I am designing a remote controlled airship. I will tune it so that the lift given by Archimedes' Principle will exactly balance the weight of all the structure. It will be propelled by brushless motors with propellers on them.

As far as I have understood, for some given velocity $v$ the drag force $D$ will be given by the air pressure, some shape dependent drag coefficient $C_D$, the surface $S$ that the airship offers to the wind, and finally the squared velocity $v^2$.

Now, to keep some speed, obviously the thrust $T$ must equal the drag $D$. Now, I need to get to some equation for the power that I need to offer such thrust at such speed, given that I am using non-ideal engines, with non-ideal propellers etc. From a barely theoretical point of view, I know that if I want to apply some force to an object that moves at some speed, I will be using some power $P \sim Fv$. Now, since the thrust is generated in some way that actually looks very dispersive, I would like to know if there exists some relationship between the power $P$, the velocity $v$ and the thrust $T$, given some specific motor and some specific propeller. In particular, what are the parameters that one needs to know to get to this relation?

On example, I can imagine that some efficiency parameter for the engine, its RPM rate, its voltage, the diameter of the propeller, the pitch of the propeller, these are all going to be relevant to the equation I am looking for, but I wouldn't know how to explicitly figure out that.

It such a relation doesn't exist in an obvious or general way, could you just give me an idea of the efficiency of the propulsion? I mean, I know that $P \geq Tv$, but how much bigger is it in general? Do these quantities share the same order of magnitude, or the dispersion is very big compared to the actual propulsion?

As you can understand, I am not an expert at all on the subject, so I would appreciate anything that could get me started.

Thank you again!


1 Answer 1


A propeller accelerates the air of density $\rho$ which is flowing through the propeller disc of diameter $d_P$. This can be idealized as a stream tube going through the propeller disc:

Section of airstream through propeller

The air speed ahead is $v_0 = v_{\infty}$ and the air speed aft of the propeller is $v_1 = v_0 + \Delta v$. The propeller effects a pressure change which sucks in the air ahead of it and pushes it out. Since the mass flow must be equal ahead and behind the propeller, the stream tube diameter is bigger ahead of the propeller and smaller downstream. In reality, there is no neat boundary between the air flowing through the propeller and that surrounding it, but for computing thrust this simplification works well if the airspeed is identical across the cross section of the propeller disc.

The mass flow (mass $m$ per unit of time $t$, written as a derivation) is: $$\frac{dm}{dt} = \pi \cdot\frac{d_P^2}{4}\cdot \rho \cdot \left( v_{\infty} + \frac{\Delta v}{2} \right)$$ The mass flow is written as the volume of air with density $\rho$ per time, moving through the propeller disk with the diameter $d_P$ at a speed that is the median between entry and exit speed. The thrust is mass flow times speed change: $$T = \pi \cdot\frac{d_P^2}{4}\cdot \rho \cdot \left( v_{\infty} + \frac{\Delta v}{2} \right) \cdot \Delta v$$ If the engine has the power P, the thrust is net power divided by airspeed in the propeller disk. To arrive at the net power, you multiply the rated engine power by the propeller efficiency $\eta_{Prop}$ and the electric efficiency $\eta_{el}$: $$T = \frac{P\cdot\eta_{Prop}\cdot\eta_{el}}{\left( v_{\infty} + \frac{\Delta v}{2} \right)}$$

A good engine will have an electric efficiency above 90%, and a good propeller will give you an efficiency between 80% and 85%. Efficiency goes up with lower $\Delta v$, so a big, slowly spinning prop is better than a small, fast one.

An airship will not move fast, so the $v_{\infty}$ is low. In the static thrust case it is zero, and the thrust equation can be simplified: $$T_0 = \frac{P\cdot\eta_{Prop}\cdot\eta_{el}}{\sqrt{\frac{2\cdot T_0}{\pi\cdot d_P^2\cdot\rho}}} = \sqrt[\LARGE{3\:}]{P^2\cdot\eta_{Prop}^2\cdot\eta_{el}^2\cdot\pi\cdot \frac{d_P^2}{2}\cdot\rho}$$ To actually know your $v_{\infty}$, you need to know the drag $D$ of your airship. The general equation is $$D = A\cdot c_D\cdot\frac{\rho}{2}\cdot v^2$$ with $A$ the frontal area of the ship and $c_D$ its drag coefficient. S. Hörner (page 14-1) gives the drag coefficient of LZ126 (later Los Angeles) as $c_D = 0.023$ for the hull alone and $c_D = 0.071$ for the complete ship, including nacelles, fins and all. A similar result is given in NACA Report 394 which documents the tests done on models of Goodyear Zeppelins in the NACA variable density wind tunnel in 1932. There is another source, NACA Report 117 by Max Munk, written in 1921 and supposedly collecting the results of German measurements on Zeppelins which can be found here.

Your model will not achieve such low values, however, because it will fly at a lower Reynolds number, which means that friction will be higher in relation to other forces. Depending on the size and speed of your model, pick a value between 0.15 and 0.3 for first calculations.

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    $\begingroup$ Never received such a clean and understandable answer in my whole stackexchange life. Thank you! $\endgroup$ Commented Sep 30, 2014 at 10:05
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    $\begingroup$ Amazing answer. I tried to calculate all this from first principles, with your guidance, in order to completely understand it. I only have one question. In the final step where you calculate thrust $T_o$, by substituting in $Δu$, i'm missing a number in my calculations. Since $ Δu=\sqrt{\frac{8T_o}{\pi d_p^2p}} $ so when you substitute in $\frac{Δu}{2}$ i get $ Δu=\sqrt{\frac{2T_o}{\pi d_p^2p}} $, instead of $ Δu=\sqrt{\frac{T_o}{\pi d_p^2p}} $ that you say.. $\endgroup$
    – KeyC0de
    Commented Aug 29, 2015 at 12:36
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    $\begingroup$ @RestlessC0bra: Thank you very much for checking this! You're right, I forgot the 2. Fixed it. $\endgroup$ Commented Aug 29, 2015 at 18:07
  • $\begingroup$ Ah no problem. I was surprised by that mistake. $\endgroup$
    – KeyC0de
    Commented Aug 29, 2015 at 18:08
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    $\begingroup$ @RestlessC0bra: Better in this context means higher efficiency. A big propeller will capture more air and will need a smaller $\Delta v$ to produce the same thrust, and will consequently spin much slower. In mathematical terms: The denominator grows with $\Delta v$, so the thrust from a given power becomes higher. Now I only hope I have understood your question correctly. $\endgroup$ Commented Apr 2, 2016 at 20:02

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