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Overunity Machines Forum



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

Started by madddann, March 26, 2014, 09:42:27 PM

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0 Members and 12 Guests are viewing this topic.

TinselKoala

The point of all this stuff about my system is just to illustrate that you cannot take the raw scopeshots at face value. Corrections must be applied, that depend on load and source parameters as well as measurement techniques and instrumentation. The QEG folks really need to tone down their proclamations of OU based on their uncorrected scope measurements. They are wondering why it's so hard to self-loop a system that gives the measurements they are getting. The answer is simple: their measurements are what they are, and must be examined carefully and corrected as necessary to give correct values. When this is done, it's easy to see why they can't make a self runner: there is in fact no OU in their system.

MileHigh

Yes indeed TK, and I will throw out a mechanical analogy.  Sometimes mechanical analogies will help people, especially people that have no experience or understanding with electronics.

I suppose the first thing to say is that this mechanical analogy is not pie-in-the-sky, it literally is a simplified representation of what is taking place in the QEG.

Think of a large church bell at ground level mounted in a frame.  If you hit it with a hammer it will ring (resonate) for a few minutes.  Let's say the note is 'C,' which is 440 Hz.

The bell represents the primary LC tank circuit.

The spinning rotor can be represented by someone holding a hand-drill that spins an octagonal block of wood.  Let's imagine that the hand drill always spins at a rate of 440/8 Hz.  Since you have 8 faces on the wooden octagon, that means that the individual faces pass a fixed point at 440 Hz.

So if someone holding the drill touches the edge of the bell with the spinning octagon, the bell will ring louder and louder over time because it is being lightly struck at its resonant frequency.

Now the analogy for the light bulb load.  On the other side of the bell there is a person with big rubber gloves on.  You know intuitively that if this person grabs the edge of the bell with their big rubber gloves, they will feel the vibrations in their hands and arms and they will make the bell ring lower and lower over time.  They are sucking stored energy out of the bell when they grab it with their hands.  That's just like the light bulb load is sucking energy out of the primary LC tank circuit.

So there you have it:

This is the QEG:   A guy is making the bell ring with a spinning wooden block.  On the other side of the bell a guy is grabbing onto the bell with big rubber gloves and muffling it because he is draining vibrational energy out of the bell.

As you can imagine using your common sense, eventually things will stabilize and the power put into the bell by the drill guy will be in balance with the power sucked out of the bell by the rubber gloves guy.

Forget about quantum anything, it's not happening, the simplified mechanical description above is deadly accurate.

Here is the current problem:  The FTW group is just measuring the amplitude of the ringing bell when it is not being touched by the rubber gloves guy and calling it "over unity."

Just use your common sense and think it out.

MileHigh

TinselKoala

OK...so let's look at my "current" set of measurements in the images above.

The DC input is easy: 12.25V x 2.78 A = 34.1 W

The output V and I are sinusoidal and phase-shifted by about 55 degrees. The average power is then given by (Vrms x Irms)cos(phase angle). The non-inductive 0.5 ohm parallel resistors I used have a special winding and according to the data sheets have inductance below 1 nH, which should result in an impedance of less than 0.010 Ohm at 793 kHz.  But when I run the numbers using that as the impedance ( or 0.26 ohm) I get crazy values, so the impedance of the resistor+the probe setup must be a lot higher than that.

But even if I use 3 ohms as the impedance of the "non inductive" CVR at 793 kHz I still get A = 6.5/3 = 2.17 A p-p, so:
(64/1.414)x(2.17/1.414)x0.573 = 39.8 W

Farmhand

Quote from: TinselKoala on July 01, 2014, 07:44:04 PM
I think his "cannabis journey" must have been right into the heart of LaLa Land. Some people should avoid those kinds of journeys at all cost... they might not be able to find their way back home.

It's not good enough that they make the Free energy searchers look bad, now they make all the pot smokers look like
hallucinating hippies on Acid. >:( >:(  ;D ;D
It's not the pot that is the problem it is the freakshow of a human that smoked it or ate it whatever his preference.
TK's right some people shouldn't drink alcohol or take drugs before getting a grip on what is real and what isn't first,
it's a recipe for Visions, hallucinations, incoherent ramblings and possibly panic attacks.

What other group will they vilify I wonder.

Certainly didn't do himself or the cause any favors. Engineer my arse, what a crock of drivel.
If he qualified as an Engineer then may God help us all.

..

P.S. This thread needs a moderator to take care of the oversized image problem, it makes the thread very difficult to read
for those with smaller screens. I am paragraphing my posts so that they can be read without scrolling sideways.

I often disregard posts that require I scroll across the screen. It annoys me so. I'm sure I'm not the only one,
so if you don't want your post read then have it spread right across the page so people need to continuously
scroll from side to side.

..

MileHigh

Okay TK:

I will get serious about your setup.  Some of these points may have already been mentioned.
I haven't poured over every detail with a fine-toothed comb so please give me some
latitude.

In your first clip there was no observable phase shift.  Would that be because of the inductive
current viewing resistor?  I am not sure.  Needless to say I found that puzzling.

Now with a proper CVR you are getting a phase shift but the numbers still apparently crunch to over unity.

Then, Isim invoked the "back door check."  The power dissipated in the loop is just Irms-squared*R.  So
presumably the resistance of the loop is much less then either CVR and so the total power dissipated
in the CVR plus the loop resistance should be quite low - under unity.

But then shouldn't that mean that your cosine multiplication resultant should be equal to the
Irms-squared*R measurement?  If they don't then something is amiss.

With the ordinary wire-wound resistor exaggerating the current measurement, I felt the echos of the
aetheric streams of you know who.  She drove a Plymouth Satellite at twice the speed of light.   8)

Anyway, not sure what you think, but I have an idea for you!

It's get out of the box time!  Are you experienced?

Why not try sensing the current phase another way?  Can you simply tack on a sensor coil to the
main coil and set up a transformer coupling?  Perhaps even just a single turn.  Connect the output
to a high-impedance scope channel.  It will show you the rate of change of the current in the main
loop.  Since you are in pure sine wave territory, a 90 phase shift of the sensor coil output will
give you the exact phase of the actual current.  Perhaps a sensor coil like this eliminates the
problems of the inductance of the CVR?

MileHigh