[Vo]:Re: [Vo]:Re: [Vo]:Re: [Vo]:Torbjörn Hartman describes power measurments

[Harry Veeder]

Among cheeses, I believe Stilton has one of the highest energy densities.

Harry

On Mon, May 27, 2013 at 7:36 PM, David Roberson  wrote:

> The report included a couple of graphs on page 27.   One was power out per
> their measurement, the other power in.  The mere fact that the power out
> versus time is clearly modulated proves that the input is not constant.
> The duty cycle can also be determined from that chart.   I am not sure that
> there is any evidence that could support their claim better.
>
>  It does no good to assume that Rossi is scamming and you guys should
> concentrate on proving that there is a problem with the measurements.  I
> assume that you understand my explanation why the DC is not important to
> the input power measurement.  That is basic electronics.
>
> Dave
>  -Original Message-
> From: Joshua Cude 
> To: vortex-l 
> Sent: Mon, May 27, 2013 6:21 pm
> Subject: [Vo]:Re: [Vo]:Re: [Vo]:Torbjörn Hartman describes power
> measurments
>
>  Check out these 2 videos. It's a clear demonstration of how full power
> can be transferred to a resistive load without registering current on
> either clamp-on or in-line ammeters. I don't know how it's done but I
> suspect high frequency, but the point is that just because I can't explain
> it, doesn't mean I must conclude that cheese can supply the power.
>
>  This switch could emulate Rossi's on/off cycling, and judging from input
> measurements one would conclude a duty cycle of 1/3, but looking at the
> resistive load, it would be 1:1.
>
>
>
>  http://www.youtube.com/watch?v=ovGXDDvc3ck
>
>  http://www.youtube.com/watch?v=Frp03muquAo
>
>
>
>
>

Re: [Vo]:Re: [Vo]:Re: [Vo]:Torbjörn Hartman describes power measurments

[David Roberson]

Andrew, I tried very hard to teach you about this subject and failed miserably. 
 If you do not understand it after my extreme effort, then it must be beyond 
your level of knowledge.

Even though I failed, you can run a spice program and see for yourself.  I did 
this for proof.  Do you want to argue with a program that does not have an 
agenda?  Apparently you are being confused by the complexity of DC versus AC 
waveforms.

Good luck,

Dave


-Original Message-
From: Andrew 
To: vortex-l 
Sent: Mon, May 27, 2013 7:08 pm
Subject: [Vo]:Re: [Vo]:Re: [Vo]:Torbjörn Hartman describes power measurments



"Power from an AC source can only be extracted by the fundamental component of 
that source, period. "
 
An uneducated and completely incorrect statement like that disqualifies you, in 
my view, from making any further comments about the EE aspects of this 
experiment. If you do, I urge anyone reading them to ignore them, because in 
all likelihood they will also be wrong.
 
Andrew
  
- Original Message - 
  
From:   David   Roberson 
  
To: vortex-l@eskimo.com 
  
Sent: Monday, May 27, 2013 1:55 PM
  
Subject: Re: [Vo]:Re: [Vo]:Torbjörn   Hartman describes power measurments
  


  
If you do not understand what I have already written then it is not going   to 
help to go over it again.   I leave this discussion by asking you   one 
pertinent question.  Where do you think the power comes from that   ends up in 
the resistor?  There is only one source and it is the AC   mains.  Power from 
an AC source can only be extracted by the fundamental   component of that 
source, period.  All others, including DC balance out   over the long run and 
can not make a long term contribution.  Once you   realize that this is true, 
which is common theory, it will become clear to you   that a measurement of 
these two waveforms is all that is required.
  
 
  
Forget the nonsense about diodes faking out good AC true RMS   instruments.  It 
don't happen.
  
 
  
Dave
  
 
  
-Original   Message-
From: Duncan Cumming 
To:   vortex-l 
Sent: Mon, May 27, 2013 4:32   pm
Subject: Re: [Vo]:Re: [Vo]:Torbjörn Hartman describes power   measurments

  
  
  
OK, I will tackle this problem head-on using the   Socratic method in stages. 

First, consider a wire carrying 100 amps of   direct current, plus one amp of 
pure sinusoidal AC current at 60Hz. What is   the AC component of the current?

Duncan

P.S. Don't worry, we   will get to the diode later.

On 5/27/2013 11:57 AM, David Roberson   wrote:

  

Duncan, I hate to keep repeating myself that the power can be measured by 
analyzing the AC components only.  When will you guys show why this is not 
true?  I suggest that you start with the simple system you proposed of a 
diode in series with a resistor driven by an AC wall socket.  Explain how 
it works as you say and I promise to show you the error of your 
calculations.

 

Dave

-Original Message-
From: Duncan Cumming 
To: vortex-l 
Sent: Mon, May 27, 2013 2:38 pm
Subject: [Vo]:Re: [Vo]:Torbjörn Hartman describes power measurments




I am not sure if I count as a skeptic, because I am not saying that any 
kind of scam was perpetrated. I am certainly not suggesting that there was 
a DC power supply hidden in the wall! My doubts are related to the 
electrical engineering skills evident in the published paper, attempting 
the notoriously difficult task of measuring three phase non sinusoidal 
power. Not only is the waveform non sinusoidal, it is a trade secret!

I am merely saying that rectification will cause a misleadingly low value 
of current to be registered using a clamp on ammeter. Since the DC is not 
smooth, there will, indeed, be a small reading from the ammeter but 
substantially lower than the actual current. This will, in turn, lead to a 
misleadingly low power measurement.

Duncan

On 5/26/2013 8:46 PM, David Roberson wrote:


  
Robin,
  
 
  
The problem at hand is that the skeptic claims that power   due to the DC 
current can be very large and not detected.  There has   been no discussion 
of the AC current reading being affected by the DC so   far.  That is a 
different issue entirely.
  
 
  
I would like for them to answer the questions because then they might   
realize that their position is invalid.  I can explain this if   required.  
No one is suggesting that Rossi actually has a DC power   supply hidden 
within the wall I hope.  This would be beyond reality   since it would be 
so easy to measure with a voltmeter or any monitor that   looks at the 
voltage.  The testers did a visual look at the voltage   from what I have 
determined.
  
 
  
So, skeptics, what say you?
  
 
  
Dave
  
-Original   Message-
From: mixent 
To:   vortex-l 
Sent:   

Re: [Vo]:Re: [Vo]:Re: [Vo]:Torbjörn Hartman describes power measurments

[David Roberson]

The report included a couple of graphs on page 27.   One was power out per 
their measurement, the other power in.  The mere fact that the power out versus 
time is clearly modulated proves that the input is not constant.   The duty 
cycle can also be determined from that chart.   I am not sure that there is any 
evidence that could support their claim better.

 It does no good to assume that Rossi is scamming and you guys should 
concentrate on proving that there is a problem with the measurements.  I assume 
that you understand my explanation why the DC is not important to the input 
power measurement.  That is basic electronics.

Dave


-Original Message-
From: Joshua Cude 
To: vortex-l 
Sent: Mon, May 27, 2013 6:21 pm
Subject: [Vo]:Re: [Vo]:Re: [Vo]:Torbjörn Hartman describes power measurments


Check out these 2 videos. It's a clear demonstration of how full power can be 
transferred to a resistive load without registering current on either clamp-on 
or in-line ammeters. I don't know how it's done but I suspect high frequency, 
but the point is that just because I can't explain it, doesn't mean I must 
conclude that cheese can supply the power.


This switch could emulate Rossi's on/off cycling, and judging from input 
measurements one would conclude a duty cycle of 1/3, but looking at the 
resistive load, it would be 1:1.






http://www.youtube.com/watch?v=ovGXDDvc3ck


http://www.youtube.com/watch?v=Frp03muquAo






On Mon, May 27, 2013 at 3:55 PM, David Roberson  wrote:

If you do not understand what I have already written then it is not going to 
help to go over it again.   I leave this discussion by asking you one pertinent 
question.  Where do you think the power comes from that ends up in the 
resistor?  There is only one source and it is the AC mains.  Power from an AC 
source can only be extracted by the fundamental component of that source, 
period.  All others, including DC balance out over the long run and can not 
make a long term contribution.  Once you realize that this is true, which is 
common theory, it will become clear to you that a measurement of these two 
waveforms is all that is required.
 
Forget the nonsense about diodes faking out good AC true RMS instruments.  It 
don't happen.
 
Dave
 

-Original Message-
From: Duncan Cumming 
To: vortex-l 


Sent: Mon, May 27, 2013 4:32 pm
Subject: Re: [Vo]:Re: [Vo]:Torbjörn Hartman describes power measurments


  

OK, I will tackle this problem head-on  using the Socratic method in 
stages. 
  
  First, consider a wire carrying 100 amps of direct current, plus  one 
amp of pure sinusoidal AC current at 60Hz. What is the AC  component of the 
current?
  
  Duncan
  
  P.S. Don't worry, we will get to the diode later.
  
  On 5/27/2013 11:57 AM, David Roberson wrote:



Duncan, I hate to keep repeatingmyself that the power can be 
measured by analyzing the ACcomponents only.  When will you guys 
show why this is nottrue?  I suggest that you start with the simple 
system youproposed of a diode in series with a resistor driven by 
anAC wall socket.  Explain how it works as you say and I
promise to show you the error of your calculations.

 

Dave

-Original Message-
  From: Duncan Cumming 
  To: vortex-l 
  Sent: Mon, May 27, 2013 2:38 pm
  Subject: [Vo]:Re: [Vo]:Torbjörn Hartman describes power  
measurments
  
  

  
I am not sure if I count as askeptic, because I am not saying 
that any kind of scamwas perpetrated. I am certainly not 
suggesting thatthere was a DC power supply hidden in the wall! 
Mydoubts are related to the electrical engineering skills   
 evident in the published paper, attempting the
notoriously difficult task of measuring three phase non
sinusoidal power. Not only is the waveform nonsinusoidal, it is 
a trade secret!

I am merely saying that rectification will cause a  
  misleadingly low value of current to be registered usinga 
clamp on ammeter. Since the DC is not smooth, therewill, 
indeed, be a small reading from the ammeter butsubstantially 
lower than the actual current. This will,in turn, lead to a 
misleadingly low power measurement.

Duncan

On 5/26/2013 8:46 PM, David Roberson wrote:
  
  
  
Robin,
  
 
  
The problem at hand is that the skeptic claimsthat power 
due to the DC current can be very large 

[Vo]:Re: [Vo]:Re: [Vo]:Torbjörn Hartman describes power measurments

[Andrew]

"Power from an AC source can only be extracted by the fundamental component of 
that source, period. "

An uneducated and completely incorrect statement like that disqualifies you, in 
my view, from making any further comments about the EE aspects of this 
experiment. If you do, I urge anyone reading them to ignore them, because in 
all likelihood they will also be wrong.

Andrew
  - Original Message - 
  From: David Roberson 
  To: vortex-l@eskimo.com 
  Sent: Monday, May 27, 2013 1:55 PM
  Subject: Re: [Vo]:Re: [Vo]:Torbjörn Hartman describes power measurments


  If you do not understand what I have already written then it is not going to 
help to go over it again.   I leave this discussion by asking you one pertinent 
question.  Where do you think the power comes from that ends up in the 
resistor?  There is only one source and it is the AC mains.  Power from an AC 
source can only be extracted by the fundamental component of that source, 
period.  All others, including DC balance out over the long run and can not 
make a long term contribution.  Once you realize that this is true, which is 
common theory, it will become clear to you that a measurement of these two 
waveforms is all that is required.

  Forget the nonsense about diodes faking out good AC true RMS instruments.  It 
don't happen.

  Dave

  -Original Message-
  From: Duncan Cumming 
  To: vortex-l 
  Sent: Mon, May 27, 2013 4:32 pm
  Subject: Re: [Vo]:Re: [Vo]:Torbjörn Hartman describes power measurments


  OK, I will tackle this problem head-on using the Socratic method in stages. 

  First, consider a wire carrying 100 amps of direct current, plus one amp of 
pure sinusoidal AC current at 60Hz. What is the AC component of the current?

  Duncan

  P.S. Don't worry, we will get to the diode later.

  On 5/27/2013 11:57 AM, David Roberson wrote:

Duncan, I hate to keep repeating myself that the power can be measured by 
analyzing the AC components only.  When will you guys show why this is not 
true?  I suggest that you start with the simple system you proposed of a diode 
in series with a resistor driven by an AC wall socket.  Explain how it works as 
you say and I promise to show you the error of your calculations.

Dave
-Original Message-
From: Duncan Cumming 
To: vortex-l 
Sent: Mon, May 27, 2013 2:38 pm
Subject: [Vo]:Re: [Vo]:Torbjörn Hartman describes power measurments


I am not sure if I count as a skeptic, because I am not saying that any 
kind of scam was perpetrated. I am certainly not suggesting that there was a DC 
power supply hidden in the wall! My doubts are related to the electrical 
engineering skills evident in the published paper, attempting the notoriously 
difficult task of measuring three phase non sinusoidal power. Not only is the 
waveform non sinusoidal, it is a trade secret!

I am merely saying that rectification will cause a misleadingly low value 
of current to be registered using a clamp on ammeter. Since the DC is not 
smooth, there will, indeed, be a small reading from the ammeter but 
substantially lower than the actual current. This will, in turn, lead to a 
misleadingly low power measurement.

Duncan

On 5/26/2013 8:46 PM, David Roberson wrote:

  Robin,

  The problem at hand is that the skeptic claims that power due to the DC 
current can be very large and not detected.  There has been no discussion of 
the AC current reading being affected by the DC so far.  That is a different 
issue entirely.

  I would like for them to answer the questions because then they might 
realize that their position is invalid.  I can explain this if required.  No 
one is suggesting that Rossi actually has a DC power supply hidden within the 
wall I hope.  This would be beyond reality since it would be so easy to measure 
with a voltmeter or any monitor that looks at the voltage.  The testers did a 
visual look at the voltage from what I have determined.

