Mechanical Design and Kinematics Analysis of a Hydraulically Actuated Manipulator Xuewen Rong1, Rui Song*,1, Hui Chai1 and Xiaolin Ma2

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The Open Mechanical Engineering Journal, 2014, 8, 457-461 457 Open Access Mechanical Design and Kinematics Analysis of a Hydraulically Actuated Manipulator Xuewen Rong1, Rui Song*,1, Hui Chai1 and Xiaolin Ma2

1School of Control Science and Engineering, Shandong University, Jinan 250061, China 2School of Mechanical and Electrical Engineering, Shandong Jianzhu University, Jinan 250101, China

Abstract: This paper gives a mechanism design of a six DOF hydraulically actuated manipulator firstly. Then its DH frames and link parameters are given. Next, its forward kinematic equations are derived according to homogeneous transformation method. Fourthly, the analytical solutions of its inverse kinematics are solved by given the position and posture of the end-effector simultaneously. The posture of the end-effector is given with three z-y-z Euler angles for they have obvious geometry meanings and are easy to be measured. In addition, the correctness of the inverse kinematic equations is verified in Simulink by comparing many sets of randomly produced joint variables in workspace and their corresponding inverse solutions. Keywords: DH frame, DOF, Euler angle, homogeneous transformation, hydraulic, manipulator.

1. INTRODUCTION the forward kinematic equations are given firstly. Then the analytical solutions of its inverse kinematics are solved by The traditional electric powered industrial manipulators given the position and posture of the end-effector have been used in all kinds of industrial productions. They simultaneously. are already very mature in technology and with high precision. But they are not suitable for several tasks in severe 2. MECHANISM DESIGN OF HydM environments for their electric drive mode and low power- The mechanical structure sketch of HydM is shown in mass ratio, such as underwater, construction, electric power Fig. (1). HydM has six rotary degrees of freedom (DOF) lines maintenance, etc. In order to make up for the which are all driven by hydraulic actuators. The first, the inadequacy of electric powered manipulators, some hydraulically actuated multi-joints manipulators have been fourth and the fifth joints are driven by helical rotary actuators. The second and the third joints are driven by developed in several countries. hydraulic linear cylinders. The sixth joint is driven by a Uninterrupted power supply has become indispensable miniature hydraulic motor through a worm reducer, so it can during the maintenance task of active electric power lines as a turn continuously. In fact, the end-effector is a gripper and result of today’s highly information-oriented society and its opening and closing movement is driven by a miniature increasing demand of electric utilities. The maintenance task has linear cylinder. The movement limits of each joint are listed the risk of electric shock and the danger of falling from high in Table 1. The movements of all the six joints and the grip- place. Therefore, it is necessary to realize an autonomous robot per are controlled by servo valves which are mounted on a system using electro-hydraulic manipulator because hydraulic manipulators have the advantage of electric insulation and high rotary actuator power-mass ratio [1, 2]. Dunnigan et al. [3-5] introduce several fourth joint fifth joint hydraulic manipulators working for underwater, forest and third joint construction tasks respectively. Taylor et al. [6, 7] introduce two hydraulic cylinder kinds of hydraulic manipulators for nuclear industries. Wang et al. [8] introduces a hydraulic manipulator developed in China rotary actuator but there is no application case given. Zhao Y. L. gives two hydraulic cylinder application cases of hydraulic manipulators using in electric hydraulic motor power line maintenance [9, 10]. sixth joint This paper gives the mechanical structure of a six DOF end-effector hydraulic manipulator in section 2, which is named HydM hydraulic cylinder and developed by Shandong University, China. In section 3, second joint rotary actuator hydraulic manifold *Address correspondence to this author at the School of Control Science first joint and Engineering, Shandong University, Jinan 250061, China; Tel: 0086- 531-88396813; Fax: 0086-531-88396813; E-mail: [email protected] Fig. (1). The mechanical structure sketch of HydM.

