Print

Please login or register.
1 Hour 1 Day 1 Week 1 Month Forever
Login with username, password and session length

Custom Search

[Magnethos]

How can you measure pure voltage?  A meter requires a small current to flow inside in order to get a voltage drop across it.  Pure voltage cannot be seen?!


D.

Maybe an electrostatic meter? A Scalar waves meter? 
I don't know

[TinselKoala]

Electrostatic sounds good. There will be an inrush of current while the plates go to their equilibrium positions, but once the voltage is on the plates and the instrument is indicating, there should be ideally no leakage current. Of course no instrument is perfect. But a good ES voltmeter won't even draw much current while it is coming up. I put up a video of an ES voltmeter hooked up to one of my mini VDG machines--its current is probably picoamperes--and you can see how quickly it comes up. This is probably the highest-impedance voltmeter type you will likely encounter, so it comes as close to a "pure voltage" measurement as you are likely to get, outside a standards laboratory.
http://www.youtube.com/watch?v=eogpGHFgV6E

[NRGFromTheVacuum]

Hi NRG,

The math looks good!  Can I ask though, and sorry if this has already been covered, you are quoting the printed capacitance values and of course they will have a manufacturing tolerance which could throw all the calculations off.  Have you measured the actual capacitance of each capacitor with a capacitance meter or perhaps used a known resistance and timed discharge method?


Thanks,

Dave.

Dave you make an excellent point,

I did indeed misinterpret the values written on the side of the supposed "1000uF capacitors".

They are .1 Farad capacitors, below is the address for the manufacturer's data sheet.

http://www.cooperbussmann.com/pdf/670fae07-c39d-4a13-852d-803aea8bc2a8.pdf

[fritz]

I can recommend vacuum tube (preamp) voltage meters.
>=100MOhm input resistance.
Maybe a little bit bulky - but - somewhat different than
what is used nowadays.

rgds.

[hartiberlin]

Looking closer:

Test #3 is:

-37.38 volts?

So #3 should be disregarded.  It was really just a preparation for test #4 (or #3 if you now relabel)

But still, #4 apparantly shows a 10% gain.... hmmm, wonder why....
Is the coil still hooked up in these tests?
And is the coil still hooked up to the battery?
And the pulse time of #4 was much shorter than #1 and #2 as evidenced by the smaller power input.
(1/2 the pulse as test #1)

P.S. It would be nice to have a wider view so as to see how the pulses are done and what leads are being used.
I see the yellow lead from the capacitor...
And the green lead.. isn't that from the battery?
etc. etc.

Okay, I see, I misinterpreted it as 373 Volts..
Surely my fault, cause I could not really see the dot on the meter.

Well, the 4th test is now with 110 % the only overunity result...

Also the new video of him with 5 x 100.000 uF caps in series are all
underunity.

http://www.youtube.com/watch?v=AYQnxDnHMiA

He  draws in the first experiment about 0.4 Joules from the 5 series caps and
gets only 0.096 Joules out at charging up the cap to 80 Volts..

It seems it depends on the cap´s inner resistance,
how good you can drive a high current pulse through the coil
at contact point.
As the 5 x 100.000 uF caps are all in series the inner resistance
is 5 times as high, so using the blue electrolyte cap in the other
video was much better, as it seemed to have less inner resistance.


So better try to use low resistance caps if you work with these systen,

Regards, Stefan.

[NRGFromTheVacuum]

Here are two pictures I captured from the video.

The first picture is the peak current draw from the blue cap before the voltage in the 30uF cap rises to 154.5 V (Minute 1:52)

The second picture is the peak current draw from the blue cap before the voltage in the 30uF cap rises to 145.7 V (Minute 1:21)

You can see that the more current I draw from the source the less energy I gain in the collector. (This is a key point in Free Energy Generation)

[CTG Labs]

Hi NRG,

Does the effect disppear if you use something other than the piece of iron on which to spark the connection?


Thanks,

Dave.

[NRGFromTheVacuum]

Hi NRG,

Does the effect disppear if you use something other than the piece of iron on which to spark the connection?


Thanks,

Dave.

Dave great question,

By "spark the connection" are you referring to the Zinc strip contacted by the two alligator leads?  Seen to the left here (Minute 0:34) http://www.youtube.com/watch?v=NhS4TxP29mM

I have used Brass, Nickle, Carbon, Zinc, Copper, and Magnesium strips in the past to connect A, B, & C wire http://www.overunity.com/index.php?topic=6734.0. None of which have an extreme negative/positive effect.

Besides the strips, I have only used the side of the 30uF capacitor case to connect AB&C wire.

I will try to include the results of using each material in future video's, which should be rather interesting.

You can find the strips here http://www.unitednuclear.com/chem.htm





 

[hartiberlin]

There is one event in the video at:
http://www.youtube.com/watch?v=4RGgrkMocCo

watch minute 1:07 and 1:09.

where the COP is at least 5 in my calculation.

Have a look at this:

http://www.youtube.com/watch?v=F3fxOqjSsyc

He calculates COP= 16,
depends on what exact voltages you read there.
Also the chargeup backside cap is only 30 uF and not 32 uF.

But this is extremly interesting and definately overunity.

My calculation got me COP= 5  with using the 
start voltage= 25.88 Volts and end voltage 25.87 Volts on
the 7500 uF supply cap C1 and the 30 uF charge cap C2 charged up to 25.62 Volts.

So lets calculate:

Before test:
EnergyC1= 0.5 x (25.88)^2 x 0.0075 Farad= 2.511654 Joules

After test:
EnergyC1= 0.5 x (25.87)^2 x 0.0075 Farad= 2.509713375 Joules

Energy lost in cap C1= 0.001940625 Joules

Energy in Cap C2= 0.5 x (25.62)^2 x 0.000030 Farad= 0.009845766 Joules

So COP = 0.009845766 Joules  / 0.001940625 Joules = 5.07

Regards, Stefan.

[PaulLowrance]

Hi Stefan,

That's definitely worth investigating!!  Another example is 1 minute and 50 seconds into the first video at -->

1:50 to 1:54
http://www.youtube.com/watch?v=4RGgrkMocCo

Blue cap: (0.5*22.07^2*7500e-6 - 0.5*20.23^2*7500e-6) = 0.29187 J
Gray cap: 0.5*155^2*30e-6 = 0.360375 J
Efficiency: 0.360375 J / 0.29187 J = 123%

That's over 100% efficiency!

PL