10
$\begingroup$

The Ingenuity helicopter that has flown on Mars is designed for roughly 0.01 bar pressure and a colder atmosphere made of mostly CO2. See links (including video) in this answer to What JPL laboratory is this exactly, and what are the functions of these amazing-looking control panels? in Space SE and sources linked within. For more about Mars' atmosphere see this answer to What could Perseverance listening to Ingenuity reveal? there as well, and sources within.

The counter-propagating rotors (driven by separate motors but synchronized) spin at about 2500 RMP. The propellors are huge by Earth standards compared to the tiny weight that they carry. But we also have to remember that it will weigh more on Earth. See answers to Does Ingenuity rotate via differential rotor speeds or differential angles of attack? and sources within for more on that.

Question: Hypothetically, could the Mars helicopter Ingenuity have been test-flown outdoors at JPL just for the heck of it? Not that they would ever do such a thing (it will have been kept very clean and very very safe at all times) but from a technical stand point would it have flown? What rotor speeds and torques would have been necessary, and would these have been possible with the light-weight motors optimized for Mars flight?

$\endgroup$
6
  • 3
  • $\begingroup$ That last one was from yesterday .. $\endgroup$
    – Mr R
    Commented May 19, 2021 at 5:23
  • 2
    $\begingroup$ I don’t think that enough information is public to say for certain, but the aircraft would need to generate more than double (2.64) times the lift that’s required on Mars. With the air being 100 times more dense than the Martian atmosphere, the rotors would experience their nominal torque at much lower speed and it’s doubtful that the motors would be able to generate the required power so far below their design speed. $\endgroup$
    – Frog
    Commented May 19, 2021 at 6:39
  • $\begingroup$ @Frog The motors are discussed in this paper found in ymb1's answer but I'm not sure if this is definitive for assumptions about a torque curve. Some kinds of motors have most of their torque all the way down to zero speed but for these I don't know. "...two brushless direct-drive propulsion motors driving the two rotors..." $\endgroup$
    – uhoh
    Commented May 19, 2021 at 7:26
  • 1
    $\begingroup$ I haven’t studied the paper, but at 1% speed you’d need 100x torque for the same power $\endgroup$
    – Frog
    Commented May 19, 2021 at 10:24

3 Answers 3

5
$\begingroup$

No, Ingenuity was too heavy and underpowered to fly while on Earth.

From 27:30 to 29:00 in the video of a 2021 May 13 press event, one of its designers, Matt Keennon, manually pilots what he called a "terrestrial version of Ingenuity," officially named Earth Copter, nicknamed Teri or Terry. (Reporters have had to guess at the spelling. The transcript of a 60 minutes segment has Terry, but that neglects I for Ingenuity.)

At 24:45 Matt explains that its only significant hardware differences compared to Ingenuity are that it is lighter (he doesn't say how much), and that its motors are taller, with twice the torque. Aerovironment hasn't published any hard numbers about Terry, or possibly even any text at all since even its name remains disputed.

But it's clear that they designed Terry to closely mimic how Ingenuity would fly on Mars (see the section "Iterations" in Inside Ingenuity With AeroVironment), so they wouldn't have lightened it and used much more powerful motors for capricious reasons. It had to actually fly. A direct quote from Matt at 24:45:

We also reduced the weight of Terry compared to Ingenuity because it's flying in Earth's gravity.

Co-designer Ben Pipenberg also says:

Terry is designed to fly here on Earth, so the motors were redesigned and are more powerful and have a higher torque to handle the denser atmosphere and the higher gravity.

