# Is my equation for calculating ducted propeller thrust correct?

The equation is 13(EfficiencyxProp diameterxoutlet diameter)^2/3

It seems really high when I use inches and really low when I use feet.

• Please edit your question so it is clear what equation you're actually using - as is, it's not clear. Also, do you have a source for this equation? A properly phrased equation will note what units are used for each term, both input and output... after all, an equation that yields thrust in ounces will have at least one different constant than an equation that gives thrust in pounds.
– Ralph J
Jan 1 at 20:02
• This is a quick tutorial on how to write equations on Stackexchange Jan 1 at 20:14
• Ralph J- No I don’t have a source. Thanks for the tutorial sophit Jan 2 at 23:06

Is my equation for calculating ducted propeller thrust correct?

Almost

It should be something like:

$$T=1.3 \eta P^⅔$$

but this is valid only on one specific case.

The easiest equation relating thrust $$T$$ and power $$P$$ is the one derived from the simple momentum theory:

$$T=\sqrt{2\rho A P²}$$

where $$\rho$$ is air density and $$A$$ is disk area i.e. $$πr²$$. In this equation you have:

• the "$$^⅔$$" applied to $$P$$, just like in your equation;
• the propeller's diameter $$d$$ (actually its radius $$r$$), just like in your equation;
• to take into account the duct, the thrust is normally increased of some 30%, the power being the same; and here you get also the 1.3 (not 13) just like in your equation;
• and to take into account all the effects which are simply disregarded in the momentum theory, the right part can be multiplied by an efficiency factor $$\eta$$ (some 0.6 to 0.8), just like in your equation.

Rearranging the terms and introducing the factors just discussed, you get:

$$T=1.3 \eta (Pd)^⅔\sqrt{½\rho π}$$

which resembles what you were actually looking for. And if you consider the density at 3km (10'000ft) height and a propeller with 1m diameter, then the equation becomes exactly what you remembered:

$$T=1.3 \eta P^⅔$$

Note that all these equations are valid only at rest (zero speed).

It seems really high when I use inches and really low when I use feet.