• rotopenguin@infosec.pub
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    1 year ago

    An AAAA cell has 200-350 mohms internal resistance. A 9v battery has 6 of them in series (many of them are literally that, others have their cells as a stack of plastic buckets). The nose ring is a short run of wire, it’s idunno a 0.2 ohm heater?

    I think the septum is going to get pretty toasty.

    https://data.energizer.com/pdfs/e96.pdf

    • enkers@sh.itjust.works
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      1 year ago

      I just tested this (for science!) with a 9V battery and an iron nail of roughly nose-ring diameter. Both the nail and the battery get unpleasantly hot after several seconds. I don’t think they’d get hot enough to burn you, though. (Don’t take my word, though, please!) I believe the internal resistance of the battery does also increase with the temperature, so it effectively somewhat self regulates itself.

      Common nose ring materials like Titanium and Stainless Steel are 4× and 7× more resistant than iron, which means they should dissipate more power than the nail, and thus get hotter. I was calculating something around 3 milliohms for a titanium 16 gauge 10mm wire, and 0.7 milliohms for an iron wire.

      Regardless of material, at 1000 milliohms internal resistance, i think the battery itself is doing most of the heat dissipation. (But also over a much bigger surface area!)

        • enkers@sh.itjust.works
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          1 year ago

          About 10-20s, I left it on until it didn’t seem to be getting much hotter. I also didn’t want the battery to overheat and fail catastrophically. I think because the “wire” is such a large gauge, there’s not enough current for it to get seriously hot. In a foam cutter, you’re passing all that current through a much smaller cross-sectional area.

          Edit: just to confirm, I did a little math. A 10cm steel wire with a tenth of the diameter would have a resistance of 5 ohms. That means that instead of 1% of the total heat dissipating in the thick wire, 80% of the heat is dissipating in the wire in foam cutter’s case, and there’s more total resistance, so more heat dissipation as well.

          This is because:

          A = π r²

          R = ρ × L / A

          So resistance is proportional to the material resistivity (ρ), the length (L), and the inverse square of the radius (r⁻²). That is to say, decreasing the radius by a factor of 10 increases resistance by a factor of 100.