1995.10.27 / Matthew Kennel / Re: Neutrinos do not have mass, Dr. Hill is the wiser, Dr. White Originally-From: mbk@I_should_put_my_domain_in_etc_NNTP_INEWS_DOMAIN (Matthew B. Kennel) Newsgroups: sci.physics,alt.sci.physics.plutonium,sci.physics.electromag sci.physics.particle,sci.physics.fusion,sci.astro Subject: Re: Neutrinos do not have mass, Dr. Hill is the wiser, Dr. White Date: 27 Oct 1995 20:10:42 GMT Organization: University of Tennessee, Knoxville Archimedes Plutonium (Archimedes.Plutonium@dartmouth.edu) wrote: : NYT tues 24 Oct 1995 page C9 says the Los Alamos Nat. Lab is reported : in the Oct 2 issue of Physical Review Letters. : Dr White who headed the team reports a neutrino mass of about 1 : millionth to 1 hundred thousandth of the mass of an electron. : Dr Hill is a maverick and says the neutrino has no mass and that the : experiment cannot say one way or the other due to contamination. In : short, the set-up is sloppy. : I believe Dr. Hill is correct and will be found to be correct. : Trouble with so much of modern day physics is that it falls into the : 'social' physics trap. We saw a case of it with Smoot where the : majority of the community wanted, prayed for, expected and demanded : that the measurements have a fluctuation. The social community put : pressure on Smoot to find fluctuations. Yet few if any said, "Hey, : there were no fluctuations, you guys just reached the precision of your : measuring devices and that in 100 years from now with more precise : instruments, we will see that the cosmic microwave radiation is uniform : down beyond and past where Smoot seemed to have erroneously found : fluctuations. Your hypothesis that the results are due to bias to find an 'expected result' results not supported by facts. People had been expecting fluctuations of larger magnitudes for many years, and yet the experimentalists did their thing and duly reported 'no evidence of fluctuations at the sensitivity of our instruments' time and time again, despite the expectation. : Social physics is pressure physics and a disregard of logic, of : reason of clean wholesome experiments. Social physics is sloppy physics : because it is the end result that is wanted and give a dam about the : sloppy work or loopholes of logic. Just get the end result and get it : printed. There is no evidence that the COBE team did this. They went over their data for a year with large groups of people. matt cudkeys: cuddy27 cudenmbk cudfnMatthew cudlnKennel cudmo10 cudqt4 cudszM cudyr1995 ------------------------------ 1995.10.27 / Barry Merriman / Re: Serious Flaws in Patterson (SOFE '95) Demo on Cold Fusion Originally-From: barry@starfire.ucsd.edu (Barry Merriman) Newsgroups: sci.physics.fusion Subject: Re: Serious Flaws in Patterson (SOFE '95) Demo on Cold Fusion Date: 27 Oct 1995 20:20:59 GMT Organization: UCSD SOE In article <46ml2c$t2a@ixnews2.ix.netcom.com> cchaynie@ix.netcom.com (Carlton Haynie) writes: > > I have 2 problems with your proposed method of determining whether the > Peterson Power Cell is producing more energy than given: > > 1) By dispersing Nitrogen or Argon through the cell, to rid the > electrolyte of any excess hydrogen, aren't you defeating the purpose of > the electrolysis, which is to saturate the electrolyte with excess > hydrogen to stimulate the 'cold-fusion' process? No. In theory, the purpose of the electrolysis is to get H _into/onto_ the palladium metal electrode, not to produce a bunch of free H2, O2 in the solution. So, no, you are not defeating the purpose of the electrolysis if you perge the inter-electrode region but allow electrolysis to continue near the electrodes. > > 2) Couldn't the introduction of an 'alien' gas to the electrolyte > interfere with the 'cold-fusion' process in an unexpected, and possibly > unknown manner? Yes, but you miss the point: Steve Jones has clearly demostrated that significant recombination can occur in cold-fusion-like electrolysis cells, that it can mimic excess heat, and that you can shut of _recombination_ with a inert gas purge of the cell. Therefore: if a cold fusion cell could _continue_ to produce excess heat after such a gas purge, it would be a clear demonstration that it was _not_ recombination. This is a major oppurtunity for the experimenter to characterize what is going on. However, if the purge stops the excess heat production, while it is not proof that it was all due to recombination, it is compatible with that known process and says to any conscientious experimenter that they had better investigate that possibility in more detail. >Wouldn't a better test of the 'cold-fusion' effect be to simply measure >the total amount of energy released from the experiment over the >*entire* run, and compare this with the total amount of energy input >during the *entire* run? Easier said than done---as S.J. has shown, the in vs out calculation often relies on assumpitons that certain rates/coefficients stay constant during an expriment, at some value that may or ay not have been accurately measured. Such assumpitons are not permissable when trying to prove the existence of a radical phenomena. A better test would probably be to directly measure the amount of recombination, via capture of the evolved gasses, and/or placing an inert barrier that prevents O and H from migrating together in the cells, as Jones did by using glass tubes in his experiments. -- Barry Merriman UCSD Fusion Energy Research Center UCLA Dept. of Math bmerriman@fusion.ucsd.edu (Internet; NeXTMail is welcome) cudkeys: cuddy27 cudenbarry cudfnBarry cudlnMerriman cudmo10 cudqt4 cudszM cudyr1995 ------------------------------ 1995.10.27 / jedrothwell@de / Re: Serious Flaws in Patterson (SOFE '95) Demo on Cold Fusion Originally-From: jedrothwell@delphi.com Newsgroups: sci.physics.fusion Subject: Re: Serious Flaws in Patterson (SOFE '95) Demo on Cold Fusion Date: Fri, 27 Oct 95 15:50:25 -0500 Organization: Delphi (info@delphi.com email, 800-695-4005 voice) John Logajan writes: "So to produce 5 watts for one hour would require a reservoir of 150 liters in which 100% of the H2 gas was eventually recombined with O2." That's a lot of water! The reservoir must be invisible. Here are two other problems, I think. 1. You would have to keep the water in a pressurized container I guess. Wouldn't the hydrogen come bubbling out spontaneously after a while, like CO2 carbonation from an open glass of Sprite? I have always wondered how that works. 