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 A1 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 A1 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. 
well, OK but you haven't addressed the Friday nights issue at all. I view that as serious.

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