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Mess with your minds: A wind powered craft going faster than a tail wind speed.
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fourfathom:

--- Quote from: IanB on December 23, 2021, 05:46:35 pm ---No. He said the power required to overcome wind resistance is the same when cycling at 60 km/h as it is when cycling at 10 km/h against a 50 km/h headwind. That is wrong. It takes much less power to go at 10 km/h.

(This has nothing to do with rolling resistance. This is only about wind effects.)
--- End quote ---

Why, when there is no rolling resistance isn't the apparent wind the determining factor?  Apparent wind speed is (the vector sum of) wind speed + ground speed.
Labrat101:
Hello guys . He is playing with you and you are falling for it .
He is side tracking you .
We are talking about can blackbird can go faster than tail wind .
He has mentioned SO Far :: Inductors . because they store energy  :-//  .Yes, Not that energy that blows a ship across the water.
Turbine on a car . Yes .. Wind turbine blade for a small blade is about 25 mtr dia .. does the blade stand still & the car rotate?
Now push bikes .
     This is even Funnier than   Monty Python's Flying Circus
Now he has realized that he has  :wtf: ed up . And drawing you All into his warped web of confusion .  :-//

And Now for something completely Different .. Reality

 key parameters in propeller design, the main ones being the power to drive the propeller,
 and the thrust that the propeller delivers.
    • The angle between blade chord and propeller plane is the geometric pitch (blade angle β
    • )
    • Angle of attack of the blade element α=β−Φ
    • The effective pitch angle is Φ=arctan(V/ω∗r)
. This is often given relative to rotational speed n = rounds per second of the propeller, and the propeller diameter R:
Φ=arctanVn⋅D∗1π⋅r/R
The advance ratio J of the propeller is
J=Vn⋅D
Induced velocity should be constant over the blade, implying that β
decreases linearly with increasing r: the propeller blade twist. Because the twist changes linearly, one point on the blade can be taken as the representative blade β
, and this is usually taken at either 70% or 75% of the radial distance.
It can be shown that for a given blade geometry, the power and thrust coefficients CP
and CT of the propeller are determined only by J and β0.75. If tip speed is below speed of sound and the blades are not stalled, Mach and Reynolds number effects are negligible.
CP=Pρ⋅n3⋅D5
CT=Tρ⋅n2⋅D4
    now look up CP
as function of J in for instance NACA reports from the NACA 

Regarding the profiles: propeller design is pretty specialist but old fashioned NACA profiles are still being used in helicopter blades for both main rotor and tail rotor. (Example) For a symmetrical profile NACA 0012 would be a representative choice, again available from the NASA .
 A reference book on this is Theory Of Wing Sections by Abbott & Von Doenhoff.

 Sorry if I am a bit rusty the last design stuff was over 45 years ago .
bdunham7:

--- Quote from: fourfathom on December 23, 2021, 06:58:11 pm ---Why, when there is no rolling resistance isn't the apparent wind the determining factor?  Apparent wind speed is (the vector sum of) wind speed + ground speed.

--- End quote ---

Yes, that determines the wind force.  The power used to move against that force is the product of the force and the speed with which you move against it, not the speed (apparent or otherwise)  of the medium. 
IanB:

--- Quote from: fourfathom on December 23, 2021, 06:58:11 pm ---Why, when there is no rolling resistance isn't the apparent wind the determining factor?  Apparent wind speed is (the vector sum of) wind speed + ground speed.

--- End quote ---

Yes, as bdunham7 said, the power required is the product of the vehicle speed and the resistance to motion. So you have:

  (power) = (speed) x (resistance)

In SI units, that is:

  [W] = [m/s] x [N]

Since 1 W = 1 Nm/s, it becomes this:

  [Nm/s] = [m/s] x [N]

So, if you are moving at 60 km/h in still air, the speed is 60 km/h, and the drag force corresponds to a wind velocity of 60 km/h.

On the other hand, if you are moving at 10 km/h against a 50 km/h headwind, the speed is 10 km/h, and the drag force corresponds to an apparent wind velocity of 10 + 50 = 60 km/h (same).

The drag force is the same in both cases, but in the second case the speed is 1/6 of the first case, so the power required is also 1/6.
Domagoj T:
Full disclosure, I skipped quite a few pages of this discussion, for obvious reasons.
In any case, how I interpret the downwind faster than wind is just as a simple lever. The wind has arbitrarily large amount of energy (size of the propeller is not specified), it's just a matter of devising a mechanism to push against the huge energy source that is the wind.
A simple demonstration is a yo-yo (yes, the spiny toy on a string).
 

Put the yo-yo on the table with the string coming from the underside of the bobbin part and pull. The yo-yo will catch up with you.
If you don't have a yo-yo at hand, substitute with a spool of solder wire.
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