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Thread: Overunity SSG - 1.23 COP! (Corrected again)

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  1. #1
    Networking Architect Aaron Murakami's Avatar
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    Thumbs up Overunity SSG - 1.23 COP! (Corrected again)







    GO DIRECTLY TO MY POST #15 FOR THE CORRECTED CALCULATIONS SHOWING 1.60 COP.










    Hi everyone,

    First of all, yes, I know the word "overunity" is an oxymoron - you can't have more than everything. But obviously this word has persisted for so long and it is known that it's intended meaning is for a device that is over 1.0 COP and that is why I used it.

    This old dinosaur, my first ever SG from 12 years ago gives a 2.38 COP on the output battery and that doesn't even include mechanical work. That is just one recent test.

    Circuit is MJl21194, both trigger and spike diodes are 1N4007 and base resistor is 60 ohms with a 1k 10 turn precision pot in series. Rotor is pink roller skate wheel from a $2 pair of roller skates from the good will with bearings in fair condition. I'm actually using neos - 3/8" thick double stacked neos...1 recessed into the wheel flush and the 2nd on top of that. Magnets are every 90 degrees and it is running in the enhanced mode described in Bedini SG - The Complete Beginner's Handbook. The power windings is 23 awg and trigger is 26 awg. The power winding is 4 ohms so that tells you how long it is. I built it originally with a MPS 8099 and followed the instructions in the original SG diagram posted on Keelynet WAY back.

    IMG_0687.jpgIMG_0685.jpg

    For this particular test, I used different batteries than are shown in the pictures. In the pictures are 12v 7ah gel cells that were from an electric scooter. I used to charge those up with a bicycle wheel trifilar SG and used to drive it down to John Bedini's shop when I worked at a pulsed light healing device company down the street about 10 years ago or so. They're not in perfect shape but it is a miracle they're even half way good considering the agony and torment that I've put them through.

    Anyway, for the COP test I did yesterday, I used 18v nicads from a Black & Decker Grasshog trimmer. They are rated at 1200 mah or 1.2 Ah.

    Looks like this:

    maximalpower-black-decker-firestorm-18v-2000mah-ni-cd-battery-for-gco18sfb-glc2500-and-more-blac.jpg

    That one is actually a 2.0 Ah model, but the ones I have are 1.2 Ah that have been heavily used for the last 5 years. 90% of the time I charged it up, I used my 1AU Tesla Charger.

    The input battery was fully charged up and so was the secondary battery. The secondary battery, I put a fixed C20 load (300 ohms) across the secondary to drain it overnight. I was going to stop it at 18.0 volts exactly but didn't catch it until it was at 17.41 volts.

    I then disconnected it and hooked up the run battery to the SSG in the pics above and got it up to the fastest speed for the least draw. With a coil this small, you don't need the neon for protection and can run it without a secondary battery hooked up.

    Anyway, after I had the base resistance where I liked it so it would be the fastest with the least draw and the lowest duty cycle, I hooked up the output battery.

    I used a Fluke Scopemeter 123 for the test. It is 2 channels. I had a 0.25 ohm calibrated current sensing resistor on the ground line of the input battery and had both channels across that resistor.

    Input battery went from 18.83 to 18.20 and I ran it with a battery on the back end for 32 minutes.

    Before I give you the real numbers, I want to give you numbers that will handicap it to show the greatest draw - more than it actually drew so I have a bigger number to beat with the draw down test on the secondary battery.

    I will use 18.83 volts (starting voltage and NOT average voltage) to calculate draw for entire running time. Voltage across the 0.25 ohm resistor was 0.0347 volts when the run started. So again, I'm using the HIGHEST numbers to show what it drew to be conservative. If I used lower numbers, it would be easier to beat so let's see what this shows us first.

    0.0347 volts / 0.25 ohm resistor = 0.1388 amps of current. 0.1388 amps X 18.83 volts = 2.614 watt seconds per second. 2.614 watts X 25.3% (using the largest duty cycle towards the end, again to handicap the results to the max) = 0.6612418 joule seconds per second "burned" from the input X 60 seconds = 39.67 joule seconds per minute X 32 minutes of running time = 1269.58 joule seconds burned from the input.

