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Overunity Machines Forum



ENERGY AMPLIFICATION

Started by Tito L. Oracion, February 06, 2009, 01:45:08 AM

Previous topic - Next topic

0 Members and 13 Guests are viewing this topic.

lancaIV

Quote from: nix85 on June 13, 2020, 06:54:35 AM
I mentioned on some other thread how some guy in Eric Dollard's lecture from late 90s i think on youtube asked the audience to ask themselves how no flux cuts transformer secondary, yet voltage is induced.

This model with etheric smoke rings and magnetic field just an axis of the ring would explain that beautifully!

http://freenrg.info/Physics/Scalar_Vector_Pot_And_Rick_Andersen/Rick_Andersen_Ortho.html
But happens not also here a " Magnus effect", the formation from a " death - Kalmen/ standing still - zone ?



http://translationportal.epo.org/emtp/translate/?ACTION=description-retrieval&COUNTRY=DE&ENGINE=google&FORMAT=docdb&KIND=A1&LOCALE=en_EP&NUMBER=3217763&OPS=ops.epo.org/3.2&SRCLANG=de&TRGLANG=en


Furthermore, the flow velocity in the vicinity of the axis is reduced to such an extent that there can be talk of a standstill in the air, which is then also accelerated towards the wrong side. The yield of the wind energy received is correspondingly low.




this eddies or whirls to understand is very important by the effective application as  "sputting process" of solar-/thermovoltaic panels :
https://www.google.com/search?q=solar+panel+sputting&rlz=1C1AVFC_enPT797PT797&oq=solar+panel+sputting&aqs=chrome..69i57.11192j0j7&sourceid=chrome&ie=UTF-8


https://en.wikipedia.org/wiki/Sputter_deposition

lancaIV

Quote from: Magluvin on June 10, 2020, 07:39:17 PM
You are not figuring in what it takes in energy to fill the cap to 40v vs 20v.  At 20v there is an amount of energy there which is 20mj for 100uf.  And 100uf at 40v is 80mj.  So are you saying it doesnt take more energy to charge the 20v cap up to 40v than it did to just charge the cap to 20v in the first place? ???
I understand what you are thinking, but it is wrong. Just like in a car audio amplifier, if the rails are +20v and -20v and the rms sine output before clipping is 14.14v, then that divided by 4ohms speaker is rms 3.535amps, then the power sent to the speaker is 49.9849w..  But if the rails are 40v + and - we calculate the same. 28.28v sine just below clipping gives us 7.07amps rms to the speaker which is 199.9396w.   

So you cant think of energy and voltage numbers as linear.  get it? ;)
Mags


https://static.weg.net/medias/downloadcenter/ha0/h5f/WEG-motors-specification-of-electric-motors-50039409-brochure-english-web.pdf


page 7 : torque = force x distance


page 11 : Effective value of voltage and current ( U and I )


It is the value of the continuous voltage and current which generate an output corresponding to that generated by the alternated current. We can identify the effective value as: U = Umáx / 2 e I = Imáx / 2 .


For example: If we connect a "resistance" to an AC circuit ( cos ϕ = 1 ) with Umax = 311 V and I max = 14. 14 A. the developed output power will be:


P = 2.200 Watts = 2.2 kW


Note: usually, when referring to voltage and current, for example, 220 V or 10 A, without mentioning any other factor, we are referring to voltage or current effective values, which are normally applied.


                                                                                                       Peak and average !


                                                                              311 V x (5/7) x 14,4 A (5/7)  to  220 V x 10 A = 2200 VA

Magluvin

Quote from: lancaIV on June 13, 2020, 08:01:11 AM

https://static.weg.net/medias/downloadcenter/ha0/h5f/WEG-motors-specification-of-electric-motors-50039409-brochure-english-web.pdf


page 7 : torque = force x distance


page 11 : Effective value of voltage and current ( U and I )


It is the value of the continuous voltage and current which generate an output corresponding to that generated by the alternated current. We can identify the effective value as: U = Umáx / 2 e I = Imáx / 2 .


For example: If we connect a "resistance" to an AC circuit ( cos ϕ = 1 ) with Umax = 311 V and I max = 14. 14 A. the developed output power will be:


P = 2.200 Watts = 2.2 kW


Note: usually, when referring to voltage and current, for example, 220 V or 10 A, without mentioning any other factor, we are referring to voltage or current effective values, which are normally applied.


                                                                                                       Peak and average !


                                                                              311 V x (5/7) x 14,4 A (5/7)  to  220 V x 10 A = 2200 VA
I get what your saying. I just didnt do the rms calc on that example. Either way it shows that energy and voltage level on a load are not a linear relationship.
Mags


lancaIV

Tito,I hope you are well !

But I have to remark : actually the use from a battery as electric source is in the 0,15-0,3 USD range ( calculating also included charger  )

This inverter feed back cycle system gives you never electricity under 25 USD-cents/KWh ,only cheaper you get this donated !

Grid-free electricity : yes ! Cheap : No !

I calculate with 75 USD/KWh storage lead battery with 300 charge cycles conventional and 750 charge cycles by pulse algorithm charging !

In the Phillipines the consumer do not pay more than 15 USD-cents/KWh,or do they ?

Nice successfull week wishing and CoVid19 beware

OCWL

p.s. : probably 2025 we get  hybrid batteries in the ≤ 5 USD-cents/KWh electric power  range comercial ,when then lithium ion batteries will be available for 0,1 USD/KWh electric power  and 100 USD/KWh electricity capaci5y  storage !