2. You have to dissolve oxygen in the water too. There is no air in the cell. You need a stoichiometric mix going into the cell. You can bubble pure H2 gas though a CF cell to scour out impurities. It comes right out the top, even when there is recombiner installed in the headspace, because there is no oxygen to combine with it. How much oxygen can you fit in the 150 liters of water? (Water already saturated with hydrogen might have a reduced carrying capacity for oxygen.) That leads to another sticky problem I mentioned earlier. Jones says that the cell stores up O2 and H2 in the water during electrolysis for 20 minutes, and then it starts to recombine the stored up gas after that (and it keeps going for weeks)! Why wouldn't it start recombining right away? Why would it build up? If the bubbles can be recombined, why should they be absorbed by the water instead? "Note however that the delivery rate is the bottleneck. The flow rate of 14.2 ml/min or 0.236 ml/sec can only deliver 0.00354 cc/sec of dissolved H2." Yup. And during the 20 minute storage period, only 280 ml from the invisible 150 liter reservoir passes through the cell. It is difficult to see how the entire reservoir can be saturated when only 0.2% of it comes through. This problem is compounded with the beads manufactured at U. Illinois, which turn on faster than the CETI beads, in 5 or 10 minutes rather than 20 or 30. I note that in order to explain the 85 MJ 7 week run, Jones requires 705,600 liters of water fully saturated with H2 and O2 gas at a flow rate of 10 liters per minute. - Jed cudkeys: cuddy27 cudenjedrothwell cudmo10 cudqt4 cudszM cudyr1995 ------------------------------ 1995.10.27 / C Harrison / Periodic Post: Cold Fusion online at sunsite.unc.edu Originally-From: harr@netcom.com (Charles (Chuck) Harrison) Newsgroups: sci.physics.fusion Subject: Periodic Post: Cold Fusion online at sunsite.unc.edu Date: Fri, 27 Oct 1995 20:45:04 GMT Organization: Fitful This message is posted periodically to inform readers about on-line data sources related to "cold fusion" which are located at the University of North Carolina SunSITE server. Two public WAIS (Wide Area Information Server) sources are online: (1) Dieter Britz's Bibliography (periodically updated), and (2) A sci.physics.fusion archive (1989 to present). WAIS provides for multiple keyword searches in these databases. It does _not_ support boolean logic in the searching :-(. 1. If you are directly connected to Internet, you can log onto a public WAIS server at the University of North Carolina: %telnet sunsite.unc.edu ... login: swais ... TERM = (unknown) vt100 It takes a minute to load ... < or use /fusion to locate the fusion-digest source> 2. If you have a "gopher" client, you can use it for WAIS access. Many university campuses provide gopher as a public information service. 2a. On most systems, you first select an option labeled "Other Systems", then from that menu select "WAIS based information". Since each gopher site creates its own menus, I can't tell you exactly where to go from there. 2b. If you can gopher to SunSITE, at UNC, navigate the menus down thru SunSITE archives..All archives..Academic..Physics..Cold-fusion. You will find the searchable databases (typically marked ), as well as the primary-literature files discussed below. 2c. If you can 'telnet' but not 'gopher', you may telnet to sunsite.unc.edu and login as 'gopher'. Then follow 2a or 2b above. 3. If you have World Wide Web (WWW) browser, such as Mosaic, Cello, or Lynx, you may use the following URL's: wais://sunsite.unc.edu/cold-fusion Britz bibliography wais://sunsite.unc.edu/fusion-digest newsgroup archive gopher://sunsite.unc.edu/11/../.pub/academic/physics/Cold-fusion 4. If you have a WAIS client on your system (the most common ones are "swais" -- character-based, and "xwais" -- for X-Windows), use it. The Britz source is called "cold-fusion" and it is listed in the directory-of-servers. If you _want_ a WAIS client program to run on your system, several are available in the public domain. Try ftp-ing to one of these sites: sunsite.unc.edu think.com There are several additional files archived at sunsite (e.g. Bollinger's Twist of Ribbon, preprints of the Fleischmann&Pons 1989 paper), which are accessible by anonymous ftp. %ftp sunsite.unc.edu . . . >cd pub/academic/physics/Cold-fusion >dir The collection (mostly primary papers) maintained by vince cate has been copied over to pub/academic/physics/Cold-fusion/vince-cate. Additional contributions are welcome; e-mail cfh@sunsite.unc.edu. cudkeys: cuddy27 cudenharr cudfnCharles cudlnHarrison cudmo10 cudqt4 cudszM cudyr1995 ------------------------------ 1995.10.27 / Bob Sullivan / Re: Serious Flaws in Patterson (SOFE '95) Demo on Cold Fusion Originally-From: bsulliva@sky.net (Bob Sullivan) Newsgroups: sci.physics.fusion Subject: Re: Serious Flaws in Patterson (SOFE '95) Demo on Cold Fusion Date: Fri, 27 Oct 95 02:02:03 GMT Organization: SkyNET Corporation In article <46o823$7fl@post.gsfc.nasa.gov>, "Lawrence E. Wharton" wrote: > I think the recombination (and now recombination/battery effect) >critics are doing the other objective skeptics a big disservice. Their >objections are not reasonable. You can't store up .06 watts for 20 >minutes (the battery effect) and deliver 5 watts for several days and >you can't take some fraction of .06 watts going into hydrogen and >oxygen, recombine it (the recombination effect), and produce 5 watts >output. No observer without some brain disfunction would think that >these objections are credable. Here are some numbers calculated from Bruce Klein's Patterson cell demos as shown on John Logajan's web page: Resistor Raw Data Calibration "Excess Heat" Efficiency Efficiency ----------- ------------- Attachment 4 34.0% 31.9% Attachment 5 34.0% 28.0% Attachment 6 30.1% 26.6% Attachmnet 7 21.7% 23.5% The first column of numbers represents the efficiency of the calibration run using a resistor to provide the heat. This is the measured heat output divided be the measured input. The second column represents the efficiency of Patterson cells supposedly producing excess heat. The efficiency calculations are identical as for the calibration run. Obviously, these calculations do not support claims of excess heat. The "excess heat" appears when the raw input and output numbers go through the magical adjustment procedure. You are arguing from ignorance, because you are using Jed's numbers at face value without taking the trouble to familiarize yourself with the calculations. You're assuming that the numbers you are working with are "measured" numbers. They are not. cudkeys: cuddy27 cudenbsulliva cudfnBob cudlnSullivan cudmo10 cudqt4 cudszM cudyr1995 ------------------------------ 1995.10.27 / John Logajan / Re: Cold Fusion Demo at SOFE '95 Worked Well Originally-From: jlogajan@skypoint.com (John Logajan) Newsgroups: sci.physics.fusion Subject: Re: Cold Fusion Demo at SOFE '95 Worked Well Date: 27 Oct 1995 03:18:31 GMT Organization: SkyPoint Communications, Inc. jonesse@plasma.byu.edu wrote: : Of course, by running at high current for some time, the flowing electrolyte : becomes loaded with dissolved hydrogen and oxygen. Then as this is passed : over the catalysts bed (Pd/Ni on beads, and the Pt anode), recombination : of H2 and O2 certainly occurs. : electrolyte makes this possible) one can keep the : PAF large for a long time -- an hour should be no problem for this battery. Well, we're all men of science here, so lets do a little ... :-) An hour of production of 5 watts requires 5J*3600S=18,000J. At 285,800 J/mole of H2+2O2 ==> H2O (one mole) we see that the production of 18,000J would also produce 0.063 moles of H2O. Therefore this would have required 0.063 moles of H2 and 0.0315 moles of O2, for a total of 0.063+0.0315=0.0945 moles combined gases. At 25C and one atmosphere pressure there are 24,500 cc per mole of gas. So 0.0945 * 24,500 = 2315 cc's of dissolved gasses are needed to produce 5 watts continuous for one hour. That's 2.3 liters of gas. Now it has to be dissolved in the water/lithium sulfate solution. I don't have handy any numbers of the soluability of such gasses in such an electrolyte. At a one-to-one ratio, you'd need a reservoir of 2.3 liters. At 1:2 you'd need 4.6 liters. At 10% you'd need 23 liters. And at 1% you'd need 230 liters. I suspect the SOFE demo didn't have a 2.3 liter reservoir. The ICCF5 demo had a 200 ml (0.2 liter) reservoir and ran almost continuosly for three days. The hypothesis that Patterson et al is nothing but conventional chemistry requires that the suggested chemistry fit the magnitudes of the observed variables. The recombination hypothesis keeps coming up short in regard to Patterson et al. -- - John Logajan -- jlogajan@skypoint.com -- 612-633-0345 - - 4248 Hamline Ave; Arden Hills, Minnesota (MN) 55112 USA - - WWW URL = http://www.skypoint.com/members/jlogajan - cudkeys: cuddy27 cudenjlogajan cudfnJohn cudlnLogajan cudmo10 cudqt4 cudszM cudyr1995 ------------------------------ 1995.10.27 / John Logajan / Re: Cold Fusion Demo at SOFE '95 Worked Well Originally-From: jlogajan@skypoint.com (John Logajan) Newsgroups: sci.physics.fusion Subject: Re: Cold Fusion Demo at SOFE '95 Worked Well Date: 27 Oct 1995 03:22:33 GMT Organization: SkyPoint Communications, Inc. John Logajan (jlogajan@skypoint.com) wrote: : At 285,800 J/mole of H2+2O2 ==> H2O (one mole) ^^^^^^ That should have been H2+0.5(O2) ==> H2O. It didn't effect the rest of the calculations. -- - John Logajan -- jlogajan@skypoint.com -- 612-633-0345 - - 4248 Hamline Ave; Arden Hills, Minnesota (MN) 55112 USA - - WWW URL = http://www.skypoint.com/members/jlogajan - cudkeys: cuddy27 cudenjlogajan cudfnJohn cudlnLogajan cudmo10 cudqt4 cudszS cudyr1995 ------------------------------ 1995.10.27 / Martin Sevior / Re: Serious Flaws in Patterson (SOFE '95) Demo on Cold Fusion Originally-From: msevior@axnd02.cern.ch (Martin Sevior) Newsgroups: sci.physics.fusion Subject: Re: Serious Flaws in Patterson (SOFE '95) Demo on Cold Fusion Date: Fri, 27 Oct 1995 04:12:32 GMT Organization: CERN European Lab for Particle Physics Steve Jones suggests that the heat produced by the Patterson Cell (TM) demonstrated at SOFE is due to recombination. It is clear that the power gain is so large (a factor of 80), that's (heat output)/(I*V), that STEADY STATE recombination could not come close to accounting for the effect. However, until the volume of water in the system became known I thought it was just possible the the heat could have come from H2 dissolved in the electrolyte before the SOFE demonstration. Clearly if the total content of dissolved H2 is less than the amount required to provide 5 watts of heat for 8 hours, the hypothesis must be rejected. I looked up the solubility of hydrogen in water (CRC handbook of Chemistry and Physics) it's 2.14 cc of Hydrogen/(cc of water). The volume of the electrolyte is 200 ml so that means we can have up to 428 cc of Hydrogen dissolved in the water. This gives us 428/22400 = 0.019 mole of Hydrogen dissolved in the water. John Logain's handy thermodynamic scorecard lists the energy released per mole of Hydrogen in the formation of H2O as 2.85*10**5 Joules. So the amount of heat that could possibly be produced though recombination by hydrogen dissolved in 200 ml of water is 0.019*2.85*10**5 = 5.4*10**3. The mount of energy released in 8 hours of operation of the SOFE demonstration is: 4.5*60*60*8 = 1.3*10**5, a factor of 24 times higher. Or alternatively the stored hydrogen could run the cell for 20 minutes as quoted by Jed. (Where did you get that number Jed?) So the Jones "stored hydrogen" recombination hypothesis fails by a huge factor assuming 200 ml of electrolyte. Martin Sevior cudkeys: cuddy27 cudenmsevior cudfnMartin cudlnSevior cudmo10 cudqt4 cudszM cudyr1995 ------------------------------ 1995.10.26 / Bill Rowe / Re: 6 = 500! Jones, the Master of Malarkey, strikes again! Originally-From: browe@netcom.com (Bill Rowe) Newsgroups: sci.physics.fusion Subject: Re: 6 = 500! Jones, the Master of Malarkey, strikes again! Date: Thu, 26 Oct 1995 21:27:06 -0700 Organization: AltNet In article , jedrothwell@delphi.com wrote: >In his continuing effort to bamboozle the public, Steve Jones >(jonesse@plasma.byu.edu) posted a long, rambling, message about the "Serious >Flaws" in the SOFE demonstration. There is no point to going over this message >point by point trying to list all of the absurdities, inaccuracies, mistakes >and distortions. Actually, the whole message boils down to two statements, >repeated over and over again: > > Recombination magically makes 0.06 watts look like 5.00 watts; and, > > You can store 5 liters of water inside of 0.2 liters of water. To reduce bandwidth and in keeping with the spirit of your first paragraph, I won't respond to your post point by point. In essence the problem is Steve Jones never makes the statements nor equivalent statements to the ones you posted above. He pointed out there was a charging time. So the issue is .06W times the charging time is energy that is later released as heat. This results in an overestimate of the output power. In no way does this imply .06W equals 5W. As far as the second statement, you do recall the water is recirculated and the source of the H2 and O2 is the same water. A correct analysis of Steve Jones post would realize there is a small loss of water since some of the H2 and O2 probably escapes. In no way is Steve Jones post equivalent to an attempt to store 5 liters in 200 mililiters. -- "Against stupidity, the Gods themselves contend in vain" cudkeys: cuddy26 cudenbrowe cudfnBill cudlnRowe cudmo10 cudqt4 cudszM cudyr1995 ------------------------------ 1995.10.27 / VCockeram / Re: 6 = 500! Jones, the Master of Malarkey, strikes again! Originally-From: vcockeram@aol.com (VCockeram) Newsgroups: sci.physics.fusion Subject: Re: 6 = 500! Jones, the Master of Malarkey, strikes again! Date: 27 Oct 1995 01:09:07 -0400 Organization: America Online, Inc. (1-800-827-6364) Professor Jones at BYU writes: >"As I mentioned in an earlier post to sci.physics.fusion, the problem > with allowing the H2 and O2 to "escape" into the atmosphere like this is > that one then has no way to preclude the possibility that recombination > is occurring in the cell, which appears as "excess" power." ---snip