    The volt reading across the resistor was done using DC Mean instead of RMS since at these relatively low speeds it is accurate. If we're running in the mhz or something, then we'd definitely want to use RMS. The Frequency was about 300hz, which is 18000 cycles per minute divided by 4 magnets every 90 degrees = 4500 RPM at the start just to give you an idea of what the wheel is doing. Anyway, 300 cycles per second is very much in the slow range to use DC Mean on a scope to measure the voltage across the current sensing resistor.

    At 32 minutes of run time, the output battery was disconnected and a C20 load was applied. 1.2 Ah C20 rate is a 60ma current draw. The battery voltage I used was 18.0 v / 0.06 amps = 300 ohm load. I used 3 X 100 ohm 10 watt power resistors with the 0.25 ohm current sensing resistor in series on the ground side of the string.

    The starting voltage was 18.7 and it took 45 minutes to go down to 17.41 v where it was drained to before it was charged up.

    AGAIN - to double handicap the numbers in favor of conservative numbers, I'm going to calculate the total draw using the voltage of 17.41 (when the battery was drained) so it will show that I drew the least amount from the output battery.

    at 17.41 volts, the voltage across the resistor was 0.015 volts. 0.015 volts / 0.25 ohms = 0.06 amps. 0.06 amps X 17.41 volts = 1.0446 watt seconds per second and of course we leave it at that since the fixed resistive load is at a 100% duty cycle.

    1.0446 watt seconds per second x 60 = 62.676 watt seconds per minute X 45 minutes until it hit the 17.41 volt goal = 2820.42 joule seconds burned from the output recovery battery, which was charged from the input.

    2820.42 joules on the output battery / 1269.58 joules on the input battery = 2.22 COP and that does NOT include any mechanical work.

    We used the highest possible numbers to show a large input and the lowest possible numbers on the recovery battery to show a small output - handicapping it in both directions for the benefit of the doubt.

    Using the real averages, averaging the average of the different geometrical ramp downs on the voltage grahps, the average voltage was 18.515 with a voltage across the current sensing resistor of 0.0346 volts. 0.0346 / 0.25 ohms = 0.1384 amps X 18.515 = 2.562476 watts X 24.90 average % duty cycle = 0.638 watt seconds per second X 60 = 38.28 watt seconds per minute X 32 minutes = actual joules burned on input of 1225.07.

    Using real averages for draw down test on output battery, considering actual averages of both geometrical ramp downs of the graph, 18.25v for 14 minutes at 0.0155 volts across resistor = 0.0155 / 0.25 = 0.062 amps X 18.25 v = 1.1315 watt seconds per second X 60 sec x 14 minutes = 950.46 joule seconds burned.

    Then the second part of the graph average is 17.6 volts with 0.015 volts across 0.25 ohm resistor = current of 0.06 amps x 17.6 volts = 1.056 watt seconds per second x 60 seconds = 63.36 watt seconds per minute X 31 minutes at this average = 1964.16 joule seconds burned.

    950.46 + 1964.16 joules burned on the output until battery got back down to 17.41 volts = 2914.62 actual joules burned on output battery.

    2914.62 divided by 1225.07 = 2.38 COP and that still doesn't include any mechanical work added to that. With mechanical work, will be about 2.8.

    Even with handicapping the input and output for the worst case scenario, the COP is still 2.22, but using the real averages, it is 2.38 COP and that is without any mechanical work added to the equation.

    The output battery obviously will recharge itself a bit and some skeptics will grip about C20 being too low of a discharge. Obvoiusly if we use a C50 rate, we could probably wind up with a COP of 5+, but C20 IS realistic. And with these numbers, I could do a C10 rate and would probably still beat 1.0 COP easily.

    Is my little SSG "overunity"?
    Aaron Murakami





    “You never change things by fighting the existing reality. To change something, build a new model that makes the existing model obsolete.” ― Richard Buckminster Fuller

  2. #2
    Networking Architect Aaron Murakami's Avatar
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    INPUT BATT
    Time Battery Volt Resistor Volt Resistor Current Amps Watts Duty Cycle Watt Sec/Sec Watt Sec per Min Running Time Min
    8:42 18.83 0.0347 0.25 0.1388 2.613604 24.30% 0.635105772 38.10634632
    8:50 18.62 0.0346 0.25 0.1384 2.577008 24.30% 0.626212944 37.57277664
    9:00 18.53 0.0346 0.25 0.1384 2.564552 24.90% 0.638573448 38.31440688
    9:14 18.2 0.0344 0.25 0.1376 2.50432 25.30% 0.63359296 38.0155776 32


