Society of Robots  Robot Forum
Mechanics and Construction => Mechanics and Construction => Topic started by: crazyengineer on March 12, 2013, 08:09:09 PM

Okay, so I'm deriving the relationship between torque and the mass. I just want to verify my derivation is correct. I'm creating a wheeled robot which it has two wheels each connected to its own motor and a ball caster at the end of the bot. The robot is traveling on a perfectly flat surface
Torque=force*wheel_radius => force=Torque/wheel_radius
Friction_force=kinetic_friction_ coefficient*normal force
Assuming a 2D plane, the normal force is equal to the weight of the robot
Friction_force=kinetic_friction_ coefficient*mass*gravity
So the summation of forces is equal to mass*acceleration. Since the friction force is opposing the force generated from the motor, and the two motors' torque will add together, I can make this assumption
mass*acceleration=2*ForceFriction_force
mass*acceleration=2*(Torque/wheel_radius)kinetic_friction_ coefficient*mass*gravity
After some algebra, I was able to determine the following
Torque=(wheel_radius/2)*(mass*acceleration+kinetic_friction_ coefficient*mass*gravity)

That would be true if you were dragging the wheel over the surface. However, the wheel is rolling, so the friction force you need to model is rolling friction (plus losses in transmission) which is generally a lot less than sliding friction.
Also, using "2" is confusing, because that's a hardcoded constant based on your number of motors (which is 2.) Using "Nmotors" or doing the calculation for the sum, and later dividing by number of motors, would be clearer IMO.

Okay! When I did a quick Google Search http://www.phy.davidson.edu/fachome/dmb/PY430/Friction/rolling.html (http://www.phy.davidson.edu/fachome/dmb/PY430/Friction/rolling.html), I found that rolling friction is
Fr=u/(r*N)
So I need to replace the friction force in my equation with the rolling friction force? Also, I'm neglecting the effects of transmission for the time being.