PHYSICS: Torque through the driveline

I am currently using engine torque to directly affect the rpm, which in turn is directly linked to the wheel speed. I want to get away from this and use torques linked from the engine through the clutch and gear system into the differential. I have been playing about with this for ages and have got nowhere! I have the torque from the engine, no problemo. The wheels need a torque to rotate. I am to get torque from the differential to do this. Questions: 1. I need to have torque at the differential. Ignoring a clutch how do I calculate this! I am going to split the torque in two for the wheels for the time being for simplicity. Does this just involve multiplying the engine torque by the gear ratio and getting an effective inertia for the whole driveline, and doing a T=Inertia*angAcc? [I know after many attempts that this is not the case, or that I am using the wrong figures!] 2. After doing this I would need the reaction from the road to be sent as a 'fighting' torque through to the engine. I asssume the system for doing this is the same, just in the opposite direction. So once I do <1.> is it straightforward to implement <2.> ? It is driving me insane! Put me out of my misery, please!!! Regards

On 24 Feb 2003 01:40:21 -0800, [email hidden] (Alex Smith) wrote:

Quoted text (11 lines)

>I am currently using engine torque to directly affect the rpm, which >in turn is directly linked to the wheel speed. I want to get away from >this and use torques linked from the engine through the clutch and >gear system into the differential. I have been playing about with this >for ages and have got nowhere! I have the torque from the engine, no >problemo. The wheels need a torque to rotate. I am to get torque from >the differential to do this. Questions: > >1. I need to have torque at the differential. Ignoring a clutch how do >I calculate this! I am going to split the torque in two for the wheels >for the time being for simplicity.

As the wheel is contrained by gears with differential rotation, i dont know how you can get ride of omega ( wheel spin ) . Sebastien TIXIER - Game Developer Dynamics and Car Physics http://www.eden-studios.fr GPLRank Normal:-44.24 Monster:-124.44 LFS GTI Online 1:30:62 LFS GTI Offline (HLVC) 1:31:14

On 24 Feb 2003 01:40:21 -0800, [email hidden] (Alex Smith) wrote:

Quoted text (11 lines)

>I am currently using engine torque to directly affect the rpm, which >in turn is directly linked to the wheel speed. I want to get away from >this and use torques linked from the engine through the clutch and >gear system into the differential. I have been playing about with this >for ages and have got nowhere! I have the torque from the engine, no >problemo. The wheels need a torque to rotate. I am to get torque from >the differential to do this. Questions: > >1. I need to have torque at the differential. Ignoring a clutch how do >I calculate this! I am going to split the torque in two for the wheels >for the time being for simplicity.

There are 3 sides to the diff; for every side you can see torques that are pushing one direction, and torques that are pushing the other direction. For example, the wheels are driven by the engine, but the road reaction forces counter their rotations. The input side of the diff (in a single-diff car) is the engine torque times the gearbox ratio. An open diff splits the incoming torque as T_left=T_right=T_input/2. (where T_input is thus your T_engine*gearBoxRatio). Also, the road reactions at the wheels are part of the system: T_input=T_engine*gearBoxRatio T_left=T_input/2+T_roadreaction_left T_right=T_input/2+T_roadreaction_right Then proceed to the quote section below. :) Note that I think most of us use a tree structure to represent the driveline. With the clutch, perhaps multiple diffs, and the accompanying ratios involved, its easier to have a tree object that can calculate its inertia. For example, I have a RDriveLine class which can do stuff like CalcInertiaOfChildren(); each component has a ratio and inertia, and you iterate through the tree, using the SQUARED ratios to add up inertia values. I use a piece of code from Gregor Veble which focuses at the diff, and from the T_input, T_left, T_right numbers is calculates the accelerations at each point, given the inertia numbers on each of the 3 ends. It's a closed system, which is why I have trouble expanding that to 3 diffs. The trick there is to see it as a symmetric acceleration combined with an asymmetric acceleration (which accelerates the 2 outputs or wheels away from eachother). Stefano seems to take another approach and calculates the torques that are transferred from one side to another; probably much better (read: simpler) in a multiple diff algorithm. Note that the road reaction forces here are implicitly passed on the engine because you calculate the accelerations of each end. The 'input' or 'engine' end doesn't accelerate as fast since the total torque for the entire system gets reduced by the road reaction forces. You might make this a separate process (engine forward, road reaction backward) but I'm not sure if you can do this, because of the differential action inbetween. A Lego differential really helps (there are 3 in their Tecnic 4x4 Offroad something car) to see how a diff works. Don't forget that there is a cage that can rotate too. >It is driving me insane! Put me out of my misery, please!!! Indeed, diffs are a pain, esp. if you want 3. :) Here's what Stefano Casillo (netKar) wrote once: --- I've tried to write a post here explaining my solution..but it was really really messy... anyway.. I try again.. U have this situation: E | _______ | | w1 w2 E, w1 and w2 are torques from engine and the 2 wheels (from the road reaction). Now, E is always splitted in 2. equally.. if u see a diff it's easy to convince yourself.. the engine can just act on the cage of the diff. Now.. w1 and w2. You need to wait to have them both to actually decide what to do. The "combined force" of w1 and w2 act on the cage. Combined force for me is the fraction of the torques acting in the same direction. Example.. if w1=100 and w2=30.. the combined force is 30*2.. the 30 N/m of w1 is acting in the same direction of the 30 of w2. so they combine and go to the cage.. Then u have the unbalanced torque.. the torque difference between w1 and w2.. this won't act on the cage but will be transfered to the weaker wheel.. in our example we'll have 100-30=70N/m going from w1 to w2... This looks crazy I know.. let's do the "wheel on ice" example.. the engine is sending say 200N/m of torque. .I split 50% and send it to the wheels.. they accellerate and get the reaction from the road.. the wheel on tarmac (say w1) will get 100 N/m of reaction force while the wheel on ice (w2) will get 0N/m reaction... Step 2.. the combined force is 0 so nothing is going back to the engine. the unbalanced force is 100 that is sent right to w2 from w1... So w1 received 100 from the engine and 100 from the road. so basically stays there... w2 got 100 from the engine plus 100 from w1 and it's spinning like a crazy... It's very easy to get this system and add stuff like internal clutches to make a limited slip (I just did it yesterday).. Whatever system u do.. there's one thing u have to check ALWAYS.. the diff cage speed MUST always be the sum of the wheel speed divided by 2.. always, no matter what kind of diff u have. In a torque based system like mine.. speed is an output not an input... and u need to check it always..I've been running with this in my code forever: if (differential->speed!=(outShaftL->speed+outShaftR->speed)/2.0f) { printf("**** DIFF SPEED CHECK FAILED! ****\n"); exit(0); } --- Hope that helps you forward, Ruud van Gaal Free car sim: http://www.racer.nl/ Pencil art : http://www.marketgraph.nl/gallery/

Thanks a lot Ruud, I totally understand the wheel on ice example which gives me confidence :) Things are starting to look a little less misty, will give it another go this week no doubt! BTW I think I accidentally replied to you and not ras. sorry! Regards