On the design of reactionless 3-DOF planar parallel mechanisms

Mechanism and Machine Theory 41 (2006) 70 82 Mechanism and Machine Theory www.elsevier.com/locate/mechmt On the design of reactionless 3-DOF planar parallel mechanisms Abbas Fattah *, Sunil K. Agrawal Mechanical Systems Laboratory, Department of Mechanical Engineering, University of Delaware, Newark, DE 19716, USA Received 22 March 2004; received in revised form 17 March 2005; accepted 26 April 2005 Available online 27 June 2005 Abstract A reactionless machine will not apply reaction forces and moments on the mounting base during motion. This article presents the theory for design of a reactionless planar parallel mechanism. The salient features of this design are: (i) it uses auxiliary parallelograms to locate the center of mass of each leg at the base, (ii) choice of geometric and inertial parameters to satisfy the reactionless conditions and/or selection of trajectory to achieve a reactionless design. We illustrate the reactionless feature of such a mechanism through computer simulations. Ó 2005 Elsevier Ltd. All rights reserved. 1. Introduction Over the years, a number of methods have been proposed for static or gravity balancing of machines through clever designs using counterweights, springs, and auxiliary parallelograms. These designs have been applied to linkages [4], serial robotic manipulators [14,16] and parallel manipulators [11,15]. Some mathematical desiptions of gravity balanced machines are (i) system center * Corresponding author. E-mail addresses: fattah@me.udel.edu (A. Fattah), agrawal@me.udel.edu (S.K. Agrawal). 0094-114X/$ - see front matter Ó 2005 Elsevier Ltd. All rights reserved. doi:10.1016/j.mechmachtheory.2005.04.005

A. Fattah, S.K. Agrawal / Mechanism and Machine Theory 41 (2006) 70 82 71 of mass remains inertially fixed during motion; (ii) potential energy remains invariant with configuration of the system; (iii) system has counter masses that balance the machine in every configuration. There are some published works on dynamic balancing of open loop systems [2,1] and the issue of reactionless control of a space robot keeping the base inertially fixed [12,18]. Also, there are some reported research on dynamic balancing of linkages, specially four-bar linkages [6,7,4]. These use counterweights or idler-loops to cancel or minimize the effects of forces and moments transmitted to the base. In addition, dynamically balanced or reactionless parallel mechanisms have been designed which use a series of reactionless four-bar linkages in their legs, and stacking these to make a reactionless leg [13,17,9]. Loop constraint equations for the four-bar linkages are used to express all dependent generalized coordinates and their time derivatives in terms of independent ones in the expression of angular momentum of the four-bar linkages. The coefficients of the new expression for the total angular momentum were then made to be zero through design choices. A problem with this design is that the dynamic balancing conditions for the four-bar linkages are very restrictive and makes the design of these mechanisms difficult [8]. Also, this design results in a large number of links for each leg and adds complexity by added masses and inertia to the system. Another design technique is to use counter-rotations with proper choice of geometric and inertia parameters to design a dynamically balanced parallel mechanisms [10,8]. The disadvantage of this approach is to add mechanical elements to the mechanism. However, the latter design is simpler compared to former one. In most research, dynamic balancing of mechanisms is attained by proper choice of geometric and inertial parameters or through proper choice of motion trajectories. In this paper, we use auxiliary parallelograms to locate the center of mass of each leg at the base. Furthermore, appropriate choices of geometric and inertial parameters and/or using proper trajectory to achieve a reactionless design for the planar parallel mechanism. The main contributions of this paper are: (i) application of auxiliary parallelograms to locate the system center of mass. This leads to a simpler design, with respect to number of links and complexity of the added masses and inertias to the system, compared to reactionless mechanisms proposed in the literature; (ii) the use of trajectories to attain a dynamically balanced design for the parallel mechanism. The organization of the paper is as follows: first, we analyze the dynamic behavior of coupled bodies in multi-loop configurations to find the conditions for design of reactionless parallel machines. Then, we apply these conditions to a 3-DOF planar parallel mechanism. Detailed mathematical model, motion planning and simulation results for a numerical example are presented towards the end of the paper. 2. Reactionless machines In this section, we first develop the equations of motion for a general system (multi-chain and/ or multi-loop) with multiple contacts with an inertial fixed frame 0, as shown in Fig. 1. The interaction of the system with inertial frame 0 is at multiple points.

