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Robotic Systems(4)

Dr Richard Crowder

School of Electronics and Computer Science

Revised Jan 2011

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Kinematic Equations

Forward Kinematics: computation of the position,orientation and velocity of the end effector, given thedisplacements and joint angles.

Reverse Kinematics: computation of the jointdisplacements and angles from the end effectors positionand velocity.

Rn

Joint space

n variables

R6

Tool space

6 Variables

Forward

Reverse

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Homogeneous Transformations

1000

10

namelyation,transformdescribematrix to44auseweroboticsIn

1

:can writewe,:spaceinpointaConsider

3,3

2,2

1,1

z

y

x

PR

PRPR

nTranslatioRotationA

c

b

a

w

cw

bw

aw

w

z

y

x

v

ckbjaiv

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3D Transformation Matrices - 1

Consider two frames F1 and F2 with origins P1 and P2.

The relative position of F1 in F2 is given by the coordinatesof P2 in F1.

F2

F1

P2

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3D Transformation Matrices - 2

The relative orientation of the frames is expressed by the cosine ofNINE angles. The orientation of the x axis of F2 in F1 is given by:

If this is repeated for the y and z axes, we can write:

Only four cases are require, translation and rotation around therespective x, y and z axes

zyxlllzyx ,,,,

222

1000

zzzz

yyyy

xxxx

pnml

pnml

pnml

H

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3D Transformation Matrices - 3

Consider the rotation around the x axis

X

Y1

Y2

Z1

Z2

Y2

10000

0

1

]Rot[x,

0hence,90z,yandx

1hence,coinsidentarexandx

112

12

zyo

x

ll

l

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3D Transformation Matrices - 4

Consider the rotation around the x axis

X

Y1

Y2

Z1

Z2

Y2

1000sin0

cos0

01

]Rot[x,

sinhence,90zandy

coshence,yandy

0hence,90isxandy

12

12

o

12

z

y

x

m

m

m

90 -

90 -

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3D Transformation Matrices - 5

Consider the rotation around the x axis

X

Y1

Y2

Z1

Z2

Y2

1000cossin0

sincos0

001

]Rot[x,

coshence,zandz

sin-hence,90yandz

0hence,90isxandz

12

12

o

12

z

y

x

n

n

n

90 +

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3D Transformation Matrices - 6

Consider the rotation around the x axis

X

Y1

Y2

Z1

Z2

Y2

1000

0cossin0

0sincos0

0001

]Rot[x,

zeroisontranslatihencecoinside,PandP12

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Other transformations

1000

0C0S-

0010

0S0C

=]Rot[y,

1000

0100

00CS

00S-C

=]Rot[z,

1000

c100

b010

a001

=c]b,Trans[a,

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Consider

uyRotzRotuzRotyRotw

vyRotw

uzRotv

oooo

o

o

90,90,90,90,

1

3

7

2

1

2

7

3

.

1000

0001

0010

0100

90,

1

2

7

3

1

2

3

7

.

1000

0100

0001

0010

90,

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Verification

X

Y

Z

uzRotyRot oo 90,90,

uyRotzRot oo 90,90,

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Movement of co-ordinate frames - 1

Y

Z

X

Z

X

Y

Z

X

Y

10007010

3001

4100

2

zRot

2

yRot]734[Trans

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Movement of co-ordinate frames - 2

Origin of the original frame [0 0 0 1]T, transform to [4 -3 7 1]T

x = [1 0 0 1]T transforms to [4 -2 7 1]T

y = [0 1 0 1]T transforms to [4 -3 8 1]T

z = [0 0 1 1]T transforms to [5 -3 7 1]T

If the four points are plotted, the new frame results, also subtracting the originposition from the new unit vectors, gives the orientation.

