[torsen] RE: Snow Driving
Keith.Maddock at trw.com
Tue Dec 17 08:50:13 EST 2002
I'll say it again.
Please reread at least the first few paragraphs of my post.
"Agreed that velocity has (nearly) nothing to do with the torque equation." !!!!
In my 4+ years of looking at data from vehicles with slipping wheels, I have never seen an instance where a wheel is slipping at a constant velocity without external control from ABS/TC/DSC. A slipping wheel is nearly always going to have some sort of acceleration.
Please tell me where the torque is coming from to accelerate my spinning wheel at 10 g's once it loses grip?
Are you saying it doesnt go through the differential?
Can you fax/scan/email these documents?
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 >>>
I think you are missing something basic here. The key is the statement in
the Chocholek paper. "Power of course is the product of torque and
rotational speed. However, since it is possible to express vehicular
traction as a reaction force acting at a given drive wheel radius, traction
considerations related to the function of power transfer to the drive axles
may be expressed in terms of torque alone."
Keith, I read exactly into this that ANY tractive forces are MEASURED as
either T1 or T2 torque, and that their sum is Trg. IOW, skinnny tires,
contact patch, tread, size, hardness, inflation, even spin, cause a
measureable torque difference at that axle. So, yes you might see slight
drop in T1 if the wheel spins up "more" as it approaches 100% wheel slip.
But the point of Chocholek's is still valid, HOW you measure that is a torque
equation, and torque alone can be used to describe the actions of the DEVICE
allocating it, REGARDLESS of what's at the end of the shaft, or it's
velocity. Whatever spin velocity you have, that speed on THAT radius wheel,
under x conditions will result in a "supported" torque measure at T1 or T2,
For more on this, read C368/88 (Chocholek - The development of a differential
for the improvement of traction control). Where you can clearly see your
math is irrelevent is in 885140 (The influence of a Torsen Centre
Differential On the Handling of Four-Wheel Drive Vehicles). Referencing
figure 5, the text reads: "With the same tractive force (max loading, going
from .9 to .4 road friction coefficient - open diff) on the surface with less
grip, the limit of adhesion is exceeded at the front wheels. The front
wheeels spin, and the tractive forces at the front and rear wheels are
reduced to 1800N which is the value that can be trasmitted with 100% wheel
slip." Drive slip on x, supported torque on y, NO reference to velocity.
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.
In a message dated 12/17/02 6:01:19 AM Central Standard Time,
Keith.Maddock at trw.com writes:
Also you must explain how a wheel spins up without requiring torque to do so!
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