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Old 05-04-2013, 06:05 AM
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Pete VanderLaan Pete VanderLaan is offline
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Element math

Element math from George Vidas

Since element math is my idea of a fun time on a Friday night:

Length of wire in a coil = revolutions * ((coil outer diameter - wire diameter) * Pi)

(Taking the circumference of the center of the wire, rather than the outside of it, will give you more accurate length.)

Revolutions in a coil = overall length / wire diameter

3/8" OD, 14GA wire, 7.5' (90") of coiled element before stretching

14GA A-1 has a diameter of 0.0641"; 90" / 0.0641" = 1404.05 revolutions.

1404.05 * ((.375" - 0.0641) * pi) = 1371.37" of wire, or 114.28'. 14GA A-1 has a resistance per foot of 0.2131 ohms, making this a 24.35 ohm element.

Watts = volts^2 * ohms; at 240V, this is a 2365 watt element. Since we know the length of the wire, and its diameter, we can imagine it as a long, skinny cylinder and take its surface area as 1371.37" * (0.0641" * pi) = 276.16 sqin, or 8.5 watts per square inch, a very reasonable surface loading for a wire melter.
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Old 05-04-2013, 06:17 PM
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Pete VanderLaan Pete VanderLaan is offline
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well, OK but you haven't addressed the Friday nights issue at all. I view that as serious.
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