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Modelling of cabling wrapping phenomenon towards an improved cable-driven mechanism Man Cheong Lei, Member, IEEE, Denny Oetomo, Senior Member, IEEE
Abstract—Traditional parallel driven cable robots relied on the fact that the cables connecting the fixed frame and the moving platform remain straight and taut under tension during operation. Most studies implicitly or explicitly modelled the cables as straight lines. As consequence, the contacts between the cables and the platform have been excluded from the admissible workspace. This practice also excludes some beneficial properties, such as higher fidelity in modelling, the ability to generate larger workspace and higher moments by allowing cable to wrap around rigid bodies. In this paper, cable wrapping phenomenon is modelled and studied. The study is limited to the modelling of a single cable wrapping around a rigid body, illustrated on a specific case of a cylindrical bar connected to the base through a ball joint. The objective of this paper is to investigate the kinematics and dynamics relationship between the cable and the resulting displacement of the cylindrical bar with the assumption of a taut wrapping. The path taken by the cable is shown to be a combined curve consisting of a helix and a straight line while it is in tension under the wrapping phenomenon. Index Terms—wrapping, cable, cable robot, shortest arclength, helix, cylindrical surface
P
I. I NTRODUCTION
ARALLEL actuated cable robots have been studied extensively in the recent years. The most common type of these robots consist only of a single rigid body connected to the stationary base via the parallel actuating cables. The early cable robots are the Texas developed by Lindemann and Tesar[1], the Robocrane from the NIST[2], the Charlotte[3] in the international space station and the FALCON [4]. Another sub-class of the parallel actuated cable robots is one where the end-effector is also connected to the stationary base via a rigid body kinematic chains in addition to the cables, such as proposed by Lau in [5]. Currently, much work is carried out on the determination and analysis of the wrenchclosure workspace [6], [7], [8], trajectory planning [9], [10] and redundancy resolution [11]. In any of these cases, the actuating cables were never allowed to collide with the rigid body (bodies) or with another cable. A cable is always assumed to go in a straight line from the point of connection on a particular rigid body to another. Wrapping of the cable on the rigid bodies of the manipulator was excluded from consideration. An algorithm was even constructed to exclude the collision between actuating cables to produce interference free workspace, as reported in [12]. Man Cheong Lei is with the Department of Mechanical Engineering, University of Mebourne, Parville, VIC 3010, Australia, e-mail:
[email protected]. Denny Oetomo is with the Department of Mechanical Engineering, University of Mebourne, Parville, VIC 3010, Australia, e-mail:
[email protected].
The exclusion of cable contact with another cable or with a rigid link of the manipulator is a very sound simplification of the problem, as it introduces modelling and computational complexity, as well as some foreseeable practical issues, for example, having moving cables in contact with each other causes rubbing between cables and potentially breaks the cables. However, allowing cable to wrap around the rigid links of the manipulator also introduces benefits, for instance: larger workspace and larger moments. Larger workspace is obtained because the workspace that used to be discarded due to cable contacting with the manipulator is now considered admissible. Larger moment is obtained as wrapping a cable around a rigid body (with non-negligible diameter or width) would produce moment that would not have existed if the actuating cables were not allowed to wrap around the rigid body of the manipulator. Furthermore, the modelling technique established here can potentially be used to model other systems, such as the human musculoskeletal systems, where muscles wrap over other muscles, joints, and bones. The technique can also be used to improve the accuracy of the model of current cable driven parallel robotics systems where an amount of wrapping occurs in the physical implementation but ignored in the modelling / computational exercise for simplification. Thus the purpose of this research is to provide the preliminary work for the