# Forward Displacement Analysis of a General 6-3 Stewart Platform Using Conformal Geometric Algebra

Authors: Feng Wei; Shimin Wei; Ying Zhang; Qizheng Liao
Journal: Mathematical Problems in Engineering
 (2017)

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## Abstract

In this paper, a new algorithm for the forward displacement analysis of a general 6-3 Stewart platform (6-3SPS) based on conformal geometric algebra (CGA) is presented. First, a 6-3SPS structure is changed into an equivalent 2RPS-2SPS structure. Then, two kinematic constraint equations are established based on the geometric characteristics, one of which is built according to the point characteristic four-ball intersection in CGA. A 16th-degree univariate polynomial equation is derived from the aforementioned two equations by the Sylvester resultant elimination. Finally, a numerical example is given to verify the algorithm.


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## Body

## 1. Introduction

Stewart platforms [1] are six-degree-of-freedom parallel manipulators that generally consist of a base platform, a moving platform, and six legs connected to each other in parallel. They were first applied to animate flight simulator platforms. In these recent decades, Stewart mechanisms have attracted wide interest from researchers and engineers due to their advantages of simplicity, high stiffness, large load capacity, quick dynamic response, and excellent accuracy [2]. The 6-3 Stewart platform (6-3SPS) is a special form of Stewart platform. The basic geometric structure of the 6-3SPS consists of a moving platform connected to the adjacent links at three distinct points Bi, i=1,2,3, by spherical kinematic pairs, which is shown in Figure 1. 6-3SPSs can be divided into either platform mechanisms or general mechanisms according to whether the six spherical points on the base platform are restricted to lie in plane or not.Figure 1
Structure of a 6-3SPS.It is well known that the forward displacement analysis (FDA) of parallel mechanisms, which is used to find the position and orientation of the moving platform given the actuator displacements, is much more challenging than the inverse displacement problem. Generally, the FDA problem involves highly nonlinear algebraic equations. There are two ways to solve these equations: numerical and closed-form solutions. Although numerical iterative approaches can obtain the forward kinematics solutions, they are not suitable for the general Stewart mechanism because they are computationally burdensome and require good initial values. A closed-form solution provides more information about the geometry and kinematic behavior over a numerical solution and the input-output closed-form univariate polynomial equation has significant theoretical values as it is fundamental to many other kinematic problems. Hence obtaining a closed-form solution to the FDA problem is more desirable in most cases.Numerous studies have been conducted regarding closed-form solutions for the FDA of the 6-3SPS. Griffis and Duffy [3] obtained a closed-form solution to the platform mechanisms of the 6-3SPS in 1989, but their method is not suitable for the solution of the general mechanisms. In 1990, Innocenti and Parenti-Castelli [4] changed the 6-3SPS structure into the 3-3RPS structure based on the 6-3SPS model. The FDA problem can be reduced to the solution of a system of three second-order nonlinear equations. In 1991, Liang and Rong [5] derived the FDA solution to a Stewart triangle platform, 6-SPS parallel manipulator. In fact, the principle of Liang’s method is the same as that in [4]. Both Nanua et al. [6] in 1990 and Akçali and Mutlu [7] in 2006 obtained the 16th-order input-output equation. However, the two methods are complex. Song and Kwon [8] obtained a closed-form direct kinematic solution of the 6-3SPS using the tetrahedron approach in 2011. The closed-form solution in [8], however, is not polynomial form. Cheng et al. [9] obtained an eighth-order input-output equation based on the orthogonal complement method. Their method introduces errors due to rounding off the coefficient of the high-order term in the process of elimination.In the process of carrying out a parallel mechanism kinematic analysis, the various components (links, joints, platforms, etc.) of the proposed parallel mechanism can be defined as three-dimensional Euclidean geometric objects such as points, lines, circles, planes, and spheres. However, it is difficult to represent and compute geometric constructions such as the intersection, extension, and contact of geometric objects. A solution for this problem is provided by conformal geometric algebra (CGA), which can represent the geometries of Euclidean space effectively by the outer product, inner product, and geometric product. At least three kinematic constraint equations, established in the previously mentioned literature, are solved based on two eliminations. This paper proposes a new process for obtaining the closed-form solution of the FDA of the 6-3SPS using CGA. Only two kinematic constraint equations are established based on the geometric characteristics and solved based on single elimination; as a result, the solution procedure is simpler and more efficient than that in the aforementioned works.The paper is arranged as follows. In Section2, we introduce the principle of CGA. In Section 3, the kinematic constraint equations are established using the CGA. In Section 4, we propose the elimination procedure. The numerical example to validate our new algorithm is given in Section 5. Conclusions are drawn in Section 6.


