机器人学导论(克雷格)第二章作业答案

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

10Trans(dx,dy,dz)Pold000023558103

01471100111Ptrans2.3 solution:

According to the constraint equations:

n•a0;n•o0;a•o0n1

Thus,the matrix should be like this:

00110001000050011003or2010100053 212.4 Solution:

PXPYPZcos=0sin0sinPn10P0

0cosPa2.7 Solution:

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

1rot(x,45)P0002222202223 22427222Pnew2.9 Solution:

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

0110ABPRot(z,90)Trans(5,3,6)Rot(x,90)P00000051300010600110051310010100490011100

Transformations relative to the reference frame

Transformations relative to the current frame2.10

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

B

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

0.527 0.369 -0.766 -0.601 T1 = -0.574 0.819 0 -2.947 0.628 0.439 0.643 -5.38 0 0 0 1

0.92 0 -0.39 -3.82 T2 -1 = 0 1 0 -6 0.39 0 0.92 -3.79 0 0 0 1 2.14

a) For spherical coordinates we have （for posihon )

1) r·cos γ·sin β = 3.1375

-1 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

units

units

Since orientation is not specified, no more information is available to check the results.

b) For case I, substifate corresponding values of sinβ , cosβ, sinγ, cosγ

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:

arctan2(ay,ax)orarctan2(ay,ax) Which lead to :

215or35

arctan2(nxSnyC,oxSoyC)0or180 arctan2(axCayS,az)50or50 2.18 Solution:

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

UTobjUTHUTRRTHUTRRTobj

Thus:

0100015001 100001UTHUTR1UTobjNo,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. x=5,y=1,z=0

According to the equations of the euler angles:

arctan2(ay,ax)orarctan2(ay,ax)0or180 arctan2(nxSnyC,oxSoyC)270or90

arctan2(axCayS,az)270or90

2.21

(a)

(b) # 0-1 1-2 2-H (c)

10UTO=00100 A=101d1d2001000θ 1 d 0 0 a d3 α 0 180 0 0

A2= A3=

(d)

UTHUTOOTHUTOA1A2A3

2.22 (a)

(b) # 0-1 1-2 = 3-H (c)

0 0 θ +90 d 0 l2 a 0 0 0 α 90 -90 0 l3 = =

= = A4=

(d)

2.23 (a)

(b) # 1 θ d 0 a 0 α 90 2 4 4 5 6 (c)

6'' 15'' 0 -90 90 -90 0 0 18'' 1 0 0 0 0 5 = = =

= A5= A6=

(d)

RTH=