On Thu, 12 Mar 1998, Allen Nugent wrote:
> >... What happens is that when you apply that brake,
> >the braking action is applied to the wheel with the least traction -
> >through the diff. In this case one wheel locks (because of the
> >differentiating action), and you enter into a dramatic spin.
>
> I can't see why this should happen. You're applying braking torque at the
> input to the diff, and you say this creates a torque imbalance within the
> diff ...?
A differential equalizes the torque that is delivered to the wheels. It
doesn't matter which direction (go or stop). If one wheel has less
traction (ie, on ice or sand), it will go/stop faster than the other...
and spin/lock up.
It takes more force to move something from rest than to keep it moving...
so if the wheel stops, it locks... and your car spins.
> It must be conservation of angular momentum that's responsible, but I'd have
> to think hard about it. Anybody want to explain it?
It's the same idea as applying the handbrake when your car is on sheet
ice. The car needs "X" amount of torque to move the wheel that's on solid
ground (you hope) and "Y" (less than X) amount of torque to spin the wheel
on ice. By applying the handbrake you increase the "Y" torque, hopefully
enough to get the other wheel to turn.
Posi-traction (or locking diffs/solid axles) don't experience this
phenomenon.
Ugh, that's enough big words for today.
-Malcolm
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