# To operate on Mars, how large would a manned helicopter's rotor blades have to be?

I asked this question at https://space.stackexchange.com/questions/53931/manned-helicopter-on-mars but there was disagreement over whether a (now deleted) answer was correct, so I've re-asked it here.

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?

• Manned is kind of ambiguous, is the only purpose to lift a single person? A group of people? Cargo? How far? How high? Jul 2, 2021 at 12:17
• You should not cross-post the same question meta.stackexchange.com/questions/64068/… Jul 2, 2021 at 12:18
• @Ron Beyer For the sake of being specific let's say 4 people to travel 100 miles and back before refuelling and high enough to travel over most terrain on Mars but not mountains.
– sno
Jul 2, 2021 at 12:36
• Perhaps a bit ambitious, nothing wrong with that in theory. Travel on Mars is like Earth at 100,000 feet with 1/3 the Gravity. This is why robotic airscouts and human ground transport might work better. Jul 2, 2021 at 12:43
• Your question focuses on the size of the rotor. Do you care about the power? A full-size, speed increased rotor on mars will be supersonic. That will generate lift, but the power requirements go through the roof. You might be able to design a rotor that could lift humans, but not anything that could power it. Jul 2, 2021 at 16:51

Ingenuity rotors need to rotate 5x (M 0.7) faster on Mars than on Earth.

5x faster rpm translates roughly to 25 times more lift (V$$^2$$). Martian atmosphere is 1/100 pressure of Earth. Gravity is 38% of Earth.

But Martian atmosphere is CO2, which has a 52% higher molecular weight than Earth N2/O2 mix.

0.38 gravity = 25x more lift × 1/100 density × 1.52 g/mole

Martian (and high altitude Earth flight) is possible with batteries or any propulsion source not requiring atmospheric O2.

So, to answer your question, your rotors need to be of 25x greater area, or spin 5x faster, or a combination of both, to lift an equal amount of mass on Mars.

• @mins they already have rated pressure. Pressure plus molecular mass give effective density. What should also be factored in is temperature, simply because the atmosphere behaves funny when at -120C, as this is actually below the freezing point of the air's main constituent, and it's really only the enthalpy of sublimation that is keeping most of the CO2 as a gas! Jul 3, 2021 at 17:07
• @mins I was thinking of the effect of lighter water vapor on density altitude. Same pressure, but less lift. Before Ingenuity ever flew, I had hoped the heavier CO2 would give it more mass to flail against. Jul 3, 2021 at 17:16
• It hurts my head to think about the many factors involved! But you may be delighted by Numerical And Experimental Aerodynamic Investigation Of A Micro-UAV For Flying On Mars and Low Reynolds Number Airfoil Evaluation for the Mars Helicopter Rotor or Predicting Aerodynamic Performance for a Mars Helicopter Rotor
– mins
Jul 3, 2021 at 17:28
• Not a good idea to let rotors rotate 5x faster on Mars, they already have tip speeds of M0.8 here on earth, since that is where the most gain in lift can be found. Best to increase the rotor size and retain the tip speed. Jul 7, 2021 at 5:09
• @Koyovis: In addition sound speed on Mars seems to average to only 240 m/s (at -63°C).
– mins
Jul 7, 2021 at 12:53

On the Space site it was determined that a manned helicopter flying on Mars would look like a Sikorsky Firefly with the rotor of the Mi-26. How much power would it require to drive that rotor, and would the co-axial configuration offer any benefits?

1. Power The Mi-26 has 17,000 kW engine power installed to lift MTOW of 56,000 kg. A Firefly has an MTOW of 930 kg = 9100 N on Earth, 3460 N on Mars. Let's allow an additional 100 Mars kilo's for the larger number of longer blades: MTOW 4500 N.

From this answer: $$C_T = \frac{T}{\rho A (\Omega R)^2}$$ and with $$\rho$$ = 0.015, A = (52.5 * 31.7) = 1,664 m$$^2$$, $$\Omega R$$ = M0.8 = 250 m/s (at -31 °C), I get $$C_T = \frac{4500}{0.015 \cdot 1,664 \cdot 250^2} = 0.0029$$ This corresponds with a $$C_P$$ of about 0.00028, and substituting that in the power equation of the earlier mentioned answer results in P = 109 kW. The Firefly has got 140 kW on board, so will be able to power this machine with no probs!

2. Co-axial On Mars, we need 31.7 times the rotor area that would be required on Earth. Yes a single rotor requires the least amount of power to drive it, but the FireFly has more power than we require for the single rotor.

If we use a co-axial rotor set-up, we don't need to drive a tail rotor and each rotor can be of a size that we have experience with on Earth, half of 1,664 = 832 m$$^2$$, a blade length of $$\sqrt{832/\pi}$$ = 16.3 m. Indeed the blade length of the Mi-26, and we only need 2 blades per rotor now.

Power to drive the co-ax rotor according to momentum theory is also given in Leishman equation 2.160: measurements have shown an increase of 22% of power required = 1.22 * 109 = 133 kW, still within the installed power of 140 kW.