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A simple exampleA simple exampleBasilio Bona ROBOTICA 03CFIOR 1

A simple exampleA simple example

Example 1.1 – vector representation

110

B

⎛ ⎞⎟⎜ ⎟⎜= ⎟⎜ ⎟⎜ ⎟⎜vA very simple manipulator

v

P

0⎟⎜ ⎟⎜⎝ ⎠

BR

v

AR1( )q t

1 1 1 1

1 1 1 1

cos sin 0 c s 0

sin cos 0 s c 0AB

q q

q q

⎛ ⎞ ⎛ ⎞− −⎟ ⎟⎜ ⎜⎟ ⎟⎜ ⎜⎟ ⎟⎜ ⎜⎟ ⎟= ≡ =⎜ ⎜⎟ ⎟⎜ ⎜⎟ ⎟R

1 1 1 1

0 0 1 0 0 1B ⎜ ⎜⎟ ⎟⎜ ⎜⎟ ⎟⎜ ⎜⎟ ⎟⎟ ⎟⎜ ⎜⎝ ⎠ ⎝ ⎠

1

1

cos

sinAB

q

q

⎛ ⎞⎟⎜ ⎟⎜ ⎟⎜ ⎟= ⎜ ⎟⎜ ⎟t1

0B

q⎜ ⎟⎜ ⎟⎜ ⎟⎟⎜⎝ ⎠Basilio Bona ROBOTICA 03CFIOR 2

Example 1.2 – vector representation

If t t i t d t th

1 11 1c sc s 0 1⎛ ⎞⎛ ⎞ ⎛ ⎞−− ⎟ ⎟ ⎟⎜ ⎜ ⎜⎟ ⎟ ⎟⎜ ⎜ ⎜

If vector represents an oriented segment, thenBv

1 1

1 11 1c ss c 0 1

00 0 1 0A

⎟ ⎟ ⎟⎜ ⎜ ⎜⎟ ⎟ ⎟⎜ ⎜ ⎜⎟ ⎟ ⎟+= =⎜ ⎜ ⎜⎟ ⎟ ⎟⎜ ⎜ ⎜⎟ ⎟ ⎟⎜ ⎜ ⎜⎟ ⎟ ⎟⎜ ⎜⎟ ⎟ ⎜ ⎟⎟⎜⎟ ⎟⎜ ⎜ ⎝ ⎠⎝ ⎠⎝ ⎠

v00 0 1 0 ⎜⎟ ⎟⎜ ⎜ ⎝ ⎠⎝ ⎠⎝ ⎠

⎛ ⎞⎛ ⎞ ⎛ ⎞ ⎛ ⎞If it represents a geometric point, then

1 1 1 1 1

1 1 1 1 1

c s 0 1 c s c

s c 0 1 c s sAA B

t

⎛ ⎞⎛ ⎞ ⎛ ⎞ ⎛ ⎞− −⎟ ⎟ ⎟ ⎟⎜ ⎜ ⎜ ⎜⎟ ⎟ ⎟ ⎟⎜ ⎜ ⎜ ⎜⎟ ⎟ ⎟ ⎟⎜ ⎜ ⎜ ⎜⎟ ⎟ ⎟ ⎟= + = + +⎜ ⎜ ⎜ ⎜⎟ ⎟ ⎟ ⎟⎜ ⎜ ⎜ ⎜⎟ ⎟ ⎟ ⎟v

1 1 1 1 1

0 0 1 0 0 0

( )

A B⎜ ⎜ ⎜ ⎜⎟ ⎟ ⎟ ⎟⎜ ⎜ ⎜ ⎜⎟ ⎟ ⎟ ⎟⎜ ⎜ ⎜ ⎜⎟ ⎟ ⎟ ⎟⎟ ⎟ ⎟ ⎟⎜ ⎜ ⎜ ⎜⎝ ⎠⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎛ ⎞

1 1

1 1

(1 ) c s

(1 ) s c

⎛ ⎞+ − ⎟⎜ ⎟⎜ ⎟⎜ ⎟= + +⎜ ⎟⎜ ⎟

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1 1( )

