Re: [Vo]:Horrace help

2009-12-29 Thread Horace Heffner


On Dec 29, 2009, at 6:42 PM, fznidar...@aol.com wrote:

In a message dated 12/29/2009 7:33:04 PM Eastern Standard Time,  
hheff...@mtaonline.net writes:
Yes, from the very little information provided above I suspect  
there might be.   I suspect you are using c as the speed of light  
instead of the speed of sound in the medium.


Best regards,

Horace Heffner
http://www.mtaonline.net/~hheffner/


No.  I tried to get an estimate of phonon frequency from the  
resonant freq formula.


It is the sq root of K/M times 1/rwo pie


By the above I take it you mean frequency f is given by:

   f = (1/(2*Pi)) (k/M)^0.5

which is the resonant frequency of a mechanical system.



For M I used 2* 1836 *9.1 *10 exp -31 KG



The mass of the deuteron is 3.3444941 x 10^-27 kg.




For K I used 1.45 exp 11 Newtons / meter

What is wrong?

Frank Z


The above k is the spring constant, given by Hooke's law:

   F = -k/x

where F is the displacement force and x is the displacement. It  
assumes a linear relationship, which does not exist with respect to  
Coulomb's law.  But, let's ignore that for now.  Palladium has a room  
temperature lattice constant of about 0.389 nm, or 3.89x10^-10 m.   
Let the deuteron move 1/100 of that distance from the zero force  
location, or let x = 3.89x10^-12 m.  Assume the deuteron is vibrating  
at a displacement +- x from the neutral force point.


From Coulomb's law:

   F1 = -C_k * q^2/(r1)^2 =  -C_k * q^2/(3.89x10^-10 m - 3.89x10^-12  
m  )^2


   F2 = +C_k * q^2/(r1)^2 =  -C_k * q^2/(3.89x10^-10 m + 3.89x10^-12  
m  )^2


where C_k * q^2 = 2.30708 J/m, and:

   F1 = -1.8x10^-9 N

   F2 =  1.49459x10^-9 N

   F = F1 + F2 = -6.099724x10^-11 N

   k = -F/x = (6.099724x10^-11 N)/(3.89x10^-12 m) = 1.568 N/m

and we obtain the frequency by:

   f = (1/(2*Pi)) (k/M)^0.5  = (1/(2*Pi)) ((1.568 N/m)/(3.3444941 x  
10^-27 kg))^0.5


   f = 3.459 x 10^12 Hz

Now that is resolved, there is another problem.  The deuterons  
actually are screened from each other by lattice electrons.  They  
exist in lattice potential wells.  They essentially move at lattice  
speeds due to lattice vibrations, except when tunneling between  
lattice sites.  These lattice vibrations might be considered to  
achieve resonance when lambda = N * 2 * (lattice constant), which for  
Pd is 7.78x10^-10 m.  The resonant frequency is given by:


   f = v / lambda

where v is the speed of sound in the medium, or 3070 m/s, so:

   f = ( 3070 m/s) / (7.78x10^-10 m) = 3.946 x 10^12 Hz

which is not far off from the other frequency I gave, which was  
computed from the deuteron 1 dimensional mechanical resonance.


In all cases the phonon energy E is quantized to:

   E = (N + 1/2)*h*f

where (1/2)*h*f is the zero point energy, and N is an integer.

Best regards,

Horace Heffner
http://www.mtaonline.net/~hheffner/






Re: [Vo]:Horrace help

2009-12-29 Thread fznidarsic
 
In a message dated 12/29/2009 7:33:04 PM Eastern Standard Time,  
hheff...@mtaonline.net writes:

Yes, from the very little information provided above I suspect there  might 
be.   I suspect you are using c as the speed of light instead of  the speed 
of sound in the medium.

Best  regards,


Horace Heffner
_http://www.mtaonline.net/~hheffner/_ (http://www.mtaonline.net/~hheffner/) 







No.  I tried to get an estimate of phonon frequency from the resonant  freq 
formula.
 
It is the sq root of K/M times 1/rwo pie
 
For M I used 2* 1836 *9.1 *10 exp -31 KG
 
For K I used 1.45 exp 11 Newtons / meter
 
What is wrong?
 
Frank Z


Re: [Vo]:JL-naudin replicates current Steorn Orbo (Dec) demo

2009-12-29 Thread Abd ul-Rahman Lomax

At 08:58 PM 12/29/2009, William Beaty wrote:

On Tue, 29 Dec 2009, Abd ul-Rahman Lomax wrote:
But there is the tantalizing middle. They find that they "almost" 
close the loop.


You're giving them the benefit of the doubt.   Count how many times 
you have to do that!  It's very telling.


Their acting very Newman-esque and using a battery?  Rather than 
using a five dollar supercapacitor?  They're either insane, or 
they're scammers.


I've already concluded that they are a variation on the latter. But 
there is a *possibility*, I'm pointing out, that they are "sincere." 
Still unethical, but that they believe they just need to get some 
more money to fix this or that, and that this justifies withholding 
the critical information. The critical information is *why* they 
believe they have overunity, or, in fact, why they believe that they 
have any evidence at all of excess energy. What they showed us, quite 
simply, didn't reveal that.


Come on, they looked at some oscilloscope traces and they looked 
okay? The amount of energy that would need to be dumped to rotation 
would be quite small compared to the heat, as if the toroids were 
resistive loads. But, as I recall, I saw some ringing.


So they think that they are actually over unity, but with losses 
that maybe with better engineering they can fix. All it takes is more money.


If they're insane, then they'll talk themselves into using a battery 
and never actually try a supercap, even in private.  They'll have 
all sorts of important reasons why they cannot ever try a 
supercap.  Oh, and by "insane," I mean the same as "fooling 
themselves."  There is a threshold past which the self-fooing 
becomes a complete break with reality.


I don't think they are simply fooling themselves, I think they got 
led into a situation where they needed to fool others. Do they know 
that the whole thing is bogus? How could they *know* that? They'd 
have to do much more careful work, and they are too busy marketing 
what they have: a concept, not engineering to *actually work*, just 
an idea that there is some anomaly here, and they want to see you the 
anomaly. You can figure out how to use it, not their business, they 
are in the business of selling you the idea and some of the equipment 
you'd use to test it. That way, they make money whether there is 
anything real here or not. Quite a business concept, actually.


I'm even doing something a *little* like it, except that I'm fully 
disclosing everything. I don't have any supersecret idea, I'm trying 
to sell kits to replicate a SPAWAR experiment. In theory, I could 
make money even if SPAWAR is bogus, though it would be more 
difficult. I could sell you the kits to show that it doesn't work. 
(But the problem is, how would I know that my kit wasn't missing some 
critical feature, some parameter that I varied, perhaps without 
realizing it?) I can say this: if I can't get the kits to work, i.e., 
to show radiation evidence, I might still sell them, but with that 
disclosure and all the associated caveats. Maybe somebody else could 
figure out the missing link. Quite simply, I have a few thousand 
dollars in this, and I could get most of it back by selling my stuff 
for other applications. I have no intention of putting myself in a 
position where I'd have to lie or deceive in order to escape with my 
shirt on. I'd rather eke it out on social security, I'd sleep better.)


