Mileage on Mars - Effect of Gravity on Mileage

[MattRob]

Hello,
For a little project I'm doing for fun, I want to know what kind of mileage would be realistic for a light, minimalistic lunar-rover type vehicle that seats six.

Using Wiki, I've found the energy density per kg of methane at 1 bar at 15*C. Unable to find it for liquid methane (my fuel of choice), I've just decided I don't need very much precision (just getting a ballpark sort of estimate), so that would do, and I ignored the loss of energy density from the gas to liquid phase, and from there I've figured that liquid methane has ~70% the energy density as liquid gasoline per volume.

First question; is that going to be a huge error, or just a small one?

But the next step has me stumped; how would the lower gravity effect mileage? Would it be proportionally greater, something more complicated, or make no difference?

 

[TheAbsoluTurk]

This video may interest you:
http://m.youtube.com/watch?v=9cUIhVEJEaM&feature=g-subs-u

 

[CWatters]

But the next step has me stumped; how would the lower gravity effect mileage? Would it be proportionally greater, something more complicated, or make no difference?

The mass will be the same. So (ignoring forces like drag and rolling resistance) the same amount of energy is required to reach a given velocity V (eg KE = 0.5mv^2). If the car doesn't have regenerative braking then that energy is lost when slowing down.

As for climbing hills... The weight will be lower so the energy needed to get to the top of a hill will be lower (eg PE = mg_marsh). Again without regenerative braking that's lost on the way down.

Atmospheric pressure is a lot lower on Mars but I suspect you won't be driving around at high speed so the difference in air resistance might not be a big factor?

What I don't know is how the rolling resistance on soft ground depends on g. I suspect the way to view this is that on soft ground a car is effectivly climbing out of it's own hole all the time. So that would scale with g ?

So overall you will get better mpg over the same terrain on Mars but by how much I wouldn't like to say.

 

[CWatters]

If you do plan on driving fast perhaps check that your brakes can dissipate the heat in the thinner atmosphere?

 

[Vanadium 50]

How exactly do you plan to burn the methane?

 

[MattRob]

This video may interest you:
http://m.youtube.com/watch?v=9cUIhVEJEaM&feature=g-subs-u

Interesting, but not exactly what I was going for...

The mass will be the same. So (ignoring forces like drag and rolling resistance) the same amount of energy is required to reach a given velocity V (eg KE = 0.5mv^2). If the car doesn't have regenerative braking then that energy is lost when slowing down.

As for climbing hills... The weight will be lower so the energy needed to get to the top of a hill will be lower (eg PE = mg_marsh). Again without regenerative braking that's lost on the way down.

Atmospheric pressure is a lot lower on Mars but I suspect you won't be driving around at high speed so the difference in air resistance might not be a big factor?

What I don't know is how the rolling resistance on soft ground depends on g. I suspect the way to view this is that on soft ground a car is effectivly climbing out of it's own hole all the time. So that would scale with g ?

So overall you will get better mpg over the same terrain on Mars but by how much I wouldn't like to say.

My real question here, I guess, is where drag force comes from on level ground, aerodynamics aside (since it's nearly a vacuum on Mars). I guess the friction would mostly come from the machinery of the engine, wheels, etc (no axle - the system I was thinking of would have an electric drive system powered by a solar panel and a methane/oxygen generator. The idea is, it's primarily driven by methane fuel, but if for some reason that ever runs dry, a solar array provides backup so the vehicle can still possibly inch along if need be. Plus, an electric system would be easier to repair in-field if the undercarriage scrapes rock or is bent than a heavy-built axle, etc).

Assuming it all comes from machinery, I know friction force is normal force multiplied by friction coefficient, creating drag on the moving parts. By that alone, however, they would be frictionless in zero-g, which simply isn't the case; close-fit parts are simply in physical contact with each other, thus, friction from that, too. So it's not entirely proportional to weight, either...

Some of the friction comes from that normal force, some of it just comes from the fact that the parts in are physical contact... So I guess all this is rather complex.

How exactly do you plan to burn the methane?

Carrying a tank of LOX to mix it with, of course, then feed it in a generator that produces electrical power to drive the vehicle (funny enough it's electric but I don't think a regenerative braking system would be worth the extra weight and complexity. To save power, just don't brake when unnecessary, and being a system designed for aerospace and Mars, I think the KISS principle applies more than ever, since it's going to necessarily be rather complicated as it is...)

 

[Vanadium 50]

OK, then you will immediately have a factor of 3 loss in mileage because you need ~2 gallons of LOX per gallon of fuel.

 

[CWatters]

Putting aside the extra weight of the LOX I can't see the range getting worse on mars. Perhaps NASA just test it on Earth and consider any extra they get on Mars to be a bonus. I wonder if it's possible to ask them?