机器人学导论(克雷格)第二章作业答案
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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:
10022(,45)0
3402
22new
P rot x P ⎡⎤⎡⎤
⎢⎥
⎢⎥⎡⎤⎢
⎢⎢⎥⎢⎢=⨯==⎢⎥⎢⎢⎢⎥⎢⎥⎢⎥⎣⎦⎢⎥⎢⎥
⎢⎥⎢⎥⎣
⎦⎣⎦
2.9 Solution:
Acording to the equations for the combined transformations ,the new coordinates are as follows:
A B
010051
00
05110
0030010310(,90)(5,3,6)(,90)001060
1004900011000111A 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
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