0⎜ ⎟⎜ ⎟⎜ ⎟⎟⎜⎝ ⎠

Example 2.1 – rotations

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Example 2.2 – rotations

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Example 3.1 – DH parameters

3( )q t2( )q t

1 11 ( ) 0 90d a

q tθ α

− 1

2

2 2

3

2 ( ) 0 903 ( ) 0 0 0

q tq t−

3 ( )q

1( )q t

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2R

2q3

1R0R

1 2

1q Denavit – Hartenberg parametersi d θ

1 1 21 ( ) 2

i i i i

q t

i d aθ απ

1 1 2

2 30 ( ) 02 tq

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01R

12R

1 1 2 10 1 1 2 11

00

0 1 0

c s cs c s⎛ ⎞⎟⎜ ⎟⎜ − ⎟⎜= ⎟⎜ ⎟⎜

T2 2 3 2

1 2 2 3 22

00

0 0 1 0

c s cs c s⎛ ⎞− ⎟⎜ ⎟⎜ ⎟⎜= ⎟⎜ ⎟⎜ ⎟

T1

10 1 00 0 0 1

⎟⎜ ⎟⎜ ⎟⎜⎝ ⎠2 0 0 0

0 0 0 1⎜ ⎟⎜ ⎟⎟⎜⎝ ⎠

⎛ ⎞+02R

02t

1 2 1 2 1 3 1 2 2 1

1 2 1 2 1 3 1 2 2 10 11 2

2 2 3 2 10

c c c s s c c cs c s s c s c ss c s

⎛ ⎞− + ⎟⎜ ⎟⎜ − − + ⎟⎜ ⎟⎜= ⎟⎜ + ⎟⎜ ⎟T T

2 2 3 2 100 0 0 1s c s +⎜ ⎟⎜ ⎟⎟⎜ ⎟⎜⎝ ⎠

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3 1 2 2 1( )

( )

x t c c c

t

= ++ E 1

3 1 2 2 1

3 2 1

( )

( )

y t s c s

z t s

= += +

Eqn. 1

c c c s s⎛ ⎞⎟⎜ ( ) ( )t tφ1 2 1 2 102 1 2 1 2 1

c c c s s

R s c s s c

⎛ ⎞− ⎟⎜ ⎟⎜ ⎟⎜ ⎟= − −⎜ ⎟⎜ ⎟⎜ ⎟

1( ) ( )

( ) 2

t q t

t

φθ π

==

2 20s c

⎜ ⎟⎜ ⎟⎟⎜⎝ ⎠

E l l

2( ) ( )t q tψ =

Euler angleseqn. (2.79) page 52

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Knowing the Euler angles, everything will be easy. Assume we do not know them.

3 1 2 2 1( )

( )

x t c c c

y t s c s

= += +

3 1 2 2 1

3 2 1

( )

( )

y t s c s

z t s

= += + 1

2

zs

−=

3

3 1 2 2 1( )

( )

x t c c c

y t s c s

= += +

Squaring and adding

( )23 1 2 2 1

3 2 1

( )

( )

y t s c s

z t s

+= + ( )22 2 2

2 3 2x y a c+ ≡ = +

2a

c−

=2

3

c

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12

3

zs

−=

2 1s z −

22

ac

−=

2 12

2 2

tan qc a

= =−

3

( )( )x t c c= + ( )1

( )x tc

c=

+( )( )3 2 2 1

3 2 2 1

( )

( )

x t c c

y t c s

= +

= +

( )

( )

3 2 2

1

( )

c

y ts

+

=( )1

3 2 2c +

1 3 2 2( ) ( )tan

s cy t y tq

+= = =

11 3 2 2

tan( ) ( )

qc c x t x t+

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Linear velocities

( ) ( ) ( )( )t t t+( ) ( ) ( )( ) ( ) ( )

( )

3 1 2 2 1 1 3 1 2 2

3 1 2 2 1 1 3 1 2 2

( )

