1. ECE 450 Introduction to Robotics Section: 50883 Instructor: Linda A. Gee 10/05/99 Lecture 10

2. 2 Geometric Approach to Inverse Kinematics Purpose: To generate an arm solution for the manipulator Example: PUMA-type Robot 6-link manipulator with rotary joints There are three configuration indicators: ARM ELBOW WRIST

3. Lecture 103 PUMA Robot *Fu, page 37

4. Lecture 104 Inverse Kinematics Solution Properties Two solutions are associated with the first three joints (i = 1, 2, 3) One solution is associated with the last three joints (i = 4, 5, 6) By solving the arm solution geometrically, a consistent solution is found

5. Lecture 105 Inverse Kinematics Solution of PUMA Robot For a 6-axis PUMA robot arm Four solutions exist for the first three joints (i = 1, 2, 3) Two solutions exist for the last three joints (i = 4, 5, 6)

6. Lecture 106 Inverse Kinematics Solution cont’d The first two indicators lead to a solution (1 of 4) for the first three joints ARM ELBOW Third indicator leads to a solution (1 of 2) for the last three joints WRIST Arm configuration indicators are specified by the user for finding the inverse transform

7. Lecture 107 Inverse Kinematic Solution cont’d Calculate the solution in two steps: Derive a position vector that points from the shoulder to the wrist –Obtain a solution to the joints (i = 1, 2, 3) by examining the projection of the position vector onto the x i-1 y i-1 plane Solve for the last three joints by using the calculated joint solution from the first three joints –Use orientation submatrices 0 T i and i-1 A i (i = 4, 5, 6) –Projection of link coordinate frames onto x i-1 y i-1 plane

8. Lecture 108 Inverse Kinematic Solution cont’d Given ref T tool, it is possible to find 0 T 6 by premultiplying ref T tool by B -1 postmultiplying ref T tool by H -1 Apply the joint-angle solution to 0 T 6 0T60T6  T = B -1 ref T tool H -1 = n x s x a x p x n y s y a y p y n z s z a z p z 0 0 0 1

9. Lecture 109 Definition of Arm Configurations RIGHT (shoulder) ARM –positive  2 moves the wrist in the positive z 0 direction while joint 3 is inactive LEFT (shoulder) ARM –positive  2 moves the wrist in the negative z 0 direction while joint 3 is inactive

10. Lecture 1010 Arm Configurations cont’d ABOVE ARM (elbow above wrist) –position of wrist of RIGHT/LEFT arm w.r.t. shoulder coordinate system has negative/positive coordinate value along the y 2 axis BELOW ARM (elbow below wrist) –position of wrist of RIGHT/LEFT arm w.r.t. shoulder coordinate system has negative/positive coordinate value along the y 2 axis

11. Lecture 1011 Arm Configurations con’td WRIST DOWN –the s unit vector of the hand coordinate system and the y 5 unit vector of (x 5, y 5, z 5 ) coordinate system have a positive dot product WRIST UP –the s unit vector of the hand coordinate system and the y 5 unit vector of (x 5, y 5, z 5 ) coordinate system have a negative dot product

12. Lecture 1012 Arm Configurations and Solutions Two indicators are defined for each arm configuration ARM ELBOW –Combine these to yield one solution of four possible for the first three joints Third indicator WRIST –gives one solution of two possible for the last three joints

13. Lecture 1013 Indicator Definitions ARM +1: RIGHT arm -1: LEFT arm ELBOW +1: ABOVE arm -1: BELOW arm WRIST +1: WRIST DOWN -1: WRIST UP FLIP +1: Flip wrist orientation -1: Remains stationary

14. Lecture 1014 Arm Configurations *Fu, Page 63

15. Lecture 1015 Arm Solution for the First Three Joints (i = 1, 2, 3) For the PUMA robot, Define a position vector, p, that points from the origin of the shoulder coordinate system (x 0, y 0, z 0 ) to the point of intersection of the last three joints p = p 6 - d 6 a = (p x, p y, p z ) T which represents the position vector 0 T 4

16. Lecture 1016 Hand Coordinate System *Fu, page 43

17. Lecture 1017 Arm Solution for the First Three Joints cont’d = C 1 (a 2 C 2 + a 3 C 23 + d 4 S 23 ) - d 2 S 1 S 1 (a 2 C 2 + a 3 C 23 + d 4 S 23 ) +d 2 C 1 d 4 C 23 - a 3 S 23 - a 2 S 2 pxpypzpxpypz Position vector 0 T 4 :

18. Lecture 1018 Joint 1 Solution Method Project p onto the x 0 y 0 plane Solve for  1 in terms of sin  1 and cos  1  1 = tan -1 (sin  1 /cos  1 )

19. Lecture 1019 Joint 1 Solution Setup (p x, p y ) B x0x0 y0y0 O     A 11 L =  -  X 1X 1 L Z 1Z 1 L radius = d2 (p x, p y ) B x0x0 y0y0 O     A X 1X 1 R Z 1Z 1 R 11 R =  +  +  Left Arm Right Arm