Graphics Reference
In-Depth Information
Figure 6.13.
Finding the intersection of a ray and a
circle.
Equation (6.12) is just a fancy way of writing the solution to this system of three equa-
tions. It is easy to check that
I
satisfies the equations. Note that
N
1
•(
N
2
¥
N
3
) is just
the determinant of that system.
The next two problems find the intersection of a ray with a circle. We shall use
the following notation:
X
will denote a
ray
from a point
p
in a direction
v
and
L
will
denote the
line
through
p
with direction vector
v
.
6.5.7 Problem.
To find the intersection
q
, if any, of the ray
X
and the circle
Y
with
center
a
and radius r.
Solution.
See Figure 6.13. Now,
q
can be written in the form
q
=
p
+ t
v
and so we
need to solve for t satisfying
pva
+-=
t
r,
or equivalently,
(
)
(
)
=
2
pa v pa v
-+
t
•
-+
t
r
.
(6.13)
Let A =
v
•
v
, B = (
p
-
a
)•
v
, and C = (
p
-
a
)•(
p
-
a
) - r
2
. Equation (6.13) can be re-
written in the form
2
At
++=.
2
Bt
C
0
(6.14)
By the quadratic formula, the roots of (6.14) are
2
-±
BB C
A
-
t
=
.
(6.15)
Note that A cannot be zero because
v
π
0
. That leaves three cases:
