Re: [Vo]:responce to the IPKat - weblog
Terry Blanton wrote: Thomis, It's 'Randell Mills'. He does not get an 'A'. Randell doesn't get an A, what does that mean? The questions are: does his reactor produce the amount of energy that he says it does, and can his processes produce novel, and useful materials. If the answer to either of these questions is yes, than his critics can go soak their heads. If his techniques contradict the laws of physics, then it's the laws of physics that have to change. The people who believe other wise have graduate degrees. --- Get FREE High Speed Internet from USFamily.Net! -- http://www.usfamily.net/mkt-freepromo.html ---
RE: [Vo]:How many volcanoes would it take...
I came across a study a few years ago that showed that the US presently has more forested land than it did in the year 1900. My personal observation verifies that. The fields around the house I grew up in, and the house I have live in now (33 yrs.) have all grown over with forest. Historical photos of the Berks county Pennsylvania area of 1900 vintage show surprising areas of cultivation that are now forest. Jeff -Original Message- From: Jed Rothwell [mailto:[EMAIL PROTECTED] Sent: Friday, April 25, 2008 1:23 PM To: vortex-L@eskimo.com Subject: Re: [Vo]:How many volcanoes would it take... thomas malloy wrote: Compared to the volcanoes, all 6,000,000,000 of us are the equivalent of a pimple on an elephant's rear end. That is incorrect, as shown by the stats Nick Palmer found. It is also obviously wrong because in North America, we burn roughly twice as much fossil fuel as all of the plants on the continent convert back into carbon and free oxygen. If volcanoes added far more CO2 to the mix then we do, than plants would have a negligible effect and the atmosphere and there would be practically no free oxygen. (By the way, decreasing levels of free oxygen have not been examined, and recent evidence shows this, too, is a threat.) This notion that people have an inherently smaller effect than natural phenomena is widespread, but it has no logical or factual basis. In North America we have stripped away most of the top soil, cut most of the trees down, destroyed the water table over large areas and paved over an area the size of Nebraska. It is inconceivable that such large scale terraforming would not have a major impact on the environment. At this rate we will destroy most of the continent in a few hundred years as effectively as people in ancient times destroyed Iraq (Mesopotamia). - Jed No virus found in this incoming message. Checked by AVG. Version: 7.5.524 / Virus Database: 269.23.5/1398 - Release Date: 4/25/2008 2:31 PM No virus found in this outgoing message. Checked by AVG. Version: 7.5.524 / Virus Database: 269.23.5/1398 - Release Date: 4/25/2008 2:31 PM
Re: [Vo]:How many volcanoes would it take...
Howdy Jeff, Same here in Texas. Before 1870 range prairie grass fires could sweep across whole counties that acted to prevent forests from gaining a foot hold. Interesting arguments for and against greenhouse effect. Al Gore and Rush Limburger cheese et al should both be proud of their ability to keep the CO2 gas balloon in the air for so long before it becomes obvious that a parallel exists.. similar to two divorce lawyers. There is money in keeping the bickering going. Meanwhile back at the ranch the whole place winds up broke and knee deep in cockle burrs and Bushes. At some point the problem becomes insoluable.. unless.. well.. err.. some kid playing with matches... Richard Jeff wrote, I came across a study a few years ago that showed that the US presently has more forested land than it did in the year 1900. My personal observation verifies that. The fields around the house I grew up in, and the house I have live in now (33 yrs.) have all grown over with forest. Historical photos of the Berks county Pennsylvania area of 1900 vintage show surprising areas of cultivation that are now forest.
Re: [Vo]:How many volcanoes would it take...
A lot of marginal farmland in the United States has been returned to forest land. It's the same throughout much of New England -- lots of woods, but it's all second growth because it all was farmland a century ago. It was terrible farmland, but in the absence of the Interstate system and cheap long distance transport, it was cheaper to grow the food locally than try to bring it in from elsewhere at great expense. I thought that was common knowledge. This is a strong argument against the Hundred Mile Diet but has nothing whatsoever to do with the supposed question of whether anthropogenic greenhouse gases are behind the now very well documented phenomenon of global warming. Note that, while the United States has regrown some forests, far more have been cut down, bulldozed, or burned elsewhere in the world. This has strong implic R C Macaulay wrote: Howdy Jeff, Same here in Texas. Before 1870 range prairie grass fires could sweep across whole counties that acted to prevent forests from gaining a foot hold. Interesting arguments for and against greenhouse effect. Al Gore and Rush Limburger cheese et al should both be proud of their ability to keep the CO2 gas balloon in the air for so long before it becomes obvious that a parallel exists.. similar to two divorce lawyers. There is money in keeping the bickering going. Meanwhile back at the ranch the whole place winds up broke and knee deep in cockle burrs and Bushes. At some point the problem becomes insoluable.. unless.. well.. err.. some kid playing with matches... Richard Jeff wrote, I came across a study a few years ago that showed that the US presently has more forested land than it did in the year 1900. My personal observation verifies that. The fields around the house I grew up in, and the house I have live in now (33 yrs.) have all grown over with forest. Historical photos of the Berks county Pennsylvania area of 1900 vintage show surprising areas of cultivation that are now forest.
