`If you look at my text you will see I wrote "catastrophic failure"`

`not just failure. This means an E-cat blows up spreading steam`

`throughout the container, injuring anyone present, and preventing`

`access to the container, causing the test to fail. I think I was`

`clear on this point. I did not refer to anything bout an E-cat`

`performance dropping. The other side of the coin to increased`

`probability of some E-cat working when multiple devices run together`

`is the increased probability of catastrophic failure.`

Best regards,

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Horace Heffner http://www.mtaonline.net/~hheffner/ On Sep 29, 2011, at 12:26 PM, Axil Axil wrote:

The failure of one module of the Rossi 1 MW reactor will not causethe entire 1 MW reactor to fail. Its performance will only degradegracefully.When the core of the module overheats or melts, the surface of thenickel nanopowder will fail before the nanopowder enclosure willfail since the enclosure will be cooled by low temperature steam orwater which would remove heat, effectively cool the enclosure, andsupport its structural strength.The failure of the nanopowder will cause the individual module tocool and be ineffective at generating thermal power.It would be analogous to a failure of one pixel of your computerscreen; if one such pixel grows dark, your screen will not fail butits performance would degrade. You would still be able to use thescreen, just the picture would not be as sharp.So too with the Rossi reactor; it would still generate heat, butnot so much as before. Its capacity would be reduced until itsperformance would eventually degrade below a certain predefinedlower threshold.When this low bound threshold is reached, the entire reactor isconsidered to have failed.On Thu, Sep 29, 2011 at 3:34 PM, Horace Heffner<hheff...@mtaonline.net> wrote:On Sep 29, 2011, at 4:02 AM, Man on Bridges wrote: Hi, On 29-9-2011 8:27, Horace Heffner wrote:Looking at the other side of the coin, the probability ofcatastrophic failure, suppose there is a 0.1% chance per hour oneof the E-cats can blow up spreading steam throughout thecontainer. There is thus a 0.999 probability of success, i.e. noexplosion for one E-cat, operating for one hour. The probabilitythat all 52 E-cats perform successfully for a 24 hour test periodis then 0.999^(52*24) = .287. That means there is a 71.3% chanceof an explosion during a 24 hour test.Me thinks you are wrong. Your statistical probability calculationis based upon the fact that the chance of a single Ecat explodingis influenced by it's behaviour earlier,This is false. The probability in each time increment is assumedto be independent. For there to be success there must be nofailures for any time increment. If there are T time increments,and the probability of failure in any time increment is p, theprobability of success q=1-p in each time increment is independentof the other time increments, and the probability of success in alltime increments is q^T (only possible if what happens in each timeincrement is independent event), and the probability of any failurehaving occurred is thus 1-(q^T).which of course is not true. Statistically each Ecat has it's ownindependent chance of explosion at any given moment which does notchange over time.The instantaneous probability of failure is zero. Zero time resultsin zero probability because lim t->0 q^t = 1 for for all 0=<q<=1and positive t. Therefore lim t->0 1-(q^t) = 0. Note that Iprovided an assumption of 0.001 percent probability of failure *perhour*.With your probability of 0,1% chance per hour this would result forthe whole of 52 Ecats then in a chance of explosion at any givenmoment of 1 - (0.999^52) = .05 or 5%.No. The probability of at least one E-cat failure in the 52 E-catsystem, based on the assumption of 0.001 probability of failure ofan individual E-cat in an hour is 1-(0.999)^52 = 0.506958 = 5%.Your number 5% is right, but your interpretation of it representingan instantaneous moment is wrong.Looking even a bit more closer again this would mean that if thechance of explosion is 0.1% per hour then the chance of explosionis 2,77e-7 per second at any given moment for a single Ecat, whichwould result for 52 Ecats into 1-((2,77e-7)^52) = 0,00001444434 or0,00144% at any time.The phrase "at any time" makes the above statement nonsensical.An hour represents 3600 seconds, which are 3600 independent eventsof 1 second duration. Let a be the probability of failure in 1second, and b=(1-a) be the probability of success in 1 second. Wehave the given probability p of failure for 3600 seconds being0.001, and the probability of success of one E-cat for one hourbeing q = 0.999. The probability of success (no failures) for the3600 1 second independent time increments isq = 0.999 = b^3600 b = q^(1/3600) = 0.999^(1/3600) a = 1 - 0.999^(1/3600) = 2.779x10^-7Note that a is the probability of failure in one second, not "atany time". This is totally consistent with the probability offailure in one E-cat in one hour being 5%. In other words, goingbackwards:p = 1-(1-a)^3600 = 1-(1-2.779x10^-7)^3600 = 1-0.999 = 0.001My calculations are therefore self consistent. The time intervalsare all treated as independent events. Your interpretation of"moment" is perhaps a conceptual problem.Kind regards, MoB Best regards, Horace Heffner http://www.mtaonline.net/~hheffner/