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Author Topic: Increase the potential energy without any energy  (Read 10479 times)

Offline Magluvin

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Re: Increase the potential energy without any energy
« Reply #90 on: September 26, 2017, 09:10:11 PM »
I must figure this out. I can't sleep at night with a puzzle like that riding my brain all night :D


Vidar
]

The cap to cap is a well known exercise that has been misconstrued for some time claiming that the 50% loss is due to heat by way of resistance.

Here is the capacitor energy calculator....
http://www.calctool.org/CALC/eng/electronics/capacitor_energy

10uf 10v is .5mJ   

10uf 5v is ,125mJ

2 10uf caps(20uf) at 5v is .25mJ  Total of 1/2 the original energy in the single 10uf cap at 10v .5mJ

We lose half the energy dumping a full cap into an identical empty cap of the same capacitance value.

Now how I started thinking about it to come to my conclusion that the resistance plays no part in the 50% energy loss as many were claiming, was when they went ahead and said that if the caps were ideal, zero resistance, superconductive per say, that we would end up with 7.07v in each cap after the cap to cap and the voltage was leveled out between the 2 caps.  If we did the electron count, it would be impossible to get 7.07v in each cap from a dump of 10v in the initial cap. Cant happen just like we can get 2 buckets of water of 7.07 gal each from a single 10 gal bucket.

1 10uf 7.07v is .2499mJ   not .25 exactly because the 7.07 is rounded off from the actual figure. But you should get the drift.

20uf(2 10uf caps in parallel) at 7.07v is .499mJ  again would be .5mj if the .707 were not rounded off

So resistance or zero resistance, we still lose 50% of the original total energy of the 10uf 10v cap by dumping half of the charge into another same value cap.

Mags

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Re: Increase the potential energy without any energy
« Reply #90 on: September 26, 2017, 09:10:11 PM »

Offline Low-Q

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Re: Increase the potential energy without any energy
« Reply #91 on: September 26, 2017, 10:01:40 PM »
]

The cap to cap is a well known exercise that has been misconstrued for some time claiming that the 50% loss is due to heat by way of resistance.

Here is the capacitor energy calculator....
http://www.calctool.org/CALC/eng/electronics/capacitor_energy

10uf 10v is .5mJ   

10uf 5v is ,125mJ

2 10uf caps(20uf) at 5v is .25mJ  Total of 1/2 the original energy in the single 10uf cap at 10v .5mJ

We lose half the energy dumping a full cap into an identical empty cap of the same capacitance value.

Now how I started thinking about it to come to my conclusion that the resistance plays no part in the 50% energy loss as many were claiming, was when they went ahead and said that if the caps were ideal, zero resistance, superconductive per say, that we would end up with 7.07v in each cap after the cap to cap and the voltage was leveled out between the 2 caps.  If we did the electron count, it would be impossible to get 7.07v in each cap from a dump of 10v in the initial cap. Cant happen just like we can get 2 buckets of water of 7.07 gal each from a single 10 gal bucket.

1 10uf 7.07v is .2499mJ   not .25 exactly because the 7.07 is rounded off from the actual figure. But you should get the drift.

20uf(2 10uf caps in parallel) at 7.07v is .499mJ  again would be .5mj if the .707 were not rounded off

So resistance or zero resistance, we still lose 50% of the original total energy of the 10uf 10v cap by dumping half of the charge into another same value cap.

Mags
You haven't lost it. You just havent spent all yet as useful energy.
I can agree that it is a confusing experiment.
1. What is the initial PE?
2. What is the loss of PE after both have the same charge?
3. What is the difference from 2. after emtying both caps?


What is the sum of the answer in question 2 and 3?
The sum is the same as initial PE in the first cap.


Energy is conserved.


Vidar (sorry for polluting your thread active25)...


Offline webby1

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Re: Increase the potential energy without any energy
« Reply #92 on: September 26, 2017, 11:12:51 PM »
No, I'm not.
 And what is the problem ? The container keeps the same volume, the number of spheres is the same, the white disk keeps constant its volume. All volumes are constant.
  The volumes are constant.

webby1: the springs have no volume, again, to simplify the calculations.


