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To achieve a matrix that describes all of the three points P C in the base frame {R} one has to
i
R
multiply P C with the rotational matrix i
iR z
The result is the matrix PC
PC = PC
⋅ i
R
i
R
z
⎡cos(
α1)(
l
=
⎢
⎢
cos( α 2)(
l
⎢⎣
cos( α3
)( l
⎡cosαi − sinα
i
=
⎢
⎢
sinα
i cosα
⎢⎣
0 0
R
i R z
i
A
A
A
cos( θ ) + R)
1
cos( θ ) + R)
2
cos( θ ) + R)
3
− ( R + l
− ( R + l
− ( R + l
______________________________________________________________________________
Public Report ELAU GmbH, Marktheidenfeld
17
A
A
A
0⎤
0
⎥
⎥
1⎥⎦
cos( θ )) sin( α )
1
cos( θ )) sin( α )
2
cos( θ )) sin( α )
3
1
2
3
− l
− l
− l
A
A
A
sin( θ1)
⎤
sin( θ )
⎥
2 ⎥
sin( θ ) ⎥ 3 ⎦
Then can three spheres be created with the forearms lengths lB as radius, and their centre in P Ci
respectively.
2
2
2 2
The equation for a sphere is ( x − x0
) + ( y − y0
) + ( z − z0
) = r which gives the three
equations
⎧
⎪
⎨
⎪
⎩
2
2
2
( x − [ cos( α1)(
lA
cos( θ1)
+ R)
] ) + ( y − [ − ( R + lA
cos( θ1))
sin( α1)
] ) + ( z − [ − lA
sin( θ1)
] )
2
2
( x − [ cos( α 2)(
lA
cos( θ2
) + R)
] ) + ( y − [ − ( R + lA
cos( θ2
)) sin( α 2)
] ) + ( z − [ − lA
sin( θ2
) ] )
2
2
( x − [ cos( α )( l cos( θ ) + R)
] ) + ( y − [ − ( R + l cos( θ )) sin( α ) ] ) + ( z − [ − l sin( θ ) ] )
2
A
3
A
Eq. 4.1
With help from computer this equation system can be solved. There will be two solutions that
describe the two intersection points of the three spheres. Then the solution that is within the
robots working area must be chosen. With the base frame {R} in this case it will lead to the
solution with negative z coordinate.
4.4 INVERSE KINEMATICS
The inverse kinematics of a parallel manipulator determines the θi angle of each actuated
revolute joint given the (x,y,z) position of the travel plate in base-frame {R}, see Figure 4.5.
3
2
A
3
2
2
= l
= l
= l
2
B
2
B
2
B