Pioneering the Applications of Interphasal Resonances http://tech.groups.yahoo.com/group/teslafy/
--- On Sat, 10/1/11, fznidar...@aol.com <fznidar...@aol.com> wrote: From: fznidar...@aol.com <fznidar...@aol.com> Subject: [Vo]:redid my video how do you like it To: vortex-l@eskimo.com, bobp...@umd.edu, georg...@aol.com, puth...@aol.com, hai...@calphysics.org Date: Saturday, October 1, 2011, 5:57 PM http://www.angelfire.com/scifi2/zpt/movies/impedance.html Frank Znidarsic In certain issues regarding air core source frequency resonance results; or tuned coupled circuits as they may be described; the interaction viewed as relevant to the noted "impedance matching" phenomenon can be analysed with the following loads; for the typical 60 hz model using 12 lb coils of 23 gauge both as sending and receiving tuned units, the receiving end displays more voltage and energy powering identical loads from the input voltage, but this effect is due to manipulations of maximum power transfer to say the least. When these same coils are used at alternator 465 hz frequencies, putting a single bulb across the receiving coils as done in the shown jpeg of 60 hz resonance effects; in the second case this will instead entirely diminish that secondary loops amperage, but as we add bulbs in series to that circuit, its voltage source according rises because of this impedance matching issue... http://www.flickr.com/photos/harvich/3338081529/ 90 Volt input to 60 hz resonances, top pair produces 1000 volts between them. Bottom air core secondary has input of .49A from primary,@ 435 volts resonant voltage rise producing 135 volts secondary volts with bulb taking 35.7 ma from coil/cap circulation of 143.1 ma. Paradoxically the meter showing the relative differences between the 435 volts and 135 volts on secondary reads 0 volts. Apparently Isolated voltage rises cannot be compared without making the autotransformer connection. http://www.flickr.com/photos/harvich/5671006273/ Input Energy for 3 ph. 266 primaries enabling 2.4H/50nf secondary coil @ 465hz to empower 5 four watt rated nite/lite consumptions @ 30 ma =20 watt output. The 266 primary has phase 3 enabling ~3.7A @10.4 volts directly underneath the 2.4H/140 ohm secondary. http://www.flickr.com/photos/harvich/5671603376/ 266 three phase triangular pancake primaries with secondary top coil receiving 62.7 ma amperage circulation between secondary L and C values, and a diversion of 35 ma across five four watt bulbs in series~ 440 volts across bulbs from ~10 volt primaries. So here apparently we have an air core transformer with the greatest power input being the layer underneath its secondary showing 10.4 volts enabling 3.7 amps. On the secondary we have ~440 volts on the 7000 ohm reactance loop, enabling a 35 ma diversion of current through load, with the source having 62 ma in circulation. The ratio of turns of secondary per phase layering are 67/1, and the ratio of resistances 140/0.5 ohms. In this case however if the loads were removed, the secondary amperage circulation would go up while its source of primary current would go down. More amp turns would be found on the secondary then the primary powering it. Operation of the (266 Primaries) with Independent Secondary loops showing household voltage and also 240 from the alternator running with unenergized field. (Note my many erroneous conclusions) http://youtu.be/ASBwp-vqhYU Dual Arc Gapped application of 266 primaries to Bipolar TC. At 0:35 both primary arc gaps are visible, but ordinarily they seem to oscillate in time. http://youtu.be/zNXtFUo4gbo Finally to conclude here, if any conclusions can be made??? Expert explains the Predicament. Wed May 11, 2011 Four months ago I posted the following to Yahoo Q&A. Can a superior amp-turn ratio made with air core primaries vs secondary lead to more heat output then inputed? This is the curiosity that seems evident in working with 465 hz AC car alternator circuits. In the cited cases the I^2R heat losses of the primary are smaller then the I^2R secondary volumes. Essentially by using a high turns ratio between primary and secondary, and then comparing heat losses between the two, more heat loss comes from the secondary then does the primary. Have I made a mistake here or is this okay? Bert K, a top contributor replied; This is absolutely okay. Power is proportional to I^2 Is = NIp R is proportional to number of turns = N Rt, Rt = resistance of one turn Pp = Ip^2 Np Rtp = Ip^2 Rp Rp = NpRtp Ps =(Ip Np/Ns)^2 Rs = Ip^2 (Np/Ns)^2 Rs Rs = NsRts = NsRts/NpRtp Rp Ps = Ip^2 (Np/Ns)^2 NsRts/NpRtp Rp Ps/Pp = (Np/Ns)(Rts/Rtp) Sp power is greater in the secondary as the turns ratio, and inversely as the resistance per turn (Np/Ns)(Rts/Rtp) For good design, the ratio is unity and the powers are equal, but in any given case, that can result in your measurement being correct. Unfortunately I have not studied his reply full enough to know what he is talking about. But here no matter how you stack things up, the secondary output seems greater then the primary one. If you use the I^2R as the primary input we found for the cited case in the video the .5 ohm primary having a current of .18A should have a true power input equal to its heat loss I^2R quantity, or .0162 watts. But applying the stame standard to the secondary we find .18144 watts with a 36 ma circulation through 140 ohms. Now we go back to some fundamental understandings. The reactive measure of the power input, or VI will ALWAYS be greater then its true power component I^2R, BUT AT PERFECT SERIES RESONANCE THEY SHOULD EQUAL EACH OTHER. At the ideal series resonance there is no timing or phase difference between the sinusoidal quantities of V and I, and the currents if ideal will come to their ohms laws value applied to AC. This is because V and I are simultaneous in time: V*I will equal I^2R, and all the power input goes into heating the conductor. Now lets go back and measure the apparent power input to the primary where we have 1.1 volts enabling .18 A or .198 watts. Since we have .1814 watts on secondary and.0162 watts on primary this gives a total of .1976 watts total, just about the same .198 watts calculated by using the apparent power. So the quandary exists as follows, even though we have been taught that the apparent power is always greater then the true power input, in this special case the apparent power is equal to the true power input, and then VI = I^2R(prim)+ I^2R(sec). Perhaps the claims of overunity are not valid then, but this particular situation we are tempted to assume that the ordinarily laws of analysis have been turned topsy turvy. Initially it would seem that the power input should only apply to the primary, not the primary and secondary as one coupled unit, but it seems that is the only way solve the paradox. HDN