## 2. Conformal Geometric Algebra

CGA [10–12] is a mathematical language that integrates various mathematical theories, such as projective geometry and quaternion, and Lie Algebra. Therefore, CGA has been spotlighted as a new method in robotics [13, 14].


### 2.1. Fundamental Theory

The fundamental algebraic operators [10] in geometric algebra are the inner product (ac·bc), the outer product (ac∧bc), and the geometric product (acbc=ac∧bc+ac·bc). CGA, which is denoted by the base space R3+1,1, has a set of orthogonal bases given by e1,e2,e3,e+, and e- with the properties(1)ei2=1,i=1,2,3,e+2=1,e-2=-1,ei·e+=ei·e-=e+·e-=0.A set of null bases{e∞,e0} orthogonal to ei can be introduced by {e+,e-}:(2)e0=12e--e+,e∞=e++e-,with the properties(3)e02=e∞2=0,e∞·e0=-1.


### 2.2. Conformal Geometric Entities

Letx∈R3 be a point expressed in the Euclidean space R3. The representation of the same point in R3+1,1 is given by(4)xc=x+12x2e∞+e0,where x=x1e1+x2e2+x3e3,x2=x12+x22+x32.InR3+1,1, the equation of a sphere of radius r centered at p∈R3 can be written as(5)Sc=p+p2-r22e∞+e0=pc-12r2e∞.


### 2.3. Conformal Transformations

Rotation, which is called a rotor, can be written as a unit quaternion as follows:(6)R=cos⁡ϕ2-nsin⁡ϕ2,R∗=cos⁡ϕ2-nsin⁡ϕ2,where n is a normalized bivector [10] representing the axis of rotation, ϕ is the rotation angle, and the superscript ∗ represents conjugation.Translation inR3+1,1can be defined as(7)T=1-12ae∞,T∗=1+12ae∞,where a=2dt, in which d denotes the displacement distance andt denotes the direction of movement. Therefore, the motion transformation of a rigid body in R3+1,1 can be denoted by(8)Qc=TRQc′R∗T∗.


### 2.4. Inner Product of CGA

InR3+1,1, the inner product of two conformal points is a direct representation of the Euclidean distance between the two points. For example, letting pc and sc be two points in R3+1,1, the inner product of two conformal points is represented by p and s inR3as follows:(9)pc·sc=-12s-p2.If the pointxcis on the surface of the sphere Sc, we can obtain(10)xc·Sc=0.A more detailed description of CGA can be found in [15].


## 2.1. Fundamental Theory

The fundamental algebraic operators [10] in geometric algebra are the inner product (ac·bc), the outer product (ac∧bc), and the geometric product (acbc=ac∧bc+ac·bc). CGA, which is denoted by the base space R3+1,1, has a set of orthogonal bases given by e1,e2,e3,e+, and e- with the properties(1)ei2=1,i=1,2,3,e+2=1,e-2=-1,ei·e+=ei·e-=e+·e-=0.A set of null bases{e∞,e0} orthogonal to ei can be introduced by {e+,e-}:(2)e0=12e--e+,e∞=e++e-,with the properties(3)e02=e∞2=0,e∞·e0=-1.


## 2.2. Conformal Geometric Entities

Letx∈R3 be a point expressed in the Euclidean space R3. The representation of the same point in R3+1,1 is given by(4)xc=x+12x2e∞+e0,where x=x1e1+x2e2+x3e3,x2=x12+x22+x32.InR3+1,1, the equation of a sphere of radius r centered at p∈R3 can be written as(5)Sc=p+p2-r22e∞+e0=pc-12r2e∞.