  So, skeptics, what say you?

  Dave
  -Original Message-
  From: mixent 
  To: vortex-l 
  Sent: Sun, May 26, 2013 11:08 pm
  Subject: Re: [Vo]:Re: [Vo]:Re: [Vo]:Re: [Vo]:Torbjörn Hartman describes 
power measurments


In reply to  David Roberson's message of Sun, 26 May 2013 22:35:09 -0400 (EDT):
Hi,

This is a little different. A full bridge rectifier will allow for both halves
of the AC current to pass, and so it should be measured as little different to a
purely resistive load. However a single diode will only allow one half to pass,
which *may* mess up magnetic field based current measurements.
(I guess whether if does or not depends on the sophistication of the device.)
>
>Assume that you have a bridge rectifier in the blue box.  This is followed by 
>a 
filtering capacitor.  The DC is then used by the electronics connected to the 
capacitor.  Are you saying that it is not possible to determine the power input 
to

[Vo]:Re: [Vo]:Re: [Vo]:Torbjörn Hartman describes power measurments

[Joshua Cude]

Check out these 2 videos. It's a clear demonstration of how full power can
be transferred to a resistive load without registering current on either
clamp-on or in-line ammeters. I don't know how it's done but I suspect high
frequency, but the point is that just because I can't explain it, doesn't
mean I must conclude that cheese can supply the power.

This switch could emulate Rossi's on/off cycling, and judging from input
measurements one would conclude a duty cycle of 1/3, but looking at the
resistive load, it would be 1:1.



http://www.youtube.com/watch?v=ovGXDDvc3ck


http://www.youtube.com/watch?v=Frp03muquAo


On Mon, May 27, 2013 at 3:55 PM, David Roberson  wrote:

> If you do not understand what I have already written then it is not going
> to help to go over it again.   I leave this discussion by asking you one
> pertinent question.  Where do you think the power comes from that ends up
> in the resistor?  There is only one source and it is the AC mains.  Power
> from an AC source can only be extracted by the fundamental component of
> that source, period.  All others, including DC balance out over the long
> run and can not make a long term contribution.  Once you realize that this
> is true, which is common theory, it will become clear to you that a
> measurement of these two waveforms is all that is required.
>
> Forget the nonsense about diodes faking out good AC true RMS instruments.
> It don't happen.
>
> Dave
>
> -Original Message-
> From: Duncan Cumming 
> To: vortex-l 
> Sent: Mon, May 27, 2013 4:32 pm
> Subject: Re: [Vo]:Re: [Vo]:Torbjörn Hartman describes power measurments
>
>  OK, I will tackle this problem head-on using the Socratic method in
> stages.
>
> First, consider a wire carrying 100 amps of direct current, plus one amp
> of pure sinusoidal AC current at 60Hz. What is the AC component of the
> current?
>
> Duncan
>
> P.S. Don't worry, we will get to the diode later.
>
> On 5/27/2013 11:57 AM, David Roberson wrote:
>
> Duncan, I hate to keep repeating myself that the power can be measured by
> analyzing the AC components only.  When will you guys show why this is not
> true?  I suggest that you start with the simple system you proposed of a
> diode in series with a resistor driven by an AC wall socket.  Explain how
> it works as you say and I promise to show you the error of your
> calculations.
>
> Dave
> -Original Message-
> From: Duncan Cumming  
> To: vortex-l  
> Sent: Mon, May 27, 2013 2:38 pm
> Subject: [Vo]:Re: [Vo]:Torbjörn Hartman describes power measurments
>
>  I am not sure if I count as a skeptic, because I am not saying that any
> kind of scam was perpetrated. I am certainly not suggesting that there was
> a DC power supply hidden in the wall! My doubts are related to the
> electrical engineering skills evident in the published paper, attempting
> the notoriously difficult task of measuring three phase non sinusoidal
> power. Not only is the waveform non sinusoidal, it is a trade secret!
>
> I am merely saying that rectification will cause a misleadingly low value
> of current to be registered using a clamp on ammeter. Since the DC is not
> smooth, there will, indeed, be a small reading from the ammeter but
> substantially lower than the actual current. This will, in turn, lead to a
> misleadingly low power measurement.
>
> Duncan
>
> On 5/26/2013 8:46 PM, David Roberson wrote:
>
> Robin,
>
> The problem at hand is that the skeptic claims that power due to the DC
> current can be very large and not detected.  There has been no discussion
> of the AC current reading being affected by the DC so far.  That is a
> different issue entirely.
>
> I would like for them to answer the questions because then they might
> realize that their position is invalid.  I can explain this if required.
> No one is suggesting that Rossi actually has a DC power supply hidden
> within the wall I hope.  This would be beyond reality since it would be so
> easy to measure with a voltmeter or any monitor that looks at the voltage.
> The testers did a visual look at the voltage from what I have determined.
>
> So, skeptics, what say you?
>
> Dave
> -Original Message-
> From: mixent  
> To: vortex-l  
> Sent: Sun, May 26, 2013 11:08 pm
> Subject: Re: [Vo]:Re: [Vo]:Re: [Vo]:Re: [Vo]:Torbjörn Hartman describes
> power measurments
>
>  In reply to  David Roberson's message of Sun, 26 May 2013 22:35:09 -0400 
> (EDT):
> Hi,
>
> This is a little different. A full bridge rectifier will allow for both halves
> of the AC current to pass, and so it should be measured as little different 
> to a
> purely resistive load. However a single diode

Re: [Vo]:Re: [Vo]:Re: [Vo]:Torbjörn Hartman describes power measurments

[Rob Dingemans]

Hi,

On 27-5-2013 20:44, Andrew wrote:
The measurement task has been made unnecessarily difficult by 
specifying 3-phase input to the control box.  Normal single-phase 
input would suffice here, given the power levels.


Not necessarily, if all three phases have a balanced load, then the 
current through Neutral is 0 Amp!

See also: http://en.wikipedia.org/wiki/Three-phase

Here is an interesting circuit: http://www.nbtv.wyenet.co.uk/6-fasen.gif
with these voltage and current 
http://www.nbtv.wyenet.co.uk/3-fasenspanning+stroom.jpg
It "converts" the 50 Hz three phases into one of 300 Hz, which is (as 
many EE knows) a lot easier due to the smaller capacitor needed to be 
directed into DC.


For those of you who can read Dutch this circuit is discussed at this 
page: http://www.circuitsonline.net/forum/view/65574/2


Kind regards,

Rob

Re: [Vo]:Re: [Vo]:Re: [Vo]:Re: [Vo]:Torbjörn Hartman describes power measurments

[Duncan Cumming]

Yes, Hall effect clamps are readily available, I am not disputing that. 
They used to suffer from drift problems, but these problems have pretty 
much been solved. The one that you show has a 3% accuracy and 8 digits 
of drift - not bad.


The only info I have about Rossi came from a single post on this list, 
his opinion may well be different than it appears - for one thing, there 
were some translation problems.


So we are in agreement on these points. Here are a few more points - can 
we agree on these as well?


1) I am not saying that Rossi is now, or has ever previously done 
anything fraudulent.
2) I am not saying that anything except regular AC voltage was present 
at the wall socket during the demo.
3) I am not claiming that the lab was in any way "re-wired" in an 
attempt to perpetrate a fraud.
4) I am not speculating that either bridge rectifiers or smoothing 
capacitors may have been used.


Duncan

On 5/27/2013 11:36 AM, Jed Rothwell wrote:

Duncan Cumming wrote:

(Of course, DC rated Hall effect clamps are available but were
not used in the demo, partially because Rossi appears to
believe that an AC outlet will only deliver AC current - this
is far from being the case).

1. People have been measuring DC amperage by measuring a magnetic 
field since 1820.


2. These are Hall effect clamps. See the specifications They are rated 
for very low DC power:


http://www.industrial-needs.com/manual/manual-clamp-meter-pce-cd3.pdf

3. Rossi played no role in this. His beliefs about AC are probably not 
as you describe them, but in any case he had no say in the matter.


- Jed

Re: [Vo]:Re: [Vo]:Re: [Vo]:Re: [Vo]:Re: [Vo]:Re: [Vo]:Torbjörn Hartman describes power measurments

[David Roberson]

You are correct Eric.  For some reason the skeptics amoung us do not want to 
understand this issue.  I suspect that it is some form of game they are playing.

Dave


-Original Message-
From: Eric Walker 
To: vortex-l 
Sent: Mon, May 27, 2013 3:38 pm
Subject: [Vo]:Re: [Vo]:Re: [Vo]:Re: [Vo]:Re: [Vo]:Re: [Vo]:Torbjörn Hartman 
describes power measurments


On Mon, May 27, 2013 at 12:33 PM, Andrew  wrote:




No separate DC power source is necessary if Duncan's diode fudge is used.



Yes, but the point is that you'd need to intentionally tamper with the mains to 
pull it off, i.e., it implies fraud, if I'm understanding the diode fudge, in 
light of the DC short to ground in the transformer.


Put another way, is there any way to draw hidden current from the mains that 
does not entail tampering between the transformer and the mains?  Earlier 
Duncan seemed to think that there was a way, but this particular approach does 
not appear to be an example.  But perhaps I'm misunderstanding an important 
detail?


Eric

Re: [Vo]:Re: [Vo]:Re: [Vo]:Re: [Vo]:Torbjörn Hartman describes power measurments

[David Roberson]

You are back to the scam by DC supply trick.  Of course, that is different than 
the other diode issue that has been shown to be in error.

I do not want to take up any more of the vortex bandwidth explaining these to 
you.  I beg forgiveness of the vortex members for making so many posts and will 
go back to answering only those that are sensible.

I might handle a couple more along the present line just to smooth the 
transition. :-)

Dave


-Original Message-
From: Andrew 
To: vortex-l 
Sent: Mon, May 27, 2013 3:28 pm
Subject: [Vo]:Re: [Vo]:Re: [Vo]:Re: [Vo]:Torbjörn Hartman describes power 
measurments



You mean an annoyance like the advance of the perihelion of Mercury? :)
 
OK, once again you furiously misunderstand. The isolation capacitor is in 
series between the grid transformer and the wall plug. Behind the wall plug, 
downstream of that capacitor, a DC power supply is connected in a T 
configuration. It's possible to do this but you can't just attach a wire. Some 
circuitry is involved to provide a DC shift without compromising the AC and 
without blowing up the DC power supply.
 
Andrew
  
- Original Message - 
  
From:   David   Roberson 
  
To: vortex-l@eskimo.com 
  
Sent: Monday, May 27, 2013 11:50 AM
  
Subject: Re: [Vo]:Re: [Vo]:Re:   [Vo]:Torbjörn Hartman describes power 
measurments
  


  
Forget the RF for now.  That is   another annoyance.
  
 
  
Please explain how much DC power will be propagated through that   "isolation 
capacitor".   Putting these in place will ensure that no   DC can find its way 
into the device.
  
 
  
Dave
  
  
  
-Original   Message-
From: Andrew 
To: vortex-l   
Sent:   Mon, May 27, 2013 2:31 pm
Subject: [Vo]:Re: [Vo]:Re: [Vo]:Torbjörn Hartman   describes power measurments

  
  
  
Dave, there are a couple of things wrong with your analysis. First off,   the 
insertion of an isolation capacitor between the main grid transformer and   the 
plug takes care of your "short circuit" problem. And then there's the   
possibility of injection of RF also, also capacitatively coupled into the plug  
 lines.
  
 
  
Andrew
  

- Original Message - 

From: David Roberson 

To: vortex-l@eskimo.com 

Sent: Monday, May 27, 2013 11:17 AM

Subject: Re: [Vo]:Re: [Vo]:Torbjörn Hartman describes power measurments




Duncan,

 

Read some of my recent posts and you will see why it will not work.  Unless 
Rossi has hidden a DC source behind the wall plug it does not matter how 
much DC flows into the control box due to rectification.  The input power 
is uniquely defined by the AC voltage and AC current waveforms leaving the 
wall.

 

You are mistaken about the DC effects since the transformer driving the 
building should present a DC short to ground.  If not, I suspect major code 
violations are present.

 

If you continue to insist that Rossi is conducting a scam by altering the 
power socket then there is no reason to continue with this discussion.  If 
you honestly believe that there is some form of DC trick that can be done 
with the control box, then we can clear up this misunderstanding.  Your 
call.

 

Dave



-Original Message-
From: Duncan Cumming 
To: vortex-l 
Sent: Mon, May 27, 2013 1:59 pm
Subject: [Vo]:Re: [Vo]:Torbjörn Hartman describes power measurments




Actually it is not beyond the bounds of possibility to set up such a 
demonstration. What exactly do you have in mind, and who would be 
interested in seeing such a demo? Do you have any contacts on the Rossi 
team?

I don't think Rossi would travel to the USA to see such a demo.
Electrical Engineers already know that a diode will convert AC to DC.
Pretty much all scientists know that an AC current clamp will not measure 
DC. (Of course, DC rated Hall effect clamps are available but were not used 
in the demo, partially because Rossi appears to believe that an AC outlet 
will only deliver AC current - this is far from being the case).

So who would your intended audience be for such a demonstration?

Duncan

On 5/26/2013 7:26 PM, David Roberson wrote:


  
Not my position.  You need to show how it was done.
  
 
  
Dave
  
-Original   Message-
From: Duncan Cumming 
To:   vortex-l 
Sent:   Sun, May 26, 2013 9:47 pm
Subject: Re: [Vo]:Re: [Vo]:Torbjörn Hartman   describes power measurments

  
  
  
So is it your position that a current clamp   without a Hall effect unit 
can measure DC? Mine is that it   cannot.

