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we have the scenario shown in Fig. 6.14, where the cylinder has been subjected to the transforms
rather than the line. From Fig. 6.14 we can predict that the ray intersection points are given by
0 8 8 and 0 12 12. Let's now confirm these predictions by transforming the ray.
Y
12
8
v
12
10
8
Z
X
Figure 6.14.
Calculate
·
=
1
2 000
0 1 2 00
00 2 0
0001
1
2 000
00 2
1 000
0 010
0
0
S 1
R 1
·
=
1
2 00
0001
100
0 001
0
and
·
=
1
2 000
00 2
1
2 00 0
00 2
100 0
010 0
001
0
5
S 1
·
R 1
·
T 1
=
0
2 00
0001
1
10
000 1
0
2 00
0001
1
If the ray is given by
p
=
t
+
v
(6.26)
where
t
=
0 and v
=
j
+
k
then
·
=
1
2 00 0
00 2
0
0
0
1
0
5
5
0
1
t =
1
2 00
0001
0
 
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