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Rec.Autos.Simulators Archive

Re: Car Physics: Clutch & Torque transfer.

9 posts by 6 people · 21 Sep 2000 - 22 Sep 2000 09:55 UTC

Quoted text (21 lines)>Hi All > >As there appears to be some very knowledgeable car dynamics people in this >newsgroup. I wonder if someone could enlighten me to a problem I have. > >I am working on a little simulation (not cars, but a clutch is envolved) and >I am a little confused on the clutch torque transfer. For example if an >engine is giving out 100Nm of torque at the flywheel, and a clutch is >attached which can transfer 400Nm of torque when fully engaged. Does this >mean if the clutch was engaged instantly that for a small period 400Nm of >torque is transferred to the drivetrain until the increased load slows the >engine? Or am I missing some important aspects. I was wondering if the speed >difference at the clutch effects the torque, or if the torque transmitted >can not be more than the engine is giving out. > >Any help to clear this up would be great. >Thanks >Mike > > >
I disagree with the other 3 posts that have been put up so far. Mike, you're exactly right about getting 400 N*M for just an instant. Anybody ever dumped a little torqueless 4 banger with a decent clutch to get the wheels to spin? I do it in my car from time to time and the engine doesn't make anywhere near enough torque by itself to do it. It's just that little period of time... If the engine's reved up and you "dump it", the full 400 N*M is sent through the drivetrain. This accelerates the drivetrain at 400 N*M no matter what the engine's torque is doing. However, this also DECELERATES the engine at a rate determined by 100 N*M - 400 N*M = 300N*M of torque. The engine snaps to a lower speed until the flywheel and clutch are coupled, THEN you'll get 100 N*M going through the drivetrain. I'll admit, I haven't given this a whole lot of thought yet, but this is the way I'm planning on testing the next version of Straightline Acceleration Simulator first :-) Todd Wasson --- Performance Simulations Drag Racing and Top Speed Prediction Software http://PerformanceSimulations.Com
Scott Spahr 21 Sep 2000#230173
This is not correct. The post by Bob Norton gives the correct answer. As he states, "The clutch is not a torque multiplier." If you want a basic description of how the clutch works check out http://www.howstuffworks.com/clutch.htm The effect Todd is describing is due to the angular momentum of the engine in transient conditions, as Bob describes. Scott Spahr > I disagree with the other 3 posts that have been put up so far. Mike,you're > exactly right about getting 400 N*M for just an instant. Anybody everdumped a > little torqueless 4 banger with a decent clutch to get the wheels to spin?I > do it in my car from time to time and the engine doesn't make anywherenear > enough torque by itself to do it. It's just that little period of time... > > If the engine's reved up and you "dump it", the full 400 N*M is sentthrough > the drivetrain. This accelerates the drivetrain at 400 N*M no matter whatthe > engine's torque is doing. However, this also DECELERATES the engine at arate > determined by 100 N*M - 400 N*M = 300N*M of torque. The engine snaps to a > lower speed until the flywheel and clutch are coupled, THEN you'll get 100N*M > going through the drivetrain. I'll admit, I haven't given this a wholelot of > thought yet, but this is the way I'm planning on testing the next versionof
Quoted text (11 lines)> Straightline Acceleration Simulator first :-) > > > > > Todd Wasson > --- > Performance Simulations > Drag Racing and Top Speed Prediction > Software > http://PerformanceSimulations.Com
Mike Cain 21 Sep 2000 14:47 UTC#230268
Hi All As there appears to be some very knowledgeable car dynamics people in this newsgroup. I wonder if someone could enlighten me to a problem I have. I am working on a little simulation (not cars, but a clutch is envolved) and I am a little confused on the clutch torque transfer. For example if an engine is giving out 100Nm of torque at the flywheel, and a clutch is attached which can transfer 400Nm of torque when fully engaged. Does this mean if the clutch was engaged instantly that for a small period 400Nm of torque is transferred to the drivetrain until the increased load slows the engine? Or am I missing some important aspects. I was wondering if the speed difference at the clutch effects the torque, or if the torque transmitted can not be more than the engine is giving out. Any help to clear this up would be great. Thanks Mike
Bob Norton 21 Sep 2000 16:33 UTC#230090
