Kinematic and Dynamic Analysis of Spatial Six Degree of Freedom ...

That’s, the variable angle α ki is in the following range:
α − α ≤ α < α + α
ij
jk
ki
This condition is met for all D 0 . Applying the same method to (2.23) solve for
angles αij or α jk , we can write the following for these angles:
ij
1 ≥
α − α ≤ α < α + α
ki
ki
α − α ≤ α < α + α
jk
jk
ij
Now that we derived and solved the equations of components of unit
o
vector e k ( Lk
, M k , Nk
) , we may find the components of the moment vector e k ( Pk
, Qk
, Rk
) .
Firstly we will transform the coordinates of the unit vector to dual coordinates of unit screw
using Kotelnikov – Shtudi transformations [38]. Using this principle, we can use the
equations of vector algebra as equations of screw algebra. The system of equations (2.22) is as
follows after transformation:
where
k
~
D
~
D
~
D
1
2
3
= D
= D
~
Lk
=
~
M k =
~
N =
k
~ ~ ~ ~ ~ ~ ~ ~ ~ −2
( L jD3
+ LiD
2 ± ( M iN
j − NiM
j ) D1
) S Aij
~ ~ ~ ~ ~ ~ ~ ~ ~ −2
( M jD3
+ M iD2
± ( NiL
j − Li
N j ) D1
) S A
~ ~ ~ ~ ~ ~ ~ ~ ~ −2
( N jD3
+ NiD2
± ( LiM
j − M iL
j ) D1
) S Aij
o
1
= D + wD
1
2
3
+ wD
+ wD
o
2
o
3
=
ij
ij
ki
2 2 2
( S A − C A − C A + 2C
A C A C A )
= C A
= C A
ij
ki
jk
− C A
− C A
jk
ki
ij
C A
C A
ij
ij
jk
jk
ki
jk
ki
ij
jk
1/
2
ij
Using the rules of screw algebra and after some arrangement, the general solution of
( L , M , N , P , Q , R )
E with real Plücker coordinates is found as follows.
k
k
k
k
k
k
(2.24)
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