Basic Free Energy Device

[Turion]

Allen,
That is absolutely true if it was a true "backspike" into the batteries that the voltage and current came out of, but it is not. It is a FORWARD spike into the batteries that are being charged. A very DIFFERENT reaction from the batteries. At least in my experience. It actually helps rejuvenate old batteries by desulfating them. In a standard setup the spike would travel the path of least resistance, which would be into the the battery it came out of. In THIS arrangement, the path of least resistance is into the battery being charged.

By the way, I kinda think the three battery system (because of this ONE FACT) clears up the conjecture as to whether the energy comes out of the negative of a battery and moves to the positive side or comes out of the positive side and moves to the negative. Just my opinion.

Dave

[Allen Burgess]

To what Matt wrote here I would like to add some observations....

If you use a stock off the shelf DC motor, you get extended runs because you are STILL hitting the battery with a pulse charge. BUT, those motors are designed to take both the generated voltage and the coil collapse and dissipate it into the unused (at that moment in time) windings so as little as possible comes out as Back EMF.

With Matt's motor run between the positives, the generated voltage and coil collapse DO come out, but down the line to the second battery, and adds to the (pulse) charge going into battery 3.

A boost module increases the voltage going into battery three.

There is one thing I have been looking at here, and that is the frequency of the pulse. It is entirely possible that an off the shelf motor pulsed SIGNIFICANTLY FASTER that what Matt's motor pulses at on 12 or 24 volts MAY be more benificial to the circuit than Matt's motor is. I haven't done enough testing to make that claim, but I thought I would throw that out there for those of you who have an off the shelf motor and a 555 timer laying around.

Dave

@Turion,

Sending the "backspike" into the batteries is the best way to kill them prematurely

[Turion]

Thoughts

To what Matt wrote here I would like to add some observations....

If you use a stock off the shelf DC motor, you get extended runs because you are STILL hitting the battery with a pulse charge. BUT, those motors are designed to take both the generated voltage and the coil collapse and dissipate it into the unused (at that moment in time) windings so as little as possible comes out as Back EMF.

With Matt's motor run between the positives, the generated voltage and coil collapse DO come out, but down the line to the second battery, and adds to the (pulse) charge going into battery 3.

A boost module increases the voltage going into battery three.

There is one thing I have been looking at here, and that is the frequency of the pulse. It is entirely possible that an off the shelf motor pulsed SIGNIFICANTLY FASTER that what Matt's motor pulses at on 12 or 24 volts MAY be more benificial to the circuit than Matt's motor is. I haven't done enough testing to make that claim, but I thought I would throw that out there for those of you who have an off the shelf motor and a 555 timer laying around.

Dave

[Matthew Jones]

Sorry Sky. I read it wrong. I apologize.

Here is how it works You do not have 2 banks of batteries although physically you can look at them that way to make sense of it, but electrically the 3 batteries wired up are one battery. If you have 2 in series and third paralleled with the load between them.
So the difference between the poles is 12 volt that is the potential running through the load. If your current is 1 amp your passing 12 watts through the motor. If your KNOWN load is 80% efficient then 9.6 watts are hitting the third battery. So if the battery is absorbing 50% percent of that 9.6 watt you are now paying 12 - (9.6* 50%) = 7.2 watts to do 12 watts of work. Your potential is NOT 24 volt it is 12. The difference between the highest and lowest point in the power system is voltage.

Now if you set a system up with a known load (One that you previously tested the efficiency on) and look at amp hours of the batteries your using you can calculate the run time. Granted the charge side has to be equal to the discharge side in abilty. You cannot take a 10 amp hour battery and discharge it into a 5 amp hour expect accurate results. But in real life if the third battery is really dead the charging efficiency can go up so the same amount can cost less until the curve of the charging cycle levels out and then absorption can come down. So you can average that curve. In cases where you would pulse the motor you can change that charging curve altogether, this is what you see in Lindemanns build with the monopole.
In most cases the load running on these types of system also raise efficiency on the load. You can go youtube and find a test done by TinMan that shows this.
There is also things you can do to keep the third battery from charging like removing the power through more work or collecting it in pulses to distribute back to the top end. But thats all neither here nor there if you are not measuring the system correctly.

The fact is the motor DOES NOT dissipate energy to do work. You only have factor in loss that come from entropy (IE heat, friction and high frequency transmission caused from brush arcing or radio transmission). You can pass current through water via hydrolysis plates and collect the current and energy on the back end of the system. Current flows through a load and entropy is the only thing that causes loss. In conventional circuit you have the same situation current runs right to ground through the load. But the ground neutralizes that current so mathematically you are at 0 when you get done. This does not happen in a potential based system.

