Yes. You do need to think about what you are doing.
I'll create a quasi-triangular region as an example.
px = [0, 1, .5, .49];
py = [0, 0, 1, 1-0.1];
plot(px([1 2 3 4 1]),py([1 2 3 4 1]),'r-')
So in the sequence I listed them, then connecting back to the beginning, the area is:
polyarea(px([1 2 3 4 1]),py([1 2 3 4 1]))
ans =
0.48
This is exactly as I would expect, just a bit less than 1/2, which would be the area of the triangle that forms the convex hull. However, in a different sequence, the included area is seen as:
polyarea(px([1 4 2 3 1]),py([1 4 2 3 1]))
ans =
0.05
plot(px([1 4 2 3 1]),py([1 4 2 3 1]),'g-')
So the same points, but in a different sequence, we got a completely different result, and that is expected.
Can you know which sequence was intended? Of course not. That is impossible, nor should a tool like polyarea be expected to know which you intended.
In the case of a convex polygon, IF the region is KNOWN to be convex, then you can just use a sort on polar angle.
t = rand(1,50)*2*pi;
px = cos(t);
py = sin(t);
polyarea(px,py)
ans =
0.968571643648727
As it turns out, I would have expected an area that was reasonably close to pi.
mux = mean(px);
muy = mean(py);
[th,r] = cart2pol(px,py); % I could have used atan2 here also
[~,ind] = sort(th);
polyarea(px(ind),py(ind))
ans =
3.11074834389343
So the area is now seen as quite close to pi. Which is correct? As I said, IF WE KNEW the region was a convex polygon, then the latter is easy. But given only the points from some general possibly non-convex domain in unsorted order, the problem is essentially impossible to solve without knowing the desired sequence of the points.
plot(px,py,'r-',px([ind,ind(1)]),py([ind,ind(1)]),'b-')
Is the area that of the blue curve in the last plot, or the area "inside" the red curve? See that when you have a Catholic polygon, that is a polygon that crosses itself multiple time,s then some parts of the polygon have positive area, some have a negative area. This effectively has some of those areas cancel each other out. (Think right hand rule, for those who know what a cross product is and what it tells you.)
