This spawns from a related question. Propeller-drive engines (piston/turboprops) seem to always list their output in horsepower or kilowatts, which are units of power.

I want to know the engine's thrust instead.

How to calculate this?

I know propeller thrust depends on airspeed. What I want is an equation where we input engine power and airspeed (and maybe other parameters), and output thrust.

  • $\begingroup$ Well, in a sense it is easy. Power = thrust · velocity. However, that is power output of the propeller. Between power output of the engine and power output of the propeller is the propeller efficiency and somebody else will have to describe that. $\endgroup$ – Jan Hudec Jun 27 '16 at 6:46

The simplest equation uses an assumed propeller efficiency $\eta_{Prop}$and converts power P into thrust T like $$T = \frac{P}{v}\cdot\eta_{Prop}$$ The tricky thing is to get $\eta_{Prop}$ right. Use 85% for lightly-loaded* propellers working at their design point. And make sure you subtract the power offtake from the engine for driving the accessories, so you only use the effective power at the propeller, not the rated engine power.

The equation only works if the propeller is in motion, so airspeed $v$ is not zero. For covering the static case, please see this answer.

*The load on a propeller can be expressed by the ratio of thrust to the propeller disc area, made dimensionless by dividing by the dynamic pressure $\frac{\rho}{2}\cdot v^2$. If $T$ is thrust and $d$ is the diameter, the dimensionless thrust loading coefficient $c_T$ is $$c_T = \frac{T}{\pi\cdot\frac{d^2}{8}\cdot\rho\cdot v^2}$$ Sometimes this is simplified to $$c_T = \frac{T}{\pi\cdot\frac{d^2}{4}\cdot v^2}$$ Note that this definition needs the propeller thrust at a specific speed. A different way of expressing the propeller loading, which lends itself to using the easy to measure static thrust, is by dividing thrust by the disc area and the dynamic pressure of the propeller tip at rest, with $n$ as the RPM of the prop: $$c_T = \frac{T}{\frac{\rho}{2}\cdot\left(\frac{n\cdot\pi\cdot d}{60}\right)^2\cdot\pi\cdot\frac{d^2}{4}}$$ and again a simplified form of this can be found in literature: $$c_T = \frac{T}{\rho\cdot\left(\frac{n}{60}\right)^2\cdot d^4}$$

Generally, a lightly loaded prop has a big diameter and flies at low to moderate speed. While WW II warbirds are still in the lightly loaded realm, all large turboprops (C-130, P-3) are firmly outside of it.

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  • $\begingroup$ Thanks but what does lightly-loaded propellers mean exactly? Are you talking about something like wing loading or disk loading for helicopters? $\endgroup$ – DrZ214 Jun 27 '16 at 7:02

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