0

0

1

0

1

7

3

4

1

7

2

4

Origin

X

This indicates that the new X axisis parallel with the original Y axis

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Relative Transformations

Up to this point we have considered all movements to bemade relative to a fixed co-ordinate frame, consider

Frame rotated around the reference Z axis

Frame rotated around the referenceY axis

Frame translated within the reference frame

2

zRot

2

yRot]734[Trans

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Relative Transformations

It is also possible to interpret the same move as:

Frame translated

Rotation 90o about the current Y axis

Rotation of 90o about the current Z axis.

The order of multiplication of the transformations is crucial toobtaining the correct final position.

If we postmultiplya transform representing a frame by a secondtransform describing a translation or rotation, then that translation orrotation is made with respect to the frame described by the firsttransformation

If we premultiplya frame transformation by a second transformdescribing a translation or rotation, then that translation or rotation ismade with respect to the base co-ordinate frame.

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Example

Execute the transformations left to rightwith respect to thereference frame postmultiply

Execute the transformations right to leftwith respect to the

reference frame premultiply

Consider Trans[a,0,0]Rot(z,45o]

X X

Y Y

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Kinematic Equations

A manipulator consists of a selection of links and joints, each with itsown co-ordinate frame, and described by four parameters.

It is possible to state: [T2]=[A1][A2]

[A1] is the position and orientation of the first link with reference to themanipulator's origin.

[A2] is the position and orientation of the second link relative to thefirst link.

Hence for a multi-jointed manipulator,

[Tn] = [A1] [A2].......[An]

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Transform Graph

A10T6

Z

E

0T6

E

Origin

X

X

X

Z

Origin

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Orientation Transformations

The matrix [T] has 16 elements, only 12 of which have anyreal meaning.

No limit on the value of px pypz

The l, m and n vectors must satisfy:

m.m = 1, n.n = 1, m.n = 0.

End effectors, two approaches:

RPY: roll, pitch and yaw

Euler angles

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Roll, Pitch, Yaw

Movement around the fixed frame (premultiply)

Rotation around the x axis is defined as yaw

Rotation around the y axis is defined as pitch

Rotation around the z axis is defined as roll

1000

0ccscs-0sc-csscc+ssscs

0ss-csccs-ssccc

xRotyRotzRotRPY

,,,,,

Z: Roll

Y: Pitch

X: Yaw

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Euler Angles

Rotation around the moving frame (postmultiply)

Rotation around the z axis

Rotation around the y axis

Rotation around the z axis

1000

0csscs-

0sscc+scsscccs

0sccs-sccssccc

zRotyRotzRotEuler

,,,,,

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Specification of the [A] matrix

Zn-1

Xn

dn

an

n

n

Zn

Joint n Joint n+1

Link n-1

Link n

Xn-1

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[A] matrix

Move sequence

Rotate around Zn-1 byn

Translate along Zn-1

by dn

Translate along Xn-1 by an

Rotate around Xn byn

1000

dcs0

assc-ccsacsscs-c

=]]Rot[x]Trans[a00]Trans[00dRot[z=]A[ n

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Application to a Manipulator

S1

S2 S3

S4

Joint 1

Joint 4

Joint 2 Joint 5

Joint 3

i 1 2 3 4 5

di S1 0 0 0 S4

i 1 2 3 4 5

ai 0 S2 S3 0 0

i 270 0 0 90 0

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Solution.

1000

d01-0

0C0S

0S-0C

=A

1

11

11

1

1000

0100

Sa0CS

Ca0S-C

=A2221

2222

2][

1000

0100

Sa0CS

Ca0S-C

=A3333

3333

3][

1000

001-0

0C0S

0S-0C

=A44

44

4][

1000

d100

00CS

00S-C

=A5

55

55

5][

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.

1000

Cd-Sa-Sa-dC-SSCS-

)Sd-Ca+Ca(SSS-CC-CS-CC-CS

)Sd-Ca+Ca(CSC-CS+SCC-SS+CC

=]T[

234523322123452345234

23452332212341512341512341

234523322123415152341512341

5

Where Cijk = Cos(qi + qj + qk)