Re: How do I work out secondary former diameter?

["Tesla list" <tesla-at-pupman-dot-com>]

Original poster: FutureT-at-aol-dot-com 

In a message dated 12/7/00 6:48:09 PM Eastern Standard Time, tesla-at-pupman-dot-com 
writes:

Matt,

The 1/4 wavelength for the wire length does not apply to Tesla coils.
It doesn't matter how long the wire is as long as the coil is tuned
properly.

Also, regarding secondary wire thickness and losses.  This has to
be considered in conjunction with the spark gap losses.  This is
very important because it's actually a tradeoff between secondary
losses and gap losses.  Perhaps surprisingly, a thinner wire on the
secondary can give much better results than a thicker wire.   More
about this at my website at:

   http://hometown.aol-dot-com/futuret/page3.html

John Freau
----

> As I recall the physics of the situation from too many years ago, the 
maximum 
> 
>  standing wave is established in the secondary (Max voltage at very top) 
when 
> 
>  the length of wire used to wind the secondary is equal to 1/4 the 
wavelength 
> 
>  of the resonant frequency. This would be length=75,000,000/f , where f is 
> the 
>  frequency in Hz and length in meters. For the metrically challenged, this 
>  would be about 246,000,000/f feet. The frequency depends in turn on the 
>  effective inductance and capacitance of the secondary & toroid as: 
>  
>  f=1/(2pi*sqrt(LC)) 
>  

>  A  very 
>  thin wire will have higher resistance meaning a lower Q for the coil, and 
a 
>  >  Thin wire is lighter and smaller than thick, but has higher resistance 
and a 
> 
>  wide coil requires more wire for a fixed number of turns than a narrow 
one. 
>  (Double the diameter means double the length of wire for the same height 
>  coil) 
>          As you can see, a good coil is a balancing act between often 
>  contradictory constraints. I can be contacted on or off list for further 
>  discussion. 
>  
>  The Light willing, 
>  Matt Deming 
>  
>