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free energy circuit setup

Started by FreeEnergy, April 01, 2006, 02:35:41 PM

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lancaIV

Hello(salute)NerzhDishual(vulgo FE),
I can read italiano i espanhol but without "sufficient ambient sphere"
I do not write or speak "foreign" !

I am not a "patent freak",
these works represent many thousands  hours of menwork and
billions of invested money units in materials,tools,machines and more !

To get the "patents" in printable version I recommend the depatisnet
web-side !

After reading the Chambrin-WO publication the study of the work
from Mr.David Hudson(Google:keelynet,ORMES) becomes a little more clear,
the relativity: iron to H2(O) derivates and element transmutation !
Electricity:What element ? Transformer,motor/generator core :
which material ?

Sincerely
            de Lanca

NerzhDishual



Salve Valveman!

Thank you very very much for answering.

What am I expecting?
First of all: setting up a suggested funny experiment, should I be allowed doing it?
Else: tying to understand something. 

Actually:
Please, just check out any (undergraduate) physics textbook.
I have one of it in hands. (I could scan it but it is in French language).

Lets take a C farad  (C1) capacitor, charged with V volts.
You get (energy) W= 1/2 * C * V^2 joules.
Example :
I used a 0.1 farad cap. charged at 12.9 volts.
The energy w is : W= 1/2 * 0.1 * 12.9 * 12.9 = 8.32 joules

Lets discharge this C1 cap into another similar C farad one (C2). With no motor(s).

What do you get?

The two capacitor are now charged at V/2 volts each.
The new energy w' becomes:
W'= 2* (1/2 * O.1 * 6.45 * 6.45)= 4.16 Joules

What the heck is this result?
Have I lost half of the Energy?
Is the energy not conserved?

Not at all, said my physics textbook!
The reasoning is (sorry for my translation, but the French text is not so limpid):

There have been production of another energy that the electric charging energy of the second cap. The wires between the two cap have been travelled by an electric current which have produced in these wires a calorific energy w''. This calorific NRG does not depends upon the resistance of the wires and represents the difference w-w'' wich equals W/2.

Ok scientists...
So, when I put a motor betwen the two caps, I guess that the calorific NRG should vanish as I have the same results (V/2 for each cap) plus a motor running.
Sixthflyingman was just asking : What about having two motors running?

Sorry, but I can not help smelling a rat.

Le bonjour chez vous...


Nolite mittere margaritas ante porcos.

NerzhDishual



Hi LancaIV

I am not sure to always understand you fully. ???

Could you tell us what  the URL of the Depatisnet web site is?
Thanks.

Best
Nolite mittere margaritas ante porcos.

lancaIV

Hello NerzhDishual,
http://depatisnet.dpma.de
I hope so !
You wrote in your  last (valveman) post about "caloric",
did you read the Chambrin or Trilles publication ? Refrigeration.
I think that we have to accumulate Lorentz with Carnot !

Sincerely
            de Lanca

p.s.: "sangue",blood:which materials ?

pg46

Hi All-

I have done the sixthflyingman experiment using two 22,000uf caps and two small motors as per diagram by NerzhDishual. I have run the test many times over.
The 2 motors spin just as long as when we only use only one motor to charge c2 as in test #1.  Sorry, as I don't have the rpm recorded or any torque figures. So seems you can spin motor #2 for free just as long as motor #1  :)
Another curious thing is I always end up with slightly more volts than I begin with. For example starting with C1 charged to exactly 14V and then discharging through the motor(s) to C2, I will end up with 7.1 or 7.2 volts in each cap??
I have to admit that my meters are not the best in the world but I have two different models getting the same results. Anyone finding the same thing or can offer any explanations?

Best,