It's well known that charging a capacitor from a voltage source wastes half the energy in the circuit resistance. The experiment can be done by using a charged capacitor to charge another capacitor of same capacitance C. At the start point we have an energy E1=1/2*C*V2. After charging the second capacitor with the first one, we are left with two capacitors charged with a voltage V/2.
The energy is E2= 2 * 1/2*C*(V/2)2 = 1/4*C*V2 = 1/2*E1. Half the energy is lost!
I found the trick to charge C2 from C1 without loss. We can either transfer the C1 charge entirely to C2, or transfer only half of its energy to C2 i.e. the start energy is conserved and equally shared in each capacitor.
First method:
(http://www.overunityresearch.com/index.php?action=dlattach;topic=1689.0;attach=9577)
In this simulation, the coil resistance is 1 ohm (to avoid unphysical conditions) and the initial voltage of C1 is 100v. At t=0 a switch (not represented) closes the circuit. Thanks to the inductance, the voltage source C1 is viewed from C2 as a current source which therefore allows for charging it without loss. It's the same as the first quater period of a resonant LC circuit, discharging for C1, charging for C2 until the voltage reaches its maximum, and then the diode prevents the charge from moving back to C1.
Second method:
(http://www.overunityresearch.com/index.php?action=dlattach;topic=1689.0;attach=9579)
C1 is discharging into the coil through D1 until 0v. A current is stored in L. Then C1 can't provide current because its voltage is nul. But the stored current in the coil doesn't reverse, and it's now the coil that is providing its current to C1 and C2 (back emf) until the minimum negative voltage is reached and the diodes become non conducting.
The final voltage is V/√2 in both capacitors, i.e an energy E2= 2 * 1/2*C*(V/√2)2 = 1/2*C*V2 = E1.
Quote from: exnihiloest on 2013.02.23, 15:24:20
It's well known that charging a capacitor from a voltage source wastes half the energy in the circuit resistance. The experiment can be done by using a charged capacitor to charge another capacitor of same capacitance C. At the start point we have an energy E1=1/2*C*V2. After charging the second capacitor with the first one, we are left with two capacitors charged with a voltage V/2.
The energy is E2= 2 * 1/2*C*(V/2)2 = 1/4*C*V2 = 1/2*E1. Half the energy is lost!
I found the trick to charge C2 from C1 without loss. We can either transfer the C1 charge entirely to C2, or transfer only half of its energy to C2 i.e. the start energy is conserved and equally shared in each capacitor.
First method:
(http://www.overunityresearch.com/index.php?action=dlattach;topic=1689.0;attach=9577)
In this simulation, the coil resistance is 1 ohm (to avoid unphysical conditions) and the initial voltage of C1 is 100v. At t=0 a switch (not represented) closes the circuit. Thanks to the inductance, the voltage source C1 is viewed from C2 as a current source which therefore allows for charging it without loss. It's the same as the first quater period of a resonant LC circuit, discharging for C1, charging for C2 until the voltage reaches its maximum, and then the diode prevents the charge from moving back to C1.
Second method:
(http://www.overunityresearch.com/index.php?action=dlattach;topic=1689.0;attach=9579)
C1 is discharging into the coil through D1 until 0v. A current is stored in L. Then C1 can't provide current because its voltage is nul. But the stored current in the coil doesn't reverse, and it's now the coil that is providing its current to C1 and C2 (back emf) until the minimum negative voltage is reached and the diodes become non conducting.
The final voltage is V/√2 in both capacitors, i.e an energy E2= 2 * 1/2*C*(V/√2)2 = 1/2*C*V2 = E1.
Will this work with larger cap's ?
Say 1000mf with multiple hf pulses
Did this back in December 2008.
34 pages explains it all fairly well I think.
Quote from: poynt99 on 2013.02.23, 17:42:33
Did this back in December 2008.
34 pages explains it all fairly well I think.
Not enough synthetic for me. Here I gave just the principle, I was surprised that it was so simple. I didn't see it in the litterature when I searched informations about the losses in charging circuits. Then I realized that a LC circuit doesn't suffer from a damping of one half the energy at each alternation. And so I came to this idea, consisting in stopping the oscillation as soon as the 1st transfer is made.
