Ok  I havnt done these calculations since mid 80s.  Im as sad story. lol
But my mind is open. 

Found a calculator on this site
http://hyperphysics.phy-astr.gsu.edu/hbase/electric/capeng.html

Im getting proper results there.

But in my example, why the loss? Is there a reverse trick?   I just love this stuff, and its killin me.  lol  Na, Im enjoying it.  I will be back in an hour to find out.
Laundromat time.  I forgot it was laundromat day.

Mags

Hmmm.  I was reading the info below the calculator on the site.

From the definition of voltage  as the energy per unit charge, one might expect that the energy stored on this ideal capacitor would be just QV. That is, all the work done on the charge in moving it from one plate to the other would appear as energy stored. But in fact, the expression above shows that just half of that work appears as energy stored in the capacitor. For a finite resistance, one can show that half of the energy supplied by the battery for the charging of the capacitor is dissipated as heat in the resistor, regardless of the size of the resistor

Im confused.  I have to really think about what is going on. This describes what is happening in that we really are getting 50% eff in exchange in woopys demo.
I find it hard to believe that we lose half due to heat on any value resistor. Well what of a super conductor? I should suppose using caps is a losing game all together?  From what this explains, when we charge a cap, we use 2 times as much energy from the source to charge the cap to the source voltage level.   

Tesla used copper ribbon or strip conductors as he said the heat was dissipated easier to keep the resistance low.  Well why worry if the heat does build up if we lose 50% no matter the resistance value?

Are we losing 50% in the Believe Circuit? When we open the switch when the cap is at 5v and the cap continues to 6.32v, did it cost twice as much to get that 5v in the cap, than what that 5v in that cap is worth?
Ok, Im calling in sick tomorrow and going for my electroshock treatment.

This can put a big dent in many things.
Something is fishy. Is this the loss that you were speaking of Loner? did you know it was a 50% loss?

BZZZZZAP    Still dont feel any better about it.  Hit me again tube sock!

magzzzzapp

Well, just did a sim with a 5v source, charged a cap through a 100ohm resistor. Then I discharged the cap into a 100 ohm resistor.
I got the same amount of power in both resistors whether charging the cap or discharging.   hmmmm

Something is fishy.  fishy fishy    The smell is bad

Mags

Quote from: Magluvin on March 30, 2011, 09:04:10 PM
Hey all

Woopy. You are doing great things.     I have to check the calculations. I think there is something we are missing in figuring this out.

I had the same thoughts as you when I just ran it thorough my head. If we have 2 10uf caps and charge one up to 20v, and we short them as you did, each cap will contain 10v, as you have wonderfully shown.
Then I thought, well what if we put the 2 caps in series now to get our 20v back. BAHH  or like you groaned in the vid, MMMMMMMM!      
I now have 20v, but in a 5uf cap!!!!   What??

BAGHHH!!  cant be!   

Then I climbed out of hell, where I thought I was for a moment and regained level ground.

This is a very important discovery, as you have experienced it and I had thought it. I like this. We work well together and communication is good.

This will be a VERY touchy subject, VERY.   Not right now. ;]

So back on earth,  we have 2 10uf caps with 5v in them.  At first we only had 10v in 47uf.  But now we have 5v in 94uf.  The difference is, we cant run a 10v device from the 5v in the caps, but we can run a 5v device on the 94uf at 5v, for just as long as a 10v device on a 47uf cap at 10v.  Get it? All being that the devices consume the same power levels. i can full describe it if you wish, but I think its good so far. So we are just not calculating nor thinking it properly the way we are.

But that loss that you calculated and I thought, well where did the power go in our thoughts and calculations?  I think its still there, we just shouldnt go about dealing with trying to get the 10v ability back after conversion to split the voltage.  You and I both know that if we use some energy into trying to put the 2 5v caps all back into one, there would be considerable LOSS =] in the work it took to do so.

