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Overunity SSG - 1.23 COP! (Corrected again)

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  • 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
    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

    Comment


    • #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?

      Comment


      • #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

        Comment


        • #5
          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

          Comment


          • #6
            Originally posted by sephiroth View Post
            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?
            Oops I just noticed your wrote "DC mean voltage" I'm assuming Aaron knows what he is doing with a scope across that resistor and is seeing a duty cycle draw rather than a constant draw.
            kind regards,
            Patrick

            Comment


            • #7
              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

              Comment


              • #8
                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, 12:23 PM.

                Comment


                • #9
                  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?

                  Comment


                  • #10
                    Overunity SSG - 2.38 COP?

                    I'll reopen the other thread after the method is validated. I don't want anyone on the wrong track if I'm in error. If it is incorrect, no point in reopening an erroneous thread.

                    Seph - that method has been around for years and is actually the only method the universities would accept for the input measurements on the Ainslie tests. You'll have to go through that old Ainslie COP 17 thread in the other forum to see all that spelled out. We even backed it up with the data dumps on the TDS3054C.

                    I'm going to do a draw on the input battery with the draw you are suggesting is happening at a 100% duty cycle. If it can sustain it for 32 minutes from 18.82 volts down to 18.20 then you're right. Please understand you are suggesting that the 18v tool battery is pulling that C8 load for a little over a half hour.
                    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

                    Comment


                    • #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

                      Comment


                      • #12
                        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, 01:29 PM. Reason: Darn auto correct

                        Comment


                        • #13
                          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

                          Comment


                          • #14
                            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

                            Comment


                            • #15
                              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

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