Quote from: tinman on 2013.02.23, 15:57:28
Will this work with larger cap's ?
Say 1000mf with multiple hf pulses
It works with any cap. Nevertheless for an efficient use in practice, the time constant must be chosen long enough to limit the current in the circuit and consequently the losses. The lower the inductance, the shorter the charging time but the higher the current.
With big capacitors, you must choose a big inductance. Some mH for some nF, some H for with some µF, and so on.
The time constant of a LC circuit is √LC.
If you estimate that t=0.1s is an acceptable time to charge 1000µF without excessive current, then you must take an inductance L=t
2/C=(10
-1)
2/10
-3=10H (in fact the charging time will be a bit longer, because the time constant is that one when about 63% of the final amplitude is reached).
If you want to charge 1000µF in 10ms, L=(10
-2)
2/10
-3=0.1H but the current will be strong. Question of compromise.
In other words, after sleeping on it:
the resistance must remain negligible in comparison with the impedance of the coil.
Good creative thinking.
Carry it a few steps further and you'll
emulate the high efficiency of the
switching (Buck/Boost) converters.
Quote from: exnihiloest on 2013.02.23, 21:08:37
It works with any cap. Nevertheless for an efficient use in practice, the time constant must be chosen long enough to limit the current in the circuit and consequently the losses. The lower the inductance, the shorter the charging time but the higher the current.
With big capacitors, you must choose a big inductance. Some mH for some nF, some H for with some µF, and so on.
The time constant of a LC circuit is √LC.
If you estimate that t=0.1s is an acceptable time to charge 1000µF without excessive current, then you must take an inductance L=t2/C=(10-1)2/10-3=10H (in fact the charging time will be a bit longer, because the time constant is that one when about 63% of the final amplitude is reached).
If you want to charge 1000µF in 10ms, L=(10-2)2/10-3=0.1H but the current will be strong. Question of compromise.
[/quote
OK this may be a dumb question,but can we use the magnetic field produced by that inductor in a way that dose not remove this effect?
Quote from: exnihiloest on 2013.02.23, 20:41:51
Not enough synthetic for me.
What does that mean?
Interesting, good work. O0
But TOO "synthetic" /theoretical for me so far. ;)
Can you set up an actual circuit and measure the voltages, before and after, then compare Efinal to Einitial? It will be informative to see how close to 100% you get in practice.
Thanks.
Not sure what I would do with those wishes.... C.C
One wants more theoretical depth. Probably because the more theory behind a concept the more real it could be.
The other wants less theory and more reality because the less theory behind a concept the more real it must be.
:D
;D
.99's paper is quite thorough and correct. The only variation you should see by building the circuits will be variations on heat losses. Even those were covered in the paper but you can't simulate physical design or assembly flaws unless you know what they will be.
I did build a couple of the more complex ones to see how results lined up with .99's paper. The variations I had were easily accounted for by considering my construction practices and the actual values of the components.
Quote from: poynt99 on 2013.02.24, 00:24:03
What does that mean?
Not enough summarizing or summing up, too many developments among which it becomes difficult to understand the elementary principles. Now this can be perfectly acceptable, it depends on our respective goals. Mine was just to give the idea with the help of a schematic, not to describe experimental methods.
(Maybe "synthetic" is a false friend of the French "synthétique" that I had in my mind, but it's not my impression after reading a dictionary).
Quote from: PhysicsProf on 2013.02.24, 01:41:15
...
Can you set up an actual circuit and measure the voltages, before and after, then compare Efinal to Einitial? It will be informative to see how close to 100% you get in practice.
...
It would not be informative except about the particular case that I would build, considering my choice of values and my choice of components, as the coil resistance, the capacitor technology, its resistance (yes, they have one, in series and even another one in parallel with high capacitances), the time constant, the diode gap, the diode capacity, its reverse current leakage and so on...
This is conventional engineering whose the results are easily predictable even from spice models. They drastically depends on the setup. Note that this thread is in the folder "Electronic Theory and Learning Center", intended to show generalities and principles that can be used in practice by everybody according to its needs, not to discuss a particular case with specific parameters.
Ok this go's against all we know.
