Graphics Reference
In-Depth Information
0
v
1
+
q
3
v
2
−
q
2
v
3
=
0
−
q
3
v
1
+
0
v
2
+
q
1
v
3
=
0
q
2
v
1
−
q
1
v
2
+
0
v
3
=
0
.
Obviously, one possible solution is
v
1
=
0, but we seek a solution for
v
in terms of
q
1
,q
2
and
q
3
. A standard technique is to relax one of the
v
terms, such
as making
v
1
=
v
2
=
v
3
=
1. Then
q
3
v
2
−
q
2
v
3
=
0
(9.16)
−
q
3
+
q
1
v
3
=
0
(9.17)
q
2
−
q
1
v
2
=
0
.
(9.18)
From (
9.18
)wehave
q
2
q
1
v
2
=
.
From (
9.17
)wehave
q
3
q
1
v
3
=
therefore, a solution is
q
1
q
1
T
q
2
q
1
q
3
q
1
v
=
which in a non-homogeneous form is
=
q
1
q
3
T
v
q
2
or in terms of the original matrix:
=
(a
23
−
a
21
)
T
v
a
32
) a
31
−
a
13
) a
12
−
(9.19)
which appears to be a rather elegant solution for finding the fixed axis of revolution.
Now let's put (
9.19
) to the test by recomputing the axis of rotation for the pure
rotations
R
α,x
,
R
β,y
and
R
γ,z
where
α
=
β
=
γ
=
90°:
⎡
⎤
10
0
⎣
⎦
R
90
°
,x
=
00
−
1
01
0
using (
9.19
)wehave
=
(
0
)
=[−
T
−
−
−
−
]
v
1
1
)
0
0
)
0
200
which is the
x
-axis.
⎡
⎤
001
010
⎣
⎦
R
90
°
,y
=
−
100
