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I am calculating the net torques and forces about the yaw axis of a quadrotor. If $I$ is the moment of inertia about the rotation axis, and $\omega$ is the angular velocity of the propeller, then the torque is given by:

$$ \tau = I \frac{d \omega}{dt} $$

This represents the net force $F$ acting on the propeller's edge at a radius $r$, giving it a net torque of $\tau = r \times F$.

From the force perspective, there is the driving force of the propeller, and the drag force of the air. For a constant angular velocity, the two forces/torques cancel each other out. Therefore, according to the formula, a constant speed propeller will contribute no torque to the body.

And yet, torque calculations in literature (see here, page 4) will express the torque of a propeller as about the yaw axis as:

$$ \tau = I \frac{d\omega}{dt} + b\cdot\omega^2 $$

Where the latter term is the drag force and $b$ is the drag coefficient. But hasn't the drag force cancelled out, leaving the residual $d\omega/dt$? Why is it being counted again? For example, when calculating net forces about the body, I don't count the weight twice like this ($v$ is velocity, $g$ is gravitational acceleration, $m$ is mass):

$$ F = m \frac{dv}{dt} + m\cdot g $$

So my question is, why is there a net yaw torque about a quadrotor body, when the propeller is spinning at constant speed?

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  • $\begingroup$ If you push against the air, the air still pushes against you just as if you were pushing off the ground when running. $\endgroup$
    – DKNguyen
    Feb 28 at 1:05

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It's important to distinguish torques on the propeller vs those on the body of the aircraft. The propeller is experiencing 0 net torque, while the aircraft is experiencing non-zero net torque. For a constant speed propeller, the drag torque and the driving torque on the propeller cancel. But in order to apply the driving torque, the motor must exert an equal and opposite torque on the aircraft. This torque remains uncancelled, leaving a net torque on the aircraft.

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  • $\begingroup$ So to elaborate in terms of forces: the motor, connected to the airframe, exerts a driving force on the propeller. But the motor also exerts a reactionary force of equal magnitude on the airframe (conservation of angular momentum). The force driving the propeller is neutralized by the drag reaction. So that prop/air interface is almost like a pushing a wall: no acceleration. But the reactionary force/torque on the airframe is unopposed and causes the net torque. And that reaction is equal in magnitude to drag, so we just use the drag formula. Yes? $\endgroup$
    – hazrmard
    Feb 28 at 17:54
  • $\begingroup$ @hazrmard yes that's right $\endgroup$ Mar 1 at 19:57

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