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 <andrew...@att.net>
To: vortex-l <vortex-l@eskimo.com>
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: vortex-l@eskimo.com
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 <andrew...@att.net>
To: vortex-l <vortex-l@eskimo.com>
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: vortex-l@eskimo.com
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 <andrew...@att.net>
To: vortex-l <vortex-l@eskimo.com>
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: vortex-l@eskimo.com
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 <andrew...@att.net>
To: vortex-l <vortex-l@eskimo.com>
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: vortex-l@eskimo.com
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: vortex-l@eskimo.com
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 <andrew...@att.net> 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
