(32)
0
-
0%
where +%
$ represents the terminal constraint(s) of the i th kinematic chain synthesized in
step 4.
8)'H#DG Synthesize the twist(s) of the i th kinematic chain with the twist basis of the ith chain,
4
$0 , obtained in step 5.
%
Suppose
4
) 4
4
4
1
2
%
$ &
$ # $
$
! $
(33)
%
0
' %0
%
0
%
0 $
(
%
where %
$ indicates the dimension of the twist basis of the i th ( % # ,
1 2 , ! , 5 ) kinematic chain.
According to linear algebra, any twist of the i th kinematic chain can be expressed as the
linear combinations of the twist basis of the chain:
4
'
4
$
# $ +
%
0
(34)
%
where + # ! -
-
! -
1
2
" "
$
.
%
Consequently, the twists of each kinematic chain can be synthesized through equation (34).
However, in order to keep the twists of the i th chain to be equivalent to the twist basis of the
chain, the rank of the total twists synthesized through equation (34) should equal the
dimension of the twist basis of the chain. .-*(#*(#6,11'?#)-'#63%()/C6)*3%#6/*)'/*3%#7#34#)-'#
4',(*K1'#L*%'0,)*6#6-,*%(.
The necessary and sufficient of this criterion can be immediately obtained from equation
(32).
According to the construction criteria 1 and 2, the required synthesis target of a mechanism
can be gradually accomplished with the above six steps. Obviously, with these six steps,
different person might synthesize different kinematic chains and different mechanisms.
However, all the end-effectors of the mechanisms synthesized with the same criteria will
surely have the identical specified free motion(s).
The next section will apply these steps to synthesize a rigid guidance mechanism that can be
utilized as a suspension of an automobile.
The synthesis target now is to use the least number of links and pure revolute joints to
design a mechanism whose end-effector has one pure translation along an exact straight
line; therefore, the mechanism must be a closed one. The reason is that it will need at least
two actuations to generate a pure straight line translation with an open chain mechanism.
And therefore, for the purpose of the suspension required, one at least needs two kinematic
chains to generate a pure straight line translation with one actuation input. According to
step 1, the specified free motion of the end-effector should be expressed in a Cartesian
Mo9ility of <patial Paral el Manipulators
485
coordinate system. Without loss of generality, the precise straight line translation of the end-
effector can be assumed to parallel 3 -axis. Therefore, the free motion can be described in
Pl cker coordinates as:
4
$#$) # !0 0 0 0 0 " "
1
(35)
So, the target now can be depicted as whether one can find two sets of screws whose pitches
represented by equation (3) are all zeros provided that they were all reciprocal to 4
$#$) of
equation (35).
According to step 2, substituting equation (35) into equation (29) yields the constraints
exerted to the end-effector, +
$#$) :
+
+
+
+
+
+
) 1
2
3
4
&
$
#
5
#$)
' $#$) $#$) $#$) $#$) $#$) $
(36)
(
%
+
where
1
$#$) # !1 0 0 0 0 0" "
represents
a
force
along
2 -axis,
+ 2
+
$
3
#$) # !0 1 0
0 0 0" " represents a force along 1 -axis, $#$) # !0 0 0 1 0 0" "
+
represents a torque about 2 -axis,
4
$#$) # !0 0 0 0 1 0" " represents a torque about 1 -
+
axis, and
5
$#$) # !0 0 0 0 0 " "
1 represents a torque about 3 -axis.
From equation (32), it is not difficult to find that the sum of the number of the independent
twists and the number of the terminal constraints of a chain is six. In order to reduce the
number of revolute joints, one might have to increase the number of the terminal constraints
of the chains as many as possible. According to equations (31) and (36), the maximum
number of the terminal constraints of a chain is five. However, if such a structure scheme is
used, one may find each kinematic chain only consists of one revolute joint, which is
unfeasible in reality. Similarly, it is not difficult to find that only when each kinematic chain
provides three terminal constraints at most, can the structure scheme is feasible.
With equation (31), one can synthesize the terminal constraints of these two kinematic
chains, individually. Selecting different (
and substituting them into
% ! % # 1 2
, , ! 5
, "
equation (31), one can synthesize three independent terminal constraints for the first
kinematic chain, for example:
)1 0 0&
'
$
'0 0 0$
Assuming
# '
1
E
0 0 0$ , one obtains
'
$
'0 1 0$
'
$
(0 0 1%
+
+
+
+
) 1
2
&
$ #
3
1
' 1
$
1
$
1
$ $
(37)
(
%
486
Parallel Manipulators, Towards New Applications
+
+
where
1
2
1
$
# !1 0 0 0 0 0" " indicates a force along 2 -axis, 1
$
# !0 0 0 0 1 0" "
+
indicates a torque about 1 -axis, and
3
1
$
# !0 0 0 0 0 " "
1 indicates a torque about 3 -
axis.
