Horace Heffner wrote:
I haven't taken the time to look into this in
detail, but my first impression . . .
With all due respect, it is a bad idea to discuss
these things without looking into them in detail,
and a person's first impressions are likely to be wrong.
is that, unless there is a typo, it makes no
sense at all to attempt to draw the 23.82 MeV line through Fig. 1 . . .
That is an expectation value. That shows how much
helium there would be if the ratio of helium to
heat was 23.82 MeV per reaction, and if every
atom of helium were recovered. Obviously, not
every atom is -- or can be -- recovered. As the
text points out a lot of the helium is stuck in
the cathode and can only be recovered after the experiment.
Perhaps I'm misreading the x axis labeling "Excess Power/Current
(mW / A)", or the intended meaning of the x axis values. To be
sensible the x axis should simply be excess energy, i.e. the integral
of mW over time.
Those are instantaneous power readings taken at
different times, arranged in ascending order. The
helium does not stay in this cell; it is open,
like the Miles cell, and the helium is collected
from the effluent gas. This is not a time graph
of the run, and that is not the integrated
energy. In other words, at one point when the
cell was producing about 70 mW the helium reading
came out 2.4 +/- 0.8, and another time when power
was ~100 mW, a helium reading came out 2.8 +/-
1.2. The points at the bottom are either
experimental error or caused by helium being
trapped in the cathode. It is difficult to say
which. Quoting the paper, p. 2 and 3:
"Figure 1 presents the results of concurrent
excess power and helium measurements performed
during open cell electrolysis using two different
Pd and Pd-alloy cathodes. In three instances
where excess power was measured at statistically
significant levels, 4He also was found to be
conveyed out of the cell in the electrolysis
gases (D2 + O2). The solid line in Figure 1 plots
the regression fit of these data to a line
passing through the origin; the dashed line is
that expected for 4He generation according to the reaction:
d + d --> 4He + 23.82 MeV (lattice) [1]
It is clear from the slopes of these two lines
that the observed 4He constitutes only 76 ± 30%
of the 4He predicted by equation [1]. A more
significant problem in Figure 1 is that three
further 4He samples, taken at times of non-zero
excess power (open diamonds), exhibited helium
concentrations only at the level of the
analytical uncertainty, as did numerous samples
taken in the apparent absence of excess power
production. Clearly if 4He is produced in
association with excess power, it is not released
to the gas phase immediately, or completely."
<http://lenr-canr.org/acrobat/McKubreMCHtheemergen.pdf>http://lenr-canr.org/acrobat/McKubreMCHtheemergen.pdf
That seems pretty clear to me. I do not
understand why people here are confused by it.
Maybe if someone took the time to look deeper into this they could
make some sense of it.
I didn't have to look very deeply.
Look folks: An author may not present data the
way you would choose to present it. I often find
that a graph shows something other than what I
assumed; i.e., it shows power rather than
integrated energy. Oops. I usually have to read a
paper several times to figure out what's what. So
let's not jump to conclusions about these things,
or assume that X or Y "doesn't make sense." You
need to cut the authors some slack. It is tough
writing papers and explaining things. Someone
once complained to Oliver Heaviside that his
papers were very difficult to read. He responded,
"That may well be -- but they were much more difficult to write."
- Jed