## 2.3. Conformal Transformations

Rotation, which is called a rotor, can be written as a unit quaternion as follows:(6)R=cos⁡ϕ2-nsin⁡ϕ2,R∗=cos⁡ϕ2-nsin⁡ϕ2,where n is a normalized bivector [10] representing the axis of rotation, ϕ is the rotation angle, and the superscript ∗ represents conjugation.Translation inR3+1,1can be defined as(7)T=1-12ae∞,T∗=1+12ae∞,where a=2dt, in which d denotes the displacement distance andt denotes the direction of movement. Therefore, the motion transformation of a rigid body in R3+1,1 can be denoted by(8)Qc=TRQc′R∗T∗.


## 2.4. Inner Product of CGA

InR3+1,1, the inner product of two conformal points is a direct representation of the Euclidean distance between the two points. For example, letting pc and sc be two points in R3+1,1, the inner product of two conformal points is represented by p and s inR3as follows:(9)pc·sc=-12s-p2.If the pointxcis on the surface of the sphere Sc, we can obtain(10)xc·Sc=0.A more detailed description of CGA can be found in [15].


## 3. Constraint Equations

A 6-3SPS has its six SPS (S, spherical joint; P, prismatic joint) legs connected at the five points in the moving platform and the five points in the base platform, shown in Figure 2. The six leg lengths provided by the prismatic joint in every leg are the six inputs to control the location and orientation of the moving platform. The six fixed spherical joints  P,Q,U,V,A3, and A4 are not restricted to lie in a plane. The three moving joints B1,B2, and B3 are composite spherical joints.Figure 2
Geometric model of a general 6-3SPS.


### 3.1. The Structure Transformation

According to [4, 5], in Figure 2 the point B1 can be only located on a circle with axis PQ. Therefore, we can replace the two SPS links B1P and B1Q by one RPS (R, revolute joint) link B1A1, where point A1 denotes the foot point on PQ with respect to point B1. The triangle  PB1Q  is equivalent to a rotation of B1 with respect to PQ, withθ1denoting the rotation angle with axis PQ. Analogously, the two SPS links B2U and B2V can be replaced by one RPS link B2A2, where point B2 denotes the foot point on UV with respect to point B2. The triangle UB2V is equivalent to a rotation of B2 with respect to UV, withθ2denoting the rotation angle with axis UV. Now clearly, the FDA of the 6-3SPS is equivalent to that of the special 4-3 parallel platform 2RPS-2SPS seen in Figure 3, where two revolute pairs are introduced in place of the pairs of legs converging at points B1 and B2.Figure 3
Equivalent 4-3 parallel platform.


