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ECE 450 Introduction to Robotics Section: 50883 Instructor: Linda A. Gee 10/07/99 Lecture 11.

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Presentation on theme: "ECE 450 Introduction to Robotics Section: 50883 Instructor: Linda A. Gee 10/07/99 Lecture 11."— Presentation transcript:

1 ECE 450 Introduction to Robotics Section: 50883 Instructor: Linda A. Gee 10/07/99 Lecture 11

2 2 Joint 2 Solution Method Project p onto the x 1 y 1 plane Use ARM and ELBOW indicators Solve for  2 in terms of sin  2 and cos  2  2 = tan -1 (sin  2 /cos  2 ) where -    2  

3 Lecture 113 Arm Configurations *Fu, Page 63

4 Lecture 114 Joint 2 Solution Setup *Fu, Page 67

5 Lecture 115 Joint 3 Solution Method Project p onto the x 2 y 2 plane Define ( 2 p 4 ) y Define  3 =  -  Solve for  3 in terms of sin  3 and cos  3  3 = tan -1 (sin  3 /cos  3 ) where -    3  

6 Lecture 116 Joint 3 Solution Setup *Fu, Page 68

7 Lecture 117 PUMA Robot *Fu, page 37

8 Lecture 118 Arm Solution for the Last Three Joints (i = 4,5,6) Having solved for the first three joint angles (  1,  2,  3 ), defines the transformation matrix 0 T 3 Set joint 4 so rotation about joint 5 aligns with axis of motion of joint 6 with the approach vector, a »z 4 =  (z 3 x a)/  z 3 x a  Set joint 5 to align axis of motion of joint 6 with a »a = z 5 Set joint 6 to align orientation vector, s = y 6 and n »s = y 6

9 Lecture 119 Joint 4 Solution Setup x3x3 44 44 x4x4 y3y3 z4z4 *wrist orientation affects sign sin  4 = - (z 4 x 3 ) cos  4 = z 4 y 3

10 Lecture 1110 Hand Coordinate System *Fu, page 43

11 Lecture 1111 Joint 5 Solution Setup 55 55 x4x4 cos  5 = - (a y 4 ) sin  5 = a x 4 y4y4 a n

12 Lecture 1112 Joint 6 Solution Setup x5x5 66 66 n y5y5 s cos  6 = s y 5 sin  6 = n y 5 *align orientation of gripper to ease picking up object: s = y 6

13 Lecture 1113 Decision Equations for Arm Configuration Indicators For the PUMA robot arm, there are 8 solutions to the inverse kinematic problem –First three joint solutions (  1,  2,  3 ) positions the arm –Last three joint solutions (  4,  5,  6 ) provides hand orientation Four solutions to the first three joints: –RIGHT shoulder –LEFT shoulder FLIP toggle yields balance of solutions

14 Lecture 1114 Decision Equations for Arm Indicators ARM = sign(-d 4 S 23 -a 3 C 23 -a 2 C 2 ) +1 RIGHT arm -1 LEFT arm ELBOW = ARMsign(d 4 C 3 - a 3 S 3 ) +1 ELBOW above wrist -1 ELBOW below wrist WRIST = sign(sz 4 )  0 = +1 WRIST DOWN sign(nz 4 )=0 = -1 WRIST UP

15 Lecture 1115 Decision Equations for Arm Indicators cont’d Arm configuration indicators can be determined from joint angles Useful for providing verification of the arm solution

16 Lecture 1116 Computer Simulation Useful to verify and validate the inverse kinematics solution Software generates the allowable locations in the workspace given the joint angle limitations Joint angles are used to produce T (arm matrix) and used to compute the decision equations which yield three arm indicators

17 Lecture 1117 System Flow for Joint Solution Direct Kinematics Decision Equations Inverse Kinematics  + - Error Joint angles Position and orientation of end-effector ARM, ELBOW, WRIST T n s a p 0 0 0 1 For T =

18 Lecture 1118 Summary General Robotic Control Object description Direct Kinematics Robot arm parameters: links, joints 4x4 Transformation Matrices Inverse Kinematics Euler angle solution Inverse transformation Geometric approach Arm indicators


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