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Renewable Energy Discussion on various alternative energy, renewable energy, & free energy technologies. Also any discussion about the environment, global warming, and other related topics are welcome here.

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  #1111 (permalink)  
Old 08-08-2012, 01:21 AM
mbrownn mbrownn is offline
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How can you be sure of something you haven't test yourself?! :/
I cant but others that are knowledgeable that I trust say that it appears to be the case. I know that overunity is possible as an airconditioner proves it and by the same principals I am sure it can be done with motors.

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despite of all clains since years and very motivated replicators, nothing occurs to be replicate-able after the claims of the first one
It is simple to verify some claims but others are more complex and exacting, the only time I got an overunity in charge I crystallized my battery so repeating it resulted in more useless batteries, not a good solution. I did get an overall overunity and that is simple with the fan. All you have to do is get it on the sweet spot and charge a lead acid battery that is in good condition and has been charged several times before by this method. By adding the mechanical power to the energy found in the charging battery we have more useable power than we put in. I know the gain is small but it is there and easily repeatable.

Quote:
For me, none of them has been able to prove honest results, I mean "infakable" devices and indisputable devices.
This is the point. build it yourself and prove it to yourself, that is the reason for the fan.

There are several ways we can get extra little bits of energy but one that is confirmed is inductive kickback in a coil. It can be collected and reused and it also adds to the motive force of a motor. This alone is two outputs for one input. The fan is weak on motive force so does not make best use of this.

Quote:
You mean using spark-gapes or fast rotating switches to create very short on and off to create picks by coupling with coils or even capacitors?
Its the pulses, they have anomalous effects on batteries.

Quote:
is not the overunity itself but the overunity in what I call mono-energy kind, conversions-like from electromagnetic energy to electromagnetic energy, or gravity to mechanical power, or caloric to caloric
.

This does seem to be significant. When we get overunity it always seems to be associated with changing the form of energy from one type to another. When we want to get the same type of energy out as we put in, it is more difficult.

Quote:
But "what"? the "mono-" or "uni-" converters I was talking about?
convert from electrical to magnetic to mechanical and back again.

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I rather thinking that most have blind themselves otherwise they would have been able to explain much more clearly why their supposed working devices were working, and replications would have bring near no problem. But yes, for me it doesn't mean they were not in the right path and have not touched pertinent points, enough to look further in their directions.
It is difficult to explain something that we don't have the words for and no one can see or conceive.

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Your wrong, Mike
Glad to here it but I haven't seen one

Quote:
No, sorry, the limit, but easily reach now, is 40%, and it's because of the "Stirling cycle" itself
OK, if out power station is 27% efficient with 73 going up the smoke stacks we could use this heat to power stirlings and recover 30% of that or 21.9% making 48.9 in all. but wait we could use lower temperature differential engines to recover the waste heat from these. so 30% of 51.1 is an additional 15.33% making 64.23 and so on. The question is when does it become no longer cost effective to add another stage. At the moment they say it isn't cost effective from the start but that is because nobody makes sterling engines of the size required. All it takes is the political will to instruct manufacturing industry to make these devices. As reliability is high and maintenance is low, Stirling engines are worth a high investment cost despite their low efficiencies.


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Of course, you can say that now after having tried to replicate with not very conclusive results, but it were not the claims of Bedini and the others who claimed effective overunity; right?
My results were conclusive although the gains were small. Conclusive and small is two different things.

So we know the problem, it is possible but it is not so simple to get substantial results. These more useful devices will be relatively expensive per Kw and at this stage we cannot say what the reliability is or the cost per Kwh but should we give up on it? I think not, remember the high cost per Kwh is offset by no input fuel requirement
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  #1112 (permalink)  
Old 08-08-2012, 03:56 PM
Khwartz Khwartz is offline
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Exclamation

Hi Mike. Thanks for your reply but looks to me your message is not complete, having been cut. So I wait a few before to answer you, wishing you have been able to have the pleasure to see that Stirling engines are indeed affordable now
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  #1113 (permalink)  
Old 08-16-2012, 09:57 AM
Khwartz Khwartz is offline
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Hi again Mike. Have you a problem with the second part of ;y before last message to not answer it? Cheer, Khwartz.
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  #1114 (permalink)  
Old 08-17-2012, 12:10 AM
mbrownn mbrownn is offline
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What part did I not answer? Ask again and I will see what I can do
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  #1115 (permalink)  
Old 08-31-2012, 10:26 PM
harctan harctan is offline
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A question for mbrownn or anyone who can answer me. I have build a fan and I want to run some tests. How do you measure the electrical and mechanical efficiency? I have been reading up on ImhotepLabs forum so I have a general idea but i would like a more detailed description. If this has been covered elsewhere please just sent me the link. Also I want to clarify something about the sweet spot. As I increase the resistance value on the potensiometer the following things happen: the fan slows down, the amp draw decreases, when no charging battery is connected the neon will light at some point and then will get brighter, when I connect a capacitor at the output the collected voltage will rise, the coils will start humming at some point and then the hum gets more high pitched. Then when I turn the pot a little more the fan will stop. Is the sweet spot at the highest resistance value just before the fan stops? Maybe even a little lower so it can work at a lower voltage as the drive battery drains?
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  #1116 (permalink)  
Old 09-02-2012, 02:34 AM
mbrownn mbrownn is offline
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Measuring the input s easy, Volts x amps.

