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Old 11-28-2021, 02:35 PM   #6
Anthony
 
Join Date: Feb 2005
Location: Berkeley, CA
Default Re: vehicles on Mars: lifting capacity

Interesting, it turns out that pretty much every variable other than molar weight can be ignored when figuring out the amount of payload a given quantity of lifting gas can lift.

Flotation boils down to one simple rule: if the weight of the fluid displaced by a vehicle equals the weight of the vehicle, the vehicle will hover; if the weight of the vehicle is less it rises, if greater it descends.

Since weight = mass*gravity, for both the vehicle and the fluid, it works out that, outside of the zero gravity case, you can replace weight with mass.

Now, according to the Ideal Gas Law, PV=nRT. This applies to both the lifting gas and the atmosphere, and therefore every mol of lifting gas displaces one mol of atmosphere; for a hydrogen balloon in a CO2 atmosphere, this means 2g of hydrogen displaces 44g of CO2, and your net lift is 21*(total mass of hydrogen). This does not actually depend on exterior temperature or pressure, unless you reach temperature where you need to treat those gases as non-ideal.

Unfortunately, GURPS Mars doesn't seem to give us mass, it gives us balloon volume. However, we can still use the same law: PV=nRT, or n=PV/RT.

P = 1200Pa (in your hypothetical mars)
V = 9.5 million cf = 269,000m^3
R = 8.314 m^3⋅Pa⋅K^−1⋅mol^−1
T = 275K? (you haven't defined it for us, let's try slightly above freezing)
n = 1200Pa * 260,000m^3 / (8.314 m^3⋅Pa⋅K^−1⋅mol^−1 * 275K)
Canceling out units
n = 1200 * 260,000 / (8.314 mol^−1 * 275) = 136,000mol.
136,000 mol of hydrogen gas = 274 kg.
136,000 mol of CO2 gas = 5984 kg.
Net lift = 5710 kg or 13,180 lbm. At 0.38g, that works out to 5,000 lbf.

Note that storage and production of lifting gas should both be dependent on the number of mols, not some theoretical volume.
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