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 -\[ L S^2 = l w t \] +\[ L S^2 = l w t \,\text{,} \] and hence -\[ L = \frac{l w t}{S^2} \] +\[ L = \frac{l w t}{S^2} \,\text{.} \] 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: -\[ L_0 = \frac{S^2 L}{S_0^2} = \frac{4 S^2 L}{\pi D_0^2} \] +\[ L_0 = \frac{S^2 L}{S_0^2} = \frac{4 S^2 L}{\pi D_0^2} \,\text{.} \] -
[This page uses MathJax for -rendering equations. It probably doesn't work if you don't enable -JavaScript.] +
[This page uses MathJax for +rendering equations. It probably doesn’t work if you don’t +enable Javascript.]