Re: [Vo]:Levi Hot Cat paper is a gem : power conditioner needed
So to continue this line of arithmetic, we have a factor of 10 gain to explain. First of all let's get rid of the Stefan Boltzmann amplification of error by taking the fourth root of 10: 10^(1/4) = 1.7782794 That means if we're looking for error as the source of the gain, we have to plausibly argue an error of 78% in the portion of the IR camera's calibration for Wein's displacement proportionality. Note, it is a proportionality -- a straight linear proportionality -- because we have removed the Stefan Boltzmann fourth power from the equation. Wein's displacement is an approximation of the Plank curve most accurate at higher frequencies -- where photons have higher energy. So if we're looking for errors in power measurement, we need to be most concerned about frequencies below the IR. The problem for those of us who want to find error in the measure is that the peak is in the camera's physical sensor bandwidth where we aren't extrapolating -- and the most likely source of error is in an area of the spectrum that not only has lower luminosity but lower energy per photon. Again, I've never seen one of these emotionally committed skeptics do so much as the simple arithmetic to come up with the factor of 10 figure for the November test let alone the 78% that results from discounting Stefan Boltzmann's sensitivity to error, let alone proceed from there to do the arithmetic to estimate what appears to be an insignificant residual error in the sensor's calibration software. That's why I laugh these people off. There's no point blather with people who refuse to do arithmetic regarding the strongest argument of their opponents. On Thu, May 23, 2013 at 2:39 PM, James Bowery jabow...@gmail.com wrote: I found the major error: The peak wavelength is in the infrared -- as it is with the sun -- and I intuitively thought that the fact that much of the surface was bright red thru yellow meant my picking dull red (700nm) was conservative. This then fed via Wien's law proportionately into the fourth power of Stefan Boltzmann's law to produce the 2MW. This arose because I simply neglected to go to the next page after page 2 -- where Figure 3 shows the temperature as 793C or 1066K. Recalculating from the substitution for Th: q=2.40137205*10^-9*pi*(Th^4-Tc^4) q=2.40137205*10^-9*pi*(1291304958736-Tc^4) ; subst(1066, Th) q=3084.152246988637*pi ; subst(289, Tc) q=9689W On Wed, May 22, 2013 at 6:58 PM, James Bowery jabow...@gmail.com wrote: I can't resist: What power level is required to get that device to barely enter the visible wavelengths (700nm), again, assuming no losses other than black body? again using http://www.ajdesigner.com/phpwien/wien_equation_t.php at 700nm: blackbody temperature (T) = 4139.6692857143 kelvin q=2.40137205*10^-9*pi*(Th^4-Tc^4) q=2.40137205*10^-9*pi*(2.9367203218388994*10^14-Tc^4) ; subst(4139.6692857143, Th) q=705199.0585641474*pi q=2.2154481E6W Yeah, Rossi had a really high frequency power supply pumping even 1/10th of that into the E-Cat HT. On Wed, May 22, 2013 at 6:40 PM, James Bowery jabow...@gmail.com wrote: One final erratum (hopefully): In the November run when the device overheated to visible wavelengths, the input power was 1kW (p2), not 360W. Therefore: 360=2.40137205*10^-9*pi*(Th^4-6975757441) 1000=2.40137205*10^-9*pi*(Th^4-6975757441) ; subst(1000, 360) Th=(59549289748750/pi+997533314063)^(1/4)/143^(1/4) ; solve(Th) Th=611.17587 Kelvin Th=338.026 Celsius using: http://www.ajdesigner.com/phpwien/wien_equation.php peak emission wavelength (λmax) = 4.741300568689E-6 meter Still deep into the infrared. On Wed, May 22, 2013 at 5:59 PM, James Bowery jabow...@gmail.comwrote: Erratum: I also left out the substitution step for room temperature: 360=2.40137205*10^-9*pi*(Th^4-6975757441) ; subst(289) On Wed, May 22, 2013 at 5:53 PM, James Bowery jabow...@gmail.comwrote: Erratum: Strike the So, what... On Wed, May 22, 2013 at 5:53 PM, James Bowery jabow...@gmail.comwrote: q=eps*s*(Th^4-Tc^4)*A q=eps*(2*pi*r^2+2*l*pi*r)*s*(Th^4-Tc^4) ; subst(2*pi*r^2+2*l*pi*r, A) q=5.6703*10^-8*eps*(2*pi*r^2+2*l*pi*r)*(Th^4-Tc^4) ; subst(5.6703e-8, s) q=5.6703*10^-8*eps*(0.11*l*pi+0.00605*pi)*(Th^4-Tc^4) ; subst(.055, r) q=2.40137205*10^-9*eps*pi*(Th^4-Tc^4) ; subst(.33, l) q=2.40137205*10^-9*pi*(Th^4-Tc^4) ; subst(1, eps) 360=2.40137205*10^-9*pi*(Th^4-Tc^4) ; subst(360, q) Th=(21437744309550/pi+997533314063)^(1/4)/143^(1/4) ; solve(Th) Th=483.6006 Kelvin Th=210.451 Celsius using: http://www.ajdesigner.com/phpwien/wien_equation.php peak emission wavelength (λmax) = 5.9920696955297E-6 meter or 6 micrometers That is with no losses other than black body radiation (ie: no convective losses). That is way into the infrared. The excursions into the visible wavelength occurred with 360W. So, what On Wed, May 22, 2013 at 4:19 PM, Jed Rothwell jedrothw...@gmail.comwrote: James
