HDTenEightyP/NetNews huggingface dataset:
From: logajan@anubis.network.com (John Logajan)
Subject: **HOT** Yellow crud (corrected calculations)
Reply-To: logajan@anubis.network.com (John Logajan)
Organization: Sci.physics.fusion/Mail Gateway
Date: Wed, 11 Nov 1992 06:08:40 GMT

***** Reposted with corrected last four paragraphs!!  Sorry ******

If we assume (guess #1) that the "yellow crud" is PtO2.4H2O, we find from
the CRC HCP that PtO2 takes 40.1 kcal/mol of Gibbs energy (heat of formation)
to form.

I couldn't find the Gibbs energy listed for PtO2.4H2O, however, but I scanned
other chemicals with .4H2O attached and found an average (guess #2) difference
in Gibbs energy between with/without .4H20 to be -290 kcal/mol.

The Gibbs energy of H2O formation is -56.7 kcal/mol.  The H2O already exists
so we want to factor out that energy from the published data for .4H2O.  Since
we have four water molecules, that is 4 * -56.7 = 227 kcal/mol.  Subtracting 
the energy of H2O formation (times 4) from the "guessed" .4H2O energy of 
formation gives -290 - (-227) = -63 kcal/mol.

So, forming PtO2 takes 40 kcal/mol of energy, but we get back -63 kcal/mol
when the .4H2O combines with it, for a net energy change of -23 kcal/mol.

That is to say, for each mole of PtO2.4H2O formed, it appears that 23 kcal
of heat energy appears in the system (is liberated by the "yellow crud.")

This is on the order of 1eV per molecule of "crud" or about 1/15eV per
constituent atom.

- John Logajan MS010, Network Systems; 7600 Boone Ave; Brooklyn Park, MN 55428
- logajan@network.com, 612-424-4888, Fax 612-424-2853
From: edgar@spectrx.saigon.com (Edgar W. Swank)
Newsgroups: sci.physics.fusion
Subject: Re:Setting up home system for internet
Date: Thu, 12 Nov 92 04:28:31 PST
Organization: SPECTROX SYSTEMS (408)252-1005 Silicon Valley, Ca

On Nov 9 Evan Simpson commented on my recommendations to Tom Droege:
 
    Note that these days you can buy a full-featured 14.4K fax/modem
    for as little as $200 (Infotel, for example, or the Gateway
    offer).  These can pay for themselves quickly if you use a pay
    dialup service.
 
I'm not familiar with either of the specific modems cited. Typical
prices for "generic" 9600 baud internal modems are $225-250. However,
I did find one ad, dated last September,
 
9600 Baud internal modem  $175
2400 "       "      "      $36
 
Micromatics
3442 De La Cruz
Santa Clara, CA 95054
Phone: 408-970-8566
Fax:   408-970-0662
 
Be aware that many "fax/modem's" floating around support a high baud
rate for FAX but a lower rate (typically 2400) for the modem function.
 
Many pay dialup services charge a higher rate for a 9600 baud dialin
port than for a 2400. (but perhaps not four times higher).  The
systems like Comp-U-Serve that charge per minute of connect time are
the most expensive.  You should be able to find a system that will
give you internet/usenet access for either free or a fixed monthly
charge.

--
edgar@spectrx.saigon.com (Edgar W. Swank)
SPECTROX SYSTEMS +1.408.252.1005  Silicon Valley, Ca

Newsgroups: sci.physics.fusion
From: ames!FNALD.FNAL.GOV!DROEGE
Subject: Just Need One More Thing
Reply-To: ames!FNALD.FNAL.GOV!DROEGE
Organization: Sci.physics.fusion/Mail Gateway
Date: Thu, 12 Nov 1992 20:44:33 GMT

Thanks to Steve Jones for postin Buehler's measurements.  Now we just need
one more thing.  The voltage and current to the Nickel cell.  

