Quote from: poynt99 on April 05, 2011, 08:55:09 PM
Magluvin,
You should know as it appears you are unaware, that not only are you off-topic, but the topic of capacitor energy transfer and the so-called "discovery" has been covered a number of times elsewhere in this forum. What you really need before you get that drink, is to obtain a clear understanding of capacitors, capacitor voltage, and the energy stored in them. THEN you will know why and how you are able to charge up these capacitors to higher voltages.
Here are a few links to get you started in your quest:
http://www.overunity.com/index.php?topic=8334.msg210138#msg210138
http://www.overunity.com/index.php?topic=8334.msg210142#msg210142 (read the attached file carefully)
http://www.overunity.com/index.php?topic=6090.msg143812#msg143812
http://www.overunity.com/index.php?topic=4419.msg148098#msg148098
http://hyperphysics.phy-astr.gsu.edu/hbase/electric/capeng.html (play with the values and see the effects)
So you may see that you are not the first to have fallen into the trap involving capacitor energy transfer, and likely you won't be the last.
.99
QuoteHey point
Actually Rose had asked me to post results here, and there are some others paying close attn also. =]
It appears you havnt been paying much attn. ;) But how could I convince you otherwise, you wont build it.
If you keep believing the formula are correct, then you are doomed.
Let me ask you something.
If you have a 10uf cap at 10v
How much energy does that cap hold? An ideal cap for simplicity.
Now connect that cap to another empty 10uf cap so that each has 5v.
Now you have 2 10uf caps at 5v = 20uf cap at 5v.
How much energy is in a 20uf cap at 5v?
If you have calculated anything less than 100% you are wrong. They are wrong. He is out of order, your all out of order!! lol
Seriously. I know ya think Im a freak. I dig it. Im Magfreakinluvin.
Now tell me how much energy is in the 10uf 10v vs 20uf 5v.
After you tell me 50% loss, I will spit on my shoe and shine it with pride, that I know what is what, and you do not.
Run some tests, spice or on the bench, tell me what you see in these questions I propose to you Point. If you dont you will never know. ;]
Run the 10uf 10v cap through a 100ohm resistor
Then the 20uf 5v cap through a 50ohm resistor
Each will start out at 1A and descending till drained. Tell me how long it took for each instance Point.
I dont know you Point. But I am just showing you the same respect you had just shown me, and I may have learned ya sumthin. ;]
Buld it, then you can tell me otherwise. Does anyone agree? Anyone?
Be cordial and so will I. We can be friends and learn new things.
It wont take you long to become a Believer when you try. I have done all the work to get you to the objective. Convert a 10uf at 10v to a 10uf cap at 13v, energy used to do the switching has not been taken from the circuit to keep the ideal function pure, so we know exactly what is in and out through the process. It is done this way all the time.
Point, if you know about what Ive been doing here the last couple weeks, Ive gone through several pains of saying, na, cant be, its all wrong. Then I climb out of the box for some fresh air, and it is all too clear. Crystal. ;) You may have to empty your cup for this one. ;]
Its easy. No complex waveforms to debate, no disrespect Rose ;], no measurements that could be denied, just what was in vs what was out. The text books will have you climbing the walls Point as you see what is going on in this circuit. Once you see that we can take a 10uf cap at 10v and end up with a 10uf cap at 13-14v, will you still question me? For what? I told ya that was the objective. Where will I have been wrong then, in any way?
We can be cool about this. Its up to you. ;]
Mags
OK, so 10uF @ 10V into 100 Ohms, and 20uF @ 5V into 50 Ohms;
both have the same tau (RC) and will "drain" in the same amount of time. So what?
.99
ok Well when we run the 10uf 1000v lets say,cap through the believe circuit, when while the cap is draining, we disconnect the cap from the circuit when it reaches down to 700v. The believe circuit continues to pump the receiver cap to 700v. Now we have a 20uf cap at 700v!
Do you see the gain? If we were to take the 10uf 1000v cap and just directly charged another 10uf cap and hold, we end up with 2 10uf caps at 500v each. 20uf at 500v vs believe circuit, 20uf at 700v!!!