Were you here when "Doctor" Stiffler was presenting his LED 
overunity device?  One of his odd behaviors was, rather than just 
sitting down and honestly demonstrating his claims, and always 
sticking with straight un-twisted discussions, he claimed to be 
making youtube postings to "mess with the heads" of skeptics.


Steorn made a claim like that about one of their prior announcements. 
It was to lead the Men in Black astray.


  In that case, nobody knew which of his videos were hoaxes 
intended to mislead "skeptics", and which were honest 
experiments.  Steorn mentioned doing something similar.


You noticed.


But this is the real and present tipoff: their development of 
extremely low-friction bearings. That is an abandonment of 
over-unity and indicates a desire to become ever more and more 
sensitive, allowing more spectacular demonstrations where a tiny 
effect is accumulated.


Definitely!  That's the Newman fallacy: pretending that a whirling 
massive flywheel represents a huge energy output.  With low-friction 
bearings, you can spin a fairly large wheel for months using just a 
few 10s of cc of battery volume.  That's how the fake PM machine 
sculpture built by David Jones of Nature journal accomplished its 
feat.  (I replaced those hidden batteries myself more than once over 
the years.)


Yeah. Classic. I've been reading Park's Voodoo Science. He makes, of 
course, some crucial errors, he fails to understand and apply his own 
advice. But he's also right about some stuff. Some of the scams he 
reports on were truly cheeky. And he 

Re: [Vo]:JL-naudin replicates current Steorn Orbo (Dec) demo

2009-12-29 Thread William Beaty

On Tue, 29 Dec 2009, Abd ul-Rahman Lomax wrote:
But there is the tantalizing middle. They find that they "almost" close the 
loop.


You're giving them the benefit of the doubt.   Count how many times you 
have to do that!  It's very telling.


Their acting very Newman-esque and using a battery?  Rather than using a 
five dollar supercapacitor?  They're either insane, or they're scammers.



So they think that they are actually over unity, but with losses that 
maybe with better engineering they can fix. All it takes is more money.


If they're insane, then they'll talk themselves into using a battery and 
never actually try a supercap, even in private.  They'll have all sorts 
of important reasons why they cannot ever try a supercap.  Oh, and by 
"insane," I mean the same as "fooling themselves."  There is a threshold 
past which the self-fooing becomes a complete break with reality.


Were you here when "Doctor" Stiffler was presenting his LED overunity 
device?  One of his odd behaviors was, rather than just sitting down and 
honestly demonstrating his claims, and always sticking with straight 
un-twisted discussions, he claimed to be making youtube postings to "mess 
with the heads" of skeptics.  In that case, nobody knew which of his 
videos were hoaxes intended to mislead "skeptics", and which were honest 
experiments.  Steorn mentioned doing something similar.


But this is the real and present tipoff: their development of extremely 
low-friction bearings. That is an abandonment of over-unity and indicates a 
desire to become ever more and more sensitive, allowing more spectacular 
demonstrations where a tiny effect is accumulated.


Definitely!  That's the Newman fallacy: pretending that a whirling massive 
flywheel represents a huge energy output.  With low-friction bearings, you 
can spin a fairly large wheel for months using just a few 10s of cc of 
battery volume.  That's how the fake PM machine sculpture built by David 
Jones of Nature journal accomplished its feat.  (I replaced those hidden 
batteries myself more than once over the years.)


Though it obviously is. They claim there is no energy going there, but that 
hasn't actually been shown except by a gross and coarse display that would 
completely miss the tiny amount of energy expenditure necessary to make that 
rotor accumulate angular momentum.


Why not just use a supercap and remove the whole battery problem?  Watch 
the capacitor voltage rise slowly as excess energy comes from nowhere? 
Stick a Zener across it to keep it from overvoltage.  There's really no 
sensible excuse for their bizarre setup, unless it's obfuscation.  Their 
setup looks sensible unless one realizes what the lack of a supercap 
implies about their collective mental state.



(( ( (  (   ((O))   )  ) ) )))
William J. BeatySCIENCE HOBBYIST website
billb at amasci com http://amasci.com
EE/programmer/sci-exhibits   amateur science, hobby projects, sci fair
Seattle, WA  206-762-3818unusual phenomena, tesla coils, weird sci



Re: [Vo]:Significant Implications - Kitamura

2009-12-29 Thread Terry Blanton
So, how does this compare to the recombination energy of atomic
hydrogen?  Here's a reference by a dubious source:

http://www.cheniere.org/misc/a_h%20reaction.htm

:-)

Terry

On Tue, Dec 29, 2009 at 4:04 PM, Jones Beene  wrote:
> OK, vorticians. This is could be an important paper and topic, so let me add
> one more point of clarification to Michel Jullian's point about the "heat of
> combustion" of hydrogen, compared to the anomalous "loading heat" of
> Kitamura's claim.
>
> Michel correctly finds that if you only look at one-half of the reaction,
> and ignore the mass of the end product, then what we have is:
>
> (294.6 / 2) / 6.02e23) * kJ = ~1.5 electron volts/amu based on hydrogen
>
> This is the energy released relative to initial hydrogen mass, but that
> might assume that oxygen is unnecessary, if you leave it out.  One should
> take the mass of O2 into consideration for the comparison with reversible
> hydride loading.
>
> ERGO. It would have been clearer back a few posts ago - if I had broken the
> comparison down this way. The steam from hydrogen combustion will have a
> molecular wt of 18 amu per hot molecule. The heat of combustion of the two
> hydrogen atoms is ~3+ eV in total. The resultant energy per amu of the
> steam, therefore, is 3/18 or .16 eV per amu of combustion end product.
>
> When we compare that energy per mass of combustion product - with the
> Kitamura reaction of hydrogen which has been reversibly loaded into a metal
> matrix, and then released, then we find that the amu of the end product is
> still about one since there is/was no permanent bond. The thermal energy
> released, according to Kitamura is ~2 eV, so the eV per amu is about a *ten
> to one ratio,* when the energy of the hydride bond is deducted - compared to
> hydrogen combustion (by mass of all non-renewable reactants).
>
> Next big issue. What is the "real" hydride bond energy for Pd? There is a
> chart here (Fig 3):
>
> http://www.iop.org/EJ/article/1742-6596/79/1/012028/jpconf7_79_012028.pdf?re
> quest-id=e4195775-a6d5-4d5f-83b9-da98912aa8c1
>
> It appears that the bond energy for Pd varies between .9 eV and a negative
> value, depending of a number of variables. The bond is field influenced,
> which could be important. From the chart - an average value appears to be
> less than .5 eV. However, the indication is that it could be much lower.
> Therefore, if Kitamura were correct on the heat energy (which I am beginning
> to doubt), then this kind of iterative recycling of hydrogen would be a
> window of opportunity for gainfulness, since the spread is very large.
>
> This is too simple and robust to be real, no?
>
> This looks like a COP of close to three. For an accurate cross-comparison
> based on all reactants - it is fair to say that we are looking an initial
> gain of almost ten to one over combustion; moreover it is an infinite gain
> if based on the renewability of the hydrogen, that is: if the COP~3 allows
> that to happen, after the conversion losses of heat back into electricity.
>
> Before we can arrive at an accurate final appraisal for the usefulness of
> the process, we must consider the net energy necessary to release the
> hydrogen from the matrix. If that were to be .5 eV as the IOP paper suggests
> (or less with an electric field) - then there is a huge potential for net
> gain from recycling the hydrogen.
>
> "IF" of course, Kitamura got the 2 eV thermal number correct. Doubts remain
> on that issue. The big "if."
>
> Jones
>
>
>
>



