with apologies for the delay had PC troubles, made blunders
http://egooutpeters.blogspot.ro/2015/03/parkhomovs-3rd-paper-text.html
Peter
--
Dr. Peter Gluck
Cluj, Romania
http://egooutpeters.blogspot.com
see it attached, the paper is at
https://yadi.sk/i/zTRxBwVofYmaE
My text is not going with Blogger huge letters appear
I am asking the colleagues skilled to combine text and images.
Peter
--
Dr. Peter Gluck
Cluj, Romania
http://egooutpeters.blogspot.com
AGP-3t.rtf
Description: RTF file
Could you have a look at conclusion 1.?
50 kWh is 180 mj.
Or 5 kWh is 18mj is 5 x 22 eurocent (electicity price) =1 euro= 0.5 liter
petrol.
Or am i missing ...
Marcus
Op 27 mrt. 2015 12:17 schreef Peter Gluck peter.gl...@gmail.com:
with apologies for the delay had PC troubles, made blunders
you are right, the error is in the original
peter
On Fri, Mar 27, 2015 at 2:51 PM, Marcus Winckers marcki...@gmail.com
wrote:
Could you have a look at conclusion 1.?
50 kWh is 180 mj.
Or 5 kWh is 18mj is 5 x 22 eurocent (electicity price) =1 euro= 0.5 liter
petrol.
Or am i missing ...
The following is overly simplistic, but also surprisingly intuitive for a
particular hypothesis, so it is worth the effort to try to get down an
explanation for the Rossi/Parkhomov effect . one containing less than a
thousand words, by using a few images.
Here is an image of the Li-7 nucleus.
Bob,
In general - the evidence says that there is an almost complete lack of high
energy radiation, or neutron activation, in the reported experiments even at
the kilowatt thermal level- this means that there are no neutrons, no fusion
and little transmutation (other than incidental).
If any of
Dear Friends
It is:
http://egooutpeters.blogspot.ro/2015/03/more-lenr-issues-at-march-27-2015.html
Peter
--
Dr. Peter Gluck
Cluj, Romania
http://egooutpeters.blogspot.com
I went into the Powerpoint file and translated the text and figures of
Alexander Parkhomov's latest presentation into English (Thank you Google
Translate). Here is a link to the translated document. Please point out
any errors and I will correct them. I took the liberty of correcting the
The spacing of the nucleons is 1.36 fermi meters. That's twice the radii. The
reciprocal of this spacing is the nuclear wave number.
The wave number and the elastic constant give a sonic velocity of 1,094,000
meters per second in the structure.
Frank Znidarsic
-Original Message-
The spacing of the nucleons is 1.36 fermi meters. That's twice their radii.
This is a constant as the density of the nucleus is constant. The reciprocal
of this spacing is the nuclear wave number. The wave number and the elastic
constant give a sonic velocity of 1,094,000 meters per
A picture is worth a thousand words, or maybe 10kJones--
The Li-6 loves neutrons and will readily change to Li-7, if one is nearby. It
may be that the Li-6 acts as a catalyst to combine the charge of a proton and
an electron to form a neutron and He-5, which in turn gives up another neutron
Great work, Bob -- very useful!
On Fri, Mar 27, 2015 at 12:35 PM, Bob Higgins rj.bob.higg...@gmail.com
wrote:
I went into the Powerpoint file and translated the text and figures of
Alexander Parkhomov's latest presentation into English (Thank you Google
Translate). Here is a link to the
The ICCF19 secretariat sent me this notice.
At the following link is now available the list of the posters and oral
presentations accepted:
http://www.iccf19.com/program_abstract.html
A weekly calendar is available at this link:
http://www.iccf19.com/program_overview.html
A more detailed
Given that the deuteron is a magnetic dipole - a quantum nuclear magnet
which interacts with its own electrons to form a magnon... we have an
interesting situation when 3 deuterons, connected at the focal point as if
one pole, and having an x,y, and z axis... oscillate at elevated
temperatures.
In reply to Bob Cook's message of Tue, 24 Mar 2015 07:57:39 -0700:
Hi,
I think the Li6 nucleus is arranged like this:-
2 neutrons paired with opposite spin.
2 protons paired with opposite spin.
1 neutron left over.
1 proton left over.
I think the negative near field of the neutron implies
Jones--
I agree with your observation that there is not radiation seen and hence
probably no neutron production, which would lead to activation and decay of
activated nuclei with their very visible gammas. My Idea of potential
activation of Al-27 is not likely.
I like your idea regarding
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