# Coefficient of Friction and Maximum Acceleration

Hey guys,
I’m the programmer of team 2609, 2nd year team.
I’m wondering if the coefficient of friction, as listed in the manual, is accurate. Does any team have a different value that provides more accurate calculations?

And for those teams with traction control, what’s the maximum acceleration of your robot driving on the crater?

Thanks everyone!
-Peter
2609 BeaverWorX

I seem to recollect that my mentor has said several times that .06G is the maximum acceleration force before wheels burnout. How accurate this is I am not sure.

now, is that the acceleration for the entire system or the angular acceleration of the wheels?
I’m trying to confirm a few calculations.

It’s for the whole system. A traction limited robot can’t accelerate faster than the coefficient of friction times one g. If the gravitational acceleration on Earth were, say, two gs, then there would be twice as much traction because traction (friction) is equal to the force due to gravitational acceleration (normal force on the wheels) times the coefficient of friction, and the robot would be able to accelerate at two gs. On the moon, of course, where gravitational force is one sixth of a g, the maximal acceleration will be a sixth of a g times the coefficient of friction. Hence the use of the slippery wheels and “regolith” in the Lunacy game to sort of simulate the traction available on the moon.

Accordingly, two traction limited robots, one light, and one heavy, will both have the same acceleration limit, so a drag race will be a tie, but in a pushing contest, the heavy robot wins.

Our open-loop traction control system maxed out (fully weighted with trailer) with our internal acceleration value set to 24 inches per second squared, which works out to 0.06g’s. The competition robot will be a little bit faster because it’ll be heavier, but we’re leaving our setting at that.

our experience was that the book values were good… for brand new wheels right out of the box on brand new regolith. After a bit of normal wear and tear, some portions of the wheel (particularly the area where the injection molding plug was) may have higher coefficients of friction.

From my understanding, mu*g will not give you the maximum possible acceleration of the robot because not all of the “system’s” weight is on the driven wheels (you need to consider the trailer!)

So, letting m be the mass of the robot and M be the mass of the robot and trailer, the maximum tractional force is mugm and so the maximum acceleration is mugm/M .