introductiontorobotics(2nded) - niku - solutionsmanual

Introduction to Robotics Analysis, Control, Applications

Solution Manual

Saeed B. Niku

Copyrighted 2010. This solution manual may not be copied, posted, made available to students, placed on

BlackBoard or any other electronic media system and the Internet without prior expressed consent of the copyright owner.

Copyrighted 2010. This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other electronic media system or the Internet without prior expressed consent of the copyright owner.

2

CHAPTER ONE Problem 1.1 Draw the approximate workspace for the following robot. Assume the dimensions of the base and other parts of the structure of the robot are as shown. Estimated student time to complete: 15-25 minutes Prerequisite knowledge required: Text Section(s) 1.14 Solution: The workspace shown is approximate.


3

Problem 1.2 Draw the approximate workspace for the following robot. Assume the dimensions of the base and other parts of the structure of the robot are as shown. Estimated student time to complete: 20-30 minutes Prerequisite knowledge required: Text Section(s) 1.14 Solution: The workspace shown is approximate.


4

Problem 1.3 Draw the approximate workspace for the following robot. Assume the dimensions of the base and other parts of the structure of the robot are as shown. Estimated student time to complete: 10-15 minutes Prerequisite knowledge required: Text Section(s) 1.14 Solution: The workspace shown is approximate.


5

CHAPTER TWO Problem 2.1 Write a unit vector in matrix form that describes the direction of the cross product of

5 3= +p i k and 3 4 5= + +q i j k . Estimated student time to complete: 5-10 minutes Prerequisite knowledge required: Text Section(s) 2.4 Solution:

( ) ( ) ( )2 2 2

5 0 3 0 12 25 9 20 0 12 16 203 4 5

144 256 400 28.28

1228.28 0.424

16 0.56628.28

0.7072028.28

x y zr r r

= = = + = + = + + = + + =

= =

i j kr p q i j k i j k

r


6

Problem 2.2 A vector p is 8 units long and is perpendicular to vectors q and r described below. Express the vector in matrix form.

0.3

0.40

yunit

q =

q 0.50.40

x

unit

r = r

Estimated student time to complete: 15-20 minutes Prerequisite knowledge required: Text Section(s) 2.4 Solution: The two vectors given are unit vectors. Therefore, each missing component can be found as:

1 0.09 0.16 0.866

1 0.25 0.16 0.768y

x

q

r

= == =

Since p is perpendicular to the other two vectors, it is in the direction of the cross product of the two. Therefore:

( ) ( ) ( )( ) ( ) ( )0.3 0.866 0.4 0.346 0.2 0.12 0.307 0.15 0.665

0.768 0.5 0.4

0.146 0.187 0.515

p = = + = +

i j ki j k

i j k

Since q and r are not perpendicular to each other, the resulting p is not a unit vector. Vector p can be found as:

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

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

2 2 2

0.146 0.187 0.515

0.146 0.187 0.515 0.567

8 14.10.567

0.146 0.187 0.515

2.06 2.64 7.27

p

p

w

w

= + = + + =

= == + = +

i j k

p i j k

p i j k


7

Problem 2.3 Will the three vectors p, q, and r in Problem 2.2 form a traditional frame? If not, find the necessary unit vector s to form a frame between p, q, and s. Estimated student time to complete: 15-20 minutes Prerequisite knowledge required: Text Section(s) 2.4 Solution: As we saw in Problem 2.2, since qr is not a unit vector, it means that q and r and not perpendicular to each other, and therefore, they cannot form a frame. However, p and q are perpendicular to each other, and we can select s to be perpendicular to those two. Of course, p is not a unit length, therefore we use the unit vector representing it.

( ) ( ) ( )

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

2 2 20.146 0.187 0.515 0.567

1 1.7640.567

0.146 0.187 0.515

0.257 0.33 0.908

p

w

w

= + + == == + = +

p i j k

p i j k

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

0.257 0.33 0.908

0.257 0.33 0.908 0.918 0.375 0.1240.3 0.866 0.4

= + = = +

p i j k

i j ks i j k


8

Problem 2.4 Suppose that instead of a frame, a point ( )3,5,7 TP = in space was translated a distance of ( )2,3,4 Td = . Find the new location of the point relative to the reference frame. Estimated student time to complete: 5 minutes Prerequisite knowledge required: Text Section(s) 2.6 Solution: As for a frame,

1 0 0 2 3 50 1 0 3 5 80 0 1 4 7 110 0 0 1 1 1

newP

= =


9

Problem 2.5 The following frame B was moved a distance of ( )5,2,6 Td = . Find the new location of the frame relative to the reference frame.

0 1 0 21 0 0 40 0 1 60 0 0 1

B

=

Estimated student time to complete: 5-10 minutes Prerequisite knowledge required: Text Section(s) 2.6 Solution: The transformation matrix representing the translation is used to find the new location as:

1 0 0 5 0 1 0 2 0 1 0 70 1 0 2 1 0 0 4 1 0 0 60 0 1 6 0 0 1 6 0 0 1 120 0 0 1 0 0 0 1 0 0 0 1

newB

= =


10

Problem 2.6 For frame F, find the values of the missing elements and complete the matrix representation of the frame.

? 0 1 5? 0 0 3? 1 0 20 0 0 1

F

=

Estimated student time to complete: 10 minutes Prerequisite knowledge required: Text Section(s) 2.4 Solution:

0 1 50 0 31 0 2

0 0 0 1

x

y

z

nn

Fn

=

From =n o a 0 0 1

x y zn n n =

i j ki

Or: ( ) ( ) ( )0y xn n + = i j k i , and therefore: 1, 0, 0y x zn n n= = =

0 0 1 51 0 0 30 1 0 20 0 0 1

F

=


11

Problem 2.7 Find the values of the missing elements of frame B and complete the matrix representation of the frame.

0.707 ? 0 2? 0 1 4? 0.707 0 50 0 0 1

B

=

Estimated student time to complete: 15-20 minutes Prerequisite knowledge required: Text Section(s) 2.4 Solution:

0.707 0 20 1 4

0.707 0 50 0 0 1

x

y

z

on

Bn

=

From =n o a 0.7070 0.707

y z

x

n no

=

i j kj

Therefore: 0.7070 0.707

y z

x

n no

=

i j kj

And ( ) ( ) ( )0.707 0.5 0y z x y x yn n o n o n + = =i j k j

From length equations: 1=n or 2 2 2

2

0.707 1 0.707

0.5 1 0.707y z z

x x

n n n

o o

+ + = = + = =

Therefore, there are two possible acceptable solutions:

0.707 0.707 0 20 0 1 4

0.707 0.707 0 50 0 0 1

B

= and

0.707 0.707 0 20 0 1 4

0.707 0.707 0 50 0 0 1

B

=


12

Problem 2.8 Derive the matrix that represents a pure rotation about the y-axis of the reference frame. Estimated student time to complete: 10 minutes Prerequisite knowledge required: Text Section(s) 2.6.2. Solution: From the figure:

cos sin

sin cos

x n a

y o

z n a

p p pp pp p p

= +== +

and 0

0 1 00

x n

y o

z a

p C S pp pp S C p

=


13

Problem 2.9 Derive the matrix that represents a pure rotation about the z-axis of the reference frame. Estimated student time to complete: 10 minutes Prerequisite knowledge required: Text Section(s) 2.6.2. Solution: From the Figure:

cos sinsin cos

x n o

y n o

z a

p p pp p pp p

= = +=

and 00

0 0 1

x n

y o

z a

p C S pp S C pp p

=


14

Problem 2.10 Verify that the rotation matrices about the reference frame axes follow the required constraint equations set by orthogonality and length requirements of directional unit vectors. Estimated student time to complete: 10 minutes. Prerequisite knowledge required: Text Section(s) 2.4.5 and 2.6.2 Solution:

For 1 0 0

( , ) 00

Rot x C SS C

=

we have:

0=n oi or nx ox + ny oy + nz oz = 0 0=n ai 0=a oi

1=n 2 2 1S C= + =o

( )2 2 1S C= + =a

( , )Rot y and ( , )Rot z will be the same.


15

Problem 2.11 Find the coordinates of point (2,3,4)TP relative to the reference frame after a rotation of 45 D about the x-axis. Estimated student time to complete: 10 minutes Prerequisite knowledge required: Text Section(s) 2.6.3. Solution:

2 1 0 0 2 2( , 45) 3 0 0.707 0.707 3 0.707

4 0 0.707 0.707 4 4.95

U P Rot x = = =

Note that the rotation is written in a 3 3 , not homogeneous form, because we are only concerned about the rotation part.


16

Problem 2.12 Find the coordinates of point (3,5,7)TP relative to the reference frame after a rotation of 30 D about the z-axis. Estimated student time to complete: 10 minutes Prerequisite knowledge required: Text Section(s) 2.6.3. Solution:

3 0.866 0.5 0 3 0.1( ,30) 5 0.5 0.866 0 5 5.83

7 0 0 1 7 7

U P Rot z = = =

Note that the rotation is written in a 3 3 form not homogeneous form, because we are only concerned about the rotation part.


