7
$\begingroup$

I'm curious about how fast a large, enclosed, manned lunar rover could feasibly travel. Toyota hasn't said yet what the top speed will be for the one that they're developing for JAXA, but that's the type of rover I'm thinking about.

I'm fairly sure it will be faster than NASA's Space Exploration Vehicle, which supposedly has a max speed of 6 mph.

So, how fast do you think a rover such as the one Toyota is designing could go?

$\endgroup$
4
  • $\begingroup$ much faster for Interplanetary Superhighway driving than for local runs $\endgroup$
    – uhoh
    Sep 30 '20 at 6:16
  • 1
    $\begingroup$ Top speed is irrelevant. How fast you can brake matters. Although given the uneven terrain, I suppose you might not want to achieve a speed at which a good-sized rock would deflect you into orbit $\endgroup$ Oct 1 '20 at 11:48
  • 1
    $\begingroup$ @CarlWitthoft Top speed isn't irrelevant to me. It's specifically what I'm asking about. I agree with the rest of what you said. $\endgroup$
    – Arbutus
    Oct 1 '20 at 17:01
  • $\begingroup$ OK, well, since there's no air resistance, so long as you stay at a constant altitude, or to be exact an iso-gravity line, you can hit any speed you want so long as the drive train can support the RPMs. That's simple physics: keep applying torque, you keep accelerating. $\endgroup$ Oct 1 '20 at 18:02
2
$\begingroup$

System Modeling

As a first approximation, we can assume traction capability is roughly proportional to flat-surface friction. For non-slipping wheels,

  1. Max Friction Force = Normal Force * Coefficient of Static Friction.

On the moon, surface gravity is 16.5% of Earth surface gravity. So, normal force would be 16.5% of an equivalent Earth value.

In a maximum performance turn of a wheeled vehicle, the maximum friction force of the wheel is used to accelerate the vehicle in circular motion.

  1. Centripetal acceleration = velocity^2 / turn radius.

Let's use maximum performance turn radius as a gravity-independent metric for the "safety" of a speed. For a certain rover of known mass and wheel/surface coefficient of friction, we can calculate the maximum speed for a target turn radius by (1) finding maximum acceleration due to friction and (2) solving for velocity.

Example

A dune buggy on Earth can be driven safely over rocky sand at 30 mph (guess).

On the Moon, there is only 16.5% of the available acceleration due to wheel friction because there is 16.5% of the gravity causing normal force.

So, for the same turning radius,

  1. velocity = sqrt(centripetal acceleration * turn radius)

A 0.165x multiplier on acceleration corresponds to a 0.406x multiplier on velocity.

The maximum safe speed of that dune buggy on similar surface on the moon is 12.2 mph.

Effects not modeled

  • Non-flat terrain
  • Lunar surface regolith having different properties than Earth regolith
  • Increased rollover risk due to lower gravity
  • Extreme consequences from collisions
$\endgroup$
2
  • $\begingroup$ Interesting. So, all vehicles would have a 0.406x multiplier on velocity, yes? But, as you've listed, there are other factors to take into account. What if there were graded tracks that served as roads? Do you have an idea of what speed a vehicle such as the JAXA Toyota rover would likely travel on such a track? Not trying to go as fast as possible, but rather, travelling at a normal speed for such a large rover on such a track. I know I'm asking for a specific speed and there are lots of unknown variables involved regarding the design of such a rover, so apologies. $\endgroup$
    – Arbutus
    Oct 15 '20 at 1:58
  • 1
    $\begingroup$ The MMSEV test vehicle from Johnson Space Center has been driven on public roads, but I don’t remember off the top of my head its maximum speed. I’m guessing 30-40 mph. Improved road surfaces improve the coefficient of friction and make the surface more regular, so you could probably trust that 0.406 number more for an improved road surface. The MMSEV has lots of its heavy components low in its body, and I think I remember it has a tilt tolerance of at least 30 degrees, maybe more like 45. So, rollover turn acceleration might not be lower than wheel slip turn acceleration? $\endgroup$ Oct 15 '20 at 2:16

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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