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Author Topic: ENERGY AMPLIFICATION  (Read 3509696 times)

XR IX

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Re: ENERGY AMPLIFICATION
« Reply #8550 on: June 08, 2020, 07:35:55 PM »
The biggest problem is making limitations and not seeing the potential from the larger picture. You see a capacitor, and that's all. But a capacitor can also a coil, and coil can be a capacitor. A plate is also a conductor.

For Example: Look at a Tesla transmitting coil and receiving coil. They are coils, but they are capacitively coupled, ergo two plates of a capacitor. You see the Earth as ground, but it also is a conductor, ergo it can maintain a standing wave (as Tesla said).

We see energy as quantity, but it most likely infinite, as the motion is in the universe. The big bang set everything in motion, and like your hand outside your car window we only harness it when we have the ability to interface the difference (i.e., slow it down).

My Theories anyway... ;)

pauldude000

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Re: ENERGY AMPLIFICATION
« Reply #8551 on: June 09, 2020, 12:27:56 AM »
If we charge a 100uf cap to 100v and connect it to a 100uf cap with 0v we end up with 50v in each cap, we lose 50% of the total energy that was in the first cap to start. Calculate the energy total in 2 100uf caps at 50v each. Where did the energy go?
Mags


Very good question, Mag. A full half of the energy is lost and no discernable work has been done, in that it is not dissipated in the form of heat or emr. Half of the energy available within the system jut goes "poof", so to speak. These are the types of questions I like as the electrical energy had to be transformed, but into what? Not electrostatic field, as that is what is being measured. Not electricity, as the voltage has lowered not raised. The available E has become 1/2E no matter how you look at the system.




For those who do not know how to do the math:



E=.5CV^2 (energy in joules stored as an electric field in a capacitor equals one half of the capacitance in farads times the square of the voltage in volts)




The first 100uF (.0001 F) is charged to 100V, so  E(joules) = (.5 X .0001) X (100 X 100) = .00005 X 10000         = .5 Joules total stored energy


The second  has two 100uF charged at 50v each, or 2((.5 X .0001) X (50 X 50)) = 2(.00005 X 2500) = 2 X .125  =  .250 Joules total stored energy




-----------


The problem you have here is that it is not a closed system. The capacitor you charged is system 1. When connected in parallel with the second capacitor thus equalizing charge you have system 2. When they are disconnected from each other to check voltage of each individual you have yet system 3. Something is happening there, no doubt, as it indicates something not identified in which the extra lost energy was stored or converted in system 2. I think the energy was probably stored, but I cannot prove that. It could have been converted into some unidentified force as well. Either way, call it system 4.





Magluvin

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Re: ENERGY AMPLIFICATION
« Reply #8552 on: June 09, 2020, 01:17:04 AM »
From capacitor/caps to cups :

You have two cups  in front,not caps  ::) ,one full with water and the other empty !
Now you fill the second = empty cup half-full by the full cup with water !


Now you have two halffull, or one halfempty and an other halffull cups :


I do not think that somebody has to explain where the water from the first cup became lost !?
100% same water content,in two cups now as 100% capacitance in Joule/Farad in 2 capacitors now !
Well that understanding is part of the key to where it went.
Think about it this way....  1 full cup and 1 empty cup. In the end we still have a total amount of water to fill 1 cup....Same with the electron charges in the caps... In this instance we should say, how many more electrons are on the negative plate of the first cap at 100v, then how many are there in total on the 2 caps negative plates?(in total excess electrons compared to a cap with 0v)

Where the work or energy happens is in the action of lifting the full cup to the brim of the empty cup and pouring the water into the empty cup till both are half full.. What useful work did we accomplish? In this case it most likely took more than 50% of the total energy of the stored water of the first cup by having to lift it in the first place, where as a cap, the higher voltage of the first cap has nothing to do with the height physically of the full cap and the empty cap. Its voltage is already at a high point electrically.  So to associate it better, with the water and containers we would need them on level surface with a hose and valve at the bottoms of each container to be equal to the cap function. With that config, we could say that if we ran a small water pump to run a small gen and the valve and hose were inline, then whatever we got out of the generator during the evening out process from one container to the other, then we could say we got some work out of the transfer rather than just moving water and losing energy in the process. So in the cap to cap process we basically wasted the 50% in the process because we did nothing with the transfer other than release the initial pressure into a container 2 times as wide..  And supposedly when we have the 2 containers half full, the amount of work that can be done is 50% less than the 1 full container can do.
Interesting, isnt it?
Mags



XR IX

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Re: ENERGY AMPLIFICATION
« Reply #8553 on: June 09, 2020, 04:01:55 AM »
Where the work or energy happens is in the action of lifting the full cup to the brim of the empty cup and pouring the water into the empty cup till both are half full..

