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I'm struggling to prove the following.

Set one ellipse in contact with a congruent one so that the minor axis of one is aligned with the major axis of the other. Now roll one round the other. The locus of the centre of the rolling ellipse is a circle centre the centre of the other, radius a + b.

Is there an obvious line of attack?

Paul Stephenson
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    Maybe this is obvious, and if so I apologize, but have you considered that an ellipse is the locus of all points such that the sum of the distances to the two foci is constant? – MJD Sep 03 '13 at 19:33
  • Or here's what might be a better idea: A point on an ellipse in standard position always has the form $(a\cos t, b\sin t)$ where $a$ and $b$ are the lengths of the vertical and horizontal semiaxes. Your locus therefore has the parametric equation $(a\cos t, b\sin t) + (a\cos (t+\frac\pi2), b\sin (t+\frac\pi 2))$, and then perhaps when you simplify that you get something like what you want. – MJD Sep 03 '13 at 19:45
  • The general problem is interesting; resulting in a not-so-simple curve.Ref1, Ref2. – Maesumi Sep 03 '13 at 20:00
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    @MJD: this parametrization does not use constant speed. Which means that the way you describe it, the two ellipses would slip at the point of contact, which is not intended here. – MvG Sep 04 '13 at 00:01
  • You're right, it's no good. Thanks for the correction. – MJD Sep 04 '13 at 00:51
  • Have you looked at similar rolling: http://mathworld.wolfram.com/images/gifs/EllipticGears.gif ? – Oleg567 Sep 05 '13 at 19:38
  • As for the circle thing, when a ellipse rolls on a line then the center describes a path which is part of a circle centered at origin! – Agile_Eagle Apr 01 '17 at 17:34

4 Answers4

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As I promised, here is animation, that show locus point-by-point.

  • Ellipses parameters: $a=2, b=1$.
  • Gray circle: circle with radius $3$.
  • Point of contact "moves" along static ellipse (and along rolling one) with constant speed.

Locus has some deviation of perfect circle.

Rolling Ellipses (GIF-animation, 120 kB)

Oleg567
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The locus is not a circle.

Counterexample.

Consider ellipses with semi-axes $a=2,b=1$. Let equation of static ellipse is $$ \dfrac{x^2}{2^2} + \dfrac{y^2}{1^2} = 1. $$

Here 3 steps are shown:

enter image description here

Suppose the locus is a circle (with radius $r = a+b =3$).

Then must be an instant/moment (see $2$nd image), when ellipses are co-directed (semimajor axes are parallel).

And 2 conditions must be true: $$KM = LM = 3/2;\tag{1}$$ (since symmetry); and $$len(BM) = len(AM),\tag{2}$$ where $len(...)$ is arc length. Yes, $len(BM) =^{\mbox{symmetry}} len(CM) =^{\mbox{rolling}} len(AM)$, since ellipse is rolling without slip.



1). Looking at condition $(1)$, let's find coordinates of point $M$.

$M$ belong to ellipse, $|KM|=3/2$.

$\left\{ \begin{array}{l} x^2+4y^2 = 4; \quad (M \mbox{belong to ellipse});\\ x^2+y^2 = 9/4;\quad (|KM| = 3/2); \end{array} \right. \quad \implies x = \sqrt{5/3},\; y = \sqrt{7/12}$.



2). Let's estimate $len(BM)$ and $len(AM)$.

Equation of arc $AB$ is $y = f(x) =\sqrt{1-x^2/4}$.

Note, that $f'(x) = \dfrac{-x/2}{2\sqrt{1-x^2/4}} = \dfrac{-x}{\sqrt{16-4x^2}}$.

$\sqrt{1 +(f'(x))^2} = \sqrt{\dfrac{16-4x^2}{16-4x^2} + \dfrac{x^2}{16-4x^2}} = \sqrt{\dfrac{16-3x^2}{16-4x^2}}$.