    SECONDARY BATT
    Time Battery Volt Resistor Volt Resistor Current Amps Watts Duty Cycle Watt Sec/Sec Watt Sec per Min
    9:16 18.7 0.016 0.25 0.064 1.1968 100% 1.1968 71.808
    9:20 18.3 0.016 0.25 0.064 1.1712 100% 1.1712 70.272
    9:25 18.14 0.016 0.25 0.064 1.16096 100% 1.16096 69.6576
    9:30 17.85 0.015 0.25 0.06 1.071 100% 1.071 64.26
    9:35 17.75 0.016 0.25 0.064 1.136 100% 1.136 68.16
    9:41 17.68 0.015 0.25 0.06 1.0608 100% 1.0608 63.648
    9:46 17.63 0.015 0.25 0.06 1.0578 100% 1.0578 63.468
    9:50 17.56 0.015 0.25 0.06 1.0536 100% 1.0536 63.216
    9:55 17.5 0.015 0.25 0.06 1.05 100% 1.05 63
    10:00 17.43 0.015 0.25 0.06 1.0458 100% 1.0458 62.748
    10:01 17.41 0.015 0.25 0.06 1.0446 100% 1.0446 62.676


    Any comments or questions are welcome. Feel free to double check the numbers or state your comments about the measurement method.

    In my opinion, the only real option that I see to be even more accurate is an "Integrated Detailed Power Analysis".

    Use a DSO and show 1 single waveform on the screen for the input measurement. The Tektronix TDS3054C for example can capture 10,000 measurements in a single screen and will give just about a perfect analysis of the draw for 1 single waveform accounting for the entire geometry. Capture that, dump it to a spreadsheet and calculate. Then mutliple that by the number of cycles for that voltage and so on until you are done. Tektronix was kind enough to loan me that $11k scope for some tests I ran in the past and it would be nice to use it again, but the Fluke Scopemeter 123 will have to do for now. I don't have the optical cable or the software to do the data dumps so I have to manually charge the data.
    Aaron Murakami





    “You never change things by fighting the existing reality. To change something, build a new model that makes the existing model obsolete.” ― Richard Buckminster Fuller

  3. #3
    Hey Aaron, I don't think I understand your measuring method of the input. You are measuring 0.0347 volts across a 0.25 ohm resistor which gives you 0.1388 amps of current. So input is 2.614 watt.

    I make that out to be 5018.88 joules as your input. Why are you factoring in the duty cycle when you were measuring the DC Mean voltage?

  4. #4
    Nice work as always Aaron!
    Duty cycle is one of the reasons it is difficult to calculate what goes into the energizer. If the energizer is only drawing current 25% of the time, that factor needs to be taken into account when calculating total joules.

    I don’t think we need a fancy scope to calculate COP however... John K. put a nice excel spread sheet together on the old forum this calculates the energy into and out of the charged battery quite nicely. If it has not already been posted in this forum, it would be a good idea.. I explained back on MP2 how to do the same on the front end (primary battery)

    However, it’s quite simple. Just take J.K.’s spread sheet for the charged battery and use the same calculations for the primary. In other words, when you charge the primary battery using conventional current from the main’s, calculate the energy into the primary battery, how many joules does it take to charge the primary battery from a wall wart for example. Then we can compare the joules that go into the system as a whole to the joules that come out of the final charged battery.
    Need to run a few cycles to make sure the numbers are not getting skewed by some anomaly.

    This method makes it simple to not have to calculate the entire wave form from a high end scope. And, really it is the whole point in the first place – how much goes in and how much goes out.
    Better yet, would be to bring a battery from the backend to the front end... All that being said

    I LIKE YOUR NUMBERS! Much faster to calculate using your method :-)

    By the way, my son has replicated your ignition coil plasma sparkplug for his science project this year. Thanks for sharing it!
    Kind regards,
    Patrick

  5. #5
    Networking Architect Aaron Murakami's Avatar
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    Quote Originally Posted by min2oly View Post
    Nice work as always Aaron!
    Duty cycle is one of the reasons it is difficult to calculate what goes into the energizer. If the energizer is only drawing current 25% of the time, that factor needs to be taken into account when calculating total joules.

