Tension Optimization for a Cable-driven Parallel

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The Open Automation and Control Systems Journal, 2015, 7, 1973-1980
1973
Open Access
Tension Optimization for a Cable-driven Parallel Robot with NonNegligible Cable Mass
Peng Liu1,2,* and Yuanying Qiu1
1
Key Laboratory of Ministry of Education for Electronic Equipment Structure Design, Xidian University, Xi’an, 710071,
P.R. China
2
School of Mechanical Engineering, Xi’an University of Science and Technology, Xi’an 710054, P.R. China
Abstract: This paper addresses the cable tension distributions of a cable-driven parallel robot with non-negligible cable
mass for large dimension mechanisms. A well-known model which describes the profile of a cable under the action of its
own weight allows us to take the cable sags into account. In addition, an approach to computing the cable tensions is presented, in which there are two major steps to obtain the cable tensions. In more detail, the first one is an iterative process
for obtaining the iterative cable tensions, while the other one is the optimization of the cable tensions obtained by the first
step. Finally, a large scale cable-driven manipulator currently under development is analyzed. The results show that cable
sags have a significant effect on the cable tensions of such manipulators.
Keywords: Cable-Driven Parallel Robot, Cable Sag, Tension Optimization.
1. INTRODUCTION
Cable-driven parallel robots, possessing several promising advantages over their rigid-link counterparts, are a special class of parallel manipulators in which the moving platform is driven by cables instead of rigid links [1-3]. Based
on the number of cables (m) and the number of degrees of
freedom of the platform (n), cable-driven parallel robots are
classified into three categories, i.e. the incompletely restrained positioning mechanisms (m<n+1), the completely
restrained positioning mechanisms (m=n+1) and the redundantly restrained positioning mechanisms (m>n+1) [2, 4]. In
most studies of cable-driven parallel robots, the cables are
considered to be straight lines, as long as cables are massless
and its tensions remain positive. Thereby, studies become
much easier and, in most cases, the model obtained is quite
close to reality [5, 6]. But, in some particular cases, more
complex cable behaviors must be taken into account to fit
well with the real manipulator. Another aspect that has to be
taken into account is sags of the long-span cables. Cables
with significant mass will tend to sag under their own weight
[1, 3, 7-13].
Determining the optimal cable tension distribution is essential for the efficient control and operation of the cabledriven parallel robots. A few researchers have worked on the
determination of the cable tensions for the cable-driven parallel robots. Generally, the cables of a parallel cable-driven
robot are considered to be massless. Fang et al. [14] developed an analytical method to optimize cable tension distributions in cable-driven parallel robots based on minimizing the
1874-4443/15
sum in of cable tensions at every pose. Their method, however, is not applicable to cable-driven parallel robots with
more than one redundant cable. Gosselin et al. [15] presented a non-iterative algorithm of cable tension determining and
proposed four optimal objective functions for the algorithm,
and the algorithm of cable tensions is just applicable to the
completely restrained cable-driven parallel robots. Liu et al.
[16] investigated optimal cable tension distributions for the
cable-based parallel robots, in which the minimum cable
tension distributions in the workspace are discussed. These
method and algorithm above, indeed, are rational for the
cable-driven parallel robots in which the cables can be modelled as the massless inextensible straight lines rather than
cable catenary model. However, this assumption, for the
cable-driven parallel robots with long-span cables, is not
rational owning to the sags caused by the non-negligible
self-mass. In addition, the effects of the cable sags on both
the kinematics and the cable tensions must be taken into account for large-span cables [8-11]. Riehl et al. [1, 3] proposed a method to determine both the kinematics and the
cable tensions of minimally constrained cable-driven manipulators. But the proposed method cannot be applied to other
cable-driven parallel robots in which the number of cables m
exceeds the number of degrees of freedom of the platform n.
Indeed, all of the above literatures, to the best of our
knowledge, do not deal with the cable tension distributions
for the redundantly restrained cable-driven parallel robots
with non-negligible cable mass.
This paper mainly focuses on the aspect of the cable sags.
