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*To*: ejfluhr@austin.ibm.com*Subject*: A simple answer here is Slip Angle*From*: QSHIPQ <QSHIPQ@aol.com>*Date*: Tue, 3 Mar 1998 15:25:03 EST*Cc*: quattro@coimbra.ans.net*Sender*: owner-quattro@coimbra.ans.net

Mr. Fluhr quieries: >How are the rear wheels turning slower than the fronts in this <spider bite> model >of cornering? As slip angle increases, traction (forward rotation of a spinning wheel) decreases, by definition (5 degrees of slip angle is considered significant, btw). Forward rotation of a spinning wheel decreases, driveshaft speed decreases, by definition. Since a front can be steered to a "lower" slip angle, (sliding with a skid lowers relative f/r slip angle, turning into the slide raises it, but you spin), you have a front and rear driveshaft at different speeds, by definition. The torsen doesn't give a care if you have a slip angle problem or a traction problem, in this case IT FUNCTIONS that the two are the same. So power (torque) is sent to the rear wheel in this case, since torsen will sense the slower rear driveshaft to "MEAN" that the front is sliding from a traction loss. Since all wheels really don't have a traction problem, but a slip angle problem, MAX Trg to the f/r diffs doesn't decrease (Dave E.'s claim of "equalization" - Page 10 - "unlikely"). Max Trg in this case decreases with Forward rotation (traction), NOT Slip Angle. So power up, add traction, the rear wheels will spin, eventually the spinning will be equal and/or faster than the front, and Torsen shifts again. As soon as it does, you have a slip problem in the rear again. Make sense? HTH Scott Justusson QSHIPQ@aol.com

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