23
$\begingroup$

Helicopter drones are awesome fun here on earth, but What if... we took one to Mars? This sounds like a good question for Randall Munroe. But I read about a proposal to send a helicopter drone (MHS) to Mars with Mars 2020.

Mars Heli-drone

Martian atmosphere is a fraction of a percent of Earth's (average 0.6% ASL). My off the cuff calculations say that such a drone would have to be 200 times more powerful than it's earth-bound cousins.

Is this a realistic proposal?

$\endgroup$
  • $\begingroup$ How did you got to that number ? Two things: We can increase lift by increasing the rotors size Gravity is significantly weaker $\endgroup$ – Antzi Jul 13 '16 at 6:24
  • 2
    $\begingroup$ You don't have to have 200 times more power. You have to spin much faster and have bigger rotor blades to generate lift, but you don't need to add much more energy into the process. $\endgroup$ – GdD Jul 13 '16 at 7:56
  • 2
    $\begingroup$ I recommend xkcd what-if: Interplanetary Cessna for a crash course (no pun intended). See also NASA's Preliminary Research Aerodynamic Design to Land on Mars, or Prandtl-m (yes, that's what it really is called). $\endgroup$ – a CVn Jul 13 '16 at 12:27
  • 1
    $\begingroup$ They have done tests proving that a small drone-like helicopter (about 1 meter across) would be able to take off and fly in Mars' thin atmosphere. $\endgroup$ – Phiteros Jul 13 '16 at 15:13
  • 2
    $\begingroup$ Increasing rotor size only works up to a point. Increase the diameter enough, and your rotor tips will go supersonic, which doesn't help lift. $\endgroup$ – Hobbes Jul 13 '16 at 15:50
26
$\begingroup$

Gravity is about a third of Earth's and competitive aerobation helicopter models have a truly excessive power surplus. Just look at this.

There won't be any manned helicopter flight. The helicopter power scales poorly with size - there's a reason we have no VTOL Jumbo Jets. But the same up-scaling issue is our friend when down-scaling. A 6kg helicopter can lift 30kg of payload, so we have equivalent of TWR=6 on Earth.

Drag and lift are both identically (quadratically) proportional to airspeed, so the same loss of lift causes loss of the primary limiting force of the helicopter - air drag on the blades. What's left is mechanical losses (friction of bearings), and mechanical durability of the construction, but these can be built with much surplus - plus larger rotor blades will offset need for extreme RPM.

Of course other things must be taken into account. These flights will be more like "hops" - a short segment of flight followed by a looong recharging period. And the drone will need to be small - and that means not much scientific equipment. Unfortunately, that also means it won't be fully autonomous as there simply won't be enough room to fit a good radio and antenna to reach Earth, or even the satellites. But it could definitely serve as a "recon vehicle" for a large rover, fetching small samples, testing firmness of ground surface for driving over, and planning the best route.

$\endgroup$
  • 1
    $\begingroup$ You need to address the factor of 100 atmospheric density difference in your answer, which is the question that was asked. To get the same lift, you need the airfoils to move 10 times as fast. $\endgroup$ – Mark Adler Jul 14 '16 at 5:20
  • $\begingroup$ @Mark: which I can, due to reduced air drag - same torque caused by air resistance. Regardless of speed and pressure, torque to lift ratio remains the same. (minus other resistances which are taken care of by the power surplus, and plus reduced lift requirements thanks to low gravity.) Air drag doesn't rise; other frictions increase by factor of ~60 (x200 density /3 gravity) but I believe they account for less than 10% of original losses, so the 6x power surplus is enough to overcome them. $\endgroup$ – SF. Jul 14 '16 at 6:27
  • 1
    $\begingroup$ No, the air drag does rise, where the dimensionless drag coefficient goes up dramatically as you approach Mach 1, starting as low as Mach 0.6. So it doesn't scale to low density as nicely as you claim. $\endgroup$ – Mark Adler Jul 14 '16 at 7:12
  • $\begingroup$ @MarkAdler: What order of increase can we expect, with the blades designed for the purpose? $\endgroup$ – SF. Jul 14 '16 at 13:47
  • $\begingroup$ It depends greatly on the details of the wing, and the other constraints you have on it. Could be a factor of 3, could be a factor of 10. See mhmaberry.files.wordpress.com/2014/04/2.jpg for an example. This sudden increase in drag is why there is a "sound barrier", since the drag drops off again above Mach 1. See this for the generic picture of the situation: history.nasa.gov/SP-367/f86.htm $\endgroup$ – Mark Adler Jul 14 '16 at 14:44
31
$\begingroup$

