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?

 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 ?

Answer 1

[John D'Errico]

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

[L K]

That 390. something is in what units ? area is calculated in what units?

And if am using polyarea over here...Is this right ? area1=polyarea(x,y)

[John D'Errico]

What units? Area has units of length^2. Only you know what the units of your variable is, but since the an angle has NO units attached, the only units you have in this problem are the units attached to radius. Again, since a radius has units of length, the area will have units of length^2.

Yes, that is the correct usage of polyarea. polyarea only makes sense when applied in Cartesian coordinates, as I responded to your other question.

[L K]

yes thank u very much :) another small doubt, my radius is actually the amplitude (in volts)...so the units volts^2 you mean ?

And according to polyarea i am getting it as 19.0449... why the difference ?

[John D'Errico]

If radius has units of volts, then yes area has units volts^2. This is just common sense.

Perhaps because you used the form I told you NOT (EXPLICITLY NOT) to use? Or perhaps you did something else completely wrong? I cannot guess. But an area of roughly 19 is nonsense in those case.

Here are the points you gave, in a cartesian plot.

plot(x,y,'.')

A box that contains the points would be roughly 90 units high, by roughly 9 units wide. So a simple bounding box would have area of roughly 810.

Since we are talking about the are of a convex hull, NOT a rectangular bounding box, the area is approximately 50% of that. 390 makes perfect sense. 19 is nonsense in this context.

19 makes no sense, except that you probably completely misused polyarea. Polyarea is not designed to compute the area of a list of random points. It computes the area contained within a POLYGON.

[L K]

This is what i had done...which fetched me 19. something as area using polyarea x and y were cartesian coordinates

%%Aluminium
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);
area1=polyarea(x(:),y(:))
k = convhull(x, y);
xch = x(k);
ych = y(k);
[thetaCH1, rhoCH1] = cart2pol(xch, ych);
%subplot(212)
polar(thetaCH1, rhoCH1, 'ro-');
DT = delaunayTriangulation(x(:),y(:));
[U,v]=convexHull(DT);

end

[John D'Errico]

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]

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]

Thank you John & Image analyst...