[argyllcms] Re: Absolute light meter calibration?

  • From: Marwan Daar <marwan.daar@xxxxxxxxx>
  • To: argyllcms@xxxxxxxxxxxxx
  • Date: Wed, 12 Aug 2015 13:25:47 -0400

This isn't a strong suit of mine, but wouldn't you need to know the
emissivity of the filament before making these calculations? If the
emissivity is 1.0 (i.e. perfect black body), then I think you can assume
that all the power "fed" into the filament is expressed as radiant flux.
But for real objects, the emissivity is lower than 1.0, which means that
not all the absorbed power is expressed as radiant flux. At least this is
my naive understanding.


On Wed, Aug 12, 2015 at 4:46 AM, Richard Kirk <richard@xxxxxxxxxxxxxxxx>

On 08/11/2015 10:47 PM, Ryan wrote:

They are likely to be independent. Temperature will be a result of power
by energy dissipation rate. The latter will depend on a light sources
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.

You can simplify the calculation a bit. The filament is very hot and very
small, so it is radiating a large amount of energy, and getting very little
back. If you had potential measuring wires going to the filament supports,
you could measure the voltage drop across the filament a bit more
accurately, but to a reasonable approximation, you could assume the
radiating power of the filament = current * voltage across the whole bulb.

What happens to the energy one it has left the filament is less
interesting. We can assume the visible wavelengths get out through the
envelope, while the infra-red is probably re-absorbed and re-emitted at 2
microns & below, but unless you are using a detector that is sensitive to
this, you can ignore it.

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.

You can calculate the total luminous flux from the power. It is harder to
get a realistic measurement of the intensity because the emission is not
uniform. If there was a single filament and the bulb was unfrosted, then
the intensity at any point would depend on the solid angle subtended by the
filament. Most sensible light bulbs have several filaments and a frosted
envelope to make the lighting more uniform, but how much is 'more uniform'?
You would have to measure the light from all directions including up from
the base to be sure of that. That may not seem a significant source of
error, but if we are wanting to get the value to a percent, it may be

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.

I have not found this to be significant. The power radiated goes as the
4th power of the temperature, so the hottest bits of the filament will
dominate the average. The filaments are usually very uniform, because if
you have a thin bit then that will get hotter, which means the resistance
goes up, so it converts more electrical power into heat, and the light-bulb
goes pop. There is some fascinating physics as it goes pop: the hottest bit
of the filament goes directly to a plasma at metal densities. This plasma
is too dense to conduct, so potential builds up until it has spread out
enough for electron avalanching, whereupon we get the pop and a brief burst
of thermal X-rays. But this is by the way: a good bulb will have a uniform
filament, and the thermal losses to the supports at the end can probably be

The best thing I can suggest is to find a bulb with a single coil filament
that goes up and down. You can measure the light flux using your instrument
from various angles up from the bulb's equator, and assume by symmetry that
the flux downward is about the same. This should give you an absolute
measurement, or about the nearest you will get to one with ordinary

Richard Kirk

Other related posts: