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Re: LC III
- To: tesla@xxxxxxxxxx
- Subject: Re: LC III
- From: "Tesla list" <tesla@xxxxxxxxxx>
- Date: Thu, 31 Mar 2005 11:48:47 -0700
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- Resent-date: Thu, 31 Mar 2005 11:51:21 -0700 (MST)
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Original poster: "Bob (R.A.) Jones" <a1accounting@xxxxxxxxxxxxx>
----- Original Message -----
From: "Tesla list" <tesla@xxxxxxxxxx>
To: <tesla@xxxxxxxxxx>
Sent: Tuesday, March 29, 2005 6:52 PM
Subject: Re: LC III
> Original poster: "Antonio Carlos M. de Queiroz" <acmdq@xxxxxxxxxx>
>
> Tesla list wrote:
>
> >Original poster: "Malcolm Watts" <m.j.watts@xxxxxxxxxxxx>
> >...
> >However as Robert Jones has pointed out, the
> >capacitance Medhurst's formula gives is *not* the static isotropic
> >capacitance of a conductive cylinder. This is easily measured and I
> >have done it. For one particular resonator I recal measuring
> >somewhere in the vicinity of 100pF where Medhurst gives (a la
> >resonant frequency calculations) somewhere around 25pF. Therefore,
> >Medhurst (and Wheeler) give us a useful tool for coil design but say
> >little about the actual working mechanism.
>
> For the coils that I have measured, or calculated, the
> Medhurst capacitance is close to one half of the capacitance
> of a hollow cylinder to "infinity".
> And there is a good reason for this, because if the distributed
> capacitance of the whole coil is split in two, one at each end
> of the coil, Each of them is at the position of the Medhurst
> capacitance when the other end is grounded. Of course, this
> assumes that the grounding at one end doesn't affect much the
> distributed capacitance at the free end of the coil, what
> appears to be true.
>
I think the ration depends if you use Medhurst C from his table or calculate
it from fr.
If I remember correctly theoretically there can be 38% difference in value
between the two, possible more for real coils.
I factor of four surprises me. I would have expected it to be closer to two
but it my be dependent on the test conditions and the above.
Bob (R. A.) Jones
A1 Accounting, Inc., Fl
407 649 6400