Graphics Reference
In-Depth Information
-
1
-
1
.
SFRC RF
=
16
In homogeneous coordinates the matrices for F
-1
and F are
1
5
2
30
2
6
1
5
2
5
Ê
ˆ
Ê
ˆ
-
0
00
Á
Á
Á
Á
Á
Á
˜
˜
˜
˜
˜
˜
Á
Á
Á
Á
Á
Á
˜
˜
˜
˜
˜
˜
2
5
1
30
1
6
2
30
1
30
5
30
-
0
-
0
and
,
5
30
1
6
2
6
1
6
1
6
0
0
-
0
Ë
¯
Ë
¯
0
0
0
1
0
0
0
1
respectively. Since the homogeneous matrix for R
-1
C
1/
R is
6
100 0
010 0
001
1
6
000 0
Ê
ˆ
Á
Á
Á
Á
˜
˜
˜
˜
,
Ë
¯
a simple multiplication of matrices gives us the same homogeneous matrix for S as
in (3.31). This shows that our new answer to the problem agrees with the two previ-
ous answers.
If all one wants is the equations for S in Cartesian coordinates, then clearly the
first solution to this problem is the simplest. The second solution was given mainly
to emphasize the fact that
any
transformation can be used in solving such problems
as long as they preserve the relevant aspects of the problem, in this case lines and
intersections of lines. Affine maps clearly do that. However, sometimes, as in com-
puter graphics applications, one is really after the equations for the central projection
in frame F coordinates. This amounts to finding the equations for the transformation
R
-1
C
1/
RF
-1
above. In that case the approach taken in the third solution needs to be
followed and this is what one must fully understand since the first two approaches
are not relevant here.
6
3.5.1.2. Example.
Use frames to find the central projection C of the x-y plane onto
the line
L
defined by the equation x - y = 2 from the point
p
= (5,1).
Solution.
See Figure 3.17. This is the same problem as in Example 3.4.3.1. We want
a frame F = (
u
1
,
u
2
,
p
) so that
u
1
is a basis for
L
and
u
2
is normal to
L
and points from
p
“to”
L
. A natural choice is
1
2
1
2
()
(
)
u
=
11
,
and
u
=-
11
,
.
1
2
The direction of
u
1
does not matter here. Since the distance from
p
to
L
is
2
, if we
let T be the translation with translation vector -
2
, then
