(Nuclear fusion engine that generates alternating current electricity by direct convrsion)
With success I ceased updating but the moment came when I turned the vacuum pump off and then threw it away. See below.
All files (C) christopher strevens 2017
For contact use: info@transfusor.co.uk
The apparatus is a coil of copper wire tuned with a capacitor to 50 Hz. Inside the winding, along the principle axis is a borosilicate glass tube containing air at reduced pressure (down to 20 Tor) with a corona wire carrying a high frequency high voltage that ionises the gas in the tube. I hope to use natural hydrogen or deuterium later.
The winding is excited by 50 Hz AC from a mains transformer the circulating current induces a current in the ionised gas that is repelled by the current in the winding to make a high vector pressure zone near the axis. The high current in the ionised gas also heats the gas.
As a result the pulsating pressure causes pulses of fusion reactions (Making pressure pulses at twice the exciter pressure) that by interacting with the magnetic field adding to the current and these photon emissions are picked up by an additional winding of thicker wire would around the outside of the inner winding. The potential ratio is the ratio of the lengths of wire so the plasma turn carrying ac induces ac in the external winding for the external load.
Experiments indicate that the pulsating neutron and neutrino source is reduced by the load on the external coil indicating that the mechanism that makes neutrons changes to an electrical current instead somehow.
The algorithm for a coil carrying ac.
See Below
Here the vector pressure is calculated on a region of ionised gas
inside the coil carrying ac because the ac induced in the plasma is
repelled by the current sheet in the winding.
The plasma current heats the gas through ohmic heating and this heat and pressure causes fusion of light nuclei within the winding.
The power depends on the plasma pressure squared but the temperature stabilises around 2 MK for deuterium.
The algorithm above predicts around a maximum of 200KW for a quite small desktop unit
The algorithm indicates that the winding has to be short and fat. This concentrates the magnetic force.
The Deuterium needs to be drawn through slowly at low pressure. This keeps the power down and makes it easy to ionise. The yield is kept down in the event of an explosion.
I think a nuclear shield could be polythene, water or high density polystyrene.
The phenomena depends on the plasma turn that is oscillating having an inductance and current at 50 Hz and therefore developing an emf. This is developed around the turn of circumference less than a mm say 0.1mm inducing an emf in the outer winding (the regulator prevents the inner winding being efected) and inducing the length of the wire (say 5000 m) as photons hit it building up the potential. If the plasma turn develops 1 volt then the terminal potential will be 50,000,000 volts so with a current capacity of the winding being 3 a the power capacity will be 150MW.
Here is my algorithm.
'Program to find the inductance of a multilayer coil by summing the
inductance of each layer and the effect of mutual inductance
'Inductance = 4.piE7.A.n^2/l
'to find the final diameter of a multilayer coil and the length of wire
'each layer adds thickness to layer so first work out turns per layer
[Start]
let Pi=3.1415926
'for room temp
Let ResistivityOfCopper=1.68E8
Let DensityOfCopper=8.94E3
'Kg per m^3
'for superconductor
'Let ResistovityOfCopper=0
'for LN temp
'Let ResistivityOfCopper=2.647E9
Let mu=4E7*Pi
Let e0=8.85E12
[InductanceLoop]
input "diameter of wire in mm " ; DiameterOfWire
'DiameterOfWire=0.170
'Print "Diameter of Wire ";DiameterOfWire
'Input "Diameter Of Lumen mm "; DiameterOfLumen
DiameterOfLumen=0
'input "Thickness of Insulation mm ";ThicknessOfInsulation
ThicknessOfInsulation=0
input "diameter of former in mm ";DiameterOfFormer
'Let DiameterOfFormer=60
'Let LengthOfFormer=60
input "length of former in mm " ;LengthOfFormer
'Print "Diameter Of Former mm ";DiameterOfFormer
'Print "Length Of Former mm ";LengthOfFormer
'Input "Resonant Frequency Hz ";ResonantFrequency
let DiameterOfWire=DiameterOfWire/1000
Let DiameterOfLumen=DiameterOfLumen/1000
Let ThicknessOfInsulation=ThicknessOfInsulation/1000
Let DiameterOfFormer=DiameterOfFormer/1000
Let LengthOfFormer=LengthOfFormer/1000
[TurnsLoop]
input "number of turns "; NumberOfTurns
'Let NumberOfTurns=13000
'Print "Number of Turns ";NumberOfTurns
let
TurnsPerLayer=LengthOfFormer/(DiameterOfWire+2*ThicknessOfInsulation)
let NumberOfLayers=NumberOfTurns/TurnsPerLayer
Let
ThicknessOfWinding=NumberOfLayers*(DiameterOfWire+2*ThicknessOfInsulation)
Let DiameterOfWinding=DiameterOfFormer+2*ThicknessOfWinding
'length of wire for each layer is the layer diameter multiplied by Pi
LayerN=0
LengthOfWire=0
'Let TotalInductance=0
'Let TotalCapacitance=0
LayerDiameter=DiameterOfFormer
[ForLoop]
LayerN=LayerN+1
Let
LayerDiameter=LayerDiameter+2*(DiameterOfWire+2*ThicknessOfInsulation)
Let LengthOfWire=Pi*LayerDiameter*TurnsPerLayer+LengthOfWire
'let
LayerInductance=mu*(Pi*(LayerDiameter/2)^2)*TurnsPerLayer^2/LengthOfFormer
'Let
LayerCapacitance=e0*Pi*LayerDiameter*((DiameterOfWire+2*ThicknessOfInsulation)/(DiameterOfWire+2*ThicknessOfInsulation))*TurnsPerLayer
'Let
InterLayerCapacitance=e0*Pi*LayerDiameter*LengthOfFormer/(DiameterOfWire+2*ThicknessOfInsulation)
'Let TotalInductance=LayerInductance+TotalInductance
'Let
TotalCapacitance=LayerCapacitance+InterlayerCapacitance+TotalCapacitance
'Let TotalCapacitance=LayerCapacitance+TotalCapacitance
if LayerN<NumberOfLayers then goto [ForLoop]
Let
ResistanceOfCoil=ResistivityOfCopper*LengthOfWire/((Pi*(DiameterOfWire/2)^2)(Pi*(DiameterOfLumen/2)^2))
'Let
TotalInductance=TotalInductance*(NumberOfLayers/2)*(NumberOfLayers1)
'Print "Inductance by sum ";TotalInductance
Print "Turns per layer ";TurnsPerLayer
Print "Number of Layers ";NumberOfLayers
Print "Diameter of coil ";DiameterOfWinding*1000;" mm"
Print "Length of Wire ";LengthOfWire; " Meters"
Print "resistance of winding ";ResistanceOfCoil; " Ohm"
'input "Resistance of Coil by measurement "; ResistanceOfCoil
'Print "Inductance of Multilayer Coil by sum ";TotalInductance;" Henry"
'Print "Capacitance of Mutilayer Coil ";TotalCapacitance*1E6; "
microfarads"
'input "1=exit 2=InductanceLoop ";OK
'if OK=2 then goto [InductanceLoop]
'to calculate dynamic impedance
'dynamic impedance is L/Cr where r is the resitance of the coil.
'let
TotalInductance=(0.8/1000)*((((DiameterOfWinding+DiameterOfFormer)/254*2)^2)*NumberOfTurns^2)/(6*(DiameterOfWinding+DiameterOfFormer)/254*2+9*LegthOfFormer/254+10*(DiameterOfWindingDiameterOfFormer)/254*2)
'print "total inductance by complex formula ";TotalInductance
'let TotalInductance=7.6
let
TotalInductance=mu*Pi*((DiameterOfFormer/2)^2)*(NumberOfTurns^2)/LengthOfFormer
Print "Inductance of winding by simple formula ";TotalInductance
input "new turns 1=y 2=n ";ok
if ok=1 then goto [TurnsLoop]
input "Inductance by measurement ";TotalInductance
'input "1 for exit 2 for new turns ";ok
'if ok=2 then goto [TurnsLoop]
'Let TotalInductance=7.6
'let TotalInductance=4.76
'Print "inductance by measurement ";TotalInductance
[ResonanceLoop]
input "Resonant Frequency Hz "; ResonantFrequency
'f=1/(2pisqr(LC) so c=1/f^2*4Pi^2L
let Capacitance=1/((ResonantFrequency^2)*4*(Pi^2)*TotalInductance)
Print "capacitance by resonance formula ";Capacitance*1E6;" uF"
'let ExternalCapacitance=1.5E6
[CapacitanceLoop]
'let VaractorCapacitance=11.9E6
'input "capacitance of Main capacitor (uF) "; ExternalCapacitance
'Let ExternalCapacitance=ExternalCapacitance/1E6
'input "Additional Series capacitance (uF)";SeriesCapacitance
'Let SeriesCapacitance=SeriesCapacitance/1E6
'input "ok1";ok
'Let Capacitance=1/(1/ExternalCapacitance + 1/SeriesCapacitance)
'Let Capacitance=TotalCapacitance+ExternalCapacitance
'Let Capacitance=ExternalCapacitance
'input "Capacitance (F) "; Capacitance
'input "ok2"; ok
Print "Capacitance ";Capacitance*1E6;" uF"
input "correct ? 1=n 2=y ";OK
if OK=1 then goto [CapacitanceLoop]
let DynamicImpedance = TotalInductance/(Capacitance*ResistanceOfCoil)
print "Dynamic impedance ";DynamicImpedance
'print "dynamic impedance  Infinite"
'input "ok3";ok
'program to calculate resonant frequency
'f=1/(2.pi.sqr(l.c))
'Print "Extermal Capacitance ";ExternalCapacitance
'Print "Additional Series Capacitance ";VaractorCapacitance
'input "Inductance of coil "; Inductance
let ResonantFrequency=1/(2*Pi*(sqr(TotalInductance*Capacitance)))
print "Resonant frequency is "; ResonantFrequency
input "1 for exit 2 for loop for another capacitance "; ok
'input "1 for exit 2 for another turns value ";ok
'if ok=2 goto [TurnsLoop]
if ok=2 then goto [ResonanceLoop]
'to calculate Q from Q= 2.Pi.f.L/R
Let Q=2*Pi*ResonantFrequency*TotalInductance/ResistanceOfCoil
Print "Q " ;Q
'Print "Q infinite"
'to calculate the total mass of copper in the winding Area times lengh
times density
let
TotalWindingMass=Pi*(((DiameterOfWire/2)^2)((DiameterOfLumen/2)^2))*LengthOfWire*DensityOfCopper
Print "Winding Mass ";TotalWindingMass;" Kg"
[PotentialLoop]
Input "Rms Potential ";RmsPotential
'to calculate slow wave
Print "Half Wavelength "; LengthOfFormer/2
Print "Velocity "; LengthOfFormer*ResonantFrequency
Print "Peak Potential "; 1.4*RmsPotential
Let
PeakCurrent=1.4*RmsPotential/(SQR((TotalInductance*2*Pi*ResonantFrequency)^2+ResistanceOfCoil^2))
Print "Peak Current ";PeakCurrent
print "Rms Current ";PeakCurrent/1.4
Let PeakAmpereTurns=PeakCurrent*NumberOfTurns
Let PeakAmpereTurnsPerMeter=PeakAmpereTurns/(LengthOfFormer/3)
Print "Peak Ampere Turns "; PeakAmpereTurns
Print "Plasma Current "; PeakAmpereTurns
Let PlasmaCurrentDensity=PeakAmpereTurns/((1.5E3)*LengthOfFormer/3)
Print "Plasma Current Density ";3*PlasmaCurrentDensity; " Amp per
meter^2, near centre"
Let PressureOnPlasma=
((3*PlasmaCurrentDensity*PeakAmpereTurnsPerMeter)*mu)/((DiameterOfFormer(6E3))/2)
Print "Ion Pressure of Plasma ";PressureOnPlasma; " Pa"
Print "Power lost through resistance ";
ResistanceOfCoil*((PeakCurrent^2)/1.4); " Watt"
'Let
ReactanceOfExternalCapacitance=1/(2*Pi*ResonantFrequency*ExternalCapacitance)
'Let
ReactanceOfSeriesCapacitance=1/(2*Pi*ResonantFrequency*SeriesCapacitance)
'Let
PotentialAcrossSeriesCapacitance=ReactanceOfSeriesCapacitance*PeakCurrent
'Let
PotentialAcrossExternalCapacitance=ReactanceOfExternalCapacitance*PeakCurrent
'Power = 0.5 watt/cc/kPa^2 10torr=133 Pa
Let
PowerGenerated=(0.5*(LengthOfFormer/3)*(3/1000)^2*2*Pi*((PressureOnPlasma)/1000)^2)*100^3
Print "Power Generated if Deuterium ";PowerGenerated;" Watt"
Print "Power Generated by natural hydrogen "; PowerGenerated*24E9;"
Watt"
'Print "Potential Across Additional Series Capacitor
";PotentialAcrossSeriesCapacitance;" Volt"
'Print "Potential Across Main Capacitor
";PotentialAcrossExternalCapacitance;" Volt"
input "1 exit 2 loop "; ok
if ok=2 then goto [PotentialLoop]
input "1 exit 2 again " ;ok
if ok=2 then goto [Start]
End
Photons emitted from the plasma turn are not absorbed by the inner
winding because it is regulated and are then collected by the outer
winding.
