# When do I transition from kinetic friction to static?

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I'm trying to figure out how to properly use the Coulomb friction model for my game. Does the transition to static friction occur when the speed is zero, or does the transition occur when normal force * static coefficient of friction / mass > speed?

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The first one. The second equation doesn't really make sense (the units don't match up).

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The second equation is close, but not quite there. You need to remember to multiply by the change in time to get velocity instead of acceleration. Once the change in speed is greater than the actual speed you just clamp it to zero.

The first method sounds like a naive idea if implemented just like that, because it would lead to oscillation about the true stopping point.

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intrest:

I goofed up while being lazy. I'm actually using an impulse based method and I manipulate the velocity vectors directly, so there isn't really any time to speak of. What I'm mostly unsure about is whether kinetic friction brings the object to a complete stop, and then static friction "digs in" so that the object requires alot of force to start again, or of the kinetic friction is "clipped" by the static friction. Which is correct?

Speed is 10 9 8 7 6 5 4 *clipped by static friction*

or

Speed is 10 9 ... 2 1 0 *static friction engaged*

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Static friction acts only when the object is not moving.

Also, remember to set null velocity when it passes zero, to prevent oscilations, since friction can't move stuff. Like this:

Acceleration: -3Time:   t1  t2  t3  t4  t5Speed:  10   7   4   1   0

If you simply apply the friction force on the object, on t5 it would still be moving, but in the opposite orientation as before.

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Quote:
 Original post by pTymNWhat I'm mostly unsure about is whether kinetic friction brings the object to a complete stop, and then static friction "digs in" so that the object requires alot of force to start again

That's correct. Under the Coulomb model, static friction applies only when the object is stationary. (The Coulomb model, of course, is an approximation, but for smooth, hard surfaces it's a surprisingly good one.)

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