Re: [Vo]:Horrace help

2009-12-29 Thread Horace Heffner


On Dec 29, 2009, at 12:08 PM, fznidar...@aol.com wrote:


How do you calculate the phonon frequency in the solid.

I get 10 exp 17 hertz..the number should be about 10 exp 12 hertz.

Is there an easy way.

Frank Z


Yes, from the very little information provided above I suspect there  
might be.   I suspect you are using c as the speed of light instead  
of the speed of sound in the medium.


Best regards,

Horace Heffner
http://www.mtaonline.net/~hheffner/






Re: [Vo]:Personal:Little help with UFO sighting?

2009-12-29 Thread Terry Blanton
Sounds like a Fastwalker.

Terry

On Tue, Dec 29, 2009 at 3:57 PM, Horace Heffner  wrote:
>
> On Dec 29, 2009, at 10:02 AM, Rick Monteverde wrote:
>
> Horace –
>
> My sighting wasn’t just after sunset, it was just after nightfall -  total
> darkness. There was just a vague hint of fading light on the horizon, but
> the sky surrounding the object, which was relatively low in the southwest,
> was already black.
>
> I did find something on the after-sunset atmospheric distortion – they say
> add 6 arc minutes to the apparent semidiameter of the sun. I’ll try to
> muddle through your figures in a little while. I sure appreciate the help,
> thanks.
>
> Yes, that accounts for distortion of the image.  There is also atmospheric
> scatter involved, and that might make an object look orange or red.   It
> makes snowy mountains here take on hot pastel colors - called Alpenglow.
> This time of year the sun sets almost due west there.  If the object was
> ever in a direction approximately due north or south of you, i.e. on a line
> perpendicular to the sunset location, then the altitude h I provided fairly
> closely applies to the object for that time t in the table.   If it was
> mainly east or west then another calculation is needed.  I would say
> anything above 100,000 feet, or 18.9 miles,  was probably not a military
> jet, and certainly not a passenger jet.   That altitude h corresponds to
> about 22 minutes after surface darkness - to whatever degree such darkness
> needs to be defined.  From experience there,  I know it gets dark pretty
> fast in Hawaii after sunset - especially compared to here - where sunsets
> can take a very long time. 8^)  If you observed the object an hour after
> sunset then I'd say it was well past the 22 minutes after darkness mark.   A
> general compass direction thus may be sufficient information for a
> definitive answer.  That far after sunset, an hour, taken even alone, is a
> pretty strong indication it was not an airplane.
> Best regards,
>
> Horace Heffner
> http://www.mtaonline.net/~hheffner/
>
>
>
>



[Vo]:Low-Energy Nuclear Reactions Sourcebook Volume 2 abstracts

2009-12-29 Thread Jed Rothwell
The abstracts for each chapter are available here:

http://pubs.acs.org/isbn/9780841224544

- Jed


Re: [Vo]:JL-naudin replicates current Steorn Orbo (Dec) demo

2009-12-29 Thread Abd ul-Rahman Lomax

At 06:00 AM 12/28/2009, William Beaty wrote:

Rather than focusing on some perhaps-unexpected measurement, just 
close the loop.   Ditch the battery.   Make a perpetual wheel.  Close the loop.

If it's real, then closing the loop should be easy.   If it's an artifact
which misleads FE-enthusiasts, then closing the loop will be impossible.


But there is the tantalizing middle. They find that they "almost" 
close the loop. So they think that they are actually over unity, but 
with losses that maybe with better engineering they can fix. All it 
takes is more money.


But this is the real and present tipoff: their development of 
extremely low-friction bearings. That is an abandonment of over-unity 
and indicates a desire to become ever more and more sensitive, 
allowing more spectacular demonstrations where a tiny effect is accumulated.


But given so much energy being dumped into heat, in the end, it only 
takes a tiny, tiny fraction of that to be coupled into rotor motion 
instead, very difficult to detect, if you have a very low-friction 
rotor which won't lose heat there. So much, though, for actually 
generating power, which will immediately dump much rotor energy into 
heat again.


Calorimetry would show the overall problem, but, of course, doing 
really accurate calorimetry is difficult. Much easier to make a roter 
spin fast and claim that the energy for that is free, that the 
battery is only generating heat, that none of this cycling of the 
magnetic field is accelerating the rotor.


Though it obviously is. They claim there is no energy going there, 
but that hasn't actually been shown except by a gross and coarse 
display that would completely miss the tiny amount of energy 
expenditure necessary to make that rotor accumulate angular momentum. 



[Vo]:Horrace help

2009-12-29 Thread fznidarsic
How do you calculate the phonon frequency in the solid.
 
I get 10 exp 17 hertz..the number should be about 10 exp 12 hertz.
 
Is there an easy way.
 
Frank Z


RE: [Vo]:Significant Implications - Kitamura

2009-12-29 Thread Jones Beene
OK, vorticians. This is could be an important paper and topic, so let me add
one more point of clarification to Michel Jullian's point about the "heat of
combustion" of hydrogen, compared to the anomalous "loading heat" of
Kitamura's claim. 

Michel correctly finds that if you only look at one-half of the reaction,
and ignore the mass of the end product, then what we have is:

(294.6 / 2) / 6.02e23) * kJ = ~1.5 electron volts/amu based on hydrogen 

This is the energy released relative to initial hydrogen mass, but that
might assume that oxygen is unnecessary, if you leave it out.  One should
take the mass of O2 into consideration for the comparison with reversible
hydride loading.

ERGO. It would have been clearer back a few posts ago - if I had broken the
comparison down this way. The steam from hydrogen combustion will have a
molecular wt of 18 amu per hot molecule. The heat of combustion of the two
hydrogen atoms is ~3+ eV in total. The resultant energy per amu of the
steam, therefore, is 3/18 or .16 eV per amu of combustion end product.

