As I suppose everyone reading this thread has already recalled, one of
Bill Beaty's "red flags of fraud" 'way back when was responding to
questions and challenges with outrage and anger, while failing to
actually address the question being raised.
We've certainly seen this sort of behavior
Russ George wrote:
> Why, because you are an armchair self-serving critic who never does
> anything but try to raise your own worth by trolling worthless comments . .
> .
>
Well, at least I have edited and published papers. You have published
little or nothing, so you have made no
compliment I can
imagine.
From: Jed Rothwell [mailto:jedrothw...@gmail.com]
Sent: Thursday, November 17, 2016 6:59 AM
To: vortex-l@eskimo.com
Subject: Re: [Vo]:How much helium?
Russ George <russ.geo...@gmail.com <mailto:russ.geo...@gmail.com> > wrote:
Yes, of course why
Russ George wrote:
Yes, of course why would anyone not do so. The methods used were all of the
> usual state of the art methods, just do your reading into the complexities
> of measuring helium in metals and you’ll see how it is done. It’s all at
> your Googling
O nuclear reactors and on the core nuclear bombs, the alpha decays makes
the material brittle. This might be part of the explanation of why cracks
are related to the efficiency of the material. Given the randomness, it
could be that the way cracks coalesce might end up making a reaction going
to
Okay, "cubic milliliters" is redundant. Like round circles.
Brian Ahern wrote:
> 1 mm3 is 1/100,000 of a leter not 1/1000,000 !
>
Nope, it is a million. There are 1000 cubic millimeters in a milliliter (10
x 10 x 10), and 1000 cubic milliliters in a liter. 1000 x 1000 = 1,000,000.
Confession: I am bad at arithmetic, so I also asked
Yes, of course why would anyone not do so. The methods used were all of the
usual state of the art methods, just do your reading into the complexities of
measuring helium in metals and you’ll see how it is done. It’s all at your
Googling fingertips.
From: Jed Rothwell
1 mm3 is 1/100,000 of a leter not 1/1000,000 !
From: Jed Rothwell
Sent: Wednesday, November 16, 2016 5:34 PM
To: vortex-l@eskimo.com
Subject: [Vo]:How much helium?
Years ago, Russ George told me that in one of his experiments he could
Russ George wrote:
Jed’s senility is showing in his recollection. In my work I have repeatedly
> shown helium bubbles, known as “loop punching” in the proper solid state
> science vernacular. These ‘bubbles’ form inside solid cold fusion metals.
>
Well, you did not
I wrote:
> Divide Avagadro's number by 10E11 gives 1.66E-12 moles, or 6.64E-12 g
> helum, which multiplied by 345,000 MJ/g gives 2.3 W. Close enough!
>
Oops. 1 mole of deuterium is 2 g. So that's 1.2 W. Even closer.
2 moles of deuterium fuse to form 1 mole of helium.
- Jed
Jed’s senility is showing in his recollection. In my work I have repeatedly
shown helium bubbles, known as “loop punching” in the proper solid state
science vernacular. These ‘bubbles’ form inside solid cold fusion metals. They
are perfectly consistent with ‘loop punching’ “bubbles” formed in
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