study of a cable-driven mechanism with wrapped cables on rigid links by conducting. In the preliminary work reported in this paper, the modelling of a single cable wrapping over a single rigid body is reported, with a specific case of a cylindrical rigid body. After the modelling of a general case cable wrapping phenomenon, the specific case of a cable wrapped around a cylindrical bar articulated by a ball joint at one end was studied for ease of presentation, analysis, observation of resulting behaviours, and the construction of the simulation. In this study, it is still necessary that the cable is taut, not having any slack, hence producing the shortest possible length of cable, between the two ends of the cable. Just like in the case of the cables in a conventional parallel actuated cable robot, all cables being in tension is a necessity in order to actuate the manipulator. As far as the authors are concerned, this is the first time cable wrapping phenomenon is studied and modelled for the purpose of improving the design and operation of parallel cable driven robot. The rest of the paper is organised as follows. In Section 2, geometry modelling of a single cable is presented, in which Lagrangian equation is utilised as an effective method in obtaining the configuration of the cable path yielding the
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shortest cable length under the given conditions (ie. the cable being taut). This is then performed on the specific case of a cable wrapping around a cylindrical rigid body, resulting in a cable path that is a combination of a helix and a straight line. For convenience, the cable length is selected as one of the constraint parameter in determining the configuration of the cable (it helps eliminate the multiple possible solutions). Section 3 presents the detail of the variations in the cable configurations (variation in possible solutions) with respect to the orientation of the cylindrical bar. Section 4 presents the simulation of the single-cable-cylinder system as set out in Section 3. Finally Section 5 concludes the work in this paper and brings out some ideas for the future direction of research. II. G EOMETRY M ODELING OF A S INGLE C ABLE W RAPPED AROUND A R IGID B ODY In this paper, an assumption was made that the cable remains in tension (no slack) during the operation considered. If the cable is not wrapped on any rigid body, it forms a straight line from one end to the other. If it is wrapped around a rigid body, the portion of the cable that is wrapped will form a path following the contour of the rigid body. As there is no slack in the cable, the arc-length of the curve formed by the cable should be the shortest among all possible curves between the starting and the end point of the length of the cable. The problem of determining the path formed by the cable wrapped around a rigid body is therefore an optimisation problem to find the curve ↵o such that the arc-length of the curve L(↵o ) is the shortest for all L(↵).
First, the set of all twice continuously differentiable functions on the interval r1 r r2 is denoted by C 2r1 ,r2 in which all the elements satisfy the boundary conditions (2) and (3). Supposed that there exists ↵o (r) = [xo (r), yo (r), zo (r)]T 2 such that ↵o (r) minimizes L on C 2r1 ,r2 . If (r) = [ 1 (r), 2 (r), 3 (r)]T 2 C 2r1 ,r2 satisfying (r1 ) = (r2 ) = 0, then for k = [k1 , k2 , k3 ] 2 R1⇥3 , ↵(r) can be defined as C 2r1 ,r2
↵(r)
Define S(k) 2 S(k)
↵(r1 )
=
[x1 , y1 , z1 ]T , and
(2)
↵(r2 )
=
[x2 , y2 , z2 ]T
(3)
with the assumptions that both x(r), y(r) and z(r) are at least twice differentiable with respect to r on the interval r1 r r2 , i.e. x0 (r), y 0 (r) and z 0 (r) are continuous. The change of the arc-length of the curve is '
x2 +
y2 +
z2
r = ↵ (r)
(4)
r
The total arc-length of the curve is therefore L(x, y, z) =
ˆ
r2
↵0 (r) dr =
r1
ˆ
r2
r1
s✓
dx dr
◆2
+
✓
dy dr
◆2
+
✓
dz dr
◆2
dr (5)
The following is to derive the necessary condition for x(r), y(r), z(r) to give the extrema to the integral, i.e. the particular xo (r), yo (r), zo (r) such that L(xo , yo , zo ) L(x, y, z). That is L(x, y, z) =
ˆ
r2
r1
F (r, x(r), x0 (r), y(r), y 0 (r), z(r), z 0 (r)) dr
(6)
1 (r), yo (r)
+ k2
2 (r), zo (r)
+ k3
3 (r)]
T
(7)
L(↵o ) L(↵o + k ).
by
C 2r1 ,r2
L(↵o + k ) ˆ r2 F (r, ↵o (r) + k (r), ↵0o (r) + k 0 (r)) dr
(8)
r1
It can be seen that the minimum of S(k) happens at k = The gradient of S(k) at k = [0, 0, 0] is
[0, 0, 0].
rS(0) =
@L @L @L , , @k1 @k2 @k3
T
(9)
.