Re: [Vo]:Levi Hot Cat paper is a gem : power conditioner needed
Erratum: luminosity should read photon flux On Fri, May 24, 2013 at 11:16 AM, James Bowery jabow...@gmail.com wrote: So to continue this line of arithmetic, we have a factor of 10 gain to explain. First of all let's get rid of the Stefan Boltzmann amplification of error by taking the fourth root of 10: 10^(1/4) = 1.7782794 That means if we're looking for error as the source of the gain, we have to plausibly argue an error of 78% in the portion of the IR camera's calibration for Wein's displacement proportionality. Note, it is a proportionality -- a straight linear proportionality -- because we have removed the Stefan Boltzmann fourth power from the equation. Wein's displacement is an approximation of the Plank curve most accurate at higher frequencies -- where photons have higher energy. So if we're looking for errors in power measurement, we need to be most concerned about frequencies below the IR. The problem for those of us who want to find error in the measure is that the peak is in the camera's physical sensor bandwidth where we aren't extrapolating -- and the most likely source of error is in an area of the spectrum that not only has lower luminosity but lower energy per photon. Again, I've never seen one of these emotionally committed skeptics do so much as the simple arithmetic to come up with the factor of 10 figure for the November test let alone the 78% that results from discounting Stefan Boltzmann's sensitivity to error, let alone proceed from there to do the arithmetic to estimate what appears to be an insignificant residual error in the sensor's calibration software. That's why I laugh these people off. There's no point blather with people who refuse to do arithmetic regarding the strongest argument of their opponents. On Thu, May 23, 2013 at 2:39 PM, James Bowery jabow...@gmail.com wrote: I found the major error: The peak wavelength is in the infrared -- as it is with the sun -- and I intuitively thought that the fact that much of the surface was bright red thru yellow meant my picking dull red (700nm) was conservative. This then fed via Wien's law proportionately into the fourth power of Stefan Boltzmann's law to produce the 2MW. This arose because I simply neglected to go to the next page after page 2 -- where Figure 3 shows the temperature as 793C or 1066K. Recalculating from the substitution for Th: q=2.40137205*10^-9*pi*(Th^4-Tc^4) q=2.40137205*10^-9*pi*(1291304958736-Tc^4) ; subst(1066, Th) q=3084.152246988637*pi ; subst(289, Tc) q=9689W On Wed, May 22, 2013 at 6:58 PM, James Bowery jabow...@gmail.com wrote: I can't resist: What power level is required to get that device to barely enter the visible wavelengths (700nm), again, assuming no losses other than black body? again using http://www.ajdesigner.com/phpwien/wien_equation_t.php at 700nm: blackbody temperature (T) = 4139.6692857143 kelvin q=2.40137205*10^-9*pi*(Th^4-Tc^4) q=2.40137205*10^-9*pi*(2.9367203218388994*10^14-Tc^4) ; subst(4139.6692857143, Th) q=705199.0585641474*pi q=2.2154481E6W Yeah, Rossi had a really high frequency power supply pumping even 1/10th of that into the E-Cat HT. On Wed, May 22, 2013 at 6:40 PM, James Bowery jabow...@gmail.comwrote: One final erratum (hopefully): In the November run when the device overheated to visible wavelengths, the input power was 1kW (p2), not 360W. Therefore: 360=2.40137205*10^-9*pi*(Th^4-6975757441) 1000=2.40137205*10^-9*pi*(Th^4-6975757441) ; subst(1000, 360) Th=(59549289748750/pi+997533314063)^(1/4)/143^(1/4) ; solve(Th) Th=611.17587 Kelvin Th=338.026 Celsius using: http://www.ajdesigner.com/phpwien/wien_equation.php peak emission wavelength (λmax) = 4.741300568689E-6 meter Still deep into the infrared. On Wed, May 22, 2013 at 5:59 PM, James Bowery jabow...@gmail.comwrote: Erratum: I also left out the substitution step for room temperature: 360=2.40137205*10^-9*pi*(Th^4-6975757441) ; subst(289) On Wed, May 22, 2013 at 5:53 PM, James Bowery jabow...@gmail.comwrote: Erratum: Strike the So, what... On Wed, May 22, 2013 at 5:53 PM, James Bowery jabow...@gmail.comwrote: q=eps*s*(Th^4-Tc^4)*A q=eps*(2*pi*r^2+2*l*pi*r)*s*(Th^4-Tc^4) ; subst(2*pi*r^2+2*l*pi*r, A) q=5.6703*10^-8*eps*(2*pi*r^2+2*l*pi*r)*(Th^4-Tc^4) ; subst(5.6703e-8, s) q=5.6703*10^-8*eps*(0.11*l*pi+0.00605*pi)*(Th^4-Tc^4) ; subst(.055, r) q=2.40137205*10^-9*eps*pi*(Th^4-Tc^4) ; subst(.33, l) q=2.40137205*10^-9*pi*(Th^4-Tc^4) ; subst(1, eps) 360=2.40137205*10^-9*pi*(Th^4-Tc^4) ; subst(360, q) Th=(21437744309550/pi+997533314063)^(1/4)/143^(1/4) ; solve(Th) Th=483.6006 Kelvin Th=210.451 Celsius using: http://www.ajdesigner.com/phpwien/wien_equation.php peak emission wavelength (λmax) = 5.9920696955297E-6 meter or 6 micrometers That is with no losses other than black body radiation (ie: no convective losses). That is way into the infrared. The excursions
Re: [Vo]:Levi Hot Cat paper is a gem : power conditioner needed
