ECE 450 Introduction to Robotics

Section: 50883

Instructor: Linda A. Gee

10/05/99

Lecture 10

Lecture 10 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

Lecture 10 3

PUMA Robot

*Fu, page 37

Lecture 10 4

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

Lecture 10 5

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)

Lecture 10 6

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

Lecture 10 7

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 xi-1 yi-1 plane

• Solve for the last three joints by using the calculated joint solution from the first three joints

– Use orientation submatrices 0Ti and i-1Ai (i = 4, 5, 6)

– Projection of link coordinate frames onto xi-1 yi-1 plane

Lecture 10 8

Inverse Kinematic Solution cont’d

• Given refTtool, it is possible to find 0T6 by

• premultiplying refTtool by B-1

• postmultiplying refTtool by H-1

• Apply the joint-angle solution to 0T6

0T6 T = B-1 refTtool H-1 =nx sx ax px

ny sy ay py

nz sz az pz

0 0 0 1

Lecture 10 9

Definition of Arm Configurations

• RIGHT (shoulder) ARM– positive 2 moves the wrist in the positive z0

direction while joint 3 is inactive

• LEFT (shoulder) ARM– positive 2 moves the wrist in the negative z0

direction while joint 3 is inactive

Lecture 10 10

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 y2 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 y2 axis

Lecture 10 11

Arm Configurations con’td

• WRIST DOWN– the s unit vector of the hand coordinate system

and the y5 unit vector of (x5, y5, z5) coordinate system have a positive dot product

• WRIST UP– the s unit vector of the hand coordinate system

and the y5 unit vector of (x5, y5, z5) coordinate system have a negative dot product

Lecture 10 12

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

Lecture 10 13

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

Lecture 10 14

Arm Configurations

*Fu, Page 63

Lecture 10 15

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 (x0, y0, z0) to the point of intersection of the last three joints

p = p6 - d6 a = (px, py, pz)T

which represents the position vector 0T4

Lecture 10 16

Hand Coordinate System

*Fu, page 43

Lecture 10 17

Arm Solution for the First Three Joints cont’d

=C1(a2C2 + a3C23 + d4S23) - d2S1

S1(a2C2 + a3C23 + d4S23) +d2C1

d4C23 - a3S23 - a2S2

px

py

pz

Position vector 0T4 :

Lecture 10 18

Joint 1 Solution Method

• Project p onto the x0y0 plane

• Solve for 1 in terms of sin 1 and cos 1

1 = tan-1 (sin 1/cos 1)

Lecture 10 19

Joint 1 Solution Setup

(px, py)B

x0

y0

O

A

1

L= -

X 1L

Z 1Lradius = d2

(px, py) B

x0

y0

O

A

X 1R

Z 1R

1

R= + +

Left Arm

Right Arm