# [torsen] RE: Snow Driving

Tue Dec 17 07:10:03 EST 2002

```OK, I'll definitely agree that you have to account for gear ratio and friction losses when calculating Trg from Teng.
Lets assume at the moment, 1:1 transmission ratio and some good fish-oil-lube with no friction :-)

However the rest I disagree with...

If you want to claim "Torque Out =Traction Torque", you're neglecting a lot.

Agreed that velocity has (nearly) nothing to do with the torque equation.

But acceleration has a LOT to do with it.

Think of what happens right when a axle slips. (at the time when input torque exceeds Mu*N)

What then happens to the slipping wheels?

They "spin up".  Yes they increase in velocity, but to do this they must accelerate.
Angular Acceleration requires  torque.

linear: Force = Mass * Acceleration
angular Torque = Inertia * AngularAcceleration (Alpha)

So now you have torque being used to accelerate the spinning wheel (and rotor and driveshaft), that is solely a function of the mass and dimensions of the spinning wheel/rotor/driveshaft.  Where does this torque come from?  It has to come from the center differential!  But, it has nothing to do with the traction.

I propose this equation
((Teng * GearRatio)-TransmissionFriction)=Trg=TractionTorque + WheelInertia*Alpha + AxleFriction + BrakingTorque

(We'll keep it simple and  assume Teng includes the effects of engine and transmission inertia)

(WheelInertia being the combined inertial of the wheel, rotor, driveshafts, and other rotating items)
(Axle friction being pretty negligible friction from the wheel bearings etc, this is related to velocity though but a very small effect we can neglect)
(Braking toque being any drag from the brake pads, whether caliper hangup,  driver, or EDL induced)

I wish I could invite you into a instrumented vehicle, we'd head to the test track and I'd show you exactly what's happening.
As a substitute, let me explain.

To keep this simple, lets examine a  RWD vehicle.
Assume the engine/trans always outputs 100Nm to the rear diff at full throttle.

One rear wheel is on asphalt (high mu wheel), and the other rear wheel is on ice.(low mu wheel)
Based on the weight of this vehicle, the tires, and the surface, the traction capability is:
High Mu Wheel: 90 Nm
Low Mu Wheel 10 Nm

First consider our favorite the locker.  Lock it up, and launch full throttle.
No problem.
Trg = 100 T1 = 90 T2= 10
Simple acceleration, no problems, no wheel spin.

Now consider a open differential.
Cautious driver

20% throttle  Trg = 20, T1=T2=10
No slip, tepid acceleration.

30% throttle Trg=30, T1=T2=15
Uh oh!  T2 has now exceeded the traction force. And the wheel spins up.
How fast does it spin up?  Simple. 5Nm/Inertia.

Assume the driver holds the throttle. Remember the same amount of gas is going to the engine.
The same amount of fuel air is exploding.  The same amount of force on the pistons is being generated
The same amount of torque is being generated on the crankshaft, so the Trg is the same!
(if anything it goes up as the engine starts to spin faster!!)

So the wheel keeps accelerating.  Higher and higher and higher.  The only limit is the driver taking his foot of the gas (from the noise of the spinning or his insane speedometer reading, or redline) or from the vehicle instability.

Regardless as long as the drive has his foot on the gas, and can stand  the insane wheelspin on the ice, he'll get that 15Nm to the high mu wheel.

Heck for a short time, he can even get the full 50Nm to the high mu wheel, but that amount of time is pretty short as he'll hit redline pretty quickly since one wheel is going 60mph. Or if he's got an automatic it'll shift all the way up! No problem.

But nobody is going to sustain wheel spin like that, so they'll back off the throttle to keep the wheel spin nominal, which means they've backed off to 20% throttle.

A torsen will do us one step better. We'll assume a max TBR of 3:

So now we can  have up to 40% throttle without slip
Teng = 40  T1=30  T1=3*T2 T2=10

Once we get more than 40% throttle, T2 will exceed the traction limit of that wheel, and it will start to spin up.

Once again we have the same problem as the open diff, the torque to the high mu wheel is limited by the driver to whatever amount of wheelspin he wants to generate on the low-mu wheel.  This wheelspin is, as explained before, limited by the speed limit of the engine and/or the balls of the driver.

Of course in either the open or torsen case, the EDL could kick in and absorb T2 to allow more T1, but that's not the scope of the conversation ;)

Anyway to sum it up, the." constant apposing force of equal summed value" that you are looking for is
"FrictionForce + Inertia*Alpha".

If this isn't sinking in, think again very carefully what happens in the RWD split-mu situation if you just FLOOR it.
Then think of what YOU would do.  My guess is "let off the throttle to reduce slip", which reduces Teng, reducing Trg.
But this reduction is from the DRIVER, not as a direct physical reaction of the slip itself.

If you want to continue your claims as below, that Trg goes down when a axle starts to slip, you must then explain where this torque goes.
Either
A) how slipping a wheel reduces engine power (without the driver lifting the accelerator)
or
B) some mystery torque accumulator.

Also you must explain how a wheel spins up without requiring torque to do so!

Cheers,
Keith

****************************************************************
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> 18:12:27 16.12.2002 >>>
Keith:
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,

>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
drivetrain.

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:
http://www.howstuffworks.com/four-wheel-drive1.htm
"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

Torsen
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.

SJ

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