23
$\begingroup$

According to this site, NASA produced a public domain painting representing what the Olympics could look like on a lunar colony. In the upper right of the picture, there are people flying by flapping wearable wings. Given Moon's gravity with Earth's atmospheric pressure in a lunar colony dome, like the one shown here, could this be possible? How much force would you have to push down on your arms to take off? Would taking off be feasible, or is just gliding?

Lunar Olmpics

Improve this question
$\endgroup$
8
  • 6
    $\begingroup$ Relevant XKCD: what-if.xkcd.com/30 $\endgroup$
    – gerrit
    Commented Jun 4, 2014 at 14:22
  • 17
    $\begingroup$ @gerrit: I just love how "there is an xkcd that might be intresting too" mutated to the standard phrase "relevant xkcd". We might want to think about "is already answered as xkcd" as a close reason $\endgroup$
    – PlasmaHH
    Commented Jun 4, 2014 at 19:31
  • 1
    $\begingroup$ I think they also addressed this in a national geographic video on colonizing the moon near the end! $\endgroup$
    – Frank
    Commented Oct 12, 2016 at 17:14
  • 1
    $\begingroup$ It would work fine as long as you don't fly too close to the Sun. ;) $\endgroup$
    – Dave
    Commented Oct 12, 2016 at 21:49
  • 2
    $\begingroup$ In the old Heinlein story "The Menace from Earth" they do this, but in a huge cavern used as the colony's air storage. I think the air pressure might have been elevated. $\endgroup$
    – Organic Marble
    Commented Oct 13, 2016 at 13:05

2 Answers 2

Reset to default
22
$\begingroup$

The lift force you would need to produce would have to be equal (stable flight) or greater (take off) than the force pulling you back towards the moon. If your mass is average for a male at 62kg, the force from the lunar gravity would be F=ma, = 62kg x 1.622m/s2 = 100N (almost exactly).

lift force = 0.5 x density x lift coefficient x area x velocity^2

This is as far as you can really calculate without specifying shape/size of the wings etc. However this site shows that the human arm can exert around 50N -60N (that's per arm). So with the correct design it seems like you have 1.1 times the required force available. This all ignores the mass of the wings themselves, the percentage of force that is applied in the appropriate direction etc, it at least implies you would be able to hover on the spot until you get tired. I suppose doing so would become comparable to the world record for treading water - after 85 hours you'd be pretty bored.

Improve this answer
$\endgroup$
8
  • 1
    $\begingroup$ With numbers that close, it seems wing design is going to be critical for the experience. I wonder how getting a running start would help. Also to consider would be the range in arm motion and muscles required and the duration of muscular exertion forces needed during the flapping motion. $\endgroup$
    – TK-421
    Commented Jun 4, 2014 at 18:36
  • 3
    $\begingroup$ That's true. Although, the values quotes for arm strength are average of a typical male. So you could expect an athlete to perform significantly better. Getting a running start would definatly help, I'd guess that you might even be able to run and take off into a glide (given the right wing profiles ofcourse). $\endgroup$
    – ThePlanMan
    Commented Jun 4, 2014 at 19:42
  • 1
    $\begingroup$ Your guy can't fly. You're showing that it takes all the guy's got to fly--but after making a flap he has to move his arm back for the next one. Is he really going to be able to reposition 10x as fast as he flapped? The athlete has a chance, though. $\endgroup$
    – Loren Pechtel
    Commented Jun 5, 2014 at 3:36
  • 13
    $\begingroup$ It looks like this assumes that the you have to be able to produce a thrust:weight ratio >= 1 in order to fly. But you don't; wings are more efficient than that. It's straightforward to create a wing with a lift:drag ratio of 10, and much higher ratios can be achieved with careful attention. Calculating the thrust available from a flapping motion is rather a complicated exercise, but the fact that human-powered sustained flight is possible (if unwieldy, and requiring extreme athletics) on Earth suggests that it would be straightforward in a 1 atm environment on the moon. $\endgroup$
    – pericynthion
    Commented Jun 5, 2014 at 5:44
  • 3
    $\begingroup$ Why the arms? Why not the legs? God Mercury did have winged calves in many ancient depictions, so it's not a new idea. And don't underestimate the power of the human ass! $\endgroup$
    – LocalFluff
    Commented Oct 12, 2016 at 19:05
2
$\begingroup$

This is a bit late but there have been claims that humans could fly by wing flapping on Titan.

https://io9.gizmodo.com/humans-might-be-able-to-fly-on-titan-if-they-use-large-15361698251

The Moon has a surface gravity of 1.62 meters per second squared, or 0.1654 gravity.

Titan has a surface gravity of 1.352 meters per second squared, or 0.14 gravity.

The surface gravity of Titan is acutally only 0.85 that of the Moon, or the surface gravity of the Moon is actually 1.176 that of Titan.

Thus at a rough guess any flying area on the Moon might have to be pressurized with 1.175 the atmospheric density of Titan's atmosphere, which is already significantly denser than Earth's, in order to be equally flyable. Thus the lunar flying area might need air too thick for humans and they might need breathing apparatus to fly by flapping their wings.

Improve this answer
$\endgroup$
1
  • $\begingroup$ Another big difference between Titan and the Moon is atmospheric density: if you use an Earthlike environment at the Moon, that's 1.25 kg/m^3; at Titan it's ~5.5 kg/m^3, and that makes a big difference. I did a back-of-envelope estimate of the power to maintain flight at Titan (not take off!), with system mass (human + wings) = 100 kg, Cd = 2 (humans aren't very streamlined!), L/D = 5 (worse even than a Cessna), projected area (human + wings) = 0.5 m^2, and got ~85 Watts at 7 MPH, not a lot of effort. The same system at the Moon takes ~235 Watts, 16 MPH, doable but considerably more work. $\endgroup$
    – Tom Spilker
    Commented May 2, 2018 at 22:58

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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