I re-read your file a few times,, and I think I missed your point, I got hung up on volumes and quantities being stated as constant for a changing geometry.


If you place the sphere back in at L2, so that you can see\use the change in energy of the spring as compared to L1 , you would need to elevate that sphere up and over the white disk raising the spring potential higher than at L1 and then allow it to reduce down to L1 as the wheel rotates.


If then you were to use that energy difference between L1 and L2 you would need to replace that energy in order to move the sphere around and back to the "place" it started from.


In short, Yes there is a change in potential between the L1,L2 positions.


What also needs to be considered is the other change in potential,, that is, if the spheres are moving from L1 to L2 then there journey must also be viewed in the same direction from L2 around to L1.


How did the sphere and spring get to the position of L1, what replaces the sphere you take out at L2, how does the replacement sphere at L2 get around the white disk.
For the volume and quantity to stay constant then there must be a full fill of spheres going up the back side of the white disk.

Offline webby1

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Re: Increase the potential energy without any energy
« Reply #93 on: September 26, 2017, 11:20:37 PM »
You haven't lost it. You just havent spent all yet as useful energy.
I can agree that it is a confusing experiment.
1. What is the initial PE?
2. What is the loss of PE after both have the same charge?
3. What is the difference from 2. after emtying both caps?


What is the sum of the answer in question 2 and 3?
The sum is the same as initial PE in the first cap.


Energy is conserved.


Vidar (sorry for polluting your thread active25)...


Those questions were answered,,
The sum is less and as such I agree that not all of what was stored has been used effectively, or maybe it would be better to say that not all that was used to create the stored potential was stored,, except for the ability to extract the energy back out of the cap by other means,,, makes you wonder doesn't it :)


Offline webby1

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Re: Increase the potential energy without any energy
« Reply #94 on: September 26, 2017, 11:24:57 PM »
P.S.


The water in the 2 tanks has the same weight as the first tank,, but now it can only fall 1\2 the distance that it could of fallen from the first tank.

Free Energy | searching for free energy and discussing free energy

Re: Increase the potential energy without any energy
« Reply #94 on: September 26, 2017, 11:24:57 PM »
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Offline activ25

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Re: Increase the potential energy without any energy
« Reply #95 on: September 27, 2017, 06:37:27 AM »

If you place the sphere back in at L2, so that you can see\use the change in energy of the spring as compared to L1 , you would need to elevate that sphere up and over the white disk raising the spring potential higher than at L1 and then allow it to reduce down to L1 as the wheel rotates.



I recover an energy from the springs because L1>L2, is it ok for you ? for all spheres I need to move in/out.

When I move out a sphere of the container, the pressure is exactly the same than the sphere I move in. Look at the identical lines of pressure, they are always perpendiculary to the springs, and the springs change their orientation.


If then you were to use that energy difference between L1 and L2 you would need to replace that energy in order to move the sphere around and back to the "place" it started from.
No, you misunderstand the potential energy stored in the spring and the energy I recover/need to move out/move in  a sphere. The length of the spring is a potential energy recovered. I recover an energy when I move out a sphere, exactly the same when I recover an energy when I move out an object from a side of a container full of water under gravity. And I lost an energy when I move in the object. But here, the pressure are symmetric.

Offline webby1

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Re: Increase the potential energy without any energy
« Reply #96 on: September 27, 2017, 07:20:30 AM »
I recover an energy from the springs because L1>L2, is it ok for you ? for all spheres I need to move in/out.
Yes,, that is fine since the spring length is not the same.
Quote

When I move out a sphere of the container, the pressure is exactly the same than the sphere I move in. Look at the identical lines of pressure, they are always perpendiculary to the springs, and the springs change their orientation.
I think you need to look into this part deeper.


Quote
No, you misunderstand the potential energy stored in the spring and the energy I recover/need to move out/move in  a sphere. The length of the spring is a potential energy recovered. I recover an energy when I move out a sphere, exactly the same when I recover an energy when I move out an object from a side of a container full of water under gravity. And I lost an energy when I move in the object. But here, the pressure are symmetric.


I understand just fine.


