The physics of flight using aerodynamic lift in extraterrestrial atmospheres

[Nilof]

This thread is largely inspired by the fact that we have a helicopter currently on its way to Mars. As a result, it feels appropriate to have a general thread for this. Feel free to insert any observations or start any new discussion within the topic.

The most surprising result to me is probably that lift at a fixed mach ratio is independent of atmospheric density, as long as we are still in the regime where the atmosphere is dense enough that it can be viewed as a fluid and that we can use thermodynamics & fluid dynamics while neglecting the corrections from statistical mechanics. One way to see this is to write:

Lift = (L/D) * Drag = (L/D) * (C_d A / 2) * ρ v^2 = (L/D) * (C_d A / 2) * M^2 * ρ c^2 = (L/D) * (C_d A / 2) * M^2 * γ

Where in the last step we have used Laplace's equation, and the previous steps were just the definitions of the lift to drag ratio, the drag coefficient, and the mach ratio.

As such, to first order the atmospheric density cancels out, and explicit dependence on it is exclusively from higher order effects as we may see differences in Lift/Drag ratio, the drag coefficient, or the specific heat ratios as a result changes in pressure, temperature, and chemical composition. Which are fundamentally not going to cause order-of-magnitude differences when the mach ratio is held constant, especially in the high Reynolds number limit.

So effectively, for Mars, it looks like level flight using electric motors is actually easier in some ways, since the lower gravity reduces lift requirements, so we just end up going faster and further for a given amount of work applied to push the craft forward. The main actual issue with low atmospheric densities is the higher takeoff & landing speed requirements for winged craft, which are partly offset by the lower gravity.

[Genial Precis]

Cross-posted from "Industrial consumables" thread in Missions to Mars subforum:

OK, if I want to make this proposal more reasonable, let me make this suggestion:

On Mars you have to get up to ridiculous speeds just to take off. Basically scale earth speeds by a factor of 10. So the aircraft which can take off subsonic are still quite lightweight. Anything that's supposed to carry useful cargo will be moving at mach several in order to generate enough lift, and the engines for such a beast will be interesting and expensive.

On the other hand, if they've got a decent glide ratio and you need to fire them at a couple km/s just to get off the ground, give up on storing and using energy in the aircraft itself and make it a glider. It's probably a lot cheaper to do it that way, since you don't need engines, and since the aircraft is supposed to be heavy relative to its size in order to carry useful cargo, it'll have such a high takeoff speed that it'll have plenty of range just from the stored energy of launch.

My baseline notion of this is a 5 km electromagnetic catapult launching gliders 1 km/s after 10 s of 10g acceleration. If the catapult is fed at 10% efficiency by a dedicated 1 GWe power plant it can launch 20 kton/day, or a 40 ton glider/container every 3 min. If I assume a glide ratio of 7 (concorde at mach 2), the range might be 500 km.

The receiving end should have multiple independent runways with redundant arresting gear, since the gliders will be coming in very hot and there's definitely not enough air resistance or ground traction to slow them down in a useful distance. Skids instead of wheels, or something, because I've got no idea how to make a tire work at 200 m/s. It was a real problem on the Concorde.

I'm skeptical that a 500 km rail link wouldn't be more practical, but that's my sketch design for Mars air travel. It's got some interesting features, I think. Like the bit about not equipping your aircraft with engines because they need so much stored kinetic energy just to get off the ground.

[sebk]

Lift = (L/D) * Drag = (L/D) * (C_d A / 2) * ρ v^2 = (L/D) * (C_d A / 2) * M^2 * ρ c^2 = (L/D) * (C_d A / 2) * M^2 * γ

This formula fails to take into account vehicle weight. Yes on Mars you could have similar lift to drag ratio than on the Earth. But since drag is tiny so is the lift. To fly on Mars you need very large lifting surfaces or very high speed to generate decent amount of both lift and drag. The problem is such surfaces are not weightless and high speeds are unwieldy.

[meberbs]

Lift = (L/D) * Drag = (L/D) * (C_d A / 2) * ρ v^2 = (L/D) * (C_d A / 2) * M^2 * ρ c^2 = (L/D) * (C_d A / 2) * M^2 * γ

This formula fails to take into account vehicle weight. Yes on Mars you could have similar lift to drag ratio than on the Earth. But since drag is tiny so is the lift. To fly on Mars you need very large lifting surfaces or very high speed to generate decent amount of both lift and drag. The problem is such surfaces are not weightless and high speeds are unwieldy.

You seem to be misreading the equation. It does say that more speed is required, but it points out that this more speed corresponds to the same Mach number, same area and same shape. (gamma would be different, not sure what it is on Mars) Since the Mach number is the same, the more speed isn't particularly more unwieldy.

