MAE263B: Dynamics of Robotic Systems Discussion Section...

MAE263B: Dynamics of Robotic Systems

Discussion Section – Week5

: Jacobian (SCARA)

Seungmin Jung

02.07.2020.

Contents

❑ Jacobian with SCARA example

• Velocity propagation

• Direct differentiation

❑ Frame of Representation

Kinematics Relations - Joint & Cartesian Spaces

• A robot is often used to manipulate object attached to its tip (end effector).

• The location of the robot tip may be specified using one of the following

descriptions:

• Joint Space

• Cartesian Space / Operational Space

{N}

=

N

2

1

=

N

N

r

PX

0

0

Euler Angles

=

10

00

0 NN

N

PRT

A minimal representation of orientation - Euler angles

=

A minimal representation of orientation - Euler angles

Kinematics Relations - Forward & Inverse

• The robot kinematic equations relate the two description of the robot tip location

=

N

2

1

=

N

N

r

PX

0

0

Tip Location in

Joint SpaceTip Location in

Cartesian Space

)(FKX =

)(XIK=

Velocity relationship

: Jacobian Matrix – Joint velocity / End-effector velocity

==

N

dt

d

2

1

][

=

==

z

y

x

z

y

x

N

Nv

v

v

vX

dt

dX

][

Tip Velocity in

Joint SpaceTip velocity in

Cartesian Space

Jacobian Matrix - Introduction

• The velocity relationship

: The relationship between

the joint angle rates ( )

and the translation and rotation velocities of the end

effector ( ).

• The relationship between

the robot joint torques ( )

and the forces and moments ( )

at the robot end effector (Static Conditions).

N

x

( ) Jx =

F

F

( ) FJT

=

Jacobian Matrix - Calculation Methods

Jacobian Matrix

Differentiation the

Forward Kinematics Eqs.

Iterative Propagation

(Velocities or Forces / Torques)


• In the field of robotics the Jacobian

matrix describe the relationship

between the joint angle rates ( )

and the translation and rotation

velocities of the end effector ( ).

This relationship is given by:

N

x

( ) Jx =

( ) xJ 1−=


• This expression can be expanded to:

• Where:

– is a 6x1 vector of the end effector linear and angular velocities

– is a 6xN Jacobian matrix

– is a Nx1 vector of the manipulator joint velocities

– is the number of joints

( )

=

Nz

y

x

Jz

y

x

2

1

N

x

( )J

N

6x1 6xN Nx1

Position Propagation

• The homogeneous transform matrix provides a complete description of the

linear and angular position relationship between adjacent links.

• These descriptions may be combined together to describe the position of a link

relative to the robot base frame {0}.

TTTT i

i

oo

i

11

21

−=

1 1

10 1

i i

i i i

i

R PT − −

−

=

Velocity Propagation

• Given: A manipulator - A chain of

rigid bodies each one capable of

moving relative to its neighbor

• Problem: Calculate the linear and

angular velocities of the link of a

robot

• Solution (Concept): Due to the robot

structure (serial mechanism) we can

compute the velocities of each link

in order starting from the base.

