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		<title>QuantumHeat.org</title>
		<description>Discuss QuantumHeat.org</description>
		<link>http://www.quantumheat.org</link>
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			<title>David Roberson says:</title>
			<link>http://www.quantumheat.org#comment-602</link>
			<description><![CDATA[@123Star This is maddening. I have used your forth order curve along with a modified quadratic and can get good fits with several different combinations. I have no idea as to how to allocate the radiation with the other possible heat escapes at this time. We need to find some method that allows us to actually measure the radiation. This is one of those times when having too many good possibilities prevents us from finding the true function. I agree, it is time to let this puppy rest for a while and maybe later something will come to us that solves the problem. Where is a good stroke of lightning when you need it?]]></description>
			<dc:creator>David Roberson</dc:creator>
			<pubDate>Tue, 20 Nov 2012 02:11:26 +0000</pubDate>
			<guid>http://www.quantumheat.org#comment-602</guid>
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			<title>123star says:</title>
			<link>http://www.quantumheat.org#comment-600</link>
			<description><![CDATA[Quadratic interpolation with gnuplot (the parameters are a bit different from yours) ------------- f(x) = a*x**2+ b*x +c a = 0.00190794 +/- 4.171e-05 (2.186%) b = -0.817727 +/- 0.03134 (3.833%) c = 75.0295 +/- 5.78 (7.704%) rms of residuals: 0.25192 ------------- Conclusions?]]></description>
			<dc:creator>123star</dc:creator>
			<pubDate>Tue, 20 Nov 2012 01:35:34 +0000</pubDate>
			<guid>http://www.quantumheat.org#comment-600</guid>
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			<title>123star says:</title>
			<link>http://www.quantumheat.org#comment-599</link>
			<description><![CDATA[Ohh I get it, "Calculated power" are just the interpolated values (it fitted TOO nicely with your function, :D )! I redo everything using the 2nd column: --------------- ------ f(x)= a*5.67E-8*(x)** 4 +c a = 0.0492754 +/- 0.0007454 (1.513%) c = -19.8548 +/- 1.19 (5.995%) RMS = 1.33571 --------- f(x)= a*5.67E-8*(x)** 4 +c + b*x a = 0.0394318 +/- 0.001125 (2.854%) b = 0.125228 +/- 0.01412 (11.28%) c = -54.3717 +/- 3.904 (7.18%) RMS = 0.328594 --------------- --- "free to move absolute zero" (parameter b) f(x)= a*5.67E-8*(x-b) **4 +c a = 0.0245443 +/- 0.00175 (7.13%) b = -100.98 +/- 11.53 (11.42%) c = -34.6504 +/- 1.727 (4.985%) RMS = 0.26854 --------------- ------------]]></description>
			<dc:creator>123star</dc:creator>
			<pubDate>Tue, 20 Nov 2012 01:29:36 +0000</pubDate>
			<guid>http://www.quantumheat.org#comment-599</guid>
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			<title>123star says:</title>
			<link>http://www.quantumheat.org#comment-598</link>
			<description><![CDATA[@David Is "Calculated power" a T_ambient corrected version of P_in? I used the third column. However you're right, with these I get: --------- f(x)= a*5.67E-8*(x)** 4 +c a = 0.0492969 +/- 0.0007695 (1.561%) c = -19.8906 +/- 1.229 (6.178%) RMS = 1.37898 --------- f(x)= a*5.67E-8*(x)** 4 +c + b*x a = 0.0392439 +/- 0.001385 (3.529%) b = 0.127891 +/- 0.01738 (13.59%) c = -55.1415 +/- 4.805 (8.714%) RMS = 0.404462 --------------- --- "free to move absolute zero" (parameter b) f(x)= a*5.67E-8*(x-b) **4 +c a = 0.0240849 +/- 0.002048 (8.502%) b = -104.123 +/- 13.83 (13.29%) c = -35.1599 +/- 2.074 (5.9%) RMS = 0.318427 Absolute 0 is way off (not zero). --------]]></description>
			<dc:creator>123star</dc:creator>
			<pubDate>Tue, 20 Nov 2012 01:19:48 +0000</pubDate>
			<guid>http://www.quantumheat.org#comment-598</guid>
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			<title>David Roberson says:</title>
			<link>http://www.quantumheat.org#comment-597</link>
			<description><![CDATA[Ok Star, I obtained mine from the viewer. The times can be found in one of my posts below but here they are: Temperature K Power_In(Watts) Calculated Power 295.6985 0 0 331.4818 13.6581 13.59494 376.827 37.9368 37.82081 401.499 54.3546 54.28809 427.494 73.7296 74.14389 439.918 84.4947 84.54173 452.173 95.622 95.37347 P(T)=.001902*T* T -.813*T +74.106 The forth order fit was not very good when these temperatures were applied. I will see how my quadratic works with yours.]]></description>