Steady-state answer (more-or-less constant rpm): The clutch is not a torque multiplier. The only way to multiply torque is by inserting gears. Note that horsepower remains the same, less loss due to fricition, so that if you multiply the torque by a factor the output shaft rpm is the same factor lower than the input shaft rpm. In your case the clutch capacity of 400 N-m is a maximum value that can be transmitted by the clutch. Since the input from the engine is 100 N-m, the clutch capacity is adequate, and the clutch will not slip (in this steady-state mode). Transient answer (sudden engagement of clutch): When the clutch is suddenly engaged with some difference in shaft speeds there will be some dynamic effects. For example, if the engine is running at 3000 rpm and the output is stationary the clutch will be asked to transmit more than the 100 N-m of steady-state torque. In a car this means that the clutch may slip, the wheels might slip, the drivetrain will wind-up, the engine may bog, etc. The relative effects will depend upon the complex dynamics. Simplifying somewhat, the torque output from the engine will the steady-state value of 100 N-m plus the moment of inertia of the engine times the angular acceleration of the engine (the angular version of F=ma). Bob On Thu, 21 Sep 2000 14:51:04 +0100, "Mike Cain" <[email hidden]> wrote:
Quoted text (20 lines)>Hi All > >As there appears to be some very knowledgeable car dynamics people in this >newsgroup. I wonder if someone could enlighten me to a problem I have. > >I am working on a little simulation (not cars, but a clutch is envolved) and >I am a little confused on the clutch torque transfer. For example if an >engine is giving out 100Nm of torque at the flywheel, and a clutch is >attached which can transfer 400Nm of torque when fully engaged. Does this >mean if the clutch was engaged instantly that for a small period 400Nm of >torque is transferred to the drivetrain until the increased load slows the >engine? Or am I missing some important aspects. I was wondering if the speed >difference at the clutch effects the torque, or if the torque transmitted >can not be more than the engine is giving out. > >Any help to clear this up would be great. >Thanks >Mike > >
Andy Laurence 21 Sep 2000 16:51 UTC#230284
> As there appears to be some very knowledgeable car dynamics people in this > newsgroup. I wonder if someone could enlighten me to a problem I have. > > I am working on a little simulation (not cars, but a clutch is envolved)and > I am a little confused on the clutch torque transfer. For example if an > engine is giving out 100Nm of torque at the flywheel, and a clutch is > attached which can transfer 400Nm of torque when fully engaged. Does this > mean if the clutch was engaged instantly that for a small period 400Nm of > torque is transferred to the drivetrain until the increased load slows the > engine? Or am I missing some important aspects. I was wondering if thespeed > difference at the clutch effects the torque, or if the torque transmitted > can not be more than the engine is giving out. Basically, if the engine only produces 100Nm, then that's all you'll get. The rating of the clutch just means that it'll be able to transfer that torque without slipping or breaking. In answer to the example you gave, 100Nm is all you'll see. You can of course get more torque, albeit with a reduction in speed. Using a reduction gearbox, you can work out how much torque or how much speed you need and optimise for that. The limiting factor is the power of the engine which is torque x speed. A reduction in speed gives and increase in torque, and vice-versa. HTH, I'm sure someone will explain it better than I have .... Andy -- Visit my site... http://mail.wm-europe.com/~andrew/ Read about my Mini... http://mail.wm-europe.com/~andrew/mini/ Use console pads on a PC ... http://www.consolepads.co.uk Music videos and DC games ... http://mail.wm-europe.com/~andrew/ftp.html
Mike Cain 21 Sep 2000 19:15 UTC#230111
Quoted text (9 lines)> Basically, if the engine only produces 100Nm, then that's all you'll get. > The rating of the clutch just means that it'll be able to transfer that > torque without slipping or breaking. In answer to the example you gave, > 100Nm is all you'll see. You can of course get more torque, albeit with a > reduction in speed. Using a reduction gearbox, you can work out how much > torque or how much speed you need and optimise for that. The limiting > factor is the power of the engine which is torque x speed. A reduction in > speed gives and increase in torque, and vice-versa. HTH, I'm sure someone > will explain it better than I have ....
Thnaks for the help Andy. The problem I have is, that I have read that fritional torque is the product of the normal force x coef_friction, so if the clutch can transfer 400Nm, then the torque at the clutch when there is a difference in speed (ie a slipping clutch) should be 400Nm. I fully see that with no slipping you will only get the torque of the engine, but under a slipping clutch will I get 400Nm?
