[Ken Leung]

After thinking about motors... I started thinking about how a drive train move the robot.

As far as I can tell... The motors used in the drive train produce a torque and angular velocity on the gears and wheel, and the wheel transform those two into linear force and linear velocity parallel to the ground.

Now, the linear force from the wheel is limited by friction, which is F = u (mu) x N (weight on powered wheels), nevertheless, there is a force from the wheel pushing against the ground. And in return, the opposite force pushes the robot forward.

Now, I understand that the speed of the robot (linear speed) can be calculated with wheel RPM and radius. But, how do the force come into play?

When the robot accelerate from 0 velocity to the linear velocity, the robot have a resisting force F = m X a, and that robot will start moving only when the linear force from the wheel got bigger than that resisting force.

Well, then... when the robot start moving, that mean the wheel force is in fact larger than resisting force of robot... and there is some extra force left, which is bigger with a more powerful motor.

I know that the left over force accelerate the robot until it reach the max speed. At that point, the robot stop accelerating. After all, the robot can't possibly go faster than the speed generated by how fast the wheel spin.

So, my question is, if there is still a force acting on the robot from the wheel, and that F = ma, then how come the robot isn't accelerating due to that force? I don't suppose drag got so big that it cancel out that pushing force... So, where did it go?

Is this left over force (wheel force - resisting force) the same as the force when robot is pushing against something (i.e. another robot)?

Do this left over force only appear when something external is pushing against it?

Or is it because the fact that we are talking about "power", and that the force is no longer a factor of velocity of the object?