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Implicit plane curve discretization

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Maxim Bogdan
Maxim Bogdan 2021 年 4 月 16 日 5:35
コメント済み: Maxim Bogdan 2021 年 4 月 17 日 8:53
I want to have a matrix with 2 lines and n columns with the coordinates of points that lie on the implicit curve where is a function. I know that the matlab function fimplicit can plot the curve. So in its subroutines it grabs a set of points from\near that curve. How can I extract those points and use them for future calculations (like the perimeter of the curve)?
Thanks!

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Matt J
Matt J 2021 年 4 月 16 日 5:59
編集済み: Matt J 2021 年 4 月 16 日 6:22
fp=fimplicit(@(x,y) x.^2 + 2*y.^2 - 1); %Example
Points=[fp.XData;fp.YData]
Points = 2×587
-0.0458 -0.0533 -0.0667 -0.0800 -0.0933 -0.1067 -0.1200 -0.1333 -0.1467 -0.1600 -0.1689 -0.1733 -0.1867 -0.2000 -0.2133 -0.2267 -0.2337 -0.2400 -0.2533 -0.2667 -0.2800 -0.2834 -0.2933 -0.3067 -0.3200 -0.3251 -0.3333 -0.3467 -0.3600 -0.3616 -0.7063 -0.7061 -0.7055 -0.7048 -0.7040 -0.7031 -0.7020 -0.7008 -0.6994 -0.6980 -0.6969 -0.6964 -0.6947 -0.6928 -0.6908 -0.6887 -0.6875 -0.6864 -0.6840 -0.6815 -0.6788 -0.6781 -0.6760 -0.6730 -0.6699 -0.6687 -0.6667 -0.6632 -0.6597 -0.6593
  3 件のコメント
Maxim Bogdan
Maxim Bogdan 2021 年 4 月 17 日 8:53
That works fine for me!
It doesn't open any figure that code for me.
Thanks a lot!

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