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 <dlrober...@aol.com> 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> <spacedr...@cumming.info> > To: vortex-l <vortex-l@eskimo.com> <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> <mix...@bigpond.com> > To: vortex-l <vortex-l@eskimo.com> <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 > > > >