The use of chain rule and integration by part on equation (6) yields @L @k1
=
= *
1 (r1 )
r2
@ F (r, x(r), x0 (r), y(r), y 0 (r), z(r), z 0 (r)) dr @k 1 r1 ˆ r2 @F @x @F @r 0 + dt @x @k1 @r 0 @k1 r ˆ 1r2 @F @F 0 (r) dr 1 (r) + @x0 1 r1 @x ✓ ◆ ˆ r2 r2 @F @F d @F (r) + 1 1 (r) dr @x0 @x dr @x0 r1 r1 ˆ
=
=
@L @k1
that is subjected to the boundary conditions:
| ↵|
= =
(1)
0
[xo (r) + k1
=
The following sub-section aims to establish the expression of the shortest-length curve ↵o , which is shown to take the form of a Lagrangian equation. Define a curve in 3D space with parametric expression,
p
↵o (r) + k (r)
=
which is also in the set C 2r1 ,r2 . Since ↵o (r) minimizes L on C 2r1 ,r2 , it follows that
A. Lagrangian equation and the shortest arc-length curve
↵(r) = [x(r), y(r), z(r)]T 2 R3 ,
=
2 (r2 )
=
ˆ
=
ˆ
r2
r1 r2 r1
Recall that ˆ
r2
r1
@L @k1
k=0
r2 @F (r) @x0 1 r1
)
=0
and thus
=0
✓ ◆ d @F dr @x0 ✓ ◆ d @F dr @x0o
@F @x @F @xo
1 (r) dr 1 (r) dr
=0 :
@F (r, ↵o , ↵0o ) @xo
d dr
✓
@F (r, ↵o , ↵0o ) @x0o
◆
1 (r) dr
=0
Then for any continuously differentiable function satisfying 1 (r1 ) = 2 (r2 ) = 0, equation (10) gives @F (r, ↵o , ↵0o ) @xo
Similarly, taking @F
@L @k2
(r, ↵o , ↵0o )
d dr
and
✓
@F (r, ↵o , ↵0o ) @x0o
@L @k3
@yo
d dr
@F (r, ↵o , ↵0o ) @zo
d dr
✓ ✓
(10) 1 (r)
◆
= 0.
(11)
◆
= 0,
(12)
= 0.
(13)
yield,
@F (r, ↵o , ↵0o ) @yo0 @F (r, ↵o , ↵0o ) @zo0
◆
Equation (11)-(13) are the Euler-Lagrange Equations. In other words, the three Lagrangian equations corresponding to the
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directions in xo, yo and zo yield the necessary conditions to determine the configuration (the curve formed by the cable) of a single cable with the shortest length while it is tautly wrapping over a rigid body. In next subsection, the equations will be utilised to obtain the geometric model of the cable on a cylindrical surface.
3) u0o , zo0 6= 0: from equation (17) & (18), zo00 u0o z 0 u00 o dzo = c3 =) duo
↵o (r) = [a cos ✓o (r), a sin ✓o (r), zo (r)]T ,
(14)
where ✓o (r) and zo (r) are the variables to be determined. The arc-length function as defined in (6), F (r, ↵(r), ↵0 (r)) becomes F (r, ✓o (r), ✓o0 (r), zo (r), zo0 (r)) and can be expressed in the form of F =
q
( a✓o0 sin ✓o )2 + (a✓o0 cos ✓o )2 + (zo0 )2 =
Let uo = a✓o , then F =
q
q
(a✓o0 )2 + (zo0 )2 (15)
(u0o )2 + (zo0 )2
(16)
Applying the Lagrangian Equation to F with respect to uo and zo , ✓
@F @uo
d dr
@F @zo
d dr
✓
@F @u0o @F @zo0
◆ ◆
=0
=)
=0
=)
zo0 (zo00 u0o
zo0 u00 o) 0 2 0 2 (uo + zo )3/2 u0o (zo00 u0o zo0 u00 o) (u0o 2 + zo0 2 )3/2
=0
(17)
=0
(18)
Note that if u0o 2 + zo0 2 = 0, it implies that u0o = zo0 = 0 and thus F = 0, which means the length of the curve equals zero. In other words, the curve is degenerate and does not exist. For all the other cases, where the curve does exist, i.e. u0o 2 +zo0 2 6= 0, there are three solutions for equation (17) and (18): 1) zo0 = 0: it implies zo = c1 , where c1 is a constant. The arc-length is L=
ˆ
r2
r1
q
(a✓o0 )2 dr =
ˆ
r2
a
r1
d✓o dr = a [✓o (r2 ) dr
✓o (r1 )] .
