uZGyY.jpgThis is a unique idea of mine. In the above image, the torque reaction propeller is rotating clockwise producing a side-thrust of 40N. With its reaction torque alone, it has to balance the moment generated by the payload on the other side. So...

vOgfF.jpg

Regarding this, I put my arguments as follows. Please choose one of these as correct ones or create your own concept to solve the issue I'm having.

Argument 1 I've also read somewhere that the yawing torque of a motor is the torque of the motor itself. If this is the case, the torque of the propeller will be: This equation (eqn.2) is what I've derived from the momentum theory. Here *tau* is the torque I've calculated.This result is based on the assumption that the relation
Rotational power=torque×angular velocity holds true in this case where a propeller involves. The torque so obtained will not be enough to balance.

Argument 2 If this torque is considered to be leveraged by the distance from the pivot upon which it acts, units on both sides of the equation doesn't balance. Refer eqn.1 in image 2 where the LHS would be in N.m and the RHS would be in N.m-squared if considered to be leveraged

Argument 3 This, eqn.3, is assuming that the thrust is directly a force acting perpendicular to the axis, on the frameIn this case, at the mentioned distance of 8.75cm, the frame will be level.

Argument 4 Here, the torque we obtained from eqn.2 in image 3 is considered as a rotational force acting at a distance, which is the propeller's mean radius, from the centre of the hub.Here, in eqn.4, the mean radius is approximated to half of the outer radius of the propellerIn this case too, like in the previous, the propeller torque reaction will be able to hold the frame level albeit at a greater distance from the pivot.

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