ICIC Express Letters Part B: Applications ICIC International c 011 ISSN 185-766 Volume, Number, April 011 pp 415 40 DYNAMIC MODEL FOR AN ARTICULATED MANIPULATOR Luis Arturo Soriano, Jose de Jesus Rubio, Salvador Rodriguez and Cesar Torres Seccion de Estudios de Posgrado e Investigacion ESIME Azcapotzalco Instituto Politecnico Nacional Av de las Granjas no 68, Col Santa Catarina, CP 050, Mexico DF, Mexico larturosoriano@gmailcom; jrubioa@ipnmx Received July 010; accepted October 010 Abstract The mathematical models of robotic arms describe the relationship between force or torque and motion The equations of motion are important to consider in the design of robotic arms, in simulation and animation of robotic arm motion and in the design of control algorithms avoiding the necessity to build a prototype of a real robotic arm The major contribution of this paper is to present an interesting method to obtain the dynamics of an articulated robotic arm Keywords: Dynamic model, Articulated manipulator 1 Introduction Now, the presition in the homeworks is required in robotics [] which is applied in the manufacture and the education [1] and in other repetitive homeworks made by humans in the past [] These homeworks are frequently with a defined trajectory [5], that is way it is an open research in robotics [4, 5, 10, 13] which present interesting results The homeworks in the education [1] and in the medicine [] are improved using a dynamic model [4, 5, 10, 13] The mathematical model of robotic arms describes the relationship between force or torque and motion There are some books that present dynamic models as are [3, 8, 9, 11, 14], the method of [6] is interesting and is different to the others, however, in [6] they do not apply their method to an articulated robotic arm, in this paper, this method is applied to an articulated robotic arm In this paper, the method of [6] is applied to obtain the dynamics of an articulated robotic arm Dynamic Model The method is based on the Lagrangian L = K P (1 where K is the total kinetic energy of the system and P is the total potential energy of the system θ i is considered for the rotatory joints and d i for the prismatic joints and τ i for the applied force moment of the joint i The equation of Lagrange-Euler is the following: ( τ i = d L dt θ i L θ i ( To get to the form (, we needed to make use of the homogeneous transformation matrix 1 Velocity of the joints arm The formulation of Lagrange-Euler knowledge needs the kinetic energy of the physical system, which in turn requires a knowledge of the velocity of every joint 415
416 L A SORIANO, J DE JESUS RUBIO, S RODRIGUEZ AND C TORRES i r i is one fixed and point at rest in the element i and expressed in homogeneous coordinates regarding the element s coordinate system i i r i = [ x i y i z i 1 ] T This coordinates of point i with respect to the system {i} The aforementioned point (any point in the link i is motionless from the system {i}, but not as i With reference to {0} the point is: 0 r i = 0 A i i r i (4 To obtain the derivative 0 r i with the time, the velocity of each link is obtained with respect base coordinates system: 0 v i = v i = d dt ( 0 r i = d dt (0 A i i r i = ( j=1 0 A i qj (3 i r i (5 The form compact of the Equation (5 is obtained because iṙ i = 0 The partial derivative of 0 A i with can be obtained easily with the help of a matrix D i, for the case of the revolution