Power is volts x amps x time

Measuring the electrical output is more difficult, again its volts times amps but often digital volt meter give false readings and when you do the measurements you will find that the output is poor. Don't worry this is normal as the fan follows all the rules of electricity.

Now we add the time factor as we charge and discharge identical batteries. Ideally you need two identical lead acid batteries whose capacity is sufficient to run the fan for 20 hours and after 20 hours they are discharged. If a battery is charged and discharged at a rate of 20 hours it is at its most efficient.

Starting with one battery fully charged ie 12.7 standing voltage and the other battery fully discharged ie 11.2v run the system until the first battery is at 11.2 volts standing voltage and the charging battery is at about 14.4v. These voltages are approximate as each battery is slightly different. You will need to measure the voltage and current drawn from the first battery throughout the discharge, I did it at 5 minute intervals.

You can also measure the combined voltage of the two batteries during the test, that is put your meter over the negative of the source and the positive of the charging battery and monitor for any overall drop or gain in voltage throughout the cycle.

Swap the batteries and repeat the measurements. Over a number of cycles you will see that the voltage is slowly dropping.

Now measure what is found in the charged battery by draining it on a known resistor or lamp that takes 20 hours to drain the battery and measure the volts and amps at intervals as before.

Divide the output batteries discharge results by the inputs batteries to get your coefficient of performance, you should get better than 0.9 or 90% efficiency.

The meter you place over the two batteries will show an initial voltage rise from about 25.4 to about 27 or 28v followed b a small drop of a few hundredths of a volt over the next 20 hours. This drop represents the loss.

In some cases it is possible to get a greater power out of the charged battery on a resistive load instead using the fan to discharge it, some have had a COP of greater than 1 when doing this.

My tests using the fan to discharge the battery varied between 95 and 97% efficiency. Using the bulb to discharge I had 97 to 110% with a margin of error of 10% if I remember correctly

As the electrical efficiency of the fan is poor it is obvious that the overunity effect is in the charging battery.

To measure the mechanical output of the fan is more difficult. you will need to remove the fan blades and rig up a pony brake using a very accurate balance to measure the torque

de Prony brake - Wikipedia, the free encyclopedia

https://www.google.com/search?q=pony...1600 &bih=778

My results gave me 20 to 25% efficiency with a margin of error of more than 10%

With these two figures added together we have an undeniable efficiency greater than 100%

The sweet spot is when you get the highest volume sound from the coils, the brightest neon or the highest voltage on the output battery. During your test runs you may need to adjust for the sweet spot from time to time as the batteries internal impedance changes.
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  #1117 (permalink)  
Old 09-02-2012, 10:31 AM
harctan harctan is offline
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I guess building it was the easy part! I know what I'll be doing now for the next month. The prony break method is what Peter Lindemann uses in his electric motor secrets video, correct? When you did that did you have to remove the plastic frame of the fan from one side so you could have the band connected to the scales in a vertical position? I was thinking maybe I could glue a plastic cylinder to the fan shaft so I can mount the band there without having to saw anything off. Also at it's sweet spot the fan has very little torque so I'm guessing the torque load should me minimal so the fan doesn't stop, therefore very accurate scales. On the electrical part of things when you cycled the batteries on the fan they would eventually get both discharged, so you would have to charge one of them. Did you use a conventional charger for that or stick to radiant charge for the duration of the measurements? What amphour rating you think would be most appropriate for C20 charge-discharge on this fan project? Lastly the meters (especially the amp meter) should they be hooked in all the time because I know that they can interfere with the circuit?

Last edited by harctan : 09-02-2012 at 10:34 AM.
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  #1118 (permalink)  
Old 09-03-2012, 09:30 AM
mbrownn mbrownn is offline
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All your assumptions are correct.