That is not to say that the higher speeds cause no problems, but from just the aerodynamics, rather than structural issues or take-off and landing, there is no need for a change. Since gravity is lower on Mars, this can be used as an offset for some of the non aerodynamic issues caused by just increasing the speed.

I was surprised by this result too and had to double check it. I would have just used lift directly rather than L/D times drag, so in the end you have Cl rather than L/D*Cd, but no real difference.

[sebk]

Mach number doesn't change appreciably with air density (or any other gas mixture reasonably close to ideal gas). It changes with temperature and with composition (the bigger molecular weight the lower speed of sound). So foe example in Mars it's actually about 3/4 than on the Earth.

So no, if more speed was needed your Mach number would change nearly proportionally.

[meberbs]

Mach number doesn't change appreciably with air density (or any other gas mixture reasonably close to ideal gas). It changes with temperature and with composition (the bigger molecular weight the lower speed of sound). So foe example in Mars it's actually about 3/4 than on the Earth.

So no, if more speed was needed your Mach number would change nearly proportionally.

Please look at the final equation Cl*A/2 *M^2*gamma

To get the same lift M is unchanged. CL is a constant. Gamma would be around 1.3 instead of 1.4 for CO2 instead of Earth air. That means to get the same lift, you want A and M to be the same, though gravity means you need a third of the lift.

This is simply the conclusion from the final equation. I went from memory to check it at first, and if there is a mistake, the only possibility I see is the last step, where p c^2 = gamma happens. Looking again, the units don't make sense, there should be a P for pressure on the right side, because gamma is unitless.

[Genial Precis]

Mach number doesn't change appreciably with air density (or any other gas mixture reasonably close to ideal gas). It changes with temperature and with composition (the bigger molecular weight the lower speed of sound). So foe example in Mars it's actually about 3/4 than on the Earth.

So no, if more speed was needed your Mach number would change nearly proportionally.

Please look at the final equation Cl*A/2 *M^2*gamma

To get the same lift M is unchanged. CL is a constant. Gamma would be around 1.3 instead of 1.4 for CO2 instead of Earth air. That means to get the same lift, you want A and M to be the same, though gravity means you need a third of the lift.

This is simply the conclusion from the final equation. I went from memory to check it at first, and if there is a mistake, the only possibility I see is the last step, where p c^2 = gamma happens. Looking again, the units don't make sense, there should be a P for pressure on the right side, because gamma is unitless.

\rho c^2 is the quantity that got renamed gamma, where \rho is the atmospheric density. You may think it a dimensionless quantity equal to Cp/Cv, but you would be wrong. Thus, your lift at any given speed is attenuated by a factor proportional to density. Have a qualitative description of flight on Mars. Hence the need for high speeds to get enough lift for an airframe with any appreciable mass on it. Hence my suggestion to eschew electric engines driving supersonic propellers, or inefficient rockets, and launch gliders around.

They need to move so fast anyway just to get off the ground that they've got a bunch of stored energy in their airspeed. Hence my suggestion to have a much cheaper aircraft by making it a glider. You can have a bunch of cheap supersonic gliders and a few expensive catapults throwing them back and forth, which seems like the right way around for flexible shipping.

[meberbs]

\rho c^2 is the quantity that got renamed gamma, where \rho is the atmospheric density. You may think it a dimensionless quantity equal to Cp/Cv, but you would be wrong.

I don't "think" that. gamma has exactly one defined meaning in this context, and it is a dimensionless quantity.
https://en.wikipedia.org/wiki/Heat_capacity_ratio

Using gamma to mean anything else would be incorrect in this context. Nilof even clearly stated what terms were present in the final equation in the first post:

As such, to first order the atmospheric density cancels out, and explicit dependence on it is exclusively from higher order effects as we may see differences in Lift/Drag ratio, the drag coefficient, or the specific heat ratios as a result changes in pressure, temperature, and chemical composition. Which are fundamentally not going to cause order-of-magnitude differences when the mach ratio is held constant, especially in the high Reynolds number limit.

Now that I have looked at it closer, it seems that he made a mistake and left out the pressure term (which is effectively proportional to density, given certain other things held constant.) But he did not try to redefine gamma as anything other than its standard meaning.

[Genial Precis]

It's not a pressure term. The following correct:
Lift=(L/D) * (Cd A / 2) * M^2 * ρ c^2
Which is proportional to \rho, which is the density.

The following is incorrect, if \gamma is supposed to be the specific heat ratio:
Lift=(L/D) * (Cd A / 2) * M^2 * γ

The true behavior is that lift is proportional to density, and you only need about .38 as much as on Earth, so any given aircraft has to move ~6.4 times as fast to achieve same ratio of lift to weight. So you either need big wings and low weight, or you need to go supersonic just to take off.

