[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
In a message dated 12/16/02 3:50:16 AM Central Standard Time,
Keith.Maddock at trw.com writes:
>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
>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.
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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