Re: Direct Down Wind Faster Than The Wind
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Reply #1581 –

Heinz poses the question: If you have a treadmill running at v_{0} = 4.0 m/s, and a cart initially keeping up with the treadmill (presumably because it has been held in place to get the wheels and prop up to speed, then released), with a braking force of F_{B} = 0.25 N, and with a transmission efficiency (from the wheels to the propeller) of τ = 0.90, what is the thrust from the propeller?

The answer depends on a bunch of other factors, of course. But having given us a figure for F_{B}, that helps narrow things down.

Let I be the moment of inertia of all the rotating parts of the cart, considered collectively, as measured at the wheels. Let r be the radius of the wheels. Let m be the mass of the cart. We don't need these things individually, but we are interested in the ratio χ = I/(m r^{2}). This ratio can be just about any positive number, although I think a case with χ > 1 would be rather weird. It will be low if the cart's mass is concentrated in the body, and somewhat higher if the mass is concentrated in the rims of the wheels and/or in the propeller.

Other things we need: Let ρ denote the density of the air, let A denote the area of the disc swept by the propeller, and let ζ denote the propeller's "figure of merit" for the case of static thrust.

The to get the thrust force F_{T}, we need to solve

F_{T}^{3/2} / (ζ v_{0} (2 ρ A)^{1/2}) + χ F_{T} = (χ + τ) F_{B}.

(Where does that come from? I'll happily go through it if someone asks, although Heinz probably won't be able to follow the math.)

So if we take (say) χ = 0.50, ζ = 0.80, ρ = 1.225 kg/m^{3}, and A = 0.126 m^{2} (corresponding to a propeller diameter of 0.40 m), then we get F_{T} = 0.41 N. Significantly greater than the braking force.

An energy accounting (in the rest frame of the room):

Rate of rotational KE gained by the cart from the treadmill: F_{B} v_{0} = 1.00 W.

Rate of rotational KE lost by the cart due to transmission inefficiency: (1 - τ) F_{B} v_{0} = 0.10 W.

Rate of rotational KE lost by the cart due to resistive torque at the prop: F_{T}^{3/2} / (ζ (2 ρ A)^{1/2}) = 0.59 W.

Net rate of rotational KE gained by the cart: 1.00 W - 0.10 W - 0.59 W = 0.31 W.

Rate of linear KE gained by the cart: F_{T} v = 0, because v = 0.

Rate of KE added to air (to create a breeze behind the prop): F_{T}^{3/2} / (2 ρ A)^{1/2} = 0.47 W.

So altogether, the 1.00 W supplied by the treadmill breaks down into:

0.31 W gained by the cart (net gain)

0.10 W in transmission losses (from wheels to prop)

0.47 W gained by the air (as a breeze)

0.12 W in other losses (at the prop)