# Friction Puzzle

This is a thought experiment I’m trying to carry out, but so far unsuccessfully.

imagine a square robot with 4 wheels , where the weight is distributed evenly between the wheels.

the wheels are in the following configuration
(lines are chassis, *** are wheels)

``````_________
|  ***   |
|*      *|
|*      *|
|__***___|
``````

now imagine you power only two of the wheels, the right-most and left-most for example.

will the robot move?

-Leav

One set of CoF numbers for the wheels seems to suggest that it would not.

However, AndyMark only has a single listing for CoF on their website which would suggest that the inline and transverse may be the same.

At least one team has already come to a similar conclusion with an empirical test.

In this case there is a slight chance your robot will drive (the strict math says it won’t be able to start). Even if it does drive, it should drive very slowly.

The one way I see that you could accomplish this would be to “deploy” the sideways pair of wheels using pneumatics when you decide to use them.

I did the math a couple of days ago. Using a simple model of friction, the robot will not move.

However, if you can get those unhelpful wheels to start sliding, it will move just fine. Just spin them quickly (perhaps alternating directions as rapidly as you feel comfortable doing) and they’ll essentially lose much of their friction.

It depends on the center of gravity and the configuration of the wheels, i.e. are the wheels co-planer or are one or more wheels raised (or lowered).

When wheels are co-planer and the CG is in the geometric center then each wheel supports an equal weight and the normal forces are equal. If the transverse friction is higher than the inline (or longitudinal) friction then the physics say the robot will not move. Spinning the wheels might not help this situation. It will lower the inline friction, but the static transverse friction still applies - not the dynamic. For the dynamic transverse friction to apply, you generally need movement in the transverse direction. (To understand why this could be so think about a grooved wheel on a deformable surface - spinning in the grove won’t help you get over the humps between the grooves ) This is true for the standard friction model but may be different with heat effects, lubrication effects, etc. are present.

If you drop (or raise) an opposing set of wheels, then the physics is quite different. The robot will be supported by at most 3 wheels. If the CG is in the geometric center, almost all of the normal force will be on the set of opposing wheels that is dropped. With a center CG the robot will teeter around the axis of the supporting wheels but both of the raised wheels will never be on the floor at the same time. To calculate the actual normal forces, you have to calculate the forces in three dimensions, accounting for the height of the CG. You also have to account for the force on the trailer hitch which can create an additional moment. Bottom line is that the robot will easily move, even with a transverse/inline ratio of 2.3. The friction force from the lower wheel will be very low because of the low normal force.

With a CG near the center, nearly all of the normal force will be on the two dropped wheels and if these are powered they can deliver the maximum force possible. If you drive those wheels in opposing directions, you will also have pretty close to the maximum possible turning moment, especially if these wheels are along the longest dimension. (Turning the orientation of the robot, however, does not necessarily turn the velocity vector).

We think it’s worth evaluating your configuration. In fact, we are currently evaluating a similar configuration with two opposing driven wheels and two opposing steerable wheels, as well as a couple of other unconventional configurations involving raisable wheels and pairs of driven wheels. Note what happens if your dropped set of opposing wheels becomes your raised set and visa versa, especially if the documented transverse/inline ratio proves to be correct.