Re: how does an open differential work (was RE: torsens & 'torque spl

[Orin Eman <orin@wolfenet.com>]

> "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. 

I disagree, as does the Bosch Automotive Handbook.  The slipping wheel
case is actually better than this.

What I am saying is that in the slipping wheel case, wheel accelerated
to its maximum speed, then there is an equal amount of torque directed
to the non-slipping wheel.  Has to be.  Moments about any point on any
bevel gear must sum to 0 or it will be accelerating.

The following are the conditions I'm using:

Force on bevel gear due to axle gear equals force on axle gear due
to bevel gear.  (...equal and opposite reaction).

Taking moments about the center of a bevel gear which eliminates
the driving torque,  you have force (axle 1) * distance from center
- force (axle 2) * distance from center = 0.

Each axle gear is the same distance from the center of the bevel.
So force (axle 1) = force (axle 2).  Or, since it's symmetrical,
the torques are equal.

What this actually means is that twice the torque at the spinning
wheel is actually making it to the ground, twice what the quoted
description indicates.  Of course, twice almost nothing is still
almost nothing!

> 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."

Depends if you are moving or not.  Power is proportional to
torque * RPM.  If you aren't moving, absolutely true, if you
are moving, there will be a little power going to the non-spinning
wheel... 

I hadn't thought of the 'power' case before, but, it gives the interesting
result that the ratio between the power going to the wheels in the open diff
case is equal to the ratio between the RPMs of each wheel...

One wheel slipping, the other stopped, the slipping wheel does get
all the power.

> 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.

Yes, but only that which is required to a) overcome the (lack of)
traction at the tire and b) friction in the axle/wheel bearings.

> > 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. 

I can see I'm going to have to draw all the pretty vector diagrams
and scan them in...

Orin.