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<title>Rolling wire-strip calculator: equations</title>
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<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
-\[ w = \sqrt{\frac{S^3}{t}} \]
+\[ w = \sqrt{\frac{S^3}{t}} \,\text{.} \]
Rearranging, we find that
-\[ S = \sqrt[3]{w^2 t} \]
+\[ S = \sqrt[3]{w^2 t} \,\text{.} \]
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.,
-\[ D = \sqrt{\frac{4 S^2}{\pi}} = \frac{2 S}{\sqrt\pi} \]
+\[ D = \sqrt{\frac{4 S^2}{\pi}} = \frac{2 S}{\sqrt\pi} \,\text{.} \]
Volume is conserved, so if the original and final wire lengths
-are $L$ and $l$ respectively, then
-\[ L S^2 = l w t \]
+are $L$ and $\ell$ respectively, then
+\[ L S^2 = \ell w t \,\text{,} \]
and hence
-\[ L = \frac{l w t}{S^2} \]
+\[ L = \frac{\ell w t}{S^2} \,\text{.} \]
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:
-\[ L_0 = \frac{S^2 L}{S_0^2} = \frac{4 S^2 L}{\pi D_0^2} \]
+\[ L_0 = \frac{S^2 L}{S_0^2} = \frac{4 S^2 L}{\pi D_0^2} \,\text{.} \]
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