# Distance traveled of the inner and outer part of the same tire

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### #1 hansforum

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Posted 19 June 2012 - 11:23

Hello!

In a corner outer part of the tire is covering greater distance then the inner part of the tire. The tire is spinning with equal speed on inner and outer part so there has to be some slip. What influence does that have on tire (car) performance? How wide the tire could be? We need differential for powered wheels to spin with different velocities in the corner so tire probably can't be as wide as a car!? I can't find anything about that topic on the internet!

### #2 GSpeedR

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Posted 19 June 2012 - 12:38

Hello!

In a corner outer part of the tire is covering greater distance then the inner part of the tire. The tire is spinning with equal speed on inner and outer part so there has to be some slip. What influence does that have on tire (car) performance? How wide the tire could be? We need differential for powered wheels to spin with different velocities in the corner so tire probably can't be as wide as a car!? I can't find anything about that topic on the internet!

This is often referred to as tire "twist" or "spin slip", caused by differences in slip speed across the tread of a tire. It can be a very significant factor in motorcycle tires at high camber; if the wheel spin axis (which is angled vertically due to camber) does not intersect the ground at that tire's turn center then there will be spin-slip, as the local slip will not cancel across the tread. Spin-slip produces both a moment and a side-force. Pacejka's "Tire and Vehicle Dynamics" provides a good description as does Cossalter's "Motorcycle Dynamics".

### #3 munks

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Posted 19 June 2012 - 16:32

I thought this was called turnslip.

### #4 GSpeedR

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Posted 19 June 2012 - 17:53

I thought this was called turnslip.

Pacejka calls it 'spin' slip in his book, Cossalter calls it a 'Twisting Moment' in his book. munks calls it 'turnslip' in his Autosport.com Bulletin Board post.

Edited by GSpeedR, 19 June 2012 - 17:54.

### #5 munks

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Posted 19 June 2012 - 20:57

I believe you are incorrect. Pacejka uses both "turn slip" and "spin slip" for two different things. From my reading, "turn slip" is what hansforum is talking about: it is due to path curvature. "Spin slip", on the other hand, is due to wheel camber.

EDIT: Actually, you may be half right. Upon re-reading and looking at the equations, I think "spin slip" includes the effects of both causes. But I'll still insist that "turn slip" is more precisely what was originally described. In any case, might want to hold off on the snark next time until you're actually correct.

Edited by munks, 19 June 2012 - 21:12.

### #6 GSpeedR

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Posted 19 June 2012 - 21:59

I believe you are incorrect. Pacejka uses both "turn slip" and "spin slip" for two different things. From my reading, "turn slip" is what hansforum is talking about: it is due to path curvature. "Spin slip", on the other hand, is due to wheel camber.

EDIT: Actually, you may be half right. Upon re-reading and looking at the equations, I think "spin slip" includes the effects of both causes. But I'll still insist that "turn slip" is more precisely what was originally described. In any case, might want to hold off on the snark next time until you're actually correct.

'Spin slip' as defined by Pacejka is 'Spin Slip' = - (Yaw_rate - omega*sin(camber))/Vel

where Yaw_rate is the vertical wheel rotation rate, omega is the angular wheel speed, and Vel is the forward velocity of the wheel center. The yaw rate term is dependent upon path curvature and obviously sin(camber) is dependent upon camber. 'Turn slip' is a simplification of 'Spin slip' neglecting the camber component. The original poster didn't specify what is causing the local slip (camber or yaw rate), only that there is a tire in a corner.

Also, I wasn't trying to be nasty above, I thought you were providing another term.

### #7 munks

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Posted 21 June 2012 - 15:21

It seems like "in a corner" implies yaw rate. You're correct that camber may be adding to/subtracting from the spin slip, but you can get that component without being in a corner (or am I having math problems today?).

No big deal, I thought you were saying that I just made up the term "turn slip", which I've dealt with previously. To be honest, I hadn't remembered the more generic "spin slip" until you made me look it up!