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