72 A. Fattah, S.K. Agrawal / Mechanism and Machine Theory 41 (2006) 70 82 Fig. 1. A closed-chain system with multiple contacts with the ground. The overall equations, with k interactions between the system and the ground, are as follows: X k F ie þ Mgn ¼ Ma C ; ð1þ d dt H C ¼ Xk M ie þ Xk r COi F ie ; where F ie and M ie are respectively the externally applied force and moment on the system by the ground O at the ith contact point O i, r COi is a vector from the system center of mass C to O i. Also n is the unit gravity vector. The conditions for reactionless machine with multi-contacts are the total forces and moments transmitted to the ground must vanish. Mathematically, these can be expressed as X k X k F ie ¼ 0; M ie þ Xk r COi F ie ¼ 0. The first condition can be satisfied if the center of mass (COM) of the machine is inertially fixed and its motion is in a plane, normal to the direction of gravity. The second condition can be satisfied by making the total angular momentum of the system about the system center of mass to be fixed, i.e., H C = constant = 0. Our strategy to satisfy these conditions for a parallel mechanism with m legs and a moving platform (MP) is as follows: (i) distribute the mass of MP at end of each chain by point masses, (ii) locate the COM of each leg at its base using an auxiliary parallelogram. Our strategy to make the angular momentum of the system to be a constant is as follows: we first write the following two identities: (i) the total angular momentum of the system is ð2þ ð3þ ð4þ

A. Fattah, S.K. Agrawal / Mechanism and Machine Theory 41 (2006) 70 82 73 H C ¼ Xm H i C ; where H i C is the angular momentum of leg i about system center of mass C, (ii) the angular angular momentum of leg H i C is related to Hi O i as H i C ¼ Hi O i þ r COi M i _r Oi C; ð6þ where M i is the total mass of leg i, (iii) the vector _r Oi C vanishes because C is inertially fixed. Upon substitution of _r Oi C ¼ 0 into Eq. (6), one obtains, H C ¼ Xm H i O i ; ð7þ (iv) if the angular momentum for each leg H i O i vanishes, then H C = 0. In the next section, we will comply these conditions to a 3-DOF planar parallel mechanism. ð5þ 3. 3-DOF planar parallel mechanism Consider a 3-DOF planar parallel mechanism as shown in Fig. 2. It consists of three legs which are connected to a thin triangular moving platform (MP). Each leg O i O i+3 O i+6, i = 1, 2, 3 comprises two links with lengths of l i,i+3 and l i+3,i+6 and masses of m i,i+3 and m i+3,i+6, connected by revolute joints. The joint angles are shown by h i and h i+3 for each leg. Cartesian space is defined by the pose of the MP, i.e., its orientation / and position of center of mass of MP given by x e and y e in the frame F 0. We will apply the reactionless conditions to this mechanism. Fig. 2. A 3-DOF planar parallel mechanism.

74 A. Fattah, S.K. Agrawal / Mechanism and Machine Theory 41 (2006) 70 82 4. Making COM inertially fixed Our approach to make the system COM inertially fixed is outlined in the following sections. 4.1. Distribution of MP using point masses We can replace the effect of mass and inertia of the MP by three point masses, one at the end of each leg, by considering the three conditions presented by Bagci [5]. These conditions are: (i) the sum of the point masses should be the same as the mass of MP, i.e., m MP ; (ii) the COM of the MP, C MP, with respect to a fixed point will be the same as COM of the all point masses; (iii) the moment of inertia of the point masses about C MP is the same as moment of inertia of MP about C MP. For example, if we replace the thin equilateral triangular MP of the mechanism by three equal point masses of m MP, it is easy to show that all three conditions are met. From now on, 3 the effect of MP is replaced by three point masses at O i+6, i =1,2,3. 4.2. Locating the center of mass of each leg at its base The center of mass of each leg can be located at its base, O i, i = 1, 2, 3, adding auxiliary parallelogram to original links O i O i+3 and O i+3 O i+6 as shown in Fig. 3 [2]. As shown, the new model of the leg 1 has four links: O 1 O 4 (link 3); P 0 1 O0 4 (link 1), parallel to link 3; O0 4 O 7 (link 2), along O 4 O 7 ; and O 1 O 0 1 (link 4), parallel to link 2. The parameters d1 1 and d1 2 are chosen such that the COM of leg 1 is located at O 1. The location of COM of leg 1 from the point P 0 1 is written as P 4 r P 0 1 O ¼ 1 m1 i r P 0 1 C1 i 1 P 4 ; ð8þ 1 m1 i Fig. 3. Modeling of leg 1 of the planar parallel mechanism.

