Game Development Reference
In-Depth Information
(c)
0.966
−.259
0.000
0.000
(d) This corresponds to a pitch of +30
o
.
4. (a) 2
(b) 5
(c) 1
(d) 3
(e) 2
(f) 1
(g) 4
(h) 6
(i) 3
5. (a) 5
(b) 2
(c) 6
(d) 1
(e) 3
(f) 5
(g) 4
(h) 2
(i) 3
6. (w
1
+ x
1
i + y
1
j + z
1
k)(w
2
+ x
2
i + y
2
j + z
2
k)
= w
1
w
2
+ w
1
x
2
i + w
1
y
2
j + w
1
z
2
k
+ x
1
w
2
i + x
1
x
2
i
2
+ x
1
y
2
ij + x
1
z
2
ik
+ y
1
w
2
j + y
1
x
2
ji + y
1
y
2
j
2
+ y
1
z
2
jk
+ z
1
w
2
k + z
1
x
2
ki + z
1
y
2
kj + z
1
z
2
k
2
= w
1
w
2
+ w
1
x
2
i + w
1
y
2
j + w
1
z
2
k
+ x
1
w
2
i + x
1
x
2
(−1) + x
1
y
2
k + x
1
z
2
(−j)
+ y
1
w
2
j + y
1
x
2
(−k) + y
1
y
2
(−1) + y
1
z
2
i
+ z
1
w
2
k + z
1
x
2
j + z
1
y
2
(−i) + z
1
z
2
(−1)
= w
1
w
2
−x
1
x
2
−y
1
y
2
−z
1
z
2
+ (w
1
x
2
+ x
1
w
2
+ y
1
z
2
−z
1
y
2
)i
+ (w
1
y
2
+ y
1
w
2
+ z
1
x
2
−x
1
z
2
)j
+ (w
1
z
2
+ z
1
w
2
+ x
1
y
2
−y
1
x
2
)k
7. First, we extract the half-angle and axis of rotation:
α = θ/2 = arccos w = arccos 0.965 ≈ 15.0
o
,
n = normalize(
0.149 −0.149
0.149
) ≈
0.577 −0.577
0.577
.
Now we form a new quaternion using the new half-angle, α
′
= 2α ≈ 30.0
o
:
2
4
cos α
′
3
5
2
4
3
5
2
4
0.867
3
5
0.867
0
@
1
A
0
@
1
A
0
@
1
A
n
x
sin α
′
n
y
sin α
′
n
z
sin α
′
0.5770.500
−0.5770.500
0.5770.500
0.288
−0.288
0.288
=
=
.
2
4
0.233
3
5
2
4
−0.752
3
5
0
@
1
A
0
@
1
A
0.060
−0.257
−0.935
0.286
0.374
0.459
8.
(a) ab =
= (0.233)(−0.752)+(0.060)(0.286)+(−0.257)(0.374)
+ (−0.935)(0.459) = −0.683
2
4
0.333
3
5
0
@
1
A
0.253
−0.015
0.906
(b) ab =
2
4
−0.752
3
5
2
4
0.233
3
5
2
4
−0.683
3
5
∗
0
@
1
A
0
@
1
A
0
@
1
A
0.286
0.374
0.459
0.060
−0.257
−0.935
0.343
−0.401
−0.500
(c) d = ba
−1
=
=
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