Robin van Spaandonk wrote:

In reply to  Edmund Storms's message of Sat, 29 Jan 2005 20:51:49 -0700:
Hi,
[snip]

For an explosion to occur, a shock wave must be produced. Simply having energy suddenly produced in a volume would only cause the temperature go up, and ionization to occur with a flash of radiation. The sudden heating would expand the gas to a higher pressure, say from 1 atm to 10 atm. This would not be enough to shatter a heavy glass vessel - blow the lid off, maybe.


Nuclear weapons essentially work on this principle, creating very little in the 
way of extra atoms compared to the size of the shock wave, which is essentially 
a result of thermal ionisation of the surrounding air.
(The actual amount of material present is only a few kg, while the shock wave 
can have an extent of many km's).

Nuclear weapons produce so much radiation that all molecules near the device are decomposed into atoms and ions, which occupy a much larger volume. In addition, the energy density is huge.

Furthermore, in the case at hand, the surrounding medium is water rather than 
air, so flash vaporization will also produce a shock wave (which the 
surrounding water will very effectively transmit to the walls of the container).

Good point. The shock wave might originate in the water as you propose.
It really all depends on just how much energy is liberated, and in what time 
frame.
[snip]

My point here was that each event adds its contribution and then is spent. The O++ catalyst is not reused.


This is actually only partly true. The reaction goes like this:

O++ + H -> O+++ + H*

followed by

O+++ + e- -> O++ + UV

where the e- comes from the plasma, or just about anything else in the 
neighbourhood that happens to have electrons attached to it. :)
So the O++ is reconstituted after use. The only problem is to reuse it before 
it captures another electron and becomes O+.


The window of time during which oxygen has the correct charge would seem to be rather short. I guess it is a matter of intuition whether the time is too short for sufficient O++ to be present.

It is not clear that the reaction its self is even capable of producing more O++. Such a replacement is only an assumption needed for your explanation.


When H[n=1/3 (or more)] is formed from H, a total of 108.8 eV is liberated.
Of this, 54.4 eV goes to the catalyst, leaving 54.4 eV either in the form of 
UV, or as kinetic energy of the hydrino. In either case, there is sufficient 
energy present to ionise O+ to O++ (which requires about 35 eV).
The UV from the reaction:

O+++ +e- -> O++ + 54.9 eV

is also sufficient to convert O+ to O++, or there is also the reaction:

O+++ + O+ -> 2 O++


However as previously mentioned, most of the time this energy won't be "spent" in this way. That means either that the UV/hydrino needs to have more initial energy so that even after losing some energy to competing processes, enough remains upon encountering O+ to ionise it to O++, or supplementary O++ needs to be formed from fusion reactions. I should point out that by the time n gets to e.g. n=1/10, a drop of 2 levels, such as would be catalyzed by O++, to n=1/12, results in an energy release of 598 eV, which with luck may even produce multiple O++ ions. Given an initial population of severely shrunken hydrinos, it should therefore be possible to reach a self sustaining (chain) reaction. (For n=1/120 -> n=1/122 this is 6582 eV according to Mills).

What I am trying to make clear here, is that once shrinkage has progressed far 
enough, the reaction can be self-sustaining, even though the production of O++ 
is not very efficient, simply because the inefficiency is out weighed by the 
energy excess from the reaction.

OK, I understand. Presumably the reaction proceeds until all of the accumulated hydrinos are used up.
It's just a matter of using hydrinos that are at such a level that O++ 
production rate exceeds consumption rate.
(I don't know what that level is, but I hope to have shown that such a level 
may well exist).
[snip]

I don't see how you get a chain reaction. A very dilute mixture of H2 and O++ is present, both of which are used up in the process. Even if O++ were replaced, this would not be expected to occur at a significant rate, i.e. in micro seconds. After all, the original concentration of O++ was accumulated only after minutes of previous electrolysis.


There was no original concentration of O++. What was accumulating over time is 
hydrinos of ever high levels of shrinkage. Once the average shrinkage level 
reaches a certain point, an explosion becomes possible (in water). It then only 
requires a trigger to set it off.
IOW the most important point in the Mizuno experiment is that fact that the 
cell had been in use for about 5 years. This gave plenty of time to cake the 
inside wall (and/or electrode(s)) with high level hydrinos.
It also means that others using the same container (or electrode(s)) for 
extended periods should also be prepared for explosions at some point.

I don't understand how the hydrinos can accumulate in the glass. Even if they were in the glass, why and how would they suddenly come out into the solution? People normally clean their equipment between runs so that material attached to the glass would not be expected. Your model needs a significant source of hydrinos that have accumulated over a period of time, which can quickly enter the water at a particular time and react. How does this occur and why the sudden release?
In a high temperature plasma containing primarily O and H, mixed with high 
energy hydrinos/UV, O++ formation would no longer be a rare occurrence.

We are not looking at a slowly accumulated supply of catalyst here, but rather 
at a situation where a more than adequate supply is created, on the fly, in 
situ.
As the reaction proceeds, the supply actually increases (because the average 
hydrino shrinkage level increases, and hence also the average energy released 
per shrinkage reaction).
[snip]

I don't understand what kind of fusion reaction you imagine using H2. In


e.g. H* + O16 -> F17 + 600 keV (as energetic electron or gamma ray)
     H* + K39 -> Ca40 + 8 MeV (as energetic electron or gamma ray)

These use severely shrunken H* not H2. The H2 is used only to produce H, which 
then further reacts with O++ to create new hydrinos and release energy.

I understand.

any case, such a reaction would release nuclear energies, which would be expected to produce visible particle and X-ray emission, unlike the cold fusion process in a solid. These are apparently not seen, or felt. (Here the "dead graduate student" effect comes in again.)


It may take a long time to produce hydrinos that are sufficiently shrunken to achieve 
fusion (i.e. "5 years").
There may have been such ionising radiation produced in the Mizuno explosion. 
Who is going to notice it when their experiment is blowing up in their face?
(In another post, someone suggested measuring the remains of the cell for 
lingering radioactivity. This was a good suggestion, particularly the remains 
of the electrode where the explosion appeared to originate.)
[snip]


Free electrons are generated by formation of ions. These ions quickly recapture their electrons so that only initially are these extra particles part of the shock wave. I don't think this would be a serious source of expansion. Heating is another matter, but not very effective.

[snip] See comments above re. nuclear weapons.
Yes, but the issue of magnitude is the essential difference. A nuclear weapon affects every molecule near it. The reactions you describe are on an local atomic scale.

Regards,
Ed

Regards,


Robin van Spaandonk

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