KeelyNet: Joel McClain: Measurement of Resonant frequencies and Harmonics

Measurement of Resonant frequencies and Harmonics

April 25, 1994

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:

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.

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.