EleMaEMCD A Theorem of Mass Being Derived From Eletrial Standing Waves (Adapted for a test by Jerry E Bayles) - by - Jerry E Bayles May 1, 000 This paper formalizes a onept presented in my book, "Eletrogravitation As A Unified Field Theory", (as well as in numerous related papers), the onept of mass being the result of standing waves More exatly, the result of eletrial standing waves First, the eletri mass equation developed in previous papers will be presented in terms related diretly to the rest mass energy of the eletron This is done so as to establish the fat that ordinary parameters suh as the permeability of free spae, the harge squared of an eletron, and the lassi eletron radius diretly establish the mass of the eletron Then we will develop an atual standing wave on a transmission line of arbitrary load resistane and line impedane Using urrents derived from harge and time related to frequeny, the mass-gain involving the urrent squared times the wavelength squared feature of the developed mass-energy equation will be presented This suggests that an eletrial mass reation may be nonlinear to the forth power by reason of the urrent squared times the length squared Then the naturally negative mass feature involving the phasor form of purely reative urrent is presented whih suggests that purely reative energy has a built-in negative mass assoiated with its field From that, we launh into a transmission line analysis involving standing waves and examine the results of a standing voltage and urrent wave given some arbitrary input voltage, line impedane, and load From the urrent derivation, we utilize the mass-energy equation to plot a resulting mass wave buildup on the transmission line Note that in Mathad, the parameters of the equations are ative and an be hanged for analysis purposes If you do not have your own Mathad 60+ or later, you an download the Mathad Explorer for free from Mathad website (Link at: http://wwweletrogravityom) This partiular analysis is speifi to Fran De Aquino's test of January 7, 000 wherein he reported a signifiant loss of weight of a test apparatus in the shape of a torus whih was supplied with a large amount of reative power His paper is loated at http://xxxlanlgov/html/gr-q/9910036 Jean Louis Naudin's website has the equation solver at: http://membersaolom/jnaudin509/systemg/html The parameters related to his page are used in the below analysis I am assuming that the test reported by Fran De Aquino is the only test with positive results sine no verifiation (to my knowledge) has been made as of this date
The Eletri Equivalent of Mass µ o 156637061 10 06 henry m 1 Magneti permeability m e 9109389700 10 31 kg Eletron rest mass q o 160177330 10 19 oul Eletron harge 99794580 10 08 m se 1 Speed of light in a vauum h 666075500 10 34 joule se Plank onstant l q 81794090 10 15 m Classial eletron radius Let: n q 1 Current multiplier for analysis h Let: t x t m x = 8093301 10 1 se eq 1 e n q q o and: i q i = t q 19796339 amp eq x i Now let: m e µ q o ( d ) Solving for d: eq 3 4 π l q has solutions) π l q m e µ o i q π l q m e µ o i q Note that the m e term is proportional to the square of the urrent term sine is a onstant Where: π l q m e = 46311 10 µ o i q 1 m eq 4 and: π l q m e = 46311 10 1 m eq 5 µ o i q Chek: d h m e d = 46311 10 1 m eq 6 (= Compton wavelength of the eletron)
Quantum mass hek of the above: i m q d x µ o 4 π l q m x = 910939 10 31 kg eq 7 (= rest mass of the eletron) Note that the mass is proportional to the urrent squared for a fixed wavelength, d Then related to urrent, the mass would inrease exponentially Let us alulate the effetive mass related to the urrent squared in an antenna with the following arbitrary parameters: Let: i ant 1 10 0 amp and f ant 1 10 04 Hz Then: d ant or, d = f ant 99795 10 4 m ant Then the effetive mass is given by: i m ant d ant ant µ o 4 π l q Note: The alulation uses real parameters sine we are onsidering the ase for antenna ation eq 8 or, m ant = 354869 10 3 kg (A signifiant mass inrease over the mass of the eletron) Note also that the above marosopi effetive mass alulation now is proportional to the square of the urrent as well as the square of the wavelength If we onsider the ase for purely indutive or apaitive urrent, then the following applies: Let: θ π and i j sw i ant e θ (+ = Indutive ase) then: i sw = 61303 10 15 + 100j amp Then the effetive mass related to purely reative urrent wave is given below as: i m sw d ant sw µ o 4 π l q This is the ase for a nonradiating or, 'antenna' that is eq 9 totally enlosed in a shield m sw = 354869 10 3 + 4345749 10 13 j kg