When charging the batteries, I always used the fan. Using a normal charger reduces the efficiency of the battery a little. The term for a battery always charged on the fan is a conditioned battery, once the battery is exposed to normal charging some of the excess capacity disappears and the battery begins the slow normal degradation. A brand new battery, when conditioned, may show capacity in excess of its specifications, but then if it is charged normally it will revert to having normal capacity. Some have experienced faster charging but as yet I am unconvinced. Gel type batteries tend to dry out on the fan so always use the good old fashioned liquid acid batteries.

To get your amp hour rating, multiply your current draw by 20, for example if your fan consumes 100ma then use a battery of 2Ah. My fans had an ignition coil slaved across the power coil, that is placed in parallel. My current draw was 350mA and my batteries 9Ah. Ok its more C20 but it was good enough.

Yes meters do drop the efficiency if left on the circuit so I remove mine. I placed a 0.2 ohm resistor in the circuit to measure voltage drop and calculate the amps.
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  #1119 (permalink)  
Old 09-09-2012, 06:44 PM
harctan harctan is offline
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I'm trying to tune my system but I'm not sure if my measurements are typical for this fan. I'm also trying to convert a pc power supply to run the fan from there but no success yet, so I'm using a new sealed lead acid battery 12V 1.3Ah. At the vicinity of the sweet spot (or at least where it should be) I have about 1.6kΩ at the pot, the neon will not flash with no output connected and when I connect a capacitor it will charge to about 80V (I don't have an oscilloscope). Meanwhile the current draw will be at the 20-30mA range. I remember reading that usually the fan draws 70-80mA. The resistance of the coils is 70-75Ω. Also before that I had it running on a 9V battery for about 5 hours (at a different sweet spot) but the charge on the second identical battery was minimal. That had me wondering if I was doing something wrong.
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  #1120 (permalink)  
Old 09-10-2012, 12:14 AM
mbrownn mbrownn is offline
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Originally Posted by harctan View Post
I'm trying to tune my system but I'm not sure if my measurements are typical for this fan. I'm also trying to convert a pc power supply to run the fan from there but no success yet, so I'm using a new sealed lead acid battery 12V 1.3Ah. At the vicinity of the sweet spot (or at least where it should be) I have about 1.6kΩ at the pot, the neon will not flash with no output connected and when I connect a capacitor it will charge to about 80V (I don't have an oscilloscope). Meanwhile the current draw will be at the 20-30mA range. I remember reading that usually the fan draws 70-80mA. The resistance of the coils is 70-75Ω. Also before that I had it running on a 9V battery for about 5 hours (at a different sweet spot) but the charge on the second identical battery was minimal. That had me wondering if I was doing something wrong.
If you are getting 80v from a 12v input you must be close to the sweet spot. you could try tuning for the highest voltage when you have a battery connected, I would not recommend doing this on a cap because the voltage may be too high for the transistor and cause it to fail.

Try lowering or raising the resistance.

If the neon is not lighting when there is no battery connected it could be that your pot is set at too high a resistance or the neon has failed or the neon is too high a voltage. I have had many neons fail, usually you get one super bright flash of white light and then the neon is useless.

Your sweet spot will depend upon the voltage and internal resistance of your source and the voltage and internal resistance of your charging battery. The internal resistance of batteries change as they charge and discharge. Usually there isn't a huge difference but it may be enough to stop the neon lighting.

Usually the sweet spot is close to where an 80 ohm fan stops, if yours will not run at the sweet spot you may need to increase your input voltage a little.

Do you here the noise from the coils when it spins? the point where the noise is loudest is also the sweet spot.
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  #1121 (permalink)  
Old 09-10-2012, 09:26 AM
harctan harctan is offline
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80V on the cap is about the highest i can go. If i raise the resistance more the fan will stop. With a battery connected to the output (instead of the capacitor) i can turn the pot just a little more but that's about it. The neon is a 110V rated one, and it works fine, i tested it. Lowering the resistance will only lower my output so i try to go as high as possible without the fan stoping. The coils do make the singing noise and it gets more high pitched as i increase the resistance but as it gets more high pitched the fan stops. I run it at 24V with 2 batteries in series and it works more like what i've read from others. The neon will flash easily from a much lower resistance (less than 1kΩ) and the amp draw close to the sweet spot is about 0.65mA. I guess this fan needs more input voltage as you suggested. Would that be because the coils have higher resistance that usual? Also i'm guessing that running it at 12V i would not get efficient charging so perhaps i shouldn't bother?
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  #1122 (permalink)  
Old 09-13-2012, 04:32 PM
mbrownn mbrownn is offline
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Yes the higher resistance is the problem, I have used similar fans and found that they only just run on a 12 volt charger that is at 14.4 volts. If you use 2 batteries in series, try putting two batteries in parallel for charging but with a separate diode going to each.
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  #1123 (permalink)  
Old 09-14-2012, 03:46 PM
El de la triste figura El de la triste figura is offline
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Hello friends of the forum energetic, very interesting topic.
excuse my intrusion ...
Please your need your help, I can Give you some queries regarding
the Bedini fan???
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  #1124 (permalink)  
Old 09-14-2012, 06:30 PM
El de la triste figura El de la triste figura is offline
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Friends if they could help me they are grateful forever.
It bedini implementing a fan, but I've only worked with old batteries and
messed up if these are not in may recover with Bedini fan ...

