the energy of impact if a bus collides heads on with a stationary car will be the exact same as if vehicle collides with a stationary buss assuming exactly the same speed in both cases.

So in order for what you say to make sense you need to chose one equation for both cases. It needs to be the same both for vehicle moving through stationary air and vehicle in moving air also anything in between.
Also the same equation when driving in to headwind and driving with tail wind.

[electrodacus]

I had the chance to converse with ChatGPT (language based AI model).
While it has many limitations it is still very impressive.

You can see part of my conversation about wind powered vehicles in the attached screenshots