So for some time we have maintained that a good motor on a good generator, one that does not bog down under load, can sustain the system, even to the point that extra work can be done at even a lower cost. And truth be told you do not need batteries at all, anything that can collect and hold current for any amount of time can do these jobs. But if you cannot measure correctly you won't ever find that to be the truth. Thats why I am up in arms and a little hot headed.
Even though I try stay out of here I do not want anyone to feed ya'll bad info.

Matt

[gotoluc]

Hi luc,

the test shows that the supposed "energy" the motor/load is "consuming" is actually not consuming at all.
the load was "Powered" by the "Charging process".
the magnetism on the motor coil can be used to do work, in which can also be for generation. in my case that was a transformer.

the concept is use the same energy more than once.

Yes ricards I've known this for over 6 years. The problem is the energy you're left with is half of what you started with as I've just explained to Dave.

Hi Dave,

not sure if this is addressed to me and related to the 3 capacitor ricards recommended?... but if it does then this is my answer:
It doesn't matter the size of a load you use between capacitors 2 and 3. The voltage will be divided equally between all capacitors and at no cost (voltage reduction) with or without the load.
I've known this for over 6 years and is kool to see.
The problem is you end up with half the energy then you began with when you do the math.

Recently Russ of RWG research has been exploring this effect to see if we could somehow get the environment like the ground to participate in a circuit using abrupt shut off of current running through a circuit like a water hammer pump effect (cavitation) to suck current out of the ground.
I'm all for it and will be testing the possibilities.

Please feel free to add anything more I may need.

Regards

Luc

[gotoluc]

Put a tiny light bulb in place of the motor, and take the SAME measurements. Are they the same? According to these new measurements, did the motor and the tiny light bulb "consume" the same amount of power to run? Same volts? Same amps? Same WATTS? Either the motor is INCREDIBLY efficient, or what is being measured is perhaps NOT what you think is being measured.

Hi Dave,

not sure if this is addressed to me and related to the 3 capacitor ricards recommended?... but if it does then this is my answer:
It doesn't matter the size of a load you use between capacitors 2 and 3. The voltage will be divided equally between all capacitors and at no cost (voltage reduction) with or without the load.
I've known this for over 6 years and is kool to see.
The problem is you end up with half the energy then you began with when you do the math.

Recently Russ of RWG research has been exploring this effect to see if we could somehow get the environment like the ground to participate in a circuit using abrupt shut off of current running through a circuit like a water hammer pump effect (cavitation) to suck current out of the ground.
I'm all for it and will be testing the possibilities.

Please feel free to add anything more I may need.

Regards

Luc

[ricards]

Hi ricards,

In all 3 tests the voltage of all 3 capacitors will be the same.
What is it you beleive this test will prove or disprove?

Regards

Luc

Hi luc,

the test shows that the supposed "energy" the motor/load is "consuming" is actually not consuming at all.
the load was "Powered" by the "Charging process".
the magnetism on the motor coil can be used to do work, in which can also be for generation. in my case that was a transformer.

the concept is use the same energy more than once.

[gotoluc]

Hi all, if matt is in disagreement with what i said gotoluc, then i guess i have some learning to do, as i'm confused now.
Though it is interesting, we have 12 volts between the positives, though obviously 24 volts is hitting the 3rd battery, or it would not charge.
I have to admit, it is a confusing setup and i need to give more thought to it or matt can enlighten me, hopefully with a more peaceful communication to all in this forum.
peace love light

Thanks SkyWatcher for your honest and peaceful reply.
There's nothing wrong or unusual with measuring current between battery 2 and 3 and see 12 to 15 volt across the 3rd battery while charging.
If the 3rd battery is uncharged you can charge it with a voltage source many times higher then its full charge voltage. The battery will convert (step down) higher voltages to current until it gets closer to being charged and or it can no longer take the current being sent to it. Ideally we shouldn't let a 12 volt battery go much above 15 volts while being charged.

I also hope we can peacefully learn as I'm truly here to do that if possible.

Kind regards

Luc

[Turion]

Info

Put a tiny light bulb in place of the motor, and take the SAME measurements. Are they the same? According to these new measurements, did the motor and the tiny light bulb "consume" the same amount of power to run? Same volts? Same amps? Same WATTS? Either the motor is INCREDIBLY efficient, or what is being measured is perhaps NOT what you think is being measured.