) '
0
0&
'
$
' .
0
0$
Assuming E # '
2
0
.
0$ , one can obtain
'
$
'0 2 ' 0$
'
$
(0
0
1%
+
+
+
+
) 1
2
&
$ #
3
2
' 2
$
2
$
2
$ $
(38)
(
%
+
where
1
1
$
# ! ' . 0 0 0 0" " denotes a force along the direction ! '
0" "
.
,
+ 2
1
$
# !0 0 0 . 2 ' 0" " denotes a torque about the direction ! . 2
0" "
'
,
+ 3
1
$
# !0 0 0 0 0 " "
1 denotes a torque about 3 -axis and '. 3 0 .
Because 4im
D
?:'$ 1
E , E 2E# 5 , the resultant terminal constraints of these 2 kinematic chains,
2
+ must be equivalent to +
$
. So the construction criterion 1 is satisfied.
%
G $
#$)
% 1
#
According to equation (32), one immediately obtains the twist bases for the two kinematic
chains with equations (37) and (38):
4
4
4
4
) 1
2
&
$
#
3
$
$
$
(39)
1
0
' 1
0
1
0
1
0 $
(
%
4
where
1
$0 # !1 0 0 0 0 0" "
represents
a
rotation
about
2 -axis,
1
4 2
4
$
3
0 # !0
0 0 0 1 0" " represents a translation along 1 -axis, $0 # !0 0 0 0 0 " "
1
1
1
represents a translation along 3 -axis, and
4
4
4
4
$ # $
$
$
(40)
0 2
B 1 2 3
0 2
0 2
0 2 C
4
where
1
$0 # !cos@ sin@ 0 0 0 0" " denotes a rotation about the direction
2
!
4
cos@ sin@
" "
0 ,
2
$0 # !0 0 0 2 sin@ cos@ 0" " denotes a translation along the
2
4
direction !2 sin@ cos@
" "
0 ,
3
$0 # !0 0 0 0 0 " "
1 denotes a translation along 3 -axis,
2
and
'
.
cos@ #
and sin@ #
.
2
2
2
2
' 6 .
' 6 .
According to step 6, one can synthesize the twists of the two kinematic chains with their
twist bases (39) and (40), individually. Considering the construction criterion 2, one can find
Mo9ility of <patial Paral el Manipulators
487
that the least number of twists in each kinematic chain is three. Therefore, the twist of the
first kinematic chain can be synthesized below with equation (34):
4'
4
4
4
$
# -
1
2
3
1
1 $0 6 - 2 $0 6 - 3 $0 # ! -
0 0 0 -
-
1
2
3 " "
(41)
1
1
1
Substituting equation (41) into equation (3) yields:
4'
> # 0
1
(42)
Equation (42) indicates that any twist having the form of equation (41) will naturally satisfy
the free motion requirements of the end-effector. The Cartesian coordinates of the joint, 1 ,
r ,
can be found from equations (7) and (9):
"
*
s s
' s
1
-
- .
0
3
2
,
r #
6
# '
1
2
//
2
s
-
- ,,
s
0
1
1 -
To make the twists of the chain be equivalent to the twist basis, there are at least three twists
indicated in the form of equation (41).
Suppose -
1
1 # and the three joints’ coordinates are
; r #
,
! ' 0 0" "
88: r #
0
! ' 10 3 " "
0
8 r #
+
! ' +
1
3+ " "
89
then, the twists of the first kinematic chain will be:
4
4
4
) ,
0
&
$
#
+
,0+
' 1
$
1
$
1
$
$
(43)
(
%
4
where
,
1
$
# !1 0 0 0 0 0" "
represents
a
rotation
about
2 -axis,
40 #
1
$
!1 0 0 0 30 2 1 " "
0
represents a rotation about a line passing through point
!
4
2
+
0
10 30 " and paralleling 2 -axis, and
#
1
$
!1 0 0 0 3+ 2 1 " "
+
represents a
rotation about a line passing through point ! 2+ 1+ 3+ " and paralleling 2 -axis.
According to equation (34), a twist of the second kinematic chain, denoted by 4'
2
$
, can be
expressed as:
4'
4
4
4
1
2
3
$0 # 1
F $0 6F2 $0 6F3 $0 # ! 1
F cos@
1
F sin@ 0 2F2sin@ F2cos@ F3" "
(44)
2
2
2
2
where F % denote real numbers and % #1 2
, 3
, .
Substituting equation (44) into equation (3) yields:
488
Parallel Manipulators, Towards New Applications
4'
>
0
2
#
(45)
Equation (45) indicates that any twist having the form of equation (44) will naturally satisfy
the free motion requirements of the end-effector.
The Cartesian coordinates of the joint, r , can be found from equations (7) and (9):
2
,
"
*
s s
. s
1F
F
F .
0