### 3.2. Basic Closure Equations

We locate a reference frameO-XYZ with its origin at random on the base platform PQUVA3A4, as shown in Figure 2. The coordinates of points P,Q,U,V,A3, and A4 are known, respectively, in fixed frame; that is to say, the coordinates of vectors P,Q,U,V,A3, and A4 are known. The distances of three points B1,B2, and B3, namely, lB1B2, lB1B3, and lB2B3, are known. The lengths of all six legs, lB1P,  lB1Q,  lB2U,  lB2V,  lB3A3, and lB3A4, are known.For the trianglePB1Q in Figure 2, the movement locus of the point B1 is a circle with its center at  A1 and lB1A1as its radius after removing the coupling of the joint B1 and the moving platform, as shown in Figure 3. Thus, the constraint of the spherical joints P and Q with respect to B1 is equivalent to that of the revolute joint A1 with respect to B1. Translating the point A1 to the origin of the frameO-XYZ, and rotating the triangle  PB1Q  with -β1, -α1, and -θ1about the axes Y,X, and Z, the result is that the point B1 locates the axis X and QP overlaps the axis Z, as shown in Figure 4.Figure 4
Position after transformation of the trianglePB1Q.Now, according to (8), the point B1 and the vector QP inR3+1,1can be written as follows:(11)B1c=M1B1c′M1∗,(12)QPc=QP+12QP2e0+e∞=Ryβ1Rxα1QPc′Rx∗α1Ry∗β1,where(13)B1c′=l1e1+12l12e∞+e0,M1=TA1Ryβ1Rxα1Rzθ1,Rzθ1=cos⁡θ12-e3sin⁡θ12,Rxα1=cos⁡α12-e1sin⁡α12,Ryβ1=cos⁡β12-e2sin⁡β12,TA1=1-A1xe1+A1ye2+A1ye3e∞,QP=QPxe1+QPye2+QPze3,QPc′=lQPe3+12lQP2e∞+e0,and l1denotes lB1A1.According to (12), α1 and β1 can be expressed as (14)a1=sin-1⁡-QPylQP,(15)β1=2tan-1⁡QPxlQPcos⁡α1+QPz.In Figure2, let γ1=∠B1PQ. According to the law of cosines, γ1 can be expressed as(16)cos⁡γ1=lB1P2+lQP2-lB1Q22lB1PlQP.Similar to the solution process ofB1c, B2c can also be obtained by(17)B2c=M2B2c′M2∗,where(18)B2c′=l2e1+12l22e∞+e0,(19)M2=TA2Ryβ2Rxα2Rzθ2,and l2 denotes lB2A2.According to (4), in R3+1,1A3 and A4 can be written:(20)A3c=A3+12A32e∞+e0,A4c=A4+12A42e∞+e0.According to (9), the distance between B1 and B2 is written:(21)B1c·B2c=-12B1-B22=-12lB1B22=-12r32,where r3 denotes lB1B2. Substituting (11) and (17) into (21) and rewriting it, we have(22)l1l2cβ1-β2cθ1cθ2+sβ1-β2sα1sθ1cθ2-sα2cθ1sθ2+cβ1-β2sα1sα2+cα1cα2sθ1sθ2+12r32-l12-l22=0,where c(β1-β2), s(β1-β2), cα1, sα1, sθ1, andcθ1 denote cos⁡(β1-β2),sin⁡(β1-β2),cos⁡α1,sin⁡α1,sin⁡θ1, and cos⁡θ1, respectively.Lettingr1=lB1B3, r2=lB2B3, l3=lB3A3, and l4=lB3A4, from Figure 5 we can see that the joint B3 is on the intersection of four spheres, that is, the sphere S1cwith its center at B1 and r1 as its radius, the sphere S2c with its center at B2 and r2 as its radius, the sphere S3c with its center at A3 and l3 as its radius, and the sphere S4c with its center at A4 and l4 as its radius. Therefore, B3 can be expressed as (23)B3c′=S1c∧S2c∧S3c∧S4c,where(24)S1c=B1c-12r12e∞,S2c=B2c-12r22e∞,S3c=A3c-12l32e∞,S4c=A4c-12l42e∞.Figure 5
Intersection of conformal geometric spheres.Expanding (23) yields(25)B3c′=W1e1+W2e2+W3e3+W4e∞+W5e0,where(26)Wi=a1sθ1sθ2i+a2sθ1cθ2i+a3cθ1sθ2i+a4cθ1cθ2i+a5sθ1i+a6sθ1i+a7sθ2i+a8cθ2i+a9i,i=1,…,5.In the above polynomials, the coefficientsa4i=0(i=1,3,5) and aji(i=1,…,5;j=1,…,9) are reported in Appendix.From (4), the expression of B3c is written as(27)B3c=B3xe1+B3ye2+B3ze3+12B3x2+B3y2+B3z2e∞+e0.According to (25), B3c can be obtained by(28)B3c=W1W5e1+W2W5e2+W3W5e3+W4W5e∞+e0.According to (9), we have(29)B3c·B3c=W12+W22+W32-2W4W5=0.Substituting for the sine and cosine the well-known expressions(30)cos⁡θi=1-ti21+ti2,sin⁡θi=2ti1+ti2,where ti=tan⁡θi/2, i=1,2, (22) and (29) can be rewritten as follows:(31)n1t12t22+n2t12t2+n3t1t22+n4t1t2+n5t12+n6t22+n7t1+n8t2+n0=0,∑i=0,…,4j=0,…,4aijt1it2j=0,where(32)n0=n1=r32-l12-l22+2l1l2cβ1-β2,n2=-n8=4l1l2sα2sβ1-β2,n3=-n7=-4l1l2sα1sβ1-β2,n4=8l1l2cα1cα2+sβ1-β2sα1sα2,n5=n6=r32-l12-l22-2l1l2cβ1-β2.and aij are real constants that depend only on aji(i=1,…,5;j=1,…,9).Equations (31) are the kinematics constraint equations in the FDA of the 6-3SPS.


