Page 97 - Dynamics of Mechanical Systems

0593_C04*_fm Page 78 Monday, May 6, 2002 2:06 PM

78 Dynamics of Mechanical Systems

n
3
n
2
N
3
n
1
R
B
FIGURE 4.2.1
N 2
A rigid body moving in a reference
frame. N 1

1 0 0   c β 0 s   c γ s − γ  0
β

S = ABC = 0 c α s − α   0 1 0   s c γ 0  
 

  γ



 
0 s α c  − s β 0 c  0 0  1 
 
β
α
(4.2.3)
 cc − c s s 
 β γ β γ β 
c c −
cs +

= ( αγ s s c ) ( α γ s s s ) −ssc 
α β
α β γ
α β γ
 ss − c s c ) ( s c + c s s ) 
α β 
 (  αγ α β γ α γ α β γ cc 
Example 4.2.1: Use of Transformation Matrices
Suppose in Eq. (4.2.3) that α, β, and γ have the values 30, 60, and 45 degrees, respectively.
Determine expressions relating the unit vector sets N and n .
i i
Solution: By substituting for α, β, and γ in Eq. (4.2.3), we obtain the transformation
matrix S as:
.
 0 354. −0 354. 0 866 
 
.
S = 0 918. 0 306 −0 250.  (4.2.4)

.
.
 −  0 176. 0 884 0 433  
From Eq. (4.2.1) we have:
N = S n and n = S N (4.2.5)
i ij j i ji j
These expressions in turn may be written in the matrix forms:
n
N = S and n = S T N (4.2.6)
where N and n are column arrays of the unit vectors N and n . Hence, by comparing Eqs.
i
i
(4.2.4) to (4.2.6), we obtain the desired relations:
0 866
0 354
N = . n − . n + . n
0 354
1 1 2 3
0 918
N = . n + . n − . n (4.2.7)
0 306
0 250
2 1 2 3
0 176
0 433
N =− . n + . n + . n
0 884
3 1 2 3