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Author Topic: A look at Arie De Geus Patent No 1032759  (Read 56754 times)

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Concepts are fine but the latter Δt marker is not placed correctly.  The time delta should be shorter than 1.07μs when measured to the beginning of the reflection.

Hi verpies, good to see you again.

With "the beginning of the reflection" you mean like in the screenshot below? (996ns would mean an electrical wavelength of 150.5m and a velocity factor of 0.60)

Quote
Also, I would use a 47Ω pot to measure the characteristic impedance more precisely (... or at least 100Ω)

Ok,  i used a 50 Ohm multi-turn potmeter so the char. impedance came out at 28.8 Ohm.

Regards Itsu
« Last Edit: 2015-04-24, 15:34:03 by Itsu »
   

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Hi Itsu,

I am reminded of the inner construction of HP wide-bandwidth baluns that I used to use.  These use the fact that a pair of twisted magnet wires create a transmission line that is close to 50 ohms.  The balun used a twisted line pair wound as a toroidal coil on a ferrite ring core.  One end of the twisted pair was an unbalanced input while the other end was the balanced output.  The image shows the construction.  As I recall these had a bandwidth covering 10 to 500MHz.

You will note that your coil using paired wires has exactly this form except your coil is solenoidal and not toroidal.  You will also note that the balanced output yields anti-phase voltages on each terminal relative to ground, which is what you are trying to achieve.  You will also note that the only difference as regards the connections is that your open ended winding is connected to ground in this balun.  This does not use resonance, but there might be some peculiarities at resonance if your function generator and load impedances differ from the line impedance.  Hope this helps.

Smudge
« Last Edit: 2015-04-24, 16:45:49 by Smudge »
   
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All,

I ran an interesting set of tests using the coil arrangement I posted previously but rather than clutter up this thread, I've posted the results on my "Magneto Electric Induction" bench in reply #53.

Regards,
partzman

 
   

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Hi Itsu,

I am reminded of the inner construction of HP wide-bandwidth baluns that I used to use.  These use the fact that a pair of twisted magnet wires create a transmission line that is close to 50 ohms.  The balun used a twisted line pair wound as a toroidal coil on a ferrite ring core.  One end of the twisted pair was an unbalanced input while the other end was the balanced output.  The image shows the construction.  As I recall these had a bandwidth covering 10 to 500MHz.

You will note that your coil using paired wires has exactly this form except your coil is solenoidal and not toroidal.  You will also note that the balanced output yields anti-phase voltages on each terminal relative to ground, which is what you are trying to achieve.  You will also note that the only difference as regards the connections is that your open ended winding is connected to ground in this balun.  This does not use resonance, but there might be some peculiarities at resonance if your function generator and load impedances differ from the line impedance.  Hope this helps.

Smudge

Hi Smudge,

it sure looks similar, however, as can be seen in this toroid version of Aries patent, the unbalanced input (FG) comes in in the middle section of the capacitor and one of the coils far end which is different.
The balanced side (Vuit = Vout) is the same.

Somehow i cannot get any 180° shift between the voltages (or currents) like shown by partzman.


All,

I tried some higher value of capacitors (2x air variable 30-350pF) but the higher the capacitance (350pF), the lower the output peaks on the known frequencies (4.7Mhs, 6.2Mhz etc.)
Also no shift was seen or induced by manipulating the one or both variable capacitors.


Regards itsu

« Last Edit: 2015-04-24, 21:27:06 by Itsu »
   

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Scanning through the patent as linked by peter in his post #1 of this thread:

http://www.overunityresearch.com/index.php?action=dlattach;topic=2469.0;attach=13892

I notice the following about the capacitors used (slightly re-translated from Dutch)

"7. A Method and Apparatus, as in (6), wherein said capacitors have the same capacitance (and thus a 180° phase difference occurs)."
"10. A Method and Equipment, as in (6), wherein the capacitance values of said capacitors are below 10 picofarads."

This "below 10 pF" was also mentioned earlier in the patent when Arie was talking about "The capacitor values are at minimum about half a picofarad. "
 
This seems to be confirmed by my using higher air variable capacitors which with increasing capacitance show lower amplitudes.

