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 translucidity when considering the radiation of heat...
- and maybe that one effect could came from changing resistors that are more or less hidden "optically"...
I propose a kind of group work,Â
I propose that people with competence, analyse goagguys arguments, and the report.
1- can someone explain first the point of goatguy on the fact that alumina is transparent...
is it noticeable ? does it change the way radiation equation are computed or is it simply emissivity change ?
what can be the order of size of the error induced ?
I did a bit of research. eg http://en.wikipedia.org/wiki/Transparent_ceramics
a) It CAN be made completely transparent to visible light
a) The kind used in the hotcat is most likely opaque to visible light
> Most ceramic materials, such as alumina and its compounds, are formed from fine powders, yielding a fine grained polycrystalline microstructure which is filled with scattering centers comparable to the wavelength of visible light.
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 independent test (which had a steel outer cylinder).
But it's proably transparent to IR , and if so I believe (without proof ... but see Jones Beene's http://www.mail-archive.com/[email protected]/msg98226.html http://www.mail-archive.com/[email protected]/msg98253.html ) that it DOES affect the power calculation.
Right now I'm changing my position from "positive" to "inconclusive". I have another post ready to send.
