How will hovercraft work on Mars?

The facts are:

1. On Mars atmosphere pressure is way much lower than on Earth.

2. To hover hovercraft blows air under itself to create air cushion. This air cushion as I understand must have enough pressure to lift hovercraft and in the same time this pressure must be higher than atmosphere pressure to create lifting force.

My specific question is: given the pressure on Mars and on Earth will the same model of hovercraft need more or less "engine power" to be lifted than on Earth? By engine power I mean power of air blowing engine not propulsion. I'm deducing that less engine power than on Earth but please give some calculation example. Also please correct if I made any mistake above:).

Parameters

The surface gravity of Mars is ~0.376 g, where g ~ 9.81 $m/s^{2}$ for Earth. The surface pressure of the atmosphere on Mars is ~0.636 kPa, which is roughly 0.63% of Earth's atmospheric pressure (i.e., ~101.325 kPa at sea level). The density of air at STP on Earth is ~1.2 $kg/m^{3}$, compared to Mars at ~0.020 $kg/m^{3}$.

Background

Typical hovercrafts make use of an impeller, or a type of axial fan. We will assume the hovercraft's altitude is in steady state, and the only differences are in the atmospheric pressure and density.

The thrust force magnitude (assume one dimensional for now) is just given by: $$F = \frac{ 1 }{ 2 } \rho \ A_{disc} \ \left( C_{in}^{2} - C_{out}^{2} \right) = \frac{ 1 }{ 2 } \rho \ A_{disc} \ \Delta C^{2}$$ where $\rho$ is the atmospheric density, $A_{disc}$ is the area of the propeller disc, $C_{in}$ is the inflow speed at air intake, and $C_{out}$ is the outflow speed at the exhaust.

Application

If everything were equal, then the thrust ratios between Earth and Mars would just be the ratio of their respective atmospheric densities (Note: $\rho_{E}/\rho_{M}$ ~ 60). The force of gravity ratio is $F_{gE}/F_{gM}$ ~ 2.67. Thus, ratio of thrust-to-weight ratios is $ratios_{E}/ratios_{M}$ ~ 22.6.

Since we cannot change $\rho_{M}$, we must increase either $A_{disc}$ or $\Delta C^{2}$ by a factor of ~ 22.6 to get an equivalent performance.

Caveats/Notes

Things I did not account for include, but are not limited to:

• differences in lift and drag on the fan blades that would arise due to the difference in pressure and density between the two atmospheres
• the efficiency of the fan
• shape of fan blades and its affect on the engine performance
• effect of thin atmosphere on fuel combustion
• etc.

This is more of an engineering question than physics, I think.

Anyway, think about it in terms of force. To hover, you have to provide a downward force to lift the craft upwards, then hold it at, say a metre above the ground. Force is mass by acceleration, and mass is density by volume. As the atmospheric density on Mars is very low, it seems unlikely, even though gravity is reduced compared to Earth, that hovercraft would be used on Mars.

1. The engine would need to be more powerful than on Earth, to spin the impeller fan faster than on Earth, so as to pull in more low density "air" per unit time.
2. The impeller fan would have to be much greater in diameter compared to Earth fans to generate sufficient volume in a given time, to increase the effective mass of working fluid sufficiently to generate any reasonable downforce. A larger impeller implies more engine power would be required.
• Yes, I was also considering that engineering aspect is vital and I agree that process of blowing air on Mars requires more engine power than on Earth. But what about pressure limits for creating air cushion? Is it more efficient on Earth or on Mars? By doing maths: size of hovercraft doesn't change (Area A is the same), force (F) is lower by factor 0.38 then required pressure (P=F/A) is also lower by factor 0.38. If what I wrote before is correct then on Earth blowed pressure must be higher than average 101 000 Pa and on Mars higher than average 600 Pa (I exclude riding on Olympus Mons;)) – Rafal Rusek Nov 9 '15 at 0:18
• To sum up. You must blow stronger but you have to achieve pressure just 0,6% of that on Earth. So overall is it more or less efficient? - Yes, it depends then on mass of hovercraft. Am I right? Small one-man hovercraft made from light materials should require less than 600 Pa, right? – Rafal Rusek Nov 9 '15 at 0:29
• Rafal, The pressure you have to achieve on Mars is not 0.6% of that on earth but $m_{craft}∗g_{Mars}/S_{craft}+0.006*1atm$ because, in order to float the craft, its weight on the red planet plus the atmospheric pressure force that pushes the hovercraft down have to be canceled by the force generated by the pressure of the gas under the hovercraft. – Energizer777 Nov 9 '15 at 3:03
• @Energizer777 So hovercraft that has for example 500 kg on Earth and bottom area of 10 m2 will need pressure on Earth = (500*9,8/10)+101325 = 101815 Pa and then on Mars = (500*3,71/10)+636= 821,5 Pa, right? Then for this particular hovercraft question about better blowing engine efficiency is if it is more efficient to blow 101815 Pa in 1 atm environment or 821,5 Pa in 0,006 atm environment, right? And here engineering points as described by count_to_10 above comes into play. – Rafal Rusek Nov 9 '15 at 8:04
• What you have to blow on earth is just $(500 kg*9.81 m/s^2/10m^2) = 490.5 Pa$. The rest is blown by the atmospheric pressure. – Energizer777 Nov 9 '15 at 18:24