Introduction to RoboticsIntroduction to Robotics

Tutorials 1-3

Technion, cs department, Introduction to Robotics 236927

Winter 2010-2011

1

Denavit-Hartenberg

• Specialized description of articulated

Reminder• Specialized description of articulated

figures

• Each joint has only one degree of

freedom

• rotate around its z-axis• rotate around its z-axis

• translate along its z-axis

http://www.cs.duke.edu/brd/Teaching/Bio/asmb/current/Papers/chap3-forward-kinematics.pdf

http://www.youtube.com/watch?v=rA9tm0gTln8 2

• Link offset di

• The distance between the origins of the coordinate

frames attached to jointi and jointi+1

•

Denavit-Hartenberg

• Measured along the axis of jointi

3

• Link rotation (joint angle) φi (In the movie - θ)

• The angle between the link lenghts αi-1 and αi

• Angle around zi-axis

Denavit-Hartenberg

•

4

Denavit-Hartenberg• Link length ai (In the movie - r)

• The perpendicular distance between the axes of

jointi and jointi+1

5

• Link twist αi

• The angle between the axes of jointi and jointi+1

• Angle around xi-axis

Denavit-Hartenberg

• Angle around xi-axis

6

Denavit-Hartenberg1.Compute the link vector ai and the link length

2.Attach coordinate frames to the joint axes

3.Compute the link twist αi

4.Compute the link offset di

5.Compute the joint angle φi

6.Compute the transformation (i-1)Ti which transforms entities from 6.Compute the transformation Ti which transforms entities from

linki to linki-1

- 2 Special Cases -

7

Denavit-Hartenberg

This transformation is done in several steps :This transformation is done in several steps :

• Rotate the link twist angle αi around the axis xi

•Translate the link length ai along the axis xi

•Translate the link offset di along the axis zi

( ) ( ) ( ) ( ) ( )ixixiziz

i

i iiiiRaTdTRT αϕ=−1

•Rotate the joint angle φi around the axis zi

8

Denavit-Hartenberg( ) ( ) ( ) ( ) ( )ixixiziz

i

i iiiiRaTdTRT αϕ=−1

( )

−

=

1000

0cossin0

0sincos0

0001

ii

ii

ixiR

αααα

α ( )

=

1000

0100

0010

001 i

ix

a

aTi

− ϕϕ

( )

=

1000

100

0010

0001

i

izd

dTi

( )

−

=

1000

0100

00cossin

00sincos

ii

ii

iziR

ϕϕϕϕ

ϕ

9

Denavit-Hartenberg( ) ( ) ( ) ( ) ( )ixixiziz

i

i iiiiRaTdTRT αϕ=−1

Multiplying the matrices :

( )

−

−

=− sinsincoscoscossin

cossinsincossincos

1 iiiiiii

iiiiiii

i

i

a

a

Tαα

ϕαϕαϕϕϕαϕαϕϕ

( ) ( ) ( ) ( )ixixizizi iiiiRaTdTRT αϕ=

In DH only φ and d are allowed to change.

=

1000

cossin0 iii

id

Tαα

10

End effector• End effector - the last coordinate system of figure

•• Located in joint N.

• But usually, we want to specify it in base

coordinates.

11

End effector

A transformation from the link N to the base :

12

End effector

• We can also express it as • We can also express it as

• three rotations (around each of the

coordinate axes)

• followed by a translation

• How can we establish a relation with

the other expression ?

13

End effector

• The origin of a coordinate frame relative • The origin of a coordinate frame relative

to some base coordinate frame is

specified by the translation :

14

End effector• Any 3D orientation relative to

some base coordinate frame

can be specified by :

three rotations, one around

each of the coordinate axes.

We do them in this order : around x, y, z. 15

End effector

16

End effector• Orientation

• The roll, pitch and yaw transformation is • The roll, pitch and yaw transformation is

then expressed :

17

End effector

• Finally, the transformation from a coordinate frame to • Finally, the transformation from a coordinate frame to

the base frame is expressed :

18

End effector

We obtain directly the translation vector :

19

End effector

We can obtain the yaw angle :

Because :

arctan is π-periodic. Let’s use our function

arctan2 to get the right angle.

20

End effector

Knowing the yaw angle, we can obtain

the pitch angle :

Because :

Again, let’s use our function arctan2 :

21

End effector

We can obtain the roll angle :

Because :

Again, let’s use our function arctan2 :

22

End effectorLet’s define the state vector

23

End effectorAs previously shown,

The state vector is composed of elements of this

matrix. It’s also a function of joint parameters :

24

DH Example - 3 revolute joints

Shown in home position

joint 1

Link 1 Link 3

Link 2

joint 1

R

joint 2 joint 3

L1 L2

25

Shown with joints in non-zero positions

DH Example - 3 revolute joints

Z0

Z2θ1

θ2 θ3

x0

x1

x2

z3

x3

Z1

θ1x0

Observe that frame i moves with link i

26

1.Compute the link vector ai and the link length

2. Attach coordinate frames to the joint axes

3.Compute the link twist αi

4.Compute the link offset di

5.Compute the joint angle φi

6.Compute the transformation (i-1)Ti

Link 2

joint 1

R

Link 1 Link 3

joint 2 joint 3

L1 L2 27

1.Compute the link vector ai and the link length

2. Attach coordinate frames to the joint axes

3.Compute the link twist αi

4.Compute the link offset di

5.Compute the joint angle φi

6.Compute the transformation (i-1)Ti

( )

−

−

=− sinsincoscoscossin

cossinsincossincos

1 iiiiiii

iiiiiii

ia

a

Tϕαϕαϕϕϕαϕαϕϕ

( )

−=−

1000

cossin0

sinsincoscoscossin1

iii

iiiiiiii

id

aT

ααϕαϕαϕϕ

28

DH Example

wherewhere

29

Example – the Stanford Arm

30

Example – the Stanford Arm

Z4

Z5

Z6

X7

Z7

Z1

X2

Z2

X3

Z3

X4

X5

X6

i ai di αi θi

1

2

3

d2

d3

X1Y1

3

4

5

6

d1

31

Pinhole Camera

d yθ

dy

x

y

ddx

ddy

θ

θ

tan

tan

=

=

32

Imaging

z

y

x

0

z

• Given a point. What is its image? (dx,dy)

• Solution:

• Convert the point’s coordinates to the • Convert the point’s coordinates to the

camera coordinate system.

• Derive θx and θy

• Calculate dx and dy.

( ) 10

0

−= TT N

N

33

Imaging

Z0

θ θz3

x30

z

y

x

x3

y3

z

y

x

0

3

0

3z

y

x

T

z

y

x

=

Z1

Z2θ1

θ2 θ3

x0

x1

x2

z3z3

x33 z

( ) ( )

( ) ( )3

3

3

3

3

3

3

3

03

tantan

tantan

x

ydddy

x

zdddx

x

y

x

z

zz

yx

yx

====

==

θθ

θθ

34

Imaging Example• Let,

904590 −=== ooo θθθ•

• What is the location of the end effector?

• Draw the image of the box with

1

0100

904590

21

321

=

===

−===

d

LLR

ooo θθθ

• Draw the image of the box with

coordinates

[ ] ( ) [ ] ( )[ ] ( ) [ ] ( )5,18,05,18,5

0,18,50,18,0

0403

0201

==

==

pp

pp

35

Imaging Example

where

0100

904590

21

321

===

−===

LLR

ooo θθθ

36

Imaging Example

( )

−

−

==−

1000

0001

102

2

2

20

02

2

2

20

10

3

3

0TT

1000

37