Brushless Gimbal motors


As some might know, some brushless gimbal motor manufacturers do not provide information on what weight their gimbal motors are ideal for (emax, for example). Contacting them is not an option because they won't reply within like half a century. My question is, is there any way to calculate the type of motor required given we know the following variables for a specific target camera payload:

Recommended voltage
Framework (Like magnets and stator numbers)

Kind Regards, Till

3 things.

  • I do not know how to calculate that from specs. If I needed a measure, I could only try it out and see.
  • I am sure that a large part of how heavy of a camera that any gimbal system will be able to handle is how close you can get the center of mass to the center of rotation.
  • Go with what works. There is a great number of people already doing this. Find the forums where they exchange ideas. Talk to them. Find someone who successfully uses a camera similar in weight to your camera. Use the same gimbal system they use.

Low KV and low resistance is what you are looking for.

A physically larger diameter will have higher torque at low speed, if all other factors are equal.

Torque scales with the rotor volume of a gearless motor with a cylindrical rotor, all else being equal
(which means magnetic field strength and maximum continuous current density in the windings).

I presume all decent gimbal motors will have their max torque listed in their data. Converting that
to a max camera weight rating is a matter of Newtonian mechanics, assuming you know the maximum
eccentricity of the camera centre-of-mass from the rotation axis. There is also the matter
of calculating the torque requirements for angular acceleration, though typically in a gimbal system
the load isn't accelerated (that's the whole point!).

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

τ * ω = I * V * E
τ = (60I * V * E) / 2pirpm

v(3s) = 11.1 V
I = 1.01 A
E (Efficiency) = 10,30,50% (We can use calculus later to maximise this)
τ = ?

So τ= 601.0111.10.1 / 2pirpm, 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 = rfsin(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)

The problem with analysis involving rpm is that a gimbal motor holds stationary, the overall efficiency
is technically zero. The torque depends on the current, and the current depends on the maximum
current rating of the motor (usually purely a thermal limit).

Speed cannot be traded for torque. That's what a gearbox does.

The motor efficiency is not constant, its a function of speed and current, and tends to zero as the speed
reduces. For most designs of motor the torque is a function of winding current and independent of speed.

Where can I find more information on torque as a function of winding current?

Thanks Till

In the datasheet for a motor.

For motors in general, there is no relationship except more current usually makes more torque.

All PMDC motors have a pretty accurate linear relationship between torque and current, its the fundamental
physics of a current carrying conductor in a magnetic field. Torque will eventually tail off at speed due
to the winding inductance.

Only when you get to wound stators does anything complicated happen, such as a series-wound DC motor
(but they tend to be in the kW range and above). With permanent magnets the relationship is simple.