Rolling Circles and Balls (Part 2)

Last time we rolled a circle on another circle the same size, and looked at the curve traced out by a point on the rolling circle:

It’s called a cardioid.

But suppose we roll a circle on another circle that’s twice as big. Then we get a nephroid:

Puzzle 1. How many times does the small circle rotate as it rolls all the way around the big one here?

By the way, the name ‘cardioid’ comes from the Greek word for ‘heart’. The name ‘nephroid’ comes from the Greek word for a less noble organ: the kidney! But the Greeks didn’t talk about cardioids or nephroids—these names were invented in modern times.

Here are my 7 favorite ways to get a nephroid:

1) The way just described: roll a circle on a circle twice as big, and track the path of a point.

2) Alternatively, take a circle that’s one and a half times big as another, fit it around that smaller one, roll it around, and let one of its points trace out a curve. Again you get a nephroid!

3) Take a semicircle, point it upwards, shine parallel rays of light straight down at it, and let those rays reflect off it. The envelope of the reflected rays will be half of a nephroid:

This was discovered by Huygens in 1678, in his work on light. He was really big on the study of curves.

As I mentioned last time, a catacaustic is a curve formed as the envelope of rays emanating from a specified point and reflecting off a given curve. We can stretch the rules a bit and let that point be a ‘point at infinity’. Then the rays will be parallel. So, we’re seeing that the nephroid is a catacaustic of a circle.

Last time we saw the cardioid is also a catacaustic of a circle, but with light emanating from a point on the circle. It’s neat that the cardioid and nephroid both show up as catacaustics of the circle. But it’s just the beginning of the fun…

4) The nephroid is the catacaustic of the cardioid, if we let the light emanate from the cardioid’s cusp!

This was discovered by Jacques Bernoulli in 1692.

5) Let two points move around a circle, starting at the same place, but with one moving 3 times as fast as the other. At each moment connect them with a line. The envelope of these lines is a nephroid!

Last time we saw that if we replace the number 3 by 2 here, we get a cardioid. So, this is yet another way these two curves are related!

6) Draw a circle in blue and draw its diameter as a vertical line. Then draw all the circles that have their center somewhere on that blue circle, and are tangent to that vertical line. You get a nephroid:

The red circle here has the red dot as its center, and it’s tangent to a point on the vertical line. Here’s a nice animation of the process, made available by the Math Images Project under a GNU Free Documentation License:

7) Finally, here’s how to draw a nephroid starting with a nephroid! Draw all the osculating circles of the nephroid—that is, circles that match the nephroid’s curvature as well as its slope at the points they touch. The centers of these circles give another nephroid:

This trick is an example of an ‘evolute’. The evolute of a curve is the set of centers of the osculating circles of that curve. Last time we saw the evolute of a cardioid is another cardioid. Now we’re seeing the nephroid shares this property!

Apparently the same is true for all curves formed by rolling one circle on another. These curves are called epicycloids. In a sense, these are the mathematical leftovers of the theory of epicycles in astronomy.

It would be nice if some of the funny relations we’ve been seeing between the cardioid and the nephroid generalize to relations between the epicycloid with k cusps and the one with k+1 cusps. But I don’t know if that’s true.

It would also be nice if the epicycloids with more and more cusps were named after increasingly disgusting organs of the body. But in fact, I don’t know any special names for them once we reach k = 3.

Puzzle 2. Use one of the 7 constructions above to get an equation for the nephroid. What is the simplest equation you can find for this curve?

Next time

Most of the pictures above are from Wikicommons, but the picture of the nephroid as a catacaustic of the cardioid is from Xah Lee’s wonderful website on plane curves. As usual, you can click on the pictures and get more informatino.

My ultimate goal is to tell you some amazing things about what happens when you roll one ball on another that’s exactly 3 times as big. These things have nothing to do with plane curves, actually. But I’ve been taking many detours, and next time I’ll talk about some curves formed by rolling one circle inside another!

Right now, thought I need a break. I need to stop thinking about all these curves. I think I’ll get a cup of coffee.

This entry was posted on Monday, September 3rd, 2012 at 3:31 am and is filed under mathematics. You can follow any responses to this entry through the RSS 2.0 feed. You can leave a response, or trackback from your own site.

