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118 Mutibody Systems Approach to Vehicle Dynamics
In a similar manner for the second column of [U] we get:
L 11 U 12 A 12 therefore U 12 A 12 /A 11
L 21 U 12 L 22 0 therefore L 22 A 21 A 12 /A 11
L 31 U 12 L 32 0 therefore L 32 A 31 A 12 /A 11
L 41 U 12 L 42 0 therefore L 42 A 41 A 12 /A 11
Moving on to the third column of [U] gives:
L 11 U 13 0 therefore U 13 0
L 21 U 13 L 22 U 23 0 therefore U 23 0
L 31 U 13 L 32 U 23 L 33 A 33 therefore L 33 A 33
L 41 U 13 L 42 U 23 L 43 A 43 therefore L 43 A 43
Finishing with the multiplication of the rows in [L] into the fourth column
of [U] gives:
L 11 U 14 A 14
L 21 U 14 L 22 U 24 A 24
L 31 U 14 L 32 U 24 L 33 U 34 0
L 41 U 14 L 42 U 24 L 43 U 34 L 44 A 44
therefore
U 14 A 14 /A 11
U 24 (A 24 A 21 A 14 /A 11 )/(A 21 A 12 /A 11 )
U 34 ((A 31 A 14 /A 11 ) (A 31 A 12 /A 11 )(A 24 A 21 A 14 /A 11 )/(A 21 A 12 /A 11 ))/A 33
L 44 A 44 A 41 A 14 /A 11 (A 41 A 12 /A 11 )(A 24 A 21 A 14 /A 11 )/(A 21 A 12 /A 11 )
A 43 ((A 31 A 14 /A 11 ) (A 31 A 12 /A 11 )(A 24 A 21 A 14 /A 11 )/
(A 21 A 12 /A 11 ))/A 13
From the preceding manipulations we can see that for this example some of
the terms in the [L] and [U] matrices come to zero. Using this we can
update (3.67) to give
⎡L11
0 0 0 ⎤ ⎡1
⎢L
21 L22
0 0 ⎥ ⎢
0
⎢
⎥ ⎢
⎢L31 L32 L33
0 ⎥ ⎢0
⎢
⎥ ⎢
⎣L41 L42 L43 L44
⎦ ⎣0
U12 0 U14
⎤ ⎡A11 A12 0 A14
⎤
1 0 U
⎥ ⎢
24 A21 0 0 A ⎥
⎥
24
⎢
⎥
0 1 U34
⎥ ⎢A31 0 A33
0 ⎥
⎥ ⎢
⎥
0 0 1 ⎦ ⎣A41 0 A43 A44
⎦
(3.68)
Consideration of (3.68) reveals that some elements in the factors [L] and [U]
are non-zero where the corresponding elements in [A] are zero. Such elements,
for example L 22 , L 32 , and L 42 here, are referred to as ‘fills’. Having completed
the decomposition process of factorizing the [A] matrix into the [L] and [U]
matrices the next step, forward–backward substitution, can commence.
The first step is a forward substitution, utilizing the now known terms in
the [L] matrix to find the terms in [y]. For this example if we expand