rec.autos.simulators

  I've been reexamining my code that handles the differential gear system in my
car simulator after finding unusual behavior with a 4 wheel drive (attempted!)
setup.  
 I'd like to clear this up as I know Ruud is working on the same thing right
now, and it's possible I might have given him incorrect information regarding
how Salisbury axles work!  That would make his debugging attempts frustrating
if he refers to my periodic "tips" on occasion, to say the least :0)

  When running an open differential (no locking), under hard acceleration in a
corner one wheel is spinning.  Theoretically, the torque bias ratio (ratio of
road reaction torques across left and right tires sharing an axle) should never
exceed 1:1.  The torques should be about equal under all circumstances from
what I've read so far.

  My concern is over ring gear (or engine) acceleration's effects on torque
bias ratio.  In my sim, with one wheel spinning wildly, the torque bias ratio
will remain about 1:1 (slight instability) as long as the ring gear is not
accelerating.  However, with the ring gear (or engine) accelerating, the torque
bias ratio can increase dramatically, and it does this in direct relation to
the ring gear acceleration.

  The code calculates the acceleration of both wheels relative to each other,
then essentially adds this to the ring gear acceleration to get the final
rotational speed of each wheel.  This "relative acceleration across the axle"
increases as I lower the polar moment of inertia of the wheels, and the further
I drop it, the closer to true 1:1 torque bias ratio behavior I'm getting.  When
implementing clutches for higher torque bias ratio, I'm seeing the same basic
thing happening.

  My question is, does the theoretical torque bias ratio of 1:1 only exist when
the ring gear is not accelerating?  

  My other question relates to four wheel drive.  I've tried a three
differential setup based on what I've got so far, and it seemed to work ok,
until I tried using front tires with 0 (or close to 0) grip.  In that
situation, with three open differentials operating (one center feeding one
front and one rear), shouldn't the car remain stationary while the front tires
accelerate wildly?  Unfortuately, this effect only happens when I've got near 0
rotational inertia at the wheels/axles.  I'm getting quite a bit of torque
transmission to the rear tires, causing hard acceleration anyway.

  Thanks for any input,

Todd Wasson
---
Performance Simulations
Drag Racing and Top Speed Prediction
Software
http://www.racesimcentral.net/

  As the rotational inertias of the wheel assemblies on one axle increase, the
torque difference across the axle doesn't allow the wheels to accelerate away
from each other as quickly as would be necessary to keep the slip ratios at
each wheel where they ought to be in order for the torque bias ratio to stay at
a constant 1:1.  When the wheel inertias are lowered on one axle, they approach
the theoretical 1:1 (or whatever) ratio because they can accelerate away from
each other more rapidly.  Perhaps this is why there is a lag, and may not
really be an error after all?  Or is it more complicated than that?

  In other words, while cornering at high lateral acceleration, and while the
ring gear is accelerating, the inside wheel needs to slow down (relative to the
ring gear) and the outside wheel speed up more quickly than it currently does.
When the wheels accelerate away from each other so slowly (while the ring gear
is speeding up, taking both wheels with it), the outside slip ratio is
increasing faster than it should, and the inside is decreasing more slowly than
it should.  

  This is the only explanation I can come up with.  Is this how it works in
real cars, or am I missing something?  

  When the ring gear is at 0 acceleration, everything is fine (regardless of
slip ratios and wheel speeds), clutch torque keeps the torque bias where it
should be (if it's used), etc....

  Any thoughts?

Todd Wasson
---
Performance Simulations
Drag Racing and Top Speed Prediction
Software
http://PerformanceSimulations.Com

Ouch! :)

Hm, this is now what I have understood (but ofcourse I may be wrong;
after all, I've never held a differential in my hand, just looked at
pretty AVIs ;-) ).
From my understanding, the torque split that is 1:1 is the dividing of
incoming torque (the engine in case of 1 diff) to the output torques
(the wheel axles). For example: Te=300Nm, then
T_wheel_left=150Nm=T_wheel_right.
However, when I look at my diff code (which was mainly contributed by
Gregor) it seems that this may not be entirely correct. Hm.

