The rover - without wings - has around 1000 kg of mass and Mars has 38% of Earths g. The density of the atmosphere is 1% of that of Earth. Assuming the surface area of the wings could be used for photoelectric cells having an efficiency of 40% at 500W/m^2, what wing dimensions, power and speed would this airborne rover need?

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    $\begingroup$ Related $\endgroup$
    – Federico
    Jan 22, 2015 at 15:26
  • $\begingroup$ There's a massive range for this, are we assuming a fixed amount of storable battery power? How long is the "runway" area available? How long should the flight be? (Time/distance?). What material is the wing made of? Are we to account for this in the weight of the lander? How much load must it handle? Should the wings be foldable for flight? This is more of a physics question, although for reference, a Cessna 172 is roughly 1000kg :) $\endgroup$
    – Jon Story
    Jan 22, 2015 at 15:59
  • $\begingroup$ Interplanetary Cessna: "X-Plane tells us that flight on Mars is difficult, but not impossible. NASA knows this, and has considered surveying Mars by airplane. The tricky thing is that with so little atmosphere, to get any lift, you have to go fast. You need to approach Mach 1 just to get off the ground, ..." $\endgroup$
    – user
    Jan 23, 2015 at 9:10
  • $\begingroup$ @MichaelKjörling have you clicked on my link? :P $\endgroup$
    – Federico
    Jan 23, 2015 at 9:25
  • $\begingroup$ @Federico No. Should I have? ;-) $\endgroup$
    – user
    Jan 23, 2015 at 9:36

1 Answer 1


Our Flight Environment

First the basics: Mars atmosphere is mainly carbon dioxide and has an average temperature of -47°C and an average pressure of 0.004 to 0.0087 bar. The speed of sound in these conditions is 238.2 m/s, the dynamic viscosity $\nu$ is 1.14$\cdot 10^{-5}$, and for my calculations I use an atmospheric density of 0.0234 kg/m³.

Now I owe all you early upvoters an apology. The earlier answer used an unsuitable methodology; now I have improved the answer with different results.

Operating Point

Next step: The electric power supply makes a propeller the right choice, and for propeller aircraft the minimum power to sustain flight can be achieved when the lift coefficient is $$c_L = \sqrt{3\cdot c_{D0}\cdot\pi\cdot\Lambda\cdot\epsilon}$$ This makes your zero-lift drag the real sizing parameter: Make the aircraft a little more draggy, and you get punished with a big size increase. Therefore, we strive to keep the zero-lift drag coefficient $c_{D0}$ down to 0.008 in order to keep the aircraft small.

This allows to get an idea which aspect ratio will be best: At minimum power the total drag is four times the zero-lift drag, and the lift coefficient should be as high as possible to keep the aircraft's size small. Realistically, our flight $c_l$ will not grow above 1.6 (which is the best that low-speed airfoils can manage at a reasonable L/D), so the aspect ratio is $$\Lambda = \frac{c_l^2}{3\cdot c_{D0}\cdot\pi\cdot\epsilon}$$ which yields an aspect ratio of 35.74 when $\epsilon$ is 0.95.

Available Power

Now we can follow the procedure in this answer to calculate power demands for your solar-electric propulsion. Solar radiation on Mars is a lot lower than on Earth, but very reliable as there are no clouds to obstruct the sun. Beware of dust storms, however!

Since you gave the Martian solar constant in your question, I will use it here: The wing will be covered to 80% with solar cells, and each square meter of them will produce your optimistic 200 W (500 W radiation times 40% efficiency). Per square meter of wing we can achieve 160 W, and with an electric efficiency of 96% and a propeller efficiency of 85% this translates into a propulsive power per square meter of 130.5 W.

Matching Thrust to Drag

Now we can equate the drag $D$ and the available thrust $T$ from the power $P$, like this: $$D = T = \frac{P}{v} => 4\cdot c_{D0}\cdot S\cdot\frac{\rho\cdot v^3}{2} = 130.5\cdot S$$ Luckily, wing area $S$ can be canceled, and we get a result for the airspeed $v$ at which the aircraft's drag can just be supported by the power the wing can generate independent of wing area: $$v = \sqrt[3]{\frac{130.5}{4\cdot c_{D0}\cdot\rho}} = 55.857 m/s$$

Now for the real size: If we assume that the payload fraction is 25%, the mass of the full aircraft will be 4 tons. To lift this at the dynamic pressure of 36.5 N/m² which corresponds to the found flight speed we need a wing area of 254 m². This can create 40.65 kW of power, 33.17 of which are transformed into thrust. The wing span will be a whopping 95.28 m. How you get this to Mars I leave to your imagination - after all, this is just a tad less than the wingspan of the Hughes H-4.

If you can achieve a higher payload fraction, the wing will become smaller: At 30% the full aircraft mass is only 3.333 tons, and the wing size will become 211.74 m² large. Its wing span will be 87 m. With these dimensions I doubt that a higher payload fraction will be possible. The closest aircraft to this design was the Boeing Condor, which had an empty mass of 3.6 tons (with a propulsion system weighing close to the mass of Curiosity) and a wing span of 61 m.

Last Checks

  1. Mach number: 56 m/s is just Mach 0.2345 on Mars, so we should not run into compressibility effects which would limit the assumed lift coefficient.

  2. Reynolds number. The bigger wing has an average chord of 2.66 m and the smaller one still 2.43 m chord, and per meter the Reynolds number in -47°C cold carbon dioxide at 56 m/s is 4,900,000 - plenty enough to make the selected lift coefficient possible. Size helps here!

  • $\begingroup$ Where does this value for the product of $c_l$ and $Ma^2$ come from? Why not use $L=c_l 0.5\rho V^2 S$? $\endgroup$
    – ROIMaison
    Jan 24, 2015 at 8:58
  • $\begingroup$ @ROIMaison: I learned it from Mark Drela. It is better to express speed in terms of $Ma$, because the Mach number limits your $c_{l max}$ at higher speeds. Just replace $v$ by $Ma \cdot a$, and your lift equation will contain all essential elements. $\endgroup$ Jan 24, 2015 at 9:19
  • $\begingroup$ I understand, and will the value of 0.4 still be valid on Mars? As there is a different 'air'-density. $\endgroup$
    – ROIMaison
    Jan 24, 2015 at 9:23
  • $\begingroup$ @ROIMaison: As long as the atmosphere behaves like an ideal gas the flow phenomena at the same Re and Ma numbers should be the same. And with its low pressure and temperature I have no doubt that the atmosphere on Mars is very similar to Earth's atmosphere 31 km up. $\endgroup$ Jan 24, 2015 at 9:26
  • $\begingroup$ I'm afraid the available solar energy is only half - 500 W impinge on the surface, but solar cell efficiency is 40% at best, so it's 200 Watt/m^2. I should have multiplied the efficiency and the raw energy in my question! $\endgroup$
    – yippy_yay
    Jan 24, 2015 at 23:56

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