[argyllcms] Re: Absolute light meter calibration?
- From: "Ryan" <uryan_@xxxxxxxxxxx>
- To: <argyllcms@xxxxxxxxxxxxx>
- Date: Wed, 12 Aug 2015 07:47:01 +1000
They are likely to be independent. Temperature will be a result of power
divided
by energy dissipation rate. The latter will depend on a light sources
geometry,
materials and the ambient conditions.
I don't think that's quite right. Once the bulb has reached steady state
(i.e. warmed up) the energy dissipation rate will equal the power input. The
steady state temperature will be roughly a function of the thermal
resistance of the bulb, bulb geometry, power input and ambient conditions.
The same applies to the filament itself which will have exactly the same
energy dissipation rate as the bulb as a whole, equal to the input power.
But its much smaller geometry will mean it has a much hotter steady state.
Anyway, since input power (and thus output power) is trivial to measure
electrically and you have an estimated spectral spread, you should be able
calculate from that the intensity of the visual portion.
A spanner in the works may be that the filament will not have a uniform
temperature, being much cooler at the ends, and near the supports which act
as heatsinks. I do not know if this would have a significant effect on the
overall result.
Other related posts:
