Testing the TK Tar Baby

Rosemary Ainslie · April 30, 2012, 04:21:00 AM

Quote from: TinselKoala on April 30, 2012, 04:14:10 AM
So now you are claiming that the oscillations are not powered by the battery.

So... we have YET ANOTHER free energy machine that needs a battery to run, but isn't powered by it, but when the battery is removed entirely, or even replaced by a capacitor... the machine won't run at all, much less make free energy.

The secret of overunity and free energy isn't in some specific circuit or pulse motor... it is IN THE BATTERIES !! As long as you have some fully charged batteries, your Free Energy machine will run just fine !! And all the time we were thinking it's the circuit. It's the BATTERIES that make a free energy machine work, every time !!

I have even referenced the PRECISE PARAGRAPH in our paper that refers to this. Indeed. The oscillation is NOT powered by the battery. It simply is advantaged by that available potential difference. That is what we're claiming. And proving. For the oscillation to occur there HAS TO BE A DISCONNECT FROM THE BATTERY SUPPLY. Therefore the battery is NOT connected during the period when the positive signal is applied to Q2 - whether it be a signal from a function generator - or a 555 with it's own battery supply - or whether the 555 switch is operated by the circuit's battery supply. And we've tested all variations of that switch. But not to our circuit apparatus - YET.

Rosemary
added

TinselKoala · April 30, 2012, 04:45:01 AM

QuoteLISTEN UP TK. We apply the probe of the function generator directly to the Gate of Q1. We apply its terminal to the Gate of Q2. And I challenge you to do this as well. Precisely to highlight our argument. I will run our demonstration of this and thereby prove what I am here stating. Therefore the applied negative or positive to the source rail is IMMATERIAL. Because then you will see that there is always an applied positive bias to afford a battery discharge during both periods of that switching cycle. The battery can take it's pick. Q1 or Q2. That would resolve in a continual discharge from the battery which would be a continual above ground voltage waveform.

LISTEN UP YOU. I ALWAYS USE THE SAME CIRCUIT HOOKUPS AS YOU except as specified, and WE ARE NOT USING THE FUNCTION GENERATOR ANY MORE. But when I do it is applied EXACTLY LIKE YOURS and your insinuations that it is not is ANOTHER LIE from you.
The rest of that paragraph is your usual word salad. Source rail? The battery can take its pick? (Without the non-English comma.) You are using the offset of your FG to provide a signal that is both positive and negative and you don't even realize it. The evidence for this is in your own scope shots in your papers.

Quote
TK said: VOLTAGE IS RELATIVE to some reference. A Negative voltage on the source means a positive voltage on the gate.
That is your assumption.

That is an absolute fact. And anyone can prove it empirically and electrical line maintenance techs prove it every day when they work on live high voltage wires.

QuoteWhat we prove is that a negative at the Gate of Q1 would be a positive at the Gate of Q2. And correspondingly a positive at the Gate of Q2 would be a negative at the Gate of Q1.

Then you are proving the obvious, because any child can see that your circuit, when fed by the FG, is wired that way. And I have never disputed that point. What you WON'T be proving is that your batteries don't discharge... because they do.

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TK said: You are turning your Q2 mosfets partially on with the negative voltage to the source and this is what causes the oscillations, as Mile High (I think) described in detail, period by period, some time ago buried under all your crap.
TK. I assure you BOTH - that there is no possibility of an oscillation if you have a continual path from either Q1 or Q2. It has NOTHING to do with whatever signal is at the source rail.

You assure us "both"? There is only one of me.
The signal at the "source rail" determines whether and which mosfets turn on. And I have ALWAYS SAID and demonstrated many times that when one or more mosfets are solidly on, the oscillations stop. I have said MANY TIMES that the main function of the FG, other than providing BIAS CURRENT FLOW, is to turn the oscillations off by turning mosfets on.
Again, you lie by distorting my actual position and by missing the point altogether.

TinselKoala · April 30, 2012, 04:50:56 AM

Quote from: Rosemary Ainslie on April 30, 2012, 04:21:00 AM
I have even referenced the PRECISE PARAGRAPH in our paper that refers to this. Indeed. The oscillation is NOT powered by the battery. It simply is advantaged by that available potential difference. That is what we're claiming. And proving. For the oscillation to occur there HAS TO BE A DISCONNECT FROM THE BATTERY SUPPLY. Therefore the battery is NOT connected during the period when the positive signal is applied to Q2 - whether it be a signal from a function generator - or a 555 with it's own battery supply - or whether the 555 switch is operated by the circuit's battery supply. And we've tested all variations of that switch. But not to our circuit apparatus - YET.

Rosemary
added

Yes Rosemary we all know what your claim is. You are wrong, though, and this has already been demonstrated. Remember the capacitor?
And you are truly pathetic. There are MILLIONS of variations of 555 timer circuits and you have not tested them all. And furthermore... The MOSFETS ARE THE SWITCH. Your 555 or function generator is only the "finger on the switch".

And even furthermore... nobody is now interested in 555 timers or function generators. You will have to show your circuit running on DC negative bias signal using its own main battery, and show that it heats a load usefully and does not discharge its batteries. THE 555 or FG are IRRELEVANT and have been superseded. Get with it. You are hopelessly behind... because Tar Baby is READY NOW and can demonstrate EVERYTHING you actually have demonstrated.. NOW... and it can do it without using a function generator or an external bias supply.

picowatt · April 30, 2012, 08:47:36 AM

Quote from: poynt99 on April 29, 2012, 09:45:19 PM
PW,

This is for you, as requested.

Note; I did not add any inductance. It oscillates nicely with the filtering across all the batteries.

Also note that 3.3W x 4 = 13.2W, precisely the same value obtained by using the power probe on one battery and multiplying by 6x.

.99.

Why does the 3.3W need to be multiplied by 4? I understood the 6X with one battery, but I am a bit confused regarding the 4X above.

I am assuming you are doing (single Vbatt x 6) x (I)= PW and (all 6 Vbatt) x (I)= PW. Should they not be the same?

When you have a moment...

Thanks,
PW

poynt99 · April 30, 2012, 09:28:22 AM

Quote from: picowatt on April 30, 2012, 08:47:36 AM
.99.

Why does the 3.3W need to be multiplied by 4? I understood the 6X with one battery, but I am a bit confused regarding the 4X above.

I am assuming you are doing (single Vbatt x 6) x (I)= PW and (all 6 Vbatt) x (I)= PW. Should they not be the same?

When you have a moment...

Thanks,
PW

These computations are in reference to the filtering across all 6 batteries. So the 6x factor was for probing the power in one battery and multiplying by 6 to get the total battery power.

The 3.3W battery power was computed by AVG[V1-V2(t) x V4-V3(t)] (V1-V2 is VBAT and V4-V3 is VCSR), but this does not account for the 0.25 Ohm CSR value. Therefore if we multiply the 3.3W by 4x, it equals the actual battery power as determined by the power probes.

This demonstrates that in order to obtain the correct total battery power, we should be treating the CSR value as 0.25 Ohms, not 1 Ohm.

If you read my analysis, you would see it explained there.