**Fun with Math** time.

My mom gave me a roll of toilet paper to put it in the bathroom, and looking at it I immediately wondered about this: is it possible, through very simple math, to calculate (with small error) the total paper length of a toilet roll?

Writing down some math, I came to this study, which I share with you because there are some questions I have in mind, and because as someone rightly said: for every problem there always are at least 3 solutions.

I started by outlining the problem in a geometrical way, namely looking only at the essential: the roll from above, identifying the salient parameters:

**Parameters**

$r = $ radius of internal circle, namely the paper tube circle;

$R = $ radius of the whole paper roll;

$b = R - r = $ "partial" radius, namely the difference of two radii as stated.

**First Point**

I treated the whole problem in the discrete way. [See the end of this question for more details about what does it mean]

**Calculation**

In a discrete way, the problem asks for the total length of the rolled paper, so the easiest way is to treat the problem by thinking about the length as the sum of the whole circumferences starting by radius $r$ and ending with radius $R$. But how many circumferences are there?

Here is one of the main points, and then I thought about introducing a new essential parameter, namely the thickness of a single sheet. Notice that it's important to have to do with measurable quantities.

Calling $h$ the thickness of a single sheet, and knowing $b$ we can give an estimate of how many sheets $N$ are rolled:

$$N = \frac{R - r}{h} = \frac{b}{h}$$

Having to compute a sum, the total length $L$ is then:

$$L = 2\pi r + 2\pi (r + h) + 2\pi (r + 2h) + \cdots + 2\pi R$$

or better:

$$L = 2\pi (r + 0h) + 2\pi (r + h) + 2\pi (r + 2h) + \cdots + 2\pi (r + Nh)$$

In which obviously $2\pi (r + 0h) = 2\pi r$ and $2\pi(r + Nh) = 2\pi R$. Writing it as a sum (and calculating it) we get:

$$ \begin{align} L = \sum_{k = 0}^N\ 2\pi(r + kh) & = 2\pi r + 2\pi R + \sum_{k = 1}^{N-1}\ 2\pi(r + kh) \\\\ & = 2\pi r + 2\pi R + 2\pi \sum_{k = 1}^{N-1} r + 2\pi h \sum_{k = 1}^{N-1} k \\\\ & = 2\pi r + 2\pi R + 2\pi r(N-1) + 2\pi h\left(\frac{1}{2}N(N-1)\right) \\\\ & = 2\pi r N + 2\pi R + \pi hN^2 - \pi h N \end{align} $$

Using now: $N = \frac{b}{h}$; $R = b - a$ and $a = R - b$ (because $R$ is easily measurable), we arrive after little algebra to

$$\boxed{L = 4\pi b + 2\pi R\left(\frac{b}{h} - 1\right) - \pi b\left(1 + \frac{b}{h}\right)}$$

**Small Example:**

$h = 0.1$ mm; $R = 75$ mm; $b = 50$ mm thence $L = 157$ meters

which might fit.

**Final Questions:**

1) Could it be a good approximation?

2) What about the $\gamma$ factor? Namely the paper compression factor?

3) Could exist a similar calculation via integration over a spiral path? Because actually it's what it is: a spiral.

Thank you so much for the time spent for this maybe tedious maybe boring maybe funny question!