# What is the apparent diameter of the sun as seen from earth?

I know this may not be physically accurate.

For my purposes, basically 3D renders, I am assuming the sun rays are parallel rays from an infinitely far lamp. If the sky is clear, what size would that lamp be?

Or in other words, how big will the sun appear if we assume it is a lamp? 1cm? 5cm?

Is there a formula for converting astronomical arc minutes / arc seconds to linear dimensions?

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"Is there a formula for converting astronomical arc minutes / arc seconds to linear dimensions?" Sure: for small angles (a few degrees or less, depends on you precision needs) convert to radians and multiple by the distance. For larger angles or for maximum precision you do a little trigonometry: $r \tan \theta$ where $r$ is the distance. –  dmckee Jan 10 '13 at 13:51
In addition to Rody Oldenhuis's answer (the first sentence of which is the most telling), I suggest you ask around in the rendering/graphics community for some succinct guide to angles and angular sizes - one must certainly exist. Often infinite distances and parallel rays are misused in this way. For an analogy, you are basically asking, "How do I convert between current and voltage if the resistance of the electrical circuit is infinite?" –  Chris White Jan 11 '13 at 23:27
as an Electrical Engineer myself, I can now relate to your analogy :-) –  Ayman Jan 13 '13 at 5:31

If you assume it is a lamp that is infinitely far away, the lamp size will also have to be infinitely large :)

What I think you want to do is let the actual light come from infinity, but place the lamp 1AU (=~150,000,000 km) away, with a radius of 695,500 km (actual values).

If this is not possible or impractical in your software for some reason, just place a lamp closer to the observer while preserving the Sun's angular diameter. So, for instance,

$\ \ \ \ \frac{\alpha}{2} = \arctan{ \left( \frac{695,500}{150,000,000} \right) } \approx 0.004633 \mathrm{\ radians}$

$\ \ \ \ \ \mathrm{new\ radius} = D \cdot \tan \left( \frac{\alpha}{2} \right) = D \cdot \frac{695 \frac12}{150,000}$

where $D$ is your desired distance in km.

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