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Re: Calculating secondary resonance of bipolar coils



Original poster: "Gavin Dingley by way of Terry Fritz <twftesla-at-uswest-dot-net>" <gavin.dingley-at-ukf-dot-net>

Hi Terry,
thanks for your advice and I will give it a try, although my intention is
not to ground the center point of the secondary coil (thus making it more
like two quarter waves). I have done some playing with bipolar coils in the
past and found something a little disturbing: it appears that the actual
length of the wire which is used to form the secondary goes along way to
determining the resonant frequency; that is to say, for instance, that a
300m length of wire wound up in to a coil will resonate close to 500kHz. Now
knowing basic TC theory this seems a bit to easy, can the distributed
capacity of a coil be that forgiving? The only thing I can think of is that
the abnormal length of such a coil makes the self capacity negligible! I
will be building such a coil to see if what I have stated is correct, but I
am pretty sure.
I had a look in the CS notes and found that Tesla did in fact find the use
of direct tx-line theory applicable (i.e resonant frequency  = speed of
light / length of wire used). He seemed to think that the lower wave
velocity encountered due to distributed capacity to be negligible and easily
tune out in most cases. However, he must have faced the problems we modern
coilers have encountered as he latter tries to find methods to remove this
inherent capacity (rather than just calculating it out, I think he saw it as
a source of loss!).
Well, I'll do my construction bit and let you know how it went; I find the
modelling of these things quite fascinating as long as the whole thing does
not go over the top!

Thanks again,

Gavin
----- Original Message -----
From: "Tesla list" <tesla-at-pupman-dot-com>
To: <tesla-at-pupman-dot-com>
Sent: Friday, January 05, 2001 1:09 AM
Subject: Re: Calculating secondary resonance of bipolar coils


> Original poster: "Terry Fritz" <twftesla-at-uswest-dot-net>
>
> Hi Gavin,
>
>         I didn't quite know what to say but since no one else did either,
I'll
> give you my guess anyway.  You can use E-Tesla6 to calculate half the coil
> assuming the center plane of the coil is a ground plane.  Since a bipolar
would
> be half wave, I would guess the Fo frequency would be the same.  So, if
you run
> half the bipolar, I think it will give you the right Fo frequency.  I
think...
>
>
> There are also supports and such that are going to mess with the external
> fields a bit too but I think it will be pretty close.
>
> http://hot-streamer-dot-com/TeslaCoils/Programs/Programs.htm
>
> When you ask a good question like this and no one answers, you know you
have
> asked a good question ;-)))
>
> Cheers,
>
>         Terry
>
>
> At 03:24 PM 1/4/2001 +0000, you wrote:
> >
> > Hi all,
> > I sent a mail to the list a couple of days ago regarding secondary
bipolar
> > coil resonance calculations. As there has not yet been a reply, I
thought I
> > should elaborate on the question I originally asked.
> > Most TCs are of the up-right quarter wave resonant type, topped with a
top
> > load such as a torroid. I have recently wound a secondary more suited
for use
> > as a bipolar or half wave resonant secondary. The primary is positioned
in
> > the middle of this secondary coil; and the there will be no earth
connection
> > at the mid point of the secondary.
> > Now, I know that it possible to calculate the resonant frequency of a
quarter
> > wave coil using medhurst and the standard LC resonance formula. It is
also
> > possible to treat the secondary as a Tx line by determining the wave
velocity
> > from the coil inductance and capacitance per unit length. Medhurst, I
> > believe, was derived from empirical measurement, while the Tx-line
method
> > relies on modelling. Either way, it seems that these methods do not work
when
> > considering half-wave resonators, the frequencies tend to be much
greater
> > than, say, the medhurst calculation. So, has there been found a method
of
> > calculating the half-wave resonance of a coil?
> >
> > Thank you in advance for your help.
> >
> > Gavin
>
>
>
>
>