## 3.1. The Structure Transformation

According to [4, 5], in Figure 2 the point B1 can be only located on a circle with axis PQ. Therefore, we can replace the two SPS links B1P and B1Q by one RPS (R, revolute joint) link B1A1, where point A1 denotes the foot point on PQ with respect to point B1. The triangle  PB1Q  is equivalent to a rotation of B1 with respect to PQ, withθ1denoting the rotation angle with axis PQ. Analogously, the two SPS links B2U and B2V can be replaced by one RPS link B2A2, where point B2 denotes the foot point on UV with respect to point B2. The triangle UB2V is equivalent to a rotation of B2 with respect to UV, withθ2denoting the rotation angle with axis UV. Now clearly, the FDA of the 6-3SPS is equivalent to that of the special 4-3 parallel platform 2RPS-2SPS seen in Figure 3, where two revolute pairs are introduced in place of the pairs of legs converging at points B1 and B2.Figure 3
Equivalent 4-3 parallel platform.


## 3.2. Basic Closure Equations

We locate a reference frameO-XYZ with its origin at random on the base platform PQUVA3A4, as shown in Figure 2. The coordinates of points P,Q,U,V,A3, and A4 are known, respectively, in fixed frame; that is to say, the coordinates of vectors P,Q,U,V,A3, and A4 are known. The distances of three points B1,B2, and B3, namely, lB1B2, lB1B3, and lB2B3, are known. The lengths of all six legs, lB1P,  lB1Q,  lB2U,  lB2V,  lB3A3, and lB3A4, are known.For the trianglePB1Q in Figure 2, the movement locus of the point B1 is a circle with its center at  A1 and lB1A1as its radius after removing the coupling of the joint B1 and the moving platform, as shown in Figure 3. Thus, the constraint of the spherical joints P and Q with respect to B1 is equivalent to that of the revolute joint A1 with respect to B1. Translating the point A1 to the origin of the frameO-XYZ, and rotating the triangle  PB1Q  with -β1, -α1, and -θ1about the axes Y,X, and Z, the result is that the point B1 locates the axis X and QP overlaps the axis Z, as shown in Figure 4.Figure 4
Position after transformation of the trianglePB1Q.Now, according to (8), the point B1 and the vector QP inR3+1,1can be written as follows:(11)B1c=M1B1c′M1∗,(12)QPc=QP+12QP2e0+e∞=Ryβ1Rxα1QPc′Rx∗α1Ry∗β1,where(13)B1c′=l1e1+12l12e∞+e0,M1=TA1Ryβ1Rxα1Rzθ1,Rzθ1=cos⁡θ12-e3sin⁡θ12,Rxα1=cos⁡α12-e1sin⁡α12,Ryβ1=cos⁡β12-e2sin⁡β12,TA1=1-A1xe1+A1ye2+A1ye3e∞,QP=QPxe1+QPye2+QPze3,QPc′=lQPe3+12lQP2e∞+e0,and l1denotes lB1A1.According