			<dc:creator>David Roberson</dc:creator>
			<pubDate>Tue, 20 Nov 2012 00:37:50 +0000</pubDate>
			<guid>http://www.quantumheat.org#comment-597</guid>
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			<title>123star says:</title>
			<link>http://www.quantumheat.org#comment-593</link>
			<description><![CDATA[@David Can you post me your data set, so that I can check too?]]></description>
			<dc:creator>123star</dc:creator>
			<pubDate>Mon, 19 Nov 2012 23:39:57 +0000</pubDate>
			<guid>http://www.quantumheat.org#comment-593</guid>
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			<title>123star says:</title>
			<link>http://www.quantumheat.org#comment-571</link>
			<description><![CDATA[@David Yes I let "b" free to test if goes to zero by itself (or - 273.15 if in Celsius). I get b = 19.576 +/- 3.88 which is incompatible with 0. If I set b=0 I get a higher rms, 0.4 W instead of 0.2. If I exclude the data points at lower temperatures (keeping only 6 of, it seems that the computed b is compatible with 0. I think this is due to the fact that convection/cond uction contribution is affecting the fit at lower temperatures.]]></description>
			<dc:creator>123star</dc:creator>
			<pubDate>Mon, 19 Nov 2012 20:29:55 +0000</pubDate>
			<guid>http://www.quantumheat.org#comment-571</guid>
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			<title>David Roberson says:</title>
			<link>http://www.quantumheat.org#comment-560</link>
			<description><![CDATA[@ 123Star I was looking over the posts and do not see the values that you are using for your constants. What does your fit want the a,b,c, etc. components to be? Sorry if you posted them and I missed finding them. Could you post your final equation so that I can evaluate it as compared to the one I have derived? Thanks!]]></description>
			<dc:creator>David Roberson</dc:creator>
			<pubDate>Mon, 19 Nov 2012 16:42:07 +0000</pubDate>
			<guid>http://www.quantumheat.org#comment-560</guid>
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			<title>David Roberson says:</title>
			<link>http://www.quantumheat.org#comment-559</link>
			<description><![CDATA[@123Star You need to eliminate the 'b' term of your equation if it is to represent the Stefan-Boltzman n function. The Temperature input is absolute in the real world and can not have an offset. If the offset is required, then your equation has a problem. Give your curve fitting routine another try without that offset and lets evaluate the match. If it still exhibits the performance you demonstrate, then you may be on to something. Of course a 4th order curve fit should be tighter than a second order one since the 4th order contains within it the second order terms. My proceedure should be entirely valid as shown. Taking the derivative of the assumed Stefan-Boltzman n ideal equation eliminates the constant term due to ambient radiation. I am confident that you agree with this. Next, taking the ratios of the derivatives at different temperatures also eliminates any unknown coefficients of the 4 th order term. Do you agree?]]></description>
			<dc:creator>David Roberson</dc:creator>
			<pubDate>Mon, 19 Nov 2012 16:31:43 +0000</pubDate>
			<guid>http://www.quantumheat.org#comment-559</guid>
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			<title>123star says:</title>
			<link>http://www.quantumheat.org#comment-554</link>
			<description><![CDATA[@David Oh sorry... I missed that you already told me your rms of residuals, about 1W. Suggestion: in your interpolation program, explicitly put the coefficient sigma (5.67E-8) in front of x^4 to help the program, if you don't for example, gnuplot fails to find the fit curve. I.e. something like f(x)=a*(5.67E-8 )*(x-b)^4+c.]]></description>
			<dc:creator>123star</dc:creator>
			<pubDate>Mon, 19 Nov 2012 12:43:53 +0000</pubDate>
			<guid>http://www.quantumheat.org#comment-554</guid>
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			<title>Rémi André says:</title>
			<link>http://www.quantumheat.org#comment-550</link>
			<description><![CDATA[Would it be possible to include the theoric impedance curves (knowing the Steinhart-Hart coefficients of the wire used) in order to evaluate the delta between measure and theory in the case of the variation of impedance with T Thank you]]></description>
			<dc:creator>Rémi André</dc:creator>
			<pubDate>Mon, 19 Nov 2012 09:25:36 +0000</pubDate>