Sky,
24 volts is not hitting battery 3. That should be apparent from the rpm of the motor. Run it on 12 volts and check the rpm. Run it on 24 volts and check the rpm. Now run it between the positives. Is the rpm closer to what it ran on 12 volts or 24 volts? What is (going through the motor) and hitting battery 3 to charge it is the DIFFERENCE between the combined voltage of battery one and two and the voltage of battery 3. 1+2-3= voltage hitting battery three. That is why the boost module is needed. It takes the 12 volt potential DIFFERENCE and boosts it up to 14.5 volts and runs THAT through the motor to hit battery 3. That gives you 2.5 volts over the battery voltage which is ENOUGH to actually charge it like it needs to be charged. The fact that Matt's motor is a pulse motor means that it is a PULSED charge, which is critical. You also get the generated output of his motor plus a coil collapse SPIKE into battery three. This is still not enough to keep up with the losses in the system, but it can extend the run times. There are ways to overcome the losses in the system, but until a pile of people are actually building this and trying it and discover what that is, this is the best you're going to get. SOME of us have done the work to figure that out, have read the patents over and over, and have actually built (like YOU Sky) models of this system to research it.

[SkyWatcher]

Hi all, if matt is in disagreement with what i said gotoluc, then i guess i have some learning to do, as i'm confused now.
Though it is interesting, we have 12 volts between the positives, though obviously 24 volts is hitting the 3rd battery, or it would not charge.
I have to admit, it is a confusing setup and i need to give more thought to it or matt can enlighten me, hopefully with a more peaceful communication to all in this forum.
peace love light

[gotoluc]

No you do get my point cause you went on and threw your bogus math and measurements out there.

So for the rest of ya'll just to some up why to gotoluc isn't worth even slightly paying attention to in this case is every person from Benitez the original inventor, to John Bedini were wrong in the way they measure. All of them measured between the potentials for voltage and current through the load for watts and were incorrect in doing so. At least that what luc would have you believe.
But he, himself has discovered the correct way to measure a set of batteries and everyone who has ever worked on and measured one of these system is wrong.
I'm not sure but something seems wrong to me...

Dear Matt,

I'm here in hopes to learn something new. Using the scenario of 1 amp was suggested by SkyWatcher and he came to the same power calculations I did but didn't include the finer details of energy losses in conversion.

Are you now going to bully SkyWatcher and accuse him of bogus math and measurements and say he's not worth paying attention since he basically came to the same power calculations I did (minus losses)?

He's scared to have an argument with me hence the reason he act confused about my statement.

I'm here aren't I?
If anyone else clearly understands why Matt wrote the below when SkyWatcher and I were talking about measuring current at 2 different points (high and low voltage potential) of the 3 battery system then please explain how the below relates to that conversation.

You do not measure a battery no matter the arrangement in 2 parts. You always measure the difference between the lowest and the highest potential. That is Voltage on a meter, period!

when you try and measure the voltage drop after the motor(because motor and 3rd battery in series) you are actually measuring the 3rd battery voltage under charging.. that is what matthew jones is saying.

Is what ricards wrote what you're trying to communicate?

Regards

Luc

[gotoluc]

Hi Luc,

take 3 capacitors of same values, preferably 4700 uf above so it will discharge slowly as you measure with a volt meter.
charge all individually up to 12v or anything close.
now connect as per 3BGS,
2 In series to have 24v and connect to that 12v.
measure the voltage write it down.
disconnect everything, discharge.. then charge all to 12v again..
do the same as per 3BGS, but this time put 1 1k ohm resistor...
measure the voltage write it down..
disconnect everything, discharge.. then charge all to 12v again..
do the same as per 3BGS, but change the load to a short piece of wire, a coil, a transformer a small motor... whatever.
now measure the voltage as it equalize.

make an interpretation of it..

Hi ricards,

In all 3 tests the voltage of all 3 capacitors will be the same.
What is it you beleive this test will prove or disprove?

Regards

Luc

[ricards]

If battery 1 & 2 are connected in series and are 12 volts each = 24 volts and the current is measured at 1 amp = 24 watts entering the motor and if battery 3 is at 12 volts and the current entering it is measured at 1 amp = 12 watts entering battery 3.

So if we have 24 watts coming out from input batteries and 12 watts going in the charge battery it means half of the input power is being used by the motor and potentially half recovered by the charge battery.
However, the motor has converted part of those 12 watts to mechanical power and at best 80% if it is available at the motor shaft which means we have about 9.6 watts in mechanical power at the motor shaft which we can recover back to electrical if we attach a generator to it and can recover at best 80% if it = 7.7 watts and add it to the charge battery which gives a potential total of 19.7 watts recovered from the 24 watts put into the system.

Hi Luc,

I think your missing the point, by seeing the motor to be consuming the "energy" like it was some kind of juice that's is being consumed. by assuming that 24v series battery connected to a motor+12v battery is using 24 watts...

when you try and measure the voltage drop after the motor(because motor and 3rd battery in series) you are actually measuring the 3rd battery voltage under charging.. that is what matthew jones is saying.

the assumption that 24 watt input that the 2 battery in series at 1 amp is false, because you already have 12v from 3rd battery opposing the 24v input. when you measure voltage and you measure current you don't just multiply, like the the conventional thinking, because the circuit itself is not "conventional".

the only reason I think why current going to the motor and the current going to the 3rd battery is showing 1 ampere is because they are in series connection not because the motor was producing "energy".

the way I see this is that the 24v series battery is charging the 12v battery passing through DC motor acting as a resistor unpulsed, when pulsed act as a charger because of radiant spikes).

from what I read here and What I saw in my experiments. the 3BGS should work as claimed. though I never built one, I had the chance to apply the concept with capacitors and you can really power a load and charge something... that is what we are doing in the other thread...

If you really are only reluctant to replicate because you are not convinced by their statement.. i can suggest simple experiments and you can convince yourself..

take 3 capacitors of same values, preferably 4700 uf above so it will discharge slowly as you measure with a volt meter.
charge all individually up to 12v or anything close.
now connect as per 3BGS,
2 In series to have 24v and connect to that 12v.
measure the voltage write it down.
disconnect everything, discharge.. then charge all to 12v again..
do the same as per 3BGS, but this time put 1 1k ohm resistor...
measure the voltage write it down..
disconnect everything, discharge.. then charge all to 12v again..
do the same as per 3BGS, but change the load to a short piece of wire, a coil, a transformer a small motor... whatever.
now measure the voltage as it equalize.

make an interpretation of it..

[Matthew Jones]

No you do get my point cause you went on and threw your bogus math and measurements out there.

Hi SkyWatcher

Yes, that is generally what I wanted to know if Satyam108 understood since he only posted in and out current measurement which may indicate he's under the impression the charge battery is receiving the same amount of power then the input batteries?
However, until he replies we won't know why he posted just the current measurements and if he understands that current is not a measurement of power.

But you do know that and I generally agree with what you posted but here is how I understand it and measured it.

If battery 1 & 2 are connected in series and are 12 volts each = 24 volts and the current is measured at 1 amp = 24 watts entering the motor and if battery 3 is at 12 volts and the current entering it is measured at 1 amp = 12 watts entering battery 3.

So if we have 24 watts coming out from input batteries and 12 watts going in the charge battery it means half of the input power is being used by the motor and potentially half recovered by the charge battery.
However, the motor has converted part of those 12 watts to mechanical power and at best 80% if it is available at the motor shaft which means we have about 9.6 watts in mechanical power at the motor shaft which we can recover back to electrical if we attach a generator to it and can recover at best 80% if it = 7.7 watts and add it to the charge battery which gives a potential total of 19.7 watts recovered from the 24 watts put into the system.

So things are not looking very good and we also have an other loss factor not yet considered which is Battery charge losses.

Example: if you load test a new fully charged 5 amp hour 12v battery you may get 60 watts hour out of it. However, to charge that battery back you will need to use about 74 watts hour which means a lead acid battery is also about 80% efficient to recover meaning the true potential to the charge battery will be 80% of the 19.7 watts = 15.76 watts of potential recovery of the 24 watts from input batteries giving a score of around 65% energy conversion from the input to the output charge battery.

One may simplify his life and get better results to just use a battery charger to charge the 3rd battery?

So what is it and can you demonstrate it that can make the reality any better then what I've explained and tested?

Regards

Luc

So for the rest of ya'll just to some up why to gotoluc isn't worth even slightly paying attention to in this case is every person from Benitez the original inventor, to John Bedini were wrong in the way they measure. All of them measured between the potentials for voltage and current through the load for watts and were incorrect in doing so. At least that what luc would have you believe.
But he, himself has discovered the correct way to measure a set of batteries and everyone who has ever worked on and measured one of these system is wrong.
I'm not sure but something seems wrong to me... He's scared to have an argument with me hence the reason he act confused about my statement.

Have a good one.

[gotoluc]

See, this is why we decided you weren't worth the time it takes to listen to you.

You do not measure a battery no matter the arrangement in 2 parts. You always measure the difference between the lowest and the highest potential. That is Voltage on a meter, period!

If you put 10 batteries in series the potential of the system is measure between the ground pole of battery 1 and the positive pole of battery 10.
If you put 10 batteries in parallel you still measure the difference between the ground pole and the hot pole.
This setup is no different except you have excluded the ground poles from measurement and your difference is measured between 2 positive poles. There is not method for knowing the voltage drop before battery 3.

Show me a textbook engineering reference that states otherwise? You cannot make up the rules as you see fit. Voltage is the difference between the highest point and the lowest point in an electrical system.

Sorry Matt, don't get your point?... my question was to get a better understanding of what the pics in Satyam108 post of input and output current meters are supposed to be demonstrating.

That's all

Regards

Luc