to (12), α1 and β1 can be expressed as (14)a1=sin-1⁡-QPylQP,(15)β1=2tan-1⁡QPxlQPcos⁡α1+QPz.In Figure2, let γ1=∠B1PQ. According to the law of cosines, γ1 can be expressed as(16)cos⁡γ1=lB1P2+lQP2-lB1Q22lB1PlQP.Similar to the solution process ofB1c, B2c can also be obtained by(17)B2c=M2B2c′M2∗,where(18)B2c′=l2e1+12l22e∞+e0,(19)M2=TA2Ryβ2Rxα2Rzθ2,and l2 denotes lB2A2.According to (4), in R3+1,1A3 and A4 can be written:(20)A3c=A3+12A32e∞+e0,A4c=A4+12A42e∞+e0.According to (9), the distance between B1 and B2 is written:(21)B1c·B2c=-12B1-B22=-12lB1B22=-12r32,where r3 denotes lB1B2. Substituting (11) and (17) into (21) and rewriting it, we have(22)l1l2cβ1-β2cθ1cθ2+sβ1-β2sα1sθ1cθ2-sα2cθ1sθ2+cβ1-β2sα1sα2+cα1cα2sθ1sθ2+12r32-l12-l22=0,where c(β1-β2), s(β1-β2), cα1, sα1, sθ1, andcθ1 denote cos⁡(β1-β2),sin⁡(β1-β2),cos⁡α1,sin⁡α1,sin⁡θ1, and cos⁡θ1, respectively.Lettingr1=lB1B3, r2=lB2B3, l3=lB3A3, and l4=lB3A4, from Figure 5 we can see that the joint B3 is on the intersection of four spheres, that is, the sphere S1cwith its center at B1 and r1 as its radius, the sphere S2c with its center at B2 and r2 as its radius, the sphere S3c with its center at A3 and l3 as its radius, and the sphere S4c with its center at A4 and l4 as its radius. Therefore, B3 can be expressed as (23)B3c′=S1c∧S2c∧S3c∧S4c,where(24)S1c=B1c-12r12e∞,S2c=B2c-12r22e∞,S3c=A3c-12l32e∞,S4c=A4c-12l42e∞.Figure 5
Intersection of conformal geometric spheres.Expanding (23) yields(25)B3c′=W1e1+W2e2+W3e3+W4e∞+W5e0,where(26)Wi=a1sθ1sθ2i+a2sθ1cθ2i+a3cθ1sθ2i+a4cθ1cθ2i+a5sθ1i+a6sθ1i+a7sθ2i+a8cθ2i+a9i,i=1,…,5.In the above polynomials, the coefficientsa4i=0(i=1,3,5) and aji(i=1,…,5;j=1,…,9) are reported in Appendix.From (4), the expression of B3c is written as(27)B3c=B3xe1+B3ye2+B3ze3+12B3x2+B3y2+B3z2e∞+e0.According to (25), B3c can be obtained by(28)B3c=W1W5e1+W2W5e2+W3W5e3+W4W5e∞+e0.According to (9), we have(29)B3c·B3c=W12+W22+W32-2W4W5=0.Substituting for the sine and cosine the well-known expressions(30)cos⁡θi=1-ti21+ti2,sin⁡θi=2ti1+ti2,where ti=tan⁡θi/2, i=1,2, (22) and (29) can be rewritten as follows:(31)n1t12t22+n2t12t2+n3t1t22+n4t1t2+n5t12+n6t22+n7t1+n8t2+n0=0,∑i=0,…,4j=0,…,4aijt1it2j=0,where(32)n0=n1=r32-l12-l22+2l1l2cβ1-β2,n2=-n8=4l1l2sα2sβ1-β2,n3=-n7=-4l1l2sα1sβ1-β2,n4=8l1l2cα1cα2+sβ1-β2sα1sα2,n5=n6=r32-l12-l22-2l1l2cβ1-β2.and aij are real constants that depend only on aji(i=1,…,5;j=1,…,9).Equations (31) are the kinematics constraint equations in the FDA of the 6-3SPS.