			<guid>http://www.quantumheat.org#comment-550</guid>
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			<title>David Roberson says:</title>
			<link>http://www.quantumheat.org#comment-549</link>
			<description><![CDATA[Continued... Then I took the ratio of each adjacent derivative to eliminate the unknown forth order coefficients. Next, the ratio of temperatures at which each pair of derivatives were obtained was cubed. I was expecting to see these pairs of numbers match, but was disappointed by the large error. A chart of the ratios of adjacent derivatives versus temperature ratios was plotted and found to be a sloped straight line instead of the expected cubic. This implied that the function I was seeking was quadratic. For this reason, I performed a second order curve fit and obtained good correlation. That is how I derived my function. It is a little complicated and the wording suffers from late night writing. P.S. The worst case error is .4 watts at one temperature, others less than .25 watts. Used 7 measurements for fit.]]></description>
			<dc:creator>David Roberson</dc:creator>
			<pubDate>Mon, 19 Nov 2012 06:32:10 +0000</pubDate>
			<guid>http://www.quantumheat.org#comment-549</guid>
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			<title>Ged says:</title>
			<link>http://www.quantumheat.org#comment-542</link>
			<description><![CDATA[@123star Ah-hah! That explains everything then :D! Looks like the equations have come together nice, I wonder if we can apply them with this data...]]></description>
			<dc:creator>Ged</dc:creator>
			<pubDate>Mon, 19 Nov 2012 03:19:55 +0000</pubDate>
			<guid>http://www.quantumheat.org#comment-542</guid>
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			<title>123star says:</title>
			<link>http://www.quantumheat.org#comment-540</link>
			<description><![CDATA[I meant 0.04 m^2 :)]]></description>
			<dc:creator>123star</dc:creator>
			<pubDate>Mon, 19 Nov 2012 01:39:04 +0000</pubDate>
			<guid>http://www.quantumheat.org#comment-540</guid>
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			<title>123star says:</title>
			<link>http://www.quantumheat.org#comment-539</link>
			<description><![CDATA[Omg! I forgot to divide by the area of the cylinder surface (which is around 0.4 m^2, can you check?) So the emissivity is about 1, comparable to a blackbody! The formulas are missing the radiation area, of course. Sorry :D]]></description>
			<dc:creator>123star</dc:creator>
			<pubDate>Mon, 19 Nov 2012 01:33:05 +0000</pubDate>
			<guid>http://www.quantumheat.org#comment-539</guid>
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			<title>JOB001 says:</title>
			<link>http://www.quantumheat.org#comment-537</link>
			<description><![CDATA[@"123star" No excess heat conclusion is possible yet because wire temperature is far below to the operate temperature 350C. Assume the heat release will double every 10F or 5.56C and will be about 18w at 350C. We would expect power to double (350-260)/5.56 = 16 times. Consequently the Excess power at 260C might be lower than 18*2^-16 = 0.0003w. The measurement error is about +/- 1w. We can remind ourselves to be patient.]]></description>
			<dc:creator>JOB001</dc:creator>
			<pubDate>Mon, 19 Nov 2012 00:36:31 +0000</pubDate>
			<guid>http://www.quantumheat.org#comment-537</guid>
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			<title>Ged says:</title>
			<link>http://www.quantumheat.org#comment-535</link>
			<description><![CDATA[@123Star I love your calculation work. Since we're measuring T_GlassOut, the calculations should be based on its emissivity and values, I would argue. What power is being lost by the wire itself and not captured in the T_GlassOut... is hard to say. A different glass type will definitely illuminate more of what's going on there. I think assuming an emissivity of 1 makes everything the most conservative for your calculations, so we could do that? Also, the cell has never been able to run at the full "activation" temperature one would expect, quite yet as far as I know. It's gotten very close, but watching for a transition from negative P_xs to positive is definitely a fingerprint sign we should be on the look out for. I agree with you fully.]]></description>
			<dc:creator>Ged</dc:creator>
			<pubDate>Mon, 19 Nov 2012 00:26:18 +0000</pubDate>
			<guid>http://www.quantumheat.org#comment-535</guid>
