Re: Need Formula for length of spiral

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

Original poster: "Jim Lux by way of Terry Fritz <twftesla-at-qwest-dot-net>" <jimlux-at-earthlink-dot-net>

Some further data on the "quarter wave being 6-15% shorter"

If you look at the actual current distribution on a halfwave dipole, fed in
the center, you can see that for the exact half wavelength, the curve
follows the expected smooth sinusoidal curve.  The calculated impedance at
the feed point is 86.45+j48.84 ohms (quite inductive...). (assuming perfect
conductors, 0.5 meter long antenna, 299.8 MHz (1 meter wavelength),
conductors 0.001 meters in diameter (a bit fat)) Shortening the antenna
reduces the reactive component, but also causes the current distribution to
get somewhat strange.  Here are some impedances:

98% 80.47+j29.71
95% 72.30+j1.169 (very close to resistive)
90% 60.51-j46.29

What's interesting is that the current distribution for the latter two
cases (5% and 10% shorter than a half wave) has a local minimum (not a big
one) at the center, with two symmetric current peaks about 5% of the way
out on either side.

The net result is that if you are building an antenna, make it a bit short
(6% is a good estimate) so that it doesn't have any reactive component, and
you'll get a better match.

If you are building a half wavelength structure, though, make it exact. 

All the same things apply for quarter wavelengths. Quarter wave ground
plane antennas essentially use the "image" of the antenna as the other
half, and are the same as a halfwavelength antenna.  This is very different
than an end fed quarter wavelength long conductor in free space.

Here's the model I used to calculate Z and current distributions

CM dipole, center fed
CE 
GW, 101, 101,-.25,0,0,.25,0,0,.001
GE
EK
EX,0,101,51,0,1,0
FR,0,1,0,0,299.8,0
XQ
FR,0,1,0,0,293.804,0
XQ
FR,0,1,0,0,284.81,0
XQ
FR,0,1,0,0,269.82,0
XQ
EN

Tesla list wrote:
> 
> Original poster: "rheidlebaugh by way of Terry Fritz
<twftesla-at-qwest-dot-net>" <rheidlebaugh-at-zialink-dot-com>
> 
> correction: The quarter wove length of a conductor is not the free space
> quarter wavelength.IT is 6 to 15% less than free space controled by Z of the
> coil or conductor.
>    Robert  H
> 
> > From: "Tesla list" <tesla-at-pupman-dot-com>
> > Date: Wed, 13 Feb 2002 21:42:18 -0700
> > To: tesla-at-pupman-dot-com
> > Subject: RE: Need Formula for length of spiral
> > Resent-From: tesla-at-pupman-dot-com
> > Resent-Date: Wed, 13 Feb 2002 21:44:56 -0700
> >
> > Original poster: "David Thomson by way of Terry Fritz <twftesla-at-qwest-dot-net>"
> > <dave-at-volantis-dot-org>
> >
> > Hi John,
> >
> > If you are going to wind a spiral coil with no inner radius, the wire
length
> > is simply the circumference at the average radius times the number of
turns.
> > If you are trying to figure the wire length according to quarter wave, you
> > make the wire length one fourth of the wave length.
> >
> > Dave
> >
> > -----Original Message-----
> > From: Tesla list [mailto:tesla-at-pupman-dot-com]
> > Sent: Wednesday, February 13, 2002 5:46 PM
> > To: tesla-at-pupman-dot-com
> > Subject: Need Formula for length of spiral
> >
> >
> > Original poster: "John Tomacic by way of Terry Fritz <twftesla-at-qwest-dot-net>"
> > <tesla_ownz_u-at-hotmail-dot-com>
> >
> > Hi everyone,
> >
> > Does anyone have a formula that I can use to calculate the length of wire
> > required in a flat spiral coil? I have the formula for inductance, however,
> > I really need the wire length.
> >
> > Thanks,
> >
> > John
> > SST coiling in Ottawa.
> >
> >
> >
> >