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[8] Z. Hu, S. E. Salcudean, and P. D. Loewen, “Robust controller design forteleoperation systems,” inProc. IEEE Int. Conf. Systems, Man, Cyber-netics, vol. 3, Vancouver, BC, Canada, Oct. 1995, pp. 2127–2132.

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An Efficient Approach to the Forward Kinematics of aPlanar Parallel Manipulator With Similar Platforms

Ping Ji and Hongtao Wu

Abstract—The forward kinematics of a parallel manipulator is so diffi-cult that the analytical solutions of only a few parallel manipulators havebeen found. The forward kinematics of a three-degree-of-freedom planarparallel manipulator, where the base and the mobile platforms are two sim-ilar triangles, is studied in this paper and its analytical solution is provided.The final solution is two independent univariate quadratic equations. Aswell, the forward kinematics solution is simplified to the parallel manipu-lator where the two triangles are equilateral and a numerical example ispresented.

Index Terms—Analytical solution, forward kinematics, parallel manipu-lator.

I. INTRODUCTION

A three-degree-of-freedom (DOF) planar parallel manipulator, ingeneral, consists of two triangles in a plane, as shown in Fig. 1. Themoving triangle�A1A2A3 is the mobile platform while three fixedpoints,B1, B2, andB3, form another triangle�B1B2B3, that is, thebase platform. An extensible limbLk links the couple verticesAk andBk (k = 1; 2; 3). The mobile platform can rotate and move in theplane if the limbs change their lengths with actuators, either prismatic[1]–[3] or revolute [2]. This manipulator was first proposed by Gosselinand Merlet [4] and discussed by some researchers, including [1]–[10].This paper only discusses the manipulator with prismatic actuators,called the RPR chains (R: revolute, P: prismatic) [5], [6]. Obviously,the parallel manipulator has three DOF. The forward kinematics tothis manipulator is to find the position and orientation of the mobileplatform with respect to the fixed base platform with the known limblengths.

This manipulator seems to have a very simple structure. However, itsforward kinematics is quite complicated. The result of its forward kine-matics is a univariate sixth polynomial [2], so it has to be solved numer-ically. Consequently, its forward kinematics has at most six solutions.The singularity problem of this parallel manipulator was discussed byCollins and McCarthy [7]. A special case of the manipulator, where twotriangles are equilateral, has been studied more deeply due to its sym-metry [8]–[10]. The purpose of this paper is to present an analyticalsolution to the forward kinematics of the planar parallel manipulatorwith two equilateral triangles. The two triangles of the manipulator tobe studied in this paper are only required to be similar, not necessarilyequilateral. Certainly, the result obtained here can be applied to theequilateral-triangular manipulator, since it is a special case of the sim-ilar triangular manipulators. This paper discusses the manipulator withtwo similar platforms in plane, as shown in Fig. 1, while the general-ization of this manipulator, that is, the manipulator with two similarplatforms in space, has been discussed by some researchers, including[11].

Manuscript received May 18, 2001; revised November 28, 2001. This paperwas recommended for publication by Associate Editor C. Gosselin and EditorI. Walker upon evaluation of the reviewers’ comments. This work was supportedby The Hong Kong Polytechnic University under Project G-T056.

P. Ji is with the Department of Industrial and Systems Engineering, TheHong Kong Polytechnic University, Hung Hom, Kowloon, Hong Kong (e-mail:mfpji@polyu.edu.hk).

H. Wu is with the Department of Mechanical Engineering, Nanjing Universityof Aeronautics and Astronautics, Jiangsu 210016, P. R. China.

Digital Object Identifier 10.1109/TRA.2002.802208

1042-296X/02$17.00 © 2002 IEEE

648 IEEE TRANSACTIONS ON ROBOTICS AND AUTOMATION, VOL. 18, NO. 4, AUGUST 2002

Fig. 1. Three-DOF planar parallel manipulator.

II. THREE-DOF PLANAR PARALLEL MANIPULATOR

Three points form a triangle if they are not in a line, so the radius ofthe circumcircle formed by pointsB1,B2, andB3 is defined asrb andthe circumcenterO is the origin of the fixed frameOXY. Since theY axis is set up through pointB2, without loss of generality, the threepoints can be denoted as

B1 =(b1x; b1y)T (1a)

B2 =(b2x; b2y)T = (0; rb)

T (1b)

B3 =(b3x; b3y)T (1c)

and their unit vectors are

B1 =B1

rb= b1x; b1y

T(2a)

B2 =B2

rb= (0; 1)T (2b)

B3 =B3

rb= b3x; b3y

T(2c)

or

Bk =rbBk (k = 1; 2; 3): (2d)

Similarly, the mobile frame is set up at the cirmcumcenterO0 ofpointsA1, A2, andA3 while theY 0 axis goes through pointA2. Thecircumradius is defined asra, then

A1 =(a1x; a1y)T (3a)

A2 =(a2x; a2y)T = (0; ra)

T (3b)

A3 =(a3x; a3y)T (3c)

and

A1 =A1

ra(4a)

A2 =A2

ra= (0; 1)T (4b)

A3 =A3

ra(4c)

or

Ak =rbAk (k = 1; 2; 3): (4d)

Obviously, the following equation exists:

Ak = Bk (k = 1; 2; 3) (5)

Therefore

AT

kBk = BT

kBk = 1 (k = 1; 2; 3): (6)

If P = (px; py)T is the position vector between the two origin points

O andO0,R is the transformation matrix between the two frames, andhere

R =cos � � sin �

sin � cos �

by (5), we have

AT

kRBk = BT

kRBk = cos � (k = 1; 2; 3): (7)

Now the limb vectors can be represented as

Lk = P+RAk �Bk (k = 1; 2; 3): (8)

Due to (2d), (4d), and (5)

Lk = P+ (raR� rbE)Bk (k = 1; 2; 3) (9)

whereE is a 2� 2 identity matrix. So, the length of each limb can beobtained as follows:

L2

k = P+ (raR� rbE)BkTP+ (raR� rbE)Bk

(k =1; 2; 3): (10)

Due to (6) and (7), (10) can be rewritten as

L2

k =PTP+ 2B

T

k raRT� rbE P+ r

2

a + r2

b � 2rarb cos �

(k =1; 2; 3): (11)

If P and� are known, the lengthLk(k = 1; 2; 3) of each limb canbe obtained easily from (11) and this is the inverse kinematics of themanipulator.