On Fri, May 24, 2013 at 9:16 AM, James Bowery jabow...@gmail.com wrote: So if we're looking for errors in power measurement, we need to be most concerned about frequencies below the IR. The problem for those of us who want to find error in the measure is that the peak is in the camera's physical sensor bandwidth where we aren't extrapolating -- and the most likely source of error is in an area of the spectrum that not only has lower luminosity but lower energy per photon. I believe Lubos Motl proposed somewhere that the E-Cat HT surface is not well-approximated by a blackbody and that the true emissivity is likely to be T^(4+d), where 0 d 1; i.e., that in the worst case scenario there will be ~T^5 relationship between temperature and power rather than T^4. I do not know what to make of this (assuming I have accurately reproduced the details). Eric
Re: [Vo]:Levi Hot Cat paper is a gem : power conditioner needed
I wrote: I believe Lubos Motl proposed somewhere that the E-Cat HT surface is not well-approximated by a blackbody and that the true emissivity is likely to be T^(4+d), where 0 d 1; i.e., that in the worst case scenario there will be ~T^5 relationship between temperature and power rather than T^4. I do not know what to make of this (assuming I have accurately reproduced the details). That it was Lubos Motl was unintentional speculation on my part, drawing upon a comment by someone else in the comments to the recent Register article [1]. The person who wanted to modify the Stefan-Boltzmann equation was HolyFreakinGhost. Elsewhere there is speculation (from the real Motl) that the emissivity of metals is 0.2 or something on that order [2]. It seems pretty clear that the E-Cat HT was well painted with black paint; I do not see how this detail could have been a point of confusion. However, if Motl's value of ~0.2 were used for the emissivity, he estimates that the calculated power would be approximately equal to the input power. Eric [1] http://forums.theregister.co.uk/forum/1/2013/05/22/e_cat_test_claims_success_yet_again/#c_1833878 [2] http://motls.blogspot.com/2013/05/tommaso-dorigo-impressed-by-cold-fusion.html
Re: [Vo]:Levi Hot Cat paper is a gem : power conditioner needed
Here's what Motl says about it: The emissivity is set to one i.e. they assume the reactor to be a black body. This choice is labeled conservative. Except that the truth seems to be going exactly in the opposite direction. The actual emissivity is lower than one and it's the coefficient multiplying the fourth power of the absolute temperature to get the power. Because they seem to calculate the power from the measured temperature (the infrared camera is claimed to give the right temperature and automatically adjust the observed radiation for emissivity etc.; see page 7 of the paper), the actual power is actually much lower than [the calculated figure] 1609 watts. The emissivity of metalshttp://www.omega.com/literature/transactions/volume1/emissivitya.html at similar reasonable temperatures seems to be 0.2 or so – something of this order – which reduces 1609 watts to something like 300 watts, pretty much equal to the consumption. Obviously, despite the fact that he cites page 7 of the paper, he didn't read it since it describes how low emissivity setting for the camera software overestimates the temperature. Hell, even Joshua Cude understood that this is a wash in the bandwidth of the camera's physical sensor. What's wrong with Motl? On Fri, May 24, 2013 at 7:23 PM, Eric Walker eric.wal...@gmail.com wrote: I wrote: I believe Lubos Motl proposed somewhere that the E-Cat HT surface is not well-approximated by a blackbody and that the true emissivity is likely to be T^(4+d), where 0 d 1; i.e., that in the worst case scenario there will be ~T^5 relationship between temperature and power rather than T^4. I do not know what to make of this (assuming I have accurately reproduced the details). That it was Lubos Motl was unintentional speculation on my part, drawing upon a comment by someone else in the comments to the recent Register article [1]. The person who wanted to modify the Stefan-Boltzmann equation was HolyFreakinGhost. Elsewhere there is speculation (from the real Motl) that the emissivity of metals is 0.2 or something on that order [2]. It seems pretty clear that the E-Cat HT was well painted with black paint; I do not see how this detail could have been a point of confusion. However, if Motl's value of ~0.2 were used for the emissivity, he estimates that the calculated power would be approximately equal to the input power. Eric [1] http://forums.theregister.co.uk/forum/1/2013/05/22/e_cat_test_claims_success_yet_again/#c_1833878 [2] http://motls.blogspot.com/2013/05/tommaso-dorigo-impressed-by-cold-fusion.html
Re: [Vo]:Levi Hot Cat paper is a gem : power conditioner needed
