Overunity SSG - 1.23 COP! (Corrected again)

[Aaron Murakami]

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]

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]

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.

[John_Koorn]

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.

[Aaron Murakami]

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.


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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

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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

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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]

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]

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.

[John_Koorn]

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.

[sephiroth]

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

[Aaron Murakami]

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.

[sephiroth]

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?

[min2oly]

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 :-)

[Aaron Murakami]

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.

[min2oly]

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

[Aaron Murakami]

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.