Hi, I am finding area enclosed by convex hull using delayunayt​​riangulat​i​on,,,i have pasted the code...I just need someone to tell me..the area i got is right according to my code?

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L K
L K 2017-3-20
评论： L K 2017-3-21
theta1=[88,89,90,91,92,94,96,94,90,89,-100,-102,-104,-105,-104,-102,-101,-100];
radius1=[5,7,11,17,26,39,46,44,32,3,0,18,34,32,33,29,28,20];
%subplot(211)
theta_rad=theta1*pi/180;
polar(theta_rad, radius1, 'b*');
hold on;
[x, y] = pol2cart(theta_rad, radius1);
k = convhull(x, y);
xch = x(k);
ych = y(k);
[thetaCH1, rhoCH1] = cart2pol(xch, ych);
%subplot(212)
polar(thetaCH1, rhoCH1, 'ro-');
DT = delaunayTriangulation(theta_rad(:),radius1(:));
[U,v]=convexHull(DT);
i got v=130.8648.... is it the right way to do it ?

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John D'Errico
John D'Errico 2017-3-20
NO. You cannot compute a convex hull of your points when they are represented in polar coordinates!!!!! If you did, the result will be nonsensical. And the area it would compute will certainly be nonsense.
Instead, convert the polar coordinates to cartesian coordinates, then compute the area of the convex hull in Cartesian coordiantes:
DT = delaunayTriangulation(x(:),y(:));
[H,A] = convexHull(DT);
A =
390.270316856299
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Image Analyst
Image Analyst 2017-3-21
Another quirk of polyarea is that if the perimeter overlaps, you can have a negative area there. For example, the area of a perfect bowtie shape is zero according to polyarea.

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