How does a propeller work?

Question

I was thinking about the force that balances the force of gravity in a drone, and I stumbled upon this question that how to mathematically describe the upthrust being generated by the propeller, as a function of the angular speed of the propeller and the angle of bend of the propellers.

Here's what I know:

1. Integral factors for the upthrust generated should be the angular speed of the propeller as well as the angle of bend and the size of the propeller, though I am not sure if there would be any other factors playing there role too.
2. I know that the propeller would be sucking the air from underneath the device from a laterally-half-toroid path around the propeller, and then creating a pressure difference, which would cause air to strike the bottom surface. Multiplying this pressure with the area of the bottom surface would give us the upward force on the device.

Here are a few links answers I referred, but couldn't get the exact answer:

1. How do propeller blades rotating through air accelerate it to provide thrust?
2. What is going on in front of and behind a fan?

Answer 1

Like anything in fluid dynamics, there's no single analytic function to describe the behavior of a propellor.

In general, the thrust of the propellor works the same way as lift on an airplane wing. Lift goes as

$$ L = C_L \frac{\rho v^2 A}{2}$$

where $C_L$ is the empirically determined coefficient for that particular wing/propellor blade.

The "angle of bend" of the propellors is known as the "Angle of Attack". For any given speed through a fluid ("advance ratio", $J$), there's an optimal angle of attack (AoA, $\beta$).

Since you must conserve momentum and kinetic energy (ignoring frictional losses), a slow/large propellor is much more efficient than a fast/small propellor for a vehicle traveling at the same speed. Sending air molecules backward at speed $v$ gives you $\frac{v}{m}$ boost in momentum but costs $\frac{1}{2}mv^2$ energy (this concept is known as "propellor slip").