Dividing ellipses

Yeah, I’m still around.

I’ve been thinking about ellipses lately. For Reasons.

Here’s a sort of silly question. Suppose I gave you a circle and asked you to mark four points on it, evenly spaced. You’d probably ask me why I don’t do such a simple thing myself, and hand me back this:


Now I ask you to do the same for eight points. You say “uh huh” and give me this:

circle8 Then I ask for four points evenly spaced around an ellipse. You sigh exasperatedly and give me this:


And then I ask for eight points on an ellipse.

Now what?

On a circle this is easy. That’s because a circle has continuous rotation symmetry; any rotation by any angle around the center maps the circle into itself. Rotate by 2π/n, 4π/n, 4π/n, … 2nπ/n radians and any point on the circle maps onto n evenly spaced points on the circle. That’s probably not how you thought about it, but that’s a way to do it.

But an ellipse is different. It has only two-fold rotation symmetry: Rotate by π radians and it maps onto itself, but any other angle (in the 0 to 2π range) doesn’t do that.

So is there any sense in which you can mark eight points evenly spaced around an ellipse? (Or seven points, or six, or even three! But let’s talk about eight.)

You know about ellipses. An ellipse centered on the origin whose semimajor axis has length a and lies along the x axis and whose semiminor axis has length b can be described by the parametric equations

x = a\sin t\\y = b\cos t

 where t ranges from 0 to 2π, from which

\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 .

 The elongatedness, or, to use the right term, the eccentricity of an ellipse is


An ellipse is the locus of points whose summed distances to two focal points is a constant, and for the above parameterization these two foci lie on the x axis at x_f = \pm a\epsilon. (They’re the magenta points in that last picture up there.)

The parameter t is not the polar angular coordinate θ. The two are related by

\theta = \tan^{-1} ((b/a)\tan t)

That’s the angle between the +x axis and a ray from the origin to the point on the ellipse. Then there’s the focal angle ɸ, between the +x axis and a ray from a focus to the point on the ellipse. Considering the focus at +aε, for points to the right of the focus — x >  — we have

\phi = \tan^{-1}((b \sin t) / (a \cos t - a\epsilon))

while for x <  — we have

\phi = \pi - \tan^{-1}((b \sin t) / (a\epsilon -a \cos t))

The length of an elliptical arc is

 a (E(t_2, \epsilon) - E(t_1, \epsilon))

where E is an elliptic integral. Not easy to calculate, in other words.

An arc and the two line segments connecting the endpoints of the arc to the origin define a central sector whose area is

A(t_1, t_2) = \frac{1}{2}ab(t_2-t_1)

And as a special case of this, the area of an ellipse is A(0,2π) = πab.

What about the area of a focal sector? This is bounded by an arc and the two line segments connecting the endpoints of the arc to a focus. In general, a focal sector can be converted to a central sector by adding one triangle and subtracting another. For instance:


Focal sector ABF is central sector ABO minus triangle BCO plus triangle ACF. In the case where point A is (a, 0) (at the end of the semimajor axis), triangle ACF disappears and triangle BCO has base  and height b sin t, so the area of that focal sector is the area of the central sector, abt/2, minus the area of the triangle, abε/2 sin t. : (ab/2)(t – ε sin t).


So back to the question. Eight evenly spaced points on an ellipse. Well, what ways are there to do it on a circle?

I talked about doing on a circle it by mapping a single point via rotations of π/4, π/2, 3π/4, … 2π radians. Equivalently, you can draw rays from the origin at angles of π/4, π/2, 3π/4, … 2π radians from the x axis and mark where the rays cross the circle.

Another way to think about it is: you know the circumference of the circle is 2πr. So divide it up into congruent arcs each of which is 2πr/8 units long.

Yet another way: Think of the circle as a pie and cut it into eight congruent slices.

Approaches like those will work for any number of points. Another, maybe somewhat odd way to think about the eight point case is to mark the points where the tangent lines have slope 0, 1, -1, or undefined.

All of these produce the same result for a circle. What if we try to adapt them for ellipses?

