Quote:
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.
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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.