The plasma loop is very small with a length of its circumference and
the length of the catcher is the length of the winding so the voltage
magnification is the ratio of these lengths. so about a million.



Transfusor 
Transfusor 
Transfusor 
13/02/2014
I a now reworking the design with a one coil system:
Having seen the Kapadnaze system on video, I realize that the single coil system is adequate. I wound 2064 T of 0.9 mm wire on to a 44 mm former 44 mm long.
diameter of wire in mm 0.9
diameter of former in mm 44
length of former in mm 44
number of turns 2064
Turns per layer 48.8888889
Number of Layers 42.2181818
Diameter of coil 119.992727 mm
Length of Wire 552.121635 Meters
resistance of winding 14.5803905 Ohm
Inductance of winding by simple formula 0.18500003
new turns 1=y 2=n 2
Inductance by measurement 273E3
Resonant Frequency Hz 50
capacitance by resonance formula 37.1139879 uF
Capacitance 37.1139879 uF
correct ? 1=n 2=y 2
Dynamic impedance 504.493843
Resonant frequency is 50.0
1 for exit 2 for loop for another capacitance 1
Q 5.88224837
Winding Mass 3.14012674 Kg
Rms Potential 100
Inner Diameter Of Tube mm 33
Half Wavelength 0.022
Velocity 2.2
Peak Potential 140.0
Peak Current 1.60926929
Rms Current 1.14947806
Peak Ampere Turns 3321.53181
Plasma Current 3321.53181
Plasma Current Density 41176014.2 Amp per meter^2, near centre
Ion Pressure of Plasma 2.13058304e9 Pa
Power lost through resistance 26.9710941 Watt
Power Generated if Deuterium 8.94470407e13 Watt
Power Generated by natural hydrogen 2146728.98 Watt
1 exit 2 loop 2
Rms Potential 300
Inner Diameter Of Tube mm 33
Half Wavelength 0.022
Velocity 2.2
Peak Potential 420.0
Peak Current 4.82780786
Rms Current 3.44843419
Peak Ampere Turns 9964.59543
Plasma Current 9964.59543
Plasma Current Density 1.23528042e8 Amp per meter^2, near centre
Ion Pressure of Plasma 1.91752474e10 Pa
Power lost through resistance 242.739847 Watt
Power Generated if Deuterium 7.24521029e15 Watt
Power Generated by natural hydrogen 1.73885047e8 Watt
1 exit 2 loop 1
1 exit 2 again 1
The low pressure hydrogen has to be ionized as before. A spark might do it from a capacitor charged to 7KV and discharged when the gas breaks down. The system is, again tuned to 50 Hz with a 37.1 uF capacitor made by series and parallel capacitors.
If 2000 kHz is used the potential across the coil rises to 4KV and so a separate exciter is no longer needed just a charger and changover system….?
diameter of wire in mm 0.9
diameter of former in mm 44
length of former in mm 44
number of turns 2064
Turns per layer 48.8888889
Number of Layers 42.2181818
Diameter of coil 119.992727 mm
Length of Wire 552.121635 Meters
resistance of winding 14.5803905 Ohm
Inductance of winding by simple formula 0.18500003
new turns 1=y 2=n 2
Inductance by measurement 273E3
Resonant Frequency Hz 2000
capacitance by resonance formula 0.23196242e1 uF
Capacitance 0.23196242e1 uF
correct ? 1=n 2=y 2
Dynamic impedance 807190.149
Resonant frequency is 2000.0
1 for exit 2 for loop for another capacitance 1
Q 235.289935
Winding Mass 3.14012674 Kg
Rms Potential 70000
Inner Diameter Of Tube mm 33
Half Wavelength 0.022
Velocity 88.0
Peak Potential 98000.0
Peak Current 28.5660143
Rms Current 20.4042959
Peak Ampere Turns 58960.2536
Plasma Current 58960.2536
Plasma Current Density 7.30912234e8 Amp per meter^2, near centre
Ion Pressure of Plasma 6.71336592e11 Pa
Power lost through resistance 8498.46359 Watt
Power Generated if Deuterium 8.88075081e18 Watt
Power Generated by natural hydrogen 2.13138019e11 Watt
1 exit 2 loop 2
Rms Potential 4000
Inner Diameter Of Tube mm 33
Half Wavelength 0.022
Velocity 88.0
Peak Potential 5600.0
Peak Current 1.63234368
Rms Current 1.16595977
Peak Ampere Turns 3369.15735
Plasma Current 3369.15735
Plasma Current Density 41766413.4 Amp per meter^2, near centre
Ion Pressure of Plasma 2.19211948e9 Pa
Power lost through resistance 27.7500852 Watt
Power Generated if Deuterium 9.4688555e13 Watt
Power Generated by natural hydrogen 2272525.32 Watt
1 exit 2 loop 2
Rms Potential 7000
Inner Diameter Of Tube mm 33
Half Wavelength 0.022
Velocity 88.0
Peak Potential 9800.0
Peak Current 2.85660143
Rms Current 2.04042959
Peak Ampere Turns 5896.02536
Plasma Current 5896.02536
Plasma Current Density 73091223.4 Amp per meter^2, near centre
Ion Pressure of Plasma 6.71336592e9 Pa
Power lost through resistance 84.9846359 Watt
Power Generated if Deuterium 8.88075081e14 Watt
Power Generated by natural hydrogen 21313801.9 Watt
1 exit 2 loop 1
1 exit 2 again 1
Winding the coil.
1422014
The 8  stage vacuum pump:
15/02/2014
In Transfusor 3 build 3 once the outer coil develops power the inner winding may be turned off as this is the exciter. The outer winding, also tuned to 50 Hz, needs to be regulated. I suggest that it is fed to the primary of a four hundred volt transformer with a secondary of 240 volt and the two back to back zeners of 200 volt shorting a 200 volt winding. This should keep the output at 240 volt. 400 volt transformers have both 240 and 110 volt windings so the regulator could be this diac at 100 volt across this winding. I do not know much about voltage regulators.
Transfusor with tuned outer winding and recirculation.
Needs regulator and offswitch.
On consideration the off switches must break the circuit, not short it, but I am unsure.
16/02/2014
Transfusor 3 radical new design.
This is a much simpler version that requires no additional supplies.
I was able to light the
fluorescent tube with this arrangement.
A simple transistor circuit would be needed to start it powered by a
battery and once started the reaction is self sustaining.
There ae 2064 turns of 0.9 mm wire on this coil and I use a 4.5 volt battery here. The former is a scrap bobbin from an electrician on a building site who sold it via B&Q for 50 pence. The wire was £270.
Its resistance is 17 ohms so to make 2 amps a 24 volt battery is needed possibly from a motorcycle. Otherwise 16 U2.
The capacitor is a 10,000 volt 0.1 µF.
The new version generates its own high tension of 4000 volts and is tuned to 2000Hz and the ends of the coil are connected to copper pipes each side of the tube so the potential is applied across along the tube to ionise the gas and then the current pulse arrives compressing and igniting the ionised gas to induce current again. I did try to separate the current and high voltage but it ligts a fluorescent tube in the coil connected like that and the quartz tube version indicated that it worked but the battery is flat now. The vacuum is down to 10 Tor. It may be low enough.
The circuit is resonant at about 1000 and generates 4000 volt to ionize the gas. It is started by one make and break of a 3 amp 30 volt supply. In the cavity is a quartz tube filled with deuterium or in my case air.
I need a way of making and then breaking the start up current. The 4.5 v battery went flat rapidly and I was unable to continue with testing.
I am working out the details.
Here is the calculation.
diameter of wire in mm 0.9
diameter of former in mm 36
length of former in mm 58
number of turns 2064
Turns per layer 64.4444444
Number of Layers 32.0275862
Diameter of coil 93.6496552 mm
Length of Wire 444.962609 Meters
resistance of winding 11.7505422 Ohm
Inductance of winding by simple formula 0.93949857e1
new turns 1=y 2=n 2
Inductance by measurement 232E3
Resonant Frequency Hz 1000
capacitance by resonance formula 0.10918231 uF
Capacitance 0.10918231 uF
correct ? 1=n 2=y 2
Dynamic impedance 180833.04
Resonant frequency is 1000
1 for exit 2 for loop for another capacitance 1
Q 124.053762
Winding Mass 2.53067241 Kg
Rms Potential 4000
Inner Diameter Of Tube mm 24
Half Wavelength 0.029
Velocity 58.0
Peak Potential 5600.0
Peak Current 3.8415463
Rms Current 2.74396164
Peak Ampere Turns 7928.95155
Plasma Current 7928.95155
Plasma Current Density 1.02529546e8 Amp per meter^2, near centre
Ion Pressure of Plasma 8.80677023e9 Pa
Power lost through resistance 123.86312 Watt
Power Generated if Deuterium 1.06554623e15 Watt
Power Generated by natural hydrogen 25573109.6 Watt
1 exit 2 loop 1
1 exit 2 again 1
Demo and test of Transfusor 3
http://youtu.be/jBpVq4r2DgA
The old design is now obsolete.
10/02/2014




vacuum of 10 Tor reached  basic unit  general view  Circuit for ignition... 
I am building this system using air as the test gas.
The device is an induction coil that is part of a Colpitts oscillator with a tube containing hydrogen or air with electrodes taking the high voltage from the induction coil along the gas column.
20/2/2014
The early success then reducing response is explained as the deuterium in the air was used up!
The batteries are suspect also but the 4.5 Volt unit was sufficient.
Since the DC resistance is 17 Ohm this battery would pass 26 mA. The bump I heard and was the deuterium igniting the electric smell (ozone and nitric oxide) came from the spark gap regulator at 4000 volt and so 3 amp was flowing at that point.
Success!
However it needs a supply of Deuterium not air.
I have just refreshed the air and it is very smelly. I need to pump it out of the house.
I suggest a mixture of deuterium and hydrogen be used to keep the power down. Recycling will make later time mixture contain more helium (inert) and lass hydrogen. The helium filter will reduce the proportion of helium.
Research to find the best mixture is needed. Unfortunately helium reacts also but at lower cross section.
In my early experiments, many years ago, we found a carbon deposit on the inside of the tube indicating that nuclear synthesis had occurred.
I am to rebuild version 2 to make it “self exciting” so the separate plasma generator is not needed. The same number of turns are needed but the Transfusor will operate at 100 Hz.
04012014 9:47
The latest version has problems with the ignition Hartley oscillator, it won't oscillate. I am seeking advice,
04/03/2014
Transfusor 3 with Hartley oscillator that won't oscillate.
10/3/2014 17:24:06
I bought a commercial circuit for a high voltage AC generator and Brian and I modified it for our setup with our transformer.
Here is the modified circuit
The experiments we did showed no signs of oscillations but the transistor became quite warm. Brian is working on the circuit.
I did try a simple resistor and switch to put a pulse of current through the 6 volt winding and caused the oscilloscope and the computer to crash with lost files and the power supply became damaged by the EMP. This may account for the circuits failure to oscillate.
I am winding off the outer coil of Transfusor 2 to be used for the 50 Hz version. I have ordered 1Kg of 0.2mm wire also for the 50Hz version.
11/03/2014 14:25:03
This is another possible ignition circuit. it puts a 5 amp current pulse through the 3 volt winding and the kickback then ignites the Transfusor.
The 71uH choke prevents high voltage spikes getting into the power supply but I will use a battery initially.
11/03/2014 16:47:05
I did some measurements and found that the 1:2 step up winding actually reduced the output potential. 20 volts in and less than one volt out.
The frequency of operation is far too low and the capacitor value it too high.
I will try the natural frequency of operation with no capacitor.
I have now cut out the capacitor. All inductors have self capacitance.
The frequency was too high to measure.
I put the capacitor back and the resonant frequency was 1KHz and the voltage gain was: 23. The output impedance of the line was 600 ohm so the current was 26mA. The output potential was 350mV
So with 5 amp the output voltage will be 67 volt.
When I removed the capacitor the coil peaked at 1.13 volt at 26 KHz and the primary read 28mV gain: 42.
So air coils are very different from coils made on ferrite. I cannot use ferrite because I have a gas core.
This coil will resonate at 1000 Hz but I suppose the coupling is poor.
At 1000 hz, top is output from 6 T coil and bottom is across the ends of the main 2064T coil. The feed is from the 3 turn primary
11/03/2014 18:44:4
So big gain at resonance. The input is less than 12 mV and output 3.33 volt. Gain 277. So at 5 amp the output may rise to 250000 volt.
That does explain the spike last time.
12/03/2014 01:07:33
Startup circuit not involving oscillator:
Here the current builds up to 1.2 A or 2.5 or 5.1 A with push button "down" then cut off by the pushbutton "up". The resulting kick back gives the high potential down the gas column, ionising ther gas and the circulating current ignites the resulting plasma. (Transfusor action). The primary does not experience high potentials and so is easy to control, the secondary turns this spike into a possible 250,0000 volt spike that ionises the gas turning it into plasma.
12/03/2014 01:07:33
Trace of switch off with 1K resistor (12 mA)
So for 12 mA it is 100 mvolt so 5 amp will give 41.6 volt. I depends on the time it takes to cut off the current as well as the current. There is a tail in as well as a tail out, here. The downward spike on the input (channel 2) may be seen as I turned the switch off.
12/03/2014 07:55:06
Trace with 10 ohm resistor in series with 12 volt supply.
About 96 volt peak across the main coil. The reading on channel 1 should be 5 volts as it was on x1.
This trace is from driver coil that connects to collector.
12/03/2014 09:07:00
Here the collector coil is taking 1 amp pulse and at cut off this output is found across the main coil.
12/03/2014 09:07:00
With the chamber pumped down to about 20 Tor (air).
12/03/2014 20:08:37
I found that the neutron count changed with air flow and pressure inside the Transfusor.
http://youtu.be/QKAvSLBSvgI
The neutron count is below, the potential across the main coil above.
Notice the peaks of the neutron count occur at the crossover point of the potential curve which is the point of maximum current.
I found it was best between 700 and 600 Tor.
13/03/2014 04:59:13
Since this "neutron signal" is still there this morning it must have another cause, like the sun or mains born interference. I will therefore return to vacuum and ionised gas techniques.
The Pulse technique was seen by the PSU as a short circuit so it "crowbard down" so to use the pulse technique I must have a battery.
13/03/2014 04:59:13
I managed to get a trace of the FB coil from the collector coil using my signal generator.
We have to set up the oscillator from this.
I suggest a Vellman mono preamplifier module that has a gain of 50 to increase the potential to the power amplifier transistor base to 4.75V
13/03/2014 08:06:51
A capacitor bank between the choke and the resistors may improve the pulse so as not to need a battery. 3 x 4700uF, 35 volt electrolytic would give 14.1 mF. This would improve the rise time, making for a higher spike potential. I will need 20 amp impulse to get 4000 volt so a resistor in a 20 am psu would not be needed, I think a car battery...
13/03/2014 08:59:54
I found the current in the coil was 34.5 mA when the FB was 20.1 mV
I used a 4 ohm resistor in the collector circuit (replaced by signal generator).
The potential across this resistor was 138 mV
Channel 2 is across the resistor in the "collector" circuit and channel 1 is across the Feed Back (FB) coil. The potential across the main coil is 11 volts. (All peak to peak).
So when the collector current is 34.5 mA the main coil has 11 volts across it: and when the collector current is 5 amp the main coil will have 1570 volts across it, It will need 12 amp in the collector current for the output to be 4000 volt across the main coil. So the base current will be 636 mA
15/03/2014 20:30:20
I was unable to set up the amplifier. I am stupid, probably as a result of psychiatric toxin forced on me.
I have survived being run through several times with a rapier and other atrocities. I was attacked for religious reasons.
And because the church has billions in petrol shares!
So far I have not had any success with these engines. There have been a series of "Just a coil" speeches that have been used since Tesla's Murder. I am sick with a rapier wound to the chest. I am violently assaulted by Christians and the British
Well it is Just a coil....
I a now looking at the ignition oscillator:
This is the amplifier with the base driven from signal generator and the collector on a 10 Ohm Load. The lower trace is the collector output.
This is with the collector connected to the driver winding. The lower trace is the collector potential.
Chris.
29/03/2014 19:51
I am recovering from a knife attack and cold and news that I have an aneurism in my aorta that will kill me suddenly and painfully when it bursts.
Transfusor 4. Another 50 Hz version with a bigger coil and smaller capacitor. The ends of the windings are at 4000 V along the deuterium filled tube that fits down the axis of the coil.
Here is the design estimate:
diameter of wire in mm 0.2
diameter of former in mm 30
length of former in mm 40
number of turns 31402
Turns per layer 200
Number of Layers 157.01
Diameter of coil 92.804 mm
Length of Wire 6135.15336 Meters
resistance of winding 3280.8384 Ohm
Inductance of winding by simple formula 21.8976176
new turns 1=y 2=n 2
Inductance by measurement 30
Resonant Frequency Hz 50
capacitance by resonance formula 0.33773729 uF
Capacitance 0.33773729 uF
correct ? 1=n 2=y 2
Dynamic impedance 27074.3102
Resonant frequency is 50.0
1 for exit 2 for loop for another capacitance 1
Q 2.87267358
Winding Mass 1.72310922 Kg
Rms Potential 4000
Half Wavelength 0.02
Velocity 2.0
Peak Potential 5600.0
Peak Current 0.56115059
Rms Current 0.40082185
Peak Ampere Turns 17621.2509
Plasma Current 17621.2509
Plasma Current Density 2.64318763e9 Amp per meter^2, near centre
Ion Pressure of Plasma 3.65809182e11 Pa
Power lost through resistance 737.93083 Watt
Power Generated if Deuterium 5.04475776e16 Watt
Power Generated by natural hydrogen 1.21074186e9 Watt
1 exit 2 loop 1
1 exit 2 again 1
The energy tube is Quatz 11 mm OD, 6 mm ID,
Winding Transfusor 4 main coil
31/03/2014 22:10:16
Transfusor 3.
I managed to get the oscillator working but with squegging. The squegging period was 60 ms and the capacitor that charged up was 470uF so the discharge capacitor is about 120 Ohm.
The oscillator squegging.
I will put a resistor of 120 ohm across the capacitor so it can discharge. The squegging period is about 60 ms and the capacitor is 470 uF so CR=60 so R = 60E3/470E6=170 Ohm.
The output voltage is the lower trace and the feedback coil is the upper trace. The output is seen as 1000 volt but it is higher than that because the oscilloscope limits the reading.
I am using a 100x probe but I need a 1000x probe.
31/03/2014 23:13:38
Transfusor 3 ignition oscillator working.
The transistor oscillator that I hope will ignite it is working. However the output potential is only 558 volt and I need 4000 volt so I will reduce the turns on the collector circuit.
When the turns in the collector are reduced to 3 from 6 the output potential drops to 408 volt. I have to match the dynamic impedance, I recall.
I calculated the dynamic impedance of the Transfusor 3 as 100,000 Ohm and the output impedance of the transistor as 1 so the ratio of the turns is about 300 so 6 turns was right. (6x300=2000 app).
The last result was 800 volts at the resonance coil terminals. However the transistor got very hot and the trimmer I used to remove squegging burned out with the dial pointing to about 600 ohms.
To enter class C the bias on the base has to be below the negative or earth rail. I suppose a capacitor resistor and diode would do that.
The turns of the collector winding were 6 and the feedback winding was 50. The resonance coil was about 2000 T and the resonance capacitor 0.1 uF. I use the Chinese capacitor supplied by Josh of a Chinese company with Josh Liu. These capacitor are of excellent quality and much lower price than similar western ones including postage from mainland china. I have also bought high voltage cable from this company. It is of a similar high quality. He trades at http://hvstuff.com/01uf10kvhighvoltagecapacitorhvteslacoilham
I use these capacitors to tune the resonance coil to the desired frequency. My calculation indicate I need 4KV to ionise the reaction gas (Deuterium) but I use air at present. Using a 10 KV rated capacitor gives a good safety margin.
02/04/2014 13:06:49
I replaced the burned out trimmer with a resistor 570 Ohm made from a 100 ohm 3W and a 470 ohm 3W resistor from Maplin.
The ignition oscillator is stable now.
I pumped the reaction chamber down but due to disturbances I only reached 40 Tor.
02/04//2014 21:59:42
I tried the ignition oscillator on the Transfusor 3 with the reaction chamber pumped down to 30 tor and the oscillator immediately squegged. The neutron detector showed bursts of neutrons after a warming up period of about 2 minutes.
07/04/2014 08:24:43
second run of Transfusor 3
It continued to generate neutrons for several minutes after the input power was cut.
Fusion power.
1/5/2014 18:50:47
Now I have succeeded in my objective I have
ceased my experiments and I am not going to build any more. I have
disposed of my tools that I bought to make the unit. I still have it
but I want to find someone to take it off me as a museum piece or
something.
video sale advert http://youtu.be/HNsaQV7rvVQ
Chris.
and nobody came..
24/05/2014 17:42
I am now reworking an old design.
25/05/2014 11:22:33
Simple step function when a 6 volt battery is switched off with only the oscilloscope probe as the load.
25/05/2014 11:54:52
Calculation for this reactor:
diameter of wire in mm 0.17
diameter of former in mm 30
length of former in mm 40
number of turns 20000
Turns per layer 235.294118
Number of Layers 85
Diameter of coil 58.9 mm
Length of Wire 2803.55724 Meters
resistance of winding 2075.06159 Ohm
Inductance of winding by simple formula 8.88264366
new turns 1=y 2=n n
Inductance by measurement 8
Resonant Frequency Hz 281
capacitance by resonance formula 0.40099379e1 uF
Capacitance 0.40099379e1 uF
correct ? 1=n 2=y 2
Dynamic impedance 96143.8134
Resonant frequency is 281
1 for exit 2 for loop for another capacitance 1
Q 6.80683426
Winding Mass 0.56889833 Kg
Rms Potential 4000
Half Wavelength 0.02
Velocity 11.24
Peak Potential 5600.0
Peak Current 0.39226093
Rms Current 0.28018638
Peak Ampere Turns 7845.21859
Plasma Current 7845.21859
Plasma Current Density 1.17678279e9 Amp per meter^2, near centre
Ion Pressure of Plasma 7.25088857e10 Pa
Power lost through resistance 228.062071 Watt
Power Generated if Deuterium 1.98204529e15 Watt
Power Generated by natural hydrogen 47569086.9 Watt
1 exit 2 loop 1
1 exit 2 again 1
I have a quartz tube for the reaction chamber.
29/05/2014 13:58:15
Experimental set up for this reconstruction.
30/05/2014 22:18:33
Feedback coil added.
Trace of oscillator. The upper trace is across the resonance coil and the lower trace from the feedback coil. The output is about 80 volt only this is very short of the 4000 volt needed. I estimate that because only 5 amp is flowing in the primary, I will need 250 amp in the primary to get the 4000 volt across the high voltage resonance coil.
That is impractical here.
