Quantum Energy Generator (QEG) Open Sourced (by HopeGirl)

[TinselKoala]

There are people who are expert spreadsheet users who may be able to help. I'm a "basic" user and I think that spreadsheets are what make computers into real computers. But.... send a message to Redmond:

Forget MicroShaft! Use LibreOffice Calc, a fully functional fully compatible free and opensource spreadsheet.  That also comes along with the rest of "Office" too, like a word processor like Word, a presentation manager like PowerPoint, and etc.

If you are familiar with Exel, you will fit into Calc seamlessly. If you are new to spreadsheets _do not waste time and money_ on Microsoft! Start right away with Calc.

http://www.libreoffice.org/download/libreoffice-fresh/

While this is primarily a Linux system they also have Mac and Windows ports available at the download site.

It is easy to import CSV or other kinds of saved data files into the spreadsheet for analysis.

Here's an example spreadsheet using Calc that I made for the Ainslie thermal efficiency runs: (You will have to unzip it as the forum won't allow .ods type files)

[isim]

@Farmhand
Good, I will do the calculus for you this evening (for me). But is the problem is what you can not scale down the big spîke(" The spike is so big the screen won't accommodate it with the current wave at any reasonable amplitude"). Because this spike is (I think) the amorcage current of your fluorescent tube. Also, to have a good measurement of the spike (pulse), you absolutely need of a purely resistive CSR...
Edit: For the graph:    Yellow = Current * 1000  with (CSR=0.1ohm and inductance=?)
                           and   Blue = InstantPower * 2
  to get visible trace on the same graph...
@+

[Farmhand]

Yes definitely isim don't go to too much effort, this is just a test to see if there is efficiency or not, just to get an idea whats
probably going on and I can choose and buy/find the correct loading resistors and CSR's for better testing.

I think it could be possible that there could be a fair amount of displacement current going on as well due to the frequency and

I still need to decide and set up with all correct ground reference and scope isolation to rule out any ground loop situation as
much as is possible with such a setup. Thank you very much isim. Looking at the wave form again I can't believe the scope
could get a proper phase angle for that wave shape and display it. But real loads are not all purely resistive. Would be good
to be able to get better accuracy with odd shaped waves.

Tinsel, I found I already have Open Office, But I've got plenty of HDD space so I'll try Libre, not too confident that I'll get the
hang of it quickly though. Like with The Gimp I gotta learn what I need as I need it, there is so much to it.

Anyway, while I await parts for further testing I can see what happens when I put the two secondaries in series with one tuned
to 420 kHz and one tuned to 840 kHz. I can also get a Lissajous pattern with the two frequencies separate. Maybe I can find
some reasonable load resistors in the parts pile. I've got about 10 roller door motor units yet to strip for motor run capacitors,
one had a 24 volt DC motor in it and a transformer. Most are small "split phase" induction motors. Busy tomorrow.

..

[isim]

@Farmland

"Yes definitely isim don't go to too much effort,..."
No problem; It's a pleasure to help somebody full of good will!
Here is the power from your capture scope:
For the full conduction i got 0.5W (blue curve) . Don't forget the CSR is (maybe) inductive.
With 12.3Vdc et 0.300A as input,
          Pout=3.69W, the efficiency is very low, but for me it's normal.
I am curious to know wich fluorescent tube you use and also can you confirme that you connect this tube between the GND and the output phase 2...

Excel, or all free programs like this one, are excellent to do calculus. It's very easy to get the power from voltage and current log of your scope, do only voltage and current multiplication of each line for the instant power;  then there is a built in function to calculate the power on one period (MEAN(instant power of one period of the voltage or currant)). You can even plot the curve easily.  I will try to find a tutorial for you( in english).
@+

[Farmhand]

Hi isim, The fluro tube is a 300 mm F10T8CW = 10 watt - type T8 - cool white. And yes it was connected between the grounded
end of the secondary and the right hand phase 2, yes definitely.

Thank you isim .

Well if power out = 0.5 Watts then there is a problem, does the video look like 0.5 watts powering the fluro tube ? Not to me.

If I can get that light with 0.5 watts power consumed by the lamp then that is incredible. The input is irrelevant if we just look
at the light produced and the calculated consumed power things do not add up.  So either there is a miracle going on and
3.2 watts of the input is wasted while the fluro is lit brightly by 0.5 watts. Or the output of the odd shaped waveform cannot
be measured accurately by my scope captures. That 0.5 Watts doesn't make much sense to me.
Did you watch the video of the circuit behavior and the input and the light ? Whats your opinion ?

If the output to the fluro cannot be calculated then we just cannot say what the efficiency actually is. 0.5 Watts is not right
it can't be. So I'm not bothered at all. Definitely not OU but it shows how difficult it is to accurately measure the power from
an odd shaped current wave form. That's a win for one.

The problem is that I can't get an accurate current wave form as yet I think.

So does the output tank look as though it gives and takes or just gives ? How much does it give and to what ?
There is that to look at.

Anyway I'll get some resistors and measure a resistive load so I have all sine waves.

..

P.S. When I look at the wave form zoomed, it does look like the majority of the current is "out of phase' except for the spike.
It looks like a sine wave until the peak at almost 90 degrees then just after is the spike, the spike goes negative from
the positive side of the Zero crossing just after the sine peak and just after 90 degrees. So The consumed power you
calculated seems it could be correct and accurate, and probably a bit less even than is calculated without the CSR
inductance taken into account. Wow.  Kinda makes sense because the primary is DC and no power can be input on the
second half of the cycle- hence the current sine below the line for the second half of the cycle is more distorted.
Kinda does the opposite of the SERPS setup.
..
EDIT: CLICK !! I think I just got it, I'll arrange some scope shot sections from tank and load to try to show where I think the
power goes and how it seems the load uses only 0.5 Watt of power and how the load is powered.
This might help explain how the reactive device OU claimed by some is actually still coming from the grid. Maybe. I'll be back.
..
Basically the tube itself is the capacitor in this case and discharges it's charge at the beginning of the second half of the cycle.
That upsets the power calculations. The first half power trace is almost symmetrical and the second half is almost all positive.

And ... If I had an AC primary driver circuit then that spike would not appear I don't think, but the lamp would still light up brightly.
Power would then be symmetrical. Wow again. I need to test that.
..