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I am working on designing a paramotor, and need to calculate estimated thrust for the propeller. Is there a calculator or equation where I can input a diameter, pitch and rpm and get a ballpark thrust estimate and power consumption. I would prefer the thrust estimate to be lower rather than higher

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  • $\begingroup$ If you can't find the formula, paramotors with decent size props driven through a reduction drive are about 5-6 lbs thrust/hp. You could use 5 lbs and be sure of being conservative. $\endgroup$
    – John K
    Dec 29, 2022 at 22:02
  • $\begingroup$ Peter Kaempf on this site will surely know the formula. $\endgroup$ Dec 29, 2022 at 23:38
  • $\begingroup$ It’s a little bit hard to understand. Can you help me summarize? $\endgroup$
    – Lelotz
    Dec 31, 2022 at 5:09

1 Answer 1

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The relation between thrust $T$ generated by a propeller and the power $P$ which has to be supplied to the propeller, is given by two distinct equations, one valid at zero flying speed and one valid at the "design speed". Let's see these two equations.


Zero speed

For this part, I gave already here exactly the needed equation:

$T=\eta (Pd)^⅔\sqrt[3]{½\rho π}$

where $\eta$ is an efficiency factor of some 0.6 to 0.8, $d$ is the propeller's diameter and $\rho$ air density.


Design speed

The propeller simply changes the engine's power $P$ in thrust $T$ pushing the airplane at a speed $V$. In this process, some power is lost in inefficiencies which are incorporated in an efficiency factor $\eta$ (different from the previous one):

$T=\eta \frac{P}{V}$

$\eta$ depends on the blade pitch as measured at ¾ of the span and on the advance ratio $J=\frac{V}{nd}$, where $n$ is the rotating speed [rev/s]. The precise trend of $\eta$ is supplied by the propeller's manufacturer but it has a typical shape like the following one¹:

propeller efficiency factor

Note that this plot becomes meaningless for $J$ smaller than some 0.2.


Let's do the math using for example the following values:

  • $\eta$@0speed, 0.6 (to be conservative); if there's a gearbox between the engine and the propeller, some power is lost there, let's say some 20%; we incorporate this loss in $\eta$ that therefore becomes $0.6 \times 0.8=0.48$.
  • $\rho$@sea level, 1kg/m³
  • Engine's power 3kW@2000rpm i.e. 33rps.
  • Propeller's diameter 1m and 40° twist@¾ span.
  • Cruise speed 60m/s.
  • $J=\frac{60}{33}=1.8$; from the plot we get a relevant $\eta=0.8$.

So:

  • at 0speed thrust is $0.48\times3'000^⅔\sqrt[3]{½π}=116N$
  • at 60m/s it is $0.8\frac{3'000}{60}=40N$

Between those two speeds thrust can be simply linearly interpolated.

Obviously these are all theoretical values which have to be tested in reality. Anyway they can be used to ballpark estimate the main geometry of the propeller and the needed power.


¹McCormick B.W. Aerodynamics, Aeronautics & Flight Mechanics. John Wiley & Sons, Inc.

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