On Oct 18, 2011, at 4:25 PM, David Roberson wrote:
Hi Horace,
Thank you for the kind welcome into vortex. I suspect that my
oversized attachment tunneled through the barrier; maybe using the
same path as Rossi's device.
It must have a long de Broglie wavelength. My guess is the mass of
paper involved is very low. 8^)
You have written an interesting description of the old ECAT version
and I plan to review it thoroughly as time allows. I guess I may
have needed a hit on the head to believe everything(or anything)
that I read on line about ECAT operation. As you know, I was just
parroting what I saw in the journal.
No problem there! Most discussion here is based on second hand
information.
My understanding of nuclear physics is lacking as my field is
electronics design. Allow me a lot of slack when I suggest
something totally whacko as sometimes I have good ideas and
approach problems from different perspectives.
New ideas are most welcome here. Slack is sometimes hard to come
by! 8^)
Now, let me ask you a few questions that I suspect you can answer
easily. You have presented some interesting calculations
concerning the penetration of gamma rays and x rays through lead.
The new ECAT design has 5 cm of lead according to reports. That is
more than twice the original thickness of 2 cm in the earlier
version. Could you help me to reverse engineer the ECAT shielding
and figure out what energy of gamma rays would be just barely
shielded enough to be undetected?
Here is a spread of results, but for 5 cm thick lead:
again using for I0:
Energy Activity (in gammas per second) for 1 kW
-------- ----------
1.00 MeV 6.24x10^15
100 keV 6.24x10^16
10.0 keV 6.24x10^17
and using:
I = I0 * exp(-mu * rho * L)
where mu is given by:
Energy mu (cm^2/gm)
-------- ----------
1.00 MeV 0.02
100 keV 1.0
10.0 keV 80
For 1 kW of MeV gammas we have:
I = (6.24x10^15 s^-1) * exp(-(0.02 cm^2/gm) * (11.34 gm/cm^3) *(5
cm))
I = 2.07x10^15 s^-1 ( was 3.7x10^15)
For 1 kW of 100 keV gammas we have:
I = (6.24x10^16 s^-1) * exp(-(1.0 cm^2/gm) * (11.34 gm/cm^3) *(5
cm))
I = 1.4x10^-8 s^-1 = ~0 s^-1 (I posted 2.9x10^5
s^-1, but this was my calc. error -
correct value was
3.04x10^4 s^-1, still readily observable with a counter)
For 1 kW of 10 keV gammas we have:
I = (6.24x10^16 s^-1) * exp(-(80 cm^2/gm) * (11.34 gm/cm^3) *(5 cm))
I = ~0 s^-1 (was ~0 s^-1)
So, the answer is the 5 cm of lead vs 2.3 cm of lead essentially
eliminates gammas in about the 100 keV range that were readily
countable with 2.3 cm of lead.
The extra lead does nothing, however, for converting more gamma
energy to heat. It still does not prevent MeV order gammas from
being readily detected, so they are still ruled out as a heat
source. The 3.04x10^4 counts/sec of 100 keV gammas eliminated
represent the unobservable 4.8x10^-10 J/s ~= one trillionth of a
watt, so no added heat is obtained there, but even close up counting
is prevented.
And, if this thickness is not adequate, is any amount of shielding
able to stop gammas to that degree?
The extra lead does nothing for 1 MeV gammas, but a lot for 100 keV
gammas, but also nothing for producing more heat. The lead can
however, add to the heat-after-death time significantly depending on
configuration. Frankly I suspect the added shielding is iron, not
lead. It can sustain heat-after-death much better and provide
stability at high temperatures that lead can not. The goal of this
test was heat-after-death.
The mu for lead at 100 keV is 0.372. For 1 kW of 100 keV gammas
through 2.7 cm of iron I get:
I = (6.24x10^16 s^-1) * exp(-(0.372 cm^2/gm) * (7.87 gm/cm^3) *(3
cm))
I = 4.66 s^-1
The counts still effectively disappear with iron on the inside and
lead on the outside or vice versa. If it is a hoax then there is of
course no need for the lead at all.
I would suspect that if you answer no amount of lead is within
reason, then you must think that the ECAT is a scam since the
shielding is arbitrary.
I see no reason to go from 2.3 cm to 5 cm, since the prior counts
were already nominal with regard to safety.
Rossi has stated that all of the energy released by the LENR
process is in the form of photons. Do you think that this is
possible?
Anything is possible. Viable prospective nuclear reactions have not
been identified. Anything that produces energy primarily from
positron emission is not viable due to the large annihilation
energy. Also, positron annihilation energies were looked for but not
found.
Do you know of any process that releases gammas or high energy x
rays but not heat directly?
That is somewhat of an inconsistent condition. If the energy output
is in the form of high energy photons then it produces heat. If the
energy is in the form of 1 MeV photons or more, then my calculations
(caveat: I make lots of mistakes) indicate it is not only readily
observable but dangerous with 2 cm or 5 cm of lead. The converse
is not inconsistent. There are many theories, including mine which
provide logical reasons why the photonic energy from LENR is in the
form of EUV or soft x-rays, which will only be observed as heat in
normal conditions. Low energy alphas or protons would produce heat
and be difficult to detect.
You answers to these simple questions would be most appreciated.
Dave
Simple but somewhat time consuming to answer. I am going to bail out
on conversation for a few days.
Best regards,
Horace Heffner
http://www.mtaonline.net/~hheffner/