The parametric equation of the curve is ↵o (r) = [a cos ✓o (r)
a sin ✓o (r)
c1 ]T .
(19) (20)
The curve is a circular arc with radius a. 2) u0o = 0: it implies uo = a✓o = c2 , where c2 is a constant. Thus L=
ˆ
r2
r1
q
zo0 2 dr =
ˆ
r2
r1
ˆ
r2
r1
In this sub-section, the ↵o (r) which minimizes the arclength of the cable when wrapped taut on a cylindrical surface is obtained using the Lagrangian equations derived above. Cylindrical surface is selected in this preliminary study as it is one of the most common shapes in the constructions of the linkages in robotics manipulators. Given the dimensions of the cylinder, i.e. the height h and the radius a, the curve can be expressed parametrically as follows
dzo dr = zo (r2 ) dr
zo (r1 ).
(21)
The parametric equation of the curve is ↵o (r) = [a cos c2 , a sin c2 , zo (r)]T .
(22)
The curve is a straight line parallel to the length of the cylinder.
=)
0 zo = ac3 ✓ + c4
(23)
where c3 and c4 are constants. The arc-length is L=
B. Shortest curve on a cylindrical surface
=
q
(a✓o0 )2 + (ac3 ✓o0 )2 dr =
q
a2 + (ac3 )2 [✓o (r2 )
✓o (r1 )] . (24)
Let v = ✓o (r), reparametrize ↵o (r) and thus the parametric equation of the curve is o (v)
= [a cos v, a sin v, ac3 v + c4 ]T ,
(25)
which is in a form of a general case helix. In summary, the shortest non-zero length of a curve on a cylindrical surface must be in the form of (1) a circular arc, (2) a straight line or (3) a general helix. To this end, the focus of the study goes to the wrapping effect of a cable on a cylinder by modelling the cable as a combination of a helix and a straight line. III. C ONFIGURATION OF A CABLE WRAPPED AROUND A CYLINDRICAL RIGID BODY
The objective of this section is to investigate the specific case of a cable wrapped around a cylindrical bar articulated at one end by a ball-joint as shown in Figure 1. One end of the cable, is fixed at Point A on the cylinder, while the other end is assumed to be attached to an arbitrary point P. A portion of the cable is assumed to be pre-wrapped on the cylinder at the start of the operation. Where the cable leaves the surface of the cylinder is denoted as Point B. Unlike A and P, the location of B changes with the displacement (in this specific case, angular displacement, as the cylinder is articulated by a ball joint) of the bar. The cable is assumed to be in tension, in other words, it cannot be pulled further if the cylinder is to remain at the same pose. The length of the cable from A to P is therefore the shortest possible in this circumstance. From the proof in Section II, the portion of cable wrapped on the cylinder, i.e. from A to B, is either a circular arc, a straight line, or a general helix, while the rest of the cable, from B to P, is a straight line. In practice, however, it can be seen that only the circular arc and the helix shapes can be achieved by a cable wrapped around a cylindrical body in the context of a parallel cable driven robot, as the occurrence of a straight line in the segment from Point A to P would require the cable to be secured to the cylinder at point P, in essence turning Point P into the start of the segment (Point A). As shown in Figure 1, three-dimensional coordinate frames are introduced, namely: Frame {Ob }, attached to the stationary frame; Frame {Oc }, attached to the cylindrical rigid body; Frame {O1 }, attached to the axis of length of the cylinder. For the sake of the future study of multi-cable mechanism, the coordinate frame with respect to cable 1 is labelled by number. In this study, the notation of a point is o P where the “o” on the top left corner represents the corresponding coordinate frame the position vector of the point is expressed in. Generally, the location of point P is known with respect to the stationary (base) coordinate frame, i.e. b P . Frame {Ob } is
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defined with its origin at the center of rotation of the ball joint (intersection of the three rotational axes of the ball joint) at the base of the cylinder. Frame {Oc } is defined with its origin coinciding with that of Frame {Ob }, with its z-axis pointing along the length of the cylinder. At zero angular displacement, Frame {Ob } and Frame {Oc } are aligned. Frame {O1 } is obtained by translating Frame {Oc } along the length of the cylinder and rotated along z-axis of Frame {Oc } by a constant angle ✓1 such that the point where the end of the cable is fixed on the cylinder lies on the x-axis of the coordinate frame. The z-axis of the helix coordinate frame remains collinear with the axis of revolution of the cylinder (and the z-axis of Frame {Oc }). See Figure 1b.