17

Problem 2.13 Find the new location of point (1, 2,3)TP relative to the reference frame after a rotation of 30D about the z-axis followed by a rotation of 60D about the y-axis. Estimated student time to complete: 10 minutes Prerequisite knowledge required: Text Section(s) 2.6.3. Solution:

( ,60) ( ,30)0.5 0 0.866 0 0.866 0.5 0 0 1 2.5310 1 0 0 0.5 0.866 0 0 2 2.232

0.866 0 0.5 0 0 0 1 0 3 1.6160 0 0 1 0 0 0 1 1 1

U

U

P Rot y Rot z P

P

= = =


18

Problem 2.14 A point P in space is defined as (5,3,4)B TP = relative to frame B which is attached to the origin of the reference frame A and is parallel to it. Apply the following transformations to frame B and find AP. Using the 3-D grid, plot the transformations and the result and verify it. Also verify graphically that you would not get the same results if you apply the transformations relative to the current frame: Rotate 90 D about x-axis, then Translate 3 units about y-axis, 6 units about z-axis, and 5 units about x-axis. Then, Rotate 90 D about z-axis. Estimated student time to complete: 15-20 minutes Prerequisite knowledge required: Text Section(s) 2.6.3. Solution:

( ,90) (5,3,6) ( ,90)0 1 0 0 1 0 0 5 1 0 0 0 5 11 0 0 0 0 1 0 3 0 0 1 0 3 100 0 1 0 0 0 1 6 0 1 0 0 4 90 0 0 1 0 0 0 1 0 0 0 1 1 1

A B

A

P Rot z Trans Rot x P

P

= = =


19

Problem 2.15 A point P in space is defined as (2,3,5)B TP = relative to frame B which is attached to the origin of the reference frame A and is parallel to it. Apply the following transformations to frame B and find AP. Using the 3-D grid, plot the transformations and the result and

verify it: Rotate 90 D about x-axis, then Rotate 90 D about local a-axis, then Translate 3 units about y-, 6 units about z-, and 5 units about x-axes. Estimated student time to complete: 15-20 minutes Prerequisite knowledge required: Text Section(s) 2.6.4. Solution:

(5,3,6) ( ,90) ( ,90)1 0 0 5 1 0 0 0 0 1 0 0 2 20 1 0 3 0 0 1 0 1 0 0 0 3 20 0 1 6 0 1 0 0 0 0 1 0 5 80 0 0 1 0 0 0 1 0 0 0 1 1 1

A B

A

P Trans Rot x Rot a P

P

= = =


20

Problem 2.16 A frame B is rotated 90D about the z-axis, then translated 3 and 5 units relative to the n- and o-axes respectively, then rotated another 90D about the n-axis, and finally, 90D about the y-axis. Find the new location and orientation of the frame.

0 1 0 11 0 0 10 0 1 10 0 0 1

B

=

Estimated student time to complete: 15-20 minutes Prerequisite knowledge required: Text Section(s) 2.6.4. Solution:

( ,90) ( ,90) (3,5,0) ( ,90)0 0 1 0 0 1 0 0 0 1 0 1 1 0 0 3 1 0 0 0 0 1 0 10 1 0 0 1 0 0 0 1 0 0 1 0 1 0 5 0 0 1 01 0 0 0 0 0 1 0 0 0 1 1 0 0 1 0 0 1 0 0

0 0 0 1 0 0 0 1 0 0 0 1 0 0 0 1 0 0 0 1

new

new

B Rot y Rot z BTrans Rot n

B

= = =

0 0 1 61 0 0 40 0 0 1


21

Problem 2.17 The frame B of problem 2.16 is rotated 90D about the a-axis, 90D about the y-axis, then translated 2 and 4 units relative to the x- and y-axes respectively, then rotated another 90D about the n-axis. Find the new location and orientation of the frame.

0 1 0 11 0 0 10 0 1 10 0 0 1

B

=

Estimated student time to complete: 15-20 minutes Prerequisite knowledge required: Text Section(s) 2.6.4. Solution:

(2,4,0) ( ,90) ( ,90) ( ,90)1 0 0 2 0 0 1 0 0 1 0 1 0 1 0 0 1 0 0 0 0 1 0 30 1 0 4 0 1 0 0 1 0 0 1 1 0 0 0 0 0 1 00 0 1 0 1 0 0 0 0 0 1 1 0 0 1 0 0 1 0 00 0 0 1 0 0 0 1 0 0 0 1 0 0 0 1 0 0 0 1

new

new

B Trans Rot y B Rot a Rot n

B

= = =

0 0 1 51 0 0 1

0 0 0 1


22

Problem 2.18 Show that rotation matrices about the y- and the z-axes are unitary. Estimated student time to complete: Prerequisite knowledge required: Text Section(s) 2.7. Solution:

0( , ) 0 1 0

0

C SA Rot y

S C

= =

2 2det 1( ) 0 1A C S = + + =

1 0 000

TA C SS C

=

1

0adj 0 1 0 /1

0

Tminor

C SA A A

S C

= = =

1 TA A =

Exact same is true for rotation about the z-axis.


23

Problem 2.19 Calculate the inverse of the following transformation matrices:

1 2

0.527 0.574 0.628 2 0.92 0 0.39 50.369 0.819 0.439 5 0 1 0 6

and0.766 0 0.643 3 0.39 0 0.92 2

0 0 0 1 0 0 0 1

T T

= =

Estimated student time to complete: 15-20 minutes. Prerequisite knowledge required: Text Section(s) 2.7. Solution: Since transformation matrices are unitary, we calculate the inverses simply by transposing the rotation part and calculating the position part by: 0.527 2 0.369 5 0.766 3 0.601

0.574 2 0.819 5 0 3 2.9470.628 2 0.439 5 0.643 3 5.38

+ = + =

+ + =

Therefore:

11

0.527 0.369 0.766 0.6010.574 0.819 0 2.947

0.628 0.439 0.643 5.380 0 0 1

T

=

And similarly,

12

0.92 0 0.39 3.820 1 0 6

0.39 0 0.92 3.790 0 0 1

T

=


24

Problem 2.20 Calculate the inverse of the matrix B of Problem 2.17. Estimated student time to complete: 5 minutes Prerequisite knowledge required: Text Section(s) 2.7. Solution:

0 1 0 11 0 0 10 0 1 10 0 0 1

B

=

1

0 1 0 11 0 0 10 0 1 10 0 0 1

B

=


25

Problem 2.21 Write the correct sequence of movements that must be made in order to restore the original orientation of the spherical coordinates and make it parallel to the reference frame. About what axes are these rotations supposed to be? Estimated student time to complete: 10 minutes Prerequisite knowledge required: Text Section(s) 2.9.3. Solution:

1 0 00 1 0

( , , ) ( , ) ( , )0 0 10 0 0 1

sph

rS CrS S

T r Rot o Rot arC

=

As shown, the rotations are relative to o- and a-axes in the sequence shown. These rotations must be relative to the current moving frame in order to prevent changing the position of the frame.


26

Problem 2.22 A spherical coordinate system is used to position the hand of a robot. In a certain situation, the hand orientation of the frame is later restored in order to be parallel to the reference frame, and the matrix representing it is described as:

1 0 0 3.13750 1 0 2.1950 0 1 3.2140 0 0 1

sphT

=

Find the necessary values of , ,r to achieve this location. Find the components of the original matrix , ,n o a vectors for the hand before the orientation was restored. Estimated student time to complete: 20 minutes Prerequisite knowledge required: Text Section(s) 2.9.3. Solution: a) The equations representing position in a spherical coordinates may be set equal to the given values as: i. 3.1375ii. 2.195iii. 3.214

rC SrS SrC

==

=

I. Assuming S is positive, from i and ii 35 = D from ii and iii 50 = D from iii r=5 units. II. If S were negative, then 215 = D 50 = D r=5 units. Since orientation is not specified, no more information is available to check the results. b) For case I, substitute corresponding values of S , C , S ,C and r in spherical coordinates to get:

0.5265 0.5735 0.6275 3.13750.3687 0.819 0.439 2.195

( , , ) (35,50,5)0.766 0 0.6428 3.214

0 0 0 1

sph sphT r T = =


27

Problem 2.23 Suppose that a robot is made of a Cartesian and RPY combination of joints. Find the necessary RPY angles to achieve the following:

0.527 0.574 0.628 40.369 0.819 0.439 60.766 0 0.643 9

0 0 0 1

T

=

Estimated student time to complete: 20 minutes Prerequisite knowledge required: Text Section(s) 2.10.1 Solution: The position is achieved by Cartesian joints. Therefore, the robot moves 4, 6, and 9 units along the x-, y-, and z-axes. The orientation is achieved by RPY rotations, therefore: 2( , ) 2(0.369,0.527) 35a y xATAN n n ATAN = = = D or 2( , ) 2( 0.369, 0.527) 215a y xATAN n n ATAN = = = D For 35a = D

( )( )2( , ( )) 2 0.766, 0.527 0.819 0.369 0.574 50o z x a y aATAN n n C n S ATAN = + = + = D For 215a = D

( )2 0.766, 0.643 130o ATAN = = D For 35a = D

( ) ( )( )

2(( ), ( ))

2

2

0.439 0.819 0.628 0.574 , 0.819 0.819 0.574 0.574

0,1 0

n y a x a y a x aATAN a C a S o C o S

ATAN

ATAN

= + ==

+ + = D

For 215a = D

( )2 0, 1 180n ATAN = = D

Either solution is acceptable: 35 21550 1300 180

a a

o o

n n

= = = = = =

D D

D D

D D


28

Problem 2.24 Suppose that a robot is made of a Cartesian and Euler combination of joints. Find the necessary Euler angles to achieve the following:

