I think I have something VERY simple producing possible over unity.
Please check out this video and let me know how I can test it further.
http://www.youtube.com/watch?v=TPZGMA6Bp5o
singerxyz
Hi Singer...
I think you are "off the mark" as you say. I was okay watching your video until the current measurement part. I couldn't exactly see what you were measuring, but if you put 1/4A through a neon bulb, it will explode! Typical neon testers like that draw micro-amps or, at the most, a few milliamps. You need to test the current and voltage at the same time. If you are shorting the current meter across the neon tester, then you will find almost zero volts and the neon will not light.
I think you have a serious measurement error. Sorry to dampen your obvious excitement!
Linda
Yup. It's a measurement error.
Srry for the bad news. If it helps, many 'experienced' do the same mistake.
Tinu
Hi Singer,
Linda is right,
you have to measure Volts and Amps at a real ohmical load.
Just try to see, how much voltage and amps you get on a 10 and/or 100 Ohm load
resistor.
(Better use a 10 Watts type, so it will not heat up too fast...)
Regards, Stefan.
Quote from: linda933 on October 08, 2007, 01:39:57 PM
Hi Singer...
I think you are "off the mark" as you say. I was okay watching your video until the current measurement part. I couldn't exactly see what you were measuring, but if you put 1/4A through a neon bulb, it will explode! Typical neon testers like that draw micro-amps or, at the most, a few milliamps. You need to test the current and voltage at the same time. If you are shorting the current meter across the neon tester, then you will find almost zero volts and the neon will not light.
I think you have a serious measurement error. Sorry to dampen your obvious excitement!
Linda
Linda,
Thanks for the response. I wasn't measuring the current with the neon bulb attached, I was measuring it open circuit. How can I test it (without 'blowin things up? :-) I don't know anything about ohms. Can someone point me in the right direction?
BTW, I really don't care too much about "the prize" (though it would be nice, and I'm sure my wife would appreciate it LOL!) I really just want to see over unity out in the public domain and I'm doing everything I can to help.
If others with better equip. could try the same type of setup it would cost about $30 US.
Hi Singer,
buy a 1 Ohm
10 Ohm,
100Ohm and
1000 Ohm
all 10Watt type resistors
and use this instead of the neon
bulb as the load.
Then connect each resistor once as the load.
Then measure the voltage across the different resistors
and post each DC voltage at each single resistor.
( Of course put only one resistor of this list to the output and then measure each one by one)
Tested current with 10 ohm 10 watt resistor and got .13 amps with slightly lower voltage.
(Batteries are close to dead- need to buy more)
I'll pick up the 100 and 1000 ohm resistors and batteries tomorrow and post results.
mramos-
the meter was shown up to .29 amps. Is that not close to .33? Mockery is not helpful except to reveal character. And as far as what "was not seen" was obviously me taking the leads off the neon bulb and putting them on the volt meter.
Quote from: singerxyz on October 08, 2007, 10:19:36 PM
Tested current with 10 ohm 10 watt resistor and got .13 amps with slightly lower voltage.
(Batteries are close to dead- need to buy more)
I'll pick up the 100 and 1000 ohm resistors and batteries tomorrow and post results.
mramos-
the meter was shown up to .29 amps. Is that not close to .33? Mockery is not helpful except to reveal character. And as far as what "was not seen" was obviously me taking the leads off the neon bulb and putting them on the volt meter.
Hi again, Singer,
One thing you said struck me and that was that you were measuring the current "open circuit". You know, that when you set your meter to any of its current scales and plug into the appropriate jacks, the meter acts as a near short circuit internally. DMMs are all actually voltmeters. To get a current reading, they just throw a fat low value resistor across the jacks inside and then measure the voltage drop across it.
Per Ohms law, if the internal shunt is 0.1 ohm and the current readout says 10A, then there will be a voltage of 1V between the meter jacks. Just remember that in Voltmeter mode, your meter will have a very high resistance (ideally infinite) and in Ammeter mode it will look like a very low resistance (dead short ideal). So when you do your resistor tests, measure voltage across (parallel) the resistor and measure current by inserting the meter in series with the test load resistor, not across it.
Probably you already knew all that, but I just wanted to clarify in case you weren't sure. When you put a current tester directly across the output of your circuit it's a virtual short circuit because of the internal shunt. I can't count how many times I've blown up the fuse in my meter by forgetting to rearrange the test leads into the right jacks and sticking the meter (still set for a current test) across some beefy power source.
Anyway...happy testing and don't be discouraged if overunity operation eludes you for a while. It's managed to hide from the mainstream of practical commercial technology for a very long time!
Linda
Linda,
Thank you for the helpful and encouraging words. I didn't know anything about resistance testing, and this helps a great deal. when you say measure 'across'' I assume you mean put + on one end of resistor and - on other and voltmeter same? And series would be + on one end of resistor and voltmeter + on the other side with - running straight through to meter?
Sorry to bother you with something so elementary as this, again I'm a little green.
No worries mramos, thanks for writing back.
Anybody know the best way to send current back to run the timer once the unit is started? it needs to replace 18V from two 9V batteries.
Hi Singer,
did you measure the 0.13 amp current at your 10 ohm
in series with the resistor or did you short out the resistor during this
with the ampmeter ?
It must be measured in series, when you take current readings.
It is probably easier for you to measure the voltage across ( in parallel) with
the resistor.
Also in parallel with the resistor must be the storage capacitor.
Please post the voltages.
Many thanks.
Again, having a little trouble understanding terminology, but what I did was connect the 10 ohm resistor to the positive out on the unit and then connected the positive terminal of the meter to the other side. the negative out of the unit I ran to the negative terminal of the meter. Is this series? I understand series/parallel when it comes to batteries, but with caps and resistors I get confused.
But after reading hartiberlin's post I think I know how to proceed now and will re-test tonight, and post with pix.
Quote from: singerxyz on October 10, 2007, 08:48:18 AM
Again, having a little trouble understanding terminology, but what I did was connect the 10 ohm resistor to the positive out on the unit and then connected the positive terminal of the meter to the other side. the negative out of the unit I ran to the negative terminal of the meter. Is this series? I understand series/parallel when it comes to batteries, but with caps and resistors I get confused. But I'm a fast learner ;-)
Yep. That's right for measuring current THROUGH a load resistor. For voltage, you measure ACROSS the load. If you know your load resistance exactly, you don't have to measure the current to figure out the power, by the way.
Power is E times I, Volts times Amps. If you know Ohm's Law, you know that I=E/R and so, if you know E and R, you can find P in Watts by E times E/R or, more often stated "E squared over R".
For accurate calculation of power, you should measure the (disconnected) actual value of the resistor using your ohm meter, rather than just using the marked value. All resistor values have a manufacturing "tolerance" associated, usually 5% or 10%, which means the actual resistance is allowed to be that much more or less than the marking.
Hope this helps. By the way, the reason Stefan suggested getting several different resistor values is because every power source or electrical system has it's own "output impedance". You will find that your circuitry puts out more power into some values of load resistor than others. When you find the value that gives you the absolute most output power, you have identified the output impedance of your circuit! My loose guess for your setup is that it will put out the most power into a several hundred ohm resistor. Just a guess!
Linda
A good way to home in on this magic number to extract maximum power is to notice the following:
As you increase the ohmic value of the load resistor, you will most certainly notice an increase in the output voltage. It will be zero if you use zero ohms (dead short) and will be maximum if you use an open load (no load...open circuit).
For most (linear) circuits, and I believe yours will probably fall into this category, the ideal load for maximum power transfer will be very near to whatever load causes the voltage to fall to half its open circuit voltage. Try to find that value by experimenting. You will then be able to tell if you have hit the maximum power point by raising and lowering the value slightly from there and observing a power falloff in both directions.
Linda
And...just in case you are not totally overwhelmed with information yet...
Please note that the load "operating point" as it is called that gives the most power into the load will not necessarily be the operating point of highest efficiency! Maximum power transfer generally occurs when load resistance is equal to the circuit's output impedance but maximum efficiency will probably be found at yet a different load!
Efficiency is the ratio of output power over input power, as you probably know. Finding that load is a bit more complicated because you must test both the input and output powers and plot them on a graph as you change your load resistor. Measuring input power WILL require a current measurement and a voltage measurement at the circuit's input terminals, since the circuit's input resistance can't be measured directly (it would have to be measured while it is running which won't work). The input power will change each time you use a new load resistor.
It is much more arduous to test for efficiency than for just the output power, as you have to measure at both ends of the circuitry and there is no simple rule for locating the max efficiency point as the "half-voltage" rule of thumb for finding the max power transfer point.
I better shut up now before you go nuts and throw the whole setup in the trash! Have fun!
Linda
Linda,
Never shut up! You're an angel. And I appreciate every word.
I bought a 1 and 100 ohm 10 watt resistor but all I could find was a 1 watt 1k ohm resistor- other than the heat, will this be sufficient to test with?
If it fries, it fries, no worries. besides, I'm only testing for short durations.
Even more than testing the Volt/Amp output, I'd like to see if it could become self running by sending current back to run timer and circuit. I imagine with a large enough capacitor it could be possible.
I'd need to send 12-18 volts and a few milliamps back.
Quote from: singerxyz on October 10, 2007, 11:08:05 AM
Linda,
Never shut up! You're an angel. And I appreciate every word.
I bought a 1 and 100 ohm 10 watt resistor but all I could find was a 1 watt 1k ohm resistor- other than the heat, will this be sufficient to test with?
If it fries, it fries, no worries. besides, I'm only testing for short durations.
Even more than testing the Volt/Amp output, I'd like to see if it could become self running by sending current back to run timer and circuit. I imagine with a large enough capacitor it could be possible.
I'd need to send 12-18 volts and a few milliamps back.
I can tell you are anxious to achieve self-running. That is a very lofty goal, you know. Certainly there are ways to "close the loop" and they would be the easy part if you truly have more output energy than input energy. First you have to find out what the ideal load is for maximum efficiency. as I have written, it's easiest to start by finding the maximum power output load resistance, but you will need more than one or two resistors to play with. I'd try the 100 ohm first. See what kind of power level you get with that, since I believe that would be much closer than your 1 ohm. The 1 ohm will probably drop your output voltage down too far.
Try the 100 ohm and tell us all what output voltage you get with that versus your output voltage with the 1000 ohm and with an open circuit (only the meter as a load). Then I can give you a good guess as to the next resistors to buy, what value and wattage you'll need to home in on this max power transfer point.
Also...be aware that as you add loading to the output side of your circuit to extract power, unless you are doing something very very special, your input power will no longer be 18Volts "at a few milliamps". You will undoubtedly observe that the current needed at the input, from the batteries, will go up rapidly as you pull more and more power from the output by adjusting your resistor value toward the maximum power value.
This is why, sadly, you have to find some operating load resistance where the output power exceeds the input power before you worry about "closing the loop" for self-running! Despite the many claims to the contrary (a.k.a. lame excuses made by people who have claimed OU but say they can't get self-running to happen because of some mysterious gobbletygook), it will be very easy to close the loop on any system that truly provides more electrical energy output than it requires at the input.
Linda
Quote from: linda933 on October 10, 2007, 11:21:05 AM
I can tell you are anxious to achieve self-running. That is a very lofty goal, you know.
Linda
Hey, if you're gonna be a bear, be a grizzly, right ;-)
I'll work on this stuff tonight,
Thanks again
SingerXYZ
Quote from: singerxyz on October 10, 2007, 11:35:51 AM
Quote from: linda933 on October 10, 2007, 11:21:05 AM
I can tell you are anxious to achieve self-running. That is a very lofty goal, you know.
Linda
Hey, if you're gonna be a bear, be a grizzly, right ;-)
I'll work on this stuff tonight,
Thanks again
SingerXYZ
"Cold grinding Grizzly bear jaws hot on your heels" a Jim Morrison Doors lyrics quote
Yes, but be prepared for a lesson in what are known as The Laws of Energy Conservation. Getting something extra, energy-wise, is a rather elusive task, stating it optimistically!
Have fun and learn at the same time...you can't beat that combination!
Linda
Doors- love it!
My results- hate it :-(
I tested the unit with the resistor as a load i.e. (positive-->resistor<--negative) and the meter connected meter positive to positive, meter neg to neg. I was using 2 nickel-cadmium 9V's because they seemed to produce more current, lasted a shorter time.
Open Circuit
600 VDC 1.6 Amps (2 Amps with fresh battery)
100 Ohm Resistor
7 VDC .8 A
10 Ohm Resistor
3.3 VDC 1.3 A
1 Ohm Resistor
.5 VDC .53 A
My batteries were running down pretty quickly, so the results may actually be slightly better but,
I guess it's back to the drawing board...
A little encouragement from an unexpected source-
Ian Lungold a researcher on the Mayan Calendar says that between Nov. 19th, 2007 - Nov. 12th, 2008
will be "The end of manufactured lack" and the arrival of free energy for all.
Interesting!
http://mayanmajix.com/cycles.html
http://www.calleman.com
http://www.mayanmajix.com/
Quote from: singerxyz on October 10, 2007, 11:43:34 PM
Doors- love it!
My results- hate it :-(
I tested the unit with the resistor as a load i.e. (positive-->resistor<--negative) and the meter connected meter positive to positive, meter neg to neg. I was using 2 nickel-cadmium 9V's because they seemed to produce more current, lasted a shorter time.
Open Circuit
600 VDC 1.6 Amps (2 Amps with fresh battery)
100 Ohm Resistor
7 VDC .8 A
10 Ohm Resistor
3.3 VDC 1.3 A
1 Ohm Resistor
.5 VDC .53 A
My batteries were running down pretty quickly, so the results may actually be slightly better but,
I guess it's back to the drawing board...
Dear Singer,
I think you are still measuring current wrong. It makes no sense to say you have 1.6A flowing in an open-circuited load! Makes me think you are still putting the current meter (with its internal low-value shunt resistor) directly across (in parallel with) your output leads. I was hoping my earlier explanations were sinking in but it looks like not yet.
Current measurement has to be done by putting the meter in series with the load, not across it. An open load (infinite resistance) leaves no way to measure current, but that's okay because it will always be zero in an open circuit!
Forget measuring the current. Measure your resistors each by itself not hooked up to anything with the meter set for ohms measurement. Write down the results for each resistor. Then measure the voltage only on each resistor as you load your circuit with it. Your numbers you give above tell me that you were doing something weird like maybe you got two meters and were trying to measure volts and current at the same time but putting the current meter right across the load. That would screw up the resistance you have because the meter has a low value internal shunt resistor when in current mode.
I was expecting you would get more voltage across your 100 ohm resistor. You also didn't test with the 1000 Ohm resistor, which would be interesting to see how much that drops the voltage as compared to open circuit. I was really hoping that if you didn't get overunity you would at least learn something about measuring basic circuits. It looks like maybe you're still not understanding yet?
Linda
OK. I just needed this diagram, now I understand. (like the saying goes, that picture could have saved a thousand words!) Basically, I tested everything like the Voltage test in the diagram. I'll re-test tonight.
Thanks again, L
Singer
I think I got it right now-
1000 ohm=8.8V .009A
100 ohm=4.2V .04A
10 ohm=2.4V .24A
Quote from: linda933 on October 10, 2007, 10:05:06 AM
A good way to home in on this magic number to extract maximum power is to notice the following:
As you increase the ohmic value of the load resistor, you will most certainly notice an increase in the output voltage. It will be zero if you use zero ohms (dead short) and will be maximum if you use an open load (no load...open circuit).
For most (linear) circuits, and I believe yours will probably fall into this category, the ideal load for maximum power transfer will be very near to whatever load causes the voltage to fall to half its open circuit voltage. Try to find that value by experimenting. You will then be able to tell if you have hit the maximum power point by raising and lowering the value slightly from there and observing a power falloff in both directions.
Linda
Somso ,,,
if i put an power load tothe 110volts line , so strom , even tat the voltage go don to 55 volts ..
than i have the maximum "power point" ...
thisway that the whole town will go to darkness ...?
Pese
(smile)
www.pese.cjb.net