where ↵0 =
d dt ↵
and c=
p
(28)
a2 + b 2
Supposed that the helix starts at ⌧ = 0 and ends at ⌧ = r, i.e. and r2 = r. Then equation (27) becomes
r1 = 0
(29)
s = cr s r= c
=)
(30)
The helix can be reparametrized as a three dimensional function of s (i.e. the length of the helix) by substituting equation (30) into equation (26). The expression of the helix becomes h ⇣s⌘ ⇣s⌘ ⇣ s ⌘iT (s) = a cos , a sin , b . c c c
(31)
The equation above expresses the helical portion of the wrapped cable around the cylindrical body parameterised over the length of the cable. B. Determination of the coefficients Referring to points A, B, P as defined in Figure 1, it can be seen that as the cable combines a helical (AB) segment and a straight line (BP) segment, the continuity of the cable at Point B must hold. In other words, the tangent at point B of AB ! must be in the same direction as BP , ie, !
0 (s)
(a) Frame {Ob }, {Oc } & {O1 } (b) Offset angle between {Oc } & {O1 }
Figure 1: The coordinate frames of the single cable wrapping system
In Section II-A, it was demonstrated that given two points on a cylinder, the shortest length of a cable is in the form of a point (zero-length), a circular arc, a straight line or a general helix. Except for the straight line, all can be expressed parametrically as (26)
↵(r) = [a cos(r), a sin(r), br],
where a, b are constants. Note that r is the angular displacement of the helix. As shown in Table 1, using different values of b and r, equation (26) can be used to represent the three spatial curves accordingly. Indeed, a straight line can also be expressed in the form of equation (26) with infinite b. Table I: Cases for shortest-length curve on a cylinder P OINT 0 0
b r
C IRCULAR A RC 0 r1 r r 2
H ELIX non-zero constant rA r rB
The general helix is then parameterised using cable length as the parameter, defined as s. Recall that the arc-length of a helix can be obtained by s=
ˆ
r2
r1
↵0 (⌧ ) d⌧ =
ˆ
r2
r1
p
a2 + b2 d⌧ = c (r2
r1 )
(27)
B
! BP ! . BP
(32)
Furthermore, 0
(s) = ↵0 (r) )
A. Parametrization of the cable
=
! 0 (s)
0
dr ↵0 (r) ↵0 (r) = ds = 0 ds |↵ (r)| dr
(s) =
|↵0 (r)| = 1. |↵0 (r)|
(33)
Hence, equation (32) becomes 0
! (s)
B
=
! BP ! BP
(34)
Defining Frame {O1 } axes as {i, j, k}, then the coordinate of P on frame {O1 } is 1
⇣ ! ! O1 P = 1 R c c R b b Ob P
1
⌘ D c = 1 xp i + 1 y p j + 1 z p k
(35)
where R denotes the rotational matrix between the coordinate frames and D is the translational vector. The location of P with respect to the base frame is assumed to be known. b Rc =c RbT represents the orientation of the cylinder and is assumed to be known. c D1 = 1 DcT is the translation from {Oc } to {O1 } along c z, and c R1 = 1 RcT is the rotational matrix to obtain Frame {O1 } by rotating Frame {Oc } by the angle ✓1 about c z. Both c D1 and c R1 can be obtained as the location where the cable 1 is fixed on the cylinder is assumed to be known, as shown in Figure 1b. The expressions of (s) and 0 (s) in Frame {O1 } are 1
! (s)
=
1 0
! (s)
=
✓ ◆ b sk c c c ⌘ ⌘ ⇣ ⇣ a s a s b sin i + sin j+ k c c c c c
a cos
⇣s⌘
i + a sin
⇣s⌘
j+
(36) (37)
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Recalling that the cable starts at s = 0, the coordinate of Point A can be computed as 1
! ! O1 A = 1 (0) = ai.