How do you not have a lose of electrical energy when fireing up an inductor?
Why will there be no loss at all through the wires them self?.
The diode has a voltage drop-why no loss there?
If this work's,then where dose the extra energy come from to overcome these losses?
Quote from: tinman on 2013.02.24, 11:37:12
Ok this go's against all we know.
Only against "all
you know", apparently. If you are not aware of a problem known as "the two capacitors paradox", I'm afraid that you should revise your knowledge for relevant discussions. See http://arxiv.org/abs/0910.5279
Quote
How do you not have a lose of electrical energy when fireing up an inductor?
Of course there are losses in practice! When I title "charging a capacitor without loss", I mean "charging a capacitor without the loss of half the energy which is a general problem of principle" as it is clearly indicated at the beginning of my first post.
The question is not about the experimental defaults that induce losses as in any setup whatever it is, but about a theoretical problem.The problem of charging a capacitor is that you put into contact two different potentials, and when the charge is made from a voltage source, for example one capacitor from another one, you face an unphysical situation: the current must be infinite.
Of course it will not be infinite in practice, the components are not ideal, any circuit except superconductors have a resistance, and so the current is limited by the circuit resistance or by the radiation from the circuit carrying a variable current but what we observe is that
half the energy is dissipated whatever the resistance.
This is not a loss due to imperfect components, this loss of half the energy is a dissipation absolutely needed for charging the capacitor this way (http://arxiv.org/abs/1210.4155), even from a thermodynamics viewpoint (http://arxiv.org/abs/1201.3890).
By changing the voltage source by a current source, the problem is solved, the capacitor is charged step by step, adiabatically: a coil is a current source after being "charged" with a current from a voltage source. The model I gave is realizing this function. It was made from an electronics viewpoint. The arXiv references that I give here link it to physics and show that it's a problem of principle.
This is the only point I have developped here, it has nothing to do with the trivial losses due to imperfect components, unavoidable in any setup.
Quote from: exnihiloest on 2013.02.24, 13:25:57
Only against "all you know", apparently. If you are not aware of a problem known as "the two capacitors paradox", I'm afraid that you should revise your knowledge for relevant discussions. See http://arxiv.org/abs/0910.5279
Of course there are losses in practice! When I title "charging a capacitor without loss", I mean "charging a capacitor without the loss of half the energy which is a general problem of principle" as it is clearly indicated at the beginning of my first post. The question is not about the experimental defaults that induce losses as in any setup whatever it is, but about a theoretical problem.
The problem of charging a capacitor is that you put into contact two different potentials, and when the charge is made from a voltage source, for example one capacitor from another one, you face an unphysical situation: the current must be infinite.
Of course it will not be infinite in practice, the components are not ideal, any circuit except superconductors have a resistance, and so the current is limited by the circuit resistance or by the radiation from the circuit carrying a variable current but what we observe is that half the energy is dissipated whatever the resistance.
This is not a loss due to imperfect components, this loss of half the energy is a dissipation absolutely needed for charging the capacitor this way (http://arxiv.org/abs/1210.4155), even from a thermodynamics viewpoint (http://arxiv.org/abs/1201.3890).
By changing the voltage source by a current source, the problem is solved, the capacitor is charged step by step, adiabatically: a coil is a current source after being "charged" with a current from a voltage source. The model I gave is realizing this function. It was made from an electronics viewpoint. The arXiv references that I give here link it to physics and show that it's a problem of principle.
This is the only point I have developped here, it has nothing to do with the trivial losses due to imperfect components, unavoidable in any setup.
My mistake
I thought you had something that may help us here in the real world when you quoted in dark black leters: I found the trick to charge C2 from C1 without loss.
But in reality this can not be done,as there will be losses.
So that statement is incorrect in reality,as there is no way to charge c2 from c1 without loss.
Quote from: exnihiloest on 2013.02.24, 11:20:47
Not enough summarizing or summing up, too many developments among which it becomes difficult to understand the elementary principles.
I thought the paper developed the same idea as yours by starting with the 50% loss problem. The ultimate conclusion in the paper is that you need to use a device which can
store the first capacitor's energy (almost without loss), then
transfer it (almost without loss) to the second capacitor. That
device of course being a large high Q factor inductance.
tinman, if we could make an ideal inductor, then the transfer could be made without loss.