In that code, the acceleration of the entire driveline part (as far as
you can see; the engine may be separated when the clutch is
disengaged) is determined by summing all the torques (engine+road/tire
reaction forces) and dividing by the total inertia of the system.
The difference between left & right wheel acceleration is then
determined by the different of the tire reactions forces
(T_reaction_right-T_reaction_left) and that difference is used as an
asymetric acceleration number which offsets the total driveline
(T_all/I_all) acceleration from the wheel's accelerations (by
(T_reaction_right-T_reaction_left)/I_both_wheels); plus at one side,
minus at the other.

So in fact, I think that both wheels still get power, only the
slipping one more so.

Still, I think torque on both axles stays the same; it's just the
resulting acceleration on both wheels that is so different (because if
one wheel can accelerate faster, the other one must accelerate
proportionally slower).
So with that in mind, I can't quite follow the rest of your
conclusions.

My intuition tells me you're looking too far, but you never know with
diffs. ;-)

Ruud van Gaal
Free car sim  : http://www.marketgraph.nl/gallery/racer/
Pencil art    : http://www.marketgraph.nl/gallery/

  That's what I've basically got too.  I tried Gregor's equation set in place
of my own yesterday and got the same behavior.

 I think that's kind of arbitrary because we aren't actually calculating output
torques to the wheels (actually, I do to check for locking, but it doesn't
directly effect wheel acceleration), but instead allowing the ring gear to
accelerate from the sum of the three torques, then letting the wheels
accelerate towards or away from each other based on their road reaction torque
difference.  What I was referring to was that the slip ratios eventually
stabilize, and those road reaction torques essentially ARE the driving torques
at each half shaft/wheel.  Theoretically (from what I've read), the two torques
(road reaction, about the same as real driving torque), should pretty much
remain identical.  When cornering with no acceleration, I'm getting the 1:1
ratio there (or whatever the clutch says), but when one wheel breaks loose
*and* there's hard driveline acceleration, I'm winding up with more torque
going to the outside wheel than the inside (not much more, but enough to
notice).  The greater the moments of inertia of the wheels, the greater the
effect.  Granted, once the driveline acceleration is returned to 0, the 1:1
ratio (road reaction torques) returns regardless of how fast one of the wheels
might be spinning.

 >Still, I think torque on both axles stays the same; it's just the

  That proportional acceleration, or as I've been calling it, the "relative"
acceleration across the axle (silly term, but it helps me think :0)) is
apparantly what is responsible for the funny acting bias ratio.  Picture this:
You've just entered a high speed corner at about max lateral g's.  With the
driveline and car at constant speed, steady state corner, the ratio of road
reaction torques across the axle should be (and is in my stuff) 1:1.  Now, you
nail the throttle, sending the inside tire up in smoke and accelerating the
driveline (ring gear) rapidly.  What I was wondering is, should the ratio of
road reaction torques remain pegged at 1:1 the whole time?  I'm finding that it
doesn't (in my stuff anyway).  In order for it to remain 1:1, the outside wheel
would have to slow down at a certain rate (relative to the ring gear, although
it probably remains at the same speed overall) while the inside tire speeds up.
 However, as the rotational inertia of the wheels is increased, the wheels
don't accelerate away from each other quickly enough, so the outside wheel
speeds up slightly right along with the ring gear, increasing slip ratio and
longitudinal force, and therefore road reaction torque.  In other words, for
the bias ratio to remain 1:1, the outside wheel would have to stay at exactly
the same speed it was at before, but with inertia in the system, it winds up
accelerating somewhat because that "proportional acceleration" doesn't get
large enough to keep the outside wheel at a constant speed.  I get more torque
at that wheel, so the theoretical 1:1 ratio falls apart.  It isn't a big
effect, but it's there.

  What I was concerned with is that I mentioned before how you might test to
see if your Salisbury diff code is working correctly.  I said that the actual
bias ratio shouldn't exceed the specified bias ratio, but I'm finding that
that's not right!  It will exceed the specified ratio when the driveline is
accelerating, but at 0 acceleration, it should pretty much stay pegged at 1.76
or whatever ratio.

  This had me thinking because with 4WD I wind up getting a lot more torque
transmitted to the rear tires than I thought there should be, especially when
the front tires have very little (if any) grip.  Anyway, I think this turns out
to be accurate behavior, I just wanted to see if anyone knew for sure.

  Maybe :0)  

Todd Wasson
---
Performance Simulations
Drag Racing and Top Speed Prediction
Software
http://PerformanceSimulations.Com

Another statement might clarify (or muddy?) the discussion??

