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2022 builders survivor board => General Builders discussion => Topic started by: floodrod on August 20, 2022, 02:45:48 AM

I have been working with the dual capacitor paradox and I was not satisfied with the conventional answer that the 50% loss was turned to heat. So I broke it all down to figure out where the other 50% went and if it could be used. Furthermore, I wondered if anything can be harvested for free.
As most know draining a capacitor to another capacitor makes both caps have equal voltage, be we lose 1/2 the usable joules. And we know if you place a motor in series between the first capacitor dumping to the 2nd capacitor, we get some work done, and catch the exact same amount. Some suggest the work done is free.. I contend that the work from the motor was done from the wasted 50% that we were not able to catch, and there is NO FREE ENERGY happening here.. Yet is is possible to recover a good portion of that wasted energy.
I used equations of Buoyancy and / or Gravity to compare what is happening. When looking at the picture below it was calculated using Buoyancy to foot pounds. The exact same logic applies to capacitors. Basically, the capacitors are going to balance the voltage.. But it uses WAY WAY more power than it needs to because the balloon is pressurized and the catch capacitor has no pressure. Basically the full 50% loss is lost during the collection phase, not the sending phase. The first sent out at FULL POWER, but to balance voltages, only 50% of the power was needed.
The good news is we can harness the lost energy by putting something between them. The catch capacitor will still catch the same amount and the waste will be used on the load between them.

Here is another simpler way to understand the dynamics..
Keep in mind there will be other small losses in the wires, cap plates, etc. Just nowhere close to 50%!!!

Last night I was wondering if indeed the load between capacitors only converted 50% to heat or the full 100% that passed through it. Logic says it should have converted the full 100% to heat and still caught 1/2 the amount of power the source sent. So today I spent a good portion of the day working with 12V batteries and boost converters. Trying to naildown exactly what's happening.. I was able to send 6 amps out of the battery through boost converters in series and collect near 95% of it, but no overunity..
I sat and looked at my math again to try to get a better grasp why I could not reproduce this effect with batteries. Then it dawned on me...
Batteries are already both pressurized... And when connected together to make a circuit, or even within itself, the internal resistance of the batteries will always be Volts / Amps.. Because they have just about full standard internal resistance, anything we "catch" will have to be sent from the source supply.
But capacitors are different!... Capacitors are FULLY DEFLATED at the start, and internal resistance builds as the capacitor fills.. I think my math in my first post is pretty correct on what's happening.. If you reference the image with my foot pound calculations, you may notice that the internal resistance of a capacitor should always calculate to 1/2 of what a battery does...
What appears to be "very inefficient" about capacitors on the first look, may just happen to be what is so special about them.. Half was lost because it actually allowed the source to send out 2X as much current as it needed to in order to balance voltages! It did this because the internal resistance of a capacitor is only 1/2 (as it's filling) than any other electronic component I know of!
This basically means it may be possible to use "catch" capacitors in a way to achieve a maximum of COP 1.5... But in reality, there will be other losses, but still room for OU..
I will be testing this by designing a circuit that does just this.. Stay tuned for updates

I did it by my old simulator.
In many different ways. The energy balance was maintained always everywhere.
Even the simulator files were saved somewhere.

I have been working on the math and comparing it to bench tests. The pattern is matching the diagram in the first post with or without a resistive load between caps..
First Cap at 20 Volts 2nd Cap empty. = You can catch up to 1/3 of the power sent.
First Cap at 20 Volts 2nd Cap at 10 Volts= You can catch 1/2 of the power sent
First Cap at 20 Volts 2nd Cap at 15 volts= You can catch 3/4 of the power sent.
When I hook a car battery at 12.8V to the capacitor through Heavy Resistance, the send and catch VOLTAGE is the exact same. You catch the FULL 12.8V in the capacitor, even through hundreds of Ohms of loads. The more resistance, the slower it goes though, because the amperage gets choked down.
This suggests VD = IxR (voltage Drop) is not happening when the negative side of the load goes into a capacitor, instead of shorting out at ground.
You can view my video of my circuit design and attempt. it's a demonstration showing you pass ALL the voltage and current through a Boost converter to recharge the same battery, AND catch some of the power in a capacitor before it gets zero'd out at ground. I have parts on order to build the circuit that will monitor the Capacitor voltage, and dump it through a 2nd boost converter back to the source battery at specified voltage levels. Also a few new different value boost converters are on the way so I can dial it all in.
>>>> https://www.youtube.com/watch?v=m7GvJ29ivg
All my math and tests are saying something is here but the window to catch more than I waste is not huge.. So it has to be done efficiently if it will work. maybe something else is happening and it won't work, but I will surely find out.

