Free Energy At Last Step By Step Must See

[El de la triste figura]

Citfta friend thank you very much for your help,
shall begin by studying the pdf ... grateful

Greetings ...

[citfta]

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

[mbrownn]

Excellent PDF, I had not seen that document before, thanks

[citfta]

Helpful Link

@El de la triste figura

Here is a link that will answer most of your questions:

http://www.rodscontracts.ws/images/projects/PulsedMotor/SSGMonopoleExperiment.pdf

Carroll

[El de la triste figura]

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???

[El de la triste figura]

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???

[mbrownn]

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.

[harctan]

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?

[mbrownn]

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.

[harctan]

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.

[mbrownn]

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.

[harctan]

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?

[mbrownn]

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+brake&hl=en&safe=off&client=firefox-a&hs=l7z&rls=org.mozilla:en-US:official&prmd=imvns&tbm=isch&tbo=u&source=univ&sa=X&ei=eLRCUMy6KeeYiAfT4YBw&ved=0CCYQsAQ&biw=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.

[harctan]

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?

[mbrownn]

What part did I not answer? Ask again and I will see what I can do