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Research Papers

# Inverse Geometrico-Static Problem of Underconstrained Cable-Driven Parallel Robots With Three Cables1

[+] Author and Article Information
Marco Carricato

Department of Industrial Engineering and Interdepartmental Center for Health Sciences and Technologies,
University of Bologna,
Viale Risorgimento 2,
Bologna40136, Italy
e-mail: marco.carricato@unibo.it

A preliminary version of this paper was presented at the 13th World Congress in Mechanism and Machine Science, Guanajuato, Mexico, 2011.

The notation $Mhij,klm$ denotes the block matrix obtained from rows h, i, and j, and columns k, l, and m, of M. When all columns are used, the corresponding subscripts are omitted.

It may happen that Eq. (3) is fulfilled because

$L1,L2,and L3$
become linearly dependent. In these situations, equilibrium is possible only if $rank(M)≤2$, since the external wrench must, however, belong to the screw subspace generated by the cable lines. Cases like these are very special and they must be studied separately.

It is worth emphasizing that having A1, A2, and A3 on a plane parallel to k, and G, B1, B2, and B3 coplanar, does not necessarily cause the 3–3 robot to behave like a planar mechanism: this is a possibility, to be considered together with the general spatial study of the manipulator. For example, when the DGP is considered, a number of feasible stable configurations may be found, for which the base and the platform are skew.

Equations (9a)(9c) coincide with those that would emerge by computing cable tensions by the first three relations in Eq. (2) and by substituting them back into the remaining ones.

A dialytic algorithm, similar to the one described in the subsequent Sec. 4, may be used as an alternative procedure with respect to the one presented in this section to solve the IGP with assigned orientation. In this way, a univariate resultant of degree 1 in x (or y) may be computed as the determinant of a $5×5$ eliminant matrix.

This is strictly related to the particularly simple form of the equations that govern this problem, especially when $Δ≠0$ (cf. Eq. (9)).

If p2 and p3 do not vanish in $(x∧,y∧),(x∧,y∧)$ is a solution of the problem only for $z→∞$.

Contributed by the Mechanisms and Robotics Committee of ASME for publication in the JOURNAL OF MECHANISMS AND ROBOTICS. Manuscript received March 24, 2012; final manuscript received April 17, 2013; published online June 10, 2013. Assoc. Editor: Vijay Kumar.

J. Mechanisms Robotics 5(3), 031002 (Jun 24, 2013) (11 pages) Paper No: JMR-12-1039; doi: 10.1115/1.4024291 History: Received March 24, 2012; Revised April 17, 2013

## Abstract

This paper studies underconstrained cable-driven parallel robots (CDPRs) with three cables. A major challenge in the study of these robots is the intrinsic coupling between kinematics and statics, which must be tackled simultaneously. Effective elimination procedures are presented which provide the complete solution sets of the inverse geometrico-static problems (IGPs) with assigned orientation or position. In the former case, the platform orientation is given, whereas the platform position and the cable lengths and tensions must be computed. In the latter case, the platform position is known, whereas the platform orientation and the cable lengths and tensions are to be calculated. The described problems are proven to admit at the most 1 and 24 real solutions, respectively.

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Topics: Robots , Cables , Polynomials

## Figures

Fig. 1

Model of a cable-driven parallel robot with three cables

Fig. 2

Conditions for positive-dimensional solution sets of the IGP with assigned orientation: (a) ri,xy∥(Ai-Ah) xy and rj,xy∥(Aj-Ah) xy; (b) the sides of the projections of the triangles A1A2A3 and B1B2B3 on the xy-plane are mutually parallel

Fig. 3

Examples of one-dimensional solution sets of the IGP with assigned orientation: (a) a2=[8;0;0],a3=[0;7;0],r1=[-0.5;-0.5;-1],r2=[1;0;-1],r3=[0;1;-1],[x,y]=[2,1.75] and (b) a2=[8;0;0],a3=[0;7;0],r1=[-0.5;-0.5;-1],r2=[1.5;-0.5;-1],r3=[-0.5;1.25;-1],[x,y]=[2,2]. In both cases, V is a vertical line and equilibrium is feasible along the entire path.

Fig. 4

Example of one-dimensional solution set of the IGP with assigned orientation: a2=[8;0;0],a3=[0;7;0],r1=[-2/3;-2/3;-3/2],r2=[14/15;-2/3;-1/2],r3=[-2/3;11/15;-1], and p23:=735x2+567xy-420y2-6062x+2534y+1960=0. V is a curve on a cylindrical surface and equilibrium configurations are feasible along the entire path.

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