How Can I derive the Parametric equation for Toilet Paper roll I was going through this wonderful article The length of toilet roll 
I have a similar problem where I wanted to find the outer Radius (R) of the Sheet metal roll which can be thought of as Toilet paper roll with thickness (h).
Is it possible for me to express same thing in Parametric equation where I calculate the outer radius (R) given length of sheet metal. If I get the parametric equation I can think of writing some python code for that using matplotlib
https://en.wikipedia.org/wiki/Parametric_equation. 
 A: Since $r$ increases by $h$ when $\theta$ increases by $2\pi$; thus, $r=\frac{h\theta}{2\pi}$.
Let $\theta=\tan(u)=\frac{2\pi r}h$, then
$$
\begin{align}
L
&=\int\sqrt{r^2+r'^2}\,\mathrm{d}\theta\\
&=\int\sqrt{\left(\frac{h\theta}{2\pi}\right)^2+\left(\frac{h}{2\pi}\right)^2}\,\mathrm{d}\theta\\
&=\frac{h}{2\pi}\int\sqrt{\theta^2+1}\,\mathrm{d}\theta\\
&=\frac{h}{2\pi}\int\sec^3(u)\,\mathrm{d}u\\
&=\frac{h}{2\pi}\int\frac1{\left(1-\sin^2(u)\right)^2}\,\mathrm{d}\sin(u)\\
&=\frac{h}{8\pi}\int\left(\frac1{(1-\sin(u))^2}+\frac1{(1+\sin(u))^2}+\frac1{1-\sin(u)}+\frac1{1+\sin(u)}\right)\mathrm{d}\sin(u)\\
&=\frac{h}{8\pi}\left(\frac1{1-\sin(u)}-\frac1{1+\sin(u)}+\log\left(\frac{1+\sin(u)}{1-\sin(u)}\right)\right)+C\\
&=\frac{h}{4\pi}\left(\tan(u)\sec(u)+\log\left(\tan(u)+\sec(u)\right)\right)+C\\
&=\frac{h}{4\pi}\left(\frac{2\pi r}h\frac{\sqrt{4\pi^2r^2+h^2}}h+\log\left(\frac{2\pi r}h+\frac{\sqrt{4\pi^2r^2+h^2}}h\right)\right)+C\tag1
\end{align}
$$
Formula $(1)$ is quite complicated, but when $h$ very small compared to $r$, $\sqrt{4\pi^2r^2+h^2}\approx2\pi r$ and $\log\left(\frac{4\pi r}h\right)$ is very small compared to $\frac{4\pi^2r^2}{h^2}$. Therefore, we can approximate $(1)$ with the much simpler
$$
L\approx\frac{\pi r^2}h+C\tag2
$$
In hindsight, $(2)$ makes a lot of sense if we think about cross-sectional area. That is, $Lh$ is the cross-sectional area of a strip of thickness $h$ and length $L$ and $\pi r^2$ is the cross-sectional area of the rolled up disk.
Therefore, we get
$$
L=\frac\pi h\left(r_\text{outer}^2-r_\text{inner}^2\right)\tag3
$$
A: I believe in practice the formula $\pi(R^2-r^2) = Lh$ is widely used. Or if you want to use $L$ as a parameter, $R(L)=\sqrt{\frac{Lh}{\pi}+r^2}$. Here $h$ is metal thickness, $r$ is inner radius of the roll. This formula is using circular approximation of the spiral roll.
