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  |  From           : KeelyNet BBS     |  DataLine        : (214) 324-3501  |
  |  Sysop          : Jerry W. Decker  |  Voice           : (214) 324-8741  |
  |  File Name      : RESMEAS.ASC      |  Online Date     : 05/22/94        |
  |  Contributed by : Joel McClain     |  Dir Category    : ENERGY          |
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  April 25, 1994

  RESMEAS.ASC  Measurement of Resonant frequencies and Harmonics

  Thanks to information  supplied  by  Gerald  O'Doucharty,  as compared with
  waveform analyses as determined by the  late Mr. Les Brown, we can begin to
  make some progress in the understanding of resonance.  I believe that these
  conclusions are in accordance with the discoveries of both  Walter Russell,
  in his resonance  based  chart  of  the  elements, as well as those of John
  Keely.

  Les Brown dropped a pebble in a bowl of water, and measured the rings which
  resulted.  From this,   he  was  able   to   determine   the   mathematical
  relationships, which he  found  to be factors of PI.  There  is  no  random
  pattern to the rings.

  The first ring  is equal to the diameter of the pebble.  The second ring is
  equal to the diameter of the first ring,  times  the  cube  root  of PI, or
  1.3313.  The third  ring  is equal to the second ring, also  times  1.3313.
  Each ring is a NATURAL HARMONIC of the previous ring.

  Comparing this with  the information supplied by Gerald, we can see that in
  the diatonic scale, the following relationships exist:

  Whole
  Note    Frequency (Hz)  1st Harmonic  (UPPER ORDER)

  C       262             F  349.33 Hz
  D       294.75          G  393.00
  F       349.33          A' 465.06
  G       393.80          C  524 (262 X 2)
  A       436.66          D  581.33  (294.75 X 2)
  C'      524             F  698.66 (349.33 X 2)

  Semi- Notes

  E       327.50          A  436.66
  B       491.25          E  654.00 (327.50 X 2)

  Energies, including radio waves, propagate  according  to  the same ratios,
  based upon the cube root of PI.  To an energy researcher, this means:

  1.  Energies are based upon the atomic model.  Therefore,  the  rings which
      surround each atom are based upon the size of the nucleus (pebble), and
      this is  the  base  frequency  of the atom.  Each ring diameter will be
      determined by the cube root of PI,  times  the  nucleus, and then times
      each ring in succession.

  2.  Once you find the resonant frequency of the nucleus,  to  resonate  the
      atom, you  will  need  to  determine the number of rings, and apply the
      correct number of harmonics at the correct frequency.  This is true for
      an element, whereas compounds will  require  each  of  the  fundamental
      frequencies, plus the harmonics.


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  Let's say, for example, that a hydrogen nucleus resonates at G, or 393 Hz.

  To resonate the aggregate hydrogen atom, you must apply G as well as C, the
  first harmonic, or 524 Hz.  If oxygen has a fundamental frequency of A, or
  436 Hz, then you must also apply D, or 589.5 Hz.

  To resonate water, you would apply 393, 436, 524, and 589.5.  However, this
  may not be  a  "hard  and  fast"  rule,  because  we  know that the highest
  frequency will contain all of the lower  frequencies at harmonic intervals,
  but at lower amplitude.

  Therefore, if a single resonating frequency is used, it will  have to be of
  high enough amplitude  so  that  the harmonics are "felt" by the element or
  compound.

  The frequencies in the above example are based upon the arbitrary values
  assigned in 1939, because they "sound" the best.  However, the multiplier
  of the cube root of PI is a constant.   So,  when  you find the fundamental
  frequency of an atom, you can apply this constant to resonate the aggregate
  mass, including the rings.

  Just multiply the constant times the fundamental freq for  the  first ring,
  and multiply the constant times the first ring for the second ring, etc.

  As Gerald pointed  out, there are inconsistencies in the man-made chart, so
  experimentation is required.  However,  Les  Brown's  constant  will  apply
  regardless, ONCE THE FUNDAMENTAL FREQUENCY has been determined.

  Referring to Russell's chart of the elements, we find that  the  number  of
  elements per octave  corresponds to the number of whole notes per octave on
  the diatonic scale.  Isotopes consist of semi-notes and harmonics.

  As such, once  we  begin to analyze elements,  we  can  quantify  Russell's
  chart, and determine the exact combination of frequencies for each element,
  isotope and inert gas.

  Russell obviously knew these correlations, or he would not have been able
  to create the chart in the first place.  We have to "reverse  engineer" the
  chart to determine  the  frequencies.   Using the cube root of PI, combined
  with experimentation, I believe that that is possible to do.
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