Sign in to follow this  
pTymN

When do I transition from kinetic friction to static?

Recommended Posts

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?

Share this post


Link to post
Share on other sites
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.

Share this post


Link to post
Share on other sites
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*

Share this post


Link to post
Share on other sites
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: -3
Time: t1 t2 t3 t4 t5
Speed: 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.

Share this post


Link to post
Share on other sites
Quote:
Original post by pTymN
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

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.)

Share this post


Link to post
Share on other sites

Create an account or sign in to comment

You need to be a member in order to leave a comment

Create an account

Sign up for a new account in our community. It's easy!

Register a new account

Sign in

Already have an account? Sign in here.

Sign In Now

Sign in to follow this