Re: torsens & 'torque split'

[Orin Eman <orin@wolfenet.com>]

> >Orin writes:
> >>
> >In summary, open diff means 50/50 torque split at the output,
> >locked diff means the split varies between 100/0 and 0/100
> >at the output.
> >>

> >I believe this statement is backwards, an open diff means 100/0
> >and 0/100. And a locked diff is set at 50/50, because both outputs 
> >are now connected as a unit and they both rotate at the same speed.

>  glen writes:
> >
> >  Torque and 'torque split' are not related to the speed or relative
> > speeds of the two shafts with an open or a locked diff. The 'torque
> > split' is purely a function
> > of the relative counterforce applied to each axle and this is a
> > function of the relative
> > traction or lack of traction available at the two road/tire interfaces
> > on the axle
> > in question. Equal traction = equal torque split. Zero traction at one
> > wheel = zero
> > total torque and zero forward vehicle acceleration/motion with an open
> > diff. Zero traction 
> > at one wheel = 0/100 'torque split' with a locked diff with the 100%
> > going to the wheel 
> > with traction - exactly what you want if you want the vehicle to
> > accelerate and move.
> >
> >  
> I thought about this last night and pretty much came to the same
> conclusion, Orins example
> of 50/50 with an open diff was based on equal 'good' traction condition
> for the wheels in question 
> and my 100/0 or 0/100 was based on a 1 wheel slip condition. And it
> stays that way, that is, 
> once that 1 wheel starts to spin it will continue to spin using 100% of
> the applied torque. 

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.

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.

Try drawing it... you can approximate the gears with square teeth
in two dimensions.

> Now for the period in between having equal 'good' traction conditions to
> the time we very low 'poor' traction 
> conditions. Orin, Steve, glen and others believe with a locked center
> diff the torque split varies between 
> 100/0 and 0/100 with 100% of the torque being applied to the wheel with
> the 'better' traction. I disagree with 
> this (look ma no buzzers). With the locked center diff there is no
> differential, that is, both output shafts rotate
> at the same speed, thereby making the wheels on the two axles rotate at
> the same speed. Now one of the 
> wheels starts to spin because the torque applied to that wheel exceeds
> its traction condition. The remaining 
> wheel which still has traction is still receiving the same amount torque
> that it was previously when both wheels
> had the same equal 'good' traction. The wheel with 'poor' traction is
> using it's torque to keep it spinning and the 

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_.

>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.

The open diff wheel spinning case is confusing because you tend
to think that the same torque is input to the system once a wheel
spins as before.  Not true.  When a wheel spins, the torque no
longer transmitted accelerates the whole drivetrain, up to the
point at which frictional losses/internal engine losses
match the torque which was previously transmitted.  A lower
total torque is now transmitted by the diff... the split IS
50/50, it's jsut that twice almost nothing is still almost nothing!

Orin.