To operate a helicopter on Mars capable of carrying people, how big would the rotor blades have to be compared to a helicopter on Earth?

This question has been reasked at https://aviation.stackexchange.com/questions/88026/manned-helicopter-on-mars

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    $\begingroup$ Maybe a better question for Aviation.SE. $\endgroup$
    – GdD
    Jun 30, 2021 at 13:05
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    $\begingroup$ really really really big :-) . No, of course, due to the Mach limit, you need multiples, as in the answer. $\endgroup$ Jul 1, 2021 at 13:52
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    $\begingroup$ Far better that you use a Glan Speeder $\endgroup$ Jul 1, 2021 at 13:55
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    $\begingroup$ You should not cross-post the same question meta.stackexchange.com/questions/64068/… $\endgroup$ Jul 2, 2021 at 12:17

1 Answer 1


If we look at order of magnitude and use momentum theory, rotor thrust can be computed as follows.

enter image description here

Above pic is fig 2.5 of Principles of Helicopter Aerodynamics by J. Gordon Leishman. Thrust T is computed as:

$$T = C_T \cdot \rho \cdot A \cdot \Omega^2 \cdot R^2$$

with: $C_T$ = thrust coefficient; $\rho$ = air density [kg/m$^3$]; A = rotor disk area [m$^2$]; $\Omega$ = angular blade speed [rad/sec]; R = blade length [m]

Scaling up. $C_T$ is an aerodynamic coefficient, mainly determined by the blade profile and -angle. Regarding the rest of the equation, we can see immediately that rotor thrust scales linearly with air density and rotor area, and quadratically with blade tip speed $\Omega R$, so the most gain in rotor thrust in any number of rotors is from increasing the tip speed. Which is why tip speed is set to the maximum that compressibility allows, between M0.7 and M0.9 depending on maximum forward speed of the heli.

Number of rotors. This answer goes into detail on how best to increase rotor area, by adding blade length or by adding more smaller rotors? The answer is: by sizing up the single rotor, requiring less engine power to generate a given amount of thrust. With possible exception of the counter-rotating dual rotor design which eliminates airframe directional torque.

Earth vs. Mars. The situation on Mars differs considerably from that on Earth.

  • A good thing is that gravity is less, Wikipedia mentions 38% of that on Earth. So we only need to generate 38% of earth rotor thrust to start flying.
  • But now the bad thing, the atmospheric density. Computing atmospheric density at h=0 from this site, I get to $\rho$ = 0.015 kg/m$^3$, 1.2% of air density on Earth which is 1.225 kg/m$^3$ at standard atmosphere sea level.
  • And with this we can compute the required rotor disk size on Mars: 0.38/0.012 = 31.7 times the rotor area on Earth.

By FlugKerl2 - Own work, CC BY-SA 3.0, https://commons.wikimedia.org/w/index.php?curid=15945118

And another bad thing: no oxygen for turbine engines on Mars, as pointed out by @RobertDiGiovanni in this answer. So the manned helicopter on Mars would be electrically powered, and we have not been able to do that on Earth really well yet. The Sikorsky Firefly has a 15 minute endurance, with the battery pack making up 57% of the MTOW. It is a modified S300-C with a rotor diameter of 8.18m, so in order to fly on Mars it would have to have a rotor diameter of 8.18 * $\sqrt{31.7}$ = 46.1m

A rotor with blades of 23m, we don't have those on earth. The largest helicopter on earth is the Mi-26, with a rotor diameter of 32 meters. It has 7 blades, and indeed adding blades does generate more thrust per rotor area, at a cost of efficiency (higher power per thrust). So perhaps a Firefly with the rotor of a Mi-26 could fly on Mars.

Note that the very long blade length on Mars does not pose much of a weight problem. The blade only needs to be able to support its own weight when at a standstill, at 38% of Earth gravity. As soon as the rotor starts turning, centrifugal forces from rotating blade inertia straighten the blades out and enable them to support the cabin weight.


This answer being long enough as it is, I had a go at computing the power required to turn this rotor on Mars on the Aviation site. The FireFly has enough installed power!

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    $\begingroup$ while the blades weigh less on mars, they have the same mass, so the stresses they encounter due to being spun are the same as on earth. $\endgroup$
    – user40799
    Jul 3, 2021 at 8:32
  • $\begingroup$ @user40799 yes indeed $\endgroup$
    – Koyovis
    Jul 3, 2021 at 11:17

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