articulation is defined as: D i = 0 1 0 1 1 0 0 1 (6 0 1 0 1 0 Therefore, i 1 A i = D i i 1 A i = We will have then for i = 1,,, n and j i: 0 1 0 1 1 0 0 1 0 1 0 1 0 i 1 A i (7 0 A i = 0 A 1 1 A j A j 1 D j j 1 A j i 1 A i (8 And for j i, the derivative will be equal to zero This equation reflects the effect of the motion of the joint j in the link i If a link j > i (eg, farther from the base than the link i in the chain of links the effect in i will be zero, because i it will not be moved due to the motion of j For simplify the notation let s define B ij = 0 A i of mode that for j i: B ij = 0 A j 1 D j 1 j A i (9 And for j > i, B ij = 0 Using the Equation (5 ( v i = B i ij qj r i (10 B ij q k = B ijk = j=1 For calculating the kinetics energy of system, also it is needded the partial derivative of B ij with respect to k : 0 A j 1 D j 1 j A k 1 D k 1 k A i i k j 0 A k 1 D k 1 k A j 1 D j 1 j A i i j k 0 i < j or i < k And for i < j or i < k the partial derivative is equal to zero This relations are because of the movements of the links j and k over the link i From the Equation (10, the velocities of the 3 links of manipulator are obtained as: v 1 = (B 11 q1 1 r 1, v = (B 11 q1 + B 1 q + B 1 q1 + B q r (11 v 3 = (B 11 q1 + B 1 q + B 13 q3 + B 1 q1 + B q + B 3 q3 + B 31 q1 + B 3 q + B 33 q3 3 r 3 (1
ICIC EXPRESS LETTERS, PART B: APPLICATIONS, VOL, NO, 011 417 Kinetic energy From the second law of Newton, the kinetic energy of a particle in movement give for e = 1 mv In this case, if K i is the kinetic energy of link i and dk i the one particle with mass dm in this link, dk i = 1 (ẋ i + ẏi + żi 1 dm = tr ( v i vi T dm (13 where tra = n a ii in others words is the sum of the diagonal elements of the matrix A, on the other hand using (10 in (13: ( dk i = 1 T ( tr B i il ql r i B i ir qr r i dm = 1 tr B i il r i dm i ri T B T irq l qr (14 B ij represents the change reason of the points i r i (in the link i related with the reference coordinate because of the change of It means that this matrix is the same for all the points in the link i (that is, it is constant in all the points and it is independent to dm At the same way the velocity dq i is the same in all the points of the link i, then to dt obtain the total kinetic energy K i of the link i by solving the integral of dk i, we can let the factors in (14 out of the integral and solving it inside of the double add ( K i = dk i = 1 ( tr B i il r i i ri T dm B T irq l qr (15 For i r i i ri T dm of the Equation (15, it is have that i r i = [ x i y i z i 1 ] T of (3 then the product of i r i i ri T is equal to: x i i r i i ri T = y i [ z i xi y i z i 1 ] x i x i y i x i z i x i = x i y i yi y i z i y i x i z i y i z i zi z i (16 1 x i y i z i 1 Each one of the elements of (16 is multiplied by the scalar dm and using the rules of integral of matrices, each element of the matrix may be solved as an integral x i dm x i y i dm x i z i dm x i dm J i = xi y i dm yi dm y i z i dm y i dm xi z i dm y i z i dm zi dm z i dm xi dm y i dm z i dm (17 dm where J i is depended of i r i Therefore, it is independent of the motion of the links and it is necessary to calculated it once And pseudo-inertia of the link is called i With (17 pseudo-inertias matrix for each link is calculated For the link 1: [ ] 03 3 0 J 1 = 1 3 (18 0 3 1 m 1 where the matrix J 1 is considered only the element J 44 because is suppose than every mass this concentrated at the source {1}, and is calculated from now on: dm = m1 For the second link is: 1 3 J = L m 0 1 m L 0 1 0 0 1 (19 m L 0 1 m where, x dm = m L 0 L x dx = m L x 3 3 0 L= m L 3 L 3 = m L 3, dm = m