However with 10 amp in the primary I should get 800 volt in the secondary.
I ain't got enough humph!
05/06/2014 20:36:24
Tesla Power reactor
18/06/2014 19:25:59
I have withdrawn some of my recent videos but here is another one. Here I put a 300 AMP pulse through a coil wich is inductively linked to a high voltage coil.
No capacitor
(image resonant burst)
Test firing of 300 amp pulse with fluorescent tube to detect high voltage
18/06/2014 22:53:57
Flow soldering to make mosfet amplifier for ignition oscillator to drive 300 amp oscillations. I also added the feedback coil to the circuit.
18/06/2014 28:18:10
A quick test with two batteries and an ammeter showed the mosfet amplifier worked.
I had a 6 volt battery connected with ve to drain and the +ve via an ammeter to the source. I put a 3 volt battery +ve to the gate and the ve to the drain and a current of 4 amp flowed round theother way the current was zero.
The feedback coil is connected between the gate and the drain and the source to the outer 300 amp coil and the drain to the power switch +ve to battery in jump starter.
19/06/2014 17:15:52
Trace from feedback oscillator
This is the trace from the feedback oscillator in response to a 300 amp pulse in the ignition oscillator coil.
20/06/2014 17:52:50
Over all view of reactor and ignition unit 
Amplifier for ignition oscillator the moset gate blew 
Another view of same 
15/08/2014 18:34
Neutron burst
Recorded with my cherenkoff detector.
Burst of Neutrons
16/08/2014 18:38
I am making two other transfusors based on my experience so far. One is designed for a much lower frequency but I won't know the frequency until I see the coil's self resonant frequency. These are based on previous versions.
This transfusor is based on a prior version but has a 400 amp winding and the capacitor has been removed as it resonates at the windings natural frequency. There are two other windings, the output winding and one with a smeter.
The 400 amp winding is to be powered by a car jump starter. A starter relay controls the current pulse. We are considering replacing thie pulse with a 400 amp oscillator using a 1000 amp MOSFET.
I will not know the frequency of oscillation until I measure it with the audio signal generator on the right.
This is a much lower frequency version ready for additional turns from another bobbin. This will also use a 400 amp winding to ignite the reaction. This will have 4 Kg of 0.170 mm copper wire eventually but now there are 3 Kg. I use solderable enamelled copper wire.
17/08/2014 09:47
Energy tubes.
The vacuum pump is very expensive system that I used to find the pressure. These will be replaced by energy tubes in small units. This is a borosilicate glass tube containing hydrogen at 5  10 Tor with conducting pins let into each end to connect to the ends of the winding.
Prototype Energy tube
19/08/2014 13:57
Another run of the transfusor showing the trigger pulse (lower) and the neutron burst (upper) 2 ms later .I used the high voltage probe on the trigger pulse to make the single sweep reliable It was set on 500 V /division.
19/08/2014 15:00
The trigger signal was taken from the push button side of the started solenoid so the relay had to close and the 400 amp current flow iducig the high voltage in the winding and this ionised the gas and then the current flowed in the coil and this generated the neutron burst.
I suppose if I remembered my elementary circuit theory I could do a prediction. However the high current point is 1/4 cycle from the high potential part of the cycle.
The circuit was previously recorded at resonating at 1.58 KHz so 1 cycle is 632.9 uS so 1/4 cycle is 158.2 uS. The neutron burst was about 40 uS long so the fusion reaction lasted for about 1/3 if the high current period. That gives the duty cycle. About 1/16 of the cycle. There was a second 1/4 cycle of high current 1/2 cycle later but that did not result in a neutron burst. This second 1/4 cycle the current pulse had decayed (see image "resonant burst" above) to about half the amplitude.
I probably need a higher Q so more turns and lower pressure. I could buy three more stages to my multisage diaphragm pump but the supplier says that they are no longer available. Another coil is more difficult as my custom made coil former was £45 and the pumps are £14 each from another similar supplier.
The starter solenoid is quite slow but probably 1/10 second. If I could look at the high voltage directly with a high voltage probe the neutron burst would be expected 158 uS after the potential peak.
It could be done...
19/08/2014 16:03
Lower trace is the secondary high voltage coil. Amplitude is out of range. atmospheric pressure
Low Pressure
Only the first cycle has sufficient amplitude to trigger the neutron burst this is out of gamut of the probe.
19/08/2014 16:59
There was a second burst probably key bounce
Second Burst
So there were three altogether and the first two means it nearly started.
More construction work on my transfusor.
Here I continue the manufacture of the low frequency transfusor.
I could not afford the energy tubes from America and the flask of Deuterium was only available to businesses.
We will have to become a business!
Brian is not responding to calls although his phone rings so I suspect he has been injected by a powerful psyche deug. This is possibly because he is out of work. He cannot work under psyche drugs.
Dr Chris
I did the calculations for this winding a while ago so I just added the last 1Kg of 0.17mm wire to the resonance coil.
Dr Chris.
Christopher Strevens, London, Best wishes.
I added 11,240 Turns of 0.170 solderable enamelled copper wire to the low frequency transfusor.
31/08/2014 06:52
Here is my estimate for the resonance coil of the low frequency transfusor.
diameter of wire in mm 0.17
diameter of former in mm 28
length of former in mm 45
number of turns 80000
Turns per layer 264.705882
Number of Layers 302.222222
Diameter of coil 130.755556 mm
Length of Wire 20077.3048 Meters
resistance of winding 14860.2795 Ohm
Inductance of winding by simple formula 110.048279
new turns 1=y 2=n 2
Inductance by measurement 110 I cannot measure this.
Resonant Frequency Hz 50 (I need to measure this)
capacitance by resonance formula 0.9211017e1 uF
Capacitance 0.9211017e1 uF
correct ? 1=n 2=y 2
Dynamic impedance 80363.367
Resonant frequency is 50
1 for exit 2 for loop for another capacitance 1
Q 2.32549588
Winding Mass 4.07409027 Kg
Rms Potential 7000
Half Wavelength 0.0225
Velocity 2.25
Peak Potential 9800.0
Peak Current 0.26051947
Rms Current 0.18608533
Peak Ampere Turns 20841.5575
Plasma Current 20841.5575
Plasma Current Density 2.77887433e9 Amp per meter^2, near centre
Ion Pressure of Plasma 4.41087743e11 Pa
Power lost through resistance 720.409295 Watt
Power Generated if Deuterium 8.25151347e16 Watt
Power Generated by natural hydrogen 1.98036323e9 Watt
1 exit 2 loop 1
1 exit 2 again 1
12/09/2014 07:09:50
Some tests on the low frequency transfusor
nothing special. I used a 100x probe with a 10 MOhm resistor. I then tried the low volage line and crashed the pc oscillosope and the pc.
Sparks and neutrons but no power.
To rewind with 80,000 T of 40 mA wire to give a higher potential at a lower frequency.
Here I add the output coil and the ignition coil
29/09/2014
A test with a 12 volt 3 amp mains transformer as the exciter. It failed during the test.
I ran another test this evening with what appears to be a strong positive result with air at normal pressure and room temp. http://youtu.be/Ycjn8ktBJV8 Not all the results described are in the video. By the time I put the video to record the neutron detector had ceased to work.
A few tests and measurements and a calculation involving changing the utility that computes the power.
fault finding
The is the power amplifier for the ignition winding on the transfusor. It is to be driven by a mains frequency (50 Hz) Wien bridge oscillator.
Latest Version of utility that I use to calculate the resonance winding.
'Inductance
= 4.piE7.A.n^2/l
'to find the final diameter of a multilayer coil and the length of wire
'each layer adds thickness to layer so first work out turns per layer
[Start]
let Pi=3.1415926
'for room temp
Let ResistivityOfCopper=1.68E8
Let DensityOfCopper=8.94E3
'Kg per m^3
'for superconductor
'Let ResistovityOfCopper=0
'for LN temp
'Let ResistivityOfCopper=2.647E9
Let mu=4E7*Pi
Let e0=8.85E12
[InductanceLoop]
input "diameter of wire in mm " ; DiameterOfWire
'DiameterOfWire=0.170
'Print "Diameter of Wire ";DiameterOfWire
'Input "Diameter Of Lumen mm "; DiameterOfLumen
DiameterOfLumen=0
'input "Thickness of Insulation mm ";ThicknessOfInsulation
ThicknessOfInsulation=0
input "diameter of former in mm ";DiameterOfFormer
'Let DiameterOfFormer=60
'Let LengthOfFormer=60
input "length of former in mm " ;LengthOfFormer
'Print "Diameter Of Former mm ";DiameterOfFormer
'Print "Length Of Former mm ";LengthOfFormer
'Input "Resonant Frequency Hz ";ResonantFrequency
input "Outside Diameter of reaction tube mm:
";OutsideDiameterOfReactionTube
input "InsideDiameter of Reaction Tube mm ";InsideDiameterOfReactionTube
Let InsideDiameterOfReactionTube=InsideDiameterOfReactionTube/1000
Let OutsideDiameterOfReactionTube=OutsideDiameterOfReactionTube/1000
let DiameterOfWire=DiameterOfWire/1000
Let DiameterOfLumen=DiameterOfLumen/1000
Let ThicknessOfInsulation=ThicknessOfInsulation/1000
Let DiameterOfFormer=DiameterOfFormer/1000
Let LengthOfFormer=LengthOfFormer/1000
[TurnsLoop]
input "number of turns "; NumberOfTurns
'Let NumberOfTurns=13000
'Print "Number of Turns ";NumberOfTurns
let
TurnsPerLayer=LengthOfFormer/(DiameterOfWire+2*ThicknessOfInsulation)
let NumberOfLayers=NumberOfTurns/TurnsPerLayer
Let
ThicknessOfWinding=NumberOfLayers*(DiameterOfWire+2*ThicknessOfInsulation)
Let DiameterOfWinding=DiameterOfFormer+2*ThicknessOfWinding
'length of wire for each layer is the layer diameter multiplied by Pi
LayerN=0
LengthOfWire=0
'Let TotalInductance=0
'Let TotalCapacitance=0
LayerDiameter=DiameterOfFormer
[ForLoop]
LayerN=LayerN+1
Let
LayerDiameter=LayerDiameter+2*(DiameterOfWire+2*ThicknessOfInsulation)
Let LengthOfWire=Pi*LayerDiameter*TurnsPerLayer+LengthOfWire
'let
LayerInductance=mu*(Pi*(LayerDiameter/2)^2)*TurnsPerLayer^2/LengthOfFormer
'Let
LayerCapacitance=e0*Pi*LayerDiameter*((DiameterOfWire+2*ThicknessOfInsulation)/(DiameterOfWire+2*ThicknessOfInsulation))*TurnsPerLayer
'Let
InterLayerCapacitance=e0*Pi*LayerDiameter*LengthOfFormer/(DiameterOfWire+2*ThicknessOfInsulation)
'Let TotalInductance=LayerInductance+TotalInductance
'Let
TotalCapacitance=LayerCapacitance+InterlayerCapacitance+TotalCapacitance
'Let TotalCapacitance=LayerCapacitance+TotalCapacitance
if LayerN<NumberOfLayers then goto [ForLoop]
Let
ResistanceOfCoil=ResistivityOfCopper*LengthOfWire/((Pi*(DiameterOfWire/2)^2)(Pi*(DiameterOfLumen/2)^2))
'Let
TotalInductance=TotalInductance*(NumberOfLayers/2)*(NumberOfLayers1)
'Print "Inductance by sum ";TotalInductance
Print "Turns per layer ";TurnsPerLayer
Print "Number of Layers ";NumberOfLayers
Print "Diameter of coil ";DiameterOfWinding*1000;" mm"
Print "Length of Wire ";LengthOfWire; " Meters"
Print "resistance of winding ";ResistanceOfCoil; " Ohm"
'input "Resistance of Coil by measurement "; ResistanceOfCoil
'Print "Inductance of Multilayer Coil by sum ";TotalInductance;" Henry"
'Print "Capacitance of Mutilayer Coil ";TotalCapacitance*1E6; "
microfarads"
'input "1=exit 2=InductanceLoop ";OK
'if OK=2 then goto [InductanceLoop]
'to calculate dynamic impedance
'dynamic impedance is L/Cr where r is the resitance of the coil.