A. Fattah, S.K. Agrawal / Mechanism and Machine Theory 41 (2006) 70 82 75 where C 1 i is COM of link i of leg 1. The supersipt 1 stands for leg 1. The vector r P 0 1 O can be written in terms of geometric parameters d 1 1 and d1 2 and the unit vectors along links 1 and 2, namely, b1 1 1 and b 1 2 as r P 0 1 O ¼ 1 d1 1 b1 1 þ d1 2 b1 2. ð9þ Also m 1 i r P of each link of leg 1 are 0 1 C1 i m 1 1 r P ¼ m 1 0 1 C1 1 1 l1 1c b1 1 ; m 1 2 r P ¼ m 1 0 1 C1 2 2 ðl1 1 b1 1 þ l1 2c b1 2Þ; ð10þ m 1 3 r P ¼ m 1 0 1 C1 3 3 ðd1 1 b1 1 þ d1 2 b1 2 þ l1 3c b1 1 Þ; m 1 4 r P ¼ m 1 0 1 C1 4 4 d 1 1 b1 1 þ d1 2 2 b1 2. Here, m 1 2 comprises the masses of m O 0 4 O, m 4 47 and m MP, 3 l1 ic is the center of mass of link i of leg 1 from the initial point of the link. Upon substitution of Eqs. (10) into Eq. (8) and the result thus obtained into Eq. (9), the parameters d 1 1 and d2 2 are determined as d 1 1 ¼ m1 1 l1 1c þ m1 2 l1 1 þ m1 3 l1 3c M 1 m 1 3 ; m1 4 m 1 2 l1 2c d 1 2 ¼ ; ð11þ M 1 m 1 3 m1 4 2 where M 1 is the total mass of the leg 1. Hence, Eqs. (11) give the expressions for the parameters d 1 1 and d 1 2 in terms of inertia parameters of the links. Note that d1 i are factors of geometry and mass distribution of the links and have been denoted by the terminology scaled lengths [3]. This method can be applied to other legs to locate the center of mass of each leg at its base as shown in Fig. 4. Therefore, the center of mass of the mechanism C will be inertially fixed as a result of fixing the center of mass of each leg. 5. Vanishing the total angular momentum of the mechanism The total angular momentum of each leg H i O i ¼ H Oi z, i = 1, 2, 3 can be written as H i O i ¼ X4 j¼1 H i j ; where z is the normal vector to the plane of the mechanism and H i j is the contribution to the angular momentum by the jth link of leg i. Its expression is H i j ¼ mi j ðri xj _ri yj _ri xj ri yj ÞþI i cj _ b j ; where r i xj and ri yj are coordinates of the position vector from point O i to the center of mass of link j and I i cj is the moment of inertia of link j about its center of mass. Also, _ b j is defined as ð12þ ð13þ

76 A. Fattah, S.K. Agrawal / Mechanism and Machine Theory 41 (2006) 70 82 Fig. 4. Modeling of the planar parallel mechanism. _b j ¼ _ h i for j ¼ 1; 3; ð14þ _b j ¼ _ h i þ _ h iþ3 for j ¼ 2; 4. ð15þ Here, _ h i is the ith joint rate. Using Eq. (13), the total angular momentum of the leg can be written as H i O i ¼ðA i þ B i cosðh iþ3 ÞÞ _ h i þðd i þ E i cosðh iþ3 ÞÞ _ h iþ3 ; ð16þ where E i ¼ m i 1 ðli 1c di 1 Þdi 2 þ mi 2 ðli 1 di 1 Þðli 2c di 2 Þ; D i ¼ I i 4 þ I i 2 þ mi 4 ðdi 2 li 4c Þ2 þ m i 2 ðli 2c di 2 Þ2 ; B i ¼ 2E i ; ð17þ ð18þ ð19þ A i ¼ m i ðl i 3 Þ2 3 2 þ Ii 3 þ mi 1 ½ðdi 2 Þ2 þðl i 1c di 1 Þ2 Šþm i 2 ðli 1 di 1 Þ2 þ D i. ð20þ It may be noted that the supersipt i stands for leg i, i = 1, 2, 3. Upon substitution of Eq. (16) into Eq. (7), the total angular momentum of the mechanism is derived as H C ¼ X3 ½ðA i þ B i cosðh iþ3 ÞÞ _ h i þðd i þ E i cosðh iþ3 ÞÞ _ h iþ3 Š. There are two approaches to make H C = 0: (i) Use loop constraint equations to express all dependent generalized coordinates and their time derivatives of Eq. (21) in terms of independent ones in joint or Cartesian space. Then try to vanish the coefficients of the new expression for H C ; (ii) Work independently on Eq. (21) to make H C = 0. The first method is more desirable because there ð21þ

are less coefficients to vanish and there is no need to apply extra conditions. However, the loop constraint equations are highly nonlinear equations and the final form of H C is very cumbersome to satisfy the conditions. Hence, we use the second approach in this work. The coefficients A i, B i, D i and E i, i = 1, 2, 3, should be vanished to have H C = 0. First we study the conditions that make E i =0. 5.1. Conditions to make E i =0 The coefficient E i as defined in Eq. (17) can be vanished by choosing appropriate values for the inertia parameters of each leg. First we locate the COM of the links 1 and 2 of the leg i as l i 1c ¼ di 1 þ li 3 ; a i 1 l i 2c ¼ di 2 þ li 2 di 2 ; a i 2 A. Fattah, S.K. Agrawal / Mechanism and Machine Theory 41 (2006) 70 82 77 where a i 1 and ai 2 are two constants. Inserting Eqs. (22) into Ei = 0 and solving for m i 1 in terms of other inertia parameters, one obtains m i 1 ¼ 2m i 2 mi 4 ai 1 ða i 2 m iþ3;iþ6 þ 2a i 2 m p 2m i 2 a 1Þ. ð23þ Here, upon substitution of B i =2E i = 0 into Eq. (21), the total angular momentum of the mechanism is written as ð22þ H C ¼ X3 ða i _ hi þ D i _ hiþ3 Þ. ð24þ Next, we present the conditions to make A i = 0 and D i =0. 5.2. Conditions to make A i = 0 and D i =0 The coefficients of A i and D i are positive values and it is impossible to vanish them by changing the inertia terms. Hence, we should add either counter-rotations at joint O i and O i+3 such that these coefficients vanish or we can make H C = 0 by appropriate trajectory planning during motion. 5.2.1. Counter-rotations We can add two counter-rotations respectively to links O i O i+3 and O i O 0 i, i = 1, 2, 3 at joints O i at each leg as shown in Fig. 5. Here, k i and k i+3 are the gear ratios of the gears that drive the counter-rotations. Using Eq. (24), the total angular momentum of the system, with adding counterrotations, is H C ¼ X3 ½ðA i I i Iiþ3 Þ h _ i þðd i I iþ3 Þ h _ iþ3 Š; ð25þ

78 A. Fattah, S.K. Agrawal / Mechanism and Machine Theory 41 (2006) 70 82 Fig. 5. Counter-rotations of leg 1 of the reactionless 3-DOF planar parallel mechanism. where I i iþ3 and I are inertia of the counter-rotations attached to links O i O i+3 and O i O 0 i, respectively. Please note that adding counter-rotations at the joints does not have any effect on the coefficients B i and E i, provided that the mass of counter-rotations attached to the links are negligible compared to masses of the links. Vanishing the coefficients of the h _ i and h _ iþ3 in Eq. (25) leads to A i I i iþ3 I ¼ 0; D i I iþ3 ¼ 0; ð26þ where A i and D i are defined in Eqs. (20) and (18), respectively. A numerical example will be presented in next section. 5.2.2. Trajectory planning This section explains how planning is obtained to have H C = 0. Upon substitution of H C =0in Eq. (24) and integration of the result, one obtains X 3 ða i h i þ D i h iþ3 Þ¼C; ð27þ where C is the integration constant. Using the inverse kinematic analysis of the mechanism, the joint angles h i and h i+3 can be derived in terms of Cartesian space, i.e., the orientation, /, and position of the COM of the MP, x e and y e,as X 3 ða i f i ðx e ; y e ; /ÞþD i g i ðx e ; y e ; /ÞÞ ¼ C. ð28þ