So i have cut my wire capacitor (11pF) in half, and it now measures 6pF at both sides.
But allthough now indeed the amplitudes of the signals are higher, no phase shift is noticed in the lower (below 10MHz) range.
Only when using an scope isolated from ground i start to see a phase shift up till 180° upwards from 23MHz, but the amplitude is way below the supplied 5Vpp, see picture.

Not sure what to think about this,  could it be that the ground loops from FG and scopes are preventing the correct displaying of the signals?
Why only phase shifts above a certain (high) frequency?

Regards Itsu




   

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because when operating above the SRF of the coil it becomes capacitive not inductive or a negative inductor.

If the frequency can be found where the current is 180 deg then it might be worth us building a standalone transistor sinewave oscillator to pump up the amplitude
   

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Hmmm,  in post #50 i showed that the both coils have a Self Resonance Frequency of ± 820 KHz, but only at above 23MHz i see this phase shift occur  (only with an isolated scope).
I will try even less capacitance to see if this phase shift frequency decreases.


Found a strange statement in the above linked patent, its both in the original Dutch patent as in the translated one:

"A-B is the input AC voltage; C-D is the output"       A typo??


Regards Itsu
 
   

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Hi Itsu,

Been thinking about this circuit and I think the junction of the two capacitors is at the hot end of the function generator.  You have to use a frequency that is resonant with the L and one of the C's so that the generator is driving a series resonant circuit.  Then the input voltage from the generator is low as its is driving a low resistance but the voltage across that L is high.  Because the other wire does not make a series resonant circuit that high voltage in the driven wire travels down the transmission line to be reflected from the open end and comes back at twice the voltage if the line is a multiple of half wavelengths.  So you get twice that high voltage across the two lines at points A and B, and from the zero reference that appears as voltages that are 180 degrees out of phase.  Getting the line and the LC to both resonate together is the key and you may have to adjust the L by putting ferrite inside it.

Smudge
   

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Hi Smudge,   great post, it makes a lot of sense, some things that are unclear with me are:

# the "zero reference" point.  With that you mean the junction of the 2 capacitors which is the hot end of the FG?
# also you say: "Because the other wire does not make a series resonant circuit.....etc."  
  Here you point to the other wire/coil with capacitor, Right?
  But why does it not make a series resonant circuit? It is an LC, right?  It does not make a LC circuit in relation to the FG is that what you mean?
# As the both wires/coils are on the same former, putting ferrite inside it does change the both wires/coils resonance points, but at the same time.
  So no adjustment between the both coils will occur.
  

As can be seen in my post #50 above i wrote:

Gr inductance 828.3uH    1st/highest resonance 823KHz @ 90Vpp, 2th resonance at 2.277MHz @ 77Vpp
Rd inductance 828.2uH    1st/highest resonance 818KHz @ 87Vpp, 2th resonance at 2.294MHz @ 77Vpp

As the capacitor was cut to be 6pF each, we have a series LC of 828uH and 6pF, see diagram.

This resonance calculator: http://www.1728.org/resfreq.htm   calculates the resonance to be at 2.258 MegaHertz which is very close to the above measured 2th resonance of 2.277Mhz
(not sure what the first resonance point of 823Khz is (3th subharmonic?), but this 2.277 is the full wavelength resonance point).

So at 2.277MHz we are at series full wave (is two ½ wavelengths) resonance having a very low (close to 0) impedance.


To confirm this theory, i would like to "catch" the both 180° out of phase signals with my scope, but need to know this "zero reference" point and if known, then i need to measure it without disrupting it with the FG/scope ground leads.


