[torsen] RE: Snow Driving

Tue Dec 17 09:12:08 EST 2002

```Agreed with a reduction of traction as a wheel slips more and more.  The Coefficient of Friction (Mu) decreases as slip increases.
Classic Mu-Slip curve.  Depending on the surface, Mu increases from a low number to it's peak from 0% to about 10% slip.  Anything more than 10% slip and you lose Mu, and since traction is Mu*N, you lose traction.

Funny how all Mu-Slip curves stop at 100% slip.  Clearly because they are usually used to document traction for braking conditions, where the most slip you can achieve *IS* 100%.   But think of positive slip (spinup) for acceleration conditions as evaluated for traction control (via diff's, engines, EDL, whatever) .  If my wheel is going 70 m/s and the vehicle is only going 5 m/s, that's a hell of a lot more than 100% slip.  Better bet that the Mu is even less there than at 100% slip.

But don't forget, I never claimed the point was velocity of slip.  It's acceleration of slip...

Wheels accelerate when they slip

Acceleration is impossible without torque

Torque must come from somewhere.

I ask again for you to tell me where the torque comes from to accelerate a slipping wheel :)

Try to think outside of the constraints of these book engineers and their SAE papers which are great... but don't forget the real world :)

Do you have MS-EXCEL to look at some data?

Keith

****************************************************************
Keith Maddock, TRW Automotive,  Koblenz, Germany
Slip Control Systems, Systems Design, Traction Control
+49 (0)261/ 895 2474     -    -    keith.maddock at trw.com

>>> <QSHIPQ at aol.com> 14:29:02 17.12.2002 >>>
(snip)
I read this to exactly mean that at ANY velocity of a given wheel, the amount
of supported torque becomes a torque number at a drive axle, period.  I also
read this to mean that that torque number will decrease (and velocity
increase) down to min xT1 = torque supported with 100% slip, then become a
constant, regardless of torque applied.

Keith, you and I don't disagree, velocity is a component of "traction".
However, these papers clearly indicate that whatever the T1 and T2 are
attached to really doesn't matter, the result of that attachment affects T1
and T2. That's all.  I don't even want to do your math, I can agree with it
or not, it really is irrelevent, I can MEASURE it as T1 or T2, and further
adding T1 and T2 gives me Trg.  With LSD attached, the mix of T1 and T2 is
affected, but the total still equates to Trg.

HTH

Scott Justusson

In a message dated 12/17/02 6:01:19 AM Central Standard Time,