I wasn't really able to follow the hyperlink. For one thing, it seems to suggest that the propeller efficiency just before takeoff is zero. I find that approach worthless.
Usually, I have found that if I need to know for sure if something it correct, at some point I will have to derive it myself, so it will usually take less time to do that then to search for what will be a non-authoritative answer anyway.
I would think a better way to handle this question would be purely from an input and output perspective, not a microscopic blade perspective. It shouldn't matter if there are 10 blades shaped like a Christmas tree. The only things you need to measure for efficiency ratios should be inputs and outputs. That's the whole idea behind such numbers.
Here, the input is power. The output is clearly thrust. So, a propeller assembly's efficiency should be the actual thrust as compared to a perfect magical conversion of power to thrust in which there was no wasted energy going elsewhere and not into thrust.
These ideas in mind, we start with $$T_0={d\over dt}mv=\dot{m}\Delta v,$$ where $T_0$ is the thrust of a perfect fan assembly, $m$ is the mass of the fluid passing through the swept area of the fan assembly, and $\Delta v=v_{\rm out}-v_{\rm plane}$ is the average difference in speed between the fluid exiting the fan and rate that the fan itself is moving through the fluid in the rest frame of the fluid.
The use of the word average here is intended to allow for a "spherical chicken," such that the effects of turbulent flow outside of the fan swept area affects the efficiency, but can cleverly be ignored in our actual calculations of efficiencies themselves. However, one could visualize the fan assembly to be a ducted fan with a constant cross-sectional area, where the air aft of the assembly could be of a higher mass density than the partial relative vacuum fore of it, just as in some automobile air intakes at high enough rpm. This way it is clear that the exhaust velocity isn't any speeds of airstreams near blades, which could be higher, but, rather, is the average velocity of all air just after the fan assembly.
Continuing, in the reference frame of the plane we have an increased kinetic power of the fluid due to the fan of $$P={\dot{m}\over2}(v_{\rm out}^2-v_{\rm plane}^2).$$ Mass flux analysis yields $$\dot{m}=\rho A v_{\rm out}$$ where $\rho$ is the mass density of the exhaust fluid excluding any fuels. This gives
$$P={\rho A v_{\rm out}\over 2}(v^2_{\rm out}-v^2_{\rm plane}).$$
This is a cubic equation. They generally have one real solution and a few imaginary ones. However, we can use a trick from relativity, with $\beta\equiv v_{\rm plane}/v_{\rm out}$, yielding $$v_{\rm out}=\left[{2P\over \rho A(1-\beta^2)}\right]^{1/3}$$ and an ideal thrust of $$T_0=\rho A v_{\rm out}(v_{\rm out}-v_{\rm plane})$$
$$=\left[4P^2\rho A{1-\beta\over(1+\beta)^2}\right]^{1/3}.$$ Using $A=\pi(D/2)^2$, this is $$T_0=\left[\pi \rho P^2 D^2{1-\beta\over(1+\beta)^2}\right]^{1/3}$$ with $\eta_{\rm P}=T/T_0$ where $T$ is the observed thrust of a fan assembly using a power $P$.
The dependence upon $\beta$ alone, as if the power were fixed, is
I think this is "How to calculate in the right way the efficiency of a propeller."
To check if this analysis is reasonable, we can compute the efficiency of one of the first propellers. I did that in the answer to Why does this calculation show Gustave Whitehead's propellers were more than 100% efficient? and arrived at 81±13%, which seems to me to be a reasonable efficiency.