The velocity of link i+1

= The velocity of link i

+ whatever new velocity components were added by joint i+1

Velocity Propagation – Intuitive Explanation

• Three Actions

– The origin of frame B moves as a function of time with respect to the origin

of frame A

– Point Q moves with respect to frame B

– Frame B rotates with respect to frame A along an axis defined by B

A

B

A Q

BP

• Linear and Rotational Velocity

– Vector Form

– Matrix Form

Q

BA

BB

A

Q

BA

BBORG

A

Q

A PRVRVV ++=

( )Q

BA

B

A

BQ

BA

BBORG

A

Q

A PRRVRVV ++=

Velocity Propagation – Intuitive Explanation

• Three Actions

– The origin of frame B moves with respect to the origin of frame A

– Point Q moves with respect to frame B

– Frame B rotates with respect to frame A about an axis defined by B

A

B

A Q

BP

Linear Velocity - Rigid Body

• Given: Consider a frame {B} attached

to a rigid body whereas frame {A} is

fixed. The orientation of frame {A}

with respect to frame {B} is not

changing as a function of time

• Problem: describe the motion of of

the vector relative to frame {A}

• Solution: Frame {B} is located

relative to frame {A} by a position

vector and the rotation matrix

(assume that the orientation is not

changing in time ) expressing

both components of the velocity in

terms of frame {A} gives

Q

BP

Q

BP

BORG

AP RA

B

0=RA

B

Q

Q

BA

BBORG

A

Q

BA

BORG

A

Q

A VRVVVV +=+= )(

0=RA

B

0=RA

B

Q

BA

BB

A

Q

BA

BBORG

A

Q

A PRVRVV ++= ( )Q

BA

B

A

BQ

BA

BBORG

A

Q

A PRRVRVV ++=

Instructor: Jacob Rosen

Advanced Robotic - MAE 263D - Department of Mechanical & Aerospace Engineering - UCLA

Frame - Velocity

• As with any vector, a velocity vector may be described in terms of any frame,

and this frame of reference is noted with a leading superscript.

• A velocity vector computed in frame {B} and represented in frame {A} would be

written

Q

BA

Q

BA Pdt

dV =)(

Q

Computed

(Measured)

Represented

(Reference Frame)

0=BORG

AV

0=RA

B

Angular Velocity - Rigid Body

Q

Q

BP

• Given: Consider a frame {B} attached

to a rigid body whereas frame {A} is

fixed. The vector is constant as

view from frame {B}

• Problem: describe the velocity of the

vector representing the the point

Q relative to frame {A}

• Solution: Even though the vector

is constant as view from frame {B} it

is clear that point Q will have a

velocity as seen from frame {A} due

to the rotational velocity

Q

BP

0=Q

BV

0=Q

BV

Q

BP

Q

BP

B

A 0=BORG

AV

Q

BA

BB

A

Q

BA

BBORG

A

Q

A PRVRVV ++= ( )Q

BA

B

A

BQ

BA

BBORG

A

Q

A PRRVRVV ++=

Angular Velocity - Rigid Body - Intuitive Approach

• The figure shows to instants of time

as the vector rotates around

This is what an observer in frame {A}

would observe.

• The Magnitude of the differential

change is

• Using a vector cross product we get

B

A

( )( )sinQ

A

B

A

Q

A PtP =

Q

A

B

A

Q

AQ

A

PVt

P==

)(tPQ

A

)( ttPQ

A +

sinQ

AP

Q

AP

Q

AP

sinQ

AP

Q

BA

BB

A

Q

BA

BBORG

A

Q

A PRVRVV ++= ( )Q

BA

B

A

BQ

BA

BBORG

A

Q

A PRRVRVV ++=

Simultaneous Linear and Rotational Velocity

• The final results for the derivative of a vector in a moving frame (linear and

rotation velocities) as seen from a stationary frame

• Vector Form

• Matrix Form

Q

BA

BB

A

Q

BA

BBORG

A

Q

A PRVRVV ++= B

A

( )Q

BA

B

A

BQ

BA

BBORG

A

Q

A PRRVRVV ++=

Q

BP

Simultaneous Linear and Rotational Velocity

• Linear and Rotational Velocity

– Vector Form

– Matrix Form

• Angular Velocity

– Vector Form

– Matrix Form

Q

BA

BB

A

Q

BA

BBORG

A

Q

A PRVRVV ++=

B

A

( )Q

BA

B

A

BQ

BA

BBORG

A

Q

A PRRVRVV ++=

Q

BP

TA

B

B

C

A

B

A

B

A

C RRRRR +=

C

BA

BB

A

C

A R +=

Velocity of Adjacent Links - Summary

• Angular Velocity

• Linear Velocity

( )

