Increase the potential energy without any energy

[Magluvin]

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.

I didnt get into it with them back them about charging a cap from a power supply, but it was said that there is the loss there and I just accepted it then as i was in awe of the 50%loss. But Im making a new connection with that idea as you posed it to me.  Here is my point..


Do you agree that we lost 50% in the 10uf to 10uf cap to cap deal?
If so do you also agree with the much smaller loss when the charged cap is 10uf and the receiving cap is 1uf? If not, you need to run the numbers again. Or you need to explain how and why the cap energy calculators are bunk.

Now if you agree with both questions, then say we have a 1Fcap at 10v and a 1uf cap at 0v. Pretty much the 1uf cap will see the 1F 10v as a huge reservoir very very similar to the power supply. If the 1F charges the 1uf to be the same voltage in the end after cap to cap, you may need a meter that reads to the 4th decimal or just thinking possibly 5 decimals to see the voltage drop of the 1F cap. Just guessing but iI think Im close. In the power supply there are output caps that are no where near 1F. So it may be possible that the 1F can charge the 1uf quicker. Just again speculating an educated guess. for simple argument sake.  Now if you do read a voltage drop in the 1F cap, of which what is left would be the same voltage the 1uf cap charged to, then calculate the original energy in the 1F at 10v and then calculate the energy left in the 1F and then the 1uf and add them. The loss would be incredibly small compared to what you are saying about the power supply charging the cap.

So now Im convincing myself right here right now that no, the cap charging from the power supply to the empty cap should not be lossy at all but very small in the least. In my 1F example, there will not be that voltage drop you speak of that would be any different than the power supply other than possibly a .001v or .ooo5v or even a .05v difference for arguments sake, of which isnt even close to a loss level as you suggest. In fact, the power supply just may dip more than the 1F cap initially in the charging of the cap. that would and could be looked at with a scope. And I happen to have a 1F 20v car audio cap on the bench and a 10 amp var supply.

So here we may have stirred another possible controversy like the bad claim that resistance was the loss in the cap to cap exercise.  I seem to remember them claiming the loss from PS to cap back then, but maybe that is all wrong also.    We may have to look at that closer now.

Mags

[activ25]

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.

Roberval is not the same device.

How the springs change ? the springs inside the container ? it is easy, each sphere has an Archimede force on it and a force from the spring attached on it, so the sum of these forces are zero. So it cost nothing to change the length of the springs WHEN the sphere is inside the container. It is like move up (or down) a molecule of water inside a glass full of water on Earth: it costs nothing.

I explained, I calculate the sum of energies for the device without the white disk, and the sum is well at zero. So the springs (of the spheres inside the container) don't need any energy nor give any energy.

[activ25]

I guess you don't know a Roberval,, they are very fun things to play with,, you can use arms and levers or pulleys and string or gears or a combination of parts,, I made one that used water.

With a Roberval the Noether's theorem is applied so the device can't create an energy.

[activ25]

The gain in energies you are measuring from L1 to L2 are the same energies needed to go from L2 back to L1.

No, the device moves from the start position to the end position. I never return to the start position, so I don't know where you find in the document I move and the lengths of the springs move from L2 to L1.

This short part describes a change in gradient caused by an external motion, that motion requires a force to go along with it.
This change in gradient is what causes the spring to change in its relative stored potential.

The device I measure the sum of energy is unstable. So I need an external device  (I call it the C_device to be clear) to control it. But this external device is perfect, no friction, etc. and this C_device recovers the energy from the device and gives energy if it is necessary, it is there only to control and in the same it counts the input/output of the energies. So, C_device counts the equations I wrote, only that.

Then by your definition as soon as the spring and sphere are removed there is no stored gain or loss since that condition is relative to the other spheres\springs\walls.


You also chose an arbitrary orientation for the observation of the spring, this is fine providing you keep using the same reference frame, which you do not when you do not also include the required changes on the spring and sphere to go back from L2 around the white wheel and back to L1.


The gain in energies you are measuring from L1 to L2 are the same energies needed to go from L2 back to L1.



This short part describes a change in gradient caused by an external motion, that motion requires a force to go along with it.
This change in gradient is what causes the spring to change in its relative stored potential.


If you are thinking about an oscillating system where the change of deformation is a natural event then the usage of the observed change in the spring stored potential will act as a damper to the system, it will slow it down,  this then would be your input.   
This could be done with a heat gradient as well,, many ways of doing it BUT the source for the gradient must be included in the energy calculations,, this does not mean that YOU as the "operator" need to pay the cost, only that the cost is shown and understood.


Calculate the energy to deform the container,, if it is a natural event then you would calculate the full energy of that event,, say geothermal heating,, how many BTU's are available then becomes the limit on what you as the operator could extract.

The device don't oscillate. I describe the start position of the device and the end position. The device don't return at its original position. I win an energy from L1->L2. I don't need to calculate the exact energy I win because with my theoretical device it becomes logical. I explain my thinking: the sum of energy is conserved without the white disk. And after, I count the differences. I don't need calculations to prove something: https://en.wikipedia.org/wiki/Propositional_calculus

[activ25]

I don't extract the energy from one spring but all the springs of the spheres that must be move out the container to let pass the white disk (and the spheres must be put inside in the same time, but the lengths of these springs are lower). The energy is extract from the springs. I move out a sphere AND in the same time I move in a sphere behind the white disk, because the volume must be constant.

I'm logic, look at the equations. And I transform a complex problem in a simple because I take simplification: no mass, no friction, volume constant, spring with a constant force. I use simple laws of physics.

"After you have taken the spring\sphere energy out it is now at a low energy condition,, you must put it back in to continue,, so now you have a higher energy state in front and a lower energy state behind,,," not at all, you forget the work of the walls. Have you understand the device without the white disk ? The potential energy lost by springs are won by the walls, the sum is well constant with the white disk, I calculated it with integrals.

And with a theoretical mechanical device, the sum of energy must be constant, it means, the sum of energy can't be negative too.