Tom Droege

P.S.  I would not be impressed if they had a hundred open cells lined up in
a row.  Too many things are uncontrolled.
From: BRITZ@kemi.aau.dk (Dieter Britz)
Newsgroups: sci.physics.fusion
Subject: RE:      Re: Fusion Digest 559
Date: 5 Nov 92 15:36:18 GMT
Article-I.D.: vms2.CD8FD715367F265768
Reply-To: Dieter Britz <BRITZ@kemi.aau.dk>
Organization: Sci.physics.fusion/Mail Gateway


Originally-From: Moishe Garfinkle <GARFINKM@DUVM.OCS.DREXEL.EDU>:

>ION IMPLANTATION AS A DEFINITIVE MEANS OF INVESTIGATING ANY POSSIBILITY OF
>INTRACRYSTALLINE NUCLEAR FUSION;  Moishe Garfinkle;  Drexel University
>FUSION TECHNOLOGY v22;(Aug 1992)p 160

Moishe, you seem concerned that this gets overlooked. Be assured that it is
not; your paper was included in the bibliography some weeks ago, and was also
presented on this list among some other updates. You have not been ignored. I
must also say, however, that you were not the first with this idea. There are
something like 10-15 papers in the bibliography, describing ion implantation,
including USA, French, Russian, Danish and perhaps other workers. One team
reckoned they achieved a local d/Pd loading as high as 9.  None of this helped
anybody, though, no hitherto unknown nuclear processes were unveiled, only
well known old self-targeting effects - most people neglect to turn off the
ion beam before measuring neutrons or whatever they measure.
----------------------------------------------------------------------------
Dieter Britz alias britz@kemi.aau.dk
Kemisk Institut, Aarhus Universitet, 8000 Aarhus C, Denmark.
----------------------------------------------------------------------------
From: rauchfuss@hpfcso.FC.HP.COM (Brian Rauchfuss)
Date: Tue, 10 Nov 1992 17:06:46 GMT
Subject: Re: LARSONIAN Physics, NO fusion in Stars
Organization: Hewlett-Packard, Fort Collins, CO, USA
Newsgroups: sci.physics.fusion

In sci.physics.fusion, pmk@prometheus.UUCP (Paul M. Koloc) writes:

> 
> Furthermore, as the star shrinks to smaller and smaller diameters
> in its later life, it leaves behind it's inverse growth rings which
> become plantets and the outer ones pick up low Z gas boil off ( i. e.
> Jupiter).  Now if the gas planet gets large enough or perhaps gets
> an ignition assist by a monster ball lightning that forms in its 
> atmosphere it will become a star and -- -   Voila!!  you have a binary
> system and WITHOUT burning IRON.  

If a star is hot enough to allow fusion, would it not just start fusing without
ignition?  If it is not hot enough, wouldn't it be impossible to ignite, no
matter how much ball lightning was around?

BDR
From: ames!FNALD.FNAL.GOV!DROEGE
Subject: Lead Loss
Reply-To: ames!FNALD.FNAL.GOV!DROEGE
Organization: Sci.physics.fusion/Mail Gateway
Date: Thu, 12 Nov 1992 20:43:40 GMT

Dieter Britz says:

"I have never had a lead just lying around that was so thin that it would
dissipate significant heat with a few amps going through it."

I have Dieter.  In the first calorimeter there were several 5" long leads 
made from #20 wire. (#20 copper wire is 0.010 ohm per foot)  At two amps, 
a 5" lead dissipates 16 mw.  This was a big problem in the first calorimeter.
I never got it right.  The potential leads were not at exactly the right place
so I had to make corrections for the heat in the leads.  Actually there was
a larger problem at first.  I had several connectors on the way into the 
calorimeter.  Connector manufacturers design for of order 50 mv drop at rated 
current.  So that is another 50 mw per amp.  Eventually I slodered everything
together.  But there were still problems with hot spots due to high resistance
regions of the lead path.  

It is not easy.  

If the measurements that have come to me on the Nagoya demonstration are 
correct, then they would have seen that silver lead glowing red had they 
turned out the lights.  From a private communication, the wire was very fine
indeed, of order #36.  It would be ohms per foot.  There could be a lot of 
heat lost where the clip lead attaches.  Easily hundreds of milliwatts into 
the air and not into the cell.  So such a demonstration requires **exactly
alike** cells.  Even then it is so hard to do this that I would never take
such a demonstration seriously.  

