tag:blogger.com,1999:blog-326167780677397310.post3643576249783551459..comments2024-03-27T21:35:04.988-07:00Comments on EGO OUT: Ed Storms' answers to 5 questions. Questions No. 6 and 7Georgina Popescuhttp://www.blogger.com/profile/04628821029016016988noreply@blogger.comBlogger4125tag:blogger.com,1999:blog-326167780677397310.post-37546060659323241262012-06-20T23:41:09.871-07:002012-06-20T23:41:09.871-07:00I have no numbers. I was just thinking if the area...I have no numbers. I was just thinking if the area there an uneven surface region touches another could create cavitys that are suiteable for NAE:s (not creating cracks in surface by the force).<br /><br />LarsAnonymousnoreply@blogger.comtag:blogger.com,1999:blog-326167780677397310.post-19185569281786603932012-06-20T04:55:24.688-07:002012-06-20T04:55:24.688-07:00Can you please give some details- how fine, which
...Can you please give some details- how fine, which<br />compressing force?<br />Thanks.<br />PeterAnonymousnoreply@blogger.comtag:blogger.com,1999:blog-326167780677397310.post-81421813458380962362012-06-20T04:48:24.276-07:002012-06-20T04:48:24.276-07:00Is it possible that NAE could form in the surface ...Is it possible that NAE could form in the surface area between particles if the powder is fine enough and it is compressed by an external force.Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-326167780677397310.post-62296166504747152592012-06-10T06:21:40.871-07:002012-06-10T06:21:40.871-07:00Question #6: I propose that a limited and relative...Question #6: I propose that a limited and relatively constant number of active cracks can form because these result from stress relief. Once all the stress is relieved, no more cracks can form. Of course, most of the cracks made this way will be too large to be active, so that only a small number of NAE sites are making the detected energy. <br /><br />The life time will be determined by variables independent of the number of active cites. For example as deuterium accumulates in the E-cat, the reaction rate will drop because the less active tritium formation reaction will start. When deuterium is used to make helium, the helium will accumulate and block access to the active sites for the deuterium. <br /><br />I do not believe that any significant transmutation takes place. All measurements of this process show that this reaction is rare, except for the claim by Rossi. <br /><br />Question #7: This question involves politics, which makes it difficult to answer. On the one hand, the Pd system has a great deal of experimental support while the Ni system can apparently produce significant power, but based on very little understanding of the process. If the crack model is correct, the metal is not important except that it be able to form active cracks and dissolve D or H as the required reactants. In fact, Ni might be a better host for the D reaction than Pd because it is cheaper and the D is more active than H because each D makes more energy than each H. So, my advice is not to focus on the metal but on understanding the process. Once the process is mastered, the claims will be accepted regardless of the metal used. In fact, I think neither Ni nor Pd are the best host for the reaction.Ed Stormsnoreply@blogger.com