Pons and Fleischman used a palladium cathode in their cold fusion
experiment of 1989 (cathode immersed in heavy water,D20)and they said
that excess energy was released from the system.
Since 1989 lots of experiments have been done and some researchers
have agreed with the findings of Pons and Fleischman.
I have seen many explanations of why cold fusion can't take place
according to known physics but here is one suggestion I have as to why
it can!
Palladium metal is very resistant to fracturing.This means that
impurity atoms or ions face a high energy barrier to get into the
crystal lattice and distort the face-centred cubic arrangement,and
that it is difficult for defects to grow larger to cause fracturing.
But when an electric current flows through palladium metal (which has
hydrogen gas on its surface and inside it - palladium absorbs 900
times its own volume of hydrogen) hydrogen molecules and hydrogen ions
(which are present in the heavy water) are given the energy to distort
the metal structure and remain in situ.As time passes the number of
distortions and hence the potential energy of the palladium cathode
increases.At some critical point, the face-centred cubic arrangement
of palladium atoms is quickly restored
and the hydrogen molecules and ions are expelled from the palladium
cathode, as potential energy is released like a mini-earthquake.In a
cylindrical-shaped cathode the released energy could be focused at its
end or in the centre of the rod.In particular,
at some locations in the rod where deuterium molecules are located,
the focused energy could be sufficient to cause helium 4 to form.The
formation of helium inside the cathode would explain why nuclear
products have rarely been detected in the heavy water in experiments
of this type.

Ditto for sonofusion I assume?
Although such back of envelope calculations might be rather complex,
depending upon what assumptions are made.

Dirk Bruere at NeoPax wrote:

Ditto for sonofusion I assume?
Although such back of envelope calculations might be rather complex,
depending upon what assumptions are made.



[Drifting the thread slightly for anecdotal reasons]
BOTE calculations have to be extremely sensitive to physical insight and
clever tricks to fit in the space allotted.

The University of Chicago used to include a handful BOTE questions as part
of their candidacy examinations. I recall one question that makes me
shudder to this day. It simply asked to calculate the height of the ocean
tides. There wasn't a single candidate who took that exam who had any idea
how to do this on the back of an envelope.

Murray Arnow wrote:
Dirk Bruere at NeoPax wrote:

Ditto for sonofusion I assume?
Although such back of envelope calculations might be rather complex,
depending upon what assumptions are made.

[Drifting the thread slightly for anecdotal reasons]
BOTE calculations have to be extremely sensitive to physical insight and
clever tricks to fit in the space allotted.

The University of Chicago used to include a handful BOTE questions as part
of their candidacy examinations. I recall one question that makes me
shudder to this day. It simply asked to calculate the height of the ocean
tides. There wasn't a single candidate who took that exam who had any idea
how to do this on the back of an envelope.

G = gravitational constant
Me = earth mass 5.97E+27 g
Mn = moon mass 7.35E+25 g
Re = earth radius 6.371315E+08 cm
Rem = distance between moon and earth 3.844E+10 cm
T = tide height
Wm = water test mass

Gravity works on a water test mass
with force from earth and moon:

G*Me*Wm/Re^2 = G*Me*Wm/(Re+T)^2 + G*Wm*Mm/(Rem)^2

cancel G and Wm

Me/Re^2 = Me/(Re+T)^2 + Mm/(Rem)^2

or

Re^2/(Re+T)^2 = 1 - (Mm/Me)*(Re/Rem)^2

Mm/Me ~ .01

Re/Rem ~ .01

Re^2/(Re+T)^2 = 1 - (.01)*(.01)^2 = 1 - 1E-6

solve ~ BOTE for T ~ 10 meters

Richard- Hide quoted text -

- Show quoted text -

Murray Arnow wrote:
Dirk Bruere at NeoPax wrote:

Ditto for sonofusion I assume?
Although such back of envelope calculations might be rather complex,
depending upon what assumptions are made.



[Drifting the thread slightly for anecdotal reasons]
BOTE calculations have to be extremely sensitive to physical insight and
clever tricks to fit in the space allotted.

The University of Chicago used to include a handful BOTE questions as part
of their candidacy examinations. I recall one question that makes me
shudder to this day. It simply asked to calculate the height of the ocean
tides. There wasn't a single candidate who took that exam who had any idea
how to do this on the back of an envelope.


G = gravitational constant
Me = earth mass 5.97E+27 g
Mn = moon mass 7.35E+25 g
Re = earth radius 6.371315E+08 cm
Rem = distance between moon and earth 3.844E+10 cm
T = tide height
Wm = water test mass

Gravity works on a water test mass
with force from earth and moon:

G*Me*Wm/Re^2 = G*Me*Wm/(Re+T)^2 + G*Wm*Mm/(Rem)^2

cancel G and Wm

Me/Re^2 = Me/(Re+T)^2 + Mm/(Rem)^2

or

Re^2/(Re+T)^2 = 1 - (Mm/Me)*(Re/Rem)^2

Mm/Me ~ .01

Re/Rem ~ .01

Re^2/(Re+T)^2 = 1 - (.01)*(.01)^2 = 1 - 1E-6

solve ~ BOTE for T ~ 10 meters

But solving energy problems should be the top priority of
science. NoEinstein

After reading this link:

 http://www.psc.edu/science/Wolf/Wolf.html

and seeing the hydrogen molecules in such close proximity to the
palladium atoms,I think that the palladium is giving electron density
to the hydrogen molecules and shortening their bond lengths,and that
the palladium atoms
having lost some electron density are repelling each other more
electrically,and increasing their bond lengths as more hydrogen enters
the system
(palladium is known to expand as hydrogen enters it).This would
explain why there is no back pressure stopping more hydrogen from
entering he palladium metal sample.
The pd-pd bond length would have to increase 2.5 times to accomodate
900 times the palladium's own volume of hydrogen ( 2.5 ^ 3 is
approximately 15, and before expansion each face-centred cubic pd unit
could contain a maximum of 60 hydrogen molecules - 15 x 60 = 900).

Murray Arnow:

Regardless of your explanation,
no one reproduced the P&F experiment.

See:
A.De Ninno, A. Frattolillo, A. Rizzo,
E. del Giudice, G. Preparata
"Experimental evidence of 4He production
in a cold fusion experiment"

Murray Arnow:

Regardless of your explanation,
no one reproduced the P&F experiment.

See:
A.De Ninno, A. Frattolillo, A. Rizzo,
E. del Giudice, G. Preparata
"Experimental evidence of 4He production
in a cold fusion experiment"
RT/2002/41/FUShttp://lenr-canr.org/acrobat/DeNinnoAexperiment.pdf

Not really.
Acceleration of ions in electric fields generated by fractures can
provide the necessary energy. IIRC there were neutron emissions detected
from (titanium?) cylinders of pressurised D2 when thermally cycled.

--
Dirk

I read this paper and I can't help thinking that the electrode
"exploded" because oxygen adsorbed into the palladium cathode reacted
with hydrogen, in the presence of an electric current.However Helium 4
was produced, or there is some major flaw in how the amount of helium
4 was measured in this experiment, but I couldn't tell what it was.