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how does an open differential work (was RE: torsens & 'torque split')

> Sorry, but just because a wheel is slipping with the open diff
> doesn't mean the other wheel isn't receiving any torque.  The
> mechanics
> and physics related to the forces involved indicate that the none
> slipping wheel receives exactly the same torque as the slipping wheel.
> I'm not even going to attempt to do an ASCII drawing of the open diff,
> but will try to do it in words.
I have been studying an exploded drawing of an open diff in Carrol
Smiths supplement to racecar engineering - the racing differential, so
no acsii art needed. BTW this supplement contains some very nice
descriptions of various differentials used in racing. Better than the
descriptions in tune/prepare to win.

> The input torque goes to the axle of the bevel gears.  Looking at just
> one of the bevel gears (they all act the same), one tooth on
> each side of this gear is engaged with a tooth on each of the drive
> axles.  Now think of pulling on the bevel gears center.  It pulls on
> both drive shafts equally.  Now as one side slips the bevel gear
> turns,
> but _it is still pulling on both axles_.  If it _isn't_ pulling on
> both axles with 50% at each, there will be a resulting moment on
> the bevel gear and it accelerates, as does the spinning wheel,
> as does the drivetrain and engine.
What you describe is just how it works for compensating for speed
differences between the output shafts. Now this is what Carrol writes
when a wheel slips.

"This rotation allows the pinions(your bevel gears) to compensate for
the difference in rotational speed of the side gears(connected to drive
axles) - while continuing to drive both side gears. This differential
action is self adjusting to any variation in axle speed - but only until
one wheel begins to slip.

When one wheel  begins to slip or to spin, the case will continue to
rotate the pinions but, instead of driving both side gears, the pinions
will take the path of least resistance, rotate on their shaft, and
"walk" around the axle connected to the wheel with greater traction. All
the torque will be directed toward the spinning wheel. 

Under "normal" conditions, the open differential is a perfectly
satisfactory device, delivering power to each driven wheel, and
providing smooth differential action when needed. However, when the
available torque exceeds the tractive capacity of the driven tyres -
either through a decrease in the frictional characteristics of the road
surface, through lateral load transfer, or through an excess of
torque(or any combination thereof) - all of the power will be delivered
to the tyre with the lesser amount of grip. This leads to wheelspin, and
the reduction of both traction and control."
> Here is where we have the problem.  Lets look at it this way.
> The wheel with poor traction tries to spin... what prevents it
> from spinning?  The other wheel of course!  One wheel is trying
> to spin, the other digs in harder and says no you can't...
> ie. _more torque at the non-spinning wheel_.
The spinning wheel is spinning, it is rotating at the same speed as the
driving wheel. And it needs torque applied to it to keep it spinning at
a constant speed.

> From Newton's laws, all the torque going into the system must
> be accounted for.  It is either balanced by the traction of
> the wheels or, if there isn't enough, the wheels will accelerate
> (spin).  Conversely, if nothing is accelerating, then wheel
> traction out is equal to torque in.  It all derives from this
> last sentance.... the 50/50 for an open diff and the 100/0
> for the locked diff.
Don't ya just love them newtons laws... I totally agree with accounting
for all the torque and the conservation of angular momentum. But I don't
agree with that last sentence.  Maybe it's time to get real nerdy and do
a numerical example...
and really bore the masses.