Graphics Reference
In-Depth Information
In that any line
L
parallel to
n
serves as a separating axis, a good choice is to have
L
go through the box center, giving
L
as
L
(
t
)
=
C
+
t
n
. The box center
C
projects
onto
L
at
t
0. Because the OBB is symmetrical about its center, the projection onto
L
results in a symmetric interval of projection
=
, centered at
C
, with
a halfwidth (or radius) of
r
. Testing if
B
intersects
P
now amounts to computing the
radius
r
of the projection interval and checking if the distance of the center point of
B
to
P
is less than
r
.
Because points of the OBB farthest away from its center are the vertices of the OBB,
the overall maximum projection radius onto any vector
n
will be realized by one of
the vertices. Thus, it is sufficient to consider only these when computing the radius
r
. The projection radii
r
i
of the eight vertices are given by
[
C
−
r
n
,
C
+
r
n
]
r
i
=
(
V
i
−
C
)
·
n
=
(
C
±
e
0
u
0
±
e
1
u
1
±
e
2
u
2
−
C
)
·
n
=
(
±
e
0
u
0
±
e
1
u
1
±
e
2
u
2
)
·
n
.
Due to the distributive properties of the dot product, this expression can be
written as
r
i
=±
(
e
0
u
0
·
n
)
±
(
e
1
u
1
·
n
)
±
(
e
2
u
2
·
n
).
The maximum positive radius
r
is obtained when all involved terms are positive,
corresponding to only positive steps along
n
, which is achieved by taking the absolute
value of the terms before adding them up:
r
= |
e
0
u
0
·
n
| + |
e
1
u
1
·
n
| + |
e
2
u
2
·
n
|
.
Because the extents are assumed positive,
r
can be written as
r
=
e
0
|
u
0
·
n
| +
e
1
|
u
1
·
n
| +
e
2
|
u
2
·
n
|
.
When the separating-axis vector
n
is
not
a unit vector,
r
instead becomes
)
r
=
(
e
0
|
u
0
·
n
| +
e
1
|
u
1
·
n
| +
e
2
|
u
2
·
n
|
n
.
The signed distance
s
of
C
from
P
is obtained by evaluating the plane equation
for
C
, giving
s
=
n
·
C
−
d
. Another way of obtaining
s
is to compute the distance
u
of
P
from
C
,
s
n
for some point
Q
on the plane. As all points on
P
project to a single point on
L
, it is sufficient to
work with the projection of
Q
onto
L
. The distance
u
can therefore be computed as
u
=−
u
. Recall that
P
is
n
·
X
=
d
, where
d
=
Q
·
=
−
·
=
·
−
·
=
−
·
(
Q
C
)
n
Q
n
C
n
d
C
n
, which up to a sign change is equivalent
