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:
Pw 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 1m
2We have vehicle at:
0m/sAvailable wind power according to your equation
0.5 * 1.2 * 1* 10
2 * |10-0| =
600WAccording to the equation I claim to be correct
0.5 * 1.2 * 1 * (10-0)
3 =
600W5m/syour equation
0.5 * 1.2 * 1* 10
2 * |10-5| =
300Wthe one I claim to be correct:
0.5 * 1.2 * 1 * (10-5)
3 =
75W9m/syour equation
0.5 * 1.2 * 1* 10
2 * |10-9| =
60Wthe one I claim to be correct:
0.5 * 1.2 * 1 * (10-9)
3 =
0.6W10m/syour equation
0.5 * 1.2 * 1* 10
2 * |10-10| =
0Wthe one I claim to be correct:
0.5 * 1.2 * 1 * (10-10)
3 =
0W15m/syour equation
0.5 * 1.2 * 1* 10
2 * |10-15| =
300Wthe one I claim to be correct:
0.5 * 1.2 * 1 * (10-15)
3 =
-75W20m/syour equation
0.5 * 1.2 * 1* 10
2 * |10-20| =
600Wthe one I claim to be correct:
0.5 * 1.2 * 1 * (10-20)
3 =
-600W30m/syour equation
0.5 * 1.2 * 1* 10
2 * |10-30| =
1200Wthe one I claim to be correct:
0.5 * 1.2 * 1 * (10-30)
3 =
-4800W50m/syour equation
0.5 * 1.2 * 1* 10
2 * |10-50| =
2400Wthe one I claim to be correct:
0.5 * 1.2 * 1 * (10-50)
3 =
-38400WCan 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.