A Theorem of Mass Being Derived From Electrical Standing Waves (As Applied to Jean Louis Naudin's Test)

Transcription

1 A Theorem of Mass Beng Derved From Eletral Standng Waves (As Appled to Jean Lous Naudn's Test) - by - Jerry E Bayles Aprl 5, 000 Ths Analyss Proposes The Neessary Changes Requred For A Workng Test Ths paper formalzes a onept presented n my book, "Eletrogravtaton As A Unfed Feld Theory", (as well as n numerous related papers), the onept of mass beng the result of standng waves More exatly, the result of eletral standng waves Frst, the eletr mass equaton developed n prevous papers wll be presented n terms related dretly to the rest mass energy of the eletron Ths s done so as to establsh the fat that ordnary parameters suh as the permeablty of free spae, the harge squared of an eletron, and the lass eletron radus dretly establsh the mass of the eletron Then we wll develop an atual standng wave on a transmsson lne of arbtrary load resstane and lne mpedane Usng urrents derved from harge and tme related to frequeny, the mass-gan nvolvng the urrent squared tmes the wavelength squared feature of the developed mass-energy equaton wll be presented Ths suggests that an eletral mass reaton may be nonlnear to the forth power by reason of the urrent squared tmes the length squared Then the naturally negatve mass feature nvolvng the phasor form of purely reatve urrent s presented whh suggests that purely reatve energy has a bult-n negatve mass assoated wth ts feld From that, we launh nto a transmsson lne analyss nvolvng standng waves and examne the results of a standng voltage and urrent wave gven some arbtrary nput voltage, lne mpedane, and load From the urrent dervaton, we utlze the mass-energy equaton to plot a resultng mass wave buldup on the transmsson lne Note that n Mathad, the parameters of the equatons are atve and an be hanged for analyss purposes If you do not have your own Mathad 60+ or later, you an download the Mathad Explorer for free from Mathad webste (Lnk at: Ths partular analyss s spef to Jean Lous Naudn's test Jean Lous Naudn's webste has the equaton solver at: The parameters related to hs page are used n the below analyss I am assumng that the test reported by Fran De Aquno s the only test wth postve results sne no verfaton (to my knowledge) has been made as of ths date

2 The Eletr Equvalent of Mass µ o henry m 1 Magnet permeablty m e kg Eletron rest mass q o oul Eletron harge m se 1 Speed of lght n a vauum h oule se Plank onstant l q m Classal eletron radus Let: n q 1 Current multpler for analyss h Let: t x t m x = se eq 1 e n q q o and: q = t q amp eq x Now let: m e µ q o ( d ) Solvng for d: eq 3 4 π l q has solutons) π l q m e µ o q π l q m e µ o q Note that the m e term s proportonal to the square of the urrent term sne s a onstant Where: and: π l q m e = µ o q π l q m e = µ o q 1 1 m eq 4 m eq 5 Chek: d h m e d = m eq 6 (= Compton wavelength of the eletron)

3 Quantum mass hek of the above: m q d x µ o 4 π l q m x = kg eq 7 (= rest mass of the eletron) Note that the mass s proportonal to the urrent squared for a fxed wavelength, d Then related to urrent, the mass would nrease exponentally Let us alulate the effetve mass related to the urrent squared n an antenna wth the followng arbtrary parameters: Let: ant amp and f ant Hz Then: d ant or, d = f ant m ant Then the effetve mass s gven by: m ant d ant ant µ o 4 π l q Note: The alulaton uses real parameters sne we are onsderng the ase for antenna aton eq 8 or, m ant = kg (A sgnfant mass nrease over the mass of the eletron) Note also that the above marosop effetve mass alulaton now s proportonal to the square of the urrent as well as the square of the wavelength If we onsder the ase for purely ndutve or apatve urrent, then the followng apples: Let: θ π and sw ant e θ (+ = Indutve ase) then: sw = amp Then the effetve mass related to purely reatve urrent wave s gven below as: m sw d ant sw µ o 4 π l q Ths s the ase for a nonradatng or, 'antenna' that s eq 9 totally nlosed n a sheld m sw = kg

