Re: Torsen defined

[Orin Eman <orin@wolfenet.com>]

> btw, your equation of what the torsen "sees" at the output shafts is wrong,
> because it does not take account of the way the torsen operates at low
> torque inputs (where the shafts are free to differentiate), and pre-bias
> ratio torque levels, where the differential's internal friction minimises
> output shaft speed differences, and after the bias ratio where the diff
> allows output shaft speed differentiation. for this reason, it is much more
> usual to describe the torsen in both frictional terms (@ the bias ratio) and
> stictional terms (pre bias ratio) as this encompasses it's operation over
> the full range of operating parameters.

NO.  That equation is always correct.  It is independant of whatever
differential you have.  You could saw off the torsen and twist the
shaft with your finger and thumb and it would still be correct.
(Though a clarification for all cases would be longitudinal traction.)

It's simply the rotational equivalent of F = ma, ie if there is an
acceleration, there must be a force causing it.  For an driveshaft
to change speed, there must be a torque causing it.  You cannot
tell the difference between tractive force and inertial reaction,
it's all torque reaction to a differential.

Note that the I*dw/dt term is ZERO for the cases where there is no
acceleration - differential 'locked' and car at a constant speed.

BTW, the I*dw/dt term is what helps when you get stuck with an open
diff and rock back and forwards to get out - as one wheel spins up,
the non-spinning wheel also gets the I*dw/dt given the 50/50
distribution of the open diff.  This little extra along with the
(almost none) traction from the spinning wheel may be enough to get
you out.  You just don't get this help for very long...

> Firstly, the claim that torque is proportioned solely on tractive
> differences is false.  What the torsen sees at the output shafts is:

> traction at wheel + I * dw/dt

> where I is the moment of inertia of that shaft and everything it's
> connected to and dw/dt is the angular acceleration of that shaft.
> That I BTW, isn't insignificant.  It's what you are reducing by
> putting lighter wheels/tires on a car.