vehicles on Mars: lifting capacity

[whswhs]

One of the vehicle written up in GURPS Mars is a Martian blimp. However, it statistics are for Mars as it is now, and the future Mars of my campaign has undergone early terraforming, increasing its atmospheric mass of carbon dioxide and thus raising its air pressure. I wanted to figure out what its lift would be under those conditions, so I took a look at data on Mars—and found that, whereas Mars's atmospheric pressure in about 0.6% of Earth's, GURPS Mars says that one cubic foot of hydrogen at Earth pressure would occupy 30 cf at Martian pressure. That doesn't seem right: PV = P'V', and the inverse of 0.006 is 167, not 30!

After some thrashing around, I decided I needed to work out the lift factor from first principles, as the differences in weights between a volume of hydrogen and a volume of Martian atmosphere.

So, to begin with, since PV = P'V', and since P' for my fictional Mars is twice that of real Mars, or 0.012 atmospheres, V' must be 83.3. That is, any given number of molecules must occupy 83.3x as much volume on Mars as on Earth. One cubic foot of Earth air turns into 83.3 cf of Earth air on Mars.

What's the mass of that volume of air? Dry air on Earth has a mass of 1.29 kg per cubic meter. The mean molecular weight of dry air is 28.97. That of hydrogen is 2.016, which gives a mass of 0.09 kg. I don't have a figure for Mars's atmosphere, but it's 95% carbon dioxide, and (by assumption) all the extra gas that goes into doubling the pressure is also carbon dioxide, raising that to 97.5%; treating it as pure carbon dioxide will come fairly close to the correct figure. Carbon dioxide has a molecular weight of 44.09, so its mass is 1.96 kg. That takes up one cubic meter on Earth, but 83.3 cubic meters on Mars. A cubic meter is 35.3 cubic feet, so 83.3 cubic meters is ~2942 cubic feet. To avoid working with ridiculously small figures, I'm going to figure masses per million cf: 30.6 kg for hydrogen and 666.2 kg for carbon dioxide.

How much do those masses weigh? On Earth, a kilogram mass weighs 2.205 pounds. Mars's gravity is 0.3794g. Taking 30.6 x 2.205 x 0.3794 gives 25.6 pounds for hydrogen; taking 666.2 x 2.205 x 0.3794 gives 557.3 pounds for carbon dioxide. The difference is 531.7 pounds (of Martian weight, of course). Dividing this into a million, I find that it takes 1880 cubic feet of hydrogen to lift one pound against Martian gravity. (For cost purposes, that's the equivalent of 22.6 cubic feet on Earth.)

The 9.5M cf lifting gas in GURPS Mars will lift a bit over 5000 pounds on Mars—if I've done the calculation right. Anyone want to check me?

If I'm right, then since the Mars blimp has a loaded weight of 1.85 tons, or 3700 pounds, it ought to need 7M cf of lifting gas.

[whswhs]

Addendum: I've looked at the idea of using a vacuum balloon; Mars's pressure of 0.012 atm is low enough so a fairly light structure can resist being crushed. It looks like it takes 1794 cf of vacuum to lift one pound. That's only a modest saving, and might well be negated by the need for DR 1 of rigid armor of some sort, which is extra weight.

On the other hand, to reuse a hydrogen airship, you need to electrolyze water to get hydrogen. To reuse a vacuum airship, you just need to pump the (Martian) air out of it, which is a fairly simple mechanical process.

On the other other hand, it won't be a "blimp"; it will need a rigid structure. That loses the ability to store it compactly. I think it's probably not worth it.

[Anthony]

You can ignore gravity in computations about lifting gas, since it affects payload weight in exactly the same way as it affects lifting ability.

As for vacuum balloons: any time a design sequence says vacuum balloons are useful, it means you've found an error in the design sequence.

[whswhs]

Originally Posted by Anthony

You can ignore gravity in computations about lifting gas, since it affects payload weight in exactly the same way as it affects lifting ability.

As for vacuum balloons: any time a design sequence says vacuum balloons are useful, it means you've found an error in the design sequence.

I ran across a page on a NASA Web site that proposed them, so I thought it worth looking at.