Thus, if you want a cargo aircraft, which should be cheap and robust and able to make long flights at high mach numbers, it makes a certain amount of sense to just accept the high takeoff and landing speeds, and indeed exploit them to avoid the need for putting any engines or fuel on the aircraft, ie a glider.

[meberbs]

It's not a pressure term. The following correct:
Lift=(L/D) * (Cd A / 2) * M^2 * ρ c^2
Which is proportional to \rho, which is the density.

The following is incorrect, if \gamma is supposed to be the specific heat ratio:
Lift=(L/D) * (Cd A / 2) * M^2 * γ

Which is what I just said 2 different ways. There needs to be a P for pressure for the units to check out. I don't know how you can say that it is not a pressure term, when pressure is literally the units needed to fix the final equation.

To repeat it again, Nilof somehow took the c^2 = gamma*P/rho equation and dropped the P. I missed this on my first pass too.

[Robotbeat]

This thread is largely inspired by the fact that we have a helicopter currently on its way to Mars. As a result, it feels appropriate to have a general thread for this. Feel free to insert any observations or start any new discussion within the topic.

The most surprising result to me is probably that lift at a fixed mach ratio is independent of atmospheric density, as long as we are still in the regime where the atmosphere is dense enough that it can be viewed as a fluid and that we can use thermodynamics & fluid dynamics while neglecting the corrections from statistical mechanics. One way to see this is to write:

Lift = (L/D) * Drag = (L/D) * (C_d A / 2) * ρ v^2 = (L/D) * (C_d A / 2) * M^2 * ρ c^2 = (L/D) * (C_d A / 2) * M^2 * γ

Where in the last step we have used Laplace's equation, and the previous steps were just the definitions of the lift to drag ratio, the drag coefficient, and the mach ratio.

As such, to first order the atmospheric density cancels out, and explicit dependence on it is exclusively from higher order effects as we may see differences in Lift/Drag ratio, the drag coefficient, or the specific heat ratios as a result changes in pressure, temperature, and chemical composition. Which are fundamentally not going to cause order-of-magnitude differences when the mach ratio is held constant, especially in the high Reynolds number limit.

So effectively, for Mars, it looks like level flight using electric motors is actually easier in some ways, since the lower gravity reduces lift requirements, so we just end up going faster and further for a given amount of work applied to push the craft forward. The main actual issue with low atmospheric densities is the higher takeoff & landing speed requirements for winged craft, which are partly offset by the lower gravity.

That's right. Other than the take-off and landing needing to be faster, you can actually go further on a charge on Mars than on Earth. Possibly even to the antipode with next-gen lithium anode batteries and glider-like lift-to-drag ratio.

But CO2 has a lower speed of sound, meaning you might get reduction in lift to drag due to sonic effects at lower speeds on Mars than on Earth, which might be a problem since you also need high speeds just to maintain lift. (EDIT:But luckily NOT supersonic--at least for typical sailplane designs--because the lower gravity means you need almost 2/3rds less lift.)


But overall, yeah, I think flight is going to be the most energy efficient way to travel long or even medium distance on Mars. The lift to drag of high performance gliders is 40-70, compared to like 5-10 on Mars' soil. Once you solve take-off and landing, electric motorgliders are going to kick butt for long distance travel on Mars... Until you can build railroads, they'll be basically unbeatable in terms of efficiency.

[Twark_Main]

My baseline notion of this is a 5 km electromagnetic catapult launching gliders 1 km/s after 10 s of 10g acceleration. If the catapult is fed at 10% efficiency by a dedicated 1 GWe power plant it can launch 20 kton/day, or a 40 ton glider/container every 3 min. If I assume a glide ratio of 7 (concorde at mach 2), the range might be 500 km.

The receiving end should have multiple independent runways with redundant arresting gear, since the gliders will be coming in very hot and there's definitely not enough air resistance or ground traction to slow them down in a useful distance. Skids instead of wheels, or something, because I've got no idea how to make a tire work at 200 m/s. It was a real problem on the Concorde.

That calls for a lot of infrastructure, and megaprojects on Mars won't be cheap.

How does this compare to "just" using Raptors at each end for propulsive take-off and landing? Delivering 2 km/s of delta-v (plus gravity loss) should be very doable.

[Genial Precis]

I calculate that each kg of payload needs a minimum of 6 MJ of fuel energy. Mass ratio is 1.70, heat of combustion of the fuel is 7.9 MJ/kg. The starship has to carry its own weight so I fudge that to 10 MJ. Making methalox from martian reagents is probably less than one third efficient, so shipping things that way costs 30 MJ/kg. Since it's a thermal process, a 1 GWe nuclear reactor would have an effective power budget of 3 GWth ish, so I can ship 50 kton/day under those assumptions. If you try to do it with solar electric it's only 15 kton/day for that same 1 GWe.