(38)
entire cable length and the z-coordinate of the point P with respect to frame {O1 }. 4) Equating (46) and (47) yields: 1 b
Assuming that the cable leaves the cylinder at s = sB , 1
! O1 B
=
(39)
Differentiating (39) with respect to s at B and comparing to
(34),
=
⇣s ⌘ ⇣s ⌘ a a b B B sin i + sin j+ k c c c c c ⇣ s ⌘i h 1 nh1 B xp a cos i + 1 yp ! c BP h ⇣ s ⌘i o B 1 zp b k c
a sin
⇣ s ⌘i B
c
j+
=
⇣s ⌘ a B cos c c
=
b c
=
a cos ! BP
sB c
1y
p
a sin ! BP
sB c
1z
b ! BP
p
(41)
(42)
sB c
(43)
and c can be calculated from equations (41) to (43): 1) Performing (41) ⇥ cos scB (42) ⇥ sin scB yields: sB
⇣s ⌘ ⇣s ⌘ B B xp cos + 1 yp sin c c1 0
a
=
1
sB c
=
sin
1
B @q
a
2
( 1 xp ) + ( 1 y p ) 2
C A
(44) + 2n⇡
(45)
where = tan 1 1 xp 1 yp and n is the number of revolutions of the helix (number of revolution the cable is wrapped around the cylinder). To determine the unique solution to the inverse sine function in (45) which provides the angular displacement of the wrapped segment of the cable forming the helix, the continuity of the cable length i.e. sB 2 C0 and (30) were considered. Since P is always located outside q the cylinder, (1 xp )2 + (1 yp )2 is always larger than the radius of the cylinder a. The inverse sine in equation (45) will therefore give one solution in quadrant I and II, and the correct solution can be verified by substituting it into equation (41). 2) (42) ⇥ cos scB ! BP
3) From (43):
(41) ⇥ sin scB : ⇣s ⌘ c h1 B = yp cos a c ! BP
=
⇣c⌘ b
1
zp
1
xp sin
sB
⇣ s ⌘i B
c
B
c
+
⌘
=
1 a
=
1
zp
,
2
6 4sin
1
(46)
(47)
Recall that the cable comprises a helix (AB) and a ! straight line (BP) with lengths sB and BP , correspondingly. Thus, (47) gives a linear relationship between the
q
0
( 1 xp ) 2 + ( 1 y p ) 2
B @q
2n⇡ +
(40)
p
1x
⇣s
(49)
a2
5) Substituting equation (45) and (49) into (48) yields: b
By comparing the coefficients of i, j and k on both sides of equation (40), ⇣s ⌘ a B sin c c
cot
(48)
1z
where
! (sB ) ⇣s ⌘ ⇣s ⌘ ⇣s ⌘ B B B a cos i + a sin j+b k c c c 1
=
⇣s ⌘i 1 h sB B + cot + c c p
=
1 a
a (1 x
q
p)
2
+
(1 y
p)
2
1 C A
( 1 xp ) 2 + ( 1 yp ) 2
+
a2
(50)
6) Finally c and sB can be calculated from (28) and (45). The configuration of the cable is therefore obtained. In a practical case, the orientation of the cylinder is unknown. Hence sB and c only depend on the rotation matrix c Rb . All the cases shown in Table II would happen during the motion of the cylinder. It should be noted however that in a practical cable robot constructed out of a cylinder articulated by a ball joint, there would be more than 1 cable utilised for the manipulator to be fully restrained. The length of the other cables will provide the necessary constraints to determine the kinematics of the manipulator. In next section, simulations are presented to demonstrate the interaction between the cylinder and the cable. It is first assumed that the motion of the cylinder is known. The configuration of the cable would change following the motion of the cylinder. Table II: Configuration of the helix with respect to different sign of b and t Growing direction of helix upwards counterclockwise upwards clockwise downwards clockwise downwards counterclockwise