( )

x t s c s q t c s q t

y t c c c q t s s q t

= − + −

= + −

( )3 2 2( )z t c q t=

Angular velocities: analytical approach

( )1( )

( ) 0

t q t

t

φθ

=

= W ll th “ l i l iti ”

( )1

0

q t⎛ ⎞⎟⎜ ⎟⎜ ⎟⎜ ⎟⎜ω( )2

( ) 0

( )

t

t q t

θ

ψ

=

=We call these “eulerian velocities”

( )2

0E

q t

⎟⎜= ⎟⎜ ⎟⎜ ⎟⎜ ⎟⎟⎜⎝ ⎠

ω

( )⎝ ⎠

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Analytic Jacobian (by differentiation)

3 1 2 2 1 3 1 2s c s c s⎛ ⎞− − − ⎟⎜ ⎟⎜ ⎟⎜

3 1 2 2 1 3 1 2

3 20

Lc c c s s

c

⎟⎜ ⎟= + −⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟⎟⎜⎝ ⎠

J Eqn. 2a

3 2⎝ ⎠

⎛ ⎞1 0

0 0

⎛ ⎞⎟⎜ ⎟⎜ ⎟⎜ ⎟= ⎜ ⎟⎜J Eqn. 2b0 0

0 1A ⎟⎜ ⎟⎜ ⎟⎜ ⎟⎟⎜⎝ ⎠

J q

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Transformation matrix (see textbook)2π

θ =

1 1

1 1

0 cos sin sin 00 sin cos sin 0E

c ss c

φ φ θφ φ θ

⎛ ⎞ ⎛ ⎞⎟ ⎟⎜ ⎜⎟ ⎟⎜ ⎜⎟ ⎟⎜ ⎜= − ≡ −⎟ ⎟⎜ ⎜M

2

1 10 sin cos sin 01 0 01 0 cos

E s cφ φ θθ

≡⎟ ⎟⎜ ⎜⎟ ⎟⎜ ⎜⎟ ⎟⎜ ⎜⎟ ⎟⎜⎜ ⎝ ⎠⎝ ⎠

M

( )0 c s s qq t⎛ ⎞⎛ ⎞ ⎛ ⎞⎟⎜⎟ ⎟⎜ ⎜( )

( )

1 1 1 21

1 1 1 2

0

0 0E E

c s s qq t

s c c q⎟⎜⎟ ⎟⎜ ⎜⎟⎟ ⎟⎜⎜ ⎜⎟⎟ ⎟⎜⎜ ⎜⎟⎟ ⎟⎜= = − = −⎜ ⎜⎟⎟ ⎟⎜⎜ ⎜⎟⎟ ⎟⎜ ⎟⎜ ⎜⎟ ⎟⎜

Mω ω

( ) 121 0 0 qq t

⎟⎜ ⎜⎟ ⎟⎜ ⎟⎜ ⎜⎟ ⎟⎟ ⎟⎜ ⎜⎟⎜⎝ ⎠ ⎝ ⎠⎝ ⎠

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From the previous slide we have the cartesian velocity in From the previous slide we have the cartesian velocity in base RF we can now compute the Jacobian matrix

1 2 1 10 0s q s s⎛ ⎞ ⎛ ⎞ ⎛ ⎞⎟ ⎟ ⎟⎜ ⎜ ⎜⎛ ⎞⎟ ⎟ ⎟⎜ ⎜ ⎜1 2 1 1

11 2 1 1

2

0 0

1 0 1 0A

qc q c c

qq

⎛ ⎞⎟ ⎟ ⎟⎜ ⎜ ⎜⎟ ⎟ ⎟⎟⎜⎜ ⎜ ⎜⎟ ⎟ ⎟⎟⎜= − = − ⇒ = −⎜ ⎜ ⎜⎟ ⎟ ⎟⎟⎜⎜ ⎜ ⎜⎟ ⎟ ⎟⎟⎟⎜⎜ ⎜ ⎜⎟ ⎟ ⎟⎝ ⎠⎜ ⎜ ⎜⎟ ⎟ ⎟⎟ ⎟ ⎟⎜ ⎜ ⎜

Jω Eq. 3

11 0 1 0q ⎝ ⎠⎜ ⎜ ⎜⎟ ⎟ ⎟⎟ ⎟ ⎟⎜ ⎜ ⎜⎝ ⎠ ⎝ ⎠ ⎝ ⎠

This is the geometric angular JacobianN i i diff Now we compute it in a different way

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Both joints are rotoidal therefore considering results at pageBoth joints are rotoidal, therefore, considering results at page.

