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Projects by members => Projects by members => J. Miller => Topic started by: jim miller on January 23, 2011, 05:08:49 am
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last modified 1 26 2010 goal clarification
last modified 1 30 2011 attached acoustic modification file
This discussion is a revisit to a concept of matching surface areas of tube sets to acoustic resonance and to look at methods of acoustic magnification for water fuel cells- See Attachment)
Basically the area of two tubes are matched in surface area while at the same time matching for acoustic resonance.
So let the area of the inner surface of the outer tube be A1 and the area of the outer surface of the smaller inner tube be A2. Now the areas will be closer in value if the gap between the two tubes is smaller. But there is a
factor that one does not want the gap to be so small as to allow arching or impediment to water and bubble flow.
So looking at the Demonstration Cell dimensions as given by Exhibit E-2
Tubular Cluster Array the distance between the two pipes is calculated by
taking the diameter of the larger pipe .75 inches minus twice the nominal wall thickness of 0.035 which yields 0.68 inches which then has the diameter of the smaller tube .50 inch subtracted from it yielding .18 inches
This result is divided by 2 giving a gap of 0.09 inches or 2.28 mm
The sympathetic vibration of the tubes might only be important in terms of removing bubbles quickly and therefore increasing effective surface area
but still bears a look at its relationship to other resonances present.
FROM POSTING IN ARCHIVE AT IONIZATIONX old archive ill update link when ionizationx is back up
"following posting is theoretical, and as such should be treated with caution.
The previous missive that I fired off on the subject of Stan Meyer tube lengths looked mainly at the negative characteristics of what has been done up to now. With this posting I am going to try and build a case for how Stan Meyer actually achieved what he did.
MY UNDERLINING JM
One of the things that puzzled me; was why did he cut slots in the outer tube in order to obtain resonance? Since it is far easier to use a lathe to take off small amounts of metal. For example, a normal metal lathe will cut down to an accuracy of 1/10 mm. (That for the USA is around 1/250 of an inch). Thus for a 25.4 mm OD tube with a wall thickness of 1.5mm, a length of 420mm gives a resonant frequency of 880 Hz, and a length of 408 mm gives a resonant frequency of 932 Hz. Thus for a length difference of 12 mm we have a frequency difference of 52 Hz. Thus the change of length per Hertz = 12 / 52 = 0.23 mm. This is easily within the tolerance of any lathe.
It has been suggested to me by various people, that they have got the best results, in terms of hydroxy LPM production, when they have ignored the resonance issue, and concentrated on getting the surface areas of the capacitive surfaces the same. What all the evidence that I now have points to, is that SM managed to combine resonance of the two tubes, with them both having the same surface area for the capacitive effect. On the face of it this is impossible, as the table given below will show, but I believe that I can illustrate how this can be done.
The following table shows the resonant frequencies of the two tubes, their respective lengths, capacitive "
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jm
The goal is to match surface area for the capacitance-impedence equation for the VIC and PPL configurations and at the same time to match the masses of the two tubes for and the sonic frequency( acoustics) to aid if bubble formation, flow and release. The are also higher order harmonics to be looked at in terms of the tube length and the keely/purharich frequencies of the water molecule ( Ie reinforcement)simultaneous solution needed - try for same area tube inner and outer tubes but both ideal tube acoustically and electrically the same