4 3 Note for a purely reatve urrent wave, the effetve mass s negatve and real Ths mples that a powerful enough wave should be able to reverse the attraton of gravty sne the effetve feld mass s negatve As an example, let us alulate the fore of repulson at the surfae of the Earth for the above mass Frst we establsh related parameters as: G newton m kg Gravtatonal onstant R E m Mean radus of the Earth M E kg Mass of the Earth Then the fore on the surfae of the earth related to the negatve effetve mass alulated above s gven by the equaton below as: G M E m sw F E R E or, eq 10 F E = newton whh s a real and negatve fore of repulson by reason of the standard equaton result s normally postve and one of attraton If we allow for a 0 or 180 degree (0 or π = half wavelength) n theta above, the fore wll be one of attraton sne the effetve mass wll be postve Insertng a theta (θ) of (π/ or -π/ = quarter wavelength) wll yeld an effetve negatve mass Then f the top of a UFO style raft had a real omponent feld whle the bottom had a reatve feld, the top would attrat and the bottom would repel other normal mass The followng s an analyss of how mass s reated by standng wave of urrent n a transmsson lne wth the parameters of Fran De Aquno's test 0f The voltage and urrents along the lne wth respet to tme are gven by the followng equatons below, whh s the sum of the forward and reverse propagatng waves V z ve V plusve e V plusve I z ve e R ω u z V negve e ω u z V negve R e ω u z ω u z z = any pont on lne eq 11 R = lne mpedene eq 1

5 where the V plusve and V negve terms are generally omplex numbers as: 4 V plusve V mplus e θ and V negve V mneg e θ Related to the above equatons, varous related parameters are defned as: f 500 Frequeny (Hz) ω π f Angular frequeny (rad/se) u Propagaton vel of transmsson lne * = See p9 ζ 160 Atual length of lne (m) R 15 ( 15 ) Charaterst Z (adusted) of transmsson lne (ohms) Z L Approx open rut Load mpedane (ohms) V S 10 ( ) Input soure voltage x (Unloaded voltage) β ω u Phase onstant where, β = rad/m Lne length as atual eletral wavelength s gven as: λ u f or, λ = (m) eq 13 The lne rato of atual length to eletral wavelength s: ζ λ = (Beomng lose to 1 / π lke Fran's test) eq 14 The refleton oeffent at the load and the nput s alulated next: Z L R Γ L Γ = Z L R L ( Load ) eq 15 Next we defne z as any pont along the lne Then: Γ ( z ) Γ L e β ( z ζ ) whh s the generalzed voltage refleton oeffent eq 16

6 5 Then the refleton oeffent at the nput to the lne s: Γ ( 0 ) = (Input) eq 17 The voltage standng wave rato (VSWR) s: 1 Γ L VSWR f Γ L 1,, 1 Γ L eq 18 VSWR = (Lne s open-ended at load) The nomnal lne nput mpedane s alulated to be: 1 Γ ( 0 ) Z n R 1 Γ ( 0 ) (Z n below s ad lose to test value) eq 19 Z n = (ohms) Next, we determne the tme doman voltage at the lne nput and at the load: Frst the soure end refleton oeffent s alulated as: Z n R Γ S Γ = Z n R S eq 0 Voltage anywhere along the lne s: Γ L e R V ( z ) V 1 Γ β ζ S Γ L e Z n R S e 1 β ( ζ z ) β z eq 1 Then the nput V(z) s: V ( 0 ) = (= V n ) eq The nput phase s: arg V ( 0 ) f V ( 0 ) 0, deg, 0 = eq 3

7 6 The absolute load voltage s: Note that the voltage at multples of a V ( ζ ) = quarter-wavelength on the lne s eq 4 equal to the nput voltage tmes the VSWR, whh ould rse to extreme The load phase s: values arg V ( ζ ) f V ( ζ ) 0, deg, 0 = eq 5 Sne we have an expresson for the voltage anywhere on the lne, (eq 1), then the urrent at the load and nput may be expressed as: V ( 0 ) (amp) I n I = Z n eq 6 n (amp) V ( ζ ) I L I = Z L eq 7 L The tme averaged load and nput power s gven below as: eq 8 1 P av ( ζ ) 1 P m ( ζ ) V ( ζ ) I L P av ( ) = V ( ζ ) I n P m ( ) = ζ eq 9 ζ The plot of phasor doman voltage and urrent s presented below npts 100 z start 0 z end ζ z 0 npts 1 z z end z start start npts 1 Mag V V z (Voltage magntude along the lne from start to end) arg V z Θ V f V z 0,, 0 (Voltage phase along lne) eq 30 deg

8 7 Voltage magntude along the 1 meter long transmsson lne s: 50 Mag V z Voltage phase along the transmsson lne 0 Θ V z Next the urrent along the transmsson lne may be gven by: Mag I Mag V 1 Γ z R 1 Γ z eq 30 V z eq 31 Equvalent to: I z Z z Where agan: I n = I L =