On Wed, May 22, 2013 at 1:33 PM, Andrew andrew...@att.net wrote: ** Since the experimenters walked up to the experiment *after* it had been turned on, we don't know for sure whether the existing cabling was used to impart the RF, or a separate kickstart cable. There were three runs. The first run (November 2012) was abortive. The second run (December 2012) was already started when they began their measurements. It seems they were present during the third run (March 2013) when the E-Cat was started (p. 15). Eric
Re: [Vo]:Levi Hot Cat paper is a gem : power conditioner needed
I found the major error: The peak wavelength is in the infrared -- as it is with the sun -- and I intuitively thought that the fact that much of the surface was bright red thru yellow meant my picking dull red (700nm) was conservative. This then fed via Wien's law proportionately into the fourth power of Stefan Boltzmann's law to produce the 2MW. This arose because I simply neglected to go to the next page after page 2 -- where Figure 3 shows the temperature as 793C or 1066K. Recalculating from the substitution for Th: q=2.40137205*10^-9*pi*(Th^4-Tc^4) q=2.40137205*10^-9*pi*(1291304958736-Tc^4) ; subst(1066, Th) q=3084.152246988637*pi ; subst(289, Tc) q=9689W On Wed, May 22, 2013 at 6:58 PM, James Bowery jabow...@gmail.com wrote: I can't resist: What power level is required to get that device to barely enter the visible wavelengths (700nm), again, assuming no losses other than black body? again using http://www.ajdesigner.com/phpwien/wien_equation_t.php at 700nm: blackbody temperature (T) = 4139.6692857143 kelvin q=2.40137205*10^-9*pi*(Th^4-Tc^4) q=2.40137205*10^-9*pi*(2.9367203218388994*10^14-Tc^4) ; subst(4139.6692857143, Th) q=705199.0585641474*pi q=2.2154481E6W Yeah, Rossi had a really high frequency power supply pumping even 1/10th of that into the E-Cat HT. On Wed, May 22, 2013 at 6:40 PM, James Bowery jabow...@gmail.com wrote: One final erratum (hopefully): In the November run when the device overheated to visible wavelengths, the input power was 1kW (p2), not 360W. Therefore: 360=2.40137205*10^-9*pi*(Th^4-6975757441) 1000=2.40137205*10^-9*pi*(Th^4-6975757441) ; subst(1000, 360) Th=(59549289748750/pi+997533314063)^(1/4)/143^(1/4) ; solve(Th) Th=611.17587 Kelvin Th=338.026 Celsius using: http://www.ajdesigner.com/phpwien/wien_equation.php peak emission wavelength (λmax) = 4.741300568689E-6 meter Still deep into the infrared. On Wed, May 22, 2013 at 5:59 PM, James Bowery jabow...@gmail.com wrote: Erratum: I also left out the substitution step for room temperature: 360=2.40137205*10^-9*pi*(Th^4-6975757441) ; subst(289) On Wed, May 22, 2013 at 5:53 PM, James Bowery jabow...@gmail.comwrote: Erratum: Strike the So, what... On Wed, May 22, 2013 at 5:53 PM, James Bowery jabow...@gmail.comwrote: q=eps*s*(Th^4-Tc^4)*A q=eps*(2*pi*r^2+2*l*pi*r)*s*(Th^4-Tc^4) ; subst(2*pi*r^2+2*l*pi*r, A) q=5.6703*10^-8*eps*(2*pi*r^2+2*l*pi*r)*(Th^4-Tc^4) ; subst(5.6703e-8, s) q=5.6703*10^-8*eps*(0.11*l*pi+0.00605*pi)*(Th^4-Tc^4) ; subst(.055, r) q=2.40137205*10^-9*eps*pi*(Th^4-Tc^4) ; subst(.33, l) q=2.40137205*10^-9*pi*(Th^4-Tc^4) ; subst(1, eps) 360=2.40137205*10^-9*pi*(Th^4-Tc^4) ; subst(360, q) Th=(21437744309550/pi+997533314063)^(1/4)/143^(1/4) ; solve(Th) Th=483.6006 Kelvin Th=210.451 Celsius using: http://www.ajdesigner.com/phpwien/wien_equation.php peak emission wavelength (λmax) = 5.9920696955297E-6 meter or 6 micrometers That is with no losses other than black body radiation (ie: no convective losses). That is way into the infrared. The excursions into the visible wavelength occurred with 360W. So, what On Wed, May 22, 2013 at 4:19 PM, Jed Rothwell jedrothw...@gmail.comwrote: James Bowery jabow...@gmail.com wrote: There is value in pursuing reductio ad absurda when they engage one of the strongest arguments that the demonstration is valid: That the power input could not conceivably have produced the radiation wavelengths observed. You have mentioned that several times. Can you please post a more detailed discussion of that, with equations and examples? That would be helpful. Please post this in a new thread so I can find it easily. You might also address the fact that the first device melted. - Jed
Re: [Vo]:Levi Hot Cat paper is a gem : power conditioner needed
Electrical INPUT is a two-edged sword. It can be measured to 6 decimal places .. IF you do it correctly, but if you don't cover ALL bases you might miss something. (eg an AC-only meter might not notice DC, or HF AC beyond its spec). I've come to the conclusion that the only way to overcome the power-side fake is to put a power conditioner between Rossi's power plug (maybe miswired per Bryce etc, or with a DC component) and his control box. I'd recommend a motor-generator, as it gives a nice sine output. Then the meter will work correctly, between the conditioner and the control box. They run at 95% + efficiency. You can probably rent one for 6 months.
