I can calculate the torque by using simple equations.

We know:

I =V/R

Pin= I*V (Consumed electrical power of motor)

Pout = τ * ω (torque * angular speed

ω = rpm * 2pi / 60 (Angular speed)

Etotal= Pin+ Pout

Thus,

τ * ω = I * V * E

τ = (60*I * V * E) / 2pi*rpm

Assuming:

v(3s) = 11.1 V

I = 1.01 A

E (Efficiency) = 10,30,50% (We can use calculus later to maximise this)

rpm=?

τ = ?

So τ= 60*1.01*11.1*0.1 / 2pi*rpm, We see that the higher the rpm, the lower the torque. It seems to be an inverse relationship. So If we wanted the maximum torque, we need the lowest RPM. As we increase the efficiency, the rpm can be higher without sacrificing as much torque. We know that brushless motors are said to have an efficiency of between 85-90%.

At 90% efficiency, we get about 70Nm of torque at 1.3 RPM, 100Nm at 0.95 RPM. Know that i can predict torque at any rpm, I can just use τ = r*f*sin(angle) to find the ideal dimensions of the gimbal arms given a certain weight, right?

Thanks Till. (Please let me know if my maths has errors in it. Also the mapping tool says 70Nm, 100Nm but the units could be, but shouldnt be off)