--- /dev/null
+<!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 4.01//EN"
+ "http://www.w3c.org/TR/html4/strict.dtd">
+<html>
+<head>
+ <title>Rolling wire-strip calculator: equations</title>
+ <script type="text/x-mathjax-config">
+ MathJax.Hub.Config({
+ tex2jax: { inlineMath: [['$', '$'], ['\\(', '\\)']] }
+ });
+ </script>
+ <script type="text/javascript"
+ src="http://cdn.mathjax.org/mathjax/latest/MathJax.js?config=TeX-AMS_HTML">
+ </script>
+ <link rel=stylesheet type="text/css" href="rolling.css">
+<head>
+<body>
+
+<h1>Rolling wire-strip calculator: equations</h1>
+
+<p>The calculations performed by the <a href="rolling.html">rolling
+wire-strip calculator</a> were derived by examining experimental data.
+We might not have considered all of the necessary variables. Anyway,
+here’s how it currently works.
+
+<p>Let’s suppose we start with square wire, with side $S$,
+and we roll it to thickness $t$. Then we find that the
+wire’s width is
+\[ w = \sqrt{\frac{S^3}{t}} \]
+Rearranging, we find that
+\[ S = \sqrt[3]{w^2 t} \]
+For round wire, we assume that the cross-section area is the important
+bit, so a round wire with diameter $D$ ought to work as well as
+square wire with side $S$ if $S^2 = \pi D^2/4$, i.e.,
+\[ D = \sqrt{\frac{4 S^2}{\pi}} = \frac{2 S}{\sqrt\pi} \]
+Volume is conserved, so if the original and final wire lengths
+are $L$ and $l$ respectively, then
+\[ L S^2 = l w t \]
+and hence
+\[ L = \frac{l w t}{S^2} \]
+Finally, determining the required initial stock length $L_0$ given
+its side $S_0$ (for square stock) or diameter $D_0$ (for
+round) again makes use of conservation of volume:
+\[ L_0 = \frac{S^2 L}{S_0^2} = \frac{4 S^2 L}{\pi D_0^2} \]
+
+<p>[This page uses <a href="http://www.mathjax.org/">MathJax</a> for
+rendering equations. It probably doesn't work if you don't enable
+JavaScript.]
+
+</body>
+</html>
~mdw / dep-ui / commitdiff