Hi all-
Had a different Capacitor Set-up for those that want to try it out themselves or for those who just want to think about it.
Take 3 capacitors all the same. I used 35V - 22,000uf caps in this example. All are empty except that I will charge only C1 to 12 volts from source...here we go..
Charge C1 to 12 Volts, hook C1 neg to C2 neg, then take C2 pos and connect to C3 neg and finally take C3 pos back over to C1 pos.
What you get is about 8V left in C1 and you get about 4Volts in each C2 and C3. Where you began with 12volts, now you have a total of 16 volts at 22,000uf - C1+C2+C3. about a 30% gain.
Now, if you then use the stored energy in C2 and C3 to do some work they will now be empty and that leaves only the 8volts of the original charge in C1. So then you charge C1 back to 12 volts(thats only adding in 4 volts of additional energy), connect the ciruit again as described above and once again you end up with about 8 volts in C1 and 4 volts each in C2 and C3 again. Only this time around you only added 4 volts to top off C1 and have gained 8volts on C2 and C3(4 volts each)-thats now a 200% gain. Use the charge in C2 and C3 to do the work again and repeat the cycle.
I have used my small dc motors to discharge the caps and carefully timed the running times for the various charges and indeed they correspond to the correct charges in each cap - more volts means more runing time.
Not likely,according to the energy formulas, that one has gained in energy but then you can run some motor much much longer than with the original 12 volt charge..hmmmm??...
best,
Hi PG46,
did you test out, if C1 alone at 12 Volts has a shorter running time
than C1 +C2 +C3 in series at 16 Volts ?
As the series capacity is then only 7333 uF with the 16 Volts
it has about 1,7 times less energy stored all in all in C1 +C2 +C3
than 12 Volts in 22000 uF.
As the stored energy is W=0.5 x C x V^2
the 12 Volts cap C1 should run longer...
If not you have found a mysterious energy multiplication...
Hi hartiberlin-
The original C1 cap charged to 12 volts will run far less time then all the caps combined after my ciruit(C1+C2+C3) After the circuit I then reconnect C1,C2 and C3 in parallel which gives me 16volts @ 22,000uf At no point do I connect C1, C2 ad C3 in series excepting between C2 and C3 in the original charging.
Using the energy formula as you posted W=0.5xCxV^2 , the original 12 V cap would have 0.5x.022x144=1.584Joules
Then after my strange circuit I end up with C1 at 8 volts and C2 at 4 volts each
So C1 now has 0.5x.022x64=0.704 Joules C2&C3 each have 0.5x.022x16=0.176joules
These 3 caps combined now only have 1.056 joules which is far less then the 1.584 joules that was in the original 12volt cap.
See, if we take a single 12 volt cap and connect in parallel with 2 others the same we end up with 3 seperate caps charged to 4 volts each. Even though these are 12 volts combined, according to the energy formula these three combined don't add up to the original 12 volt charge (3 x 0.176=0.528) The original 12 volt cap has 1.584 joules
So, does my little ciruit have any value, I don't know. You surely can run a motor much much longer than when using only the original 12 volts, although I am not testing under any load. Its real interesting after the first round if you use the energy in C2 and C3 and then top off C1 with just 4 additional volts to have 4 volts to use in each C2 and C3 again.
Best,
Then after my strange circuit I end up with C1 at 8 volts and C2 at 4 volts each
oops, this should read- Then after my strange circuit I end up with C1 at 8 volts and C2 and C3 at 4 volts each
Another way one could look at this circuit is where with one 12 volt cap one normally can get only 3 - 4 volt caps when hooked in parallel with 2 other caps.
Now with my circuit one can begin with a 12 volt cap and end up with 4 seperate 4 volt caps :)
Let me ask so I know... these caps are good either way, so the neg and pos doesn't matter really, right?
No, all the caps I am using are clearly marked pos and neg. One must be very careful when charging the caps to connect properly and not to exceed the rated voltage capacity neither as otherwise they can explode or at the minumum ruin your capacitor.
so they are polarized caps?
<After the circuit I then reconnect C1,C2 and C3 in parallel which gives me 16volts @ 22,000uf
......
How do you get 16 Volts, when you put all 3 caps in parallel again ??? Then you have 66000 uF at 16 Volts ???
