Power to space use ratio

[PierreB]

@blimpyway

I can not read that paper only the abstract, so I cannot confirm or deny what it states

It doesn’t matter.

@blimpyway

Tests on parachutes exhibit coefficients up to 1.7

On a publication I know, and according to a specified Reynolds number. Other publications that I have cited give other values.

@blimpyway

because it is proven wrong

Please read all the links and comments I provide on this issue before making an opinion.

The 4/27 limit for yo-yo mode is logic: the swept area goes downwind, the apparent wind speed being 2/3 real wind speed. The paper proves it. My demonstration ( I put again below) proves it, using the formula of force multiplied by reel-out speed to match the formula of power at Betz limit. This has nothing to do with the drag coefficient. This is related to the loss of apparent wind speed, given that the swept area is moving downwind.

That said when the Cd increases beyond 1, we can reasonably assume that the area swept increases also (see my previous comment), being the parachute (or autogyro) area plus the area of ​​air deflected by the parachute and which increases if the Cd increases.

Thus, the demonstration remains valid even if I use a Cd of 2 (the disk area becoming 200 m² by taking account of additional air deflection), which I do below, with a temporary conclusion.

@PierreB

Let us try to check this with an example, wind speed = 10 m/s, air density = 1.2.
The power of a HAWT is determined by the disk swept by the rotor. Supposing the disk area is 100 m².
Power at Betz limit: 1/2 x 100 x 1.2 x 1000 x 16/27 = 35555 W.
Now let us take a surface of 100 m². Traction (force): 1/2 x 100 x 1.2 x 100 = 6000 N. Let us reel-out this surface in the optimal speed of 1/3 wind speed, considering the wind speed on the surface is now only 2/3 wind speed, leading to the result: 20000 x 4/9 = 8888 W, so exactly 1/4 of 35555 W.
Checked!

Note: 4/9 is (2/3)². It is because the traction force goes with squared wind speed.

Now, with a Cd of 2: the traction force is 12000 N instead of 6000 N. The result is: 40000 x 4/9 = 17776 W, so exactly (to within decimal places) 1/4 of 71104 W, by taking account of the doubled swept area, which becomes 200 m² due to the additional deflection of air around the parachute or autogyro.

In practice this would mean that the higher the Cd is beyond 1 or even 2 (a value much higher than that from tests on a parasail that was not a scale model linked here), the more the units should be spaced out to maintain their aerodynamic efficiency.