When we compare that energy per mass of combustion product - with the
Kitamura reaction of hydrogen which has been reversibly loaded into a metal
matrix, and then released, then we find that the amu of the end product is
still about one since there is/was no permanent bond. The thermal energy
released, according to Kitamura is ~2 eV, so the eV per amu is about a *ten
to one ratio,* when the energy of the hydride bond is deducted - compared to
hydrogen combustion (by mass of all non-renewable reactants). 

Next big issue. What is the "real" hydride bond energy for Pd? There is a
chart here (Fig 3):

http://www.iop.org/EJ/article/1742-6596/79/1/012028/jpconf7_79_012028.pdf?re
quest-id=e4195775-a6d5-4d5f-83b9-da98912aa8c1

It appears that the bond energy for Pd varies between .9 eV and a negative
value, depending of a number of variables. The bond is field influenced,
which could be important. From the chart - an average value appears to be
less than .5 eV. However, the indication is that it could be much lower.
Therefore, if Kitamura were correct on the heat energy (which I am beginning
to doubt), then this kind of iterative recycling of hydrogen would be a
window of opportunity for gainfulness, since the spread is very large.

This is too simple and robust to be real, no?

This looks like a COP of close to three. For an accurate cross-comparison
based on all reactants - it is fair to say that we are looking an initial
gain of almost ten to one over combustion; moreover it is an infinite gain
if based on the renewability of the hydrogen, that is: if the COP~3 allows
that to happen, after the conversion losses of heat back into electricity. 

Before we can arrive at an accurate final appraisal for the usefulness of
the process, we must consider the net energy necessary to release the
hydrogen from the matrix. If that were to be .5 eV as the IOP paper suggests
(or less with an electric field) - then there is a huge potential for net
gain from recycling the hydrogen.

"IF" of course, Kitamura got the 2 eV thermal number correct. Doubts remain
on that issue. The big "if."

Jones





Re: [Vo]:Personal:Little help with UFO sighting?

2009-12-29 Thread Horace Heffner


On Dec 29, 2009, at 10:02 AM, Rick Monteverde wrote:


Horace –

My sighting wasn’t just after sunset, it was just after nightfall  
-  total darkness. There was just a vague hint of fading light on  
the horizon, but the sky surrounding the object, which was  
relatively low in the southwest, was already black.


I did find something on the after-sunset atmospheric distortion –  
they say add 6 arc minutes to the apparent semidiameter of the sun.  
I’ll try to muddle through your figures in a little while. I sure  
appreciate the help, thanks.


Yes, that accounts for distortion of the image.  There is also  
atmospheric scatter involved, and that might make an object look  
orange or red.   It makes snowy mountains here take on hot pastel  
colors - called Alpenglow.


This time of year the sun sets almost due west there.  If the object  
was ever in a direction approximately due north or south of you, i.e.  
on a line perpendicular to the sunset location, then the altitude h I  
provided fairly closely applies to the object for that time t in the  
table.   If it was mainly east or west then another calculation is  
needed.  I would say anything above 100,000 feet, or 18.9 miles,  was  
probably not a military jet, and certainly not a passenger jet.
That altitude h corresponds to about 22 minutes after surface  
darkness - to whatever degree such darkness needs to be defined.   
From experience there,  I know it gets dark pretty fast in Hawaii  
after sunset - especially compared to here - where sunsets can take a  
very long time. 8^)  If you observed the object an hour after sunset  
then I'd say it was well past the 22 minutes after darkness mark.   A  
general compass direction thus may be sufficient information for a  
definitive answer.  That far after sunset, an hour, taken even alone,  
is a pretty strong indication it was not an airplane.


Best regards,

Horace Heffner
http://www.mtaonline.net/~hheffner/






Re: [Vo]:OT: Change Metal Surface to Reflct any Color (Not plating or Paint)

2009-12-29 Thread Harry Veeder
Heat from welding can tint steel too, but I guess with lasers the effect can be 
controlled.

This topic reminds of Chromoskedasic printing, where black&white photographic 
paper is chemically
treated so that it will scatter light and generate abstract/psychedelic colour 
effects.

http://chemigramist.com/chromoskedasic.html

example
http://www.loreal.com/_en/_ww/loreal-art-science/bronzefigure-2005.aspx

Harry



- Original Message 
> From: Jed Rothwell 
> To: vortex-l@eskimo.com
> Sent: Tue, December 29, 2009 9:24:51 AM
> Subject: RE: [Vo]:OT: Change Metal Surface to Reflct any Color (Not  plating 
> or Paint)
> 
> This is great stuff! Amazing.
> 
> Summary: they are changing the color of metals by changing the surface 
> structure.
> 
> - Jed



  __
Get a sneak peak at messages with a handy reading pane with All new Yahoo! 
Mail: http://ca.promos.yahoo.com/newmail/overview2/



[Vo]:Digital Quantum Battery Could Boost Energy Density Tenfold

2009-12-29 Thread Mark Iverson
Digital Quantum Battery Could Boost Energy Density Tenfold
http://www.physorg.com/news180704455.html
 
In their study, Alfred Hubler and Onyeama Osuagwu, both of the University of 
Illinois at
Urbana-Champaign, have investigated energy storage capacity in arrays of nano 
vacuum tubes, which
contain little or no gas. When the tubes' gap size - or the distance between 
electrodes - is about
10 nanometers wide, electric arcing is suppressed, preventing energy loss. 
Further, each tube can be
addressed individually, making the technology digital and offering the 
possibility for data storage
in conjunction with energy storage.
 
17pg paper at UIUC...
https://netfiles.uiuc.edu/a-hubler/www/digitalquantumbatteries.pdf
 
Happy Holidays, 
Mark 
 


RE: [Vo]:Personal:Little help with UFO sighting?

2009-12-29 Thread Rick Monteverde
Sunset at my location in Honolulu that day was 5:59 PM, sighting was at 6:58
PM.  I'll look for a star chart to get the sighting angle for the object.

 

-  R.



RE: [Vo]:Personal:Little help with UFO sighting?

2009-12-29 Thread Rick Monteverde
Horace -

 

My sighting wasn't just after sunset, it was just after nightfall -  total
darkness. There was just a vague hint of fading light on the horizon, but
the sky surrounding the object, which was relatively low in the southwest,
was already black. 

 

I did find something on the after-sunset atmospheric distortion - they say
add 6 arc minutes to the apparent semidiameter of the sun. I'll try to
muddle through your figures in a little while. I sure appreciate the help,
thanks.

 

-  R.

 

From: Horace Heffner [mailto:hheff...@mtaonline.net] 
Sent: Tuesday, December 29, 2009 8:08 AM
To: vortex-l@eskimo.com
Subject: Re: [Vo]:Personal:Little help with UFO sighting?

