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