The interesting question is why were they using a silver lead?  No significant
difference as a conductor - a few percent.  But perhaps they did not want to 
get copper ions into their cell.  Perhaps the silver was the cathode/anode
material?

Tom Droege

Also, perhaps it was not silver at all, but "silver colored".  Then it was 
likely platinum.  
Newsgroups: sci.physics.fusion
From: bijal@fusion.Berkeley.EDU (Bijal C. Modi)
Newsgroups: sci.physics.fusion
Subject: Question on Langmuir Probe
Date: 12 Nov 1992 08:21:48 GMT
Organization: University of California, Berkeley
Distribution: world
NNTP-Posting-Host: fusion.berkeley.edu


Hi,

   Can somebody provide me with the I-V (current-voltage) relationship
for a Langmuir Probe? The one I got by simple derivation is

                    1 - J
      exp(eV/T) =   _____     , V = voltage, J = current
                  
                    1 + J

     My suspect is that instead of exp(eV/T) it is exp(-eV/T).


thanks,

-Bijal.
 bijal@garnet.berkeley.edu
Newsgroups: sci.physics.fusion
From: Dieter Britz <BRITZ@kemi.aau.dk>
Subject: D/Pd Loading
Reply-To: Dieter Britz <BRITZ@kemi.aau.dk>
Organization: Sci.physics.fusion/Mail Gateway
Date: Wed, 11 Nov 1992 15:23:00 GMT


Tom Droege and Todd Green have discussed the problem of measuring the loading.
I still like the excess oxygen method Tom used, even with the interference by
the yellow crud. You could largely account for the effect of that. For one
thing, you can weigh the Pt anode before and after, which gives you the Pt
content of the crud. At the end, carefully collect as much crud as you can,
and have it analysed quantitatively for Pt, O and D (or H; Bockris tells us a
lot of H gets in there! {:] ) and you can then estimate the total amount formed
from the actual Pt lost. Then calculate how much oxygen was swallowed by that
and correct the excess.
 There is a problem here. If the crud were to eat oxygen, Tom actually
underestimated the excess oxygen - i.e. there would have been more, if not for
the crud. The way I see this happening (and I do wonder at so much Pt
dissolving) is the corrosion reaction, in a strongly alkaline solution:

Pt + 4OD-  ---> Pt(OD)4 + 4e-                  (1)

(Never mind the extra D2O's attached to the stuff). This reaction competes
with the predominant one taking place at the Pt,

4OD- ---> O2  + 2D2O + 4e-                     (2)

so while D2 is being produced at the Pd cathode, reaction (1) robs us of a
bit of O2 supposed to be made at the Pt anode. I hate to say it - because
frankly I don't believe in loadings far above 0.8 - but it looks to me as if
you might have underestimated the loadings. Please correct me, somebody, if
I have this wrong (Todd?).

I believe you start with filling the cell headspace with D2 gas - and all D2
formed should either  burn at the catalyst or go into the Pd. If the crud
reaction (1) takes away oxygen, though, could you be getting D2 coming out,
along with the excess O2? At first, while you are charging the Pd, there
would be excess oxygen; later, when the Pd is fully charged, no more gas
should come out, all being recombined, but crud formation would now produce
a small excess of D2 gas, with no O2 there to burn it.
----------------------------------------------------------------------------
Dieter Britz alias britz@kemi.aau.dk
Kemisk Institut, Aarhus Universitet, 8000 Aarhus C, Denmark.
----------------------------------------------------------------------------
From: ames!FNALD.FNAL.GOV!DROEGE
Subject: Reply to Jim Batka
Reply-To: ames!FNALD.FNAL.GOV!DROEGE
Organization: Sci.physics.fusion/Mail Gateway
Date: Fri, 13 Nov 1992 00:35:03 GMT

Jim Batka wonders why we did not respond to his initial request for cel
temperature studies.  I saw it Jim, and questions like this are appreciated,
but I just do not get around to everything. 

I can't wait to study things like this.  We have been collecting literature
and prices on mineature IR video cameras.  But first I have to think that I
see heat.  And I have to be able to repeat the experiment.  If we ever get 
to that state you can bet that everything but the kitchen sink will be 
attached to the cell!!