Re: [Vo]:STEORN Musings
Other possibilities based on my personal experience with these things: 1) Their counsel told them to STFU and quit bragging about things for which they yet have no patent protection. and 2) They encountered challenges of scale when they tried to create a commercial application. Or, as one of my engineers puts it, I don't understand, it worked in the Powerpoint presentation. Terry On Sat, Apr 26, 2008 at 3:06 PM, OrionWorks [EMAIL PROTECTED] wrote: A follow-up on my recent STEORN musings. There is another possibility that comes to mine, a remote scenario that involves a bit of drama. It assumes STEORN still has something of value that they eventually plan on revealing. However, based on the fact that there have been no official announcements since October 2007, it is admittedly a stretch, even a leap of faith for me to assume that the Dublin based company may still be in the running. Assuming for the sake of argument that it has taken STEORN's engineers this long to fix the embarrassing ORBO problem. If so, I could see how in order to generate maximum effect the marketing portions of the company might be planning on presenting their comeback on the 1st year anniversary of the original Kinetica failed demonstration. Again, it's a stretch, but I will be noting the date on my calendar. Meanwhile, a more realistic and prosaic conclusion to reach under the circumstances is that STEORN's engineers may have discovered much to their dismay that the ORBO PM configuration is not as robust as their original tests indicated. If so I would well imagine there's a lot of internal hand wringing and what-to-dos being bandied about. Back on April 17, 2008, OrionWorks sed: Speaking of alternative energy companies... It's been a long dry spell since we've heard anything from STEORN, particularly since their spectacularly failed July 2007 demo debacle. My gut reaction would be to assume, sadly, that things are probably not being going well for them. Common sense would suggest to me that STEORN's engineers would have been able to by now correct the kinks so embarrassingly revealed in the failed demo. Surely they would have by now presented a sequel: The new-and-improved ORBO. Still waiting. In absence of hard data, speculation runs rampant. One of my favorite STEORN conspiracies can be found out on Wikipedia where: * * * * * * Eric Berger, writing on the Houston Chronicle website, commented that: Recall that Steorn is a former e-business company that saw its market vanish during the dot.com bust. It stands to reason that Steorn has re-tooled as a Web marketing company, and is using the free energy promotion as a platform to show future clients how it can leverage print advertising and a slick Web site to promote their products and ideas. If so, it's a pretty brilliant strategy.[33] http://en.wikipedia.org/wiki/Steorn * * * * * * Taking my own advice to heart, where it is advisable to choose the conspiracy one wishes to believe in wisely, I have to admit that I actually gave Mr. Berger's theory serious consideration. In the end, however, I discarded it on the premise that Berger's theory violated my personal understanding of the principals of Occam's Razor. The theory personally strikes me as possessing too many complicated assumptions that would have to be set in place for the final payoff to eventually be realized. ...and just when is that payoff supposed to occur? Of course, this leave me once again clueless as to what might really be going on. I've therefore decided appeal to the collective intelligence of Vort's membership, particularly to all those entrepreneurs and former CEOs who have suffered their own personal stories - the slings and arrows of outrageous misfortune while running their own companies and start-ups. What say you all to the STEORN saga? What do your own gut reactions suggest? Regards Steven Vincent Johnson www.OrionWorks.com www.zazzle.com/orionworks
Re: [Vo]:Ethanol Al
On Sat, Apr 26, 2008 at 6:24 PM, OrionWorks [EMAIL PROTECTED] wrote: Not all are ducking There is absolutely nothing wrong with making ethanol from corn. Our forefathers certainly did it. They were not so foolish as to use it for transportation, however. Terry Alcohol and driving don't mix.
Re: [Vo]:responce to the IPKat - weblog
slaps head Terry On Sun, Apr 27, 2008 at 1:20 AM, thomas malloy [EMAIL PROTECTED] wrote: Terry Blanton wrote: Thomis, It's 'Randell Mills'. He does not get an 'A'. Randell doesn't get an A, what does that mean? The questions are: does his reactor produce the amount of energy that he says it does, and can his processes produce novel, and useful materials. If the answer to either of these questions is yes, than his critics can go soak their heads. If his techniques contradict the laws of physics, then it's the laws of physics that have to change. The people who believe other wise have graduate degrees. --- Get FREE High Speed Internet from USFamily.Net! -- http://www.usfamily.net/mkt-freepromo.html ---
[Vo]:Capturing CO2 ?
