Andrew_M

Well it would seem nobody here knows the answer to this problem. I will try to find other explanatory texts on luminance measurement and calculation. I will also check if the difference in magnitude between my answer and the Wikipedia figure is the same as the difference in magnitude between the area subtended by the pupil and the area subtended by the sun. Perhaps someone has used the wrong solid angle in the calculation, or the wrong area.

Hello everyone, I have been reading about photometric optics and photometric units recently. I keep seeing several documents saying that the luminance of the sun is 1.6×109 cd/m2 at noon, eg Wikipedia. I tried to use the definition of luminance and some facts about the sun to see if I could calculate what the luminance should appear to be to someone on Earth, just to see if I got the same number as in Wikipedia. Well not only did I not get the same number, but my answer was off by a factor of 1030 too big! I am puzzled because I think I am using the right formulae and all the correct numbers, but still it does not match, so I am wondering where I have gone wrong. So generally the formula is Lum = Flux / SolidAngle / Area Wikipedia includes a cosine theta term there too, but in this case we are looking at the sun and the sun is looking at us, we're all in the same orbital plane. The Flux will be half of the total flux of the sun, because we can only see one half at a time. From WP, this is 3.75×1028 / 2 lumens. According to the setup for a luminance photometer, the Solid Angle is the angle subtended by the receiving area as seen by a point on the source. Since the receiving area is going to be the pupil of an observer's eye (hopefully he is wearing powerful sunglasses) then the solid angle can be found by formula for solid angle of the base of a cone with full angle theta: sr = 2*pi*(1 - cos(theta)) I'll assume the pupil radius is 1mm. The distance of 1AU is 1.496×108 km or 1.496×1011 m sr = 2*pi*(1 - cos( 2*arctan(pupil radius / 1 AU) )) = (2π×(1−cos(2× tan⁻¹(0.001÷(1.496×10¹¹))))) This is going to be rather a small solid angle, obviously. The Area is the area of the source being measured, which here is half the total area of the sun, so 3×1012 km2 which is 3×1018 m2 So putting it altogether... Lumsun = (1.8×10²⁸) ÷ (2π×(1−cos(2× tan⁻¹(0.001÷(1.496×10¹¹)) ))) ÷ (3×10¹⁸) = 1.068573927×10³⁷ Hah, quite a bit different to 1.6x109, so where did I go wrong?