418 L A SORIANO, J DE JESUS RUBIO, S RODRIGUEZ AND C TORRES Similarly form for the link 3: 1 3 J 3 = L m 3 0 1 m 3L 0 1 0 0 1 (0 m 3L 0 1 m 3 Now, the total kinetic energy of arm K is: ( K i = K i = 1 tr Developing the Equation (1, we have: K = n Ki = 1 [( Bil J i Bir T ql ] qr (1 tr ( B il J i Bir T ql qr ( For n = 1,, 3, l = 1,, 3 and r = 1,, 3: tr ( B 11 J 1 B T 11 θ1 θ1 + tr ( B 1 J B T 1 θ1 θ1 + tr ( B 1 J B T θ1 θ K = 1 + tr ( B J B T 1 θ θ1 + tr ( B J B T θ θ + tr ( B 31 J 3 B T 31 θ1 θ1 + tr ( B 31 J 3 B3 T θ1 θ + tr ( B 31 J 3 B33 T θ1 θ3 + tr ( B 3 J 3 B31 T θ θ1 + tr ( B 3 J 3 B3 T θ θ + tr ( B 3 J 3 B33 T θ θ3 + tr ( B 33 J 3 B31 T θ3 θ1 + tr ( B 33 J 3 B T 3 θ3 θ + tr ( B 3 J 3 B T 33 θ3 θ3 3 Potential energy The potential energy of joint is P i and the total manipulator P we have: P i = m i g 0 r i = m i g ( 0 A i i r i (4 In this equation, g is the vector of gravity, g = [ g x g y g z 0 ] is the system of the base The specific value of the vector g depends of how is the contact of the robot If the base is over the floor and the axe Z 0 is normal to it, g = [ 0 0 981 0 ] The total potential energy is the sum of all the energy of each link: P = P i = m i g ( 0 A i i r i (5 The potential energy of arm for n = 3: P = m 1 g ( 0 A 1 1 r 1 m g ( 0 A r m3 g ( 0 A 3 3 r 3 4 The lagrangian and the dynamics model Now, they have the necessary data to obtain the Lagrangian, L, and the equation of Euler-Lagrange Replacing ( and (5 in (1 gives: L = 1 tr ( B il J i Bir T ql qr + m i g ( 0 A i i r i Remember that l > i, B il = 0, of mode that the Lagrangian is simplified considerably Now, using (7 in ( and remembering the definitions of the B The derivation of L L whit respect to q x is: tr ( B ix J i Bik T qx If B il = 0 for l > i, and the q x = 1 l=1k=1 derivative with respect to t, we have d dt ( L q x = n l=ik=1 l tr ( B ik J l Bli T (3 (6 (7 q k The factor 1
ICIC EXPRESS LETTERS, PART B: APPLICATIONS, VOL, NO, 011 419 disappear since it has to derive partially with respect to i and to k By means of similar considerations for L q x gives the expression: { l τ i = tr ( B ip J l Bli T l l q p + tr ( } B ipq J l Bli T qp q q m l gb l li r l (8 l=i p=1 p=i q=1 For the arm of tree degree of freedom 3 equations are required for the moments, this equations can denoted in matrix form and in form more compact using the next few definitions: τ(t is the vector of 3 1 the moments applied in the joints q(t is the vector of 3 1 the variables of link, q(t, q(t, fist and second derivative of the q(t with respect to time M(q is the symmetric matrix of 3 3 that depends of the inertias and with elements gives: M ik = j=max(i,k tr ( B jk J j Bji T i, k = 1,, 3,, n (9 C(q, q is the 3 1 no lineal vector that depends of the Coriolis and Centrifuge force, we have: C i = c ikm q k q m i = 1,, 3,, n (30 where, c ikm = j=max(1,k,m k=1m=1 tr ( B jkm J j Bji T i, k, m = 1,, 3,, n (31 G(q is the 3 1 vector that is obtained by severely exited force on the robot Each of the elements is given as: G i = m j g B j ji r j i = 1,, 3,, n (3 j=i using this definitions in (8 the equation τ i gives τ i = n developing the Equation (9 gives M ik k=1 q k + c i ( q k q m + G i (q i or: τ i = M(q q(t + C(q, q + G(q (33 M(θ = [M ij ] R 3 3, i = 1, 3, j = 1, 3 (34 substituting values gives M 11 = 1 6 L (m +4m 3 +3m 3 cos θ 3 +3m 3 cos (θ + θ 3 +m cos θ +3m 3 cos θ + m 3 cos (θ + θ 3, M 1 = M 13 = M 1 = M 31 = 0, M = 1 3 L (m + 4m 3 +3m 3 cos θ 3, M 3 = M 3 = 1 6 L m 3 (3 cos θ 3 +, M 33 = 1 3 L m 3 On the other hand develop the Equation (30 we have: C(θ, θ = [ C 1 C C 3 ] T (35 where C = 1 6 L (6m 3 sin (θ + θ 3 + m sin θ + 6m 3 sin θ + m 3 sin (θ + θ 3 θ 1 θ, C 1 = 0 and C 3 = 1 3 L m 3 (3 sin θ + sin (θ + θ 3 + 3 sin (θ + θ 3 θ 1 θ 1 48 L m 3 (16 sin (θ + θ 3 + 4 sin θ 3 + 4 sin (θ + θ 3 θ 1 θ3 Finally, developing the Equation (3 we have: G(θ = [ G 1 G G 3 ] T (36 where G 1 = 0, G = 1 Lg(m 3 cos (θ + θ 3 + m cos θ + m 3 cos θ and G 3 = 1 Lgm 3 cos (θ + θ 3
40 L A SORIANO, J DE JESUS RUBIO, S RODRIGUEZ AND C TORRES 3 Conclusion In this paper, we obtain the equations for the robot s dynamic Model articulated through the formulation of [6], the equations are very useful for understanding the functioning of the robot and can be compared with existing ones Acknowledgment The authors thank the Secretaria de Investigación y Posgrado and the Comisión de Operación y Fomento de Actividades Académicas del IPN and the Consejo Nacional de Ciencia y Tecnologia for their help in this research REFERENCES [1] G Capi, Application of recurrent neural controllers for robot complex task performance, International Journal of Innovative Computing, Information and Control, vol5, no5, pp1171-1178, 009 [] M-K Chang and T-H Yuan, Experimental implementations of adaptive self-organizing fuzzy sliding mode control to 3-DOF rehabilitation robot, International Journal of Innovative Computing, Information and Control, vol5, no10(b, pp3391-3404, 009 [3] J J Craig, Robótics, Prentice Hall [4] Y Dai, M Konishi and J Imai, Rnn-based cooperative motion control of -DOF robot arms, International Journal of Innovative Computing, Information and Control, vol3, no4, pp937-95, 007 [5] M M Fateh and M R Soltanpour, Robust task-space control of robot manipulators under imperfect transformation of control space, International Journal of Innovative Computing, Information and Control, vol5, no11(a, pp3949-3960, 009 [6] K S Fu, R C González and C S G Lee, Robotics: Control, Detection, Vision and Intelligence, McGraw-Hill [7] P Groover, Robótica Industrial, McGraw-Hill, Madrid, 1989 [8] R Kelly and V Santibañez, Movement Control Manipulator Robots, Prentice Hall, 003 [9] F L Lewis, D M Dawson and C T Abdallah, Robot Manipulator Control Theory and Practice, nd Edition, Marcel Dekker [10] K Najim, E Ikonen and E Gomez-Ramirez, Trejectory tracking control based on a general geological decision tree controller for robot manipulators, International Journal of Innovative Computing, Information and Control, vol4, no1, pp53-6, 008 [11] R M Murray, Z Li and S S Sastry, A Mathematic Robot Modeling and Control [1] R Radharamanan and H E Jenkins, Laboratory learning modules on CAD/CAM and robotics engineering education, International Journal of Innovative Computing, Information and Control, vol4, no, pp433-443, 008 [13] J J Rubio and L A Soriano, An asymptotic stable proportional derivative control with sliding mode compensation and with a high gain observer for robotic arms, International Journal of Innovative Computing, Information and Control, vol6, no10, pp4513-456, 010 [14] M W Spong and M Vidyasagar, Robot Dynamics and Control, John Wiley & Sons