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