Anyway, none of this is new. Design engineers have known about this "problem", and the "solution" for eons.
Quote from: poynt99 on 2013.02.24, 14:38:02
I thought the paper developed the same idea as yours by starting with the 50% loss problem. The ultimate conclusion in the paper is that you need to use a device which can store the first capacitor's energy (almost without loss), then transfer it (almost without loss) to the second capacitor. That device of course being a large high Q factor inductance.
tinman, if we could make an ideal inductor, then the transfer could be made without loss.
Anyway, none of this is new. Design engineers have known about this "problem", and the "solution" for eons.
Hi Poynt99
I was asumeing this could be done in the real world,which could open the doors to some interesting systems.
Your statement-if we could make an idead inductor-shows that we can not yet do this,unless in a super conductor situation.
Also the diode would have to be ideal,along with a switch that has no arcing at all.
You know me and how i feel about simulation's.
tinman,
Yes the diode would have to be ideal as well, but we can approach ideal with a MOSFET switch in place of the diode.
In terms of the simulation, it predicts very well what happens in the real world in this case.
Quote from: tinman on 2013.02.24, 14:22:49
My mistake
I thought you had something that may help us here in the real world when you quoted in dark black leters: I found the trick to charge C2 from C1 without loss.
But in reality this can not be done,as there will be losses.
So that statement is incorrect in reality,as there is no way to charge c2 from c1 without loss.
I give a method to save half the energy that otherwise is wasted when you charge a capacitor from a voltage source, and you say that this can't "help us here in the real world" under the pretext that much weaker losses remain? C.C It's completely crazy.
Sorry that the real reality doesn't fit your dreamed reality, and that an elementary knowledge of the real reality seems outside of your interest, because I deal only with the real reality.
The reality is that you can charge a capacitor without wasting half the energy, if you do it properly.
The reality is that to do it, you need a current source instead of a voltage source which alas is the usual option.
The reality is that this practical method is especially interesting and can help us when, for example, it is question to loop an OU device.
It explains also many things, like the efficiency of the back emf recharging a capacitor, often more than 90%.
What I provided here is a practical method to avoid real losses not coming from imperfect components. The fact that I try to explain the theory sustaining this method doesn't make it theoretical, it's just a supplementary information for those with a sufficient background in electronics or physics, who want understand and are able to understand instead of just applying a method as unaware robots.
If you don't yet see how this method can help us, moreover when it can be easily implemented according to a schematic which I took time to present here, it's your problem.
Quote from: poynt99 on 2013.02.24, 14:38:02
I thought the paper developed the same idea as yours by starting with the 50% loss problem. The ultimate conclusion in the paper is that you need to use a device which can store the first capacitor's energy (almost without loss), then transfer it (almost without loss) to the second capacitor.
...
I agree. But the relevant point is that the intermediate energy storage must be a current source, not a voltage source. So an intermediate coil will do the job, not a third capacitor. Once we know that, the problem is solved.
Quote from: exnihiloest on 2013.02.24, 16:43:41
I give a method to save half the energy that otherwise is wasted when you charge a capacitor from a voltage source, and you say that this can't "help us here in the real world" under the pretext that much weaker losses remain? C.C It's completely crazy.
Sorry that the real reality doesn't fit your dreamed reality, and that an elementary knowledge of the real reality seems outside of your interest, because I deal only with the real reality.
The reality is that you can charge a capacitor without wasting half the energy, if you do it properly.
The reality is that to do it, you need a current source instead of a voltage source which alas is the usual option.
The reality is that this practical method is especially interesting and can help us when, for example, it is question to loop an OU device.
It explains also many things, like the efficiency of the back emf recharging a capacitor, often more than 90%.
What I provided here is a practical method to avoid real losses not coming from imperfect components. The fact that I try to explain the theory sustaining this method doesn't make it theoretical, it's just a supplementary information for those with a sufficient background in electronics or physics, who want understand and are able to understand instead of just applying a method as unaware robots.
If you don't yet see how this method can help us, moreover when it can be easily implemented according to a schematic which I took time to present here, it's your problem.
