Wikipedia:Reference desk/Archives/Mathematics/2008 September 30

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September 30

calculating thickness of plated metal

I have a certain geometric problem that is causing a hiccup to my lab writeups. Pennies contain 2.5% copper, weigh 2.5 grams, are 1.0 mm thick, have a radius of 8.50 mm, etc. (not measured values but ones I wish to use for this inquiry). So now I have to find the thickness of the copper plating on top of the zinc core. Just how do I calculate this? I know how to find the mass of copper, the total volume of the penny, and given that the density of copper is 8.96 grams per cubic centimetre, I know the total volume of copper the copper has to occupy (the zinc occupies the rest). But I don't know how to find the thickness of the layer on top of the core, since the surface area of the coin is in two dimensions, not three.

I suppose it could be a purely abstract problem for optimisation -- how do I find the minimal uniform thickness such that I can layer 0.0085 cubic centimetres of stuff A over stuff B, given that the total volume of the coin is 23 cubic centimetres? John Riemann Soong (talk)

Imagine a whole coin and its zinc core as cylinders. You know a radius and height of the coin, so you know its volume: $V=\pi \,r^{2}h$. As the copper layer has uniform thickness $t$, the zinc core is a cylinder with radius $r_{c}=r-t\,$ and height $h_{c}=h-t\,$, so its volume $V_{c}=\pi \,(r-t)^{2}(h-t)$. The volume of copper layer $V_{l}$ is a difference of the total volume and a core volume: $V_{l}=V-V_{c}\,$. Plug relations above into the last equation and you'll get a cubic equation with unknown variable $t$. --CiaPan (talk) 05:54, 30 September 2008 (UTC)

Hmm. Is it that $h_{c}=h-2t\,$? I was wondering if the small volume ratio (3.7%?) made the contribution of the thickness of the coating towards the whole coin's radius negligible as far as affecting surface area, so I tried an alternate means of dividing the known volume of the copper by the surface area of the coin, yielding something like approximately 17 microns. If the height of the cylinder = h - t, then my value for thickness becomes 19.6 microns. But trying to compensate that thickness would "eat" height both ways reduces that value to barely above a micron when I use height = h - 2t. John Riemann Soong (talk) 10:39, 30 September 2008 (UTC)

Of course, you're right – the copper layer covers the core on both sides, so $h_{c}=h-2t\,$ and $V_{c}=\pi \,(r-t)^{2}(h-2t)$.
Thank you for catching my mistake. --CiaPan (talk) 12:10, 30 September 2008 (UTC)