In order to meet that capacity with starships carrying 200 tons from A to B 10 times a day, I would need 8 or 25 starships for the link.

If I assume 100% efficiency of charging the catapult from the power plant, the catapult launch scheme costs only 500 kJ/kg, 1 MJ/kg if the glider has the same mass as the cargo. Thus 86 kton/day capacity.

If your shipping is limited by energy efficiency, winged aircraft and rockets are much worse than ground rail. This is true when the power plant is much more constraining than the other stuff.

If your shipping is limited by the capital cost of either 1) a bunch of high supersonic winged aircraft with engines 2) A bunch of gliders and a 5 km electric catapult 3) 8-25 Starships 4) 500 km rail link, it is not completely clear to me as I write this which wins out, or even how they stack up against the 1 GW power plant I have assumed to be driving it all. 1 GW power plant is a 1E9 USD object even on Earth, so maybe the energy efficiency is so important that you're better off with rail.

[edzieba]

Rail has the up-front energy cost of constructing the tracks before you can start running energy efficient trains. That now adds a distance-dependant factor and a overall usage period (amortisation over time) factor to the total system energy calculation.

[Genial Precis]

For the upfront costs of rail, I assume that scrap steel costs 500 USD/ton on Earth, and that primary steel has an embodied energy of 50 MJ/kg. Thus the effective price for avoiding the energy input of making primary steel is 1E8 J/USD. I note the price to small consumers of electricity in my area is 0.1 USD/kWh or 3.5E7 J/USD. I'll take the higher energy figure since railroad building is big industry

Thus, on only that basis, I assume that the roughly 1E6 USD/km cost of constructing new rail track (including earth moving etc) represents an input energy of 1E6 USD/km *1E8 J/USD = 1E14 J/km. Thus the 500 km rail link will require resources equivalent to 5E16 J to construct, or 580 days of output from the 1 GWe power plant.

So it seems to me that for this 500 km point to point link, if you would be running for multiple years at a cargo capacity 20 kton/day or more and more capacity would be useful, you should consider building a railroad. So if you have 2E7 or more tons of cargo to take from point A to point B and it can be used as fast as it's received.

This 20 kton/day figure, which represents 8-25 Starships running from the dedicated output of a 1 GW power plant, is a throughput of 7 MT/yr. Most major highways in the USA have more throughput than that, and I think essentially all freight rail links.

So I suspect the decision of whether to go rail or road or else some more generalist but energy-intensive vehicle such as Starship, comes down to the total amount of mass you want to move. The point at which you make the decision one way or another is probably not exactly represented by these back-of-the-napkin calculations, but I would guess they are within an order of magnitude at least.

Now, whether or not my catapult glider scheme has a niche depends on just how much goes into the catapult, which I don't have a good estimate for at the moment.

[Vahe231991]

On April 19, 2021, the Ingenuity helicopter launched aboard the Perseverance rover made its first flight on Mars, becoming the first aircraft to fly on an extraterrestrial body. Since then, the Ingenuity helicopter has made a total of 52 flights, the most recent being on April 27, 2023. In your opinion, how does the flight performance of the Ingenuity helicopter stack up against your formulas for aerodynamic lift in extraterrestrial atmospheres?

[redneck]

The Ekranoplan (Caspian Sea Monster) effect might be considered for well defined runs. High speed in ground effect might allow routes that were just more or less graded to road conditions. Though 400 knots and altitude 3 feet might be a bit sporty.

[Vahe231991]

The Ekranoplan (Caspian Sea Monster) effect might be considered for well defined runs. High speed in ground effect might allow routes that were just more or less graded to road conditions. Though 400 knots and altitude 3 feet might be a bit sporty.

The highest altitude achieved by the Ingenuity exceeds the maximum height above the water at which ekranoplans past and present can travel (for starters, the ekranoplan concept was perfected in the USSR during the Cold War to help create an ocean-going vehicle combining the strengths of airplanes and ships by flying a few feet over the water on a cushion of air).

[redneck]

The Ekranoplan (Caspian Sea Monster) effect might be considered for well defined runs. High speed in ground effect might allow routes that were just more or less graded to road conditions. Though 400 knots and altitude 3 feet might be a bit sporty.

The highest altitude achieved by the Ingenuity exceeds the maximum height above the water at which ekranoplans past and present can travel (for starters, the ekranoplan concept was perfected in the USSR during the Cold War to help create an ocean-going vehicle combining the strengths of airplanes and ships by flying a few feet over the water on a cushion of air).

My suggestion was only for heavy loads on routes in frequent use. That was the route grading such that the vehicle could do the cushion of air. The thought was in the context of railroad and suborbital hopper suggestions. The altitude limitations on Ekranoplan would make it useless for exploration and unprepared destinations..