Re: [Vo]:Levi Hot Cat paper is a gem : power conditioner needed
Alan Fletcher a...@well.com wrote: I've come to the conclusion that the only way to overcome the power-side fake is to put a power conditioner between Rossi's power plug (maybe miswired per Bryce etc, or with a DC component) and his control box. That would do it. But the fact is, any $20 watt meter would also do it. Experts tell me there is no way you can fool one. They are better than meters costing thousands of dollars were 20 years ago. Levi has one of those things. I expect he used it. He did in previous tests. I suggest you should stop fantasizing about this. Rossi did not take apart the wall and install secret equipment that he turned on and then turned off during the calibration. He did not find a way to send so much power through an ordinary electric cord that he melted steel and ceramic. That is not possible. You can dismiss it from your mind. The electric cord would have burned. The other gadgets such as the computers plugged into that circuit would have been roached. - Jed
RE: [Vo]:Levi Hot Cat paper is a gem : power conditioner needed
-Original Message- From: Alan Fletcher [mailto:a...@well.com] Sent: mercredi 22 mai 2013 22:19 To: vortex-l@eskimo.com Subject: Re: [Vo]:Levi Hot Cat paper is a gem : power conditioner needed Electrical INPUT is a two-edged sword. It can be measured to 6 decimal places .. IF you do it correctly, but if you don't cover ALL bases you might miss something. (eg an AC-only meter might not notice DC, or HF AC beyond its spec). I've come to the conclusion that the only way to overcome the power-side fake is to put a power conditioner between Rossi's power plug (maybe miswired per Bryce etc, or with a DC component) and his control box. I'd recommend a motor-generator, as it gives a nice sine output. Then the meter will work correctly, between the conditioner and the control box. A basic quality control check of the power-side will be at first a good step. The idea to put conditioner between Rossi's plug and the wall power socket will remove any doubt for a miswired error (Voluntary or not)
Re: [Vo]:Levi Hot Cat paper is a gem : power conditioner needed
I doubt that Rossi would allow a power conditioner, because he himself states that there is some initial RF powering going on to kickstart the device. Since the experimenters walked up to the experiment after it had been turned on, we don't know for sure whether the existing cabling was used to impart the RF, or a separate kickstart cable. Were I to guess, I would assume that the existing cabling was used, and that the RF generator resides in the control box. There's a whole lot of detail about the input side that would benefit from the light of day. What's required is an interview with the Swedes from someone who understands the issues. Andrew - Original Message - From: Jed Rothwell To: vortex-l@eskimo.com Sent: Wednesday, May 22, 2013 1:27 PM Subject: Re: [Vo]:Levi Hot Cat paper is a gem : power conditioner needed Alan Fletcher a...@well.com wrote: I've come to the conclusion that the only way to overcome the power-side fake is to put a power conditioner between Rossi's power plug (maybe miswired per Bryce etc, or with a DC component) and his control box. That would do it. But the fact is, any $20 watt meter would also do it. Experts tell me there is no way you can fool one. They are better than meters costing thousands of dollars were 20 years ago. Levi has one of those things. I expect he used it. He did in previous tests. I suggest you should stop fantasizing about this. Rossi did not take apart the wall and install secret equipment that he turned on and then turned off during the calibration. He did not find a way to send so much power through an ordinary electric cord that he melted steel and ceramic. That is not possible. You can dismiss it from your mind. The electric cord would have burned. The other gadgets such as the computers plugged into that circuit would have been roached. - Jed
Re: [Vo]:Levi Hot Cat paper is a gem : power conditioner needed
Andrew andrew...@att.net wrote: There's a whole lot of detail about the input side that would benefit from the light of day. What's required is an interview with the Swedes from someone who understands the issues. And who understands Swedish. Any volunteers? - Jed
Re: [Vo]:Levi Hot Cat paper is a gem : power conditioner needed
Talar ni Svenska. Not much, anyway. - Original Message - From: Jed Rothwell To: vortex-l@eskimo.com Sent: Wednesday, May 22, 2013 1:46 PM Subject: Re: [Vo]:Levi Hot Cat paper is a gem : power conditioner needed Andrew andrew...@att.net wrote: There's a whole lot of detail about the input side that would benefit from the light of day. What's required is an interview with the Swedes from someone who understands the issues. And who understands Swedish. Any volunteers? - Jed
Re: [Vo]:Levi Hot Cat paper is a gem : power conditioner needed
On Wed, May 22, 2013 at 3:33 PM, Andrew andrew...@att.net wrote: ** I doubt that Rossi would allow a power conditioner, because he himself states that there is some initial RF powering going on to kickstart the device. You misunderstand: The power conditioner would be placed between the wall socket and any other equipment in the room, including the RF generator.
Re: [Vo]:Levi Hot Cat paper is a gem : power conditioner needed
On Wed, May 22, 2013 at 3:27 PM, Jed Rothwell jedrothw...@gmail.com wrote: ... I suggest you should stop fantasizing about this. Rossi did not take apart the wall and install secret equipment that he turned on and then turned off during the calibration. He did not find a way to send so much power through an ordinary electric cord that he melted steel and ceramic. That is not possible. You can dismiss it from your mind. The electric cord would have burned. The other gadgets such as the computers plugged into that circuit would have been roached. There is value in pursuing reductio ad absurda when they engage one of the strongest arguments that the demonstration is valid: That the power input could not conceivably have produced the radiation wavelengths observed.
