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Newton's third law problem.
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:
--- Quote from: Kleinstein on December 09, 2022, 07:20:13 am ---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.
--- End quote ---
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
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