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MOHAMMED AND LI: DYNAMIC NEURAL NETWORKS FOR KINEMATIC REDUNDANCY RESOLUTION 3
Fig. 1. Schematic of the Stewart platform.
Fig. 1, a fixed base, and six independent driving legs connect-
ing the aforementioned two parts. The two ends of each leg are
fixed on the mobile platform and the fixed based, respectively,
using universal joints. Each leg can be actuated to change
its length for the adjustment of the distance between the two
fixed points on the platform and the base. All together, the six
legs collaborates to adjust the orientation and position of the
mobile platform by changing their lengths.
A. Geometric Relation
For the Stewart platform, the global coordinate is fixed on
the base and the platform coordinate is fixed on the mobile
platform. a
i
∈ R
3
for i = 1, 2,...,6 represents the position
in global coordinates of the ith connection point on the base.
b
i
∈ R
3
for i = 1, 2,...,6 represents the position in platform
coordinates of the ith connection point on the platform. We use
b
i
to represents its position in the global coordinate, as shown
in Fig. 1. d
i
= b
i
− a
i
for i = 1, 2,...,6 represents the vector
corresponding to the ith leg, which points from the base to
the platform. For a point x
∈ R
3
in the platform coordinate,
its position x ∈ R
3
in global coordinate is obtained after a
rotational and translational transformation
x = p + Qx
(7)
where p = [x
p
, y
p
, z
p
]
T
∈ R
3
is the global coordinate of the
zero position in the platform coordinate, and it corresponds
to the translational transformation, Q ∈ R
3×3
is the rota-
tional matrix, which is uniquely defined by the Euler angles
θ = [θ
x
,θ
y
,θ
z
]
T
∈ R
3
Q = Q
z
Q
y
Q
x
Q
x
=
⎡
⎣
10 0
0 cos θ
x
sin θ
x
0 − sin θ
x
cos θ
x
⎤
⎦
Q
y
=
⎡
⎣
cos θ
y
0 − sin θ
y
01 0
sin θ
y
0 cos θ
y
⎤
⎦
Q
z
=
⎡
⎣
cos θ
z
sin θ
z
0
− sin θ
y
cos θ
y
0
001
⎤
⎦
. (8)
Following (7), as to the ith connection point on the platform,
i.e., the ones with x = b
i
in the global coordinates or the ones
with x
= b
i
in the platform coordinates, we have:
b
i
= p + Qb
i
. (9)
Therefore, the ith leg vector can be further expressed as
d
i
= b
i
− a
i
= p + Qb
i
− a
i
. (10)
For the vector d
i
, we define r
i
=d
i
to represent its length.
Accordingly, we have
r
i
=
p + Qb
i
− a
i
. (11)
Notice that both a
i
and b
i
are constants and are determined
by the geometric structure. p = [x
p
, y
p
, z
p
]
T
defines the trans-
lation of the platform, and Q as a function of the Euler angles
θ = [θ
x
,θ
y
,θ
z
]
T
, defines the rotation of the platform. Overall,
the right-hand side of (11) depends on the pose variables of
the platform π = [x
p
, y
p
, z
p
,θ
x
,θ
y
,θ
z
]
T
∈ R
6
while the left-
hand side of (11) is the length of the leg, which is controlled
for actuation. In this sense, (11)fori = 1, 2,...,6 defines the
kinematic relation between the actuation variables and the pose
variables. For a six-dimensional reference pose, the desired leg
length r
i
can be directly obtained from (11). However, in real
applications, the reference are usually not six-dimensional. For
example, for surgical applications of Stewart platform, peo-
ple may only care about the position of an end-effector on
the platform, instead of its orientation. In this situation, the
reference is 3-D and we have three additional DOF as redun-
dancy. For such a situation, we usually have infinite number
of feasible solutions of r
i
for i = 1, 2,...,6 to reach the
reference. Among the feasible solutions, we may be able to
identify one, which outperforms others in terms of certain opti-
mization criteria. This intuitive analysis motivate us to model
it as an optimization problem and identify the optimal one
for improved performance. However, due to the nonlinear-
ity of (11), direct treatment of (11) is technically prohibitive.
Instead of direct solution in position space, we turn to solve
the problem in its velocity space to exploit the approximate
linearity.
B. Velocity Space Resolution
For easy treatment, (11) is rewritten as
r
2
i
=
p + Qb
i
− a
i
T
p + Qb
i
− a
i
. (12)
To obtain the velocity space relations, we first compute time
derivative on both sides of (12), which yields
r
i
˙r
i
=
p + Qb
i
− a
i
T
˙p +
˙
Qb
i
+ Q
˙
b
i
−˙a
i
=
p + Qb
i
− a
i
T
˙p +
˙
Qb
i
. (13)
Recall that both a
i
and b
i
are constants and their time deriva-
tives, ˙a
i
and
˙
b
i
, equal to zero. For the rotational matrix Q,
according to the preliminary equations (1) and (6), it has the
following property for its time derivative:
˙
QQ
T
=
⎡
⎣
0 −
˙
θ
z
˙
θ
y
˙
θ
z
0 −
˙
θ
x
−
˙
θ
y
˙
θ
x
0
⎤
⎦
=
⎡
⎣
˙
θ
x
˙
θ
y
˙
θ
z
⎤
⎦
×
=
˙
θ
×
. (14)
Therefore,
˙
Q writes as follows:
˙
Q =
˙
θ
×
Q
T
−1
=
˙
θ
×
Q. (15)