Among the most brain-dead of all the environmental proposals being floated around to combat greenhouse gases, is the kind of carbon sequestration where CO2 is pumped into the earth... this is beyond silly for so many reasons... But here is an interesting development: http://www.sciencedaily.com/releases/2008/04/080424103217.htm ... which is a highly energy-efficient method of converting carbon dioxide (CO2) into marketable chemical compounds- such as cyclic carbonates. The technique relies upon the use of a catalyst to force a chemical reaction between CO2 and an epoxide http://en.wikipedia.org/wiki/Epoxide Epoxides are cheap, but not that cheap. However, this kind of process could be a method which is an actual money-maker for the grid plant, and not an added cost to the consumer. Furhtermore, coal itself is much cheaper than epoxides, even after the recent run-up in price, and if the power-plant is burning coal anyway, then the prospect of removing some of most of the CO2 with *more coal* is intriguing ...it sounds crazy at first; since more coal could be required for removing CO2 than what is burned, until you think about the possible advantages. Many plastics, those high tonnage varieties currently being made from petroleum, are costly, about 15 times more costly per pound than is coal - and could possibly be made from CO2+coal at a huge profit to the power plant- which would be a win-win situation... since it would reduce the demand for petroleum and reduce the emissions of CO2 at the same time. I wonder if organic chemists, those who are not on the payroll of the PetroMafia, are already looking seriously at this possibility, especially for the ether-based polymers, or those from polyethylene glycol http://en.wikipedia.org/wiki/Polyethylene_glycol It is so obvious (as a concept) that it must be already RD in progress, or already proved not to be feasible, even with the advanced catalysts. Jones
[Vo]:Re: HUP-spread-out electron feels (and thus Coulomb-screens?) like a point charge... - T.GIF
- Original Message - From: Robin van Spaandonk [EMAIL PROTECTED] Sent: Saturday, April 26, 2008 11:51 PM Indeed momentum p is the denominator of the De Broglie wavelength h/p, my mistake. So in fact the more immobile the target deuteron, the less close the incident deuteron needs to get in order to fuse... This is also the way I used to think, however it is somewhat problematic. In order to tunnel, a particle has to make an attempt, i.e. it has to move at least a little. Furthermore, the more frequent the attempts, the more likely tunneling is in any given period of time. Indeed two immobile d's wouldn't attempt much I don't think, but isn't it ok if _one_ d in the pair, namely the incident one, makes the attempt, as in the Desorbing vs Incident Excess Surface Electron Catalyzed Fusion (DIESECF) scenario we are discussing? (However maybe the zero point motion is enough to count as an attempt?) - Anyone know how to calculate the frequency? 1/ Wouldn't it therefore dramatically improve things if we threw (by electrolysis, gas discharge or whatever) the incident deuterons onto a deep-cooled deuteron desorbing cathode? (liquid deuterium in a back side chamber could provide the deuterium, the low temp and the pushing pressure maybe) IIRC, there were a few early attempts using loaded cathodes rapidly immersed in liquid nitrogen (with a few neutrons detected if I'm not mistaken). I was thinking of permanently cooling the back (loading) side and exposing the front side to more energetic incident deuterons. This way we would benefit from the cooling-induced wider quantum spread (larger De Broglie wavelength h/p) of the target deuteron, allowing tunneling to it from a larger distance, and of the energy of the incident one to get as close as possible to the target, does this make sense? 2/ Another thought triggered by your correction, forced cooling or not, isn't a deuteron about to desorb (for Jones's entertainment: stuck half way through the surface Pd sphincter) particularly immobile due to its squeezed condition, and therefore an easier fusion target? I wouldn't think *anything* is particularly stuck at the atomic level. Not really stuck of course, but particularly slow, think of the motion of an internal d traveling between successive octahedral sites on its way out, you will certainly agree it slows down while passing bottlenecks: in terms of potential, it's a succession of hills and valleys where the bottlenecks are the mountain passes where the d almost comes to a halt before gathering speed again on the next downhill stretch. As a result the bottlenecks are the most favorable places for fusion to occur, with the output sphincter at the cathode surface the most favorable of all since it features surface electron screening, plus initially relatively energetic incident deuterons (a few to a few hundred eV depending on the CF experiment type) in frontal collision course, agreed? 