45 Responses to Rolling Circles and Balls (Part 2)

  1. Douglas Summers-Stay says:
    3 September, 2012 at 9:53 am

    I thought your comment about “increasingly disgusting organs” was funny. The one with six lobes could be an elephant brain:
    http://www.smithsonianmag.com/science-nature/The-Social-Brain.html

    Reply
  2. Greg Egan says:
    3 September, 2012 at 10:45 am

    The easiest parameterisation to derive is:

    (x, y) = 3a (\cos \theta, \sin \theta) - a (\cos 3\theta, \sin 3\theta)

    where a is the radius of the small circle, which moves in a circular path of radius 3a, and the point that inscribes the curve needs to counter-rotate three times faster so that the tangential velocity of that point where it makes contact with the large circle is zero.

    So the answer to puzzle 1 is “three times”.

    We can simplify the parameterisation a little to:

    (x, y) = 2a (3\cos \theta - 2\cos^3 \theta, 2\sin^3 \theta)

    It’s not hard to solve this to express x in terms of y, though the result’s not especially simple. But I can massage it into this peculiar sixth-degree equation:

    \left(x^2+y^2-4 a^2\right)^3-108 a^4 y^2 = 0

    Reply
    • John Baez says:
      3 September, 2012 at 11:31 am

      Great! Did you get that sextic equation without peeking at the back of the book, Greg? The same equation, complete with that peculiar factor of 108, appears here:

      • Wolfram MathWorld, Nephroid.

      If you got it independently, I think I can trust it.

      My own favorite equation for the nephroid is a parametric equation, but it’s a wee bit simpler-looking than the ones you wrote down. Can anyone guess the rather easy trick I used?

      For beginners, it’s probably worth emphasizing that the \theta in your equation is not one of the polar coordinates (r,\theta) of the point on the small circle that’s tracing out the nephroid we’re trying to describe. Instead, it’s one the polar coordinates of the center of the small circle.

      The main reason I want to emphasize this is as follows. Suppose we take your first formula for the nephroid and call the parameter t instead of \theta to avoid the possible confusion I just mentioned, which will become dangerous in a minute or two:

      (x, y) = 3a (\cos t, \sin t) - a (\cos 3 t, \sin 3 t)

      Then let’s choose the radius a of the small circle to be 1, just to keep things simple:

      (x, y) = 3 (\cos t, \sin t) - (\cos 3 t, \sin 3 t)

      We can write down a very similar parametric equation for the cardioid. The only difference is that now the little circle rotates twice as fast as it revolves, instead of three times:

      (x, y) = (\cos t, \sin t) - (\cos 2 t, \sin 2 t)

      Then if we let

      r^2 = x^2 + y^2

      in the usual way, we get

      \begin{array}{ccl} r^2 &=& (\cos t - \cos 2 t)^2 + (\sin t - \sin 2 t)^2 \\ &=& \cos^2 t + \sin^2 t + \cos^2 2t + \sin^2 2t - 2 (\cos t \cos 2t + \sin t \sin 2 t) \\ &=& 2 + 2 \cos t \end{array}

      where in the last step we used a famous trig identity for \cos(\alpha - \beta). (This looks mildly clever, but I originally did the whole calculation based on my simpler parametrization, and then no cleverness at all is required!)

      So we get a cute fact:

      r^2 = 2 (1 + \cos t)

      This becomes even simpler if we use another trig identity. We get

      r^2 = 4 \cos^2 (t/2)

      or

      r = 2 |\cos(t/2)|

      But what’s funny is that this equation

      r^2 = 2 (1 + \cos t)

      vaguely resembles the formula for a cardioid in polar coordinates:

      r = 1 + \cos \theta

      where the parameter \theta is one of the polar coordinates of the point tracing out the cardioid! They’re both true facts about the cardioid, but I don’t see a quick way to get from one to the other.

      Reply
    • Michael Habeck says:
      4 September, 2012 at 1:44 pm

      An easier answer to puzzle 1: As the smaller circle travels around the larger one, it covers a distance of L=2\pi (3a). Because one rotation of the small circle covers a distance of l=2\pi a, the small circle has rotated L/l = 3 times in total.

      Reply
  3. Greg Egan says:
    3 September, 2012 at 12:38 pm

    I didn’t peek at the MathWorld article, but I did use Mathematica to help with the algebra, so I guess the answers aren’t 100% independent (a fiendish Mathematica bug could be generating this result whenever someone does the calculation).

