I have often been trying to capture personal time in this blog through old and new photographs. What you see above, is a recent (like 20 minutes ago) photo of an old toy of mine. I received this early edition of *Spirograph* when I was maybe 9 years old. You can now purchase a 50th anniversary edition (without the tasty pins …).

This is something that I (and my daughter) have used intermittently over all these years, and it has acquired a meaning for me way beyond its mere presence. Already back then I cherished it so much that I kept the *products* in the box. So this is, well, an *ancient* artifact:

The lavishly illustrated instruction manual promised perfection that I never achieved. Too often one of the wheels started sliding instead of just rolling, or the pins didn’t quite hold. What counts, however, is the process. We are, truly, not interested in the ideal, the mathematical perfect curve, but in the process of getting there.

The curves that one can make with *Spirograph* are called *Cycloids*. You can get them abstractly by tracing a point on a wheel that is rolling along a curve. In its simplest form, you roll a circle along a line,

and you learn that these curves can be found on the icy Saturn moon Europa, or as geodesics in the upper half plane when using the Riemannian metric 1/y ds (which is not quite the hyperbolic metric, of course). The ancient ones used them to model planetary orbits when popular belief pinned *man* into the center of the universe.

As my early *Spirograph* experiments show, the results make nice designs. Using contemporary software like *Mathematica* allows you to create these to perfection, you think? Unfortunately, plotting the true cycloids will result in images that are either inaccurate (not enough anchor points) or difficult to manipulate in Adobe Illustrator (too many anchor points). So, to make this:-:,

I replaced the cycloidal arcs between intersections by cubic Bezier splines that have the same curvature as the cycloids at their end points. Again, this was just to find satisfaction in the process to approximate the ideal.

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How

doyou fit a cubic to a given pair of curvatures (as well as tangents)? By deriving the curvature from the cubic function and correcting it with Newton’s method or the like?LikeLike

Good question. To determine the spline for a given arc, we know the first and last control point as end points of the arc, and I set up the remaining two control points to lie on the tangents through the end points of the arc, using two variables to be determined. Because the curvatures at the end points are also known, this gives two algebraic equations in the two unknowns. Mathematica solves these for me numerically, most of the solutions are complex. So the problem is not particularly well posed, and I don’t know for what given data (two points with tangents and curvatures) there is a unique cubic spline through these points.

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