4569846498

Modelling and analysis of suspension systems 219
We can now apply the triangle law of vector addition to equate the expression
for {A DH } 1 in equation (4.196) with {A D } 1 in equation (4.181) and
{A H } 1 in equation (4.193):
{A DH } 1 {A D } 1 {A H } 1 (4.197)
⎡128.
269⎤
⎡ 44 4z
514
⎢
1.
121
⎥ ⎢
144
⎢ ⎥ ⎢ 4z
51 4
⎣⎢
45.
535 ⎦⎥
⎢
⎣
144 y 44
⎤
⎥ 2
⎥
mm/s
⎦⎥
(4.198)
Rearranging (4.198) yields the next three equations required to solve the
analysis:
Equation 4 3346f 3 51 4y 44 4z 8 5y 228 5z 32.229 (4.199)
Equation 5 1840f 3 51 4x 144 4z 8 5x 1539.678 (4.200)
Equation 6 54 970f 3 44 4x 144 4y 228 5x 73.753 (4.201)
We can now proceed to set up the next set of three equations working from
point G to point P and using the triangle law of vector addition:
{A PG } 1 {A P } 1 {A G } 1 (4.202)
Determining an expression for the relative acceleration {A PG } 1 of point P
relative to point G using values for { 4 } 1 and {V PG } 1 found from the earlier
velocity analysis gives
{A PG } 1 { 4 } 1 {V PG } 1 { 4 } 1 {R PG } 1 (4.203)
⎡A
⎢
⎢
A
⎣⎢
A
PGx
PGy
PGz
y
x
4 4x
⎤ ⎡ 96.030
⎤ ⎡3346
f
⎥
⎢
⎥ 47 281.651
⎥
⎢
1840
⎢
⎥ ⎢
f
⎥
⎦ ⎣⎢
1576.804 ⎦⎥
⎣⎢
54 970 f
3
3
3
⎡9.
602
10 ⎤ ⎡228 z 8
⎢
⎥
45 743.
094
⎢
⎢
⎥ ⎢
85x
⎢
⎣
1605.
022 ⎥
⎦ ⎣⎢
2285x
3
⎤ ⎡ 0 1.446 10
0.
945
⎥ ⎢
⎥
⎢1. 446 10 0 8.
774 10
⎦⎥
⎢
3
⎣
0. 945 8.
774 10
0
3 3
3
⎤
⎥
⎥
⎦⎥
5 5y
⎤ ⎡166.468⎤
⎥ ⎢
⎥ 1.535
⎥
⎢ ⎥
⎥
⎦ ⎣⎢
7.150⎦⎥
⎡ 0
⎢
⎢
⎢
⎣
4z
4y
4z
4x
0
4y
4x
0
⎤ ⎡ 7
⎤
⎥ ⎢
⎥ 58
⎥
mm/s
⎢ ⎥
2
⎥
⎦ ⎣⎢
176⎦⎥
(4.204)
⎡A
⎢
⎢
A
⎣⎢
A
PGx
PGy
PGz
⎤ ⎡ 6.
755 ⎤ ⎡58 176
⎥ ⎢ ⎥ ⎢
⎥
0.303
⎢ ⎥ ⎢ 7 176
⎦⎥
⎣⎢
157.326⎦⎥
⎢
⎣
74y
584x
4z
4y
4z
4x
⎤
⎥
⎥ mm/s
⎥
⎦
(4.205)
We already have an expression for {A G } 1 in equation (4.177) and we can
define the vector {A P } 1 in terms of the known vertical acceleration component
A Pz and the unknown components A Px and A Py :
2