## 4. Solution Procedure

Equations (31) represent a system of two equations in two unknown variables, t1 and t2. t2 can be eliminated between them obtaining one equation in the only unknown, t1. Equations (31) can be rearranged as follows:(33)M1t22+M2t2+M3=0,G1t24+G2t23+G3t22+G4t2+G5=0,where(34)M1=n1t12+n3t1+n6,M2=n2t12+n4t1+n8,M3=n5t12+n7t1+n0,G1=∑i=0,…,4ai4t1i,G2=∑i=0,…,4ai3t1i,G3=∑i=0,…,4ai2t1i,G4=∑i=0,…,4ai1t1i,G5=∑i=0,…,4ai0t1i.Constructing the Sylvester resultant, the eliminant of (33) is the following:(35)0G1G2G3G4G5G1G2G3G4G50000M1M2M300M1M2M300M1M2M300M1M2M3000=0.Expanding (35), a 16th-order polynomial equation with the variable t1 can be obtained as follows:(36)∑i=016bit1i=0,where bi are real constants that depend only on input data. Therefore 16 real and complex solutions for t1 are possible.Fort=t1, (33) are indeed algebraic equations in the only unknown t2, and these have a common root, t2, whose value can be found by equating to zero the first-degree greatest common divisor (GCD) of the polynomials on the left-hand sides of (33).Hence, a unique solution (t1,t2) of (31) is derived for every solution t1 of (36) and, consequently, through (11), (17), and (28) a unique location of B1,B2, and B3 is obtained. Therefore, the FDA of the 6-3SPS provides 16 solutions in the field of complex numbers.


## 5. Numerical Verification

In order to validate the new algorithm, the input data of a numerical example are given in Table1. The data are equivalent to the data of the example discussed in [4]. All 16 sets of output data obtained by the proposed method agree with those given in [4]; the outputs of four real solutions are listed in Table 2.Table 1
Input data.
W a P Q U V A 3 A 4 X 50 −25 80 −50 48 36 Y 0 −2 20 −20 15 31 Z 100 40 50 70 68 −93 Distance between three points ofBi (i=1,2,3) lB1B3=190,lB2B3=135,lB1B2=141 Length of six legs lB1P=76,lB1Q=160,lB2U=139,lB2V=55,lB3A3=128,lB3A4=217Table 2
Four real solutions from 16 sets of output solutions.
i t 1 t 2  B 1 B 2 B 3 1 −0.6258 −1.0743 X 79.5353 −26.0942 −70.9222 Y −45.8809 −68.9457 54.3104 Z 152.9018 62.3955 94.3853  2 −0.3717 0.442 X 68.8676 −47.0215 21.2493 Y −33.0062 21.0886 137.0612 Z 165.8073 106.4396 95.7395  3 0.9609 0.240 X 82.5389 −48.8261 14.0067 Y 51.0783 24.7276 −108.359 Z 145.9154 101.9852 71.8847  4 0.7363 0.5102 X 90.9016 −40.7763 −6.7822 Y 53.3944 6.2851 −100.5172 Z 135.3847 117.4237 74.218


## 6. Conclusions

This paper presents a closed-form solution for the FDA of the 6-3SPS using the CGA. The 16th-order input-output equation is derived from two equations with two variables by using the Sylvester resultant method. The two equations are derived by using a computer algebra symbolic calculation. The number of solutions agrees with that of [4]. The procedure is unique in that the final resultant matrix is in the symbolic form of geometric parameters. This symbolic formulation not only can provide insights into the nature of the problem, but also greatly simplifies the implementation of computer programming. To compute solutions for a new set of structural parameter and limb actuation data, one only needs to substitute numerical values of these parameters into the symbolic resultant matrix without requiring any algebraic manipulation. Compared with the previously reported method, the novelty of the proposed method lies in that the problem is modeled based on the geometric characteristics and is solved based on single elimination; as a result, the solution procedure is simpler, more efficient, and more readily programmed. The computation time using the new solution presented in this paper is about 0.15 s in Maple 16 running on a PC with Intel Pentium Dual-Core E5800 3.2 GHz and 2 GB Ram. However, employing the same computer system and software, the computation time using the method developed in [4] is 0.3 s.



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*Source: 4687638-2017-02-13.xml*