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Re: Tesla coil formula



Original poster: Steve Conner <steve@xxxxxxxxxxxx>

>you have to effectively introduce a fiddle factor

I agree. I spent a long time staring at Jared's wire length equations trying to figure out if they were saying something new. Eventually I came to the same conclusion as Malcolm did, about the fiddle factor.

What Jared does is to work out the quarter wave resonance frequency of a given length of wire. Of course, when he makes it into a Tesla resonator, it doesn't resonate anywhere near that frequency, so he adds terminal capacitance until it does. He then seems to claim that there is something magic about that terminal capacitance, and the alignment of the operating frequency with the quarter wave frequency of the straight wire, that makes good performance.

Both theory and experiment are against him on this count. We predict- and find in practice- that the best terminal capacitance to use is just the biggest that you can drive to about its breakout voltage, u sing the tank capacitors and power supply you have available. However, if you make the toroid too big compared to the secondary coil, the secondary will be overstressed by voltage and die from racing sparks and flashovers before you achieve the spark length you should be getting.

Finally, even a straight piece of wire doesn't resonate at exactly the frequency predicted by the "rope resonance" equation, using the speed of light as the wave velocity. When you make a wire antenna, it turns out the effective speed of "light" is slightly lower than "c" and you have to cut it slightly shorter for resonance. IIRC the required length of a half-wave dipole is about 468/f, for length in feet and frequency in megahertz.

Steve Conner
http://www.scopeboy.com/