1874-155X/14 2014 Bentham Open 458 The Open Mechanical Engineering Journal, 2014, Volume 8 Rong et al. hydraulic manifold. HydM does not integrate with a p = s (a c + a c + a c − d s ) − d (s c s − c c ), hydraulic power unit, so it must be supplied hydraulic power y 1 2 2 3 23 4 234 5 234 6 1 234 5 1 5 from an external hydraulic power unit through high pressure p = a s + d + a s + a s + d c − d s s , hose. z 2 2 1 3 23 4 234 5 234 6 234 5

a3 a4 d6 where, z x y y 5 6 3 4 y5 si = sinθi , ci = cosθi , sij = sin(θi +θ j ), cij = cos(θi +θ j ), z 5 6 x3 x4 d θ a2 6 s = sin(θ +θ +θ ), c = cos(θ +θ +θ ). ijk i j k ijk i j k θ3 θ4 θ5 Table 1. Link parameters of HydM.

z1 y2 x2 Link No. ai-1 di Joint x1 αi-1 θi Limits θ2 i (mm) (mm) Variable

1 1 0 0 d1 θ1 θ1 (-90°, 90°) d θ 1 2 0 90° 0 θ2 θ2 (0°, 120°) z0 x0 3 a2 0 0 θ3 θ3 (-125°, -5°)

4 a3 0 0 θ4 θ4 (-50°, 70°) Fig. (2). The DH frames of HydM. 5 a4 -90° d5 θ5 θ5 (-150°, -30°) 3. KINEMATICS ANALYSIS OF HydM 6 0 -90° d6 θ6 θ6 (-180°, 180°) 3.1. The DH Frames and Link Parameters of HydM According to the DH rules [11], the global frame {O0} 3.3. Inverse Kinematics of HydM and local frames {Oi} fixed to each link can be build as shown in Fig. (2). For manipulators, inverse kinematics analysis is very important since it is the base of trajectory planning and real- Then the link parameters in DH frames of each link can time control. It is the best result to obtain analytical solutions be gained as listed in Table 1. The open intervals in last of the inverse kinematic problem for it can well meet the column show the motion range of each joint. requirements of real-time control. In the case of a serial 3.2. Forward Kinematics of HydM manipulator with six degrees of freedom, it must meet one of the following two conditions to have analytical solutions of The homogeneous transformation method based DH its inverse kinematic problem. One is that it has three rules is the most commonly used method to solve the adjacent parallel joints. The other is that it has three adjacent forward kinematics of manipulators or other complex serial joints whose axes intersect at one point. HydM meets the mechanisms. According to Fig. (1) and Table 1, it is easy to first condition, so we can obtain analytical solutions of its get the homogeneous transformation matrix between two inverse kinematics. adjacent links. The HydM manipulator has six joints and seven links including the mounting base, so there have six Generally, the position and posture of the end-effector homogeneous transformation matrices. The successive are given with homogeneous transformation matrix shown as product of these matrices describes the position and posture Eq. (2). And the vector [p! , p! , p!]T gives the position while of the end-effector in global frame. And it is also the forward x y z kinematic equation of the manipulator shown as Eq. (1). the 3×3 matrix at the upper left corner of 0 T! gives the 6 posture of the end-effector in global frame {O }. " n o a p % 0 $ x x x x ' " n o a p % 0 0 1 2 3 4 5 $ n o a p ' !x !x !x !x T = T ⋅ T ⋅ T ⋅ T ⋅ T ⋅ T = $ y y y y ' (1) $ ' 6 1 2 3 4 5 6 $ ' 0 n!y o!y a!y p!y $ nz oz az pz ' T6! = $ ' (2) $ ' n o a p # 0 0 0 1 & $ !z !z !z !z ' $ ' # 0 0 0 1 & n = c c c c − s s c + c s s , n = s c c c + c s c + s s s , x 1 234 5 6 1 5 6 1 234 6 y 1 234 5 6 1 5 6 1 234 6 According to the homogeneous transformation rules, the n = s c c − c s , o = −c c c s + s s s + c s c , given position and posture of the end-effector described in z 234 5 6 234 6 x 1 234 5 6 1 5 6 1 234 6 frame {O1} can be shown with Eq. (3). Simultaneously, the o = −s c c s − c s s + s s c , o = −s c s − c c , same position and posture of the end-effector can be shown y 1 234 5 6 1 5 6 1 234 6 z 234 5 6 234 6 with Eq. (4) due to the forward kinematics. ax = −c1c234s5 − s1c5 , ay = −s1c234s5 + c1c5 , $ n!c + n!s o!c + o!s a!c + a!s p!c + p!s ' & x 1 y 1 x 1 y 1 x 1 y 1 x 1 y 1 ) & ) (3) 1 0 −1 0 −n!s + n!c −o!s + o!c −a!s + a!c − p!s + p!c az = −s234s5 , px = c1(a2c2 + a3c23 + a4c234 − d5s234 ) − d6 (c1c234s5 + s1c5 ), T T T x 1 y 1 x 1 y 1 x 1 y 1 x 1 y 1 6! = 1 ⋅ 6! = & ) & n! o! a! p! − d ) & z z z z 1 ) 0 0 0 1 % ( Kinematics Analysis of a Hydraulically Actuated Manipulator The Open Mechanical Engineering Journal, 2014, Volume 8 459