0.527 0.574 0.628 40.369 0.819 0.439 60.766 0 0.643 9

0 0 0 1

T

=

Estimated student time to complete: 20 minutes Prerequisite knowledge required: Text Section(s) 2.10.2 Solution: The position is achieved by Cartesian joints. Therefore, the robot moves 4, 6, and 9 units along the x-, y-, and z-axes. The orientation is achieved by Euler rotations, therefore:

2( , ) 2( , )y x y xATAN a a or ATAN a a = = or

2(0.439,0.628) 352( 0.439, 0.628) 215

ATANATAN

= = = =

D

D

2[( ), ( )]x y x yATAN n S n C o S o C = + +

or 2[0,1] 02[0, 1] 180

ATANATAN

= = = =

D

D

2[( ), )]x y zATAN a C a S a = +

or ( ) ( )2[(0.628 0.819 0.439 0.573),0.643)] 502[(0.628 0.819 0.439 0.573 ),0.643)] 50

ATANATAN

= + = = + =

D

D

Then 35 2150 180

50 50

= = = = = =

D D

D D

D D


29

Problem 2.25 Assume that the three Euler angles used with a robot are 30 ,40 ,50D D D respectively. Determine what angles should be used to achieve the same result if RPY is used instead. Estimated student time to complete: 20-25 minutes Prerequisite knowledge required: Text Section(s) 2.10 Solution: For Euler angles: Euler( , , ) ( , ) ( , ), ( , ) ( ,30) ( , 40), ( ,50)Rot a Rot o Rot a Rot a Rot o Rot a = =

=

0.0435 0.8296 0.5568 00.9096 0.2635 0.3215 00.4134 0.4925 0.766 0

0 0 0 1

2( , ) 2(0.9096,0.0435) 87.26a y xATAN n n ATAN = = = D

or 2( , ) 2( 0.9096, 0.0435) 267.26a y xATAN n n ATAN = = = D

( )2( , ( )) 2 0.4134,0.9106 24.5o z x a y aATAN n n C n S ATAN = + = = D or ( )2 0.4134, 0.9106 155.5o ATAN = = D

( )2(( ), ( ))

2 0.5408,0.8412 32.7n y a x a y a x aATAN a C a S o C o S

ATAN

= + = = D

or 2( 0.5408, 0.8412) 212.7n ATAN = = D


30

Problem 2.26 A frame UB was moved along its own o-axis a distance of 6 units, then rotated about its n-axis an angle of 60D , then translated about the z-axis for 3 units, followed by a rotation of 60D about the z-axis, and finally rotated about x-axis for 45D . Calculate the total transformation performed. What angles and movements would we have to make if we were to create the same location and orientation using Cartesian and Euler configurations? Estimated student time to complete: 25-30 minutes Prerequisite knowledge required: Text Section(s) 2.9- 2.10 Solution:

( , 45) ( ,60) (0,0,3) (0,6,0) ( ,60)0.5 0.433 0.75 5.196

0.6123 0.4355 0.6596 00.6123 0.789 0.0474 4.242

0 0 0 1

UB

new

T Rot x Rot z Trans Trans Rot n

B

= =

The Cartesian translations are: 5.196, 0, 4.242x x xp p p= = = . Euler angles are:

2( , ) 2( , )y x y xATAN a a or ATAN a a = = or

2( 0.6596,0.75) 41.332(0.6596, 0.75) 138.67

ATANATAN

= = = =

D

D

2[( ), ( )]x y x yATAN n S n C o S o C = + +

or 2[0.79, 0.613] 127.82[ 0.79,0.613] 52.2

ATANATAN

= = = =

D

D

2[( ), )]x y zATAN a C a S a = +

or 2[0.9988,0.0474] 87.282[ 0.9988,0.0474] 87.28

ATANATAN

= = = =

D

D


31

Problem 2.27 A frame UF was moved along its own n-axis a distance of 5 units, then rotated about its o-axis an angle of 60D , followed by a rotation of 60D about the z-axis, then translated about its a-axis for 3 units, and finally rotated 45D about the x-axis. Calculate the total transformation performed. What angles and movements would we have to make if we were to create the same location and orientation using Cartesian and RPY configurations? Estimated student time to complete: 20-25 minutes Prerequisite knowledge required: Text Section(s) 2.11 Solution:

( , 45) ( ,60) (5,0,0) ( ,60) (0,0,3)0.25 0.866 0.433 3.8

0.918 0.354 0.177 3.590.306 0.354 0.884 5.71

0 0 0 1

UB

new

T Rot x Rot z Trans Rot o Trans

B

= =

For Cartesian coordinates: 3.8, 3.59, 5.71x y zp p p= = = For RPY angles:

2( , ) 2(0.918,0.25) 74.8a y xATAN n n ATAN = = = D or 2( , ) 254.8a y xATAN n n = = D

( )2( , ( )) 2 0.306,0.951 17.8o z x a y aATAN n n C n S ATAN = + = = D ( )2 0.306, 0.951 162.2o ATAN = = D

[ ]2(( ), ( ))

2 0.372,0.928 21.8n y a x a y a x aATAN a C a S o C o S

ATAN

= + = = D

( )2 0.372, 0.928 201.8n ATAN = = D


32

Problem 2.28 Frames describing the base of a robot and an object are given relative to the Universe frame. Find a transformation RTH of the robot configuration if the hand of the robot is to be placed on the object. By inspection, show whether this robot can be a 3-axis spherical robot, and if so, find

, , r . Assuming that the robot is a 6-axis robot with Cartesian and Euler coordinates, find

, , , , ,x y zp p p .

1 0 0 1 0 1 0 20 0 1 4 1 0 0 10 1 0 0 0 0 1 00 0 0 1 0 0 0 1

U Uobj RT T

= =

Estimated student time to complete: 20-25 minutes Prerequisite knowledge required: Text Section(s) 2.11 Solution: a. We want to place the hand on the part. Therefore, 1R U UH H objT T T

=

1U U R U R R U Uobj R H R obj H R objT T T T T T T T

= = =

Substitute:

0 1 0 1 1 0 0 1 0 0 1 51 0 0 2 0 0 1 4 1 0 0 1

0 0 1 0 0 1 0 0 0 1 0 00 0 0 1 0 0 0 1 0 0 0 1

RHT

= =

b. No. The 3,2 element of spherical coordinate transformation matrix should be zero. This is not. c. 5, 1, 0x y zp p p= = =

2[0, 1] 180 or 2[0,1] 0ATAN ATAN = = = =D D 2[ 1( 1),0] 90 or 2[ 1(1),0] 270ATAN ATAN = = = =D D 2[ 1( 1),0] 90 or 2[ 1(1),0] 270ATAN ATAN = = = =D D

Then 180 090 27090 270

= = = = = =

D D

D D

D D


33

Problem 2.29 A 3-DOF robot arm has been designed for applying paint on flat walls, as shown. Assign coordinate frames as necessary based on the D-H representation. Fill out the parameters table. Find the UTH matrix.

Estimated student time to complete: 30 minutes Prerequisite knowledge required: Text Section(s) 2.12 Solution: In this solution, the locations of the origins of some of the frames are arbitrary. Therefore, intermediate matrices might be different for each case. However, the final answer should be the same. Please also note that no particular reset position is specified.

# d a

0-1 1 l4 0 0 1-2 0 0 l5 -90 2-H 90 l6 0 0

1 2 3

1 1 1 5

2 1 1

3 4 6

1 1 1 6 1 5 1

1 1 1 6 1 5 2

4 3

1 0 0 0 0 1 0 0 0 1 0 00 1 0 0 0 0 0 1 0 1 0 0 00 0 1 0 0 1 0 1 0 0 0 0 10 0 0 1 0 0 0 1 0 0 0 1 0 0 0 1

001 0 0

0 0 0 1

U U R UH R H RT T T T A A A

l C S ll S Cl l l

C S S l C l lS C C l S l l

l l

= = =

+ + + += +


34

Problem 2.30 In the 2-DOF robot shown, the transformation matrix 0TH is given in symbolic form, as well as in numerical form for a specific location. The length of each link l1 and l2 is 1 ft. Calculate the values of 1 2 and for the given location.

12 12 2 12 1 1

12 12 2 12 1 10

0 0.2924 0.9563 0 0.69780 0.9563 0.2924 0 0.8172

0 0 1 0 0 0 1 00 0 0 1 0 0 0 1

H

C S l C l CS C l S l S

T

+ + = =

z

x

y

x0

z1

x1

zH

xH21

z0

Figure P.2.30

Estimated student time to complete: 20 minutes Prerequisite knowledge required: Text Section(s) 2.13 Solution: This example is very simple, with only two degrees of freedom. Therefore, there is really no need to use the traditional method of inverse kinematic mentioned in Section 2.13. The joint angles can be determined as follows:

( )( ) ( ) ( )( )

1 2 12 12

1 2 12 12 12 1 11 2 12 1 2 12 1

2 12 1 1 1 2 12 1

2( , ) 2(0.9563, 0.2924) 107

/2 / , /

/xx

y xy y

ATAN S C ATAN

C p l C ll C l C pATAN p l S l p l C l

l S l S p S p l S l

+ = = == + = = + = =

D

Substitute:

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

1 2 12 1 2 12 1

1

2

2 / , /

2 0.8172 0.9563 , 0.6978 0.2924 2 0.1391,0.9902 8

107 ( 8) 115

y xATAN p l S l p l C l

ATAN ATAN

= = + = = = =

D

D


35

Problem 2.31 For the following SCARA-type robot: Assign the coordinate frames based on the D-H representation. Fill out the parameters table. Write all the A matrices. Write the UTH matrix in terms of the A matrices. Estimated student time to complete: 20-30 minutes Prerequisite knowledge required: Text Section(s) 2.12 Solution: In this solution, the locations of the origins of some of the frames are arbitrary. Therefore, intermediate matrices might be different for each case. However, the final answer should be the same. Please also note that no particular reset position is specified.

