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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 ?

John D'Errico
on 20 Mar 2017

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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