There's nothing like seeing it flying in a Mars-density chamber to answer your question:

Crazy Engineering: Mars Helicopter

I have a really nice video of it in controlled flight in the chamber, but I can't find that one on the interwebs yet.

Update two years later:

Thanks SF for this link to nice video.

$\endgroup$
6
$\begingroup$

You can calculate the hover power required out of ground effect by using the following formulas:

Given m, the mass of the helicopter, the required lift force is $L = g_{mars}*m$

The required shaft power is:

$Power (Watts) = (L^{3/2} / R * \sqrt{( 2 / (\pi * density) )})/FM$

where $R$ is rotor diameter and FM is the "figure of Merit". For a small helicopter the FM is smaller than 0.66, say 0.55.

The density at low altitude on Mars is 0.0152 kg/m^3

The gravitational acceleration $g_{mars}$ is 3.8 m/sec^2.

Example

  • mass = 2 kg
  • Lift = 7.6 Newton
  • FM = 0.55
  • Rotor diameter = 1 meter
  • density = 0.0152 kg/m^3
  • pi = 3.1416

The result is 264 Watts.

The fact that the counter rotating rotors are coaxial does not significantly reduce the hover power. The effective diameter is nearly the same. Note that, when the mass is reduced to 1 kg the required power would be only 87 Watts !

In summary: Power required is proportional to (Lift to the 3/2 power exponent), inversely proportional to rotor diameter and inversely proportional to the square root of the density.

When flying forward instead of hovering the required power goes down significantly for a well shaped streamlined fuselage. In 1929 Glauert came up with an approximate formula (the solution of a quartic) which is still used today. A good reference text book is: B W McCormick: Aerodynamics of V/STOL Flight

$\endgroup$
1
$\begingroup$

I do not object to Mark's answer, but I am concerned with the rotor blades rotating so fast that they will be entirely within a supersonic flow. Maintaining helicopter blades within a subsonic flow is a limiting factor for helicopter speeds here on earth. How would the rotor's lift be affected if it were entirely behind a shockwave? On earth most subsonic wings placed in such conditions simply stall. Also because Mars' atmosphere is so cold, the local sound velocity should be quite low, hence a rotor for Mars better be designed to function within transonic and supersonic regimes. A good example of this problem: the U2 spy plane (granted not a helicopter) at its highest altitude it was at the combined limits of not flying fast enough to generate enough lift (in atmosphere), or flying too fast and going supersonic (low local sound velocity) which was not good for the type of plane it was.

$\endgroup$
  • $\begingroup$ This is called the "Coffin corner". See en.wikipedia.org/wiki/Coffin_corner_(aerodynamics) . $\endgroup$ – Tom Spilker May 14 '18 at 4:23
  • $\begingroup$ Wit the right airfoil, a rotor can work at supersonic speeds. The reason this isn't done on Earth is the huge amount of noise generated by supersonic propellers. $\endgroup$ – Hobbes Nov 1 '18 at 16:03
-3
$\begingroup$

mars has thin atmosphere so, when there is rotation of blades there is less impact on ground so we need atleast 250 miles an hour to lift or takeoff so it is impossible to land on rockey ground also

$\endgroup$

protected by Community Feb 24 '18 at 9:44

Thank you for your interest in this question. Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site (the association bonus does not count).

Would you like to answer one of these unanswered questions instead?

Not the answer you're looking for? Browse other questions tagged or ask your own question.