Ha ha ha, I would love to see a physics whiz try to work the maths out on this one, it will take a month of Sundays. Scoff all you want.
Ok,
For the past 4 days (96 hours solid). I have had three lead acid car batteries hooked together in a mix of parallel and series connections in a closed circuit, using the same resistance principle I referred to in the previous topic. Guess what. One of the batteries is charging up and there is no loss or degradation in the other 2 batteries. You REALLY have to think outside the square for this one. It's really happening lol.
I'll keep updating as time goes on.
Hi dandman, could you post a wiring diagram or is this top secret because you believe fake money makes the world go around, thanks.
peace love light
tyson :)
Hi Skywatcher, I still don't believe it myself yet but as soon as I work out what is going on I'll be posting one. btw, you got something against making a buck, grin
Hey Dand
Are you using inductors? Just wondering considering the losses involved in direct discharge/charge.
Mags
Hi Mags, nope, no inductors, just resistors, something I have just twigged to in this set up is that two of the batteries appear to be swapping values of charge state, each swing in state takes them in turn to higher state of charge, wierd ey. I will keep posting as things become clearer Cheers
Again, you must not have understood what I said. I will try a different way
You may show the same voltage but you will not show the same amperage. You must know how many amps are in the charged battery. Once you know that, then charge the other batterys with that battery and find out how many amps they have. I bet you will loose amps. You can not go on volts alone, volts mean nothing if there are no amps. They all may show the same volts and even differ slightly but the amps are what is need to know to know if you have a gain or a loss.
Dear Nightlife, I get it, I really do. THE WATT IS KING. Voltage to me is only a very unreliable indication of watts or amps as everyone likes to call them, available. I load tested the batteries with a 30amp draw for 15 seconds before, the battery dropped to 10.5v and returned to its original voltage of 12.64. I repeated this after 2 discharges and recharges. Same result. I presume we are talking about my previous post here?.
This post I havent quite got my head around at all yet. Currently waiting for it to run its full cycle before I do any load testing. So far its been running for just over 4 days. Thanks for the input. I am not averse to learning and I do want ppl to pull holes out of these things as I am averse to looking like a prat. grin
Lets try this again. You must know how many amps. I have yet to read how many amps you started with and how many you ended with. Once we know the amps, we then can find out how many watts becuase you have already gave us the volts. Volts X amps = watts.
Ok, I must be missing something here. Please be patient.
1 amp at 12v = 12w. 1 amp hr at 12v is the constant supply of 12w for an hour. If I start at voltage x and discharge to voltage y for 1 hour at 1 amp. I have discharged 1 amp hour. If I then return the state of charge back to voltage x and discharge to voltage y again for 1 hour at 1 amp. I only us the voltage as a reference point. Is it not true that I must have replaced the 1 amp hour of charge I used back in the battery. Or is there another reading or calculation you would have me do, if so, what?. Thanks
I cant say because we dont have the starting amps and the ending amps. The car battery you have problably has 700 to 1000 amps at 12 volts. You charge another battery with it, you mostlikely will still see the same voltage but with lower amps. The battery that was charged may also show the same voltage but it too may have lower amps. You have to know what the watts are in each battery before and then add them up and then find out the watts in each battery after and then add them up. If the total of the start is lower then the total after, you have a gain, if not, you have a loss. We can't tell you based on the lack of information you have gave us. We have to know how many amps are in them at the start and how many after before we can say wether or not you have a gain or a loss.
Now I know what Im missing. how do I calculate the current amp holding of a battery. The only reference I have found to battery amperage holding is new at full charge divide the cca rating by 6.5 and that will give the as close you can get to the total ah available to dead.
I'm not sure. I have an automtive battery tester I use but it is not as accurate as I would like it to be.
Okaaay, mate it's been very informative. thanks for the interaction.
DOES ANYONE OUT THERE KNOW HOW TO CALCULATE THE CURRENT AMP HOUR HOLDING OF BATTERY AT ANY GIVEN POINT PLEASE?
pls give details / data of you battery !
is this an car battery ??
example 84Ah 12volt = about 1 Kw/h = 1 Kilowatt-hour
or
an array of AAA NiCad batteries at 1 Ah ???
if you unload your battery with
30 AMP ! , for only 15 second
so ist this (to see the capacity (energy) per hour
ONLY 0.125 Ampere/hours = 1/8 Ah.
SO ... IT is nothing unload -in relation- to an strong battery !
So the battery will return to his (near) fully voltage after unloading
Pese
Hi Pese, these are lead acid car batteries. the load test is deliberate to determine the condition of the battery. If the battery is charged above 12.64v hitting it at 5 percent of its cca rating in ah for 15 seconds at 80 degrees f. If in ok condition it should return to its original voltage if not it is deteriorating.
