In a previouis post I discussed the physics of hovering drones and helicopters. My interest in the physics of propellers was reignited by video discussing the possibility of a wind-powered vehicle to move faster than the wind (in the downwind direction).

The video describes a type a vehicle (Blackbird) and the physics involved.

After searching the web a while I found quite a few places where the propeller equation is derived and discussed, like this one. Unfortunately none of them are simple enough for me. They helped me understand that my analysis was quite wrong. Fortunately my conclusions were not so bad.

Using the momentum change of the air under the action of the blades of size \(R\) rotating at frequency \(f\) and advancing with \(V\), simple estimate of the thrust \(T\) can be obtained as follows:

\[T= K M (V_{displaced}-V_{relative}).\]

Here \(M = \rho_{air} f R^2 p,\) is the mass of the displaced air while \(V_{displaced}=f p,\) \(V_{relative}=V-V_{wind}\) and and \(K\) is some constant. The ‘pitch’ \(p\) is the linear amount of air displaced by the blades rotating once. For better agreement with data
some assume \(K=K\left(\frac{R}{p}\right)\) in order to take into account details of the aerodynamics.

For the case of the hovering drone/helicopter we have the following relations for thrust and power:

\[T=K\left(\frac{R}{p}\right) \rho_{air} f^2 R^2 p^2, \quad P= L\left(\frac{R}{p}\right) \rho_{air} f^3 R^3 p^2\]

For simplicity we can consider that the structure factor \(\frac{R}{p}\) is fixed and defines a class of propellers. Among those propellers developing a certain level of thrust (vertical force) $T$ requires a power

\[P \sim \frac{\sqrt{T^3}}{R}\]

thus implying that the most efficient hovering is done with large propellers driven at low frequency. It should also be a large pitch propeller, given that we have assumed \(\frac{R}{p}\) is constant.

Getting back to the Blackbird driving downwind we have the equation:

\[T=K\left(\frac{R}{p}\right)\rho_{air} f R^2 p (f p - V + V_{wind})\]

It is tempting to assume that the coupling between the propeller and the wheels which \(f*d=V\) (\(d\) is also the length linking torque on the propeller and linear force on the wheels) will reduce to \(f*p=V\) in some reference frame. In this case

\[F=K\left(\frac{R}{p}\right) \rho_{air} R^2 V V_{wind}\]

which shows that it is the wind which drives the vehicle in front of it. The relation \(f*p=V\) could be the link to a geometrical interpretation of the “Down Wind Faster than Wind” mode of propulsion. As in the video above the propeller could be viewed as a wheel rolling on an imaginary track moving with the wind. We use the symbol \(F\) because this force driving the vehicle is not exactly the propeller thrust, and describes the action of the wind and the reaction of the wheels.

If I understand correctly the Blackbird can move both downwind and upwind. In the upwind direction the propeller acts as a wind turbine driving the wheels. The question is what happens if the land-yacht is stationary \(V=0\). Which way does it go? The last formula implies that the force is zero (small?) for stationary vehicles. Thus the \(V=0\) state is an equilibrium one, but non-stationary. Small perturbations will start the motion. My guess is that the relationship between the pitch of the propeller and the mechanics of the wheel-propeller coupling will determine which is the more probable mode.