Increase the potential energy without any energy

[activ25]

Low-Q, webby1, Magluvin: please, speak about my device because if you mix electricity and mechanics (or other example in mechanics) it becomes impossible to read. I'm agree the energy is conserved with standard capacitors and Magluvin didn't understand something. But anyway, come back to my device, I count all the energies. And again, the theorem of Noether cannot be applied here because the function is not continuous. Could you be precise, where in my thinking, I'm wrong ? The energies X, Y and W are known and without a white disk the energy is conserved, the springs lost a potential energy but the walls give an energy, I calculate with integrals and the sum of energy is conserved without the white disk. But with the white disk, the sum of energy is not zero, it is 'e'. Now, be specific, and try to explain where I'm wrong please:

1/ move in/out spheres need an energy ?
2/ rotate the disk around the magenta point needs an energy ?
3/ I can't recover the energy from the difference of length L1/L2 ?
4/ the potential energy is not decrease by the same value 'd' at start and at final ?

Because, there are no other possibility.

Like there is no mass, there is no delay to transmit the pressure, so for you if the question 1/ and 2/ need and energy, could you explain ?
I don't know why the question 3/ could be true.
I calculate the sum of length for all springs at start and at final and it is the same (even it is logical with geometry), so I don't see why the question 4/ could be true.

NB: change the length of a spring don't need nor give an energy because it is like a molecule of water inside a glass of water on Earth, the molecule is attracted by gravity but others molecules around give Archimede force, so the sum if zero. It is the same with the springs and spheres. It is verified by calculation, without the white disk the sum of energy is well conserved and I take in account only the potential energy of the springs and the work from walls.

[activ25]

You might be surprised at the connection between mechanical and electrical.

No, I'm not.

This is wrong,, you keep the same quantity of spheres. 

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.

Same quantity within a smaller volume simply means more pressure,, same quantity larger volume less pressure.

The volumes are constant.

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

[Magluvin]

I must figure this out. I can't sleep at night with a puzzle like that riding my brain all night


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

[Low-Q]

]

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)...

[activ25]

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.