    I don’t think we need a fancy scope to calculate COP however... John K. put a nice excel spread sheet together on the old forum this calculates the energy into and out of the charged battery quite nicely. If it has not already been posted in this forum, it would be a good idea.. I explained back on MP2 how to do the same on the front end (primary battery)

    However, it’s quite simple. Just take J.K.’s spread sheet for the charged battery and use the same calculations for the primary. In other words, when you charge the primary battery using conventional current from the main’s, calculate the energy into the primary battery, how many joules does it take to charge the primary battery from a wall wart for example. Then we can compare the joules that go into the system as a whole to the joules that come out of the final charged battery.
    Need to run a few cycles to make sure the numbers are not getting skewed by some anomaly.

    This method makes it simple to not have to calculate the entire wave form from a high end scope. And, really it is the whole point in the first place – how much goes in and how much goes out.
    Better yet, would be to bring a battery from the backend to the front end... All that being said

    I LIKE YOUR NUMBERS! Much faster to calculate using your method :-)

    By the way, my son has replicated your ignition coil plasma sparkplug for his science project this year. Thanks for sharing it!
    Kind regards,
    Patrick
    Hi Patrick,

    You can see I actually didn't calculate the whole waveform, but that would be my preference if I had the optical cable and software for this Fluke. But you're right, that isn't needed. I just used the scope for the duty cycle, resistor voltage and frequency.

    I tried to put the battery graphs in my post but it came out as jumbled code. Anyone should be able to copy that cell data and paste it into a spreadsheet to see the graphs.

    The 100% duty cycle draw that Seph is suggesting would be very close to a C8 draw on that 18v 1.2Ah battery. It would take a 122.45 ohm load across the battery if I did the numbers right.

    That means that at 100% duty cycle at that draw, when the battery hits 18.83 volts and goes down to 18.2, it should be able to sustain that draw for 32 minutes.

    Would be interesting to see if this input battery could actually do that.

    That's awesome about your son - I hope he wins! Take lots of pics. Showing the normal CDI discharge then connecting the diode to get the plasma blast should have them scratching their head since we're using the same input.
    Aaron Murakami





    “You never change things by fighting the existing reality. To change something, build a new model that makes the existing model obsolete.” ― Richard Buckminster Fuller

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    Quote Originally Posted by sephiroth View Post
    Why are you factoring in the duty cycle when you were measuring the DC Mean voltage?
    Seph,

    Did you ever do any of those Ainslie tests? I don't recall. But did you use the protocols that the universities were so adamant about? The DC Mean is just the net voltage of a single waveform on a per waveform basis. You still have to account for duty cycle - at least according to those professors.

    At these low frequencies, a simple volt meter on a cheap multimeter will show you the same average dc voltage across the resistor - don't need a scope for that. But even with the 25% duty cycle at these speeds, that is still too fast to see the volt meter fall in voltage. The volt meter shows the same reading as the scope. You still have to account for the duty cycle.

    Can you show me a reference where you have to do the calculation at 100% duty cycle? You could be right and all those professors could be wrong. I'm open to correction.
    Aaron Murakami





    “You never change things by fighting the existing reality. To change something, build a new model that makes the existing model obsolete.” ― Richard Buckminster Fuller

  7. #7
    Quote Originally Posted by Aaron Murakami View Post
    Seph,

    Did you ever do any of those Ainslie tests? I don't recall. But did you use the protocols that the universities were so adamant about? The DC Mean is just the net voltage of a single waveform on a per waveform basis. You still have to account for duty cycle - at least according to those professors.

    At these low frequencies, a simple volt meter on a cheap multimeter will show you the same average dc voltage across the resistor - don't need a scope for that. But even with the 25% duty cycle at these speeds, that is still too fast to see the volt meter fall in voltage. The volt meter shows the same reading as the scope. You still have to account for the duty cycle.

    Can you show me a reference where you have to do the calculation at 100% duty cycle? You could be right and all those professors could be wrong. I'm open to correction.
    Seph,
    guess I should learn how to read more thoroughly before typing. I'll wait to hear from Aaron. sorry for jumping in on this... my method of slowly calculating the joules into the primary still rocks :-)
    kind regards,
    Patrick

    looks like we typing at the same time, I'll stop :-)
    Last edited by min2oly; 12-11-2012 at 12:23 PM.

