intelligent robotic chapter 4

7/31/2019 Intelligent Robotic CHAPTER 4

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Between two frames we have a kinematic relationship -basically a translation and a rotation.

This relationship is mathematically represented by a 4 4Homogeneous Transformation Matrix.

z

y

x


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x

y 3 1 Translation

z

r1 r2 r3

r4 r5 r6

r7 r8 r9

3

3 RotationalMatrix

0 0 0

1 3 Perspective

1 Global Scale

Homogeneous transformations:


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4

Base

Frame

Tool Frame

x

Goal Frame

Link Frame

Camera Frame


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We are interested in two kinematics topics:

Forward Kinematics -angles to position (See Chapters 2 & 3)What you are given: The length of each link AND the angle of each joint

What you can find: The position of any point (i.e. its (x, y, z) coordinates)OR

(Given the joint angles for the robot, what is the orientation and position of the endeffector)

Inverse Kinematics - position to angles

What you are given: The length of each link and the position of some point on the robotWhat you can find: The angles of each joint needed to obtain that position.

OR

(Given a desired end effector position what are the joint angles to achieve this)


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For a robot system the inverse kinematic problem isone of the most difficult to solve.

The robot controller must solve a set of non-linearsimultaneous equations.

The problems can be summarised as:

The existence of multiple solutions.

The possible non-existence of a solution.

Singularities.


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Goal

This two link planar manipulator has two possible solutions.This problem gets worse with more Degrees of Freedom.

Redundancy of movement.


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Goal

A goal outside the workspace of the robot has no

solution.

An unreachable point can also be within the

workspace of the manipulator - physical constraints.

A singularity is a place of acceleration - trajectory

tracking.


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2

1

(x , y)

l2

l1

Given: l1, l2,x , y

Find:1, 2

(x , y)

Inverse Kinematics of a Two Link Manipulator

Redundancy: A unique solution to this problem doesnot exist. Notice, that using the givens two solutions are

possible. Sometimes no solution is possible.


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The Geometric Solution

l1

l2

(x , y)

2

1

Using the Law of Cosines:

21

2

2

2

1

221

2

21

2

2

2

1

22

21

2

2

2

1

22

222

2cos

2)cos(

)cos()180cos(

)180cos(2)(cos2

ll

llyx

ll

llyx

llllyxCabbac

2

22

2

Redundant since 2 could be in the first or

fourth quadrant.

xy

yxl

l

cClbBcC

bB

1

22

2211

1

11

22

2

22

2

2

1

22221

tan)sin(sin;xytan

;yx

)sin(

yx

)sin(180sin

yx;180;;;sinsin

Redundancy caused since 2

has two

possible values

Using the Law of Cosines:


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12

21

2

2

2

1

221

2

221

2

2

2

1

21121121

2

2

2

1

21121

2

21

2

2

2

1

2

121121

2

21

2

2

2

1

2

1

2222

2

yx

cos

c2

)(ss)(cc2)(ss2)(ss)(cc2)(cc

yx)()(

ll

ll

llll

llllllllllll

iii

The Algebraic Solution

21

21211

21211

2121

11

)(

ssy)(

ccx)(

)cos(c

cosc

iii

llii

lli

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

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

:

abbaba

bababa

Note

l1

l2

(x , y)

1

2

Only unknown


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)c(s)s(c

cscss

sinsy

)()c(c

ccc

ccx

2211221

12221211

21211

2212211

21221211

21211

lll

lll

ll

slsll

sslll

ll

We know what 2 is from the previous slide. We

need to solve for 1 . Now we have two equations

and two unknowns (sin 1 and cos 1 )

22

222211

2212

22

1122221

221122

221

221

221

2211

yx

x)c(ys

)c2(sx)c(

1

)c(s)s()c(

)(xy

)c(

)(xc

slll

llllslll

lllll

sls

ll

sls

Substituting for c1 and simplifying

many times

Notice this is the law of cosinesand can be replaced by x2+ y2

22

222211

yx

x)c(yarcsin

slll

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

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

:

abbaba

bababa

Note

intelligent robotic chapter 4

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