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.
On Sep 29, 2011, at 12:26 PM, Axil Axil wrote:
The failure of one module of the Rossi 1 MW reactor will not cause
the entire 1 MW reactor to fail. Its performance will only degrade
When the core of the module overheats or melts, the surface of the
nickel nanopowder will fail before the nanopowder enclosure will
fail since the enclosure will be cooled by low temperature steam or
water which would remove heat, effectively cool the enclosure, and
support its structural strength.
The failure of the nanopowder will cause the individual module to
cool and be ineffective at generating thermal power.
It would be analogous to a failure of one pixel of your computer
screen; if one such pixel grows dark, your screen will not fail but
its performance would degrade. You would still be able to use the
screen, just the picture would not be as sharp.
So too with the Rossi reactor; it would still generate heat, but
not so much as before. Its capacity would be reduced until its
performance would eventually degrade below a certain predefined
When this low bound threshold is reached, the entire reactor is
considered to have failed.
On Thu, Sep 29, 2011 at 3:34 PM, Horace Heffner
On Sep 29, 2011, at 4:02 AM, Man on Bridges wrote:
On 29-9-2011 8:27, Horace Heffner wrote:
Looking at the other side of the coin, the probability of
catastrophic failure, suppose there is a 0.1% chance per hour one
of the E-cats can blow up spreading steam throughout the
container. There is thus a 0.999 probability of success, i.e. no
explosion for one E-cat, operating for one hour. The probability
that all 52 E-cats perform successfully for a 24 hour test period
is then 0.999^(52*24) = .287. That means there is a 71.3% chance
of an explosion during a 24 hour test.
Me thinks you are wrong. Your statistical probability calculation
is based upon the fact that the chance of a single Ecat exploding
is influenced by it's behaviour earlier,
This is false. The probability in each time increment is assumed
to be independent. For there to be success there must be no
failures for any time increment. If there are T time increments,
and the probability of failure in any time increment is p, the
probability of success q=1-p in each time increment is independent
of the other time increments, and the probability of success in all
time increments is q^T (only possible if what happens in each time
increment is independent event), and the probability of any failure
having occurred is thus 1-(q^T).
which of course is not true. Statistically each Ecat has it's own
independent chance of explosion at any given moment which does not
change over time.
The instantaneous probability of failure is zero. Zero time results
in zero probability because lim t->0 q^t = 1 for for all 0=<q<=1
and positive t. Therefore lim t->0 1-(q^t) = 0. Note that I
provided an assumption of 0.001 percent probability of failure *per
With your probability of 0,1% chance per hour this would result for
the whole of 52 Ecats then in a chance of explosion at any given
moment of 1 - (0.999^52) = .05 or 5%.
No. The probability of at least one E-cat failure in the 52 E-cat
system, based on the assumption of 0.001 probability of failure of
an individual E-cat in an hour is 1-(0.999)^52 = 0.506958 = 5%.
Your number 5% is right, but your interpretation of it representing
an instantaneous moment is wrong.
Looking even a bit more closer again this would mean that if the
chance of explosion is 0.1% per hour then the chance of explosion
is 2,77e-7 per second at any given moment for a single Ecat, which
would result for 52 Ecats into 1-((2,77e-7)^52) = 0,00001444434 or
0,00144% at any time.
The phrase "at any time" makes the above statement nonsensical.
An hour represents 3600 seconds, which are 3600 independent events
of 1 second duration. Let a be the probability of failure in 1
second, and b=(1-a) be the probability of success in 1 second. We
have the given probability p of failure for 3600 seconds being
0.001, and the probability of success of one E-cat for one hour
being q = 0.999. The probability of success (no failures) for the
3600 1 second independent time increments is
q = 0.999 = b^3600
b = q^(1/3600) = 0.999^(1/3600)
a = 1 - 0.999^(1/3600) = 2.779x10^-7
Note that a is the probability of failure in one second, not "at
any time". This is totally consistent with the probability of
failure in one E-cat in one hour being 5%. In other words, going
p = 1-(1-a)^3600 = 1-(1-2.779x10^-7)^3600 = 1-0.999 = 0.001
My calculations are therefore self consistent. The time intervals
are all treated as independent events. Your interpretation of
"moment" is perhaps a conceptual problem.