My question is this:

What is the method to retrieve the defective batteries for example if
are sulfated??
As when a battery is fully charged???
I can charge a 12V battery with 55A bedini fan???
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  #1125 (permalink)  
Old 09-14-2012, 06:39 PM
citfta citfta is offline
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Helpful Link

@El de la triste figura

Here is a link that will answer most of your questions:
http://www.rodscontracts.ws/images/p...Experiment.pdf

Carroll
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  #1126 (permalink)  
Old 09-15-2012, 02:11 AM
mbrownn mbrownn is offline
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Excellent PDF, I had not seen that document before, thanks
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  #1127 (permalink)  
Old 09-15-2012, 02:28 AM
citfta citfta is offline
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Actually it was wrtner (Paul-R) that posted that link in the thread about the scooter wheel energizer. I just reposted the link to this thread. You are right it is a good pdf. It answers a lot of questions for those people interested in the Bedini energizers.

Carroll
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  #1128 (permalink)  
Old 09-15-2012, 06:16 AM
El de la triste figura El de la triste figura is offline
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Citfta friend thank you very much for your help,
shall begin by studying the pdf ... grateful

Greetings ...
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  #1129 (permalink)  
Old 09-17-2012, 11:25 PM
harctan harctan is offline
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So I've been playing with my fan for a few days but I'm still uncertain about the sweet spot. I run it with 24V from a pc power supply and charge a 12V 1.3Ah SLA battery. The charging voltage peaks at about 820 Ohms, thats about where the coils start their "singing", at about 62 mA current draw. But the peak charge voltage stays there all the way to 6000 Ohms at 20mA. The sound pitch of the coils changes so I can't really tell when it's louder, and without the battery the neon flashes an orange colour which at first gets brighter but then the only thing I notice is it's frequency droping as the fan slows down. Should I just aim for the lowest amp draw at 6000 Ohms? Or would that result to inefficient charging due to low frequency of the spikes? Perhaps there is some other way to find out? (meaning other than spending many days/weeks charging and discharging the battery at different base resistance values)
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  #1130 (permalink)  
Old 09-18-2012, 10:04 AM
mbrownn mbrownn is offline
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tune for the highest charging voltage on your battery.
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  #1131 (permalink)  
Old 09-18-2012, 10:27 AM
harctan harctan is offline
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Originally Posted by mbrownn View Post
tune for the highest charging voltage on your battery.
As I mentioned earlier the highest charging voltage remains the same from 800 Ohms all the way to 6000 Ohms, hence my confusion.
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  #1132 (permalink)  
Old 09-20-2012, 12:41 AM
mbrownn mbrownn is offline
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As I mentioned earlier the highest charging voltage remains the same from 800 Ohms all the way to 6000 Ohms, hence my confusion.
Are you sure? the difference may only be one hundredth of a volt or so if the battery is big.

if you get in the range of the loudest sounds, the highest charging voltage and when no battery is connected the brightest neon you are in the sweet spot.

Any lead acid battery, be it new or old, responds better after a number of charges provided it id always charged on the fan; Capacity increases even on a brand new battery and charge time drops. Do not overcharge gel type batteries they will rapidly deteriorate by drying out, in this condition they act much like a capacitor charging very quickly to high voltages but have little capacity. This may be a cause of some of the very spectacular charging results that some people have obtained. I always use the simple cheap lead acid batteries now, the type that you have to add water. They are also better in the fact you can check each cell with a hydrometer to make sure that the acid is the same strength in each cell.
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  #1133 (permalink)  
Old 09-20-2012, 11:45 PM
harctan harctan is offline
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I'm using 2 new 12v 1.3Ah SLA batteries which I'm currently conditioning. My multimeter has an accuracy to one hundreth of a volt (at least on paper) and it will not register any change in charging voltage for a wide range in base resistance. I run the fan at 24V (due to high coil resistance at 70-75 Ohms) from a computer power supply. I tried again tuning the fan with sound and neon. This way I settled at around 5.000 Ohms. The sound from the coils, although I can't determine if it's the highest in volume, it is the highest in pitch. The neon does light the brightest but the fan has slowed down enough so that the blinking is easily noticable. At that point the fan draws 23mA and charges the battery very very slow (it took about 1 hour to get from 12.25V to 12.27V). After a lot of delay I finally recieved a coil of AWG #32 wire that I had ordered which I used to rewire another striped fan that I had from a previous failed attempt. This one seems to run nicely at 12V so I will run it for the next days and see if I will get better results.