snip--- Professor, please, please give it up! You cant input .06 watts and get out more than that for the length of time these were run even assuming 100 percent recombination! It just wont work. This recombination wagon your on is not working. They are not charging up the cell at high power and the reducing input power to measure temps. They are running at a continious power input of .06 watts and getting continious power out of around 5 watts. Total recombination would yield power out of .06 watts. Right?? Back to lurking in Las Vegas. Vince cudkeys: cuddy27 cudenvcockeram cudlnVCockeram cudmo10 cudqt4 cudszM cudyr1995 ------------------------------ 1995.10.26 / Barry Merriman / Re: Serious Flaws in Patterson (SOFE '95) Demo on Cold Fusion Originally-From: barry@starfire.ucsd.edu (Barry Merriman) Newsgroups: sci.physics.fusion Subject: Re: Serious Flaws in Patterson (SOFE '95) Demo on Cold Fusion Date: 26 Oct 1995 22:59:38 GMT Organization: UCSD SOE In article <46o823$7fl@post.gsfc.nasa.gov> "Lawrence E. Wharton" writes: > I think the recombination > critics are doing the other objective skeptics a big disservice... > you can't take some fraction of .06 watts going into hydrogen and > oxygen, recombine it (the recombination effect), and produce 5 watts > output. No observer without some brain disfunction would think that > these objections are credable. > Then why do you think people are suggesting it? If you see rational folks making blatantly irrational suggestions, it seems you should conclude you are missinterpreting their statements. I'm just guessing, but I suggest that at least some of these folks may be implying that recombination precipitated on the thermocouples is causing a local rise in temp, that fools the calorimeter into thinking the entire bulk has raised as much. So, in other words, there would not really be 5 watts out. Rather, a few 0.1's of watts, carefully placed, can mimic the effect of 5 spatially distributed watts. -- Barry Merriman UCSD Fusion Energy Research Center UCLA Dept. of Math bmerriman@fusion.ucsd.edu (Internet; NeXTMail is welcome) cudkeys: cuddy26 cudenbarry cudfnBarry cudlnMerriman cudmo10 cudqt4 cudszM cudyr1995 ------------------------------ 1995.10.27 / Wigger Othmar / Re: Neutrinos do not have mass, Dr. Hill is the wiser, Dr. White Originally-From: owigger@psw205.psi.ch (Wigger Othmar) Newsgroups: sci.physics,sci.physics.electromag,sci.physics.particle,sci. hysics.fusion,sci.astro Subject: Re: Neutrinos do not have mass, Dr. Hill is the wiser, Dr. White Date: 27 Oct 1995 08:06:07 GMT Organization: Paul Scherrer Institute (PSI), Switzerland On Oct 27 01:31:41 MET 1995, wrote: > forgive my ignorance as a non-scientist but , > if E=mc^2 > then it can be said that E is proportional to M > in other words with energy you have mass ? > > is this not correct or does some other law apply ? > > is it also not true that the only things that we can observe are the > movements of mass , ie other particles are undetectable ? This concept of mass (proportional to energy) is outdated and not used any more in the modern language of physics. Read for example the article "The Concept of Mass" in Physics Today, June 1989. Othmar Wigger -- Othmar Wigger PSI LG/E19 CERN 580-R-004 Othmar.Wigger@psi.ch CH-5232 Villigen CH-1211 Geneva PS-195 (CPLEAR) +41 56 993279 +41 22 7673088 geek code: GS d? -p+ c+++ l++ u+ e+++(*) m s !n h--- f? g+ w- t- r- y+ cudkeys: cuddy27 cudenowigger cudfnWigger cudlnOthmar cudmo10 cudqt4 cudszM cudyr1995 ------------------------------ 1995.10.27 / mitchell swartz / Serious Flaws in Patterson (SOFE '95) Demo on Cold Fusion Originally-From: mica@world.std.com (mitchell swartz) Newsgroups: sci.physics.fusion Subject: Serious Flaws in Patterson (SOFE '95) Demo on Cold Fusion Date: Fri, 27 Oct 1995 12:34:07 GMT Organization: The World Public Access UNIX, Brookline, MA In Subject: Re: Serious Flaws in Patterson (SOFE '95) Demo on Cold Fusion Message-ID: <46p3sq$q3u@soenews.ucsd.edu> barry@starfire.ucsd.edu (Barry Merriman) writes: "I'm just guessing, but I suggest that at least some of these folks may be implying that recombination precipitated on the thermocouples is causing a local rise in temp, that fools the calorimeter into thinking the entire bulk has raised as much. So, in other words, there would not really be 5 watts out. Rather, a few 0.1's of watts, carefully placed, can mimic the effect of 5 spatially distributed watts." That is really cute. Most people think you need a catalyst to produce recombination to form water from diatomic hydrogen and oxygen. In fact, here is a hypothesis that is really amazing because it circumvents all knowledge of Group VIII metal catalysis by claiming that recombination occurs on copper constantan thermocouples. Too bad this is a just a joke because a lot of money might be saved on catalytic systems by substituting copper for platinum. Best wishes. Mitchell Swartz (mica@world.std.com) cudkeys: cuddy27 cudenmica cudfnmitchell cudlnswartz cudmo10 cudqt4 cudszM cudyr1995 ------------------------------ 1995.10.27 / jedrothwell@de / Incredibly stupid comments from Jones & Sullivan Originally-From: jedrothwell@delphi.com Newsgroups: sci.physics.fusion Subject: Incredibly stupid comments from Jones & Sullivan Date: Fri, 27 Oct 95 09:10:17 -0500 Organization: Delphi (info@delphi.com email, 800-695-4005 voice) Steve Jones and Bob Sullivan are either incredibly stupid or they are gluttons for punishment. Steve Jones AGAIN writes: "By charging a large quantity of electrolyte with H2 and O2 initially, (having a flowing electrolyte makes this possible) one can keep the PAF large for a long time -- an hour should be no problem for this battery." Yeah, right. Sure Steve! You have a great grasp of physics. And your arithmetic, wow! Okay, we charge the cell for 20 minutes at 60 milliwatts. That's a total of 72 joules. Half of that is lost to joule heating; that leaves 36 joules. Of course, we can see the rest leaving in the bubbles, but we will pretend that every single bubble is tucked away in the electrolyte. Say, if the cell is such a good recombiner, how come it doesn't start recombining right away Steve? I mean, why should it selectively save up the gas at one time during electrolysis, and release it later on? Oh, well, so what, let's move on. So we have this mighty 36 joules sitting in this gigantic ocean of electrolyte (200 ml; a coffee cup full). Okay, a 5 watt reaction begins. And you say an hour should be no problem? At the rate of 300 joules a minute? Hey Steve, can you do third grade arithmetic? 5 watts times 60 seconds is 300 joules. In one hour we get 18,000 joules. Does 36 = 18,000? Can you tell which is the BIGGER number, and which is the SMALLER number? Are you completely brain dead, or what? Suppose we had an ocean of flowing electrolyte. The total input energy in the first 20 minutes of the experiment is far less than you get out every minute during the excess heat run, and the excess heat run has gone on for SEVEN WEEKS!!! Which is longer, 7 weeks or 7.2 seconds? Can you tell? No, of course you can't. An idiot who thinks 6 = 500 cannot tell the difference between 4,233,600 and 7. Hey, do they actually let you teach college courses there at BYU? Maybe you should go for some adult education remedial classes. Your local YMCA may offer something. They will teach you the difference between 4 million and 7. Let me give you a hint: If I give you $7, you and your friend can buy lunch at McDonalds here in Chamblee. If I give you $4 million dollars, you can buy the restaurant and have plenty of money left over. Furthermore, as I said before, in order to store the energy released during that seven week run in the form of H2 and O2 gas, you would have to put 5 liters of water into a 200 milliliter container. That's a tight squeeze! And Bob Sullivan, not be outdone in the stupidity department, writes: "Pssst, Jed, go to John Logajan's web page, look up the Bruce Klein demo, and see how these calculations are actually made. If you don't understand it, just say so, and we'll help you out." Yo! Earth to Bob. Earth to Bob! Come in please! That was a DIFFERENT EXPERIMENT. Got it? We are not talking about the same experiment. Bruce Klein is not George Miley. See, the names sound different, the initials are different. It's B.K. and G.M. Totally different. Chicago is not in Florida. Patterson's lab is not located in the Midwest Ballroom of The Chancellor Hotel in Champaign, Illinois. Bruce used his method of computing excess heat because the cell he observed did not produce much excess, and it was running uninsulated at a much higher temperature so most of the heat was not accounted for. The cells shown at ICCF5 and SOFE were running in completely different calorimeters, at different temperatures, with new, improved, more effective beads. The performance of the two cells was different, and the methods used to compute excess heat was different. Got it? No of course you don't! Naturally, you will repeat this inane comment of yours a dozen more times. - Jed cudkeys: cuddy27 cudenjedrothwell cudmo10 cudqt4 cudszM cudyr1995 ------------------------------ 1995.10.27 / John Logajan / Re: Serious Flaws in Patterson (SOFE '95) Demo on Cold Fusion Originally-From: jlogajan@skypoint.com (John Logajan) Newsgroups: sci.physics.fusion Subject: Re: Serious Flaws in Patterson (SOFE '95) Demo on Cold Fusion Date: 27 Oct 1995 14:18:00 GMT Organization: SkyPoint Communications, Inc. Bob Sullivan (bsulliva@sky.net) wrote: : You are missing a critical point. The "efficiency" calculations for the : Patterson cell involve adjustments to REDUCE the measured input and INCREASE : the measured output. : Everyone seems to agree that water is dissociated in the cell. The TBs say : that the energy to dissociate the water escapes in the form of gases. They : argue that it is inappropriate to consider this as part of the input, so they : reduce the measured input to reflect the presumed loss: : (measured output) * (a cell constant) : -------------------------------------- : (measured input) - (dissociation loss) I'm not sure why you keep repeating this, but it simply is not true. The ICCF5 demo (and apparently the SOFE demo) did not use any gas loss correctiont term, so this recombination explanation does not apply. Even Steven Jones has had to rescue the recombination explanation by appealing to pre-dissolved H2 and O2 storage in the electrolyte -- he apparently has gotten the hint that the gas loss correction was not taken in these demos. -- - John Logajan -- jlogajan@skypoint.com -- 612-633-0345 - - 4248 Hamline Ave; Arden Hills, Minnesota (MN) 55112 USA - - WWW URL = http://www.skypoint.com/members/jlogajan - cudkeys: cuddy27 cudenjlogajan cudfnJohn cudlnLogajan cudmo10 cudqt4 cudszM cudyr1995 ------------------------------ 1995.10.27 / John Logajan / Re: Serious Flaws in Patterson (SOFE '95) Demo on Cold Fusion Originally-From: jlogajan@skypoint.com (John Logajan) Newsgroups: sci.physics.fusion Subject: Re: Serious Flaws in Patterson (SOFE '95) Demo on Cold Fusion Date: 27 Oct 1995 15:14:26 GMT Organization: SkyPoint Communications, Inc. Martin Sevior (msevior@axnd02.cern.ch) wrote: : I looked up the solubility of hydrogen in water (CRC handbook of Chemistry : and Physics) it's 2.14 cc of Hydrogen/(cc of water). Thanks for suggesting the CRC but you have it slightly wrong! :-) Actually it is 2.14 cc per 100 cc of cold water, or 0.85 cc per 100 cc of hot water. We probably have warm water, or about 1.5 cc H2 per 100 cc warm water. So to produce 5 watts for one hour would require a reservoir of 150 liters in which 100% of the H2 gas was eventually recombined with O2. Note however that the delivery rate is the bottleneck. The flow rate of 14.2 ml/min or 0.236 ml/sec can only deliver 0.00354 cc/sec of dissolved H2. Now there are 24,500 cc per mole of H2 at one atmosphere and 25C. So we have 0.00345/24,500 = 0.00000014 moles of H2 per second. At 285800 J per mole of H2O produced from one mole of H2 and half a mole of O2, we see that 285800*0.00000014=0.040 J/sec. So the *MOST* power possible to be seen from the Steven Jones Memorial Dissolved H2 Recombination Process is 40 milliwatts at 14.2 ml/min electrolyte flow. Not quite 5 watts, no matter how you slice it. RIP -- - John Logajan -- jlogajan@skypoint.com -- 612-633-0345 - - 4248 Hamline Ave; Arden Hills, Minnesota (MN) 55112 USA - - WWW URL = http://www.skypoint.com/members/jlogajan - cudkeys: cuddy27 cudenjlogajan cudfnJohn cudlnLogajan cudmo10 cudqt4 cudszM cudyr1995 ------------------------------ 1995.10.27 / jedrothwell@de / Re: Serious Flaws in Patterson (SOFE '95) Demo on Cold Fusion Originally-From: jedrothwell@delphi.com Newsgroups: sci.physics.fusion Subject: Re: Serious Flaws in Patterson (SOFE '95) Demo on Cold Fusion Date: Fri, 27 Oct 95 13:51:31 -0500 Organization: Delphi (info@delphi.com email, 800-695-4005 voice) mitchell swartz writes: > That is really cute. > Most people think you need a catalyst >to produce recombination to form water from diatomic >hydrogen and oxygen. In fact, here is a hypothesis >that is really amazing because it circumvents all knowledge >of Group VIII metal catalysis by claiming that recombination >occurs on copper constantan thermocouples. Too bad It *is* cute. It is even cuter than you realize, because Dennis covers the ends of his thermocouple probes with inert plastic. (Not sure what; I think he uses that shrink wrap teflon stuff.) Also, in some experiments, he positioned a thermometer below the gas trap. I mean the test tube where the liquid electrolyte falls, and the bubbles go out the top and through the gas flowmeter. So it is mighty hard to see how those bubbles could reverse course, hop through that test tube and recombine on the thermometer. They are cute little devils though, you never know what