OverUnity Demonstration ... by Introvertebrate

[nul-points]

@Tinu

i think you'll find the 12V initial charge making two charges of 6+V was just showing you what happens when you connect the two capacitors together manually without the 'charge-switching' circuit

the title text before the vid clearly states "12V + 0V = 15V ?"

it is not necessary to lose the initial 50% energy, as i mentioned in my previous post, if you modify the circuit to make use of the charging energy - my circuit does this and converts at least 20% more energy than the total energy used from the input capacitor

this is not being 'amazed by an apparent 200% efficiency' just by preventing energy loss - this is real energy gained over the total supplied

let's look at the charge: in one of my first experiments i started with a total initial charge of 2 Coulombs - and finished with a total charge of 2.9 Coulombs - this is not an imaginary gain! 

ok, now let's look at the energy:  the total initial energy stored in the input capacitor is 9.57 Joules, the final energy stored in the input capacitor is 7.33 Joules

therefore, total energy input to the circuit: 2.24 Joules

energy converted by the circuit:
  0.25 Joules used by the switching oscillator
  1.22 Joules passed thro' load resistor to charge up output capacitor
  1.22 Joules discharged thro load-resistor from output capacitor at end

Total energy converted by circuit: 2.69 Joules

Energy Quotient of circuit: 2.69/2.24 = 1.2

and the charge anomaly that accompanied this result:
   start charge: 2.39 Coulombs
   final charge:  3.0 Coulombs

...this anomaly can provide very real overunity!

have a nice day
sandy
Doc Ringwood's Free Energy site  http://ringcomps.co.uk/doc

[tinu]

let's look at the charge: in one of my first experiments i started with a total initial charge of 2 Coulombs - and finished with a total charge of 2.9 Coulombs - this is not an imaginary gain! 

Total charge is zero. It always was and it will remain so. Charge conserves until otherwise proved.

If you want to talk about charge on ONE plate of a capacitor, that?s a different story and it is not relevant as long as you decouple it from potential in respect to the other plate. So, you have to decide if you want to connect the final capacitors in series (in which case you?d have indeed 15V but only half ?charge?) or in parallel (in which case you obviously don?t have 15V any longer).

ok, now let's look at the energy:  the total initial energy stored in the input capacitor is 9.57 Joules, the final energy stored in the input capacitor is 7.33 Joules

therefore, total energy input to the circuit: 2.24 Joules

energy converted by the circuit:
  0.25 Joules used by the switching oscillator
  1.22 Joules passed thro' load resistor to charge up output capacitor
  1.22 Joules discharged thro load-resistor from output capacitor at end

Total energy converted by circuit: 2.69 Joules

Energy Quotient of circuit: 2.69/2.24 = 1.2

I can?t comment on energy estimations but how can you be sure that figures are correct?

and the charge anomaly that accompanied this result:
   start charge: 2.39 Coulombs
   final charge:  3.0 Coulombs

...this anomaly can provide very real overunity!

I wish it also.
Electrolytics have memory and recover a large amount (several good percents and often much more!) of initial energy when discharged.  In fact, I think both apparent surplus charge as well as energy as in the above quotes are due to this memory effect. It?s not free energy neither OU: they reside in your power supply (or battery).

Cheers,
Tinu

[Omega_0]

...this anomaly can provide very real overunity!

Isn't it ironical ?
You know the secret of free energy and you are still paying electricity bills and running your car on oil.

[nul-points]

Total charge is zero. It always was and it will remain so. Charge conserves until otherwise proved.

If you want to talk about charge on ONE plate of a capacitor, that's a different story and it is not relevant as long as you decouple it from potential in respect to the other plate

i think most forum members will be comfortable with the concept of 'charging' a capacitor by applying a voltage across its terminals

so when i report that the total initial 'charge' in the capacitors was 2 Coulombs, and the total final charge was 2.9 Coulombs, people can come to their own conclusions about the implication of that

my observation - from actually doing the experiment, not just talking about it - was that the total amount of Coulombs stored in the capacitors had increased without any further application of voltage across the capacitors from external circuitry

I can't comment on energy estimations but how can you be sure that figures are correct?

anyone who is interested enough can see the measurement readings on my website which i linked above (and which is also given at the end of this post) - there are also some FAQs on the site answering questions i've received about possible ways in which errors could have been introduced into the results

Electrolytics have memory and recover a large amount (several good percents and often much more!) of initial energy when discharged.  In fact, I think both apparent surplus charge as well as energy as in the above quotes are due to this memory effect. It's not free energy neither OU: they reside in your power supply (or battery).

i agree, there is a well known 'memory' effect easily seen with larger valued capacitors where, after a capacitor has been discharged, it self-recharges to some extent

this effect is not confined to modern Electrolytics - there are reports dating back to the 1920s recording self-recharge on capacitors using Rochelle salt crystal as a dielectric

i've noted in my description of the experiments that self-recharge was observed - however, this effect takes place after the discharge of the input & output capacitors - and by this time the experiment has finished and the measurements already taken

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!

if there IS a difference between the two cases then how can we ever hope to calculate the energy stored in a capacitor?

i'm not aware of ever having seen TWO equations for calculating the Coulombs stored in a capacitor - one for battery/PSU applied voltages and a different one for self-recharged voltages

It's not free energy neither OU: they reside in your power supply (or battery).

i have to disagree - the additional energy does not "reside in my power supply (or battery)" - it enters the circuit from some external source (energetic medium of space? aether? zero-point energy?) during the operation of the circuit which is not connected to a power supply or battery

all i can say is that once my experiment starts running the only energy which i supplied is contained in the input capacitor - and when the experiment stops more energy has been converted than is drawn from that capacitor

it's not rocket-science to confirm this anomaly - i'm aware of several people (Introvertebrate included) who have put together a handful of simple components and seen the charge anomaly effect for themselves

Isn't it ironical ?
You know the secret of free energy and you are still paying electricity bills and running your car on oil.

you can make fun of me all you like Omega_0, that doesn't alter the fact that this anomaly exists and people are starting to pick up on it

i started these experiments with the switched capacitors in March of this year - i think i've been very fortunate to get to a point where i can record Energy Quotients greater than 1 in that time

i'm following it through now with further investigations of the effect with a view to scaling up the power, looking for practical confirmation of excess energy and developing my basic test circuit to make it a continuous process

but then, that's me - i'm just a 'get on and do it, kinda guy', rather than a 'tell other people it's impossible' kinda guy

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

[ORION]

Sorry no OU here. 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.

(remember the formula for two equal value capacitors in series ? Look it up or just divide one of them by two.)

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.  Energy (Joules) =1/2 CV^2.

Many starters make this fundamental error.