3 Note for a purely reative urrent wave, the effetive mass is negative and real This implies that a powerful enough wave should be able to reverse the attration of gravity sine the effetive field mass is negative As an example, let us alulate the fore of repulsion at the surfae of the Earth for the above mass First we establish related parameters as: G 667590000 10 11 newton m kg Gravitational onstant R E 637 10 6 m Mean radius of the Earth M E 598 10 4 kg Mass of the Earth Then the fore on the surfae of the earth related to the negative effetive mass alulated above is given by the equation below as: G M E m sw F E R E or, eq 10 F E = 3489674 10 4 + 473477 10 1 j newton whih is a real and negative fore of repulsion by reason of the standard equation result is normally positive and one of attration If we allow for a 0 or 180 degree (0 or π = half wavelength) in theta above, the fore will be one of attration sine the effetive mass will be positive Inserting a theta (θ) of (π/ or -π/ = quarter wavelength) will yield an effetive negative mass Then if the top of a UFO style raft had a real omponent field while the bottom had a reative field, the top would attrat and the bottom would repel other normal mass The following is an analysis of how mass is reated by standing wave of urrent in a transmission line with the parameters of Fran De Aquino's test 0f 01-7-000 The voltage and urrents along the line with respet to time are given by the following equations below, whih is the sum of the forward and reverse propagating waves V z ve V plusve e V plusve I z ve e R j j ω u z V negve e ω u z V negve R e j j ω u z ω u z z = any point on line eq 11 R = line impedene eq 1
where the V plusve and V negve terms are generally omplex numbers as: 4 V j plusve V mplus e θ V j negve V mneg e θ Proposal for new test by Jerry E Bayles: The test below is based on a bifilar wound oil onsisting of parallel elements wound a total of N turns on a 1/ in diameter iron rod for a distane of 6 feet The parameters are developed in appendix 1,, and 3 E in 03 Input volts (loaded) I in 4666 Element urrent, amps R loss 6415394 10 4 Element d resistane, ohms P kva E in I in P kva = 13998 Power Input, VA (loaded) Z S 1 10 06 j 649 10 04 Xfmr se Z (Appendix ) R in 13057 Transmission line harateristi Z (ohms) C line 0033747 Assumed total line apaitane (farads) L line 511085 10 4 Assumed total line indutane (henry) Now using standard formulae for indutive and apaitive reatane, the reatane related to the above L line and C line is alulated to be: X L π f L line or, X L = 019653 ohm eq 13 1 X C or, X π f C = 007860 C line ohms eq 14 The resonant frequeny related to the alulated indutane and apaitane is: 1 f r f r = 3834808 Hz eq 15 π L line C or, line 1 f f r = 1565566 eq 16 This frequeny is lower than the test frequeny of 60 Hz whih implies that it would be able to provide a kik every yle to the standing wave in the line Note that: atan 1 f f r = 100373 radian eq 17
5 The phase veloity and Z is alulated next where: u p 1 or, u p = 95760873 m/se eq 18 L line ζ C line ζ where: L line = 511085 10 4 henry C line = 0033747 farad L line and: Z' ζ C line ζ 1 Z' = 013057 ohms eq 19 Related to the above equations, various related test parameters are defined as: f 600 Frequeny (Hz) (Global delaration) ω π f Angular frequeny (rad/se) u u p Propagation vel of transmission line * = See above ζ 39768 Atual length of line (m)(global delaration) R R in Charateristi Z of transmission line (ohms) Z L 1 10 09 j 0 Approx open iruit Load impedane (ohms) V S 03 Input soure voltage β ω u Phase onstant where, β = 0393675 rad/m Line length as atual eletrial wavelength is given as: λ u f or, λ = 15960348 (m) eq 0 The line ratio of atual length to eletrial wavelength is: ζ λ = 049168 (Very lose to 1/4 wavelength) eq 1 The refletion oeffiient at the load and the input is alulated next: Z L R Γ L Γ = Z L R L 1 ( Load ) eq