III. A NALYTICAL SOLUTION

In the forward kinematics, the three limb lengthsLk(k = 1; 2; 3) in(11) are given, whileP and� are unknown. Thus, (11) is a set of threenonlinear equations with three unknowns. By defining

W = (wx; wy)T = raR

T� rbE P (12)

then (11) can be expanded as

PTP+ 2B

T

1W =L2

1 � r2

a + r2

b + 2rarb cos � (13a)

PTP+ 2B

T

2W =L2

2 � r2

a + r2

b + 2rarb cos � (13b)

PTP+ 2B

T

3W =L2

3 � r2

a + r2

b + 2rarb cos �: (13c)

Obviously, we have

2 BT

1 �BT

2 W =L2

1 � L2

2 (14a)

2 BT

3 �BT

2 W =L2

3 � L2

2: (14b)

That is

b1x b1y � 1

b3x b3y � 1

wx

wy

=

(L �L )2

(L �L )2

: (15)

Therefore

wx

wy

=b1x b1y � 1

b3x b3y � 1

�1 (L �L )2

(L �L )2

: (16)

In order to get�, let

�2 = r

2

a + r2

b � 2rarb cos �: (17)

Here,�, ra, andrb form a triangle and, obviously,� � 0. From (12),we have

wx

wy

=ra cos � � rb ra sin �

�ra sin � ra cos � � rb

px

py: (18)

IEEE TRANSACTIONS ON ROBOTICS AND AUTOMATION, VOL. 18, NO. 4, AUGUST 2002 649

Therefore

px

py=

1

�2ra cos � � rb �ra sin �

ra sin � ra cos � � rb

wx

wy

: (19)

Also, by (19), we have

PTP = p

2

x + p2

y =w2

x + w2

y

�2: (20)

Now (13b) becomes

w2

x + w2

y

�2+ 2wy = L

2

2 � �2: (21)

Equation (21) is a quadratic equation in terms of�2 and its solution is

�2=

L2

2

2� wy � L4

2

4� wyL

2

2� w2

x: (22)

So far, we have solved the forward kinematics problem of thismanipulator. The procedure is to getwx, wy first by use of (16), thento get�2 from (22) and� from (17), and finally to obtainpx, py from(19). Equation (22) may yield two real solutions of�2 and each ofthem leads two�’s, so there are at most four solutions to the forwardkinematics of the parallel manipulator.

The advantage of this method is obvious. Equation (22) is aunivariate quadratic equation and (17) can be considered as anotherunivariate quadratic equation, so only two independent univariatequadratic equations are required to solve and, certainly, the manip-ulator has an analytical solution to its forward kinematics withoutnumerical iterations. Recently, Kong and Gosselin [5] also obtainedan analytical solution to the forward kinematics of this manipulator.Compared with their solution procedure, the approach presented inthis paper is more elegant and concise.

The general three-DOF planar parallel manipulator has at most sixsolutions to its forward kinematics [2]. However, as we derived ear-lier, the special three-DOF planar parallel manipulator with two sim-ilar triangles has at most four solutions to its forward kinematics. If thesolution procedure [2] for the general planar parallel manipulator, thatis, the case where the two platforms are not similar, is applied to thismanipulator with two similar platforms, it can be proved that the othertwo solutions are extraneous. They are

T2= � ra � rb

ra + rb

2

whereT is defined as

cos � =1� T 2

1 + T 2and sin � =

2T

1 + T 2:

IV. EXAMPLE

Since the three-DOF planar parallel manipulator with two equilateraltriangles has been investigated deeply, a numerical example is providedhere for this type of manipulators. Before the example is presented, theprevious solution is simplified for this equilateral triangle case. Due toequilateral triangles, we have

B1 =B1

rb= b1x; b1y

T=

p3

2;�1

2

T

B3 =B3

rb= b3x; b3y

T= �

p3

2;�1

2

T

:

Thus, from (16), we have

wx =L2

1 � L2

3

2p3

(23)

wy =2L2

2 � L2

1 � L2

3

6: (24)

TABLE IFINAL SOLUTIONS

Now (22) becomes

�2

=L2

1 + L2

2 + L2

3

6� 2L2

1L2

2+ 2L2

1L2

3+ 2L2

2L2

3� L4

1� L4

2� L4

3

12:

(25)

Suppose the parameters of the manipulator arera=30 andrb=40.If px = 12, py = 8, and� = 40

�, then from (11) we can obtainL1 =

33:919134, L2=11:592566, andL3=36:380912.Now, if the aboveL1, L2, andL3 are known, in order to getpx, py,

and�, we can getwx=�49:9550 andwy=�367:5529 from (23) and(24), and�21=661:4933 and�22=208:0 from (25). The four solutionsto the forward kinematics are listed in Table I.

V. CONCLUSION

The forward kinematics of the three-DOF planar parallelmanipulator with two similar triangles was studied in this paper andits analytical solution with only two independent univariate quadraticequations was presented. One of its special cases, where the twotriangles are both equilateral, was also discussed and its simplifiedsolution was provided.

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