 

In case there is any doubt, the following is my final answer - unless of
course I find other mistakes!  8^)

 

 

On Dec 29, 2009, at 8:44 AM, Horace Heffner wrote:

 

 

Hi Rick,

 

Coincidentally, I saw something similar yesterday  (Dec 28, 2009) around
noon AKST, (about 11 orbits later) west of Palmer AK, but heading SW.  It
was one small finger width at arms length above the horizon.  It had a
periodic (about 10 second) flash to it, so I assumed it might be a  booster,
but strange it was heading SW, not SE or NE, or just S.  Of course a U-turn
is not a typical satellite maneuver, nor did I see that! 

 

The altitude h to the directly overhead sun midline is given by:

 

   h = r_earth * ( SQRT(1 + sin^2 theta) -1)

 

Given time after sunset t we have:

 

   theta = (t/(8.64x10^4 s))*(2*Pi) radians = (t/(1440 min))*(2*Pi) radians

 

Earth radius, r_earth, at Hawaii is about 3951 mi.  Here are some numbers:

 

t (min) theta (radians)   h (miles)



1 0.00436331944 0.03760073165

5 0.02181659722 0.93976780755

100.04363319444 3.75594358

200.08726638889 14.973936498

300.13089958333 33.506081478

600.26179916667 130.1553394

900.39269875  279.3533269

 

 

Since the above is time after total sunset, you don't have to correct for
the angular width of the sun.  However, even total sunset is not good enough
to black out an object though, due to light diffraction.   Clearly not
enough time, i.e. "shortly after sunset", passed to rule out an airplane. 

 

Best regards,

 

Horace Heffner

http://www.mtaonline.net/~hheffner/

 





 



[Vo]:Personal:Little help with UFO sighting?

2009-12-29 Thread Horace Heffner
For background information, the attached is a graphic showing my  
derivation of the formula used below.  The umbra line, shown as  
horizontal, and the line segment x directly above the viewer is  
roughly the same as the altitude to an object on the umbra line,  
depending on the latitudinal distance from the viewer.


The altitude h to the directly overhead sun midline, with r factored  
out, is given by:


   h = r_earth * ( SQRT(1 + sin^2 theta) -1)

Given time after sunset t we have:

   theta = (t/(8.64x10^4 s))*(2*Pi) radians = (t/(1440 min))*(2*Pi)  
radians


Earth radius, r_earth, at Hawaii is about 3951 mi.  Here are some  
numbers:


t (min) theta (radians) h (miles)

1   0.00436331944   0.03760073165
5   0.02181659722   0.93976780755
10  0.04363319444   3.75594358
20  0.08726638889   14.973936498
30  0.13089958333   33.506081478
60  0.26179916667   130.1553394
90  0.39269875  279.3533269


Since the above is time after total sunset, you don't have to correct  
for the angular width of the sun.  However, even total sunset is not  
good enough to black out an object though, due to light  
diffraction.   Clearly not enough time, i.e. "shortly after sunset",  
passed to rule out an airplane.


<>


Best regards,

Horace Heffner
http://www.mtaonline.net/~hheffner/






RE: [Vo]:Significant Implications - Kitamura

2009-12-29 Thread Jones Beene
-Original Message-
From: Michel Jullian [

> Or rather, as it turns out, exactly right. Physics works, contrary to
your suggestions  :)

It works of course, but not as perfectly as you suggest, in real world
applications.

> Besides, you don't have to take my word, see

http://en.wikipedia.org/wiki/Heat_of_combustion

Hydrogen: 140 kJ/g, which is about 1.5eV per atom.

Yes, but once again your reference is NOT to "burning hydrogen in air". At
the very top of the page you site, it clearly says "The heat of combustion
is the energy released as heat when one mole of a compound undergoes
complete combustion with oxygen" 

Burning H2 in air is not "complete combustion with oxygen" and in fact H2
can be leaned-out sufficiently in air so that it will not burn at all.
Contrary to what you state, parasitic loses cannot be ignored - unless you
are merely trying to prove a pedantic point, which seems to be the case.

> The important result here is that the 2 eV you get by letting an
hydrogen atom bond to the _surface_  of a Pd nanoparticle are
comparable with the chemical energy you get by letting it bond to an
oxygen atom  (starting from molecular gas phase in both cases)

NO! Absolutely not a relevant comparison, nor an accurate value. Otherwise
metal hydrides could not be used for hydrogen storage, and palladium could
not be used as a filter to separate H2 from other gases, both of which
applications are common. 

Imagine having to apply 2 eV of thermal energy to a metal hydride in order
to release the stored hydrogen gas for use in an engine. That is absurd.

Jones



Re: [Vo]:Personal:Little help with UFO sighting?

2009-12-29 Thread Horace Heffner
In case there is any doubt, the following is my final answer - unless  
of course I find other mistakes!  8^)



On Dec 29, 2009, at 8:44 AM, Horace Heffner wrote:



Hi Rick,

Coincidentally, I saw something similar yesterday  (Dec 28, 2009)  
around noon AKST, (about 11 orbits later) west of Palmer AK, but  
heading SW.  It was one small finger width at arms length above the  
horizon.  It had a periodic (about 10 second) flash to it, so I  
assumed it might be a  booster, but strange it was heading SW, not  
SE or NE, or just S.  Of course a U-turn is not a typical satellite  
maneuver, nor did I see that!


The altitude h to the directly overhead sun midline is given by:

   h = r_earth * ( SQRT(1 + sin^2 theta) -1)

Given time after sunset t we have:

   theta = (t/(8.64x10^4 s))*(2*Pi) radians = (t/(1440 min))*(2*Pi)  
radians


Earth radius, r_earth, at Hawaii is about 3951 mi.  Here are some  
numbers:


t (min) theta (radians) h (miles)

1   0.00436331944   0.03760073165
5   0.02181659722   0.93976780755
10  0.04363319444   3.75594358
20  0.08726638889   14.973936498
30  0.13089958333   33.506081478
60  0.26179916667   130.1553394
90  0.39269875  279.3533269


Since the above is time after total sunset, you don't have to  
correct for the angular width of the sun.  However, even total  
sunset is not good enough to black out an object though, due to  
light diffraction.   Clearly not enough time, i.e. "shortly after  
sunset", passed to rule out an airplane.


Best regards,

Horace Heffner
http://www.mtaonline.net/~hheffner/






[Vo]:Personal:Little help with UFO sighting?

2009-12-29 Thread Horace Heffner


On Dec 28, 2009, at 2:09 PM, Rick Monteverde wrote:

Saw an orange fire colored UFO last night just after nightfall. The  
path was that of an object flying in a curved path at high altitude  
(a u-turn, basically), definitely not a satellite, and a bit  
brighter than a good space station sighting. Even through 8x  
binoculars it appeared as a point source. The time of day, the  
sighting angle, and the variable characteristics of the light over  
the two minutes or so the object was visible suggests it was  
reflected light from the setting sun on the lower surface of a  
solid object, but given that the sky was almost completely dark at  
my location at that point, my guess is that its altitude could have  
been above the atmosphere. I’ve seen conventional aircraft  
reflecting sunset’s light after local sunset, and though the  
appearance was similar, in those cases it was much closer to sunset  
and sky was still quite light. Once it gets dark, I think such  
reflections tend to be in satellite territory.


Anyway, since I can find the sight angle because of its passage  
near identifiable stars, the time, and my location on that date, it  
should be possible to calculate the earth’s shadow line from the  
setting sun and see where my sight line crosses it. That would tell  
me if it was just an airplane at very high altitude, or something  
maneuvering up a bit higher than conventional aircraft can reach.  
Anybody have an idea how I would go about that (umbra?) line?


Thanks,

-  Rick


Hi Rick,

Coincidentally, I saw something similar yesterday  (Dec 28, 2009)  
around noon AKST, (about 11 orbits later) west of Palmer AK, but  
heading SW.  It was one small finger width at arms length above the  
horizon.  It had a periodic (about 10 second) flash to it, so I  
assumed it might be a  booster, but strange it was heading SW, not SE  
or NE, or just S.  Of course a U-turn is not a typical satellite  
maneuver, nor did I see that!


The altitude h to the directly overhead sun midline is given by:

   h = r_earth * ( SQRT(1 + sin^2 theta) -1)

Given time after sunset t we have:

   theta = (t/(8.64x10^4 s))*(2*Pi) radians = (t/(1440 min))*(2*Pi)  
radians


Earth radius, r_earth, at Hawaii is about 3951 mi.  Here are some  
numbers:


t (min) theta (radians) h (miles)

1   0.00436331944   0.03760073165
5   0.02181659722   0.93976780755
10  0.04363319444   3.75594358
20  0.08726638889   14.973936498
30  0.13089958333   33.506081478
60  0.26179916667   130.1553394
90  0.39269875  279.3533269


Since the above is time after total sunset, you don't have to correct  
for the angular width of the sun.  However, even total sunset is not  
good enough to black out an object though, due to light  
diffraction.   Clearly not enough time, i.e. "shortly after sunset",  
passed to rule out an airplane.


Best regards,

Horace Heffner
http://www.mtaonline.net/~hheffner/






Re: [Vo]:Significant Implications - Kitamura

2009-12-29 Thread Michel Jullian
2009/12/29 Stephen A. Lawrence :
>
>
> On 12/29/2009 11:19 AM, Jones Beene wrote:
>> Since water can be split into H2 and O2 with 1.23 volts - does it stand to
>> reason that one could get 1.5 eV in return ? That was rhetorical; and of
>> course this one of nature's built-in cases of "systemic overunity" -
>>
>
> Now you're neglecting the splitting cost of H2->2H and O2->2H.

No he isn't, that's comprised in the price (if you use the correct
value of 1.48V that is). What's the energy needed to go from water  to
the gases? 1.48V, times the charge of the transferred electrons (1
electron per hydrogen atom). Of course, you get the same energy when
going the other way.

Michel



[Vo]:Personal:Little help with UFO sighting?

2009-12-29 Thread Horace Heffner


On Dec 28, 2009, at 2:09 PM, Rick Monteverde wrote:

Saw an orange fire colored UFO last night just after nightfall. The  
path was that of an object flying in a curved path at high altitude  
(a u-turn, basically), definitely not a satellite, and a bit  
brighter than a good space station sighting. Even through 8x  
binoculars it appeared as a point source. The time of day, the  
sighting angle, and the variable characteristics of the light over  
the two minutes or so the object was visible suggests it was  
reflected light from the setting sun on the lower surface of a  
solid object, but given that the sky was almost completely dark at  
my location at that point, my guess is that its altitude could have  
been above the atmosphere. I’ve seen conventional aircraft  
reflecting sunset’s light after local sunset, and though the  
appearance was similar, in those cases it was much closer to sunset  
and sky was still quite light. Once it gets dark, I think such  
reflections tend to be in satellite territory.


Anyway, since I can find the sight angle because of its passage  
near identifiable stars, the time, and my location on that date, it  
should be possible to calculate the earth’s shadow line from the  
setting sun and see where my sight line crosses it. That would tell  
me if it was just an airplane at very high altitude, or something  
maneuvering up a bit higher than conventional aircraft can reach.  
Anybody have an idea how I would go about that (umbra?) line?


Thanks,

-  Rick


Hi Rick,

Coincidentally, I saw something similar yesterday  (Dec 28, 2009)  
around noon AKST, (about 11 orbits later) west of Palmer AK, but  
heading SW.  It was one small finger width at arms length above the  
horizon.  It had a periodic (about 10 second) flash to it, so I  
assumed it might be a  booster, but strange it was heading SW, not SE  
or NE, or just S.  Of course a U-turn is not a typical satellite  
maneuver, nor did I see that!


The altitude h to the directly overhead sun midline is given by:

   h = r_earth * ( SQRT(1 + sin^2 theta) -1)

Given time after sunset t we have:

   theta = (t/(8.64x10^4 s))*(2*Pi) radians = (t/(1440 min))*(2*Pi)  
radians


Earth radius, r_earth, at Hawaii is about 3951 mi.  Here are some  
numbers:


t (min) theta (radians) h (miles)

1   0.00436331944   0.03760073165
5   0.02181659722   0.93976780755
10  0.04363319444   3.75594358
20  0.08726638889   14.973936498
30  0.13089958333   33.506081478
60  0.26179916667   130.1553394
90  0.39269875  279.3533269


The above only provides mid-line information.  The angular radius of  
the sun is 31.65 minutes, so the total sunset line follows the above  
by about 15 minutes and 50 seconds.  Even total sunset is not good  
enough to black out an object though, due to light diffraction.
Clearly not enough time passed from sunset to rule out an airplane.


Best regards,

Horace Heffner
http://www.mtaonline.net/~hheffner/






[Vo]:Personal:Little help with UFO sighting?

2009-12-29 Thread Horace Heffner


On Dec 28, 2009, at 2:09 PM, Rick Monteverde wrote:

Saw an orange fire colored UFO last night just after nightfall. The  
path was that of an object flying in a curved path at high altitude  
(a u-turn, basically), definitely not a satellite, and a bit  
brighter than a good space station sighting. Even through 8x  
binoculars it appeared as a point source. The time of day, the  
sighting angle, and the variable characteristics of the light over  
the two minutes or so the object was visible suggests it was  
reflected light from the setting sun on the lower surface of a  
solid object, but given that the sky was almost completely dark at  
my location at that point, my guess is that its altitude could have  
been above the atmosphere. I’ve seen conventional aircraft  
reflecting sunset’s light after local sunset, and though the  
appearance was similar, in those cases it was much closer to sunset  
and sky was still quite light. Once it gets dark, I think such  
reflections tend to be in satellite territory.


Anyway, since I can find the sight angle because of its passage  
near identifiable stars, the time, and my location on that date, it  
should be possible to calculate the earth’s shadow line from the  
setting sun and see where my sight line crosses it. That would tell  
me if it was just an airplane at very high altitude, or something  
maneuvering up a bit higher than conventional aircraft can reach.  
Anybody have an idea how I would go about that (umbra?) line?