You are not using the correct reference frame,, you do not have all of the energy in and out accounted for,, mainly that that is being put into your system, the deformation and or rotation is an input.




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Re: Increase the potential energy without any energy
« Reply #96 on: September 27, 2017, 07:20:30 AM »
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Offline activ25

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Re: Increase the potential energy without any energy
« Reply #97 on: September 27, 2017, 07:33:02 AM »
I think you need to look into this part deeper.
Explain how you see the pressure please, because the springs simulate gravity. So, the lines of pressure are perpendiculary to the force of attraction with gravity, tou are ok with that ? So, with the springs it is the same.

You are not using the correct reference frame,, you do not have all of the energy in and out accounted for,, mainly that that is being put into your system, the deformation and or rotation is an input.
Your sentence is like, you're wrong, I don't know where, but you're wrong.


Offline Magluvin

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Re: Increase the potential energy without any energy
« Reply #98 on: September 27, 2017, 07:34:34 AM »
You haven't lost it. You just havent spent all yet as useful energy.
I can agree that it is a confusing experiment.
1. What is the initial PE?
2. What is the loss of PE after both have the same charge?
3. What is the difference from 2. after emtying both caps?


What is the sum of the answer in question 2 and 3?
The sum is the same as initial PE in the first cap.


Energy is conserved.


Vidar (sorry for polluting your thread active25)...

You tell me. ;) Ive laid out what I claim. I dont have 2 months to convince you like I had to do when I last made the claim. It took that long for that session to sink in with other top guys here. Once they actually took notice they finally agreed and it has been settled here. I made the claim several times before that and was ignored over bout a 2 year stint. Reread what I wrote here about this and do some tests and calculations yourself. If you find fault in what I claimed then just simply prove me wrong. I put it to you here in a more detailed manner than I had done to convince others before because its ingrained in my mind and can describe it in better detail now than then. So put your thinking cap on and work it out, and if you can, provide convincing argument that what I said is not correct. Then Ill respond with a rebuttal.

You want to question me on Pe without giving me your version of that. Show me how and what I said is wrong and give me your numbers and how you came to arrive at those numbers and I will respond.

Mags

Offline Magluvin

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Re: Increase the potential energy without any energy
« Reply #99 on: September 27, 2017, 07:41:09 AM »
P.S.


The water in the 2 tanks has the same weight as the first tank,, but now it can only fall 1\2 the distance that it could of fallen from the first tank.

To correctly use the water tank analogy for the cap to cap, the tanks need to be on the same level, not one above the other. Doing so with tanks at different levels offsets the balance that would be inherent with 2 like caps. The water needs to balance to equal parts by leveling out just like the caps. Having one water tank at a different height, that divided balance cant be had.

mags

Free Energy | searching for free energy and discussing free energy

Re: Increase the potential energy without any energy
« Reply #99 on: September 27, 2017, 07:41:09 AM »
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Offline Magluvin

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Re: Increase the potential energy without any energy
« Reply #100 on: September 27, 2017, 09:19:58 AM »
You tell me. ;) Ive laid out what I claim. I dont have 2 months to convince you like I had to do when I last made the claim. It took that long for that session to sink in with other top guys here. Once they actually took notice they finally agreed and it has been settled here. I made the claim several times before that and was ignored over bout a 2 year stint. Reread what I wrote here about this and do some tests and calculations yourself. If you find fault in what I claimed then just simply prove me wrong. I put it to you here in a more detailed manner than I had done to convince others before because its ingrained in my mind and can describe it in better detail now than then. So put your thinking cap on and work it out, and if you can, provide convincing argument that what I said is not correct. Then Ill respond with a rebuttal.

You want to question me on Pe without giving me your version of that. Show me how and what I said is wrong and give me your numbers and how you came to arrive at those numbers and I will respond.