'let
TotalInductance=(0.8/1000)*((((DiameterOfWinding+DiameterOfFormer)/254*2)^2)*NumberOfTurns^2)/(6*(DiameterOfWinding+DiameterOfFormer)/254*2+9*LegthOfFormer/254+10*(DiameterOfWindingDiameterOfFormer)/254*2)
'print "total inductance by complex formula ";TotalInductance
'let TotalInductance=7.6
let
TotalInductance=mu*Pi*((DiameterOfFormer/2)^2)*(NumberOfTurns^2)/LengthOfFormer
Print "Inductance of winding by simple formula ";TotalInductance
input "new turns 1=y 2=n ";ok
if ok=1 then goto [TurnsLoop]
input "Inductance by measurement ";TotalInductance
'input "1 for exit 2 for new turns ";ok
'if ok=2 then goto [TurnsLoop]
'Let TotalInductance=7.6
'let TotalInductance=4.76
'Print "inductance by measurement ";TotalInductance
[ResonanceLoop]
input "Resonant Frequency Hz "; ResonantFrequency
'f=1/(2pisqr(LC) so c=1/f^2*4Pi^2L
let Capacitance=1/((ResonantFrequency^2)*4*(Pi^2)*TotalInductance)
Print "capacitance by resonance formula ";Capacitance*1E6;" uF"
'let ExternalCapacitance=1.5E6
[CapacitanceLoop]
'let VaractorCapacitance=11.9E6
'input "capacitance of Main capacitor (uF) "; ExternalCapacitance
'Let ExternalCapacitance=ExternalCapacitance/1E6
'input "Additional Series capacitance (uF)";SeriesCapacitance
'Let SeriesCapacitance=SeriesCapacitance/1E6
'input "ok1";ok
'Let Capacitance=1/(1/ExternalCapacitance + 1/SeriesCapacitance)
'Let Capacitance=TotalCapacitance+ExternalCapacitance
'Let Capacitance=ExternalCapacitance
'input "Capacitance (F) "; Capacitance
'input "ok2"; ok
Print "Capacitance ";Capacitance*1E6;" uF"
input "correct ? 1=n 2=y ";OK
if OK=1 then goto [CapacitanceLoop]
let DynamicImpedance = TotalInductance/(Capacitance*ResistanceOfCoil)
print "Dynamic impedance ";DynamicImpedance
'print "dynamic impedance  Infinite"
'input "ok3";ok
'program to calculate resonant frequency
'f=1/(2.pi.sqr(l.c))
'Print "Extermal Capacitance ";ExternalCapacitance
'Print "Additional Series Capacitance ";VaractorCapacitance
'input "Inductance of coil "; Inductance
let ResonantFrequency=1/(2*Pi*(sqr(TotalInductance*Capacitance)))
print "Resonant frequency is "; ResonantFrequency
input "1 for exit 2 for loop for another capacitance "; ok
'input "1 for exit 2 for another turns value ";ok
'if ok=2 goto [TurnsLoop]
if ok=2 then goto [ResonanceLoop]
'to calculate Q from Q= 2.Pi.f.L/R
Let Q=2*Pi*ResonantFrequency*TotalInductance/ResistanceOfCoil
Print "Q " ;Q
'Print "Q infinite"
'to calculate the total mass of copper in the winding Area times lengh
times density
'let
TotalWindingMass=Pi*(((DiameterOfWire/2)^2)((DiameterOfLumen/2)^2))*LengthOfWire*DensityOfCopper
let
TotalWindingMass=Pi*((DiameterOfWire/2)^2)*LengthOfWire*DensityOfCopper
Print "Winding Mass ";TotalWindingMass;" Kg"
[PotentialLoop]
Input "Rms Potential ";RmsPotential
'to calculate slow wave
Print "Half Wavelength "; LengthOfFormer/2
Print "Velocity "; LengthOfFormer*ResonantFrequency
Print "Peak Potential "; 1.4*RmsPotential
Let
PeakCurrent=1.4*RmsPotential/(SQR((TotalInductance*2*Pi*ResonantFrequency)^2+ResistanceOfCoil^2))
Print "Peak Current ";PeakCurrent
print "Rms Current ";PeakCurrent/1.4
Let PeakAmpereTurns=PeakCurrent*NumberOfTurns
Let PeakAmpereTurnsPerMeter=PeakAmpereTurns/(LengthOfFormer/3)
Print "Peak Ampere Turns "; PeakAmpereTurns
Print "Plasma Current "; PeakAmpereTurns
Let PlasmaCurrentDensity=PeakAmpereTurns/((1.5E3)*LengthOfFormer/3)
Print "Plasma Current Density ";3*PlasmaCurrentDensity; " Amp per
meter^2, near centre"
Let PressureOnPlasma=
((3*PlasmaCurrentDensity*PeakAmpereTurnsPerMeter)*mu)/((DiameterOfFormer(6E3))/2)
Print "Ion Pressure of Plasma ";PressureOnPlasma; " Pa"
Print "Power lost through resistance ";
ResistanceOfCoil*((PeakCurrent^2)/1.4); " Watt"
'Let
ReactanceOfExternalCapacitance=1/(2*Pi*ResonantFrequency*ExternalCapacitance)
'Let
ReactanceOfSeriesCapacitance=1/(2*Pi*ResonantFrequency*SeriesCapacitance)
'Let
PotentialAcrossSeriesCapacitance=ReactanceOfSeriesCapacitance*PeakCurrent
'Let
PotentialAcrossExternalCapacitance=ReactanceOfExternalCapacitance*PeakCurrent
'Power = 0.5 watt/cc/kPa^2 10torr=133 Pa
Let
PowerGenerated=(0.5*(LengthOfFormer*100/3)*(InsideDiameterOfReactionTube*100)^2*2*Pi*((PressureOnPlasma)/1000)^2)
Print "Power Generated if Deuterium ";PowerGenerated;" Watt"
Print "Power Generated by natural hydrogen "; PowerGenerated*24E9;"
Watt"
'Print "Potential Across Additional Series Capacitor
";PotentialAcrossSeriesCapacitance;" Volt"
'Print "Potential Across Main Capacitor
";PotentialAcrossExternalCapacitance;" Volt"
input "1 exit 2 loop "; ok
if ok=2 then goto [PotentialLoop]
input "1 exit 2 again " ;ok
if ok=2 then goto [Start]
End
Calculation from above:
diameter of wire in mm 0.17
diameter of former in mm 28
length of former in mm 40
Outside Diameter of reaction tube mm: 13
InsideDiameter of Reaction Tube mm 11
number of turns 50000
Turns per layer 235.294118
Number of Layers 212.5
Diameter of coil 100.25 mm
Length of Wire 10136.5814 Meters
resistance of winding 7502.62217 Ohm
Inductance of winding by simple formula 48.3610599
new turns 1=y 2=n 2
Inductance by measurement 49
Resonant Frequency Hz 50
capacitance by resonance formula 0.20677793 uF
Capacitance 0.20677793 uF
correct ? 1=n 2=y 2
Dynamic impedance 31584.8497
Resonant frequency is 50.0
1 for exit 2 for loop for another capacitance 1
Q 2.05178981
Winding Mass 2.0569169 Kg
Rms Potential 7000
Half Wavelength 0.02
Velocity 2.0
Peak Potential 9800.0
Peak Current 0.57226969
Rms Current 0.40876407
Peak Ampere Turns 28613.4847
Plasma Current 28613.4847
Plasma Current Density 4.2920227e9 Amp per meter^2, near centre
Ion Pressure of Plasma 1.05223125e12 Pa
Power lost through resistance 1755.03804 Watt
Power Generated if Deuterium 5.6117248e18 Watt
Power Generated by natural hydrogen 1.34681395e11 Watt
1 exit 2 loop 1
1 exit 2 again 1
I think it will start at a lower continuous
current than 400 amp. I might get 10 amp out of this amplifier that
will be driven by a Wien bridge oscillator. This should give a 140 volt
transfusor potential. Once it starts the potential will jump to 400
volt then to 7000 volt.
If I use a mains transformer as the ignition
oscillator the started transfusor will put electric current back into
the mains.
The local electrician reported a 400 volt surge
when I tried a transformer. I think we need a power engineer.
The mains transformer I was using failed but the output rose from 12 volt to 20 volt just before.
I wonder ...
24/10/2014 10:56:05
I have now built the ignition oscillator and power amplifier for the ignition winding. The Wien bridge oscillator does not work It is possible the lm 741 has failed through static.
This is a sketch of the device. There are three windings, the inner or transfusor winding of 50,000T tuned by a 0.2 uF capacitor and the ignition winding of 50T and the output winding that depends on application.
Ignition oscillator The Wien bridge oscillator does not work yet, it looks like the lm 741 cn has failed.
Image of unfinished ignition oscillator
4/3/2015 06:19
The transfusor winding needs a regulator to prevent over voltage and I used a spark gap set at about 2 cm in air for about 10000 v.
I put an additional winding (the regulator winding with a triac a1=a2 across it and the gate to a potentiometer attached to the sense winding.
The ignition oscillator is now a car jump starter powering a 1500 watt 220 volt inverter.
This is being updated.
25/06/2017 21:40:15
Hi, they just found me not guilty of murder and discharged me.
When I turned the vacuum pump off, the same AC flowed but without ionisation. It was displacement current instead. I made a control box where a relay energised when the output reached 12 volt and cut off the input power and then the transfusor powered a 3 watt 12 volt lamp for four minutes. The pc fan was not connected at that stage.
I made a Cherenkoff counter to detect neutrons and found that the thing gave of sine waves of them.
I ceased experimenting at that point because reported illness next door.
I have put up some videos of my experiments on utube. Look up the channel by Christopher Strevens.
Chris,
London, SW.
25/06/2017
21:40:15
25/06/2017
25/06/2017 21:40:15
I did further experiments and wound other coils similar to above but a fire bomb cut my work short. I'm now in a care home being poisoned to age me and quieten me.
This calculation is another design.
diameter of wire in mm 0.17
diameter of former in mm 25
length of former in mm 40
Outside Diameter of reaction tube mm: 9
InsideDiameter of Reaction Tube mm 7
number of turns 70000
Turns per layer 235.294118
Number of Layers 297.5
Diameter of coil 126.15 mm
Length of Wire 16703.9144 Meters
resistance of winding 12363.4541 Ohm
Inductance of winding by simple formula 75.5641561
new turns 1=y 2=n 2
Inductance by measurement 75.56
Resonant Frequency Hz 50
capacitance by resonance formula 0.13409368 uF
Capacitance 0.13409368 uF
correct ? 1=n 2=y 2
Dynamic impedance 45576.7978
Resonant frequency is 50.0
1 for exit 2 for loop for another capacitance 1
Q 1.92000338
Winding Mass 3.38956129 Kg
Rms Potential 7000
Efield = 175000 Volts/meter
Half Wavelength 0.02
Velocity 2.0
Peak Potential 9800.0
Peak Current 0.36615585
Rms Current 0.26153989
Peak Ampere Turns 25630.9094
Plasma Current 25630.9094
Plasma Current Density 3.84463642e9 Amp per meter^2, near centre
Ion Pressure of Plasma 9.77612635e11 Pa
Power lost through resistance 1183.97828 Watt
Power Generated if Deuterium 4.90408854e11 Watt
Power Generated by natural hydrogen 11769.8125 Watt
1 exit 2 loop 1
1 exit 2 again 1
Here is the update of the transfusor designer. it includes the other windings data to generate.