Look at RCVD Figure 20.2.  The "levers" at the bottom are meant to be a
"snapshot" of the spider (differential) gears, balancing the torque
between the two side gears.

I think what you are saying below is that, dynamically, the "levers" also
have some inertia about the center (pivot), so the output torques aren't
exactly equal under dynamic conditions.  The inertia comes from the parts
outboard of the diff (mostly the wheels/tires/brake-rotors).

Helpful or not??

-- Doug Milliken

  Perfect :0)  I thought about this a lot today and came to the conclusion that
it must be right after all (well, at least *that* part :-)).  I must say I
tried to track down this "bug" for probably 6 hours the other evening!  Thanks
for confirming that the inertial effect indeed does throw off the torque bias
sometimes :0)  I figure now the only way the bias ratio could truly be 1:1 in a
situation where the ring gear is accelerating is if there was no rotational
inertia in the rotating assemblies.  Maddening, but glad to know my basic
thinking was right :-)  

  Thanks,

Todd Wasson
---
Performance Simulations
Drag Racing and Top Speed Prediction
Software
http://PerformanceSimulations.Com

...

Could it even? If you have quite a high CG height, so the car tips
over slightly, getting the inner wheel to lose touch with the surface,
it would get nearly 0 road reaction torque (just some friction). How
could you get a 1:1 ratio then? The outer wheel would have to get
nearly no road reaction torque as well. You end up with no torque on
either side, but where then would you leave the incoming engine
torque?
It seems that you can't reach that point. Therefore, I'd say the 1:1
ratio of the reaction torques is a RESULT, rather than the GOAL. A
result because the steady state will go there. Inevitably, as both
axles get 1/2*T_engine torque, so if this endures into a steady state,
the road reaction ratio *can only be* the same, as the diff doesn't
know how to distribute torque otherwise. So *if* you have a steady
state, you *must* have the ratio for the reaction torques to be the
same as well (otherwise one wheel is still accelerating away/toward
the other and you haven't reached a stable state yet).

Obviously, a suddenly spinning wheel (or getting of the ground) is not
by any means a steady state and very dynamic, so then your road
reaction ratio goes berzerk. The only thing you can count on I think
is that the torque on each axle will be 1/2*T_engine.

Perhaps a comparison with electricity may be useful (not only for you,
but for me as well ;-)). If you look at a system like this:

   ___   ____
  /   \ /    \
 R1    B      R2

Where R1=resistance 1, B=battery and R2=resistance 2 (which may be
infinite). If the Battery has a some high voltage (relate
voltage=torque), then electricity will flow through R1 and R2 (relate
resistance=inertia) in a specific ratio. Damn, this has been a long
time ago. How was that again?
Anyway, depending on the ratio of R1/R2, electricity will flow through
R1 and R2 but with different power running through them. I'll have to
look this up, sigh, all I can remember is V=I*R (V=voltage, I=ampere,
R=resistance).
It seems the differential can be viewed of much the same way, so in
principle you'd have a similar system to a parallel row of lamps and
batteries (for which the formula are well-documented).

I may be on to something there. ;-)
I must have a look at some of these pages to see what formulae they
used again.

The heisenberg compensators will probably result automatically from
there. ;-)

Hm, in my view the ratio could be anything, and the diff just starts
locking more or less (upto a maximum defined by the locking ratio)
based on the ratio.
Suppose I have a max bias ratio of 3:1 and the road reaction torques
are 400:100 (Nm). Then I'd have (starting from the low-power side):
- maxTorque_on_other_side = 100 * (3:1) = 300Nm
- torque_overflow = 400-300 = 100Nm.

This 100Nm will be applied to both wheels to accelerate away from
eachother.

I think that should be the case. It's not as if torque suddenly
disappears everywhere and really flows to one wheel, but the ratio is
so big that the car becomes undrivable and one wheel gets spinning.
But I think the other 3 wheels should get a bit of power anyhow.

Perhaps Doug has the formulae right there in an upcoming book?
("Racing Car Dynamics for the mathematically disabled") ;-)

Ruud van Gaal
Free car sim  : http://www.marketgraph.nl/gallery/racer/
Pencil art    : http://www.marketgraph.nl/gallery/

  The same thing as happens when one wheel is on ice.  The wheel with no
traction spins, and the other side just sits there without pushing the car at
all.  The road reaction torques stay equal, at about 0 and 0, a 1:1 ratio.  