Duncan

On 5/26/2013 5:34 PM, David Roberson   wrote:

  

How do we know that your diode trick will actually do what you think?  
You need to prove that this is possible, otherwise anyone can make the 
assumptio

[Vo]:Re: [Vo]:Re: [Vo]:Re: [Vo]:Re: [Vo]:Re: [Vo]:Torbjörn Hartman describes power measurments

[Eric Walker]

On Mon, May 27, 2013 at 12:33 PM, Andrew  wrote:

**
> No separate DC power source is necessary if Duncan's diode fudge is used.
>

Yes, but the point is that you'd need to intentionally tamper with the
mains to pull it off, i.e., it implies fraud, if I'm understanding the
diode fudge, in light of the DC short to ground in the transformer.

Put another way, is there any way to draw hidden current from the mains
that does not entail tampering between the transformer and the mains?
 Earlier Duncan seemed to think that there was a way, but this particular
approach does not appear to be an example.  But perhaps I'm
misunderstanding an important detail?

Eric

[Vo]:Re: [Vo]:Re: [Vo]:Re: [Vo]:Re: [Vo]:Torbjörn Hartman describes power measurments

[Andrew]

No separate DC power source is necessary if Duncan's diode fudge is used.

Andrew
  - Original Message - 
  From: Eric Walker 
  To: vortex-l@eskimo.com 
  Sent: Monday, May 27, 2013 12:09 PM
  Subject: [Vo]:Re: [Vo]:Re: [Vo]:Re: [Vo]:Torbjörn Hartman describes power 
measurments


  On Mon, May 27, 2013 at 11:31 AM, Andrew  wrote:


Dave, there are a couple of things wrong with your analysis. First off, the 
insertion of an isolation capacitor between the main grid transformer and the 
plug takes care of your "short circuit" problem. And then there's the 
possibility of injection of RF also, also capacitatively coupled into the plug 
lines.


  Assume that the three-phase power coming into the transformer has not been 
tampered with (seems like a safe assumption, but you never know ;).  Assume as 
well that Dave is correct that at the transformer there will be a DC short to 
ground.  With these assumptions, am I correct in drawing one of the two 
following conclusions?
a.. No hidden DC to the E-Cat resistors exceeded the power measured at the 
mains; or
b.. Rossi or an associate intentionally added an isolation capacitor (i.e., 
tampered with the mains). 
  Eric

[Vo]:Re: [Vo]:Re: [Vo]:Re: [Vo]:Torbjörn Hartman describes power measurments

[Andrew]

You mean an annoyance like the advance of the perihelion of Mercury? :)

OK, once again you furiously misunderstand. The isolation capacitor is in 
series between the grid transformer and the wall plug. Behind the wall plug, 
downstream of that capacitor, a DC power supply is connected in a T 
configuration. It's possible to do this but you can't just attach a wire. Some 
circuitry is involved to provide a DC shift without compromising the AC and 
without blowing up the DC power supply.

Andrew
  - Original Message - 
  From: David Roberson 
  To: vortex-l@eskimo.com 
  Sent: Monday, May 27, 2013 11:50 AM
  Subject: Re: [Vo]:Re: [Vo]:Re: [Vo]:Torbjörn Hartman describes power 
measurments


  Forget the RF for now.  That is another annoyance.

  Please explain how much DC power will be propagated through that "isolation 
capacitor".   Putting these in place will ensure that no DC can find its way 
into the device.

  Dave
  -Original Message-
  From: Andrew 
  To: vortex-l 
  Sent: Mon, May 27, 2013 2:31 pm
  Subject: [Vo]:Re: [Vo]:Re: [Vo]:Torbjörn Hartman describes power measurments


  Dave, there are a couple of things wrong with your analysis. First off, the 
insertion of an isolation capacitor between the main grid transformer and the 
plug takes care of your "short circuit" problem. And then there's the 
possibility of injection of RF also, also capacitatively coupled into the plug 
lines.

  Andrew
- Original Message - 
From: David Roberson 
To: vortex-l@eskimo.com 
Sent: Monday, May 27, 2013 11:17 AM
Subject: Re: [Vo]:Re: [Vo]:Torbjörn Hartman describes power measurments


Duncan,

Read some of my recent posts and you will see why it will not work.  Unless 
Rossi has hidden a DC source behind the wall plug it does not matter how much 
DC flows into the control box due to rectification.  The input power is 
uniquely defined by the AC voltage and AC current waveforms leaving the wall.

You are mistaken about the DC effects since the transformer driving the 
building should present a DC short to ground.  If not, I suspect major code 
violations are present.

If you continue to insist that Rossi is conducting a scam by altering the 
power socket then there is no reason to continue with this discussion.  If you 
honestly believe that there is some form of DC trick that can be done with the 
control box, then we can clear up this misunderstanding.  Your call.

Dave
-Original Message-
From: Duncan Cumming 
To: vortex-l 
Sent: Mon, May 27, 2013 1:59 pm
Subject: [Vo]:Re: [Vo]:Torbjörn Hartman describes power measurments


Actually it is not beyond the bounds of possibility to set up such a 
demonstration. What exactly do you have in mind, and who would be interested in 
seeing such a demo? Do you have any contacts on the Rossi team?

I don't think Rossi would travel to the USA to see such a demo.
Electrical Engineers already know that a diode will convert AC to DC.
Pretty much all scientists know that an AC current clamp will not measure 
DC. (Of course, DC rated Hall effect clamps are available but were not used in 
the demo, partially because Rossi appears to believe that an AC outlet will 
only deliver AC current - this is far from being the case).

So who would your intended audience be for such a demonstration?

Duncan

On 5/26/2013 7:26 PM, David Roberson wrote:

  Not my position.  You need to show how it was done.

  Dave
  -Original Message-
  From: Duncan Cumming 
  To: vortex-l 
  Sent: Sun, May 26, 2013 9:47 pm
  Subject: Re: [Vo]:Re: [Vo]:Torbjörn Hartman describes power measurments


  So is it your position that a current clamp without a Hall effect unit 
can measure DC? Mine is that it cannot.

  Duncan

  On 5/26/2013 5:34 PM, David Roberson wrote:

How do we know that your diode trick will actually do what you think?  
You need to prove that this is possible, otherwise anyone can make the 
assumption that it might not work just as with the ECAT tests.  If you do not 
prove that this will work, then why should we accept it as a possibility?

A lot of time and energy is being wasted trying to see if bull frogs 
can fly.  Some might actually be born with wings.  Have we proven that none of 
them can fly?

Rossi and the testers have done a lot to prove that the ECAT works.   
No one has proven that it does not.  The only offers from the other side of the 
table assume fraud.  Is this a valid position for them to take?

Dave
-Original Message-
From: Duncan Cumming 
To: vortex-l 
Sent: Sun, May 26, 2013 8:18 pm
Subject: Re: [Vo]:Re: [Vo]:Torbjörn Hartman describes power measurments


I am not trying to assert anything as fact. I am merely pointing out 
that a simple diode inside the controller box 

Re: [Vo]:Re: [Vo]:Re: [Vo]:Re: [Vo]:Torbjörn Hartman describes power measurments

[David Roberson]

Eric, the isolation capacitor does not serve a purpose in this discussion.  It 
would ensure that no DC gets through.

You assumption that no DC power exceeds the input power measured at the mains 
should be accurate.  It would be very difficult to keep excess DC flowing at a 
higher level for the long periods associated with the PWM.

Dave


-Original Message-
From: Eric Walker 
To: vortex-l 
Sent: Mon, May 27, 2013 3:09 pm
Subject: [Vo]:Re: [Vo]:Re: [Vo]:Re: [Vo]:Torbjörn Hartman describes power 
measurments


On Mon, May 27, 2013 at 11:31 AM, Andrew  wrote:




Dave, there are a couple of things wrong with your analysis. First off, the 
insertion of an isolation capacitor between the main grid transformer and the 
plug takes care of your "short circuit" problem. And then there's the 
possibility of injection of RF also, also capacitatively coupled into the plug 
lines.





Assume that the three-phase power coming into the transformer has not been 
tampered with (seems like a safe assumption, but you never know ;).  Assume as 
well that Dave is correct that at the transformer there will be a DC short to 
ground.  With these assumptions, am I correct in drawing one of the two 
following conclusions?

No hidden DC to the E-Cat resistors exceeded the power measured at the mains; or
Rossi or an associate intentionally added an isolation capacitor (i.e., 
tampered with the mains).

Eric

[Vo]:Re: [Vo]:Re: [Vo]:Re: [Vo]:Torbjörn Hartman describes power measurments

[Eric Walker]

On Mon, May 27, 2013 at 11:31 AM, Andrew  wrote:

**
> Dave, there are a couple of things wrong with your analysis. First off,
> the insertion of an isolation capacitor between the main grid transformer
> and the plug takes care of your "short circuit" problem. And then there's
> the possibility of injection of RF also, also capacitatively coupled into
> the plug lines.
>

Assume that the three-phase power coming into the transformer has not been
tampered with (seems like a safe assumption, but you never know ;).  Assume
as well that Dave is correct that at the transformer there will be a DC
short to ground.  With these assumptions, am I correct in drawing one of
the two following conclusions?

   - No hidden DC to the E-Cat resistors exceeded the power measured at the
   mains; or
   - Rossi or an associate intentionally added an isolation capacitor
   (i.e., tampered with the mains).

Eric

Re: [Vo]:Re: [Vo]:Re: [Vo]:Re: [Vo]:Torbjörn Hartman describes power measurments

[David Roberson]

OK, you are mistaken with this analysis.  The input power is determined by the 
AC 50/60 hertz fundamental and the fundamental component of the current flowing 
from the wall socket.  The DC just comes along for the ride since it is 
converted from some of the input AC power.

And yes, you can measure accurately the input power with a good instrument that 
does not measure the DC component.  I can explain further if you like.

Dave


-Original Message-
From: Duncan Cumming 
To: vortex-l 
Sent: Mon, May 27, 2013 2:11 pm
Subject: Re: [Vo]:Re: [Vo]:Re: [Vo]:Re: [Vo]:Torbjörn Hartman describes power 
measurments


  

What I am proposing is a lot simpler  than that. No bridge rectifier, no 
capacitor, just a simple diode.  I am saying that given a diode in series 
with a resistor, it is  not possible to measure the power using a clamp on 
ammeter.
  
  I am not suggesting that anybody has performed a scam. I am  
suggesting that the equipment used would not have measured the  power 
consumed by the resistor if rectification were present in  the controller 
box.
  
  Is there anybody reading this that can do SPICE simulations? Might  
it be possible to simulate a resistor in series with a diode and  determine 
the actual and apparent power if an AC coupled current  meter is used?
  
  Duncan
  
  P.S. I never mentioned either bridge rectifiers or capacitors. In  
the case of a bridge rectifier type power supply, then a clamp on  ammeter 
will work OK. I do not suspect such a thing in the demo.
  
  On 5/26/2013 7:35 PM, David Roberson wrote:



Assume that you have a bridgerectifier in the blue box.  This is 
followed by a filteringcapacitor.  The DC is then used by the 
electronics connectedto the capacitor.  Are you saying that it is 
not possible todetermine the power input to this type of network by 
   measuring the input AC voltage and current?  Or are you
saying that someone has performed a scam and put a DC supplyin 
series with the normal AC voltage?

 

You do know that this could easily be measured by a simple  DC 
voltmeter, right?

 

Dave

-Original Message-
  From: Duncan Cumming 
  To: vortex-l 
  Sent: Sun, May 26, 2013 10:01 pm
  Subject: Re: [Vo]:Re: [Vo]:Re: [Vo]:Re: [Vo]:Torbjörn Hartman 
 describes power measurments
  
  

  
Almost. The power being fedto the heater exceeds that measured 
at the wall, becausethe sensor used (an AC current clamp) 
cannot sense thedirect current being drawn from the wall socket.

Some people find the difference between current and 
   voltage confusing. What I am saying here is that if you
connect a resistor in series with a diode to a wallsocket, then 
the CURRENT drawn is direct even though theVOLTAGE at the 
socket is alternating. (Rossi does notseem to understand this 
concept judging by his messagethat got posted today). So unless 
you use a DC ratedcurrent meter (such as a shunt) you will not 
sense allof the current, and hence power, drawn from the wall   
 socket. 

The electrical power meter in your house certainloy IS  
  rated for DC, so you will certainly be BILLED for the
power even though you didn't measure it yourself!

V = IR
Power = Voltage * Current * Power Factor

Duncan

On 5/26/2013 5:57 PM, Eric Walker wrote:
  
  

I wrote:
  


  

On Sun, May 26, 2013 at 5:18 PM,  Duncan Cumming 
  wrote:   
   
 

  

  

I am not trying to assert  anything as 
fact. I am merely  pointing out that a 
simple diode  inside the controller box (to 
 which access was forbidden by  
Rossi) COULD HAVE given the 
 observed results. I am NOT saying  
that it, in fact, did, merely  speculating 
that it could

[Vo]:Re: [Vo]:Re: [Vo]:Torbjörn Hartman describes power measurments

[Eric Walker]

On Mon, May 27, 2013 at 10:59 AM, Duncan Cumming wrote:

 I don't think Rossi would travel to the USA to see such a demo.
>

How about a YouTube video?

Eric

Re: [Vo]:Re: [Vo]:Re: [Vo]:Torbjörn Hartman describes power measurments

[David Roberson]

Forget the RF for now.  That is another annoyance.

Please explain how much DC power will be propagated through that "isolation 
capacitor".   Putting these in place will ensure that no DC can find its way 
into the device.

Dave


-Original Message-
From: Andrew 
To: vortex-l 
Sent: Mon, May 27, 2013 2:31 pm
Subject: [Vo]:Re: [Vo]:Re: [Vo]:Torbjörn Hartman describes power measurments



Dave, there are a couple of things wrong with your analysis. First off, the 
insertion of an isolation capacitor between the main grid transformer and the 
plug takes care of your "short circuit" problem. And then there's the 
possibility of injection of RF also, also capacitatively coupled into the plug 
lines.
 
Andrew
  
- Original Message - 
  
From:   David   Roberson 
  
To: vortex-l@eskimo.com 
  
Sent: Monday, May 27, 2013 11:17 AM
  
Subject: Re: [Vo]:Re: [Vo]:Torbjörn   Hartman describes power measurments
  


  
Duncan,
  
 
  
Read some of my recent posts and you will see why it will not   work.  Unless 
Rossi has hidden a DC source behind the wall plug it does   not matter how much 
DC flows into the control box due to rectification.The input power is 
uniquely defined by the AC voltage and AC current waveforms   leaving the wall.
  
 
  
You are mistaken about the DC effects since the transformer driving the   
building should present a DC short to ground.  If not, I suspect major   code 
violations are present.
  
 
  
If you continue to insist that Rossi is conducting a scam by altering the   
power socket then there is no reason to continue with this   discussion.  If 
you honestly believe that there is some form of DC trick   that can be done 
with the control box, then we can clear up this   misunderstanding.  Your call.
  
 
  
Dave
  
  
  
-Original   Message-
From: Duncan Cumming 
To:   vortex-l 
Sent: Mon, May 27, 2013 1:59   pm
Subject: [Vo]:Re: [Vo]:Torbjörn Hartman describes power   measurments

  
  
  
Actually it is not beyond the bounds of possibility   to set up such a 
demonstration. What exactly do you have in mind, and who   would be interested 
in seeing such a demo? Do you have any contacts on the   Rossi team?

I don't think Rossi would travel to the USA to see such a   demo.
Electrical Engineers already know that a diode will convert AC to   DC.
Pretty much all scientists know that an AC current clamp will not   measure DC. 
(Of course, DC rated Hall effect clamps are available but were not   used in 
the demo, partially because Rossi appears to believe that an AC outlet   will 
only deliver AC current - this is far from being the case).

So who   would your intended audience be for such a   demonstration?

Duncan

On 5/26/2013 7:26 PM, David Roberson   wrote:

  

Not my position.  You need to show how it was done.

 

Dave

-Original Message-
From: Duncan Cumming 
To: vortex-l 
Sent: Sun, May 26, 2013 9:47 pm
Subject: Re: [Vo]:Re: [Vo]:Torbjörn Hartman describes power measurments




So is it your position that a current clamp without a Hall effect unit can 
measure DC? Mine is that it cannot.

Duncan

On 5/26/2013 5:34 PM, David Roberson wrote:


  
How do we know that your diode trick   will actually do what you think?  
You need to prove that this is   possible, otherwise anyone can make the 
assumption that it might not work   just as with the ECAT tests.  If you do 
not prove that this will   work, then why should we accept it as a 
possibility?
  
 
  
A lot of time and energy is being wasted trying to see if bull frogs   can 
fly.  Some might actually be born with wings.  Have we   proven that none 
of them can fly?
  
 
  
Rossi and the testers have done a lot to prove that the ECAT   works.   No 
one has proven that it does not.  The only   offers from the other side of 
the table assume fraud.  Is this a   valid position for them to take?
  
 
  
Dave
  
-Original   Message-
From: Duncan Cumming 
To:   vortex-l 
Sent:   Sun, May 26, 2013 8:18 pm
Subject: Re: [Vo]:Re: [Vo]:Torbjörn Hartman   describes power measurments

  
  
  
I am not trying to assert anything as fact. I   am merely pointing out that 
a simple diode inside the controller box (to   which access was forbidden 
by Rossi) COULD HAVE given the observed   results. I am NOT saying that it, 
in fact, did, merely speculating that it   could have.

For any scientific experiment, the onus is on the   experimenters to 
produce the result. The best way to do this is to provide   sufficient 
information for others to replicate the   experiment.

Duncan

On 5/26/2013 5:07 PM, David Roberson   wrote:

  

Perhaps you should build one of these scam machines and prove that it 
will work without being detected.  That would be the best w

[Vo]:Re: [Vo]:Re: [Vo]:Torbjörn Hartman describes power measurments

[Andrew]

The measurement task has been made unnecessarily difficult by specifying 
3-phase input to the control box.  Normal single-phase input would suffice 
here, given the power levels.

They redesigned the control box between the December and March tests, changing 
the output from 3-phase to single-phase. I would suggest that they do the same 
sort of thing on the input side. Measurement ambiguity would be reduced as a 
consequence.

Just a suggestion to the Rossi team (on the off-chance they are reading any of 
this).

Andrew
  - Original Message - 
  From: Duncan Cumming 
  To: vortex-l@eskimo.com 
  Sent: Monday, May 27, 2013 11:38 AM
  Subject: [Vo]:Re: [Vo]:Torbjörn Hartman describes power measurments


  I am not sure if I count as a skeptic, because I am not saying that any kind 
of scam was perpetrated. I am certainly not suggesting that there was a DC 
power supply hidden in the wall! My doubts are related to the electrical 
engineering skills evident in the published paper, attempting the notoriously 
difficult task of measuring three phase non sinusoidal power. Not only is the 
waveform non sinusoidal, it is a trade secret!

  I am merely saying that rectification will cause a misleadingly low value of 
current to be registered using a clamp on ammeter. Since the DC is not smooth, 
there will, indeed, be a small reading from the ammeter but substantially lower 
than the actual current. This will, in turn, lead to a misleadingly low power 
measurement.

  Duncan

  On 5/26/2013 8:46 PM, David Roberson wrote:

Robin,

The problem at hand is that the skeptic claims that power due to the DC 
current can be very large and not detected.  There has been no discussion of 
the AC current reading being affected by the DC so far.  That is a different 
issue entirely.

I would like for them to answer the questions because then they might 
realize that their position is invalid.  I can explain this if required.  No 
one is suggesting that Rossi actually has a DC power supply hidden within the 
wall I hope.  This would be beyond reality since it would be so easy to measure 
with a voltmeter or any monitor that looks at the voltage.  The testers did a 
visual look at the voltage from what I have determined.

So, skeptics, what say you?

Dave
-Original Message-
From: mixent 
To: vortex-l 
Sent: Sun, May 26, 2013 11:08 pm
Subject: Re: [Vo]:Re: [Vo]:Re: [Vo]:Re: [Vo]:Torbjörn Hartman describes 
power measurments


In reply to  David Roberson's message of Sun, 26 May 2013 22:35:09 -0400 (EDT):
Hi,

This is a little different. A full bridge rectifier will allow for both halves
of the AC current to pass, and so it should be measured as little different to a
purely resistive load. However a single diode will only allow one half to pass,
which *may* mess up magnetic field based current measurements.
(I guess whether if does or not depends on the sophistication of the device.)
>
>Assume that you have a bridge rectifier in the blue box.  This is followed by 
>a 
filtering capacitor.  The DC is then used by the electronics connected to the 
capacitor.  Are you saying that it is not possible to determine the power input 
to this type of network by measuring the input AC voltage and current?  Or are 
you saying that someone has performed a scam and put a DC supply in series with 
the normal AC voltage?
>
>You do know that this could easily be measured by a simple DC voltmeter, right?
>
>Dave
[snip]
Regards,

Robin van Spaandonk

http://rvanspaa.freehostia.com/project.html

[Vo]:Re: [Vo]:Re: [Vo]:Re: [Vo]:Torbjörn Hartman describes power measurments

[Jed Rothwell]

Duncan Cumming wrote:

> (Of course, DC rated Hall effect clamps are available but were not used in
> the demo, partially because Rossi appears to believe that an AC outlet will
> only deliver AC current - this is far from being the case).
>
> 1. People have been measuring DC amperage by measuring a magnetic field
since 1820.

2. These are Hall effect clamps. See the specifications They are rated for
very low DC power:

http://www.industrial-needs.com/manual/manual-clamp-meter-pce-cd3.pdf

3. Rossi played no role in this. His beliefs about AC are probably not as
you describe them, but in any case he had no say in the matter.

- Jed

[Vo]:Re: [Vo]:Re: [Vo]:Torbjörn Hartman describes power measurments

[Andrew]

Dave, there are a couple of things wrong with your analysis. First off, the 
insertion of an isolation capacitor between the main grid transformer and the 
plug takes care of your "short circuit" problem. And then there's the 
possibility of injection of RF also, also capacitatively coupled into the plug 
lines.

Andrew
  - Original Message - 
  From: David Roberson 
  To: vortex-l@eskimo.com 
  Sent: Monday, May 27, 2013 11:17 AM
  Subject: Re: [Vo]:Re: [Vo]:Torbjörn Hartman describes power measurments


  Duncan,

  Read some of my recent posts and you will see why it will not work.  Unless 
Rossi has hidden a DC source behind the wall plug it does not matter how much 
DC flows into the control box due to rectification.  The input power is 
uniquely defined by the AC voltage and AC current waveforms leaving the wall.

  You are mistaken about the DC effects since the transformer driving the 
building should present a DC short to ground.  If not, I suspect major code 
violations are present.

  If you continue to insist that Rossi is conducting a scam by altering the 
power socket then there is no reason to continue with this discussion.  If you 
honestly believe that there is some form of DC trick that can be done with the 
control box, then we can clear up this misunderstanding.  Your call.

  Dave
  -Original Message-
  From: Duncan Cumming 
  To: vortex-l 
  Sent: Mon, May 27, 2013 1:59 pm
  Subject: [Vo]:Re: [Vo]:Torbjörn Hartman describes power measurments


  Actually it is not beyond the bounds of possibility to set up such a 
demonstration. What exactly do you have in mind, and who would be interested in 
seeing such a demo? Do you have any contacts on the Rossi team?

  I don't think Rossi would travel to the USA to see such a demo.
  Electrical Engineers already know that a diode will convert AC to DC.
  Pretty much all scientists know that an AC current clamp will not measure DC. 
(Of course, DC rated Hall effect clamps are available but were not used in the 
demo, partially because Rossi appears to believe that an AC outlet will only 
deliver AC current - this is far from being the case).

  So who would your intended audience be for such a demonstration?

  Duncan

  On 5/26/2013 7:26 PM, David Roberson wrote:

Not my position.  You need to show how it was done.

Dave
-Original Message-
From: Duncan Cumming 
To: vortex-l 
Sent: Sun, May 26, 2013 9:47 pm
Subject: Re: [Vo]:Re: [Vo]:Torbjörn Hartman describes power measurments


So is it your position that a current clamp without a Hall effect unit can 
measure DC? Mine is that it cannot.

Duncan

On 5/26/2013 5:34 PM, David Roberson wrote:

  How do we know that your diode trick will actually do what you think?  
You need to prove that this is possible, otherwise anyone can make the 
assumption that it might not work just as with the ECAT tests.  If you do not 
prove that this will work, then why should we accept it as a possibility?

  A lot of time and energy is being wasted trying to see if bull frogs can 
fly.  Some might actually be born with wings.  Have we proven that none of them 
can fly?

  Rossi and the testers have done a lot to prove that the ECAT works.   No 
one has proven that it does not.  The only offers from the other side of the 
table assume fraud.  Is this a valid position for them to take?

  Dave
  -Original Message-
  From: Duncan Cumming 
  To: vortex-l 
  Sent: Sun, May 26, 2013 8:18 pm
  Subject: Re: [Vo]:Re: [Vo]:Torbjörn Hartman describes power measurments


  I am not trying to assert anything as fact. I am merely pointing out that 
a simple diode inside the controller box (to which access was forbidden by 
Rossi) COULD HAVE given the observed results. I am NOT saying that it, in fact, 
did, merely speculating that it could have.

  For any scientific experiment, the onus is on the experimenters to 
produce the result. The best way to do this is to provide sufficient 
information for others to replicate the experiment.

  Duncan

  On 5/26/2013 5:07 PM, David Roberson wrote:

Perhaps you should build one of these scam machines and prove that it 
will work without being detected.  That would be the best way to show that it 
is possible.  Why should we accept this assertion as fact any more than 
believing that the testers missed finding the scam?

We can spend an equal amount of time knocking down any theory that is 
put forth as others can spend assuming they are real.

Dave
-Original Message-
From: Duncan Cumming 
To: vortex-l 
Sent: Sun, May 26, 2013 7:59 pm
Subject: [Vo]:Re: [Vo]:Torbjörn Hartman describes power measurments


"The only possibility to fool the power-meter then is to raise the DC 
voltage on all the four lines"

This turns out not to be the case. You could also draw DC current 
through any of the lines,

Re: [Vo]:Re: [Vo]:Re: [Vo]:Re: [Vo]:Re: [Vo]:Re: [Vo]:Torbjörn Hartman describes power measurments

[David Roberson]

Half wave rectifiers are not the way to go.  They have been all but abandoned 
in the electronic world because of the issues you have found.  Full wave 
bridges eliminate the DC component from the mix and should be used.

This does not suggest that accurate power measurements can not be obtained from 
the AC waveforms.  This can be done and is the reason for much of the 
discussion taking place.

Dave


-Original Message-
From: Alain Sepeda 
To: vortex-l 
Sent: Mon, May 27, 2013 2:19 pm
Subject: [Vo]:Re: [Vo]:Re: [Vo]:Re: [Vo]:Re: [Vo]:Re: [Vo]:Torbjörn Hartman 
describes power measurments




http://www.circuitstoday.com/half-wave-rectifiers



(ii)Disadvantages:1. The output current in the load contains, in addition to dc 
component, ac components of basic frequency equal to that of the input voltage 
frequency. Ripple factor is high and an elaborate filtering is, therefore, 
required to give steady dc output.
(iii)2.The power output and, therefore, rectification efficiency is quite low. 
This is due to the fact that power is delivered only half the time.
(iv)3.Transformer utilization factor is low.
(v)4.DC saturation of transformer core resulting in magnetizing current and 
hysteresis losses and generation of harmonics.
transformer are not perfect and saturation solve the DC component problem.


