[torsen] Re: [V8] differential/torque se

Andres Kroonmaa andre at online.ee
Fri Mar 21 08:00:54 EST 2003

On 20 Mar 2003, at 8:47, QSHIPQ at aol.com <QSHIPQ at aol.com> wrote:

> I believe the answer can be found below.  To understand it, think of 
> reversing input direction (engine braking torque force = reverse drive mode)
> From Zexel Gleason themselves (an SAE paper, I lost the first page):
> "1.1
> An interaxle center differential operates in the following 4 basic modes show 
> in fig 10.  In reverse, DRIVE becomes COAST and vise versa.
> In DRIVE mode, a Torque Sensing differential will distribute the higher 
> driving torque to the axle that tends to turn slower than the other one.
> In COAST mode the higher braking torque will be distributed to the axle which 
> tends to turn faster than the other one."
> That indicates to me that in reverse, more torque goes to the axle spinning 
> faster.  The torque split (TBR) doesn't change, just the allocation 
> properties of reversing the input torque.  IOW, engine braking is the same as 
> driving in reverse.  Under engine braking the effects of it in terms of 
> tractive force is nil.  Under engine torque (reverse), the effects of the 
> drive mode greatly affect tractive force.  

 Nonono. Now you confused few things. In coast mode here, speeding axle
 "comes towards" opposing force quicker, therefore higher counterforce.
 In drive mode, slower axle is encountered first while quicker axle
 "speeds away" from driving force - higher driving force to slower axle.

 In reverse gear, difference is that faster axle does not offer force
 as implied above. In reverse we still should look at "drive" mode, only
 gearing inside torsen has forces against opposite sides of toothing.

> Let's not confuse this with TBR in forward or reverse axle rotation.  It's 
> still 75/25/75.  

 Please take a look at section 4.3 and figure 7, and fig.5

 "The side gears within the differential are designed with the same hand
  of helix angle. When engine power is applied to the differential (i.e.,
  drive mode), both side gears are thrust against the same end of the
  differential housing. Alternatively, when the engine is used to brake
  the drive wheels (i.e., coast mode), the side gears are thrust against
  the opposite end of the housing. This feature provides an opportunity
  to vary frictional characteristics between opposite ends of the housing
  to vary bias ratios between the opposite directions of power transfer
  through the differential."

 I'm about this. Its so easy to do, even accidentally. Even upto simple
 fact that reverse torque is so rarely used that surfaces might be covered
 with slippy shit, causing no friction during casual usage..

 Hey, maybe indeed THAT is the reason? Maybe after some time in rear on
 torque, it would start "working" again?

> >Engine has some finite rotational acceleration. Rotational acceleration of
> >wheel is called only to compensate for that rpm difference. When engine rpm
> >is limited, torque goes to 0. Torque=Inertia*acceleration. Its quite easy
> >to derive max torque from rotational inertia if we know max rotational
> >acceleration of engine, divided by final drive.
> Keith is thinking oh no.  I was here at one point too Andre.  Torque at the 
> ring gear doesn't change, hence torque at the torsen = engine torque.  
> Tractive torque (supported by wheels on ground) + inertial torque (slip) + 
> frictional loss = torque at the torsen/engine torque.  Add more engine torque 
> beyond what's supported by wheels on the ground, that torque goes to slip.

 eh :) I'm tough guy, so it'll take you some effort to gun me down,
 even if 2 against 1 ;) I'm still suspicious about inertial torque.

 Take a tooth-pick, apply some torque with your hands. You feel it. Apply
 more, until it brakes. What happens? You feel complete loss of torque.
 There is some "acceleration", but only so much so. If you continued to
 apply "torque", you would still feel nothing. There is no torque, even
 though your hands have plenty to offer. Your hands acceleration is limited
 not by any counterforce, but by inertia of your own hands. Mass of the
 toothpick is nil.

 Have you ever tried to calculate acceleration of wheel in air when ring
 gear supposedly has same torque applied as before liftoff? Its quite
 educating. Torque = inertia x acceleration (T=Ia). Inertia of wheel is
 I=(mR^2)/2. Lets take rwd car, rear axle, one wheel liftoff, open diff.
 Inertia of wheel with 20kg mass, 0.3m radius is 0.9kgm^2. Say, engine
 gives off 200nm of torque. Drivetrain total ratio is 12 (ring gear is
 final drive). At axle, we'd have 200*12/2=1200nm of torque. That makes
 angular acceleration of 20kg wheel to be 1333rad/s^2, or 12729 rpm/s^2,
 AFTER total drivetrain. Now translate that to required engine rpm change,
 x12, thats 152750rpm/s^2! Excuse me, thats 15000rpm change in 0.1 sec,
 including overcoming all of the drivetrain friction. OUCH.

 Have we such engines?! Also, wheel with 0.3m radius has 1.8m circle, 
 so with 13krpm/s^2 change, thats 23km/s^2! You could slingshot your
 beercan to the Moon with such acceleration. Thats nuts. Something
 has to give. Engine is NOT able to give same torque at such amount
 of acceleration. Friction? Where is it larger, in drivertrain or last
 wheel bearing?

 What is realistic? 13000rpm change in 1sec? Thats >100 times less torque
 against same wheel inertia!
 So, from our torque of 1200nm at axle, all thats left is 12nm! Engine
 is simply unable to revv up faster. There is not enough of combustion
 speed and energy to do that.
 Yes, from theoretical point all that 200nm engine torque goes into spin,
 but we live in world of finites. That torque is impossible to sustain
 at such crazy accelerations.

 Conclusion: when wheel looses traction and spin develops, there cannot
 be same amount of torque at ring gear. It drops, dramatically.

> >but no, this is not the case. Reality is that if you engage full engine 
> torque,
> >then 90% of it will go to accelerate low-mu wheel, 2% on traction from it
> >and 8% to the gripping one. Or more correctly, you really can't engage all
> >of the engine torque, only 10% of it. Too quick wheel acceleration will
> >prohibit any further torque increase.
> Tractive force vs wheel spin.  Add them up, you get Teng = Tring gear

 I don't buy that. Real engines can't do that. Only tractive force can
 offer any support, even during wheel spin.

> >Take open diff. We could say it has bias ratio 1.0:1, 50/50 split at most.
> >Could we say that with one axle in air, we can apply 50% of engine torque
> >onto gripping wheel?
> An open diff allocates 50/50 ALWAYS T1=T2.  If torque is 1 then .5 goes 
> forward, .5 goes back.  If one axle is in the air, T1 = 0, then T2=0.  No 
> engine torque can go to gripping wheel.

 exactly. And same happens with torsen. 80/20, if T1=0, then T2=0.
 If T1=0.01, T2=0.04

> >"Torsen is AT MOST capable of transmitting 80% of current engine torque 
> output to the
> >axle with most traction and current torque output is limited to 5 times 
> traction
> >capacity of the axle with the least traction or limited by the drivers 
> desire not to
> >spin the low traction wheels at very high speeds.." 
> >Yes ;) thats correct. If one wants to nag further, it must be noted that
> >5 times traction of worst axle cannot exceed sum of traction of worst and
> >best axles. Or, in other words, this applies only after 80/20 max split
> >is reached, and does not apply to highgrip torque shifts due to turn radius.
> Sure it does.

 I was just pointing out that 5 times 40% of torque would result 200%
 of torque, which is error. Construct is valid when least tractive axle
 can support 20% or less of torque. Shifts due to turn radius are not
 due to _real_ traction capacity. I specially put that reservation for
 you Scott ;) Torsen behaviour in turn is peculiarity, not result of real

> Remember a Torsen isn't really sensing torque.  A torsen 
> senses axle speed diferentiation at a very low angle, that's all.  You could 
> build more slop into it so that it's more like a VC, but then you'd have 
> abrupt allocation and lose many of the straight line advantages.

 I know where you are coming from, Scott ;) This is true, and at the
 same time, it is not. You are right in that torsen is sensitive to
 forced speed difference.
 But you are wrong saying that it isn't really sensing torque. Problem
 is that torque is function of inertia times acceleration, ie. has
 velocity component by definition. F=ma or T=ia. mass being car mass,
 inertia or traction, acceleration being change in rpm, torque being
 product to be sensed. Change either - traction, or angular velocity,
 and torque product changes. We can as well say that torsen is "angular
 acceleration sensing", or angsen or accsen device. But that'd be still
 abit misleaing.

 Reason why torsen still is torque sensing, is that it works the same
 even if there is practically zero rotation. If ring gear rotates 1 rev
 per year, and applies torque of 5000nm, and there are 2 axles with
 different traction, then torsen will sense that difference. axles will
 move at different speeds, one at lets say 0.1 rev per year, other at 0.2.
 But they will have different torques in them as per TBR. So, torsen is
 insensitive to _magnitudes_ of rotational speeds, that cancels out for
 it, it sense torques no matter what rpms. Yet it still is sensitive to
 changes to ratios of speeds.

 But I can think of no way to build more slop in it..

 Andres Kroonmaa <andre at online.ee>
 CTO, Microlink Data AS
 Tel: 6501 731, Fax: 6501 725
 Pärnu mnt. 158, Tallinn
 11317 Estonia

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