++=

+

+

+

+

+

1

1

1

1

1 0

0

i

i

i

i

iii

ii

i

d

vPRv

+=

+

+

+

+

1

1

1

1 0

0

i

i

ii

ii

i R

0 - Prismatic Joint

0 - Revolute Joint

The velocity of link i+1

= The velocity of link i

+ whatever new velocity components were added by joint i+1

Jacobian: Velocity propagation

• Therefore the recursive expressions for the adjacent joint linear and angular

velocities can be used to determine the Jacobian in the end effector frame

• This equation can be expanded to:

( ) JX NN =

( )

=

=

=

N

N

N

N

N

N

N

N Jv

z

y

x

z

y

x

X

2

1

Velocity of Adjacent Links - Angular Velocity 5/5

• The result is a recursive equation that shows the angular velocity of one link in

terms of the angular velocity of the previous link plus the relative motion of the

two links.

• Since the term depends on all previous links through this recursion, the

angular velocity is said to propagate from the base to subsequent links.

+=

+

+

+

+

1

1

1

1 0

0

i

i

ii

ii

i R

1

1

+

+

i

i


• From the relationship developed previously

• we can re-assign link names to calculate the velocity of any link i relative to the

base frame {0}

• By pre-multiplying both sides of the equation by ,we can convert the frame

of reference for the base {0} to frame {i+1}

C

BA

BB

A

C

A R +=

+→

→

→

1

0

iC

iB

A

1

00

1

0

++ += i

i

iii R

Ri 1

0

+


• Using the recently defined notation, we have

- Angular velocity of frame {i+1} measured relative to the robot base, and

expressed in frame {i+1} - Recall the car example

- Angular velocity of frame {i} measured relative to the robot base, and

expressed in frame {i+1}

- Angular velocity of frame {i+1} measured relative to frame {i} and

expressed in frame {i+1}

1

01

0

01

01

01

0 +

++

+

+ += i

i

i

i

i

i

i

i RRRR

1

11

1

1

+

++

+

+ += i

ii

ii

i

i

i R

1

1

+

+

i

i

i

i 1+

1

1

+

+ i

ii

i R

c

c

c

wcvV =


• Angular velocity of frame {i} measured relative to the robot base, expressed in

frame {i+1}

1

11

1

1

+

++

+

+ += i

ii

ii

i

i

i R

i

ii

ii

i R 11 ++ =


• Angular velocity of frame {i+1} measured (differentiate) in frame {i} and

represented (expressed) in frame {i+1}

• Assuming that a joint has only 1 DOF. The joint configuration can be either

revolute joint (angular velocity) or prismatic joint (Linear velocity).

• Based on the frame attachment convention in which we assign the Z axis

pointing along the i+1 joint axis such that the two are coincide (rotations of a link

is preformed only along its Z- axis) we can rewrite this term as follows:

=

+

+

+

1

1

1 0

0

i

i

ii

i R

1

2

3

i+1

i

1

11

1

1

+

++

+

+ += i

ii

ii

i

i

i R

SCARA – RRRP – DH Parameter (Modified form)

SCARA – RRRP

SCARA – RRRP – Forward Kinematics

Simplify Function

SCARA example _ Jacobian: Velocity propagation

• The recursive equation for the Angular Velocity is

𝑖+1𝜔𝑖+1 = 𝑖𝑖+1𝑅 𝑖𝜔𝑖 + 𝜌

00ሶ𝜃𝑖+1

,𝜌 = 0 𝑖𝑛 𝑡ℎ𝑒 𝑝𝑟𝑖𝑠𝑚𝑎𝑡𝑖𝑐 𝑗𝑜𝑖𝑛𝑡𝝆 = 𝟏 𝒊𝒏 𝒕𝒉𝒆 𝒓𝒆𝒗𝒐𝒍𝒖𝒕𝒆 𝒋𝒐𝒊𝒏𝒕

𝑖𝑖+1𝑅 is the transpose of 𝑖+1

𝑖𝑅 ( 𝑖𝑖+1𝑅 = 𝑖+1

𝑖𝑅𝑇) and 𝑖+1

𝑖𝑅 = 𝑖+1𝑖𝑇(1: 3,1: 3)

𝑖+1𝑖𝑅 can be obtained from the transformation matrix in the forward kinematics.