2013/5/27 David Roberson 

The concept mentioned below by Duncan is not correct.  The DC current that 
flows into the resistor from the wall socket finds a short circuit to ground in 
the power transformer center tap in most cases.
 

[Vo]:Re: [Vo]:Re: [Vo]:Torbjörn Hartman describes power measurments

[Andrew]

Word up Duncan - Rossi currently resides in Florida!
You could call it "The Power Sneaker".

Andrew
  - Original Message - 
  From: Duncan Cumming 
  To: vortex-l@eskimo.com 
  Sent: Monday, May 27, 2013 10:59 AM
  Subject: [Vo]:Re: [Vo]:Torbjörn Hartman describes power measurments


  Actually it is not beyond the bounds of possibility to set up such a 
demonstration. What exactly do you have in mind, and who would be interested in 
seeing such a demo? Do you have any contacts on the Rossi team?

  I don't think Rossi would travel to the USA to see such a demo.
  Electrical Engineers already know that a diode will convert AC to DC.
  Pretty much all scientists know that an AC current clamp will not measure DC. 
(Of course, DC rated Hall effect clamps are available but were not used in the 
demo, partially because Rossi appears to believe that an AC outlet will only 
deliver AC current - this is far from being the case).

  So who would your intended audience be for such a demonstration?

  Duncan

  On 5/26/2013 7:26 PM, David Roberson wrote:

Not my position.  You need to show how it was done.

Dave
-Original Message-
From: Duncan Cumming 
To: vortex-l 
Sent: Sun, May 26, 2013 9:47 pm
Subject: Re: [Vo]:Re: [Vo]:Torbjörn Hartman describes power measurments


So is it your position that a current clamp without a Hall effect unit can 
measure DC? Mine is that it cannot.

Duncan

On 5/26/2013 5:34 PM, David Roberson wrote:

  How do we know that your diode trick will actually do what you think?  
You need to prove that this is possible, otherwise anyone can make the 
assumption that it might not work just as with the ECAT tests.  If you do not 
prove that this will work, then why should we accept it as a possibility?

  A lot of time and energy is being wasted trying to see if bull frogs can 
fly.  Some might actually be born with wings.  Have we proven that none of them 
can fly?

  Rossi and the testers have done a lot to prove that the ECAT works.   No 
one has proven that it does not.  The only offers from the other side of the 
table assume fraud.  Is this a valid position for them to take?

  Dave
  -Original Message-
  From: Duncan Cumming 
  To: vortex-l 
  Sent: Sun, May 26, 2013 8:18 pm
  Subject: Re: [Vo]:Re: [Vo]:Torbjörn Hartman describes power measurments


  I am not trying to assert anything as fact. I am merely pointing out that 
a simple diode inside the controller box (to which access was forbidden by 
Rossi) COULD HAVE given the observed results. I am NOT saying that it, in fact, 
did, merely speculating that it could have.

  For any scientific experiment, the onus is on the experimenters to 
produce the result. The best way to do this is to provide sufficient 
information for others to replicate the experiment.

  Duncan

  On 5/26/2013 5:07 PM, David Roberson wrote:

Perhaps you should build one of these scam machines and prove that it 
will work without being detected.  That would be the best way to show that it 
is possible.  Why should we accept this assertion as fact any more than 
believing that the testers missed finding the scam?

We can spend an equal amount of time knocking down any theory that is 
put forth as others can spend assuming they are real.

Dave
-Original Message-
From: Duncan Cumming 
To: vortex-l 
Sent: Sun, May 26, 2013 7:59 pm
Subject: [Vo]:Re: [Vo]:Torbjörn Hartman describes power measurments


"The only possibility to fool the power-meter then is to raise the DC 
voltage on all the four lines"

This turns out not to be the case. You could also draw DC current 
through any of the lines, which current would not register on the 
clamps. The simplest way to do this would be just to use a diode in 
series with the heating element.

Since power = current x voltage x pf, it is NOT necessary to change the 
voltage in order to change the power.

Duncan

On 5/26/2013 2:21 PM, Jed Rothwell wrote:
> A Swedish correspondent sent me this link:
>
> http://www.energikatalysatorn.se/forum/viewtopic.php?f=2&t=560&sid=5450c28dab532569dee72f88a43a56f0&start=330
>
> This is a discussion in Swedish, which Google does a good job 
> translating. Before you translate it, you will see that in the middle 
> of it is a message from one of the authors, Torbjörn Hartman, in 
> English. Here it is, with a few typos corrected.
>
> QUOTE:
>
> Remember that there were not only three clamps to measure the 
> current on three phases but also four connectors to measure the 
> voltage on the three phases and the zero/ground line. The protective 
> ground line was not used and laid curled up on the bench. The only 
> possibility to fool the power-meter then is to raise the DC voltage on 
> all the four lines but that also means that the current must have an 
> other way to leave the system and I tried to find suc

Re: [Vo]:Re: [Vo]:Re: [Vo]:Re: [Vo]:Torbjörn Hartman describes power measurments

[Duncan Cumming]

Yes, Robin is correct.
Duncan

On 5/26/2013 8:08 PM, mix...@bigpond.com wrote:

In reply to  David Roberson's message of Sun, 26 May 2013 22:35:09 -0400 (EDT):
Hi,

This is a little different. A full bridge rectifier will allow for both halves
of the AC current to pass, and so it should be measured as little different to a
purely resistive load. However a single diode will only allow one half to pass,
which *may* mess up magnetic field based current measurements.
(I guess whether if does or not depends on the sophistication of the device.)

Assume that you have a bridge rectifier in the blue box.  This is followed by a 
filtering capacitor.  The DC is then used by the electronics connected to the 
capacitor.  Are you saying that it is not possible to determine the power input 
to this type of network by measuring the input AC voltage and current?  Or are 
you saying that someone has performed a scam and put a DC supply in series with 
the normal AC voltage?

You do know that this could easily be measured by a simple DC voltmeter, right?

Dave

[snip]
Regards,

Robin van Spaandonk

http://rvanspaa.freehostia.com/project.html

Re: [Vo]:RE: [Vo]:Re: [Vo]:Torbjörn Hartman describes power measurments

[Duncan Cumming]

My source was Hanno Essen, one of the authors. He answered a question 
asked by email by one Sterling D. Allan 

/ of Pure Energy Systems News/, reported earlier in this list.
>/4. Have you tried to test the output of the power supply to exclude 
that/ /also a DC current is supplied to the device, which clamp 
amperometers/ /could not detect?/ No, we did not think of that. The 
power came from a normal wall socket and there did not seem to be any 
reason to suspect that it was manipulated in some special way. Now that 
the point is raised we can check this in future tests.


The PCE clamp to which you link is, indeed, DC rated. But Essen does 
seem to believe, erroneously, that it is not possible to draw direct 
current from an ordinary AC outlet. In fact, a simple diode does enable 
one to take fluctuating DC from an AC outlet, which outlet has not been 
manipulated in any special way. Maybe Essen does not have an EE background?


On 5/26/2013 7:53 PM, Jones Beene wrote:


*From:*Duncan Cumming

So is it your position that a current clamp without a Hall effect unit 
can measure DC? Mine is that it cannot.



Did you actually check the PCE site?

It looks to me like all the current clamps on the PCE power analyzer 
site measure both AC and DC


http://www.industrial-needs.com/technical-data/current-detector-PCE-DC-3.htm

[Vo]:Re: [Vo]:Re: [Vo]:Re: [Vo]:Re: [Vo]:Re: [Vo]:Torbjörn Hartman describes power measurments

[Alain Sepeda]

http://www.circuitstoday.com/half-wave-rectifiers

*(ii)**Disadvantages:1.* The output current in the load contains,* in
addition to dc component*, *ac components of basic frequency equal to that
of the input voltage frequency*. Ripple factor is high and an elaborate
filtering is, therefore, required to give steady dc output.

*(iii)*2.The power output and, therefore, rectification efficiency is quite
low. This is due to the fact that power is delivered only half the time.

*(iv)*3.Transformer utilization factor is low.

*(v)*4.*DC saturation of transformer core resulting in magnetizing current
and hysteresis losses and generation of harmonics.*

transformer are not perfect and saturation solve the DC component problem.

2013/5/27 David Roberson 

>  The concept mentioned below by Duncan is not correct.  The DC current
> that flows into the resistor from the wall socket finds a short circuit to
> ground in the power transformer center tap in most cases.
>
>

Re: [Vo]:Re: [Vo]:Re: [Vo]:Re: [Vo]:Torbjörn Hartman describes power measurments

[Duncan Cumming]

What I am proposing is a lot simpler than that. No bridge rectifier, no 
capacitor, just a simple diode. I am saying that given a diode in series 
with a resistor, it is not possible to measure the power using a clamp 
on ammeter.


I am not suggesting that anybody has performed a scam. I am suggesting 
that the equipment used would not have measured the power consumed by 
the resistor if rectification were present in the controller box.


Is there anybody reading this that can do SPICE simulations? Might it be 
possible to simulate a resistor in series with a diode and determine the 
actual and apparent power if an AC coupled current meter is used?


Duncan

P.S. I never mentioned either bridge rectifiers or capacitors. In the 
case of a bridge rectifier type power supply, then a clamp on ammeter 
will work OK. I do not suspect such a thing in the demo.


On 5/26/2013 7:35 PM, David Roberson wrote:
Assume that you have a bridge rectifier in the blue box.  This is 
followed by a filtering capacitor.  The DC is then used by the 
electronics connected to the capacitor.  Are you saying that it is not 
possible to determine the power input to this type of network by 
measuring the input AC voltage and current?  Or are you saying that 
someone has performed a scam and put a DC supply in series with the 
normal AC voltage?
You do know that this could easily be measured by a simple DC 
voltmeter, right?

Dave
-Original Message-
From: Duncan Cumming 
To: vortex-l 
Sent: Sun, May 26, 2013 10:01 pm
Subject: Re: [Vo]:Re: [Vo]:Re: [Vo]:Re: [Vo]:Torbjörn Hartman 
describes power measurments


Almost. The power being fed to the heater exceeds that measured at the 
wall, because the sensor used (an AC current clamp) cannot sense the 
direct current being drawn from the wall socket.


Some people find the difference between current and voltage confusing. 
What I am saying here is that if you connect a resistor in series with 
a diode to a wall socket, then the CURRENT drawn is direct even though 
the VOLTAGE at the socket is alternating. (Rossi does not seem to 
understand this concept judging by his message that got posted today). 
So unless you use a DC rated current meter (such as a shunt) you will 
not sense all of the current, and hence power, drawn from the wall 
socket.


The electrical power meter in your house certainloy IS rated for DC, 
so you will certainly be BILLED for the power even though you didn't 
measure it yourself!


V = IR
Power = Voltage * Current * Power Factor

Duncan

On 5/26/2013 5:57 PM, Eric Walker wrote:

I wrote:

On Sun, May 26, 2013 at 5:18 PM, Duncan Cumming
mailto:spacedr...@cumming.info>> wrote:

I am not trying to assert anything as fact. I am merely
pointing out that a simple diode inside the controller box
(to which access was forbidden by Rossi) COULD HAVE given the
observed results. I am NOT saying that it, in fact, did,
merely speculating that it could have.


Am I right in understanding that this line of reasoning requires
tampering with the mains itself, where the electrical
measurements were made, in addition to any sly customizations
that might have been made at the controller?


I think I'm starting to understand.  This is a separate line of 
reasoning to the one about the possibility of hidden DC and RF 
passing undetected through the clamp meters at the mains.  In this 
line of reasoning, the duty cycle (35 percent ON) is misunderstood, 
and there is a hidden DC component from the controller delivering 
power to the E-Cat, but not above what was read from the wall -- am I 
describing this right?


Eric

Re: [Vo]:Re: [Vo]:Re: [Vo]:Re: [Vo]:Re: [Vo]:Torbjörn Hartman describes power measurments

[David Roberson]

The concept mentioned below by Duncan is not correct.  The DC current that 
flows into the resistor from the wall socket finds a short circuit to ground in 
the power transformer center tap in most cases.

All of the power being delivered into the resistor from the wall socket can be 
determined by taking the AC voltage which is a sine wave and multiplying it by 
the fundamental frequency of the AC current(also a sine wave).  This must be 
adjusted by multiplication by the cosine of the phase angle between the supply 
voltage and fundamental current.

There can be no power associated with this imaginary DC source since its drive 
value is 0.  This is difficult to understand and has lead to a lot of confusion 
about the power input.

Harmonic currents can not deliver power from the line source either.  This is 
also confusing.  This fact leads to interesting measurements such as that the 
pf can be .5 in the case at hand.  Harmonic currents due to distortion show up 
in the RMS reading of the power meter since they are real.  They can be 
significant if a rectifier or phase modulation triac control is used but they 
do not contribute to power being delivered from the mains.

I hope this clears up some of the confusion that has been rampart.

Dave


-Original Message-
From: Alain Sepeda 
To: vortex-l 
Sent: Mon, May 27, 2013 8:25 am
Subject: [Vo]:Re: [Vo]:Re: [Vo]:Re: [Vo]:Re: [Vo]:Torbjörn Hartman describes 
power measurments


Right, that was a know problems with simple rectification and transformers that 
get magnetized.

however you can clearly see it on the waveform.

You see the asymetry of shape. impossible to miss.

second point is that mixing two voltage,  it will kill or trouble other 
instruments plugged (peak too high), if not using transformer.

Rossi would have to know no classic switch power supply will be plugged.

this is why the only question if how far Rossi controlled the installation.
This is the only important point about fraud : was there enough freedom for 
testers to make fraud too risky.