Mags

One more thing.  My actual claim is that resistance is not the cause for the loss. But you are arguing the basic fact that there is no loss in the cap to cap. So you have a ways to go with understanding the basic point of the exercise. We lost it by reducing pressure that resulted in no work being done during the cap to cap connection of a charged cap into an identical cap at 0v.  A very similar case would be to have 2 12v car batteries. It may not be a 50% loss as a battery is a bit different than a cap.
If we have a fully charged 12v battery and a fully drained battery of the same make and model then we can calculate how much work can be done with the fully charged battery and use it till it is all the way drained to get the result, then we take the fully charged battery and connect it to the fully drained battery, you will not end up with the same amount of total energy of the 2 batteries combined as you could get out of the single fully charged battery.

There are a lot of sites and articles that will give you the same results for the cap to cap 50% loss. Most all will say the "unanimously agreed upon loss" is is the fault of resistance. On your point of there being no loss, that is an argument that just about anyone that works out the numbers will argue against you till you break. And if you dont end up realizing it for what it really is, then you must not agree with the capacitor energy calculators, nor any of the equations of formula that all arrive at the real and actual lossy conclusion. When we first discussed this when it was brought up in a big discussion about caps, most of us were just like you are now. Denial without full understanding. But once actual test were done, and the energy calculations were made over and over, the loss became very real. So like I said, show me your numbers and how exactly you arrived at those numbers that you say prove your reasons for believing that there is no loss and I will respond with where you are making your mistakes and correct them for you.

in the real world, doing the cap to cap as described is not a function used in any electronic devices. Its not a useful circuit in the least. it is a losing proposition. It is only a learning tool.  We do not want this circuit in any products because of the inherent loss.  One more time.......



If we have a 10uf cap at 10v and we use that to do work till its 0v, then we can measure and calculate the total energy used and the amount of work that can be done.



1)  If we used the energy from the 10uf 10v cap only till it is drained down to 5v and we disconnect out load, then we did a certain amount of work with that usage. Now we are left with a 10uf cap at 5v.

2)  So now say we have 2 10uf caps and each has 5v. Now lets just use one of them to do work till it is 0v and leave the other one at 5v.

In both cases 1 and 2, we end up with 1 cap with 5v. Are you going to tell me that the work we did with the first cap beginning at 10v till it was drained to the 5v level then disconnect the load, that the amount work that the cap provided during that drain down to 5v is not more work done than the second case where we did work from 10uf cap at 5v starting and drained down to 0v???  If that is your argument then you need to work on your basics of V*A=P.   If the load were a resistor and we repeated the work done tests 1 and 2 and we measured the heat generated by the resistor in each case, do you believe that in both cases the resistor produces the same 'amount' of heat for each case???  P=V*A is not a credible equation here? The higher voltage during the case 1 drain is not always throughout the test more than the voltage in case 2 during its drain into the resistor? ??? Think man think!

Mags
« Last Edit: September 27, 2017, 01:04:23 PM by Magluvin »

Offline webby1

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Re: Increase the potential energy without any energy
« Reply #101 on: September 27, 2017, 03:13:10 PM »
Explain how you see the pressure please, because the springs simulate gravity. So, the lines of pressure are perpendiculary to the force of attraction with gravity, tou are ok with that ? So, with the springs it is the same.
Your sentence is like, you're wrong, I don't know where, but you're wrong.


I have a fairly good idea of parts that are not being considered, I have a reasonable idea of the path those forces take,, I did suggest that you look into a Roberval system to help in being able to see some of these interactions.


You have not spoken about HOW the spring gets changed,, you have not spoken about how the changing shape that does not change in volume makes things move, and the input it takes to make the shape change,, whether that is from the container morphing or from the wheel spinning or both.


So ask yourself "HOW" does the spring get changed.  Ask yourself if there is motion needed for that change in spring to happen, then link the motion used and the change in the spring so you can follow the path of interaction.


Offline webby1

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Re: Increase the potential energy without any energy
« Reply #102 on: September 27, 2017, 03:25:03 PM »
To correctly use the water tank analogy for the cap to cap, the tanks need to be on the same level, not one above the other. Doing so with tanks at different levels offsets the balance that would be inherent with 2 like caps. The water needs to balance to equal parts by leveling out just like the caps. Having one water tank at a different height, that divided balance cant be had.

mags


Well,, I did not say they were,, BUT the WATER is!


mgh,,, interesting formula,, CoM of the water column is, when uniformly shaped, at the midpoint in the height of the column,, so a 1m tall tank has its CoM at 0.5m,, if you take the water above the 0.5m mark and dump it into the second tank :)


The P.S. part is after all is said and done the CoM is now lower and if it is mgh the same m with less h at the same g means less stored.


