Graphics Reference
In-Depth Information
v
,” and use the previous
solution; that's an excellent choice, because the cross-product computation can be
performed once, and stored for later reuse. The point here is that computing a ray
intersection with the implicit form is computationally far easier than doing so with
the parametric form of the plane.
An alternative is to say, “Let's just compute
n
=
u
×
Once again, suppose we have a ray whose points are of the form
P
+
t
d
, with
t
0. We want to compute its intersection with a sphere given by its center,
Q
,
and its radius,
r
.
The implicit description of this sphere is that the point
X
is on the sphere if the
distance from
X
to
Q
is exactly
r
; this is equivalent to the
squared
distance (which
is almost always easier to work with) being
r
2
, that is,
≥
Q
)=
r
2
,
(
X
−
Q
)
·
(
X
−
(7.103)
where we've used the fact that the squared length of a vector
v
is just
v
v
.
Now let's ask, “Under what conditions on
t
is
P
+
t
d
on the sphere?” It must
satisfy
·
Q
)=
r
2
.
((
P
+
t
d
)
−
Q
)
·
((
P
+
t
d
)
−
(7.104)
Because the vector
P
Q
is going to come up a lot, we'll give it a name,
v
;now
we can simplify the expression above:
−
Q
)=
r
2
((
P
+
t
d
)
−
Q
)
·
((
P
+
t
d
)
−
(7.105)
Q
)+
t
d
)=
r
2
((
P
−
Q
)+
t
d
)
·
((
P
−
(7.106)
(
v
+
t
d
)=
r
2
(
v
+
t
d
)
·
(because
v
=
P
−
Q
)
(7.107)
t
d
)=
r
2
(
v
·
v
)+
2
(
t
d
·
v
)+(
t
d
·
(7.108)
r
2
)+
t
(
2
d
v
)+
t
2
(
d
(
v
·
v
−
·
·
d
)=
0.
(7.109)
This last equation is a quadratic in
t
, which therefore has zero, one, or two real
solutions.
Inline Exercise 7.13:
Describe geometrically the conditions under which this
equation will have zero, one, or two solutions, as in “If the ray doesn't intersect
the sphere, there will be no solutions; if ...”
Fortunately, it's easy to tell whether a quadratic
c
+
bt
+
at
2
=
0 has zero, one,
or two real solutions. The quadratic formula tells us that the solutions are
±
√
b
2
t
=
−
b
−
4
ac
,
(7.110)
2
a
which are real if
b
2
0; the two solutions are identical precisely if the
square root is zero, that is, if
b
2
=
4
ac
. In our case, this turns out to mean that
there are two solutions if
−
4
ac
≥