This is fun topic. So I think we have to recognize the difference in potential to move to a neutral state. The two 50v caps produced at discharge, doesn't retain the potential the single cap had at 100v. The word "discharge" is the relevant idea. Though I used the example of water, you have to understand the relationships that are and aren't the same. Returning to a lower state is probably flows like water, raising it to a higher level is like winding a spring. If you ever popped a cap, you know what happens when you wind the spring beyond it's breakdown voltage. The capacity is reached when the charges accumulate on the plates, and the plates get full. The charge rate is not linear, because as you place more like charges, they resist being placed there - the action is like compression.

When you discharged the first cap, into the 2nd cap, the energy to bring it to the higher potential was lost, as you can't return the the energy from the 2nd cap into the 1st cap because they are at equal states. You need to wind the spring (build the compression) of either cap to allow it to flow to a lower state capacity.


lancaIV

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Re: ENERGY AMPLIFICATION
« Reply #8554 on: June 09, 2020, 01:56:37 PM »
#8552 :
Peak and average !
You have a bottle full with water ,open without cap :
you turn the bottle slowly : water comes out
you turn the bottle fast : no(very low) water comes out
action/reaction time !

Water with 80° ,60°,20°,0° C,0° F,0°K degree(s) : temperature and movement freedom degrees linear/rotatory/translatory( viscosity,Engeler units)

This is in a capacitor(spring : Lord Kelvin analogon) similar

Scaling f.e. Richter or Beaufort : never linear,ever dynamical : scale down plenitude(-x),scale up magnitude(+x)

The (non expected accelerated) movement in nano-tubes !

Magluvin

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Re: ENERGY AMPLIFICATION
« Reply #8555 on: June 09, 2020, 04:01:15 PM »
There is more to the story.  Will post it later today after work
Mags

Tito L. Oracion

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Re: ENERGY AMPLIFICATION
« Reply #8556 on: June 10, 2020, 04:23:15 AM »
Ok sorry if i got here again ok?,
Don't worry it's not my plan to  indulge here more ok?
I just want to spice and add some salt into the topic of
Capacitor ok?.
Here it is:
First is we always have to pour or discharge the cap
Into a lot lot lot lot lot of coil and the very thing that we
Should know is the controller of discharger, what i mean is
As much as possible we only need an electron movement or a very
Very very short movement is all it takes to energized the bunch of coils
And collectors.
A very huge of magnetic field is created in just a very very small movement
An electron point of move or push is all we need everyone. Imagine that.
Offcourse the strenght of magnetic field will  still depends on the primary voltage
,size of coils and the faster we can sudden stop the movement.
Ok.
Thank you and goodbye
Covid is getting nearer and nearer in my vicinity i'm really  in danger
Maybe and i think i'm going to discharge the capacitor in me to kill the virus if
I got infected ;D


Bye friends probably this is my really last post. :)
By the way just to give hopè to new comers
There is one free energy solution in dC and one in AC and  in capacitor, tesla switch and radiant energy receiver ok?
Currently i'm thinking about and will try to combine different kinds of metals with americium 231 and a magnet finding a natural pouring of electron movement continously. In that sense
We only need a vibrating relay to make power.


Bye everyone
GOD Bless all of you :)




WhatIsIt

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Re: ENERGY AMPLIFICATION
« Reply #8557 on: June 10, 2020, 04:56:15 PM »
100uF , 20V discharge parallel into another 100uF.
Now you have 2 caps of 100uF , 10V each.

Connect them in series, 2x100uF, 10V and you got output of 100uF, 20V.
There is no energy loss.

WhatIsIt

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Re: ENERGY AMPLIFICATION
« Reply #8558 on: June 10, 2020, 06:31:11 PM »
As far as I know, there was 2 members trying to exploit cap conundrum.
Both were nailed to the cross by the members eventually.
Aldo, they both gave interesting solutions.

One was to use load while caps equalize,
first cap discharge trough resistor, load into second and
do some work, until both caps equalize in voltage.
And then with the help of serial law, the energy is returned to source.
It involved complicating??? switching of 3 switch.
Both members was buried before they had chance to present their project all the way through.
I stayed in contact with them for a while, and did some experiments
which showed me that in some cases source was spent just a little and result was positive.

Anyway, it was hard to do it with mosfets, and there was lots of circuitry involved,
and in time I abandon idea and left the project to sit in corner for some future events.


Void

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Re: ENERGY AMPLIFICATION
« Reply #8559 on: June 10, 2020, 07:16:30 PM »
100uF , 20V discharge parallel into another 100uF.
Now you have 2 caps of 100uF , 10V each.

Connect them in series, 2x100uF, 10V and you got output of 100uF, 20V.
There is no energy loss.

Sorry, but that is completely wrong.
Analyzing this only requires very basic electronics knowledge.