1Ai i−=J k

1 1,Li i i p− −= ×J k r

First we compute the angular Jacobian

0⎛ ⎞ 0⎛ ⎞ ⎛ ⎞

1 0

0

0A

⎛ ⎞⎟⎜ ⎟⎜ ⎟⎜ ⎟= = ⎜ ⎟⎜ ⎟⎜J k

10

2 1 1 1

0

0A

s

c

⎛ ⎞ ⎛ ⎞⎟ ⎟⎜ ⎜⎟ ⎟⎜ ⎜⎟ ⎟⎜ ⎜⎟ ⎟= = = −⎜ ⎜⎟ ⎟⎜ ⎜⎟ ⎟⎜ ⎜J k R

1 0

1A ⎜ ⎟⎜ ⎟⎜ ⎟⎟⎜⎝ ⎠

2 1 1 1

1 0A ⎜ ⎜⎟ ⎟⎜ ⎜⎟ ⎟⎜ ⎜⎟ ⎟⎟ ⎟⎜ ⎜⎝ ⎠ ⎝ ⎠

These two columns are equal to those in Eqn. 3

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⎛ ⎞

0 p

x

y

⎛ ⎞⎟⎜ ⎟⎜ ⎟⎜ ⎟= ⎜ ⎟⎜ ⎟r This relation is

b d f d0,py

z⎜ ⎟⎜ ⎟⎜ ⎟⎟⎜⎝ ⎠

⎛ ⎞⎛ ⎞

obtained from direct KF – Eqn. 1

( )1 0 0 0 0

0 1 0

1 0 0L p p

x

y

⎛ ⎞⎛ ⎞− ⎟ ⎟⎜ ⎜⎟ ⎟⎜ ⎜⎟ ⎟⎜ ⎜⎟ ⎟= × = = ⎜ ⎜⎟ ⎟⎜ ⎜⎟ ⎟J k r S k r( )1 0 0, 0 0,

0 0 0L p p

y

z⎜ ⎜⎟ ⎟⎜ ⎜⎟ ⎟⎜ ⎜⎟ ⎟⎟ ⎟⎜ ⎜⎝ ⎠⎝ ⎠

⎛ ⎞ ⎛ ⎞3 1 2 2 1

3 1 2 2 1

y s c s

x c c c

⎛ ⎞ ⎛ ⎞− − −⎟ ⎟⎜ ⎜⎟ ⎟⎜ ⎜⎟ ⎟⎜ ⎜⎟ ⎟= = +⎜ ⎜⎟ ⎟⎜ ⎜⎟ ⎟3 1 2 2 1

0 0⎜ ⎜⎟ ⎟⎜ ⎜⎟ ⎟⎜ ⎜⎟ ⎟⎟ ⎟⎜ ⎜⎝ ⎠ ⎝ ⎠

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Then we compute3 2c

s

⎛ ⎞⎟⎜ ⎟⎜ ⎟⎜ ⎟= ⎜ ⎟r

1

1, 3 2

0p

R

s= ⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟⎟⎜⎝ ⎠

r

1R

Instead of transforming it in RF 0 and after making the vector product,we make the vector product in RF 1 and then we transform the result

( )2 1 1 1 1L⎡ ⎤⎡ ⎤⎡ ⎤ ⎡ ⎤= × =⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦J k r S k r

to express it in RF 0

( )1 1 1 1

2 1 1, 1 1,

3 2 3 20 1 0

L p pR R R R

c s

⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦⎛ ⎞⎛ ⎞ ⎛ ⎞− −⎟ ⎟ ⎟⎜ ⎜ ⎜⎟ ⎟ ⎟⎜ ⎜ ⎜

