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

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Let Ei ( Li
, M i,
Ni
) and Ek ( Lk
, M k , Nk
) be two unit screws positioned arbitrarily in
~ ~ ~
space. A third unit screw E j ( L j , M j , N j ) , perpendicular to both E i and E k , thus E j lies on
the line of shortest distance between E i and E k as shown in figure 2.5.
Figure 2.5 – Three recursive screws in space
In figure 2.5, the dual angles between E i , j E and Ek are given by 2 / π
α = + = A ij ij waij
( a = 0),
A jk = α jk + wa jk = π / 2 ( a jk = 0),
A ki = α ki + waki
where αki is the twist angle and
ij
e k
// e i
// n
ei
n
αki
αkn
a ki is the shortest distance between screw axis E i and E k .
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Now, the recurrent screw equations to find E k ( Lk
, M k , Nk
) from E i ( Li
, M i,
Ni
) and
~ ~ ~
E j ( L j,
M j , N j ) will be created. First of all we will describe the recurrent equations for unit
vector ek ( Lk
, M k , Nk
) using known vectors ei ( Li
, Mi
, Ni
) and e j ( L j,
M j,
N j ) . To have only
e j
aki
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