Re: [Vo]:If Mizuno is correct, this design is likely tobetheprecursor to all future devices

[mixent]

In reply to  bobcook39...@hotmail.com's message of Wed, 24 Jul 2019 13:19:02
+:
Hi,
[snip]
>For example, spin energy transitions within a coupled “coherent” system may 
>not entail any radiation at all, if there is a perfect conservation of angular 
>momentum during the LENR event.  Of course radiant heat may be emitted in a 
>follow-up reaction involving the decay of the phonic energy of the coherent 
>system’s lattice.

Can you explain in detail what you have in mind?

Regards,


Robin van Spaandonk

local asymmetry = temporary success

Re: [Vo]:If Mizuno is correct, this design is likely tobetheprecursor to all future devices

[mixent]

In reply to  Jürg Wyttenbach's message of Wed, 24 Jul 2019 14:15:44 +0200:
Hi,
[snip]
>K shells are not usually vacant, so such an electron would still upset 
>things. Regards,
>
>You miss the point! If you increase the nuclear charge by +2 then 
>exactly 2 k-shell electrons are missing!

True, I did miss that point, but your statement raises another. In that case,
you are only supplying a single extra electron from the neutron of the D, so the
other K shell vacancy remains unfilled, and will cause a higher level electron
to drop into the vacancy releasing an x-ray.
Of course, for light elements this will only be a soft x-ray, but for mid-range
or heavy atoms this can be quite energetic.
Regards,


Robin van Spaandonk

local asymmetry = temporary success

[Vo]:Hardness of nickel wrt palladium and also Johnson-Matthey "Type A"

[Jones Beene]

Given there is conflicting information floating around concerning the relative 
hardness of nickel vs palladium - perhaps more attention should be directed to 
this detail.
Apparently all of the direct comparisons agree that nickel is indeed softer 
than palladium unless it has been work hardened (as when it is drawn into 
wire). It is indeed drawn into wire to make mesh, normally so it should be 
harder than Pd... end of story.

Catch-22 when drawn nickel is to be woven to make the wire mesh then it is 
almost always first annealed as it is too hard to weave, otherwise. When 
annealed it is softer. 

Opps. Nickel does not re-harden after a heat treatment and quench so the normal 
mesh should, on paper, be too soft for burnishing with Pd. In short - it should 
NOT be possible to use a palladium rod to coat nickel mesh unless the nickel 
has been work hardened, which it has been in order to make wire - BUT when wire 
is woven into mesh it is most often but not always annealed to make it softer. 
So the bottom line is that nickel wire must hardened and not annealed in order 
to coat it and yet this detail is not mentioned... yet there is more. 

One exception to this hardness issue would be if the rod being used to apply 
the Pd was made from J-M Type A palladium, which is considerably softer than 
pure. I double checked and nowhere could I find the composition of the 
palladium rod. There are several relevant papers and I may have missed it. Does 
anyone know?

BTW - Some of this detail about Type A goes back a decade or more to when BARC 
in India discovered that the alloy used in palladium filters (which is Type A) 
was testing dramatically better at excess heat than pure Pd. Later in France 
IIRC, Type A was used for the hero results. Normally it would be specified by 
anyone following P protocol.

Prior to BARC, it was thought that Silver prevents full deuterium loading, but 
there is scant evidence for that, and anyway - in the new Mizuno technique, 
high loading is to be avoided so it makes sense that the rod would be Type A or 
else the nickel was not annealed before weaving.

Given the cost of palladium these days, I suspect it could be a rod that Mizuno 
has owned for some time and he may not have been fully aware that it was Type A 
alloy.
Hopefully Jed will have the answer to this ...

RE: [Vo]:If Mizuno is correct, this design is likely tobetheprecursor to all future devices

[bobcook39...@hotmail.com]

Robin—



If the electron is slow enough, I would guess it very well may react with other 
atomic electrons and be absorbed producing a charge on the system and a small 
increase in phonic energy.  For all practical purposes it would not be observed 
except for a small charge , accumulated with repeated like LENR reactions.



Bob Cook



Sent from Mail for Windows 10




From: mix...@bigpond.com 
Sent: Tuesday, July 23, 2019 6:46:04 PM
To: vortex-l@eskimo.com 
Subject: Re: [Vo]:If Mizuno is correct, this design is likely tobetheprecursor 
to all future devices

In reply to  Jürg Wyttenbach's message of Sun, 21 Jul 2019 18:39:48 +0200:
Hi,
[snip]
>Bob
>
>One reason why the D* path is working like adding +2p/2e could be that
>the internal electron from the neutron only needs to do a little push to
>get to the k-shell. Thus no need to emit an electron!

K shells are not usually vacant, so such an electron would still upset things.
Regards,


Robin van Spaandonk

local asymmetry = temporary success

RE: [Vo]:If Mizuno is correct, this design is likely tobetheprecursor to all future devices

[bobcook39...@hotmail.com]

Axil—

Gamma radiation is by definition generated by a nuclear transition from one 
energy state to another.   I is not necessarily of Mev intensity.  In contrast 
x-rays are by definition  are produced by atomic electronic transitions from 
one energy state to another.

However a nuclear transition may happen within a coherent system of coupled 
particles with no gammas being produced. Not even low intensity irradiation as 
is produced in NMR events  may not happen in some LENR phenomena.

For example, spin energy transitions within a coupled “coherent” system may not 
entail any radiation at all, if there is a perfect conservation of angular 
momentum during the LENR event.  Of course radiant heat may be emitted in a 
follow-up reaction involving the decay of the phonic energy of the coherent 
system’s lattice.

Bob Cook

From: Axil Axil 
Sent: Tuesday, July 23, 2019 12:39:57 PM
To: vortex-l 
Subject: Re: [Vo]:If Mizuno is correct, this design is likely tobetheprecursor 
to all future devices

If a nuclear reaction (fusion) was responsible for the transmutation, wouldn't 
gamma radiation be produced?

On Tue, Jul 23, 2019 at 4:35 PM mailto:mix...@bigpond.com>> 
wrote:
In reply to  Axil Axil's message of Mon, 22 Jul 2019 00:41:51 -0400:
Hi,
[snip]
>http://www.jmcchina.org/html/2019/1/20190101.htm
>
>Replication of biologic transmutation using a chemical reaction.
>
>The productivity of the transmutation was a function of the ambient
>temperature of the solution. 75C produced the most transmutation. Note that
>there was no report of a heating effect or other energy release that
>accompanied the transmutation.

The actual reported change was in the ppm range, so you should be able to
calculate whether or not any normal nuclear reaction energy release would have
been noticeable.
[snip]
Regards,


Robin van Spaandonk

local asymmetry = temporary success

Re: [Vo]:If Mizuno is correct, this design is likely tobetheprecursor to all future devices

[Jürg Wyttenbach]

K shells are not usually vacant, so such an electron would still upset 
things. Regards,


You miss the point! If you increase the nuclear charge by +2 then 
exactly 2 k-shell electrons are missing!

(If you understand the energy levels ...)

Jürg Wyttenbach

Am 24.07.19 um 04:46 schrieb mix...@bigpond.com:

In reply to  Jürg Wyttenbach's message of Sun, 21 Jul 2019 18:39:48 +0200:
Hi,
[snip]

Bob

One reason why the D* path is working like adding +2p/2e could be that
the internal electron from the neutron only needs to do a little push to
get to the k-shell. Thus no need to emit an electron!

K shells are not usually vacant, so such an electron would still upset things.
Regards,


Robin van Spaandonk

local asymmetry = temporary success





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