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Re: The Torsen Mentality of Slip Angle

>>i understand what you are saying about slip angles, i just have no
>>about how they would affect the operation of the torsen.  by definiton they
>>affect the torque reaction of the wheels, and so would affect torque
>>distribution.  but by how much?  on dry roads, not much at all (relative f:r
>>split), although on snow, it could be more.  the same factor affects the open
>>or locked centre diff...
>NO.  Not correct.  Slip Angle to a fixed diff is 50/50, to an open diff, it
>becomes 100 front.  Both defined as understeer.  Remember, no wheels have left
>the ground, so a fixed diff is fixed at 50/50.  Why? Because of the same
>"equalizing" effect you describe with wheel lift.  If cf is constant on all 4
>wheels and they are grounded, you will have 50/50.  With an open diff, the
>fronts will spin before the rears, and more torque sent forward (lift and
>absolute traction are the same thing).  So, we can say that understeer will
>stay with the above two scenarios, both over and understeer will occur in the
>Torsen Center.  

but you've just said that the 'cf' *isn't* constant, because you've got slip
front or rear...

orin's point is that 50:50 does not occur with slip and a locked centre.

>Understand Dave, the difference between your "loss of traction" definition and
>the Torsens interpretation.  The Center Torsen is looking ONLY for torque
>indicators in the front and rear driveshaft, it isn't looking at traction per


>say.  Slip Angle is traction to a Torsen.  You are looking for forward
>rotation, doesn't need to be.  To us lay folk, a tire sliding sideways has
>"lost traction".  To a Center Torsen, a tire sliding side is a tire that has
>GAINED traction.  Less Slide to relative to opposing wheels (slip angle) =

no, i disagree.  a tyre sliding sidewards has lost traction; it's slipping, and
the torsen recognises this by it's *decreased* torque reaction ie. through the
slipping wheels.

why do you say that a tyre slipping sidewards loses forward movement when a car
is cornering?  when a wheel has a large slip angle, it still has forward
rotation, and if a tyre has lost forward movement, it is skidding.  to put this
another way: how does a wheel lose forward rotation - what is happening to that
[virtual axle] to rob the axle of forward rotation?

if a skid (slip) occurs under braking, then the torsen is not involved (and the
abs is at work to prevent it with torsen).  if a tyre is slippng under
acceleration, then traction loss is involved and the torque reaction
communicates this fact to the torsen.

>Less Traction.  Absolutely opposite of reality.  Specifically, to a Center
>Torsen,  if the front is sliding sideways less than the back (like steering
>with a slide around an apex), the rear has more traction.  Reread that.  To
>your definition that is false, to the Torsen it will transfer torque Tmax to
>exactly your axle with LESS traction (as you defined it).  See any problems
>with that?  Are you sure you want to put that in the lower percentile?  Want
>to look at cf first?  I don't think you need to based on my experiences...

no, in this case, if you're steering with a slide around the apex
(oversteering), the front is travelling *less* distance around the corner than
the rear, by definition.

because it is travelling less, then the torsen sees a better torque reaction
and feeds the power to the *correct* [front] axle.  this is what you want.

if you're understeering around the corner, your front wheels are going to be
travelling *more* distance around the corner than the rears, and agin the
torsen will make the reight decisions...

this is perfectly illustrated in the behaviour of the 'mb' and 'rr' torsen
ur-quattros and their relative lack of understeer compared to the 'wr'
generation 1 ur-quattros.

why do *you* think that the torsen ur-quattro's understeer less?

perhaps i should suggest a test session with the appropriate cars on a track. 
volunteers anyone? (have to the be uk, i guess).

'95 rs2
'90 ur-q