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:

?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:

2

?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:

3

?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

4

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 -1 T2 = 0 1 0 -6 -1 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

5