机器人学导论(克雷格)第二章作业答案

机器人学导论(克雷格)第二

章作业答案

-标准化文件发布号：（9456-EUATWK-MWUB-WUNN-INNUL-DDQTY-KII

2.1 solution:

According to the equation of pure transition transformation,the new point after transition is as follows:

1002350

10358(,,)00147110

00111trans

x y z old P Trans d d d P ⎡⎤⎡⎤⎡⎤

⎢⎥⎢⎥⎢⎥⎢

⎥⎢⎥⎢⎥=⨯==⎢⎥⎢⎥⎢⎥

⎢⎥⎢⎥⎢⎥

⎣⎦⎣⎦⎣⎦

2.3 solution:

According to the constraint equations:

0;0;01

n a n o a o n •=•=•==

Thus,the matrix should be like this:

0015001

51003100301020102000

1000

1or --⎡⎤⎡⎤⎢⎥⎢⎥-⎢

⎥⎢

⎥⎢⎥⎢⎥--⎢

⎥⎢

⎥⎣⎦⎣⎦

2.4 Solution:

X Y Z

P P P ⎛⎫ ⎪ ⎪ ⎪⎝⎭=cos 0sin 010sin 0cos θθθθ⎛⎫ ⎪ ⎪ ⎪-⎝

⎭0n a P P P ⎛⎫

⎪

⎪ ⎪⎝⎭

2.7 Solution:

According to the equation of pure rotation transformation , the new coordinates are as follows:

100222

22(,45)0

34227202

22new

P rot x P ⎡⎤⎡⎤

⎢⎥

⎢⎥⎡⎤⎢

⎢⎢⎥⎢⎢=⨯==⎢⎥⎢⎢⎢⎥⎢⎥⎢⎥⎣⎦⎢⎥⎢⎥

⎢⎥⎢⎥⎣

⎦⎣⎦

2.9 Solution:

Acording to the equations for the combined transformations ,the new coordinates are as follows:

B 01

005100051

100030010310 (,90)(5,3,6)(,90)

00106010049

00011000111 A B

P Rot z Trans Rot x P

-

⎡⎤⎡⎤⎡⎤⎡⎤⎡⎤

⎢⎥⎢⎥⎢⎥⎢⎥⎢⎥

-

⎢⎥⎢⎥⎢⎥⎢⎥⎢⎥=⨯⨯⨯==

⎢⎥⎢⎥⎢⎥⎢⎥⎢⎥

⎢⎥⎢⎥⎢⎥⎢⎥⎢⎥

⎣⎦⎣⎦⎣⎦⎣⎦⎣⎦Transformations relative to the reference frame

Transformations relative to the current frame

2.10

P=Trans(5,3,6)Rot(x,90)Rot(a,90) P

A

1 0 0 5 1 0 0 0 0 -1 0 0

2 = 0 1 0

3 0 0 -1 0 1 0 0 0 3 0 0 1 6 0 1 0 0 0 0 1 0 5 0 0 0 1 0 0 0 1 0 0 0 1 1 2 = -2 8 1 2.12

2.14

a) For spherical coordinates we have （for posihon )

1) r ·cos γ·sin β = 3.1375

units

units

2) r ·sin γ·sin β = 2.195 3) r ·cos β = 3.214

I) Assuming sin β is posihve, from a and b → γ=35°

from b and c → β=50° from c → r=5

II) If sin β were negative. Then γ=35°

β=50° r=5 Since orientation is not specified, no more information is available to check the results.

b) For case I, substifate corresponding values of sin β , cos β, sin γ, cos γ

and r in sperical coordinates to get: 0.5265 -0.5735 0.6275 3.1375

Tsph(r,β,γ)=Tsph(35,50,5)= 0.3687 0.819 0.439 2.195

-0.766 0 0.6428 3.214 0 0 0 1

2.16 Solution:

According to the equations given in the text book, we can get the Euler angles as follows:

arctan 2(,)arctan 2(,)y x y x a a or a a Φ=-- Which lead to :

21535or Φ=

②arctan 2(,)0180x y x y n S n C o S o C or ψ=-Φ+Φ-Φ+Φ= ③arctan 2(,)5050x y z a C a S a or θ=Φ+Φ=- 2.18 Solution:

①Since the hand will be placed on the object, we can obtain this:

U

U U R U R obj H R H R obj T T T T T T ===

Thus:

10015100101000

001U

U U H R obj

T T T --⎡⎤⎢⎥-⎢⎥==⎢⎥

⎢⎥

⎣⎦

②No,it can ’t.

If so,the element at the position of the third row and the second column should be 0.However, it isn ’t. ③x=5,y=1,z=0

According to the equations of the euler angles:

arctan 2(,)arctan 2(,)0180y x y x a a or a a or Φ=--= arctan 2(,)27090x y x y n S n C o S o C or ψ=-Φ+Φ-Φ+Φ=

arctan 2(,)27090x y z a C a S a or θ=Φ+Φ=

2.21