Funny thing that I can balance things together that are not even close to being in the same system except for the shared force to balance,,,,,  :)


Do you realize there is the same kind of loss if you charge the cap from a fixed DC voltage?  unless you charge it my way anyhow :)

Offline Magluvin

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Re: Increase the potential energy without any energy
« Reply #103 on: September 27, 2017, 05:14:09 PM »

Well,, I did not say they were,, BUT the WATER is!


mgh,,, interesting formula,, CoM of the water column is, when uniformly shaped, at the midpoint in the height of the column,, so a 1m tall tank has its CoM at 0.5m,, if you take the water above the 0.5m mark and dump it into the second tank :)


The P.S. part is after all is said and done the CoM is now lower and if it is mgh the same m with less h at the same g means less stored.


Funny thing that I can balance things together that are not even close to being in the same system except for the shared force to balance,,,,,  :)


Do you realize there is the same kind of loss if you charge the cap from a fixed DC voltage?  unless you charge it my way anyhow :)

"Do you realize there is the same kind of loss if you charge the cap from a fixed DC voltage?  unless you charge it my way anyhow"

Well i might not agree there. If the 2 caps are the same value, then there is a total 50% loss. If 1 cap is say 10uf 10v and the other is 1uf at 0v and then we do cap to cap. We get 9v each. Now calculate the energy left in the 10uf 9v and calculate the energy in the 1 uf 9v then add them together. The loss is significantly less. So if we think about the power supply as a very very large cap 10v and we charge a 1uf 0v to the ps 10v, wouldnt that in comparison to what I just explained be very similar with not that much of a loss? Would seem to me that the power supply lost nearly as much as the cap gained.  The loss happens hard in the identical cap to cap because the source cap lost a lot of pressure(V) resulting in part of the whole loss, where the power supply if solid would not. The 10u 10v to 1uf 0v cap to cap, the 10uf did not lose a lot of pressure so its energy level is still quite high, and then we add that to what energy level in in the 1uf 9v to find the end total.

I really cant see 50% loss in charging a cap from a power supply that simulates basically an infinitely large cap compared to the receiving empty cap. And I might say that the loss in charging a very tiny cap in comparison that the losses would be very minuscule.

Mags


Offline webby1

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Re: Increase the potential energy without any energy
« Reply #104 on: September 27, 2017, 06:12:19 PM »
Your 10V PSU will provide its current at 10V,, how much voltage is needed to move charge carriers onto a plate at 0V,, it takes 0V+ some infinitesimal amount.


As long as the charging voltage is that infinitesimal amount of voltage above the plate voltage charge will move into the cap.
It takes Coulombs of charge to raise the voltage, the PSU will provide those Coulombs at the voltage it is set to,, so 10V @ 1A to charge a 1F cap from 0V to 1V , another 10V @ 1A to take that cap and raise it from 1V to 2V,, and so on. <<== it is the Coulombs that "fill" the cap.


What is the resistance of a cap with no charge to current flow from the source.
What is the resistance of a cap with 99% charge to current flow from the source.


If your source uses 10V @ 1A to charge the cap it is then wasting energy,, if this were water you would have this large fountain happening while filling until the water reached the top,, what a waste unless the fountain is what you want :)


Plot the energy per step while charging both for the cap and that supplied by the source,, use some value of base resistance for the cap so as to not hit that infinite current flow issue,,


To not have this "loss of unused energy" you can either ramp the voltage up from 0 to charge voltage or you can start with a very small capacitance and increase the capacitance until charge quantity is reached.
A coil storing some of that current "rush" in the magnetic field uses what would be wasted and then returns the field potential into the cap as the cap charges and its resistance to current flow from the source goes up,, is this the best way??? not sure but it does work fairly well.

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Re: Increase the potential energy without any energy
« Reply #104 on: September 27, 2017, 06:12:19 PM »

 

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