4
$\begingroup$

I am currently working on a basic simulation of the Mars Helicopter Ingenuity, and I am having trouble interpreting the outputs I am receiving from the lift equation. The equation reads as follows:

L = Cl * A * .5 * r * V²

L is lift in newtons, Cl is the coefficient of lift, A is the area of the circle traced by a blade in one rotation, r is atmospheric density, and V is the velocity of the blade. Ingenuity has two rotors 1.2 meters across, each with two blades. For this example, let's assume a rotation rate of 2400 rpm. The coefficient of lift is 0.4, the average for small helicopters. Atmospheric density on Mars is 0.020 kg/m^3. The velocity is taken 1/2 way down the blade and is found by using ((1.2/2)π * rpm / 60). Plugging in the values for Ingenuity gives:

102.871 = 0.4 * π(0.6)² *.5 * 0.020 * 5684.858 * 4

The 4 at the end is added because Ingenuity has four blades, each generating its own lift in accordance with the lift equation.

Dividing the lift by the mass of Ingenuity (1.8 kilograms) gives an upward acceleration of 57.15 m/s². Unfortunately, this number is too large to be accurate. Martian gravity is only 3.711 m/s², and whatever drag that will be generated by the thin atmosphere is likely to be insignificant in relation to such a large upwards acceleration.

So, the question is, how should lift be calculated in this case? Am I using the lift equation incorrectly, or is it simply not applicable to small drones like Ingenuity? If you have any resources for calculating lift more accurately in this case, please share them.

Thank you.

$\endgroup$
2
  • $\begingroup$ Welcome to AV Stack Exchange. I'm not familiar with the martian atmosphere, but that velocity looks like you're going much faster than the speed of sound. At that point, the lift calculation becomes different. I think it would also be useful to calculate the Knudsen number for this phase of flight (en.wikipedia.org/wiki/Knudsen_number) to see if continuum mechanics is applicable. $\endgroup$ Aug 28 '20 at 14:15
  • $\begingroup$ @StuartBuckingham actually, the ends of the rotor blades only reach ~0.7 times the speed of sound, and the spacecraft is never planned to exceed 3 m/s vertically. Although, if the output of my equation was correct, Ingenuity would reach the speed of sound very quickly! $\endgroup$
    – marsman
    Aug 28 '20 at 15:37
6
$\begingroup$

The formula is correct, but you're measuring the wrong area.
A is the area of the wing, i.e., rotor blade, not the area of the rotor disc.
When π(0.6 m)² is replaced with the correct value, about 0.6 m * 0.05 m, then the final number for lift drops by about 40 times, as does the upward acceleration, to a more plausible 1.4 m/s².

(This error is also committed by the first google hit I got for lift equation rotor.)

The next step to improve accuracy would be to not ignore that the root of the blade moves much more slowly than the tip, and conversely that (for Ingenuity at least) the root has greater chord and greater angle of attack. But the math for that is much more involved.

$\endgroup$
2
  • 2
    $\begingroup$ What a great response! It seems like the area misconception resulted from a mix-up between different sets of equations. This source appears to show that the full disk area is used in the more complicated airflow-based Navier-Stokes Equations. Perhaps the author of the article you linked confused this with the area for the lift equation. Also, it seems like you have answered many questions like mine on this forum. Helpful folks like you make the internet a much more useful tool, and keep communities like this one alive, so thanks for your work. $\endgroup$
    – marsman
    Aug 28 '20 at 15:44
  • $\begingroup$ Heh, answering such questions is what keeps me sharp! $\endgroup$ Aug 28 '20 at 15:56
1
$\begingroup$

Your calculation is wrong. For a first approximation, you may take 0,7R as the representative station, but the area should be the area of the blade, not that of the swept surface.

Let's take 2400 rpm. That's 251 rad/s. And let's take the airspeed V at 0,7R => 0,7 * 0,6 = 0,43 0,43 * 251 = 105 m/s

Let's assume that the blade chord is 0,1m. The blade area A would be 0,1 * 0,6 = 0,06 m2

L = Cl * A * .5 * r * V²

Takin Cl as 0,4

Lift = L = 0,4 * 0,06 * 0,5 * 0,02 * 105^2 = 2,7 N

But we have four blades, so we may take 4 * 2,7 = 10,8N as the total lift.

Gravity acceleration in Mars is 3,7 m/s2, so that ingenuity's weight will be 1,8 * 3,7 = 6,6 N

You have plenty of lift...!

Edited in order to add an elementary derivation that results in a differential expression that, being a polynomial, can be easily integrated:

enter image description here

$\endgroup$
1
  • $\begingroup$ Awesome! I actually already have a setup for calculating lift for individual blade segments--I just wanted to keep things simple in this post. $\endgroup$
    – marsman
    Aug 28 '20 at 15:51
1
$\begingroup$

Two ways of calculating helicopter rotor thrust: with the simplified momentum theory, and with finite element methods looking at the lift/drag equations of rotating blades.

  1. Momentum theory. The simple way is to use momentum theory, where the reference area is the rotor disk area. The relevant dimensionless aerodynamic coefficient is $C_T$ rather than $C_L$, as in: $$T = C_T \cdot \rho \cdot A \cdot \Omega^2 \cdot R^2$$ with: $C_T$ = thrust coefficient; $\rho$ = air density [kg/m$^3$]; A = rotor disk area [m$^2$]; $\Omega$ = angular blade speed [rad/sec]; R = blade length [m]

    The $\Omega$R in this equation is the blade tip speed. $C_T$ would be in order of magnitude of 0.004 - 0.008 in normal circumstances, depending on number of blades, twist etc. Some more information in this answer, loads more in J. Gordon Leishman, Principles of Helicopter Aerodynamics.

  2. Blade element theory. With this method the reference area is the total blade area - we compute the lift/drag at every section of the blades, then integrate over the blade length, as per example for propellers in this answer. The helicopter blade will have:

    • twist: variable angle-of-attack
    • sometimes variable blade cord;
    • variable speed as a function of distance to the rotational centre and forward/sideways speed;
    • variable boundary layer and Reynolds effects.
    • a large amount of tip loss,
    • a bit of blade lacking around the rotor centre, so blade length integration boundary does not start from zero.

SO the method OP used was the blade element method, but with a couple of things not entirely right:

  • Computing lift per blade, the reference area is the blade area, not the rotor area.
  • Velocity should be taken as an integral over the blade length, not as a constant at half blade length (the lift generated at the tip is four times as much!)
  • The tip/root losses should be taken into account.

For simple birds eye view considerations, nothing beats the momentum method.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.