Quoted text (9 lines)>This is not correct. The post by Bob Norton gives the correct answer. As >he states, "The clutch is not a torque multiplier." If you want a basic >description of how the clutch works check out >http://www.howstuffworks.com/clutch.htm > >The effect Todd is describing is due to the angular momentum of the engine >in transient conditions, as Bob describes. > >Scott Spahr
Right. I disagreed with Bob's and the others' posts without carefully reading what Bob stated here.: (For a sudden engagement with a difference in rotating speed between flywheel and clutch) "Simplifying somewhat, the torque output from the engine will the steady-state value of 100 N-m plus the moment of inertia of the engine times the angular acceleration of the engine (the angular version of F=ma)." What I was really trying to show was HOW to calculate the angular acceleration of the engine given Mike's data. If it was already known, then yes, this quote is a good explanation. However, Mike Cain's question was: "For example if an engine is giving out 100Nm of torque at the flywheel, and a clutch is attached which can transfer 400Nm of torque when fully engaged. Does this mean if the clutch was engaged instantly that for **>a small period 400Nm of torque is transferred to the drivetrain until the increased load slows the engine?<**" The answers everybody (except Bob) gave was that 100 Nm would be transmitted during the "small period" of time where the flywheel and clutch were not spinning at the same speed (instant engagement). I disagreed with this because if it were true, dumping a 400Nm, 700Nm or 10,000Nm capacity clutch at high rpm with a stationary vehicle would result in **0** change in engine rpm until the vehicle "caught up" to the engine's speed. (100-100=0 torque and angular acceleration). The engine revs would merely "hang" at the rpm it was at when the clutch was engaged, no matter *HOW* strong the clutch plate is pressed against the flywheel. Yes, it IS a transfer of angular momentum as Bob stated, but I think Mike wants to know HOW to find this and which (100 or 400) torque to use to do it. In order for Mike to calculate the angular acceleration of the engine (or drivetrain for that matter), he'd need the 100Nm the engine is creating, and the -400Nm the clutch can transfer, giving a -300Nm net torque decelerating the engine and a 400 Nm torque accelerating the drivetrain. We're not talking about a long period of time here, only during the period where the clutch and flywheel haven't matched speeds. Once they have, then of course, "the clutch is not a torque multiplier" and he'll have 100Nm accelerating the drivetrain. Of course, Mike said he isn't simulating a car, so maybe this is all just... ;-P Todd Wasson --- Performance Simulations Drag Racing and Top Speed Prediction Software http://PerformanceSimulations.Com
Matthew V. Jessick 22 Sep 2000 06:38 UTC#230051
For a simulation, the torques involved in a racing engine are so high and the inertias can be so low that you are unlikely to be able to run the simulation in real time for all conditions. So be prepared to make appropriate compromises. After all, if you aren't including detailed and complicated things like thermal effects, taking minute care of the transients would seem to be a bit of overkill... - Matt
Andy Laurence 22 Sep 2000 09:55 UTC#230072
> > Basically, if the engine only produces 100Nm, then that's all you'llget. > > The rating of the clutch just means that it'll be able to transfer that > > torque without slipping or breaking. In answer to the example you gave, > > 100Nm is all you'll see. You can of course get more torque, albeit witha > > reduction in speed. Using a reduction gearbox, you can work out howmuch > > torque or how much speed you need and optimise for that. The limiting > > factor is the power of the engine which is torque x speed. A reductionin > > speed gives and increase in torque, and vice-versa. HTH, I'm suresomeone > > will explain it better than I have .... > > Thnaks for the help Andy. The problem I have is, that I have read that > fritional torque is the product of the normal force x coef_friction, so if > the clutch can transfer 400Nm, then the torque at the clutch when there isa > difference in speed (ie a slipping clutch) should be 400Nm. I fully seethat > with no slipping you will only get the torque of the engine, but under a > slipping clutch will I get 400Nm? By slipping the clutch, you are letting the torque through gradually, ie. the clutch lets a percentage between 0 and 100 through depending on how much you use it. If the clutch had a linear pedal (which it by no means does), then when the pedal is fully depressed, the clutch would transfer no torque (0%), if you pulled the pedal back half way, then you'd get half torque (50% of 100Nm - 50Nm), and so on. A bit busy to explain this now ... maybe someone else can take over? Andy -- Visit my site... http://mail.wm-europe.com/~andrew/ Read about my Mini... http://mail.wm-europe.com/~andrew/mini/ Use console pads on a PC ... http://www.consolepads.co.uk Music videos and DC games ... http://mail.wm-europe.com/~andrew/ftp.html

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