Stored energy of a 100uF cap at 20V: 20mJ
Stored energy of 100uF cap at 10V: 5mJ
Total energy of two 100uF caps at 10V: 5mJ + 5mJ = 10 mJ (Yes, 50% of the energy is already lost)
Stored energy of two 100uF caps in series, each cap at 10V (Note: two 100uF caps in series equals 50uF, so we have 50uF at 20V): 10mJ

Clearly, 50% of the energy was already lost before ever having connected a load,
so there would only be 50% of the energy left for powering a load in your above scenario.
This type of approach very obviously does not work.

This is the same reason why Bedini's 4 battery ("Tesla Switch") or 3 battery setup system can not work.
Basic analysis indicates the batteries will run down.

WhatIsIt

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Re: ENERGY AMPLIFICATION
« Reply #8560 on: June 10, 2020, 07:33:17 PM »
Note: two 100uF caps in series equals 50uF, so we have 50uF at 20V): 10mJ


2x100uF = 200uf = in series 200/2 = 100uF, 20V

How you came from 200uF to 50uF?
Serial law states that in case 2 caps, their capacitance will be halved (200uF/2) and voltage doubled.

I don't wanna go down into this rabbit hole just as those members did and had nailed to cross.

So, for me this story is over, you are right and let it stays that way.

Magluvin

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Re: ENERGY AMPLIFICATION
« Reply #8561 on: June 10, 2020, 09:24:02 PM »
2x100uF = 200uf = in series 200/2 = 100uF, 20V

How you came from 200uF to 50uF?
Serial law states that in case 2 caps, their capacitance will be halved (200uF/2) and voltage doubled.

I don't wanna go down into this rabbit hole just as those members did and had nailed to cross.

So, for me this story is over, you are right and let it stays that way.

(2) 100uf caps in parallel = 200uf
(2) 100uf caps in series =50uf

You need more study time before you nail the story to the cross. ::)

(2) 100uf caps 50v each in parallel = 200uf at 50v
(2) 100uf caps 50v each in series = 50uf at 100v

Now, as you may need to, find an online calculator for energy in a capacitor and run these figures...

(1) 100uf cap at 100v = ?
(1) 200uf cap at 50v = ?

Are they the same?

For a Sr. Member, you should know this.... ;)

Mags

NickZ

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Re: ENERGY AMPLIFICATION
« Reply #8562 on: June 10, 2020, 09:38:43 PM »
   Mags:    So now you are  :) making us work for  it...    Thanks for the reminder.

Magluvin

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Re: ENERGY AMPLIFICATION
« Reply #8563 on: June 10, 2020, 10:14:18 PM »
The story continues...

When I was told this by Sr. members years ago, I was in denial at first. And once the calculations were made, I was still in disbelief a bit. It was said that the 50% was lost due to resistance(no matter how small or large the resistance) and heat during the action of connecting the say 100v cap to the 0v cap. Seemed plausible, but something just didnt seem right still. 

Then the Sr. members gave an example that enabled me to argue that they were wrong and it was not the resistance that caused the loss at all. They said if we looked at it in an ideal world of no resistance, only then would we not experience the loss in the cap to cap experiment.  Well that had me going even more into disbelief. My mind was running like a full steam locomotive.  So doing some calculations, in the supposed ideal world, having 1 cap at 100v connected to a 0v cap, we would need to end up with 70.7v in each cap to still have 100% of the original energy of the first cap we started off with of 100v.

Now I was on the right track.  There was something terribly wrong here. Even more wrong than I had initially thought. I made analogies in my head. Then it hit me..  'Hmm. How could I take 10gal of water in 1 bucket and end up with 2 buckets that contain 7.07gal each????  We cannot.

There is a specific law in electricity that proves we cannot end up with 2 caps of 70.7v each from 1 cap of 100v in the experiment described, neither real world nor ideal world of no resistance.  ???? ;)

If we have a capacitor at 0v that was very precise as to its value of capacitance, and we were able to count the number of electrons pulled from the + plate and stored to the - plate, we should be able to calculate the exact voltage across that cap with extreme accuracy.

So.  with all this, I was able to argue that it was not the resistance and heat during the transfer of cap to cap that caused the losses. In the real world or ideal, we still lost 50% mostly because we didnt do anything with the transfer action. Whether we did anything with it like run a motor, a light bulb, heat from a resistor, or even ideally where no energy was expelled beyond what was stored in the caps, there will always be 50% of the initial energy gone in the end of the experiment.

So. How would we explain where the energy went in the ideal no resistance world??  Where did it convert to??? ;)

In the real world where resistance is applied, did we get that heat for free????   ;D

Mags

NickZ

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Re: ENERGY AMPLIFICATION
« Reply #8564 on: June 10, 2020, 10:57:44 PM »
  No...  No free lunch in the real world. As free energy is killed by each and every component used. The more of them the less for us. Right?  So how do we get to OU. Remember OU...   As an old timer here, you should know how, by now. But,...