Game Development Reference
In-Depth Information
2
2
2
ma
I
ma
ma
2
1
1
3
4
4
2
2
11
ma
+
I
11
ma
I
0
=
0
.
(14.83)
12
12
2
2
2
1
ma
1
ma
2
ma
I
4
4
3
2
2
Factoring
11
12 ma
I
out of the second row and setting
b
=
1
4
ma
gives us
−− −
−− =
−− −
8
3
bI
b
b
(
)
bI
11 0 0
.
(14.84)
11
3
b
b
8
3
b
I
Evaluating the resulting determinant, we have
(
) (
)
2
(
)
2
0
=−
11
3
bI
8
bI
−− −−
bbI
8
2
b
3
3
) (
)
(
=−
bII
2
−+
13
bI
b
2
11
22
3
3
9
(
)(
)(
)
=−
11
bI
11
bI bI
2
.
(14.85)
3
3
3
The principal moments of inertia I , I , and I are thus given by
2
I
==
==
==
b
a
11
11
1
3
12
2
I
11
b a
11
2
3
12
I
b
a
2
.
(14.86)
2
1
3
3
6
To find the principal axis of inertia corresponding to the eigenvalue I , we need
to solve the homogeneous linear system
2
2
2
2
ma
I
1
ma
1
ma
 
x
3
3
4
4
 
2
2
2
1
ma
2
ma
I
1
ma
y
=
0 .
(14.87)
 
3
4
3
4
 
ma
2
ma
2
ma
2
I
z
1
1
2
 
3
4
4
3
2
Again using the constant
b
=
1
4
ma
and substituting the value
, we have
I
=
2
b
3
3
2
bbbx
bbby
bbbz
−−
 
 
2
=
0 .
(14.88)
 
−−
2
 
 
The reduced form of this system is