  Exactly.  This happens because the wheel on ice won't turn and develops no
slip ratio.  Actually, the wheel *would* have to develop a "slip ratio" in
order for the wheel on ice to accelerate.  The two wheels are accelerating away
from each other (relative to the ring gear) in opposite directions and speeds,
so there *must* be a torque difference across the axle.  Therefore, the wheel
with traction really IS generating force, but only when there's an acceleration
at the ring gear.  How else can left_torque - right_torque be greater than 0 if
right_torque is 0?  (In order for the wheels to accelerate away from each
other, left_torque - right_torque must be greater than 0, otherwise the two
wheels just follow the ring gear's acceleration)  This is what the wierd
behavior due to wheel inertia I saw is really caused by.

  It accelerates the ring gear, along with both wheels' road reaction torques.

  We're in agreement, but we're thinking about the same thing slightly
differently.  I never think about the engine torque except when it comes to
figuring the ring gear's acceleration.  I don't think about how much engine
torque is going to each wheel, although you're absolutely right here and are
definately answering the question in a way that backs up what I've been
thinking, so thank you :0)

  Stability isn't what I meant.  By steady state, I was referring to a
situation where the ring gear and both wheels were not accelerating.  By
dynamic, I meant "something somewhere is accelerating".  I was wondering if an
open differential really exhibited a 1:1 torque bias ratio in both situations.
I know now that it doesn't, but in a dynamic situation (something's
accelerating, bad choice of phrase perhaps?), the actual torque bias ratio ends
up being dependent on the wheel inertias quite a bit.  

 > The only thing you can count on I think

  I'm not too sure about that, but with the equation set Gregor sent you, it
wouldn't matter anyway because the engine torque is never considered for each
wheel.  In my system (which works pretty much like Gregor's set but has a
clutch and is very messy looking), there are only three things looked at.
Engine torque at the ring gear, and each wheel's road reaction torque.  Those
three torques accelerate everything automatically (yours does too if you're
using Gregor's stuff) without ever looking at how much torque goes to each
wheel.  

  Perhaps the torque on each axle really is 1/2*T_engine (not that it matters
anyway).  I said I'm not too sure about that because it seems that the road
reaction torque winds up *being* the engine torque that's applied to that axle,
and when an open diff's ring gear is accelerating in situations like you and I
described, this doesn't appear to be what happens at all.  But, on the other
hand, maybe this road reaction torque isn't really the same as the engine
torque, it shouldn't be, because otherwise there would be no wheel
acceleration.... I'm dizzy now and will stop..... :-P  My point there is that
the engine torque to each axle doesn't really matter until you want to check
for brake locking.  In that case, I'll have to look again for what I did and
see if it's really right.

<snip on electrical>

  Very cool way to look at it.  You might be on to something there after all
:0)  I saw in an engine book how an electrical formula (Helmholtz or
something?) was used to calculate the engine speed where resonance would occur
in the intake system.  

  As you pointed out, we'll have to check the heisenberg compensators to make
sure :0)

  Yes, the ratio could be anything of course.  1.76 was just an example :0)
But you're right, the torque_overflow accelerates the wheels away from each
other.  Now you just need to calculate the clutch torque and add that in.

Todd Wasson
---
Performance Simulations
Drag Racing and Top Speed Prediction
Software
http://PerformanceSimulations.Com

And now for a differential pop-quiz:

Rear drive car.  Stuck with one wheel on wet ice with an open diff.
Other wheel has some traction, but of course you are not moving...
What do you do to get the car to move?

In my mandatory one year military service I was driving a belt-driven
sypply vehicle, a BV 206. Fun stuff, could go almost anywhere.
However, they had no locking option on the diff. One day when we got
this *** undercooled rain on a road that were already covered with
hard-packed snow and one of the vehicles went slightly off the road.
Seeing this nice go-anywhere, nothing will stop us vehicle helplessy
stranded with two belts onto a perfectly flat road was kinda ironic.