    I think there might be a fiendish bug in your parametric equation for the cardioid: unless I’m confused, you haven’t placed the centre of the outer circle far enough from the origin. For a cardioid, I get:

    (x,y) = 2(\cos t, \sin t) - (\cos 2t, \sin 2t)

    and I end up with:

    r^2 = 5 - 4 \cos t

    For a nephroid, with:

    (x,y) = 3(\cos t, \sin t) - (\cos 3t, \sin 3t)

    I get:

    r^2 = 10 - 6 \cos 2t

    And for a general epicycloid, with:

    (x,y) = (k+1)(\cos t, \sin t) - (\cos (k+1)t, \sin (k+1)t)

    I get:

    r^2 = k (k+2)+2 - 2 (k+1) \cos (k t)

    It would be nice if some of the funny relations we’ve been seeing between the cardioid and the nephroid generalize to relations between the epicycloid with k cusps and the one with k+1 cusps.

    I made some pictures of the catacaustics to check, and alas that pattern doesn’t continue.

    Reply
    • John Baez says:
      3 September, 2012 at 3:49 pm

      Here are Greg’s pictures of catacaustics of epicycloids, formed by letting light rays emanate from a cusp and then reflect:

      The catacaustic of the 1-cusped epicycloid (the cardioid) is the 2-cusped one (the nephroid):

      The catacaustic of the 2-cusped epicycloid (the nephroid) is, alas, not the 3-cusped one:

      The catacaustic of the 3-cusped epicycloid is not the 4-cusped one:

      The catacaustic of the 4-cusped epicycloid is not the 5-cusped one:

      And so on, no doubt. But they’re certainly charming curves!

      Reply
    • John Baez says:
      3 September, 2012 at 4:04 pm

      Greg wrote:

      I think there might be a fiendish bug in your parametric equation for the cardioid: unless I’m confused, you haven’t placed the centre of the outer circle far enough from the origin.

      I actually made at least two mistakes. I left out a 2 in my parametric equation for the cardioid, which should have been

      (x, y) = 2 (\cos t, \sin t) - (\cos 2 t, \sin 2 t)

      in true analogy with your nephroid

      (x, y) = 3 (\cos t, \sin t) - (\cos 3 t, \sin 3 t)

      I also grabbed this equation for a cardioid in polar coordinates off the shelf:

      r = 1 + \cos \theta

      but this is a different cardioid, with its cusp at the origin instead of at (1,0).

      I’ll fix these sometime when I’m more awake…

      Reply
      • romain says:
        3 September, 2012 at 5:00 pm

        It puzzles me that the catacaustic of the cardioid, which has just one symmetry axis, is the nephroid, which has two. It looks as if the two cusps of the nephroid were conjugate (though not in the usual optical system meaning).

        I wonder whether, in order to understand the relationship between the n-cusped epicycloids, it might be worth trying to understand how catacaustics change the symmetry of the figure depending on the location of the source.

        It may be that to obtain the 3-cusped epicycloid requires taking the ray source of the nephroid at a place different than the cusp.

        Reply
      • Greg Egan says:
        3 September, 2012 at 11:33 pm

        Romain, if I switch the light source to the coordinate origin I get a much more symmetrical figure, which is almost the 2k-cusped epicycloid, but not quite!

        In these images, the blue curve is a true 2k-cusped epicycloid, scaled so that its cusps coincide with those of the original k-cusped epicycloid.

        Reply
        • John Baez says:
          4 September, 2012 at 2:47 am

          Wow, that’s really ‘close but no cigar’!

        • John Baez says:
          4 September, 2012 at 8:57 am

          Hmm. Maybe these curves you’re getting as catacaustics are given by polynomial equations with the same degree as the desired epicycloids, but with different coefficients.

          Unfortunately I’m not in the mood right now to think about the formulas for the reflected rays and then their envelopes. Envelopes are calculated like this. There should at least be a simple upper bound on the degree of the catacaustic as a function of the degree of the original curve… this would be a basic part of that ‘general theory of curve constructions’ I’d like to exist but haven’t seen yet.

        • Greg Egan says:
          4 September, 2012 at 11:13 am

          I managed to derive an explicit formula for the catacaustic of an epicycloid with the source at the origin. For the epicycloid given by:

          z = (k+1)e^{i t} - e^{i (k+1) t}

          the catacaustic for reflections from the source at (0,0) is:

          z = \frac{k \left((k+1) (2 k+1) e^{i t}-(k-1) e^{i (k+1) t}-(k+1) \left(e^{i (1-k) t}+e^{i (2 k+1) t}\right)\right)}{2 \left(k^2+k+1-(k+1) \cos (k t)\right)}

          Of course that might not be the nicest possible parameterisation; it’s the parameter of the point on the original curve that produces the reflected ray tangent to the catacaustic at the coordinates given by this expression. I don’t have the energy right now to figure out what the order of this would be as a polynomial in the Cartesian coordinates.

  4. Allan E says:
    3 September, 2012 at 2:55 pm

    Be careful not to drink that disgusting nephroid!

    Reply
  5. Todd Trimble says:
    3 September, 2012 at 4:33 pm

    The title of the post put me in mind of this, which you may remember from long ago, John. :-)

    Reply
  6. A.M. says:
    3 September, 2012 at 6:40 pm

    Another interesting fact is that these epicycloids match the main bodies of the generalized Mandelbrot fractals one gets using k as the exponent:

    z_{n+1} = z_{n}^k + c

    Reply
    • A.M. says:
      3 September, 2012 at 6:42 pm

      I guess this could count as yet another weird way of getting these curves.

      Reply
    • John Baez says:
      4 September, 2012 at 2:45 am

      Good point! That’ll count as another way of getting these curves if we can give a nice characterization of those ‘main bodies’, like this one.

      By the way, I fixed up your LaTeX. The correct syntax for LaTeX on this blog is described at top right of this blog:

      You can include math in your comments using LaTeX, but you need to do it this way:

      $latex E = mc^2$

      You need the word ‘latex’ right after the first dollar sign, and it needs a space after it. Double dollar signs don’t work, and other limitations apply, some described here. You can’t preview comments here, but I’m happy to fix errors.

      Reply
  7. chrisnamastephys11uca says:
    3 September, 2012 at 8:30 pm

    Hi John!

    Interesting post, as usual! I learned some new things from your blog post and they are quite cool!

    The answer I obtained for Puzzle 1 is that the small circle rotates three times as it rolls around the big circle.

    The answer to your second puzzle I found by investigating the general parametric equations for an epicycloid in a calculus textbook–Calculus with Analytic Geometry by Robert Ellis and Denny Gulick. So, the parametric equations that constitute the nephroid are:

    x = (3/2)b \cos(t) - (1/2)b \cos(3t)
    y = (3/2)b \sin(t) - (1/2)b \sin(3t),

    with r=(1/2)b.

    The variables r and b refer to the radii of the small and big circles, respectively.

    So, an example of a nephroid, with b=2, r=1, we have:

    x = 3\cos(t) - \cos(3t)
    y = 3\sin(t) - \sin(3t)

    I’m not sure that this is the simplest equation you are looking for, though, John.

    Reply
    • John Baez says:
      4 September, 2012 at 2:34 am

      Nice, Chris Namaste! These equations look good. I prefer a plus sign where you have a minus sign… you get a slightly different version of the epicycloid that way, and I always deduct one point from the ‘beauty score’ of an equation if it has a minus sign where it could have had a plus.

      But that’s a tiny nitpick. The real way to pretty up the equations you wrote is not to use the real numbers, but some other number system, which is perfectly designed for studying rotations.

      Reply
  8. Ilya Surdin says:
    4 September, 2012 at 11:13 am

    I made a small observation, it’s pretty trivial, but I think it gives a different light to all these calculations.
    If you write the coordinates of the nephroid equation as complex numbers you get
    3e^{it}-e^{3it} = e^{it} (3 + e^{2it})
    So, what we have a point turning around (3,0), which is on a turning plane. And we also get the r equation for r^2 for free.

    Reply
    • John Baez says:
      4 September, 2012 at 2:15 pm

      Great! This is the form of the equation that I was hinting at Puzzle 2 and my later comments.

      If the fixed circle has radius 2 and the rolling circle has radius 1, the center of the rolling circle traces out the path

      3 e^{it}

      and since it rotates 3 times each time it revolves (by Puzzle 1), a point on its edge traces out a path like

      3 e^{it} + e^{3 i t}

      If you want the curve here instead of a version rotated 90 degrees:

      you should use

      3 e^{it} - e^{3 i t}

      but the minus sign offends me.

