Having spent a number of years as a councilor at a BSA
summer-camp in Kansas, I'm sorry to say that your
common-sense approach will not work.
Thirst has very little to do with the actual level of
body fluids in your system.
At camp drinking water is the prescribed treatment for
all ailments
From: John.Rudiger
Here in Perth, Western Australia they are powering 3 Public Busses using
Hydrogen Fuel Cells as a trial with more to come in the near future.
http://www.dpi.wa.gov.au/fuelcells/
Thanks, John. Any idea what those buses cost?
Here in Georgia, upon examining the cost of
Hey, Robin and George... wait a minute
Naudin apparently incorporated the low duty cycle in his
calculations (in a non-standard way) by figuring the average
value by multiplying the peak voltage (from a 12 volt power DC
battery power supply) by the duty cycle to give an average voltage
The Auto ionization reaction in water:
2 H2O ---( H3O +) + OH - requires a free energy of 335,000 joules per mole
or about 3.4 eV per H3O + formed.
This and similar phenomena strongly suggests that Zero Point Energy (ZPE) pumping is
providing the energy.
The ~ 0.1 picosecond lifetime of
Fred,
The Auto ionization reaction in water:
2 H2O --- ( H3O +)+ OH - requires a free energy of 335,000
joules per mole or about 3.4 eV per H3O + formed.
Hmmm3.4 eV the signature of ZPE, it would seem.
kinda like the smile of the Cosmic-Cheshire-Cat... eh, Lewis?
See:
http://www.new-harvest.org/default.php
http://www.newscientist.com/article.ns?id=dn3208
This technology is coming along faster than I thought it would. This could
be very good news. See my book, chapter 16.
- Jed
Jones.
I wrote:
The Auto ionization reaction in water:
2 H2O ---( H3O +) + OH - requires a free energy of 335,000 joules per mole
or about 3.4 eV per H3O + formed.
This and similar phenomena strongly suggests that Zero Point Energy (ZPE) pumping is
providing the energy.
The ~ 0.1
Hi Jones,
You wrote:
Hey, Robin and George... wait a minute
Naudin apparently incorporated the low duty cycle in his
calculations (in a non-standard way) by figuring the average
value by multiplying the peak voltage (from a 12 volt power DC
battery power supply) by the duty cycle to
Soylent green is . . .
:-)
George,
Actually 166 A (peak)* 12 V(peak) * .05 (duty factor) gives
about
100 watts input at an average current of about 8 A
and an average voltage of .6 V.
OK, now I see what you are saying but how does an auto battery
provide 166 amps, even if it is peak? Is this what you really
think
Here is a good semi-technical introduction to the field:
http://www.new-harvest.org/img/files/Invitro.pdf
Many practical problems must be overcome, but it sounds like they are
making progress. For example: Benjaminson and others succeeded in using a
serum-free medium made from maitake
Jones,
George,
Actually 166 A (peak)* 12 V(peak) * .05 (duty factor) gives
about
100 watts input at an average current of about 8 A
and an average voltage of .6 V.
OK, now I see what you are saying but how does an auto battery
provide 166 amps, even if it is peak? Is this what you
Jones Beene wrote:
George,
Actually 166 A (peak)* 12 V(peak) * .05 (duty factor) gives about
100 watts input at an average current of about 8 A
and an average voltage of .6 V.
OK, now I see what you are saying but how does an auto battery provide
166 amps, even if it is peak?
The
Terry Blanton wrote:
Soylent green is . . .
:-)
Hey, what's your objection here? Recycling is a _good_ thing, right?
;-)
The site has been updated with radiation measurements:
http://jlnlabs.imars.com/mahg/tests/mahg2e.htm
Exec summary: NADA so it is doubtful that anything nuclear is
going on in the sense of LENR.
However, I hope that we can get some kind of quick confirmation
that his P-in is accurate...
Hi George,
One more thing...
OK, now I see what you are saying but how does an auto battery
provide 166 amps, even if it is peak? Is this what you really
think is really happening ?