I have found you can diode both sides of the capture circuit into a cap and use an isolated buck/boost converter to send it back to the input. Just have a diode as well on the source or it may well try to backfeed to the bat.

I performed many many tests to determine if all the energy can be harvested while passing to charge a copacitor or battery. Sadly, the answer I found is No..
The entire "recycling" concept is nothing more than robbing from peter to pay paul.
Here are some of the tests and results I have done.. If there are any disagreements, I can repeat and record each test and show the results..
Take a 12V Source going through 2 resistors then finally to a 6V catch battery.
The amperage that flows is determined by the total ohms of the resistors multiplied by the potential difference between the source and catch. So in this example, the potential differential is 6V. Now take the total Ohms of the resistors (lets say 10 Ohms) , the current will be 0.6 Amps.
The question bow becomes, is the source really putting out the full 12 Volts, or only the potential difference? So lets find out..
Take the multimeter and probe from the hot terminal of the source to inbetween the resistors. Each resistor is dropping the voltage by 3 Volts, and 6 are coiming out the tail end to the battery.. 6+3+3 = 12..
Lets verify it again.. Plug a 12V AC to DC pack into a wall watt meter. Put it across the resistors and record watts pulled from the wall. Put it on the battery alone and record watts pulled. Now put the resistors in series + the catch battery and record watts.. Make sure to record all the amperages, then crunch the numbers.. The above proves correct.
Now lets do another test to lock it in concrete..
Take a 10,000uf Capacitor, fill it to 20V.. Take an identical capacitor that is empty and transfer from 1 to the other. You will sens 1.5 Joules and only catch 0.5 Joules. You lost 66% of your power. But you know you sent out 1.5 joules because it's measurable.
Next, fill the same capacitor to 17 Volts, which is 1.5 joules.. Dump it right into a motor and record spin time..
Now repeat the first part. fill Capacitor to 20V so when we dump to an empty one, voltages balance and exactly 1.5 joules sends out. Now put your motor between and record spin time..
Both times we sent out 1.5 Joules. With motor directly to cap I got 4.76 seconds spin time. When sending the same amount and catching to a 2nd cap, I get 3.14 seconds spin time. The motor loses exactly 33% of it's spin time when we catch 33%..
I made a video explaining how the capacitor to capacitor dual paradox works if it is unclear. https://www.youtube.com/watch?v=yL9FI9tTDJU
Anyway, the point is, my conclusion in the whole recycling matter is, there is absolutely no way to get more back.. When you try to catch some on the tail end, your device runs slower and weaker in the exact percentage that you caught. The voltage drop is able to be seen with a multimeter. And the only way to make the load between the source and catch run as efficiently is by jacking up the source voltage and hoping the source really isn't sending that voltage out.. But all the tests show it certainly is..
Amperage will be determined by Load resistance multiplied by potential difference. But true source wattage is really volts sent out to create that potential difference multiplied by the amperage in the circuit.
I also tested parallel loads between source and catch. Parallel loads follows conventional electrical theory. The amperage divides between each parallel leg and all pulls from the source. And the parallel loads run at the differential.
To precombat rebuttals, I did NOT say you do not catch! YES you catch.. But you just took that same amount of power away from the load in the process.