Thanks,  regards itsu
   

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Hi Smudge,   great post, it makes a lot of sense, some things that are unclear with me are:

# the "zero reference" point.  With that you mean the junction of the 2 capacitors which is the hot end of the FG?
No, it is the ground point you show in your diagram smudge1.
Quote
# also you say: "Because the other wire does not make a series resonant circuit.....etc."  
  Here you point to the other wire/coil with capacitor, Right?
  But why does it not make a series resonant circuit? It is an LC, right?  It does not make a LC circuit in relation to the FG is that what you mean?
Yes.  It can't be a series resonant circuit because the other end is open so you can't pass current through the L.
Quote
# As the both wires/coils are on the same former, putting ferrite inside it does change the both wires/coils resonance points, but at the same time.
  So no adjustment between the both coils will occur.
It is not adjustment between the two coils that matters.  You have to realize that there are two distinct resonances.  One is the simple LC series resonance.  This has no connection with the time delay down the transmission line.  The other resonance is that transmission line when it is a multiple of a half wavelength.  So you have two different resonant frequencies that must coincide.  You therefore need to measure both (which you have effectively done, you quote the L and C values and you have done a TDR measurement on the transmission line).  That should tell you what value capacitor you need to sync those two together and I only suggested ferrite as a fine adjustment assuming you would have a fixed value capacitor.

The output from A and B is at high voltage and high impedance so you must use high value load resistors.  An with such low value capacitors you can't hang 10pF scope probes there, that will screw things up.  You need potential dividers as the load so you can connect scope probes at a lower impedance point, see my modification of your diagram.
Quote
As can be seen in my post #50 above i wrote:

Gr inductance 828.3uH    1st/highest resonance 823KHz @ 90Vpp, 2th resonance at 2.277MHz @ 77Vpp
Rd inductance 828.2uH    1st/highest resonance 818KHz @ 87Vpp, 2th resonance at 2.294MHz @ 77Vpp

As the capacitor was cut to be 6pF each, we have a series LC of 828uH and 6pF, see diagram.

This resonance calculator: http://www.1728.org/resfreq.htm   calculates the resonance to be at 2.258 MegaHertz which is very close to the above measured 2th resonance of 2.277Mhz
(not sure what the first resonance point of 823Khz is (3th subharmonic?), but this 2.277 is the full wavelength resonance point).

So at 2.277MHz we are at series full wave (is two ½ wavelengths) resonance having a very low (close to 0) impedance.

That 2.277MHz is the LC resonance and has nothing to do with the wavelength down the transmission line.  You must adjust the frequency (by changing capacitor values) so that it coincides with the transmission line resonance.  Why don't you try driving the transmission line itself from just one end to find its resonant points, that would help in getting you to the right point.

Quote

To confirm this theory, i would like to "catch" the both 180° out of phase signals with my scope, but need to know this "zero reference" point and if known, then i need to measure it without disrupting it with the FG/scope ground leads.
Answered above.

Regards
Smudge
   

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Very nice Smudge,  that clears up many things.

As usual for me, this simple circuit has much more to it then i realized.

I have to treat those 2 distinct resonances in a different way ("simple LC series resonance" and "transmission line when it is a multiple of a half wavelength").

Thanks,   regards Itsu
   

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OK, I have just looked back at your earlier posts.  Your transmission line is 90.24m long.  You measured 20nF.  That is a distributed capacitance of 222pF/m.  You also measured its characteristic impedance Z as 30 ohms.  Now since Z=sqrt(L/C) where the L and C are the distributed values we can get the L as 0.2uH/m.  The time delay per metre t=sqrt(LC) comes out at 6.66nS which for the total line length of 90.24m is a one way delay of 0.61uS.  Two way delay is 1.22uS which is close to the 1.07uS you measured.  Thus the first resonant point for the transmission line calculates as 832KHz.  That explains your mystery first resonance that you measured at 823KHz.  So you have to get your LC resonance to coincide with multiples of 823KHz which you are close to with your 6pF capacitors but not quite there.  Keep trying.

Smudge
   

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Smudge, that is clear all the way to the 823KHz  :(

Still puzzled about that as it seems to me having no relation to either my measured 2-way reflection pulse of 1.07us or your calculated 2-way delay of 1.22us:

full wavelength 140.1m (from the 1.07us reflection pulse) which is a frequency of 2.141MHz
full wavelength 122.8m (from the 1.22us reflection pulse) which is a frequency of 2.442MHz

What is this 823KHz resonance point?

