Introduction to Robotics Mechanics and Control 4th Edition Craig Solutions Manual

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Chapter 2 Solutions for Introduction to Robotics

1. a) Use (2.3) to obtain

A B

R

1   = 0 0

0 0 −1

0  1 0

b) Use (2.74) to get

α  = 90 degrees 90  degrees β  =  = 90 degrees 90  degrees γ  =  =

90 degrees 90  degrees

−

2. a) Use (2.64) to obtain

A B

R

 330   =  908 259 . .

−

.

.770

−

.418 .483

 547  0396 837

. . .

b) Answer is the same as in (a) according to (2.71) 3. Use (2.19) to obtain the transformat transformation ion matrices. matrices. The rotation rotation is X-Y-Z fixed angles, so use (2.64) for that 3×3 submatrix, with angles

γ  = 0  degrees β  = −sin−1





  tripod_height  = distance_along_optical_axis

αC  = 0  degrees αD  = 120 degrees 120  degrees αE  = 240 degrees 240  degrees

The position vectors to the camera-frame origins are

−

sin−1

1 5 .

5

 =

−

107 degrees 107  degrees

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horizontal_distance 4 77   =  0  0  =  1 50 tripod_height horizontal_distance cos   2 39      =  horizontal_distance sin  =  4 13  15 tripod_height horizontal_distance cos   2 38      =  horizontal_distance sin  =  4 13 1 50 tripod_height  distance_along_optical_axis)) where horizontal_distance = (distance_along_optical_axis .

B

P CORG CORG

.

×

B

P DORG DORG

×

×

B

P EORG EORG

×

αD αD

−

αE  αE 

−

.

. .

. .

−

,

.

2

−

(tripod_height) tripod_height)2 .

Combining the rotation and translation yields the transformation matrices via (2.19) as

 300 0 954 4 77   0 1 00 0 0   =  9054 00 0300 1 150  150 866 477 2 39   260 500 826 4 13   =  9054 00 0300 1 150   150 866 477 2 39   260 500 826 4 13  =  9054 00 0300 1 150  .

−

B C 

T 

T 

B E 

T 

.

.

.

.

.

−

.

B D

.

−

. −.

.

−

.

−

. −.

. . .

.

−

. .

.

. .

−

.

.

−

. .

− −

.

.

−

− 2 5]T . Use 4. The camera-frame origin is located at B P CORG Use (2.19 (2.19)) to get get the CORG = [7 B transformation, C T . The rotation is Z-Y-X Euler angles, so use (2.71) with

α  = 0  degrees β  = −110 degrees 110 degrees γ  = −20 degrees 20 degrees

to get

 342   9400 =  0 .

−

B C 

T 

.

.321 .940 .117

0

.883

−

.342 −.321

0

 7 00  2 00 5 00  1 .

.

−

.

Full file at https://testbankuniv.eu/Int https:// testbankuniv.eu/Introduction-to-Roboti roduction-to-Robotics-Mechanics-and-C cs-Mechanics-and-Control-4th-Editi ontrol-4th-Edition-Craig-Solutions on-Craig-Solutions-Manual -Manual 5. Let B

B

P 1  =

P 0 + 5 B V 0  = [9.5

1.00

−

1.50]T

The object’s position in { A} is A

P 1  =

A B

2.11 3.60]T

T  B P 1  = [−4.89

6.   (2.1)  ˆ , θ ) R  = rot  =  rot((Yˆ , φ) rot( rot(Z

 c 0 s  c s 0    c 0 =  0 1 0  s 0 0 1 s 0 c c c c s s    = s 0 c s c s s c φ

φ

φ

−

θ θ

−

θ θ

φ

φ θ θ − φ θ

−

φ θ θ φ θ

φ

φ

7.   (2.2)  ˆ , 60) rot( R  = rot  =  rot((X rot(Yˆ ,

1 0 0  45)707   = 0 500 866  0 0 866 500 707  707 0 707   612 =  612 500 353 866 353 −

.

. .

−

.

.

.

.

0 1 0

 707  0  707

−

.

.

. −. . −

.

. .

−

.

8.   (2.12) Velocity (2.12)  Velocity is a “free vector” and only will be affected by rotation, and not by translation:

 707  =  612 353 .

A

A B

B

V  = R V 

−

.

.

= [−14.1

−

0 .500 .866

29.0 62.9]T

   707 30 0   612 40 0 353 50 0

. −. .

. . .

−

9.   (2.31)

C  B

T 

 0   500866 =  0 .

.

−

0 −.866 −.500 0

1 0 0 0

−

2  0 0 1

10.   (2.37) Using (2.37)  Using (2.45) we get that B

P AORG AORG  =

A − B

T A

R

P AORG AORG  =

−

 25  43 86 . .

.87 −.50

00

. .

     43 50 94      75   4 0 =  6 4 50 30 28 .

−

.

.

−

.