Medhurst self resonant frequency and nodality (fwd)

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---------- Forwarded message ----------
Date: Sun, 14 Oct 2007 20:44:25 -0400
From: Jared Dwarshuis <jdwarshuis@xxxxxxxxx>
To: Pupman <tesla@xxxxxxxxxx>
Subject: Medhurst self resonant frequency and nodality

               Medhurst self resonant frequency and nodality





When Medhurst wrote the self resonance equations he based his curve fit on
the curios fact that a coil self resonates at: (wave length/ 2)  when the
coil is long and slender and has many turns.



So the self capacitance can be written as Cself = 4 e H/ Pi

 (for a long slender coil with many turns)



We will demonstrate how this works with nodality where:



n = 1/2, 2/2, 3/2, ,….. corresponds to wire length = Wavelength/4,
Wavelength/2,  3 Wavelength/4……..



Starting with:



F = 1 / 2pi sqrt LC            we place 2 pi inside the sqrt as:



F = 1 / sqrt ( 4 (pi)sqrd LC )



We will use L = (wire length/  C)sqrd / 4 pi e H (2n)      (we inserted
inductance nodality here as (2n) )



Thus frequency = 1 / sqrt( 4(pi)sqrd 1/2n  (Wire length/C)sqrd  x  pi/ eH  x
capacitance )



Capacitance is also nodal in behavior so we will write:



Frequency = 1 / sqrt( 4(pi)sqrd 1/2n  (Wire length/C)sqrd  x  pi/ eH  x  [ 4
e H/ 2n pi ] )

(we insert Capacitance nodality here as (2n) )



After canceling like terms we get:



F = sqrt ( (n)sqrd  (C/ wire length)sqrd  )



F =  n   C/ wire    ( again as a reminder: where n = 1/2, 2/2,
3/2,………corresponds to 1/4 wave, 1/2 wave, 3/4 wave etc….  )



This not only applies to an inductor but to a straight wire as well. In both
instances the results will be closest to ideal when the solenoid (or wire)
is long and many nodes are present.



Let us examine the straight wire LC resonance capacitance



Using: L =  u (Wire length)sqrd / 4 pi H 2n

When Wire length = H:

This can be simplified to:



L straight wire =   u Wire length/ 4 pi



The self capacitance for a straight wire at resonance already defined as:
Wire length = Wavelength/2 can now be written as:



Capacitance straight wire = 4 e Wire length / pi



Then LC frequency for a wire equals:



F =  1/ 2pi sqrt ( (Wire/C)sqrd   x  1/ (pi)sqrd ) = 2 C/ wire = 4 C /
wavelength



We can see that frequency is set by the quarter wave region when the ideal
self capacitance and self inductance formula are adjusted to incorporate
nodality.



Observations:



The self resonant frequency of Wavelength/2 for a wire is solely a function
of its length. Thus a given wire will have the same resonant frequency
whether it is straight or wound in a solenoid. This stands to reason since
node spacing is entirely a function of wavelength, or in our case, wire
length.



These equations tell us that a stretched out wire or the same wire wound
around a solenoid will have a natural LC distribution such that waves at the
resonant frequency travel at the speed of light.



When a wire is wrapped around a solenoid the inductance increases but so
does the proportion of self capacitance such that waves at the resonant
frequency will still travel at C.



Perhaps equations are complicated, but the bottom line is rather straight
forward:



Waves go a fixed velocity at resonance. If they have 1/2  as long to travel
they will travel 1/2 as far. This means the nodes are then  half as far
apart and we need twice as many to cover the same distance.



We found some of our information about Medhurst equations from the web site
G3YNH



Jared Dwarshuis and Lawrence Morris



Oct. 07