Tom Droege
Newsgroups: sci.physics.fusion
From: ames!FNALD.FNAL.GOV!DROEGE
Subject: Gas Loading
Reply-To: ames!FNALD.FNAL.GOV!DROEGE
Organization: Sci.physics.fusion/Mail Gateway
Date: Tue, 10 Nov 1992 22:10:50 GMT

Dick Blue says that he does not see a good explanation as to how high gas
loadings are measured.  I don't know how to do it either.  

Tod Green gives a good discussion and picks the four point resistance method.
I too used this for a while.  I now measure the excess oxygen.  As far as I
can see neither method is very good.  But they are both better than anything
else that I can figure out.  

I watch the skyscraper and count the number of people coming in and out the
front door to estimate how many people are inside.  Tod monitors the elevator
traffic.  My method is bad because there is a subway under the building, and
also a back door.  I think Tod's method fails because we really don't know 
the relation between elevator traffic and people in the building when the 
building is occupied by more people than the fire department allows.  

But this does not stop us from learning some things.  One can still make 
comparative measurements.  With plain rods I get to 0.7 to 0.8 by the 
excess oxygen method.  With a polished cathode, I have achieved 0.94 of the 
first load three times in a row.   Later in the cycle I appear to get to 
much higher loading, but by then I no longer have a good count on the 
skyscraper occupancy.  

Tom Droege
Newsgroups: sci.physics.fusion
From: barry@arnold.math.ucla.edu (Barry Merriman)
Subject: Re: LARSONIAN Physics, NO fusion in Stars
Organization: UCLA, Mathematics Department
Date: Thu, 12 Nov 92 13:35:13 GMT

In article <7600011@hpfcso.FC.HP.COM> rauchfuss@hpfcso.FC.HP.COM (Brian  
Rauchfuss) writes:
> If a star is hot enough to allow fusion, would it not just start fusing  
without
> ignition?  If it is not hot enough, wouldn't it be impossible to ignite, no
> matter how much ball lightning was around?
> 

Have you ever lit a match?


--
Barry Merriman
UCLA Dept. of Math
UCLA Inst. for Fusion and Plasma Research
barry@math.ucla.edu (Internet; NeXTMail is welcome)
Newsgroups: sci.physics.fusion
Subject: Re: the Notoya Demo
From: jonesse@physc1.byu.edu
Date: 11 Nov 92 16:28:57 -0700
Distribution: world
Organization: Brigham Young University

The following measurements were recorded by David Buehler (BYU graduate
student) on the Notoya cells.  I post these to meet the request by D. Britz,
without comment.
Cell with Ni cathode AND control cell with resistor, initially:
3.4 volts,  0.6 amps.
Buehler moved alligator clip from power supply about 5 cm closer to the cell
on very thin lead entering the control cell, result:  
3.0 volts,  0.6 amps (constant current supply).
Adjusting both clips to points on two wires close to entry into cell,
presumably by Dr. Notoya after Buehler pointed out his concern (control cell):
2.57 volts, 0.72 amps (current evidently adjusted also).
(Don't know why current was adjusted on control cell.)

This is all the data I collected from D. Buehler and recorded in my logbook.
It may suffice for a mock-up of the Notoya demonstration if anyone is 
interested, at least to test the effects of the higher resistance of the
lead wires entering the control cell.  One more item, the wires entering the
Ni cell were nearly the thickness of a dime (U.S.), while the wires entering
the control cell were much thinner (one was very fine).
I mean no offense in offering these data.  David Buehler was correct, I think,
to examine the Hokkaido University demonstration very carefully, and to report
his findings to Dr. Notoya at the conference in a straightforward manner.
Best Regards,
Steven Jones

Newsgroups: sci.physics.fusion
From: Bob.Clarke%bbs@actcnews.res.utc.com (Bob Clarke)
Date: Mon, 09 Nov 92 20:18:32 EDT
Organization: United Technologies Research Center
Subject: Who is Funding Pons?
Distribution: world
X-Posting-Software: UnixBBS 1.02a S/N 190921