      The same sort of formula works for the epicycloid formed by a circle of radius 1 rolling on a circle of radius k-1:

      k e^{it} + e^{k i t}

      I’d say this is as elegant as we can make it. But your interpretation in terms of a turning plane makes the derivation even clearer.

      Reply
      • Ilya Surdin says:
        5 September, 2012 at 9:23 am

        The thing is that, it really seems to me just like a cycloid that goes on a circle instead of a line (you could say that this is a generalization of a cycloid, where the cycloid is for k\to\infty , but you need to put a point of the big circle in the center for it to work.

        Reply
  9. P. Bloom (@pbloemesquire) says:
    4 September, 2012 at 9:57 pm

    Since the last post had a comment mentioning the cardoid in the Mandelbrot set, I feel we should mention the Multibrots here:

    Just change the exponent of the core function, and you’ll get all the epicycloids you can match organs to. I’m guessing that the negative exponents will relate to the upcoming post.

    Reply
    • John Baez says:
      8 September, 2012 at 6:05 am

      Nice! Sorry to take a while, but I converted your URLs to a link and a picture.

      So epicycloids are related to Multibrot sets!

      Now I’m wondering if someone has tried to define the catacaustic of the Mandelbrot set. This sounds tough, because light reflects off a curve in a way that makes equal angles, and that’s not defined for a nondifferentiable curve… but there still might be some trick to get around this, some sort of limiting process. And people have definitely studied eigenfunctions of the Laplacian on fractals, and these are related to the propagation of light waves, which reduce to light rays in a certain limit: the geometrical optics limit. So one could at least study what happens when light waves are emitted from a point source in the Mandelbrot set, and then what happens in the limit where the frequency of this light gets higher and higher.

      Reply
  10. Greg Egan says:
    7 September, 2012 at 10:52 pm

    The Cartesian equation for an epicycloid with 4 cusps is:

    \left(r^2-6\right)^5-375 \left(r^2-\frac{64}{9}\right)^3+\frac{50000}{243} \left(189 r^2-2230\right)-50000 x^2 y^2 = 0

    i.e. a polynomial of degree 10. This is clearly symmetrical under an exchange of x and y.

    The Cartesian equation for the catacaustic of a nephroid is also of degree 10! But it’s a lot messier:

    375 \left(1893 x^2-880\right) y^8 +16384 \left(x^2-4\right)^3 \left(3 x^2+4\right)^2 \\ +\left(1433331 x^4-2364960 x^2-1995200\right) y^6 \\ +3 \left(482331 x^6-1705392 x^4-307712 x^2+1568768\right) y^4 \\ +768 \left(951 x^8-5804 x^6+4816 x^4-63296 x^2+9216\right) y^2+140625 y^{10} = 0

    Reply
    • John Baez says:
      8 September, 2012 at 5:56 am

      Wow, that’s impressive.. .even with the help of Mathematica.

      So, on the basis of this, I think we can conjecture that the catacaustic of the k-cusped epicycloid has the same degree as the (k+1)-cusped epicycloid.

      Let’s see… the 1-cusped epicycloid has degree 4, the 2-cusped one has degree 6, the 3-cusped one has degree… well, I’m not sure we talked about that, and the 4-cusped one has degree 10. So, the k-cusped epicycloid should have degree 2k+2. This should be easy to check, since a nice formula for it is known.

      So, we can restate the conjecture by saying the catacaustic of the epicycloid with k cusps has degree 2k+2.

      And I’ll be happy to go out on a limb and guess without much evidence that the epicycloids are ‘generic’ in this respect, meaning ‘no better than average’. Namely, I’ll guess that the catacaustic of a curve of degree d has at most degree d+2.

      (By the way, when I say ‘the’ catacaustic I really should say which point we’re using as our light source. For the epicycloids it makes some sense to use a cusp, but I actually bet the degree of the catacaustic is independent of the light source, except in some degenerate special cases where it drops to a smaller number.)

      Reply
    • Greg Egan says:
      8 September, 2012 at 7:39 am

      I agree that the epicycloid with k cusps has degree 2k+2, but I think the process of taking the catacaustic will (at least for epicycloids) produce a curve with double the number of cusps.

      So taking the catacaustic looks like it maps degree 2k+2 to degree 2(2k)+2, or in other words d to 2d-2.