Last edited by harctan : 09-21-2012 at 09:11 PM.
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  #1134 (permalink)  
Old 09-21-2012, 12:18 AM
mbrownn mbrownn is offline
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It sounds like you are on the sweet spot. Unfortunately it is true that the high resistance fans run poorly and it is better to rewind them. I use #27wire for this, the lower the resistance we have the better it works but we still need some turns so that the fan will run.

What amp hour rating of battery do you use?
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  #1135 (permalink)  
Old 09-21-2012, 09:25 PM
harctan harctan is offline
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My battery is 1.3Ah rating. I had tried in the past to rewire with #26 and #28 wire but it was a no go since the induced current wasn't enough to trigger the transistor. The #32 works fine for me but only when I use smaller fan blades (80mm diameter) instead of the original (120mm) as shown in the attached pictures. I've been playing around with this fan today and it is definately easier to tune. I haven't pinpointed the sweet spot exactly yet but I'm closing in. I have it running at 1KΩ,130mA,12v and the battery seems to be charging at an acceptable rate, rising from 12.42v to 12.50v within an hour.

Last edited by harctan : 07-14-2013 at 03:12 PM.
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  #1136 (permalink)  
Old 09-22-2012, 06:45 AM
mbrownn mbrownn is offline
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Originally Posted by harctan View Post
My battery is 1.3Ah rating. I had tried in the past to rewire with #26 and #28 wire but it was a no go since the induced current wasn't enough to trigger the transistor. The #32 works fine for me but only when I use smaller fan blades (80mm diameter) instead of the original (120mm) as shown in the attached pictures. I've been playing around with this fan today and it is definately easier to tune. I haven't pinpointed the sweet spot exactly yet but I'm closing in. I have it running at 1KΩ,130mA,12v and the battery seems to be charging at an acceptable rate, rising from 12.42v to 12.50v within an hour.
130mA is a C10 charge rate which is twice as fast as the ideal but you should still get good results.

Using #27 wire does not have many turns on the stator but is lower resistance so your current is more. My pot is set at around 68 ohms but gets hot. I guess different devices have different settings.
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  #1137 (permalink)  
Old 09-25-2012, 03:46 PM
harctan harctan is offline
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So I'm in the proccess of conditioning my new batteries using a C10 charge-discharge rate (1.24Ah rated at C10), running the fan from a pc power supply at 12V and using as load either 12V bulb, standard fan or bedini fan to charge smaller batteries. I charge them to 14.5-15V and discharge them to 11.2V. I have concluded 3 full charge-discharge cycles until now. The first battery takes about 7 hours to discharge and about twice as long to charge. The second battery charges a little faster but when I discharge it at about 2 hours the voltage goes down to 10V. At first it looks like it drains at a normal rate but in the last 10-20 minutes it plummets fast. Does that mean that it is sulfated? Perhaps it was sitting on the self for a long time before I bought it? If so should I continue charging as usual and see what happens or would it be beneficial if I used more amps and drive it more than 15V?

Last edited by harctan : 09-25-2012 at 10:22 PM.
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  #1138 (permalink)  
Old 09-25-2012, 05:03 PM
mbrownn mbrownn is offline
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With sealed batteries I would not charge above the recommended voltage, Your second battery could be sulfated or drying out. Use a slightly smaller load so you are at least C10 on your discharge, C20 is better.
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  #1139 (permalink)  
Old 09-25-2012, 10:21 PM
harctan harctan is offline
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Ok, the manufacturer gives 15V as top charge voltage so I'll stick to that. The load I use is 120-130mA so it's at C10. I bought these batteries when I was experimenting with my first fan which had a lower current draw and would give me at least a C20 charge. C10 will have to do for now. I guess I'll keep cycling the battery and see if there will be any improvement.
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  #1140 (permalink)  
Old 09-26-2012, 12:17 AM
mbrownn mbrownn is offline
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Yes, after 6 to 10 cycles you wont see much more improvement. To be honest I have never had good experiences with sealed gel batteries. I stick to the old fashioned ones you have to top up with distilled water.
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