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

  1. #11
    Sorry Aaron. I didn't mean to cause trouble. I have been satisfied with my measurements using RMS, and haven't come across the method you described. If RMS is inaccurate and your way of measuring this kind of power is correct then all my tests in the past have been with inaccurate data so I would like to know

  2. #12
    Senior Member John_Koorn's Avatar
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    Hey Aaron,

    i was also wondering why you calculated the joules from the primary at 25.3% duty cycle.

    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.
    I have not measured it this way in the past, so I'm not sure on this but if you are reading 0.0347v over the resistor isn't the meter showing you the average voltage?

    I agree with Sephiroth that the 0.1388 amp draw should be factored at 100% duty cycle since I've always seen this as the average draw current.

    Perhaps another way to measure the draw energy from the battery is to measure the area under the curve from a current probe reading. Even this is difficult as when the transistor turns on the current does not go from zero to max instantly, it goes up in a non linear curve. I've also noticed that when the transistor switches off there is a current spike that goes back to the primary battery.

    If you would like I'm happy to plug your numbers into my COP calculator spreadsheet and see how they compare.

    John K.
    Last edited by John_Koorn; 12-11-2012 at 01:29 PM. Reason: Darn auto correct

  3. #13
    Networking Architect Aaron Murakami's Avatar
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    Quote Originally Posted by sephiroth View Post
    Sorry Aaron. I didn't mean to cause trouble. I have been satisfied with my measurements using RMS, and haven't come across the method you described. If RMS is inaccurate and your way of measuring this kind of power is correct then all my tests in the past have been with inaccurate data so I would like to know
    No trouble, I don't think I was supposed to multiply it by duty cycle either. Yes, use 100% duty cycle. I found an old Ainslie test and I did not multiply it by duty cycle...but that was a single waveform measurement to see the net draw and to show that the net draw was negative (more below the line than above) so it really isn't the same kind of test but same protocols can be used. Sorry for the confusion - it's been about 4-5 years since I did those measurements.

    Here is an example of the waveform analysis fyi: https://www.youtube.com/watch?v=sAsydnawSpA You can do that for very high accuracy waveform analysis and simply multiply that by however many of those waveforms occur per second then x the minutes. Obviously the voltage of the battery changes over time so you have to account for that, but that is the idea. There is no method that is more accurate that I've seen.

    You can do RMS too but not necessary at these low frequencies.

    What this does show however, is a very valid test protocol.

    I did the draw down on the input battery with the fixed 120 ohm load and it powered it from 18.82 to 18.20 and it was close to the time that it took the SSG input run so it is close enough in the ballpark. What that validates is that the method I proposed is a valid protocol.

    Also, this tiny coil is way to small to push that 18v battery anywhere and I know a good % of the input draw was wasted in vain so the secondary battery wasn't getting charged up from the last x minutes of the run. A 6v 1.7ah gel cell is probably about the biggest battery I'd want to use on that machine.
    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

  4. #14
    Networking Architect Aaron Murakami's Avatar
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    Quote Originally Posted by John_Koorn View Post
    Perhaps another way to measure the draw energy from the battery is to measure the area under the curve from a current probe reading. Even this is difficult as when the transistor turns on the current does not go from zero to max instantly, it goes up in a non linear curve. I've also noticed that when the transistor switches off there is a current spike that goes back to the primary battery.
    I don't have a current probe for either of my scopes.

    I just need to get some serial cables to charge the batteries on one of my laptops. Sure will make it easier than having to hand chart it.

    The real cop I think is closer to 0.8 if we consider the following:

    1. The real average draw. The draw suggested by Seph is using the highest voltage and current from the beginning and using that to calculate the draw for the entire 32 minutes. The real average input would be about 4900 joules - so that alone is a COP of 0.6. But - #2 below is the primary loss.

    2. A % of the end of the running time was wasted because it wasn't doing anything to the secondary battery since the small coil is not pushing. That is a 23 awg power winding of 4 ohms. It is about the size of a 35mm film canister - actually smaller. I would have gotten the same draw from the input if I would have ended it several minutes sooner (maybe). Without this loss, would be closer to 0.8.

    3. With mechanical, it is probably about 1.0 COP. Which is still incredible for a bad bearing roller skate with neo magnets.

    Here is the thing that is still incredible even for a low COP SSG. The conventional science says it takes x to power the input to charge the coil x amount of time. If you recover ANYTHING at all, conventional science is automatically defeated no matter how low of a COP an SSG is. That is what most people don't realize. There isn't supposed to be any electrical recovery at all and every single SSG is already violating conventional science because of this.

    If there is a COP of 0.8, then it actually only is a 20% net cost on the front - that already beats the books.
    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

  5. #15
    Networking Architect Aaron Murakami's Avatar
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    Ok, correction to my numbers...

    Because the current sensing resistor says 0.25 ohms doesn't mean it is so.

    I used a Fluke 123 Scopemeter and it says 0.65 ohms, a Harbor Freight Centech Mutlimeter says 0.7 and a 79III Fluke Multimeter says 0.7 - that averages to 0.683 ohms!
    I'm not sure why it is so far off from 0.25 ohms (I'll post a youtube vid showing these measurements - the resistor really is 0.7 ohms and NOT 0.25.
    Using the actual resistance for the calculation AND at 100% duty cycle shows 1800.86 joules burned from the input using real averages.


