Car Physics: Anyone?

Hi all, I've just been missing some physics discussions lately I guess. Anyone, any questions, explanations etc. :)? -Gregor

"Gregor Veble" <[email hidden]> skrev i en meddelelse news:[email hidden]... > Hi all, > > I've just been missing some physics discussions lately I guess.Anyone, > any questions, explanations etc. :)? >Has there ever been a discussion about modelling the real non-linearity of springs? ;-) Or the rigidity of the chassis? /Jens

Physics. Gregor Veble wrote: > Hi all, > > I've just been missing some physics discussions lately I guess. Anyone, > any questions, explanations etc. :)? > > -Gregor

Not really been working on physics for the last month. Been working on the graphics side to give my brain a rest .... But here's a small point I was wondering about. For the springs in my suspension system I generate forces using F = k_s * extension + k_d * v_diff where k_s and k_d are spring and damper constatnts, and v_diff is the velocity of the sprung mass relative to the unsprung mass. In other sims (eg Racer) I've noticed they use damper bump rate and rebound rate rather than just a damper constant. Can anybody tell me how this works? Cheers, Mike. "Gregor Veble" <[email hidden]> wrote in message news:[email hidden]...

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> Hi all, > > I've just been missing some physics discussions lately I guess. Anyone, > any questions, explanations etc. :)? > > -Gregor > > >

I have a long line of questions, but need to work through them a bit myself before burdening the RAS population quite yet :). I have been too lazy recently ;) Ash

If the spring is being compressed, use k_bump, if it's being extended, use k_rebound, if it's not being compressed or extended then it doesn't matter anyway as v_diff = 0. Hope that's right (!) Jim "Mike Stanley" <[email hidden]> wrote in message news:_h8N7.2486$[email hidden]...

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> Not really been working on physics for the last month. Been working on the > graphics side to give my brain a rest .... > > But here's a small point I was wondering about. For the springs in my > suspension system I generate forces using > > F = k_s * extension + k_d * v_diff > > where k_s and k_d are spring and damper constatnts, and v_diff is the > velocity of the sprung mass relative to the unsprung mass. > > In other sims (eg Racer) I've noticed they use damper bump rate and

rebound > rate rather than just a damper constant. Can anybody tell me how thisworks?

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> > Cheers, > > Mike. > > "Gregor Veble" <[email hidden]> wrote in message > news:[email hidden]... > > Hi all, > > > > I've just been missing some physics discussions lately I guess. Anyone, > > any questions, explanations etc. :)? > > > > -Gregor > > > > > > > >

On Wed, 28 Nov 2001, Mike Stanley wrote:

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> > F = k_s * extension + k_d * v_diff > > where k_s and k_d are spring and damper constatnts, and v_diff is the > velocity of the sprung mass relative to the unsprung mass. > > In other sims (eg Racer) I've noticed they use damper bump rate and rebound > rate rather than just a damper constant. Can anybody tell me how this works? >

That one's not too difficult. It just means that if v_diff>0, then k_d has a different value (say k_d=k_bump) than when v_diff<0 (k_d=k_rebound, depending on how you define the v_diff). In general, though, you might even want to generalize the damping force to not be F_d= k_d * v_diff but F_d=F_damp(v_diff) which is now a general function of velocity which can then separate high speed from low speed behaviour. But that's a new topic altogether. Now, I'll think about the chassis rigidity question and post tomorrow.... :) Thanks to all, I needed that ;) -Gregor

On Wed, 28 Nov 2001, andrew wrote: > Physics. > Thanks, I needed that :). -Gregor

On Wed, 28 Nov 2001 18:20:50 +0100, Gregor Veble <[email hidden]> wrote:

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>On Wed, 28 Nov 2001, Mike Stanley wrote: >> >> F = k_s * extension + k_d * v_diff >> >> where k_s and k_d are spring and damper constatnts, and v_diff is the >> velocity of the sprung mass relative to the unsprung mass. >> >> In other sims (eg Racer) I've noticed they use damper bump rate and rebound >> rate rather than just a damper constant. Can anybody tell me how this works? > >That one's not too difficult. It just means that if v_diff>0, then k_d has >a different value (say k_d=k_bump) than when v_diff<0 (k_d=k_rebound, >depending on how you define the v_diff).

