It could just as well be that the resistive wires are what are bright and
the gaps between them are where it gets darker.
If this were the case, won't there be a double dark shadow cast on either
side of the wire with the bright wire in between.
On Sat, Oct 11, 2014 at 1:54 AM, Eric Walker
The dark wire is thinner than the bright shadows so I think that the wire
is casting the shadow.
On Sat, Oct 11, 2014 at 2:03 AM, Axil Axil janap...@gmail.com wrote:
It could just as well be that the resistive wires are what are bright and
the gaps between them are where it gets darker.
If
On Sat, Oct 11, 2014 at 12:15 AM, Axil Axil janap...@gmail.com wrote:
The dark wire is thinner than the bright shadows so I think that the wire
is casting the shadow.
Maybe. Do you have a closeup that you're looking at? The details in the
image I see in the writeup are hard to make out. The
The two pictures on page 25 of the 54 page report can be zoomed to a
high resolution by using the control key of your keyboard and the wheel on
your mouse if you are using a new windows computer running with high screen
resolution.
You can see the dark wires as clear as day.
On Sat, Oct 11, 2014
On Sat, Oct 11, 2014 at 9:02 AM, Axil Axil janap...@gmail.com wrote:
You can see the dark wires as clear as day.
Yes. And now where does it say in the report that the team conducting the
trial determined that current was flowing through them?
Eric
Page 25:
The resistors appear to glow intensely in the parts lying outside the caps,
whereas inside the reactor body they seem to shade an underlying emission
of light. This may be explained if we consider that the main source of
energy inside the reactor body is actually the charge, and that it
If it has a COP 1 you might expect that, right
On Saturday, October 11, 2014, Axil Axil janap...@gmail.com wrote:
Page 25:
The resistors appear to glow intensely in the parts lying outside the
caps, whereas inside the reactor body they seem to shade an underlying
emission of light. This
Right...
On Sat, Oct 11, 2014 at 12:31 PM, ChemE Stewart cheme...@gmail.com wrote:
If it has a COP 1 you might expect that, right
On Saturday, October 11, 2014, Axil Axil janap...@gmail.com wrote:
Page 25:
The resistors appear to glow intensely in the parts lying outside the
caps,
On Sat, Oct 11, 2014 at 9:23 AM, Axil Axil janap...@gmail.com wrote:
Page 25:
The resistors appear to glow intensely in the parts lying outside the
caps, whereas inside the reactor body they seem to shade an underlying
emission of light.
What this sentence says to me is that the team assumed
When talking about the resistor heaters... Remember that Rossi repeats
that his E-Cat requires AC and can't run (directly) with DC. The
current on the three phases of electricity going in is different. But
it sounded like the phase and frequency going into the reactor matches
that from the mains.
Especially if they switch to a pulse mode where they are not really heating
directly anymore, the pulses are working like an induction stovetop where
the quickly changing magnetic fields are inducing arcs/currents in the
secret sauce
http://www.finecooking.com/videos/induction-cooktop-action.aspx
On Sat, Oct 11, 2014 at 9:48 AM, ChemE Stewart cheme...@gmail.com wrote:
Especially if they switch to a pulse mode where they are not really heating
directly anymore, the pulses are working like an induction stovetop where
the quickly changing magnetic fields are inducing arcs/currents in the
It basically means goat guys theory might be goat F'd...
On Saturday, October 11, 2014, Eric Walker eric.wal...@gmail.com wrote:
On Sat, Oct 11, 2014 at 9:48 AM, ChemE Stewart cheme...@gmail.com
javascript:_e(%7B%7D,'cvml','cheme...@gmail.com'); wrote:
Especially if they switch to a pulse
On Sat, Oct 11, 2014 at 9:48 AM, ChemE Stewart cheme...@gmail.com wrote:
Especially if they switch to a pulse mode where they are not really heating
directly anymore, the pulses are working like an induction stovetop
On page 6 there's a photo of the power and harmonic analyzer. I don't know
At 09:02 AM 10/11/2014, Axil Axil wrote:
The two pictures on page 25 of the 54 page report can be zoomed to a
high resolution by using the control key of your keyboard and the
wheel on your mouse if you are using a new windows computer running
with high screen resolution.
