3DOF

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KINEMATICS

ANALYSISOF

ROBOTS

(Part 3)

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This lecture continues the discussion on the analysis of theforward and inverse kinematics of robots.

After this lecture, the student should be able to:

Solve problems of robot kinematics analysis using transformation

matrices

Kinematics Analysis of Robots III

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Example: A 3 DOF RRR Robot

Link and Joint Assignment

Link (2) Link (3)Link (1)

Revolute joint

Link (0)

Revolute joint

Revolute joint

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Frame Assignment

Z1

Z1

Y1

Y1

X1

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Frame Assignment

Y0, Y1

X0, X1

Z0, Z1

Z2

Z2

X2

Y2

Y2

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Frame Assignment

Y0, Y1

X0, X1

Z0, Z1

Z2

X2

Y2

Z3

Z3

X3

Y3

Y3

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Frame Assignment

Y0, Y1

X0, X1

Z0, Z1

Z2

X2

Y2

Z3

X3

Y3

1

2 3

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Tabulation of D-H parameters

Y0, Y1

X0, X1

Z0, Z1

Z2

X2

Y2

Z3

X3

Y3

A

B

0= (angle from Z0to Z1measured along X0) = 0

a0= (distance from Z0to Z1measured along X0) = 0

d1= (distance from X0to X1measured along Z1)= 0

1= variable (angle from X0to X1measured along Z1)

1= 0(at home position) but 1can change as the arm moves

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Y0, Y1

X0, X1

Z0, Z1

Z2

X2

Y2

Z3

X3

Y3

A

B


a1= (distance from Z1to Z2measured along X1) = Ad2= (distance from X1to X2measured along Z2) = 0


2= 0(at home position) but 2can change as the arm moves

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Y0, Y1

X0, X1

Z0, Z1

Z2

X2

Y2

Z3

X3

Y3

A

B


a2= (distance from Z2to Z3measured along X2) = Bd3= (distance from X2 to X3measured along Z3) = 0


3= 0(at home position) but 3 can change as the arm moves

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Link i Twist i Linklength aiLink offsetdi

Joint angle i

i=0 0 0

i=1 90 A 0 1

(1=0at home

position)

i=2 0 B 0 2

(2=-0at home

position)

i=3 0 3

(3=-0at home

position)

Summary of D-H parameters

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Tabulation of Transformation Matrices from the D-H table

1000

)cos()cos()cos()sin()sin()sin(

)sin()sin()cos()cos()sin()cos(0)sin()cos(

1111

1111

1

1

iiiiiii

iiiiiii

iii

i

id

da

T

1000

010000)cos()sin(

00)sin()cos(

11

11

01

T

0,0,0 100 da

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0,,90 211 dAa

1000

00)cos()sin(0100

0)sin()cos(

22

22

12

A

T

1000



1111

1111

1

1

iiiiiii

iiiiiii

iii

i

id

da

T

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0,,0 322 dBa

1000

0100

00)cos()sin(

0)sin()cos(

33

33

2

3

B

T

1000



1111

1111

1

1

iiiiiii

iiiiiii

iii

i

id

da

T

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Forward Kinematics

1000

00)cos()sin(

)sin()cos()sin()sin()cos()sin(

)cos()sin()sin()cos()cos()cos(

22

112121

112121

1

2

0

1

0

2

A

A

TTT

1000

)sin(0)cos()sin(

)sin())cos(()cos()sin()sin()cos()sin(

)cos())cos(()sin()sin()cos()cos()cos(

23232

121321321

121321321

2

3

0

2

0

3

B

BA

BA

TTT

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Y0, Y1

X0, X1

Z0, Z1

Z2

X2

Y2

Z3

X3

Y3

A=3B=2 C=1

P


What is the position

of point P at the

home position?

Solution:

1

0

0

1

1

3p

11

30

3

0 pT

p

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1

00

1

1000

)sin(0)cos()sin()sin())cos(()cos()sin()sin()cos()sin(


123232

121321321

121321321

3

0

3

B

BA

BA

pT

1= 2= 3=0, A=3, and B=2:

1

0

0

6

1

0

0

1

1000

0010

0100

5001

11

3

0

3

0 pT

p

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Inverse Kinematics

Given the orientation and position of point P:

TTTTpaon

paon

paon

zzzz

yyyy

xxxx

03

23

12

01

1000

1

0

0

1

1000

)sin(0)cos()sin(

)sin())cos(()cos()sin()sin()cos()sin(


123232

121321321

121321321

3

0

3

B

BA

BA

pT

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Inverse Kinematics

Equate elements (1,3) and (2,3):

y

x

y

x

a

a

a

a1

1

1

1tan

)cos(

)sin(

Provided that 1)( 22 yx aa


Ap

BBAp

ApB

BAp

y

y

xx

)cos(

1)sin()sin())cos((

)cos(1)cos()cos())cos((

1

212

1

212

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Inverse Kinematics

If cos(1)0, then use pxto find cos(2). Afterwards, find

Otherwise use pyto find sin(2) and then solve using

)cos(

)sin(

tan

)(cos1)sin(

2

21

2

2

2

2

)cos(

)sin(tan

)(sin1)cos(

2

21

2

2

2

2

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Inverse Kinematics


z

z

z

z

o

n

o

n1

32

32

32tan)(

)cos(

)sin(

2323

Now find 1, 2, and 3given the orientation and position of point P:

1000

1001

01005010

1000

zzzz

yyyy

xxxx

paon

paonpaon

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Inverse Kinematics

0tan1

0 11

y

x

y

x

a

aaa

Now cos(1)=10. We use pxto find cos(2):

1)cos(

1)cos(

2,3,5

1

2

Ap

B

BAp

x

x

0)cos(

)sin(tan

0)(cos1)sin(

2

21

2

2

2

2

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Inverse Kinematics90tan)(

0

11

32

z

z

z

z

o

n

o

n

900902323

Y0

X0

Z0

Z3

X3

Y3

1000

1001

0100

5010

1000

zzzz

yyyy

xxxx

paon

paon

paonA=3 B=2

PC=1

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Forward & Inverse Kinematics Issues

Given a set of joint variables, the forward kinematics will always

produce an unique solution giving the robot global position and

orientation.

On the other hand, there may be no solution to the inverse

kinematics problem. The reasons include:

The given global position of the arm may be beyond the robotwork space

The given global orientation of the gripper may not be possible

given that the gripper frame must be a right hand frame

For the inverse kinematics problem, there may also exist multiple

solutions, i.e. the solution may not be unique.

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Forward & Inverse Kinematics Issues

Example of multiple solutions given the same gripper global

position and orientation:

First solution

Second solution

First solution

Second solution

Some solutions may not be feasible due to obstacles in the workspace

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Summary

This lecture continues the the discussion on the analysis of the

forward and inverse kinematics of robots.

The following were covered:

Problems of robot kinematics analysis using transformation

matrices

3DOF

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