Re: [Vo]:Levi Hot Cat paper is a gem : power conditioner needed
James Bowery jabow...@gmail.com wrote: There is value in pursuing reductio ad absurda when they engage one of the strongest arguments that the demonstration is valid: That the power input could not conceivably have produced the radiation wavelengths observed. You have mentioned that several times. Can you please post a more detailed discussion of that, with equations and examples? That would be helpful. Please post this in a new thread so I can find it easily. You might also address the fact that the first device melted. - Jed
Re: [Vo]:Levi Hot Cat paper is a gem : power conditioner needed
I wrote: That the power input could not conceivably have produced the radiation wavelengths observed. You have mentioned that several times. Can you please post a more detailed discussion of that, with equations and examples? I realize you challenged Mary Yugo and other skeptics to do this analysis. That is a forlorn hope. They will not do it. So, why don't you do it? I would appreciate that. Others here can check your work. - Jed
Re: [Vo]:Levi Hot Cat paper is a gem : power conditioner needed
Did anyone scope the the power in for 50Hz? Or allow the researchers to choose any outlet? I imagine anything on the same heater circuit would fry if someone tried to insert an extra 500 watts. A light bulb added to the circuit would have detected additional power... or any decent UPS will include power line conditioning which will deliever a pure sine wave AC voltage.. Besides a circuit diagram showing all of the inputs, outputs, and test equipment, are there some other notes that should be added to the report? Anyone know if peer review is in the works? I think its exciting to think that this test was done accurately and we can anticipate that there will be more positive tests like this to follow in the very near future. - Brad On Wed, May 22, 2013 at 2:22 PM, Jed Rothwell jedrothw...@gmail.com wrote: I wrote: That the power input could not conceivably have produced the radiation wavelengths observed. You have mentioned that several times. Can you please post a more detailed discussion of that, with equations and examples? I realize you challenged Mary Yugo and other skeptics to do this analysis. That is a forlorn hope. They will not do it. So, why don't you do it? I would appreciate that. Others here can check your work. - Jed
Re: [Vo]:Levi Hot Cat paper is a gem : power conditioner needed
Realistic speaking, to get a respectable scientists or engineers doing formal peer review for a magazine is an impossible task right now. So, this is a catch 22 problem to begin with. 2013/5/22 Brad Lowe ecatbuil...@gmail.com Besides a circuit diagram showing all of the inputs, outputs, and test equipment, are there some other notes that should be added to the report? Anyone know if peer review is in the works? -- Daniel Rocha - RJ danieldi...@gmail.com
Re: [Vo]:Levi Hot Cat paper is a gem : power conditioner needed
q=eps*s*(Th^4-Tc^4)*A q=eps*(2*pi*r^2+2*l*pi*r)*s*(Th^4-Tc^4) ; subst(2*pi*r^2+2*l*pi*r, A) q=5.6703*10^-8*eps*(2*pi*r^2+2*l*pi*r)*(Th^4-Tc^4) ; subst(5.6703e-8, s) q=5.6703*10^-8*eps*(0.11*l*pi+0.00605*pi)*(Th^4-Tc^4) ; subst(.055, r) q=2.40137205*10^-9*eps*pi*(Th^4-Tc^4) ; subst(.33, l) q=2.40137205*10^-9*pi*(Th^4-Tc^4) ; subst(1, eps) 360=2.40137205*10^-9*pi*(Th^4-Tc^4) ; subst(360, q) Th=(21437744309550/pi+997533314063)^(1/4)/143^(1/4) ; solve(Th) Th=483.6006 Kelvin Th=210.451 Celsius using: http://www.ajdesigner.com/phpwien/wien_equation.php peak emission wavelength (λmax) = 5.9920696955297E-6 meter or 6 micrometers That is with no losses other than black body radiation (ie: no convective losses). That is way into the infrared. The excursions into the visible wavelength occurred with 360W. So, what On Wed, May 22, 2013 at 4:19 PM, Jed Rothwell jedrothw...@gmail.com wrote: James Bowery jabow...@gmail.com wrote: There is value in pursuing reductio ad absurda when they engage one of the strongest arguments that the demonstration is valid: That the power input could not conceivably have produced the radiation wavelengths observed. You have mentioned that several times. Can you please post a more detailed discussion of that, with equations and examples? That would be helpful. Please post this in a new thread so I can find it easily. You might also address the fact that the first device melted. - Jed
Re: [Vo]:Levi Hot Cat paper is a gem : power conditioner needed
