Re: Surprise! (TDC image brightness)

[P3D John W Roberts <roberts@xxxxxxxxxxxxxxxxx>]

>Date: Thu, 19 Dec 1996 00:19:36 -0600
>From: "P3D  Dr. George A. Themelis"  <DrT-3d@xxxxxxxxxxxxxxxxxxxxxxxxxxx>
>Subject: Surprise! (TDC image brightness)

>Well, I did my measurements and concluded that the 750W bulbs
>are *not* brighter than the 500W bulbs in my TDC 116 projector....

>...At the end I do not know why the two sets of bulbs are giving
>the same light output even though one is rated higher than the
>other...  This being the case, it is better to use the lower
>wattage bulbs which are less demanding on the projector, slide
>and electric bill!

Reflector and lens issues aside, I think the point about effective surface
area of the filaments could be the reason. [Disclaimer: I've never seen
either of these bulbs.]

For a given wattage, a filament with a larger surface area will be able to
radiate more effectively, and will therefore have a lower surface temperature.
As a result, a larger percentage of its output would be in the infrared, and
therefore it would put out less visible light, and what light was put out
would be less white and more red. Even if two bulbs have different wattage,
if the higher wattage one has much greater effective filament surface area,
it could still put out less visible and redder light. [Note that the previous
post commented on the very white color of the lower-wattage light.]
Consider an extreme example - an electric clothes iron can be regarded as
an incandescent lamp with an enormous "filament" surface area - it puts
out plenty of "thermal" infrared and some "near infrared" [I've seen a
photograph taken with IR film and an iron as a "light" source], but
no visible light.

I have a 500W 120V halogen light which puts out light that's much redder
than some lower-wattage (and lower voltage) halogen lights I have, which
adds plausibility to the line of reasoning.

I don't know how to figure "effective surface area" - a tightly coiled
filament obviously interferes with itself in terms of radiation, so
[pi * filament diameter * total filament length] would give a very bad
approximation.

John R


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