Right. Also note that Racer v0.4.9a3.1 (geez, what a name!) now uses Gregor's excellent implicit integration proposition. What I mean is you look at the integration formula, newWheelVel=oldWheelVel+... where ... is some F=m*a formulae that is normally like: F_total/m * dt but as damper forces really made my cars unstable, it changed into 2 separate integrations; 1 for the spring (which is ok for a 100-200Hz integrator mostly), and 1 for the damper. Look up Baraff's paper on Implicit Integration. That explains the idea. Weird formulae that come out of it but it works. :) Also note, as for the original question, nonlinearity in springs, you can do: F_spring = k1*d + k2*d^3 'k2' can be negative or positive, giving increasingly tight or decreasingly tight springs. Near the equilibrium position (and probably if k2<<k1) things will be linear enough but progressively non-linear when 'd' gets bigger (the extension). Now, for the next issue then, ;), how does turbo work? Compressed air into the cylinders, right? But when does it kick in? And how much is its effect in mathematical terms? And why does an intercooler improve the turbo effect? Ruud van Gaal Free car sim : http://www.marketgraph.nl/gallery/racer/ Pencil art : http://www.marketgraph.nl/gallery/

"Gregor Veble" <[email hidden]> wrote in message news:[email hidden]... > That one's not too difficult. ok, here's a trickier one then :) I've been thinking about upgrading my integrator from Euler to midpoint or RK4. Although the formulas are quite simple I'm a bit confused as to the implementation. If I take as an example the integration of velocity to produce position, then the problem is to solve ds(t)/dt = v(t) which for euler is s(t+dt) = s(t) + dt*v(t). I've been reading about using midpoint integration, and if I understand correctly the formula for this is s(t+dt) = s(t) + dt * v(t+dt/2) My question is, how do I evaluate v at the half time step? I do not have an expression for v, just a discrete set of values, so presumably I have to do it numerically. But wouldn't this affect the accuracy of the integrator, and reduce it back to 1st order? Has anybody tried implementing integrators other than Euler yet? Cheers, Mike.

"Ruud van Gaal" <[email hidden]> wrote in message news:[email hidden]... > integrator mostly), and 1 for the damper. Look up Baraff's paper on > Implicit Integration. That explains the idea. Weird formulae that come > out of it but it works. :) > What's the title of that paper? I had it once but can't seem to find it now (he's done so many!). Cheers, Mike.

It's ok I've found it. "Large Steps in Cloth Simulation", right? "Mike Stanley" <[email hidden]> wrote in message news:5UqN7.225$[email hidden]...

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> > "Ruud van Gaal" <[email hidden]> wrote in message > news:[email hidden]... > > integrator mostly), and 1 for the damper. Look up Baraff's paper on > > Implicit Integration. That explains the idea. Weird formulae that come > > out of it but it works. :) > > > > What's the title of that paper? I had it once but can't seem to find it

now

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> (he's done so many!). > > Cheers, > > Mike. > > >

Ruud van Gaal <[email hidden]> wrote in message news:[email hidden]... > Now, for the next issue then, ;), how does turbo work? Compressed air > into the cylinders, right? But when does it kick in? And how much is > its effect in mathematical terms? As a first approximation, boost pressure is proportional to exhaust flow (read ~engine power) with a lag; first-order should do okay. Flow rate through the throttle is then going to be proportional to (atmosphere + boost pressure), and engine power should be proportional to intake flow. Now, precisely how you determine intake flow (as a combination of throttle opening and engine speed) is another question - as is whether back-pressure is significant. My guess to the latter is "no" assuming mild turbocharging (perhaps up to ~0.7bar gauge). > And why does an intercooler improve the turbo effect? It increases the mass flow of air for a given volume flow, hence allowing you to burn more fuel at stoichiometric A:F ratio and thus produce more power. HTH! Jonny

"Jonny Hodgson" <[email hidden]> wrote in message news:tcvN7.1097$[email hidden]... > > Ruud van Gaal <[email hidden]> wrote in messagenews:[email hidden]... > > > And why does an intercooler improve the turbo effect? > > It increases the mass flow of air for a given volume flow, > hence allowing you to burn more fuel at stoichiometric A:F > ratio and thus produce more power. Compressing the air charge also heats it, making detonation an issue (at least with a gasoline engine). Adding an intercooler allows higher boost pressures without inducing detonation. Randy

