Why this hangup about fundamental components? I can extract current from
an AC waveform any way I want. Switched mode power supplies usually do
this at 20kHz or so, even though the fundamental component is 60Hz.
But you are right about one thing - we may as well end this discussion.
It is like trying to explain the purpose and function of a Dewar vessel
to an ant!
Duncan
On 5/27/2013 1:55 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 <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