6 Next we define z as any point along the line Then: Γ ( z ) Γ L e j β ( z ζ ) whih is the generalized voltage refletion oeffiient eq 3 Then the refletion oeffiient at the input to the line is: Γ ( 0 ) = 0999945 0010461j (Input) The voltage standing wave ratio (VSWR) is: 1 Γ L VSWR if Γ L 1,, 1 Γ L eq 4 VSWR = 816315 10 9 (Line is open-ended at load) The nominal line input impedane is alulated to be: 1 Γ ( 0 ) Z in R 1 Γ ( 0 ) (Z in below is lose to stated test value) eq 5 Z in = 1514344 10 11 6436856 10 4 j (ohms) Next, we determine the time domain voltage at the line input and at the load: First the soure end refletion oeffiient is alulated as: Z S R Γ S Γ = Z S R S 099999 0010448j eq 6 Voltage anywhere along the line is: Γ L e R V ( z ) V 1 Γ j β ζ S Γ L e Z S R S e 1 j β ( ζ z ) j β z eq 7 Then the nominal input V(z) is: V ( 0 ) = 0300183 33314 10 4 j (= V in ) The input phase is: if V ( 0 ) 0, arg V ( 0 ) deg, 0 = 55156 deg eq 8
7 The absolute load voltage is: The omplex load voltage is: V ( ζ ) = 57388478 V ( ) = ζ 5738846 0044606j The load phase is: arg V ( ζ ) if V ( ζ ) 0, deg, 0 = 551603 deg eq 9 Sine we have an expression for the voltage anywhere on the line, (eq 7), then the urrent at the input and load may be expressed as: V ( 0 ) (+ = urrent leading voltage = apaitive) I in Z I = in in 036477 + 466350351j (amp) eq 30 V ( ζ ) I L I = Z L 5738846 10 8 446059 10 11 j (amp) eq 31 L The input and load power is given below as:(volts times onjugated urrent) P in V ( 0 ) I in P in = 393437 10 6 139990531j (= Negative imaginary power) eq 3 P ld V ( ζ ) I L P ld = 393437 10 6 + 310103 10 5 j eq 33 The plot of phasor domain voltage and urrent is presented below npts 100 z start 0 z end ζ z i 0 npts 1 z i z end z start start i npts 1 eq 34 Mag Vi V z i (Voltage magnitude along the line from start to end) arg V z i Θ Vi if V z 0 i,, 0 (Voltage phase along line) eq 35 deg
Voltage magnitude along the transmission line is: 8 60 Mag Vi 40 0 0 0 05 1 15 5 3 35 4 z i Voltage phase along the transmission line 0044535 Θ Vi 0044533 00445335 0044534 0 05 1 15 5 3 35 4 z i Next the urrent along the transmission line may be given by: Mag Ii Mag Vi 1 Γ z R i 1 Γ z i Equivalent to: I z i V z i Z z i eq 36 Where again: I in = 036477 + 466350351j I L = 5738846 10 8 446059 10 11 j
Plot of the reative Mag I urrent along line: 9 500 Im Mag Ii 0 500 0 1 3 4 z i arg Mag Ii The urrent phase is derived as: Θ Ii if Mag 0 Ii,, 0 eq 37 deg The urrent phase plot is provided below as: 100 Θ Ii 50 0 50 0 1 3 4 z i Based on the line amps alulated above at the given line length, the effetive field mass in the transmission line an be alulated as: Mag Ii m ζ i µ iron µ o 4 π l q Where: µ iron 808 µ o 4 π 1 10 07 l q 81794090 10 15 99794580 10 08 (= negative mass in kg) eq 38 05 0 m i 05 1 15 0 1 3 4 z i
10 The average negative field mass in kg of the line is given by: m avg m i npts 1 m avg = 0555519 + 863569 10 4 j eq 39 i The final result in the above equation demonstrates that if the torus geometry is not treated as an antenna, but rather as a transmission line, then negative mass ours naturally via the reative terms as shown above APPENDIX1: It is of interest that a quarter wavelength presents a series resonant ondition whih would appear as a very low Z iruit to a soure at the transmission line input Note that all of this is also for a line that is open iruit at the load end Conlusions: Treating Fran De Aquino's test torus as a transmission line yields a negative field mass diretly without resorting to the assumption that the iron atoms in the shield must absorb energy To absorb energy, the atoms must onvert that energy to vibratory motion whih translates into heat build whih means that they will be subjet to temperature rise Fran De Aquino assures us that the torus and shield do not get hot Therefore the power is reative and the transmission line equations above use that property to arrive at the negative field mass results in a straightforward manner I suggest that this analysis more aurately desribes the results of Fran De Aquino's test muh better than assuming that the ation of antenna radiation ours The mehanis of antenna radiation involve non-reative radiation, ie, real power, whih is alulated in free spae by use of the Poynting vetor equation for power See the next page for ontinuation of this paper Ω NOTE: The original work that supports this onept is hapter 7 of my book, "Eletrogravitation As A Unified Field Theory"