they might do! - Jed cudkeys: cuddy27 cudenjedrothwell cudmo10 cudqt4 cudszM cudyr1995 ------------------------------ 1995.10.27 / MARSHALL DUDLEY / Re: Serious Flaws in Patterson (SOFE '95) Demo on Cold Fusion Originally-From: mdudley@brbbs.brbbs.com (MARSHALL DUDLEY) Newsgroups: sci.physics.fusion Subject: Re: Serious Flaws in Patterson (SOFE '95) Demo on Cold Fusion Date: Fri, 27 Oct 1995 10:28 -0500 (EST) janowsky@gaston.ma.utexas.edu (Steven Janowsky) writes: -> This is absolute nonsense. -> -> Take an everyday fridge; repackage the motor so that it's inside the -> insulated region. Then most of the fridge gets slightly warm, but the -> outside coils get hot. In that case the total energy that leaves the refrigerator including what leaves the coils would equal the energy in. No more, no less. To the outside world you no longer have a heat pump, although heat would still be getting pumped within the refrigerator, it would have no effect on the heat balance between power in and heat loss to the ambient. The heat pump would be totally within a closed system, having no effect on total heat flow to the outside world. I understand the point you are trying to make, but this example does not do it. Marshall cudkeys: cuddy27 cudenmdudley cudfnMARSHALL cudlnDUDLEY cudmo10 cudqt4 cudszM cudyr1995 ------------------------------ 1995.10.27 / Jacques Distler / Re: Neutrinos do not have mass, Dr. Hill is the wiser, Dr. White Originally-From: distler@golem.ph.utexas.edu (Jacques Distler) Newsgroups: sci.physics,alt.sci.physics.plutonium,sci.physics.electromag sci.physics.particle,sci.physics.fusion,sci.astro Subject: Re: Neutrinos do not have mass, Dr. Hill is the wiser, Dr. White Date: Fri, 27 Oct 1995 15:14:45 -0600 Organization: Physics Department, University of Texas at Austin In article <46p99d$q9c@numbat.devetwa.edu.au>, Clinton Smith wrote: >re : neutrinos > >forgive my ignorance as a non-scientist but , >if E=mc^2 >then it can be said that E is proportional to M >in other words with energy you have mass ? Once upon a time, people liked to talk about a "velocity-dependent" mass, for which the above formula would hold. The sensible, modern approach is to abandon this (rather useless) notion, and use the formula ______________ E = \/ p^2 c^2 + m^2 c^4 This reduces to the familiar formula when the particle is at rest (the momentum p=0). When the mass of the particle vanishes (m=0), the formula gives, instead, E = |p| c which is the appropriate relation for a particle which travels at the speed of light. >is it also not true that the only things that we can observe are the >movements of mass , ie other particles are undetectable ? Uh, can you SEE your computer screen? You are, at this very moment, detecting photons, particles with zero mass, using nothing but your eyes. Detecting, say, neutrinos, requires more sophisticated equipment. Jacques Distler cudkeys: cuddy27 cudendistler cudfnJacques cudlnDistler cudmo10 cudqt4 cudszM cudyr1995 ------------------------------ 1995.10.27 / John White / Re: Hilarious 'skeptical' handwaving awards Originally-From: jnw@lys.vnet.net (John N. White) Newsgroups: sci.physics.fusion Subject: Re: Hilarious 'skeptical' handwaving awards Date: 27 Oct 1995 15:09:05 -0500 Organization: Vnet Internet Access, Inc. - Charlotte, NC. (704) 374-0779 jedrothwell@delphi.com writes: > In any case, the > J. White theory holds that the heat from the motor crossed the shaft, went > through the plastic housing and into water. I did not specify exactly *how* the heat gets into the fluid. There are many possible ways, but I was thinking mostly in terms of loss in the pump mechanism itself. > And not only that, the pump > heat increased and decreased with the level of electrolysis, even though > nobody touched the pump motor speed controls. The crystals can only dissolve until they are all gone. The crystals form at the nickel cathode in the presence of an overpotential, and thus the rate of formation will be current dependent. Note that these cells show diminishing returns at high currents, which is consistent with my proposal. > It's magic! Magic is neat, and there is no magic more wonderful than the extraordinary magical metal coated beads made by the great archmage Patterson and his apprentices. This amazing spell component of great power and potency will, when used with the proper incantation, generate excess heat by a mystic process that is totally beyond the limited comprehension of primitive races, such as humans. Note that in order to invoke this awesome power in its purest form, the measurement must be made in a flowing-electrolyte calorimeter. Otherwise only low levels of this mystic energy will be observed. It is not for mere humans to ask why this is so. -- jnw@vnet.net cudkeys: cuddy27 cudenjnw cudfnJohn cudlnWhite cudmo10 cudqt4 cudszM cudyr1995 ------------------------------ 1995.10.27 / John White / Re: Hilarious 'skeptical' handwaving awards Originally-From: jnw@lys.vnet.net (John N. White) Newsgroups: sci.physics.fusion Subject: Re: Hilarious 'skeptical' handwaving awards Date: 27 Oct 1995 15:11:48 -0500 Organization: Vnet Internet Access, Inc. - Charlotte, NC. (704) 374-0779 21cenlogic@i-link.net (Mitchell Jones) writes: > ***{John, we are talking about a *pump* here, not about a complex > electronic gadget with an onboard computer and a host of temperature and > flow sensors, plus fluid mixing capability, etc. The Elmeco E-50 is just a pump, motor, and controls. There is no onboard computer, no temperature or flow sensors, no mixing capability. It has a toggle-style on-off switch, a little red power-on indicator light, and a motor speed control knob. That's it. > As a matter of interest, before I posted the message in which I claimed > you couldn't buy a pump with the characteristics you needed, I actually > looked at a few pumps. From my pocket notebook, here is an example. The > Penguin 660 (which is just a pump, nothing more) draws .16 amp at 120 > volts, and pumps 170 gallons per hour. This pump would be totally unsuitable for pumping electrolyte. It would quickly contaminate and destroy the cathode. This sort of experiment has to be *extremely* clean. That is the reason for using a peristaltic pump. This sort of pump has a rotating drum with rollers on the outside. These rollers press against a piece of tubing. As the drum rotates, fluid in the tubing is pushed along by the rollers. In terms of energy this is incredibly inefficient, but it avoids contamination because the fluid never leaves the tubing. > Are you, perhaps, a glutton for punishment? --Mitchell Jones}*** I must admit that trying to teach elementary concepts to someone who is determined not to understand them is a particularly fruitless endeavor. -- jnw@vnet.net cudkeys: cuddy27 cudenjnw cudfnJohn cudlnWhite cudmo10 cudqt4 cudszM cudyr1995 ------------------------------ 1995.10.27 / Mitchell Jones / Re: Fitting vs physics (was Re: Testing