Measures of the vertical and horizontal vectors of athletes running shows that once they get up to speed the ratio of vertical to horizontal exertion is about 4:1. This paper by Jean-Benoit Morin Ph.D. gave data using the latest measurement techniques:

force vectors during sprinting

So, imagining that on the Moon, it would take you 1/6 of the force you need to apply vertically on the Earth to achieve the same vertical vector. Then the question is how efficiently you can apply force horizontally - how much of the effort you save on the vertical element you can put into the horizontal element. This must imply a different running stance, which implies an entirely different biomechanical model.

Let's assume this can be done fairly efficiently, albeit with a very different running stance which may be supported by the use of different gear - different shoes, perhaps poles somewhat analogous to ski poles.

Air friction needs to be considered, for a model for within a habitat . I've looked at the formula $F_D\, =\, \tfrac12\, \rho\, v^2\, C_D\, A$ . I don't know how to properly use this formula for this case, though I've looked up the figures. Indoors, in fairness a lower air pressure than on Earth would likely be chosen. Let us say half the air pressure with twice the oxygen.

Traction would be an issue. How does one incorporate that into a calculation? Also, if the environment was designed for running, some solution for lack of traction could surely be found through the use of a custom surface and shoes with cleats.

Put together, what top speeds could a person achieve INDOORS in 0.5 atmospheres, and OUTDOORS, in a vacuum?

  • 1
    $\begingroup$ Couldn't this be answered as a simple energy question? $\endgroup$
    – ThePlanMan
    Commented Dec 31, 2014 at 9:59
  • 2
    $\begingroup$ Scientists are researching (also here) on this topic. $\endgroup$
    – mins
    Commented Dec 31, 2014 at 11:53
  • 4
    $\begingroup$ Seems related to this question on Physics:SE "What is our estimated running speed on Moon's surface?" $\endgroup$
    – mins
    Commented Dec 31, 2014 at 12:13
  • 1
    $\begingroup$ @mins I took a look, especially at the 3rd answer. I doubt a jumping technique would be the best solution. The longer you are in contact with the ground, the longer you can apply acceleration, which also means you can get more acceleration with a given coefficient of friction. The less of your force is wasted on the vertical vector, the more can go to the horizontal vector. The videos linked to in the 2nd answer are interesting, but the subject had to deal with the restrictions of the test format. I am going to edit the question so it seeks an answer that assumes the situation proposed. $\endgroup$
    – kim holder
    Commented Dec 31, 2014 at 16:01
  • $\begingroup$ The subject in the video has to "lean in " a lot to achieve his horizontal acceleration. I doubt he can go much faster. And this was apparently on a solid flat surface. On a dust-covered lunar surface, he would likely slip. $\endgroup$
    – oefe
    Commented Jan 2, 2015 at 18:51

2 Answers 2


Because the biomechanics of running would change a lot in the moon environment, any answer is a preliminary one, pending actual experimentation on the moon. :D

That said, the indoor case can still be calculated pretty accurately due to the fact air resistance is by far the most important factor. This paper calculated that Usain Bolt was using 92.5% of his effort to overcome drag when he was at his top speed of 12.2 m/s, or 44 km/h. Since the force required to overcome drag increases with the cube of velocity, even in 0.5 atmospheres he could only go slightly faster.

The case in a vacuum depends entirely on how much of the savings on vertical force can be applied as horizontal force, and how well a runner can manage the increased speed. The graph in the question suggests that if all the saved vertical effort could be applied horizontally, that force would increase by a factor of 3.3. Bolt would then be able to run at 40 m/s, or 144 km/h. You may already see the problem, but before getting into that, let us note that since there is no air resistance, if there wasn't that other problem, he could run even faster. Nothing is slowing his forward momentum, and he only needs to apply a small fraction of his force to overcome gravity, so as long as he could continue to apply the remaining force against the ground, he could continue to accelerate.

But he has to be able to swing his leg faster than the ground is already passing under him in order to increase his speed. The force that represents the difference between how fast he can swing his leg, and how fast he is already moving, would be applied as acceleration.

I took a little video of me swinging my leg as fast as i could, just above the ground, above a measuring tape. It took very close to two frames of the video for my foot to cover 50 cm. That is a speed of 15 m/s, 54 km/h. My leg measures 88 cm. I can't find a figure for Bolt's legs, but he is 13 inches taller than me and has a much longer leg relative to the length of his torso. I'm going to estimate his leg is 108 cm. If he can also swing his leg through the air at my top speed, he would be moving it at 66 km/h. He can probably do it faster, but maybe not by much since this is not a simple question of strength. I'll call it 70 km/h.

If he is moving forward faster than he can move his legs, he is going to fall. But moving at the top speed his legs can move is only going to occupy about 60 % of his strength. It seems reasonable to think that he could do that despite the biomechanical problems. For one thing, he could slightly increase the force he is applying vertically so that he is hopping high enough that he has time to position himself exactly how he wants for the next push against the ground. Also, one could surely maintain a pace that only requires 60 % of your strength for a long time. You could probably run a whole marathon at your top speed.

His main problem would be learning not to overshoot the speed his legs can handle, and having adequate protection for when he does. Falling at over 70 km/h is no fun.

(Notes - This would be on the hard-packed lunar soil 10 to 15 cm under the surface - a track would have to be prepared. It is assumed that through the use of adequate cleats the lack of traction due to low weight could be overcome. Matters not covered: how to turn.)