We all know how a child on a swing cycles back and forth till they stop.
The Tesla coil/automobile spark is made from power in to a primary coil and back emf back to the capacitor of greater voltage than the power in and then repeated till there is not enough voltage to make a spark at the spark plug.
What this does is once started it repeats and thus makes numerous sparks as the air is ionized enough for another spark thus causing what appears to us as a wider spark and thus making a more complete burn of the fuel. I know this from experience when the condenser in the distributor went bad the car would barely go up hill but run fine on the level because of the low vacuum retarded spark when going up a hill.
If you jumper across the condenser there will be a faint spark and less power from a partial liquid burn.
The Bedini School Girl motor was similar in that it ran the motor and captured back emf charging the battery thus making the motor run longer than with no back emf. So it would be more energy efficient.
My point is that if you can find a way to use the secondary output from a given process then you can increase its efficiency. For example.
A Sterling heat engine could run from ICE exhaust heat and charge the battery and maybe run some electrical devices.
Norman

We all know how a child on a swing cycles back and forth till they stop.
The Tesla coil/automobile spark is made from power in to a primary coil and back emf back to the capacitor of greater voltage than the power in and then repeated till there is not enough voltage to make a spark at the spark plug.
What this does is once started it repeats and thus makes numerous sparks as the air is ionized enough for another spark thus causing what appears to us as a wider spark and thus making a more complete burn of the fuel. I know this from experience when the condenser in the distributor went bad the car would barely go up hill but run fine on the level because of the low vacuum retarded spark when going up a hill.
If you jumper across the condenser there will be a faint spark and less power from a partial liquid burn.
The Bedini School Girl motor was similar in that it ran the motor and captured back emf charging the battery thus making the motor run longer than with no back emf. So it would be more energy efficient.
My point is that if you can find a way to use the secondary output from a given process then you can increase its efficiency. For example.
A Sterling heat engine could run from ICE exhaust heat and charge the battery and maybe run some electrical devices.
Norman
Yes all true...
I have setup a recycling circuit and am witnessing overall losses at the moment. But there very well may be more to this than meets the eye.
*** I retract the word "Conclusion".. *****
My brain says there should be something here but the bench is not cooperating so far. lol
My investigation continues in this matter..

Thanks Flood. I do not recall hearing of the Capacitor paradox before. Running a motor and still catching the same amount of charge... That is interesting.

floodrod
Many people misunderstand the two capacitor paradox because like many things they treat it like a math problem. The problem has nothing to do with math or equations more so charges(Electron, Proton), motion from one place to another and surface area.
Where is the missing energy in the paradox, why do we lose half the energy?...
Suppose we force 20 electrons onto a metal plate giving it a negative charge. Now we attach another equal sized plate and the electrons disperse to cover both plates equally because electrons have a like charge and repel each other. We now have 10 electrons on each plate where before we had 20 electrons on one plate. On energy, we end up with 1/2 the energy because the plate surface area doubled (two plates versus one plate). When the surface area doubled the electron density dropped to 1/2 because we have the same number of electrons covering twice the plate surface area. We lost energy because the energy in the system was coulomb forces or electron repulsion which can do work to move electrons in a circuit.
Same as two balloons, one full one empty. The air pressure will drop to 1/2 because the same air occupies the volume of two balloons instead of one. The energy is in the air pressure and when it dropped so did the energy.
Then we can examine cause and effect, the electrons in a charged capacitor move to the uncharged one because the electron density (electrical pressure) is greater in the charged capacitor. The energy gained in the electron current between capacitors is equal to the energy loss as the electron density in the charged capacitor falls. Here we can say the electron density falling is the true cause of the energy loss and the electron current is simply an effect. It doesn't matter if the electron current between capacitors did useful work or not we still lose energy because the electrons now cover twice the surface area and have 1/2 the electron density.
Here's a neat concept, if we place a resistor between the two capacitors as a current flows we measure a "voltage drop" across the resistance. However what is a voltage drop and why does it occur?. We measure a voltage drop and voltage is a "difference in potential" thus the potential must be changing in this area. What is potential?, the potential relates to the electric field density which is determined by the charge density ie. electron density. So the voltage drop tells us the electrical pressure is falling across/within the resistor as the current flows.
On free energy, some clever inventors understood what I just explained and found a solution to work around the falling electrical pressure due to an increasing area/volume problem. They found the total energy in every electrical system is much greater than most are aware of. Not recycling energy but transforming it into another form which can do more work then reforming the energy back to it's original state. So the obvious rules still apply, we cannot create or destroy energy but we can transform it.
Regards
AC

