On Sep 13, 2011, at 10:20 PM, Peter Heckert wrote:
Am 14.09.2011 01:20, schrieb Horace Heffner:
On Sep 13, 2011, at 12:55 PM, Peter Heckert wrote:
Am 13.09.2011 22:47, schrieb Man on Bridges:
Hi,
On 13-9-2011 20:44, Horace Heffner wrote:
<snip calculation of lead shielding>
Hmmm, is there a way to start and stop a gamma radiation source,
as it may be used only to trigger the process?
There is no other way than shielding or increasing the distance.
Rossi could inside use a shield that is moved electrically or by
heat (bimetal).
Or he could control the distance to the gamma source.
If it is a very small point source the /local/ intensity of
radiation could be changed by factor 10^2 or 10^3.
Peter
The above is incorrect. A 2 cm thick lead shield will only reduce
Co-60 gammas by 75%.
I = I0 * exp (-0.694 * x)
So we want I/Io = 0.01 to achieve 1/100 reduction factor.
I/I0 = exp (-0.694 * x)
0.01 = exp (-0.694 * x)
ln(0.01) = -0.694*x
x = ln(0.01)/(-0.694) = 6.63
It takes 6.6 cm of lead to divide Co-60 gamma intensity by 100.
Similarly, it takes about 10 cm of lead (on all sides) to
attenuate CO60 gammas by a factor of 1/1000.
Maybe I am in error.
The error I am pointing out is that it does not matter at all how
small the source inside the device is - assuming it is centrally
located. It could be microscopic, or a couple cm in diameter and
this would make no difference at all to the gamma flux measured at
the surface of the Rossi device. If the source were even the size
of one atom, vs a few mm, or cm, it would make no difference to the
intensity measured at the surface of the Rossi device. The intensity
is proportional to surface area divided by total counts per minute.
The size of the source inside the device is of no consequence
provided it is centrally located.
Rossi used two counters right up against the device, primarily in
coincidence mode, but they would saturate at the count rate expected,
so coincindece mode would be irrelevant. Celani measured radiation
right near the device before turn on as near background, using two
different types of counters. It is not possible to put enough lead
in the device to suppress the 1.33 MeV gammas from cobalt to even a
non-lethal level - provided there is enough cobalt to sustain a 15 kW
reaction at one gamma per LENR reaction. Celani also measured the
counts a few meters away. A few meters away, where Celani also
measured, is not enough to suppress the counts to background.
If a 2 cm thick lead shielded source has even a very modest amount of
Co-60 then detectors nearby will detect the gammas - at all times. I
showed it would take at least 6x10^11 gammas a second to account for
a 12 kW LENR reaction, even assuming 10 MeV per reaction, which is
high. Even if Rossi could stuff his source behind a blanket of 6.6
cm of lead on all sides, giving a device radius of 13 cm, leaving no
room for water or fuel, that would only reduce the count by a factor
of 100, thus outside the reactor a 6x10^9 count per minute (cpm)
source would be manifest. At a distance of 6.6 meters, the flux would
be reduced by a factor of 6.6/660 = 10^-4, or to 6x10^5 cpm. Celani
could not miss this.
I understand it this way:
A shield cannot alter the wavelength and so it cannot alter the
photon energy respective frequency.
Yes.
Only the amount or density of gamma photons can be changed by
photon absorption.
That is in practical terms true. Some of the gammas cause positron
emission which results in a lower energy gamma, but at CO60 energy
levels this is not important.
Now, lets assume the gamma radiator has a volume of 1mm. Then the
photon density in 100mm distance must be 40000 times weaker as the
density directly measured in 0.5 mm distance at the surface of the
gamma source. (Inverse square law as in optics)
This is where the conceptual error occurs. The source is not measured
at its radius. It is measured at the radius of the Rossi device, and
further.
Even without shield we can get a large attentuation factor purely
from distance, if the diameter of the source is small.
This is irrelevant because the distances at which measurement
actually occurred are fixed.
So if the gamma source is in direct contact with nickel, the photon
density must be 100 times larger than in 10 mm distance.
Is this wrong?
You are mixing apples and oranges. There is a difference between how
the radiation affects the Ni and determining the amount of radiation
by counting outside the device. If the Co60 were a nano-sized
particle it would provide a high intensity radiation to nano-sized
nickel particles at nano-distances from it, but not to all the fuel.
A point source does not provide a means to irradiate the entire fuel
at the point source flux level. What counts in irradiating the fuel
is achieving as nearly as possible a 1-1 relationship between gammas
emitted and LENR reactions produced. As I noted, even at optimal 1-1
conditions, this requires a huge amount of gammas a second. It is not
possible to shield such a large source located within a Rossi sized
device, no matter if it is entirely composed of lead. For counting
purposes, it does not matter whether the source is the size of an
atom or a grape.
An ideal arrangement would probably consist of a fine mixture of the
radioactive stimulator with the fuel. However, it would not then be
feasible to hide the stimulator when post experiment fuel analysis
occurs.
Another thought:
I think Rossi is naive and will loose if he think he can
commercialize a discovery of this magnitude and eternal history
changing importance and keep it secret. This is impossible to do,
he must go the scientific route, not the commercial route.
Also his fans and investors are naive to believe this.
I agree the commercial route makes little business sense at this
point, unless maybe he can build power plants and operate using trade
secrets for as long as possible - which likely would not be long. The
reason for not going commercial is the lack of strength of the patent
(s) involved. On the other hand, if he does have a breakthrough in
energy production, it could be one of the greatest scientific
achievements of mankind. If he is looking for a financial reward,
the worst thing he could do, for himself and everyone else, is sit on
the discovery, or struggle in great competition to survive as a
business - likely until he dies or goes broke. He said he wants to
do good with his invention. It could avoid needless war, save
millions of lives, and bring water, food and even prosperity to
billions of people. This can not happen any time soon if he hangs on
to it as a trade secret. Making such a great discovery is
undoubtedly worthy of substantial awards.
As soon as it is totally and unmistakenly clear, this is a nuclear
reaction that produces large amounts of energy, law will stop him.
And international scientific research will start.
You cannot discover the stone of philosophers and commercialize
this and keep it secret, this is impossible.
This must be done in a scientific way.
As soon as large amounts of energy are produced, it must be also
scientifically investigated, if this can be abused to build bombs
and so on. Rossi says no, this is not possible, but as long as it
is a secret he cannot proof it is without dangers.
I think no government can tolerate something like this going on and
reaching very large dimensions unsupervised.
The unknown potential of danger is too high.
Only if his customer is NASA or another large scientific and
trusted organisation he could have luck selling this.
Best,
Peter
All very real concerns.
Best regards,
Horace Heffner
http://www.mtaonline.net/~hheffner/
