# 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. Commented Aug 28, 2020 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! Commented Aug 28, 2020 at 15:37
• THERE IS A BIT OF MISUNDERSTANDING HERE, THAT EQUATION CANNOT BE DIRECTLY USED FOR HELICOPTERS, wherever they fly --- The correct equation for rotary-wing is: $T = ½ \rho (\omega R)^2 \pi R^2 C_T$ where $\omega$ is the rotating speed of the rotor and $C_T$ is the thrust coefficient ("thrust" since rotors generate no lift) which cannot be more than some 0.03 for very aerodynamically optimised rotors. Please have a look at my answer for its correct application. Commented Sep 26, 2022 at 20:37
• I'm not sure, but maybe this should be moved from aviation to physics? Commented Nov 10, 2022 at 19:29
• @JosephDoggie very very hopefully not. It has 4 good answers and numerous useful comments. The subject is if far greater interest to this community than to Physics where it may well get minimal attention. It's 2+ years old . Commented Nov 10, 2022 at 23:29

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: see sophit's or xxavier's answers, which are fancier than what the OP needed, and which lack this answer's "Add citations from reputable sources" post notice.

• 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. Commented Aug 28, 2020 at 15:44
• Heh, answering such questions is what keeps me sharp! Commented Aug 28, 2020 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. Commented Nov 29, 2021 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. Commented Nov 30, 2021 at 18:20
• THERE IS A BIT OF MISUNDERSTANDING HERE, THAT EQUATION CANNOT BE DIRECTLY USED FOR HELICOPTERS, wherever they fly --- The correct equation for rotary-wing is: $T = ½ \rho (\omega R)^2 \pi R^2 C_T$ where $\omega$ is the rotating speed of the rotor and $C_T$ is the thrust coefficient ("thrust" since rotors generate no lift) which cannot be more than some 0.03 for very aerodynamically optimised rotors. Please have a look at my answer for its correct application. Commented Sep 26, 2022 at 20:33

## There is a bit of misunderstanding here, the "lift" equation cannot be directly used for helicopters, wherever they fly and whatever the chosen answer says.

The correct equation for rotors and propellers is the "thrust" equation which is formally similar to the lift equation but uses the speed due to the rotation and the disk surface instead:

$$T = ½ \rho (\omega R)^2 \pi R^2 C_T$$

where $$\omega$$ is the rotating speed of the rotor, $$R$$ its radius and $$C_T$$ is the thrust coefficient ("thrust" since rotors generate thrust, not lift) which cannot be more than some 0.03 for very aerodynamically optimised rotors.

The equation given in the question is used to calculate lift generated by a wing (or, more generally, an aerodynamic body) with surface $$S$$ when it moves in a fluid of density $$\rho$$ with a well defined speed $$V$$:

$$L = ½ \rho V^2 S C_L$$

Why is this equation not suitable for rotary-wing applications? The speed $$V$$ seen by an helicopter blade is basically given by the sum of 3 different terms:

1. the speed due to its rotation around the rotor shaft; this speed varies linearly along the bladespan from 0 at the rotor head to $$\omega R$$ at the tip, where $$\omega$$ is the rotating speed and $$R$$ the bladespan;
2. the speed at which the whole helicopter is flying;
3. during its rotation, the blade bumps into the wake shed by the previous blade; and after one complete rotation, it bumps into its own wake; these wakes also modify the local speed seen by the blade.

The sum of these 3 terms gives a total speed which continuously changes in space and time and that definitely cannot be represented by one single constant value $$V$$.

So, can the lift equation be used for an helicopter? No, because the blade does not move within a uniform and well defined airflow with steady speed $$V$$.

That equation can be used for airplane, cars, cyclists on bike, bridges and so on, but not for helicopter or, more generally, not for rotary-wing aircraft.

Is it possible to use an equation as easy as that one for rotary-wing applications? Yes it is, but in the helicopter world nomenclature is a bit different: lift, per definition, is the component of the aerodynamic force perpendicular to the airflow. As discussed, the airflow seen by a rotor is the sum of 1.+2.+3. and it is definitely not constant, it changes continuously. Therefore defining a lift for a rotor is quite difficult simply because the airflow (and it's perpendicular component) changes continuously. Therefore for rotor not lift rather thrust is defined: it is the aerodynamic force parallel to the rotor shaft and positive upward.

Analogously to the lift equation, also the thrust can be represented as the product of a density, a velocity (squared) and a surface but with the following distinctions:

• as a velocity, the velocity of the rotating blade tip is used, that is $$\omega R$$
• as a surface, the area $$A$$ of the rotor disk is used, that is $$\pi R^2$$

Therefore for rotary-wing application, the following thrust equation is used:

$$T = ½ \rho (\omega R)^2 \pi R^2 C_T$$

(Note: in USA the $$½$$ is normally dropped)

$$C_T$$, a bit like $$C_L$$, depends on several geometrical factors and trade-off, but in general its maximum value is within the range 0.008 and 0.028, according to how much optimised the rotor is.

