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Multiple press sources (eg this one) are reporting the planned experimental helicopter on Mars 2020, which is really cool. It seems this vehicle will only be able to fly for 90 seconds or so on its stored power, however it demonstrates the basic aerodynamics and rotor design.

Suppose a future Mars helicopter were to be built big enough to carry astronauts (or comparable-sized scientific payloads), how much stored power would it need to carry to fly for a few hours, and is there a feasible way of storing that within the mass budget of the copter?

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    $\begingroup$ I'm not sure helicopters would scale up well for mars. Square-cube law makes it hard to scale up things and have them still fly, like how the largest birds are all flightless. Mars choppers need either very light rotors, or very rapidly rotating ones (or both), neither scales up very well. $\endgroup$
    – Blake Walsh
    Commented May 13, 2018 at 9:51
  • $\begingroup$ related: space.stackexchange.com/questions/17176/… $\endgroup$
    – Organic Marble
    Commented May 13, 2018 at 12:07
  • $\begingroup$ I guess one way to make it work would be nuclear ramjet. In 1950-60 several designs of ramjet-powered helicopters were being designed; the designs were mothballed due to accidents resulting from combustion instability. Nuclear heating could replace combustion, low air density isn't much of a problem, and the problem of excessive fuel usage would be gone too. Though leaving a trail of radioactive exhaust could be considered unwelcome... $\endgroup$
    – SF.
    Commented May 13, 2018 at 12:46
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    $\begingroup$ @Dragongeek not the question I asked, also it can lift a maximum of $0.015 kg/m^3$ (kilograms of structure and payload, per cubic meter of gas). So a 100m radius spherical balloon (full of vacuum, hydrogen, or helium, it makes little difference) can lift about 60 tons, which has to include the 100 000 $m^2$ of gasbag. It's probably feasible, but it's going to be unwieldy and light gases are rare on Mars $\endgroup$
    – Steve Linton
    Commented May 13, 2018 at 17:00
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    $\begingroup$ @SF. Helicopters at Mars are all about specific power & specific energy of the power source - with a nod to rotor size. Fission thermal power sources have large numbers for both, as long as there are no passengers, semiconductor-based avionics, or polymers involved. All those things are sensitive to the huge neutron and gamma fluxes from the (unshielded) reactor. As soon as you start adding shielding, especially for components you can't put far from the reactor(s), those numbers fall precipitously. Spec. energy is still OK, but spec. power isn't. I seriously doubt it would work. $\endgroup$
    – Tom Spilker
    Commented May 14, 2018 at 3:28

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Thanks to @Organic_Marble for the related question. One of the answers there gives a formula written out in words which could be

$$P = {F^{3/2}\over d FM}\sqrt{\pi \rho}$$

or

$$P = {F^{3/2}\over d FM\sqrt{\pi \rho}}$$

depending on how you read the English. The second one is dimensionally correct, and therefore probably actually correct.

where $P$ is the power needed (at the shaft), $F$ the lift (force) needed, $d$ the rotor diameter, $FM$ is a measure of rotor efficiency called the figure of merit and $\rho$ the atmospheric density. $FM$ is quotes about typically 0.55 - 0.6 for helicopters on Earth.

Using known values for atmospheric density and $g$ we find that, to hover, a 1 metric ton helicopter with $10m$ rotor diameter would need just over $100kW$ of power.

If this is right (and I'm suspicious because the blades will surely be supersonic) then a Tesla car engine and battery pack with a total mass of about 500kg would have enough power for liftoff, and almost an hour of flight. Alternatively, a methane/oxygen fuel cell and compressed or liquidised gas fuel might do it.

Can anyone tell me whether this is a reasonable extrapolation, or have I missed something?

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    $\begingroup$ So on the Moon, where $\rho \rightarrow 0$, a helicopter wold work great! Hmm... $\endgroup$
    – uhoh
    Commented May 13, 2018 at 13:51
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    $\begingroup$ I had misread the formula in words. Answer now edited. $\endgroup$
    – Steve Linton
    Commented May 13, 2018 at 14:42
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    $\begingroup$ This is why it's better to only cite equations when you can cite the source at the same time. I'll bet a quick browse through Aviation SE will find some answers with similar equations that are well sourced. $\endgroup$
    – uhoh
    Commented May 13, 2018 at 14:46
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    $\begingroup$ 3.8 $ms^{-2}$ so thrust needs to be 3800N $\endgroup$
    – Steve Linton
    Commented May 13, 2018 at 16:15
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    $\begingroup$ @SteveLinton I translated that word formula and got your #2, but I have a √2 in the numerator. With that √2 I get the numbers they (he? she?) do, except I think they transposed two digits. In the 2-kg example I get 246 W, not 264. $\endgroup$
    – Tom Spilker
    Commented May 15, 2018 at 6:09

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