rec.autos.simulators

I recently read some more about SMD systems; spring-mass-dampers. In
there, the energy of the system was defined as:

E=Ep+Ek=constant (Ep=potential, Ek=kinetic energy).

Ep is defined as the amount of work done to get the
compression/decompression, so Ep is the integral over the spring
offset:

Ep=1/2*k*x^2 (k=spring rate, x=offset w.r.t. rest length)

Ek is defined as the usual 1/2*m*v^2.
I want to get a look at the energy in the system, since I've read
(from FastCar) that with Euler integration, when the car is in the
air, the suspensions and car can add energy into the system quite
quickly (again, with Euler integration), as the dampers don't really
do a lot of work once airborne.

The question is, how much is 'm' in the Ek formula? A regular SMD
system has 1 mass attached to it, but a suspension has both mass on
one end (the tire) and on the other end (the car). I'm not sure
whether to see one of both as the 'ground' and use the mass of the
other, as that doesn't seem right, since both can move considerably
(although the tire mass is very small compared to the car's mass).

Does anybody have an idea about this?

The reason for all this is to check my integrations; if I integrate,
and Ep+Ek becomes larger, I can reduce the suspension movement until I
get sort of an average new Ep+Ek that will keep things constant.

Thanks for any ideas,

Ruud van Gaal
Free car sim  : http://www.racesimcentral.net/
Pencil art    : http://www.racesimcentral.net/

Ummmm.......Can I call you back ?

(what the ***y hell is he talking about ?)

E = Ep + Ek(s) + Ek(us)

OTOH I'm not sure why you are bothering.  Best just to use F=mA, IMO.  I
did some drag race simulation code in the 80's for a project I was
working on, and that was all I used.  The results came out quite
accurate as I recall.  Obviously the key is breaking down the F term
into as much detail as possible, and then simulating it in as tiny steps
as possible (I think I calculated every foot, but of course that is not
very convenient for real time code).  But when you start adding in
energy balance and Euler integration and all that other stuff it just
gets confusing.

I agree with that - each of the masses influences the spring and should
therefore make a contribution to the energy in the system. But like Ruud
said, the wheel's mass will be small compared to the vehicle, so the
contribution may be negligible. However if you decided to simulate a tractor
or a monster truck it would be a different story.

I agree with the energy thing. Personally I've never had to directly track
energy in a system in any physics sims I've written, but OTOH I'm not sure I
completely understand what Ruud's trying to do. However I don't see how you
could do a physical simulation without doing some form of numerical
integration - you need to somehow get from acceleration to velocity to
position. Could you explain how you avoided doing this?

I think I just got lost in the terminology.  I've been in a product
releasing position for the past 7 years or so, and so my knowledge of
theory is getting rusty.  ;o)  Yes I did do some form of integration,
you are absolutely right.  And I just looked up Euler integration to
refresh my memory, and that is of course one way to do it.  I think I
added some other methods to minimize error accumulation, but I don't
have the code here at home so I can't tell you for sure.  Sorry for the
confusion.

That's perfectly alright :)

Just thought there might have been some other method I've never heard of. I
must admit that when I first started games programming, I was using Euler
integration for a while without even knowing it was a form of numerical
integration :o)

Mike

Hm yes, that may be the right one. :)

Well, if you've tried*** an undamped wheel to a car which is
floating in air, you can see that Euler integrators (in math English;
pos+=vel*t, vel+=acc*t) can generate quite some error which allows the
suspension to not just bounce up & down indefinitely, but the
amplitude gets bigger and bigger and BIGGER until it all explodes. ;-)
There are multiple ways of doing rigid body physics; Newtonian is
known to most of us, but you also have LaGrangian mechanics, which
writes everything down as energy. The nice thing is (I think) that you
can check for constant energy in systems with the latter, where with
forces you just apply and see what happens.
The energy could *check* whether a given force and resulting
displacement is too much and would increase the energy. WSC uses it
for example, and it seems to make things <quote> 'rock-solid'
</quote>. So it's good to look into. :)

Thanks for the tip; I'll see how that would work out.

Ruud van Gaal
Free car sim  : http://www.racesimcentral.net/
Pencil art    : http://www.racesimcentral.net/

I missed the thread earlier, obviously. :)

The energy indeed needs to be written for all components of the system.
Therefore, for each rigid body in the system you need to introduce 1/2 m
v^2 for translational, 1/2 omega^T J omega for rotational (omega is a
vector, J the inertia tensor and T means the transpose), and 1/2 k x ^2
for any spring in the system (including the tyre 'spring' I suppose).

However, if you indeed use a small enough timestep and have damping in
the system, the energy should by all means not increase, or something
wrong is happening in the system. The Euler integrators (once the
timestep is smaller than the one suggested by the highest frequency of
the system) do increase energy, but only after multiple oscillations.
The moment you introduce a sufficient damper in your system, the energy
should strictly decrease. Only if the timestep is too large this isn't
true anymore, and in that case keeping energy in check may provide
stability but most likely not accuracy.

-Gregor (who is by now a Dr. - cue congratulations ;) )

Hi Gregor!

(audience warmup man raises his hand)
*clap clap* Hurrah! Congratulations!
Jeez, a Dr.
Dr. Veble, hehe. But you deserve it, you're brilliant.

Back to the topic at hand then. If I read you correctly, this really
means I can mostly check the energy of the entire car (4 springs, 1
sprung mass, 4 tire masses) rather than isolate things to 1 suspension
and do a local check there?

*virtual beer* <= have one on me. ;-)

Ruud
PS With J you don't mean the inverse inertia tensor? (in other words,
I'll need the normal inertia tensor after all?)

Hi Gregor

Do you feel that large weight lifting off your shoulders?  I certainly did
at the end of mine.

Congratulations
Stephen

p.s. don't worry, there another even heavier weight just around the
corner...

Todd Wasson
---
Performance Simulations
Drag Racing and Top Speed Prediction
Software
http://PerformanceSimulations.Com

thanks for the congrats! I may not be as brilliant as you say, but at
least I worked long enough on the Ph.D. to deserve it :).

Now, to the trick question you pose. In the large scheme of things,
energy gets converted all the time, so looking at the overall picture is
the correct way. However, since the frequencies are very much different,
looking at each individual wheel, perhaps even relative to the car (not
taking into account the overall translation due to car velocity), should
work very well, especially since it is the quick divergencies that you
are trying to take into account. Furthermore, the rotation of the wheel
is mostly uncoupled from the suspension movement, so if you just use the
1/2 mv^2 of the wheel and the spring 1/2kx^2 you should get a good
estimate of the energy stored in the suspension, and should clamp your
variables accordingly. Only if it doesn't work should you try to
consider other contributions (the KISS principle applies well here).

On the other hand, the rotational energy does contain the inertia tensor
and not its inverse; however, since you might not need it the point is
moot.

Greetings,

-Gregor

thanks for the congrats!

-Gregor