15
$\begingroup$

Imagine a decreasing sequence of (positive) radii $r_1 > r_2 > r_3 > \cdots$ and a series of nested circles $C_1 \supset C_2 \supset C_3 \supset \cdots$ with these radii, initially each resting on the $x$-axis tangent at $(0,0)$. Each is assigned a rolling speed $s_1, s_2, s_3, \ldots$, the amount of arc length that $C_i$ rolls on the inside of $C_{i-1}$ per unit time. (Positive $s_i$ represents clockwise spinning of $C_i$; negative, counterclockwise.) $C_1$ rolls on the $x$-axis, which can be considered $C_0$.

Here is an example, with $(r_1, r_2, r_3)=(1, \frac{1}{2}, \frac{1}{4})$ and $(s_1, s_2, s_3)=(1,2,3)$, with the track of a point on the third circle highlighted:

Call the curve that is the track of the $n$-th circle a rolling epicycle curve, or just a rolling curve. My question is:

Q1. What is the class of functions on some interval $[0,X]$ that can be approximated by some rolling curve?

Say that a function $f(x)$ is approximated by a rolling curve if, for any $\epsilon > 0$, a curve may be found that remains within an $\epsilon$-tube around $f$. (One can easily substitute other reasonable definitions of approximation.) To be specific, we could insist that $f(0)=0$ and the rolling curve tracks the innermost circle's point that initially touches $(0,0)$.

Here is a more random example of four circles of radii $(1, \frac{3}{4}, \frac{2}{3}, \frac{1}{2})$, with green tracking the fourth circle:

There is considerable flexibility, but it seems difficult to control. To pose a more specific version of Q1:

Q2. Can a straight line through $(0,0)$ be approximated on a given interval $[0,X]$?

Wondering about the power of Ptolemaic epicycles led me to this question (although I realize the rolling constraint renders my question different). Thanks for insights!

Addendum. As per J.M.'s request, here is an animated GIF for the three-circle example (which may or may not animate, depending on your browser settings):
         a3 animation

$\endgroup$
  • $\begingroup$ Very interesting! By any chance, did you make animations of these? $\endgroup$ – J. M. is not a mathematician Sep 13 '11 at 10:13
  • 1
    $\begingroup$ @J.M.: No, but I should! When I get a chance... $\endgroup$ – Joseph O'Rourke Sep 13 '11 at 11:25
  • 2
    $\begingroup$ In Q1, it might be necessary for f(x) to have derivative 0 at the origin. (It looks so from the figure, but I do not know if it is true.) If this is true, then the answer to Q2 will be clearly No. $\endgroup$ – Tsuyoshi Ito Sep 13 '11 at 21:15
  • 1
    $\begingroup$ The curves which can be approximated by rolling are not different from the curves which can be approximated by sliding since you can add an $n+1$st innermost microscopic circle whose rolling approximates any rotation speed of the $n$th circle. $\endgroup$ – Douglas Zare Sep 23 '11 at 5:15
  • 1
    $\begingroup$ If some curve has a nonzero finite slope at the origin, then you can approximate a line with that slope by rescaling. However, $y(t)$ is an even power series in $t$ while $x(t)$ is an odd power series, and they can't have a nonzero finite ratio as $t \to 0$. This doesn't rule out the possibility of approximating a line other than a vertical or horizontal one, but it will be much more complicated if it is possible. $\endgroup$ – Douglas Zare Sep 25 '11 at 22:17

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.