Overunity.com Archives is Temporarily on Read Mode Only!



Free Energy will change the World - Free Energy will stop Climate Change - Free Energy will give us hope
and we will not surrender until free energy will be enabled all over the world, to power planes, cars, ships and trains.
Free energy will help the poor to become independent of needing expensive fuels.
So all in all Free energy will bring far more peace to the world than any other invention has already brought to the world.
Those beautiful words were written by Stefan Hartmann/Owner/Admin at overunity.com
Unfortunately now, Stefan Hartmann is very ill and He needs our help
Stefan wanted that I have all these massive data to get it back online
even being as ill as Stefan is, he transferred all databases and folders
that without his help, this Forum Archives would have never been published here
so, please, as the Webmaster and Creator of these Archives, I am asking that you help him
by making a donation on the Paypal Button above.
You can visit us or register at my main site at:
Overunity Machines Forum



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

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

Previous topic - Next topic

0 Members and 86 Guests are viewing this topic.

MarkE

Quote from: Farmhand on August 22, 2014, 05:56:45 AM
To me the FTW claim and the SERPS claim are more or less the same, they are saying that the VAR can be utilized not just continuously but also without paying for it in generation costs.

So it's all relevant.

If I have a 10 klm distribution line (not reality just thinking out loud) just say the line has 10 Ohms DC resistance,
then if we connect a load with a power factor of 0.5 and there is 240 VRMS and 10 amps of current to the load.

So we then have 2400 VA x 0.5 = 1200 Watts and 1200 VAR, so the current supplied is double that which is used.
Now all the current is still there 10 Amps. both ways with AC, so the losses associated with reactive power are not
from the reactive power itself
but from the excess current need in the first place. What is dropped is voltage
across the load. The current remains the same. So the potential energy in the volts x amps in phase is all the energy
that gets transferred to the load.

Therefore if a resistive load like a filament bulb is, through some miracle actually made to act "reactive" the filament would
have to pass more current than it should. And the out of phase current would not be "allowed" to heat the filament.



..
This is why industrial users get penalized for low PF, because it imposes reals costs for the power company.  In the 1980s the local power utility:  PG&E engineered the AC power distribution in a new building for HP in Silicon Valley.  They failed to account for the fact that  HP would install computers everywhere.  This was before power factor correctors, PFCs, became mandatory above 500W.  The AC wiring caught fire because of the reactive current sloshing back and forth between all the non PFC power supplies in the computers and the utility.  The red-faced PG&E had to pay for the building repairs.  Rather than rewire the place, they paid for PFCs installed in front of every computer.

Farmhand

Yeah, I forgot Thane Heins because he's a has been but still has a strong following. All these scammers need skinning.  ;)
Just like they're trying to do to others and they use places like this to make better their exposure. With no regard for the
safety or the expense it causes all the unwitting but very good intentioned experimenters. To me it looks like they should be
prosecuted in some way and jailed in some cases.

Anyone not trying to perpetuate a ruse or not actively and knowingly supporting one should have nothing to be concerned about.

I made a comment to one person that if they want to use their house wiring for central heating then using a Barbosa and Leal
device might work.
hahahhahahha

I think Thane Heins started this virus.

Someone should keep a tally of the reactive power OU claims since Thane first made his, just for curiosity. Quite a few.
..

Farmhand

Quote from: MarkE on August 22, 2014, 06:04:48 AM
The relation between VA, VAR, and Real power, S, Q, and P is:  VA2 = VAR2 + P2 as you applied to derive P = (VA2 - VAR2)0.5.  So, since Farmhand isn't observing anything close to 30W of real power dissipation, then probably the 78.4W was obtained incorrectly due to a phase angle error or other measurement anomaly.  If his dissipation is in the 5W range and the 83.5VA measurement is correct (should be with good true RMS measurements), then the reactive power should be more like 79.7W.   Just 2.7 deg phase error would account for the difference.

Can't be 5 Watts dissipated Mark, most would be 3.8 Watts which is the input. If we do not consider any input is coming from anywhere else then 3.5 Watts Max under the best possible conditions or so I would say. I get ya though.

So whatever that means to the measurement errors I'm not sure, and don't want to confuse myself at this point with it.
But I do think that 2.7 Watts is reasonable and so would be a bit more but I doubt it's less than 2.7 Watts. And regardless
of what the output power is the light is good for the input power.

Another thing I will do is make a checklist for everything I must take into account regarding ground loops ect. for better
measurements. Scope settings ect. ie. turn off the bandwidth filters and whatnot.

Umm, wouldn't the bandwidth filters tend to make the wave form smaller ?

PCB

QuoteSo we then have 2400 VA x 0.5 = 1200 Watts and 1200 VAR, so the current supplied is double that which is used. Now all the current is still there 10 Amps. both ways with AC, so the losses associated with reactive power are not from the reactive power itself but from the excess current need in the first place. What is dropped is voltage across the load. The current remains the same. So the potential energy in the volts x amps in phase is all the energy that gets transferred to the load.
Well said. This is also why transformers power ratings are quoted in VA and not watts.

MarkE

Quote from: Farmhand on August 22, 2014, 07:00:26 AM
Can't be 5 Watts dissipated Mark, most would be 3.8 Watts which is the input. If we do not consider any input is coming from anywhere else then 3.5 Watts Max under the best possible conditions or so I would say. I get ya though.
It is the scale from 25W to 5W that I am concerned with.
Quote

So whatever that means to the measurement errors I'm not sure, and don't want to confuse myself at this point with it.
But I do think that 2.7 Watts is reasonable and so would be a bit more but I doubt it's less than 2.7 Watts. And regardless
of what the output power is the light is good for the input power.

Another thing I will do is make a checklist for everything I must take into account regarding ground loops ect. for better
measurements. Scope settings ect. ie. turn off the bandwidth filters and whatnot.

Umm, wouldn't the bandwidth filters tend to make the wave form smaller ?
Yes, but if you have a filter set at 20MHz and are measuring at 300kHz the difference is pretty much imperceptible.