IG this unloading (for 15 seconds) = 0,125 W/h are only an small part from the energy of ean BIG car lead battery (with 1000W hour (1kw/h, so the relation of the lost energy (unload 15 seconds" it it NOTHING. (a 1 /10000 part !)
If you have used (example) 12V 2,2AH batteries . I will wonder, and think about it.
Pese
Update. 3 batteries, closed circuit, a mix of series and parallel connections using resistance, passing .8 of an amp for 5 days. Still maintaining a swing in voltage in turn of .05 of a volt within the same range. WOW
Quote from: dandman on April 20, 2011, 05:10:06 PM
Update. 3 batteries, closed circuit, a mix of series and parallel connections using resistance, passing .8 of an amp for 5 days. Still maintaining a swing in voltage in turn of .05 of a volt within the same range. WOW
Yes, with higher voltage, you can charge one battery,
but you unload 2 batteries.
So you compensate the lasses (mor than 20% that needs !!)
in SUMMARY, anytime you find LOSSES!
Your IDEA is "fine",
but infortunatly it give not an practical, usuable solution to find (this way) :overunity.
Pese
This is probably how he is doing it.
It is claimed by some that this is the method Tesla used to run his electric car continuously without charging.
.
this ideas, are over years in the net.
this cant work.
unfortunatly it produce only LOSSES !
Tesla car system (also made from others in 1930+odd
and some years ago (in eastern europe) have nothing to do
with recharging DC. Tesla used an special AC Motor, made by Westinghouse.
Pese
Since when did we find out anything about Tesla's mysterious car? All I have found is heresay. It was said to be powered by a A/C motor, battery, antenna, and a certain amount of tubes. It was said that when one tube was installed, it would then be ready to drive anywhere at speeds of over 80 miles an hour I believe.
But again, this is heresay and there is no proof of such a car nor is there any documentation from Tesla about this mysterious car. I believe it was his grandson or nephew that was said to have witnesses this car. If anyone has any information about this car, please post your findings.
This damn car has been haunting me for quite some time and it has took me away from all my ideas I had about other types of energy producing projects I was working on. I am sure there is a way to collect natural flowing energy and that is about all I try to figure out any more.
Another is the TV that was supposed to have imploded killing a person. It was said to have created a magnetic suction that attracted metal objects and was drawn in to the TV killing a person within the path. Again there is no proof of this but it too has my attention. to me, both are connected in some way.
Sorry Dandman, I don't mean to get off topic here but in way, we really aren't. There is ways to charge battery's for just an antenna and some circuitry but the way you have stated is impossible. You are actually losing energy but you can't see it. You are loosing amps which you will not be able to see untill you have the right equiptment to find out how many amps you are starting with and how many you are ending with. Mine that I use at my shop is a gauge and needle setup and I wouldn't think it was acurate enough. You would need a digital load testing meter to get the most accurate reading. I sold mine a few years back and I am kicking myself in the butt today for selling it.
Fatbird you are soooooo close mate. Wondered when someone would work out the dead battery thing. Havent heard about Tesla's car but believe me a few mods to your drawing and thing happens.
Nightlife the energy swing appears to be a natural phenomenon of the setup. This morning, having found that the energy swing slows I kick started it by closing the circuits in a different way and the energy used is being replaced to original levels. YOWZA.
I will give Fatbirds drawing of Tesla's setup a go.
Stay tuned
Oh yeah, Nightlife. I did a bit of checking. Load testing is apparently not the way to go. The only accurate way to find current starting and finishing amp hours is converting hydrometer readings. Peukerts law cannot be used reliably unless you duplicate exactly his starting values and loads. The theorists proved it.
Tesla's drawing has left out quite a few small but significant pieces of the puzzle. Just in case your all giving it a go.
Dandman, you are correct that there is no real accurate way to check amps in a battery becuase the load test actually takes amps away as does anything used to check the ampreage. You also need to know that transferring travels through resistance and resistance takes away amps. The voltage change you are expiriencing is not a gain, it is still a loss but the loss is by way of the amps. Just becuase you show 12 volts at the start and 12.6 volts or higher in the end, doesn't mean you have a gain, you still need to know how many amps are present at the 12 volt start and at the 12.6 volt end. 12 x 500 amps = 6000 watts and 12.6 x 470 = 5875 watts which shows that it may show 12 volts at the start and 12.6 volts at the end but there is still a loss of 125 watts.
Thanks Nightlife, understand that. Also understand the relationship between surface voltage and floating which is why as far as Im aware the old fashioned diesel electric submariners way, using hydrometer specific gravitiy readings and converting them is the only accurate way to know whether you have lost energy in a battery (lead acid)
Question. If a battery charger forces energy the wrong way into a battery to charge it. Is the equalization that appears to happen when you put two different strength batteries to equalize in energy or voltage or both. If both, how?.
Update, I think I know what is going on. The energy being used is being stored and returned to the discharged batteries in the conventional charging style. So far Ive used bulbs across it, inductive motors and even charged another battery across it and every time the batteries return to the same state of charge. This is doing my head in...... Ill be back
Wow its gone quiet in here. Anyone "killed" their good batteries yet?
Anyone find anything unusual about the statistics below???
They represent the discharge of a 3 lead acid battery configuration at a constant 12v 1.1 amp with a resistor (car bulb) and the charge states at 24 hour (approx) intervals. It should be noted that at no point during discharge were the batteries attached to any kind of charging device, mechanical or electrical. Prior to beginning discharge, the batteries were charged to their start state with a standard 2.5amp car charger in parallel and the first readings are prior to disconnection. The remainder of the readings are at the end of the 24hr periods, after which the configuration was changed systematically. The same results can be achieved with different discharge rates and with induction ie attaching an inverter to run a.c. applications.
Day 1 Midday 11deg c 5/9/11
Battery Hydrometer Reading Voltage Reading
B1 1.245 14.25v
B2 1.240 14.25v
B3 1.240 14.25v
Day 2 9am 8deg c 6/9/11
B1 1.235 12.54v
B2 1.250 12.74v
B3 1.250 14.89v
Day 3 7:30am 7deg c 7/9/11
B1 1.245 14.21v
B2 1.235 12.43v
B3 1.240 12.69v
Day 4 7:30am 10deg c 8/9/11
B1 1.240 12.50v
B2 1.255 13.25v
B3 1.245 12.40v
Day 5 8:45am 10deg c 9/9/11
B1 1.245 13.12v
B2 1.250 12.33v
B3 1.245 12.40v
Day 6 7:30am 6deg c 10/9/11
B1 1.230 10.75v
B2 1.250 12.52v
B3 1.250 12.67v
Day 7 9am 9deg c 11/9/11
B1 1.240 12.35v
B2 1.250 12.61v
B3 1.225 12.17v
The problem is you are using batteries.
Hey cutting edge, they are fickle things arent they. Check day2 b3, day3 b1,day4 b2 etc. As 2 of the batteries discharge 1 charges, the hydrometer readings are there to confirm that the electrolyte density is changing and not just battery voltage. Therefore proving that charging is actually taking place, grin
Quote from: dandman on September 26, 2011, 05:35:50 PM
Hey cutting edge, they are fickle things arent they. Check day2 b3, day3 b1,day4 b2 etc. As 2 of the batteries discharge 1 charges, the hydrometer readings are there to confirm that the electrolyte density is changing and not just battery voltage. Therefore proving that charging is actually taking place, grin
I agree with your analysis but batteries wear out over time.
AT LAST, thanks cutting edge. a battery does have a finite life, agreed, however, if over the period of that life one can produce a benefit (energy) in another battery, as a consequence of the use, be it light, heat or motive power surely it is worthy of further investigation?. As far as I am aware, the stats provided prove that there is a consequencial benefit available over and above its use?.
Quote from: dandman on September 26, 2011, 11:41:30 PM
AT LAST, thanks cutting edge. a battery does have a finite life, agreed, however, if over the period of that life one can produce a benefit (energy) in another battery, as a consequence of the use, be it light, heat or motive power surely it is worthy of further investigation?. As far as I am aware, the stats provided prove that there is a consequencial benefit available over and above its use?.
Are you aware of how much energy goes into building a battery? does one get more energy out of the battery then what it took to make it? processing, refinement, element purity and so forth, what is the total energy spent on making it compared to its out put?
I am aware that the same amount of energy that went into making a battery yesterday, before an additional benefit was apparent, is the same as today, after there is an additional benefit available, therefore, any benefit is an additional gain, is it not?.
Quote from: dandman on September 27, 2011, 12:16:33 AM
I am aware that the same amount of energy that went into making a battery yesterday, before an additional benefit was apparent, is the same as today, after there is an additional benefit available, therefore, any benefit is an additional gain, is it not?.
I'll give you that but it is still not over unity compared to all that was required to make the process occur. that includes any energy you put into it yourself to make it happen and Humans do use a lot of energy inefficiently.
Hi Cutting edge, Are you just sorry you agreed with my analysis or is there some point to this diversion on energy spent making electorchemical devices. As far as Im aware, if you make a rowing boat, the more times you can use it the more it was worth making it??
Quote from: dandman on September 27, 2011, 12:31:17 AM
Hi Cutting edge, Are you just sorry you agreed with my analysis or is there some point to this diversion on energy spent making electorchemical devices. As far as Im aware, if you make a rowing boat, the more times you can use it the more it was worth making it??
Now you are talking, I love that analogy. I am not here to screw up your enjoyment of the ride, I am just here to remind you that it took a lot of effort in getting that ride. enjoy and I do look forward in your next update.
thanks.
Jerry
Jerry, it's been a pleasure. Thank you for the interaction. I got the message. I aint here for the enjoyment of the ride though. I got nearly 6 years invested in this search. Problem is getting to someone open minded enough, without the restrictions of academic learning, to accept that it can happen. The beginning of these post were more to try to find that kind of mind.
Cheers buddy
Danny
Hi All, If the following is a correct modern day description of overunity.
hypothetical machines that produce more work or energy than they consume, whether they might operate indefinitely or not.
Then does it follow that if one can cause 1 amp of electro chemical energy to pass across a battery with a charging effect for every 1 amp of electro chemical energy used from another battery would be considered overunity?.
This is a serious question, so please, refrain from scoffing, sarcasm, or any derogatory replies.
Thanks
Quote from: dandman on September 28, 2011, 11:47:47 PM
Hi All, If the following is a correct modern day description of overunity.
hypothetical machines that produce more work or energy than they consume, whether they might operate indefinitely or not.
Then does it follow that if one can cause 1 amp of electro chemical energy to pass across a battery with a charging effect for every 1 amp of electro chemical energy used from another battery would be considered overunity?.
This is a serious question, so please, refrain from scoffing, sarcasm, or any derogatory replies.
Thanks
Hi Dandman
In the search for the truth, one must first of all be truthful and the truth is sometime hard to take...however I shall be gentle and not derogatory.
Probably the first thing wrong with your question is the description of OU. You either have OU or no OU...there is no half way. Either the glass is overflowing or it is not, it is that simple so therefore when you stated "whether they (the device) might operate indefinitely or not", I have a big problem with that! If the device is not running indefinitely how can it be claimed as OU? I guess someone pedantic here might reply and say " I calculated all the horsepower worked as opposed to the horsepower/joules stored in battery/capacitor and its only because the output is not routed back to input that the device does not run indefinitely" Yes...on paper that might be ok but considering there is so much talk in forums and theories ad nauseum, a self running device with load attached is always going to be proof of pudding. Besides, if you truly have excess COP then it is nothing to plug output into input.
Now to your experiments. I have personally run these in the past so I understand what you are attempting to do. Yes I agree that you end up with better EFFICIENCY incorporating a load into the charging circuit, but unless you can show definitive proof of COP OU and a self running device I would not be so hasty to claim OU. There are losses in all the transfers. Losses on the wires, loss at the load, loss in the charging and hence this is the reason why you need more than just COP 1.5 for a self running device.
Back to you at the news desk.
Quote from: highvoltage on September 29, 2011, 02:29:20 AM
If the device is not running indefinitely how can it be claimed as OU?
how long is indefinitely? (rhetorical question...) and whom shall be there to verify that yes, it did run 'indefinitely'?
Quote from: highvoltage on September 29, 2011, 02:29:20 AM
Besides, if you truly have excess COP then it is nothing to plug output into input.
LMFAO... how asinine of me not to realize... i'll just go and plug the output of my heatpump into the input. why didn't i think of this before!?! ::)
back to you at the news desk...
Hi, Me from the News Desk here, I guess there are differing interpretations of OU. Purists etc. The wording I posted was the definition from google. By the way, thanks for "being gentle" grin. Methinks Ill go and borrow a video recorder and post it in colour and leave you all to ponder it.
Back soon
Great idea danny ;)
With a 4 lead acid battery configuration, 2 in series loading 2 in parallel via a joule thief circuit , like shown in the attachment , it may be possible ,to charge the batteries by themself.
Notifying the ringing effekt , that causes energy be drawn from the ambient.
Ossi's circuit (experienced guy)
Imho it's not advantageous to apply a load during the charging process, because the dipole is destroyed then.
So remove the parallel stack , apply a load to it and write down volts amps and duration
(Energy consumed).
Then put the paralell stack back into the charging circuit and charge it full.
Exchange series and parallel stack
(prior parallel batteries now in series, and series batteries now parallel) simple
Charge the new parallel stack. When it's loaded , remove it from the charging circuit and apply
the load. ... And so on. If the summary of the energy beeing drawn by the load exceeds
the capacity V * Ah of all 4 batteries you have your OU circuit.
http://www.overunity.com/downloads/sa/view/down/487/
http://www.overunity.com/downloads/sa/view/down/486/