Intermission

Let's do the math for Ingenuity making the following assumptions

• $$\rho = 0.01 kg/m^3$$ Note: density on Mars changes according to season from 0.01 to 0.02; we use the lowest value in order to be conservative;
• $$\omega$$ = 2400$$rpm$$ = 250$$rad/s$$
• $$C_T$$, we suppose that the rotors are of a very optimised design = 0.028.

Then we get, remembering that Ingenuity has 2 rotors:

$$T =2 \cdot ½ \cdot0.01(250\cdot0.6)^2 \pi 0.6^2 \cdot 0.028 = 7.125N$$

Since $$g_{mars} = 3.72m/s^2$$ that corresponds to 2.17kg thrust, which is enough to hover and manoeuvre Ingenuity with its 1.8kg weight.

Bonus material: what if we do use the lift equation (in the correct way)?

Is, in the helicopter world, the lift equation as given in the question really useless? If the hypothesis is done that the blade is composed by a continuous sequence of adjacent airfoils, then that equation can be locally used for each of these airfoils in the sequence.

So, for each slice of blade that equation can be used to get the local lift (and the local drag); afterward the local lift (and drag) is summed up (integrated) along the bladespan, from the rotor head till the tip, to get the total lift and drag; and finally the total lift and drag is decomposed in 1) a force parallel to the rotor shaft, which is the thrust generated by the rotor; and 2) a force perpendicular to it, which gives the torque needed to make the rotor spin.

Obviously that equation can be used if it is know the local speed $$V$$ on each slice of blade as given by 1.+2.+3. As seen, 1. and 2. are know (since $$\omega$$, $$R$$ and the helicopter speed are known); the speed 3. due to the wake depends itself on the lift and gives rise to a vicious circle: the lift depends on the wake and the wake depends on the lift! This vicious circle is broken by either using simplified models of the wake and/or wind tunnel measurements and/or CFD simulations.

Let's break the vicious circle via a simplified model that we will later test again on Ingenuity. If the simplification is done that:

• the helicopter is hovering, i.e. term 2. is null;
• the wake in/upon the rotor is constant everywhere, i.e. term 3. has a constant value;
• the blades have a hyperbolic twist (this is normally true for a propeller blade, helicopter blades have a more linear twist but using a hyperbolic one the math is easier);
• and each airfoil is operating at its $$\alpha$$ of maximum efficiency i.e. maximum $$C_l/C_d$$;

then we get that:

$$C_T=¼ \frac{N_b c_{tip}}{\pi R} C_{l_{\alpha}} \alpha_{@maxC_l/C_d}$$

where:

• $$N_b$$ is the number of blades;
• $$c_{tip}$$ is the chord at the tip of the blade;
• $$C_{l_{\alpha}}$$ is the slope of the airfoil's lift coefficient;
• and $$\alpha_{@maxC_l/C_d}$$ is the airfoil's $$\alpha$$ for maximum efficiency.

For Ingenuity we have:

• $$N_b=4$$;
• $$c_{tip}=0.06m$$ (ballpark estimate)
• $$C_{l_{\alpha}}=2\pi$$ (typical thin-airfoils value);
• and $$\alpha_{maxC_l/C_d}=8°$$ (this value really depends on the airfoil; I have chosen a standard NACA 2412 at a very low Reynold number). This gives:

$$C_T=¼ \frac{4\cdot0.06}{\pi\cdot0.6} 2\pi \cdot0.14=0.028$$ Yeah

Yet another bonus material: momentum theory

A first approximation of the thrust can be obtained also using the simple momentum theory which gives:

$$T = \sqrt[3]{2 \rho A P^2}$$

Where:

• $$A$$ is the rotor area;
• $$\rho$$ is the density;
• $$P$$ is the power needed to generate the thrust.

To see the simple momentum theory in action for Ingenuity, please have a look at this answer.

Useful references:

• Excellent answer! I notice that to account for there being two rotors, you simply multiply by two. As they are coaxial, I imagine that further complicates the vicious circle of interaction, so my first thought that multiplying by two was leaving something out. My second guess is that these effects are captured by the coefficient of thrust for the assembly differing from that of each rotor considered individually. Am I making any sense here? Commented Sep 25, 2022 at 12:42
• @sdenham: yes you're right, two coaxial rotors normally need some 15% more power than two isolated rotors, the total thrust being the same. And for the thrust I made the indirect simplification that all the complications are incorporated in the $C_T$ Commented Sep 25, 2022 at 18:35
• This is such a great and detailed answer. I don't understand it at all, but kudos ;)
– Jamiec
Commented Nov 1, 2022 at 15:22
• @Jamiec it's not that difficult: for airplane speed is simply V, for rotating blades it's V + rotation speed + wake speed ;) Commented Nov 1, 2022 at 16:36
• In the USA the 1/2 is dropped? Could it be factored into CT? Going to try your equation for an air plane prop. Commented Jan 12, 2023 at 2:42

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. Commented Aug 28, 2020 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
• 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.