Question: How many caps charged to 4 volts could one get from a single cap charged at 12 volts?
Hi - Sorry have been away travelling and so am with little computer access for a few days.
Elvis-
Yes, the caps I use are polarized.
hartiberlin -
Sorry about the voltage confusion. I will try to reword it a bit better. In my example from my 1st posting I might say instead that I have gained an extra cap charged at 4 volts than I would have otherwise.
How is that so?
Well lets see, if in set up #1 one takes a cap charged to 12 volts and hooked it in parallel with two other empty caps then one ends up with 3 caps charged at 4 volts each which is standard since the voltage is divided evenly amongst the caps. If in setup #2, one takes a 12 volt charged cap but uses my circuit instead you'll wind up with 4 caps charged at 4 volts each. 8) there, thats better
Sorry however, as I haven't been able to get to 16 volts with the same capacitance I stared out with. :-[
Nevertheless, where most folks will want to answer the first question of "how many 4 volt caps can one get from a 12 volt cap?" with a 3, I can answer it with "4 at the very least" ;)
pg46 please post picture(s)
but you claim you get more energy alltogether in all caps.. so do you put then all 3 or 4 caps in series or what ? I am now totally confused and don?t know what you are doing exactly... please describe again in detail how you get the motor to run longer compared to the standard 12 volts single cap. Thanks.
Hi All-
Sorry for the delayed responses, as like I said before I haven't steady access to internet just now since I am currently travelling.
hartiberlin-
If we put the energy formulas aside for the moment.
Let me say here first that it is far easier to actually do the circuit than to write about it :D Let's run through it again here anyhow-
1)Take 3 identical caps 2) lets call them C1,C2 and C3 3)charge C1 to 12 volts 4)connect C1 neg to C2 neg 5) connect C2 pos to C3 neg 6) finally connect C3 pos back over to C1 pos
You will end up with approx. 8 volts in C1 and 4 volts in each C2 and C3 - thats it !
Best,
there is a saying - energy cannot be destroyed or created, it can only change form - somewhere in quantum physics
Hi pg46
Is that what you really mean?
Sounds not 'conventional'. :D
Anyway, I will try it.
Best
(https://overunityarchives.com/proxy.php?request=http%3A%2F%2Fjeanyves.hervouet.free.fr%2F4overun2%2Fcaps_1.gif&hash=ac39d2799468530f2dcc2046f51d5814333690c0)
(https://overunityarchives.com/proxy.php?request=http%3A%2F%2Fjeanyves.hervouet.free.fr%2F4overun2%2Fcaps_2.gif&hash=ecb1d14be51c7dd0f79b84306b7506482c9d109b)
(https://overunityarchives.com/proxy.php?request=http%3A%2F%2Fjeanyves.hervouet.free.fr%2F4overun2%2Fcaps_3.gif&hash=c36ddee347962e94de0eb3b3ccba527eb1f389cc)
(https://overunityarchives.com/proxy.php?request=http%3A%2F%2Fjeanyves.hervouet.free.fr%2F4overun2%2Fcaps_4.gif&hash=fa68e95ff7c9eb3188402722321ed16654ec5b4d)
(https://overunityarchives.com/proxy.php?request=http%3A%2F%2Fjeanyves.hervouet.free.fr%2F4overun2%2Fcaps_5.gif&hash=9ab04b0969bf8597a8514a3834b7776d14d6193a)
(https://overunityarchives.com/proxy.php?request=http%3A%2F%2Fjeanyves.hervouet.free.fr%2F4overun2%2Fcaps_6.gif&hash=7c2977c8121b6b2b5badfa4ff3914e0eed820768)
Hi everybody and especially pg48
Should I be understood when craming my own stuff into this one:
a quote from pg48 QuoteSee, if we take a single 12 volt cap and connect in parallel with 2 others the same we end up with 3 seperate caps charged to 4 volts each. Even though these are 12 volts combined, according to the energy formula these three combined don't add up to the original 12 volt charge (3 x 0.176=0.528) The original 12 volt cap has 1.584 joules.
That is exactly what I was saying in another post. When you parallel-connect 2 or more caps,
it appears that you are loosing NRG, as pg46 is pointing out with his 3 caps (1.584 joules vs 0.528 joules).
Should I be allowed to 'self-quote' my May 16, 2006 post to "
free energy circuit setup"? (It was about 2 caps and the "lost NRG".)
