rec.autos.simulators

Does anyone know the equations for the forces affecting a vehicle? I'm
writing a program where a car drives around a track and I wanted to add
physics to it. Eventually maybe I'll make a game out of it.

--
Peter Vieth

IGZ Handle: Fitek
ICQ UIN: 3660410
Web page: http://www.racesimcentral.net/~Fitek/index.html

Well, it depends on how fancy you want to get.  Do you want true
four-wheel suspension?  Or are you willing to assume that the surface
is perfectly flat and fudge things like body roll?  Will each tire be
modeled separately?  The basic equation for the linear motion dynamics
of any rigid body is, of course, the classic F=MA (it has a
corresponding equation for angular motion).  This is generally
implemented as a set of differential equations:

dV = F / M
dP = V

where V is velocity and P is position, expressed as 3-tuples.  These
are solved using some sort of numerical integration technique, such as
Euler or Runga-Kutta.

So the problem then becomes one of figuring out the forces acting on
the vehicle.  Not counting collisions and the like, the forces
involved fall into one of two classes: vertical forces deriving from
the suspension and the force of gravity; and horizontal forces
deriving from the interaction of the tires with the road surface.  For
simplicities sake, most games have mostly ignored vertical forces up
until now, since it GREATLY complicates things.  Even dealing with the
forces generated by the tires is complex enough that Nascar 1 and 2
(supposedly) really only model ONE tire, although of course they do
keep track of separate tire temperatures.  The problem further breaks
down into longitudinal (acceleration/braking) and lateral (cornering)
forces.  Both usually involve modeling tire "slip" and the resulting
friction; for longitudinal forces a simple function which returns a
friction value of 0 for no slip, rising rapidly to the tires maximum
coeff. of friction as slip increases and then trailing off
asymptotically to constant value (representing a slipping tire's
coeff. of friction) works pretty well.  The force being generated is
then proportional to the force pushing DOWN on the tire (usually the
car's weight divided by the acceleration of gravity and divided again
by the number of tires you are modeling) multiplied by the calculated
coeff. of friction.  Lateral forces are more complicated; one must
take into account not only the amount of slip but the angle of slip as
well, and that's all I can really say about that.

I hope this has been helpful and that I didn't say too much that was
obvious or already understood.  I should say that it helps to have a
thorough grounding in linear algebra and, to some extent, calculus
(for the differential equations), although there is a lot of
public-domain software around for doing the numerical integration
stuff (I highly recommend "Numerical Recipes in C" if you don't
already have it).  Good luck!

Todd

But in reality not everything is so cut and dried. This model suits
hard surfaces which don't deform at all. The nature of *** means
that it's surface molds to fit the surface of the road (or anything
else) that it touches, on a molecular level, giving extra sideways
grip. It is this extra sideways grip which gives more traction with
more contact area. Of course, I'm just theorising here, and so it's
likely that this is not the right reason for *** gripping more:)

Grant.

Huh?
Please explain this physical impossibility......!

Both of you could be right. It depends on the exact parameters in the
problem:

Any surface that is in contact with another surface has a certain maximum
grip (Maximum friction-force). This maximum is some kind of function of
the pressure on the surface, lets say Grip per unit area = (Pressure)^a,
with a some exponential (this must be true, at least in first order
approximation).

When the surface decreases with an factor b (e.g. new surface =  (old
surface)/b, for b > 1), the presure increases with the same factor b

In that case the Grip/area      = (presure)^a
                                = (old presure * b)^a
                                = (old presure)^a * b^a
                                = (Old Grip/area) * b^a

The total grip becomes: (Grip/Area) * Area

                        = (Old Grip/Area) * Area * b^a
                        = (Old Grip/Area) * (Old Area)/b  * b^a
                        = (Old Grip) * b^(a-1)

Conclusion: if the grip/area is proportional to the pressure (thus: a=1),
then the grip wil be constant.

If the grip/area is sublineair with the pressure (thus; a<1) the grip will
increase when the area becomes less.

Personally I think the latter case is most likely.

Greetings, Han Monsees

----------------------------------------------------------------------------

        Unless that was already mentioned, and I'm a stooge, or my
newsreader sucks, or both :)

What we know:

1. Soft tyres has more grip than hard tyres.
2. Hard tyres last longer than soft tyres.
3. The friction force stays the same no matter what the contact size is.

Do I have to continue, or have you figured it out already :o)? Suppose I add:

4. The more pressure per square centimeter there is on a tyre, the more the tyre
wears.

This leads to the conclusions that size dont matter on grip, but a larger size
will give us a lower pressure per square centimeter, thus lower tyre wear, thus
allowing us to have softer tyre, which gives us more grip :o).

The grooves in the F1 tyres decrease the contact size, thus force the teams to
have harder tyres to get the same tyre wear as before.

/Christer, now I could be completely wrong, but I dont think so :o)

Of course, I didn't think this through enough in my post :o). With a smaller
width tyre you have a smaller tyre to absorb the same energy (the weight of the
car is the same), so it get's heated faster and probably has a higher running
temperature. You will probably need a harder tyre just to cope with the heat.

/Christer, I think I now understand why I want a tyre to have a contact area as
large as possible :o).

And to whoever says grooves affect heat dissipation is wrong also. I
believe they said it was because slicks have more surface area to
dissipate the heat. Think again on that one. Grooves add surface area. But
the tread squirm (the true technical term) caused by the grooves generates
heat, causing the tire to fail if pushed hard.

One last comment. It pains me to think of all the people who believe the
false answers they read on these newgroups and pass them on to other naive
people. Everyone would benefit if unsure people would check up one the
facts instead of just guessing. But then again maybe I'm the only one who
cares about correct answers. Maybe it's my engineering degree, my work in
a tribological lab, and my racing cars and motorcycles that lets me know
the right answer to these questions. Not just listening to others
guessing.

Jer

Todd

How much does one have to increase the width to
get a 10% increase in friction? (Double it? ;)

- Matt

-Mike

PS- Regarding N1 & N2, all 4 tires are modeled, not just one.  Think of
those cars as a brick with 4 wheels.(as opposed to a multiple rigid body
system as in GPL)

I had second thoughts as soon as I hit the "Post" button.  A little
too late, unfortunately.  I tried to qualify my statements as being my
"what I'd heard," not any sort of last word.  Thanks for setting me
straight!

Todd

Todd

Just to make this stuff even more complicated...

Since the tires are narrower, they'll also cool-off faster!!!

Best,
Edwin Solheim
Check out The Paddock -  a legendary site...
for some cool Grand Prix Legends stuff and tips!
(http://home.c2i.net/thepaddock)