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>Now, for the next issue then, ;), how does turbo work? Compressed air >into the cylinders, right? But when does it kick in? And how much is >its effect in mathematical terms? >And why does an intercooler improve the turbo effect? > > >Ruud van Gaal

AAARGGH!!! $%$%!!! Sorry, I got infected with a virus called IMAGES.DOC.pif... Doh!! Should have known better than to run that one :0) This is now my third attempt at responding to this. Better type fast and save a lot! lol Here is my extremely scientific explanation ;-): A turbo compresses air into the cylinders, right. It's basically just a spinning shaft with a turbine on one end that the exhaust pipe runs through, and a compressor on the other end that the intake air runs through. Essentially, it's like two windmills on a stick.. Hmmm... Giant candy? Har har har, Green Giant.. Ok, I'm goofy today, the virus has infected me through the keyboard and I'm losing my mind. Anyway, the exhaust gases are going really fast and are at high pressure, this pushes against the (where the heck did my cursor just go??) exhaust turbine "paddle", provided the paddle is moving more slowly than the exhaust gas itself, I suppose. (Cursor's back now, everything's fine. Time to save this real quick. Ok, done.) This speeds up the shaft, turbine and compressor (since they're all connected.) The compressor "paddle", which is hopefully spinning faster than the air in the intake system, forces the intake air to a higher speed, which pushes it into the air ahead of it, and ahead of it, and ahead of it, compressing it and raising it's pressure in a sonic wave as it goes, all the way into the cylinder. This additional pressure is the turbo boost pressure you read on the dashboard of Gregor Veble's Lingenfelter Twin Turbo Corvette, if he's ever taken you for a ride in it ;-) To put it more simply, the exhaust gases are pushing on one end of the turbo's shaft, the intake is pushing on the other. Usually, the exhaust gases are trying to speed up the turbo, while the intake gases are trying to slow it down. This isn't always the case, but usually it is. A turbo could probably be done by declaring a moment of inertia for the turbo, then letting a combination of gas pressure and velocity on each side of the turbo give it two "torques", one in the exhaust and one in the intake, which speed it up and slow it down. You could probably generate a volumetric efficiency curve (which is basically how full the cylinders are of air, expressed as a % of what mass would fill the cylinder at atmospheric pressure) from your normalized engine torque curve, which might be moved up or down slightly, and maybe shifted a bit down the rpm scale. This is because the "cylinder filling", or volumetric efficiency is closely related to the engine torque. As I currently understand it, they deviate from each other primarily because of changes in fuel mixture quality at certain rpm ranges and increased engine friction as rpm goes up. Anyway, given the engine displacement, power/torque, and the volumetric efficiency at whatever engine speed you're looking at, you could calculate the air flow rate through the engine. If you knew the cross sectional areas at the intake and exhaust sections running through the turbo, you could estimate the velocities. If you want a formula for that I'd be happy to oblidge. Looking at a paddle in one of the air flow streams, there would basically be two forces I think that could take care of things. One would be a collision force. Actually, come to think of it, the gas momentum could probably be used directly... So you might be able to do it with just the mass flow rate, although it seems to me that if the... Well, I guess maybe I haven't quite figured it out yet! lol Sorry :-( Still, if you can come up with a way to vary the boost pressure realistically, it's a pretty safe assumption that the power will scale right along with the change in boost pressure. You could have a factor to multiply the fixed torque curve by according to boost pressure. For instance, if standard air pressure was 1 bar, and the boost pressure was .2 bar above that (1.2 bar on Gregor's guage, or .2 bar??) you'd get about a 20% increase in torque. That's not set in stone of course, 'cause things can happen to spark, burn duration, ignition timing may no longer be right, valve timing might be off optimum now, etc., but it's a pretty close assumption provided everything's dialed in right, I think. (I'm back again for #4 attempt, good thing I just saved this!) Anyway, it seems that basically the speed of the turbo in relation to the engine speed determines the boost pressure, so maybe you could say something like: BoostPressure = (turbospeed * factor) / enginespeed You could then use BoostPressure as a multiplier for the fixed engine torque curve. The problem is, how do you alter the turbo speed without calculating all the gas movement stuff and keep it simple? I think the "fixed" engine torque as it's modified by the throttle percentage (without the turbo added in), could be used to adjust the turbo speed in combination with the engine rpm, since that would pretty much get the basic effect of exhaust flow rate to speed up the turbo... Before I mentioned two torques. How about for the first torque, use something like: TorqueExhaust = EngineTorque * throttle * rpm * some_factor Then, maybe you could calculate boost pressure: BoostPressure = (turbospeed * some_other_factor) / enginespeed Then get a new engine torque with turbo added in: EngineTorqueWithTurbo = EngineTorque * BoostPressure Maybe this new engine torque with the turbo added in could be used for the torque at the compressor (intake), since the cylinder pressure with the turbo boost added is what tries to slow down the turbo: TorqueIntake = EngineTorqueWithTurbo * throttle * rpm * some_factor Then maybe you could add the two torques together and accelerate the turbo? Just an idea anyway to get us thinking, probably wouldn't really work that way. As far as intercoolers go, they are used to cool the intake air. Cold air is denser than warm air, so you wind up with more air and fuel in the cylinder than you otherwise would. There's a rule of thumb (here in the US anyway, sorry about the English units), that each 11 degree Fahrenheit drop in temperature increases torque/power by 1%. Air under the hood around the engine can get rather hot, which means full power isn't available once the thing's running awhile. To solve this, many folks duct outside air in from the front of the car, or through a ram-air hood or scoop of some sort. This can make a noticable difference in some cars. Intercoolers cool the air a lot more than this though, so it's better (but more expensive of course than some hoses and plastic.) Well, I think that's about it on my end on the subject of turbos :0) I haven't studied them at all nor given them much though yet. If you'd like, I'll try to make a longer, more rambling post next time though! Todd Wasson --- Performance Simulations Drag Racing and Top Speed Prediction Software http://PerformanceSimulations.Com