Sign of b + + -
Direction of t increases decrease decrease increase
IV. S IMULATION OF THE INTERACTION BETWEEN ONE CABLE AND A CYLINDRICAL BAR ARTICULATED BY A BALL JOINT Based on the information in the previous section, the configuration of the cable will change according to the orientation of the cylindrical bar. In order to observe the variation of the cable with respect to the moving cylinder, four simulations are made by considering different cases. The first three cases would have the cylindrical bar stay parallel to the zb -axis initially and the forth case would have it remain a non-zero constant angular displacements on both xb and yb -axis. Each case is distinguished as a different trajectory on the cylinder. Therefore the configuration of the cable is subjected to the change of the rotation matrix c Rb , i.e. the transformation from the base coordinate frame to the cylinder coordinate frame. In this section, the location of P is assumed to be fixed at (40, ! 20, 0) with the unit of cm, i.e. b Ob P = [40, 20, 0]T . The radii of
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the cylinder and the ball-joint are 2cm and 5cm respectively. That is, a = 2cm. One end of the cable is mounted at the height of h = 28cm measured along the zc -axis. The cable is pre-wrapped on the cylinder with two revolutions initially. The following provides the simulations of the cylindrical platform performing unidirectional rotation under different conditions. Figure 2-4 simulate the cylindrical rigid body performing uni-directional rotations at a constant speed about the xb , yb and zb -axis respectively. Sub-figure (a) in each figure shows the varying configuration of the cable while the cylinder is rotating. The black curve is the cable and the red star denotes where the cable is mounted on the cylinder, i.e. point A on Figure 1a. Meanwhile, the blue dashed line indicates the direction from P to A. The purpose of it is to illustrate that the cable cannot be connected directly between the two points in some circumstances. Besides, although practically such cylindrical bar with a ball-joint is unlikely to rotate 360 along xb or yb -axis, the simulation is still feasible as the cylinder can be assembled reversely. On the other hand, subfigure (b) plots the cable length with respect to the rotation angle. The solid line represents the entire cable length L. The dashdot line and dashed line represent the length of the helical segment (AB) and the length of the straight segment (BP ) correspondingly as the cylinder is rotated from 0 to 360 about the respective axis. Figure 5 simulates the motion of the cylinder rotating about the zb -axis with non-zero constant angular displacements on the other two directions. Here ✓x = ✓y = 30 were set for this simulation. By comparing Figure 4 and Figure 5, although both cases performed the rotation about the zb -axis, the cable length of the former changes linearly with respect to the rotation angle while the latter has a sinusoidal-like relationship. It is because the zb -axis is not colinear with the zc -axis. Hence 1 zp is no longer a constant as the case in Figure 4.