Thanks,

-  Rick


Hi Rick,

Coincidentally, I saw something similar yesterday  (Dec 28, 2009)  
around noon AKST, (about 11 orbits later) west of Palmer AK, but  
heading SW.  It was one small finger width at arms length above the  
horizon.  It had a periodic (about 10 second) flash to it, so I  
assumed it might be a  booster, but strange it was heading SW, not SE  
or NE, or just S.  Of course a U-turn is not a typical satellite  
maneuver, nor did I see that!


The altitude h to the directly overhead sun midline is given by:

   h = r_earth * ( SQRT(1 + sin^2 theta) -1)

Given time after sunset t we have:

   theta = (t/(8.64x10^4 s))*(2*Pi) radians = (t/(1440 min))*(2*Pi)  
radians


Earth radius, r_earth, at Hawaii is about 3951 mi.  Here are some  
numbers:


t (min) theta (radians) h (miles)

1   0.00436331944   0.03760073165
5   0.02181659722   0.93976780755
10  0.04363319444   3.75594358
20  0.08726638889   14.973936498
30  0.13089958333   33.506081478
60  0.26179916667   130.1553394
90  0.39269875  279.3533269


The above only provides mid-line information.  The angular radius of  
the sun is 31.65 minutes, so the total sunset line follows the above  
by about 31 minutes and 39 seconds.  Even total sunset is not good  
enough to black out an object though, due to light diffraction.
Clearly not enough time passed from sunset to rule out an airplane.


Best regards,

Horace Heffner
http://www.mtaonline.net/~hheffner/






Re: [Vo]:Significant Implications - Kitamura

2009-12-29 Thread Michel Jullian
2009/12/29 Jones Beene :
> -Original Message-
> From: Michel Jullian
>
>> - but the 2 eV available
>> from loading alone without deuterium (contrast that to about .5 eV if the
>> hydrogen were burned in air) is a huge surprise -
>
> MJ: Jones, where did you get that  .5 eV figure? I did the maths and found
> about 1.5 eV instead, here is the Google calculator result;
>
> ((294.6 / 2) / 6.02e23) * kJ = 1.52719998 electron volts
>
>
> Michel, the half-eV figure is the common 'real world' estimate based on the
> maximum average temperature of the resultant steam - but even so, it appears
> you did not first deduct the dissociation energy of O2 and H2

Their formation enthalpy is zero, by convention

> and then later
> deduct the parasitic losses of NOx, peroxides etc. and the other losses that
> are expected in actual practice, for combustion in air?

Negligible

> IOW there are lies, damn lies, and theoretical calculations ;) when trying
> to go from 'paper numbers' to actual practice. Kitamura's numbers were
> indicated to be actual practice (if they can be trusted) so it is fair to
> contrast those numbers with that which would happen if one were to actually
> burn H2 in air - and .5 eV is a fair estimate

No (see below)

> even if you discount the 80%
> of air which is nearly inert.

why would you not discount them???

> Since water can be split into H2 and O2 with 1.23 volts - does it stand to
> reason that one could get 1.5 eV in return ? That was rhetorical; and of
> course this one of nature's built-in cases of "systemic overunity" -

This was not rhetorical at all actually, I hadn't made the connexion
but yes, the combustion energy per D atom in eV should be, of course,
exactly equal to the thermoneutral electrolysis voltage... and it is,
as a matter of fact: the thermoneutral voltage for electrolysis of D2O
is 1.54V, which confirms my 1.53V calculation. And BTW, it's 1.48V for
H2O, not 1.23V.

> ... except for the damn lie that it simply does not work out that way in
> practice - but it does serve to contrast the large disparity of the "actual
> with the calculated".
>
>> Did I get it wrong?
>
> Well, let's say that you got it partly right and mostly wrong

Or rather, as it turns out, exactly right. Physics works, contrary to
your suggestions  :)
Besides, you don't have to take my word, see
http://en.wikipedia.org/wiki/Heat_of_combustion
Hydrogen: 140 kJ/g, which is about 1.5eV per atom.

The important result here is that the 2 eV you get by letting an
hydrogen atom bond to the _surface_  of a Pd nanoparticle are
comparable with the chemical energy you get by letting it bond to an
oxygen atom  (starting from molecular gas phase in both cases)

Michel



Re: [Vo]:Significant Implications - Kitamura

2009-12-29 Thread Stephen A. Lawrence


On 12/29/2009 11:19 AM, Jones Beene wrote:
> -Original Message-
> From: Michel Jullian 
>
>   
>> - but the 2 eV available
>> from loading alone without deuterium (contrast that to about .5 eV if the
>> hydrogen were burned in air) is a huge surprise -
>> 
> MJ: Jones, where did you get that  .5 eV figure? I did the maths and found
> about 1.5 eV instead, here is the Google calculator result;
>
> ((294.6 / 2) / 6.02e23) * kJ = 1.52719998 electron volts
>
>
> Michel, the half-eV figure is the common 'real world' estimate based on the
> maximum average temperature of the resultant steam

Isn't combustion of hydrogen in air rather different from the situation
we've got here?


>  - but even so, it appears
> you did not first deduct the dissociation energy of O2 and H2 and then later
> deduct the parasitic losses of NOx, peroxides etc. and the other losses that
> are expected in actual practice, for combustion in air?

Parasitic losses, in particular, would not seem to apply in the present
case.


> IOW there are lies, damn lies, and theoretical calculations ;) when trying
> to go from 'paper numbers' to actual practice. Kitamura's numbers were
> indicated to be actual practice (if they can be trusted) so it is fair to
> contrast those numbers with that which would happen if one were to actually
> burn H2 in air - and .5 eV is a fair estimate even if you discount the 80%
> of air which is nearly inert.
>
>
> Since water can be split into H2 and O2 with 1.23 volts - does it stand to
> reason that one could get 1.5 eV in return ? That was rhetorical; and of
> course this one of nature's built-in cases of "systemic overunity" - 
>   

Now you're neglecting the splitting cost of H2->2H and O2->2H.


> ... except for the damn lie that it simply does not work out that way in
> practice - but it does serve to contrast the large disparity of the "actual
> with the calculated".
>
>   
>> Did I get it wrong?
>> 
> Well, let's say that you got it partly right and mostly wrong - if your
> intent was to suggest that hydrogen can be burned in air with resultant
> steam being formed at about 17,000 degrees K. 
>
> Jones
>
>   



RE: [Vo]:Significant Implications - Kitamura

2009-12-29 Thread Jones Beene
-Original Message-
From: Michel Jullian 

> - but the 2 eV available
> from loading alone without deuterium (contrast that to about .5 eV if the
> hydrogen were burned in air) is a huge surprise -

MJ: Jones, where did you get that  .5 eV figure? I did the maths and found
about 1.5 eV instead, here is the Google calculator result;

((294.6 / 2) / 6.02e23) * kJ = 1.52719998 electron volts


Michel, the half-eV figure is the common 'real world' estimate based on the
maximum average temperature of the resultant steam - but even so, it appears
you did not first deduct the dissociation energy of O2 and H2 and then later
deduct the parasitic losses of NOx, peroxides etc. and the other losses that
are expected in actual practice, for combustion in air?