A. Fattah, S.K. Agrawal / Mechanism and Machine Theory 41 (2006) 70 82 79 Using trajectories that satisfy this constraint leads to H C = 0. Eq. (28) is a scalar constraint equation and thereby provides only one constraint on the possible trajectories of the 3-DOF system. The method is outlined in Section 6 with a numerical example. 6. Numerical examples This section desibes numerical examples for the reactionless mechanism. The length and mass of the original links are given in Table 1. Using the conditions as mentioned in Eqs. (11), (22) and (23) to locate the COM of each leg to its base and also to make E i = 0, one can obtain the detailed design of the mechanism as: d i 1 ¼ 0.3782 m, di 2 ¼ 0.0276 m, mi 4 ¼ 0.6 kg, ai 1 ¼ 10, ai 2 ¼ 2, mi 1 ¼ 1.365 kg; i =1,2,3. We can choose different designs which satisfy these conditions since there are free variables. Now, one can use either the counter-rotation method or trajectory planning to make a reactionless mechanism. 6.1. Reactionless mechanism example using counter-rotations Using counter-rotation method, one can use Eq. (26) to obtain the counter-rotations at joints as I i ¼ 0.0642 kg m2 ; i ¼ 1; 2; 3; ð29þ I iþ3 ¼ 0.0287 kg m 2 ; i ¼ 1; 2; 3. ð30þ Therefore, the mechanism with the above-mentioned inertia and geometric parameters is reactionless for all motion of the MP of the mechanism. 6.2. Reactionless mechanism example through motion planning The following Cartesian space trajectory is chosen for the position of the COM of the MP y e ¼ 0.42857x e þ 0.17687 ð31þ with the initial point x e = 0.2831 m, y e = 0.2982 m and the final point x e = 0.4231 m, y e = 0.3582 m. Upon substitution of the position trajectory in Eq. (28), one obtains the orientation of the MP, i.e., / that is shown in Fig. 6(a). Also, the initial and final configuration of the mechanism is shown in Fig. 6(b). Having the Cartesian space trajectory and using the inverse kinematic analysis, the joint angle motions can be obtained. Table 1 The geometric and inertia parameters for 3-DOF parallel mechanism Original link Dimension (m) Mass (kg) O 1 O 4 = O 2 O 5 = O 3 O 6 0.15 0.1908 O 4 O 7 = O 5 O 8 = O 6 O 9 0.2 0.1430 MP(O 7 O 8 O 9 ) 0.3464 0.6

80 A. Fattah, S.K. Agrawal / Mechanism and Machine Theory 41 (2006) 70 82 Fig. 6. (a) Orientation of the MP versus x e and (b) initial and final configuration of the mechanism. Fig. 7. Time history of the joint angles (a) and their time derivatives (b) of the reactionless mechanism through special trajectory. In order to show that the mechanism is reactionless, the total moment transmitted to the base for the reactionless mechanism using special trajectory is compared to a normal design. To this end, the time history of the joint angles for the above-mentioned motion was approximated by a polynomial during a period of 1 s as shown in Fig. 7(a). Their time rate of changes are also given in Fig. 7(b). Likewise, the time history of the joint angles and their time derivatives for the normal one are shown in Fig. 8(a) and (b). Using Eq. (2), the total moment transmitted to the base with this planning and the normal design are computed and shown in Fig. 9. As shown, the total moment transmitted to the base for this planning (solid line) is quite small and close to zero compared to normal design (dotted line). The results are not identically zero for this planning because the joint angles were approximated by the polynomials using only 10 points. We can get better results using more points. However, this comparison shows the effectiveness of reactionless mechanism through motion planning.

A. Fattah, S.K. Agrawal / Mechanism and Machine Theory 41 (2006) 70 82 81 Fig. 8. Time history of the joint angles (a) and their time derivatives (b) of the mechanism through normal motion planning. Fig. 9. Total moments transmitted to the base for the reactionless mechanism through special trajectory (solid line) and for the normal design (dotted line). 7. Conclusions The paper provided the design and theory for a reactionless planar parallel mechanism. The system center of mass was inertially fixed using auxiliary parallelograms. The total angular momentum of the system vanishes using two approaches: (i) through proper choice of inertia and geometric parameters, (ii) using appropriate motion planning. The results verified that the mechanism is reactionless and there are no forces or moments transmitted to the base during the motion of the moving platform. The first approach requires the use of counter-rotations which inease the mass and inertia of the system. Therefore, the second approach is recommended if we

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