APPENDIX : 11 The Q related to the indutive property of the transmission line may be used as a transfer parameter to design transmission lines of different geometry The same Q (or greater) will insure that an effiient mass loss would also our The Q related to Fran De Aquino's test was derived on the bottom of page 11 of EleMaBMCD and was equal to 450999105 This is the ratio of XL / Rloss of Fran De Aquino's test where the R loss is the atual element ohmi loss alulated from a opper wire table for 4/0 opper wire 1139 meters long The angle related to Q was 8879789 degrees The Q and phase related to this paper's alulations is given below as: Q X L or, Q = 3009833 R loss X L = 019653 R loss = 6415394 10 4 eq 40 θ Q atan( Q ) θ Q = 8980905 deg eq 41 APPENDIX 3: It is possible that a small amount of power, properly applied, may yield large results For instane, a soldering gun delivers a substantial amount of urrent on its seondary sine it supplies a load of very small resistane Further, it is the urrent squared that is most important if the line length is held onstant Therefore, if we design a quarter wave open ended transmission line properly, it will offer a series resonant iruit of very low impedane and thus offer only ohmi resistane to the transformer power soure Simply mathing that seondary transformer Z with the proper impedane will assure maximum transfer of power and thus urrent to the load, in this ase the transmission line Sine it is the standing wave urrent that generates the mass effet, having as muh reative urrent as possible in the transmission line will generate a maximum amount of negative field mass The following nominal parameters apply to a soldering gun that I happen to have on hand P max 140 V pri 130 V se 03 watts volts volts, loaded A pri T r P max A = V pri 107693 Primary amps pri V pri T = V r 433333333 Transformer turns ratio se
A se A pri T r A se = 466666667 Seondary amps 1 R se V se R se = 648571 10 4 Seondary impedane, ohms A se The following parameters are stated for the design of a transmission line that is formed as a bifilar winding on an iron rod 1/ inh in diameter and approximately 13 feet long The number of turns will be derived that will yield the same Q as above The wire used is #4/0 AWG opper wire Permeability of iron rod Length of ore (= 5 in) Ohms per 1000 ft of #4/0 AWG wire µ rod µ iron ore lngth 0635 meters R kft 04901 ohms R kft R meter 38084 or, R = 1000 meter 160794 10 4 ohms Then the total physial length in meters of the 4/0 wire is given by dividing the transformer seondary impedane by the per meter resistane, R meter This will math the seondary impedane of the transformer to the transmission line ohmi resistane This must also be lose to the quarter-wave atual physial length R se Then: L wire L = R wire 3998018 meters meter ore radius 635 10 03 meters (= 5 in) N, 1 100 Then: µ rod µ o N π ore radius XL ( N ) π f eq 4 ore lngth ohms / meter Iron rod 4/0 ore rad wire rad And: R ( N ) R meter N π ore radius 635 10 03 eq 43 In terms of N, the Q of the line is: Q ( N ) XL ( N ) R ( N )
Then the intersetion of the fixed Q and the Q based on N turns is plotted as: 13 800 600 Q ( N ) Q 400 00 0 0 0 40 60 80 100 N The number of turns is 50 for a Q of 300 as the graph above shows The indutive reatane in terms of N is: 1 XL ( N ) 05 0 0 0 40 60 80 100 The ohmi resistane in terms of N is: N 00015 R ( N ) 0001 5 10 4 0 0 0 40 60 80 100 N
14 We now state the parameters based on the above harts involving N, XL and R to ompute the input Z of our line whih is really the new resistive R (The value of XL fix below must equal the starting X L above to yield 1/4 λ in the final analysis) N fix 49834 Q = 3009833 X C = 007860 X L = 019653 µ rod µ o N fix π ore radius XL fix π f XL fix = 019643 eq 44 ore lngth R fix R meter N fix π ore radius 635 10 03 = R fix 6394095 10 4 eq45 XL fix L line Chek: = 3018796 Z R line Z = fix C line 013057 eq 46 line and: N fix π ore radius 635 10 03 = 3976576 meters eq 47 The physial weight of 1305 ft of opper is: (4/0 is rated at 1561 ft per pound) N fix π ore radius 635 10 03 ( 38084 ) 1561 The total 4/0 opper wire length in feet is: = 8357791 lbs eq 48 N fix π ore radius 635 10 03 38084 = 13046511 feet total eq 49 The bifilar winding will allow for design below whih is driven from the opposite ends by a variable frequeny power amplifier apable of driving the soldering gun primary at 140 watts and 130 volts at a variable frequeny entered on the nominal design frequeny of 60 Hz The length of wire is ut in half and the iron ore is shared by both windings Figure 1 below is a basi design onept of the transmission line FIG 1 < ---------------------------- meters ---------------------------- >
