Re: flat coil ?

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

Original poster: "Barton B. Anderson by way of Terry Fritz <twftesla-at-qwest-dot-net>" <tesla123-at-pacbell-dot-net>

Hi Chris,

Since you mentioned programs and flat coil specs, here's what I do. I use
JavaTc and use only the
primary area (normally for flat primarys). I simply input (in inches) the
I.D. (0.125), the wire
diameter (0.063), the wire spacing (0.0034, this is the insulation
thickness between closewound turns
if magnet wire), the angle (0, for flat coil), and zero out the cap and
height boxes. Then just change
the turns until you hit ~2000 feet of wire (339 turns). Many outputs will
not calc due to other missing
elements, however, the OD, Length, and Inductance will be shown. No
capacitance unfortunately.
Anyway, OD is 45.02", length is 2003 ft, and inductance is 43.4mH. Just for
giggles, I popped these
numbers into Acmi and inductance showed 45.95mH. Of course, the inputs used
are the same as those I
used in JavaTC. Probably best to wait until you actually wind it. After
that point, make OD and ID
measurements, and count the turns, then let us know. These are still new
and I for one would be
interested in doing these exersizes again after we know the actual dimensions.

Take care,
Bart


Tesla list wrote:

> Original poster: "Chris Swinson by way of Terry Fritz
<twftesla-at-qwest-dot-net>" <exxos-at-cps-games.co.uk>
>
> Hi all,
>
> Thanks for everyone's help. I cut the coil form out today and hope to wind
> it tomorrow.  Will be interesting to see what inductance it will actually
> be. I guessed it about about 8mH but could end up anything. Does anyone know
> the self C of the coil ?  Most tesla programs will not work with this type
> of flat coil so its hard to tell on specs of this coil :-\
>
> Thanks,
> chris
>
> ----- Original Message -----
> From: "Tesla list" <tesla-at-pupman-dot-com>
> To: <tesla-at-pupman-dot-com>
> Sent: Wednesday, April 24, 2002 4:14 AM
> Subject: Re: flat coil ?
>
> > 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
> >
> >
> >
> >