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To a Torsen: What is Traction
More posts come to my mailbox, regarding torque, traction, slip, and track.
Go to page 10, the "unlikely" page. Let's compare this paragraph to a fwd/rwd
torsen vs a center torsen, maybe that will help.
(Trg = Engine torque - for KISS, let's assume you can transmit 100% after
friction losses etc.)
This paragraph says that as you go thru a turn with fwd/rwd, and you exceed
traction (forward rotation) on either wheel, the total (collective) Trg
(engine torque) will reduce in relation to that traction, equalizing the
torque to each axle. This reduces the likelyhood of 'spin up', as you lift or
are adding "power". So here, traction is considered forward spinning of a
wheel in relation to another. Power is added to the "outside" wheel, until it
slips, then collective Trg reduces equalizing traction. The paper indicates
that Collective traction can't exceed the combined traction supported by
either wheel. Ok let's go center Torsen. That changes the statement above to
really read: Collective traction can't exceed the combined traction supported
by ALL wheels. Let's explore what that means.
Go thru a turn, the slip angle of the rear increases to the point where the
rear driveshaft slows. You have 100% Trg. Why? Because you don't have a
"traction" (torsen defines traction as forward rotation of a sliding wheel,
power up) issue. The forward rotation is lower on the rear. So from a
physics standpoint, the torsen senses that the rear has more "traction"
ability, specifically, in this same turn as the fwd/rwd car, the torsen is
sensing front wheel slip (front wheels spinning, remember in a torsen center,
it doesn't matter which, if any, really are), when that is exactly what is NOT
happening. So torque is sent to the rear to compensate for that "sensed"
front wheel slip. Since rear traction (forward rotation) is maximum (you
haven't spun any wheel yet), Trg remains 100%. So ALL of the torsen torque
split of Trg goes to the back, since the torsen "thinks" the back wheel has
more available traction. It goes to maximum, because, you haven't done
anything with power to make the torsen think the back is slipping. In fact,
the disparity is large enough that you need to really power up to get a wheel
As soon as you power up, the back wheels eventually will spin faster than the
fronts. Understand, from a center torsen physics, you can be in a turn, with
that "unlikely" power up happening. Why? Because, in order for the Torsen to
sense too much power in the rear, you HAVE TO BE spinning a rear wheel, by
definition. Given the right slip angle, you can spin the rear tires all the
way thru a turn, with MAXIMUM Trg. "Unlikely"? By design, you can do it.
However, since torsen is immediate, that line is pretty well balanced. Why?
Because a change in apex angle, too much throttle input, or a change in
steering, will upset that balance (spinning the rear axle + track + slip angle
to equal the non spinning front) as you go thru the turn. So what happens?
Well, let's say you added too much power. Now, Torsen senses rear wheel
slipping more than the fronts, so the Trg is sent to the front wheels. Since
weight shift is to the rear, and you have an open front diff, all the transfer
goes to the inside front wheel, effectively wanting to spin it. In the
meantime, the back slows again, because you are back to slip angle (not a
traction problem you tried to equalize with power up) slower rotation. So,
torsen senses front inside wheel slip (faster front driveshaft/slower rear),
and torque goes to the back again. The total Trg doesn't need to be anything
but 100% thruout this process. Where is the Trg equalization? I say the
"unlikely" is EXACTLY what happens in a audi Center Torsen car.
Lower cf? Ok, now we can talk about a lower Trg. But as you lower cf, you
haven't changed the scenario involved above, you just changed the amount of
Trg that is significant. I have experienced the significance of that total
Trg in a NA '88 90q. You don't change what happens, only at what Trg and what
slip angle it will happen at. The higher the cf > the higher the slip angle>
the higher the Trg. Does the same hold true in reverse? I argue so. The the
lower the cf > the lower the slip angle > the lower the Trg. Both scenarios
significant. Why? Because you exactly don't have a "traction" problem, as
is defined in this article, you have a track and slip problem. The torsen is
defined in terms of traction only, in a Center diff application. At a certain
track and slip, the torsen becomes fooled, by the very same dyanamics that
makes it a great traction device.
So, I look at page 10, and see the definition of "the bite". Really looking
at it, or experiencing it, makes you wonder why traction was dictated over
chassis dynamics of that chassis. Is that the 7/10ths world we were given in
1988? "It's unlikely that you will see that 3/10ths. I object, btdt. I
wonder given the change from 1998>, whether some others did as well.
Referencing article found at: