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Author Topic: Increase the potential energy without any energy  (Read 9752 times)

Offline Magluvin

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Re: Increase the potential energy without any energy
« Reply #75 on: September 26, 2017, 11:13:20 AM »
And that is how you do the electron count. 100gal of water div into 2 containers, you end with 50gal in each. The resistance is the valve and it only changes the time of the transfer to leveling out, the valve does not consume energy nor take or add water to the system. Same with the electron count, we end up with the same electron count and the resistance only changes the time it takes for each cap to be equal and the resistance doesnt add or take away electrons from the system.

So if we use the energy stored in the 10uf 10v cap only till it is down to 5v, we got more done with that first half of the 10v down to 5v than we can get from either pair of caps at 5v each


10uf 10v use energy till you are down to 5v.   So we still have 5v left

Now have 2 10uf 5v ea  and use the energy of 1 of them.  We still have 1 10uf 5v left just like above.


So the energy from the 10uf at 10v will do more work down to the 5v cutoff than 10uf 5v can do and in both situations we are left with 10uf 5v.

Mags

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Re: Increase the potential energy without any energy
« Reply #75 on: September 26, 2017, 11:13:20 AM »

Offline Low-Q

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Re: Increase the potential energy without any energy
« Reply #76 on: September 26, 2017, 12:02:58 PM »
And that is how you do the electron count. 100gal of water div into 2 containers, you end with 50gal in each. The resistance is the valve and it only changes the time of the transfer to leveling out, the valve does not consume energy nor take or add water to the system. Same with the electron count, we end up with the same electron count and the resistance only changes the time it takes for each cap to be equal and the resistance doesnt add or take away electrons from the system.

So if we use the energy stored in the 10uf 10v cap only till it is down to 5v, we got more done with that first half of the 10v down to 5v than we can get from either pair of caps at 5v each


10uf 10v use energy till you are down to 5v.   So we still have 5v left

Now have 2 10uf 5v ea  and use the energy of 1 of them.  We still have 1 10uf 5v left just like above.


So the energy from the 10uf at 10v will do more work down to the 5v cutoff than 10uf 5v can do and in both situations we are left with 10uf 5v.

Mags
Actually not. From 10V to 5V, or 5V to 0V in the same capacitor provides the same amount of energy. The difference in both is 5V at the same capacity.
The difference is time.
If you load the capacitor with a 1kOhm resistor, it takes shorter time to discharge from 10 to 5V, and longer time to discharge from 5V to 0V - actually, it takes forever to discharge from 5V to 0V if the capacitor is perfect. The energy in both cases is equal. When you charge the capacitor through the same resistor, it takes shorter time to charge from 0V to 5V, than from 5V to 10V. In the meantime you have lost energy through the resistor during charging. The energy input is therfor higher than the potential energy in the fully charged capacitor.


Lets say for the cause of simplicity, you have a mass of 1kg you want to lift 1 meter. Lets say you spend 1 second the first half meter. Then 1 hour the last half meter. What is the potential energy between 0m and 0.5m, and between 0.5m and 1m? Not surprisingly they are equal.
In this particular case, the altitude is the charge, and the mass is the capacity.


Vidar


Offline activ25

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Re: Increase the potential energy without any energy
« Reply #77 on: September 26, 2017, 12:09:37 PM »
Mags: look at this link: http://www.smpstech.com/charge.htm

But I'm agree with you Mags, it is possible to create the energy (or destroy).

Low-Q: my document is not clear ?

Offline Magluvin

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Re: Increase the potential energy without any energy
« Reply #78 on: September 26, 2017, 01:07:21 PM »
Actually not. From 10V to 5V, or 5V to 0V in the same capacitor provides the same amount of energy. The difference in both is 5V at the same capacity.
The difference is time.
If you load the capacitor with a 1kOhm resistor, it takes shorter time to discharge from 10 to 5V, and longer time to discharge from 5V to 0V - actually, it takes forever to discharge from 5V to 0V if the capacitor is perfect. The energy in both cases is equal. When you charge the capacitor through the same resistor, it takes shorter time to charge from 0V to 5V, than from 5V to 10V. In the meantime you have lost energy through the resistor during charging. The energy input is therfor higher than the potential energy in the fully charged capacitor.


Lets say for the cause of simplicity, you have a mass of 1kg you want to lift 1 meter. Lets say you spend 1 second the first half meter. Then 1 hour the last half meter. What is the potential energy between 0m and 0.5m, and between 0.5m and 1m? Not surprisingly they are equal.
In this particular case, the altitude is the charge, and the mass is the capacity.


Vidar

"Actually not. From 10V to 5V, or 5V to 0V in the same capacitor provides the same amount of energy. The difference in both is 5V at the same capacity.
The difference is time."

Absolutely not.  You are missing it.

The 10v down to 5v has a great advantage over 5v down to 0v. 

The 10v to 5v still has the potential of the beginning output of the 5v when it gets down to 5v, And the 10v to 5v is working on potential above 5v throughout the use, where the 5v only has 5v down to 0v potential during its use. Run the numbers.  10uf 10v holds 2 times the energy of a 20uf cap at 5v. Yep. ;)

In the cap to cap, we lost 50% doing the cap to cap.  There are many online calculators for caps. Run the calculator to calc the energy of a 10uf cap at 10v, then run it for 20uf at 5v(equal to total 2 10uf at 5v) You will see the loss. You can do more work with a 10uf at 10v than 20uf at 5v. Check it out for yourself. Its not what you think. Its the same for water tanks and gravity and the same as air pressure tanks. You would need to end up with 7.07v in each cap for the 2 caps combined energy to equal the energy of 1 cap at 10v. Run that on the energy in a cap calculator online or work out the equation yourself. Well by the electron imbalance count we cannot end up with 7.07v each. Just like we cannot end up with 2 buckets of water with 7.07gal from 1 bucket that had 10gal.

Its been a big discussion on and off here. It is what it is.  Even Woopy said Opah! Couldnt believe the loss. Its been through the ringer.  In the back of my mind I couldnt see the resistance as the loss, considering the electron count evidence.  If you could count how many electrons you take from one plate of the cap and shove it into the other plate, you can determine the resulting voltage charge of a particular cap value, every time. More electrons taken from one plate and forced into the other plate is stored pressure(voltage) If you know the exact capacitance and the voltage charge on it, you can use Coulombs law to figure the electron imbalance of the plates. And that number will be the exact same if the cap is charged to exactly 10v after discharging it to 0 and charge back to exactly to 10v. Each cap value will have its own electron count imbalance across its plates for 10v charge.  A tiny cap will need less electron count differential than a large cap, just like an air tank, less air molecules to get a small tank to 100lb than a large tank to get to 100lb.

Some claim that charging a cap say in a output of a power supply has this loss. Its not true. If it were then we would not have power supplies that approach 100%eff. We are mostly just topping off an output cap in a supply, not replenishing from 0v.  Only at start up of the supply where the cap is actually 0v is there a possibility of these losses, and even then the new slow start supplies take care of that issue with step charging till optimum output is achieved.

Mags


Offline Magluvin

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Re: Increase the potential energy without any energy
« Reply #79 on: September 26, 2017, 01:18:46 PM »
Mags: look at this link: http://www.smpstech.com/charge.htm

But I'm agree with you Mags, it is possible to create the energy (or destroy).

Low-Q: my document is not clear ?

The energy loss in the cap to cap scenario can be associated to the energy it took to move the electrons/water/air from 1 container to the other one till leveled out. That is the only thing I can think of as to where the energy went. But, we could have used that energy in the transfer from cap to cap to do something and still end with 5v in each cap.

But the saying that the energy is lost in the cap to cap is due to resistance and heat is not true. It is only a valve set to a particular position between the 2 tanks to slow the transfer down, but in the end all of the electrons leveled out just like water or air and we lost the energy by not using the energy that occurred in the transfer to do other work in the process. We just released potential pressure into a tank that is basically 2 times as large. We lost it stupidly as I say, unless the goal was to divide that pressure/water/air into 2 containers for what ever reason. That is a lossy goal. ;)

Mags

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Re: Increase the potential energy without any energy
« Reply #79 on: September 26, 2017, 01:18:46 PM »
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Offline Low-Q

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Re: Increase the potential energy without any energy
« Reply #80 on: September 26, 2017, 02:55:22 PM »
The energy loss in the cap to cap scenario can be associated to the energy it took to move the electrons/water/air from 1 container to the other one till leveled out. That is the only thing I can think of as to where the energy went. But, we could have used that energy in the transfer from cap to cap to do something and still end with 5v in each cap.

But the saying that the energy is lost in the cap to cap is due to resistance and heat is not true. It is only a valve set to a particular position between the 2 tanks to slow the transfer down, but in the end all of the electrons leveled out just like water or air and we lost the energy by not using the energy that occurred in the transfer to do other work in the process. We just released potential pressure into a tank that is basically 2 times as large. We lost it stupidly as I say, unless the goal was to divide that pressure/water/air into 2 containers for what ever reason. That is a lossy goal. ;)

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


Vidar

Offline webby1

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Re: Increase the potential energy without any energy
« Reply #81 on: September 26, 2017, 03:05:52 PM »
webby1:  my device needs to have a function discontinuous, it is very important, in the contrary, the theorem of Noether is applicable and the energy is conserved, so I need to have something that moves outside the device. With your example the function is continuous (or I didn't understand your example).

Thanks to try to help me :)


I was trying to point out that something does indeed change,, it must change,, and that is where the "energy" change comes from.


Example:  If you have 100 cubes stacked up so they are 10 cubes high and 10 cubes wide, the containment "box" will have 4 sides each 10 cubes long,, so the containment "box" will be 40 cubes in length,, now take those 100 cubes and have them 1 cube high and 100 cubes wide,, now your containment "box" has 2 sides 1 cube high and 2 sides 100 cubes long for a total "length" of 202 cubes.


Example:  If you have a containment "box" that is 40 cubes in length and arrange it so that there are 2 sides 1 cube long you then have 2 sides that are 19 cubes long and you can fit 19 cubes within that containment box, now arrange the "box" so that you have 4 sides 10 cubes long and fill it with 19 cubes, you now have a volume deficit of 81 cubes.


The change in energy then comes from either stretching the containment length or changing the distance of separation of the cubes.


Energy enters into the system.



Well, that is not completely true. If you charge a 1000 l tank with water. The tank is a capacitor. This tank is connected to an equal tank at the same level. The connection is a tube with a ball valve at the bottom.
Now, open the ball valve, and let the water flow into the other tank. After a while there is 500 liters in each. Now you have the same total storage of potential water.
What you suggest, is that both tanks should have 707 liters in each, and asks where did the lost 414 liters of water go, and conclude with the following state: There is lost energy on the way.




The same example applies to a capacitor. It is not only the Voltage (pressure at the bottom of the tank), but also the potential energy storage. You end up with 20uF capacitors with 5V in each. That is the exact same potential energy storage as one 10uF capacitor at 10V. No energy is lost.




Vidar


Energy stored in a cap is 0.5*(C*V^2),, and you also see this relationship within a tank of water,, due the math and it shows that twice the volume with the same quantity at 1\2 the pressure has 1\2 the stored potential,, it is that X^2 part :) when dealing with a quantity.  It would be best if you transferred 1\2 the quantity into a "vessel" that had 1\2 the volume while decreasing the volume of the first vessel to 1\2.


There is insufficient  supplies of either pressure or quantity or both,, depends on how you choose to look at it,, to deal with twice the volume that needs to be occupied.


The final decision was to just teach everyone that the lost energy is in the resistance of the connection, friction of the water through the pipe or the resistance to charge flow  resulting in the energy dissipated as heat, it is not the correct answer but for how we use things it works,, some of us however prefer to find out what really is happening :)

Free Energy | searching for free energy and discussing free energy

Re: Increase the potential energy without any energy
« Reply #81 on: September 26, 2017, 03:05:52 PM »
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Offline Low-Q

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Re: Increase the potential energy without any energy
« Reply #82 on: September 26, 2017, 05:22:47 PM »
So if we have two tanks. 1x1x1m each. One filled with 1000 l of fluid.
That full tank have a potential energy of 500 units with reference to the ground because center of the mass is 0.5 meter above ground.
Then we open the valve so 500 liter fills the other tank.
As the waterlevel rises in the other tank, the remaining potential energy in the first tank drop both because the level is decreasing, and the initial reference level for the other tank will rise.


In reality, the potential energy in the first tank is only 250 units because we choose to fill the other tank, and not poor the fluid on the ground.


The remaining potential energy in each tank will finally reach 125 units each. 250 units combined. The same as the initial potential energy in the first tank. Energy is conserved.


Does this make sense? Will this apply to the capacitors as well?


Vidar.

Offline webby1

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Re: Increase the potential energy without any energy
« Reply #83 on: September 26, 2017, 06:06:14 PM »
What is the PE in the first tank?


If you choose to call it 500 units only if you pour it out of the bottom and through your device and onto the floor,, that is fine.


How does using the fall to fill change the PE of the first tank?  it does not.


When you are done the first tank has 1\2 the height and 1\2 the quantity of liquid,, 1\2 times 1\2 is 1\4 and guess how much you have in the other tank,, the same,, and unless math has changed 1\4+1\4 is 1\2,, so you have 1\2 the stored energy in the 2 tanks as you started with.


NOTE, you can USE the water to both fill the other tank AND do something else,, simultaneously,,  :)


It goes to show that just because something CAN do something does not mean it must,, here you CAN fill the other tank WITHOUT using the energy in the flow while filling, not "destroying" energy just throwing it away.

Offline webby1

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Re: Increase the potential energy without any energy
« Reply #84 on: September 26, 2017, 06:17:45 PM »
By the way,, silly thing really.


If you take and make a hole in the first tank 1\2 way up and use a hose from that hole to fill the other tank you do not do anything with the lower half of the first tank.
Now the water has moved down from 1m down to 0m, then from .9m down to .1m,, .8 to .2 and so on,,,,


How much energy is thrown away letting the water free fall into the other tank and how does that change the initial conditions?
This energy is also thrown away in a capacitor to capacitor dump,, which is why you can use something else in the way to use up some of that disposed of energy.


Plot the energy exchange,, you exchange the highest PE with the lowest PE first,,  if you CHOOSE to not use that difference then you choose to be wasteful.

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Re: Increase the potential energy without any energy
« Reply #84 on: September 26, 2017, 06:17:45 PM »
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Offline Low-Q

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Re: Increase the potential energy without any energy
« Reply #85 on: September 26, 2017, 06:36:52 PM »
I just claimed that no energy is lost during the filling of the other tank.
The initial scenario is filling up a 2.nd cap with voltage.
While doing that from a 10V charged cap, the PE of the first cap will decrease. Not only because the voltage drops in that cap, but also because the reference voltage is rising in the other cap that is being charged.
The average PE of the first cap is therefor 1/2 of the PE that could exist if we discharged that 10V cap to zero. But we are not. It is discharged to 5V while the second cap is charged to 5V. The sum of discharged PE and the remaining PE is the same as the fully 10V charged single cap.


I just confirmed that energy is conserved.


Vidar

Offline activ25

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Re: Increase the potential energy without any energy
« Reply #86 on: September 26, 2017, 07:13:06 PM »
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.
 


Offline webby1

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Re: Increase the potential energy without any energy
« Reply #87 on: September 26, 2017, 08:07:10 PM »
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.


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

Note, it is possible to keep
constant all the volumes, because I use spheres:
This is wrong,, you keep the same quantity of spheres.  In my second example I kept the same quantity of cubes but the volume ended up being 81 cubes short of full.  What happens to the length of the angled side legs?  to keep the top at a constant height those legs get shorter,, so does the bottom and top grow longer?

Offline webby1

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Re: Increase the potential energy without any energy
« Reply #88 on: September 26, 2017, 08:26:46 PM »
I just claimed that no energy is lost during the filling of the other tank.
The initial scenario is filling up a 2.nd cap with voltage.
While doing that from a 10V charged cap, the PE of the first cap will decrease. Not only because the voltage drops in that cap, but also because the reference voltage is rising in the other cap that is being charged.
The average PE of the first cap is therefor 1/2 of the PE that could exist if we discharged that 10V cap to zero. But we are not. It is discharged to 5V while the second cap is charged to 5V. The sum of discharged PE and the remaining PE is the same as the fully 10V charged single cap.


I just confirmed that energy is conserved.


Vidar


0.5*CV^2
0.5×(.005×10^2) = 0.25
0.5×(.005×5^2) =  0.0625
0.0625×2 =  0.125


You did not confirm energy is conserved.


This "condition" is well documented to be a loss of energy and as such that loss has been attributed to a loss due to resistance of the connections and dissipated as heat.
When that loss is NOT accepted then one needs to find what part of the energy used to create the initial condition is not being used when transferring the PE.


This "condition" is also well documented when using water,, same thing with the volume to quantity to pressure,, as the volume changes or the quantity that occupies that volume changes the pressure changes,, add energy into or take energy out of the system and these relationships change.


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


In respect to this thread,, what happens to the space between the spheres that is, I assume, occupied by the springs?  When you keep the volume and quantity the same there is no change in PE.


Offline activ25

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Re: Increase the potential energy without any energy
« Reply #89 on: September 26, 2017, 08:30:31 PM »
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.


Free Energy | searching for free energy and discussing free energy

Re: Increase the potential energy without any energy
« Reply #89 on: September 26, 2017, 08:30:31 PM »

 

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