rec.autos.simulators

Car Physics: Anyone?

Gregor Vebl

Car Physics: Anyone?

by Gregor Vebl » Thu, 29 Nov 2001 20:32:58

Hi all,

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

-Gregor

Jens H. Kruus

Car Physics: Anyone?

by Jens H. Kruus » Thu, 29 Nov 2001 23:16:42



Has there ever been a discussion about modelling the real non-linearity
of springs? ;-)
Or the rigidity of the chassis?

/Jens

andre

Car Physics: Anyone?

by andre » Fri, 30 Nov 2001 00:35:24

Physics.

> Hi all,

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

> -Gregor

Mike Stanle

Car Physics: Anyone?

by Mike Stanle » Fri, 30 Nov 2001 01:31:41

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.


amcconnel

Car Physics: Anyone?

by amcconnel » Fri, 30 Nov 2001 01:43:44

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

Jim Seamu

Car Physics: Anyone?

by Jim Seamu » Fri, 30 Nov 2001 02:01:56

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


> 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.



> > Hi all,

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

> > -Gregor

Gregor Vebl

Car Physics: Anyone?

by Gregor Vebl » Fri, 30 Nov 2001 02:20:50


>     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

Gregor Vebl

Car Physics: Anyone?

by Gregor Vebl » Fri, 30 Nov 2001 02:21:30


> Physics.

Thanks, I needed that :).

-Gregor

Ruud van Ga

Car Physics: Anyone?

by Ruud van Ga » Fri, 30 Nov 2001 22:32:27

On Wed, 28 Nov 2001 18:20:50 +0100, Gregor Veble



>>     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/

Mike Stanle

Car Physics: Anyone?

by Mike Stanle » Fri, 30 Nov 2001 22:36:39


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.

Mike Stanle

Car Physics: Anyone?

by Mike Stanle » Fri, 30 Nov 2001 22:41:11



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.

Mike Stanle

Car Physics: Anyone?

by Mike Stanle » Fri, 30 Nov 2001 22:52:52

It's ok I've found it. "Large Steps in Cloth Simulation", right?




> > 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.

Jonny Hodgso

Car Physics: Anyone?

by Jonny Hodgso » Sat, 01 Dec 2001 03:12:38


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).

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

Randy Cassid

Car Physics: Anyone?

by Randy Cassid » Sat, 01 Dec 2001 06:05:44





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

J. Todd Wass

Car Physics: Anyone?

by J. Todd Wass » Sat, 01 Dec 2001 11:24:41

  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


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