'Inductance = 4.piE7.A.n^2/l
'to find the final diameter of a multilayer coil and the length of
wire
'each layer adds thickness to layer so first work out turns per layer
[Start]
let Pi=3.1415926
'for room temp
Let ResistivityOfCopper=1.68E8
Let DensityOfCopper=8.94E3
'Kg per m^3
'for superconductor
'Let ResistovityOfCopper=0
'for LN temp
'Let ResistivityOfCopper=2.647E9
Let mu=4E7*Pi
Let e0=8.85E12
[InductanceLoop]
input "diameter of wire in mm " ; DiameterOfWire
'DiameterOfWire=0.170
'Print "Diameter of Wire ";DiameterOfWire
'Input "Diameter Of Lumen mm "; DiameterOfLumen
DiameterOfLumen=0
'input "Thickness of Insulation mm ";ThicknessOfInsulation
ThicknessOfInsulation=0
input "diameter of former in mm ";DiameterOfFormer
'Let DiameterOfFormer=60
'Let LengthOfFormer=60
input "length of former in mm " ;LengthOfFormer
'Print "Diameter Of Former mm ";DiameterOfFormer
'Print "Length Of Former mm ";LengthOfFormer
'Input "Resonant Frequency Hz ";ResonantFrequency
input "Outside Diameter of reaction tube mm:
";OutsideDiameterOfReactionTube
input "InsideDiameter of Reaction Tube mm ";InsideDiameterOfReactionTube
Let InsideDiameterOfReactionTube=InsideDiameterOfReactionTube/1000
Let OutsideDiameterOfReactionTube=OutsideDiameterOfReactionTube/1000
let DiameterOfWire=DiameterOfWire/1000
Let DiameterOfLumen=DiameterOfLumen/1000
Let ThicknessOfInsulation=ThicknessOfInsulation/1000
Let DiameterOfFormer=DiameterOfFormer/1000
Let LengthOfFormer=LengthOfFormer/1000
[TurnsLoop]
input "number of turns "; NumberOfTurns
'Let NumberOfTurns=13000
'Print "Number of Turns ";NumberOfTurns
let
TurnsPerLayer=LengthOfFormer/(DiameterOfWire+2*ThicknessOfInsulation)
let NumberOfLayers=NumberOfTurns/TurnsPerLayer
Let
ThicknessOfWinding=NumberOfLayers*(DiameterOfWire+2*ThicknessOfInsulation)
Let DiameterOfWinding=DiameterOfFormer+2*ThicknessOfWinding
'length of wire for each layer is the layer diameter multiplied by Pi
LayerN=0
LengthOfWire=0
'Let TotalInductance=0
'Let TotalCapacitance=0
LayerDiameter=DiameterOfFormer
[ForLoop]
LayerN=LayerN+1
Let
LayerDiameter=LayerDiameter+2*(DiameterOfWire+2*ThicknessOfInsulation)
Let LengthOfWire=Pi*LayerDiameter*TurnsPerLayer+LengthOfWire
'let
LayerInductance=mu*(Pi*(LayerDiameter/2)^2)*TurnsPerLayer^2/LengthOfFormer
'Let
LayerCapacitance=e0*Pi*LayerDiameter*((DiameterOfWire+2*ThicknessOfInsulation)/(DiameterOfWire+2*ThicknessOfInsulation))*TurnsPerLayer
'Let
InterLayerCapacitance=e0*Pi*LayerDiameter*LengthOfFormer/(DiameterOfWire+2*ThicknessOfInsulation)
'Let TotalInductance=LayerInductance+TotalInductance
'Let
TotalCapacitance=LayerCapacitance+InterlayerCapacitance+TotalCapacitance
'Let TotalCapacitance=LayerCapacitance+TotalCapacitance
if LayerN<NumberOfLayers then goto [ForLoop]
Let
ResistanceOfCoil=ResistivityOfCopper*LengthOfWire/((Pi*(DiameterOfWire/2)^2)(Pi*(DiameterOfLumen/2)^2))
'Let
TotalInductance=TotalInductance*(NumberOfLayers/2)*(NumberOfLayers1)
'Print "Inductance by sum ";TotalInductance
Print "Turns per layer ";TurnsPerLayer
Print "Number of Layers ";NumberOfLayers
Print "Diameter of coil ";DiameterOfWinding*1000;" mm"
Print "Length of Wire ";LengthOfWire; " Meters"
Print "resistance of winding ";ResistanceOfCoil; " Ohm"
'input "Resistance of Coil by measurement "; ResistanceOfCoil
'Print "Inductance of Multilayer Coil by sum ";TotalInductance;" Henry"
'Print "Capacitance of Mutilayer Coil ";TotalCapacitance*1E6; "
microfarads"
'input "1=exit 2=InductanceLoop ";OK
'if OK=2 then goto [InductanceLoop]
'to calculate dynamic impedance
'dynamic impedance is L/Cr where r is the resitance of the coil.
'let
TotalInductance=(0.l8/1000)*((((DiameterOfWinding+DiameterOfFormer)/254*2)^2)*NumberOfTurns^2)/(6*(DiameterOfWinding+DiameterOfFormer)/254*2+9*LegthOfFormer/254+10*(DiameterOfWindingDiameterOfFormer)/254*2)
'print "total inductance by complex formula ";TotalInductance
'let TotalInductance=7.6
let
TotalInductance=mu*Pi*((DiameterOfFormer/2)^2)*(NumberOfTurns^2)/LengthOfFormer
Print "Inductance of winding by simple formula ";TotalInductance
input "new turns 1=y 2=n ";ok
if ok=1 then goto [TurnsLoop]
input "Inductance by measurement ";TotalInductance
'input "1 for exit 2 for new turns ";ok
'if ok=2 then goto [TurnsLoop]
'Let TotalInductance=7.6
'let TotalInductance=4.76
'Print "inductance by measurement ";TotalInductance
[ResonanceLoop]
input "Resonant Frequency Hz "; ResonantFrequency
'f=1/(2pisqr(LC) so c=1/f^2*4Pi^2L
let Capacitance=1/((ResonantFrequency^2)*4*(Pi^2)*TotalInductance)
Print "capacitance by resonance formula ";Capacitance*1E6;" uF"
'let ExternalCapacitance=1.5E6
[CapacitanceLoop]
'let VaractorCapacitance=11.9E6
'input "capacitance of Main capacitor (uF) "; ExternalCapacitance
'Let ExternalCapacitance=ExternalCapacitance/1E6
'input "Additional Series capacitance (uF)";SeriesCapacitance
'Let SeriesCapacitance=SeriesCapacitance/1E6
'input "ok1";ok
'Let Capacitance=1/(1/ExternalCapacitance + 1/SeriesCapacitance)
'Let Capacitance=TotalCapacitance+ExternalCapacitance
'Let Capacitance=ExternalCapacitance
'input "Capacitance (F) "; Capacitance
'input "ok2"; ok
Print "Capacitance ";Capacitance*1E6;" uF"
input "correct ? 1=n 2=y ";OK
if OK=1 then goto [TurnsLoop]
let DynamicImpedance = TotalInductance/(Capacitance*ResistanceOfCoil)
print "Dynamic impedance ";DynamicImpedance
'print "dynamic impedance  Infinite"
'input "ok3";ok
'program to calculate resonant frequency
'f=1/(2.pi.sqr(l.c))
'Print "Extermal Capacitance ";ExternalCapacitance
'Print "Additional Series Capacitance ";VaractorCapacitance
'input "Inductance of coil "; Inductance
let ResonantFrequency=1/(2*Pi*(sqr(TotalInductance*Capacitance)))
print "Resonant frequency is "; ResonantFrequency
input "1 for exit 2 for loop for another capacitance "; ok
'input "1 for exit 2 for another turns value ";ok
'if ok=2 goto [TurnsLoop]
if ok=2 then goto [ResonanceLoop]
'to calculate Q from Q= 2.Pi.f.L/R
Let Q=2*Pi*ResonantFrequency*TotalInductance/ResistanceOfCoil
Print "Q " ;Q
'Print "Q infinite"
'to calculate the total mass of copper in the winding Area times lengh
times density
let
TotalWindingMass=Pi*(((DiameterOfWire/2)^2)((DiameterOfLumen/2)^2))*LengthOfWire*DensityOfCopper
Print "Winding Mass ";TotalWindingMass;" Kg"
[PotentialLoop]
Input "Rms Potential ";RmsPotential
'to calculate efield
Let Efield=RmsPotential/LengthOfFormer
Print "Efield = ";Efield; " Volts/meter"
'to calculate slow wave
Print "Half Wavelength "; LengthOfFormer/2
Print "Velocity "; LengthOfFormer*ResonantFrequency
Print "Peak Potential "; 1.4*RmsPotential
Let
PeakCurrent=1.4*RmsPotential/(SQR((TotalInductance*2*Pi*ResonantFrequency)^2+ResistanceOfCoil^2))
Print "Peak Current ";PeakCurrent
print "Rms Current ";PeakCurrent/1.4
Let PeakAmpereTurns=PeakCurrent*NumberOfTurns
Let PeakAmpereTurnsPerMeter=PeakAmpereTurns/(LengthOfFormer/3)
Print "Peak Ampere Turns "; PeakAmpereTurns
Print "Plasma Current "; PeakAmpereTurns
Let PlasmaCurrentDensity=PeakAmpereTurns/((1.5E3)*LengthOfFormer/3)
Print "Plasma Current Density ";3*PlasmaCurrentDensity; " Amp per
meter^2, near centre"
Let PressureOnPlasma=
((3*PlasmaCurrentDensity*PeakAmpereTurnsPerMeter)*mu)/((DiameterOfFormer(6E3))/2)
Print "Ion Pressure of Plasma ";PressureOnPlasma; " Pa"
Print "Power lost through resistance ";
ResistanceOfCoil*((PeakCurrent^2)/1.4); " Watt"
'Let
ReactanceOfExternalCapacitance=1/(2*Pi*ResonantFrequency*ExternalCapacitance)
'Let
ReactanceOfSeriesCapacitance=1/(2*Pi*ResonantFrequency*SeriesCapacitance)
'Let
PotentialAcrossSeriesCapacitance=ReactanceOfSeriesCapacitance*PeakCurrent
'Let
PotentialAcrossExternalCapacitance=ReactanceOfExternalCapacitance*PeakCurrent
'Power = 0.5 watt/cc/kPa^2 10torr=133 Pa
Let
PowerGenerated=(0.5*(LengthOfFormer/3)*((InsideDiameterOfReactionTube/2)/1000)^2*2*Pi*((PressureOnPlasma)/1000)^2)*100^3
Print "Power Generated if Deuterium ";PowerGenerated;" Watt"
Print "Power Generated by natural hydrogen "; PowerGenerated*24E9;"
Watt"
'Print "Potential Across Additional Series Capacitor
";PotentialAcrossSeriesCapacitance;" Volt"
'Print "Potential Across Main Capacitor
";PotentialAcrossExternalCapacitance;" Volt"
input "1 exit 2 loop "; ok
if ok=2 then goto [PotentialLoop]
'ignition winding, 2 mm wire rms potential/12 x number of turns in
ignition winding/number of turns in the resonance winding=1
'Let IgnitionPotential=12
Input "Ignition Potential? ";IgnitionPotential
let IgnitionWindingTurns=(NumberOfTurns/RmsPotential)*IgnitionPotential
Print "Ignition winding turns ";IgnitionWindingTurns; " 2 mm wire"
Print "Sense winding ";IgnitionWindingTurns;" Turns of 0.9 mm wire"
Print "Local Power Winding ";IgnitionWindingTurns;" Turns of 0.9 mm
wire"
Print "Brake Winding ";IgnitionWindingTurns*2;" Turns of 0.9 mm wire"
input "1 exit 2 again " ;ok
if ok=2 then goto [Start]
End
End of just basic program
Output for short transfusor
diameter of wire in mm 0.17
diameter of former in mm 25
length of former in mm 10
Outside Diameter of reaction tube mm: 9
InsideDiameter of Reaction Tube mm 7
number of turns 15000
Turns per layer 58.8235294
Number of Layers 255.0
Diameter of coil 111.7 mm
Length of Wire 3249.63382 Meters
resistance of winding 2405.22656 Ohm
Inductance of winding by simple formula 13.8791307
new turns 1=y 2=n 2
Inductance by measurement 18
Resonant Frequency Hz 50
capacitance by resonance formula 0.56289548 uF
Capacitance 0.56289548 uF
correct ? 1=n 2=y 2
Dynamic impedance 13295.0125
Resonant frequency is 50
1 for exit 2 for loop for another capacitance 2
Resonant Frequency Hz 50
capacitance by resonance formula 0.56289548 uF
Capacitance 0.56289548 uF
correct ? 1=n 2=y 2
Dynamic impedance 13295.0125
Resonant frequency is 50
1 for exit 2 for loop for another capacitance 1
Q 2.35107444
Winding Mass 0.65941627 Kg
Rms Potential 1750
Efield = 175000 Volts/meter
Half Wavelength 0.005
Velocity 0.5
Peak Potential 2450.0
Peak Current 0.39868965
Rms Current 0.28477832
Peak Ampere Turns 5980.34468
Plasma Current 5980.34468
Plasma Current Density 3.58820681e9 Amp per meter^2, near centre
Ion Pressure of Plasma 8.51552027e11 Pa
Power lost through resistance 273.085014 Watt
Power Generated if Deuterium 9.30223001e10 Watt
Power Generated by natural hydrogen 2232.5352 Watt
1 exit 2 loop 1
Ignition Potential? 12
Ignition winding turns 102.857143 2 mm wire
Sense winding 102.857143 Turns of 0.9 mm wire
Local Power Winding 102.857143 Turns of 0.9 mm wire
Brake Winding 205.714286 Turns of 0.9 mm wire
1 exit 2 again 1
I hope to get permission to make one here in the care home but
I'm hoping to be able to move out back into the community when te
poisoning stops.