Is there a gain? Would you concur that it would be about 130%eff in transfer of energy? Think! its hard to fathom that it can be real. But it is. ;]
Does this not compute or what? ;]
Woopy and I have had a hard time with calculations as they did not figure what we have, on the sim, and an the bench.
Woopy has gone off the edge the last day or so. He is being strained between what he has on the bench and what the books say.
Im over it now that I have seen the ability to go from cap to same value cap and reach a higher voltage in the end. An only the only energy that went into the circuit was a single 10uf cap charged to 1000v, goes through 4 stages(new tests show that is my goal for 2kv out on a 10 uf cap) no other input, and end up with 2kv on a 10 uf cap. Im working on the switching for the end game, for putting the output to the input. Switches. ;]
Mags
Quote from: poynt99 on April 05, 2011, 10:37:54 PM
OK, so 10uF @ 10V into 100 Ohms, and 20uF @ 5V into 50 Ohms;
both have the same tau (RC) and will "drain" in the same amount of time. So what?
.99
Well, a 10uf at whatever voltage, with a resistor that initially pulls 1Amp, vs a 20uf at whatever voltage, with a resistor that initially pulls 1Amp, you say they will be drained at equal time periods? Hmmm, show me.
Mags
Quote from: Magluvin on April 05, 2011, 11:07:51 PM
ok Well when we run the 10uf 1000v lets say,cap through the believe circuit, when while the cap is draining, we disconnect the cap from the circuit when it reaches down to 700v. The believe circuit continues to pump the receiver cap to 700v. Now we have a 20uf cap at 700v!
Do you see the gain? If we were to take the 10uf 1000v cap and just directly charged another 10uf cap and hold, we end up with 2 10uf caps at 500v each. 20uf at 500v vs believe circuit, 20uf at 700v!!!
Is there a gain? Would you concur that it would be about 130%eff in transfer of energy? Think! its hard to fathom that it can be real. But it is. ;]
Does this not compute or what? ;]
Mags
I do not see any gain. Gain in
what exactly? Please be specific.
.99
Quote from: Magluvin on April 05, 2011, 11:13:04 PM
Well, a 10uf at whatever voltage, with a resistor that initially pulls 1Amp, vs a 20uf at whatever voltage, with a resistor that initially pulls 1Amp, you say they will be drained at equal time periods? Hmmm, show me.
Mags
Here was your scenario again:
QuoteOK, so 10uF @ 10V into 100 Ohms, and 20uF @ 5V into 50 Ohms;
First of all, in this case the initial current in each is 0.1A, not 1A. But that does not matter for this question.
Second, what exactly do you want me to show you? The time constant (tau) for both cases is 1ms, i.e. R x C = 1ms.
.99
Quote from: poynt99 on April 05, 2011, 11:34:17 PM
I do not see any gain. Gain in what exactly? Please be specific.
.99
Well, if we agree that 10uf at 10v has the same energy as a 20uf at 5v, then 20uf at 700v is? ;)
mags
After a period of 5Ï,, (5*tau), each capacitor will be discharged to about 1% of its initial voltage.
So after 5ms, the 10uF will be at about 0.1V, and the 20uF will be at about 0.05V.
.99
Quote from: Magluvin on April 05, 2011, 11:48:34 PM
Well, if we agree that 10uf at 10v has the same energy as a 20uf at 5v, then 20uf at 700v is? ;)
mags
That did not answer my question. You asked another question. Here it is again:
QuoteGain in what exactly? Please be specific.
.99
Hmm maybe im not the best at explaining what I mean. hmm
Let me ask you this, if we used a 10000uf cap at 100v as a source that we had charged and disconnected the charger.
So at the input of our circuit we have this cap 10000uf 100v as the only input to our circuit, from start to finish. Get it?
Would you agree that if when we run my circuit, using this cap as the source, and when all is said and done, we end up with a cap that is 10000uf at 130v, would I have a gain?
I have found that it is most peoples goal is to replace the battery with a cap, mostly to relieve any doubts about batteries, and hey, chuck the battery if not needed. ;]
And when we know that a cap at a particular voltage contains a for certain, amount of energy, an amount we can rely on each time.
