Real OU-Effect to Share with everyone!!!

[PaulLowrance]

Your source cap is 49000 uF, and it had 25 V

Input energy = 0.5 * 49 000 * (10^(-6)) * (25^2) = 15.3125 Joules (half C times V square)


Load cap is 10 uF, and it got 196 V max

Output energy = 0.5 * 10 * (10^(-6)) * (196^2) = 0.19208 Joules

Efficiency = Output/Input = 0.19208 / 15.3125 = 0.012544 that is way underunity.

Remember, all newbies - volts dont mean energy.

Sorry no OU here, I must move on.

My calculations of one of NRG's measurements in his video shows 123% efficiency. Even though this has not been cleared up, the reason I now believe this is not "free energy" is because of seeing further details of his core, and for other very good reasons that I'm not going to openly discuss at the moment. Legit researchers should know what's happening in this "free energy" case. It's a classical case.

PL

[gyulasun]

Scotty,

Thanks a lot for the clear vid. I wish everyone makes such clear vids, we can reach conclusions fast.

Your source cap is 49000 uF, and it had 25 V

Input energy = 0.5 * 49 000 * (10^(-6)) * (25^2) = 15.3125 Joules (half C times V square)


Load cap is 10 uF, and it got 196 V max

Output energy = 0.5 * 10 * (10^(-6)) * (196^2) = 0.19208 Joules

Efficiency = Output/Input = 0.19208 / 15.3125 = 0.012544 that is way underunity.

Remember, all newbies - volts dont mean energy.

Sorry no OU here, I must move on.

@Omega_0

Hi,  do not be so fast...

Scotty mentioned in his video and also wrote here that the voltage level he consumed from the 25V was between 1 to 3V, is that right?  So there was 22-24V still left in the 49000uF capacitor.  It is true he furnished in 15.31 Joule when he charged it up but consumed only much less from it, the rest remained in his capacitor, he did not use it for getting more output within one particular test.  Do you agree with this?

Now if we figure out the input power consumed from the capacitor, we receive the followings:

1) when 1V was used up, the used energy hence the input energy was 0.049/2=0.0245J   and even if we consider only 150V was collected in the 10uF,  the output energy was 0.5 * 10 * (10^(-6)) * (150^2) = 0.1125 J   Efficiency or rather let's call it coefficience of performance, COP: 0.1125/0.0245=4.59

2) when 2V was consumed from the 25V, the input energy used was (0.049*2*2)/2=0.098 J and still using the conservative 150V received in the 10uF, the output energy was 0.1125 J, the COP was 0.1125/0.098=1.14

3) when 3V was consumed from the 25V, the input energy used was (0.049*3*3)/2=0.2205 J  and now using the 196V (but remember Scotty mentioned higher than this voltage received when he lost 2-3V from the 25V) the output energy in the 10uF was like you calculated already:  0.5 * 10 * (10^(-6)) * (196^2) = 0.19208 J   the COP was 0.19208/0.2205=0.871  this is indeed underunity.
If the output voltage in the 10uF capacitor reaches 210V or higher (instead of the 196V) while only 3V is consumed from the 25V in the 49000uF capacitor, then a COP of 1 or higher is involved.  Scotty mentioned getting 300V in cases? when the voltage loss was 3V or around that?

I know the best case would be to find that when the 49000uF input capacitor is charged up to 3 or max 3.5V you could still receive over 230-280V in the 10uF;   or maybe continue with several 10uF capacitors to collect further 200V+ voltages in as many capacitors as possible to consume fully the 25V input voltage in the input capacitor, one after the other.

So is it really underunity, I wonder?

regards, Gyula

[gotoluc]

Hi everyone,

I could not resist to test this

http://www.energetictube.com/play/Vacuum_Radiant_Energy/NRG%20From%20The%20Vacuum%20replication%20test

I did not find any effects that favored over unity ... but that would not mean that NRG's claim is not correct!... I just was not able to replicate what he claims.

Luc

[Omega_0]

@Omega_0

Hi,  do not be so fast...

Scotty mentioned in his video and also wrote here that the voltage level he consumed from the 25V was between 1 to 3V, is that right?  So there was 22-24V still left in the 49000uF capacitor.  It is true he furnished in 15.31 Joule when he charged it up but consumed only much less from it, the rest remained in his capacitor, he did not use it for getting more output within one particular test.  Do you agree with this?