      But I should stare a bit longer at those polynomials and their derivatives, because I think I’m fairly close to getting a precise answer from that.

      Reply
    • Greg Egan says:
      8 September, 2012 at 8:17 am

      Thinking about the circle giving the cardioid as its catacaustic, that’s a case of a curve of degree 2 producing a curve of degree 4, so the upper bound must be at least 2d, not 2d-2 as I was guessing.

      So either the 2d-2 cases (such as the degree 6 nephroid giving a degree 10 catacaustic) or the 2d cases (such as the degree 2 circle giving a degree 4 catacaustic) must be non-generic.

      Reply
  11. Greg Egan says:
    8 September, 2012 at 7:30 am

    Suppose we have an algebraic plane curve defined implicitly by P(x,y)=0. Wlog, we form a catacaustic by reflecting rays from the origin to the curve. The direction of the reflected ray through the point (x,y) is given by:

    F(x,y) = (x,y) - \frac{2(x,y)\cdot\nabla P}{\nabla P\cdot\nabla P}\nabla P

    It will be useful to define two polynomials:

    A(x,y) = \left(\frac{\partial P}{\partial y}\right)^2-\left(\frac{\partial P}{\partial x}\right)^2 \\ B(x,y) = 2 \frac{\partial P}{\partial x} \frac{\partial P}{\partial y}

    If the degree of P is n, the degree of A and B will be 2(n-1).

    Points (X,Y) on the reflected ray satisfy the equation G(x,y,X,Y)=0, where:

    G(x,y,X,Y)\\ = \left(A(x,y) (y, x) + B(x,y)(x, -y)\right)\cdot (X-x, Y-y) \\ = A(x,y) (x (Y-2 y)+X y) + B(x,y) (x (X-x)+y(y-Y))

    The envelope of the reflected rays is given by (X,Y) that satisfy, simultaneously, G(x,y,X,Y)=0 and \partial_t G(x(t),y(t),X,Y)=0, where (x(t), y(t)) is some
    parameterisation of the curve.

    We can solve these conditions as two linear equations in X and Y, along with the requirement that \partial_t P(x(t),y(t))=0, to obtain the solutions:

    X = \frac{1}{D} \left(\frac{\partial P}{\partial y} \left(B \left(\left(y^2-x^2\right) A+2 x y B+r^2 x \frac{\partial A}{\partial x}\right)-r^2 x \frac{\partial B}{\partial x} A\right)+\frac{\partial P}{\partial x} \left(r^2 B \left(B-x \frac{\partial A}{\partial y}\right)+x A \left(r^2 \frac{\partial B}{\partial y}-2 y B\right)+2 x^2 A^2\right)\right)

    Y = \frac{1}{D} \left(\frac{\partial P}{\partial x} \left(B \left(\left(y^2-x^2\right) A+2 x y B - r^2 y \frac{\partial A}{\partial y}\right)+r^2 y \frac{\partial B}{\partial y} A\right)+\frac{\partial P}{\partial y} \left(r^2 B \left(B+y \frac{\partial A}{\partial x}\right)+y A \left(2 x B-r^2 \frac{\partial B}{\partial x}\right)+2 y^2 A^2\right)\right)

    D = \frac{\partial P}{\partial x} \left(r^2 \frac{\partial B}{\partial y} A+B \left(x B-r^2 \frac{\partial A}{\partial y}\right)+x A^2\right)+\frac{\partial P}{\partial y} \left(B \left(y B+r^2 \frac{\partial A}{\partial x}\right)+y A^2 - r^2 \frac{\partial B}{\partial x} A\right)

    If we form an arbitrary polynomial Q(X,Y) and use these solutions to express it as a polynomial Q'(x,y), the condition that P(x,y) divides Q'(x,y) will be sufficient to make Q(X,Y)=0.

    What we know about the degrees of P, A and B ought to be enough to put an upper bound on the degree of Q(X,Y) needed to ensure that P(x,y) divides Q'(x,y). But I’ve run out of time now, so I might tackle that later …

    Reply
  12. Greg Egan says:
    8 September, 2012 at 7:53 am

    I wrote:

    If we form an arbitrary polynomial Q(X,Y) and use these solutions to express it as a polynomial Q'(x,y) …

    I should have said that Q(X,Y) will become a rational function Q'(x,y), rather than a polynomial, and the condition we’re seeking is that P(x,y) divides the numerator of that rational function.