# d a

0-1 1 0 d3 0 1-2 2 0 d4 180 2-H 3 d5 0 0

0

0 0 1 2 3

1 1 3 1 2 2 4 2 3 3

1 1 3 1 2 2 4 2 3 3

1 2 5

1 0 0 0 0 0 0 00 1 0 0 0 0 0 00 0 1 0 0 1 0 0 0 1 0 0 0 10 0 0 1 0 0 0 1 0 0 0 1 0 0 0 1

U U UH HT T T T A A A

C S d C C S d C C SS C d S S C d S S C

d d d

= = = +

Multiplying these matrices will yield the total transformation.


36

Problem 2.32 A special 3-DOF spraying robot has been designed as shown: Assign the coordinate frames based on the D-H representation. Fill out the parameters table. Write all the A matrices. Write the UTH matrix in terms of the A matrices.

Estimated student time to complete: 20-30 minutes Prerequisite knowledge required: Text Section(s) 2.12 Solution: Please note that no particular reset position is specified for this robot.

# d a

0-1 190 + 0 0 90 1-2 0 l2 0 -90 2-3 3 0 0 90 3-H 0 l3 0 0

1 1 1

1 10 1

3 3

3 32 3 4

2 3

1 0 0 0 00 1 0 0 0 00 0 1 0 0 1 0 00 0 0 1 0 0 1 1

1 0 0 0 0 0 1 0 0 00 0 1 0 0 0 0 1 0 00 1 0 0 1 0 0 0 0 10 0 0 1 0 0 0 1 0 0 0 1

U

l S CC S

T A

C SS C

A A Al l

= = = = =

00 0 1 2 3 4

U U UH HT T T T A A A A= =

The total transformation can be summarized by multiplying these matrices.


37

Problem 2.33 For the Unimation Puma 562, 6-axis robot shown, Assign the coordinate frames based on the D-H representation. Fill out the parameters table. Write all the A matrices. Find the RTH matrix for the following values: Base height=27 in, d2=6 in, a2=15 in, a3=1 in, d4=18 in 1 2 3 4 5 60 , 45 , 0 , 0 , 45 , 0 = = = = = =D D D D D D

Estimated student time to complete: 30-40 minutes Prerequisite knowledge required: Text Section(s) 2.12 Solution: In this solution, the locations of the origins of some frames are arbitrary. Therefore, intermediate matrices might be different for each case. However, the final answer should be the same. To express the height from the table, the 27-in base height must be added to the zp value. Please also note that no particular reset position is specified. Therefore, the results relate to a position and orientation that would correspond to the given transformations from a reset position. We assume at reset, there is 90 degrees between 0x and 1x . Please note how frames 2, 3, and 4 relate to each other by referring to the side view of the arm.

(Reprinted with permission from Staubli Robotics.)


38

1 2 3

4 5

0 0 1 0 0.707 0.707 0 10.61 1 0 0 01 0 0 0 0.707 0.707 0 10.61 0 0 1 00 1 0 0 0 0 1 6 0 1 0 00 0 0 1 0 0 0 1 0 0 0 1

1 0 0 0 0.707 0 0.707 00 0 1 0 0.707 0 0.707 00 1 0 18 0 1 0 00 0 0 1 0 0 0 1

A A A

A A

= = = = =

6

06

1 0 0 00 1 0 00 0 1 00 0 0 1

0 1 0 61 0 0 1.410 0 1 240 0 0 1

A

T

= =


39

Problem 2.34 For the given 4-DOF robot: Assign appropriate frames for the Denavit-Hartenberg representation. Fill out the parameters table. Write an equation in terms of A-matrices that shows how U HT can be calculated. Estimated student time to complete: 20-30 minutes Prerequisite knowledge required: Text Section(s) 2.12 Solution: In this solution, the locations of the origins of some of the frames are arbitrary. Therefore, intermediate matrices might be different for each case. However, the final answer should be the same. Please also note that no particular reset position is specified.

1

200 0 1 2 3 4 1 2 3 4

3

1 0 00 1 00 0 10 0 0 1

U U UH H

ll

T T T T A A A A A A A Al

= = =


40

Problem 2.35 For the given 4-DOF robot designed for a specific operation: Assign appropriate frames for the Denavit-Hartenberg representation. Fill out the parameters table. Write an equation in terms of A-matrices that shows how U HT can be calculated. Estimated student time to complete: 20-30 minutes Prerequisite knowledge required: Text Section(s) 2.12 Solution: In this solution, the locations of the origins of some of the frames are arbitrary. Therefore, intermediate matrices might be different for each case. However, the final answer should be the same. Please also note that no particular reset position is specified.

* ( )4 5 / sinL l l = +

The total transformation for the robot can be determined from: 1

00 0 1 2 3 4 1 2 3 4

1 0 00 1 0 00 0 1 00 0 0 1

U U UH H

l

T T T T A A A A A A A A

= = =


41

Problem 2.36 For the given specialty-designed 4-DOF robot: Assign appropriate frames for the Denavit-Hartenberg representation. Fill out the parameters table. Write an equation in terms of A-matrices that shows how U HT can be calculated. Estimated student time to complete: 20-30 minutes Prerequisite knowledge required: Text Section(s) 2.12 Solution: In this solution, the locations of the origins of some of the frames are arbitrary. Therefore, intermediate matrices might be different for each case. However, the final answer should be the same. Please also note that no particular reset position is specified.

1

200 0 1 2 3 1 2 3

1 0 00 1 00 0 1 00 0 0 1

U U UH H

ll

T T T T A A A A A A

= = =


42

Problem 2.37 For the given 3-DOF robot: Assign appropriate frames for the Denavit-Hartenberg representation. Fill out the parameters table. Write an equation in terms of A-matrices that shows how U HT can be calculated. Estimated student time to complete: 20-30 minutes Prerequisite knowledge required: Text Section(s) 2.12 Solution: In this solution, the locations of the origins of some of the frames are arbitrary. Therefore, intermediate matrices might be different for each case. However, the final answer should be the same. Please also note that no particular reset position is specified. Please also note that there can be an additional frame between frames 1 and H. In that case, the two joint variables will be represented by two separate matrices.

# d a 0-1 190 + 3 4l l 0 -90 1-H 2 5 6 7l l l+ + 0 0

1

200 0 1 2 1 2

1 0 00 1 00 0 1 00 0 0 1

U U UH H

ll

T T T T A A A A

= = =


43

Problem 2.38 For the given 4-DOF robot: Assign appropriate frames for the Denavit-Hartenberg representation. Fill out the parameters table. Write an equation in terms of A matrices that shows how U HT can be calculated. Estimated student time to complete: 20-30 minutes Prerequisite knowledge required: Text Section(s) 2.12 Solution: In this solution, the locations of the origins of some of the frames are arbitrary. Therefore, intermediate matrices might be different for each case. However, the final answer should be the same. Please also note that no particular reset position is specified.

# d a 0-1 1 3l 4l 90 1-2 290 + 5l 0 90 2-H 3 6l 0 0

1

200 0 1 2 3 1 2 3

1 0 00 1 00 0 1 00 0 0 1

U U UH H

ll

T T T T A A A A A A

= = =


44

Problem 2.39 Derive the inverse kinematic equations for the robot of Problem 2.36. Estimated student time to complete: 20-30 minutes if Problem 2.36 is already solved Prerequisite knowledge required: Text Section(s) 2.13 Solution: We assume all positions are made relative to the base of the robot, and therefore, 0

UT is not included in the solution. Using the Denavit-Hartenberg representation and the joint parameters shown, we get:

1 1 2 2 5 2 3 3

1 1 2 2 5 2 3 31 2 3

3 4 6 7

0 1 01 2 3 1 2 3

1 1

3 4

1 1

0 0 0 0 00 0 0 0 0

0 1 0 0 1 0 0 0 0 10 0 0 1 0 0 0 1 0 0 0 1

0 00 0 1

0 00 0 0 1

H H

x x x x

y y y

C S C S l C C SS C S C l S S C

A A Al l l l

T A A A A T A A

C S n o a pl l n o a

S C

= = = + +

= =

2 2 5 2 3 3

2 2 5 2 3 3

6 7

0 0 00 0 0

0 1 0 0 0 0 10 0 0 1 0 0 0 1 0 0 0 1

y

z z z z

C S l C C Sp S C l S S C

n o a p l l

= + Multiplying these matrices we get:


45

1 1 1 1 1 1 1 1

3 4

1 1 1 1 1 1 1 1

2 3 2 3 2 2 6 7 5 2

2 3 2 3 2 2 6 7 5 2

3 3

0 0 0 1

( )( )

0 00 0 0 1

x y x y x y x y

z z z z

x y x y x y x y

C n S n C o S o C a S a C p S pn o a p l l

LHSS n C n S o C o S a C a S p C p

C C C S S S l l l CS C S S C C l l l S

RHSS C

+ + + + =

+ + + + =

From 3,3 or 3,4 elements we get:

11 1 1

11 1 1

0 tan

0 tan

yx y

x

yx y

x

aS a C a

a

pS p C p

p

= = = =

From 1,3 and 2,3 elements we

get: ( )2 1 1 2 1 12

( )2 ,x y x y z

z

S C a S aATAN C a S a a

C a= + = =

From 2,1 and 2,2 elements we get:

( )3 1 1 3 1 1 1 13 1 1

2 ,x y x y x yx y

S S n C nATAN S n C n S o C o

C S o C o= = =


46

Problem 2.40 Derive the inverse kinematic equations for the robot of Problem 2.37. Estimated student time to complete: 20-30 minutes if Problem 2.37 is already solved Prerequisite knowledge required: Text Section(s) 2.13 Solution: We assume all positions are made relative to the base of the robot, and therefore, 0

UT is not included in the solution. Using the Denavit-Hartenberg representation and the joint parameters shown, we get:

# d a 0-1 190 + 3 4l l 0 -90 1-H 2 5 6 7l l l+ + 0 0

1 1 2 2

1 1 2 21 2

3 4 6 6 7

01 2

0 0 0 00 0 0 0

0 1 0 0 0 10 0 0 1 0 0 0 1

H

S C C SC S S C

A Al l l l l

T A A

= = + +

=


47

1 1 2 2

1 1 2 2

3 4 5 6 7

1 2 1 2 1 1 5 6 7

1 2 2 1 1 1 5 6 7

2

0 0 0 00 0 0 0

0 1 0 0 0 10 0 0 1 0 0 0 1 0 0 0 1

( )( )

x x x x

y y y y

z z z z

n o a p S C C Sn o a p C S S Cn o a p l l l l l

S C S S C C l l lC C S C S S l l l

S

= + + + +

+ += 2 3 400 0 0 1

C l l

From 1,3 and 2,3 elements we get: ( )1 11

2 ,y y xx

S aATAN a a

C a= = =

From 3,1 and 3,2 elements we get: ( )2 22

2 ,z z zz

S nATAN n o

C o= = =

From 1,4 elements we get: 1 5 6 7 5 6 71

( ) xxpC l l l p l l lC

+ + = + + =


48

CHAPTER THREE Problem 3.1 Suppose the location and orientation of a hand frame is expressed by the following matrix. What is the effect of a differential rotation of 0.15 radians about the z-axis, followed by a differential translation of [0.1, 0.1, 0.3]? Find the new location of the hand.

0 0 1 21 0 0 70 1 0 50 0 0 1

RHT

=

Estimated student time to complete: 15-20 minutes Prerequisite knowledge required: Text Section(s) 3.6, 3.7. Solution: For 0.15, 0.1, 0.1, 0.3,z dx dy dz = = = = we get:

0 0.15 0 0.10.15 0 0 0.1

0 0 0 0.30 0 0 0

=

[ ]0 0.15 0 0.1 0 0 1 2 0.15 0 0 0.95

0.15 0 0 0.1 1 0 0 7 0 0 0.15 0.40 0 0 0.3 0 1 0 5 0 0 0 0.30 0 0 0 0 0 0 1 0 0 0 0

0.15 0 1 1.051 0 0.15 7.40 1 0 5.30 0 0 1

R RH H

R R RH H Hnew old

d T T

T T d T

= = = = + =


49

Problem 3.2 As a result of applying a set of differential motions to frame T shown, it has changed an amount dT as shown. Find the magnitude of the differential changes made ( , , , , ,dx dy dz x y z ) and the differential operator with respect to frame T.

1 0 0 5 0 0.1 0.1 0.60 0 1 3 0.1 0 0 0.50 1 0 8 0.1 0 0 0.50 0 0 1 0 0 0 0

T dT

= =

Estimated student time to complete: 15-20 minutes Prerequisite knowledge required: Text Section(s) 3.6-3.8. Solution: [ ] [ ][ ] [ ] [ ][ ]

[ ]

1

0 0.1 0.1 0.6 1 0 0 5 0 0.1 0.1 0.10.1 0 0 0.5 0 0 1 8 0.1 0 0 00.1 0 0 0.5 0 1 0 3 0.1 0 0 00 0 0 0 0 0 0 1 0 0 0 0

dT T dT T = = = =

By inspection: d = [0.1,0,0], = [0,0.1,0.1]. [ ] [ ] [ ] [ ]1

1 0 0 5 0 0.1 0.1 0.6 0 0.1 0.1 0.60 0 1 8 0.1 0 0 0.5 0.1 0 0 0.50 1 0 3 0.1 0 0 0.5 0.1 0 0 0.50 0 0 1 0 0 0 0 0 0 0 0

T T

T

dT T T dT = = = =


50

Problem 3.3 Suppose the following frame was subjected to a differential translation of [1 0 0.5]d = units and a differential rotation of [0 0.1 0] = . a. What is the differential operator relative to the reference frame? b. What is the differential operator relative to the frame A?

0 0 1 101 0 0 50 1 0 00 0 0 1

A

=

Estimated student time to complete: 10 minutes Prerequisite knowledge required: Text Section(s) 3.6-3.8 Solution: a.

0 0 0.1 10 0 0 00.1 0 0 0.50 0 0 0

=

b.

[ ] [ ][ ]10 1 0 5 0 0 0.1 1 0 0 1 100 0 1 0 0 0 0 0 1 0 0 51 0 0 10 0.1 0 0 0.5 0 1 0 00 0 0 1 0 0 0 0 0 0 0 1

0 0 0 00 0 0.1 0.50 0.1 0 10 0 0 0

A A A = =

=


51

Problem 3.4 The initial location and orientation of a robots hand are given by T1, and its new location and orientation after a change are given by T2. a. Find a transformation matrix Q that will accomplish this transform (in the Universe frame). b. Assuming the change is small, find a differential operator that will do the same. c. By inspection, find a differential translation and a differential rotation that constitute this operator.

1 2

1 0 0 5 1 0 0.1 4.80 0 1 3 0.1 0 1 3.50 1 0 6 0 1 0 6.20 0 0 1 0 0 0 1

T T

= =

Estimated student time to complete: 15-20 minutes Prerequisite knowledge required: Text Section(s) 3.6-3.8 Solution: a.

12 1 2 1

1 0 0.1 4.8 1 0 0 5 1 0.1 0 0.10.1 0 1 3.5 0 0 1 6 0.1 1 0 00 1 0 6.2 0 1 0 3 0 0 1 0.20 0 0 1 0 0 0 1 0 0 0 1

T QT Q T T

= = = =

b. ( )2 1 1 1 1 1T T dT T T I T QT= + = + = + =

Therefore: Q I =

or

0 0.1 0 0.10.1 0 0 00 0 0 0.20 0 0 0

=

c. By inspection: d = [0.1, 0, 0.2] and = [0, 0, 0.1].


52

Problem 3.5 The hand frame of a robot and the corresponding Jacobian are given. For the given differential changes of the joints, compute the change in the hand frame, its new location, and corresponding .

66

8 0 0 0 0 0 00 1 0 10 3 0 1 0 0 0 0.11 0 0 5 0 10 0 0 0 0 0.10 0 1 0 0 1 0 0 1 0 0.20 0 0 1 0 0 0 1 0 0 0.2

1 0 0 0 0 1 0

TT J D

= = =

Estimated student time to complete: 30 minutes Prerequisite knowledge required: Text Section(s) 3.6-3.11 Solution:

[ ]

6

6

66 6

6

6

6

8 0 0 0 0 0 0 03 0 1 0 0 0 0.1 0.1

0 10 0 0 0 0 0.1 10 1 0 0 1 0 0.2 0.30 0 0 1 0 0 0.2 0.21 0 0 0 0 1 0 0

T

T

TT T

T

T

T

dxdydz

D J Dxyz

= = = =

Substitute in 6T to get: 60 0 0.2 00 0 0.3 0.10.2 0.3 0 10 0 0 0

T

=

Then

[ ] 66 60 1 0 10 0 0 0.2 0 0 0 0.3 0.11 0 0 5 0 0 0.3 0.1 0 0 0.2 00 0 1 0 0.2 0.3 0 1 0.2 0.3 0 10 0 0 1 0 0 0 0 0 0 0 0

TdT T

= = =

[ ] [ ] [ ]6 6 60 1 0.3 9.91 0 0.2 5

0.2 0.3 1 10 0 0 1

new oldT T dT

= + =


53

[ ] [ ]6 16 60 1 0 10 0 0 0.2 0 0 1 0 51 0 0 5 0 0 0.3 0.1 1 0 0 100 0 1 0 0.2 0.3 0 1 0 0 1 00 0 0 1 0 0 0 0 0 0 0 1

0 0 0.3 0.10 0 0.2 00.3 0.2 0 10 0 0 0

TT T = =

=


54

Problem 3.6 Two consecutive frames describe the old (T1) and new (T2) positions and orientations of the end of a 3-DOF robot. The corresponding Jacobian relative to T1, relating to

1 1 1, ,T T Tdz x z is also given. Find values of joint differential motions 1 2 3, ,ds d d of the robot that caused the given frame change.