So I am presenting a circuit that can do just that.
If the 4th stage gives me 2000+volts, we have a self runner with 100% output to boot. Would this not interest you?
If we were to forget the caps, or the circuit, if we get out 2 times what we put in, is that not a good thing? Lets say my capacitor ideas and theories just plain suck, but I get out more than in, would that interest you? Would ya push me aside and take over the project(if real in your mind ;]) and not think that I may be of value still?
I didnt come up with this by accident. I studied. I absorbed. I did it.
Im an odd thinker. Im weird, just like everyone else. Im a kook. lol
But the better question you should be asking is how am I doing it.
its all on the thread next door. 5 pages of happiness and sadness. Really. But we got it. Got it in 2 weeks.
While your wanting to compete for king capacitance here, we gunna be ou'ing next door this week.
Join us. I know you have a good mind. Just try to see if you can make sense of it. I really dont want any arguements and fuss. I had seen what some of the guys at our were crackin about me.
They dont even know. But maybe you will. Im sharing, and Its all original to me. And it works.
It seems strange that we gain while voltages go down in each stage, but as I have shown you, im not just spliting the one cap into 2 equal halves, we are above average each time 700v vs 500v, and collecting interest compound. Once all is said and done, we have an 80uf cap at 350v, from a 10u at 1000v. And I can make that 80uf cap produce a 10uf cap at 1320v in the end game circuit. Still in design. Just switches.
With the 4th stage we should be at 10uf 2000v, then we dump half of that back into the believe circuit and we get as much for free.
It sounds like fantasy. Well put on your 3d goggles and see.
you will forget this thread quickly. ;]
mags
I'm pretty sure I understand what you are saying mags, and you are doing a good job explaining yourself. I also am not the best at understanding what folks write. But that aside...
I must slightly correct myself from before; I do see a gain by using your circuit, but it is not the type of gain I think you are assuming you have.
I would still appreciate an answer to my simple question though if you wouldn't mind (in ten words or less):
Gain in what exactly? Please be specific.
.99
I have read your thread actually, and as I said, I believe I understand where you guys are coming from.
What type of test using a simulation would you like to see to prove your theory? You seem to have faith in SPICE simulation and so I am asking and offering to do a sim.
I have one that I can suggest:
Let's use the two pairs of RC components you mentioned before, i.e. a 10uF cap at 10V, with a 100 Ohm load, and a 20uF cap at 5V with a 50 Ohm load. From this, we can use SPICE to show us the total energy that will be discharged from the capacitors and dissipated in the resistor in each pair.
What do you think?
.99
thanks for understanding , even if it is not fully understood. ;]
The question. The gain is that in the end, we have a same uf value cap as we started with, but more voltage than we started with..
I just came up with how to do the end game switching that made things very easy as compared to what i had thought earlier. And i made a mistake in my 3rd stage calc that worked out in our favor. ;]
Im tired Point. Glad you see where im going with it. And if you have any easier solutions, I welcome all. It can only get better with a collective of good minds working together.
See ya tomorrow point. I dont think you will be disaPOINTed ;]
thanks
mags
Quote from: Magluvin on April 06, 2011, 01:55:41 AM
The question. The gain is that in the end, we have a same uf value cap as we started with, but more voltage than we started with..
Yes, if you can end up with a higher voltage on a same value capacitor, you would have a gain. Aside from the obvious voltage increase, what would the element of gain be exactly? (I'm still trying to get an answer to my original question).
Have you already obtained the result of a higher voltage on the same value capacitor? Where?
Quote
Im tired Point. Glad you see where im going with it. And if you have any easier solutions, I welcome all. It can only get better with a collective of good minds working together.
I have already offered what I think is a quick and solid solution to checking your theory. Would you like me to perform that simulation I mentioned, but in this case use your present values of 10uF at 1000V, and 80uF at 350V? What loads would you like to see on these two capacitors?
Since you are generating much of this data using a simulator, I am assuming you would be content to see what a SPICE simulator shows in terms of the energy content of each of the aforementioned capacitors.
.99