Re: Space winding question

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

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

Hi Malcolm,

I agree the difference is greater than suggested but due to a top load
capacitance. When I replied about
a 10kHz difference, I had modeled both cases using a top load size picked
out of the air but probably
appropriate. I didn't model without a top load since for all practical
purposes, the coils wouldn't be
used without one. Medhurst was used in my calcs.

Bart A.

Tesla list wrote:

> Original poster: "Malcolm Watts by way of Terry Fritz
<twftesla-at-uswest-dot-net>" <m.j.watts-at-massey.ac.nz>
>
> On 11 Mar 01, at 12:53, Tesla list wrote:
>
> > Original poster: "by way of Terry Fritz <twftesla-at-uswest-dot-net>"
> > <Mddeming-at-aol-dot-com>
> >
> > In a message dated 3/10/01 4:45:00 PM Eastern Standard Time,
> > tesla-at-pupman-dot-com writes:
> >
> > Terry & All,
> >
> > It has been said that nothing ruins a good discussion like someone
> > with facts. However, I for one appreciate the fact that Terry has
> > again "Done the Math" before answering the question. IMVHO this
> > approach enhances both the List's and his personal credibility.
> >
> > Matt D.
>
> In that case I should apologize for my first reply to your question
> since I did a mental calc to arrive at the aanswer. However, I do
> believe that there is a greater difference between the capacitances
> than the figures calculated below suggest. Medhurst's tabular
> approach certainly shows it.
>
> Regards,
> Malcolm
>
> > >
> > > Your questions is straight forward and we can figure it out.
> > >
> > > The coil's resonant frequency is determined basically by it's
> > > inductance and effective capacitance.  "Sort of" like a simple LC
> > > circuit.  "I" don't think wire length has anything at all to do with
> > > it and wire gauge does not have much affect either.
> >
> >
> >
> >
> >
> > >
> > > So we have two 1000 turn coils 12 x 54 and 12 x 60.
> > >
> > > The inductance of a coil is given by the following to with about 1%
> > > accuracy:
> > >
> > > L = (N x R)^2 / (9 x R + 10 x H)
> > >
> > >
> > > L = inductance of coil in microhenrys (µH)
> > > N = number of turns
> > > R = radius of coil in inches (Measure from the center of the coil to
> > > the middle of the wire.) H = height of coil in inches
> > >
> > > Putting in our numbers:
> > >
> > > L = (1000 x 6)^2 / (9 x 6 + 10 x 54) == 60600uH = 60.6mH
> > >
> > > L = (1000 x 6)^2 / (9 x 6 + 10 x 60) == 55000uH = 55.0mH
> > >
> > > So now we know the inductance of both coils...
> > >
> > > For the capacitance, we can use the famous Medhurst equation also at
> > > the site above that is about 1% accurate:
> > >
> > > C = 0.29 x L +0.41 x R + 1.94 x SQRT(R^3/L)
> > >
> > > C = self capacitance in picofarads
> > > R = radius of secondary coil in inches
> > > L = length of secondary coil in inches
> > >
> > > for the first coil I get C = 22.0pF
> > > for the second coil I get C = 23.54pF
> > >
> > > So now we know the effective capacitances for both coils.
> > >
> > > Using the resonant circuit formula (also at the site above):
> > >
> > > F = 1 / (2 x pi x SQRT(L x C)
> > >
> > > F = frequency in hertz
> > > L = inductance in henrys
> > > C = capacitance in farads
> > >
> > > The first coil gives:
> > >
> > > F = 1 / (2 x 3.14159 x SQRT(0.0606 x 22 x 10^-12) = 137.84KhZ
> > >
> > > and the second coil gives:
> > >
> > > F = 1 / (2 x 3.14159 x SQRT(0.0550 x 23.54 x 10^-12) = 139.87KhZ
> > >
> > > Thus, the two coils have very close to the same frequency around
> > > 139kHz.
> > >
> > > See how easy that is :-))  Ok, there are a bunch of computer
> > > programs around that will do all this easily but this is the "stuff"
> > > behind how those programs work.
> >
> >
> >
> >
> >
> >
> >