<lurk off> just felt an urge to throw some random extras into the mix :) "J. Todd Wasson" wrote: > To put it more simply, the exhaust gases are pushing on one end of the > turbo's shaft, the intake is pushing on the other. Usually, the exhaust gases > are trying to speed up the turbo, while the intake gases are trying to slow it > down. This isn't always the case, but usually it is. > Here's a common example of when the exhaust gases are not driving the turbo, and neither are the intake gases. When you accelerate at full throttle, then lift off the gas to change gears. Suddenly the exhaust gas speed falls away, but there's still a full load of compressed air in the intake. Combine that with the backpressure wave from the intake air hitting the closed throttle plate and you get nasty intake backpressure waves crashing into the compressor wheel. This isn't very healthy to the long term life of a turbo, so there is usually some kind of pressure relief valve on the intake just after the turbo designed to open up when there's a large pressure differential between the intake and the manifold pressure to let this air out and save your turbo. The term I'm used to for naming this device is the blow-off valve, (BOV), and this is the source of all the weird whistle/whoosh noises during shifts or when letting off the throttle from the tasteless boy racer turbo charged honda civic with neon lights you see downtown (or my very tasteful (no neon lights) MR2 :) ). > Still, if you can come up with a way to vary the boost pressure > realistically, it's a pretty safe assumption that the power will scale right > along with the change in boost pressure. You could have a factor to multiply > the fixed torque curve by according to boost pressure. For instance, if > standard air pressure was 1 bar, and the boost pressure was .2 bar above that > (1.2 bar on Gregor's guage, or .2 bar??) that would be .2 bar on the gauge. Gauges read zero at atmospheric pressure, and positive for greater than atmospheric pressure amounts. They usually use units of vacuum for pressures from atmospheric down to zero. 0.2 bar is a pathetic amount of boost :) > Then maybe you could add the two torques together and accelerate the turbo? Remember to make sure to model lots of turbo lag! If you're modelling a turbo and you don't get pissed off at the bad throttle response compared to a normally aspirated engine, you're probably missing something :) The turbo does have it's own momentum, but the major source of turbo lag is simply the amount of space it has to pressurise in the intake plumbing. For instance, if you stuck an intercooler in the front grill of a mid- engined car, you'd have about 6 miles of intake piping from the back of the car up to the front and back to pressurise, so it would take half an hour to boost up! Of course this is an extreme example.

