[torsen] RE: Snow Driving

QSHIPQ at aol.com QSHIPQ at aol.com
Mon Dec 16 12:31:04 EST 2002

Comments inserted, I'll respond to this in the context of the torsen in a 
center application:
In a message dated 12/16/02 3:50:16 AM Central Standard Time, 
Keith.Maddock at trw.com writes:

>Scott - 
>Trg has to equal Teng.
Never happens for two reasons.  The first is strictly math, you might be able 
to argue that Trg = Teng + gear ratio.  That isn't technically correct 
either, cuz in actuality, Trg = Teng + gear ratio - friction loss.  The 
reason we need to speak of Trg in terms of what the torsen allocates, is that 
the torsen is exactly transferring the torque available at the input of the 
device, IOW plus the gear ratio and minus the friction loss between Teng and 
torsen.  In either case, Trg or Teng can't exceed available traction.

>Don't forget that torque going to the axles  has can go to either or both of 
the >following
>A) Providing torque to the tire/road interface
>B) providing power to accelerate the rotational speed of the rotating 

Hmm.  Torque is a variable, not a constant.  Torque can never exceed 
availalbe traction.  Hence Trg is reduced as a wheel slips.  For more on the 
concept, run the video and read the print at:
"The interesting thing about torque is that in low-traction situations, the 
maximum amount of torque that can be created is determined by the amount of 
traction, not by the engine."   
OR for a more simple explanation (SAE 885140)
"Engine torque applied to the ring gear (Trg) is substantially equal in 
magnitude to the sum of the reaction torques which are developed at each 
drive axle."  I read that to exactly translate into:
Trg = T1 +T2
As such, if you decrease traction of the front axle (T1), or the rear axle 
(T2) by slipping, Trg must decrease so the exact formula stays the same, 
which supports my claim that Trg (Defined here as Trg = (Teng + gear 
mutiplier) - frictional loss)) = "torsen input" (since audis don't 
technically have a "ring gear") can never generate more torque than the 
available traction.  Teng can never generate more torque than the available 
traction plus gear ratio minus frictional loss.  

The torsen can *shift* the mix of T1 and T2, the effect of which is to:  
"....support a torque imbalance between drive axles which contributes to the 
total amount of torque which can be transferred from the engine to the drive 
axles when the amount of torque which can be supported in one of the drive 
axles is limited by available traction."  IOW, if T1 is slipping, and T2 
isn't, your Trg isn't reduced until you've exceeded the Bias Ratio.

A better idea might be to plug and play here.  Let's say you have Teng of 
75lb/ft at 5krpm.  Gear ratio is 1.50, and frictional loss of 12.5%.  So
Trg = 100lb/ft
A awd car will allocate that with no wheel slip as T1 + T2 = Trg.  So, if no 
wheels are slipping at all, a torsen, locker or open will allocate the above 
100lb/ft = T1(50lb/ft) + T2(50lb/ft).  Simple.  Ok, let's slip a drive axle.

Front starts slipping, let's assume the rears aren't.  In an open diff:
24lb/ft + 24lb/ft = Trg, Trg = 48lb/ft.  This is because if either axle 
starts slipping, the torque supported by T1 and T2 are equal. always.

Locked diff:
24lb/ft + 76lb/ft = Trg, Trg = 100lb/ft

24lb/ft + (24 * 3TBR=) = Trg, Trg = 96lb/ft

>In the case that input torque to an axle exceeds it's friction capability at 
the >tire/road interface (SurfaceMu*NormalForce), the torque "T1" or "T2" 
doesn't >decrease, the excess over Mu*N just goes into making the wheels spin 
faster >rather than pushing the vehicle.  Eventually Teng will decrease 
because the driver >lifts the gas pedal because they see their speedo pegged 
at 180mph while the >vehicle is sitting still (depending on what the speedo 
uses to detect vehicle >speed), or they hear the spinning wheels  (depends on 
the skill level of the driver :->D  )

I'm not buying into the argument that Trg is a constant, cuz it can never be 
unless no wheels are slipping.  Dynos can't calculate slip, all dynos go to 
maximum traction (tires on a chassis dyno, or brake torque in a engine dyno) 
and that's the measure.  It will error (or just break), because it can't 
measure loss as a function of slip velocity.  

Please feel free to point me in the right direction Keith.  This is the first 
time I've heard that Teng or Trg is a constant.  I just don't see that at 
all, no without a constant apposing force of equal summed value.


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