Graphics Reference
In-Depth Information
Fig. 11.12
The point
(
0
,
1
,
1
)
is rotated 90° about
the vector
v
to
(
1
,
0
,
1
)
0
.
5 should compute a mid-way position for an interpolated
quaternion, with its vector at 45° between the
x
- an
d
y
-axes, as shown in Fig.
11.12
.
We already know that
θ
Using (
11.7
) with
t
=
=
√
3
/
2, and using (
11.7
)
=
60°, therefore sin
θ
√
2
2
+
√
2
2
√
2
2
+
√
2
2
j
i
sin
2
60°
sin 60°
sin
2
60°
sin 60°
q
=
+
√
2
2
+
√
2
2
√
2
2
+
√
2
2
j
i
1
√
3
1
√
3
=
+
√
2
2
√
3
+
√
2
2
√
3
√
2
2
√
3
+
√
2
2
√
3
=
j
+
i
√
2
√
3
+
1
√
6
i
1
√
6
j
=
+
therefore,
√
2
√
3
,x
1
√
6
,y
1
√
6
,z
s
=
=
=
=
0
which when substituted in (
11.3
)gives
⎡
⎤
⎡
⎤
(
6
)
2
(
6
)
2
(
3
)
1
−
x
u
y
u
z
u
⎣
⎦
qpq
−
1
⎣
⎦
2
(
6
)
2
(
6
)
1
=
1
−
2
(
−
3
)
1
2
(
3
)
2
(
6
+
1
2
(
−
3
)
1
−
6
)
and
⎡
⎣
⎤
⎦
⎡
⎤
2
3
1
3
2
3
x
u
y
u
z
u
qpq
−
1
1
3
2
3
2
3
⎣
⎦
.
=
−
2
3
2
3
1
3
−
Substituting the coordinates
(
0
,
1
,
1
)
gives
⎡
⎣
⎡
⎣
⎤
⎦
⎤
⎦
=
⎡
⎣
⎤
⎦
2
3
1
3
2
3
1
0
1
0
1
1
1
3
2
3
2
3
−
(11.9)
2
3
2
3
1
3
−
which gives the point
(
1
,
0
,
1
)
.
