Come on Andrew. Bite the bullet. Review the pictures in the report regarding the time domain response of the device in both the on and off state. The writers make a big deal about the difference between the behavior of the ECAT and a regular resistor. I see the effects of positive feedback.
Dave -----Original Message----- From: Andrew <[email protected]> To: vortex-l <[email protected]> Sent: Mon, May 27, 2013 1:39 pm Subject: Re: [Vo]: About the March test I am not acknowledging any such thing - yet :). That's because I don't know what's going on during the pulse OFF time, which is 66% of the total time. Certainly the temperature drops a little during that time, as the report shows. The question is whether there truly is no power delivered during OFF time. It seems clear that during ON time the device behaves just like an electrical resistor. Andrew ----- Original Message ----- From: David Roberson To: [email protected] Sent: Monday, May 27, 2013 10:26 AM Subject: Re: [Vo]: About the March test Yes, what you say in bold type is true but not a problem in this case. Why do you think that energy must be radiated and convected at a level that is greater than the input throughout the entire cycle? Consider energy storage within the device as the place where some of the generated energy is deposited. You are over simplifying the system and leaving out important details. If you are now acknowledging that the COP might be greater than one, then we are making some headway. :-) Dave -----Original Message----- From: Andrew <[email protected]> To: vortex-l <[email protected]> Sent: Mon, May 27, 2013 1:06 pm Subject: Re: [Vo]: About the March test Sure, I completely understand that the calculated COP in the report is wholly due to the 35% duty cycle. But this misses my point. Let me say it again: If input and output power are equal, then there is no energy generation by the device itself. Andrew ----- Original Message ----- From: David Roberson To: [email protected] Sent: Monday, May 27, 2013 7:03 AM Subject: Re: [Vo]: About the March test A little humor never hurts! The bottom line is that the average power being emitted by the ECAT must be equal to the peak duty cycled drive when the COP is 3 and the duty cycle is 33%. This is by definition. Dave -----Original Message----- From: Andrew <[email protected]> To: vortex-l <[email protected]> Sent: Mon, May 27, 2013 1:39 am Subject: Re: [Vo]: About the March test You have stopped processing information and now are talking about bullfrogs. When you return from bullfrog land, we might be able to resume a serious dialogue. Until then, have a hoppingly great time. ----- Original Message ----- From: David Roberson To: [email protected] Sent: Sunday, May 26, 2013 6:29 PM Subject: Re: [Vo]: About the March test I read that section and found that this is not a problem. The input is applied for 1/3 of the time while the average output is roughly equal to that value. The calculation shows that the COP is therefore approximately 3. This is what they say in the report. The maximum instantaneous peak power output should be greater than the peak input. This is consistent. Operation at low temperatures and therefore COP are limited. I prefer to see them run her at full warp, but control issues make this difficult for long duration tests. Dave -----Original Message----- From: Andrew <[email protected]> To: vortex-l <[email protected]> Sent: Sun, May 26, 2013 8:20 pm Subject: Re: [Vo]: About the March test p22. Emitted Power E-Cat HT2 = (741.3 + 17 + 58) [W] = (816.3± 2%) [W] = (816±16) [W] (24) Instantaneous Power Consumption E-Cat HT2 = (920 110) [W ]= 810 [W] (25) ----- Original Message ----- From: David Roberson To: [email protected] Sent: Sunday, May 26, 2013 5:17 PM Subject: Re: [Vo]: About the March test Where does this statement appear? I suspect that you are misreading. Dave -----Original Message----- From: Andrew <[email protected]> To: vortex-l <[email protected]> Sent: Sun, May 26, 2013 8:12 pm Subject: Re: [Vo]: About the March test I continue to be worried about the fact that the input and output power are measured equal in the report in the pulse ON state. One would have thought that, if the device truly is generating its own energy, that this should not be the case. Andrew ----- Original Message ----- From: Andrew To: [email protected] Sent: Sunday, May 26, 2013 4:23 PM Subject: Re: [Vo]: About the March test Eric, The idea here is that the extras (DC and/or RF) are undetectable to the meter using clamp ammeters (we know this for a fact), and when this extra gets passed on to the control box, it's able to pass them on to the device, perhaps with some customisation. The device, being chiefly ohmic, will dissipate DC and will likely also dissipate RF. So no customisation by the control box of the extras is in principle necessary - the power simply gets passed along to the device, which consumes it and generates heat as a result. Now, as I've described, the shenanigans chiefly occur during the pulse OFF state, so there will have to be some customisation in the control box. The idea here is to dissipate the extras during pulse ON and pass them along during pulse OFF. The mains doesn't know about the pulse schedule, so cannot itself switch the extras in or out (actually, a Byzantine arrangement could be made to work in this way, but I'm not going that far out). Since no type of electronics control circuitry could survive colocated with the device, the implication is that the control box has to dissipate significant power continuously. That raises a question about the control box temperature. Since it's a sealed unit, and we're talking a couple hundred watts at least, it would have to get bloody hot. There's another data point we don't have. But you'd think they would have mentioned it. I'm talking myself out of this, aren't I? :) Andrew ----- Original Message ----- From: Eric Walker To: [email protected] Sent: Sunday, May 26, 2013 4:00 PM Subject: Re: [Vo]: About the March test On Sun, May 26, 2013 at 3:45 PM, Andrew <[email protected]> wrote: B) seems unlikely because it would require batteries, and Hartman states that it was much lighter than that. Battery technology does not exist that could be that light, and/or occupy so little volume, and make up that total energy difference as measured over 100+ hours. Therefore, it seems that the only workable theory of possible deception is A). I recall Hartman clarifying that measurements were taken on the mains side (from Jed's post). I am not too familiar with circuitry. I assume that either (1) the measurement equipment (including the laptop) will need some kind of single-phase conversion in order to work off of the same mains, or (2) they will have to be routed to a separate source (in the case where the mains side has been tampered with). Assuming (1) for the moment, how easy or hard would it be to filter out hidden DC or AC when constructing the single phase conversion in order to protect the measurement equipment? Would you need a heavy transformer? Eric