3 2 3 21 0 0

0 0 0 0 0

s c⎟ ⎟ ⎟⎜ ⎜ ⎜⎟ ⎟ ⎟⎜ ⎜ ⎜⎟ ⎟ ⎟= =⎜ ⎜ ⎜⎟ ⎟ ⎟⎜ ⎜ ⎜⎟ ⎟ ⎟⎜ ⎜ ⎜⎟ ⎟ ⎟⎜ ⎜ ⎜⎟ ⎟ ⎟⎟ ⎟ ⎟⎜ ⎜ ⎜

1

0 0 0 0 0R

⎟ ⎟ ⎟⎜ ⎜ ⎜⎝ ⎠⎝ ⎠ ⎝ ⎠

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Now we transform from RF 1 to RF 0

3 2 1 1 3 2 3 1 20s c s s c s⎛ ⎞ ⎛ ⎞⎛ ⎞ ⎛ ⎞− − − −⎟ ⎟ ⎟ ⎟⎜ ⎜ ⎜ ⎜⎟ ⎟ ⎟ ⎟⎜ ⎜ ⎜ ⎜

0

3 2 1 1 3 2 3 1 20

2 1 3 2 1 1 3 2 3 1 20

0 0 1 0 0L R

c s c c s s

c

⎟ ⎟ ⎟ ⎟⎜ ⎜ ⎜ ⎜⎟ ⎟ ⎟ ⎟⎜ ⎜ ⎜ ⎜⎡ ⎤ ⎟ ⎟ ⎟ ⎟= = = −⎜ ⎜ ⎜ ⎜⎟ ⎟ ⎟ ⎟⎢ ⎥ ⎜ ⎜ ⎜ ⎜⎣ ⎦ ⎟ ⎟ ⎟ ⎟⎜ ⎜ ⎜ ⎜⎟ ⎟ ⎟ ⎟⎜ ⎜ ⎜ ⎜⎟ ⎟ ⎟ ⎟⎟ ⎟ ⎟ ⎟⎜ ⎜ ⎜ ⎜

J R

In conclusions the two Jacobians are

1 1 03 2

0 0 1 0 0R R R

c⎟ ⎟ ⎟ ⎟⎜ ⎜ ⎜ ⎜⎝ ⎠ ⎝ ⎠⎝ ⎠ ⎝ ⎠

In conclusions, the two Jacobians are

3 1 2 2 1 3 1 2s c s c s⎛ ⎞− − − ⎟⎜ ⎟

( )3 1 2 2 1 3 1 2

1 2 3 1 2 2 1 3 1 2

0L L L

c c c s s⎜ ⎟⎜ ⎟⎜ ⎟= = + −⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟

J J J

It coincides with the results of Eqn. 2a

3 20 c⎜ ⎟⎟⎜⎝ ⎠

q

Basilio Bona ROBOTICA 03CFIOR 19

Linear Jacobians are independent from the methods used to compute them(since we use always Cartesian representation)

Instead, angular Jacobians, depends on the conventions used to expressthe TCP orientation

s c s c s⎛ ⎞⎟⎜In conclusions:

3 1 2 2 1 3 1 2

3 1 2 2 1 3 1 2L

s c s c s

c c c s s

− − − ⎟⎜ ⎟⎜ ⎟⎜ ⎟= + −⎜ ⎟⎜ ⎟⎜ ⎟J

3 20 c

⎜ ⎟⎜ ⎟⎟⎜⎝ ⎠

1 0

0 1

⎛ ⎞⎟⎜ ⎟⎜ ⎟⎜ ⎟= ⎜ ⎟J1

0

0

s

c

⎛ ⎞⎟⎜ ⎟⎜ ⎟⎜ ⎟= ⎜ ⎟J0 1

0 0A= ⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟⎟⎜⎝ ⎠

J1

0

1 0A

c= −⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟⎟⎜⎝ ⎠

J

Analytical Jacobian Geometric Jacobian

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