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 <andrew...@att.net>
To: vortex-l <vortex-l@eskimo.com>
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 <spacedr...@cumming.info>
To: vortex-l <vortex-l@eskimo.com>
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 <spacedr...@cumming.info>
To: vortex-l <vortex-l@eskimo.com>
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 <mix...@bigpond.com>
To: vortex-l <vortex-l@eskimo.com>
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
