Locating the midpoint of a quarter-ellipse

I’m doing some CAD work (it’s by no means my strength, but necessary for this research task), and I find myself needing to locate the midpoint of a quarter-elliptical segment. The CAD package I’m using, to my knowledge, won’t select the midpoint automatically like it does for lines and circular arcs.

So I’m left with having to create a geometric construction or apply a formula to find this midpoint. I may need to do this repeatedly, so I’m hoping for something not too tedious. I know that calculating elliptical arc length is difficult, but luckily the program does that automatically.

Anyone know how to do this? Is there a relation based on the dimensions of the semi-axes that will make the midpoint easy to find?

aerodave, when you say a “quarter-elliptical segment”, what exactly do you mean?


And by midpoint, do you mean the middle of the segment (the edge), the middle of the area if it was ‘squared off’ or the focal point?

This should work:

Assume that your ellipse is centered at the origin, and that the axes intersect it at points A and B. Draw the line segment AB, and draw the ray from the origin to the midpoint of AB. That ray should intersect the ellipse at a point Q such that the arc from A to Q is the same length as the arc from B to Q.

(I’m assuming that’s the sort of midpoint you’re looking for. If you mean something else, please clarify.)

Calculation of elliptical arc length involves using the elliptical integrals. It’s been some time since I did this, but I do recall it being a mess. I don’t think there’s a simple relation that will tell you the midpoint of an arc length of 1/4 of the ellipse, bounded by the major and minor axes, except in special cases like the circle.



Unfortunately, this won’t work; the ratios arc lengths aren’t preserved under affine transformations. (The figure you describe can be obtained from a quarter-circle with a 45-degree line drawn through it, and “squashing” it in an affine transformation.) The two “slices” of the ellipse you make this way will have equal areas, but that’s not much help.

As CalMeacham noted, there may not be a way to do this without elliptic integrals. Basically, you would need to find a solution to the equation

E([symbol]q[/symbol], e) = 1/2 E(e)

where e is the eccentricity of the ellipse, E([symbol]q[/symbol], e) is the incomplete elliptic integral of the second kind, and E(e) is the complete elliptic integral of the second kind. If you can solve for [symbol]q[/symbol], then the point on the ellipse which marks the halfway point is given by (x,y) = (a sin([symbol]q[/symbol]), b cos([symbol]q[/symbol])).

I did a little tinkering with Mathematica, and discovered that [symbol]q[/symbol] goes smoothly from pi/4 for e = 0 (as expected) to pi/6 for e = 1 (somewhat surprising, but OK.) It probably wouldn’t be too hard to get approximate expansions for nearly circular or highly eccentric ellipses, but beyond that I’m afraid you may just have to suck it up and solve the above equation numerically.

Well guys, thanks for the input. I guess it is going to be a little messy.

And for the record, the midpont of the arc is what I am interested in. I realize using the word “segment” was misleading, because the means something different.

Thanks again.

Just how precise does your midpoint have to be? If it needs to be exact, then you will have to use the integral, but in real life you’re working with limited precision.

My first thought, since your software calculates arc length, would be to try successive approximation. (I’m assuming you have a scripting language in the program.) Guess a value for x (say the mean of the min and max values), calculate y, get the length of the partial arc, and use that to adjust x toward the long side of the arc. Loop until close enough to half the original.

What definition of eccentricity are you using? Under the standard definition, an eccentricity of 1 is a parabola, and the problem becomes ill-defined.

I suppose this was a little sloppy; I should have said the eccentricity “approaches 1”. For a non-degenerate ellipse, the eccentricity is just sqrt(a[sup]2[/sup] - b[sup]2[/sup])/a, where a and b are the semi-major and semi-minor axes. The “e = 1” case I was talking about is then the case where b goes to zero while a is held constant. If you prefer, you can talk about the angular eccentricity going to pi/2.

Since I’m posting, I’ll also note that after a little more thought, the value of [symbol]q[/symbol] = pi /6 makes perfect sense in this limit: the quarter-ellipse becomes a straight line running from (0,0) to (a,0), and the “midway point” is (a* sin(pi/6), 0) = (a/2,0).