    Reply
  13. Greg Egan says:
    9 September, 2012 at 5:25 am

    Continuing the argument from my earlier comment …

    The upper bound on the degree of the numerator polynomials in our solution for X and Y is 5n-3, where n is the degree of P. The upper bound on the degree of the denominator D is 5n-4.

    Suppose we take a polynomial Q(X,Y) of degree at most d, substitute for X and Y, and clear all fractions by multiplying through by D^d. The result will be a polynomial S(x,y) of degree at most s=d(5n-3). The bivariate polynomials of degree at most s form a real vector space of dimension (s+1)(s+2).

    We then construct R(x,y), the reduction of S(x,y) modulo P(x,y). This process gives a linear map whose kernel is spanned by the products of P(x,y) with monomials of degree at most s-n, of which there are (s-n+1)(s-n+2). This means the range of the map will have dimension (s+1)(s+2)-(s-n+1)(s-n+2).

    Overall, we’ve gone from Q(X,Y), in a vector space of polynomials of degree at most d, of dimension (d+1)(d+2), to R(X,Y), in a vector space of dimension (s+1)(s+2)-(s-n+1)(s-n+2). As soon as the dimension of the domain exceeds that of the range, the map must have a non-trivial kernel, and so there will be some non-zero Q(X,Y) that evaluates to zero whenever (X,Y) lies on the catacaustic.

    Unfortunately, this gives absurdly large bounds! For n=2, it’s not until d=25 that the space we take Q(X,Y) from (of dimension (d+1)(d+2)=702) is as large as the space that R(x,y) belongs to (s=d(5n-3)=175 and (s+1)(s+2)-(s-n+1)(s-n+2)=702).

    There must be something I’m missing that would make it possible to tighten this enormously.

    Reply
  14. Greg Egan says:
    9 September, 2012 at 5:53 am

    I can get lower-degree expressions for X and Y by expanding out the definitions of A(x,y) and B(x,y). The numerator polynomials are then:

    2 \left(-2 x \frac{\partial P}{\partial x} \frac{\partial P}{\partial y} \left(r^2 \frac{\partial ^2P}{\partial x\, \partial y}+y \frac{\partial P}{\partial x}\right)+\left(\frac{\partial P}{\partial y}\right)^2 \left(r^2 x \frac{\partial ^2P}{\partial x^2}-y^2 \frac{\partial P}{\partial x}\right)+x \left(\frac{\partial P}{\partial x}\right)^2 \left(r^2 \frac{\partial ^2P}{\partial y^2}-x \frac{\partial P}{\partial x}\right)\right)

    and

    2 \left(-\frac{\partial P}{\partial x} \frac{\partial P}{\partial y} \left(2 r^2 y \frac{\partial ^2P}{\partial x\, \partial y}+x^2 \frac{\partial P}{\partial x}\right)+y \left(\frac{\partial P}{\partial y}\right)^2 \left(r^2 \frac{\partial ^2P}{\partial x^2}-2 x \frac{\partial P}{\partial x}\right)+r^2 y \left(\frac{\partial P}{\partial x}\right)^2 \frac{\partial ^2P}{\partial y^2}-y^2 \left(\frac{\partial P}{\partial y}\right)^3\right)

    and the denominator is:

    -\frac{\partial P}{\partial x} \frac{\partial P}{\partial y} \left(4 r^2 \frac{\partial ^2P}{\partial x\, \partial y}+y \frac{\partial P}{\partial x}\right)+\left(\frac{\partial P}{\partial y}\right)^2 \left(2 r^2 \frac{\partial ^2P}{\partial x^2}-x \frac{\partial P}{\partial x}\right)+\left(\frac{\partial P}{\partial x}\right)^2 \left(2 r^2 \frac{\partial ^2P}{\partial y^2}-x \frac{\partial P}{\partial x}\right)-y \left(\frac{\partial P}{\partial y}\right)^3

    The degrees of the numerators are now bound by 3n-1, and the denominator by 3n-2, so we now have s=d(3n-1). But that still only bounds the degree of the catacaustic of a quadratic to be at most 18.

    Reply
  15. Greg Egan says:
    9 September, 2012 at 7:53 am

    Though the bounds I’ve derived above are much too high, my earlier guess that a curve of degree d would have a catacaustic of degree 2d is much too low. A generic catacaustic for an ellipse has degree 6!

