## 4 May 2015

### String around the rod

I've decided to feed my blogs with puzzles, sometimes. There are already two examples packed into one post in the other blog. I will use this blog to post about puzzles when I decide not to show how to tackle them with a computer language as a tool.

Usually you can find a solution online which is written better, explained better, shown better, and so on. This is why these posts must be read still as pointless ramblings and why I consider relevant to show also whatever could be described as suboptimal paths or even, maybe, wrong solutions. Warned: don't be fooled!

So, in this post the problem is the Sunday puzzle n. 28, that «stumped 96% of America's top math students»: String around a rod.

To a second look, this is the kind of problem that should have reminded me Martin Gardner's column “Mathematical Games” in Scientific American. I have two books collecting several games of this kind1, and among them there are two problems which are somewhat useful: the problem of the spider and the fly, and the problem of the fly and the honey (both are Dudeney's problems).

It is interesting to note that although Dudeney had little familiarity with topology, then in its infancy, he frequently used clever topological tricks for solving various route and counter-moving puzzles. He called it his "buttons and string method."2

We can borrow an interesting idea from the solution of those problems: “manipulate” the object so that we have to deal with a flat surface. In our case, this means to unroll the cylinder (the rod) and see the drawing made by the “lines” (the string around the rod when in 3D)… Now, it's easier to figure out a solution. This is one of the possible solution, and likely the best one.

There's no reason to talk about it further—it's well explained elsewhere. Instead, let's see whatelse we can do.

## Approach 1: compute the length of the curve

This is indeed the first thing I've thought. It does not use a clever “trick” and the math you need is a little bit harder (not too much, hopefully), but it is more straightforward to me; conceptually trivial: how do you find the length of something? By “measuring” it… i.e. by computing the length of the curve (the string) along its path.

In fact in its winding around the rod, the string is nothing but a curve in a 3D space, outlined by a point rotating in circle while advancing—namely a helix. We can describe the curve through a parametric set of equations, where the parameter t can be called time.

\begin{align*} x(t) & = r\cos s(t) \\ y(t) & = r\sin s(t) \\ z(t) & = Lf(t) \end{align*}

Let t be between 0 (when we are at the beginning of the string, hence z(0) = 0) and 1 (when we are at the end of the string, hence z(1) = L); moreover, we know that as t goes from 0 to 1, the “point” must complete four “rounds”. Summarizing:

\begin{align*} f(t) & = t \\ s(t) & = 4\cdot 2\pi t \end{align*}

Now we must compute a line integral which is trivial3:

$l = \int_{0}^{1}dt\; \sqrt{ (8\pi r)^2 + L^2 }$

The parameter t does not appear, so the square root can be put outside the integral, which has value 1. Then we also notice that the problem gives us the circumference, instead of the radius. But c = 2πr; and all we have to compute is:

$l = \sqrt{ (4c)^2 + L^2 }$

If we put the numbers in (rod circumference is 4cm, the rod is 12cm long), we obtain:

$l = \sqrt{ 256 + 144 } = \sqrt{ 400 } = 20$

So, the length of the string is 20cm.

## Approach 2 (wrong result)

Why the string is so smoothly wrapped around the rod? What kind of magic made it possible? Are there other ways to wind the string? Consider several cylindrical sections: each one is an ellipse. Consider just a half of the ellipse. The other half has a different slope and shares with the previous half ellipse a single “point”. This way we can “link” several ellipses and them, altogether, are our whole strings.

Since the strings wrap fully four times, we have 8 half-ellipses. If we decide that all the half-ellipses are equal, then all we need to know it's the length of a single half-ellipse. Computing this length isn't trivial, since we have to deal with a complete elliptic integral of the second kind—we can use approximation formulae, or evaluate it numerically. My choice (with GNU/Octave).

With E(k), namely the complete elliptic integral of the second kind, we can compute the circumference of an ellipse: C = 2aE(k). We need the major axis of the ellipse, a, and its eccentricity k (so, we need also the minor axis).

The minor axis is simply the diameter of the cylinder (2r, i.e. 2c/2π, while Pythagora helps us for the major axis:

$a = \sqrt{\left(\frac{L}{8}\right)^2 + (2r)^2}$

If we feed our calculator —GNU/Octave or whatever— with all this, we obtain 18.938. It's not 20. Why? We can speculate that each sharp turn can be done using less of the twine, while if we want smooth rounds, then we must “squander” more of the string. (We have seven turns, so each smooth turn eats about 0.152 cm that we can save if we decide to bend abruptly each half a round).

However this doesn't fit the image we have (the smooth rounds are overt); moreover, there's a simmetry, but it's not the one you'd expect reading that something «is wound symmetrically» (if we rotate the cylinder along it's axis, we would see slight different “line patterns”)4. ■

(should I add a picture here?)

1. One is “The Second Scientific American Book of Mathematical Puzzles and Diversions”; I can't find the other book right now (likely it hides behind stacks in a shelf, while for some reason the second was in sight).

2. If you know already how to compute derivatives of sine, cosine and composite functions: if you remember that sin2α + cos2α = 1; if you remember basic arithmetic facts, e.g. the distributive law of multiplication

3. Homework: check if I am right when I try to explain why I am wrong (given the statements of the problem).