These points have been covered in open discussion - but they really need to be held in full focus. We claim that we are measuring a negative wattage on our tests - at a variety of different settings to the applied frequency - and with a scheduled list of recorded data that would stretch from here to the moon. Effectively we have positioned those two transistors at a kind of 'mirror image' to each other and then applied the probe from the signal generator to Q1 and its ground to Q2. As shown in both the following circuits.
The red pencilled lines are intended to show the commonality of the switches. Therefore, in effect, one has those transistors arranged that the battery can access either Q1 or Q2 - REGARDLESS. Which argument then claims that there is ALWAYS a path for the positive flow of current from the battery supply. And therefore too - the battery is NEVER disconnected. Which, by default - means that when we measure energy being delivered from the supply - which is that voltage measured above zero - then correctly it IS INDEED being discharged by the battery. This conclusion makes not the slightest difference to our claim. We still measure that negative wattage number. There is still, evidentially, more energy being delivered back to the battery than was first supply by the battery. But there's a nicety that needs to be factored in which goes to the real anomaly and not to standard assumptions about anything at all. Lest we lose the significance of this data - I'm taking the trouble to show this here - as it's been argued on the forum.
This may be a better way to explain the anomalies and it may also get to the heart of Bubba's objection. The oscilloscope probes are placed directly across the batteries that ground is at the source rail and the probe is at the drain. Which is standard convention. Then. During the period when the oscillation is greater than zero - in other words - when the battery is DISCHARGING - then it's voltage it falls. And it SERIOUSLY falls. It goes from + 12 volts to + 0.5. Given a 6 battery bank, for example, then it goes from + 72 volts to + 3 volts. At which point the oscillation reaches its peak positive voltage. And this voltage increase is during the period when the applied signal at Q1, is negative. WE KNOW that this FAR EXCEEDS THE BATTERY RATING. In order for that battery to drop its voltage from + 12V to + 0.5V then it must have discharged A SERIOUS AMOUNT OF CURRENT. Effectively it would have had to discharge virtually it's ENTIRE potential as this relates to its watt hour rating. We EXPECT the battery voltage to fall during the discharge cycle. But we CERTAINLY DO NOT expect it to fall to such a ridiculous level in such a small fraction of a moment AND SO REPEATEDLY - WITH EACH OSCILLATION.
Now. If we take in the amount of energy that it has discharged during this moment - bearing in mind that it has virtually discharged ALL its potential - in a single fraction of a second. And then let's assume that we have your average - say 20 watt hour battery. For it to discharge it's entire potential then that means that in that small fraction of second - during this 'discharge' phase of the oscillation it would have to deliver a current measured at 20 amps x 60 seconds x 60 minutes giving a total potential energy delivery capacity - given in AMPS - of 72 000 AMPS. IN A MOMENT? That's hardly likely. And what then must that battery discharge if it's rating is even more than 60 watt hours? As are ours? And we use banks of them - up to and including 6 - at any one time. DO THE MATH. It beggars belief. In fact it's positively ABSURD to even try and argue this.
NOW. You'll recall that Poynty went to some considerable lengths to explain that the battery voltage DID NOT discharge that much voltage. Effectively he was saying 'IGNORE THE FACT THAT THE BATTERY VOLTAGE ALSO MEASURES THAT RATHER EXTREME VOLTAGE COLLAPSE'. JUST ASSUME THAT IT STAYS AT ITS AVERAGE 12 VOLTS. Well. It's CRITICAL - that he asks you all to co-operate on this. And in a way he's right. There is NO WAY that the battery can discharge that much energy. SO? What gives? Our oscilloscope measures that battery voltage collapse. His own simulation software measures it. Yet the actual amount of current that is being DISCHARGED at that moment is PATENTLY - NOT IN SYNCH.
But science is science. And if we're going to ignore measurements - then we're on a hiding to nowhere. So. How to explain it? How does that voltage at the battery DROP to +0.5V from +12.0V? Very obviously the only way that we can COMPUTE a voltage that corresponds to that voltage measured across the battery - is by ASSUMING that there is some voltage at the probe of that oscilloscope - that OPPOSES the voltage measured across the battery supply. Therefore, for example, IF that probe at the drain - was reading a voltage of +12 V from the battery and SIMULTANEOUSLY it was reading a negative or -11.5 volts from a voltage potential measured on the 'other side' of that probe - STILL ON THE DRAIN - then it would compute the available potential difference on that rail +0.5V. Therefore, the only REASONABLE explanation is to assume that while the battery was discharging its energy, then simultaneously it was transposing an opposite potential difference over the circuit material. WHICH IS REASONABLE. Because, essentially, this conforms to the measured waveforms. And it most certainly conforms to the laws of induction.
OR DOES IT? If, under standard applications, I apply a load in series with a battery supply - then I can safely predict that the battery voltage will still apply that opposing potential difference - that opposite voltage across the load. Over time. In fact over the duration. It most certainly will NOT reduce its own measured voltage other than in line with its capacity related to its watt hour rating. It will NOT drop to that 0.5V level EVER. Not even under fully discharged conditions. So? Again. WHAT GIVES? Clearly something else is coming into the equation. Because here, during this phase of the oscillation, during the period when the current is apparently flowing from the battery - then the battery voltage LITERALLY drops to something that FAR exceeds it's limit to discharge anything at all. And we can discount measurement errors because we're ASSURED - actually WE'RE GUARANTEED - that those oscilloscopes are MEASURING CORRECTLY. Well within their capabilities.
SO. BACK TO THE QUESTION? WHAT GIVES? We know that the probe from the oscilloscope is placed ACROSS the battery supply. BUT. By the same token it is ALSO placed across the LOAD and across the switches. It's at the Drain rail. And its ground is on the negative or Source rail. And we've got all those complicated switches and inductive load resistors between IT and its ground. Could it be that the probe is NOT ABLE to read the battery voltage UNLESS IT'S DISCHARGING? UNLESS it's CONNECTED to the circuit? Unless the switch is CLOSED. IF there's a NEGATIVE signal applied to the GATE then it effectively becomes DISCONNECTED? In which case? Would it not then pick up the reading of that potential difference that IS available and connected in series - in that circuit? IF so. Then it would be giving the value of the voltage potential that is still applicable to that circuit. It may not be able to read the voltage potential at the battery because the battery is DISCONNECTED. It would, however, be able to read the DYNAMIC voltage that is available across those circuit components that are STILL CONNECTED to the circuit? In which case? We now have a COMPLETE explanation for that voltage reading during that period of the cycle when the voltage apparently RAMPS UP. What it is actually recording is the measure of a voltage in the process of DISCHARGING its potential difference from those circuit components. Which ONLY makes sense IF that material has now become an energy supply source.
It is this that is argued in the second part of that 2 part paper - as I keep reminding you. Sorry this took so long. It needs all those words to explain this. The worst of it is that there's more to come.
Kindest regards as ever,