rec.autos.simulators

Simulating

Sebastien Tixie

Simulating

by Sebastien Tixie » Tue, 16 Mar 2004 20:03:02

Hi everybody, i'm sure this question has been already asked, but i didnt
find anything about this, so maybe i can post it all the same :)

Please somebody help me or point me on some ras archive i didnt find.

So i have probleme modeling differential.

Lets start with an open differential :)

lets assume those datas

TorqueIn  .. torque given to the diff
TorqueOut[2] .. torque out of the diff to the both axles
ResistiveTorque[2].. torque that come back from axles

When reading books i can read that an open diff split  torque_in in 2 to
the TorqueOut[i].

So TorqueOut[i] = TorqueIn / 2

Its also say ( Race Car Vehicules Dynamics page 734 ) "if one wheel
loses traction the total drive force available will be just twice the
force available from the tire with least grip". i.e. ResistiveTorque is
weak ?

So i modify to

TorqueMaxPossible = Min( ResistiveTorque[0], ResistiveTorque[1] )
Torque = Min( TorqueIn, TorqueMaxPossible )
TorqueOut[i] = Torque / 2

The probleme is when the driving wheels are not sliding at all ( i.e.
slip ratio ~0 ) the ResistiveTorque is almot zero , so TorqueOut[i]
become slower, so slipratio become slower, etc... the car doesnt accell
anymore :(

So what should i do ?

1) Always use Torque = TorqueIn ?
2) Use Torque = Min ( TorqueIn, TorqueMaxPossible ) only if
Abs(SlipRatio ) > 1 ( if we assume SlipRatio is normalize with Optimum
Slipratio ).
3) Maybe i should use a ResistiveBiasInstead ? TorqueBias =
ResistiveTorque[Grippy] / ResistiveTorque[Slippy] ? How should i use it
then ?

Limited Slip Diff

Ok, one i'll get Open Diff working , i think limited Diff are easy.

lets assume

OmegaOut[2] are angle velocity of axles
OmegaIn are anglular velocity Engine side
OmegaS , angulare velocity of internal diff

OmegaIn = ( OmegaOut[0] + OmegaIn[1] ) / 2
OmegaS = OmegaOut[1] - OmegaOut[2]

Td = Differential torque.

Torque = ?? , depend of how its works with open diff lets assume we have
2 grippy wheels
Torque = TorqueIn

Td = OmegaS * DiffFriction
TorqueOut[0] = ( Torque - Td ) / 2
TorqueOut[1] = ( Torque + Td ) / 2

Is that right ?

Now, should i clamp Td so Abs( Td ) is not > to Torque ??

thanks in advance for any help.

Seb.

Ruud van Ga

Simulating

by Ruud van Ga » Sat, 20 Mar 2004 03:28:35

Hi Sebastien!

On Mon, 15 Mar 2004 12:03:02 +0100, Sebastien Tixier


>Hi everybody, i'm sure this question has been already asked, but i didnt
>find anything about this, so maybe i can post it all the same :)

>Please somebody help me or point me on some ras archive i didnt find.
>So i have probleme modeling differential.

There are some good posts on that. A year or 2, 3 ago Gregor Veble
explained his system; using asymmetrical torque for resistive torque
on an axle that doesn't point the same way as the other axle.
Also, Todd Wasson has at some point explained bias ratio quite
thoroughly, being the resistive torque bias, influencing locking.
And Stefano Casillo from netKar has explained on RAS how he did it.
Some more searching to do then. :)

Hard to imagine what you did in VRally3 considering your question! ;-)

Read that carefully; 'driving' force means the effective force being
used to accelerate the car. This does NOT mean forces/torque just
disappears! Just that the excess force is used to spin the wheels.

So this is not the case at all. TorqueOut[i]=torque/2. That's it.

Consider the example of one tire on the road, one on perfect ice.
torqueIn=torqueEngine*gearing
torque[i]=torqueIn/2

Suppose torqueIn=100Nm (engine), torqueResistive[0]=50Nm (road) and
torqueResistive[1]=0Nm (ice). What would happen?
The sum of resistive torques add up to torquein, so on the engine
side, no accelerating will take place; the engine will keep rotating
at the same speed.
The diff will split torque 50/50: 50Nm on both sides. The [0] side
will stay still. However, the [1] side gets 50Nm from the engine
(torque[1]=50Nm), plus torque transfer happens from [0] to [1].
Totalling 50Nm of torque transfer means 100Nm is used on the [1] side
to spin the wheel on ice.

The transfer is a bit odd, and this example evens up too nicely. A
different example:
torqueResist[0]=100, torqueRestist[1]=30, torqueEngine=200.
The torque that transfers back to the engine is the resistive torque
pointing in the same direction; in this case 30 from [0] and 30 from
[1]. The remaining torque transfer (this is the diff acting) of
100-30=70Nm goes to spin up [1].
So [0] gets 200/2=100Nm engine torque, road reaction 100Nm, summing to
0Nm = no motion.
And [1] gets 200/2=100Nm engine torque, road reaction 30Nm, torque
transfer from [0] = 70Nm. 100+70-30=140Nm spinning the wheel.

Not sure if that last sum is totally correct. :) Would be good to read
up on Casillo's post. Search for 'Ruud van Gaal physics torque through
the driveline' for a post of mine quoting Stefano (otherwise I have it
in my archives here as well).

LSD is then just adding a torque resisting the motion of the axles, I
think. ;-)

Hope that helps,

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

Sebastien Tixie

Simulating

by Sebastien Tixie » Sat, 20 Mar 2004 04:24:36

hey hey, well, in fact, i only modeled LDS in VRally3, and the diff were
never open. The minimum slip was 50% ;)
In VR3 i split torque equaly all the time plus the resistorque from
diff. I know its good on very low grip :(

arf, lack of English for me :)))

Hey, great explaination :) and i'm quite happy, because,  as i didnt saw
this answer i try to figure out myslef looking to the mechanical scheme
of an open diff :) and i almost find the good answer except for the
torque going back to the engine :)
Thanks a lot for this good explanation :)

Well i didnt find any archive no where, so it would be so great if u

Again thanks u.


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