Re: flat coil ?

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

Original poster: "Bert Hickman by way of Terry Fritz <twftesla-at-qwest-dot-net>" <bert.hickman-at-aquila-dot-net>

Hi Chris,

You can estimate the wire length and outer diameter of the spiral coil
as follows:
 Let:
  Starting diameter = Do  ("inside" diameter)
  Number of Turns   = N
  Wire Diameter     = Dw

 Then:
  Total Length of wire  = Pi[(N+1)*D + Dw*(N^2 + N)]  
  Outer spiral Diameter = Do+(2*N*Dw)

Plugging in Do = 0, N = 500, and Dw = 1.6mm and converting units, I end
up with a coil about 5.25 feet in diameter, requiring over 4100 feet of
wire.

Working backwards, a 4 foot diameter coil would require about 381 turns
and 2400 feet of wire. 

Hope this helps!

-- Bert --
-- 
Bert Hickman
Stoneridge Engineering
Coins Shrunk Electromagnetically!
http://www.teslamania-dot-com  

Tesla list wrote:
> 
> Original poster: "Chris Swinson by way of Terry Fritz
<twftesla-at-qwest-dot-net>" <exxos-at-cps-games.co.uk>
> 
> Hey all,
> 
> I am just about to build a large flat secondary coil.  The wire length will
> be 2000ft, the wire dia is 1.6mm. I *think* this will be 500 turns at about
> 4foot dia.  Anyone got any ideas if that sounds about righ before I wind the
> thing ?
> 
> Thanks,
> chris