  • $\begingroup$ When swinging your leg as fast as you could, wouldn't 92.5% of the energy be lost to drag? So in a vacuum wouldn't you be able to swing it faster? $\endgroup$
    – ThePlanMan
    Commented Feb 8, 2015 at 17:33
  • $\begingroup$ Swinging your leg as fast as possible once is easier than maintaining a rhythm. Try running on a treadmill and cranking up the speed until you can't keep up (and try not to faceplant). Youtube has various videos of people doing this. $\endgroup$
    – Hobbes
    Commented Feb 10, 2015 at 9:01
  • $\begingroup$ @FraserOfSmeg I thought about that, but i didn't feel any sense of drag... maybe because of my leg's profile? $\endgroup$
    – kim holder
    Commented Feb 11, 2015 at 15:35
  • 1
    $\begingroup$ You can work out how fast Usain Bolt moves his legs, because each leg is moving half the time. So his legs actually each moved 100 meters in 4.79 secs. That's 75.16 km / hour. He can probably swing his legs a bit faster than that as that also includes stopping and starting at the end of each stride. If this way of thinking about it is right, then good athletes in superb spacesuits on the Moon could go a bit faster than double their speed on Earth.. $\endgroup$ Commented Apr 14, 2016 at 16:35
  • 2
    $\begingroup$ If you assume you are in vacuum, you are either dead or with a spacesuit. In both cases, I think it is difficult for you to move your legs at 70 km/h ;) $\endgroup$
    – frarugi87
    Commented Apr 14, 2017 at 7:26

This is a 3 part question imo.

Part 1: Force

Based on the graphic in the question it looks (as you've surmised) as if the steady state top speed is based on avoiding the ground more than overcoming drag. So the question then becomes do large hops lead to higher speed? Intuitively I'd guess no. Larger hopes mean less time in contact with the ground and hence less power (energy per second) going into forward motion. You want to be in the air long enough for your leg to get in position for the next step, no longer. So you'd probably want a similar step frequency as that on Earth (looks like 4hz-ish). This means you want less energy going into upward motion and hence more going into forward motion. The 2000N vertical force in the graphic could be reduced to 1/6th of the value (333N) and the remaining 1667N could be put into forward motion.

Part 2: Drag The drag equation you've stated is the appropriate equation. We can even calculate the drag experienced by the runner in the graphic easily from equilibrium conditions. The running speed is approximately constant towards the end of the graphic, with a horizontal force of about 500N. Therefore the drag must be equivalent. If we can estimate the horizontal force per time this should give a value of drag per time. To me it looks like it about 500N for about a quarter of the period of one step. So lets say an average of 125N/step. We said earlier there's about 4 steps per second so that's about 500N per second. We now know our drag must be 500N each second.

If we look at the drag equation we can make a few assumptions. First lets say the drag area ($A$) of the runner stays the same. Lets also say the coefficient of drag ($C_d$) stays the same (this may not quite be the case). We know the 1/2 stays the same. So we're left with the $V$ and the $ \rho\ $. On Earth, at sea level $ \rho\ $ is approximately $1.225\text{ kg/m}^3$. The top speed of the Earth runner looks like its around $6.5\text{ m/s}$. We can now get a rough value for our $C_d \times A$. $$C_d \times A = F/( \rho\ \times V^2 \times 1/2) = 19.3$$ (units here would be $\text{m}^2$ but they're somewhat meaningless).

Part 3: Lunar air density You've said the habitat would have a lower air pressure than Earth, but you've not specified a value. Lets assume an ISS style habitat, air pressure of 101.3kpa. The air density can be calculated from $\text{rho} = p /(R \times T)$ where $R$ is specific gas constant and $T$ is temperature (Kelvin). So we have a density of $1.17\text{ kg/m}^3$.

The results:

Based on the above it looks as if running on the moon will have a similar drag profile to that on Earth (at least in my chosen habitat). So it becomes a question of calculating the equilibrium where the horizontal force (of 500 + 1667 = 2167N) is equal to the drag force. So simply $$V = \sqrt{2167/(1.17 \times 1/2 \times 19.3)} = *13.85\text{ m/s}*$$

This is based on a lot of estimations though, so anywhere from 10 - 15 m/s wouldn't be unreasonable.

  • 1
    $\begingroup$ I've enjoyed your answer, but there is a flaw, unfortunately - the graph in the question comes from a study that used a specially adapted treadmill to make measurements. There was no drag. And the next thought is - hey, yeah, why did the force level off? I'm tempted to conclude the runner (who was a world-class sprinter) simply couldn't move his legs faster. Which would then give a separate way of answering the question. Obviously if your forward motion is faster than you can move your legs, you are about to fall. That would then be the limiting factor. Hah! $\endgroup$
    – kim holder
    Commented Feb 7, 2015 at 16:24
  • $\begingroup$ @briligg dam it, I should have flicked through the paper you linked to! No drag does cause quite a major flaw in my calculations. I'd guess it's probably the maximum speed of the runners legs that are the limiting factor here then as you've said. $\endgroup$
    – ThePlanMan
    Commented Feb 7, 2015 at 16:36
  • $\begingroup$ The question now does state "Let us say half the air pressure with twice the oxygen.". $\endgroup$
    – user
    Commented Feb 9, 2015 at 9:56

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.