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

one meaning of orientation for unit vector k e wrt. plane ( e i, e j ), we describe a new vector n as
n = ei
× e j → n = ei
e j S αij
= 1.
To determine the direction of vector e k ( α kn = π/2-α ki ) we
have:
e n = e n C α = C( π / 2 −α
) = Sα
k
k
e
k
i
kn
k
i
ki
ki
ki
ki
e e = e e C α = Cα
(2.26)
k
e
j = ek
e j kj
C α = 0
Re-writing equation (2.26) in coordinate form:
Lk(MiNj – NiMj) + Mk(NiLj – LiNj) + Nk(LiMj – MiLj) = Sαki
LkLi + MkMi + NkNi = Cαki (2.27)
LkLj + MkMj + NkNj = 0
Solving the system of linear equations (2.27) for Lk, Mk and Nk using Crammer’s
method, ∆=1 and the components of vector ek ( Lk
, M k , Nk
) are found as:
Lk = (MiNj – NiMj) Sαki + Li Cαki
Mk = (NiLj – LiNj) Sαki + Mi Cαki (2.28)
Nk = (LiMj – MiLj) Sαki + Ni Cαki
Using Kotelnikov - Shtudi transformations on (2.28) to find the equations of screw
~ ~ ~
E k ( Lk
, M k , Nk
) we change the form of equation (2.28) to:
~ ~ ~ ~ ~ ~
L = ( M N − M N ) S A + L C A
k
k
i
i
j
j
j
j
i
i
ki
ki
i
~ ~ ~ ~ ~ ~
M = ( N L − N L ) S A + M C A
(2.29)
~ ~ ~ ~ ~ ~
N = ( L M − L M ) S A + N C A
k
i
j
j
i
From equations (2.29), we can find the components of the moment of unit vector
o
e k ( Pk
, Qk
, Rk
) as:
Pk = (MiRj + NjQi – NiQj – MjRi – akiLi) Sαki + [(MiNj – NiMj)aki + Pi] Cαki
ki
i
i
ki
ki
ki
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