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.