Turn back while you still can! ;)
Modelling
a car as simply as possible would be removing the third
dimension. You might still model the third dimension but might
ignore any torques that cause the car to pitch or roll. If you
don't allow pitch and roll it won't behave so very realistically,
but it's simple.
The trick is how tires
behave. They have variable friction, basically. They have
partially static friction and partially kinetic friction as they
roll along with their non-solid bodies, deforming with contact
with the road and partially sliding and partially sticking. Go
search up "Physics of Racing".
Simple-ish
summary:
You could look at a tire as a three-dimensional
spring. If it's deformed from its zero-force point it will exert
a force between the road and the wheel. The problem is that as it
rolls along this "relaxes" the deformation. This is
easier to visualize in the lateral direction. If the bottom of
the tire is displaced to one side, it's obviously pulling the
wheel to that side just like a normal spring. But even if we keep
the wheel from moving laterally, as the wheel rolls the tire will
relax, moving back to directly under the wheel. You can see that
this must be true in that the top of the tire not in contact with
the ground has zero lateral displacement and without anything
providing a sideways force and movement it will have no sideways
displacement when it gets to the ground. Longitudinal
displacement works the same. Thankfully, axial displacement
(toward the axis) is just air pressure and can be measured as a
spring just from the center of the wheel and that doesn't change
as the wheel rotates (although the tires do expand as they rotate
as seen in drag racing).
Maybe that's already
too complex even if it's just theory. :)
The
short summary is that the tire "spring" provides no
force without a displacement, and rolling along the ground
relaxes this displacement. To keep a constant force the tire must
have a difference in velocity with the ground to keep "pulling"
the spring away from the zero point as the rolling relaxing
it.
Aha, maybe you can relate this to skiing.
You have to turn the ski sideways to the velocity of the ground
to have any force. Similarly for tires. To get a lateral force
you have to turn the wheel sideways to the velocity of the
ground. For a longitudinal force you must speed them up
(acceleration) or slow them down (braking).
The
upshot is that mathematically near the zero-force point the force
from the tire behaves as the sine of its velocity relative to the
ground. It's nearly linear near zero and "flattens" off
to maximum grip. Past that it decays down to completely kinetic
friction (locking the wheels under breaking, car completely
sideways, smoking burnout). The lateral velocity component is
proportional to the angle at which the wheel is turned, and the
longitudinal the ratio between the rate the wheel is spinning at
and the linear velocity of the wheel.
Oh, and
that's really how you determine the tires coefficient of grip.
The force is then the "weight" on the tire multiplied
by that.
Now since the ground exerts a force
on the tire (and the tire on the ground, but the ground isn't
going to experience any noticable acceleration from that), and
the tire on the wheel (and wheel on the tire, but we consider the
tire more or less massless), and the wheel on the car, if you
don't want to model the physics of the rotating wheel then you
have to make some shortcuts.
The linear and
rotational velocities of the car determine the range of forces
the tires could possibly provide. At zero braking/engine force
(torque on the wheel) you'll have only lateral force as given by
the slip angle from the steering angle of the wheels. You can
apply more longitudinal force up until the maximum total force
from the tire is reached, after which the tire is slipping more
than it's gripping, and the coefficient of "grip"
decreases.
So in a practical example: let's
begin with a car travelling at a constant velocity and heading
(orientation) such that it's pointed in its direction of travel,
zero angular velocity, and wheel rotation rates such that the
linear velocity of the tire's contact patch matches groundspeed
(i.e. the wheels are neither rotating too fast nor too slow). If
you steer the front wheels to the right then you've introduced an
angle between the wheel and the ground velocity. This causes a
lateral force at the front end of the car which results in both
linear and rotational acceleration for the car. As the car
rotates this introduces a slip angle for the rear tires, and they
provide a lateral force which increases the cars linear
acceleration and decreases its angular acceleration (to zero once
you've reached a constant turn).
So as the car
travels in a constant circle the tires at it's four corners are
not pointed in the direction of their travel (think skiier's
skiis). In a gentle turn they'd all be pointed at a small angle
(say one degree), while near maximum grip they'd be at more like
ten degrees.
Well anyway, the four tires
provide separate forces which must be summed to give total torque
and force. There's a maximal force vector (direction and
magnitude) they can provide for any given steering angle. You
could keep track of the wheel's rotational velocity and
acceleration as given by the torque both the tire and the car
(engine/brakes), but to keep things more simple you might just
say that if matching the torque from the car with torque from the
tire would exceed the maximum grip of the tire then you just
increase the slip ratio (longitudinal difference in velocity with
the ground) of the tire at a rate proportional to the difference
in the two torques. Hmm...I guess that is just distilling out the
angular acceleration on the wheel.
Anyway, the
tangential-to-the-road forces from the four tires depend
primarily on how fast the wheels are rotating, their steering
angle, and the linear and angular velocity of the car, and the
normal force between the tire and road (which is a constant if we
neglect the third dimension).
"Physics of
Racing"
http://phors.locost7.info/contents.htm
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