Author Topic: Charging Batteries from Batteries part 2. Free Energy? (Read 36946 times)

dandman · « **on:** April 20, 2011, 04:21:33 AM »

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.

SkyWatcher123 · « **Reply #1 on:** April 20, 2011, 05:13:11 AM »

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

dandman · « **Reply #2 on:** April 20, 2011, 05:29:49 AM »

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

Magluvin · « **Reply #3 on:** April 20, 2011, 05:37:24 AM »

Hey Dand

Are you using inductors? Just wondering considering the losses involved in direct discharge/charge.

Mags

dandman · « **Reply #4 on:** April 20, 2011, 05:46:02 AM »

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

nightlife · « **Reply #5 on:** April 20, 2011, 06:01:03 AM »

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.

dandman · « **Reply #6 on:** April 20, 2011, 06:26:31 AM »

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

nightlife · « **Reply #7 on:** April 20, 2011, 06:36:58 AM »

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.

dandman · « **Reply #8 on:** April 20, 2011, 07:05:17 AM »

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

nightlife · « **Reply #9 on:** April 20, 2011, 07:21:50 AM »

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.

dandman · « **Reply #10 on:** April 20, 2011, 07:28:36 AM »

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.

nightlife · « **Reply #11 on:** April 20, 2011, 07:38:44 AM »

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.

dandman · « **Reply #12 on:** April 20, 2011, 07:52:36 AM »

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?

pese · « **Reply #13 on:** April 20, 2011, 08:13:49 AM »

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

dandman · « **Reply #14 on:** April 20, 2011, 08:21:38 AM »

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.

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Author Topic: Charging Batteries from Batteries part 2. Free Energy? (Read 36946 times)

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