# Can the lift equation be used for the Ingenuity Mars Helicopter?

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.

• 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. Aug 28 '20 at 14:15
• @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! Aug 28 '20 at 15:37

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.

• 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. Aug 28 '20 at 15:44
• Heh, answering such questions is what keeps me sharp! Aug 28 '20 at 15:56
• Isn't $C_L$ and $A$ dependent of each other? ie. if you choose to use area of entire rotor, the lift coefficient also describes entire rotor. For approximation like this that might be more suitable approach as you cannot add blades indefinitely to increase lift of the rotor. Nov 29 at 10:53
• Yes, the OP's formula ignores interactions between blades. But ignoring how many blades each rotor has may matter less than ignoring the inter-blade effect of two coaxial rotors. Nov 30 at 18:20
• Why did you do $0.6 \times 0.05$? is 0.05 the width of the blade? Dec 1 at 4:28

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...!

We can try to be more accurate, by integrating the lift along the blades. A elementary derivation/calculation follows:

$$L=\frac{1}{2}\cdot C_{L}\cdot v^{2}\cdot S\cdot \rho$$

Let the radius being r, the chord c, and the pitch p. Then, we take a rectangular element of blade surface:

$$dS=c\cdot dr$$

For that blade element, and in differential form,

$$dL=\frac{1}{2}\cdot C_{L}\cdot v^{2}\cdot c\cdot dr\cdot \rho$$

Taking v as a function of rotor angular speed omega at radius r in air of density rho and an inflow w (blowing from below the disk) $$v^{2}=\Omega ^{2}r^{2}+w^{2}$$

Taking CL as a linear function of the AoA, itself determined by the pitch p and $$\arctan (w/\Omega r)$$ then we have $$AoA=p-\arctan (w/\Omega r)$$

We may take the airfoil NACA0012. From the NACA graph, and with the AoA in radians, $$C_{L}=5,72\cdot AoA$$

We end up with $$C_{L}=5,72\cdot ((p-\arctan (w/\Omega r))$$

The angles being small, they may be taken as the tangents, in radians.

Inserting CL and v2 in $$dL=\frac{1}{2}\cdot C_{L}\cdot v^{2}\cdot c\cdot dr\cdot \rho$$

We arrive to $$dL=2,86\cdot c\cdot \rho \cdot (p\cdot \Omega ^{2}\cdot r^{2}+w\cdot \Omega r+p\cdot w^{2}-w^{3}/\Omega r)\cdot dr$$

For constant values of chord, air density, inflow, rotor angular velocity and blade pitch, the whole expression may be integrated, but if we assume that inflow is zero, and taking only the first term, that is the important one, we have that, for one blade:

$$L=2,86\cdot c\cdot \rho \cdot p\cdot \Omega ^{2} \int_{0}^{r}r^{2} dr$$

Inserting values for the Mars helicopter:

rotor radius 0,6m, Omega = 2400 rpm = 251 rad/s rho = 0,02 kg/m3 chord = 6 cm blade pitch = 9º = 0,157 rad

$$L=2,86\cdot 0,06\cdot 0,02 \cdot 0,157\cdot \ 251 ^{2} \int_{0}^{0,6}r^{2} dr$$

Integrating…

$$L=2,86\cdot 0,06\cdot 0,02 \cdot 0,157\cdot 251 ^{2}\cdot 0,6^{3}/3$$

We end up with a lift of 2,4 newton per blade...

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

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