I zoomed and did
To me, the width/continuity of the dark lines seems much more consistent
then the light colored areas so I would say the dark areas are wires
On Saturday, October 11, 2014, Alan Fletcher a...@well.com wrote:
At 09:02 AM 10/11/2014, Axil Axil wrote:
The two pictures on page 25 of the 54 page
I think you all made the job (respect to Jed BTW, as usual)
1- the window of transparency can be real for some alumina materials, but
not in the wavelength that the IRcam use (7um)
2- if the IRcam was troubled by the white light, the bright zone would be
much hotter for the IR cam. the IRcam
On Sat, Oct 11, 2014 at 1:04 PM, Eric Walker eric.wal...@gmail.com wrote:
On Sat, Oct 11, 2014 at 9:48 AM, ChemE Stewart cheme...@gmail.com wrote:
Especially if they switch to a pulse mode where they are not really
heating directly anymore, the pulses are working like an induction stovetop
Hi,
among the skeptic argument one of the only that is not laughable is the one
of goatguy...
maybe is it because I don't understand it well...
He seems to say
- that alumina is not a grey body, but transparent, and that emissivity
must be mixed with translucidity when considering the radiation
I find it funny that anonymous GoatGuy is literally one of the best-read
skeptics out there and get's so much play, but in my view he deserves it
because he's pretty good and the skeptical community generally sucks.
Still don't think his objections discredit the report, but I wouldn't mind
seeing
The 7 professors who wrote the TIP report are supposed to be answering such
criticisms. They should have set up a website for just that purpose.
Rossi did.
On Fri, Oct 10, 2014 at 2:31 PM, Foks0904 . foks0...@gmail.com wrote:
I find it funny that anonymous GoatGuy is literally one of the
At 02:22 PM 10/10/2014, Alain Sepeda wrote:
Hi,
among the skeptic argument one of the only that is not laughable is the
one of goatguy...
maybe is it because I don't understand it well...
He seems to sayÂ
- that alumina is not a grey body, but transparent, and that emissivity
must be mixed with
Alain,
There are several answers to your question.
1. Alumina is not completely transparent and so heats to equilibrium.
2. The run with the dummy unfueled E-Cat takes care of any IR
measurement error.
3. I believe they did use calibrated dots at some point.
Adrian Ashfield
Again how serious this is depends on the temperature difference between the
inner and outer shell no. If that was serious you would expect
the top edge of a picture of the hot cat to have unsharp color shade
because the top edge should represent the heat of the outer shell. I have
not find such an
Not scientific -- but a search of google images for alumina
transmission indicates that you can get pretty much any profile you
want (Include transparent sapphires, of course), and that the actual
profiles vary wildly.
One would thus have to characterize the ceramic actually used, and
then
Yes and the thickness of the alumina and the time constants of heat
transfer dTouter/dt = K(Tinner - Touter) or similare suitable equation.
On Sat, Oct 11, 2014 at 12:44 AM, Alan Fletcher a...@well.com wrote:
Not scientific -- but a search of google images for alumina transmission
indicates
At 03:48 PM 10/10/2014, you wrote:
Yes and the thickness of the alumina and the time constants of
heat transfer dTouter/dt = K(Tinner - Touter) or similare suitable equation.
Fundamentals of Ceramics
Michael Barsoom
About 600 pages.
I found a probably bootleg copy on the web, but you'll have
Jones is right...
If the reactor material is transparent to infrared to any degree, the
remote temperature sensor would be looking at the temperature somewhere
inside the ceramic tube. Since the amount of radiate heat is proportional
to the surface area of the radiating body at the air boundary,
http://digital.csic.es/bitstream/10261/83021/1/Sintering%20to%20transparency.pdf
See page 528
Al2O3 is transparent to mid range infrared between the 2 and 5 micron
wavelengths. That is the operating temperature of the E-Cat.