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			<title>123star says:</title>
			<link>http://www.quantumheat.org#comment-533</link>
			<description><![CDATA[I'm noting the fact that we are not finding any triggering temperature for P_xs (t_glassout vs p_in doesn't make any jump, it's always smooth). This is an indication that there isn't any excess heat, or if there is, there is at any temperature (unlikely). No?]]></description>
			<dc:creator>123star</dc:creator>
			<pubDate>Mon, 19 Nov 2012 00:01:42 +0000</pubDate>
			<guid>http://www.quantumheat.org#comment-533</guid>
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			<title>wookie says:</title>
			<link>http://www.quantumheat.org#comment-532</link>
			<description><![CDATA[i see u guys heat with both wires now. i m curious how that works. how many steps up to which power in per wire ? greetz]]></description>
			<dc:creator>wookie</dc:creator>
			<pubDate>Sun, 18 Nov 2012 23:56:28 +0000</pubDate>
			<guid>http://www.quantumheat.org#comment-532</guid>
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			<title>Matt says:</title>
			<link>http://www.quantumheat.org#comment-530</link>
			<description><![CDATA[So can anyone suggest the current potential conclusions from the data so far? Is there a good sign of excess heat during that run with hydrogen at 4.5 Bar or was the data just too dirty to make any reasonable conclusions?]]></description>
			<dc:creator>Matt</dc:creator>
			<pubDate>Sun, 18 Nov 2012 23:46:39 +0000</pubDate>
			<guid>http://www.quantumheat.org#comment-530</guid>
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			<title>wookie says:</title>
			<link>http://www.quantumheat.org#comment-529</link>
			<description><![CDATA[@Ged thanks for the sum up. absolutely!! cheers]]></description>
			<dc:creator>wookie</dc:creator>
			<pubDate>Sun, 18 Nov 2012 23:35:51 +0000</pubDate>
			<guid>http://www.quantumheat.org#comment-529</guid>
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			<title>Ged says:</title>
			<link>http://www.quantumheat.org#comment-524</link>
			<description><![CDATA[@123star, The glass should have an emissivity from 0.85 to 0.95, depending on the type. Take a look at http://www.engineeringtoolbox.com/emissivity-coefficients-d_447.html for the emissivity of common materials. The emissivity you are calculating is definitely not right. A higher emissivity means the easier it is to radiate away power via radiation (IR at these temperatures). Celani assumes an emissivity of 1, a perfect black body, which means he is -underestimatin g- the heat energy out of the cell if he uses black body radiation only to calculate the power produced; because glass has an emissivity below 1 and is not perfectly radiating (i.e. power is lost by convection/cond uction more, so if you only use radiation to calculate, you are missing a lot of the power out).]]></description>
			<dc:creator>Ged</dc:creator>
			<pubDate>Sun, 18 Nov 2012 23:14:03 +0000</pubDate>
			<guid>http://www.quantumheat.org#comment-524</guid>
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			<title>123star says:</title>
			<link>http://www.quantumheat.org#comment-523</link>
			<description><![CDATA[If there is thermal conduction the formula for the dissipated should read P = Pconduction + Pradiation Pconduction should be (roughly?) linear in T_glassout since the heat conduction follows the rule P = k*A*(T_hot - T_cold)/L (See http://en.wikipedia.org/wiki/Thermal_conductivity) Thus it should have a linear relationship with T_hot (the other parameters are the same). I don't think we can easily distinguish conduction from radiation from the fit, but feel free to correct me.]]></description>
			<dc:creator>123star</dc:creator>
			<pubDate>Sun, 18 Nov 2012 23:07:34 +0000</pubDate>
			<guid>http://www.quantumheat.org#comment-523</guid>
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			<title>123star says:</title>
			<link>http://www.quantumheat.org#comment-522</link>
			<description><![CDATA[I tried to fit the T_glassout(x) with P_In(y) using the Stefan-Boltzman n equation using gnuplot P_radiation = (emissivity * sigma * T_glass_out ^ 4) - (room_air_emiss ivity * sigma * T_ambient^4). See Celani paper too (he neglects emissivity altogether?!?) The second part (room part) can be taken as a constant. The interpolation formula I used is: f(x)= a*5.67E-8*(x+b) **4 +c It fist nicely with a rms of about 0.2 W, but the arguments a,b,c are strongly correlated. I get emissivity ~ 0.04, can anyone confirm this? I'm not sure if this value is reliable (I used RunHe2_USA and T_rise_glassout but I should have used T_glassout maybe).]]></description>
