On Sat, Jun 1, 2013 at 10:10 AM, Jones Beene jone...@pacbell.net wrote:
Anyway the Farnsworth Fusor is a fusion reactor that many high school level
students have built, including Conrad.
It involves adding electrical energy in order to achieve LENR reactions.
Sound familiar, Joshua?
You
On Sat, Jun 1, 2013 at 11:25 AM, Edmund Storms stor...@ix.netcom.comwrote:
We are taking about two different phenomenon of nature. Trying to use the
same concepts and words to describe both results in confusion. Those of us
who have studied cold fusion for the last 23 years have a definition
Jones,
Did he make the background measure and the active run measure with the
detector in the same place and same orientation?
If he did, then the dip recorded during the active run would mean an
_active_ ecat can reduce background radiation.
Harry
On Sun, Jun 2, 2013 at 12:08 AM,
From: David Roberson
Robin, how would Rossi prevent the lead from melting at the
elevated temperatures? Do you suspect that he has it confined within a
closed shell of some kind? I do not recall seeing any place for it to hide.
OK, Jones, let me try to summarize what you propose.
You believe CF is like the Mills effect even though CF is known to
produce nuclear products and the Mills effect does not.
You believe that Rossi made the Ni-H2 system create energy using the
Mills effect while everyone else who explored
From: Edmund Storms
OK, Jones, let me try to summarize what you propose You
believe CF is like the Mills effect even though CF is known to produce
nuclear products and the Mills effect does not.
Not even close, Ed.
I specifically said that I do not address
Jones Beene jone...@pacbell.net wrote:
No. That is not correct. Tritium would have already have been detected by
Bianchini if it was there . . .
I do not think so. Tritium would be trapped inside the cell. The decay
product is a low energy beta. If a little tritium leaks out of the cell it
Jed is correct. Tritium can not be detected by an ordinary detector
because the beta is too weak. Unless the required special detector is
used, tritium would be totally missed no matter how much is present.
That is why tritium is dangerous. Nevertheless, modern methods can
detect tritium
You do not need to remove the gas.
I know you have heard of Bremsstrahlung, even if the word is almost
unspellable to Anglos. Thank heavens for spell checkers and Wiki vids. Here
is a little video that tells you why Bianchini would see tritium, if it was
there.
Jones, you are simply wrong. I have worked with tritium and I know how
it behaves. It cannot be detected using its Bremsstrahlund unless a
huge amount is present because this radiation is produced at only a
small fraction of the beta and is absorbed very quickly by only a
small amount of
On Jun 2, 2013, at 10:05 AM, Jones Beene wrote:
From: Edmund Storms
OK, Jones, let me try to summarize what you propose You
believe CF is like the Mills effect even though CF is known to produce
nuclear products and the Mills effect does not.
Not even
From: Edmund Storms
Jones, you are simply wrong. I have worked with tritium and I know how it
behaves.
You apparently have not worked with tritium very intuitively, if you cannot
understand this simple video.
It cannot be detected using its Bremsstrahlund unless a huge
Apparently Jones, I have to be clearer and more emphatic. Tritium can
not be detected when it is in a container as massive as the E-cat.
THIS IS A FACT. Please at least acknowledge that I might know
something about tritium that you do not. The video only shows that
some unknown amount of
Ed,
You are not very good at misdirection, try hard as you might - and you are
fighting a losing battle in trying to wedge an incorrect theory into the
most important LENR experiment out there at present.
My advice is to quit before you are completely embarrassed. You theory works
in some
Jones Beene jone...@pacbell.net wrote:
. . . a signal should show up above background on his meter - especially
when the Rossi device is disassembled, as it is in the Penon report.
They disassemble it by cutting it in half with a saw, don't they? There is
no way you could capture tritium by
OK Jones, useful discussion has come to an end. I will wait until the
proper measurements are made . Then we will talk again.
Ed Storms
On Jun 2, 2013, at 12:59 PM, Jones Beene wrote:
Ed,
You are not very good at misdirection, try hard as you might - and
you are fighting a losing battle
Tritium is preferentially absorbed into nickel. Most of it would be retained
in the nickel powder, if it were present.
