Designing a Mars-Ready RC Plane: What Factors Must Be Considered?

[Batmoosemike]

So, I'm building an rc plane for school that has to theoretically be able to fly on Mars. What differences/limitations are there compared to designing a regular remote controlled plane? This is for a terraformation project, so at this point Mars would have an atmospheric pressure of 50kPa.

 

[David Lewis]

The gravitational constant and atmospheric density. Atmospheric viscosity might also play a role. Then make sure your airplane can withstand the temperature extremes and atmospheric instability it will encounter.

 

[Batmoosemike]

The gravitational constant and atmospheric density. Atmospheric viscosity might also play a role. Then make sure your airplane can withstand the temperature extremes and atmospheric instability it will encounter.

Thank you for the quick reply! As the atmospheric pressure would be roughly half of earth's, would I need to double the thrust to weight ratio?

 

[mrspeedybob]

Gravity on Mars on only 3.7 m/s² vs. 9.8 m/s² on earth, so you need a lot less lift.
Unless your terraformation project involves adding significant mass to the planet, thereby increasing its gravity. It would probably have to in order to retain the more dense atmosphere that you're talking about.
You are aware that Mars's current atmosphere is more like 0.5 kpa right? Not 50 kpa. I'm assuming the 50 kpa atmosphere that you're talking about is the result of the terraformation project.

 

[Batmoosemike]

Gravity on Mars on only 3.7 m/s² vs. 9.8 m/s² on earth, so you need a lot less lift.
Unless your terraformation project involves adding significant mass to the planet, thereby increasing its gravity. It would probably have to in order to retain the more dense atmosphere that you're talking about.
You are aware that Mars's current atmosphere is more like 0.5 kpa right? Not 50 kpa. I'm assuming the 50 kpa atmosphere that you're talking about is the result of the terraformation project.

Wait so I would need less lift? Interesting.

Also, yeah after terraformation it should be around 50kPa.

 

[David Lewis]

You may not have to worry about barometric pressure directly because the density of the atmosphere automatically takes that into account. Without knowing your model airplane's mission, in the preliminary design phase it often comes down to how high of a stall speed you can tolerate. The slower the airplane can fly, the easier it will be to control, the more endurance it will have, the less likely it will crash or be damaged in a crash, and the easier it will be to land and take off. Most of the time it's assumed the airplane will be operated by moderate-to-low skill pilots.

 

[mrspeedybob]

Atmospheric density is going to be more relevant then pressure. If you've changed the Martian atmospheric composition to match Earth's then 50% pressure should equal 50% density. If the composition is different then figure out what the density will be.

Half the air density will mean that the same wings will only generate half the lift. 38% of Earth's gravity will mean that the same plane only needs 38% of the lift. If that's as deep as you need to go with your project then you can say that any aircraft that can fly on Earth, would fly even more easily on New Mars.

 

[DaveC426913]

Also, yeah after terraformation it should be around 50kPa.

This is still only 5% of Earth sea level pressure, not 50%.

 

[Batmoosemike]

This is still only 5% of Earth sea level pressure, not 50%.

I was under the impression that Earth's atmospheric pressure at sea level was 101325 pascals, so 101 kPa?

 

[SteamKing]

This is still only 5% of Earth sea level pressure, not 50%.

It's that danged metric system again, keeping people confused.

S.L. barometric pressure on Earth is 101.325 kPa, so 50% pressure would be ≈ 50 kPa. (or about 7.3 psi)

 

[mfb]

If the terraformed Mars has the same chemical composition temperature as Earth, it is easy. We have a region of half the atmospheric density (and roughly half its pressure) already, about 5000 to 6000 meters above sea level. Commercial airplanes fly way above that height - to have an even lower pressure, which allows faster flight. Planes on Mars would do the same. There is no commercial airport at this altitude, but we can extrapolate from those located ~4000 m above sea level: you would need very long runways. The lower required lift makes it easier, however.

Note that Mars has significantly taller features on its surface - Olympus Mons' peak is 22-26 kilometers above the surrounding terrain. You would have regions where flight is significantly easier than elsewhere, and regions where airports are not reasonable at all.

It's that danged metric system again, keeping people confused.

