How many mandrel wraps are required to get a desired strip length I have a machine that wraps a strip of aluminum with thickness $t$ on a mandrel. If I want a specified length $x$, how many mandrel (diameter $d$) revolutions $n$ are required?
That is:
$$x=\pi(d+(d+2t)+(d+4t)+\cdots+(d+2nt))$$
Solving for $n$.  My brain went tilt.
 A: Not sure if I understood your question. However, assuming that your expression is right, you can find $n$ in terms of the other quantities. Notice that
$$
x
= \pi \left( \sum_{j=0}^n (d + 2tj)\right)
= \pi \left( d(n+1) + 2t \frac{n(n+1)}{2}\right)
= \pi (tn^2 + (t+d)n + d),
$$
a quadratic expression on $n$. Just solve for $n$.
A: Notice you have $n+1$ total $d$'s. Now using the fact that $1+2+\cdots+n= \dfrac{n(n+1)}{2}$ (this is not obvious but is well known, for example here), we have
$$
\begin{aligned}
x&= \pi \,\big(d+(d+2t)+(d+4t)+\cdots+(d+2nt)\big)\\
x&= \pi \big( (d+d+\cdots+d) + (2t+4t+\cdots+2nt) \big) \\
x&= \pi \big( (n+1)d + 2t(1+2+\cdots+n) \big) \\
x&= \pi \left((n+1)d + 2t \cdot \dfrac{n(n+1)}{2} \right) \\
\frac{x}{\pi}&= (n+1)d + 2t \cdot \dfrac{n(n+1)}{2} \\
\frac{x}{\pi}- (n+1)d &= 2t \cdot \dfrac{n(n+1)}{2} \\
\dfrac{\frac{x}{\pi}- (n+1)d}{t}&= n(n+1)
\end{aligned}
$$
We have this odd (but straightforwardly calculated term) $\dfrac{\frac{x}{\pi}- (n+1)d}{t}$. Let $C=\dfrac{\frac{x}{\pi}- (n+1)d}{t}$. Then 
$$
\begin{aligned}
C&= n(n+1) \\
C&= n^2+n \\
n^2+n-C&=0
\end{aligned}
$$
which is a quadratic eqation, so $n= \dfrac{-1 \pm \sqrt{1+4C}}{2}$. The problem  now is that this term is probably not an exact integer (one probably could show that this is always the case) but $n$ is a count so it must have an integer value. But being a real world problem just round down or up, depending on the context called by this scenario - I do not know anything about mandrels.