    You can see pictures at MathWorld. That article doesn’t give any formulas for the generic case, but the picture clearly shows two cusps (like a nephroid, which is degree 6), and working through the algebra it’s possible to show that the degree really is 6.

    Reply
  16. Greg Egan says:
    9 September, 2012 at 9:40 am

    For the record, all my dimensions of polynomial spaces above are too large by a factor of 2. The actual dimension of the space of bivariate polynomials of degree at most d is (d+1)(d+2)/2.

    Unfortunately, this has no effect on the results! Hopefully the next error I find will actually improve the bounds …

    Reply
  17. John Baez says:
    10 September, 2012 at 5:03 am

    All your comments are very interesting to me, Greg, but also rather intimidating…. especially because I feel some algebraic geometers out there must have been studying this question for at least a century, and I’m reluctant to dive in and compete with them without even the help of Mathematica.

    I hadn’t known the catacaustic of an ellipse could have degree 6, or I wouldn’t have made some of my more optimistic guesses! Here’s a fun site where you can use an applet to move around through the ‘moduli space’ of ellipses and their catacaustics, but only in the case of parallel rays:

    • Irina Boyadzhiev, GeoGebra Applet Constructing the Catacaustic of an Ellipse – Parallel Rays.

    You’ve got to crank up the number of rays to see anything interesting.

    It’s wonderful what computers have done to help explain classical topics in 2d geometry like this! It makes me want to dream up some new questions that’d help revitalize these subjects among professional mathematicians.

    Reply
  18. Greg Egan says:
    10 September, 2012 at 8:10 am

    That applet’s great! And there are links from that page to other versions for all the conics, for both parallel and radial rays.

    I think the bounds I’ve derived are correct, but still pretty naive; they ignore the correlations between the original polynomial and the various intermediate polynomials derived from it, treating those intermediates as if they were free to be any polynomial of the same degree.

    Here’s one more nice example. For the quartic:

    x^4 + y^4 =1

    the catacaustic with the source at the origin has degree 28.

    Reply
    • John Baez says:
      11 September, 2012 at 11:29 am

      Wow! That’s a great picture and a scarily high degree. I had guessed that epicycloids are ‘generic’ when it comes to the degree of their catacaustics—mainly because if you guess that things are generic, then generically you’re correct. But it seems like I was way off, at least in the degrees you’ve investigated. So that means some potentially interesting mechanism is at work, to make their catacaustics have lower degree than average.

      Reply
  19. arch1 says:
    10 September, 2012 at 9:05 pm

    It’s likely redundant w/ comments already made, but I find this stuff easier to think about if I add in a non-black (say, blue) radius from the center of the rolling cirle to the point of tangency. This more graphically distinguishes rotating reference frame from what’s happening relative to that frame: In general, the black radius crosses the blue one n times as the blue radius executes a single rotation, thus the black radius executes n+1 rotations.

    When the rolling circle rolls on the inside of another circle (rather than on the outside), the two rotations cancel rather than add (I managed to puzzle myself for awhile trying to visualize what happens when that cancellation becomes perfect:-)

    Reply
  20. Rolling Circles and Balls (Part 3) « Azimuth says:
    11 September, 2012 at 9:18 am

    In Part 1 and Part 2 we looked at the delightful curves you get by rolling one circle on another. Now let’s see what happens when you roll one circle inside another!

    Reply
  21. Rolling Circles and Balls (Part 4) « Azimuth says:
    2 January, 2013 at 12:38 am

    In Part 2 we rolled a circle on a circle that’s twice as big […]

    Reply
  22. Zane Dobler says:
    28 May, 2013 at 3:45 am


    As you can see, a cardioid can be made by starting with a circle, choosing one point on that circle, and making every other point a circle center. This collection of circles makes a cardioid. I did this same process, but instead of a cicrcle, I started with an equilateral triangle. This created a nephroid, which I put here on ImageShack:
    http://imageshack.us/photo/my-images/837/nephroid.jpg/

    Reply
  23. Ariel Hoffman says:
    25 November, 2013 at 12:56 am

    Something wrong with example 2, in my mind. One circle has r=2, and the other r=3. How is one “twice” the other? I think it should be 3/2 times as “big” as the other.

    Reply

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