In reply to Edmund Storms's message of Sun, 30 Jan 2005 10:05:41 -0700: Hi, [snip] >Nuclear weapons produce so much radiation that all molecules near the >device are decomposed into atoms and ions, which occupy a much larger >volume. In addition, the energy density is huge. [snip] Precisely. >> So the O++ is reconstituted after use. The only problem is to reuse it >> before it captures another electron and becomes O+. >> >> >The window of time during which oxygen has the correct charge would seem >to be rather short. I guess it is a matter of intuition whether the time >is too short for sufficient O++ to be present.
I think it's more a matter of what else is present that it can collide with before it comes into contact with H, and what the result of that collision will be. In a stoichiometric mix of H and O, there will be twice as many H as O atoms, so a lone O++ is twice as likely to come in contact with H as it is with an O atom. Of course there is also the competing reaction: H + O++ -> H+ + O+ and it's anybody's guess what the ratio of the two reaction rates is. Of course pre-existing hydrinos in the plasma will shift the balance in favour of a shrinkage reaction, because the O percentage is decreased, and also because when O++ reacts with a hydrino rather than with H, there is no competing reaction. Shrinkage is the only game in town. This means that once shrinkage has started, there is practically speaking no real way back. [snip] >> What I am trying to make clear here, is that once shrinkage has progressed >> far enough, the reaction can be self-sustaining, even though the production >> of O++ is not very efficient, simply because the inefficiency is out weighed >> by the energy excess from the reaction. >> >OK, I understand. Presumably the reaction proceeds until all of the >accumulated hydrinos are used up. Yes, or the cell blows itself apart, and puts and end to the process. In which case, there should still be a supply of severely shrunken hydrinos bound to the walls/electrodes, which is why I suggested that it might be possible to replicate using the remains of the shattered cell/electrodes. [snip] >I don't understand how the hydrinos can accumulate in the glass. Hydrinos can bind an extra electron to become hydrinohydride (H*-). This is essentially a very small negative ion. The second electron can be very tightly bound to the hydrino (up to 70 eV binding energy according to Mills). Because this ion is very small, it can snuggle up very close to a positive ion, which in turn implies a high binding energy between the two. To give an idea of what this means, O-- ions bind very tightly to metal ions because they are relatively small, which is why oxides generally have high melting points. The H*- ion if much smaller than O--, and hence should sit closer to a metal ion than even O--, implying a much stronger bond. These substances could have melting points of tens to hundreds of thousands of degrees. Consequently H*- could easily be bound to Si++++ or Na+ in the glass, displacing O--. This bond would be so strong that no amount of scrubbing and no solvent would remove it. Essentially it would be stronger than the glass itself. This same reasoning applies equally to the electrodes. [snip] >Even >if they were in the glass, why and how would they suddenly come out into >the solution? The extraction process requires a threshold energy. Below the threshold, nothing happens, which is why cleaning has no effect. Because of the strength of the bond, it takes a very energetic process to free them, however hydrino shrinkage provides just such energies. IOW shrinkage reactions taking place in the plasma can supply the energy required to free the H*- from its bound position in the lattice. Once free, O+++ will remove the electron from H*-, provided that the binding energy of the second electron doesn't exceed 54 eV. The H* thus provided, is then free to undergo further shrinkage. [snip] > material attached to the glass would not be expected. Your model >needs a significant source of hydrinos that have accumulated over a >period of time, which can quickly enter the water at a particular time >and react. How does this occur and why the sudden release? Please see above. However the plasma required has to start somewhere. The initial trigger may be a cosmic ray or a random fusion event occurring in the lattice, between an "embedded" H*- and the metal atom to which it is bound. Because of the mass and size of H*-, it's even possible that these particles actually orbit the nucleus of the metal atoms inside the K shell, effectively displacing a K shell electron during the binding process. This is a closer analogy to the muonic molecule. From such an orbit, it is only a matter of time before a fusion reaction occurs. Naturally such reactions would have a characteristic half life, depending on the metal atom in question, and the shrinkage level of the H*-. The fusion reaction probably produces ionising radiation. Ionising radiation can create a small localized plasma, which can grow according to the model previously described, provided that it is adjacent to a surface containing bound H*-. [snip] >> See comments above re. nuclear weapons. >Yes, but the issue of magnitude is the essential difference. A nuclear >weapon affects every molecule near it. The reactions you describe are on >an local atomic scale. True, but the energy release can be several orders of magnitude greater than normal chemical reactions, and the products of the reaction are high energy products (i.e. tens to thousands of eV each), so massive disruption of water molecules is certain. IOW the same MO, but on a smaller scale. We are looking at a plasma temperature possibly as high as 100 eV, which is equivalent to nearly 800,000 K. All of this leads me to infer that O++ is an excellent Mills catalyst, but that it operates best in a temperature range that is rarely attained in our everyday world. However, once a high energy trigger event occurs that locally elevates the temperature into that range, it really shines. ;) One last comment I would make. Improving the containment strength of the cell may just result in a more powerful bomb. Next time, the results may be deadly. I would opt for a weak cell surrounded by a strong containment with a large air gap between the two. Regards, Robin van Spaandonk All SPAM goes in the trash unread.