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    Networking Architect Aaron Murakami's Avatar
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    This shows three meters showing the same thing - current sensing resistor is actually 0.65 to 0.7 ohms:



    Certainly NOT 0.25 like I assumed just because it is written on the resistor.
    Aaron Murakami





    “You never change things by fighting the existing reality. To change something, build a new model that makes the existing model obsolete.” ― Richard Buckminster Fuller

  9. #9
    Quote Originally Posted by Aaron Murakami View Post
    Seph,

    Did you ever do any of those Ainslie tests? I don't recall. But did you use the protocols that the universities were so adamant about? The DC Mean is just the net voltage of a single waveform on a per waveform basis. You still have to account for duty cycle - at least according to those professors.

    At these low frequencies, a simple volt meter on a cheap multimeter will show you the same average dc voltage across the resistor - don't need a scope for that. But even with the 25% duty cycle at these speeds, that is still too fast to see the volt meter fall in voltage. The volt meter shows the same reading as the scope. You still have to account for the duty cycle.

    Can you show me a reference where you have to do the calculation at 100% duty cycle? You could be right and all those professors could be wrong. I'm open to correction.
    As I said, I don't understand. Do you have a link showing an explaination of this measuring method? How do your calculations compare with other methods of measuring the input?

  10. #10
    Networking Architect Aaron Murakami's Avatar
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    Ok everyone, am correcting the readings again.

    Was brought up in a youtube post about considering the resistance of the leads themselves that could be added to the current sensing resistor.

    I'm not going to consider anything that Cen Tech Harbor Freight meter says and will only rely on the Fluke's because of their quality.

    So - the 123 Scopemeter showed me 0.65 ohms and the Fluke multimeter shows me 0.7. Averaging those shows me 0.675 ohms.

    The leads of the Fluke multimeter have a resistance that bounces back and forth between 0.1 and 0.2 but hangs at 0.1 for about 75% but I'll just call it 0.15 ohms resistance to be conservative even though it is slightly less on average.

    So if the direct current sensing resistor reading of that is 0.7 and we subtract lead resistance of 0.15 ohms that leaves 0.55 ohms as the actual reading of the current sensing resistor according to this meter.

    The 123 Scope Meter accounting for using ground connected to the probe cable or the COM ground between both A & B ports, it comes out to be right at 0.5 ohms as the real reading and not 0.65 which doesn't account for lead resistance or other possible interferences.

    So between both meters - lead resistance, etc... I'm going to pin the resistor at 0.55 + 0.5 / 2 = 0.525 ohms.

    Using 0.525 ohms to calculate the input draw, here are the new numbers:

    Time Battery Volt Resistor Volt Resistor Current Amps Watts Duty Cycle Watt Sec/Sec Watt Sec per Min Running Time Min Total Watt Sec
    8:42 18.83 0.0347 0.525 0.066095238 1.2445733 100.00% 1.244573333 74.6744
    8:50 18.62 0.0346 0.525 0.065904762 1.2271467 100.00% 1.227146667 73.6288
    9:00 18.53 0.0346 0.525 0.065904762 1.2212152 100.00% 1.221215238 73.27291429
    9:14 18.2 0.0344 0.525 0.06552381 1.1925333 100.00% 1.192533333 71.552 32 2289.664
    To be conservative, we will use the starting voltage and current to calculate draw so there will be a higher number to beat AND will use the higher duty cycle at the end!
    18.83 0.0347 0.525 0.066095238 1.2445733 100.00% 1.244573333 74.6744 32 2389.5808
    Estimate of actual averages
    8:42~9:14 18.515 0.0346 0.525 0.065904762 1.2202267 100.00% 1.220226667 73.2136 32 2342.84

    That means the draw is 2342.84 joules from the input.

    The output still remains at 2883.96 joules - that is pretty dead on and won't change. I'm using the resistance of 3 X 100 ohm 10 watt power resistors and a current sensing resistor and I can tell you at that resistance at about 300 ohms plus or minus a couple ohms, that isn't going to change the reading at any significant level.

    2883.96 / 2342.84 = COP 1.23
    If estimating 25% in mechanical, then the estimated total COP is 1.54 but I'll leave it at 1.23 since I didn't measure the mechanical work of that wheel spinning at 4500 RPM.

    So, forget what the resistor is rated at. Measure the resistance of your probes by making good contact with each other and keep them shorted. Whatever that is, subtract that from the reading you get when you measure the resistor because lead resistance will add to that resistor.

    Keep in mind this is a one pass test and not multiple passes. I'll post something else later.

    Any comments, questions or suggestions are welcome.
    Aaron Murakami





    “You never change things by fighting the existing reality. To change something, build a new model that makes the existing model obsolete.” ― Richard Buckminster Fuller

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