Now if we figure out the input power consumed from the capacitor, we receive the followings:

1) when 1V was used up, the used energy hence the input energy was 0.049/2=0.0245J   and even if we consider only 150V was collected in the 10uF,  the output energy was 0.5 * 10 * (10^(-6)) * (150^2) = 0.1125 J   Efficiency or rather let's call it coefficience of performance, COP: 0.1125/0.0245=4.59

2) when 2V was consumed from the 25V, the input energy used was (0.049*2*2)/2=0.098 J and still using the conservative 150V received in the 10uF, the output energy was 0.1125 J, the COP was 0.1125/0.098=1.14

3) when 3V was consumed from the 25V, the input energy used was (0.049*3*3)/2=0.2205 J  and now using the 196V (but remember Scotty mentioned higher than this voltage received when he lost 2-3V from the 25V) the output energy in the 10uF was like you calculated already:  0.5 * 10 * (10^(-6)) * (196^2) = 0.19208 J   the COP was 0.19208/0.2205=0.871  this is indeed underunity.
If the output voltage in the 10uF capacitor reaches 210V or higher (instead of the 196V) while only 3V is consumed from the 25V in the 49000uF capacitor, then a COP of 1 or higher is involved.  Scotty mentioned getting 300V in cases? when the voltage loss was 3V or around that?

I know the best case would be to find that when the 49000uF input capacitor is charged up to 3 or max 3.5V you could still receive over 230-280V in the 10uF;   or maybe continue with several 10uF capacitors to collect further 200V+ voltages in as many capacitors as possible to consume fully the 25V input voltage in the input capacitor, one after the other.

So is it really underunity, I wonder?

regards, Gyula

Gyula, you managed to confuse me there and for a second I thought its indeed overunity

Now, your mistake (as you have noted in the last para) is to take that 1 V drop as the INPUT. The real input is the energy generated by his 'magnet motor' or whatever it is. He could have used a battery or any other source, and the energy consumed out of that source is the real input.

To show real OU, scotty (or anyone who cares) needs to demo a setup which starts from 1V and still manages to get the other cap to 150 V.
Or he can try using identical caps and also measure the capacitance after the experiment not relying on their printed value.

So again , please see my calculation and I shall word it differently this time, from a wider perspective:

Energy in the system before shorting the connectors : 15.3 Joules
Energy in the system after shorting the connectors : 0.19208 Joules + (0.5 * 49 000 * (10^(-6)) * 24 * 24) = 14.30408 Joules
Where the last term is energy remaining in the 49000 uF cap with a voltage of say 24 V....
No gain.....

This circuit just re-distributes the energy and in the process loses a lot of it. Its like connecting two dis-similar charged capacitors together (through a coil) and disconnecting them before steady state is reached. Thats all it is.

[PaulLowrance]

You people enjoy using conventional capacitor equations. Don't limit yourself. Conventional magnetic equations show a gain of energy per cycle by means of magnetic viscosity and a change in *effective* permeability -->

http://globalfreeenergy.info

It's my hope that any legit people who post at this site will begin to realize that the odds of creating your own custom physics theory that can outdo present conventional physics (predict almost all known effects from the macro down to the sub-atomic), and that predicts more than conventional physics, perhaps some cool aether theory, is slim and none. I should know because in my early twenties I formulated two fantasy aether theories. Oh, they looked awesome at the time, LOL. I used some math, a lot of arm waving. People don't have a clue how many hundreds of thousands of effects and unique situations such theories must predict to even be equal to a conventional physics.

Conventional mathematics is all that's required to successfully produce a "free energy" machine. It's by far the quickest path, and I've already discovered it, and posted the details on my website. And no, I don't sell advertisement. I have always and will continue to refuse to accept any amount of money for this research. Just one remaining goal needs to be worked out-- searching for the exact materials and final design that will overcome all of the losses involved to make a self-running machine.

Conventional physics is not bad. It's not perfect, but no theory is perfect. The problem has been that conventional scientists have falsely believed 2LoT is a law. My diode research proved that 2LoT is a tendency. My trapdoor simulations consisting of particles clearly showed me how 2LoT is a tendency.

You don't need a college degree to do this technology. You don't need good spelling skills. You don't need good grammar. You don't need to know political science, geology, chemistry, biology, sociology, economics, history, philosophy, psychology, etc. etc. It should take someone less than a 1/2 year to learn nearly everything to do this research. It is guaranteed path!!

PL