11 2

0 0 1 8 0 0.01 1 8.15 10 0

1 0 0 5 1 0.05 0 53 0 0

0 1 0 2 0.05 1 0.01 20 1 1

0 0 0 1 0 0 0 1

TT T J

= = =


[ ][ ] [ ]

[ ]

1

1

2 1 1 1

11

0 0.01 0 0.10 0.05 0 0

0.05 0 0.01 00 0 0 0

0 1 0 5 0 0.01 0 0.1 0 0.05 0 00 0 1 2 0 0.05 0 0 0.05 0 0.01 01 0 0 8 0.05 0 0.01 0 0 0.01 0 0.10 0 0 1 0 0 0 0 0 0 0 0

T

T

T T dT T T

T dT

= = = = = = =

Therefore,

1

1 1

1

0.05 0.10.01 0.01

0.050.1

Tz

T Tx

Tz

Dd

= = = =

And: [ ] [ ]1 1 1 11T T T TD J D D J D = = Please see Appendix A for methods to calculate the inverse of a non-unitary matrix. Substitute to get:

[ ] 1 11

15 10 0 0.1 0.00333 0 0 0.01 0.008330 1 1 0.05 0.0417

T TD J D

= = =


55

Problem 3.7 Two consecutive frames describe the old (T1) and new (T2) positions and orientations of the end of a 3-DOF robot. The corresponding Jacobian, relating to , ,dz x z is also given. Find values of joint differential motions 1 2 3, ,ds d d of the robot that caused the given frame change.

1 2

0 0 1 10 0.05 0 1 9.755 10 0

1 0 0 5 1 0.1 0.05 5.23 0 0

0 1 0 3 0.1 1 0 3.70 1 1

0 0 0 1 0 0 0 1

T T J

= = =


2 1

0.05 0 0 0.250 0.1 0.05 0.2

0.1 0 0 0.70 0 0 0

T T dT

= =

[ ] [ ][ ] [ ] [ ][ ] 11 10.05 0 0 0.25 0 1 0 50 0.1 0.05 0.2 0 0 1 3

0.1 0 0 0.7 1 0 0 100 0 0 0 0 0 0 1

0 0.05 0 00.05 0 0.1 0

0 0.1 0 0.20 0 0 0

dT T dT T = = =

=

Therefore, 0.2 0.20.1 0.1

0.050.05

z

x

z

dD

= = = =

Substitute to get: [ ] [ ] [ ]1

15 10 0 0.2 0.03343 0 0 0.1 0.003340 1 1 0.05 0.0467

D J D

= = =


56

Problem 3.8 A camera is attached to the hand frame T of a robot as given. The corresponding inverse Jacobian of the robot at this location is also given. The robot makes a differential motion, as a result of which, the change in the frame dT is recorded as given. a. Find the new location of the camera after the differential motion. b. Find the differential operator. c. Find the joint differential motion values associated with this move. d. Find how much the differential motions of the hand-frame ( T D ) should have been

instead, if measured relative to frame T, to move the robot to the same new location as in part a.

1

1 0 0 0 0 00 1 0 3 2 0 1 0 0 01 0 0 2 0 0.2 0 0 0 00 0 1 8 0 1 0 0 1 00 0 0 1 0 0 0 1 0 0

1 0 0 0 0 1

T J

= =

0.03 0 0.1 0.790 0.03 0 0.090 0.1 0 0.40 0 0 0

dT

=

Estimated student time to complete: 30-40 minutes Prerequisite knowledge required: Text Section(s) 3.6-3.11 Solution:

a.

0.03 1 0.1 3.791 0.03 0 2.090 0.1 1 7.60 0 0 1

new oldT T dT

= + =

b.

1

0.03 1 0.1 3.79 0 1 0 2 0 0.03 0.1 0.051 0.03 0 2.09 1 0 0 3 0.03 0 0 00 0.1 1 7.6 0 0 1 8 0.1 0 0 0.10 0 0 1 0 0 0 1 0 0 0 0

dT T

= = =

c. [ ]0.05 0 0.1 0 0.1 0.03 TD =

Therefore: 1

1 0 0 0 0 0 0.05 0.052 0 1 0 0 0 0 0.20 0.2 0 0 0 0 0.1 00 1 0 0 1 0 0 0.10 0 0 1 0 0 0.1 01 0 0 0 0 1 0.03 0.08

D J D

= = =


57

d.

1

0 1 0 2 0.03 0 0.1 0.791 0 0 3 0 0.03 0 0.090 0 1 8 0 0.1 0 0.40 0 0 1 0 0 0 0

0 0.03 0 0.090.03 0 0.1 0.790 0.1 0 0.40 0 0 0

T TdT T T dT

= = = =

[ ]0.09 0.79 0.4 0.1 0 0.03 TT D =


58

Problem 3.9 A camera is attached to the hand frame T of a robot as given. The corresponding inverse Jacobian of the robot relative to the frame at this location is also given. The robot makes a differential motion, as a result of which, the change dT in the frame is recorded as given. a. Find the new location of the camera after the differential motion. b. Find the differential operator. c. Find the joint differential motion values D associated with this move.

1

1 0 0 0 0 00 1 0 3 2 0 1 0 0 01 0 0 2 0 0.1 0 0 0 00 0 1 8 0 1 0 0 1 00 0 0 1 0 0 0 1 0 0

1 0 0 0 0 1

TT J

= = 0.02 0 0.1 0.70 0.02 0 0.080 0.1 0 0.30 0 0 0

dT

=

Estimated student time to complete: 30-40 minutes Prerequisite knowledge required: Text Section(s) 3.6-3.11 Solution: a.

0.02 1 0.1 3.71 0.02 0 2.080 0.1 1 7.70 0 0 1

new oldT T dT

= + =

b.

1

0.02 0 0.1 0.7 0 1 0 2 0 0.02 0.1 0.060 0.02 0 0.08 1 0 0 3 0.02 0 0 0.020 0.1 0 0.3 0 0 1 8 0.1 0 0 00 0 0 0 0 0 0 1 0 0 0 0

dT T

= = = c.


59

[ ][ ] [ ]1 10 1 0 2 0.02 0 0.1 0.7 0 0.02 0 0.081 0 0 3 0 0.02 0 0.08 0.02 0 0.1 0.70 0 1 8 0 0.1 0 0.3 0 0.1 0 0.30 0 0 1 0 0 0 0 0 0 0 0

T T T T dT = = = = =

Therefore: [ ]0.08 0.7 0.3 0.1 0 0.02 TT D =

1

1 0 0 0 0 0 0.08 0.082 0 1 0 0 0 0.7 0.140 0.1 0 0 0 0 0.3 0.070 1 0 0 1 0 0.1 0.70 0 0 1 0 0 0 0.11 0 0 0 0 1 0.02 0.06

T T T TD J D D J D

= = = =


60

Problem 3.10 The hand frame TH of a robot is given. The corresponding inverse Jacobian of the robot at this location relative to this frame is also shown. The robot makes a differential motion relative to this frame described as

[ ]0.05 0 0.1 0 0.1 0.1H TT D = . a. Find which joints must make a differential motion, and by how much, in order to create the indicated differential motions. b. Find the change in the frame. c. Find the new location of the frame after the differential motion. d. Find how much the differential motions (given above) should have been if measured relative to the Universe, to move the robot to the same new location as in Part c.

1

5 0 0 0 0 00 1 0 3 2 0 1 0 0 01 0 0 3 0 0.2 0 0 0 00 0 1 8 0 1 0 0 1 00 0 0 1 0 0 0 1 0 0

1 0 0 0 0 1

HTHT J

= =


[ ] 15 0 0 0 0 0 0.05 0.252 0 1 0 0 0 0 0.20 0.2 0 0 0 0 0.1 00 1 0 0 1 0 0 0.10 0 0 1 0 0 0.1 01 0 0 0 0 1 0.1 0.15

H HT TD J D

= = =

Therefore, joints 1, 2, 4, and 6 must move as shown. b. Substituting values from HT D to form HT , we get:

0 1 0 3 0 0.1 0.1 0.05 0.1 0 0 01 0 0 3 0.1 0 0 0 0 0.1 0.1 0.050 0 1 8 0.1 0 0 0.1 0.1 0 0 0.10 0 0 1 0 0 0 0 0 0 0 0

HTHdT T

= = =

c.


61

0 1 0 3 0.1 0 0 0 0.1 1 0 31 0 0 3 0 0.1 0.1 0.05 1 0.1 0.1 3.050 0 1 8 0.1 0 0 0.1 0.1 0 1 8.10 0 0 1 0 0 0 0 01 0 0 1

new oldH HT T dT

= + = + =

d.

1

0.1 0 0 0 0 1 0 3 0 0.1 0 0.30 0.1 0.1 0.05 1 0 0 3 0.1 0 0.1 1.15

0.1 0 0 0.1 0 0 1 8 0 0.1 0 0.20 0 0 0 0 0 0 1 0 0 0 0

HTH H

H

dT T T

dT T

= = = = =

Therefore, [ ]0.3 1.15 0.2 0.1 0 0.1 TD =


62

Problem 3.11 The hand frame T of a robot is given. The corresponding inverse Jacobian of the robot at this location is also shown. The robot makes a differential motion described as

[ ]0.05 0 0.1 0 0.1 0.1 TD = . a. Find which joints must make a differential motion, and by how much, in order to create the indicated differential motions. b. Find the change in the frame. c. Find the new location of the frame after the differential motion. d. Find how much the differential motions (given above) should have been, if measured relative to Frame T, to move the robot to the same new location as in Part c.

