On Feb 25, 2010, at 5:40 AM, Jed Rothwell wrote:
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.
If I were afraid of being wrong it would destroy my creativity, I
would learn little, and I would post nothing at all. That of course
might be a good thing from your perspective, but not mine. 8^) In
this particular case, for example, I would have no excuse for
procrastinating on doing my tax return!
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.
Here you have missed the point entirely. There is no such "expected
value" of energy per helium atom as a function of excess heat
power. There is an expected value of energy per helium atom as a
function of excess heat *energy*.
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.
Apparently it does not. It shows a ratio of helium to excess power,
not excess heat.
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.
.
This makes the graph seem nonsensical.
.
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.
.
Then the green line makes no sense at all without a further explanation.
.
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.
These numbers do not relate to
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).
This makes total sense.
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]
This is the part that needs clarification. There is no clear link
established between helium concentration and power produced.
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].
The helium concentration is not predicted by equation 1. Equation 1
only establishes a relationship between helium atoms created and
excess *energy* produced. It has nothing to do with power.
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
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.
And you didn't make any sense.
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
As an ineffective writer I have to emphatically agree with the above
paragraph.
I think I can make some sense of this, at last in my head. The
purpose of dividing by current I is to roughly account for the flow
rate of the effluent gas. In an imaginary ideal case the excess heat
would be produced at a constant COP, and the He would be produced and
ejected into the effluent gas at a constant rate. If the voltage
is held constant then there is, in the ideal case, a constant
concentration of helium produced in the effluent per unit of time,
and the amount of non-helium effluent produced is proportional to the
current (ignoring entrained water content of the effluent, and
recombination). Therefore, provided the current is head steady
throughout a run, dividing the excess heat by the current is a rough
method to normalize the data for the effluent flow rate. If you know
the flow rate and helium production rate, and the system is in a
steady state, you can then predict the helium concentration in the
effluent, and produce the green line. I don't think this model was
expressed in the article, but was assumed. There are of course many
problems with this model, but in my head this makes some sense of
the approach and the graph.
Best regards,
Horace Heffner
http://www.mtaonline.net/~hheffner/
