4569846498

224 Multibody Systems Approach to Vehicle Dynamics
{V C6C7 } 1 . We can also introduce a scale factor A s to simplify the sliding
acceleration calculation giving
{A c C6C7} 1 2{ 6 } 1 {V C6C7 } 1 (4.238)
{A s C6C7} 1 A s {R CI } 1 (4.239)
Combining these components of acceleration gives {A C6C7 } 1 as
{A C6C7 } 1 2{ 6 } 1 {V C6C7 } 1 A s {R CI } 1 (4.240)
⎡A
⎢
⎢
A
⎣⎢
A
C6C7x
C6C7y
C6C7z
⎡A
⎢
⎢
A
⎣⎢
A
C6C7x
C6C7y
C6C7z
⎤ ⎡ 0 1.159 0.245 ⎤ ⎡ 13.682 ⎤ ⎡ 3 ⎤
⎥
2
⎥
2
⎢
1.159 0 0.904
⎥ ⎢
41.045
⎥
A
⎢
s 9
⎥
mm/s
⎢
⎥ ⎢ ⎥ ⎢ ⎥
⎦⎥
⎣⎢
0.245 0.904 0 ⎦⎥
⎣⎢
1988.378⎦⎥
⎣⎢
436⎦⎥
(4.241)
(4.242)
Applying the triangle law of vector addition yields
{A C6C7 } 1 {A C6 } 1 {A C7 } 1 (4.243)
⎡879.162⎤
⎢
3626.702
⎥
⎢ ⎥
⎣⎢
80.457
⎦⎥
⎤ ⎡879.162⎤
⎡ 3 ⎤
⎥
⎥
⎢
3626.702
⎥
A
⎢
s 9
⎥
mm/s
⎢ ⎥ ⎢ ⎥
⎦⎥
⎣⎢
80.457
⎦⎥
⎣⎢
436⎦⎥
⎡ 3 ⎤ ⎡420.212⎤
⎡9 436
⎢
9
⎥
⎢
3742.479
⎥ ⎢
3 436
⎢ ⎥ ⎢ ⎥ ⎢ z
⎣⎢
436⎦⎥
⎣⎢
463.361⎦⎥
⎢
⎣
3 y 9
6z 6y
A s
6 6x
⎡ 210.614
⎤
⎢
28 476.707
⎥
mm/s
⎢
⎥
2
⎣⎢
3455.690 ⎦⎥
6 6x
(4.244)
Rearranging (4.244) yields three equations that can be used to solve this
part of the analysis:
Equation 1 3A s 436 6y 9 6z 669.564 (4.245)
Equation 2 9A s 436 6x 3 6z 28 592.484 (4.246)
Equation 3 436A s 9 6x 3 6y 3838.594 (4.247)
This leaves us with four unknowns, 6x , 6y , 6z and A s , but only three
equations. We can use the same approach here as used in the preceding
velocity analysis. Since the spin degree of freedom of Body 6 about the
axis C–I has no bearing on the overall solution we can again use the vector
dot product to enforce perpendicularity of { 6 } 1 to {R CI } 1 . This will yield
the fourth equation as follows:
{ 6 } 1 • {R CI } 1 0 (4.248)
2
⎤
⎥
⎥
⎥
⎦
⎡ 3 ⎤
[ 6 6 6 ]
⎢
9
⎥
x y z 0 mm/s
⎢ ⎥
⎣⎢
436⎦⎥
(4.249)