From: Jed Rothwell
. . . a signal should show up above background on his meter - especially
when the Rossi device is disassembled, as it is in the Penon report.
They
Jones Beene jone...@pacbell.net wrote:
Tritium is preferentially absorbed into nickel. Most of it would be
retained in the nickel powder, if it were present.
Good point. Still, if you were doing a serious study you would not cut it
in half.
McKubre devised a complicated way to puncture the
Let me say that almost everyone concerned, other than Andrea Rossi himself -
would be delighted if tritium had been found in the spent fuel of the
HotCat. If tritium were found in proportion to thermal gain - this would
explain the mechanism in accordance with Ed Storm's theory - and not only
Jones, I agree with your conclusion about Rossi. However, tritium is
not his only problem. His patent will probably not reveal how the Ni
can be treated to make it active. Simply adding Ni62 is obviously not
the only thing he does to the Ni. Without the ability to replicate the
patent by a
In reply to Jones Beene's message of Sun, 2 Jun 2013 06:15:39 -0700:
Hi,
[snip]
The fact that there is no radiation at all detectable (at kW thermal output)
from Rossi's device (above a threshold of tens of keV) is rather conclusive
that there is no fusion, and essentially no nuclear reaction of
Jones, please do not confuse hot fusion with cold fusion. The
difference is in the products. Cold fusion does not produce neutrons
and energetic radiation. Hot fusion produce neutrons and radiation
because the conditions require the nuclear product to fragment. This
fragmentation does not
-Original Message-
From: Edmund Storms
Jones, please do not confuse hot fusion with cold fusion. The difference
is in the products.
Not necessarily. Perhaps that is your definition, but as I stated - the
Farnsworth Fusor is LENR on the input side. Same with sonofusion - it is the
We are taking about two different phenomenon of nature. Trying to use
the same concepts and words to describe both results in confusion.
Those of us who have studied cold fusion for the last 23 years have a
definition of CF that is not up for discussion. Please try to
understand what I'm
I thought we agreed to call Muon assisted fusion warm fusion.
On Sat, Jun 1, 2013 at 12:25 PM, Edmund Storms stor...@ix.netcom.com wrote:
We are taking about two different phenomenon of nature. Trying to use the
same concepts and words to describe both results in confusion. Those of us
who
-Original Message-
From: Edmund Storms
We are taking about two different phenomenon of nature. Trying to use
the same concepts and words to describe both results in confusion.
Those of us who have studied cold fusion for the last 23 years have a
definition of CF that is not up for
You can call it what you want. Jones called the muon reaction cold
fusion before he applied the term was applied to the F-P effect.
Nevertheless, the products are those that result from hot fusion, i.e.
equal amounts of neutron and tritium that result from fragmentation of
the resulting
On Jun 1, 2013, at 11:16 AM, Jones Beene wrote:
-Original Message-
From: Edmund Storms
We are taking about two different phenomenon of nature. Trying to use
the same concepts and words to describe both results in confusion.
Those of us who have studied cold fusion for the last 23
-Original Message-
From: Edmund Storms
The Mills effect is a different phenomenon all together. His effect is not
nuclear, as he admits.
Yes, but that is not relevant to understanding Rossi. Many other
researchers, including Miley have incorporated major parts of Mills theory
into a
Jones Beene jone...@pacbell.net wrote:
Bianchini finds zero radiation over hundreds of hours of careful radiation
testing.
Most cold fusion experiments produce no measurable radiation over hundreds
of hours, including Pd-D ones.
Essen finds no radioactivity in the ash. No excess deuterium
From: Jed Rothwell wrote:
Bianchini finds zero radiation over hundreds of hours of careful radiation
testing.
Most cold fusion experiments produce no measurable radiation over hundreds
of hours, including Pd-D ones.
Most cold fusion experiments have been milliwatt level and do
Jones Beene jone...@pacbell.net wrote:
Most cold fusion experiments have been milliwatt level and do not use the
very sophisticated setup of Bianchini . . .