1T = 1T 2 T 3T 4 T 5 T 6 2 3 4 5 6 a c a s a!z !x 1 + !y 1 " c c c + s s −c c s + s c −c s a c + a c + a c − d s − d c s % a s + a s = p! − d + a + d − d a! (18) $ 234 5 6 234 6 234 5 6 234 6 234 5 2 2 3 23 4 234 5 234 6 234 5 ' (4) 2 2 3 23 z 1 4 s 5 s 6 z $ ' 5 5 s5c6 −s5s6 c5 d6c5 = $ ' $ s234c5c6 − c234s6 −s234c5s6 − c234c6 −s234s5 a2s2 + a3s23 + a4s234 + d5c234 − d6s234s5 ' $ ' Define 0 0 0 1 # & a!c + a!s a! The Eqs. (3) and (4) are all describe the position and m = p!c + p!s + a x 1 y 1 − d z − d (a!c + a!s ) x 1 y 1 4 s 5 s 6 x 1 y 1 posture of the end-effector in frame {O1}, so the 5 5 corresponding elements of the two matrices should be equal a! a!c + a!s to each other. The three elements (1, 4), (2, 4) and (3, 4) in n = p! − d + a z + d x 1 y 1 − d a! z 1 4 s 5 s 6 z homogeneous matrix decide the position of end-effector, 5 5 while as elements (1, 3), (2, 3) and (3, 3) decide the posture of z axis of the end-effector in a given frame. But the three Then, Eqs. (5) and (7) can be re-written as elements (1, 3), (2, 3) and (3, 3) are not independent, so an a2c2 + a3c23 = m (19) element in other columns should be selected to decide the posture of x- or y-axis of the end-effector. The seven pairs of a2s2 + a3s23 = n (20) corresponding elements (1, 4), (2, 4), (3, 4), (1, 3), (2, 3), (3, 1 1 3) and (2, 1) in matrices T and T! are selected and listed We can get an equation only included c3 by adding the 6 6 as following: two sides square of Eq. (19) to those of Eq. (20) respectively. 2 2 2 2 a c + a c + a c − d s − d c s = p"c + p"s (5) 2a2a3c3 = m + n − a2 − a3 (21) 2 2 3 23 4 234 5 234 6 234 5 x 1 y 1 then d6c5 = − p"xs1 + p"yc1 (6) 2 2 2 2 m + n − a2 − a3 a s + a s + a s + d c − d s s = p" − d (7) θ3 = −arccos (22) 2 2 3 23 4 234 5 234 6 234 5 z 1 2a a 2 3 −c s = a"c + a"s (8) 234 5 x 1 y 1 Eqs. (19) and (20) can be transformed to the following two equations due to the trigonometric function algorithm. c5 = −a"xs1 + a"yc1 (9) (a + a c )c − a s s = m (23) 2 3 3 2 3 3 2 −s s = a" (10) 234 5 z a s c +(a + a c )s = n (24) 3 3 2 2 3 3 2 s5c6 = −n"xs1 + n"yc1 (11) From Eqs. (23) and (24)，we can eliminate s2 and get From Eqs. (6) and (9), we have !(a + a c )2 +(a s )2 #c = m(a + a c )+ na s (25) " 2 3 3 3 3 $ 2 2 3 3 3 3 (−a"xs1 + a"yc1)d6 = − p"xs1 + p"yc1 (12) then then m(a + a c )+ na s arccos 2 3 3 3 3 (26) p d a θ2 = 2 2 "y − 6 "y (a + a c ) +(a s ) θ = arctan (13) 2 3 3 3 3 1 p" − d a" x 6 x According to the homogeneous transformation method, From Eq. (9), we have the position and posture of the end-effector described in frame {O3} can be derived as the following matrix: θ =−arccos( −as′′+ ac ) 511xy (14) 3 3 4 5 T6 = T4 T5 T6 = Since sinθ ≠ 0 ，so Eqs. (8) and (10) can be re-written " c c c + s s −c c s + s c −c s a + a c − d s − d c s % 5 $ 4 5 6 4 6 4 5 6 4 6 4 5 3 4 4 5 4 6 4 5 ' (27) as $ s c c − c s −s c s − c c −s s a s + d c − d s s ' $ 4 5 6 4 6 4 5 6 4 6 4 5 4 4 5 4 6 4 5 ' s c s s c d c a"c + a"s $ − 5 6 5 6 − 5 − 6 5 ' c x 1 y 1 $ ' 234 = − (15) 0 0 0 1 s # & 5 3 p! = 2 T−1 ⋅ 1T−1 ⋅ 0 T−1 ⋅ 0 p! = a" 6 3 2 1 6 s = − z (16) 234 s $ −a c + p!c c + p!s c + p!s − d s ' 5 & 2 3 x 1 23 y 1 23 z 23 1 23 ) Substitute Eqs. (15) and (16) into Eqs. (5) and (7), then & a s − p!c s − p!s s + p!c − d c ) (28) & 2 3 x 1 23 y 1 23 z 23 1 23 ) & ) a!c + a!s a! p!xs1 − p!yc1 x 1 y 1 z (17) a2c2 + a3c23 = p!xc1 + p!ys1 + a4 − d5 − d6 (a!xc1 + a!ys1) & ) s5 s5 & 1 ) % ( 460 The Open Mechanical Engineering Journal, 2014, Volume 8 Rong et al.

