# how to calculate the intersection area of two ellipses

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amal Mbarki 2016 年 2 月 18 日
コメント済み: Image Analyst 2018 年 12 月 2 日
hello, how can i calculte the intersection area of two ellipses, each ellipse is characterized by(x,y,a,b,w).I thought of solving the system of equations of the two ellipses ,then calculate the area bounded by the points found as solution. would you please help me ?
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Hongchu Yu 2018 年 12 月 1 日
Had you solved the problem? I have the same problem now. Could you tell me the solution?
Image Analyst 2018 年 12 月 2 日
The only solution I've seen is a numerical one, not an analytical one.

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### 回答 (2 件)

Kevin Claytor 2016 年 2 月 18 日
Here is an arxiv paper on ellipse intersection algorithms.
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Image Analyst 2016 年 2 月 18 日
To create an ellipse numerically in 2-D is very easy - just see the FAQ. http://matlab.wikia.com/wiki/FAQ#How_do_I_create_an_ellipse.3F To rotate it, just multiply the coordinates by the rotation matrix [cos(theta), -sin(theta); cos(theta), sin(theta)]. Then just AND the two ellipse images
intersectionArea = ellipsoid1 & ellipsoid2;
pixelArea = sum(intersectionArea (:)); % Compute the area in pixels.
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Image Analyst 2016 年 2 月 19 日
what do you consider the "real" area? Do you have anlytical formulas for your ellipses?
amal Mbarki 2016 年 2 月 19 日

In fact each ellipse is characterized by (x,y)its center coordinate, it big and small axes(a,b) and it's orientation w. I’ve read a code which calculate the intersection area of two circles analytically .the equation is: when (d(i,j)> abs(ri-rj)) & (d(i,j)<(ri+rj), d is the distance between the centers of the two objects) then the intersection area is M(i,j) = f(xi,yi,ri,xj,yj,rj) = ri^2*arctan2(yk,xk)+rj^2*arctan2(yk,d(i,j)-xd(i,j)*yk where xk = (ri^2-rj^2+d(i,j)^2)/(2*d(i,j)) and yk = sqrt(ri^2-xk^2). So I thought why not i calculate the intersection area of two ellipse with the same way.But i don't know how to extract the analytical formula?

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