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> As far as intercoolers go, they are used to cool the intake air. Cold air is > denser than warm air, so you wind up with more air and fuel in the cylinder > than you otherwise would. There's a rule of thumb (here in the US anyway, > sorry about the English units), that each 11 degree Fahrenheit drop in > temperature increases torque/power by 1%. Air under the hood around the engine > can get rather hot, which means full power isn't available once the thing's > running awhile. To solve this, many folks duct outside air in from the front > of the car, or through a ram-air hood or scoop of some sort. This can make a > noticable difference in some cars. Intercoolers cool the air a lot more than > this though, so it's better (but more expensive of course than some hoses and > plastic.)

Intercoolers are great in this respect, especially on a nice cool chilly day :) However as Randy mentioned earlier, they are also extremely important if you're trying to run anything like .4 bar or more to prevent your engine from blowing up a few pistons. As the piston is compressing the air in the cylinder, the cylinder pressure and temperature build. At a certain combination of both factors, the fuel can (and will!) ignite by itself. This of course is really bad if the piston is still going up and the resulting explosion is trying to push the piston back down. You can avoid it by either running less boost (which is the no-fun solution), or using an intercooler to reduces the intake charge temperature. (Or running a richer fuel mixture - the charge won't ignite so easily if it's overly rich) You can also push the self ignition point further away by using higher octane fuel. Running 87 octane gas in my MR2 at 1 bar boost would be a guaranteed blown motor. 93-94 octane is OK. Full on racing gas is vastly superior than pump gas in this respect, which is why you hear of CART (or 80's F1) running at 40psi (2.5 bar) or more. Or those cars Rod Millen runs at Pike's Peak at 80psi. This is also why most turbo cars use some form of boost pressure regulation to prevent the boost from going over a certain limit. This is done by bypassing excess exhaust gas around the turbo's exhaust turbine once the desire intake pressure is reached, via a wastegate. <lurk on>

Grant, WONDERFUL post! :0) >that would be .2 bar on the gauge. Gauges read zero at atmospheric >pressure, >and positive for greater than atmospheric pressure amounts. >They usually use units of vacuum for pressures from atmospheric down to >zero. 0.2 bar is a pathetic amount of boost :) Ah, ok. I thought so, but wasn't sure... .2 bar IS a pathetic amount of boost :0) >Remember to make sure to model lots of turbo lag! If you're modelling a >turbo and you don't get pissed off at the bad throttle response compared >to a normally aspirated engine, you're probably missing something :) >The turbo does have it's own momentum, but the major source of turbo lag >is simply the amount of space it has to pressurise in the intake >plumbing. Good point, hadn't thought of the space issue.

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>Intercoolers are great in this respect, especially on a nice cool chilly >day :) However as Randy mentioned earlier, they are also extremely >important >if you're trying to run anything like .4 bar or more to prevent your >engine >from blowing up a few pistons. As the piston is compressing the air in >the >cylinder, the cylinder pressure and temperature build. At a certain >combination of both factors, the fuel can (and will!) ignite by itself.

Ah ha! I knew there had to be another reason besides increasing density :-) Hadn't thought of the temperature change from the work to the gas during compression. Todd Wasson --- Performance Simulations Drag Racing and Top Speed Prediction Software http://PerformanceSimulations.Com

Mike Stanley wrote:

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> > > ok, here's a trickier one then :) > > I've been thinking about upgrading my integrator from Euler to midpoint or > RK4. Although the formulas are quite simple I'm a bit confused as to the > implementation. If I take as an example the integration of velocity to > produce position, then the problem is to solve > > ds(t)/dt = v(t) > > which for euler is > > s(t+dt) = s(t) + dt*v(t). > > I've been reading about using midpoint integration, and if I understand > correctly the formula for this is > > s(t+dt) = s(t) + dt * v(t+dt/2) > > My question is, how do I evaluate v at the half time step? I do not have an > expression for v, just a discrete set of values, so presumably I have to do > it numerically. But wouldn't this affect the accuracy of the integrator, and > reduce it back to 1st order? Has anybody tried implementing integrators > other than Euler yet?