So it can be done today-sounds great
Could you build the system and show us this effect?
I remember doing this type of thing some time ago useing a mosfet triggered pulse motor-but as an unaware robot.
Quote from: tinman on 2013.02.24, 22:47:00
...
Could you build the system and show us this effect?
...
Certainly not. Imagine the difficulties! Two capacitors, a coil and a diode, it is as difficult as showing that U=RI, well beyond my lab ;D.
I have done this. It is possible to transfer energy without loss. For example, 1/2CV^2. If you have 10V 1mF cap, you can actually convert it into .32V 1F cap.
Now I have read Faraday law of electrolysis. It basically say that the amount of hydrogen produce is proportional to the electrical quantity Q(charge). Does this means .32V 1F cap would produce more hydrogen than 10V 1mF cap?
Ex
Quote:
Certainly not. Imagine the difficulties! Two capacitors, a coil and a diode, it is as difficult as showing that U=RI, well beyond my lab .
--------------------
That's OK Exnihiloest
Don't feel to bad we all have our limitations.
O0
Some folks just have a problem building things..........
Thx
Chet
I went through this a while back.
If we had 2 air tanks, say 1 full, 1 empty, connected together via hose to each tanks valve, valves closed, then we open the valves till the pressure levels out in the 2 tanks from the full one.
Now, if we were to measure the amount of energy that 1 full tank could provide to say an air driven motor from full till empty, and compared that to measuring the output of the air motor running on the 2 half full tanks, measuring the total for 1 tank and then the other and adding, do we get the same amount of energy from our air motor using the full tank till empty, and from the total of the 2 half empty tanks? ;)
Did we lose half of the energy stored in the full tank, when we connected the empty tank to the full tank and leveled their pressures?
And when we level out the 2 tanks, is there a loss in the system because of heat produced by the transfer in the system? ;)
That heat was lost back when the full tank was filled. ;) All we did was release pressure into a container that is twice the volume of the original pressured chamber. In fact, the tanks will reduce in temperature compared the when 1 was full and 1 was empty. After leveling, once the tanks sit half full for a while, they will come up to room temperature by absorbing heat from the environment. So we produced heat at one time in the system and lost it, and we took it back when the air pressure was released and expanded. Weird thought.
But we didnt lose half of the energy in releasing the full tank into the empty tank because of heat loss when we opened the valve. We lost it because we released a level of air pressure into a larger container(or caps ;) ) without 'using' the energy transfer to power something till the tanks are equal levels. To me it adds up to a form of waste that doesnt transfer to another form. The pressure left in the 2 half full tanks is still pressure, we just chose a stupid way to dilute it. ;)
So, are we really losing half the energy in the cap transfer, 'HALF' due to heat losses, because of resistance? Or are we just carelessly wasting energy by connecting a full cap to an empty cap, of the same values, till they have half the original voltage(pressure), by unleashing 'pressures' into larger containers, without using the energy in the transfer from one cap the the next.
If we had 2 large caps, 1 full and the other empty, and we put a light bulb in line, we would get light and heat, and the caps will still be half the original voltage, but we made use of the transfer. Or a motor. Or a JT. Lol, wouldnt you think that if we lost half in the cap to cap transfer, and we make use of all that energy transfer, and still have half the voltage left in each cap, that the loss we are talking about are not even real. If we add a light bulb to the system during transfer, which is a 'resistance', and producing a lot more heat compared to a very low resistance transfer, and we get light to boot, but we still get half the voltage in each cap when we are done!!! :D Lol, we used a light bulb, a resistor, to obtain more energy output from the system to make use of an inherent loss in diluting a pressure.
Now, it is said, that if it were all supper conducting, that the caps will be more than half of the original voltage. But I dont know.
Like in a sim, if you dont add a resistance value to the caps or circuit, would we still get only 5v in each cap, from 1 with 10v? Or would it be more in each? ;) I would like to see it. ;)
Mags
Mags what can I say, you have said it all O0
Mike 8)
Quote from: Magluvin on 2013.02.26, 07:14:01
I went through this a while back.