Based on a well-known model of a cable with non-negligible
mass, described in section 2, the numerical computations of
the cable tensions for the redundantly restrained cable-driven
parallel robots are presented in section 3, in which an iterative optimization algorithm to solve the cable tensions, for a
given pose, is proposed. This paper studies the effects of the
2015 Bentham Open
1974
The Open Automation and Control Systems Journal, 2015, Volume 7
Liu and Qiu
zic
Ai
oic
xic
li
ci
di
Li
Bi
Hi
Vi
γi
Ti
Fig. (1). Catenary cable in its vertical plane.
A2
A1
B2
ρ2 d 2
·∙·∙·∙
Bi
o
X
ρmdm
P
BASE
B1
Y′
X′
Z
ai
O′
bi
ρi di
Ai
ρ1d1
Z′
PLATFORM
Bm
Am
Y
Fig. (2). Schematic of a cable-driven parallel robot.
cable sags on the cable tensions. And furthermore, the differences between the massless and the sags models regarding
the determination of the cable tensions are discussed in section 4. Finally, section 5 concludes this paper.
2. CABLE MODEL AND OPTIMAL CABLE TENSIONS
In order to make the cable-driven parallel robots move
stably and reliably, a uniform motion of the end-effector is a
requirement. Indeed, in many situations, an inextensible
catenary of non-negligible mass is likely to be an appropriate
modeling of the cables of the robots with long-span cables
[13]. As the result, the cables are modeled as inextensible
catenary, while the inertias of the cables and platform can be
ignored. It is well known that the profile of the cable
catenary and the cable tensions are coupled. Contrary to the
case of massless cable modeling, the cable tensions depend
on the external forces acting on the cables but also on their
own mass of the cables [1, 3]. Cable structures have also
complex mechanics behaviors because of their nonlinearity
in a geometric sense. Thus, complete modeling for the cabledriven parallel robots with the catenary cables must be done.
2.1. Cable Catenary Model
The first challenge for analyzing a cable-driven robot
with sagging cables is to mathematically describe the shape
of the cables under the influence of gravity. One end of the
homogeneous cable, in our model introduced here, is connected to the motor while the other end is connected to the
end-effector. As shown in Fig. (1), a local cable frame, noted
{oicxiczic}, is attached to Bi in Fig. (2), in which the zc axis of
the cable frame points in the same direction with the global z
axis. The non-elastic cables with significant mass will tend
to sag under their own weights. Therefore, the inextensible
catenary equation can be applied for describing the profile of
the cables under their sags [7, 9]. As a consequence, the profile of the cable i is the catenary within the ocxczc plane,
which can be expressed as follows [9]:
zic =
⎛ 2βi xic
⎞⎤
Hi ⎡
− αi ⎟⎥ , i = 1, 2,L , m
⎢cosh αi − cosh ⎜
ρg ⎣
l
⎝ i
⎠⎦
(1)
where xic and zic are x and z-direction coordinates in
{oicxiczic}; ρ is linear density of the inextensible cable;
g=9.8m/s2 is the gravitational acceleration with the direction
Tension Optimization for a Cable-driven Parallel Robot
along
the
negative
direction
$ # (c l ) '
" gli
! i = sinh "1 & i i i ) + # , ! i =
i
2H i
sinh
#
&%
i )
(
of
the
The Open Automation and Control Systems Journal, 2015, Volume 7
z
axis;
;Hi and Vi are the horizontal
and vertical components of each cable tension at the cable
end Bi; li and ci are the horizontal and vertical spans of each
cable respectively.
It must be noted from Eq. (1) that this kinematics model
with inextensible catenary cables differs from the traditional
kinematics with the straight line cables for the cable-driven
robots. As we can see from the number of the unknowns, we
must take into account both the cable tensions and the cable
lengths simultaneously because, in this case, we cannot separate the both aspects, the cable lengths depending directly on
the cable tensions according to the cable sagging model described in Eq. (1). Furthermore, the slope of catenary at the
last node Bi can be obtained from Eq. (1) as follows:
tan! i =
"z c i
2$ i x c i
=
#sinh(
# % i ), i = 1, 2, ! , m
li
"x c i
(2)
where γi is the angle between the tangent and the horizontal
plane at the last node of the catenary.
