how to model motor+gearbox

[James Critchley]

Interesting question.

A flat mechanical efficiency (single number such as 90%) is a Coulomb friction model. It represents the load dependent binding of the gears and shafts due to tolerances in the bearings and the relative sliding of gear tooth surfaces under load.

The "speed reduction" through a gear box is always the gear ratio (unless you start skipping teeth) so you must enforce the constant ratio of lost power through the torque by multiplying it by 0.9. That is, if I have 100 RPM and 10 ft-lbs on the input side of a mesh ratio "R" with an efficiency of 0.9, then I must have 100/R RPM on the output side, and that means the torque must be 0.9xRx10 ft-lbs.

By definition there is no load at free speed. At no load, there is no friction. Your approximation will achieve the exact free speed indicated by your gear ratio (1:10). It'll take you a little longer to get there, but the "model" will get there.

Simillarly, stall torque would be 90% of the motor torque times your (1:10).

Interestingly, backdriving the same gear box would give you a holding torque at stall of 111% of the motor torque times your (1:10).

In reality, the "flat efficiency" number is obtained as the average of common operating points and thus incorporates gear windage (speed dependent forces from spinning the bearings and squishing grease out from between the meshing teeth). There might be splits for various types of gear boxes and meshes, but I would expect it to have more to do with the intended operating speed and load of the gearbox.

To approximate the combined free speed you need to know (guess or measure) the windage component of the resistance and it should be a function of speed. Subtracting this from the motor torque curve (straight line in KOP) would give you a new motor curve... the zero point of which is the new free speed.

I know, I know... You were looking for something simple.