			<dc:creator>123star</dc:creator>
			<pubDate>Sun, 18 Nov 2012 22:55:38 +0000</pubDate>
			<guid>http://www.quantumheat.org#comment-522</guid>
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			<title>David Roberson says:</title>
			<link>http://www.quantumheat.org#comment-517</link>
			<description><![CDATA[I was reviewing the recent data and saw that the power output versus outside glass temperature fits very nicely to a second order curve. I was looking for a forth order match since I was expecting to see the heat being removed from the device by radiation. I recall seeing a function in your earlier documentation that also showed the quadratic relationship and was curious about your explanation. Do you assume that most of the heat escapes by convection or conduction from the cell? If this is true, how do we take into account the IR loss by radiation mechanisms? I am not convinced that the present calibration technique is flawed. I just wonder what would happen if the energy released by the LENR effects is burst IR radiation that escapes capture? Perhaps these questions have been explored earlier and if so please forgive my post.]]></description>
			<dc:creator>David Roberson</dc:creator>
			<pubDate>Sun, 18 Nov 2012 20:55:11 +0000</pubDate>
			<guid>http://www.quantumheat.org#comment-517</guid>
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			<title>Ryan Hunt says:</title>
			<link>http://www.quantumheat.org#comment-516</link>
			<description><![CDATA[@ Grieg - Good point. There is a copper/celani and copper/NiCr junction. Two, in fact. One on each end of the cell. Everything else is either constant temperature or copper-copper junction. We originally didn't worry about that effect because we expected to have an excess power measurement that was unambiguous. I think it may be a good idea to carefully address it now. Thanks for the reminder. @ Ged - Do you see anything different between those runs? I will take a look at it soon. First I want to post all the raw data in a list of all the phases of the experiment. That's my afternoon work in between setting up the next phase. @ Rats - At the moment we are using the calibration based on the T_Glass_out from the initial runs in Helium. They were extremely close together even though one was at 3.5 bar and the other at 0.5 bar. Both these runs were higher than the previous runs before we put the Celani wire in the cell. I am attributing that to the fact that the wire was wrapped loosely and came up one wrap short on the end, thereby concentrating the heat towards the center a significant amount. Since we are running in Hydrogen instead of helium, the negative excess power means that the glass is not as hot as it was during the Helium run. At this moment, we are running a couple watts lower than at the same power yesterday, which I cannot explain. @ Ecco - Are you saying resistance is futile? ;)]]></description>
			<dc:creator>Ryan Hunt</dc:creator>
			<pubDate>Sun, 18 Nov 2012 20:51:42 +0000</pubDate>
			<guid>http://www.quantumheat.org#comment-516</guid>
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			<title>Rats says:</title>
			<link>http://www.quantumheat.org#comment-515</link>
			<description><![CDATA[Can someone please explain what it means when P_xs is negative?]]></description>
			<dc:creator>Rats</dc:creator>
			<pubDate>Sun, 18 Nov 2012 20:21:52 +0000</pubDate>
			<guid>http://www.quantumheat.org#comment-515</guid>
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			<title>Ged says:</title>
			<link>http://www.quantumheat.org#comment-514</link>
			<description><![CDATA[It may be interesting to compare the current run behavior with the first run that was tried. It seems like it's not behaving the same way now, after the helium run and when we were worried something might have happened to the wire. Hopefully the wire is still functional.]]></description>
			<dc:creator>Ged</dc:creator>
			<pubDate>Sun, 18 Nov 2012 19:42:14 +0000</pubDate>
			<guid>http://www.quantumheat.org#comment-514</guid>
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