3/ I have found this 2002 paper Study on Physical Foundation of Cold Fusion : http://www.swip.ac.cn/cfs/english/Information/nb2002/024.2.pdf The English is very poor but the physics seem quite understandable, even to this QM ignoramus. I find the volcano section view shaped potential curve quite helpful: positive hill shape is Coulomb repulsion potential (hill is lowered and narrowed by any screening negative charge density I guess), narrow central pit going down to very negative values is nuclear force attraction potential. Summary of my understanding of this paper: in order to fuse i.e. fall into the pit an incident deuteron doesn't have to classically go all the way up the hill, instead it can tunnel through it if it gets closer than the target's De Broglie wavelength. It seems the incident deuteron can be treated as a classical point charge loosing KE and gaining PE to find how high on the hill --and therefore how close to the target deuteron-- it gets. Your comments on the paper or my summary welcome. 4/ Do you have a ref for your factor 10 to 20 (0.035 to 0.07 Å instead of D2's 0.7 Å separation) required for practical D-D fusion power production? Sorry, but that's based upon my own (possibly wildly inaccurate) calculations. (See attached gif, which contains one of my various attempts to get this right). Z = atomic number of the target nucleus d = initial distance before tunneling m = mass of tunneling particle (assuming that tunneling particle has an atomic number of 1). I appreciate, thanks also for your other posts. What do you mean by initial distance before tunneling, is this the distance at which the incident (projectile) deuteron comes to a halt? I don't see where the De Broglie wavelength or the energy of the particle(s) comes into play BTW, it should matter as we discussed above. Surely there must be a standard way to compute this, could the approach used in the Chinese paper I quoted above be a standard one? See also the paper:-
Re: [Vo]:responce to the IPKat - weblog
- Original Message - From: thomas malloy [EMAIL PROTECTED] To: vortex-l@eskimo.com Sent: Sunday, April 27, 2008 2:20 AM Subject: Re: [Vo]:responce to the IPKat - weblog Terry Blanton wrote: Thomis, It's 'Randell Mills'. He does not get an 'A'. Randell doesn't get an A, what does that mean? The questions are: does his reactor produce the amount of energy that he says it does, and can his processes produce novel, and useful materials. If the answer to either of these questions is yes, than his critics can go soak their heads. If his techniques contradict the laws of physics, then it's the laws of physics that have to change. The people who believe other wise have graduate degrees. Well said. There is much misunderstanding. Some points: a) H exists in a 'ground' state *isolated*, e.g. in a vacuum with nothing *nearby* b) H can have a lower state in chemical compounds, with other atoms nearby c) Non-photon resonant transfer of energy occurs in conpunds, e.g. in phosphors d) BLP catalysis is resonant transfer of energy from H to an *ion* *nearby* e) At the instant of transfer, the H is *not isolated in a vacuum* as in a), it is transiently in a compound, as b) f) Ergo, no laws of physics have been violated. What is missing from Mills' publications, and apparaently the scientific literature as well, is data on the effective range of the resonant transfer effect, or the 'cross section'. Mills may have some theory or data, which is unpublished. Or he may not. I don't know. Experiments with microwave plasmas use gas pressures about 1 Torr and mixure of 5% H and 95% catalyst gas to increase the chance of close encounters of the resonant kind. The solid fuel apparently emits H and catalyst when heated, atoms in proximity, at high densities, and thus appears to be a key to commercial viability. Mike Carrell --- Get FREE High Speed Internet from USFamily.Net! -- http://www.usfamily.net/mkt-freepromo.html --- This Email has been scanned for all viruses by Medford Leas I.T. Department.
Re: [Vo]:responce to the IPKat - weblog
In reply to Mike Carrell's message of Sun, 27 Apr 2008 16:40:29 -0400: Hi, [snip] What is missing from Mills' publications, and apparaently the scientific literature as well, is data on the effective range of the resonant transfer effect, or the 'cross section'. [snip] My bet is that actual physical contact is necessary, in order to transfer angular momentum. Regards, Robin van Spaandonk The shrub is a plant.
Re: [Vo]:responce to the IPKat - weblog
Tomis, It's 'RandEll' not 'RandAll'. sigh T On Sun, Apr 27, 2008 at 10:26 AM, Terry Blanton [EMAIL PROTECTED] wrote: slaps head Terry On Sun, Apr 27, 2008 at 1:20 AM, thomas malloy [EMAIL PROTECTED] wrote: Terry Blanton wrote: Thomis, It's 'Randell Mills'. He does not get an 'A'. Randell doesn't get an A, what does that mean? The questions are: does his reactor produce the amount of energy that he says it does, and can his processes produce novel, and useful materials. If the answer to either of these questions is yes, than his critics can go soak their heads. If his techniques contradict the laws of physics, then it's the laws of physics that have to change. The people who believe other wise have graduate degrees. --- Get FREE High Speed Internet from USFamily.Net! -- http://www.usfamily.net/mkt-freepromo.html ---
Re: [Vo]:Capturing CO2 ?