    ---------------------------------------------------------------------------------------------------------------------------------------

    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 0.683 0.050805271 0.9566633 100.00% 0.95666325 57.39979502
    8:50 18.62 0.0346 0.683 0.050658858 0.9432679 100.00% 0.943267936 56.59607613
    9:00 18.53 0.0346 0.683 0.050658858 0.9387086 100.00% 0.938708638 56.3225183
    9:14 18.2 0.0344 0.683 0.050366032 0.9166618 100.00% 0.916661786 54.99970717 32
    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 0.683 0.050805271 0.9566633 100.00% 0.95666325 57.39979502 32 1836.793441
    Estimate of actual averages
    8:42~9:14 18.515 0.0346 0.683 0.050658858 0.9379488 100.00% 0.937948755 56.27692533 32 1800.86

    ---------------------------------------------------------------------------------------------------------------------------------------

    On the draw down of the secondary battery - I'm not going to use the current sensing resistor at all. 0.25 ohms or even a couple ohms from a fixed 300 ohm load won't significantly change the results. The C20 for that 1.2 Ah battery is 60ma or 0.06 Amps - the load is 300 ohms and that was the load used to draw it down. Over time, the voltage will go down and
    so with the amps drawn but since I'm starting above 18v (18.7) and going below 18v (17.41), it will be very close to accurate using 0.06 amps as the current for the calculations. Conservative estimate is 2820 joules burned but using real averages considering the voltage, it is closer to 2883 joules burned from the secondary.



    Time Battery Volt Resistor Volt Resistor Current Amps Watts Duty Cycle Watt Sec/Sec Watt Sec per Min Running Time Minutes Total Watt Sec
    9:16 18.7 0.06 1.122 100% 1.122 67.32 0
    9:20 18.3 0.06 1.098 100% 1.098 65.88
    9:25 18.14 0.06 1.0884 100% 1.0884 65.304
    9:30 17.85 0.06 1.071 100% 1.071 64.26 0
    9:35 17.75 0.06 1.065 100% 1.065 63.9
    9:41 17.68 0.06 1.0608 100% 1.0608 63.648 0
    9:46 17.63 0.06 1.0578 100% 1.0578 63.468
    9:50 17.56 0.06 1.0536 100% 1.0536 63.216 0
    9:55 17.5 0.06 1.05 100% 1.05 63 0
    10:00 17.43 0.06 1.0458 100% 1.0458 62.748 0
    10:01 17.41 0.06 1.0446 100% 1.0446 62.676 45 2820.42
    Since we're conservative on the input calculations, we'll be conservative here and use the weakest draw using the last numbers for the whole time!
    17.41 0.06 1.0446 100% 1.0446 62.676 45 2820.42
    Estimate of Actual Averages
    9:16 ~ 9:30 18.25 0.06 1.095 100% 1.095 65.7 14 919.8
    9:30 ~ 10:01 17.6 0.06 1.056 100% 1.056 63.36 31 1964.16
    2883.96

    ---------------------------------------------------------------------------------------------------------------------------------------

    Input 1800 joules
    Output 2883 joules

    2883 / 1800 = 1.6 COP

    And that doesn't include mechanical work.


    Again, I'll post a video showing what I mean about the current sensing resistor actually being .7 ohms instead of 0.25 like I wrongly assumed.

    I was willing to accept 0.6 COP if that is what it was but I still had to check it out. I did quite a few different tests all day scratching my head at how the results could be that poor. It is almost impossible to get a COP that bad even if you tried with these circuits. 1.6 COP is much closer to what is expected running in the better mode explained in the book.

    Now it all makes sense.
    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

  6. #16
    Senior Member John_Koorn's Avatar
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    Hey Aaron, those numbers look a lot better. Thanks heaps for posting the correction.

    If you have time could you do the same test with the coil in repulsion mode? It would be great for people to see what the performance difference is between the two modes.

    John K.

  7. #17
    Networking Architect Aaron Murakami's Avatar
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    Quote Originally Posted by John_Koorn View Post
    Hey Aaron, those numbers look a lot better. Thanks heaps for posting the correction.

    If you have time could you do the same test with the coil in repulsion mode? It would be great for people to see what the performance difference is between the two modes.

    John K.
    Hi John,

    I'm anxious to do that when the kits are ready. I want to run the same machine and change out the wheels with a different magnetic configuration for comparison with an ABA test.
    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

  8. #18
    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. #19
    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

  10. #20
    Networking Architect Aaron Murakami's Avatar
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    One thought I have is to use a high enough resistance current sensing resistor so that any invariance in lead resistance, etc... will have no significant relevance. And, will be most practical for anyone that doesn't have a $300 precision micro-ohm meter, etc...

    If the current sensing resistor is 10 ohms, a 0.1 or 0.2 ohm lead resistance won't make a difference that is big enough to take you out of the ballpark.

    1.23 cop from 1.6 is about a 23% difference because of the lead difference factor I experienced.

    With a 10 ohm resistor, a 0.1 or 0.2 ohm lead resistance difference is insignificant like 1-2% difference.

    Using super small resistors for current sensing is so that you don't limit the overall current significantly to effect the machine's operation. So perhaps using the largest resistor you can but small enough to allow as much current as you're ever going to need may be a good idea. Using a larger resistor will simply make it easier, more accurate and more practical for the average builder who can use this method with any good quality multimeter to see exactly what the input vs output is.
    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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