classical mechanics - 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:).

Answer

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 $kg/m^{3}$.

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.

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.

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.