floodrod
Many people misunderstand the two capacitor paradox because like many things they treat it like a math problem. The problem has nothing to do with math or equations more so charges(Electron, Proton), motion from one place to another and surface area.
Where is the missing energy in the paradox, why do we lose half the energy?...
Suppose we force 20 electrons onto a metal plate giving it a negative charge. Now we attach another equal sized plate and the electrons disperse to cover both plates equally because electrons have a like charge and repel each other. We now have 10 electrons on each plate where before we had 20 electrons on one plate. On energy, we end up with 1/2 the energy because the plate surface area doubled (two plates versus one plate). When the surface area doubled the electron density dropped to 1/2 because we have the same number of electrons covering twice the plate surface area. We lost energy because the energy in the system was coulomb forces or electron repulsion which can do work to move electrons in a circuit.
Same as two balloons, one full one empty. The air pressure will drop to 1/2 because the same air occupies the volume of two balloons instead of one. The energy is in the air pressure and when it dropped so did the energy.
Then we can examine cause and effect, the electrons in a charged capacitor move to the uncharged one because the electron density (electrical pressure) is greater in the charged capacitor. The energy gained in the electron current between capacitors is equal to the energy loss as the electron density in the charged capacitor falls. Here we can say the electron density falling is the true cause of the energy loss and the electron current is simply an effect. It doesn't matter if the electron current between capacitors did useful work or not we still lose energy because the electrons now cover twice the surface area and have 1/2 the electron density.
Here's a neat concept, if we place a resistor between the two capacitors as a current flows we measure a "voltage drop" across the resistance. However what is a voltage drop and why does it occur?. We measure a voltage drop and voltage is a "difference in potential" thus the potential must be changing in this area. What is potential?, the potential relates to the electric field density which is determined by the charge density ie. electron density. So the voltage drop tells us the electrical pressure is falling across/within the resistor as the current flows.
On free energy, some clever inventors understood what I just explained and found a solution to work around the falling electrical pressure due to an increasing area/volume problem. They found the total energy in every electrical system is much greater than most are aware of. Not recycling energy but transforming it into another form which can do more work then reforming the energy back to it's original state. So the obvious rules still apply, we cannot create or destroy energy but we can transform it.
Regards
AC
Thank you for the response... You ask some questions, but I don't think you expect answers. But ... Lol.
What is voltage drop? Well there is no such thing as voltage really. It's just a measuring scale of how much pressure is on either side of a resistive load. Picture it like a reducing valve in a water pipe. Once side has much more pressure than the other. That's all the voltage reading really is..
What is difference in potential? Pretty much the same thing. One side has more pressure than the other. The difference between them is the voltage end or the difference in potential that we read on a multimeter.
I know this goes against everything we are taught.
I am making gains everyday and getting closer and closer.

When the surface area doubled the electron density dropped to 1/2 because we have the same number of electrons covering twice the plate surface area. We lost energy because the energy in the system was coulomb forces or electron repulsion which can do work to move electrons in a circuit.
Great explanation.
Therefore, we do not need to increase the area or volume of a substance that carries a charge, but rather decrease it. Well, you understand what I mean. Only here we will interfere with the Coulomb forces. But let's turn our attention to the hollow metal sphere. We can add charge inside, through the hole. Without large resistance. Thus increasing the charge density on the outside.
Van der Graf's machine is OU? ;)

