Implicit plane curve discretization

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Maxim Bogdan
Maxim Bogdan 2021년 4월 16일
댓글: Maxim Bogdan 2021년 4월 17일
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 2021년 4월 16일
편집: Matt J 2021년 4월 16일
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
  댓글 수: 3
Walter Roberson
Walter Roberson 2021년 4월 17일
fp = fimplicit(@(x,y) x.^2 + 2*y.^2 - 1, 'visible', 'off'); %Example
Points = [fp.XData;fp.YData]
delete(fp)
However this creates a graphic object -- so for example if you have no existing figure then one will be created with an axes in it (that just will not have any visible content.)
Maxim Bogdan
Maxim Bogdan 2021년 4월 17일
That works fine for me!
It doesn't open any figure that code for me.
Thanks a lot!

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