Area of two identical overlapping ellipses

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laoliu102
laoliu102 2017-12-19
评论： John D'Errico 2017-12-20
Hi, I have two identical overlapping ellipses (one is moved by 0.2 in the y direction). How can I calculate the overlapping area?
I know how to analytically solve the overlapping area of two identical circles, but the analytical solution for two identical overlapping ellipses is very hard to find...Maybe Matlab can do it numerically or by comparing pixels?
My two ellipses are generated using the following code:
a = 1;
b = sqrt(2);
x0=0;
y0=0;
t=-pi:0.01:pi;
x=x0+a*cos(t);
y=y0+b*sin(t);
m1=fill(x,y,'b')
m1.FaceAlpha=0.2;
y01=y0+0.2;
y1=y01+b*sin(t);
hold on
m2=fill(x,y1,'r')
m2.FaceAlpha=0.2;

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John D'Errico
John D'Errico 2017-12-20
编辑：John D'Errico 2017-12-20
Pretty simple actually as just a transformation of variables.
You claim to have the analytical solution for a pair of circles, with a simple y offset.
Solve the problem for a pair of circles with unit radius, where one circle is offset in y by a delta of 0.2/b. (.2/b is important here.) Get the area as A_c. (Thus the area of intersection for two circles of radius 1.)
Now, if you implicitly transform the variables such that
X_e = x_c*a
Y_e = Y_c*b
The area of the ellipse intersection will be a*b*A_c.
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laoliu102
laoliu102 2017-12-20
The ellipses are not rotated. The second ellipse only moves a distance in the y direction on the plane where the ellipse sits.
John D'Errico
John D'Errico 2017-12-20
Exactly. Were one of the ellipses rotated, it gets a bit messy. But this case is a simple one.

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