1

5 0 0 0 0 00 1 0 3 2 0 1 0 0 01 0 0 3 0 0.2 0 0 0 00 0 1 8 0 1 0 0 1 00 0 0 1 0 0 0 1 0 0

1 0 0 0 0 1

T J

= =


1

5 0 0 0 0 0 0.05 0.252 0 1 0 0 0 0 0.20 0.2 0 0 0 0 0.1 00 1 0 0 1 0 0 0.10 0 0 1 0 0 0.1 01 0 0 0 0 1 0.1 0.15

D J D

= = =

Therefore, joints 1, 2, 4, and 6 must move as shown. b. Using D given above to form ,

0 0.1 0.1 0.05 0 1 0 3 0.1 0 0.1 0.550.1 0 0 0 1 0 0 3 0 0.1 0 0.30.1 0 0 0.1 0 0 1 8 0 0.1 0 0.40 0 0 0 0 0 0 1 0 0 0 0

dT T

= = =


63

c. 0.1 1 0.1 3.551 0.1 0 3.30 0.1 1 7.60 0 0 1

new oldT T dT

= + =

d.

[ ][ ] [ ]1 10 1 0 3 0.1 0 0.1 0.551 0 0 3 0 0.1 0 0.30 0 1 8 0 0.1 0 0.40 0 0 1 0 0 0 0

0 0.1 0 0.30.1 0 0.1 0.550 0.1 0 0.40 0 0 0

T T T T dT

= = = =

The differential motions should have been:

[ ]0.3 0.55 0.4 0.1 0 0.1T D =


64

Problem 3.12 Calculate the 6 21

T J element of the Jacobian for the revolute robot of Example 2.25. Estimated student time to complete: 20 minutes Prerequisite knowledge required: Text Section(s) 3.10 Solution: The robot has 6 revolute joints. From Equation 3.26: ( )6 2T i x y y xJ o p o p= + where for i=1 we use 6 1 2 3 4 5 6oT A A A A A A= given in Equation 2.59. Substitute to get:

[ ] [ ][ ] [ ]

621 1 234 5 6 234 6 1 5 6 1 234 4 23 3 2 2

1 234 5 6 234 6 1 5 6 1 234 4 23 3 2 2

( ) ( )

( ) ( )

T J C C C C S C S S S S C a C a C a

S C C C S C C S S C C a C a C a

= + + ++ + +


65

Problem 3.13 Calculate the 6 16

T J element of the Jacobian for the revolute robot of Example 2.25. Estimated student time to complete: 20 minutes Prerequisite knowledge required: Text Section(s) 3.10 Solution: The robot has 6 revolute joints. From Equation 3.26: ( )6 1T i x y y xJ n p n p= + where for i=6 we use 5 6 6T A= given in Equation 2.57. Substitute to get:

( )( ) ( )( )6 21 6 60 0 0T J C S= + =


66

Problem 3.14 Using Equation (2.34), differentiate proper elements of the matrix to develop a set of symbolic equations for joint differential motions of a cylindrical robot and write the corresponding Jacobian. Estimated student time to complete: 10 minutes Prerequisite knowledge required: Text Section(s) 3.10-3.11 Solution: For a cylindrical coordinate (without additional rotation) there can only be 3 variables. The Jacobian will be a 3 3 matrix.

00

( , , )0 0 10 0 0 1

cyl

C S rCS C rS

T r ll

=

Then x

y

z

p rCp rSp l

= = = and

x

y

z

dp drC rS ddp drS rC ddp dl

= = + =

Therefore, 00

0 0 1

x

y

z

dp C rS drdp S rC ddp dl

=


67

Problem 3.15 Using Equation (2.36), differentiate proper elements of the matrix to develop a set of symbolic equations for joint differential motions of a spherical robot and write the corresponding Jacobian. Estimated student time to complete: 10 minutes Prerequisite knowledge required: Text Section(s) 3.10-3.11 Solution: For a spherical coordinate (without additional rotation) there can only be 3 variables. The Jacobian will be a 3 3 matrix.

( , , )0

0 0 0 1

sph

C C S S C rS CC S C S S rS S

T rS C rC

=

Then ..

x

y

z

p rS Cp rS Sp rC

= = = and

. . .

. . .x

y

z

dp drS C rC C d rS S ddp drS S rC S d rS C ddp drC rS d

= + = + + =

Therefore, . . .. . .

0

x

y

z

dp S C rC C rS S drdp S S rC S rS C ddp C rS d

=


68

Problem 3.16 For a cylindrical robot, the three joint velocities are given for a corresponding location. Find the three components of the velocity of the hand frame. r=0.1 in/sec, =0.05 rad/sec, l=0.2 in/sec, r=15 in, 30 = D , l=10 in. Estimated student time to complete: 10 or 20 minutes Prerequisite knowledge required: Text Section(s) 3.10-3.11 Solution: Referring to the solution of Problem 3.14, and dividing both sides by dt, substitute the given values into the equation to get:

0 0.866 7.5 0 0.1 0.2880 0.5 13 0 0.05 0.7

0 0 1 0 0 1 0.2 0.2

x

y

z

p C rS rp S rCp l

= = =

in/sec


69

Problem 3.17 For a spherical robot, the three joint velocities are given for a corresponding location. Find the three components of the velocity of the hand frame. r=2 in/sec, =0.05 rad/sec, =0.1 rad/sec, r=20 in, 60 = D , 30 = D . Estimated student time to complete: 10 or 20 minutes Prerequisite knowledge required: Text Section(s) 3.10-3.11 Solution: Referring to the solution of Problem 3.15, and dividing both sides by dt, substitute the given values into the equation to get:

0.75 8.66 8.66 2 1.0670.433 5 15 0.05 2.616

0 0.5 17.32 0 0.1 0.134

x

y

z

p S C rC C rS S rp S S rC S rS Cp C rS

= = =


70

Problem 3.18 For a spherical robot, the three joint velocities are given for a corresponding location. Find the three components of the velocity of the hand frame. r=1 unit/sec, =1 rad/sec, =1 rad/sec, r=5 units, 45 = D , 45 = D . Estimated student time to complete: 10 or 20 minutes Prerequisite knowledge required: Text Section(s) 3.10-3.11 Solution: Referring to the solution of Problem 3.15, and dividing both sides by dt, substitute the given values into the equation to get:

0.5 2.5 3.535 1 0.5350.5 2.5 3.535 1 6.535

0 0.707 3.535 0 1 2.828

x

y

z

p S C rC C rS S rp S S rC S rS Cp C rS

= = =


71

Problem 3.19 For a cylindrical robot, the three components of the velocity of the hand frame are given for a corresponding location. Find the required three joint velocities that will generate the given hand frame velocity. 1 in/sec, 3 in/sec, 5 in/sec, 45 , 20 in, 25 inx y z r l= = = = = =D Estimated student time to complete: 10 or 25 minutes Prerequisite knowledge required: Text Section(s) 3.12 Solution: Referring to the solution of Problem 3.16, substitute to get:

0 1 0.707 14.14 00 2 0.707 14.14 0

0 0 1 3 0 0 1

x

y

z

p C rS r rp S rCp l l

= =

2.83 in/sec0.707 14.14 1

0.707 14.14 3 =0.071 /sec

5 5 in/sec

rrr

l l

= = + = = = D

Please note that this problem could have been solved by calculating the inverse of the Jacobian as well.


72

Problem 3.20 For a spherical robot, the three components of the velocity of the hand frame are given for a corresponding location. Find the required three joint velocities that will generate the given hand frame velocity. 5 in/sec, 9 in/sec, 6 in/sec, 60 , 20 in, =30x y z r = = = = =D D Estimated student time to complete: 15 or 30 minutes Prerequisite knowledge required: Text Section(s) 3.12 Solution:

Referring to the solution of Problem 3.17, substitute to get:

5 0.75 8.66 8.669 0.433 5 15

0 6 0.5 17.32 0

x

y

z

p S C rC C rS S r rp S S rC S rS Cp C rS l

= =

0.75 8.66 8.66 5 10.647 in/sec

0.433 5 15 9 = 0.039 /sec0.306 /sec0.5 17.32 6

r rr

r

+ = = + + = = = D

D

Please note that this problem could have been solved by calculating the inverse of the Jacobian as well.


73

CHAPTER FOUR Problem 4.1 Using Lagrangian mechanics derive the equations of motion of a cart with two tires under the cart as shown:

km

F

xr, I r, I

Figure P.4.1

Estimated student time to complete: 15-20 minutes Prerequisite knowledge required: Text Section(s) 4.2 Solution: I. Kinematics: Since the rollers are in contact with the block at the top, the contact-point velocity is x . Therefore:

22xx rr

= = xvc

II. Dynamics:

2 2 2 2 2 2 22

1 1 1 122 2 2 2 4 4

w cw c c eff

m IK mx m v I mx x x m xr

= + + = + + =

where 212 4 4

w ceff

m Im mr

+ + = 21

2P kx=

Therefore, 2 212eff

L K P m x kx= =

2 2eff effL d Lm x m xx dt x

= = L kxx

= 2 effF m x kx= +


74

Problem 4.2 Calculate the total kinetic energy of the link AB, attached to a roller with negligible mass, as shown.

l, I

m

Vo

A

B

Cx

y

Figure P.4.2

Estimated student time to complete: 20-30 minutes Prerequisite knowledge required: Text Section(s) 4.2 Solution: First, find the velocity of point C, assuming that the roller has negligible mass too:

/ ( ) cos sin2 2 2sin cos( ) cos sin

2 2 2 2

C A C A o o

o o

l lv

l l l lv j v

= + = + = + + = + = +

lv v v v i k i j

i i i j

2 2 2 2 2

2 2 2 2 2 2sin cossin sin4 4 4C o o o o

l l lv v v l v v l = + + = + 2

2 2 2 2 21 1 1 1sin2 2 2 4 2C o o

lK mv I m v v l I = + = + +


75

Problem 4.3 Derive the equations of motion for the 2-link mechanism with distributed mass, as shown.

x0

y0

m1

m2

x1

y1

l1 , I1

l2 , I

22

1

O

C

AD

B

Figure P.4.3

Estimated student time to complete: 1.5-2 hours Prerequisite knowledge required: Text Section(s) 4.2 Solution: I. Kinematic: First we write equations describing the kinematic relationships.