It is noted from Eq. (2) that the slope of a catenary is relevant to the parameter βi being a function of the horizontal
component Hi of each cable tension. Thus the profiles of the
cable catenaries, as mentioned before, are completely coupled to the forces applied to the cables, and therefore, these
variables (cable lengths and cable tensions) have to be determined at the same time.
2.2. Kinetostatic Model of a Cable-Driven Parallel Robot
In order to make a cable-driven parallel robot completely
controllable, it must be actuated redundantly due to the unilateral constraints nature of cables. The cable-driven parallel
robot, as shown in Fig. (2), consists mainly of a moving platform connected to the fixed pulleys by the cables and the
fixed base. The each pulley can be attached to the ceiling of
a mast. As a result, the moving platform moves freely in
every direction because the cables can be shortened and
lengthened controlled by servo motors mounted to the fixed
base.
When the cable sags are non-negligible, the cable tension
distributions of cable-driven parallel robot become quite
different from the ones with straight lines. The cable tension
distributions must be solved as a problem solving the statics
and the kinematics simultaneously rather than as a pure statics problem. In this section, the cable tension distributions
for a cable-driven parallel robot are posed, and a solution
method is presented. The cable tension distributions can be
regarded as an iterative optimization problem. With regard to
the completely restrained cable-driven parallel robots, the
equations for this type of system consist of a set of statics
equations and a set of kinematic-static equations. The position and orientation of the end-effector are specified, and
therefore, the unknown variables are the cable lengths and
the cable tensions. Thus, for a manipulator with m cables,
there will be a total of 2m unknowns which are cable length
1975
and cable tension for each cable. And furthermore, there are
m kinematic-static equations which come directly from Eq.
(1), while there are n equations for static equilibrium, and
they can be written as:
$m
&! Ri "Ti + mg+Fe = 0
& i=1
%m
& b # R T +M = 0
i
i i
e
&'!
i=1
( )
(3)
where Ri is the rotation matrix that maps the force vector Ti
from the cable frame to the fixed/manipulator frame; bi is the
position vector of the end point of the ith cable attached to
the platform, and expressed in the frame attached to the center of gravity of the platform; and g is the acceleration of
gravity.
Combining with the Eqs. (1) and (2), Eq. (3) can be written in the following matrix form:
JH=W
(4)
where J is the structure matrix associated the driving cable;
H=[H1, H2, ···, Hm]T is the vector describing horizontal components of all the cable tensions at their last node; W=fg+We,
fg=[0 0 mg 0 0 0]T∈R6 1 in which m is the mass of the moving platform, and We=[Fe Me]T∈R6 1 is the external wrench.
It should be noted that the structure matrix J is relevant
to the vector H through its dependency in tanγi being a function of the horizontal component Hi of each cable tension.
Thus, Eq. (4), formally, seems to be a linear equation but it is
not substantially, since the relationship between the coefficient matrix J and the vector H is not straightforward. Actually, Eq. (4) is a non-linear equation in terms of the vector H
describing horizontal components of all the cable tensions
because the structure matrix J is relevant to the vector H, in
general, which is solved using an iterative algorithm employing the cable tensions and lengths obtained by massless
straight line model as the initial iterative values here. And in
more detail, the iterative algorithm can be shown as follows:
Firstly, giving the end-effector platform position Xk in the
workspace (the kth platform position is denoted by the subscript k), the initial cable tensions, H(0), can be obtained with
the massless straight line model, while the slope of catenary
at the first node, tanγi(0), can be also obtained using Eq. (2).
Secondly, substituting tanγi(0) into the structure matrix J of
the equation (4), the new vector H(1) can be calculated using
the equation (4). Then, substituting the new vector H(1) into
Eq. (2), the associated slope of cable catenary, tanγi(1), can be
obtained. And moreover, substituting tanγi(1) into the equation (4), the new vector H(2) can be calculated using the
equation (4). Repeating the process until the vector H(j) satisfies the convergence condition ( H ( j ) − H ( j −1) ≤ α , α is the
iterative stop threshold). The vector H(j) is considered to be
the solution to the horizontal components of the cable tensions.
There may be a group of the vectors satisfying the above
convergence condition because the number of cables exceeds
1976
The Open Automation and Control Systems Journal, 2015, Volume 7
the degree of freedom of the platform for a completely restrained cable-driven parallel robot. Thus, an optimization
algorithm is required to obtain a unique solution to the vector
H which can confirm that the cable-driven parallel robot
operation as stable as possible.
