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Kinematics and dynamics of rigid bodies 67
⎡1
[ T]
T [ T]
⎢
0
⎢
⎣⎢
0
(2.204)
This allows us to further develop (2.203) to give
[I 2 ] 2/2 { 2 } 1/2 [T 3 ] 2 [I 2 ] 2/3 [T 3 ] T 2[T 3 ] 2 { 2 } 1/3 (2.205)
Since we also know that { 2 } 1/2 [T 3 ] 2 { 2 } 1/3 we can therefore write
[I 2 ] 2/2 [T 3 ] 2 [I 2 ] 2/3 [T 3 ] T 2 (2.206)
In this case we require a transformation matrix [T 3 ] 2 which is orthogonal
and will lead to a diagonal matrix [I 2 ] 2/2 . If we now take (2.206) and premultiply
by [T 3 ] T 2 we get
[T 3 ] T 2[I 2 ] 2/2 [I 2 ] 2/3 [T 3 ] T 2
⎡T T T
⎢
⎢
T T T
⎣⎢
T T T
11 21 31
12 22 32
13 23 33
0 0⎤
1 0
⎥
⎥
0 1⎦⎥
⎡T11I1 T21I2 T31I3
⎤
⎢
T I T I T I
⎥
⎢ 12 1 22 2 32 3⎥
⎣⎢
T13I1 T23I2 T33I3⎦⎥
⎡IxxT11 IxyT12 IxzT13 IxxT21 IxyT22 IxzT23 IxxT31 IxyT32 IxzT33⎤
⎢
⎥
⎢IxyT11 IyyT12
IxzT13 IxyT21 IyyT22 IyzT23 IxyT31 IyyT32 IyzT33⎥
⎢ I T I T I T I T I T I T I T I T I T ⎥
⎣ xz 11 yz 12 zz 13 xz 21 yz 22 zz 23 xz 31 yz 32 zz 33 ⎦
(2.207)
Equating the first columns from the matrices on either side of (2.207) leads
to the eigenvalue equation in (2.208):
⎡T11⎤
⎡Ixx Ixy Ixz
⎤ ⎡T11⎤
I
⎢
T
⎥ ⎢
⎥
1 ⎢ 12 ⎥
I I I
⎢
T
⎥
⎢ xy yy yz ⎥ ⎢ 12 ⎥
(2.208)
⎣⎢
T13⎦⎥
⎢I I I ⎥
⎣ xz yz zz ⎦ ⎣⎢
T13⎦⎥
This equation can be rearranged to give
⎡Ixx I1
Ixy Ixz
⎤ ⎡T11⎤
⎡0⎤
⎢
⎥
⎢ Ixy Iyy I1
I
⎢
yz ⎥ ⎢
T
⎥
12 ⎥
⎢
0
⎥
⎢ ⎥
(2.209)
⎢
⎣
Ixz Iyz Izz I ⎥
1⎦
⎣⎢
T13⎦⎥
⎣⎢
0⎦⎥
For a non-trivial solution to (2.209) we require the determinant of the square
matrix containing I 1 to be zero leading to the characteristic equation:
Ixx I1
Ixy Ixz
Ixy Iyy I1
Iyz
I I I I
⎤ ⎡I1
0 0⎤
⎡I I I
⎥ ⎢
0 I2
0
⎥ ⎢
⎥ ⎢ ⎥
⎢I I I
⎦⎥
⎣⎢
0 0 I3⎦⎥
⎢
⎣
I I I
0
xz yz zz 1
xx xy xz
xy yy yz
xz yz zz
⎤ ⎡T T T
⎥ ⎢
⎥ ⎢
T T T
⎥
⎦ ⎣⎢
T T T
11 21 31
12 22 32
13 23 33
⎤
⎥
⎥
⎦⎥
(2.210)