IOW there are lies, damn lies, and theoretical calculations ;) when trying
to go from 'paper numbers' to actual practice. Kitamura's numbers were
indicated to be actual practice (if they can be trusted) so it is fair to
contrast those numbers with that which would happen if one were to actually
burn H2 in air - and .5 eV is a fair estimate even if you discount the 80%
of air which is nearly inert.

Since water can be split into H2 and O2 with 1.23 volts - does it stand to
reason that one could get 1.5 eV in return ? That was rhetorical; and of
course this one of nature's built-in cases of "systemic overunity" - 

... except for the damn lie that it simply does not work out that way in
practice - but it does serve to contrast the large disparity of the "actual
with the calculated".

> Did I get it wrong?

Well, let's say that you got it partly right and mostly wrong - if your
intent was to suggest that hydrogen can be burned in air with resultant
steam being formed at about 17,000 degrees K. 

Jones



Re: [Vo]:Significant Implications - Kitamura

2009-12-29 Thread Michel Jullian
2009/12/28 Jones Beene :

> - but the 2 eV available
> from loading alone without deuterium (contrast that to about .5 eV if the
> hydrogen were burned in air) is a huge surprise -

Jones, where did you get that  .5 eV figure? I did the maths and found
about 1.5 eV instead, here is the Google calculator result;

((294.6 / 2) / 6.02e23) * kJ = 1.52719998 electron volts

294.6 kJ/mol is the energy released per mole of D2O formed (=minus the
enthalpy of formation of D2O), which I divided by 2 (2 D per D2O) and
by Avogadro's number and then converted to eV to find the burning
energy in eV per D atom. Did I get it wrong?

Michel



RE: [Vo]:Significant Implications - Kitamura

2009-12-29 Thread froarty572


On Mon, 28 Dec 2009 10:22:14 Jones Beene said 





  

[snip] Others here on Vo - have mentioned or debated the fact that the gravity 



force must grow exponentially at close dimensions - IF - "grand unification" 



is accurate. I think it is accurate. Dufour puts some numbers to that 



hypothesis. He may be onto something.[End Snip] 





  

[snip] But here is an irony. We have often asked the rhetorical question: "if 
the 



Casimir 'force' is essentially negative, then how can it produce a net 



energy gain?"  And now, with pico-gravity in the picture, we seems to 



have a tantalizing clue, in a reversed solution, so to speak. [End Snip] 





  

[reply] How about if the gravity is also a relativistic effect? If Naudts is 
correct then the hydrino can be explained relativistically. The Casimir plates 
are a negative energy sink reshaping longer vacuum fluctuations to fit between 
the plates meaning the Casimir cavity represents a different inertial frame. 
Any matter diffused inside the cavity is redrawn on these reshaped vacuum 
fluctuations which also modify gravity from our perspective outside the cavity 
because gravity is defined as distance/time^2. Gamma is changing inside the 
cavity in the same way as the Twin approaching C see his twin back on earth 
except the boundary is abrupt and the accumulating dv results from a difference 
in equivalent accelerations between the ambient gravitational field outside the 
cavity and reduced field inside. The cavity maintains the zones spatially 
stationary to each other but the reshaped / restricted vacuum flux open a novel 
relativistic solution for the dv. We pull away in our ambient inertial frame 
while the cavity contents fall behind in time proportional to the Casimir 
force/ plate spacing at their locality, different inertial frames forming a 
gradient for different spacing until finally reaching the plate boundaries and 
restoring the normal ambient energy levels for vacuum fluctuations. The 
gradient represents different levels of deceleration (or negative energy/sink), 
the energy would be conservative upon exiting the cavity unless we somehow pin 
the reshaped atoms into their new shape making the vacuum flux do work to 
restore the atoms on the way out such as forming a compound or molecule. I 
don't think the world will ever see a steep fractional hydrogen outside of a 
Casimir cavity / skeletal catalyst and I think this property can be exploited 
in conjunction with natures preference for diatomic states. The black light 
plasma could well be decelerated hydrogen oscillating between H1/H2 as it is 
drug back up to speed exiting the cavity.[end reply] 





  

Fran 

animation - long vacuum fluctuations reshaped instead of displaced 
http://www.byzipp.com/finished2.swf  

RE: [Vo]:OT: Change Metal Surface to Reflct any Color (Not plating or Paint)

2009-12-29 Thread Jed Rothwell

This is great stuff! Amazing.

Summary: they are changing the color of metals by changing the 
surface structure.


- Jed



Re: [Vo]:steorn 2009 patent: "System and method for measuring energy in magnetic interactions"

2009-12-29 Thread Terry Blanton
You might note that these are patent applications.  The patents are pending.

Terry

On Tue, Dec 29, 2009 at 5:27 AM, Esa Ruoho  wrote:
> http://www.google.com/patents?id=1dOyEBAJ&zoom=4&pg=PA1#v=onepage&q=&f=true



Re: [Vo]:steorn 2009 patent: "System and method for measuring energy in magnetic interactions"

2009-12-29 Thread Esa Ruoho
http://www.google.com/patents?id=1dOyEBAJ&zoom=4&pg=PA1#v=onepage&q=&f=true

On Tue, Dec 29, 2009 at 12:11 PM, Esa Ruoho  wrote:

> (gleaned from overunity.com)
>
> "US 2009/0009157 A1 : SYSTEM AND METHOD FOR MEASURING ENERGY IN
> MAGNETIC INTERACTIONS : An apparatus and method is provided for
> measuring magnetic force response time due to the magnetic viscosity
> of materials and for measuring total energy exchanged due to relative
> motion of magnetic materials. Voltage and current versus time through
> an electromagnet is measured and recorded
> Inventors: Sean David McCarthy, Alan Simpson, Martin Flood, Maxime Sorin"
>
> and
>
> http://sites.google.com/site/steornlab/home
>
>


[Vo]:steorn 2009 patent: "System and method for measuring energy in magnetic interactions"

2009-12-29 Thread Esa Ruoho
(gleaned from overunity.com)

"US 2009/0009157 A1 : SYSTEM AND METHOD FOR MEASURING ENERGY IN
MAGNETIC INTERACTIONS : An apparatus and method is provided for
measuring magnetic force response time due to the magnetic viscosity
of materials and for measuring total energy exchanged due to relative
motion of magnetic materials. Voltage and current versus time through
an electromagnet is measured and recorded
Inventors: Sean David McCarthy, Alan Simpson, Martin Flood, Maxime Sorin"

and

http://sites.google.com/site/steornlab/home