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Overunity Machines Forum



Russian Overunity Resonance Transformer

Started by synchro1, May 09, 2014, 03:46:33 PM

Previous topic - Next topic

0 Members and 1 Guest are viewing this topic.

avalon

Wrong patents. Look further.

Now for the most educated, look what I found on the Web: a SGS TUV certificate.
No mention of the second law of thermodynamics there, but a good read nevertheless. They do mention 9.4 times power increase though...

~A

synchro1

@Avalon,


Thanks! It's very refreshing to get some fine help for a change instead of ten tons of horseshit static:

Farmhand

Quote from: synchro1 on May 10, 2014, 12:57:01 PM
@Avalon,


Thanks! It's very refreshing to get some fine help for a change instead of ten tons of horseshit static:

The other thread keeps locking up with bad scripts or something

Anyway I'm not here to argue for ever over things that cannot be proved. I've already given my opinion. I only see pieces of documents.

What would you guys have me say ? Something like = OK then it's real I believe it, he can produce more energy out than in.

Thing is, what good does it do anyone unless it can be replicated.

Avalon, when was I previously looking at the wrong patent ? You write it like I do it all the time. I looked at the patent that was linked by number.

If I'm looking at the wrong patent and you want me to look at the right patent then post a link. I'll be glad to look at it. I'm not the one claiming someone else's stuff is OU so I ain't looking for it.

..


Farmhand

Where exactly does the snapshot below come from, it doesn't make much sense to me.

It says there is 400 Watts real power consumed at the input with 17.5 kVA and 9.8 kVAR. so if we subtract the real power 0.4 kW and the reactive power 9.8 kVAR from the applied power 17.5 kVA we get 7.3 kVAR missing. If that reactive power does not go back to the supply it gets paid for.

At the output it says 4.7 kW real power and 4.7 kW total power. so the output is 4.7 kW and the applied input power is 17.5 kVA and the reactive power is 9.8 kVAR with 400 Watts consumed, then where is the rest of the applied power ? 7.3 kVAR of it ?

Basically the reactive power back to the supply should almost equal the (applied power minus the output power).

..

synchro1

@Farmhand,


Here's the PDF once again posted by member avalon in his post number 119 from above. Wake up man!