• The recursive equation for the Angular Velocity is

𝑖+1𝜔𝑖+1 = 𝑖𝑖+1𝑅 𝑖𝜔𝑖 + 𝜌

00ሶ𝜃𝑖+1

,𝜌 = 0 𝑖𝑛 𝑡ℎ𝑒 𝑝𝑟𝑖𝑠𝑚𝑎𝑡𝑖𝑐 𝑗𝑜𝑖𝑛𝑡𝝆 = 𝟏 𝒊𝒏 𝒕𝒉𝒆 𝒓𝒆𝒗𝒐𝒍𝒖𝒕𝒆 𝒋𝒐𝒊𝒏𝒕

• The base frame does not move

• Three revolute joints

• One prismatic joint

1𝜔1 = 01𝑅 0𝜔0 +

00ሶ𝜃1

2𝜔2 = 12𝑅 1𝜔1 +

00ሶ𝜃2

3𝜔3 = 23𝑅 2𝜔2 +

00ሶ𝜃3

4𝜔4 = 34𝑅 3𝜔3 +

00

ሶ𝜃4 = 0

0𝜔0 =000

(The base frame does not move)



• The recursive equation for Linear Velocity is

𝑖+1𝑣𝑖+1 = 𝑖𝑖+1𝑅 𝑖𝜔𝑖 ×

𝑖𝑃𝑖+1 +𝑖𝑣𝑖 + 𝜌

00ሶ𝑑,𝜌 = 1 𝑖𝑛 𝑡ℎ𝑒 𝑝𝑟𝑖𝑠𝑚𝑎𝑡𝑖𝑐 𝑗𝑜𝑖𝑛𝑡𝝆 = 𝟎 𝒊𝒏 𝒕𝒉𝒆 𝒓𝒆𝒗𝒐𝒍𝒖𝒕𝒆 𝒋𝒐𝒊𝒏𝒕

• The base frame does not move

• Three revolute joints

• One prismatic joint

0𝑣0 =000

1𝑣1 = 01𝑅 0𝜔0 ×

0𝑃1 +0𝑣0

2𝑣2 = 12𝑅 1𝜔1 ×

1𝑃2 +1𝑣1

3𝑣3 = 23𝑅 2𝜔2 ×

2𝑃3 +2𝑣2

4𝑣4 = 34𝑅 3𝜔3 ×

3𝑃4 +3𝑣3 +

00ሶ𝑑4

SCARA example

Jacobian: Velocity propagation


• The Jacobian 4𝐽4is defined as 4𝑣44𝜔4

= 4𝐽4

ሶ𝜃1ሶ𝜃2ሶ𝜃3ሶ𝑑4

• According to the definition 4𝑣44𝜔4

=

4𝑣4,14𝑣4,24𝑣4,34𝑤4,14𝑤4,24𝑤4,3

,

4𝑣4,14𝑣4,24𝑣4,34𝑤4,14𝑤4,24𝑤4,3

= 4𝐽4

ሶ𝜃1ሶ𝜃2ሶ𝜃3ሶ𝑑4


Jacobian: Frame of Representation

• The Jacobian provides the relationship between the end effector’s Cartesian

velocity measured relative to the robot base frame {0}

• For velocity expressed in frame {N}

• For velocity expressed in frame {0}

( ) JX NN =

( ) JX 00 =




• Consider the velocities in a different frame {B}

• We may use the rotation matrix to find the velocities in frame {A}:

• The Jacobian transformation is given by a rotation matrix

)(Jv

X B

N

B

N

B

B =

=

=

=

N

BA

B

N

BA

B

N

A

N

A

A

R

vRvX

( ) ( ) JRJX B

J

A

B

AA ==




• where is given by

or equivalently,

J

A

B R

=

R

R

R

A

B

A

B

J

A

B

000

000

000

000

000

000

( )

( ) J

R

R

J B

A

B

A

B

A

=

000

000

000

000

000

000



Jacobian: Frame of Representation - 3R Example

( ) ( )

0

4

0 4

4 4

0

4

0 0 0

0 0 0

0 0 0

0 0 0

0 0 0

0 0 0

R

J J

R

=




Jacobian: Direct Differentiation

• This expression can be expanded to:

( )

=

Nz

y

x

Jz

y

x

2

1

6x1 6xN Nx1



=

10

00

0 NN

N

PRT

2 1 2 1 1

2 1 2 1 1

4

2 1 2 1 1 1 2 1 2 2

2 1 2 1 1 1 2 1 2 2

4

cos( ) cos( )

sin( ) sin( )

[ sin( ) sin( )] [ sin( )]

[ cos( ) cos( )] [ cos( )]

x

y

z

x

y

z

p l t t l t

p l t t l t

p d

p l t t l t t l t t t


p d

+ +

= + + −

− + − + − +

= + − + + −

Jacobian: Direct Differentiation (Jp)

1

2

3

4

x

y p

z

x

y p

z

v

v J

v

tp

tp J

tp

d

=

=

2 1 2 1 1

2 1 2 1 1

4

2 1 2 1 1 1 2 1 2 2

2 1 2 1 1 1 2 1 2 2

4

cos( ) cos( )

sin( ) sin( )

[ sin( ) sin( )] [ sin( )]

[ cos( ) cos( )] [ cos( )]

x

y

z

x

y

z

p l t t l t

p l t t l t

p d



p d

+ +

= + + −

− + − + − +

= + − + + −

1

2 1 2 1 1 2 1 2

2

2 1 2 1 1 2 1 2

3

4

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

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

0 0 0 1

x

y

z

tp l t t l t l t t

tp l t t l t l t t

tp

d

− + − − +

= + − + −

Jacobian: Direct Differentiation (Jo)

1

2

3

4

x

y

z

x

y

z

w

w Jo

w

tw

tw Jo

tw

d

=

=

1 2 3

1

2

3

4

0

0

0

0 0 0 0

0 0 0 0

1 1 1 0

x

y

z

x

y

z

w

w

w

w

w

wd

= + + +

=







𝐽𝑃1= 𝑧0_1 × (𝑝𝑒 − 𝑝0_1)

= ×( - ) =

𝐽𝑃2= 𝑧0_2 × (𝑝𝑒 − 𝑝0_2)

= ×( - ) =

𝐽𝑃3= 𝑧0_3 × (𝑝𝑒 − 𝑝0_3)

= ×( - ) =

𝐽𝑃4= 𝑧0_4= =


𝐽𝑃1= 𝑧0_1 × (𝑝𝑒 − 𝑝0_1)

= ×( - ) =

𝐽𝑃2= 𝑧0_2 × (𝑝𝑒 − 𝑝0_2)

= ×( - ) =

𝐽𝑃3= 𝑧0_3 × (𝑝𝑒 − 𝑝0_3)

= ×( - ) =

𝐽𝑃4= 𝑧0_4= =


Jacobian: Direct Differentiation 𝑱 =

𝑱𝑷𝟏 𝑱𝑷𝟐 𝑱𝑷𝟑 𝑱𝑷𝟒𝑱𝑶𝟏

𝑱𝑶𝟐𝑱𝑶𝟑

𝑱𝑶𝟒

Jacobian - Comparison


Summary

✓ Jacobian with SCARA example

• Velocity propagation

• Direct differentiation

✓ Frame of Representation

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