Hi, I am finding area enclosed by convex hull using delayunaytriangulation,,,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 2017-3-20

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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 2017-3-20

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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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John D'Errico 2017-3-21

编辑：John D'Errico 2017-3-21

在 MATLAB Online 中打开

I made your code readable.

I see that you tried to use the "{} Code" button. The trick is to paste in your code.Then select the entire code block, and only then click on the "{} Code" button.

I suspected what you had done. I was right.

polyarea does NOT take a random set of points,and then compute the area of the convex hull.

polyarea is designed to compute the area of a POLYGON, so a list or ordered pairs of points that enclose a region in the (x,y) plane.

So, you computed the area contained in the polygonal region enclosed by these points:

plot(x,y,'-o')

Even worse, that polygon crosses over itself, which makes things even worse. It is no surprise at all that polyarea computed a very small area.

But IF you want to compute the area of the region contained in the convex hull of those points, it is 390.

E = convhull(x,y);
polyarea(x(E),y(E))
ans =
          390.270316856299

We can see the difference in this plot.

plot(x,y,'b-o',x(E),y(E),'r-')

The area inside the red polygon is 390. The area "inside" the blue polygonal thing that crosses over itself is 19.

As I said, you cannot just throw random points at polyarea. polyarea is designed to compute the area of a polygon.

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.

L K 2017-3-21