Erratum: Strike the So, what... On Wed, May 22, 2013 at 5:53 PM, James Bowery jabow...@gmail.com wrote: q=eps*s*(Th^4-Tc^4)*A q=eps*(2*pi*r^2+2*l*pi*r)*s*(Th^4-Tc^4) ; subst(2*pi*r^2+2*l*pi*r, A) q=5.6703*10^-8*eps*(2*pi*r^2+2*l*pi*r)*(Th^4-Tc^4) ; subst(5.6703e-8, s) q=5.6703*10^-8*eps*(0.11*l*pi+0.00605*pi)*(Th^4-Tc^4) ; subst(.055, r) q=2.40137205*10^-9*eps*pi*(Th^4-Tc^4) ; subst(.33, l) q=2.40137205*10^-9*pi*(Th^4-Tc^4) ; subst(1, eps) 360=2.40137205*10^-9*pi*(Th^4-Tc^4) ; subst(360, q) Th=(21437744309550/pi+997533314063)^(1/4)/143^(1/4) ; solve(Th) Th=483.6006 Kelvin Th=210.451 Celsius using: http://www.ajdesigner.com/phpwien/wien_equation.php peak emission wavelength (λmax) = 5.9920696955297E-6 meter or 6 micrometers That is with no losses other than black body radiation (ie: no convective losses). That is way into the infrared. The excursions into the visible wavelength occurred with 360W. So, what On Wed, May 22, 2013 at 4:19 PM, Jed Rothwell jedrothw...@gmail.comwrote: James Bowery jabow...@gmail.com wrote: There is value in pursuing reductio ad absurda when they engage one of the strongest arguments that the demonstration is valid: That the power input could not conceivably have produced the radiation wavelengths observed. You have mentioned that several times. Can you please post a more detailed discussion of that, with equations and examples? That would be helpful. Please post this in a new thread so I can find it easily. You might also address the fact that the first device melted. - Jed
Re: [Vo]:Levi Hot Cat paper is a gem : power conditioner needed
Erratum: I also left out the substitution step for room temperature: 360=2.40137205*10^-9*pi*(Th^4-6975757441) ; subst(289) On Wed, May 22, 2013 at 5:53 PM, James Bowery jabow...@gmail.com wrote: Erratum: Strike the So, what... On Wed, May 22, 2013 at 5:53 PM, James Bowery jabow...@gmail.com wrote: q=eps*s*(Th^4-Tc^4)*A q=eps*(2*pi*r^2+2*l*pi*r)*s*(Th^4-Tc^4) ; subst(2*pi*r^2+2*l*pi*r, A) q=5.6703*10^-8*eps*(2*pi*r^2+2*l*pi*r)*(Th^4-Tc^4) ; subst(5.6703e-8, s) q=5.6703*10^-8*eps*(0.11*l*pi+0.00605*pi)*(Th^4-Tc^4) ; subst(.055, r) q=2.40137205*10^-9*eps*pi*(Th^4-Tc^4) ; subst(.33, l) q=2.40137205*10^-9*pi*(Th^4-Tc^4) ; subst(1, eps) 360=2.40137205*10^-9*pi*(Th^4-Tc^4) ; subst(360, q) Th=(21437744309550/pi+997533314063)^(1/4)/143^(1/4) ; solve(Th) Th=483.6006 Kelvin Th=210.451 Celsius using: http://www.ajdesigner.com/phpwien/wien_equation.php peak emission wavelength (λmax) = 5.9920696955297E-6 meter or 6 micrometers That is with no losses other than black body radiation (ie: no convective losses). That is way into the infrared. The excursions into the visible wavelength occurred with 360W. So, what On Wed, May 22, 2013 at 4:19 PM, Jed Rothwell jedrothw...@gmail.comwrote: James Bowery jabow...@gmail.com wrote: There is value in pursuing reductio ad absurda when they engage one of the strongest arguments that the demonstration is valid: That the power input could not conceivably have produced the radiation wavelengths observed. You have mentioned that several times. Can you please post a more detailed discussion of that, with equations and examples? That would be helpful. Please post this in a new thread so I can find it easily. You might also address the fact that the first device melted. - Jed
Re: [Vo]:Levi Hot Cat paper is a gem : power conditioner needed
peak emission wavelength (λmax) = 5.9920696955297E-6 meter or 6 micrometers That is about the diameter of the Rossi micro-powder, could there be a dipole blackbody resonant condition at work here? Of course there is! On Wed, May 22, 2013 at 6:59 PM, James Bowery jabow...@gmail.com wrote: Erratum: I also left out the substitution step for room temperature: 360=2.40137205*10^-9*pi*(Th^4-6975757441) ; subst(289) On Wed, May 22, 2013 at 5:53 PM, James Bowery jabow...@gmail.com wrote: Erratum: Strike the So, what... On Wed, May 22, 2013 at 5:53 PM, James Bowery jabow...@gmail.com wrote: q=eps*s*(Th^4-Tc^4)*A q=eps*(2*pi*r^2+2*l*pi*r)*s*(Th^4-Tc^4) ; subst(2*pi*r^2+2*l*pi*r, A) q=5.6703*10^-8*eps*(2*pi*r^2+2*l*pi*r)*(Th^4-Tc^4) ; subst(5.6703e-8, s) q=5.6703*10^-8*eps*(0.11*l*pi+0.00605*pi)*(Th^4-Tc^4) ; subst(.055, r) q=2.40137205*10^-9*eps*pi*(Th^4-Tc^4) ; subst(.33, l) q=2.40137205*10^-9*pi*(Th^4-Tc^4) ; subst(1, eps) 360=2.40137205*10^-9*pi*(Th^4-Tc^4) ; subst(360, q) Th=(21437744309550/pi+997533314063)^(1/4)/143^(1/4) ; solve(Th) Th=483.6006 Kelvin Th=210.451 Celsius using: http://www.ajdesigner.com/phpwien/wien_equation.php peak emission wavelength (λmax) = 5.9920696955297E-6 meter or 6 micrometers That is with no losses other than black body radiation (ie: no convective losses). That is way into the infrared. The excursions into the visible wavelength occurred with 360W. So, what On Wed, May 22, 2013 at 4:19 PM, Jed Rothwell jedrothw...@gmail.comwrote: James Bowery jabow...@gmail.com wrote: There is value in pursuing reductio ad absurda when they engage one of the strongest arguments that the demonstration is valid: That the power input could not conceivably have produced the radiation wavelengths observed. You have mentioned that several times. Can you please post a more detailed discussion of that, with equations and examples? That would be helpful. Please post this in a new thread so I can find it easily. You might also address the fact that the first device melted. - Jed
Re: [Vo]:Levi Hot Cat paper is a gem : power conditioner needed
