机器人学导论克雷格作业答案

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^he new point after transition is as follows:
2.3 solution:
According to the constraint equations: 〃 • d = 0;〃 • o = 0；d • o = 0
n = 1
Thus,the matrix should be like this:
'o
0 -1 5"
0 -1 5" 1 0 0 3 or -1
0 0 3 0 -1 0
2 0 -1 0 2 0
1
_0
1
2.4
= Trans (d x , d y , ) x P ()ld
Solution:
Solution:
According to the equation of pure rotation transformation , the new coordinates are as follows:
2.9 Solution:
Acording to the equations for the combined transformations ,the new coordinates are as follows:
2.7
1
P new = rot(xA5 )xP = 0
V2
2
返
V2
2 .
7A /2
Transformations relative to the current frame 2.10
此T「ans(536)Rot(x，90)Rot(a，90) P
1
0_丁'1 '
0 3 10
0 4 9
1 1 1
0 -1
t
. 1 0
A P = Rot(z,90 )X Trans(5y3,6)x Rot(x.90°)x P= ° Q
0 0 Tran sformatio ns relative to the ref ere nee frame
(/O 5 1
0 10 3 0 0-1 0 10 0 0
0 10 0 0 0 10
0\Q 0
2.12
0. 0.369 -0.766 -0.601
Tl = 0.574 0.819 0 -2.947
0.6 28 0.439 0.643 -5.38
OJ0
0 0 -0.39 -3.82
0 -6
0.3 0.92 ・3.79
\0 0
2.14
a) For spherical coordinates we have (for posihon )
1)r cos y sin 0 二 3.1375
2)r sin ysin 0 二 2.195
3)r cosp = 3.214
I) Assuming sin P is posihve, from a and b —> y二35°
from b and c t 0二50°
units
II) If sin p were negative. Then
Y二35。

p=50°「二5 units
Since orientation is not specified, no more information is available to
check the results・ b) For case substifate corresp on ding values of sin0 , cos 色sin” cosy
and r in sperical coordinates to get:
0.5265 -0.5735 56275 3.1375
Tsph(r,P,Y)=Tsph(35,50,5)= 3687 0.819 0.439 2.195
-0.766 0 0.6428 3.214
2.16
Solution:
According to the equations given in the text book, we can get the Euler angles as follows:
①①=arctan 2(-a y, —a x )or arctan 2(a>., a x)
Which lead to :
①= 215 "35
②W = arctan 2（-qS① + n y C^-o x S①+o、C①）=0 orl 80
③0 = arctim2（“ C① + yS①,a.） =50 "-50
2.18
Soluti on: ©Since the hand will be placed on the object, we can obtain this:
Thus:
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.
③ x=5,y 二 1忆二 0
According to the equations of the euler angles:
①=arctan2（-ci,,—a v）or arctan2（g,a t） = 0 or\80
T = arctan 2（-n x S①+n v C^-o x S①+o、C①）=270。

皿
6 = arctan 2（yC① + a、.S0«.） = 270 or90
2.21
(b) #
e
d a a 0-1 q 0 £ 0 1-2
&2
£ 180 2-H
0、、
〃、
(c)
j 0 0 0
7】
0 d 3c \
0 1 0
A {=
q 0 心
0 0 1 4+d? 0
0 1 0 0 0 0 1
1 ■
(d)
T H^T O°T H^T O A X A2A3
2.22
⑻
c2 s2 0 d4c2
0 0-10
0 0 0 1 0
11
# e
d a a
0-1 q+90 0 0 90 1-2 0
/,
-90
二
3-H 0
0 0
(c)
j 0 0
"1 0 0
o'
A 2 0
1 0
人二
0 ~c \ 0 A 二
0 1 0 0
0 -1 0 l 2 0 1
0 0 0 0 1 h
0 0 0 1
0 0
1
0 0 0 1
(d)
IO
U T O =
10 0/,
0 5| 0 _$1
(b)
(c)
A}='1
1
o"
0 A2=
「0.707
0.707
-0.707
0.707
10.6f
10.61 去二
"1
1
1
0 0 -1 0 0 0 0 1 6 0 -1 0 0 _0 0 0 1 _ 0 0 0 1 0 0 0 1
■1 0 0 o-_ 0.707 0 0.707 0_'1 0 0 o-
0 0 -1 0 -0.707 0 0.707 0 0 1 0 0
A.=人二
r 0 1 0 18 J 0 -1 0 0 0 0 1 5
0 0 0 1 . 0 0 0 1. _0 0 0 1
(d)
"1 0 0 -1.414
Rqr 0 -1 0 6
0 0 -1 -19
_0 0 0 1
13。