Robin--
I am reluctant to think you are joking about dispropotionation reactions for
hydrinos and catalysis of hydrinos and regular hydrogen by water, but
that crossed my mind. ( :
Bob
-Original Message-
From: mix...@bigpond.com
Sent: Saturday, July 04, 2015 8:38 PM
To: vorte
In reply to mix...@bigpond.com's message of Sun, 05 Jul 2015 13:38:34 +1000:
Hi,
BTW, I would guess that Hydrinos in this size category may not last long, as
they would probably undergo nuclear reactions fairly readily (unless well
separated from other matter), thus the spectral line seen is prob
In reply to Jones Beene's message of Sat, 4 Jul 2015 19:47:40 -0700:
Hi Jones,
[snip]
Ok, I found the reason. It lies in the disproportionation reactions.
If you start with a mixture of p = 16 & p = 4, you get:- (16^2)/2 + 4 = 132.
(Formula derivation available on request).
Note that water mole
-Original Message-
From: mix...@bigpond.com
> Guessing that the observed value might match a different transition, I
created a little table for p = 120-136 ...As you can see, p=132->133 is a
good match
Interesting. Nothing obvious pops up at first glance - as to why this
132/133 lev
In reply to Jones Beene's message of Sat, 4 Jul 2015 16:54:51 -0700:
Hi Jones,
Actually it should be 13.598, rather than 13.6, and for a transition from 136 to
137 a catalyst with m=1 is required, which absorbs 27.2 eV first, so the actual
amount remaining to be emitted as a photon would be 3658
Robin,
Yes this Rydberg calculation is close, but probably not close enough. Red
shift could change that assessment and make it exact.
As you know, Mills addressed this issue years ago (that dark matter is
composed of hydrinos) and he concocted a formula that unfortunately provides
a value which
On Sat, Jul 4, 2015 at 6:19 PM, Bob Higgins
wrote:
> The Li and Al are going to be present in equal amounts in the fuel, but
> only the Al will show in the XRD. XRD has a pretty small spot and you can
> be pretty sure that the measurement diameter will not include much that is
> not the Ni parti
In reply to Jones Beene's message of Sat, 4 Jul 2015 15:46:57 -0700:
Hi,
[snip]
For Mills, the difference between any two adjacent states is (2p-1)*13.6 eV
where p is the smaller of the two numbers.
Thus the difference between 136 & 137 would be:
((2*136)-1)*13.6 eV = 3686 eV.
Regards,
Robin
152 went out on the first day of promotion
Sent from my iPad
Bob,
The 510 keV of Maly & Vavra is almost certainly incorrect, but there are a
number of values in the range of several hundred keV which represent the total
energy which can be released in 136 steps. Robin has mentioned his value around
255207+eV but that is almost certainly wrong if FQHE
In reply to Roarty, Francis X's message of Sat, 4 Jul 2015 11:57:36 +:
Hi Fran,
[snip]
>Robin said [snip] This is perhaps because it's the electron the shrinks, while
>the assumption is
>made that the proton is constant.
>This would result in a p value for the maximum energy release in my mod
The Li and Al are going to be present in equal amounts in the fuel, but
only the Al will show in the XRD. XRD has a pretty small spot and you can
be pretty sure that the measurement diameter will not include much that is
not the Ni particle. OTOH, SIMS is a much bigger spot size and it would be
e
The Lugano report said that the nickel particle varient was the same
particle kind and the fuel was very find grained with a gray color.
"All of the Ni becomes quickly surface coated with liquid Li-Al-H. "
Not so. The nickel particles are covered by pure lithium from the fuel (see
below). No alu
Note that there are many optimizations of carbonyl processing designed to
produce, in particular, long strand connected particles with high surface
area optimized for nickel metal hydride battery performance. I suspect
that Rossi used a standard variant of this process that is available COTS.
It i
However, Maly & Vavra, and Naudts predict the lowest DDL state as giving up
510 keV to be reached, not 3.56 keV. That is 2 orders of magnitude lower
energy for their DDL solution than what you are describing. Where has all
the energy gone in this calculation?
On Fri, Jul 3, 2015 at 5:52 PM, Jone
Please see:
;
http://egooutpeters.blogspot.ro/2015/07/my-own-declaration-of-independence-in.html
Peter
--
Dr. Peter Gluck
Cluj, Romania
http://egooutpeters.blogspot.com
http://www.amazon.com/Energy-Fusion-Antigravity-Znidarsic-Science-ebook/dp/B00AD6ARD6/ref=sr_1_1?s=digital-text&ie=UTF8&qid=1436020667&sr=1-1&pebp=1436020670808&perid=00Z27HTS9TA9SD3SYB8E
Robin said [snip] This is perhaps because it's the electron the shrinks, while
the assumption is
made that the proton is constant.
This would result in a p value for the maximum energy release in my model of 119
and a matching energy of 102 keV. [/snip]
Robin... and what value if both electron an
18 matches
Mail list logo