In reply to Jones Beene's message of Sun, 27 Apr 2008 08:54:28 -0700 (PDT): Hi, [snip] ...it sounds crazy at first; since more coal could be required for removing CO2 than what is burned, until you think about the possible advantages. [snip] What happens to the cyclic carbonates when they reach the end of their useful life? Regards, Robin van Spaandonk The shrub is a plant.
[Vo]:Re: responce to the IPKat - weblog
Thomas won't get an A then? ;) Michel - Original Message - From: Terry Blanton [EMAIL PROTECTED] To: vortex-l@eskimo.com Sent: Sunday, April 27, 2008 11:24 PM Subject: Re: [Vo]:responce to the IPKat - weblog Tomis, It's 'RandEll' not 'RandAll'. sigh T On Sun, Apr 27, 2008 at 10:26 AM, Terry Blanton [EMAIL PROTECTED] wrote: slaps head Terry On Sun, Apr 27, 2008 at 1:20 AM, thomas malloy [EMAIL PROTECTED] wrote: Terry Blanton wrote: Thomis, It's 'Randell Mills'. He does not get an 'A'. Randell doesn't get an A, what does that mean? The questions are: does his reactor produce the amount of energy that he says it does, and can his processes produce novel, and useful materials. If the answer to either of these questions is yes, than his critics can go soak their heads. If his techniques contradict the laws of physics, then it's the laws of physics that have to change. The people who believe other wise have graduate degrees. --- Get FREE High Speed Internet from USFamily.Net! -- http://www.usfamily.net/mkt-freepromo.html ---
Re: [Vo]:Re: HUP-spread-out electron feels (and thus Coulomb-screens?) like a point charge... - T.GIF
In reply to Michel Jullian's message of Sun, 27 Apr 2008 18:05:33 +0200: Hi, [snip] Indeed two immobile d's wouldn't attempt much I don't think, but isn't it ok if _one_ d in the pair, namely the incident one, makes the attempt, as in the Desorbing vs Incident Excess Surface Electron Catalyzed Fusion (DIESECF) scenario we are discussing? [snip] I was thinking of permanently cooling the back (loading) side and exposing the front side to more energetic incident deuterons. This way we would benefit from the cooling-induced wider quantum spread (larger De Broglie wavelength h/p) of the target deuteron, allowing tunneling to it from a larger distance, and of the energy of the incident one to get as close as possible to the target, does this make sense? The De Broglie wavelength is based on the *relative* velocity between the particles. In fact one should probably calculate this in the common centre of mass frame of reference (in which case both particles have the same De Broglie wavelength - my thanks to Charles Cagle). What this means is that you can't have one fast particle and one slow one. They both need to be slow. This also implies that tunneling probably only takes place when the approaching particle has used up all its kinetic energy in overcoming the electrostatic potential of the other particle, and is just on the verge of reversing course. I.e. it's usually a one shot affair. Consequently, glancing blows may not contribute, it may need to be a head on collision. [snip] I appreciate, thanks also for your other posts. What do you mean by initial distance before tunneling, is this the distance at which the incident (projectile) deuteron comes to a halt? Yes. I don't see where the De Broglie wavelength or the energy of the particle(s) comes into play BTW, it should matter as we discussed above. The De Broglie wavelength is only a guide - a rule of thumb if you will. Note that when the particles are stationary relative to one another, the De Broglie wavelength is infinite, hence no longer relevant. When two particles are within the De Broglie wavelength of one another tunneling is possible, but not guaranteed, and the chance that it will happen is strongly related to the separation distance. That chance is what I have attempted to calculate in the gif file I attached to my previous post. Note also that even when tunneling does take place, fusion is not guaranteed. Whether or not it happens, depends also on the nuclear cross section of the reaction. Some reactions are more likely than others. A good example of a poor reaction is the p-p reaction. Tunneling probably happens quite often, yet a nuclear reaction seldom ensues. OTOH, a reaction with a good cross section is the p-B reaction, but this is limited by the reduced tunneling probability due to the high charge on the B nucleus. Surely there must be a standard way to compute this, could the approach used in the Chinese paper I quoted above be a standard one? Yes, in fact the essence of it is the only method I have seen employed. If you look closely, you will see it is also the method I used. If you feel like working on this, and you come up with something that works well, I would very much appreciate it, if you would pass it on. See also the paper:- Catalysis of Nuclear Reactions between Hydrogen Isotopes by mu- Mesons by J.D. Jackson, Physical Review, Vol. 106, Number 2, April 15 1957, page 330. Thanks, do you have a pdf version by any chance? I'm afraid it's copyrighted, but you can purchase a version on line, as I did, or visit your local technical/university library, and read it for free. (BTW Jackson essentially uses the same basic concept). Regards, Robin van Spaandonk The shrub is a plant.