3. OPTIMAL MODEL OF THE CABLE TENSION
DISTRIBUTION
3.1. Cable Tension Equation
Because the cable-driven parallel robots are redundant
systems having more actuators than the task dimensions,
infinitely many solutions for the distribution of cable tensions can be obtained for a given external load. One of the
issues in the operation of the cable-driven parallel robots is
resolving the actuation redundancy and determining the optimum cable forces distribution. Therefore, Eq. (4) from
which the vector H can be obtained may exist infinite solutions. In addition, the vector H(j) satisfying the convergence
condition is obtained with the iterative algorithm above, and
furthermore, the structure matrix J is constant values substituting the vector H(j) into the equation (4). And therefore,
with the introduction of Moore-Penrose generalized inverse
matrix (J(H(j)))+ of the structure matrix J(H(j)), the vector H
can be expressed as follows [2, 9]:
H=Hs+Hh
Liu and Qiu
subject to
(6)
( j)
(8)
1.
Input the end-effector platform position Xk in the
workspace and the cable density ρ, and obtain li, ci
andθi (θi is the angle between xc axis of ocxczc and X
axis of OXYZ; the ith cable is denoted by the subscript
i); k=1 (the kth end-effector platform position is denoted by the subscript k).
2.
Calculate the initial value tanγi(0) =ci/li , and obtain the
initial value J(0)=[J(0)1, J(0)2 ,···, J(0)m]T; and then, calculate the initial value of the vector H using Eq. (4),
H(0)=(J(0))+W; j=1 (the jth iterative step vector H is
denoted by the superscript j).
3.
Obtain the associated slope of cable catenary tanγi(j)
by substituting the vector H(j-1) into Eq. (2).
4.
Substitute tanγi(j) into the equation (4) and update the
new structure matrix J(j) , and then calculate the new
vector H(j) using the equation (4), H(j)=(J(j))+W.
5.
Judge
In order to ensure the normal work of the cable-driven
parallel robot, the tension vector T must satisfy the following
condition:
Ts,min≤T≤Ts,max
) H = Q(H )
As mentioned above, particular importance will be attached to the fact that Eq. (4) is a highly non-linear equation.
In order to solve the nonlinear equation, we present an iterative optimization algorithm to solve the vector H, as mentioned previously, which employs the cable tensions and
lengths obtained by massless straight line model as the initial
iterative values. Here, the iterative optimization algorithm
can be summarized as follows:
The cable tension of cable i at the cable node Bi is obtained from the following equation:
Ti = H i 1 + tan 2 γ i (i = 1, 2,L m )
( j)
where E(H)=(H1+H2+···+Hm)/m is the arithmetic mean value
of the vector H.
+
where Hs=(J(H )) W is the special solution to the vector H;
Hh =N(J(H(j))λ is the homogeneous solution to the vector H;
H(j) is the solution to the horizontal components of the cable
tensions using the iterative algorithm; λ is an arbitrary scalar.
(
J H
T s min ≤ T ≤ T s max
(5)
(j)
⎛1⎡m
2⎤⎞
min ⎜ ⎢ ∑ ( H i − E ( H ) ) ⎥ ⎟
m
i
=
1
⎦⎠
⎝ ⎣
Object
that
the
predetermined
condition
j
H ( j+1) ! H ( ) " # (α is the iterative stop threshold) is
satisfied or not. If it is, stop the iterative process and
record the vector H(j) satisfying the nonlinear equation
(4). If not, go to the 4th step, and repeat the process
from the steps (3)-(5), j=j+1.
(7)
where T=[T1, T2, ···, Tm] ∈ is the vector consisting of all cable tensions; the lower bound of the cable tension
Ts,min=[T1,min, T2,min, ···, Tm,min ]T is required to keep cables
taut; while upper bound of the cable tension Ts,max=[T1,max,
T2,max, ···, Tm,max]T is limited by the output torques of the servo
motors and the maximum tension the cable can withstand
without breaking.
6.
