Car physics --More on open diffs

Hello, 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://PerformanceSimulations.Com

After some more messing around just now, maybe I can offer my own little theory on this :0) If someone can verify this behavior, that would be great. 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

Hi Todd, On 28 Oct 2001 06:02:24 GMT, [email hidden] (J. Todd Wasson) wrote:

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> Hello, > > 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)

Ouch! :) > 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. 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/

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>> 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. > >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. >

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. >So in fact, I think that both wheels still get power, only the >slipping one more so. 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 >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. 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. >My intuition tells me you're looking too far, but you never know with >diffs. ;-) 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 On 29 Oct 2001, J. Todd Wasson wrote:

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> >> 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. > > > >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. > > > > > 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. > > > >So in fact, I think that both wheels still get power, only the > >slipping one more so. > > > 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 > >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. > > 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. > > > >My intuition tells me you're looking too far, but you never know with > >diffs. ;-) > > Maybe :0) > > > > > Todd Wasson > --- > Performance Simulations > Drag Racing and Top Speed Prediction > Software > http://PerformanceSimulations.Com > >

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>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

On 29 Oct 2001 22:55:30 GMT, [email hidden] (J. Todd Wasson) wrote: ...

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>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?

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. ;-) > 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. 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. > 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. 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/

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>>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? > >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 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. >The outer wheel would have to get >nearly no road reaction torque as well. 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. > You end up with no torque on >either side, but where then would you leave the incoming engine >torque? It accelerates the ring gear, along with both wheels' road reaction torques.

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> 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).

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) >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. 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 >is that the torque on each axle will be 1/2*T_engine. 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)

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>> 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. > >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.

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.

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>> 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. > >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") ;-) > >

Hehe :0) I'd buy it. I'll have to try this with very low inertias and high frequencies to see if it's just the inertial effect. I'm pretty sure it is. 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? On 31 Oct 2001, J. Todd Wasson wrote:

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> >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 > >

Doug Milliken <[email hidden]> writes: > 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? Get out and push. 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 nasty 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

On 31 Oct 2001 23:13:21 GMT, [email hidden] (J. Todd Wasson) wrote: > 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. 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? :) >> You end up with no torque on >>either side, but where then would you leave the incoming engine >>torque? > > It accelerates the ring gear, along with both wheels' road reaction torques. 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. ... > 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) You're most welcome. :) I'm glad to give another perspective that turns out the same way. ;-)

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> > The only thing you can count on I think >>is that the torque on each axle will be 1/2*T_engine. > > 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.

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.

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> 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

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... ;-) ><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. 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. ;-) > As you pointed out, we'll have to check the heisenberg compensators to make >sure :0) Without them, you couldn't call a sim a sim. GPL surely has them.

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>>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. > > Yes, the ratio could be anything of course. 1.76 was just an example :0)

Ah yes, I didn't read careful enough; I thought you meant it would not exceed this ratio ever then. >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. 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). >>Perhaps Doug has the formulae right there in an upcoming book? >>("Racing Car Dynamics for the mathematically disabled") ;-) > > > Hehe :0) I'd buy it. I'll have to try this with very low inertias and high >frequencies to see if it's just the inertial effect. I'm pretty sure it is. 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/

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>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? > >

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

I think that "pull the hand-brake" is the more more canonical answer, given the forward brake bias of most street cars... unless your front wheel were also on ice. Luckily/unluckily I'll probably have to test this out in a few months when Minnesota stops screwing around and actually gets down to some real winter, especially if I don't put on my Blizzaks in time! "J. Todd Wasson" <[email hidden]> wrote in message news:[email hidden]...

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> >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? > > > > > > 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. On Thu, 1 Nov 2001, Dave Pollatsek wrote:

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> I think that "pull the hand-brake" is the more more canonical answer, given > the forward brake bias of most street cars... unless your front wheel were > also on ice. Luckily/unluckily I'll probably have to test this out in a few > months when Minnesota stops screwing around and actually gets down to some > real winter, especially if I don't put on my Blizzaks in time! > > "J. Todd Wasson" <[email hidden]> wrote: > > >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? > > Push the brake a little along with the gas. What do I win? :-) > > Todd Wasson

You're in Minnesota too, Dave?? I'm in Richfield, about one mile from the Mall of America. Today was a real scorcher, what was it, 60 degrees? It ought to be about 30 below next week, then back up to 90 the week after.... :0) Todd Wasson --- Performance Simulations Drag Racing and Top Speed Prediction Software http://PerformanceSimulations.Com

Yup, MGI is located in bucolic Northfield (although a bunch of us live in Minneapolis)--which explains why if you look at all the track (.trk) files in Viper Racing, they are named things like "kenyon.trk", "bemidji.trk", etc. The weather's crazy now, but I'm not complaining! "J. Todd Wasson" <[email hidden]> wrote in message news:[email hidden]... > You're in Minnesota too, Dave?? I'm in Richfield, about one mile fromthe > Mall of America. Today was a real scorcher, what was it, 60 degrees? Itought

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> to be about 30 below next week, then back up to 90 the week after.... :0) > > > > Todd Wasson > --- > Performance Simulations > Drag Racing and Top Speed Prediction > Software > http://PerformanceSimulations.Com

Push in the button to activate the locking diff. :-) SLG [email hidden] (J. Todd Wasson) wrote:

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>>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? >> >> > > 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

(All spelling errors are intentional and are there to show new and improved ways of spelling old words. Grammatical errors are due to too many English classes/teachers)

> >>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? Walk without gloves or hat in the freezing -20 C temperatures along the railroad track that parallels the road until you find some small stones that you can chip free and bring back to wedge under the leading edge of the tires. Ask me how I know this. <grin> ...Klinn

In article <8aAF7.866$[email hidden]>, "Colin French" <[email hidden]> writes:

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>> >>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? > >Walk without gloves or hat in the freezing -20 C temperatures along the >railroad track that parallels the road until you find some small stones that >you can chip free and bring back to wedge under the leading edge of the >tires. Ask me how I know this. <grin> >

Ooh, THAT sucks.... Eldred -- Dale Earnhardt, Sr. R.I.P. 1951-2001 Homepage - http://www.umich.edu/~epickett GPLRank - under construction... Never argue with an idiot. He brings you down to his level, then beats you with experience... Remove SPAM-OFF to reply.

Colin French <[email hidden]> wrote in message news:8aAF7.866$[email hidden]...

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> > >>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? > > Walk without gloves or hat in the freezing -20 C temperatures along the > railroad track that parallels the road until you find some small stones

that > you can chip free and bring back to wedge under the leading edge of the > tires. Ask me how I know this. <grin> > Too bad the handbrake answer didn't come up earlier :-) Stephen

"Stephen Ferguson" <[email hidden]> wrote > Too bad the handbrake answer didn't come up earlier :-) > ...heh-heh, quite true, although I don't think it would have helped in my case - *both* rear wheels were on a glass-smooth sheet of ice. ...Klinn