Gearbox friction model

[Chris is me]

Doesn't your OutputSpeed equation not factor in efficiency at all?

OutputSpeed = (1 - %Torque) * FreeSpeed
OutputSpeed = (1 - InputTorque/StallTorque) * FreeSpeed

InputTorque is the variable you multiply by efficiency to get OutputTorque - i.e. torque before efficiency is factored in. At a system's free speed, the InputTorque would only be enough torque to overcome kinetic friction in the system (no external load).

It's very possible I'm misunderstanding.

[IKE]

If you think about what is going on in a gearbox, it can help you with the model immensely.
With planetary and spur gears, the teeth of the gear are ment to roll relative to one another. The better the tooth, the lower the friction. This rolling will have some friction which will be proportional to load, and thus a simple efficiency factor works well espeically for moderate torques. Speeds are merely a function of the ratio.

Thus
output torque=eff*input torque*gear_ratio
Outputspeed=inputspeed/gear_ratio

While this model works pretty well, as other have pointed out, it falls apart towards "top speed". IE, if my motor has a free speed of 15,000 rpm and my planetary gearbox has a ratio on 3:1, why doesn't my gearbox freespeed = 5,000 rpm? Instead I get about 4,000 rpm!

So the other element tends to be drag. For example when riding a bicycle, you have to overcome the grade, rolling resistance, and wind resistance. On a still day, on flat ground, the initial torque is dependent on acceleration and rolling resistance, but as you speed up, wind resistance becomes a bigger factor.

For gearboxes, the "wind resistance" is often the oil or grease in the gearbox. These drag forces tend to have a torque to speed relationship of:
drag torque = C*speed^2 with C being a constant.

C is a combination of the thickness of the grease and bearings and....It is often temperature sensitive for many gearboxes.

A good way to get a value for this is to measure motor free speed, then motor gearbox free speed. Use the gear ratio and motor curve to find the torque for that speed, and then...
Set dragT=c*speed^2

For your model, you will then have:
Torque out=eff*ratio*input_torqu-c*speed^2

If you use this method, and have the ability to measure stall torque, I think you can make a very accurate model, and you will likely see a much better efficiency number than using the other methods.

For instance in high power oiled gearboxes, we might use as little a 1%-1.5% efficiency loss per gear mesh and 0.1% to 0.15% for bearings. Of course, this is with gearboxes with really good gear geometry, properly weighted oils, and high quality bearings. Overloaded gears, bushings, bearings, and really thick greases tend to perform at lower efficiencies with higher "windage", but they tend to work pretty good for FRC.

[Ether]

Originally Posted by IKE

For your model, you will then have:
Torque out=eff*ratio*input_torque-c*speed^2

The above is essentially what I used in the Drivetrain Acceleration Model, except that instead of c*speed^2 I used Kro+Krv*speed (line 85 in the C code). This one line of C code is easily changed to make it whatever function of speed you like. In FRC drivetrains, the Kro constant term is probably needed since even at very low speeds and output torque there is friction in the drivetrain (due to tension in the chains/belts, misalignment, etc).

[IKE]

Originally Posted by Ether

The above is essentially what I used in the Drivetrain Acceleration Model, except that instead of c*speed^2 I used Kro+Krv*speed (line 85 in the C code). This one line of C code is easily changed to make it whatever function of speed you like. In FRC drivetrains, the Kro constant term is probably needed since even at very low speeds and output torque there is friction in the drivetrain (due to tension in the chains/belts, misalignment, etc).

While I agree that a certain amount of friction there initially, I think that value should go away once it is overcome, much like the delta between static and dynamic coefficients of friction. If you are pushing a block on a smooth surface, you need a certain amount of push to get it going, but once it is going, the term would reduce.
I will need to think about that one a bit more, and how it would effect an acceleration model.

[Ether]

Originally Posted by IKE

While I agree that a certain amount of friction there initially, I think that value should go away once it is overcome, much like the delta between static and dynamic coefficients of friction.

I wasn't talking about static friction. Note how I carefully worded it:

at very low speeds and output torque there is friction in the drivetrain (due to tension in the chains/belts, misalignment, etc).

Low speeds, not zero speed. The tension in the chain combined with any misalignment can create a non-negligible loss even at very low speeds even if there is no load on the output. This can be modeled with a constant factor. You may also need a speed-dependent factor, as in your post.

[IKE]

I will have to look into other acceleration models to see what all they have, and how to do some simple test in order to get an accurate model of what is going on.

It would be really neat to come up with a VI or something similar that could be run with an FRC robot and output some parameters.

[Ether]

Originally Posted by IKE

I will have to look into other acceleration models to see what all they have, and how to do some simple test in order to get an accurate model of what is going on.

Consider the following scenarios:

Scenario1:

1) put the robot up on blocks, so there is no external torque load on the drivetrain.

2) apply enough voltage to the motors to get them moving

3) slowly reduce the voltage to the lowest value that still sustains a constant (very slow) speed. Since the wheels are not accelerating, there is no net torque on the wheels.


For Scenario1 I assert the following:

A) at that same voltage, a free motor (i.e. motor not connected to gearbox and drivetrain) would be spinning faster and drawing less current. In other words, the motors in the drivetrain are producing torque... but that torque gets lost through the friction of the gearbox and drivetrain.

B) using the model:

wheel_torque=eff*ratio*motor_torque- (Kro+Krv*speed)

... with wheel_torque = 0 and wheel_speed ~ 0, you get:

motor_torque = Kro/(eff*ratio)

... which allows you to choose a value for Kro which reflects the real-world situation that the motor_torque is not zero under these conditions.

C) This effect can be non-negligible for a poorly-built drivetrain (chains/belts too tight; misalignment; inadequate lubrication, etc).

D) since the net wheel_torque is zero and the wheel_speed ~0, the model:

wheel_torque=eff*ratio*motor_torque-c*speed^2

... says the motor_torque is ~zero, which does not allow the model to take into account the losses mentioned above.


Scenario2:

Same as Scenario1, except place the robot on a straight, flat, level, carpeted surface


For Scenario2 I assert the following:

A) The motor current (and thus torque) will be even higher than in Scenario1.

B) The reason is the extra losses due to rolling resistance (carpet compression) plus additional work done on the carpet due to misalignment of the wheels (or in the case of mec or omni, the stretching of the carpet due to the sideways component of the reaction force of the roller on the carpet.

C) These extra losses will include a constant term (which can be added to Kro) and a speed dependent term. Whether the speed dependent term is better approximated as linear or quadratic I do not know at this point.

[Nemo]

Originally Posted by Chris is me

Doesn't your OutputSpeed equation not factor in efficiency at all?

OutputSpeed = (1 - %Torque) * FreeSpeed
OutputSpeed = (1 - InputTorque/StallTorque) * FreeSpeed

InputTorque is the variable you multiply by efficiency to get OutputTorque - i.e. torque before efficiency is factored in. At a system's free speed, the InputTorque would only be enough torque to overcome kinetic friction in the system (no external load).

It's very possible I'm misunderstanding.

The input torque depends on the output torque. Your output torque requirement is higher if the efficiency is lower. Or, your input torque is higher for the same output torque.