Substitute the vector H(j) satisfying the nonlinear
equation (4) into the right of Eq. (5), obtain
(J(H(j)))+W+N(J(H(j))λ, and then, calculate the optimized vector Ho,k (the optimized vector is denoted by
the subscript o) according to the optimal model Eq.
(8).
7.
Calculate the cable tension of the ith cable Ti substituting the optimal vector Ho,k into Eq. (6).
3.2. Optimal Model of the Cable Tension Distribution
8.
Judge the camera platform position Xk is whether the
last position. If it does not, go to the step (1) and
solve the cable tensions of the next position, k=k+1; If
it does, record and output the optimized vector of H
and the cable tension Ti of the ith cable, stop the calculation.
T
It can be shown from Eq. (5) that the homogeneous solution Hh is infinite due to the arbitrariness of λ. Therefore, in
order to obtain a unique solution, the minimum variance having the least differences among all cable tensions and the
arithmetic mean value of them is used to optimize while using Eq. (5) and (7) as the constraint conditions. Mathematically, the determination of the vector H can be formulated as
follows:
It is noted that, observing the iterative optimization algorithm above, the steps (1)-(5) are employed to obtain the
Tension Optimization for a Cable-driven Parallel Robot
solution of the cable tensions using an iterative method, in
which the steps (1)-(2) are used to calculate the initial values
for the iteration process. And moreover, the structure matrix
J in Eq. (4) is determined by substituting the solution of the
cable tensions using the iterative method into the left of Eq.
(3), and therefore, Eq. (4) can be considered as a linear equation. The vector H, in this case, can be optimized because
Eq. (4) is an indeterminate equation having infinitely many
solutions. While the steps (6)-(7) are used to obtain the optimized solution to the vector H at the present end-effector
platform position. From above, it can be seen that there are
two major steps to obtain the horizontal components of the
cable tensions here. The first one is the step being used to
solve the nonlinear equation (4) with an iterative algorithm
employing the cable tensions and lengths obtained by massless straight line model as the initial iterative values, and
therefore, the solution of the equation (4) can be obtained.
And the other one is the optimization of the cable tension
solutions obtained by the first step, and furthermore, the optimized solution of the cable tensions can be received. Generally speaking, the horizontal components of the cable tensions, H, can be obtained with the iterative algorithm in the
first step without the optimization of it. However, the optimization procedure with which the optimized vector H can
be obtained, in our manuscript, is required to confirm that
the cable-driven parallel robot operations as stable as possible. It should be pointed that the optimized vector Ho,k, in
general, can satisfy the predetermined condition
j
( H ! H ( ) " # , δ is a very small quantity), particularly in
o,k
instances where the cable-driven parallel robot locates at the
symmetric geometric position, and therefore satisfy the nonlinear equation (4).
In the context of the design of large workspace cabledriven parallel robots, besides the cable tension distributions,
one of the main aspects to take into account is the cable capabilities. Actually, the cables have to be able to support the
maximal cable tension all over the desired workspace. The
maximal tension is also involved in the proper determination
of the cable. The cables employed here can be chosen directly from the maximal tension. In fact, for most cables, the
maximal supported tension has to satisfy the equation (7).
When the cable model with non-negligible cable mass is
used, the determination of the appropriate cable can be done
by determining the maximal tension that can occur in the
cable over the whole workspace.
As far as I am concerned, when cable mass is taken into
account, the cable sags appears. Indeed, contrary to straight
cables, the cable tensions with non-negligible masses are not
constant along the profile. Kozak [7] has shown that the cable sags have effects on the positioning of the platform. But,
the cable tensions are also significantly affected by the mass
of the cables [1]. Actually, the actuators and the cables have
to support a part of the platform mass, but also the mass (or a
part of it) of the hanging cable. Thus, while taking cable
mass into account, the maximal tension Timax will be higher
than with the simple model of cable, because:
The Open Automation and Control Systems Journal, 2015, Volume 7
Timax = H i2 + (Vi − ρ gLi )
2
1977
(9)
It is noted that Vi , in the paper, is consider to be negative
shown in Fig. (1). The problem here is that the cable tensions
in the cables are directly dependent on the cable masses and
meanwhile have to satisfy the equation (7). Therefore, when
obtaining the cable tensions with the presented iterative optimization algorithm, the maximal tension Timax is required to
satisfy the equation (7), and furthermore, the obtained cable
tensions, in the case, are rational.
4. SIMULATION EXAMPLES
4.1. Description of the Cable-Driven Parallel Robot Studied
The cable-driven parallel robot considered in the study is
a point mass one, that is to say, a 4 cables / 3 DOF spatial
robot which is used to realize the aerial panoramic photographing [9]. In more detail, the 3 DOF are the three translations along the x, y and z axes. The cable-driven parallel robot considered here is called camera robot. The 4 cable exit
points are located at the 4 extremities of a rectangle. And the
density of the cable used here is 0.188Kg/m. The end effector mass has been chosen to be 50Kg. The boundary conditions of the cable tensions are presented as follow:
Ts,max=[10000 10000 10000 10000]T N, Ts,min=[10 10 10
10]T N. The dimensions of the structure have been chosen to
allow us to see the effect of cable mass deliberately. Thus,
the positions of the exit points of each cable, expressed in the
fixed frame, are given in Table 1. The dimensions of the
structure have been chosen to allow us to see the effect of
cable mass deliberately.
A spatial sloping straight line is selected to depict the
significant effects of the cable sags on the tensions in the
cables. It should be pointed that the starting position of the
trajectory is (25, 25, 5)T m; while the ending position of the
trajectory is (65, 53, 18)T m. It takes ten seconds from the
starting position to the ending position of the trajectory
above.
In order to better illustrate and understand the significant
effects of cable sags on the cable tensions, the relative differences denoted by ε between the cable tensions with the catenary model and the ones with the massless straight line model can be used to assess the divergences of the cable tensions,
and it can be expressed as follows:
ε=
T c i -T s i
×100% i = 1, 2,3, 4
T si
(10)
where Tci is the tension in the cable i at cable end Bi obtained
by the catenary model and Tsi the tension in the cable i obtained by the massless straight line model.
And furthermore, the relative differences of the cable
tensions denoted by ω between the first node and the last
node with the catenary model can be used to evaluate the
divergences of the cable tensions at the two nodes of the
cable catenary, and it can be expressed as follows:
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The Open Automation and Control Systems Journal, 2015, Volume 7
Liu and Qiu
Table 1. Exit point positions.
x (m)
y (m)
z (m)
Position of 1# pulley
0
0
23
Position of 2# pulley
100
0
23
Position of 3# pulley
100
90
23
Position of 4# pulley
0
90
23
2000
1800
Cable 3
Tension/N
1600
Cable 2
1400
1200
1000
800
Cable 1
600
Cable 4
400
200
0
1
2
3
4
5
t/s
6
7
8
9
10
(a) Tensions with the straight line model
5000
Cable 3
4500
4000
Cable 2
Tension/N
3500
3000
2500
Cable 1
2000
1500
Cable 4
1000
500
0
1
2
3
4
5
t/s
6
7
8
9
10
(b) Tensions with the catenary model
190
185
Cable 3
180
175
Cable 1
ε /%
170
165
160
155
Cable 4
145
0
B
A
150
Cable 2
1
2
3
4
5
t/s
6
7
8
9
10
(c) Relative differences ε
Fig. (3). Cable tensions along the sloping straight line with the two models and their Relative differences ε.
ω=
Timax - Ti
×100% i = 1, 2,3, 4
Ti
(11)
where Timax is the tension in the cable i at the last node (the 4
cable exit points) obtained by the catenary model and Ti the
tension in the cable i at the last node Bi obtained by the catenary model.
4.2. Results and Discuss
As shown in Fig. (3), the cable tensions of the four cables
along the spatial sloping straight line with the catenary and
straight line models are shown to demonstrate the significant
effects of cable sags on the cable tensions. As it is clearly
seen in this figure, the cable tensions computed by the two
models, indeed, are quite different. It should be pointed in
Tension Optimization for a Cable-driven Parallel Robot
The Open Automation and Control Systems Journal, 2015, Volume 7
1979
5500
5000
Cable 3
4500
Cable 2
Tension/N
4000
3500
3000
2500
Cable 1
2000
1500
Cable 4
1000
500
0
1
2
3
4
5
t/s
6
7
8
9
10
9
10
(a) Maximal tensions with the catenary model
30
Cable 4
25
Cable 2
ω /%
20
Cable 3
15
10
Cable 1
5
0
0
1
2
3
4
5
t/s
6
7
8
(b) Relative differences ω
Fig. (4). Cable tensions along the sloping straight line with the two models.
the Fig. (3a) and (3b) that the cable tensions obtained by the
catenary model are about 2 times larger than the ones by the
straight line model, and this is because the non-negligible
cable mass has an important effect on the cable tensions.
