[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 
the same axle speeds, wheels would have to be straight ahead, and the loading 
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 
reverse, not about traction.

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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