17/04/20
18/04/2017 09:35:43
26/04/2017 04:17:37
13/05/2017 07:16:25
Below I recalculate the integral and show it is a central force and also calculate the pressure on the plasma collumn inside thw winding.
14/05/2017 16:48:00
The tunnelling threshold is at 13 nm and that is 3*10^9 meters so the calculationshows that fusion by pressure alone is possible.
14/05/2017 17:23:47
diameter of wire in mm 2.76
diameter of former in mm 25
length of former in mm 40
Outside Diameter of reaction tube mm: 10
InsideDiameter of Reaction Tube mm 7
number of turns 70000
Turns per layer 14.4927536
Number of Layers 4830.0
Diameter of coil 26686.6 mm
Length of Wire 2937700.73 Meters
resistance of winding 8249.14228 Ohm
Inductance of winding by simple formula 75.5641561
new turns 1=y 2=n 2
Inductance by measurement 75.5
Resonant Frequency Hz 50
capacitance by resonance formula 0.13420025 uF
Capacitance 0.13420025 uF
correct ? 1=n 2=y 2
Dynamic impedance 68200.073
Resonant frequency is 50
1 for exit 2 for loop for another capacitance 1
Q 2.87533216
Winding Mass 157127.828 Kg
Rms Potential 120000000
Efield = 3.0e9 Volts/meter
Half Wavelength 0.02
Velocity 2.0
Peak Potential 1.68e8
Peak Current 6689.88063
Rms Current 4778.48616
Peak Ampere Turns 4.68291644e8
Plasma Current 4.68291644e8
Plasma Current Density 7.02437466e13 Amp per meter^2, near centre
Ion Pressure of Plasma 3.26340963e20 Pa
Power lost through resistance 2.63704473e11 Watt
Power Generated if Deuterium 5.46471947e28 Watt
Power Generated by natural hydrogen 1.31153267e21 Watt
1 exit 2 loop 1
19/05/2017 00:43:24
This is the particular instance the first transfusor was made to except tat I had 50,000 T the new ones had 70,000T.
diameter of wire in mm 0.17
diameter of former in mm 25
length of former in mm 40
Outside Diameter of reaction tube mm: 10
InsideDiameter of Reaction Tube mm 7
number of turns 70000
Turns per layer 235.294118
Number of Layers 297.5
Diameter of coil 126.15 mm
Length of Wire 16703.9144 Meters
resistance of winding 12363.4541 Ohm
Inductance of winding by simple formula 75.5641561
new turns 1=y 2=n 2
Inductance by measurement 75.5
Resonant Frequency Hz 50
capacitance by resonance formula 0.13420025 uF
Capacitance 0.13420025 uF
correct ? 1=n 2=y 2
Dynamic impedance 45504.4441
Resonant frequency is 50
1 for exit 2 for loop for another capacitance 1
Q 1.91847877
Winding Mass 3.38956129 Kg
Rms Potential 7000
Efield = 175000 Volts/meter
Half Wavelength 0.02
Velocity 2.0
Peak Potential 9800.0
Peak Current 0.36638468
Rms Current 0.26170335
Peak Ampere Turns 25646.9279
Plasma Current 25646.9279
Plasma Current Density 3.84703919e9 Amp per meter^2, near centre
Ion Pressure of Plasma 9.78834969e11 Pa
Power lost through resistance 1185.45864 Watt
Power Generated if Deuterium 4.91635961e11 Watt
Power Generated by natural hydrogen 11799.2631 Watt
1 exit 2 loop 1
Ignition Potential? 12
Ignition winding turns 120 2 mm wire
Sense winding 120 Turns of 0.9 mm wire
Local Power Winding 120 Turns of 0.9 mm wire
Brake Winding 240 Turns of 0.9 mm wire
Sense Winding 120 Turns of 0.3 mm wire
1 exit 2 again 1
using original utility software.
19/05/2017 01:21:57
These incorrect calculations removed
3KW
I think this is right...I cannot work out the construction.
21/05/2017 14:46:54
Ion Pressure
This is more than 10^4 so cold fusion will take place.
29/05/2017 15:44:38
Large Transfusor
.
I cannot be sure of the correctness of this work but I sat down and reasoned again and did a hand calculation on a calculator and pen and paper.
Diagram:
below is a new version of the just basic agorithm.
The estimate of the power below is too high.
'Inductance = 4.piE7.A.n^2/l
'to find the final diameter of a multilayer coil and the length of wire
'each layer adds thickness to layer so first work out turns per layer
[Start]
let Pi=3.1415926
'for room temp
Let ResistivityOfCopper=1.68E8
Let DensityOfCopper=8.94E3
'Kg per m^3
'for superconductor
'Let ResistovityOfCopper=0
'for LN temp
'Let ResistivityOfCopper=2.647E9
Let mu=4E7*Pi
Let e0=8.85E12
[InductanceLoop]
input "diameter of wire in mm " ; DiameterOfWire
'DiameterOfWire=0.170
'Print "Diameter of Wire ";DiameterOfWire
'Input "Diameter Of Lumen mm "; DiameterOfLumen
DiameterOfLumen=0
'input "Thickness of Insulation mm ";ThicknessOfInsulation
ThicknessOfInsulation=0
input "diameter of former in mm ";DiameterOfFormer
'Let DiameterOfFormer=60
'Let LengthOfFormer=60
input "length of former in mm " ;LengthOfFormer
'Print "Diameter Of Former mm ";DiameterOfFormer
'Print "Length Of Former mm ";LengthOfFormer
'Input "Resonant Frequency Hz ";ResonantFrequency
Input "Ignition Potential? ";IgnitionPotential
input "Outside Diameter of reaction tube mm:
";OutsideDiameterOfReactionTube
input "InsideDiameter of Reaction Tube mm ";InsideDiameterOfReactionTube
Let InsideDiameterOfReactionTube=InsideDiameterOfReactionTube/1000
Let OutsideDiameterOfReactionTube=OutsideDiameterOfReactionTube/1000
let DiameterOfWire=DiameterOfWire/1000
Let DiameterOfLumen=DiameterOfLumen/1000
Let ThicknessOfInsulation=ThicknessOfInsulation/1000
Let DiameterOfFormer=DiameterOfFormer/1000
Let LengthOfFormer=LengthOfFormer/1000
[TurnsLoop]
input "number of turns "; NumberOfTurns
'Let NumberOfTurns=13000
'Print "Number of Turns ";NumberOfTurns
let
TurnsPerLayer=LengthOfFormer/(DiameterOfWire+2*ThicknessOfInsulation)
let NumberOfLayers=NumberOfTurns/TurnsPerLayer
Let
ThicknessOfWinding=NumberOfLayers*(DiameterOfWire+2*ThicknessOfInsulation)
Let DiameterOfWinding=DiameterOfFormer+2*ThicknessOfWinding
'length of wire for each layer is the layer diameter multiplied by Pi
LayerN=0
LengthOfWire=0
'Let TotalInductance=0
'Let TotalCapacitance=0
LayerDiameter=DiameterOfFormer
[ForLoop]
LayerN=LayerN+1
Let
LayerDiameter=LayerDiameter+2*(DiameterOfWire+2*ThicknessOfInsulation)
Let LengthOfWire=Pi*LayerDiameter*TurnsPerLayer+LengthOfWire
'let
LayerInductance=mu*(Pi*(LayerDiameter/2)^2)*TurnsPerLayer^2/LengthOfFormer
'Let
LayerCapacitance=e0*Pi*LayerDiameter*((DiameterOfWire+2*ThicknessOfInsulation)/(DiameterOfWire+2*ThicknessOfInsulation))*TurnsPerLayer
'Let
InterLayerCapacitance=e0*Pi*LayerDiameter*LengthOfFormer/(DiameterOfWire+2*ThicknessOfInsulation)
'Let TotalInductance=LayerInductance+TotalInductance
'Let
TotalCapacitance=LayerCapacitance+InterlayerCapacitance+TotalCapacitance
'Let TotalCapacitance=LayerCapacitance+TotalCapacitance
if LayerN<NumberOfLayers then goto [ForLoop]
Let
ResistanceOfCoil=ResistivityOfCopper*LengthOfWire/((Pi*(DiameterOfWire/2)^2)(Pi*(DiameterOfLumen/2)^2))
'Let
TotalInductance=TotalInductance*(NumberOfLayers/2)*(NumberOfLayers1)
'Print "Inductance by sum ";TotalInductance
Print "Turns per layer ";TurnsPerLayer
Print "Number of Layers ";NumberOfLayers
Print "Diameter of coil ";DiameterOfWinding*1000;" mm"
Print "Length of Wire ";LengthOfWire; " Meters"
Print "resistance of winding ";ResistanceOfCoil; " Ohm"
'input "Resistance of Coil by measurement "; ResistanceOfCoil
'Print "Inductance of Multilayer Coil by sum ";TotalInductance;" Henry"
'Print "Capacitance of Mutilayer Coil ";TotalCapacitance*1E6; "
microfarads"
'input "1=exit 2=InductanceLoop ";OK
'if OK=2 then goto [InductanceLoop]
'to calculate dynamic impedance
'dynamic impedance is L/Cr where r is the resitance of the coil.