9 Plot of the reatve Mag I urrent along lne: Im Mag I z arg Mag I The urrent phase s derved as: Θ I f Mag 0 I,, 0 eq 3 deg The urrent phase plot s provded below as: 100 Θ I z Based on the lne amps alulated above at the gven lne length, the effetve feld mass n the transmsson lne an be alulated as: Where the adusted permeablty s: µ ron and µ o Mag I m ζ µ ron µ o 4 π l q 4 π l q (= negatve mass n kg) eq 33 5 m The adusted permeablty s the average permeablty of the ron powder + sheld z

10 The average negatve feld mass n kg along the lne s gven by: eq 34 9 m avg m npts 1 m avg = (In kg) The fnal result n equaton (34) demonstrates that f the torus geometry s not treated as an antenna, but rather as a transmsson lne, then negatve mass ours naturally va the reatve terms as shown above APPENDIX: The followng presents further ratonale of why the torus test an be analyzed as f t s a transmsson lne Below are the new parameters that would allow Jean Lous Naudn's test to approah the workng phase of Fran De Aquno's test V n 10 Input voltage R loss 030 (Old value = ohm) Element ohm resstane, ohms I n 80 Element urrent, amps P n V n I n P n = Power Input, VA The angle related to the VA nput and the resstve loss s: eq 35 θ aos I n R loss θ = deg P n Ths phase angle s lose ompared to the deg of De Aquno test Now that we have the angle above, we an alulate the ndutve reatane related to the nput Z (Ths new phase angle s also lose to 1 / π) Z n 15 Assumed nput mpedane X L Z n sn( θ ) X L = ohm eq 36 L lne X Ḷ π f L lne = henry eq 37 The above ndutane s qute reasonable onsderng the entre element assembly s surrounded wth powdered and pure ron Then we alulate the apatane related to the ndutane by usng standard transmsson lne equatons on the next page (The ndutane may be rased by wrappng the wre around an ron rod)

11 10 C lne L lne Z n C lne = farad eq 38 The resonant frequeny related to the alulated ndutane and apatane s: 1 f r f r = Hz eq 39 π L lne C lne (Frans' test was Hz: very lose to 60 Hz) The phase veloty s alulated next u p 1 L lne ζ C lne ζ u p = m/se eq 40 * NOTE: The above value of phase veloty s the value used on page 4 prevous The large value of apatane s also possble sne n Fran De Aquno's test, the very thn flm of pant around the elements allows for the E feld to transfer to the ondutve ron powder and thus to the nearby opposte element so that t would appear as f the elements were very lose to eah other The lne R on page 4 was adusted to reflet the lne Z of the elements so they would represent an atual transmsson lne Ths s also a very possble value Fnally, the nput voltage on page 4 represents the unloaded value and s halved when onneted to the lne whh yelds the same nput voltage as gven by the orgnal torus test parameters Ths s shown by the voltage plot on the top of page 7 Conlusons: Treatng Fran De Aquno's test torus as a transmsson lne yelds a negatve feld mass dretly wthout resortng to the assumpton that the ron atoms n the sheld must absorb energy To absorb energy, the atoms must onvert that energy to vbratory moton whh translates nto heat buld whh means that they wll be subet to temperature rse Fran De Aquno assures us that the torus and sheld do not get hot Therefore the power s reatve and the transmsson lne equatons above use that property to arrve at the negatve feld mass results n a straghtforward manner I suggest that ths analyss more aurately desrbes the results of Fran De Aquno's test muh better than assumng that the aton of antenna radaton ours The mehans of antenna radaton nvolve non-reatve radaton, e, real power NOTE: The orgnal work that supports ths onept s hapter 7 of my book, "Eletrogravtaton As A Unfed Feld Theory" Ω

12 ADDITIONAL COMMENTS: The reason that Jean Lous Naudn's test dd not replate Fran De Aquno's test may be that the osne of the phase angle between the volt-amperes and the resstve power should be lose to the rato of 1 / (p) Ths would ensure that the alulated frequeny related to the ndutane and apatane parameters for ordnary resonane are lose to the atual nput frequeny Then lke the aton of a pendulum, the mass-wave wll be "kked" nto ts next yle at the proper tme I am remnded of when the orbts of the atoms were found to have standng waves related to the alulated DeBrogle wavelength and that those wavelengths had to be a whole number multple of some number n It was found to be m v r = n h / p, or the angular momentum was equal to n h / p The mass wave lkely ded out n Jean Lous test sne the phase angle was not lose to the 1 / p whh would allow t to ft wthn the 'yle' of operaton properly Lowerng the opper loss wll boost the nput urrent whle at the same tme t wll sustan the standng wave as shown above -- Jerry E Bayles