Whenever I talk about math in this blog, you know it’s going to be something that is simultaneously incredibly important and incredibly useless. Today is no exception — I want to tackle one of the most important (but no worries, also useless) examples of polar coordinates and path-length integrals around, even if you don’t realize it yet:

When you have a roll of stuff — think paper towels, aluminum foil, Saran wrap, use your imagination — how much of that stuff is left, based on the diameter of the roll? How many more chicken dinners can you cover and stick in the fridge? How many more spilled beers can you wipe off of the table?

(This question, for the record, came up at work, albeit with some industrial polymer examples instead of household chores. The times I do real math as a mechanical engineer are surprisingly few and far between, so apologies if I come off as over-excited.)

On one level, this question isn’t that tricky. Your roll of stuff — I’ll stick with aluminum foil in these examples — is wound around a core, and you can measure the core diameter, measure the roll diameter, measure the thickness of the foil and chug-chug-chug along until a number comes out.

Which is, of course, exactly what we’re going to do first. (Spoiler alert: there’s a massive shortcut and a cool arithmetic trick at the end, but we’re going to slog through some calculus because I learned all this shit and never get to use it, goddammit.)

The foil is wound around the core in a spiral — an Archimedean spiral, to be precise. Each layer sits directly on top of the last layer, around and around and around, until it reaches the point you pull.

In polar coordinates, the equation for an Archimedean spiral is *r = c θ*. (Polar coordinate recap: angle,

*θ*,

*and distance from the origin,*

*r*,

*are just as valid a way to define a point’s location as*

*x*and

*y*, and the bears are white instead of brown.) The constant

*c*defines how much space is between each turn of the spiral; in our case this is directly related to the thickness of the foil, which I’ll call

*t*. Every winding around the core, a distance of 2

*π*radians, increase the roll’s radius by

*t*, so we have:

*c = t */ (2*π*)

Therefore the equation of our aluminum foil spiral is:

*r* = *t**θ** */ (2*π*)

Without deriving it here (I let Stephen derive it here instead), the length of a curve — any curve — in polar coordinates is given by:

Our starting and ending coordinates, *θ*_{0} and *θ*_{1}, are given by the radius of the core the foil is wound around (*r*_{0}) and the outer radius of the roll (*r*_{1}), as well as our Archimedean spiral equation. We get:

*θ*_{0} = 2*π**r*_{0} / *t*

*θ*_{1} = 2*π**r*_{1} / *t*

Putting this all together, we get:

This is, uh, not a nice integral to solve. But we’re really good at calculus (and by “we” I mean “Wolfram|Alpha”), and so we know this evaluates to:

Holy dammit Christmas, we’ve gone hyperbolic. If it makes you feel better, we can write the inverse sinh function as a logarithm, and so this soup of advanced math classes reduces down to:

Okay, fine, it’s still not pretty. But it is precise.

Standard household aluminum foil is about 0.016 mm thick. If it’s wrapped around a 12 mm radius core to a final outer radius of 16 mm, this formula tells us that it should have a total overall length of 21.99 m — enough for a whole lot of Chipotle burrito swaddling, and just about exactly what we’d expect. Two points calculus.

• • • • • • •

The engineer-graduate-degree half of my brain is done at this point (we have, after all, the most exact answer possible, though it took a decent amount of computation to get there). But the undergraduate-physics-degree half of my brain, which was at one point instructed that pi is “approximately 1” and prides itself on Fermi estimation, doesn’t love all the work we had to go through to pull it off. I mean — path integrals? Hyperbolic trig functions? Really?

What if we assume the foil is wound in concentric circles instead of a spiral? This way simpler geometrically (ignore the fact that it’s physically impossible, please — this is a physics approximation after all), and since aluminum foil or paper towels or whatever are so thin the answer should be very similar. In this case, the radius of each subsequent winding increases by exactly the thickness of the foil.

So if we add up the circumferences of all the concentric circles from the first to the *N*th, we get the overall length of the foil:

*s* = 2*π**r*_{0} + 2*π*(*r*_{0}* + t*) + 2*π*(*r*_{0}* + *2*t*) + 2*π*(*r*_{0}* + *3*t*) + … + 2*π*(*r*_{0}* + *(*N – *1)*t*)

Or, rearranging:

*s* = 2*π*(*r*_{0} + *r*_{0} + *t* + *r*_{0} + 2*t* + *r*_{0} + 3*t* + … + *r*_{0} + (*N – *1)*t*)

*s* = 2*π*(*N**r*_{0} + (1 + 2 + 3 + … + (*N – *1))*t*)

Lurking in here is a really neat arithmetic identity — the sum of every number between 1 and *x *(here, *x* is played by *N *– 1). There’s an apocryphal story about a young Carl Friedrich Gauss, whose 18th century elementary school teacher tasked his class with summing every number from 1 to 100 — presumably to shut the little bastards up while the teacher nursed a massive Oktoberfest hangover. While every other student began to assiduously add numbers up, our wunderkind Gauss thought about it and came up with a much more elegant solution: each pair of numbers — 1 and 100, 2 and 99, 3 and 98, etc. adds up to exactly the same value, and there are exactly 100/2 = 50 of those pairs. Therefore the sum of all numbers from 1 to 100 is:

(100 / 2)(1 + 100) = 5050

That Gauss guy was a smart dude. There’s a reason literally everything in mathematics is named after him (well, him and Euler). In algebraic form, the sum of all numbers from 1 to *x* is:

*(x* / 2)(1 + *x*)

And for us, where *x = N – *1:

(1 + 2 + 3 + … + (*N* – 1)) = ((*N* – 1)/2)(1 + (*N* – 1)) = (*N*^{2} – *N*) / 2

So that means the length of our roll is approximately:

*s* = 2*π*(*N r*

_{0}+ ((

*N*

^{2}–

*N*) / 2)

*t*) =

*πN*(2

*r*

_{0}+ (

*N*– 1)

*t*)

We still need to figure out how many turns are in our foil coil, but that’s just based on the inner and outer radius:

*N* = (*r*_{1} – *r*_{0}) / *t*

Putting everything together, we have:

*s* = *π*((*r*_{1} – *r*_{0}) / *t*)* *(2*r*_{0} + (((*r*_{1} – *r*_{0}) / *t*) – 1)*t*)

*s = *(*π */ *t*)(*r*_{1}^{2} – *r*_{0}^{2} + *r*_{0}*t* – *r*_{1}*t*)

Somehow this just looks much nicer than the solution with a hyperbolic trig function in it. And when you plug the same numbers in, you get — wait for it — 21.98 m. That’s a 0.01 m difference over more than 20 m of length… or a discrepancy of less than 0.05%.

So yeah, I’ll stick with Gauss on this one. You can call it intelligence or you can call it indolence, but either way I know how many dinners I can wrap up and shove in the fridge.