QuoteNot 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 calorificNRG 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. [...]
Sorry, but I can not help smelling a rat.
That is what I was saying a couple of weeks ago.
I can say that again.
The caps's behavior seems to be strange.
But, the experimenter creativity is infinite.
Best
Quote from: pg46 on June 15, 2006, 02:00:29 AM
1)Take 3 identical caps 2) lets call them C1,C2 and C3 3)charge C1 to 12 volts 4)connect C1 neg to C2 neg 5) connect C2 pos to C3 neg 6) finally connect C3 pos back over to C1 pos
You will end up with approx. 8 volts in C1 and 4 volts in each C2 and C3 - thats it !
Best,
Okay, you end up with 3 x 22.000uF caps in series which have due to the series circuit now
about 7333 uF with a total voltage of 16 Volts= 8 + 4 + 4 Volts.
So does this 16 Volts at 7.333 uF run your motor longer than the 12 Volts at 22.000 uF ?
If yes, it is probably due to the motor you are using, that it has a better efficiency at
higher voltages or something like this...
If you would use just a load resistor you would see, that it will get hotter
at the 12 Volts at 22.000 uF cap.
Salve a tutti
Very sorry.
But, I am just wondering.
According to pg46 descriptions and my following pics (should these pics being accurate).
In that case, the most important question could be: how can you charge a cap with a
single wire?
Or, I am mistaken something?
Best
@NerzhDishual
he is just placing C2+C3 in parallel with C1.
Then C1 discharges until it has 8 Volts and also C2+C3 have
8 Volts, so each C2= 4 Volts and C3= 4 Volts.
Then he puts alls 3 caps in series getting 16 Volts but at a lower
total capacity...
So I don?t think, he can run the motor longer on the 3 caps in series...
at least not in normal theory....
Thanks NerzhDishual -
Very nice diagrams, thanks! Yes indeed, that's the way I connected them.
Please note that the last connection from C3 pos back to C1 pos need only be momentary. Thanks also for your comments.
hartiberlin- You can see by the connections that my circuit is neither in series or parallel or else they are both! or maybe you could say that C1 and C3 are hooked in parallel but with a cap(C2) "in-line" between them ::)
I do not connect any caps in series after my circuit as although that would be more volts, there would be less capacitance and so it will not run a motor any longer than the original C1 cap charged to 12 volts. So, before and after my circuit I do all testing on caps individually or else in parallel only.
In this group of tests and examples I used all the same caps, 35volt 22,000uf caps and a single DC motor for any discharge timings.
SET UP #1 - I charge C1 to 12 Volts, 22,000uf. If I connect C1 in parallel to empty C2 and C3 caps, I will end up with 4 volts and 66,000uf. When I discharge this through my motor, it will take 30 seconds to empty the charge.
SETUP #2, if I charge C1 to 12 volts at 22,000uf but then this time I run the pg46 circuit with C2 and C3 , I can get 1 cap at approx. 8 volts(C1) and two caps(C2&C3)at 4 volts each. So, then if I introduce another cap C4 and reconnect the 4 caps in parallel I will end up with 4 volts @ 88,000uf. Now, when I discharge this through the motor, this time it will take 40 seconds to empty the charge, approx. a 30% gain over the "normal" circuit with the same 12 volt input charge.
The reason I use 4 volts to time the discharges is because "4" is a common denominator in both the original 12 volt charge and the resulting voltage from the pg46 ciruit(16v) The motor will run faster or slower depending on the voltage so I wanted to make sure I did the timings all using the same voltage.
Another way I can say all this is that in setup#1, I can produce 3 caps of 4 volts each from a single 12 volt charge. With my circuit I can produce 4 caps at 4 volts each with the same single 12 volt charge.
If one thinks of capacitors as batteries, one person might offer 3- 4 volt batteries from a single 12 volt charge whereas with my circuit one could offer 4 - 4volt batteries from a single 12 volt charge :)
Best,
Okay, PG46,
now I see, what you are doing...
But you did not test,how long the motor will
run on the 12 Volts C1 cap only ?
The only thing you do with your circuit is, you
reduce the losses a bit..
