INTRODUCTION TO DYNAMICS ANALYSIS OF ROBOTS (Part 3)

INTRODUCTION TO

DYNAMICS ANALYSIS

OF ROBOTS(Part 3)

This lecture continues the discussion on the analysis of the instantaneous motion of a rigid body, i.e. the velocities and accelerations associated with a rigid body as it moves from one configuration to another.

After this lecture, the student should be able to:•Derive the acceleration tensor and angular acceleration tensor•Derive the principles of relative motion between bodies in terms of acceleration analysis

Introduction to Dynamics Analysis of Robots (3)

Summary of previous lectures

333231

232221

131211

)()(

tRRt T

Velocity tensor and angular velocity vector

12

31

23

21

13

32

3

2

1

)(

t

)()()()()( tPtQttvtv PQ

)()()()()( tPtQttvtv PQ

bQabaobaQ PRPP ///

Velocity and moving FORs

bQabbQ

ababaobaQ vRPRvv /////

Relative Angular Velocity

Consider 3 FORs {a}, {b} and {c}. is the rotation of frame {b} w.r.t. frame {a}. Let

Rab

= relative angular velocity of frame {b} w.r.t. frame {a}aba

/

= relative angular velocity of frame {c} relative to frame {b} w.r.t. frame {a}bc

a/

= relative angular velocity of frame {c} w.r.t. frame {a}aca

/

)( // bcabbc

a R

)( ///

///

bcababac

bca

aba

aca

R

Example: Relative Angular Velocity

Example: The 3 DOF RRR Robot:

Y0, Y1

X0, X1

Z0, Z1

Z2

X2

Y2

Z3

X3

Y3

A=3 B=2 C=1

P

What is after 1 second if all the joints are rotating at

3,2,1,6

it

i

0/3

0

5236.0

0

1/2

100

0866.05.0

05.0866.001R

5236.0

0

0

0/1


5236.0

0

0

2/3

0866.05.0

100

05.0866.012R

100

0866.05.0

05.0866.023R

Solution: We re-used the following data obtained from the previous lecture

0

0472.1

0

5236.0

0

0

0866.05.0

100

05.0866.0

0

5236.0

0

)(

1/3

2/3121/21/3

R


5236.0

9068.0

5236.0

0

0472.1

0

100

0866.05.0

05.0866.0

5236.0

0

0

)(

0/3

1/3010/10/3

R

)(

)(

1/3010/10/3

2/3121/21/3

R

R


You should get the same answer from the overall rotational matrix and its derivative, i.e.

)1,2(

)3,1(

)2,3(

0/3

0/3

0/3

0/303

030/3

23

12

01

23

12

01

23

12

01

03

23

12

01

03

TRR

RRRRRRRRRR

RRRR

5236.0

9068.0

5236.0

0/3

05.0866.0

866.0433.025.0

5.075.0433.023

12

01

03 RRRR

09069.05236.0

2618.06545.02267.0

4534.02267.09162.023

12

01

23

12

01

23

12

01

03 RRRRRRRRRR


5236.0

9068.0

5236.0

05236.09068.0

5236.005235.0

9068.05236.00

05.0866.0

866.0433.025.0

5.075.0433.0

09069.05236.0

2618.06545.02267.0

4534.02267.09162.0

0/30/3

0/3

T

Acceleration tensor

Consider 2 points “P” and “Q” of a rigid body:

)()()()()()(

})()()(){()()()()()(

)()()()()()()()(

)()()()()(

2 tPtQtttPtQ

tPtQtttPtQttPtQ

tPtQttPtQttPtQ

tPtQttPtQ

Rearranging:

)()()()()( tPtQtAtata PQ

where)()()( 2 tttA

A(t) is called the acceleration tensor

00

00

000

)()(

tRRt T

Example: Acceleration tensor

Given

Find the acceleration tensor if =t2

Solution:

020

200

000

00

00

000

)(

t

222 tt

Example: Acceleration tensor

22

222

2

22

400

040

000

)(

00

00

000

00

00

000

00

00

000

)(

t

tt

t

22

22

22

222

420

240

000

)(

400

040

000

020

200

000

)(

t

ttA

t

ttA

Angular Acceleration vector

)()()()()()()()(

)()()()()()()()(

tvtvttPtQttPtQ

tPtQttPtQttPtQ

PQ

where

12

31

23

21

13

32

3

2

1

)(

t Angular

velocity vector

12

31

23

21

13

32

3

2

1

)(

t

Similarly:

Angular acceleration vector

Example: Angular Acceleration vector

00

00

000

)()(

tRRt T

Given

Find the angular acceleration vector if =t2

Solution:

020

200

000

00

00

000

)(

t

222 tt

0

0

2

)(

21

13

32

t

Acceleration and moving FORs

RRR

RRRRR

RRRR

PRPRPRPRPRvv

PRPRvv

T

TT

TT

bQabbQ

abbQ

abbQ

abbQ

abaobaQ

bQabbQ

abaobaQ

1

1111

1

///////

////

)()()(

)(

bQabbQ

abab

bQabababbQ

ababaobaQ

bQabbQ

abbQ

abbQ

abaobaQ

bQabbQ

abbQ

abbQ

abbQ

abaobaQ

bQabbQ

abbQ

abbQ

abbQ

abaobaQ

PRPR

PRPRaa

PRPRPRPRvv

PRPRPRPRPRvv

PRPRPRPRPRvv

///

///////

//////

///////

///////

)(2

)(

)(2)(

)()()(

)()()(

Acceleration and moving FORs

bQ

abbQ

abab

bQabababbQ

ababaobaQ

PRPR

PRPRaa

///

///////

)(2

)(

Let

bQabrel

bQabrel

bQabrel

PRa

PRV

PRP

/

/

/

)(

relrelabrelababrelabaobaQ aVPPaa )(2)( //////

Example: Acceleration and moving FORs

Example: The 3 DOF RRR Robot:

Y0, Y1

X0, X1

Z0, Z1

Z2

X2

Y2

Z3

X3

Y3

A=3 B=2 C=1

P

What is the acceleration of point “P” after 1 second if all the joints are rotating at

3,2,1,6

it

i


We know from the previous lecture that at t=1

100

0866.05.0

05.0866.001R

000

05.0866.0

0866.05.0

11

1101

R

1000

0100

00)cos()sin(

00)sin()cos(

11

11

01

T

0000

0000

00)sin()cos(

00)cos()sin(

1111

1111

01

T

61

0000

0000

00)sin()cos(

00)cos()sin(

1111

1111

01

T


0000

0000

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

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

112

11112

11

112

11112

11

01

T

000

0866.05.0

05.0866.02

12

1

21

21

01

R


Similarly at t=162

1000

00)cos()sin(

0100

0)sin()cos(

22

22

12

A

T

0866.05.0

100

05.0866.012R

05.0866.0

000

0866.05.0

22

2212

R

0000

00)sin()cos(

0000

00)cos()sin(

2222

2222

12

T


0000

00)sin()cos(

0000

00)cos()sin(

2222

2222

12

T

0000


0000


2222222

22

2222222

222

12

T

0866.05.0

000

05.0866.0

22

22

22

22

12

R

At t=1,63

1000

0100

00)cos()sin(

0)sin()cos(

33

33

23

B

T

000

05.0866.0

0866.05.0

33

3323

R

100

0866.05.0

05.0866.023R

0000

0000

00)sin()cos(

00)cos()sin(

3333

3333

23

T


0000

0000

00)sin()cos(

00)cos()sin(

3333

3333

23

T


0000

0000



3323333

233

3323333

233

23

T

000

0866.05.0

05.0866.023

23

23

23

23

R

TTT RRRRRRt )(

00 0/101

01

01

010/1

TT RRRR


Substitute the matrices given into the equation, we get:

Similarly

00

00

2/323

23

23

232/3

1/212

12

12

121/2

TT

TT

RRRR

RRRR

TPP 001;6 3/321

We need to find 0/Pa

With

5236.0

0

0

0/1

5236.0

0

0

2/3

0

5236.0

0

1/2

For the data given, the following were determined in the previous lecture:


0

4534.0

2618.0

2/Pv

4304.1

0

4304.1

1/Pv

4304.1

6571.1

6084.2

0/Pv

02/31/20/1


3/

233/

232/3

3/232/32/33/

232/32/32/

)(2

)(

PP

PPoP

aRvR

PRPRaa

0

0

0

3/

3/

2/3

P

P

o

v

a

a

There is no translation acceleration between frames {3} and {2} and no translation velocity and acceleration of point “P” in frame {3}

3/232/32/32/ PP PRa

0

1371.0

2374.0

0

0

1

100

0866.05.0

05.0866.0

5236.0

0

0

5236.0

0

0

2/Pa


2/

122/

121/2

2/121/21/22/

121/21/21/

)(2

)(

PP

PPoP

aRvR

PRPRaa

01/2 oa There is no translation acceleration between

frames {2} and {1}

TPoP PRPP 05.0866.23/232/32/

TPa 01371.02374.02/

01/2

TPP PRv 04534.02618.03/232/32/

2237.1

0

023.1

0

1371.0

2374.0

0866.05.0

100

05.0866.0

0

4534.0

2618.0

0866.05.0

100

05.0866.0

0

5236.0

0

2

0

5.0

866.2

0866.05.0

100

05.0866.0

0

5236.0

0

0

5236.0

0

1/

1/

P

P

a

a


Substituting the values into the equation:

2/122/

121/22/

121/21/21/ )(2 PPPP aRvRPRa


1/

011/

010/1

1/010/10/11/

010/10/10/

)(2

)(

PP

PPoP

aRvR

PRPRaa

00/1 oa There is no translation acceleration between

frames {1} and {0}

TPa 2237.10023.11/

0/ ab

TPoP PRPP 866.10232.52/121/21/

TPPP vRPRv 4304.104304.12/122/

121/21/


Substituting the values into the equation:

1/011/

010/11/

010/10/10/ )(2 PPPP aRvRPRa

22.1

53.2

38.1

2237.1

0

023.1

100

0866.05.0

05.0866.0

4304.1

0

4304.1

100

0866.05.0

05.0866.0

5236.0

0

0

2

866.1

0

232.5

100

0866.05.0

05.0866.0

5236.0

0

0

5236.0

0

0

0/

0/

P

P

a

a


We should get the same answer if we use transformation matrix method.

Try it at home and we’ll discuss this in the next lecture!

Summary

This lecture continues the discussion on the analysis of the instantaneous motion of a rigid body, i.e. the velocities and accelerations associated with a rigid body as it moves from one configuration to another.

The following were covered:•The acceleration tensor and angular acceleration tensor•The principles of relative motion between bodies in terms of acceleration analysis

INTRODUCTION TO DYNAMICS ANALYSIS OF ROBOTS (Part 3)

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