If we had 2 air tanks, say 1 full, 1 empty, connected together via hose to each tanks valve, valves closed, then we open the valves till the pressure levels out in the 2 tanks from the full one.
Now, if we were to measure the amount of energy that 1 full tank could provide to say an air driven motor from full till empty, and compared that to measuring the output of the air motor running on the 2 half full tanks, measuring the total for 1 tank and then the other and adding, do we get the same amount of energy from our air motor using the full tank till empty, and from the total of the 2 half empty tanks? ;)
Did we lose half of the energy stored in the full tank, when we connected the empty tank to the full tank and leveled their pressures?
And when we level out the 2 tanks, is there a loss in the system because of heat produced by the transfer in the system? ;)
The parallel between the two capacitors paradox and the two tanks problem is interesting and there are real similarities nevertheless we are not exactly in the same conditions.
Firstly you must give the way according which you fill one tank with the other: is it adiabatic or not? You may have noticed that I used the term "adiabatically" to qualify the charge of the capacitor from a current source. This meant that we "filled" the capacitor step by step, electron by electron, without creating a wide potential difference. In this case we avoid heat and entropy increase, and this is the interest of the above mentioned method. For pressure, it's the same. If your filling is not adiabatic, as the charge of a capacitor from a voltage source like from another capacitor, you will produce heat and so, loss energy.
The second point is that you suppose that your experiment is made in an environment which is neither at the absolute zero nor at a zero pressure. You must firstly obtain the pressure difference with a work exerted from a start point including the ambient temperature and pressure. The initial energy is this work. At the end of your experiment, the two tanks would be at the same pressure, but the possibility to extract energy from this state depends on the difference from the ambient pressure unlike the case of the capacitors that can be fully discharged in any case whatever the ambient potential because they are dipoles while your tank is like a monopole.
Quote
...
If we had 2 large caps, 1 full and the other empty, and we put a light bulb in line, we would get light and heat, and the caps will still be half the original voltage, but we made use of the transfer. Or a motor. Or a JT.
That's correct. When we charge a capacitor from a voltage source we could use half the energy that otherwise would be wasted. It's an other option than the goal of the proposed method, which is to move without loss the energy from a capacitor to another one.
Quote
...
Now, it is said, that if it were all supper conducting, that the caps will be more than half of the original voltage. But I dont know.
The caps wouldn't be more than half of the original voltage (see the arXiv papers, especially this one involving thermodynamics).
If you led the experiment with surperconductors, there would be an infinite current, incompatible with an experimental setup. So what happens in reality?
In fact in this case, if the conductors don't fuse and the strong magnetic field doesn't collapse the superconducting state, then you have to take into account the inductance of the circuit, because it becomes no more negligible in comparison with the circuit resistance. In the real life, all circuits have an inductance. The time constant due to the inductance will limit the current, working exactly as the circuit with the coil, the coil being replace by the natural inductance of the circuit.
Quote from: GibbsHelmholtz on 2013.02.25, 23:16:11
I have done this. It is possible to transfer energy without loss. For example, 1/2CV^2. If you have 10V 1mF cap, you can actually convert it into .32V 1F cap.
Now I have read Faraday law of electrolysis. It basically say that the amount of hydrogen produce is proportional to the electrical quantity Q(charge). Does this means .32V 1F cap would produce more hydrogen than 10V 1mF cap?
Ok, so you started off with 50m joules of energy-and you ended up with 51.2m joules of energy.
And this was transfering from one cap to another?
So not only didnt you have any loss-you had a gain in energy at the end?
Was the system a closed loop,or was there an out side energy source aswell?
May we have the schematic to the circuit that achieved this?
hi Tinman,
You can see the extra 1.2 mJ of energy as experimental error or uncertainty. :)
Quote from: tinman on 2013.02.26, 10:36:58
Ok, so you started off with 50m joules of energy-and you ended up with 51.2m joules of energy.
And this was transfering from one cap to another?
So not only didnt you have any loss-you had a gain in energy at the end?
Was the system a closed loop,or was there an out side energy source aswell?
May we have the schematic to the circuit that achieved this?
Looks to me like there is not enough precision in the answer of 0.32V, it was probably rounded up error, should have been 0.3162V.