( )( )

2 21 1 12 1 1 1 1 2 12

2 21 1 12 1 1 1 1 2 12

2 2

2 2

D D

D D

l lx l C C x l S S

l ly l S S y l C C

= + = + = + = + +

The velocity of point D will be:

( ) ( )( ) ( )

( ) ( )

22 2 2 2 2 2 2 2 22

1 1 1 1 2 1 2 12 1 2 1 1 2 1 12

22 2 2 2 2 22

1 1 1 1 2 1 2 12 1 2 1 1 2 1 12

22 2 2 22

1 1 1 2 1 2 1 2 1 1 2 2

24

24

24

D D Dlv x y l S S l l S S

ll C C l l C C

ll l l C

= + = + + + + +

+ + + + + +

= + + + + +

Rearrange: 2 2 2

2 2 2 22 2 21 1 1 2 2 2 1 2 1 2 24 4 2D

l l lv l l l C l l C = + + + + +


76

II: Kinetics: Kinetic Energy:

( )( )

22 21 2 1 1 2 2

22 2 2 21 1 1 2 2 1 2 2

2 2 2 2 2 2 21 1 1 2 2 2 1 2 1 2 2 2 2 2 1 2 2 2 2 1 2 2

1 1 12 2 2

1 1 1 1 12 3 2 12 2

1 1 1 1 1 1 16 6 2 2 6 3 2

O D D

D

K K K I I m v

m l m l m v

m l m l m l m l l C m l m l m l l C

= + = + + + = + + + = + + + + + +

Potential Energy:

( ) ( ) ( )1 21 2 1 1 2 1 1 2 121 1 12 2l lP P P m g C m gl C m g C= + = + +

Lagrangian:

2 2 2 2 2 2 21 1 1 2 2 2 1 2 1 2 2 2 2 2 1 2 2 2 2 1 2 2

1 2 1 21 2 1 1 2 12 1 2 1 2

1 1 1 1 1 1 16 6 2 2 6 3 2

2 2 2 2

L K P m l m l m l m l l C m l m l m l l C

l l l lm g m gl C m g C m g m gl m g

= = + + + + + + + + +

Derivatives and equations of motion:

2 2 2 21 1 1 2 2 2 1 2 1 2 2 2 2 2 2 1 2 2

1

2 2 2 21 1 1 2 2 2 1 2 1 2 2 2 2 2 2 1 2 2

1

21 2 2 1 2 2 2 2 1 2 2

1 1 1 1 1 126 6 2 2 3 2

1 1 1 1 1 126 6 2 2 3 2

12

L m l m l m l m l l C m l m l l C

d L m l m l m l m l l C m l m l l Cdt

m l l S m l l S

= + + + + + = + + + + +

1 21 2 1 1 2 12

1 2 2l lL m g m gl S m g S

= + Substitute to get:


77

2 2 2 21 1 1 2 2 2 1 2 1 2 2 1 2 2 2 1 2 2 2

22 1 2 2 1 2 2 1 2 2 2 1 2 1 1 2 2 12

1 1 1 13 3 3 2

1 1 12 2 2

T m l m l m l m l l C m l m l l C

m l l S m l l S m m gl S m gl S

= + + + + + + + +

For the second variable:

2 22 2 2 1 2 2 2 1 2 2

2

2 22 2 2 1 2 2 2 1 2 2 2 1 2 2 1 2

2

1 1 13 3 2

1 1 1 13 3 2 2

L m l m l m l l C

d L m l m l m l l C m l l Sdt

= + + = + +

22 1 2 2 1 2 1 2 2 1 2 2 2 12

2

1 1 12 2 2

L m l l S m l l S m gl S =

Substitute to get:

2 2 22 2 2 2 1 2 2 1 2 2 2 2 1 2 2 1 2 2 12

1 1 1 1 13 2 3 2 2

T m l m l l C m l m l l S m gl S = + + + +

The result can be written in a matrix form too.


78

Problem 4.4 Write the equations that express U62, U35, U53, U623, and U534 for a 6-axis cylindrical-RPY robot in terms of the A and Q matrices. Estimated student time to complete: 30 minutes Prerequisite knowledge required: Text Section(s) 4.4.1. Solution:

For joints 1 and 3: ( )

0 0 0 0

0 0 0 0

0 0 0 1

0 0 0 0

prismatic PQ Q

= =

For joints 2,4,5,6: ( )

0 1 0 01 0 0 00 0 0 00 0 0 0

revolute RQ Q

= =

0

1 2.. ..iij j j ij

TU A A Q A A j iq

= = ijk ij kU U q=

( )( )( )( )( )

01 2 3 4 5 66

62 1 2 3 4 5 6

01 2 33

350 0

01 2 3 4 55

53 1 2 3 4 5

1 2 3 4 5 662623 1 2 3 4 5 6

1 2 3 4 553534 1 2 3 4 5

0 ( )

R

P

RR P

PP R

a a

A A A A A ATU AQ A A A A A

A A ATU i j

A A A A ATU A A Q A A Al l

AQ A A A A AUU AQ A Q A A A Al l

A A Q A A AUU A A Q A Q A Aq q

= = = = = = = = = = = = = = =


79

Problem 4.5 Using Equations 4.49 to 4.54, write the equations of motion for a 3-DOF revolute robot and describe each term. Estimated student time to complete: 15-20 minutes Prerequisite knowledge required: Text Section(s) 4.4.4. Solution: We have:

2 2 21 11 1 12 2 13 3 1( ) 1 111 1 122 2 133 3

112 1 2 113 1 3 123 2 3 1

2 2 22 21 1 22 2 23 3 2( ) 2 211 1 222 2 233 3

212 1 2 213 1 3 223

2 2 2

2 2 2

act

act

T D D D I D D D

D D D D

T D D D I D D D

D D D

= + + + + + ++ + + +

= + + + + + ++ + +

2 3 2

2 2 23 31 1 32 2 33 3 3( ) 3 311 1 322 2 333 3

312 1 2 313 1 3 323 2 3 32 2 2act

D

T D D D I D D D

D D D D

+= + + + + + ++ + + +

( ) iD terms are caused by angular accelerations. ( ) 2iD terms represent centripetal accelerations. ( ) i jD terms are related to the Coriolis accelerations. ( )iD terms are gravity terms.


80

Problem 4.6 Expand the D134 and D15 terms of Equation 4.49 in terms of their constituent matrices. Estimated student time to complete: 1 hour Prerequisite knowledge required: Text Section(s) 4.4 Solution:

a. From Equation 4.44: max( , , )

( )n

Tijk pjk p pi

p i j kD Trace U J U

==

For D134 we have i =1, j =3, k =4, p =4, n =6. Then:

( ) ( ) ( )6134 34 1 434 4 41 534 5 51 634 6 614

( )T T T Tp p pD Trace U J U Trace U J U Trace U J U Trace U J U= = + + ( )( )( )

01 2 3 44

43 1 2 3 3 43 3

1 2 3 3 443434 1 2 3 3 4 4

4 4

01 2 3 44

41 1 1 2 3 41 1

A A A ATU A A Q A Aq q

A A Q A AUU A A Q A Q Aq q

A A A ATU Q A A A Aq q

= = = = = = = = =

From Equation A.11 of Appendix A: 41 4 3 2 1 1T T T T T TU A A A A Q=

( )( )( )( )

1 2 3 4 553 1 2 3 3 4 5

3

1 2 3 3 4 5534 1 2 3 3 4 4 5

4

1 2 3 4 551 1 1 2 3 4 5 51 5 4 3 2 1 1

1

1 2 3 4 5 663 1 2 3 3 4 5 6

3

63634 1 2 3 3 4 4 5 6

4

61 6 5 4

T T T T T T T

T T T T

A A A A AU A A Q A A A

qA A Q A A A

U A A Q A Q A Aq

A A A A AU Q A A A A A U A A A A A Q

qA A A A A A

U A A Q A A A Aq

UU A A Q A Q A A Aq

U A A A

= == == = == == =

= 3 2 1 1T T T TA A A Q

Substitute to get: ( )

( )( )

134 1 2 3 3 4 4 4 4 3 2 1 1

1 2 3 3 4 4 5 5 5 4 3 2 1 1

1 2 3 3 4 4 5 6 6 6 5 4 3 2 1 1

T T T T T

T T T T T T

T T T T T T T

D Trace A A Q A Q A J A A A A Q

Trace A A Q A Q A A J A A A A A Q

Trace A A Q A Q A A A J A A A A A A Q

=++


81

b. From Equation 4.43: max( , )

( )n

Tij pj p pi

p i jD Trace U J U

==

For D15 we have i =1, j =5, p =5, n =6. Then:

( ) ( )615 5 1 55 5 51 65 6 615

( )T T Tp p pD Trace U J U

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