Re: [Vo]:responce to the IPKat - weblog
- Original Message - From: Robin van Spaandonk [EMAIL PROTECTED] To: vortex-l@eskimo.com Sent: Sunday, April 27, 2008 5:17 PM Subject: Re: [Vo]:responce to the IPKat - weblog In reply to Mike Carrell's message of Sun, 27 Apr 2008 16:40:29 -0400: Hi, [snip] What is missing from Mills' publications, and apparaently the scientific literature as well, is data on the effective range of the resonant transfer effect, or the 'cross section'. [snip] My bet is that actual physical contact is necessary, in order to transfer angular momentum. === Peter Zimmerman advanced the 'contact' agument on another forum some time ago. The radii of a H atom and typical catalyst, such as Ar+, are so small that if contact were required, the probability of reaction would be vanishingly small. Since Ar+ experimetally catalyzes H, that argument is void. It is more likely that the transfer is a near-field phenomenon that can be approached from antenna theory, but I do not know of any detailed work on such. Mike Carrell Regards, Robin van Spaandonk The shrub is a plant. This Email has been scanned for all viruses by Medford Leas I.T. Department.
Re: [Vo]:responce to the IPKat - weblog
In reply to Mike Carrell's message of Sun, 27 Apr 2008 18:52:51 -0400: Hi, [snip] Peter Zimmerman advanced the 'contact' agument on another forum some time ago. The radii of a H atom and typical catalyst, such as Ar+, are so small that if contact were required, the probability of reaction would be vanishingly small. The probability of reaction is quite small, in fact Ar+ is not a very good catalyst. Nevertheless, contacts happen all the time, otherwise gas pressure wouldn't exist. Since Ar+ experimetally catalyzes H, that argument is void. That remains to be seen. It is more likely that the transfer is a near-field phenomenon that can be approached from antenna theory, but I do not know of any detailed work on such. Would such allow for transfer of angular momentum? Regards, Robin van Spaandonk The shrub is a plant.
[Vo]:Re: HUP-spread-out electron feels (and thus Coulomb-screens?) like a point charge... - T.GIF
Robin, Although it seems to make sense, something doesn't fit in your center of mass frame of reference and therefore equal De Broglie wavelengths (DBW) paradigm: in that frame, as you say, when the particles are stationary relative to one another, the DBW is infinite, hence no longer relevant, whereas that distance r1 where the incident d has lost all its initial kinetic energy is precisely where the Li et al paper compares the distance by which it missed (r1-r0) with the DBW, which they don't find infinite but equal to 0.78 Å... But on the other hand, how can the DBW not be infinite if momentum is zero?? On yet another hand, the DBW seems the right parameter to define the spread of a particle and therefore its capacity to tunnel or be tunneled to... if it's infinite, it's all over the place so tunneling should have 100% probability! This point is definitely unclear to me, any enlightening welcome. Michel - Original Message - From: Robin van Spaandonk [EMAIL PROTECTED] To: vortex-l@eskimo.com Sent: Monday, April 28, 2008 12:13 AM Subject: Re: [Vo]:Re: HUP-spread-out electron feels (and thus Coulomb-screens?) like a point charge... - T.GIF In reply to Michel Jullian's message of Sun, 27 Apr 2008 18:05:33 +0200: Hi, [snip] Indeed two immobile d's wouldn't attempt much I don't think, but isn't it ok if _one_ d in the pair, namely the incident one, makes the attempt, as in the Desorbing vs Incident Excess Surface Electron Catalyzed Fusion (DIESECF) scenario we are discussing? [snip] I was thinking of permanently cooling the back (loading) side and exposing the front side to more energetic incident deuterons. This way we would benefit from the cooling-induced wider quantum spread (larger De Broglie wavelength h/p) of the target deuteron, allowing tunneling to it from a larger distance, and of the energy of the incident one to get as close as possible to the target, does this make sense? The De Broglie wavelength is based on the *relative* velocity between the particles. In fact one should probably calculate this in the common centre of mass frame of reference (in which case both particles have the same De Broglie wavelength - my thanks to Charles Cagle). What this means is that you can't have one fast particle and one slow one. They both need to be slow. This also implies that tunneling probably only takes place when the approaching particle has used up all its kinetic energy in overcoming the electrostatic potential of the other particle, and is just on the verge of reversing course. I.e. it's usually a one shot affair. Consequently, glancing blows may not contribute, it may need to be a head on collision. [snip] I appreciate, thanks also for your other posts. What do you mean by initial distance before tunneling, is this the distance at which the incident (projectile) deuteron comes to a halt? Yes. I don't see where the De Broglie wavelength or the energy of the particle(s) comes into