Particular emphasis was placed on the cable tensions at the
starting and the ending position, and it is perfectly clear that
the cable tension at the ending position is bigger than the one
at the starting position this is because the elevation at the
ending position is higher than the one at the starting position.
Indeed, increasing the elevation leads to an increase in the
angles between the tensions Ti and their vertical components
Vi, so the cable tensions must increase to withstand the
weights of the cables and the platform. Consequently, the
tensions of the four cables at the ending point are larger than
the ones at the starting point. In addition, the cable tensions,
without exception, tend to increase with increasing height
along z-direction using the two models.
Further enhancing the attraction to the relative difference,
it is clearly seen from Fig. (3c) that the maximal relative
difference is up to 186%. It can be seen that the relative differences of the cable tensions from the point A (x=48.96,
y=41.772, z=12.787) to point B (x=58.96, y=48.772,
z=16.037) are same in the four cables, and this is because
these platform positions locate at the geometric symmetrical
position of the workspace, and therefore, there is a same effect of the cable sags on the cable tensions. Note that presented Fig. (3c) the relative differences of cable tensions in
cable 1 and cable 3 at the ending position are less than the
one at the starting position, while are more in cable 2 and
cable 4, this is because the starting position of the sloping
straight line is away from the first and third towers, and
therefore, there is an obvious effect of the cable sags on the
cable tensions.
Referring in Fig. (4), the maximal cable tensions and the
relative differences ω along the spatial sloping straight line
with the catenary model are shown to demonstrate the difference of cable tensions along the profile of the cable catenary.
It can be seen that the maximal cable tensions along the profile of the cable catenary are more than the ones shown in
Fig. (3b). And furthermore, it is clear from Fig. (4b) that the
maximal relative difference ω is about 26%. Thus, taking the
maximal cable tensions of the cable catenary model into
consider is required to confirm that the cables do not exceed
the restriction of the cable tensions.
From the results it can be concluded that the tension solutions obtained from the proposed algorithm are continuous
and smooth. According to these results, cable tensions seem
to be the parameter that is mostly affected by the nonnegligible cable mass model. It can be noticed that the mean
difference between the tensions in the two models is about
155% more for the model including cable mass. Using this
model turns out to be essential to an appropriate design of
the robot, i.e. to the choice of the motors able to supply
enough torques, but also to the dimensioning of the structure
to prevent from deformations, or breakdown.
1980
The Open Automation and Control Systems Journal, 2015, Volume 7
Liu and Qiu
CONCLUSION
This paper addresses a main issue related to the cabledriven parallel robots: the computation of the optimal cable
tensions for a cable-driven parallel robot due to the sags in
its cables. The process for solving the problem begins from
the basic analysis of a single cable, which, in turn, is based
on the static displacement of that cable. A cable-driven parallel robot has infinite tension solutions for a particular pose
due to actuation redundancy. In this paper, a tension optimization algorithm, using an iterative optimization method, is
proposed to obtain the optimal cable tensions. This research
can readily be applied to other manipulators for which cable
sag is significant to better analyze those manipulators. Furthermore, this research can potentially be extended to address other important issues, such as kinematics problem to
account for cable sags and incorporating cable-sags predictions into the manipulator design process.
[4]
CONFLICT OF INTEREST
[10]
The authors confirm that this article content has no conflict of interest.
[5]
[6]
[7]
[8]
[9]
[11]
[12]
ACKNOWLEDGEMENTS
This work was financially supported by the National Science Foundation of China (51175397, 51105290).
[13]
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Revised: August 23, 2015
Accepted: September 31, 2015
© Liu and Qiu; Licensee Bentham Open.
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