2013/5/27 Duncan Cumming 

  

Some people find the difference between current and voltage  confusing. 
What I am saying here is that if you connect a resistor  in series with a 
diode to a wall socket, then the CURRENT drawn is  direct even though the 
VOLTAGE at the socket is alternating.  (Rossi does not seem to understand 
this concept judging by his  message that got posted today). So unless you 
use a DC rated  current meter (such as a shunt) you will not sense all of 
the  current, and hence power, drawn from the wall socket. 

  

Re: [Vo]:Re: [Vo]:Re: [Vo]:Re: [Vo]:Re: [Vo]:Torbjörn Hartman describes power measurments

[David Roberson]

Do you think that Rossi would be stupid enough to have DC voltage at the power 
socket pins which would be so easy to check?  The testers could have looked at 
this at any time and his gig would have been up.  This is not reasonable to 
assume as he was not around to prevent this from happening.

Dave


-Original Message-
From: Andrew 
To: vortex-l 
Sent: Mon, May 27, 2013 1:57 am
Subject: [Vo]:Re: [Vo]:Re: [Vo]:Re: [Vo]:Re: [Vo]:Torbjörn Hartman describes 
power measurments



By direct admission of the team, posted here, it did not occur to them to check 
for a DC level change.
  
- Original Message - 
  
From:   David   Roberson 
  
To: vortex-l@eskimo.com 
  
Sent: Sunday, May 26, 2013 8:46 PM
  
Subject: Re: [Vo]:Re: [Vo]:Re: [Vo]:Re:   [Vo]:Torbjörn Hartman describes power 
measurments
  


  
Robin,
  
 
  
The problem at hand is that the skeptic claims that power due   to the DC 
current can be very large and not detected.  There has been no   discussion of 
the AC current reading being affected by the DC so far.That is a different 
issue entirely.
  
 
  
I would like for them to answer the questions because then they might   realize 
that their position is invalid.  I can explain this if   required.  No one is 
suggesting that Rossi actually has a DC power supply   hidden within the wall I 
hope.  This would be beyond reality since it   would be so easy to measure with 
a voltmeter or any monitor that looks at the   voltage.  The testers did a 
visual look at the voltage from what I have   determined.
  
 
  
So, skeptics, what say you?
  
 
  
Dave
  
  
  
-Original   Message-
From: mixent 
To: vortex-l   
Sent: Sun, May 26, 2013 11:08 pm
Subject:   Re: [Vo]:Re: [Vo]:Re: [Vo]:Re: [Vo]:Torbjörn Hartman describes power 
  measurments

  
In reply to  David Roberson's message of Sun, 26 May 2013 22:35:09 -0400 (EDT):
Hi,

This is a little different. A full bridge rectifier will allow for both halves
of the AC current to pass, and so it should be measured as little different to a
purely resistive load. However a single diode will only allow one half to pass,
which *may* mess up magnetic field based current measurements.
(I guess whether if does or not depends on the sophistication of the device.)
>
>Assume that you have a bridge rectifier in the blue box.  This is followed by 
>a 
filtering capacitor.  The DC is then used by the electronics connected to the 
capacitor.  Are you saying that it is not possible to determine the power input 
to this type of network by measuring the input AC voltage and current?  Or are 
you saying that someone has performed a scam and put a DC supply in series with 
the normal AC voltage?
>
>You do know that this could easily be measured by a simple DC voltmeter, right?
>
>Dave
[snip]
Regards,

Robin van Spaandonk

http://rvanspaa.freehostia.com/project.html

[Vo]:Re: [Vo]:Re: [Vo]:Re: [Vo]:Re: [Vo]:Torbjörn Hartman describes power measurments

[Alain Sepeda]

by the way, not a bridge but a single wave (one diode) rectification...

usual bridge does not cause asymetry, neither double wave.


2013/5/27 David Roberson 

> Assume that you have a bridge rectifier in the blue box.  This is followed
> by a filtering capacitor.  The DC is then used by the electronics connected
> to the capacitor.  Are you saying that it is not possible to determine the
> power input to this type of network by measuring the input AC voltage and
> current?  Or are you saying that someone has performed a scam and put a DC
> supply in series with the normal AC voltage?
>
>

[Vo]:Re: [Vo]:Re: [Vo]:Re: [Vo]:Re: [Vo]:Torbjörn Hartman describes power measurments

[Alain Sepeda]

Right, that was a know problems with simple rectification and transformers
that get magnetized.

however you can clearly see it on the waveform.

You see the asymetry of shape. impossible to miss.

second point is that mixing two voltage,  it will kill or trouble other
instruments plugged (peak too high), if not using transformer.

Rossi would have to know no classic switch power supply will be plugged.

this is why the only question if how far Rossi controlled the installation.
This is the only important point about fraud : was there enough freedom for
testers to make fraud too risky.

2013/5/27 Duncan Cumming 

>  Some people find the difference between current and voltage confusing.
> What I am saying here is that if you connect a resistor in series with a
> diode to a wall socket, then the CURRENT drawn is direct even though the
> VOLTAGE at the socket is alternating. (Rossi does not seem to understand
> this concept judging by his message that got posted today). So unless you
> use a DC rated current meter (such as a shunt) you will not sense all of
> the current, and hence power, drawn from the wall socket.
>
>

RE: [Vo]:Re: [Vo]:Re: [Vo]:Re: [Vo]:Torbjörn Hartman describes power measurments

[Sunil Shah]

or a scope and LOOK at the damn thing : D

> Anyway... next piece of equipment on the shopping list (for the next test) 
> ought to be http://www.tortech.com.au/category/3-phase-isolation/
> Plug everything into an isolation transformer. That should do it.
>
> .s  

RE: [Vo]:Re: [Vo]:Re: [Vo]:Re: [Vo]:Torbjörn Hartman describes power measurments

[Sunil Shah]

If you want to sneak DC into a system, you'd never get it passed a clamp meter, 
if you just use some diodes.  You'll need serious decoupling. I say serious, 
because the load is substantial and will quickly drain the reservoir 
capacitors. Any ripple on the DC will generate a varying magnetic field, albeit 
small if your ripple is small, and this variation is what a clamp meter picks 
up.

Anyway... next piece of equipment on the shopping list (for the next test) 
ought to be http://www.tortech.com.au/category/3-phase-isolation/
Plug everything into an isolation transformer. That should do it.

.s

[Vo]:Re: [Vo]:Re: [Vo]:Re: [Vo]:Re: [Vo]:Torbjörn Hartman describes power measurments

[Andrew]

By direct admission of the team, posted here, it did not occur to them to check 
for a DC level change.
  - Original Message - 
  From: David Roberson 
  To: vortex-l@eskimo.com 
  Sent: Sunday, May 26, 2013 8:46 PM
  Subject: Re: [Vo]:Re: [Vo]:Re: [Vo]:Re: [Vo]:Torbjörn Hartman describes power 
measurments


  Robin,

  The problem at hand is that the skeptic claims that power due to the DC 
current can be very large and not detected.  There has been no discussion of 
the AC current reading being affected by the DC so far.  That is a different 
issue entirely.

  I would like for them to answer the questions because then they might realize 
that their position is invalid.  I can explain this if required.  No one is 
suggesting that Rossi actually has a DC power supply hidden within the wall I 
hope.  This would be beyond reality since it would be so easy to measure with a 
voltmeter or any monitor that looks at the voltage.  The testers did a visual 
look at the voltage from what I have determined.

  So, skeptics, what say you?

  Dave
  -Original Message-
  From: mixent 
  To: vortex-l 
  Sent: Sun, May 26, 2013 11:08 pm
  Subject: Re: [Vo]:Re: [Vo]:Re: [Vo]:Re: [Vo]:Torbjörn Hartman describes power 
measurments


In reply to  David Roberson's message of Sun, 26 May 2013 22:35:09 -0400 (EDT):
Hi,

This is a little different. A full bridge rectifier will allow for both halves
of the AC current to pass, and so it should be measured as little different to a
purely resistive load. However a single diode will only allow one half to pass,
which *may* mess up magnetic field based current measurements.
(I guess whether if does or not depends on the sophistication of the device.)
>
>Assume that you have a bridge rectifier in the blue box.  This is followed by 
>a 
filtering capacitor.  The DC is then used by the electronics connected to the 
capacitor.  Are you saying that it is not possible to determine the power input 
to this type of network by measuring the input AC voltage and current?  Or are 
you saying that someone has performed a scam and put a DC supply in series with 
the normal AC voltage?
>
>You do know that this could easily be measured by a simple DC voltmeter, right?
>
>Dave
[snip]
Regards,

Robin van Spaandonk

http://rvanspaa.freehostia.com/project.html

Re: [Vo]:RE: [Vo]:Re: [Vo]:Torbjörn Hartman describes power measurments

[Alan Goldwater]

That is a different instrument. The one used in the tests  (PCE-830 
) 
does not measure DC.


On 5/26/2013 7:53 PM, Jones Beene wrote:


Did you actually check the PCE site?

It looks to me like all the current clamps on the PCE power analyzer 
site measure both AC and DC


http://www.industrial-needs.com/technical-data/current-detector-PCE-DC-3.htm

Re: [Vo]:Re: [Vo]:Re: [Vo]:Re: [Vo]:Torbjörn Hartman describes power measurments

[David Roberson]

Robin,

The problem at hand is that the skeptic claims that power due to the DC current 
can be very large and not detected.  There has been no discussion of the AC 
current reading being affected by the DC so far.  That is a different issue 
entirely.

I would like for them to answer the questions because then they might realize 
that their position is invalid.  I can explain this if required.  No one is 
suggesting that Rossi actually has a DC power supply hidden within the wall I 
hope.  This would be beyond reality since it would be so easy to measure with a 
voltmeter or any monitor that looks at the voltage.  The testers did a visual 
look at the voltage from what I have determined.

So, skeptics, what say you?

Dave


-Original Message-
From: mixent 
To: vortex-l 
Sent: Sun, May 26, 2013 11:08 pm
Subject: Re: [Vo]:Re: [Vo]:Re: [Vo]:Re: [Vo]:Torbjörn Hartman describes power 
measurments


In reply to  David Roberson's message of Sun, 26 May 2013 22:35:09 -0400 (EDT):
Hi,

This is a little different. A full bridge rectifier will allow for both halves
of the AC current to pass, and so it should be measured as little different to a
purely resistive load. However a single diode will only allow one half to pass,
which *may* mess up magnetic field based current measurements.
(I guess whether if does or not depends on the sophistication of the device.)
>
>Assume that you have a bridge rectifier in the blue box.  This is followed by 
>a 
filtering capacitor.  The DC is then used by the electronics connected to the 
capacitor.  Are you saying that it is not possible to determine the power input 
to this type of network by measuring the input AC voltage and current?  Or are 
you saying that someone has performed a scam and put a DC supply in series with 
the normal AC voltage?
>
>You do know that this could easily be measured by a simple DC voltmeter, right?
>
>Dave
[snip]
Regards,

Robin van Spaandonk

http://rvanspaa.freehostia.com/project.html


 

Re: [Vo]:Re: [Vo]:Re: [Vo]:Re: [Vo]:Torbjörn Hartman describes power measurments

[mixent]

In reply to  David Roberson's message of Sun, 26 May 2013 22:35:09 -0400 (EDT):
Hi,

This is a little different. A full bridge rectifier will allow for both halves
of the AC current to pass, and so it should be measured as little different to a
purely resistive load. However a single diode will only allow one half to pass,
which *may* mess up magnetic field based current measurements.
(I guess whether if does or not depends on the sophistication of the device.)
>
>Assume that you have a bridge rectifier in the blue box.  This is followed by 
>a filtering capacitor.  The DC is then used by the electronics connected to 
>the capacitor.  Are you saying that it is not possible to determine the power 
>input to this type of network by measuring the input AC voltage and current?  
>Or are you saying that someone has performed a scam and put a DC supply in 
>series with the normal AC voltage?
>
>You do know that this could easily be measured by a simple DC voltmeter, right?
>
>Dave
[snip]
Regards,

Robin van Spaandonk

http://rvanspaa.freehostia.com/project.html

[Vo]:RE: [Vo]:Re: [Vo]:Torbjörn Hartman describes power measurments

[Jones Beene]

 

 

From: Duncan Cumming 

 

 

So is it your position that a current clamp without a Hall effect unit can 
measure DC? Mine is that it cannot.




Did you actually check the PCE site?

 

It looks to me like all the current clamps on the PCE power analyzer site 
measure both AC and DC 

 

http://www.industrial-needs.com/technical-data/current-detector-PCE-DC-3.htm

 

 

Re: [Vo]:Re: [Vo]:Re: [Vo]:Re: [Vo]:Torbjörn Hartman describes power measurments

[David Roberson]

Assume that you have a bridge rectifier in the blue box.  This is followed by a 
filtering capacitor.  The DC is then used by the electronics connected to the 
capacitor.  Are you saying that it is not possible to determine the power input 
to this type of network by measuring the input AC voltage and current?  Or are 
you saying that someone has performed a scam and put a DC supply in series with 
the normal AC voltage?

You do know that this could easily be measured by a simple DC voltmeter, right?

Dave


-Original Message-
From: Duncan Cumming 
To: vortex-l 
Sent: Sun, May 26, 2013 10:01 pm
Subject: Re: [Vo]:Re: [Vo]:Re: [Vo]:Re: [Vo]:Torbjörn Hartman describes power 
measurments


  

Almost. The power being fed to the  heater exceeds that measured at the 
wall, because the sensor used  (an AC current clamp) cannot sense the 
direct current being drawn  from the wall socket.
  
  Some people find the difference between current and voltage  
confusing. What I am saying here is that if you connect a resistor  in 
series with a diode to a wall socket, then the CURRENT drawn is  direct 
even though the VOLTAGE at the socket is alternating.  (Rossi does not seem 
to understand this concept judging by his  message that got posted today). 
So unless you use a DC rated  current meter (such as a shunt) you will not 
sense all of the  current, and hence power, drawn from the wall socket. 
  
  The electrical power meter in your house certainloy IS rated for  DC, 
so you will certainly be BILLED for the power even though you  didn't 
measure it yourself!
  
  V = IR
  Power = Voltage * Current * Power Factor
  
  Duncan
  
  On 5/26/2013 5:57 PM, Eric Walker wrote:


  
I wrote:

  


  
On Sun, May 26, 2013 at 5:18 PM, DuncanCumming 
wrote:
  
  

  

  
I am not trying to assert anything asfact. I am 
merely pointing out that a simplediode inside the 
controller box (to whichaccess was forbidden by 
Rossi) COULD HAVEgiven the observed results. I am 
NOT sayingthat it, in fact, did, merely speculating 
   that it could have.
  

  


  


Am I right in understanding  that this line of reasoning 
requires tampering with  the mains itself, where the electrical 
measurements  were made, in addition to any sly customizations 
that  might have been made at the controller?
  

  
  


I think I'm starting to  understand.  This is a separate line of 
reasoning to the one  about the possibility of hidden DC and RF passing 
undetected  through the clamp meters at the mains.  In this line of 
 reasoning, the duty cycle (35 percent ON) is misunderstood,  and 
there is a hidden DC component from the controller  delivering power to 
the E-Cat, but not above what was read  from the wall -- am I 
describing this right?




Eric



  


  

Re: [Vo]:Re: [Vo]:Re: [Vo]:Re: [Vo]:Torbjörn Hartman describes power measurments

[Duncan Cumming]

Almost. The power being fed to the heater exceeds that measured at the 
wall, because the sensor used (an AC current clamp) cannot sense the 
direct current being drawn from the wall socket.


Some people find the difference between current and voltage confusing. 
What I am saying here is that if you connect a resistor in series with a 
diode to a wall socket, then the CURRENT drawn is direct even though the 
VOLTAGE at the socket is alternating. (Rossi does not seem to understand 
this concept judging by his message that got posted today). So unless 
you use a DC rated current meter (such as a shunt) you will not sense 
all of the current, and hence power, drawn from the wall socket.


The electrical power meter in your house certainloy IS rated for DC, so 
you will certainly be BILLED for the power even though you didn't 
measure it yourself!


V = IR
Power = Voltage * Current * Power Factor

Duncan

On 5/26/2013 5:57 PM, Eric Walker wrote:

I wrote:

On Sun, May 26, 2013 at 5:18 PM, Duncan Cumming
mailto:spacedr...@cumming.info>> wrote:

I am not trying to assert anything as fact. I am merely
pointing out that a simple diode inside the controller box (to
which access was forbidden by Rossi) COULD HAVE given the
observed results. I am NOT saying that it, in fact, did,
merely speculating that it could have.


Am I right in understanding that this line of reasoning requires
tampering with the mains itself, where the electrical measurements
were made, in addition to any sly customizations that might have
been made at the controller?


I think I'm starting to understand.  This is a separate line of 
reasoning to the one about the possibility of hidden DC and RF passing 
undetected through the clamp meters at the mains.  In this line of 
reasoning, the duty cycle (35 percent ON) is misunderstood, and there 
is a hidden DC component from the controller delivering power to the 
E-Cat, but not above what was read from the wall -- am I describing 
this right?


Eric

Re: [Vo]:Re: [Vo]:Re: [Vo]:Torbjörn Hartman describes power measurments

[Duncan Cumming]

No, it does not. What happens is that the diode rectifies the mains to 
DC, and the DC is not sensed by the clamp-type current meter.

http://en.wikipedia.org/wiki/Current_clamp
No tampering with the mains itself would be necessary if one were to use 
a diode.


Duncan

On 5/26/2013 5:34 PM, Eric Walker wrote:
On Sun, May 26, 2013 at 5:18 PM, Duncan Cumming 
mailto:spacedr...@cumming.info>> wrote:


I am not trying to assert anything as fact. I am merely pointing
out that a simple diode inside the controller box (to which access
was forbidden by Rossi) COULD HAVE given the observed results. I
am NOT saying that it, in fact, did, merely speculating that it
could have.


Am I right in understanding that this line of reasoning requires 
tampering with the mains itself, where the electrical measurements 
were made, in addition to any sly customizations that might have been 
made at the controller?


Eric

[Vo]:Re: [Vo]:Re: [Vo]:Re: [Vo]:Torbjörn Hartman describes power measurments

[ChemE Stewart]

It looks like some flying/gliding frogs have flaps of skin, which are like
wings...
http://en.m.wikipedia.org/wiki/Flying_frog

It is hard for me to comment on the rest of the unfalsifiable
falsifications of Rossi's results.  As for me, I think he has something.  I
am not much for conspiracy theories involving multiple Scientists although
I suppose you can find anything under the sun and I certainly don't think a
comet is a snowball

Stewart

On Sunday, May 26, 2013, David Roberson wrote:

> That is funny.  I guess they can fly, but do they use their wings?
>
> If you want to get technical about this you would say that the people who
> saw them fly were all confused by a scammer.  Perhaps someone down the
> street had a catapult and used it to throw a large number of them to the
> observers.  How do we know that this did not happen?  Also, maybe all of
> the observers were in the scam together.  Their reward was to keep all of
> the frog legs they could capture as long as the king. who loves frog
> legs dearly, does not find out and slay the entire bunch.
>
> I am confident that many silly stories can be concocted just as with the
> "CAT scam".
>
> Dave
>  -Original Message-
> From: ChemE Stewart  'cheme...@gmail.com');>>
> To: vortex-l  'vortex-l@eskimo.com');>>
> Sent: Sun, May 26, 2013 8:39 pm
> Subject: [Vo]:Re: [Vo]:Re: [Vo]:Torbjörn Hartman describes power
> measurments
>
>  It has already been proven that bullfrogs can fly, most likely caused
> from strings of vacuum energy.
>
>  http://en.m.wikipedia.org/wiki/Raining_animals
>
> On Sunday, May 26, 2013, David Roberson wrote:
>
> How do we know that your diode trick will actually do what you think?  You
> need to prove that this is possible, otherwise anyone can make the
> assumption that it might not work just as with the ECAT tests.  If you do
> not prove that this will work, then why should we accept it as a
> possibility?
>
> A lot of time and energy is being wasted trying to see if bull frogs can
> fly.  Some might actually be born with wings.  Have we proven that none of
> them can fly?
>
> Rossi and the testers have done a lot to prove that the ECAT works.   No
> one has proven that it does not.  The only offers from the other side of
> the table assume fraud.  Is this a valid position for them to take?
>
> Dave
>  -Original Message-
> From: Duncan Cumming 
> To: vortex-l 
> Sent: Sun, May 26, 2013 8:18 pm
> Subject: Re: [Vo]:Re: [Vo]:Torbjörn Hartman describes power measurments
>
>  I am not trying to assert anything as fact. I am merely pointing out
> that a simple diode inside the controller box (to which access was
> forbidden by Rossi) COULD HAVE given the observed results. I am NOT saying
> that it, in fact, did, merely speculating that it could have.
>
> For any scientific experiment, the onus is on the experimenters to produce
> the result. The best way to do this is to provide sufficient information
> for others to replicate the experiment.
>
> Duncan
>
> On 5/26/2013 5:07 PM, David Roberson wrote:
>
> Perhaps you should build one of these scam machines and prove that it will
> work without being detected.  That would be the best way to show that it is
> possible.  Why should we accept this assertion as fact any more than
> believing that the testers missed finding the scam?
>
> We can spend an equal amount of time knocking down any theory that is put
> forth as others can spend assuming they are real.
>
> Dave
> -Original Message-
> From: Duncan Cumming 
> To: vortex-l 
> Sent: Sun, May 26, 2013 7:59 pm
> Subject: [Vo]:Re: [Vo]:Torbjörn Hartman describes power measurments
>
>  "The only possibility to fool the power-meter then is to raise the DC
> voltage on all the four lines"
>
> This turns out not to be the case. You could also draw DC current
> through any of the lines, which current would not register on the
> clamps. The simplest way to do this would be just to use a diode in
> series with the heating element.
>
> Since power = current x voltage x pf, it is NOT necessary to change the
> voltage in order to change the power.
>
> Duncan
>
> On 5/26/2013 2:21 PM, Jed Rothwell wrote:
> > A Swedish correspondent sent me this link:
> >
> > http://www.energikatalysatorn.se/forum/viewtopic.php?f=2&t=560&sid=5450c28dab532569dee72f88a43a56f0&start=330
> >
> > This is a discussion in Swedish, which Google does a good job
> > translating. Before you translate it, you will see that in the middle
> > of it is a message from one of the authors, Torbjörn Hartman, in
> > English. Here it is, with a few typos corrected.
>
>

[Vo]:RE: [Vo]:Re: [Vo]:Re: [Vo]:Torbjörn Hartman describes power measurments

[MarkI-ZeroPoint]

“I took a picture every day of the connectors and cables to the powermeter in 
case anyone would tamper with them when we were out.”

 

Its comments like this one and several others made by Torbjörn Hartman which 
indicates that these guys came at this with the idea of possible cheating, and 
tried to cover all the bases.  Lifting the Control box to look for hidden 
wires, the metal frame (which was ‘freestanding’) so no hidden wires there, 
etc.  


He did mention that the AC waveform (for each phase?) could be viewed on the 
power analyzer… 

 

So the list of possible AND REASONABLE ways for Rossi to fake this are 
diminishing with each interview of the team…

 

-Mark

 

 

From: Andrew [mailto:andrew...@att.net] 
Sent: Sunday, May 26, 2013 2:50 PM
To: vortex-l@eskimo.com
Subject: [Vo]:Re: [Vo]:Re: [Vo]:Torbjörn Hartman describes power measurments

 

Indeed it has Dave. That's heartening.

 

Andrew

- Original Message - 

From: David Roberson <mailto:dlrober...@aol.com>  

To: vortex-l@eskimo.com 

Sent: Sunday, May 26, 2013 2:43 PM

Subject: Re: [Vo]:Re: [Vo]:Torbjörn Hartman describes power measurments

 

I assume that your opinion of the test guys has improved according to your 
latest statement.

 

Dave

-Original Message-
From: Andrew 
To: vortex-l 
Sent: Sun, May 26, 2013 5:29 pm
Subject: [Vo]:Re: [Vo]:Torbjörn Hartman describes power measurments

My position is nicely summarised in that final paragraph. So if you attack me, 
you attack by proxy one of the authors of the paper.

 

I'm gratified that at least one of the testers had his head screwed on. I woke 
up this morning thinking about a wire through the bench into the control box. 
Hartman is my kind of guy.

 

Andrew

- Original Message - 

From: Jed Rothwell <mailto:jedrothw...@gmail.com>  

To: vortex-l@eskimo.com 

Sent: Sunday, May 26, 2013 2:21 PM

Subject: [Vo]:Torbjörn Hartman describes power measurments

 

A Swedish correspondent sent me this link: 

 

http://www.energikatalysatorn.se/forum/viewtopic.php?f=2 
<http://www.energikatalysatorn.se/forum/viewtopic.php?f=2&t=560&sid=5450c28dab532569dee72f88a43a56f0&start=330>
 &t=560&sid=5450c28dab532569dee72f88a43a56f0&start=330

This is a discussion in Swedish, which Google does a good job translating. 
Before you translate it, you will see that in the middle of it is a message 
from one of the authors, Torbjörn Hartman, in English. Here it is, with a few 
typos corrected.

 

QUOTE:

 

Remember that there were not only three clamps to measure the current on three 
phases but also four connectors to measure the voltage on the three phases and 
the zero/ground line. The protective ground line was not used and laid curled 
up on the bench. The only possibility to fool the power-meter then is to raise 
the DC voltage on all the four lines but that also means that the current must 
have an other way to leave the system and I tried to find such hidden 
connections when we were there. The control box had no connections through the 
wood on the table. All cables in and out were accounted for. The E-cat was just 
lying on the metal frame that was only free-standing on the floor with no 
cables going to it. The little socket, where the mains cables from the wall 
connector where connected with the cables to the box and where we had the 
clamps, was screwed to the wood of the bench but there was no screws going 
through the metal sheet under the bench. The sheet showed no marks on it under 
the interesting parts (or elsewhere as I remember it). Of course, if the white 
little socket was rigged inside and the metal screws was long enough to go just 
through the wood, touching the metal sheet underneath, then the bench itself 
could lead current. I do not remember if I actually checked the bench frame for 
cables connected to it but I probably did. However, I have a close-up picture 
of the socket and it looks normal and the screws appear to be of normal size. I 
also have pictures of all the connectors going to the powermeter and of the 
frame on the floor. I took a picture every day of the connectors and cables to 
the powermeter in case anyone would tamper with them when we were out.

I lifted the control box to check what was under it and when doing so I tried 
to measure the weight and it is muck lighter than a car battery. The box itself 
has a weight, of course, and what is in it can not be much.

All these observations take away a number of ways to tamper with our 
measurements but there can still be things that we "didn't think of" and that 
is the reason why we only can claim "indications of" and not "proof of" 
anomalous heat production. We must have more control over the whole situation 
before we can talk about proof.

