[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.

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