What is going wrong with my mesh command?
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I am trying to visualize this function and eventually its contour lines:
clear all
[x,y]=meshgrid(-10:0.1:10)
d=max([sqrt(x.^2+y.^2);sqrt((x-5).^2+(y+1).^2);sqrt((x-4).^2+(y-6).^2);...
sqrt((x-1).^2+(y-3).^2)])+eps;
mesh(d)
I get an error message saying Z should be a matrix. I looked at the example here http://www.mathworks.com/help/matlab/ref/mesh.html, and I don't understand why their function generates a matrix and mine don't. Altough mine is a 'weird' function with a max it shouldn't be conceptually too different, what am I missing here?
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採用された回答
Iman Ansari
2013 年 5 月 6 日
編集済み: Iman Ansari
2013 年 5 月 6 日
Hi.
clear
[x,y]=meshgrid(-10:0.1:10)
d=max(sqrt(x.^2+y.^2),sqrt((x-5).^2+(y+1).^2));
d=max(d,sqrt((x-4).^2+(y-6).^2));
d=max(d,sqrt((x-1).^2+(y-3).^2))+eps;
mesh(x,y,d)
OR
meshc(x,y,d)
2 件のコメント
Iman Ansari
2013 年 5 月 6 日
編集済み: Iman Ansari
2013 年 5 月 6 日
From Help: If A is a matrix, max(A) treats the columns of A as vectors, returning a row vector containing the maximum element from each column.
Your input in max function is 804*201, so output become 1*201.
No.
その他の回答 (1 件)
Walter Roberson
2013 年 5 月 6 日
d=max( cat(3, sqrt(x.^2+y.^2), sqrt((x-5).^2+(y+1).^2), sqrt((x-4).^2+(y-6).^2), sqrt((x-1).^2+(y-3).^2)), 3) + eps;
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