As for ignoring gravity, that seemed intuitively right, but I wanted to go back to first principles, by comparing the weights of the lifting gas, the equivalent volume of atmospheric gas, and the payload. My intuition isn't infallible, after all.

[johndallman]

Originally Posted by whswhs

I ran across a page on a NASA Web site that proposed them, so I thought it worth looking at.

Sadly, that is not reliable. There is, for example, a NASA team that's reported positively on the EmDrive.

[Anthony]

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.

[whswhs]

Originally Posted by Anthony

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.

Indeed, and that was what I relied on in figuring the relative weights of hydrogen and the hypothetical Martian atmosphere. You'll note that my answer for 9.5M cf of lifting gas was "a bit over 5000 pounds," which is quite close to your answer of 5000 pounds exactly, so we're working on the same basic logic and getting comparable results.

I would attribute the discrepancy to my not following your suggestion that "storage and production of lifting gas should both be dependent on the number of mols, not some theoretical volume." GURPS doesn't use this approach; it uses a fixed number of cubic feet of various lifting gases per pound of lift, disregarding differences in temperature. Since I'm looking for an order-of-magnitude answer to "Can this thing get off the ground?" I went along with that as a gamable abstraction. I don't plan to be looking up exact altitudes, figuring exact local pressure, estimating exact local temperature, and so on during play; rather, I'm going to assume that managing gas volume and ballast falls under the skill of Piloting (Lighter-than-Air, Mars).

For what it's worth, my estimate was that artificial greenhouse gases had given Mars a mean temperature that worked out to a few degrees below freezing (at least by Earth standards of freezing; Mars's lower pressure may change that). The difference is probably not great enough to change your result substantially.

[whswhs]

Here's another aspect of lifting capacity: GURPS Vehicles says that the best helicopter drivetrain has lift of 12 lbs. per kW, and weight of 0.3 lbs. per kW + 15 lbs. But that's on Earth, at 1 atm! What happens on Mars, at 0.012 atm?

The power required to move air at a given velocity v is proportional to the density of the air, and to the cube of the velocity. What's the density of Mars's atmosphere? Its volume is 83.3 times that of Earth's atmosphere. Its molar weight is that of carbon dioxide, 44.09, as opposed to that of air, 28.97. So I get a density of 0.018. That gives a velocity of 3.8x that for a similar system on Earth.

Since power is dimensionally equal to force x velocity, I can divide lift (vertical thrust) by that same factor of 3.8 to get 3.16 pounds of lift per kW.

A 10-kW system will have lift of 31.6 lbs., and will weigh 18 lbs. A 100-kW system will have lift of 316 lbs., and will weigh 45 lbs.

An advanced battery weighing 13.6 pounds would store 13,600 kJ, enough for 1360 seconds of flight, or just over 20 minutes. One weighing 271 lbs. would be sufficient for 2710 seconds of flight, or about 45 minutes. When you factor in structure, (minimal) armor, and control systems, it looks as if even a simple robotic drone with no payload would be marginal at best.

On the other hand, I'm a little surprised to find that it looks like even a marginal possibility.

[Anthony]

Originally Posted by whswhs

Here's another aspect of lifting capacity: GURPS Vehicles says that the best helicopter drivetrain has lift of 12 lbs. per kW, and weight of 0.3 lbs. per kW + 15 lbs. But that's on Earth, at 1 atm! What happens on Mars, at 0.012 atm?

The approximation is even more wrong than it is on Earth, as helicopter lift isn't linear in power to start with -- it scales with the 2/3 power of rotor power, the 1/3 power of rotor area, and the 1/3 power of atmospheric density.

I would point out Mars Ingenuity.

[whswhs]

Originally Posted by Anthony

The approximation is even more wrong than it is on Earth, as helicopter lift isn't linear in power to start with -- it scales with the 2/3 power of rotor power, the 1/3 power of rotor area, and the 1/3 power of atmospheric density.

It seems to me that I just scaled it with the 1/3 power of atmospheric density.

I was pretty sure the formulae in Vehicles were oversimplified; they seem to be approximation some sort of curve with, first, a simple proportionality, and second, a linear equation with a nonzero intercept. But I'm not knowledgeable enough about vehicular engineering to come up with more accurate formulae.