During wartime they use big studs on the belts, but that was not
allowed because of the wear on the roads .
--
  -asbjxrn

Lol, hm, my head's close to exploding when trying to think of the ring
gear accelerating, and than figuring the 2 wheels accelerating away
from eachother, only to make one wheel stand in place! :)
But you probably have a point. I recently saw a nice animation (on
how-things-work.com) which showed the ring gear and the axles
rotating, so I can picture it a bit. Still hard to get the velocities
right in my imagination though. :)

Where's Gregor to express his feelings about his code and why he did
it like that? :)

Hm, right. So then the inertia of the ring gear becomes important too.
Or it might be in there already as the 'differential inertia', just
like the gearbox inertia with their (in)famous squared influences.

...

You're most welcome. :) I'm glad to give another perspective that
turns out the same way. ;-)

How about that. Having source code and still figuring out what it
does. BTW the set I have (and use) also includes wheel locking. It
uses a matrix (2x2) that was derived from Lagrangian mechanics in case
one or both of the wheels lock(s) up. You might want to take a look at
that; I haven't been able to decypher where Gregor derived them from.

Hm yes. Weird. Perhaps it's just lagging, as the road reaction will
only get there once the wheel starts rotating. I don't know. It's too
late now also. Need... food... ;-)

Right. I refreshed my electronics yesterday; I used

http://webhome.idirect.com/~jadams/electronics/combination.htm

Nice to see is the first thing they refer to (at the start page) is
hydraulics to explain electricity symbols (voltage, current and
resistance). Now if some hydraulics page then states 'to explain
hydraulics, look at a car's differential'... ;-)

It struck me that Ohm's Law, V=I*R is not far from F=m*a;
V=voltage=force, I=current=acceleration (velocity rather though), and
R=resistance=inertia (mass). Hehe, perhaps searching too far here.
But still, a rear axle is like a parallel circuit; with electricity,
if you have 2 parallel resistors, the total resistance goes down:

1/Rt = 1/R1+1/R2.

So if you have R1=100 (the tire still having grip) and R2=1 (the
spinning tire), the total resistance becomes: 1/Rt=1/100+1/1 =>
Rt=1/1.01=0.99.
And lots of current then flows through the low resisting part. Seems
to fit a differential's bill very nicely. :) I'll have to look more
into it to be sure though, but it seems to have some parallels, and
the good thing is that more formulae are available for that. I mean,
combined circuits (including serial and parallel resistors) can be
calculated quite easily, and it is like describing the driveline as a
combined circuit (starting with some serial resistors at the
clutch->gearbox->diff input) into a parallel system (diff->axles).
More interesting when combining 3 differentials and trying to figure
things out. Just like engine power going to the wheel with the worst
grip, electricity wants to go through the path with the least
resistance as well, and it is calculatable how much so with
high-school physics formulae.

Seems to good to be true, and you know what they say about that. ;-)

Without them, you couldn't call a sim a sim. GPL surely has them.

Ah yes, I didn't read careful enough; I thought you meant it would not
exceed this ratio ever then.

The diff clutch torque in the above case is just the 300Nm (the
non-overflow part), if I'm not mistaken?

BTW From what I've read, the ratio is more important than the number
of clutches. So when setting up a car, you first decide on the spider
gear angles, then finetune that with adding/removing clutches. Does
that correspond with your experience? Seems to me that clutches would
have a LOT to do with the total locking (perhaps more than some
ratio/angle tweaking).

Let us know the results. :) If anybody is still following this, lol.

Onto simpler things now for a change tonight; replays. Aww, sometimes
just work and no thought is so nice. :)

Ruud van Gaal
Free car sim  : http://www.marketgraph.nl/gallery/racer/
Pencil art    : http://www.marketgraph.nl/gallery/

  Push the brake a little along with the gas.  What do I win?  :-)  
Todd Wasson
---
Performance Simulations
Drag Racing and Top Speed Prediction
Software
http://PerformanceSimulations.Com

Handbrake is the correct answer -- and why I used a rear drive car for the
example.  Sorry, no prizes this time, but like driving practice with sims,
it might help you on the road some day...

Some farm tractors have open diffs and individual rear wheel brakes (two
pedals side-by-side), then the driver can just apply the brake on the
spinning wheel -- very effective once learned.

Todd Wasson
---
Performance Simulations
Drag Racing and Top Speed Prediction
Software
http://PerformanceSimulations.Com