Thanks,  Itsu 
   

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Smudge, that is clear all the way to the 823KHz  :(

Still puzzled about that as it seems to me having no relation to either my measured 2-way reflection pulse of 1.07us or your calculated 2-way delay of 1.22us:

full wavelength 140.1m (from the 1.07us reflection pulse) which is a frequency of 2.141MHz
full wavelength 122.8m (from the 1.22us reflection pulse) which is a frequency of 2.442MHz

What is this 823KHz resonance point?

Thanks,  Itsu  

Don't know how you calculated those wavelengths.  The 1.07us reflection represents a waveform of period 1.07uS for 1 cycle and that is 935KHz.  The 1.22us is 1 cycle period of 820KHz.  When you are dealing with times you don't need to go into wavelengths.

But just to get this right the calculated time delay per metre was 6.66nS which is a velocity of 1.5x108m/s.  At that velocity the wavelength of an 820KHz waveform is 183m and that is twice the length of your line or near enough.  Methinks you have got in a muddle with your calculations.

Edit.  And how can an increased time delay (going from 1.07 to 1.22uS) result in a reduced length (from 140.1 down to 122.8 )?  Something wrong in the formula used here.

Smudge
   

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Ok,  i screwed up the math from time to wavelength somehow.

But what about my measured 1.07us pulse.  This is the forth and back time of my pulse, so the single route time would be half of it = 0.535us.
So the waveform of period 0.535us for 1 cycle would be 1.869MHz, not 935Khz.          (or do i see a half wave (half cycle) as resonance?).

Regards Itsu
 
   

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Yes the open ended line is resonant when it is a half wavelength long.  So the one way time is only half a cycle.
Smudge
   

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Let's be clear about this.  The transmission line is resonant when the reflected wave comes back to reach the input in phase with the signal there, hence it adds to the signal and creates twice the voltage.  Thus the resonant line acts like the familiar parallel tuned circuit offering a resonant peak (not a trough although that can be arranged).  To achieve this the two way time has to be exactly one cycle period (for the first resonant peak or integer multiples of that time for other peaks at higher frequency).  That means the one way time is exactly a half cycle, the line is electrically one half wavelength long.

If the half wavelength line were terminated at the far end with a short circuit, the reflected wave would come back in anti-phase and would subtract from the input voltage.  The line would then look like a series resonant circuit yielding a voltage trough.  To get the peak resonance with a shorted line you need a quarter wavelength line or any integer multiple of quarter wavelengths.  Actually a shorted line is better because you can get a definite short there, whereas an open circuit suffers from stray capacitance and the open circuit is not so well defined.

Smudge
   

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Quote
With an arrangement of the embodiment of Fig . 3b and a cylinder length of the double winding of 33 cm and a winding-diameter of 16,5 cm - windings , wherein wire diameter ( with insulation
included , so called magnet wire ) of 1.1 mm was used, it was in the application of a square-wave
voltage of 21 volts, with a frequency measured at or near 8.5 MHz, a power increase of close to 8%
. ( measurement done with instruments that are accurate to 8.5 MHz ) .
If one has  input voltages at higher frequencies or could have there is therefore the possibility to
generate free electric power , and at any scale level . It is possible by properly choosing the length
of the wires , inductances and capacities with respect to the input frequency with a simple circuit ,
as described above,to bring frequencies at a high level with very low losses

I wonder why he uses such a high frequency for a coil that is 16.5cm diameter and 33cm long, the resonant frequency has to be way lower than 8.5MHz
   

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OK i think i have worked things out to a degree.

There are only 2 things we need to concern ourselves with to get this working.

1 we need as large an amplitude as possible, this is achieved by adjusting wire length, capacitance values.

2 This is the most important thing see quote below from the patent.
Quote
The zero point energy absorption process does not work if electrons flows move next to each other
in the same direction .

Forget about 1 lets concentrate on 2, so how can we guarantee that our electron flow in adjacent bifilar coils is not the same?
If we can answer that then all we need do is achieve a large drive amplitude efficiently using a transistor square wave generator.

I would have thought that we could do this by driving both coils 180 degrees out of phase, but this may use twice the power so maybe Arie has a more efficient way of doing this by driving only one and having the second open.

How do we monitor the electron flow phase is this our current or voltage.
OK so our electron phase is our voltage phase not current.
   

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Hi Peter,  i agree, point 2 is the first condition we need to match.