The 12 Volt Cap has 1584 units energy,
the 3 caps at 4 Volts have 528 units energy
and the
4 caps at 4 volts have 704 units of energy, so
you reduce the losses from about 66 % to just 50 % ,
but in total you have lost about 50 % of the used input energy all in all.
You put 1584 units of energy into the C1 cap, when you had charged it up to 12 Volts
and end up with 4 caps at 4 Volts , that is just left only 704 units of energy !
So just let the motor run on the 12 Volts C1 cap and it will
probably run for about 50 seconds or more...
so the energy equation is still valid with Wcap= 0.5 x C x V^2
Hi hartiberlin-
Yes, when looking at the energy formulas there doesn't seem to be any gains. Where does the energy go when connecting a cap thats charged to 12 volts and connected in parallel with 2 others I wonder?
The interesting thing is that when you discharge the 12 volt cap alone through the motor, it will only run it for about 20 seconds!! not like the 50 sec you expected Hartiberlin. So whats going on then? Well of course I am not counting RPMs and nor testing with additional loads on. So its very likely the motor simply spins faster at 12 volt than at 4 volts and so discharges the stored energy in the cap quicker. This is why I even out all the caps to 4 volt charges before and after my circuit to do the tests, so that the motor will spin at the same RPMs.
I didn't make any claims of "free energy". I just thought it was quite interesting that out of one cap charged to 12 volts, you normally would expect to only be able to get 3 caps charged to 4 volts each when connected in parallel, whereas with my circuit I can get 4 caps charged to 4 volts each from that same single cap charged to 12 volts. In this particular experiment thats an increase of 33.3 percent over the first circuit.
I just wanted to report what I observed and thought that this perhaps could be a useful circuit for someone trying to use electricity more efficiently.
Best,
Hi PG46,
many thanks anyway for running your experiment.
This is good, as it can show sometimes strange things
other people have not yet observed...
Anyway, that the 12 Volt cap is only running 20 seconds is
kind of strange, but okay, as the motor is probably a lower voltage
type it does not generate so much counter EMF voltage, so that
the inout current into the motor will be much higher
and thus the cap will be faster discharged...
yes, it is hard to explain, where all this energy went, when ou just
load up a parallel cap from a charged cap...
It is not like filling water from one glas to the other, but you loose always almost
50 % of the energy.
It comes probably mostly from the work you have todo to chargeup the dielectrium
inside the other cap.
Regards, Stefan,
HELLO !
DO KEEP TRYING ON !
THE ELEMENTARY IDEA IS WELL FOUNDED !
Your suggestions are about , how in fact that caps inter-charges ,
e.g. the well-known (1/2)*(caps-value)*((caps-voltage)^2) to be the "caps-charged-energy" !
Your suggestions may seemingly , but just seemingly , appear to be "stopped" by the in-fact-caps-value-when-connected-in-parallel-serial ..... !
BUT DO CONTINUE , FOR YOUR EFFORTS WILL PROVIDE SOME REAL
FUNCTIONING POSSIBILITIES FOR DISCUSSIONS ABOUT , WHAT ENERGY IN FACT MAY HAPPEN TO BE !
You have made a "usefull conceipt" for further science investigations !
Ideas like your here written are in fact of the usefull kind making well foundings for getting the real facts working set-ups !
How about moveable C-plates/topological-structures , dielectrica to move , etc !
Your experimental layout I self emperically/theoretically by "modified networks theories/rules" did some calculations about some 15-20 Y ago !
(many such layouts/ideas do work and important: the steady going-on-and-aim-for discussions do improve Science in Theory & Practice)
Your idea/suggested conceipt appear somehow like the Classical-Mechanical-"integral of Force*way" , but and significantly marking this very point :
Electro-dynamics do not just "copy" all of "Classical Mechanics"-Force/Force-between-interacting-objects/acceleration/speed....etc;
THERE ARE SURDEN POSSIBILITIES IN ELECTRO-DYNAMICS !
Well , also to discuss/include "self-inductances" when having changing currents-to-plates/plates-charging-alterations !
May conventionel practical-electronics-120 Vac->lower-supply-voltages-layouts depict the tricky circumstances , that inductances
in some form/way have to be included when trying to directly-charge a capacitor !
Have a nice day and results-filled studies e.g. among "electrical networks/components/how to use/connect....etc" !
And saluting greetings also meant for other "topics"-writing authors i this overunity-forum !