the Mitchell Jones Hypothesis) Originally-From: 21cenlogic@i-link.net (Mitchell Jones) Newsgroups: sci.physics.fusion Subject: Re: Fitting vs physics (was Re: Testing the Mitchell Jones Hypothesis) Date: Fri, 27 Oct 1995 14:44:22 -0500 Organization: 21st Century Logic In article <46ogtm$gdg@soenews.ucsd.edu>, barry@starfire.ucsd.edu (Barry Merriman) wrote: > In article <21cenlogic-2410951824030001@austin-2-12.i-link.net> > 21cenlogic@i-link.net (Mitchell Jones) writes: > > In article <46gmvo$bti@soenews.ucsd.edu>, barry@starfire.ucsd.edu (Barry > > Merriman) wrote: > > > > > > In article <21cenlogic-1910951340010001@austin-2-6.i-link.net>, > > > > 21cenlogic@i-link.net (Mitchell Jones) wrote: > > > > > > > > > > > >Result: an > > > > >electron orbiting in a cavitation channel does no work, loses no energy, > > > > >and thus cannot possibly radiate. It's simple! --Mitchell Jones}*** > > > > > > > > > > MJ: normally, charges radiate when they are accelerated---they need > > > not be doing any obvious ``work''. .. why does it not radiate away power > > > by standard radiation from an accelerated charge? > > > > > > Your question makes it sound like you are disturbed by the violation of > > the man-made rule of thumb which says that an accelerated charge must > > radiate. The proper rule, however, is that when a charge is accelerated > > *in the ether*, it must radiate. > > So, in your model, radiated power = energy lost to collsions > with etherons. ***{No. Two things are operating here. First, as I have noted several times in the past, E1 is an electronic cloud, not a kinetic gas. The etherons are repellent of one another and also of the particles of ordinary matter. Second, as I have also noted repeatedly, pure etherons are particles of virtually zero mass. Thus the energy required to brush them aside is very small even in those rare cases where surface contact collisions occur. These two considerations, taken together, imply that surface contact collisions are rare, and require extremely high velocities. (I would add, however, that etherons are capable, under certain conditions, of acting as centers of accumulation of other material. When this happens, the resulting impure etherons can pack a wallop.) --Mitchell Jones}*** > > That's conceptually lovely, but does not fit the reality of > E&M radiation: > > (1) a charge moving at a constant speed through the vacuum > does not radiate...but it is contantly knocking your etherons > out of the way. Contradiction. ***{Not true. See above. --MJ}*** > (2) radiation from an accelerated charge scales like (accel)^2. > Work done by collsions against a background field of > etherons could not scale that way---it has to scale with velocity or > distance traversed or whateever, but not instantaneous acceleration. ***{As noted above, we are not talking about work done by collisions. Such a model is crudely simplified, by comparison to what is actually going on within atomic systems. I brought up the picture of a satellite in high earth orbit as an analogy, not as an exact representation of what is going on when an electron orbits. What you need here is a visual model of the actual causation that is taking place when photons are emitted and when they are not emitted. Unfortunately, I don't know how to convey such a model without a book length writeup accompanied by visual aids. The source of the underlying problem is simply this: if we get into this area, we leave the world of established physics, where you and I have common ground, and enter my world, where everything is guaranteed to be alien to you. In spite of that, I will now attempt to convey an image that will assist you in answering, in an approximate way, your question. Suppose that you are driving a dump truck that is filled to overflowing with fine dust. In that case, the faster you drive, the more dust blows off. Thus, dust loss correlates with linear velocity and with distance traveled. However, that is because air is a kinetic gas--i.e., because air molecules actually make contact collisions with the dust particles, and knock them away from the truck. But suppose, just for the sake of argument, that air was not a kinetic gas but an electronic cloud. Suppose, in other words, that the molecules of air repel one another, and that they repel the particles of your truck and the dust particles in the bed of the truck. In that case, it might be possible for you to drive very fast, so long as you drove at a constant velocity, and yet the dust would remain in place. However, the fact that there would be no contact collisions to knock the dust off would *not* mean you couldn't get it off. Acceleration could be used to get it off. Slam on your brakes, and some of the dust would fly off. Or tromp on the accelerator: that would throw off some dust as well. Every time the truck accelerated strongly enough, in any direction, to overcome attractive forces, some of the surface dust would fly off. Now, in a parallel way, let's take the admittedly simplified view that an electron is a tiny ball covered with a deep layer of dust. By the above analogy, every time the electron experiences an acceleration that is strong enough to overcome attractive forces, it is going to throw off some dust. Does that mean that the steady centripetal acceleration that it encounters as it sweeps around its orbit is going to be enough? The answer: no, because the dust is very deep, and that toward the core is more tightly bound by attractive forces. Thus what will happen is that the dust will be shaken free down to the depth where the acceleration is too weak to remove it, and no further. Since the orbital acceleration is greater for the inner orbits, this implies that the mass of the dust will be less for electrons in inner orbits than in outer orbits. Indeed, by this model, predictable amounts of mass will be lost as an electron "jumps" to a lower orbit. As I am sure you are aware, this is in fact the case. Concerning cavitation channels, note that an electron that has just been captured and has made one pass around a circular orbit is going to encounter an acceleration as it enters its second orbit, if its cross section doesn't match the shape of the cavitation channel that it created on the first orbit. This will be a different acceleration from the centripetal one that characterizes the orbit itself, and will be additive to it--which means: on one side of the electron or the other, some more dust is going to be hurled off. Now, if the dust that is thrown off is called "energy," then we have fashioned, in a very rough way, an image of how it may be that cavitation channel characteristics can be instrumental in determining whether an orbiting electron does or does not radiate, and how the quantity of such radiation may be related to charge acceleration rather than to the sorts of things you mentioned: velocity, distance traversed, etc. Naturally, this image is