Game Development Reference
In-Depth Information
The center of the circle that the car is traveling on is located at the intersection of lines
drawn perpendicular to the front and rear right wheels. The radius of the circle can be found
from trigonometric relations. The distance from the centers of the front and back wheels,
l
, is
known as the
wheelbase
. The ratio of the wheelbase to the circle radius is equal to the sine of
the wheel angle,
d
. Rearranging this relation results in an equation for the circle radius.
l
r
=
(8.36)
c
sin
δ
Another important quantity to determine is the rate that a car will make the turn—that is,
the angular velocity of the car during its turn. If the wheels are rolling without friction, the
angular turn velocity,
w
t
, is equal to the translational velocity magnitude,
v
, of the car divided
by the turn radius.
v
r
ω
=
(8.37)
t
c
Using Equation (8.36), the angular turn velocity can be expressed in terms of the wheelbase
and wheel angle.
v
sin
l
δ
ω
=
(8.38)
t
Equations (8.36) and (8.38) provide all the information needed to model a low-speed turn.
The turn radius can be determined from the wheelbase and wheel angle. The car then travels
along this circle at an angular velocity determined from Equation (8.38). Let's look at an example
to see how it all fits together. Let's say the driver of a Boxster S car wants to perform a low-speed
90-degree turn at a translational velocity of 10
m/s
(36
km/hr
). To make this turn, the wheels are
turned at an angle of 10 degrees. The wheelbase of the Boxster S is 2.41
m
. What is the radius of
the turn, and how long will it take the car to make the turn?
The turn radius can be computed from Equation (8.36).
2.41
m
r
=
=
13.9
⎛
⎟
10
π
(8.39)
⎜
sin
⎟
⎜
⎝
⎟
⎠
180
The time required to make a 90-degree turn is equal to the number of radians in the turn,
p
/2, divided by the angular turn velocity.
π
π
l
π
* 2.41
s
t
==
=
=
2.2
⎛
⎟
2
ω
2 sin
v
δ
10
π
(8.40)
⎜
2 *10 *sin
t
⎟
⎜
⎝
⎟
⎠
180
High-Speed Turns
A general model for describing high-speed car turns is complicated by several factors. For one
thing, as the car goes around the curve, the centripetal force experienced by the car may cause
the tires to slide outward. In other words, the tires will have a velocity component normal to the
direction in which they are rotating. The normal force component can also generate a torque
