How many turns are needed to get a length L of the steel blade roll?

Question

I'm doing a college project and I came across the need to calculate how many turns it takes to get a length $L$ of a steel blade roll. I am not able to develop a formula for this.
The roll length formula I'm using is:$$L=\frac{\pi}{T}\cdot \left (\left (\frac{d_2}{2}\right )^2-\left (\frac{d_1}{2}\right )^2\right ).$$And here is a sketch of the roll:

Answer 1

Your formula is equivalent to $$ L \cdot T = \pi\, {r_2}^2 - \pi\, {r_1}^2, \tag{$\dagger$} \label{length} $$ where $L$ is the length, $T$ is the thickness, and $r_1$ and $r_2$ are the inner and outer radii, respectively. This is clearly a correct approximation, as each is a description of the area of the cross-section region: an annulus, which is a circle with a smaller circle removed.

Imagine that the annulus is filled in with concentric thickened circles (rather than the actual more subtle shape: a thickened spiral). The innermost circle has radius $r_1$, then next circle has radius $r_1 + T$ the next $r_1 + 2T$, etc. The outermost circle has radius $r_2 = r_1 + n T$, where
$$ n = \frac{r_2 - r_1}{T}. $$ Again, we're approximating things by assuming that $T$ is very small compared to the radii, so $n$ is quite large and be assumed to be an integer.

Now the total length is the sum of the circumferences of the circles: \begin{align} L &= 2\pi r_1 + 2\pi (r_1 + T) + 2\pi (r_1 + 2T) + \cdots + 2\pi (r_1 + nT) \\ &= 2\pi \biggl( (n+1) r_1 + (0 + 1 + 2 + \cdots + n)T \biggr), \end{align} and using the well-known formula for the triangle number $$ 1 + 2 + \cdots + n = \frac{n(n+1)}{2}, $$ we have $$ L = 2\pi(n+1)\, r_1 + \pi n(n+1)\, T = \pi(n+1)(2r_1 + nT). $$ Recall that $r_2 = r_1 + nT$, so $$ L = \pi(n+1)(r_1 + r_2) $$ Let's multiply by $T$ and equate this formula with the original formula {\ref{length}) for the length of the roll: $$ \pi(n+1)(r_1 + r_2)\, T = \pi\, {r_2}^2 - \pi\, {r_1}^2 $$ or $$ (n+1) T = r_2 - r_1 $$ Finally, \begin{align} n = \frac{r_2 - r_1}{T} - 1. \end{align} Converting back to diameters, and discarding the ${}-1$, as it's just an approximation of a large number anyway, $$ n = \frac{d_2 - d_1}{2T} $$