IF he is getting a reading at the battery itself - of about 1/3+
amp average current - then you would
From: Stephen A. Lawrence
Terry Blanton wrote:
Soylent green is . . .
:-)
Hey, what's your objection here?
Recycling is a _good_ thing, right?
;-)
Maybe it's time to go home.
E. G. Robinson really did have a great send off.
Regards,
Steven Vincent Johnson
Hi Jones,
OK, now I see what you are saying but how does an auto battery
provide 166 amps, even if it is peak? Is this what you really
think is really happening ?
It certainly looks possible. Batteries are capable of surprisingly
high peak currents and the pulse width here is only 1 ms.
From: Jones Beene
This is something that should be pretty easy to ascertain right
away...
I was under the impression that the values on his chart were rms (root mean
square) since that's what the Fluke measures.
From: [EMAIL PROTECTED]
From: Stephen A. Lawrence
Terry Blanton wrote:
Soylent green is . . .
:-)
Hey, what's your objection here?
Recycling is a _good_ thing, right?
;-)
Maybe it's time to go home.
E. G. Robinson really did have a great send off.
Hi George,
Not wanting to waste anymore bandwidth on this
device than is necessary, if it isnot OU,I sent the following
message to intereseted parties, which I hope gets to JNL or to Nicholas Moller
very soon - as they are difficult to contact directly. COPY of
posting:
A serious question
From: Terry Blanton
From: Jones Beene
This is something that should be pretty easy to ascertain right
away...
I was under the impression that the values on his chart were rms (root mean
square) since that's what the Fluke measures.
http://jlnlabs.imars.com/mahg/tests/index.htm
This test does not prove that LENR is not occurring. LENR does not
produce detectable radiation outside of the cell. The only radiation of
any significance has only been detected within the cell. In addition,
such a counter would not detect radiation from tritium nor neutron
emission.
See the mechanical drawings:
http://jlnlabs.imars.com/mahg/diagram.htm
which show the assembly. Note that this is what we EEs call a two port
(terminal) device. The tungsten grid is connected to one port and the center
rod appears to be connected to the other. But is the upper grid support
Ed,
Your are correct and I was too hasty in saying that LENR had been
eliminated as a factor. I'm having a low batting average today it
seems.
This test does not prove that LENR is not occurring. LENR does
not produce detectable radiation outside of the cell. The only
radiation of any
FWIW, another reality check for the input power consumed by the MAHG is to
first calculate the energy in Joules per pulse and multiply by the rep rate.
IE, the input energy is 166 * 12 * (1/50) * .05 = 1.992J per pulse.
Therefore, the input power = 1.992 * 50 = 99.6 watts. With the equipment JLN
Jon,
FWIW, another reality check for the input
power consumed by the MAHG is to first calculate the energy in Joules
per pulse and multiply by the rep rate. IE, the input energy is 166 * 12
* (1/50) * .05 = 1.992J per pulse. Therefore, the input power = 1.992 *
50 = 99.6 watts. With the
Jones is making good points, but
snip
You (Robin, at least) is willing to accept Mills equally
surprising claims - whereas Mills gives almost zero detail, and
often bases his P-out claims on guess-timates of what the power
would be IF the photon radiation were converted, and yet in
From: [EMAIL PROTECTED]
Date: 7/8/2005 1:28:11 PM
Subject: [BOBPARKS-WHATSNEW] What's New Friday July 8, 2005
WHATÂ’S NEW Robert L. Park Friday, 8 Jul 05Washington, DC
1. POLITICAL SCIENCE: IS THE CONGRESSMAN DOING CLIMATE STUDIES? Who among
us has not engaged in disputes over
I've been trying to organise my ideas on
the inverse of temperature (which I have
called Compreture) seen as a differential
pressure and tying it in with volume so
that the classic PV/T is a constant can
be viewed as a hierarchical string of
pressures - or more fundamentally
hierarchical
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