Making a transformation matrix in matrix format

6 ビュー (過去 30 日間)
C K Chan
C K Chan 2021 年 3 月 30 日
コメント済み: C K Chan 2021 年 4 月 3 日
Hi, I've a matrix shows below
syms th d alph l
A = [cos(th) -cos(alph)*sin(th) sin(alph)*sin(th) l*cos(th);
sin(th) cos(alph)*cos(th) -sin(alph)*cos(th) l*sin(th);
0 sin(alph) cos(alph) d;
0 0 0 1]
i end up getting 4 matrices
A =
[cos(th), -cos(alph)*sin(th), sin(alph)*sin(th), l*cos(th)]
[sin(th), cos(alph)*cos(th), -sin(alph)*cos(th), l*sin(th)]
[ 0, sin(alph), cos(alph), d]
[ 0, 0, 0, 1]
but I want it in one matrix format, what can I do with it?
  1 件のコメント
C K Chan
C K Chan 2021 年 3 月 30 日
syms th d alph l
A = [ cos(th) -cos(alph)*sin(th) sin(alph)*sin(th) l*cos(th);
sin(th) cos(alph)*cos(th) -sin(alph)*cos(th) l*sin(th);
0 sin(alph) cos(alph) d;
0 0 0 1 ];
syms th1
A1 = subs(A,{l,alph,d,th},{0,pi/2,20,th1})
or I use subs numbers in the the matrix, I end uo getting this 4 matrices
A1 =
[cos(th1), 0, sin(th1), 0]
[sin(th1), 0, -cos(th1), 0]
[ 0, 1, 0, 20]
[ 0, 0, 0, 1]
I want to get it in one matrix like below
A1 =
cos(th1), 0, sin(th1), 0;
sin(th1), 0, -cos(th1), 0;
0, 1, 0, 20;
0, 0, 0, 1;
Thank you!

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採用された回答

Paul
Paul 2021 年 3 月 30 日
You're not getting 4, 1 x 4 matrices. It just looks like that because of the formatting to the diplay.
>> syms th d alph l
A = [ cos(th) -cos(alph)*sin(th) sin(alph)*sin(th) l*cos(th);
sin(th) cos(alph)*cos(th) -sin(alph)*cos(th) l*sin(th);
0 sin(alph) cos(alph) d;
0 0 0 1 ];
>> size(A)
ans =
4 4
As shown, A is, in fact, one 4 x 4 sym matrix.
  5 件のコメント
Paul
Paul 2021 年 3 月 31 日
So everything works as you need it to work?
C K Chan
C K Chan 2021 年 4 月 3 日
Thanks a lot.
I fixed it.

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