First the angular approach. Divide 2π into eighths… but which 2π? There are really three angles in play here for any point on the ellipse: The parameter angle t, the polar angle θ, and the focal angle ɸ. Well, two focal angles, but they’re symmetric: let’s only consider the +aε focus. For the circle t = θ and there are no foci (or they coincide with the origin and θ = ɸ, if you prefer) but for general ellipses these all differ.

Then the arc approach. You can’t divide the perimeter up into eight congruent arcs, but you can at least make eight arcs of equal length.

Next, the sector approach. Again, you can’t cut your elliptical pie up into congruent slices, but you can at least make your slices the same area. But are you cutting central sector slices or focal sector slices?

Finally the dy/dx approach.

Okay, seven ways to do it: Parameter angles, polar angles, focal angles, arcs, central sectors, focal sectors, and dy/dx. Let’s do this. For each approach compute eight values of t and mark the corresponding points. Ready?

1. Parameter angles. This is easy. t = π/4, π/2, 3π/4, … 2π radians. On an ellipse with eccentricity of 0.8 it looks like this:


I haven’t drawn in any angles because there’s nothing to draw; t doesn’t correspond with any geometrical angles here. On an ellipse with eccentricity 0.99 it looks like this:


That’s probably not what you mean by evenly spaced.

2. Polar angles. This is almost as easy. θ = π/4, π/2, 3π/4, … 2π radians and t=\tan^{-1}((a/b)\tan\theta). For the four off-axis points you just get t=\tan^{-1}(\pm a/b). On our 0.8 ellipse, with the polar rays drawn in:


And on our 0.99 ellipse:ellipse_poangle_99Okay, that’s really not what you mean.

3. Focal angles. Just because we can, though it’s not so easy. To get the ɸ = π/4 and ɸ = 3π/4 points you need to solve

\frac{b\sin t}{a\cos t-a\epsilon}=\pm 1

(do that by squaring and substituting 1-\cos^2t for \sin t, giving you a quadratic in \cos t) and for the ɸ = π/2 point it turns out

t=\cos^{-1}\epsilon .

The ɸ = 0 and ɸ = π points I hope are obvious and the others you get by reflection. Result, on the 0.8 ellipse, with the focal rays drawn:


You don’t even want to see the 0.99 ellipse here. Trust me.

4. Arcs. On the whole this seems likely to be the most reasonable way to interpret “evenly spaced”. So how’s it work out? Well, all you have to do is to solve

E(t_{n+1}, \epsilon) - E(t_{n}, \epsilon)=E(2\pi,\epsilon)/8

for t1, t2, and t3 (t0 we’ll set to 0, and the other tn we’ll get from symmetry). Which is… hard. Let’s skip that.

5. Central sectors. Back to easy again; use the central sector area formula. Here it is for 0.8:

ellipse_posect and for 0.99:ellipse_posect_99 Those are equitable slices of elliptical pie. May still not be what you mean by evenly spaced, though.

6. Focal sectors. Uh oh. The area formula has (t – ε sin t) in it; how’re you going to solve that for t? I asked wolframalpha.com and it just laughed at me. Good thing this is guaranteed to be a silly way to interpret “evenly spaced” anyway.

7. dy/dx. Just for grins.The point spacings may not be regular, but their tangent line orientations are! The derivative is

\frac{dy}{dx}=\frac{(b/a)}{\tan t}

so for the off axis points, where the slope is to be +1 or –1, t=\tan^{-1}(\pm b/a). (Compare that to the polar angles method, where we got t=\tan^{-1}(\pm a/b)!) For 0.8:

ellipse_dydxand for comedy relief, 0.99:ellipse_dydx_99

Let’s pretend I never drew that.




4 thoughts on “Dividing ellipses

  1. How are number 1.- and number 5.- any different? If you divide the area in equal parts and then solve t2 you get:
    t2 = t1 + (1/parts)*2*PI

  2. Any chance we could revisit the arc length one? As you said, that really seems like the only one that truly matches the definition of ‘evenly spaced’ 😛

  3. I am on the edge of my seat here and so disappointed there is no conclusion! I prepared an outline last night that looks very similar to the proposed arc length method. My intent is to approximate a means to distribute points on an ellipse so the linear distance between them is same. I realized before arriving here that equal arclength segments does not facilitate this.

    Is there a way to solve this problem that does not require programming? If not, has anyone attempted it?

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