In a message dated 7/26/04 5:38:10 PM Pacific Daylight Time, [EMAIL PROTECTED] writes:
Ironically, the reason that any of this can happen at all, as Fred has been suggesting, is that a high voltage gradient serves to oscillate the ever-present neutrino flux to the degree that they will interact w
Hi Robin,
> Actually I was trying to come up with a way of measuring the ratio of "stripping"
> reactions to fusion reactions, but upon further consideration I realize that since
> both would produce lone neutrons, this isn't a good measure anyway.
That measurement can be done with a neutron s
In reply to Jones Beene's message of Mon, 26 Jul 2004 07:26:26 -0700:
Hi,
[snip]
>> So there are lots of 700 keV electrons detected?
>
>How would you detect these in a Fusor?
By putting a small instrument portal in the wall?
>
>If the question is, "are secondary gammas detected" the answer is ye
In reply to Jones Beene's message of Mon, 26 Jul 2004 07:22:10 -0700:
Hi,
[snip]
>> But fusors don't depend on Maxwellian tails. All the deuterons get accelerated to
>> thousands of eV, which equates to a temperature of approx. 1E8 K.
>
>NO!! You are confusing kv with KeV. All Fusore depend on M
> So there are lots of 700 keV electrons detected?
How would you detect these in a Fusor?
If the question is, "are secondary gammas detected" the answer is yes.
Robin,
> As they approach, the first bond grows weaker as the second bond grows stronger:- N
> P NP followed by N PNP, and 2 D has become N + He3. The force required to tear
> the D apart is supplied by the growing force between the P and the other D. The net
> reaction is exothermic, and
Robin,
> I still don't see how this can happen without supplying the 2.2 MeV binding energy
> of the deuteron.
The neutrino oscillation, which is elevated and instigated by the electric field,
provides whatever energy is needed, but it is in the KeV range
> >Whoa, Robin! You are missing the m
In reply to Jones Beene's message of Sat, 17 Jul 2004 12:48:30 -0700:
Hi,
[snip]
>That is to say, in stripping, the energy deficit that appears to prohibit the
>reaction from happening in the first place comes from the energy that should have
>been left over once it happened.
Of course it does,
In reply to Jones Beene's message of Sat, 17 Jul 2004 12:48:30 -0700:
Hi,
[snip]
>The two physicists found that, when a deuteron is fired into a target atom even
>weakly, the neutron of that atom can be stripped off the proton and penetrate the
>nucleus of the target.
[snip]
This is actually the
In reply to Jones Beene's message of Sat, 17 Jul 2004 12:48:30 -0700:
Hi,
[snip]
>5) The neutrons measured in a Fusor are indeed mostly fusion neutrons, but they
>represent a small proportion of the total neutrons which have been produced
A 2.5 MeV neutron/proton has enough energy to break apart
In reply to Jones Beene's message of Fri, 16 Jul 2004 07:49:44 -0700:
Hi,
[snip]
>a variety of beta decay - not markedly different from the situation where the neutron
>became partially disengaged and decayed. There are QM reason why the neutron cannot
>itself decay within the confines of the D
In a message dated 7/17/04 12:55:41 PM Pacific Daylight Time, [EMAIL PROTECTED] writes:
Executive summary:
1) High voltage has the demonstrated property of forcing a 'neutrino oscillation.'
"There is also a one-sided emphasis on entropic (explosive, radiative, centrifugal) phenomena in physics
From: "explorecraft"
> Allegedly
> The Farnsworth type Fusor centers around kinetic ions
> which apparently free the neutrons by collision.
Not exactly.
The Farnswoth Fusor is an Inertial Confinement Fusion device which beneifts from
spherical convergence, but the nuetrons are not collisiona
> -Original Message-
> From: Jones Beene [mailto:[EMAIL PROTECTED]
> Sent: Thursday, 2004 July 15 00:59
> To: vortex
> Subject: "free-energy" mechanism with H2O
>...X...<
> In a Farnsworth type Fusor,
> it has been proven beyond any doubt that
> a non-static electric field of 10,000 v
From: "Robin van Spaandonk"
> How should D decay?
a variety of beta decay - not markedly different from the situation where the neutron
became partially disengaged and decayed. There are QM reason why the neutron cannot
itself decay within the confines of the D nucleus; which is, I suppose, wha
In reply to Jones Beene's message of Wed, 14 Jul 2004 10:59:15 -0700:
Hi,
[snip]
>two added neutrons (tritium) are rapidly unstable. The yield on D decay is over 1.4
>MeV. As with the neutron, there still exists a considerable measure of uncertainty as
>to the precise value of deuterium half-lif
At 10:59 am 14-07-04 -0700, you wrote:
>For whatever reason, the subject of "accelerated radioactive decay" may be the most
>neglected concept in free energy research. At the same time, accelerated decay may
>offer an immediate solution to the problem of finding an acceptable transportation
>fue
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