Analysis of Leon Dragone's paper by William Alek - Intalek

Analysis of Leon Dragone's,"Energetics of Ferromagnetism"By William S. AlekINTALEK, INC.November 2002Version 5a

Given:Tstart :=0 ⋅ secTend :=0.1 ⋅ secT :=Tstart , 0.0005 ⋅ sec .. TendV :=950 ⋅ voltR :=20000 ⋅ ohmLc :=2000 ⋅ HIstart :=0 ⋅ amp⎛ Tend1 − exp⎛−R⋅ ⎞⎞⎝ ⎝ Lc ⎠⎠Iend := V ⋅Iend = 0.03AR1t := 0.050 ⋅ secf := f = 10Hz2t⎛t1 − exp⎛−R⋅ ⎞⎞⎝ ⎝ Lc⎠⎠i := V ⋅i = 0.019AR

Compute Current Through Coil:( ) := Vi T⋅⎛⎝T1 − exp⎛−R⋅⎝ LcR⎞⎞⎠⎠0.0350.030.0250.02Ampsi ( T)0.0150.010.0050 0.02 0.04 0.06 0.08TSecFigure 2

Compute Voltage Across Resistive Portion of Coil:( ) := R ⋅ i ( T)Vr T700600500400VoltsVr( T)3002001000 0.02 0.04 0.06 0.08TSecFigure 3

Compute Power Dissipated By Resistive Portion of Coil:( ) := R ⋅ ( i ( T)) 2Pr T2015WattsPr ( T)1050 0.02 0.04 0.06 0.08TSec( ( )) 2Figure 4Pr := R ⋅ i TendPr = 18.031W

Compute Voltage Across Coil:IDOT( T)( )d:=dT i T( ) := Lc⋅IDOT( T)Vc T1000800VoltsVc( T)6004002000 0.02 0.04 0.06 0.08TSecFigure 5( ) 2IendEc := Lc⋅Ec = 0.902 J2

Given Atomic Current, Compute Time-Reversed Mutual Energy:Ia := 100000 ⋅ IendIa = 3002.573Ak := 1.0Lm := 0.0001 ⋅ µH( )2Em := Lm ⋅IaEm = 0 J2( ) 1 2M := k ⋅ Lc⋅LmM = 0HEmutual := M ⋅ Ia ⋅ IendEmutual = 0.04JCompute Time-Reversed Mutual Voltage:Vrev T( ) := M ⋅ Ia( )( )IDOT T⋅i T1000800600VoltsVrev( T)4002000 0.02 0.04 0.06 0.08TSecFigure 6

Compute Total Voltage Across Coil:Vin T( ) := Vc( T) Vr( T)( )+ − Vrev T1000800600VoltsVin( T)4002000 0.02 0.04 0.06 0.08TSecFigure 7

Compute Total Power Dissipated Coil:t :=0.000 ⋅ sec( )( )( ( )) 2Pin := Lc ⋅ i t ⋅ IDOT t + R ⋅ i t − M ⋅ Ia ⋅ IDOT tPin = −0.638W( )t :=0.001404⋅sec( )( )( ( )) 2Pin := Lc ⋅ i t ⋅ IDOT t + R ⋅ i t − M ⋅ Ia ⋅ IDOT tPin = 0WPin T( ) := Lc⋅i( T)IDOT( T)( ( )) 2⋅ + R ⋅ i T − M ⋅ Ia ⋅( )IDOT( T)302520WattsPin ( T)151050 0.02 0.04 0.06 0.08TSecFigure 8

Compute Total Resistance of Coil:t :=0.0010⋅sec( ( )) 2( )Vin tRin := Rin = −816888.855ΩPin tRin( T)( Vin( T)) 2:=Pin T( )5 . 10 5 Sec4 . 10 53 . 10 5OhmsRin( T)2 . 10 51 . 10 50 0.02 0.04 0.06 0.081 . 10 5TFigure 9

Compute Total Energy of Coil:t :=0.050 ⋅ sect⌠Ein := ⎮ Pin T dTEin = 0.456 J⌡Tstart( )Ein( T)⌠:= ⎮⌡TTstartPin( T)dT0.0080.006JoulesEin ( T)0.0040.0020 0.002 0.004 0.006 0.0080.002TSecFigure 10Vr := R ⋅ IendVr = 600.515V( )IDOTend :=ddTend i TendIDOTend = 0.175 A sIDOTendVrev := M ⋅ Ia ⋅Vrev = 7.815VIendVc := Lc ⋅ IDOTendVc = 349.485V

Vin := Vc + Vr − VrevVin = 942.185V( ) 2Pin := Lc ⋅ Iend ⋅ IDOTend + R ⋅ Iend − M ⋅ Ia ⋅ IDOTendPin = 28.29W( ) 2VinRin := Rin = 31379.266ΩPin

Compute Time-Reversed Impulse Current:Vrev T( ) := M ⋅ Iai_neg( T):=⎡⎢⎣( )( )IDOT T⋅i T( V + Vrev( T))R⎤⎥⎦0.090.08Ampsi_neg ( T)0.070.060.050.040 0.02 0.04 0.06 0.08TSecFigure 11

Compute Time-Forward Current:i_pos T( ) := V⋅⎛⎝T1 − exp⎛−R⋅⎝ LcR⎞⎞⎠⎠0.0250.02Ampsi_pos( T)0.0150.010.0050 0.02 0.04 0.06 0.08TSecFigure 12

Add Time-Reversed Impulse Current to Time-Forward Current:( ) := i_neg( T) + i_pos( T)i T0.0750.07Ampsi ( T)0.0650.060.0550 0.02 0.04 0.06 0.08TSecFigure 13___________COMMENTS:Referring to figure 1, the moment S1 closes (t = 0 sec), a "negative energy" impulse function occurs.This is also shown in figures 6 and 7 as a voltage impulse function."Negative power" is produced during the time duration as shown in figure 8."Negative energy" is sent back to the source as shown in figure 10.Because the coil is resistive, it will dissipate "positive energy" as shown in figure 5. However, thetotal power (Pin) is negative as shown in figure 8.Figure 9 is of special interest because it exhibits a property I call "supernegativity".

lse function occurs.