rec.autos.simulators

An example of a situation where the torque fed back from each wheel
(F cross R) is different is in low speed cornering. The longitudinal slip
ratios will be noticibly different on the inside and outside wheels from
the geometry of the turning situation.

The tranmission efficiencies are much higher than 50%.  (Closer to 1
than 50% ;). Gillespie gives some examples in "Fundamentals of
Vehicle Dynamics". I've heard estimates of the torque lost in the
total drive train for Trans-Am cars as something like 7% to 11%.

I believe Trans-Am cars are now also allowed a differential
(rear end) temperature gauge in***pit, so presumably they can be
concerned about heating it.

- Matt

Ash - watching and learning intently :)

| Hi Ruud,
|
| the differential is definitely maybe the hardest part to comprehend. :)
|
| You correctly identified the locked axle and the fully open differential
| as the two extreme cases. The main difference between the two, though,
| is that any (semi-)open differential introduces an additional degree of
| freedom in the movement with respect to the locked axle.
|
| The RPM of the engine and the output RPM on each of the wheels are
| always in a direct relationship with a locked axle, determined by the
| gearing with a locked axle.
|
| In an open differential, it is only the average of the RPMs on the
| driven wheels that is proportional to the engine RPM in the same
| fashion, but the difference in RPM on both wheels can be arbitrary.
|
| In my sim, I therefore do not deal directly with the RPMs of the engine,
| the left and the right wheel independently, but put the focus of my
| calculations on the differential. Therefore I introduce the 'symmetric'
| RPM (which matches the engine RPM as seen through the gearing and also
| equals the average of the wheel RPMs), and an 'antisymmetric' RPM (half
| the difference between the wheel RPMs). For both of these it is possible
| to calculate the 'symmetric' torques (the properly scaled torque on the
| engine plus the torques on both wheels) and inertias (in a similar
| fashion), while for the 'antisymmetric' case I took half the difference
| between the torques on the left and right wheel to be the
| 'antisymmetric' torque (with modifications for semi-locked axles, as
| will be noted below), and for the 'antisymmetric' inertia the average of
| the inertias of the left and right wheel assemblies. Taking that into
| account, both the symmetric and antisymmetric RPM can be updated as per
| any rotating assembly.
|
| The actual RPMS on the wheels can then be had by adding or subtracting
| the antisymmetric RPM to / from the symmetric one, depending on which
| wheel you are updating, while, as noted before, the engine RPM matches
| the symmetric one.
|
| When the differential is not fully open, the antisymmetric torque also
| acquires an additional component that is a function of the antisymmetric
| RPM (directly proportional to it in a simplified viscous differential
| which my sim currently uses) and the input engine torque.
|
| Well, that's how I do it. Perhaps someone else might have another
| approach to share.
|
| -Gregor

the differential is definitely maybe the hardest part to comprehend. :)

You correctly identified the locked axle and the fully open differential
as the two extreme cases. The main difference between the two, though,
is that any (semi-)open differential introduces an additional degree of
freedom in the movement with respect to the locked axle.

The RPM of the engine and the output RPM on each of the wheels are
always in a direct relationship with a locked axle, determined by the
gearing with a locked axle.

In an open differential, it is only the average of the RPMs on the
driven wheels that is proportional to the engine RPM in the same
fashion, but the difference in RPM on both wheels can be arbitrary.

In my sim, I therefore do not deal directly with the RPMs of the engine,
the left and the right wheel independently, but put the focus of my
calculations on the differential. Therefore I introduce the 'symmetric'
RPM (which matches the engine RPM as seen through the gearing and also
equals the average of the wheel RPMs), and an 'antisymmetric' RPM (half
the difference between the wheel RPMs). For both of these it is possible
to calculate the 'symmetric' torques (the properly scaled torque on the
engine plus the torques on both wheels) and inertias (in a similar
fashion), while for the 'antisymmetric' case I took half the difference
between the torques on the left and right wheel to be the
'antisymmetric' torque (with modifications for semi-locked axles, as
will be noted below), and for the 'antisymmetric' inertia the average of
the inertias of the left and right wheel assemblies. Taking that into
account, both the symmetric and antisymmetric RPM can be updated as per
any rotating assembly.

The actual RPMS on the wheels can then be had by adding or subtracting
the antisymmetric RPM to / from the symmetric one, depending on which
wheel you are updating, while, as noted before, the engine RPM matches
the symmetric one.

When the differential is not fully open, the antisymmetric torque also
acquires an additional component that is a function of the antisymmetric
RPM (directly proportional to it in a simplified viscous differential
which my sim currently uses) and the input engine torque.

Well, that's how I do it. Perhaps someone else might have another
approach to share.

-Gregor

Talk to you later!

Ruud van Gaal, GPL Rank +53.25
Pencil art    : http://www.marketgraph.nl/gallery/
Free car sim  : http://www.marketgraph.nl/gallery/racer/

Does it let one wheel spin-up freely if that wheel is off the ground?  And,
if it's "perfect", then there should be no driving force from the wheel
that stays on the ground...

Ah, there's where I go wrong I guess. I apply torque to both ends
equally. So even if the right rear wheel spins ***ly, the left
rear wheel still gets the same torque. Should I bias the torque
instead with the ratio leftRPM/rightRPM? So if the right wheel has
twice as much rotation as the left wheel, the right wheel will also
get twice as much torque (talking open differential here)?
Even though I may do it wrong now, it seems silly simulating this, as
a Torsen differential seems much better by putting power on the ground
wheel (left in my example) instead (the inverse of the open
differential). So a very little hunch says to me that open diffs
aren't used at all without locking (when my car breaks out, it is
hardly stoppable until it flips ends).
But ofcourse, I lack practical knowledge here; do you happen to know
what diffs they normally use in F1, Sports Cars, Nascar and normal
passenger cars?

Ruud van Gaal, GPL Rank +53.25
Pencil art    : http://www.racesimcentral.net/
Free car sim  : http://www.racesimcentral.net/

To get it right, you have to apply torque to the spider gears -- which act
like little levers balancing the torque between the wheels.  The engine
pushes on the center pivot of the "lever".  RCVD Fig 20.2....

Real open diffs (not perfect ones) have some friction in them, so even if
one wheel is off the ground (and engine rpm is constant), the friction will
apply some small amount of drive torque to the wheel that is still on the
ground.

Same if one wheel is on glare ice, the internal friction will still
apply some torque to the other wheel so the car *might* be able to move.

Open diffs are great at tracks where the car never lifts a wheel and rarely
spins a wheel.  They have little friction, so they absorb little power,
thus they are ideal for super speedways and any track where the car is
power limited.

-- Doug

                Milliken Research Associates Inc.

-Gregor