# Irradiance due to a sphere and circular surface light sources

Hi Andrei,

While it may seem confusing, it is the projected solid angle that matters to irradiance, and this corresponds to the visible disc of the light source rather than the total surface area, which is why you get the same answer for a ring as a sphere light source in Radiance. If you think about it, the emitted radiance of a diffuse source does not change with position or inclination on the emitter, so all that matters is the shape as visible from the point on the surface where you are calculating irradiance. In either case, that point sees a disk off in the distance -- it can't tell if it's a circle or a sphere. The light that arrives from either is the same.

What *is* different is the total light emitted from the two sources. The sphere will emit 4 times as much light because it will have the same effective brightness in all directions, where the total light from the ring will fall off as the cosine of the angle to its normal, and will be zero for the back-side.

Cheers,
-Greg

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From: "Kolomenski, Andrei (JSC-SF311)[WYLE LABORATORIES, INC.]" <[email protected]>
Subject: [Radiance-general] Irradiance due to a sphere and circular surface light sources
Date: January 6, 2016 8:05:59 AM PST

I need help understanding the rtrace/rpict irradiance output for a sphere light source. The sphere has a radius of 1 mm. and it is projecting light onto a planar surface, 1 m. away. For this geometry, a point source approximation can be used to represent the light source.

The irradiance falling onto the planar surface is governed by the analytical relation:

E(theta) = I(theta=0)*cos(theta)*(1/d^2).

I(theta=0) is the radiant intensity falling perpendicular to the surface. I(theta=0) = L*A, where L=10, is the radiance of the light source and A=4*pi*r^2, is the surface area of the sphere light source. Plugging the numbers into the irradiance formula for sphere at theta=0 (normal projection) yields:

E_sphere(theta=0) = (L * A * 1)/(1) = 10 * 4 * pi * (0.01)^2 = 0.012566 W/m^2

I ran Radiance with this geometry, both rpict(-i) and rtrace (-i –I) produced the irradiance of 0.003141 W/m^2 for theta=0. This result is consistent with the analytical solution if a circle surface is used for the light source, instead of a sphere.

E_circle(theta=0) = (L * A * 1)/(1) = 10 * pi * (0.01)^2 = 0.003141 W/m^2

I changed my Radiance simulation to use a circle (ring with no inner radius) for the light source. For this case, rpict & rtrace both produced that same answer (0.003141), equal to the case of a sphere light source..

Why is the irradiance the same for both light source geometries? The sphere emanates light in all directions (4pi sr.), unlike the circular surface which emanates light into a hemisphere (2pi sr.). Since both light geometries were given the same radiance definition (Lr=10, Lg=10, Lb=10), why do they yield the same result in Radiance? While their analytical solutions are different, by a factor of 4.

I did not use the brighdata primitive to assign a beam pattern to the sphere or circle light sources. I’m assuming a Lambertian light distribution is default, if no beam pattern is provided. Anyway, I would really appreciate your explanation on why these two different geometries produce the same irradiance.

Thank you very much for your insight,

Andrei Kolomenski