OverUnity Demonstration ... by Introvertebrate

[nul-points]

Sorry no OU here.

the only thing not happening 'here'... is a considered inspection of other people's experimental results! 


in the case of Introvertebrate's test, he clearly states that he doesn't have a net increase in energy - but the increase in voltage is in violation of conservation of charge in his closed circuit

in the case of my experiments, i measure a clear excess (about 50%, in one test) of total Coulombs stored in the circuit after a test run

if anybody bothered to do the Coulomb calcs for this anomaly they would find that the total Coulombs stored on the two caps at the end of the test are greater than the total Coulombs stored at the start of the test

charge conservation has been violated

Voltage alone is not a measure of an increase in energy, you must consider the capacitance which is now one half since your capacitors are now in series.

i agree that it's of no value to connect the two capacitors together at the end and say, 'look, more voltage' - that's why i don't do it

it's the increased total charge in the system which is important (i'm sure Introvertebrate understands that)

this is not about capacitors in series doubling the voltage & halving the capacitance - the capacitors are connected by a switched circuit in these tests - not a piece of wire


in order to store Coulombs of charge into a capacitor, work must be done - if work is done, then energy is expended

if you've stored more Coulombs than you expected, you must have used more energy than you expected

a charge gain anomaly will always be accompanied by an energy gain anomaly

this has happened!  on a workbench near you!  to Introvertebrate - to NerzhDishual - to me - to people in other OU forums

these are real measured events

- we could deny they ever happened, and hope they go away

- or we could investigate further  ...because we've found something unexpected  ...because it could be useful


so - the big question is... is it possible for this 'unexpected' charge gain to be sufficiently high to enable the total energy use of the circuit to go overunity?

the results indicate: yes, it is!

in the case of my test circuit (which uses inductance and a resistive load to capture the usual 50% energy loss - a technique sometimes used in switched-mode power supply design) i'm measuring similar gains of Coulombs stored - plus excess energy converted by the circuit over total energy supplied

this is not theory - these are measured results

if anyone has any issue with this it will have to be on the basis of the experimental setup and measurements

** health warning: this will involve some reading, thinking and calculating **   

Your original charged cap started out with more energy (not voltage) than the final two capacitors in series. Even though their final combined voltage was higher their effective capacitance was halved by the act of putting them in series

Many starters make this fundamental error.

i'm sure they do

i don't connect my capacitors in series - i use a switched-charge circuit


this is not a time for the textbook - this is a time for the workbench!

...we can amend the textbook later

all the best
sandy
Doc Ringwood's Free Energy site  http://ringcomps.co.uk/doc

[tinu]

in the case of my experiments, i measure a clear excess (about 50%, in one test) of total Coulombs stored in the circuit after a test run

if anybody bothered to do the Coulomb calcs for this anomaly they would find that the total Coulombs stored on the two caps at the end of the test are greater than the total Coulombs stored at the start of the test

charge conservation has been violated

Everything is normal in your description exactly the way it should be less the last line, which is false. In a general setup if you step down the voltage, you?ll certainly increase the current, hence the ?charge?.
No offense, but you do not understand what charge conservation is. Please read more about it, otherwise you may create a lot of confusion.

if you've stored more Coulombs than you expected, you must have used more energy than you expected

Not necessarily at all!
See also the above explanation.

Cheers,
Tinu

[ORION]

I am interested to know what is the difference in energy between that in a capacitor that i have 'charged' to voltage 'V' from a battery and the same capacitor which has self-recharged to voltage 'V' after a full discharge

i ask this because if there is NO difference then even 'self-recharge' is free energy!

Dielectric absorption, the electret effect and PVDF  piezo films are probably worthy of further investigation.

[nul-points]

No offense, but you do not understand what charge conservation is.

no offense taken 

charge conservation can be viewed as an empirical law by considering the 'continuity equation' in electromagnetic theory

the continuity equation provides a relationship between the electric current density (Amps / square metre) and the charge density (Coulombs / cubic metre)

the equation basically states that if charge is leaving (or entering) a certain enclosed space then the rate of change of the remaining charge density varies with the opposite sign - ie. remaining charge density decreases with a removal of charge, and it increases with an addition of charge

- if charge density has increased in a closed system it must be as a result of charge entering the system from 'outside';
- if charge density has decreased in a closed system it must be as a result of charge leaving the system and flowing to something 'outside'

hence charge conservation: if no external flow of charge (in or out) occurs then the total charge density in a closed system must remain constant

what's not to understand? 

Please read more about it, otherwise you may create a lot of confusion.

i believe that other forum members, who are sufficiently interested in these anomalies, are perfectly capable of verifying for themselves any statements which i - or you - make

Quote from: nul-points on Today at 07:42:42 AM
if you've stored more Coulombs than you expected, you must have used more energy than you expected

Not necessarily at all!

absolutely at all! 

Q (Coulombs charge) = V (volts) * C (capacitance)

more voltage on the capacitor -> more Coulombs charge stored


E (Joules energy) = 1/2 * C * V * V = 1/2 * Q * V

more Coulombs charge stored -> more Joules energy stored


E (Joules energy) = W (Watts work) * t (time)

more Joules energy stored -> more work done


so - as i said above, if you've stored more Coulombs than you expected, you've also used more energy than you expected (more stored plus more as work done)

a charge gain -> an energy gain

a charge gain anomaly -> an energy gain anomaly

Dielectric absorption, the electret effect and PVDF  piezo films are probably worthy of further investigation.

i agree


Quote from: nul-points on August 08, 2008, 07:57:20 PM

i am interested to know what is the difference in energy between that in a capacitor that i have 'charged' to voltage 'V' from a battery and the same capacitor which has self-recharged to voltage 'V' after a full discharge

i ask this because if there is NO difference then even 'self-recharge' is free energy!

it appears that no-one can provide any difference...

...in that case, i suggest that capacitor 'self-recharge' (aka Dielectric absorption) can be considered as free energy also


all the best
sandy
Doc Ringwood's Free Energy site  http://ringcomps.co.uk/doc

[ORION]

Assuming  this is for real, what is the difficulty in arranging a series of switches, mechanical or solid state, that can make the anomaly into a continuous process rather than a one shot affair?

Has anyone gotten down to building such a device to harvest the energy on a continuous basis?

Shuttling the capacitors around so that the energy or charge keeps building?

What is the roadblock here?