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achieve ANSWERS and
First Examiner: Paul Aljabar
Second Examiner: Daniel Rueckert
In order, the transformation can be achieved in four steps:
Paper: C317  Graphics
Question: 1
Page 2 of 16
i) Translate the object to the origin
ii) Perform a shrink relative to the origin
iii) Perform a rotaion around the zaxis
iv) Translate the object back to its original centre
In fact the order of the shrink and the rotation, in this case, do not matter and
reversing them would give the same result.
Relative to the origin, the matrix for the scaling is the same as in part a:
0
1
0.9
0
00
B 0 0.9
0 0C
B
C
@0
0 0.9 0A
0
0
01 A rotation around 21
Graphics Lecture 11: Slidethe! zaxis by an angle q is given by
0
1
cos q
sin q 0 0
B
C 0 Transformations" 1 0.9
0
00
B 0 0.9
0 0C
B
C
@0
0 0.9 0A
0
0
01 A rotation around the zaxis by an angle q is given by
0
1
cos q
sin q 0 0
B sin q
cos q 0 0C
B
C
@
0
0 1 0A
0
001 So the complete sequence of transformations in matrix form is
0
10
10
1
1 0 0 10
0.9
0
00
0.98
0.17 0 0
B0 1 0 0C B 0 0.9
0 0C B0.17
0.98 0 0C
B
CB
CB
C
@0 0 1 10A @ 0
A@ 0
0 0.9 0
0 1 0A
000 1
0
0
01
0
001
which evalutes to Marks: 0 0.8820
B0.1530
B
@
0
0 0.1530
0
0.8820
0
0 0.9000
0
0 0 1
B0
B
@0
0 0
1
0
0 0
0
1
0 1
10
0C
C
10A
1 1 1.1800
1.5300C
C
1.0000A
1.0000 4 0 0.8820
B0.1530
B
@
0
0 Transformations"
Marks:
c 0.1530
0
0.8820
0
0 0.9000
0
0 1 1.1800
1.5300C
C
1.0000A
1.0000 For another sequence the object is to shrink, as deﬁned in part a, and to drop
vertically downwards by 1 unit each frame.
If the animation sequence is made up of a number of frames, numbered
consecutively from zero, what is the transformation matrix that should be
applied at frame n?
Frame n is obtained after n translations downwards by one unit. This means the
centre of the object is now at (10, n, 10). 4 vertically downwards by 1 unit each frame.
If the animation sequence is made up of a number of frames, numbered
Conﬁdential
consecutively from zero, what is the transformation matrix that should be
Tapplied at frame n? ANSWERS and MARKING SCHEME
ransformations"
MODEL Department of Computing Examinations – 2011  2012 Session
First Examiner: Paul Aljabar Second Examiner: Daniel Rueckert Frame n is obtained after n translations downwards by one unit. This Page 3 of 16
means the
Question: 1
centre of the object is now at (10, n, 10). Paper: C317  Graphics The matrix for the transformation can therefore be obtained by multiplying:
0
10
10
10
1
1 0 0 10
1 0 0 10
0.9
0
00
100
10
B0 1 0
0 0C B0 1 0
nC
1C B0 1 0
nC B 0 0.9
B
CB
CB
CB
C
@0 0 1 10A @0 0 1 10A @ 0
A @0 0 1
0 0.9 0
10A
000
1
000
1
0
0
01
000
1
or, equivalently,
0
10
B0 1
B
@0 0
00 0
0
1
0 10 10 100
10
0.9
0
00
n 1C B 0 0.9
0 0C B0 1 0
CB
CB
10A @ 0
0 0.9 0A @0 0 1
1
0
0
01
000 Working from the right:
0
100 10 10 0.9 0 0 9 1 1 10
nC
C
10A
1 0 1
B0
B
@0
0 0
1
0
0 0
0
1
0 10 10 1 100
10
0.9
0
00
n 1C B 0 0.9
0 0C B0 1 0
CB
CB
A@ 0
10
0 0.9 0A @0 0 1
1
0
0
01
000 Transformations"
Working from the right:
0
10
B0 1
B
@0 0
00 0
0
1
0 10 10
nC
C
10A
1 1 10
0.9
0
0
9
0 0.9nC
n 1C B 0 0.9
CB
C
A@ 0
10
0 0...
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 Spring '14

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