Fleischmann and Pons ran hundreds of tests with boiling cells, at 20 to 100
W. They has sophisticated detectors. They found nothing as
On Sat, Jun 1, 2013 at 2:02 PM, Jones Beene jone...@pacbell.net wrote:
The ultimate source of energy cannot be determined as of now but Rossi’s
hundreds of hours of operation at kilowatt levels with no gammas clearly
indicates NO fusion.
I don't exclude the possibility that there's
Eric,
I have dined on crow before and prefer mine well-charred with a nice Pinot Noir…
The ultimate source of energy cannot be determined as of now but Rossi’s
hundreds of hours of operation at kilowatt levels with no gammas clearly
indicates NO fusion.
I don't exclude the
On Sat, Jun 1, 2013 at 5:33 PM, Jones Beene jone...@pacbell.net wrote:
I have dined on crow before and prefer mine well-charred with a nice Pinot
Noir…
Foul! Fowl demands a white, say chardonnay,
In reply to Jones Beene's message of Sat, 1 Jun 2013 14:33:22 -0700:
Hi,
[snip]
Eric,
I have dined on crow before and prefer mine well-charred with a nice Pinot
Noir
The ultimate source of energy cannot be determined as of now but Rossis
hundreds of hours of operation at kilowatt
-Original Message-
From: mix...@bigpond.com
Hi Robin,
The H2 is of course f/H molecules.
Still three body reactions - no way
Nevertheless, I suspect that indeed the primary source of energy in his
reactor is the formation of f/H.
Yup. By a large factor.
There is actually an easy
On Sat, Jun 1, 2013 at 5:10 PM, Eric Walker eric.wal...@gmail.com wrote:
On Sat, Jun 1, 2013 at 2:02 PM, Jones Beene jone...@pacbell.net wrote:
The ultimate source of energy cannot be determined as of now but
Rossi’s hundreds of hours of operation at kilowatt levels with no gammas
clearly
In reply to Jones Beene's message of Sat, 1 Jun 2013 17:27:32 -0700:
Hi,
[snip]
-Original Message-
From: mix...@bigpond.com
Hi Robin,
The H2 is of course f/H molecules.
Still three body reactions - no way
No, these are all two body reactions, because the f/H is bound in a
-Original Message-
From: mix...@bigpond.com
Hi Robin,
The H2 is of course f/H molecules.
Still three body reactions - no way
No, these are all two body reactions, because the f/H is bound in a
pico/femto molecule, and approaches the target nucleus as a single
(composite) entity.
Jones,
Interesting concept..[snip] No, these are all two body
reactions, because the f/H is bound in a
pico/femto molecule,[/snip] how about combining it with Naudt's paper on
relativistiv hydrogen, the hydrogen has an equivalent negative acceleration
of relativistic proportion
Let me add that in the appendix to the Penon report, David Bianchini finds
not only no significant radiation over background, but actually the peak
radiation counts are slightly less during the experiment than background,
indicating the apparatus shields the detector from cosmic rays slightly.
In reply to Jones Beene's message of Sat, 1 Jun 2013 19:35:11 -0700:
Hi,
[snip]
-Original Message-
From: mix...@bigpond.com
Hi Robin,
The H2 is of course f/H molecules.
Still three body reactions - no way
No, these are all two body reactions, because the f/H is bound in a
In reply to Jones Beene's message of Sat, 1 Jun 2013 19:59:52 -0700:
Hi,
[snip]
Let me add that in the appendix to the Penon report, David Bianchini finds
not only no significant radiation over background, but actually the peak
radiation counts are slightly less during the experiment than
-l@eskimo.com
Sent: Sun, Jun 2, 2013 12:08 am
Subject: Re: [Vo]:Adding Energy to get Energy
In reply to Jones Beene's message of Sat, 1 Jun 2013 19:59:52 -0700:
Hi,
[snip]
Let me add that in the appendix to the Penon report, David Bianchini finds
not only no significant radiation over background
for it to hide.
Dave
You are correct. :) I was confused with the earlier versions that used lead
shielding. However any solid will provide *some* shielding.
-Original Message-
From: mixent mix...@bigpond.com
To: vortex-l vortex-l@eskimo.com
Sent: Sun, Jun 2, 2013 12:08 am
Subject: Re: [Vo]:Adding
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