When vehicle speed equal wind speed there is plenty available wind power.
You just need to power your propeller with your wheels, and get more force out of it than you took from the wheels so that your wheels turn faster.
The air is accelerated by the propeller. So in the Earth reference frame, wind decreased behind the car: the kinetic energy of the wind is now in the car.
That's how it works and if you can't understand you should read about Newton laws and stop thinking in terms of energy until you understand them.
The amount of power available to a wind powered vehicle (any design) is 1/2*mass flow*wind speed^2. If the vehicle seed is twice the wind speed, you get mass flow=density*area*wind speed.
Your computations with the wind turbine are quite funny. What if there is wind not for 1ms, but for 1 week?
What if I replace the wind turbine by a water one? There you go: take a dam (1 million ton of concrete, say), and it produces 1 GW for 10 years. What is the speed of the dam?
How much wind power do you think is available when vehicle speed equal wind speed and what is the equation describing that ?
What you say sounds exactly like "generator supplies the motor and the motor supplies the generator" not only that but there is also plenty of extra energy.
Power and energy is all you need to use the forces and speed (voltage and current) are not relevant when you try to do a energy balance calculation.
Do you agree that a typical propeller is just around 70% efficient for propulsion in air while a wheel is well over 90 to 95% efficient.
So when you are at wind speed what is the difference between using the energy generate at the back wheels in to the front wheel/wheels instead of putting it in a less efficient propeller ?
If you take 1000W from the wheels for 1ms that will be 1Ws when vehicle is at wind speed vehicle kinetic energy will be reduced by this exact amount 1Ws as there is no wind power available.
Then when you take this 1Ws and put in to the 70% efficient propeller the kinetic energy of the vehicle increases by 0.7Ws so vehicle ends up with a deficit of 0.3Ws meaning vehicle has lower speed after you do this.
The amount of power available to a wind powered vehicle (any design) is 1/2*mass flow*wind speed^2. If the vehicle seed is twice the wind speed, you get mass flow=density*area*wind speed.
We are making progress as now you say that wind power available to vehicle is the same as I always say but just with the equation presented a bit different
You wrote:
0.5 * (air density * area * wind speed) * wind speed
2 That is correct and the same with
0.5 * air density * area * wind speed
3All you need to clarify now is that wind speed in this equation is wind speed relative to vehicle and not to the ground.
The wind speed relative to vehicle = (wind speed relative to ground - vehicle speed relative to ground) and now you will have the equation I provided you with many times.
Now when vehicle is at 2x the wind speed you have
0.5 * air density * area * (wind speed - (2 * wind speed))
3 = 0.5 * air density * area * (-wind speed)
3Notice the negative sign in front of wind speed meaning negative power relative to vehicle meaning vehicle will be slowed down / decelerated at that rate and not accelerated.
The vehicle hits the air not the air hits the vehicle when vehicle speed is 2x wind speed.
If you replace air (compressible fluid) with water (incompressible fluid) you can no longer store energy in pressure differential thus you can no longer demonstrate 2x or 3x fluid flow speed for that vehicle.