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

J is the 6x6 Jacobian matrix that’s the partial derivations (5.7) of six outputs to six
input parameters. Solving equation (5.7), we can find
.
.
.
x , y,
z , wx,wy,wz. Thus the problem of
determination of the partial derivatives in (5.7) comes to kinematic analysis of mechanism
having 1-DOF. After substituting the values of output parameters (5.7) into (5.5) and
straightforward transformations, we obtain the equation of kinetic energy for the spatial 6-
DOF parallel manipulator with three angular and three linear actuators as:
where
2T
=
[ I w w + I w V + I V V ]
3
∑ k,
i
k,
i=
1
3
∑
k,
i=
1
k
i
k,
i+
3
k
i+
3
k+
3,
i+
3
o ⎡o
o
o
⎤
2T
= ⎢I
k,
i wk
wi
+ I k,
i+
3 wkVi
+ 3 + I k+
3,
i+
3 VkVi
+ 3 ⎥ (5.8)
⎣
⎦
I k,
i = [ m(
J1,
k J1,
i + J 2,
k J 2,
i + J 3,
k J 3,
i ) + I x J 4 , k J 4,
i + I y J5,
k J5,
i + I z J 6,
k J 6,
i + ζ k,
i ]ξ k,
i
o
I k,
i = [ m(
J1,
k J1,
i + J 2,
k J 2,
i + J 3,
k J 3,
i ) + I x J 4 , k J 4,
i + I y J 5,
k J 5,
i + I z J 6,
k J 6,
i + ζ k,
i ]ξ k,
i
ζ = 1 if (i = k or i < 4) else ζ , ξ = 2
k, i = I k , ξk
, i
o
o
k,
i = k , k,
i
k,
i = 0 k,
i
o o
k,
i = 0 k,
i
ζ I ξ = 1 if (i = k or i > 3) else ζ , ξ = 2
Differentiation of kinetic energy equations (5.8) wrt. six input parameters gives:
∂T
1
=
∂ϕ
2
∂
∂s
i
3
∑
k=
1
⎛ ∂I
k,
i ∂I
k,
i+
3
⎜
wk
wi
+ w
⎝ ∂ϕi
∂ϕi
⎛
k
V
i+
3
∂I
+
∂ϕ
k+
3,
i+
3
i
k
V
i+
3
k+
3
V
i+
3
+
+ +
∑ +
+ +
+ = +
+
+
⎟ ⎟⎟
o
o
o
o
3 T 1 ⎜ ∂ I k,
i ∂ I k,
i 3 ∂ I k 3,
i 3
= ⎜ wk
wi
+ wkVi
3 + Vk
3Vi
3
i 3 2 k 1 ∂si
3 ∂si
3
∂si
3
⎜
⎝
⎞
⎟
⎠
⎞
⎠
o
o
(5.9)
Note that the inertia factors in six equation (5.9) depends on the Jacobian matrix (5.7)
which, in one’s turn, depends on input angular and linear displacements. The partial
derivatives for inertia forces can easily be found from equations (5.8).
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