I understand Pons is working in France with a staff of 32 scientists and
other staff.  Who is funding this effort?
--
Bob Clarke, user of the UnixBBS System @ actcnews.res.utc.com
E-Mail: Bob.Clarke%bbs@actcnews.res.utc.com
========================================================================
   United Technologies Corporation
   Opinions expressed do not necessarily represent UTC.
From: vnoninski@fscvax.fsc.mass.edu
Subject: Re: S.Jones' posting of Notoya's data
Reply-To: vnoninski@fscvax.fsc.mass.edu
Organization: Sci.physics.fusion/Mail Gateway
Date: Fri, 13 Nov 1992 05:51:02 GMT

Dear Colleagues,

S.Jones should be advised that, if we are to believe the figures he himself 
posted today, the *heating* power of Notoya's electrolysis cell 
(3.4V-1.48V)x0.6A=1.15 W remains lower than the input power in the control 
cell, 3.0Vx0.6A=1.8W, even after the student "moved clips from power supply 
about 5cm closer to the cell on very thin lead entering the control cell ...".

Thus, even after the "correction" the temperature of the electrolysis cell 
should be expected to be lower than the temperature of the control cell. This 
is contrary to what was reported to have been observed by many in Nagoya !

Therefore, the easy solution S.Jones offers to the Notoya puzzle is invalid.

Truly yours,

Vesselin Noninski                                November 12, 1992
Newsgroups: sci.physics.fusion
From: Jed Rothwell <72240.1256@compuserve.com>
Subject: The Delights of Nature
Reply-To: Jed Rothwell <72240.1256@compuserve.com>
Organization: Sci.physics.fusion/Mail Gateway
Date: Fri, 6 Nov 1992 02:33:22 GMT

To: >INTERNET:fusion@zorch.SF-Bay.ORG

I thank Tarl Neustaedter for posting the latest hilarious piece from Nature
magazine. Their reporter, David Swinbanks, is apparently based in Tokyo. I
surmise two things about him:

*    He does not speak Japanese, since he could not get a simple abstract
     out of the PR department of NTT two weeks after Business Week and the
     Wall Street Journal evidently got it.

*    He must be on a very short allowance, since he could not afford the
     $160 round trip train ticket to Nagoya, where he would have found the
     abstract on the page 92 of the program, along with 3 references to
     Yamaguchi's earlier work.

Mr. Neustaedter aptly describes the Nature article as "a word from the
establishment about that Nagoya conference," but I quibble with the definite
article. "An establishment" might be a better choice of words. There is more
than one "establishment" in this world, after all. I myself think of the
Japanese Telephone Company, the Ministry of International Trade and
Industry, the New Hydrogen Energy Panel, and the five sponsors of the Nagoya
conference as "The Establishment." Compared to these organizations, Nature
magazine is a flea. Nature's challenge of the Japanese establishment's
scientific prowess is comical and absurd fun. It reminds me of a recent
foray by their editor, Maddox, in which he gave advice to Japan about how to
improve their education system, science and technology! You have to admire
the chutzpa of this panhandler who is so presumptuous he gives investment
tips to a millionaire.

- Jed

Newsgroups: sci.physics.fusion
From: chuck@coplex.com (Chuck Sites)
Subject: Re: New Scientist: Lukewarm reception for Japanese cold fusion
Organization: Copper Electronics, Inc.
Distribution: sci
Date: Mon, 9 Nov 1992 01:44:07 GMT

webb+@CS.CMU.EDU (Jon Webb) writes:

>Yamaguchi claims the palladium atoms fuse as they migrate towards the
>palladium oxide.  One critic at the conference claimed the He-4
>probably came from the glass in the apparatus.

>NTT has repeated the experiment five times.

>David Williams, head of the department of chemistry at University
>College London, suggests the rise in temperature came from oxidation
>of the deuterium, and the He-4 came from the glass.