play BTW, it should matter as we discussed above. The De Broglie wavelength is only a guide - a rule of thumb if you will. Note that when the particles are stationary relative to one another, the De Broglie wavelength is infinite, hence no longer relevant. When two particles are within the De Broglie wavelength of one another tunneling is possible, but not guaranteed, and the chance that it will happen is strongly related to the separation distance. That chance is what I have attempted to calculate in the gif file I attached to my previous post. Note also that even when tunneling does take place, fusion is not guaranteed. Whether or not it happens, depends also on the nuclear cross section of the reaction. Some reactions are more likely than others. A good example of a poor reaction is the p-p reaction. Tunneling probably happens quite often, yet a nuclear reaction seldom ensues. OTOH, a reaction with a good cross section is the p-B reaction, but this is limited by the reduced tunneling probability due to the high charge on the B nucleus. Surely there must be a standard way to compute this, could the approach used in the Chinese paper I quoted above be a standard one? Yes, in fact the essence of it is the only method I have seen employed. If you look closely, you will see it is also the method I used. If you feel like working on this, and you come up with something that works well, I would very much appreciate it, if you would pass it on. See also the paper:- Catalysis of Nuclear Reactions between Hydrogen Isotopes by mu- Mesons by J.D. Jackson, Physical Review, Vol. 106, Number 2, April 15 1957, page 330. Thanks, do you have a pdf version by any chance? I'm afraid it's copyrighted, but you can purchase a version on line, as I did, or visit your local technical/university library, and read it for free. (BTW Jackson essentially uses the same basic concept). Regards, Robin van Spaandonk The shrub is a plant.
Re: [Vo]:Re: responce to the IPKat - weblog
You have a far greater command of the language. Terry On Sun, Apr 27, 2008 at 5:11 PM, Michel Jullian [EMAIL PROTECTED] wrote: Thomas won't get an A then? ;) Michel - Original Message - From: Terry Blanton [EMAIL PROTECTED] To: vortex-l@eskimo.com Sent: Sunday, April 27, 2008 11:24 PM Subject: Re: [Vo]:responce to the IPKat - weblog Tomis, It's 'RandEll' not 'RandAll'. sigh T On Sun, Apr 27, 2008 at 10:26 AM, Terry Blanton [EMAIL PROTECTED] wrote: slaps head Terry On Sun, Apr 27, 2008 at 1:20 AM, thomas malloy [EMAIL PROTECTED] wrote: Terry Blanton wrote: Thomis, It's 'Randell Mills'. He does not get an 'A'. Randell doesn't get an A, what does that mean? The questions are: does his reactor produce the amount of energy that he says it does, and can his processes produce novel, and useful materials. If the answer to either of these questions is yes, than his critics can go soak their heads. If his techniques contradict the laws of physics, then it's the laws of physics that have to change. The people who believe other wise have graduate degrees. --- Get FREE High Speed Internet from USFamily.Net! -- http://www.usfamily.net/mkt-freepromo.html ---
Re: [Vo]:responce to the IPKat - weblog
- Original Message - From: Robin van Spaandonk [EMAIL PROTECTED] snip The probability of reaction is quite small, in fact Ar+ is not a very good catalyst. Nevertheless, contacts happen all the time, otherwise gas pressure wouldn't exist. MC: Gas pressure results for impacts with the container, more than other atoms/molecules. Much bigger target. What is the basis for your statement that Ar+ is not a very good catalyst? In the water bath calorimeter studies, He, Ar, and O give approximately the same energy yield. Each are as effective on a per-atom basis. The problem is optimizing the reactor parameters to get the highest energy yield against competting processes. The is one reason the 'solid fuel' is significant. Since Ar+ experimetally catalyzes H, that argument is void. That remains to be seen. MC: You are familiar with the water bath calorimetry? Have you another explanation for the observed excess heat? It is more likely that the transfer is a near-field phenomenon that can be approached from antenna theory, but I do not know of any detailed work on such. Would such allow for transfer of angular momentum? MC: Why do you enter angular momentum into the discussion? The 'form' of the energy transfered is not stated. The H atoms exhibit very high kinetic energy/ temperature, manifested as Balmer line broadening. Regardds, Mike Carrell
Re: [Vo]:Re: HUP-spread-out electron feels (and thus Coulomb-screens?) like a point charge... - T.GIF
