# [torsen] RE: differential/torque sensing

QSHIPQ at aol.com QSHIPQ at aol.com
Sat Mar 29 13:52:17 EST 2003

```Andre:
Hang in there...  I believe I see part of the problem.  Do you agree with the
statement:
In straight line acceleration, front rear torque in a center torsen/locker
follows exactly weight distribution up to the TBR?

ANY tire slip causes axle slip = torsen reallocation.  I appreciate your
paradoxes, and will address both.  In order for a 4wheel drift to accomplish
on front and rear wheels exactly identical.  I have a hard time
conceptualizing that scenario on any 60/40 quattro.  Technically under
conditions of 4 wheel lockup, you'd have Trg=0 anyway (either car is stalled,
or clutch is in)= no torque to allocate.

Reverse/coasting.  I'm not coming to the same conclusion wrt reverse engine
torque.  If you have reverse rotation at a low engine torque (coast), you
will already have maxed the gear to the side housing when starting in
reverse.

Andre, from my perspective 2 more statements appear to support my conclusion
that reverse torque is driving in reverse or engine brake torque:
3.2 in c498/30/14
"In case of too low a locking effect in mode 3 (Drive:  Reverse driving or
COAST:  rear high alxe torque), the consequences are not relevent as far as
vehicle stability is concerned."  A true statement about stability in

In SAE 885140 fig 8 (bevel/open vs torsen center) you can see that the engine
torque in the bevel is allocated 50f/50r under maximum braking.  In the
torsen under maximum braking 25% of the engine torque is at the front wheels,
75% is at the rear wheels.  Problem supported in text:  Front wheels are
slower than rears.

Andre,  I assure you I read your posts in great detail.  I believe that you
might find that this reverse torque and the friction forces are not "equal
and uniform" between forward and reverse rotation.  As I inversely impose the
frictional forces on the helix gear in reverse, I see gears going to the
opposing side of the basket vs forward rotation.  Isn't it indeed true that
in order for the forward and reverse torque allocations to be identical, you
would need to change the helix gear angle so that it ramps in the same
fashion?  Specifically picturing fig 5, the helix gear would need to have a
reverse gear cut down (like an inverted "V") top to bottom with the same
frictional properties?

I would propose that the way you have presented your theory, at maximum brake
force of 80f/20r you should have 75f/25r engine retardation force, when it's
actually exactly opposite according to 885140.

Still willing to learn, just not seeing your light yet.  WRT engine torque
causing wheel spin up, if this is not a truism, why would the Keith's of the
world add engine output control systems to EDL with torsens?  C498/30/14
"LSD can proportion but not limit torque."  That statement alone indicates
that Trg = T1+T2.  T1 or T2 = sum of internal (acceleration) and external
torque (traction) to a given differential axle.  ANY slip decreases external
torque as it increases the internal torque.  The rate is exactly inversely
proportional.

SJ

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