But now for the super nasty thoughts. lol   

If we have 20v in a 10uf cap and another 10uf that is empty, then we connect them to get 10v in each 10uf cap.  If we put the caps in series and we only have 20v in a 5uf cap, we DID lose, a lot. We could never get the 20v 10uf amount of energy from a 20v 5uf cap. That is nuts woop. How dose that kind of loss happen?
What if it is a trick that is a real trick, the effect of energy actually disappeared in thin air and we were left with only 2.5 level in our water jugs instead of 5?
Here is the nasty good part, I hope. If we can make that energy just disappear, gone, nada, neva commin back, cant have it, well maybe just maybe we can double it by another way. It only seems logical that if the energy could just vanish into thin air, that there could be a way to increase it, out of thin air.  Lol, that is just mind boggling.  If we had never seen this issue, maybe we would never think of a possible way to accomplish the same trick but in reverse.

I may be very wrong here an we may have suffered a HUGE loss. But freekin how?  I know you are felling this right now woops. I feel ya.

We have to investigate this. At least its not boring. 

Like in your vid, you said very low resistance to connect the caps, so little loss, try this woops, do the same thing but with a resistor, and check to see if we lost anything. If the resistor is high value, it will take some time for the caps to level out. Tell me how much we lost.  hehe =]  If we lost anything, then heat is a loss. If you end up with 5v in each cap, what can we say about heat? Was it free?  I havnt tried this as Im really being spontaneous with the thoughts here and you have the test setup with all the same parts as before.

If the caps have less than 5v after then I submit that the heat was energy taken and converted into heat . But what if those caps level out to 5v woops? Do we have something here or what?   

I have said that I wasnt completely sure about heat not being a loss of energy from the source, but if you would, entertain me.
I will still maintain the flywheel as free till we discover otherwise.

Imagine a battery and we have a resistor that is getting hot, emitting heat, do less electrons make it to the positive side of the batt than what came out of the negative? Or are we just discharging the battery and heat is a free byproduct?  Remember, the hotter it gets, the less we pull from the battery. Would we not think that the hotter, the more current? =]
I dunno. Maybe Im just nuts and should be sent out for an afternoon of electroshock treatments. 

I just may be eating both of my feet tomorrow for lunch. I will clean them well in the morning, and bring some A1 sauce.  ;]

But Im ok with it.  I just cant see why we lost 50% woops. Something is up.

If you could do the 2 caps and resistor test, we will at least nailed down something that is on the list. I just cant imagine a 50% loss in heat here. Where did it go. We must be calculating something wrong.  Will think on it.

Well, this is more fun than titos puzzles.  No offense Teets.  We just seem to have plain objectives to work with and interesting results.

Mags

I'm not an EE or a mathematician. But either the math or the theorem has to be wrong. I would tend to think that a simple experiment could be devised to test. I thought about using a low voltage circuit like a 1.5v quartz movement and see how long it runs but it is some what voltage dependent. How bout using a resistor as a load submerged in a calorimeter to test series V. parallel cap charge?  This seems to me as a definitive, quantitative experiment. 

hey Tek

Either Im an idiot and still cant see it yet like a fool, or what Im seeing and reading today has just put me into the 4th dimension of twilight zone.

Something is fishy.  This 50% eff is just nuts.  How do we get anywhere near cop with this in mind.

From the explanation, it takes work to charge the cap.  I always thought that as long as the caps voltage level is lower than the source, that it would gladly accept additional charge to equal the source. Where is the work in that. The source sees an empty or even partially empty cup. I would think the leveling out is a release of unbalanced tensions. The source has to do 2 times the work to get the 1 thing done from what they say.  What da heck. I do not buy that.  We better have cop8 or we aint gunna do nuthin but drain our batteries using caps..
Im exaggerating a bit. But im angry.  Again  Fishy fishy.

Anyone remember the Barney Miller episode where the asian guy ate the marijuana brownies by mistake?  lol  Remember what he kept saying?  Mushy Mushy.    That was his way of saying fishy fishy. =]

Woopy,  this is very awkward, no?  How can these things make no sense, unless there is something wrong with what we know and understand.  I have never run into something like this in all my experience. 

We have gone from having what we might think of as 130%eff, to "we dont really know nuthin at all" all in one day? We found that capacitors are just beyond our comprehension.
Have you ever heard before that it is a 50%eff transfer from batt to cap?


Fishy fishy fishy freekin fishy




Mags marina del fishy