In reply to Michel Jullian's message of Mon, 28 Apr 2008 01:33:07 +0200: Hi, Robin, Although it seems to make sense, something doesn't fit in your center of mass frame of reference and therefore equal De Broglie wavelengths (DBW) paradigm: in that frame, as you say, when the particles are stationary relative to one another, the DBW is infinite, hence no longer relevant, whereas that distance r1 where the incident d has lost all its initial kinetic energy is precisely where the Li et al paper compares the distance by which it missed (r1-r0) with the DBW, which they don't find infinite but equal to 0.78 Å... But on the other hand, how can the DBW not be infinite if momentum is zero?? Without re-reading their paper, I think you will find that the DBW they calculate is based upon thermal energy. That's fine for a first rough guess, which is why I said it's a rule of thumb. What they mean is that the DBW is *at least* that big, ergo tunneling is possible. It's also possible that they are simply guilty of sloppy thinking. On yet another hand, the DBW seems the right parameter to define the spread of a particle and therefore its capacity to tunnel or be tunneled to... if it's infinite, it's all over the place so tunneling should have 100% probability! No, just possible, not necessarily probable. It's only infinite for a very short period of time. Probability is also determined by confinement time, and at least in the literature, by the cross section of the nuclear reaction. (However IMO, QM probably compounds tunneling probability with the cross section). IOW while I have tried to separate the two, QM usually doesn't. This point is definitely unclear to me, any enlightening welcome. [snip] Regards, Robin van Spaandonk The shrub is a plant.
Re: [Vo]:responce to the IPKat - weblog
In reply to Mike Carrell's message of Sun, 27 Apr 2008 22:26:06 -0400: Hi, [snip] The probability of reaction is quite small, in fact Ar+ is not a very good catalyst. Nevertheless, contacts happen all the time, otherwise gas pressure wouldn't exist. MC: Gas pressure results for impacts with the container, more than other atoms/molecules. Much bigger target. The atomic radius of Argon is about 0.98 Angstrom. Thus the diameter is about 2 A. When an electron is removed, it will shrink a bit, but I doubt it will shrink by more than half, so lets say about 1 A. For a gas at 1 torr, and a temperature of 1000 K with a particle diameter of 1 A, the mean free path is about 2.3 mm. That's a pretty small container. Even if we assume that the total collision cross section is supplied by the Ar+, and that the Hydrinos have 0 size, this still works out to 9 mm. Still a small container. Furthermore, it's only a *mean* free path, that doesn't mean collisions aren't going to happen in less distance occasionally. What is the basis for your statement that Ar+ is not a very good catalyst? In one of Mills early experiments he compares various catalysts, and Ar+ doesn't exactly shine. In the water bath calorimeter studies, He, Ar, and O give approximately the same energy yield. Each are as effective on a per-atom basis. The problem is optimizing the reactor parameters to get the highest energy yield against competting processes. The is one reason the 'solid fuel' is significant. I already sent Mills a wash list of my doubts about the water bath experiments years ago. To summarize I suspect the actual Hydrino energy contribution was more on the order of a couple of watts (not 10's of watts), and that the small differences in energy output were primarily due to the differences in efficacy of the different catalysts. [snip] MC: You are familiar with the water bath calorimetry? Have you another explanation for the observed excess heat? I am very suspicious of any experiment that purports to put out 60 W of heat, when that also just happens to be the nameplate power output of the microwave generator used. Suffice it to say, that I strongly suspect measurement error in those experiments. It is more likely that the transfer is a near-field phenomenon that can be approached from antenna theory, but I do not know of any detailed work on such. Would such allow for transfer of angular momentum? MC: Why do you enter angular momentum into the discussion? The 'form' of the energy transfered is not stated. The H atoms exhibit very high kinetic energy/ temperature, manifested as Balmer line broadening. ...because IMO, the hydrino can't shrink without shedding angular momentum. In fact I think that's the primary reason for the existence of the normal ground state of the Hydrogen atom. In order to shrink it would need to shed angular momentum, and it can't do this via photon emission, because the creation of a circularly polarized photon requires more angular momentum than the electron can supply. Hence H can't shrink via photon emission, hence the ground state. The Hause condition appears to be a necessary, but not sufficient condition, according to the skeptics, and I think they may be right about that. However the angular momentum condition is sufficient. Thus if shrinkage requires shedding angular momentum, and it can't be done via photon emission, then that only leaves passing it off to another particle, which in turn implies a physical collision. BTW I think you will find that Mills assumes that the angular momentum of the electron is a constant during shrinkage. One of various points upon which I disagree with him. Regards, Robin van Spaandonk The shrub is a plant.
