another small breakthrough on our NERD technology.

[CuriousChris]

The answer to this argument is simple.

place some very large capacitors in parallel with the batteries
place a switch in series with the batteries such that when the switch is open the batteries are disconnected from the circuit, but the capacitors are still connected to the circuit.

Start the oscillator. When you are happy the oscillator is running stably. turn the switch off. the capacitors are now supplying the current.

If your circuit is indeed OU then the capacitors should remain charged and the circuit should continue to oscillate. if instead the circuit is UU then the voltage across the caps will quickly dissipate and the cct will stop oscillating.

If the cct continues to oscillate congratulations. If not the cct is not OU.

If you believe that capacitors don't support the oscillation due to fundamental differences between caps and batteries you must be able to put forward a cogent explanation of why. Once you have that fundamental explanation then you can alter the cct to allow for those differences and then make it work using the caps.

If you cannot make it work using the caps. then it serves no useful purpose. as long as it needs batteries it will never be considered OU.

CC

P.S.

I have often wondered if I crack the OU puzzle how would I get the message out. For me the answer is simple. Create a kit, sell it on Ebay with a say 60 day warranty. This lets others test it and validate it for you. If the kit doesn't work you will quickly learn about it in negative feedback and paypal will refund the peoples money.

If your kit works you wont need to worry about refunds.

[Rosemary Ainslie]

Hello Chris

I wonder if you could perhaps take the trouble to read the papers.  We cover that point about capacitors.

Regards,
Rosemary

[CuriousChris]

I had read it but I wanted you to cover it again for your own sake

Therefore, to test whether this retained
potential difference is  a  required condition to enable the
oscillation, capacitors were applied to the circuit during
operation when the oscillation was fully established. The
batteries were then disconnected leaving the capacitors in
series with the circuit and the oscillation then collapsed to  a
zero  voltage. This evidence may support the conclusion that
the retained potential difference at the primary supply source
is required, if not entirely responsible,  for driving this
oscillation.

The conclusion is correct. In that one paragraph you have proved the device is not OU.

If any current was SOURCED from the device it would have recharged the capacitor(s) and provided the necessary potential difference to keep the cct running.

This simple test showed that any current you were seeing "flowing back into the battery" was little more than leakage current caused by the breakdown of the Zener diodes. Because you used an inductive load, when Q2 was switched off by the signal generator, the flux around the inductor collapses and causes a voltage spike (cemf). The voltage quickly exceeds the zeners breakdown voltage of 1000V and current flows back to the battery, because the voltage is quite high it 'recharges' the battery, but only by a very small amount.

I won't enter the discussion on the signal generator being the source of energy because I could not find any details about it. In any real test it must be factored into it. it sources current into the system so that MUST be taken into account. In general signal generators are quite low impedance as well, some I have seen as low as 50 ohms, which means that current can flow through the generator in ways that needs to be accounted for.

If you still fail to see your own test as proof the system is UU. The next test is not so much harder.

Supply a large source of liquid (preferably repleneshing i.e. from a tap) place your heater element in the liquid (flow).
Calculate the watt hours the battery can give you
Calculate the wattage used by the heater element (remember to use Vrms or determine your duty cycle and use that to calculate the watt hours your load consumes)
Properly heatsink your mosfets so they don't fail during the test. (perhaps use the same water supply? you can buy liquid cooled heat sinks. just look up liquid cooled PC's)

Turn your device on

Wait n hours (till the batteries have consumed their calculated watt hours). Smile you are 1/3 of the way there.

Wait another n*2 hours (if you have true OU it won't matter how long you wait, but it should be this at a minimum)

If its still running return to the capacitor problem and try and work out a "COGENT" explanation for why it failed. If you can't explain it in a simple scientific way don't try to make Shit Up. Just accept you don't know why the capacitor test failed and let the physicists determine why the caps failed.


If you reach this point then do as I suggest. market your device in kit form. it will both generate an income for you and silence your critics.

I will be the first to buy one, provided it is suitably guaranteed of course.

On to the flame wars you are having with poynt99. It is doing you no favours, You are behaving in such a condescending manner, and your verboseness indicates you seem to relish in it. What does that say about you as a person?

Just agree to disagree with Poynt99 and leave it at that.


CC

[Bubba1]

... POWER IS NOT WATTAGE....  Power is ALWAYS REPRESENTED AS JOULES which is vi dt. 

Rosemary

OMG!
Rose, how many times can you get this wrong?
JOULES IS NOT POWER!!!!!

[hartiberlin]

Dear Harti,

Thank you for getting back to me.  And compliments of the season.  I trust you had a good holiday.
If I understand this correctly you're asking where we measured the power FROM the generator in the application of that signal at the transistors.  If this is right we explained this in the paper.  We measured the amount of power and found it to not only be negligible but to have a negative value in relation to the supply.  Therefore it would served to DEPLETE rather than to ADD to the energy coming from the battery.  In any event the current value is that negligible that it can be comfortably factored into the margins for error. 

No, I meant to measure the power into the function generator at the grid input of the function generator, so at the 230 Volts AC side with a digital power meter.

We used high wattage resistors PRECISELY because we were generating HIGH CURRENT.  No doubt it would be preferred to use those highly calibrated shunt resistors but, unfortunately, they were and are outside our budget.  HOWEVER - the problems associated with the small inductances on those resistors are only relevant if our measurements of energy are marginal.  This is not the case in any of the examples included in that paper. 
Margins for error has been factored in and most certainly IS referenced in that paper.

You can build yourself very cheaply NONINDUCTIVE Shunts for high power by using a parallel and serial circuit of
SMD shunt resistors. These are noninductive then.

We tested this on a 555 switch.  Reference again in that paper.  The results are the same.  And our results were measured using a Tektronix oscilloscope meter in conjunction with the LeCroy.  They give precisely equivalent results.  The Tektronix is not grounded.  Therefore the LeCroy results are confirmed not be influenced by ground.  We only used the LeCroy screen downloads for the paper because they are clearer and gives a fuller account of the circuit values.

Lets exactly see the circuit diagram then on this and also a new video with this.
How is the 555 circuit powered ?

Will the circuit then also put out these power levels without any scope
or measurement gear connected ?
Just the 12 Volt car batteries and the circuit alone ?

Also you should use a professional battery capacity meter so see the
remaining energy still stored inside the batteries.