Yeah, I can never remember the conversion factors.
Sometimes they are 1000, sometimes they are 1000, sometimes they are 1000. So confusing.
3 feet in a yard, 12 inches in a feet, 16 ounces in a pound, 32.174049 pounds in a slug, 231 cubic inches in a gallon, 1760 yards in a mile, and approximately 1.000002 miles are a US survey mile. So much easier.

 

[David Lewis]

mfb wrote: "There is no commercial airport at this altitude (5 km to 6 km pressure altitude), but we can extrapolate from those located ~4000 m above sea level: you would need very long runways."

Correct. That is true if your model airplane needs to fly fast, as commercial airliners are required to do. However, if high speed flight is not a concern then short takeoff and landing (STOL) capability is simply achieved with low wing loading and high lift coefficient.

 

[SteamKing]

Yeah, I can never remember the conversion factors.
Sometimes they are 1000, sometimes they are 1000, sometimes they are 1000. So confusing.
3 feet in a yard, 12 inches in a feet, 16 ounces in a pound, 32.174049 pounds in a slug, 231 cubic inches in a gallon, 1760 yards in a mile, and approximately 1.000002 miles are a US survey mile. So much easier.

I don't know about you, but I've notice people at PF who have had their share of difficulties with the metric system. Like saying there are 100 cm³ in a cubic meter and such.

At least the USCS keeps you on your toes. I can switch back and forth between measurement systems without too much difficulty. Can you?

 

[DaveC426913]

It's that danged metric system again, keeping people confused.
S.L. barometric pressure on Earth is 101.325 kPa, so 50% pressure would be ≈ 50 kPa. (or about 7.3 psi)

My bad. I was thinking of millibars.

 

[mfb]

mfb wrote: "There is no commercial airport at this altitude (5 km to 6 km pressure altitude), but we can extrapolate from those located ~4000 m above sea level: you would need very long runways."

Correct. That is true if your model airplane needs to fly fast, as commercial airliners are required to do. However, if high speed flight is not a concern then short takeoff and landing (STOL) capability is simply achieved with low wing loading and high lift coefficient.

You are right, the lower required lift makes it easier if we are not interested in high speeds.

I don't know about you, but I've notice people at PF who have had their share of difficulties with the metric system. Like saying there are 100 cm³ in a cubic meter and such.

There are also people struggling with quadratic equations, that doesn't mean it would help to introduce them to quantum field theory.

At least the USCS keeps you on your toes. I can switch back and forth between measurement systems without too much difficulty. Can you?

Why should I remember tens of conversion factors for a (basically) US-specific unit system? There are about 200 countries in the world, and the US just represents ~5% of the world population. And science is done in SI units anyway - for a good reason.

 

[SteamKing]

There are also people struggling with quadratic equations, that doesn't mean it would help to introduce them to quantum field theory.

No one is talking about quadratic equations or quantum theory here.

Measurement systems are merely conventions, and conventions have some good features and some not so good features, often.

Why should I remember tens of conversion factors for a (basically) US-specific unit system?

No one is asking you to.

There are about 200 countries in the world, and the US just represents ~5% of the world population. And science is done in SI units anyway - for a good reason.

The output of the US economy represents about 22% of the gross world product. Not bad for 5% of the world population.

Just because science is done in SI units doesn't mean that SI units can't be confusing or that mistakes can't be made if SI units are carelessly applied.

Take pressure measurements, for example. You've got bars, pascals, torrs, etc. That seems like a lot of units to keep up with, and that's just for pressure. And that's not even mentioning units like kg/cm², which were once in common use.

 

[nasu]

Take pressure measurements, for example. You've got bars, pascals, torrs, etc. That seems like a lot of units to keep up with, and that's just for pressure. And that's not even mentioning units like kg/cm², which were once in common use.

You are right. Mixing units from different systems may be confusing.
Bars and tors are not SI units. If you want to make conversions simple you need to stick to SI.
However the other units ("tolerated" units) are found convenient for people in various fields and they are willing to put up with the "confusion".
Same as in US the traditional units are found convenient enough for practical use if not for science.
In common use there are very few conversions to do.

 

[A.T.]

At least the USCS keeps you on your toes.

That's the most positive way to say "It's a mess" that I have ever heard.

 

[David Lewis]

If the maximum wind speed in which the airplane must be able to fly is part of your design specification then you must make sure the airplane can fly faster than that speed, plus a margin for gusts.