very rough and deliberately simplified. We will have to go through many iterations before we will have an accurate visualization of the phenomena we are considering. --Mitchell Jones}*** > > > Your etheron model simply does not come close to yielding the known > properties of E&M radiation ***{Wrong. You do not yet understand the model, Barry. --Mitchell Jones}*** , especially if you define > radiation as work done by collisons against etherons. ***{As noted above, I don't. --MJ}*** > > (And while you are at it, explain why a circular current flow in a > > superconductor doesn't radiate away power, either! My version of the rule > > works for superconductors. Yours does not.) --Mitchell Jones}*** > > I would say they do radiate---but since electrons in a macroscopic > superconduting circular current undergo negligible accelerations, > the radiated power would be incredible small---to get significant > radiated power from a circular orbit, the electron needs to have > speed near c, and the radius needs to be on the order of centimeters > or less. ***{An interesting answer. I think I will respond to it later, after I have had a chance to explain some of the concepts that I will need. --Mitchell Jones}*** > -- > Barry Merriman > UCSD Fusion Energy Research Center > UCLA Dept. of Math > bmerriman@fusion.ucsd.edu (Internet; NeXTMail is welcome) =========================================================== cudkeys: cuddy27 cuden21cenlogic cudfnMitchell cudlnJones cudmo10 cudqt4 cudszL cudyr1995 ------------------------------ 1995.10.27 / jedrothwell@de / Re: Water Replenishment in Open Cell? Originally-From: jedrothwell@delphi.com Newsgroups: sci.physics.fusion Subject: Re: Water Replenishment in Open Cell? Date: Fri, 27 Oct 95 23:10:55 -0500 Organization: Delphi (info@delphi.com email, 800-695-4005 voice) writes: >In Jed's report dated Oct. 9, 1995, and titled "Cold Fusion Demo at SOFE '95 >Worked Well," carried in Scott Hazen Mueller's Fusion Digest No. 4455, dated >Oct. 10, Jed said that the CETI cell demonstrated at SOFE was an open cell >and that half of the input energy was carried off by free hydrogen and >oxygen. > >Doesn't one have to add water occasionally in those circumstances? Yes. During the SOFE panel discussion Dennis Cravens said that during high power tests he has to replenish at the rate of about 10 ml per week with pure distilled water. You do not have to add salts (lithium) again. 10 ml is not much compared to the 200 ml reservior, so the lithum concentration does not change much and resistance stays about the same. That is good. With a small cell, with less electrolyte, you can get large changes because of this, which complicates things. - Jed cudkeys: cuddy27 cudenjedrothwell cudmo10 cudqt4 cudszM cudyr1995 ------------------------------ 1995.10.27 / Mitchell Jones / Re: Cold Fusion Demo at SOFE '95 Worked Well Originally-From: 21cenlogic@i-link.net (Mitchell Jones) Newsgroups: sci.physics.fusion Subject: Re: Cold Fusion Demo at SOFE '95 Worked Well Date: Fri, 27 Oct 1995 19:31:16 -0500 Organization: 21st Century Logic In article <46pj27$kup@stratus.skypoint.net>, jlogajan@skypoint.com wrote: > jonesse@plasma.byu.edu wrote: > : Of course, by running at high current for some time, the flowing electrolyte > : becomes loaded with dissolved hydrogen and oxygen. Then as this is passed > : over the catalysts bed (Pd/Ni on beads, and the Pt anode), recombination > : of H2 and O2 certainly occurs. > > : electrolyte makes this possible) one can keep the > : PAF large for a long time -- an hour should be no problem for this battery. > > > Well, we're all men of science here, so lets do a little ... :-) > > > An hour of production of 5 watts requires 5J*3600S=18,000J. > > At 285,800 J/mole of H2+2O2 ==> H2O (one mole) we see that the production > of 18,000J would also produce 0.063 moles of H2O. Therefore this would > have required 0.063 moles of H2 and 0.0315 moles of O2, for a total of > 0.063+0.0315=0.0945 moles combined gases. > > At 25C and one atmosphere pressure there are 24,500 cc per mole of gas. > > So 0.0945 * 24,500 = 2315 cc's of dissolved gasses are needed to produce > 5 watts continuous for one hour. > > That's 2.3 liters of gas. Now it has to be dissolved in the water/lithium > sulfate solution. I don't have handy any numbers of the soluability of > such gasses in such an electrolyte. > > At a one-to-one ratio, you'd need a reservoir of 2.3 liters. At 1:2 > you'd need 4.6 liters. At 10% you'd need 23 liters. And at 1% you'd > need 230 liters. ***{Since the solubility of gases in water decreases as the temperature increases, the highest solubility of a gas is generally in cold water, and that is explicitly the case for both H2 and O2. From my Handbook of Chemistry and Physics, 63rd edition, it appears that the highest solubility of H2 gas in cold water is 2.14 cc of H2 per 100 cc of water, and the highest solubility of O2 is 4.89 cc per 100 cc of water. This means that the H2 requires *at a minimum* 47 times its volume of water in order to dissolve, and the O2 requires at least 20 times its volume. Since the .063 moles of H2 equates to .063 times 24,500 = 1544 cc of H2, and the .0315 moles of O2 equates to .0315 times 24500 = 772 cc of O2, this means 47 times 1544 = 72568 cc of H20 is required to dissolve the H2, and 20 times 772 = 15440 cc of H2O is required to dissolve the O2. Total amount of electrolyte required: 72568 + 15440 = 88008 cc, or 88 liters. This, of course, falsely assumes that the water is near freezing, that the dissolved oxygen does not interfere with the solubility of the dissolved hydrogen and vice versa, and that the electrolyte is distilled water containing no dissolved lithium sulphate. By those assumptions, we need at least 88 liters of electrolyte, and, due to the falsity of the assumptions, that is itself probably a tiny fraction of what would actually be required. Since it is crudely obvious that no such amounts of electrolyte were involved, it is equally clear that the recombination explanation does not work. --Mitchell Jones}*** > > I suspect the SOFE demo didn't have a 2.3 liter reservoir. The ICCF5 > demo had a 200 ml (0.2 liter) reservoir and ran almost continuosly > for three days. > > > The hypothesis that Patterson et al is nothing but conventional chemistry > requires that the suggested chemistry fit the magnitudes of the > observed variables. The recombination hypothesis keeps coming up > short in regard to Patterson et al. > > -- > - John Logajan -- jlogajan@skypoint.com -- 612-633-0345 - > - 4248 Hamline Ave; Arden Hills, Minnesota (MN) 55112 USA - > - WWW URL = http://www.skypoint.com/members/jlogajan - =========================================================== cudkeys: cuddy27 cuden21cenlogic cudfnMitchell cudlnJones cudmo10 cudqt4 cudszM cudyr1995 ------------------------------ processed by cud.pl ver. 0.5 Sat Oct 28 04:37:08 EDT 1995 ------------------------------