[CAR PHYSICS] Simulating

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.

Hi Sebastien! On Mon, 15 Mar 2004 12:03:02 +0100, Sebastien Tixier <[email hidden]> wrote: >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! ;-)

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>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 ?

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

> Hard to imagine what you did in VRally3 considering your question! ;-) 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 :( > 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. arf, lack of English for me :)))

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> 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.

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 :) > 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). Well i didnt find any archive no where, so it would be so great if u could send it to me at [email hidden] :) Again thanks u.