If memory serves correctly, propeller thrust for a given propulsive horsepower is inversely proportional to airspeed, assuming features like constant RPM prop, and being outside Mach Effects

The problem is, such a relation falls apart at zero airspeed by virtue of dividing by zero, and even before that, it would predict thrust sufficient to rip the propeller blades clean off the aircraft.

Obviously, there must be something more to this situation, but I don't know what the rules are in that. How do we know the thrust of an ideal propeller at zero airspeed if we know the thrust for given power at some higher airspeed?


2 Answers 2


The trick to avoid a division by zero is to use the speed at the plane of the propeller disk and not the speed at infinite distance. Then you can use the same equation for a moving and a static propeller.

Air of density $\rho$ flows through the propeller disc of diameter $d_P$, creating a thrust force $T$ by being accelerated by $\Delta v$. The air speed ahead is $v_0 = v_{\infty}$ and the air speed aft of the propeller is $v_1 = v_0 + \Delta v$. In the plane of the propeller it can be assumed that air speed is $v_{\infty} + \frac{\Delta v}{2}$:

$$T = \pi \cdot\frac{d_P^2}{4}\cdot \rho \cdot \left( v_{\infty} + \frac{\Delta v}{2} \right) \cdot \Delta v$$

In the static case $v_{\infty}$ is zero:

$$T = \pi \cdot\frac{d_P^2}{4}\cdot \rho \cdot \frac{\left(\Delta v\right)^2}{2}$$

Without knowing $\Delta v$ it is impossible to solve. Here the static thrust equation which uses engine power $P$ helps:

$$T_0 = \frac{P\cdot\eta_{Prop}}{\sqrt{\frac{2\cdot T_0}{\pi\cdot d_P^2\cdot\rho}}} = \sqrt[\LARGE{3\:}]{P^2\cdot\eta_{Prop}^2\cdot\pi\cdot \frac{d_P^2}{2}\cdot\rho}$$

Still, for a solution you need to know the propeller efficiency $\eta_{Prop}$. For the static case this parameter can at most become 0.5. For practical results use 0.4 with constant speed props and - depending on propeller pitch - for fixed-pitch propellers lower values if the propeller is optimized for higher speeds.


the relation falls apart at zero airspeed by virtue of dividing by zero

With props, thrust is affected by airspeed, but it is important to realize engine power output only works against the drag of spinning the propeller (and of course its own internal friction).

Engine Brake Horsepower = prop Drag torque/time

The thrust output of the propeller depends on the amount of air mass accelerated and it's angle of attack to the relative wind and its coefficient of lift (as an airfoil).

this is known as prop efficiency

Props are a bit more of an art because, while they lose efficiency as airspeed increases from an air mass acceleration view, efficiency greatly increases if the prop blade angle of attack is properly aligned with the relative wind.

Prop propulsors generate thrust far more efficiently as spinning airfoils than they do as mass movers. This is why variable pitch prop blades enable much higher airspeeds.

In summary, the power output of an engine is only related to its fuel burn. The load on the engine is prop drag. No need to lose your mind with division by zero.

thrust is derived by multiplying in the efficiency factor, which can be positive, zero, or negative (reverse thrust)

Finally, it is important to consider whether or not the given situation is transient or steady state. Unless brakes are applied, an aircraft generating thrust will begin accelerating until thrust = drag.


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