4569846498

Kinematics and dynamics of rigid bodies 39
Multiplying out these three equations leads to:
2
2 2
2 2 2 2
| RDA | ( Dx 2 DxAx Ax ) ( Dy 2 DyAy Ay
)
2 2
( Dz 2 DzAz Az
)
2 2 2
| RDB | ( Dx 2 DxBx Bx ) ( Dy 2 DyBy By
)
2 2
( Dz 2 DzBz Bz
)
(2.76)
(2.77)
2 2
| RDC | ( Dx 2 DxCx Cx ) ( Dy 2 DyCy Cy
)
2 2
( Dz 2 DzCz Cz
)
(2.78)
At this stage we need to introduce some numerical data to demonstrate how
a solution can be obtained. This will also demonstrate how cumbersome
the algebra will become and the need to utilize computer software to solve
these problems. As an example we can take the following co-ordinates for
points A, B and C:
{R A } 1 [103 350 142] mm
{R B } 1 [127 350 128] mm
{R C } 1 [15 500 540] mm
The lengths of the three rigid links can be taken as:
|R DA | 172.064 mm
|R DB | 183.527 mm
|R DC | 401.103 mm
Substituting these numerical values into the equations above we get
172.064 2 (Dx 103) 2 (Dy 350) 2 (Dz 142) 2
183.527 2 (Dx 127) 2 (Dy 350) 2 (Dz 128) 2
401.103 2 (Dx 15) 2 (Dy 500) 2 (Dz 540) 2
Multiplying out gives
2 2
29 606.02 ( Dx 206 Dx 10 609) ( Dy 700 Dy 122 500)
2
( Dz 284 Dz 20 164)
2 2
33 682.16 ( Dx 254 Dx 16 129) ( Dy 700 Dy 122 500)
2
( Dz 256 Dz 16 384)
2 2
160 883.617 ( Dx 30 Dx 225) ( Dy 1000 Dy 250 000)
2
( Dz 1080 Dz 291600)
Combining the terms in these three equations leads to
2 2
2 2 2
123 667 ( Dx 206 Dx) ( Dy 700 Dy)
2
( Dz 284 Dz)
(2.79)