On Fri, Oct 10, 2014 at 7:34 PM, Axil Axil janap...@gmail.com wrote:
This transparency to infrared photons must be why Rossi uses this ceramic
material to get heat unencumbered to his powder. Rossi is clever.
On Fri, Oct 10, 2014 at 7:55 PM, Axil Axil janap...@gmail.com wrote:
http://digital.csic.es/bitstream/10261/83021/1/Sintering%20to%20transparency.pdf
At 04:34 PM 10/10/2014, Axil Axil wrote:
Jones is right...
Fundamentals of Ceramics
Michael Barsoom
The chapter on optics is mostly concerned with transparent ceramics. But
it does point out that ceramics are mostly transparent, and that they
become opaque by scattering from point sources
At 05:15 PM 10/10/2014, Alan Fletcher wrote:
b) If it were perfectly transparent, then we can treat the outside
of the inner cylinder as the source.
The energy per square can be calculated, but the area is smaller (as r^2)
But what's the emissivity of the inner cylinder? Or can we
The issue of translucency would alter the absolute power calculations but
wouldn't the relative difference between input and output power remain
roughly the same and therefore the COP too?
Harry
On Fri, Oct 10, 2014 at 7:08 PM, Alan Fletcher a...@well.com wrote:
At 03:48 PM 10/10/2014, you
On Fri, Oct 10, 2014 at 7:58 PM, Axil Axil janap...@gmail.com wrote:
This transparency to infrared photons must be why Rossi uses this ceramic
material to get heat unencumbered to his powder. Rossi is clever.
Or maybe it allows more infrared photons to escape unencumbered once the
reactor
At 06:14 PM 10/10/2014, H Veeder wrote:
The issue of translucency would alter the absolute power
calculations but wouldn't the relative difference between input and
output power remain roughly the same and therefore the COP too?
No -- the input power calculation is correct as it is. The
No -- the input power calculation is correct as it is. The output
power -- and hence COP (output/input+output) -- may change.
Ooops COP = (input+output)/input
On Fri, Oct 10, 2014 at 9:35 PM, Alan Fletcher a...@well.com wrote:
At 06:14 PM 10/10/2014, H Veeder wrote:
The issue of translucency would alter the absolute power calculations but
wouldn't the relative difference between input and output power remain
roughly the same and therefore the COP
No, its very laughable. He uses phrases like, well know that. as in, we
should all know this. but... he gives no sources, no numbers, and has
failed to notice that there are DIFFERENT types of sintered alumina, some
of which are DESIGNED to be transparent (sapphire shielding), and some
which
Mistakes happen, NASA crashed a Mars probe because they mixed up metric and
standard measurements.
On Fri, Oct 10, 2014 at 10:19 PM, leaking pen itsat...@gmail.com wrote:
No, its very laughable. He uses phrases like, well know that. as in, we
should all know this. but... he gives no
Rossi would nave used alumina that is transparent to infrared in his
reactor design because he wants the heat from his primary heater that is
imbedded in the alumina to get to the nickel powder. An infrared insulator
is not good reactor design.
On Fri, Oct 10, 2014 at 10:19 PM, leaking pen
At 07:42 PM 10/10/2014, you wrote:
Rossi would nave used alumina that is transparent to infrared in his
reactor design because he wants the heat from his primary heater
that is imbedded in the alumina to get to the nickel powder. An
infrared insulator is not a good reactor design.
The report
the alumina is outside the resistors and the reactor.
On Fri, Oct 10, 2014 at 7:42 PM, Axil Axil janap...@gmail.com wrote:
Rossi would nave used alumina that is transparent to infrared in his
reactor design because he wants the heat from his primary heater that is
imbedded in the alumina to
I discount Goat's hypothesis for the following reasons:
As shown in figure 10 they split the reactor IR camera image into 10
segments plus the ends. They record the temperature for each segment. As
shown in the photograph, some segments were incandescent and others were
not. If incandescent
On Fri, Oct 10, 2014 at 2:52 PM, Alan Fletcher a...@well.com wrote:
The shadows of the wires in figs 12 are problematic ... but we don't have
enough information to figure out if they are actually the result of light,
or if they represent zones of different thermal conductivity, as in the
first
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