I have been having difficulty determining for certainty what the source is actually outputting when doing these tests on recycling, so today I found a way to verify it all. (as far as I am concerned anyway)
Because the catch battery is at 12V, we have to set the power supply higher than the positive terminal of the battery.. So the amps show correctly on the power supply, but it leaves us guessing how many Volts it's actually using.. And Ohms law works both ways, depending if we calculate the battery resistance ohms or not.. I had my hunches, but could never verify with certainty what was leaving the source..
So today I hooked a boost converter to my power supply and locked it at 6V.. Just the amps could fluctuate on the PS so I can tweedle the volts with the boost adjuster to see exactly what is leaving the source and where it is all going..
First I verified efficiency of the boost converter by hooking it right up to resistors (no battery).. The booster was showing 79% efficiency.. Fair enough at least now I had some sort of number to reference for later..
Next I put 17 ohms of resistors across the supply and let them burn. as expected, 6V @ .35 amps (as ohms law states)
Next I put 17 ohms of resistors between the positive out of the battery and positive out of the booster.. Tweedled the booster adjuster till exactly 6V was going through the resistors. Power supply read 6V 1.45 Amps.. (8.7 Watts). I then measured amps inside the resistor circuit. It was 0.35 amps...
2.1 watts were going through the resistors. (6V at .35a).. So where are the last 6.6 watts? I took the battery voltage, multiplied by .35 amps and landed at about 4.6 Watts. Giving me a total of 6.7 watts..
6.7 watts is what percent of 8.7? ===== 77%.. Now remember when I measured the boost converter efficiency at 79%? Yep we are right in the margin there...
Conclusion I am not saying there is no way yo make gains by collecting scraps or reusing byproducts.. There may be other angles to explore.. BUT.... As far as just putting loads between 2 hot terminals and hoping to catch some it is an absolute fail... Everything that is "caught" comes from the source, and the loads between tack right on to the amount the source needs to send..
As far as all the posts about moving batteries around to charge themselves, I am calling it bunk... Sure you will see the middle and end battery charge, but as much charge as they took, it all came out of the source battery.. Playing with parallel / series / and fiddling the amp hours may give the illusion it is a total gain in the short term but you are really not gaining any. Just shuffling it around..

Do you know other ways to increase the charge density without spending energy for this?

Do you know other ways to increase the charge density without spending energy for this?
Not Yet. But I did draw some interesting conclusions from all this that may end up being very useful for future endeavors. Viewing electricity as "pressure" and the amplitude being the difference between 2 pressure units (instead of Volts and Amps) proves to be true.
Here is another thing very interesting in all this... They say "Electricity wants to get to ground". This is NOT true... I drove an 8' ground rod in with a 10 gauge wire straight to my bench.. Under no circumstances can I get any electricity out of a positive side of a battery, capacitor, or even my power supply by connecting the Earth Ground.. Positive pressure has absolutely no desire to "get to ground"... The only thing it want's to do is to get into the negative side of the battery... (or coil, or negative side of the capacitor).... It just want's to balance, and the only way current will flow is if the power source can balance by going through it..
What goes out MUST come in...

Under no circumstances can I get any electricity out of a positive side of a battery, capacitor, or even my power supply by connecting the Earth Ground..
Only if you ground the negative plate too.
It is possible connect to another grounding rod at some distance.

Only if you ground the negative plate too.
It is possible connect to another grounding rod at some distance.
Yes, if we bond the negative terminal to the ground it will flow because all amperage gets back to the negative side still.
And between ground rods "could" work as long as 1 rod is bonded to the negative terminal and parasitic capacitance of the ground allows the current back. But in both scenarios, the electric found it's way back to the negative side of the battery.

When you run a motor as a ‘load’ (with no load on the motor),
Almost all of the energy passing through the motor’s coil is recoverable
after it gets up to speed.
The amount of energy that comes out (to charge a capacitor or battery),
will decrease as the motor encounters resistance. (conversion to heat and work)
This has fooled many experimenters with battery to battery setups.