15 In the above drawing, (figure 1), the toroidal winding is employed verses the normal onfiguration whih will allow for the same resistane as alulated above The normal line would have twie the loss due to the line having to be twie as long to obtain the same indutane Also, the toroidal line indutane is additive sine the urrent is flowing in the same diretion in the adjaent line windings Lastly, the normal line would require separate line ores sine the indutanes would be subtrative The number of apaitors and their values would have to hosen so that if 10 apaitors were used, their sum total apaitane value would add up to the C line total value given on page 4 above C line Or, = 10 33747 10 3 farad eah eq 50 The voltage rating should be at least 80 volts and bi-polarity, or a rated (Capaitors of double the value of apaitane and polarity sensitive in series bak to bak ould also be used) They would be onneted between the two bifilar windings every: ζ ( 10 ) = 019884 linear wrap meters or 788 inhes eq51 The atual length of 4/0 wire wrapped 50 turns on a 1/ inh diameter ore 5 inhes long is omputed below This must be lose to the atual wavelength and fit the parameters for the indutane alulation above The ore length is: ore lngth = 0635 meter = 50 in The ore radius is: ore radius = 635 10 3 meter = 05 in N fix = 49834 The distane between the enter of eah turn is: Dis T ore lngth Dis = N T 00174 meter = 050 in fix The diameter of eah turn is: Dia T 054 meter = 100 in (Center to outside of eah wire normal to ore axis + ore= x 05 + 05 = 1 in) The atal elongated (spiral distane) length of wire used for eah turn is:
16 Dis T L T Dia T π or, L T = 00868 meter Finally, the total length of wire used for N fix turns is: Length N fix L T Length = 4099766 meter This distane is very lose to the ζ requirement above Wrapping the turns of opper around a plasti pipe that fits over the 1/" diameter iron ore will allow for the iron rod to be moved in or out of the oiled transmission line for fine tuning the line for maximum performane (Even though the plasti pipe will inrease the turns radius, the oil's indutive reatane depends mostly on the diameter of the iron ore whih has a muh higher permeability than air) Also, driving the primary of the soldering gun transformer with a 00 W power amplifier at 130 volts and 60 Hz will allow for a variane of the input frequeny if the amplifier is driven by a suitable audio osillator and this will also serve as a means of fine tuning the transmission line for maximum effiieny A final note: It is not the power that is important here but the urrent and effetive length of the line that will determine the field mass amplitude Both the urrent and the length of the line have a squared effet on the final field mass result NOTE: It is of interest that the Tesla oil operates on a quarter-wavelength priniple in that the atual physial length of the seondary winding is 1/4 wavelength (or 3/4, 1 1/4, et) related to the resonant seondary operating frequeny, usually around 100 to 00 KKz This is equivalent to 1500 to 3000 meters This length squared times the urrent squared (average) along the seondary length will determine the theoretial negative mass buildup in the seondary Has anyone atually measured the weight of a Tesla oil before and during the oil being operated? I suggest that it may be of interest to the experimenter who is testing for the negative mass effet to test a Tesla oil for a mass derease during its operation relative to when it is not operating The urrent and voltage in the seondary are purely reative and fulfill the test requirements of the above presentation If anyone does this test, please let me know how it turns out Author Jerry E Bayles May 1, 000