'let
TotalInductance=(0.l8/1000)*((((DiameterOfWinding+DiameterOfFormer)/254*2)^2)*NumberOfTurns^2)/(6*(DiameterOfWinding+DiameterOfFormer)/254*2+9*LegthOfFormer/254+10*(DiameterOfWindingDiameterOfFormer)/254*2)
'print "total inductance by complex formula ";TotalInductance
'let TotalInductance=7.6
let
TotalInductance=mu*Pi*((DiameterOfFormer/2)^2)*(NumberOfTurns^2)/LengthOfFormer
Print "Inductance of winding by simple formula ";TotalInductance;"
Henry"
input "new turns 1=y 2=n ";ok
if ok=1 then goto [TurnsLoop]
input "Inductance by measurement H ";TotalInductance
'input "1 for exit 2 for new turns ";ok
'if ok=2 then goto [TurnsLoop]
'Let TotalInductance=7.6
'let TotalInductance=4.76
'Print "inductance by measurement ";TotalInductance
[ResonanceLoop]
input "Resonant Frequency Hz "; ResonantFrequency
'f=1/(2pisqr(LC) so c=1/f^2*4Pi^2L
let Capacitance=1/((ResonantFrequency^2)*4*(Pi^2)*TotalInductance)
Print "capacitance by resonance formula ";Capacitance*1E6;" uF"
'let ExternalCapacitance=1.5E6
[CapacitanceLoop]
'let VaractorCapacitance=11.9E6
'input "capacitance of Main capacitor (uF) "; ExternalCapacitance
'Let ExternalCapacitance=ExternalCapacitance/1E6
'input "Additional Series capacitance (uF)";SeriesCapacitance
'Let SeriesCapacitance=SeriesCapacitance/1E6
'input "ok1";ok
'Let Capacitance=1/(1/ExternalCapacitance + 1/SeriesCapacitance)
'Let Capacitance=TotalCapacitance+ExternalCapacitance
'Let Capacitance=ExternalCapacitance
'input "Capacitance (F) "; Capacitance
'input "ok2"; ok
Print "Capacitance ";Capacitance*1E6;" uF"
input "correct ? 1=n 2=y ";OK
if OK=1 then goto [TurnsLoop]
let DynamicImpedance = TotalInductance/(Capacitance*ResistanceOfCoil)
print "Dynamic impedance ";DynamicImpedance
'print "dynamic impedance  Infinite"
'input "ok3";ok
'program to calculate resonant frequency
'f=1/(2.pi.sqr(l.c))
'Print "Extermal Capacitance ";ExternalCapacitance
'Print "Additional Series Capacitance ";VaractorCapacitance
'input "Inductance of coil "; Inductance
let ResonantFrequency=1/(2*Pi*(sqr(TotalInductance*Capacitance)))
print "Resonant frequency is "; ResonantFrequency
input "1 for exit 2 for loop for another capacitance "; ok
'input "1 for exit 2 for another turns value ";ok
'if ok=2 goto [TurnsLoop]
if ok=2 then goto [ResonanceLoop]
'to calculate Q from Q= 2.Pi.f.L/R
Let Q=2*Pi*ResonantFrequency*TotalInductance/ResistanceOfCoil
Print "Q " ;Q
'Print "Q infinite"
'to calculate the total mass of copper in the winding Area times lengh
times density
let
TotalWindingMass=Pi*(((DiameterOfWire/2)^2)((DiameterOfLumen/2)^2))*LengthOfWire*DensityOfCopper
Print "Winding Mass ";TotalWindingMass;" Kg"
[PotentialLoop]
Input "Rms Potential ";RmsPotential
'to calculate efield
Let Efield=RmsPotential/LengthOfFormer
Print "Efield = ";Efield; " Volts/meter"
'to calculate slow wave
Print "Half Wavelength "; LengthOfFormer/2
Print "Velocity "; LengthOfFormer*ResonantFrequency
Print "Peak Potential "; 1.4*RmsPotential
Let
PeakCurrent=1.4*RmsPotential/(SQR((TotalInductance*2*Pi*ResonantFrequency)^2+ResistanceOfCoil^2))
Print "Peak Current ";PeakCurrent
print "Rms Current ";PeakCurrent/1.4
Let PeakAmpereTurns=PeakCurrent*NumberOfTurns
Let PeakAmpereTurnsPerMeter=PeakAmpereTurns/(LengthOfFormer/3)
Print "Peak Ampere Turns "; PeakAmpereTurns
Print "Plasma Current "; PeakAmpereTurns
Let
PlasmaCurrentDensity=PeakAmpereTurns/((InsideDiameterOfReactionTube/2)*LengthOfFormer)
Print "Plasma Current Density ";PlasmaCurrentDensity; " Amp per
meter^2, near centre"
Let PressureOnPlasma=
((PlasmaCurrentDensity*PeakAmpereTurnsPerMeter)*mu)/((DiameterOfFormer(InsideDiameterOfReactionTube))/2)
Print "Ion Pressure of Plasma ";PressureOnPlasma; " Pa"
Print "Power lost through resistance ";
ResistanceOfCoil*((PeakCurrent^2)/1.4); " Watt"
'Let
ReactanceOfExternalCapacitance=1/(2*Pi*ResonantFrequency*ExternalCapacitance)
'Let
ReactanceOfSeriesCapacitance=1/(2*Pi*ResonantFrequency*SeriesCapacitance)
'Let
PotentialAcrossSeriesCapacitance=ReactanceOfSeriesCapacitance*PeakCurrent
'Let
PotentialAcrossExternalCapacitance=ReactanceOfExternalCapacitance*PeakCurrent
'Power = 0.5 watt/cc/kPa^2 10torr=133 Pa
'Let
PowerGenerated=(0.5*(LengthOfFormer/3)*((InsideDiameterOfReactionTube/2)/1000)^2*2*Pi*((PressureOnPlasma)/1000)^2)*100^3
Let
PowerGenerated=(0.5*(LengthOfFormer)*((InsideDiameterOfReactionTube/2))^2*2*Pi*((PressureOnPlasma)/1000)^2)*100^3
Print "Power Generated if Deuterium ";PowerGenerated;" Watt"
Print "Power Generated by natural hydrogen "; PowerGenerated*24E9;"
Watt"
'Print "Potential Across Additional Series Capacitor
";PotentialAcrossSeriesCapacitance;" Volt"
'Print "Potential Across Main Capacitor
";PotentialAcrossExternalCapacitance;" Volt"
input "1 exit 2 loop "; ok
if ok=2 then goto [PotentialLoop]
'ignition winding, 2 mm wire rms potential/12 x number of turns in
ignition winding/number of turns in the resonance winding=1
'Let IgnitionPotential=12
'Input "Ignition Potential? ";IgnitionPotential
let IgnitionWindingTurns=(NumberOfTurns/RmsPotential)*IgnitionPotential
Print "Ignition winding turns ";IgnitionWindingTurns; "Turns wire"
Print "Sense winding ";IgnitionWindingTurns;" Turns wire"
Print "Local Power Winding ";IgnitionWindingTurns;" Turns wire"
Print "Brake Winding ";IgnitionWindingTurns*2;" Turns wire"
Print "Sense Winding ";IgnitionWindingTurns;" Turns wire"
input "1 exit 2 again " ;ok
if ok=2 then goto [Start]
End
03/06/2017 22:25:27
Transfusor diagram
25/06/2017 21:40:15
A linked topic is the Electric Jet Motor that may run from the
transfusor output.
Christopher
A more suitable material is ceramic with quartz glass as the reaction
tube.
End
25/07/2017 04:18:
info@transfusor.co.uk
The transfusor is dangerous because it gives off neutrons and a cat and a dog were killed when this apparatus was used. The winding fused and the internal organs were destroyed. Another test a 1/4 HP motor was powered by this apparatus. There is also the phenomina a secondary radiaton from activation by neutron bombardment of the materials of the appartus and the containment vessel. It cannot be used in any domestic environment such as cars or homes. It my be used well away from people and in well built shielded buildings as district power supplies.
A 1.5 volt pen cell connected to give a current pulse trough the winding was required to ignite thereactor. To control it a gas discharge tube (bourne) that conducts at a standard potential is connected across the winding.
I suggest a variable transformer be used here to alter the potential and to turn the device off.
To reduce the radiation hazard shielding must be used. I suggest graphite to moderate the neutron energy. Hydrogen may be used to absorb the nutrons and to create deuterium which may then be used as fuel for the transfusor.
292018 00:01
Calculations indicate a winding of 600T of 2 mm wire with a 0.1 uF 10000 volt capacitor gives the ionic presure needed to bring the deuterons close enough for the strong nuclear force to overcome the coloumb repulsion so cold fusion takes place.
Unfortunately the theory depends on the gaussian theory of magnetic shells is wrong and the theory of biotsavart is right. Measurements with a plotting compass used as a tangent galvanometer showed that the value of the field at the centre is as given by Ampere but rises linearly towards the wire reaching a maximum at the wire'ssurface this supports biotsavart.
The deuterium gas is made into a plasma by a dicharge along the tube and the current sheet induces an antiparallel current in the plasma. The plasma current is repelled by the winding current sheet and forced towards the axix and as the length gets smaller so the area is reduced and since the force is not changed the pressure risis towards infinity. At some point the Deuterons fuse to form helium3 with the emmission of a neutron and neutrino.
20190627 04:28
It is possible that very high ionic pressure could lead to the transformation of matter into electrical output...
The Power will depend on how long the plasma remains in the active zone so the duty cycle will be much smaller than 0.1.
1072019 06:08
the basic concept
The engine needs an enclosure, a radiation screen (neutrons) and a hydrogen blanket to soak up the neutrons and create deuterium. The engine runs on deuterium so it is a breeder. This D2 system needs designing. The output is alternating current at 5500 Hz. (c) Christopher Strevens 2019. I call this the ion compressor. This one had an ID of 28.1 mm, 40 mm wide with 937 turns of 1.6 mm wire.
completed skeleton transformer, it needs enclosure and output winding.
this shows the graphite screen and hydrogen blanket to convert hydrogen into deuterium for fuel  a breeder..
He3 that is made is converted to He4 by an additional Neutron with energy.
output transformer fits concentricly over the ion compessor and will depend on the application.
an example calculation for 200 amp bus is as follows:
I ran a test and got the odour of NO2.
Transfusor running in air with the production of NO2, N2O and electrity at 5500Hz.
the clamp meter may not work at 5500 Hz
output winding
control circuit placed in box
.
loop within loop energy tube
Two views of completed prototype transfusor just before I donated it to the nation. The startup oscillator is in the foreground
.
The outer tube contains hydrogen and the inner tube deuterium, The gas in the inner tube is ionised by the output of the ion compresor and neutrons from the fusion reactions are absorbed by the hydrogen which is transmuted to deuterium. Helium exits from the inner tube.
This is an outline diagram of proposed nuclear alternator
.
Circuit Diagram of proposed transfusor with brake, startup oscillator and output winding. The brake winding is connected to a triac circuit designed to short the winding above a preset level. An additional relay cuts off the startup oscillator once power is developed and turns on an indicator lamp powered by the reactor. The ion compressor winding is designed to oscillate at 50 Hz with a terminal potential of 6000V and current of 100mA.
Disclaimer: I used dimentional analysis to get the power to pressure relationship. The temperature term is missing because I thought that
Pressure alone would bring the nuclei close enough for fusion at absolute zero K. There should be a term with the Boltzmann distribution. If someone knows the relation between the power and temerature and pressure perhaps they would contact me.
We are now regularly and reliably transmuting hydrogen into helium.
We would like venture capital to continue as we are exceedingly poor..
info@transfusor.co.uk
Thank you,
Transfusor conference nearly 1 hour zoom video aparrently showing transmutation Deuterium to Helium .
videohttps://1drv.ms/v/s!ApOqJXw2X2WJjat1J9UqdFus0zhU1g?e=J7SDsq
Conference 20210331 Video about 45 min
We talk about the latest engine. The yellow line was caused by sodium vapour.
Anotated photograph of transfusor there is no brake or output winding
Transfusor conference 20210404
video about 1 hour where we discuss lates developments
Transfusor conference 742021.
Video 80 mins approximately where our test rig is described. Luke describes how the engine ran for some time with no exciter power. The test sample of deuterium now
has the green line of helium (Not shown in this video)
Transfusor conference 20210411.
Video app 40 Minutes we try different frequencie, potentials and currents without neutron detections
This is a mathcad calculation on the ion compressor winding. To replace the BASIC coding.
Another mathcad calculation where I attempt to insert the temperature dependance. A is arbitary and B is the threshold energy from a graph
A 900T Ion compressor at 60KHz.
A mathcad calculation for the 900 turn winding.
It should be impossible to make the pressure high enough but eperimentally it works.
Here I reason aloud.
New calculation using the gamow energy
Here I try to use the Gamow enegy to calculate the power possible from the reactor. I calculate the energy available from the work function
caused by the pressure forcing the ions together. This high pressure is only available near the axis of the reaction tube.
Transfusor conference 20210425
We discuss our present situation and that we have no success. About 40 minutes.
I think my equation is fundimentally wrong with the term 0.5 watt per cc. at NTP. It will be far less but not zero. However the ionisd gas in the reaction tube gets hot by ohmic heating and could reach mega kelvins in a few seconds. This is because the current is 7 kilo Amp and the ionised gas has a very low specific heat. So the reactor is a normal thermonuclear device after all. The pressures we anticipate are not high enough for cold fusion so the available energy to coagulate the nuclei is the energy through pressure plus the energy due to heat.
I hope we can continure experimenting to find the solution.
This assumes that the factor is very small but that the ionised gas reaches 1 billion K. The output is limited by the conductor
Although the theory predicts that it cannot work, experimentally it does. I think the pressure and the copper rings contain the ionised gas for a short time the induced current heats the ionised has to ten million K to induce deuterium coagulation.
(C) Christopher Strevens 2019.
(Cheshire)