q=eps*s*(Th^4-Tc^4)*A q=eps*(2*pi*r^2+2*l*pi*r)*s*(Th^4-Tc^4) ; subst(2*pi*r^2+2*l*pi*r, A) q=5.6703*10^-8*eps*(2*pi*r^2+2*l*pi*r)*(Th^4-Tc^4) ; subst(5.6703e-8, s) q=5.6703*10^-8*eps*(0.11*l*pi+0.00605*pi)*(Th^4-Tc^4) ; subst(.055, r) q=2.40137205*10^-9*eps*pi*(Th^4-Tc^4) ; subst(.33, l) q=2.40137205*10^-9*pi*(Th^4-Tc^4) ; subst(1, eps) 360=2.40137205*10^-9*pi*(Th^4-Tc^4) ; subst(360, q) Th=(21437744309550/pi+997533314063)^(1/4)/143^(1/4) ; solve(Th) Th=483.6006 Kelvin Th=210.451 Celsius
using: http://www.ajdesigner.com/phpwien/wien_equation.php peak emission wavelength (λmax) = 5.9920696955297E-6 meter or 6 micrometers That is with no losses other than black body radiation (ie: no convective losses). That is way into the infrared. The excursions into the visible wavelength occurred with 360W. So, what On Wed, May 22, 2013 at 4:19 PM, Jed Rothwell <[email protected]> wrote: > James Bowery <[email protected]> wrote: > > >> There is value in pursuing reductio ad absurda when they engage one of >> the strongest arguments that the demonstration is valid: >> >> That the power input could not conceivably have produced the radiation >> wavelengths observed. >> > > You have mentioned that several times. Can you please post a more detailed > discussion of that, with equations and examples? That would be helpful. > Please post this in a new thread so I can find it easily. > > You might also address the fact that the first device melted. > > - Jed > >