(a)
(b)
Figure 3: Rotation about yb -axis from 0 to 360 (a) simulation; (b) cable length vs. rotating angle ✓y
(a)
(b)
Figure 4: Rotation about zb -axis from 0 to 360 (a) simulation; (b) cable length vs. rotating angle ✓z
(a)
(b)
Figure 5: Rotation about zb -axis from 0 to 360 with non-zero angular displacement in xb and yb direction (a) simulation; (b) cable length vs. rotating angle ✓z V. C ONCLUSION AND F UTURE W ORK (a)
(b)
Figure 2: Rotation about xb -axis from 0 to 360 (a) simulation; (b) cable length vs. rotating angle ✓x
This paper investigates and models the wrapping phenomenon of a cable mechanism. As each cable of a cable robot has to be in tension during any motion, the length of it should be the shortest among all the possible configurations
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at that moment. Euler Lagrange equation was demonstrated an effective tool to obtain the configuration of the cable subjected to minimum length. For the platform with a cylindrical contour, the portion of a cable wrapped on it would be a standard helix. Thus the entire cable is a combination of a helix and a straight line. This study utilizes the arc-length of the helix as the parameter to express the configuration of the entire cable. The helical portion of the cable varies with respect to the orientation of the platform. Four simulations are given to illustrate the variance of the length of two segments and the entire cable with respect to different motion and initial orientation of the cylindrical bar. Since a cable robot with m degrees of freedom requires at least m + 1 cables to operate, essentially multi-cables are expected to wrap around the platform. Prior to dealing with multi-cables, the next step of future work is to extend the formulation on different contours, number of pre-wrapped revolution of the cable and direction of wrapping. As all the above would result in different configuration of the cables in a mechanism. It may lead to the study on optimizing the design of the mechanism giving some constraints and conditions. R EFERENCES [1] R. Lindemann and D. Tesar, “Construction and Demonstration of A 9-String 6-DOF Force Reflecting Joystick for Telerobotics,” NASA International Conference on Space Telerobotics, pp. 55–63, Jan. 1989. [2] J. Albus, R. Bostelman, and N. Dagalakis, “The NIST ROBOCRANE,” Journal of Robotic Systems, vol. 10, no. 5, pp. 709–724, Jul. 1993. [3] P. D. Campbell, P. L. Swaim, and C. J. Thompson, “Charlotte Robot Technology for Space and Terrestrial Applications,” SAE, International Conference on Environmental Systems, Jul. 1995. [4] S. Kawamura, W. Choe, S. Tanaka, and S. R. Pandian, “Development of an Ultrahigh Speed Robot FALCON using Wire Drive System,” Proceedings of the IEEE International Conference on Robotics and Automation, pp. 215–220, May 1995. [5] D. Lau, T. Hawke, L. Kempton, D. Oetomo, and S. Halgamuge, “Design and Analysis of 4-DOF Cable-Driven Parallel Mechanism,” Proceedings of the Australasian Conference on Robotics and Automation, Dec. 2010. [6] D. Lau, D. Oetomo, and S. K. Halgamuge, “Wrench-Closure Workspace Generation for Cable Driven Parallel Manipulators Using a Hybrid Analytical-Numerical Approach,” Journal of Mechanical Design, vol. 133, no. 7, pp. 071 004–1–10, 2011. [7] W. Bin Lim, G. Yang, S. H. Yeo, and S. K. Mustafa, “A generic force-closure analysis algorithm for cable-driven parallel manipulators,” MAMT, vol. 46, no. 9, pp. 1265–1275, Sep. 2011. [8] M. Gouttefarde, D. Daney, and J.-P. Merlet, “Interval-Analysis-Based Determination of the Wrench-Feasible Workspace of Parallel CableDriven Robots,” IEEE Transactions on Robotics, vol. 27, no. 1, pp. 1–13, Feb. 2011. [9] D. McColl and L. Notash, “Workspace formulation of planar wireactuated parallel manipulators,” Robotica, vol. 29, no. 04, pp. 607–617, Aug. 2010. [10] Alberto Trevisani, “Underconstrained planar cable-direct-driven robots: A trajectory planning method ensuring positive and bounded cable tensions,” Mechatronics, vol. 20, no. 1, pp. 113–127, Feb. 2010. [11] H. R. Fahham, M. Farid, and M. Khooran, “Time Optimal Trajectory Tracking of Redundant Planar Cable-Suspended Robots Considering Both Tension and Velocity Constraints,” Journal of Dynamic Systems, Measurement, and Control, vol. 133, no. 1, p. 011004, 2011. [12] J. Merlet and D. Daney, “Legs Interference Checking of Parallel Robots over A Given Workspace or Trajectory,” Proceedings of the IEEE International Conference on Robotics and Automation, pp. 757–762, May 2006.