Therefor my quest to have this 180° phase shift realized.

The trick to realize this is in the open ended coil, which should present a mirror of the signal on the other coil.
But as Smudge tried to explain, this will only happen under certain specific conditions.
The transmission line coil (open ended one) should match with its length and therefor with its standing waves the series resonance
frequency of the other coil/cap LC.

I have set up a voltage divider of 10Mohm/1Mohm on each A and B point to ground (see Smudge his drawing in post #60) and measure with my probes
across the 1Mohm resistors to ground.
But still the signals are in phase, and i have trimmed my capacitors from 11pf up till 1pF now.
The only effect noticed is that the resonance points frequencies slowly increases, low 1st frequency was 823KHz, now 1200KHz, 2nd was 2770kHz, now 3600Khz.

Regards Itsu
 


   

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It just occurred to me that if the coils were wired in anti series then our inductance would be small, thus our resonant frequency much higher than at the moment, this would mean that the output indicated on the drawing is actually our input, which you stumbled on the other day in the patent.
   

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Peter,

not sure what you mean by "anti series", do you mean "counterwound" as mentioned by partzman in his post #36?
If so, i did some tests with that config (swapped 1 coil leads) and sure we get 180° phase shifts then.
Not sure i understand your input / output swap statement, could you please draw a diagram?

Regards Itsu
   

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Ok,  i did screw up the math of the time/wavelength calculations as stated in post #30 and to be seen in the video there.
I did ask to check it   ;D

Here a retry.
As can be seen in the screenshot below, the forth and back reflection pulse is 1.07us depending on where the 2th marker is positioned.   
I think it should be in the middle like shown (1.07us), but others mentioned that it should be on the start of the pulse which then would measure 996ns.   
For the time being i keep my 1.07us which i have measured severall times now.  (why does my 3ns wide pulse get smeared out to this ugly 200ns or so reflection pluse?  Capacitance?)

So the 1.07us is the forth and back reflection on my 90.24m (measured) long transmisstion line.
The single length time reflection time then would be 1.07us / 2 =  0.535us.

As the speed of light is 299792458m/s = 299.792458m/us the pulse traveled 299.792458m * 0.535 = 160.38m.
So the electrical length of my transmission line coil is 160.38m

As my transmission line is only 90.24m long, the calculated velocity factor of it is 90.24/160.38 = 0.56.


So we have:

a  physical transmission line length of    90.24m
an electrical transmission line length of 160.38m
a velocity factor of                                 0.56
char. impedance of                                 30 Ohm
capacitance measured                            20nF
Inductance of each coil                          828uH

According to http://www.wavelengthcalculator.com/ a wavelength of 160.38m corresponds with a frequency of 1.870MHz (1 cycle), so half a wavelength resonance would be at  935KHz.

This 935KHz is what i found as my 1st reonance points, but this is creeping up in frequency while trimming my capacitors (823KHz with 11pf, 935KHz with 6pf, and presently 1180KHz or so with about 1pf)

My idea was that this transmissionline frequency (open ended coil) should stay fixed no matter what capacitor value i connect.
What should (and does) change when trimming the cap is the LC resonance of the series LC circuit (not open ended coil)  :D

I made a 1Mohm/10Mohm voltage divider to probe the both A and B outputs as suggested by Smudge,  see diagram.
Works fine, but no phase shifts noticed yet.

Regards Itsu
   

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What do the two output voltages look like?  Is there a difference between their amplitudes?   Is one voltage 3 times the amplitude of the other so that the difference is 2 times.  What I am getting at is that although A and B are in phase, if B is 3 times A then you have exactly what Arie is saying, there is twice the A voltage across the transmission line.  From the transmission line perspective that is like a balanced line relative to ground with positive and negative voltages, whereas in fact it is not relative to ground.  Just a thought.  It's what goes on inside the transmission line that is important.

Smudge
   

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Smudge,

see the screenshot, a video of the setup and connections used is being uploaded right now.

Video here: https://www.youtube.com/watch?v=hBuyy3VK95I&feature=youtu.be

Regards Itsu
« Last Edit: 2015-04-27, 18:19:22 by Itsu »
   
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