Small errors become large errors when the square function is invoked. Why precise measurement is required.
A low cost DMM will not give the degree of precision nor accuracy required.
I've just made a setup to test Faraday electrolysis.
First test:
- A 4700uF is charged to about 30 volts and discharge into water. Gas level is observed.
Second test:
- Four 4700uF in series is charged to 30 volts each (30x4= 120V) and discharge into water. Same gas level observed.
In conclusion, Faraday law of electrolysis is correct that gas is proportional to the charge Q. It is also interesting to note that even the energy ratio is 1:4, the gas ratio is still 1:1. According to this, hydrogen can be generated in unlimited quantity.
Quote from: Centraflow on 2013.02.26, 08:19:22
Mags what can I say, you have said it all O0
Mike 8)
Thanks. I say things that I regret later when it comes to these things, but I stick to this one like glue for some time now.
But what does it mean, if it means anything? If Im right, did 'they' get it wrong? Or is it that things are not what they seem on purpose? ;)
Today I read an article about how the milk industry wants to be able to use aspartame and such in milk products, WITH OUT HAVING IT ON THE LABEL!! Why? Does milk and butter need sugar? Let alone an artificial sweetener? Does gmo milk not taste so good without an added sweetener? The Yogurt companies surly want it omitted from their labels, as the people flock to more brands that just use sugar.
Is aspartame cheaper than sugar? Is it safer? If its safe, then why all that magic acts to hide it? Why use it at all? Subsidies?
If this is what is happening to milk, then why is it so hard to believe that electronics theory may have a few secrets that 'they' dont want everyone to know about? ;)
Dats what me thinks. ;D
Mags
Quote from: Magluvin
Today I read an article about how the milk industry wants to be able to use aspartame and such in milk products, WITH OUT HAVING IT ON THE LABEL!! Why? Does milk and butter need sugar? Let alone an artificial sweetener? Does gmo milk not taste so good without an added sweetener? The Yogurt companies surly want it omitted from their labels, as the people flock to more brands that just use sugar.
Is aspartame cheaper than sugar? Is it safer? If its safe, then why all that magic acts to hide it? Why use it at all? Subsidies?
It could be all of the above, except the "safer"
possibility. Milk already contains proprietary
(non-revealed) ingredients which are needed to
extend its shelf life, to keep it from souring
too early.
I suspect the aspartame would tend to mask the
souring in its early stages. Milk loses its sweetness
as it begins to sour - before it begins to smell.
One thing we know for a certainty - it has nothing
to do with benefiting the health of the consumers.
Quote from: Magluvin
If this is what is happening to milk, then why is it so hard to believe that electronics theory may have a few secrets that 'they' dont want everyone to know about? ;)
Dats what me thinks. ;D
Mags
Aye, there are quite a few "secrets" having to do
with electronics and electronic devices which "they'd"
rather keep hidden from the people.
Additional experiment shown that hydrogen production is independent of the circuit resistance. Adding resistance to the circuit only slow down the discharge rate and not limit the amount of charge passing through water. This is in line with hydrogen production is independent of energy. I would say that it depends on the amount of charge passing through the water. If resistance is what taking away the circuit's energy, then hydrogen is the product of reactive power.
Couple more test:
- When four 4700uF in parallel (30V) and discharge to water, the gas level is much higher than in series (120). Again, the same energy level in both cases but different amount of gas.
- Additional water cell in series seems to double the production. I would say the more cells in series, the more gas produced for the same amount of discharge.
Next step is to find out how much hydrogen needed to set unity.
Some numbers scavenged from the net.
240 kJ/mol of hydrogen
Faraday constant = 96000 C/mol
There are two grams in a mole of hydrogen so we probably need 2 moles of electrons. That equates to about 192,000 Coulombs. That would gives 1.25 Joules/Coulombs.
1.25 = .5V
V = 2.5 volts
Double check:
If we have 1 Farad 2.5V, energy = 3.125 J
Q = 2.5, E = 2.5 x 1.25 = 3.125 J
If we have 5F 2.5V, energy = 15.625 J
Q = 12.5, E = 12.5 x 1.25 = 15.625 J
So there seems to be a threshold to each cell's voltage.