Best regards,
Torbjörn

END QUOTE

 

- Jed

 

[Vo]:Re: [Vo]:Re: [Vo]:Re: [Vo]:Torbjörn Hartman describes power measurments

[Eric Walker]

I wrote:

On Sun, May 26, 2013 at 5:18 PM, Duncan Cumming wrote:
>
>  I am not trying to assert anything as fact. I am merely pointing out
>> that a simple diode inside the controller box (to which access was
>> forbidden by Rossi) COULD HAVE given the observed results. I am NOT saying
>> that it, in fact, did, merely speculating that it could have.
>>
>
> Am I right in understanding that this line of reasoning requires tampering
> with the mains itself, where the electrical measurements were made, in
> addition to any sly customizations that might have been made at the
> controller?
>

I think I'm starting to understand.  This is a separate line of reasoning
to the one about the possibility of hidden DC and RF passing undetected
through the clamp meters at the mains.  In this line of reasoning, the duty
cycle (35 percent ON) is misunderstood, and there is a hidden DC component
from the controller delivering power to the E-Cat, but not above what was
read from the wall -- am I describing this right?

Eric

Re: [Vo]:Re: [Vo]:Re: [Vo]:Torbjörn Hartman describes power measurments

[David Roberson]

That is funny.  I guess they can fly, but do they use their wings?

If you want to get technical about this you would say that the people who saw 
them fly were all confused by a scammer.  Perhaps someone down the street had a 
catapult and used it to throw a large number of them to the observers.  How do 
we know that this did not happen?  Also, maybe all of the observers were in the 
scam together.  Their reward was to keep all of the frog legs they could 
capture as long as the king. who loves frog legs dearly, does not find out and 
slay the entire bunch.

I am confident that many silly stories can be concocted just as with the "CAT 
scam".

Dave


-Original Message-
From: ChemE Stewart 
To: vortex-l 
Sent: Sun, May 26, 2013 8:39 pm
Subject: [Vo]:Re: [Vo]:Re: [Vo]:Torbjörn Hartman describes power measurments


It has already been proven that bullfrogs can fly, most likely caused from 
strings of vacuum energy.


http://en.m.wikipedia.org/wiki/Raining_animals

On Sunday, May 26, 2013, David Roberson  wrote:

How do we know that your diode trick will actually do what you think?  You need 
to prove that this is possible, otherwise anyone can make the assumption that 
it might not work just as with the ECAT tests.  If you do not prove that this 
will work, then why should we accept it as a possibility?
 
A lot of time and energy is being wasted trying to see if bull frogs can fly.  
Some might actually be born with wings.  Have we proven that none of them can 
fly?
 
Rossi and the testers have done a lot to prove that the ECAT works.   No one 
has proven that it does not.  The only offers from the other side of the table 
assume fraud.  Is this a valid position for them to take?
 
Dave


-Original Message-
From: Duncan Cumming 
To: vortex-l 
Sent: Sun, May 26, 2013 8:18 pm
Subject: Re: [Vo]:Re: [Vo]:Torbjörn Hartman describes power measurments


  

I am not trying to assert anything as  fact. I am merely pointing out that 
a simple diode inside the  controller box (to which access was forbidden by 
Rossi) COULD HAVE  given the observed results. I am NOT saying that it, in 
fact, did,  merely speculating that it could have.
  
  For any scientific experiment, the onus is on the experimenters to  
produce the result. The best way to do this is to provide  sufficient 
information for others to replicate the experiment.
  
  Duncan
  
  On 5/26/2013 5:07 PM, David Roberson wrote:



Perhaps you should build one of thesescam machines and prove that 
it will work without beingdetected.  That would be the best way to 
show that it ispossible.  Why should we accept this assertion as 
fact anymore than believing that the testers missed finding the 
   scam?

 

We can spend an equal amount of time knocking down any  theory that is 
put forth as others can spend assuming they are  real.

 

Dave

-Original Message-
  From: Duncan Cumming 
  To: vortex-l 
  Sent: Sun, May 26, 2013 7:59 pm
  Subject: [Vo]:Re: [Vo]:Torbjörn Hartman describes power  
measurments
  
  

"The only possibility to fool the power-meter then is to raise the DC 
voltage on all the four lines"

This turns out not to be the case. You could also draw DC current 
through any of the lines, which current would not register on the 
clamps. The simplest way to do this would be just to use a diode in 
series with the heating element.

Since power = current x voltage x pf, it is NOT necessary to change the 
voltage in order to change the power.

Duncan

On 5/26/2013 2:21 PM, Jed Rothwell wrote:
> A Swedish correspondent sent me this link:
>
> http://www.energikatalysatorn.se/forum/viewtopic.php?f=2&t=560&sid=5450c28dab532569dee72f88a43a56f0&start=330
>
> This is a discussion in Swedish, which Google does a good job 
> translating. Before you translate it, you will see that in the middle 
> of it is a message from one of the authors, Torbjörn Hartman, in 
> English. Here it is, with a few typos corrected.
>
> QUOTE:
>
> Remember that there were not only three clamps to measure the 
> current on three phases but also four connectors to measure the 
> voltage on the three phases and the zero/ground line. The protective 
> ground line was not used and laid curled up on the bench. The only 
> possibility to fool the power-meter then is to raise the DC voltage on 
> all the four lines but that also means that the current must have an 
> other way to leave the system and I tried to find such hidden 
> connections when we were there. The control box had no connections 
> through the wood on the table. All cables in and out were 
> accounted for. The E-cat was just lying on the metal fra

[Vo]:Re: [Vo]:Re: [Vo]:Torbjörn Hartman describes power measurments

[ChemE Stewart]

It has already been proven that bullfrogs can fly, most likely caused from
strings of vacuum energy.

http://en.m.wikipedia.org/wiki/Raining_animals

On Sunday, May 26, 2013, David Roberson wrote:

> How do we know that your diode trick will actually do what you think?  You
> need to prove that this is possible, otherwise anyone can make the
> assumption that it might not work just as with the ECAT tests.  If you do
> not prove that this will work, then why should we accept it as a
> possibility?
>
> A lot of time and energy is being wasted trying to see if bull frogs can
> fly.  Some might actually be born with wings.  Have we proven that none of
> them can fly?
>
> Rossi and the testers have done a lot to prove that the ECAT works.   No
> one has proven that it does not.  The only offers from the other side of
> the table assume fraud.  Is this a valid position for them to take?
>
> Dave
>  -Original Message-
> From: Duncan Cumming  'spacedr...@cumming.info');>>
> To: vortex-l  'vortex-l@eskimo.com');>>
> Sent: Sun, May 26, 2013 8:18 pm
> Subject: Re: [Vo]:Re: [Vo]:Torbjörn Hartman describes power measurments
>
>  I am not trying to assert anything as fact. I am merely pointing out
> that a simple diode inside the controller box (to which access was
> forbidden by Rossi) COULD HAVE given the observed results. I am NOT saying
> that it, in fact, did, merely speculating that it could have.
>
> For any scientific experiment, the onus is on the experimenters to produce
> the result. The best way to do this is to provide sufficient information
> for others to replicate the experiment.
>
> Duncan
>
> On 5/26/2013 5:07 PM, David Roberson wrote:
>
> Perhaps you should build one of these scam machines and prove that it will
> work without being detected.  That would be the best way to show that it is
> possible.  Why should we accept this assertion as fact any more than
> believing that the testers missed finding the scam?
>
> We can spend an equal amount of time knocking down any theory that is put
> forth as others can spend assuming they are real.
>
> Dave
> -Original Message-
> From: Duncan Cumming 
> To: vortex-l 
> Sent: Sun, May 26, 2013 7:59 pm
> Subject: [Vo]:Re: [Vo]:Torbjörn Hartman describes power measurments
>
>  "The only possibility to fool the power-meter then is to raise the DC
> voltage on all the four lines"
>
> This turns out not to be the case. You could also draw DC current
> through any of the lines, which current would not register on the
> clamps. The simplest way to do this would be just to use a diode in
> series with the heating element.
>
> Since power = current x voltage x pf, it is NOT necessary to change the
> voltage in order to change the power.
>
> Duncan
>
> On 5/26/2013 2:21 PM, Jed Rothwell wrote:
> > A Swedish correspondent sent me this link:
> >
> > http://www.energikatalysatorn.se/forum/viewtopic.php?f=2&t=560&sid=5450c28dab532569dee72f88a43a56f0&start=330
> >
> > This is a discussion in Swedish, which Google does a good job
> > translating. Before you translate it, you will see that in the middle
> > of it is a message from one of the authors, Torbjörn Hartman, in
> > English. Here it is, with a few typos corrected.
> >
> > QUOTE:
> >
> > Remember that there were not only three clamps to measure the
> > current on three phases but also four connectors to measure the
> > voltage on the three phases and the zero/ground line. The protective
> > ground line was not used and laid curled up on the bench. The only
> > possibility to fool the power-meter then is to raise the DC voltage on
> > all the four lines but that also means that the current must have an
> > other way to leave the system and I tried to find such hidden
> > connections when we were there. The control box had no connections
> > through the wood on the table. All cables in and out were
> > accounted for. The E-cat was just lying on the metal frame that was
> > only free-standing on the floor with no cables going to it. The little
> > socket, where the mains cables from the wall connector where connected
> > with the cables to the box and where we had the clamps, was screwed to
> > the wood of the bench but there was no screws going through the metal
> > sheet under the bench. The sheet showed no marks on it under the
> > interesting parts (or elsewhere as I remember it). Of course, if the
> > white little socket was rigged inside and the metal screws was long
> > enough to go just through the wood, touching the metal sheet
> > underneath, then the bench itself could lead current. I do
> > not remember if I actually checked the bench frame for cables
> > connected to it but I probably did. However, I have a close-up picture
> > of the socket and it looks normal and the screws appear to be of
> > normal size. I also have pictures of all the connectors going to th
>
>

[Vo]:Re: [Vo]:Re: [Vo]:Torbjörn Hartman describes power measurments

[Eric Walker]

On Sun, May 26, 2013 at 5:18 PM, Duncan Cumming wrote:

 I am not trying to assert anything as fact. I am merely pointing out that
> a simple diode inside the controller box (to which access was forbidden by
> Rossi) COULD HAVE given the observed results. I am NOT saying that it, in
> fact, did, merely speculating that it could have.
>

Am I right in understanding that this line of reasoning requires tampering
with the mains itself, where the electrical measurements were made, in
addition to any sly customizations that might have been made at the
controller?

Eric

[Vo]:Re: [Vo]:Re: [Vo]:Torbjörn Hartman describes power measurments

[Andrew]

Indeed it has Dave. That's heartening.

Andrew
  - Original Message - 
  From: David Roberson 
  To: vortex-l@eskimo.com 
  Sent: Sunday, May 26, 2013 2:43 PM
  Subject: Re: [Vo]:Re: [Vo]:Torbjörn Hartman describes power measurments


  I assume that your opinion of the test guys has improved according to your 
latest statement.

  Dave
  -Original Message-
  From: Andrew 
  To: vortex-l 
  Sent: Sun, May 26, 2013 5:29 pm
  Subject: [Vo]:Re: [Vo]:Torbjörn Hartman describes power measurments


  My position is nicely summarised in that final paragraph. So if you attack 
me, you attack by proxy one of the authors of the paper.

  I'm gratified that at least one of the testers had his head screwed on. I 
woke up this morning thinking about a wire through the bench into the control 
box. Hartman is my kind of guy.

  Andrew
- Original Message - 
From: Jed Rothwell 
To: vortex-l@eskimo.com 
Sent: Sunday, May 26, 2013 2:21 PM
Subject: [Vo]:Torbjörn Hartman describes power measurments


A Swedish correspondent sent me this link: 



http://www.energikatalysatorn.se/forum/viewtopic.php?f=2&t=560&sid=5450c28dab532569dee72f88a43a56f0&start=330


This is a discussion in Swedish, which Google does a good job translating. 
Before you translate it, you will see that in the middle of it is a message 
from one of the authors, Torbjörn Hartman, in English. Here it is, with a few 
typos corrected.


QUOTE:


Remember that there were not only three clamps to measure the current on 
three phases but also four connectors to measure the voltage on the three 
phases and the zero/ground line. The protective ground line was not used and 
laid curled up on the bench. The only possibility to fool the power-meter then 
is to raise the DC voltage on all the four lines but that also means that the 
current must have an other way to leave the system and I tried to find such 
hidden connections when we were there. The control box had no connections 
through the wood on the table. All cables in and out were accounted for. The 
E-cat was just lying on the metal frame that was only free-standing on the 
floor with no cables going to it. The little socket, where the mains cables 
from the wall connector where connected with the cables to the box and where we 
had the clamps, was screwed to the wood of the bench but there was no screws 
going through the metal sheet under the bench. The sheet showed no marks on it 
under the interesting parts (or elsewhere as I remember it). Of course, if the 
white little socket was rigged inside and the metal screws was long enough to 
go just through the wood, touching the metal sheet underneath, then the bench 
itself could lead current. I do not remember if I actually checked the bench 
frame for cables connected to it but I probably did. However, I have a close-up 
picture of the socket and it looks normal and the screws appear to be of normal 
size. I also have pictures of all the connectors going to the powermeter and of 
the frame on the floor. I took a picture every day of the connectors and cables 
to the powermeter in case anyone would tamper with them when we were out.

I lifted the control box to check what was under it and when doing so I 
tried to measure the weight and it is muck lighter than a car battery. The box 
itself has a weight, of course, and what is in it can not be much.

All these observations take away a number of ways to tamper with our 
measurements but there can still be things that we "didn't think of" and that 
is the reason why we only can claim "indications of" and not "proof of" 
anomalous heat production. We must have more control over the whole situation 
before we can talk about proof.

Best regards,
Torbjörn

END QUOTE


- Jed

[Vo]:Re: [Vo]:Re: [Vo]:Torbjörn Hartman describes power measurments

[Jed Rothwell]

Arrgh! This thread seems to have the "Vo: Re Vo:" problem, induced by the
server.

It seems that any character outside of the normal ascii set 0 - 9 a - Z
triggers this.

(This is a test message to see if the problem happens, responding to David
Roberson.)

- Jed