Eq. (27) shows the given position of the end-effector to the position of the objective in global frame {O0}. The described in frame {O3}. It should be equal to the fourth posture of the end-effector can be given with z-y-z Euler column of the position and posture matrix deduced from the angles shown in Fig. (3) [11]. In fact, α and β have forward kinematics shown as Eq. (28). Therefore, we can get obvious geometric meanings and are easy to measure. α is two equations as below: the included angle between axis x and the projection line in 0 a3 +(a4 − d6s5 )c4 − d5s4 = −a2c3 + p"xc1c23 + p"ys1c23 + p"zs23 − d1s23 x0 y0 plane of axis z6 . β is the included angle between axes (29) z0 and z6 . (a − d s )s + d c = a s − p"c s − p"s s + p"c − d c (30) 4 6 5 4 5 4 2 3 x 1 23 y 1 23 z 23 1 23 From Fig. (3), it can be concluded that the frame {O6} is Define transformed from the original posture duplicated with the u = −a c − a + p"c c + p"s c + p"s − d s global frame {O0} through three rotation transformations. 2 3 3 x 1 23 y 1 23 z 23 1 23 The first one makes {O6} to rotate α around axis z to 6 v = a s − p"c s − p"s s + p"c − d c 2 3 x 1 23 y 1 23 z 23 1 23 {O!}. The second one makes {O!} to rotate β around axis 6 6 Then, Eqs. (29) and (30) can be re-written as y! to {O!!}. The last one makes {O!!} to rotate γ around z!! 6 6 6 6 (a4 − d6s5 )c4 − d5s4 = u (31) to {O!!!}. The posture of {O!!!} is the given posture of the end- 6 6 (a − d s )s + d c = v (32) effector which is shown as Eq. (37). 4 6 5 4 5 4 y From Eqs. (31) and (32), we can eliminate c and get zz,  6 y, y 4 06 66 2 2 zz,  "(a − d s ) + d $s = v(a − d s ) −ud (33)  66 # 4 6 5 5 % 4 4 6 5 5 y Then 0 v(a − d s ) −ud  θ = arcsin 4 6 5 5 (34) 4 (a − d s )2 + d 2 4 6 5 5 x0  x6 From Eq. (11), we can get θ as  6 x x6 6

−n#xs1 + n#yc1 Fig. (3). z-y-z Euler angles. θ6 = ±arccos (35) s5 0 R6! = Rot(z,α)Rot( y,β)Rot(z,γ) = According to the DH rules, the plus sign should be & cαcβcγ − sαsγ −cαcβsγ − sαcγ cαsβ ) selected for the right side of Eq. (35) if cos(x ⋅ z ) is ( + (37) 6 5 ( cαsβcγ + cαsγ −sαcβsγ + cαcγ sαsβ + nonpositive, otherwise the minus sign should be selected. ( + −sβcγ sβsγ cβ And cos(x ⋅ z ) can be derived from the following equation: '( *+ 6 5 " % 3.5. The Correctness Verification of the Inverse Solution " cos(x ⋅ x ) % n( $ 6 5 ' $ x ' The block diagram in Simulink for verifying the $ ' 4 T 3 T 2 T 1 T 0 T $ n ' cos(x6 ⋅ y5 ) = R5 R4 R3 R2 R1 (y = correctness of the inverse solution is shown in Fig. (4). $ ' $ ' cos(x ⋅ z ) $ n( ' Firstly, a certain number of sets of joint variables are #$ 6 5 &' # z & (36) produced randomly in their limits. Then the corresponding

" n( (c c c − s s )+ n( (s c c + c s )+ n(s c % $ x 1 234 5 1 5 y 1 234 5 1 5 z 234 5 ' $ −n( (c c s + s c ) − n( (s c s − c c ) − n(s s ' $ x 1 234 5 1 5 y 1 234 5 1 5 z 234 5 ' $ ' −n(xc1s234 − n(ys1s234 + n(zc234 # & 3.4. The Given Position and Posture of the End-Effector The end-effector should approach to the objective in an optional posture and a series of planned posture in processing procedure. So, the position and posture of the end-effector are known and they are generally given with a homogeneous transformation matrix shown as Eq. (2). The position vector [p! , p! , p!]T can be given directly according x y z Fig. (4). Block diagram for inverse solution verification.

Kinematics Analysis of a Hydraulically Actuated Manipulator The Open Mechanical Engineering Journal, 2014, Volume 8 461 positions and postures of the end-effector are calculated. The REFERENCES next procedure is to obtain the corresponding inverse [1] K. Ahn, and S. Yokota, “Application of discrete event control to solutions through the inverse kinematic equations. The last the insertion task of electric line using 6-link electro-hydraulic procedure is to compare the inverse solutions and the manipulators with dual arm”, JSME International Journal, Series random joint variables. If they are equal to each other within C: Mechanical Systems, Machine Elements and Manufacturing, a certain precision, it can be concluded that the inverse vol. 46, no. 1, pp. 263-269, 2003. [2] L. Q. Cai, “Hydraulic manipulator system for live electric line solutions are correct. operation”, Construction Machinery for Hydraulic Engineering & Power Stations, vol. 16, no. 2, pp. 42-48, 1994. (In Chinese) CONCLUSION [3] M. W. Dunnigan, D. M. Lane, A. C. Clegg, and I. Edwards, This paper gives a mechanism design of a six DOF “Hybrid position/force control of a hydraulic underwater manipulator”, IEE Proceedings: Control Theory and Applications, hydraulically actuated manipulator which has some obvious vol. 143, no. 2, pp. 145-151, 1996. advantages compared to traditional electric powered [4] P. M. La Hera, U. Mettin, S. Westerberg, and A. S. Shiriaev, manipulators. Those are with high power-mass ratio, electric “Modeling and control of hydraulic rotary actuators used in insulation, suitable for severe tasks, etc. So it can be used in forestry cranes”, In: Proceedings of the IEEE International underwater, electric power line maintenance, forest, Conference on Robotics and Automation, 2009, pp. 1315-1320. [5] R. F. Abo-Shanab, and N. Sepehri, “On dynamic stability of construction, nuclear and other industries. Then the manipulators mounted on mobile platforms”, Robotica, vol. 19, analytical solutions of its inverse kinematics are solved by no. 4, pp. 439-449, 2001. given the position and posture of the end-effector [6] C. J. Taylor, and D. Robertson, “State-dependent control of a simultaneously. The posture of the end-effector is given hydraulically actuated nuclear decommissioning robot”, Control Engineering Practice, vol. 21, no. 12, pp. 1716-1725, 2013. through three z-y-z Euler angles for they have obvious [7] M. J. Bakari, K. M. Zied, and D. W. Seward, “Development of a geometry meanings and are easy to be measured. The multi-arm mobile robot for nuclear decommissioning tasks”, solving method of inverse kinematics and the posture given International Journal of Advanced Robotic Systems, vol. 4, no. 4, method of the end-effector can provide important references pp. 387-406, 2007. to those with similar configuration to it. [8] X. Y. Wang, R. Liu, and G. Z. Jia, “On hydraulic manipulator control with generalization pulse code modulation”, Chinese CONFLICT OF INTEREST Journal of Mechanical Engineering, vol. 38 (supp.), pp. 203-206, 2002. (In Chinese) The authors confirm that this article content has no [9] Y. L. Zhao，S. Y. Lu，X. C. Lv，J. Li，and Z. L. Wang, conflict of interest. “Design of force feedback hydraulic arm for live working robot”, Chinese Hydraulics & Pneumatics, no. 8, pp. 31-34, 2013. (In ACKNOWLEDGEMENTS Chinese) [10] Y. L. Zhao，Q. Zhang，H. Qi，J. Li，and Y. C. Li, “Design of This work is supported by the High Technology Research non-force feedback hydraulic arm for live working”, Machinery and Development Program of China under grant no. Design & Manufacture, no. 9, pp. 117-119, 2012. (In Chinese) [11] J. J. Craig, Introduction to Robotics: Mechanics and Control (3rd 2012AA041506. Edition), Prentice Hall Press, 2004.

Received: March 5, 2014 Revised: May 2, 2014 Accepted: May 6, 2014

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