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The Lambertian BRDF is defined as $R / \pi$, where the division by $\pi$ is there to maintain conservation of energy. The derivation is provided here.

The maths makes sense, and I can understand where the BRDF comes from in a mechanical sense. But I'm struggling to understand what it represents; how does division by $\pi$ relate to reflectance within the unit hemisphere?

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A perfect Lambert reflector actually reflects light in a cosine distribution - that is, the amount of light per unit area reflected in any given direction $R$ is proportional to $N.R$. The reason the radiance appears constant for all angles is that as the view direction moves away from the normal, the reflected light per unit area decreases, but the surface area per projected beam area (and hence per pixel) increases by the same amount - the cosine factors cancel out.

Given this distribution, to be energy conserving, the probability density for the reflected light in all possible directions cannot sum to more than 1. If you integrate the cosine function over the whole hemisphere around the normal, you get $\pi$, so this is the normalization factor you need.

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  • $\begingroup$ Would it be reasonable to say that the cosine distribution integrates to $\pi$ because $cos$ is a one-dimensional function? So integrating it over a hemispherical domain gives you the solid angle consumed by the disk passing through the equator? $\endgroup$ – Paul Ferris Jan 25 '18 at 7:15
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    $\begingroup$ Yeah that's an interesting way of looking at it. Intuitively, any infinitesimal patch of the hemisphere's surface will have its projected area onto the equatorial disk scaled by N.L, so the sum of the projected areas is just the area of the disk. Spherical calculus is not really my strong suit though ;) $\endgroup$ – russ Jan 25 '18 at 10:02
  • $\begingroup$ I guess that's understandable :P. Thanks for the answers. $\endgroup$ – Paul Ferris Jan 25 '18 at 19:39

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