One final erratum (hopefully): In the November run when the device overheated to visible wavelengths, the input power was 1kW (p2), not 360W. Therefore: 360=2.40137205*10^-9*pi*(Th^4-6975757441) 1000=2.40137205*10^-9*pi*(Th^4-6975757441) ; subst(1000, 360) Th=(59549289748750/pi+997533314063)^(1/4)/143^(1/4) ; solve(Th) Th=611.17587 Kelvin Th=338.026 Celsius using: http://www.ajdesigner.com/phpwien/wien_equation.php peak emission wavelength (λmax) = 4.741300568689E-6 meter Still deep into the infrared. On Wed, May 22, 2013 at 5:59 PM, James Bowery jabow...@gmail.com wrote: Erratum: I also left out the substitution step for room temperature: 360=2.40137205*10^-9*pi*(Th^4-6975757441) ; subst(289) On Wed, May 22, 2013 at 5:53 PM, James Bowery jabow...@gmail.com wrote: Erratum: Strike the So, what... On Wed, May 22, 2013 at 5:53 PM, James Bowery jabow...@gmail.com wrote: q=eps*s*(Th^4-Tc^4)*A q=eps*(2*pi*r^2+2*l*pi*r)*s*(Th^4-Tc^4) ; subst(2*pi*r^2+2*l*pi*r, A) q=5.6703*10^-8*eps*(2*pi*r^2+2*l*pi*r)*(Th^4-Tc^4) ; subst(5.6703e-8, s) q=5.6703*10^-8*eps*(0.11*l*pi+0.00605*pi)*(Th^4-Tc^4) ; subst(.055, r) q=2.40137205*10^-9*eps*pi*(Th^4-Tc^4) ; subst(.33, l) q=2.40137205*10^-9*pi*(Th^4-Tc^4) ; subst(1, eps) 360=2.40137205*10^-9*pi*(Th^4-Tc^4) ; subst(360, q) Th=(21437744309550/pi+997533314063)^(1/4)/143^(1/4) ; solve(Th) Th=483.6006 Kelvin Th=210.451 Celsius using: http://www.ajdesigner.com/phpwien/wien_equation.php peak emission wavelength (λmax) = 5.9920696955297E-6 meter or 6 micrometers That is with no losses other than black body radiation (ie: no convective losses). That is way into the infrared. The excursions into the visible wavelength occurred with 360W. So, what On Wed, May 22, 2013 at 4:19 PM, Jed Rothwell jedrothw...@gmail.comwrote: James Bowery jabow...@gmail.com wrote: There is value in pursuing reductio ad absurda when they engage one of the strongest arguments that the demonstration is valid: That the power input could not conceivably have produced the radiation wavelengths observed. You have mentioned that several times. Can you please post a more detailed discussion of that, with equations and examples? That would be helpful. Please post this in a new thread so I can find it easily. You might also address the fact that the first device melted. - Jed
Re: [Vo]:Levi Hot Cat paper is a gem : power conditioner needed
I can't resist: What power level is required to get that device to barely enter the visible wavelengths (700nm), again, assuming no losses other than black body? again using http://www.ajdesigner.com/phpwien/wien_equation_t.php at 700nm: blackbody temperature (T) = 4139.6692857143 kelvin q=2.40137205*10^-9*pi*(Th^4-Tc^4) q=2.40137205*10^-9*pi*(2.9367203218388994*10^14-Tc^4) ; subst(4139.6692857143, Th) q=705199.0585641474*pi q=2.2154481E6W Yeah, Rossi had a really high frequency power supply pumping even 1/10th of that into the E-Cat HT. On Wed, May 22, 2013 at 6:40 PM, James Bowery jabow...@gmail.com wrote: One final erratum (hopefully): In the November run when the device overheated to visible wavelengths, the input power was 1kW (p2), not 360W. Therefore: 360=2.40137205*10^-9*pi*(Th^4-6975757441) 1000=2.40137205*10^-9*pi*(Th^4-6975757441) ; subst(1000, 360) Th=(59549289748750/pi+997533314063)^(1/4)/143^(1/4) ; solve(Th) Th=611.17587 Kelvin Th=338.026 Celsius using: http://www.ajdesigner.com/phpwien/wien_equation.php peak emission wavelength (λmax) = 4.741300568689E-6 meter Still deep into the infrared. On Wed, May 22, 2013 at 5:59 PM, James Bowery jabow...@gmail.com wrote: Erratum: I also left out the substitution step for room temperature: 360=2.40137205*10^-9*pi*(Th^4-6975757441) ; subst(289) On Wed, May 22, 2013 at 5:53 PM, James Bowery jabow...@gmail.com wrote: Erratum: Strike the So, what... On Wed, May 22, 2013 at 5:53 PM, James Bowery jabow...@gmail.comwrote: q=eps*s*(Th^4-Tc^4)*A q=eps*(2*pi*r^2+2*l*pi*r)*s*(Th^4-Tc^4) ; subst(2*pi*r^2+2*l*pi*r, A) q=5.6703*10^-8*eps*(2*pi*r^2+2*l*pi*r)*(Th^4-Tc^4) ; subst(5.6703e-8, s) q=5.6703*10^-8*eps*(0.11*l*pi+0.00605*pi)*(Th^4-Tc^4) ; subst(.055, r) q=2.40137205*10^-9*eps*pi*(Th^4-Tc^4) ; subst(.33, l) q=2.40137205*10^-9*pi*(Th^4-Tc^4) ; subst(1, eps) 360=2.40137205*10^-9*pi*(Th^4-Tc^4) ; subst(360, q) Th=(21437744309550/pi+997533314063)^(1/4)/143^(1/4) ; solve(Th) Th=483.6006 Kelvin Th=210.451 Celsius using: http://www.ajdesigner.com/phpwien/wien_equation.php peak emission wavelength (λmax) = 5.9920696955297E-6 meter or 6 micrometers That is with no losses other than black body radiation (ie: no convective losses). That is way into the infrared. The excursions into the visible wavelength occurred with 360W. So, what On Wed, May 22, 2013 at 4:19 PM, Jed Rothwell jedrothw...@gmail.comwrote: James Bowery jabow...@gmail.com wrote: There is value in pursuing reductio ad absurda when they engage one of the strongest arguments that the demonstration is valid: That the power input could not conceivably have produced the radiation wavelengths observed. You have mentioned that several times. Can you please post a more detailed discussion of that, with equations and examples? That would be helpful. Please post this in a new thread so I can find it easily. You might also address the fact that the first device melted. - Jed