$\endgroup$
2
  • $\begingroup$ This is very helpful, thanks! Since the density is of order 100 x difference and these changes are of order 3-5 we can guess that the scaling will scale as the cube root of density. Hopefully we'll get the math at some point and can then find out. Thanks! $\endgroup$
    – uhoh
    Commented May 19, 2021 at 20:24
  • 1
    $\begingroup$ AeroVironment has the math, of course, but asking them to publish prematurely would be as effective as asking SpaceX to! Matt does respond courteously when asked for very specific things that obviously only he knows, but I'm careful to not overstay my welcome. $\endgroup$ Commented May 19, 2021 at 21:40
2
$\begingroup$

...test flown outdoors at JPL just for the heck of it

Perhaps if they wanted to lose their jobs as engineers, but if one were to try it:

Ingenuity is a constant speed/variable pitch rotor design. Earth gravity is 2.64x greater than Martian. Rotor speed is 2500 rpm. Rotor diameter is 4 feet. Rotor tip velocity is:

2500 rpm × 4 feet × 3.14 × 60 min/hr × 1 mile/5280 feet = 340 mph

Why so slow?

Speed of sound in CO2 is slower (78%) than air. In helium it travels almost 3x faster! Speed is sound is also slower at colder temperatures.

These factors affect the Martian design. The rotors will work fine on Earth but ...

On Earth they will have to lift 2.64 times more weight.

Lift = Coefficient × Area × Velocity$^2$ x Density

If air density is 100 times greater rotor pitch control would be much more challenging. Assuming you don't burn out the motors trying to get 2500 rpm, the helicopter would probably crash from excessive control inputs.

Forging ahead, we could reduce the RPM spec for Earth flight:

Square Root 2.64 x 2500 rpm/100 density factor = 40 rpm!

or employ smaller rotors, making it much more controllable. Adding a bit of buoyancy with a helium dirigible attachment would also be possible. Then, it may be flyable.

A final consideration would be the effect of Reynolds number on the performance of the rotor blades. Here the Martian atmosphere and higher blade velocity would offer an advantage.

Reynolds number = chord × Velocity/Kinematic Viscosity

In summary, a re-design would be in order for the Earth Ingenuity, but, conceptually, the contra-rotating, variable pitch, constant speed approach (with or without gasbag) would be of interest.

$\endgroup$
2
  • $\begingroup$ Nice answer, thanks! It's operated at at least a roughly constant speed and likely optimized for it, but are we sure they can't just instruct a different speed? I don't think we should assume the speed is permanently fixed. Alas I still can't find a description of the exact motor beyond "...two brushless direct-drive propulsion motors driving the two rotors..." but I think we should allow for the probability that the speed can be changed $\endgroup$
    – uhoh
    Commented May 20, 2021 at 22:27
  • 1
    $\begingroup$ @uhoh judging from airfoil polar data, those rotors would need to be spun fast enough to get most of them above a Reynolds of 10$^6$, where L/D ratio is much better. But, from a marketing point of view, how many people will run out and get an Earth modified Ingenuity drone from the hobby store. $\endgroup$ Commented May 21, 2021 at 1:33
0
$\begingroup$

Given the higher gravity on Earth, the effective mass of the craft on Earth is higher than on Mars. The drag is also far higher because of the higher atmospheric density.

Both are factors causing more lift and power to be required, in the form of larger rotors and stronger engines.

Luckily the denser air also means more lift per unit of area of the rotors, reducing the required rotor size, somewhat offsetting the requirement for larger rotors.

I don't have the math or the data to make the required calculations, but it's unlikely that a craft designed to be as small and light weight as possible to work in a Martian environment and be able to be launched on the available cheapest launch vehicle available (cost...) would be well suited for use on earth.

It's possible but unlikely.

$\endgroup$
2
  • 3
    $\begingroup$ I'm not exactly taking an opinion pole here, "it's unlikely" without any support based on anything besides "optimized for A means can't work for B" is a comment, but not really a Stack Exchange answer. Someone could write a similar sounding answer with the exact opposite conclusion and there'd be no way for readers to tell which was right. $\endgroup$
    – uhoh
    Commented May 19, 2021 at 12:22
  • 3
    $\begingroup$ The effective mass will be identical on Mars or anywhere. The weight will be different. $\endgroup$
    – Frog
    Commented May 19, 2021 at 19:49

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .