Interesting inductance script. However, either my calc's are wrong or the script calc's are wrong for the inductance.
For multilayer, I'm using the following: LuH = (31.6 x n^2 x ri^2) / (6 x ri + 9 x L + 10 x (ro-ri)) n = total turns ri = inside coil radius ro = outside coil radius L = coil length (all in meters)Quite a discrepancy between the script inductance result and the above. For the calculators inputs at 5H, he realizes 51.18 layers. If I use 24 awg (.0213" in his code) and adjust my layers the same, everything works out fine (id, od, dcr, turns, wire length, etc..) except I show 1.75H where he shows 5H (2"D, 4.75"L, 24G). Somethings odd here... and can't put my finger on it.
Anyone else have a known working multilayer calculation for L? Take care, Bart Jim Harvey (UDN) wrote:
For what its worth, I wondered "What would Tesla do?" If he needed an inductor, he didn't worry too much about space in his lab or the cost of wire. My thought was, hey, just make it bigger in diameter since L goes up (approximately, for large diameters) by the square of the diameter. My initial calculations tell me that larger diameter will break the bank in terms of total wire length. In my search, I found this: http://www.pronine.ca/multind.htm Any way, just thoughts for a snowing morning here in Utah. Jim Harvey, W7YV
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