As per the capacitor “paradox”, its not really a paradox,
Just a mathematical conundrum in how we measure energy in a capacitor.
The time variance of the discharge curve is not taken into account.
We relate everything to 1 second of time (or some fraction of this)
First thing to note is the original capacitor was only allowed to discharge for half of its’ time.
The other half is in that both capacitors are only charged half way, which allows them to discharge
more quickly (higher current).
We calculate each at 25%,
But the energy didn’t actually go anywhere.
The 2 in conjunction return 2x the expected value in half the time.
This can even be performed under infrared analysis and it is observed that heat is not a major factor.

When you run a motor as a ‘load’ (with no load on the motor),
Almost all of the energy passing through the motor’s coil is recoverable
after it gets up to speed.
The amount of energy that comes out (to charge a capacitor or battery),
will decrease as the motor encounters resistance. (conversion to heat and work)
Incorrect..
Regardless if the motor is spinning or not, the motor will only run on the difference in potential between the send battery and the catch device. (in this case the capacitor). And the catch device will only collect it's nominal voltage at that exact time.
Example...
12V @ 1 amp coming out of the source through a motor to a 6 volt battery catch... 12 Watts comes out of the source.. Catch battery catches 6 watts (6V X 1A) and Motor runs on 6 watts (potential difference X amps).
Not 1 fraction of a watt is "recoverable" at any time in this setup.. Motor at 10,000 RPM or standing still. Amperage will change, but every drop of energy that was caught by the catch battery came from the source. In addition to every drop of wattage that the motor consumed.
Experimenters are "fooled" because the amperage that gets sent from the source calculates to make it seem the source is only outputting the differential voltage. And it's very hard to confirm what exactly is coming from the source. But the way I accomplished it leaves little doubt

When you run a motor as a ‘load’ (with no load on the motor),
Almost all of the energy passing through the motor’s coil is recoverable
after it gets up to speed.
If our motor would have a flywheel on its shaft,
capacitors could been recharge with reverse polarity.
right?
Calculate the total energy after that, maybe there will be an extra one? ;)

No. There will be a heavier pull spinning the flywheel up,
some of which is recoverable by slowing it down,
But overall the motor will drawn slightly more current,
you will not ‘gain’ from this, but if you connect a load the flywheel
can help with stability, and consistency of the electrical load.

@floodrod
What i meant by this is what is ‘passing through’, as at optimal speed with no load,
the motor isn’t itself drawing very much power.
However, the coil CAN still conduct electricity, beyond the magnetic saturation of the coil.

@floodrod
What i meant by this is what is ‘passing through’, as at optimal speed with no load,
the motor isn’t itself drawing very much power.
However, the coil CAN still conduct electricity, beyond the magnetic saturation of the coil.
Lets say we have an unloaded motor at absolute full speed and we have a 6 V "Catch Battery" after the coil to "recapture"..
Motor is FULL speed at 12V @ 0.01 Amps.. Now we unhook the 6 Volt catch battery and hook the motor driver up standard fashion.. Now when it is Full Speed again, the power supply will read 6V @ 0.01 amps. And the motor will be going the exact same speed.
In the first way we caught 6V @ 0.01 amps... But we only caught that because we fed that exact same amount more into the motor.
Many get confused at this because when we are "catching", they think the power supply is only putting out the 6V differential, even though the supply reads 12V. It,s easy to make this assumption because the amperage flows exactly as it would with 6V. They don't get the whole "pressure" part. As Onepower stated (and I stated in my first 2 posts), take 1 balloon fully blown up and transfer half the air into an empty balloon. You still have all the air, but you lost a bunch of pressure power. Putting a motor or a a load between will allow you to recapture some of that lost pressure, but that is all. You will not end up with more power after..
Is there another way?? I don't know.. You can convert the lost pressure to flux in between, but you will still need an overunity device that can convert that flux to more power than was sent to it. At the moment, I see no advantage in catching....
BUT..... I do think performance of a generator can be improved by using a capacitor in such a way to always keep the volts the generator is producing HIGH and the amps low.. And continually dumping the top part of the cap into a battery without depleting the cap. But the battery and the generator will always need to be isolated.