Myabe. Myabe not. I tend to believe Yamaguchi et al. Their technique
drives the point home. If CF occurs in solid state system, what is 
the best way exagerate the effect to make it measurable. To me, its
a further demonstration of the hydrogen band model during an explosive
release of H through the Pd lattice.  Back in the days when I was 
proposing "Heavy Heat", one of the points that became clear was,
that if H moves through the Pd lattic rapidily, almost explosivly,
a significant heating effect would occur.  Dieter was better than 
I at describing the effect, but it was basically the heat of rapid
diffusion of H from a charged H absorbing metal. My argument
was that such H movements would cause a band effect that could lead
to a secondary EM conduction path which could be hidden in ordinary
electrical measurements is still valid.  But also the band model lends
itself to an increased fusion rate in experimental circunnstances like
Yamaguchi's.  Do not rule his work out because of presumptions 
of CNF.  His work is very sofisticated indeed and deserves a little 
head scratching.       
      
>-- J
Have fun,
------------------------------------------------------------------------------
. . . \Chuck Sites uunet!coplex!chuck Copper Electronics |Cold Fusion  / . . .
o o o o \chuck@coplex.com Ph: 502-454-7218 Wrk: 968-8495 |as real as / o o o o
O O O O O \   Klein bottle for sale.  Inquire within.    |it gets. / O O O O O
------------------------------------------------------------------------------
From: webb+@CS.CMU.EDU (Jon Webb)
Newsgroups: sci.physics.fusion
Subject: Re: New Scientist: Lukewarm reception for Japanese cold fusion
Date: 9 Nov 92 03:13:24 GMT
Article-I.D.: cs.BxFIAC.KLA.1
Distribution: sci
Organization: School of Computer Science, Carnegie Mellon University
Originator: webb@DUCK.WARP.CS.CMU.EDU

In article <7989@transfer.stratus.com> tarl@sw.stratus.com (Tarl Neustaedter) writes:

   In article <BxB0ou.1uF.1@cs.cmu.edu>, webb+@CS.CMU.EDU (Jon Webb) writes:
   > [...palladium plate...] Yamaguchi claims it absorbs 760 times its own 
   > weight of deuterium.

   I presume that's a typo of some kind. That would be 9 kg/cc, which could
   only be some variant of degenerate matter.

Not on my part.  That is a direct quote from New Scientist.  It does
sound unlikely, though.  But your calculation of the density is
probably wrong; it's well known that palladium expands as it absorbs
deuterium (in electrolisys, at least.) -- J
From: tiq@fennel.cc.uwa.oz.au (Todd Green)
Newsgroups: sci.physics.fusion
Subject: Re: Missing Oxygen
Date: 9 Nov 92 13:33:37 +0800
Organization: University of Western Australia
NNTP-Posting-Host: fennel.cc.uwa.edu.au

In article <921106162911.208010ce@FNALD.FNAL.GOV>, ames!FNALD.FNAL.GOV!DROEGE writes:

> This leaves us with no good way to measure the cathode loading.  I think 
> the resistance is not well known - or even meaningful at high loadings.  
> Expansion of the cathode is hard to do, and depends on the shape of the
> cathode.  I cannot imagine how to weigh the cathode while it is under 
> electrolysis.  I have all along assumed that a cell can not be turned off 
> and the cathode removed and weighed.  It it is, then at least sometimes 
> a lot of gas is lost.  So what to do????

Tom, I reckon you should get back to the resistance technique for measuring
loading values. Having mucked about with most of the alternatives, I decided
that this was the best procedure. A reasonable calibration of resistance
versus loading can be constructed from the literature (see McKubre's paper
for the refs) although I estimate an error of ca. 3% up to 0.90 and maybe as
high as 5% for > 0.90. If there is any interest I can post the calibration
data to this group, or maybe even a postscript file of the actual graphs.

A 4 wire resistance measurements is mandatory, of course, given the low
resistance of the cathodes. Also, the electrolysis current produces a
voltage offset, but this can be corrected for by pulsing the excitation
currents in the forward and reverse directions. In a typical run the initial
resistance might be 2 milliohms but upon charging it will increase to about
4 milliohms and then start to drop as you pass through the resistance maxima
at around PdD0.73. Usually the resistance will stabilise at about 3.8 milliohms 
corresponding to PdD0.85 but one sample that was annealed and acid etched got
to PdD0.91.

Anyway, I think resistance is probably the best and easiest method to use.
There are other problems which I haven't mentioned, but they can all be
solved rather easily.


----
Todd

Newsgroups: sci.physics.fusion