So,

$\displaystyle len(BM) = \int\limits_0^{\sqrt{5/3}} \sqrt{1+(f'(x))^2} dx = \int\limits_0^{\sqrt{5/3}} \sqrt{\dfrac{16-3x^2}{16-4x^2}}dx \approx 1.32081$ (wolfram alpha).

$\displaystyle len(AM) = \int\limits_{\sqrt{5/3}}^{2} \sqrt{1+(f'(x))^2} dx = \int\limits_{\sqrt{5/3}}^2 \sqrt{\dfrac{16-3x^2}{16-4x^2}}dx \approx 1.10131$ (wolfram alpha).

When condition $(1)$ is true, condition $(2)$ isn't true.

So, the locus isn't circle.

Oleg567
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  • Thank you Oleg 567. I found two typos in your text but they were only that. From a physical experiment I did, I was sure I had a perfect circle. The animation at vimeo.com/20944110 seemed to support this but I realise there could be a lot of play between the teeth of the gears. – Paul Stephenson Sep 04 '13 at 17:52
  • @Paul, tell me please about my typos. I am ready to correct them. I'll try to draw this locus later and compare it with perfect circle. – Oleg567 Sep 04 '13 at 18:48
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For the arc length, the middle expression should have a plus sign between the two terms. In the last integral the limits should be root-5-over-3 to 2.

On the circle hypothesis, my crude experiment with cheese boxes wrapped in velcro, a = 69+or-1 mm, b = 43+or-1 mm, gave a diameter of 225 mm with a standard deviation of 5 mm.

The experiments form part of an article for a U.K. maths magazine targeted at secondary teachers (Mathematics in School). Since your posting I'd like to make teachers of older students aware of Mathstackexchange. (Apart from the final integrals there is nothing in your proof which would not be in an A level further maths syllabus.)

Paul Stephenson
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  • Thank you very much for answer. I corrected typos now. Very interesting real experiment. In general, cheese boxes may be not an ellipses, but other kind of oval. Maybe, there exist 2 congruent ovals (not perfect ellipses), that derive circle, when rolling one around the other. – Oleg567 Sep 05 '13 at 18:50
  • Beautiful! (I had argued in my article that the figure must have 4 symmetry axes and your animation seems to bear this out.) – Paul Stephenson Sep 07 '13 at 10:54
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Just another style of counterexample.

We consider moment/instant, when ellipses are co-directed (image $2$) (see previous post).

1) Assume condition $(1)$ is true. So, find point $M$, that divides arc $AB$ into 2 parts of equal length.

$$ \displaystyle \int\limits_0^{x_M} \sqrt{\dfrac{16-3x^2}{16-4x^2}}dx = \int\limits_{x_M}^2 \sqrt{\dfrac{16-3x^2}{16-4x^2}}dx = \dfrac{1}{2} \int\limits_0^{2} \sqrt{\dfrac{16-3x^2}{16-4x^2}}dx. $$

Coordinates of $M$: $x_M\approx 1.18894378$, because $$ \displaystyle \int\limits_0^{x_M} \sqrt{\dfrac{16-3x^2}{16-4x^2}}dx \approx 1.211056028, $$ $$ \displaystyle \int\limits_{x_M}^2 \sqrt{\dfrac{16-3x^2}{16-4x^2}}dx \approx 1.211056028. $$ Then $y_M = \sqrt{1-x_M^2/4} \approx 0.80411639$.

2) But condition $(2)$ is false now. $KL = 2KM = 2\sqrt{x_M^2+y_M^2} \approx 2.8706727$. It is far from $3$. Deviation is $4.31 \%$.

When ellipses will have $a=5,b=1$, you'll find $KL \approx 5.5108044$, instead of $6$.
Deviation is $8.15 \%$ now.

Oleg567
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  • I'll try to find out from the box manufacturer what the specification is for the 'forme' ( I believe they're called) used to cut out the shapes. – Paul Stephenson Sep 06 '13 at 09:47