An efficient second order integrator for real time systems that use second order systems (like your velocity and position set above) is a combination of a second order Adams for velocity and a second order Moulton Corrector for position. Here, the _prev variables are the previous value dt ago and a is acceleration v = v_prev + (dt/2)(3 * a - a_prev) ' 2nd order Adams p = p_prev + (dt/2)(v + v_prev) ' 2nd order Moulton Corrector Generally you can't use a corrector like that in a true real time system using measurements because you don't have the future measurements yet. However, in this case you can use the Adams integrator to generate the data required by the second integrator. This gives you second order integration with a cost of only one derivative evaluation each. Note that you now have to maintain the previous value of the derivative of velocity (a_prev) A Runge-Kutta 4th order integrator will require 4 derivative calculations. If you push the integration step high enough to flirt with instability, you may find that the second order integrator above is actually more efficient than the Runge-Kutta even though you may have to run it at a 3 times smaller time step. (3 derivatives in the same duration that the RK4 would need 4 derivatives.) I'm not considering accuracy in this case, rather the frequency response that most efficiently gives "decent" results without immediate instability. - Matt Ref: "Math Toolkit for Real-Time Programming," Jack W. Crenshaw, CMP Books, Lawrence, Kansas, 2000. ISBN 1-929629-09-5

On 30 Nov 2001 02:24:41 GMT, [email hidden] (J. Todd Wasson) wrote: >>Now, for the next issue then, ;), how does turbo work? Compressed air >>into the cylinders, right? But when does it kick in? And how much is >>its effect in mathematical terms? >>And why does an intercooler improve the turbo effect?... >Here is my extremely scientific explanation ;-): > > A turbo compresses air into the cylinders, right.... Thanks for the elaborate reply, Todd! Also thanks to Jonny, Grant and Randy (nice to see Randy still pop up here). I now finally understand a little about turbo. Will have to deal with all these other deadlines here though before I can actively implement stuff, but at least my right brain can work on that subconsciously. ;-) To be continued in the future somewhere... Ruud van Gaal Free car sim : http://www.marketgraph.nl/gallery/racer/ Pencil art : http://www.marketgraph.nl/gallery/

On Sat, 01 Dec 2001 04:51:16 GMT, "Matthew V. Jessick" <[email hidden]> wrote: >Mike Stanley wrote: >> I've been thinking about upgrading my integrator from Euler to midpoint or >> RK4.... >> ds(t)/dt = v(t) >> >> which for euler is >> >> s(t+dt) = s(t) + dt*v(t). Also take note that for F=m*a type integrations, you start out with 'a', which is the 2nd derivate. So you can expand to: s(t+dt)=s(t)+dt*v(t)+0.5*dt^2*a(t) Which is already more stable, and nearly free. :) And ofcourse, precalculate t^2. Matt's combined integrator seems very interesting too, I'd try that too if I were you (and test with big time steps to see how it's doing). Ruud van Gaal Free car sim : http://www.marketgraph.nl/gallery/racer/ Pencil art : http://www.marketgraph.nl/gallery/

OK, thanks to you both. I'm currently adding in code to allow me extend the integrator (ie storing velocities and accelerations from the previous n timesteps) after which I'll be in a position to test lots of different integrators. My main aim is to improve on the stability of the Euler integrator (mainly for spring forces) so I may attempt an implicit integrator as well. I'll post when I've done a few tests, and let you all know my conclusions. Cheers, Mike. "Ruud van Gaal" <[email hidden]> wrote in message news:[email hidden]... > On Sat, 01 Dec 2001 04:51:16 GMT, "Matthew V. Jessick" > <[email hidden]> wrote: > > >Mike Stanley wrote: > >> I've been thinking about upgrading my integrator from Euler to midpointor

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> >> RK4. > ... > >> ds(t)/dt = v(t) > >> > >> which for euler is > >> > >> s(t+dt) = s(t) + dt*v(t). > > Also take note that for F=m*a type integrations, you start out with > 'a', which is the 2nd derivate. So you can expand to: > > s(t+dt)=s(t)+dt*v(t)+0.5*dt^2*a(t) > > Which is already more stable, and nearly free. :) > And ofcourse, precalculate t^2. > > Matt's combined integrator seems very interesting too, I'd try that > too if I were you (and test with big time steps to see how it's > doing). > > > Ruud van Gaal > Free car sim : http://www.marketgraph.nl/gallery/racer/ > Pencil art : http://www.marketgraph.nl/gallery/