Units for Light, and Radiosity

CS 481 Lecture, Dr. Lawlor

So in reality, everything's a light source. Yes, the sun is a light source. But the sky is also a light source, filling in areas not illuminated by direct sunlight with a soft blue glow. Your pants are a light source, lighting up the ground in front of you with pants-colored light.

Does this matter? Well, does this look like OpenGL?
Soft-edged radiosity image

This is an image computed using "radiosity".

Radiosity Units

The terms we'll be using to talk about light power are:

Power and Energy carried by light (e.g., power received at a 100% efficient photocell)

It's not energy, it's energy per unit time.
Measured in Watts
Also measured in "lumens"-- power as perceived by the human eye. At a wavelength of 555nm, one lumen is 1/680 watt.
AKA flux, radiant flux

Irradiance: light power per unit surface area leaving or arriving at a surface

Measured in watts per square meter [of receiver or transmitter area].
So multiply by the receiver's area to get watts. (Assuming everything's facing dead-on.)
E.g., the Sun delivers an irradiance of 700W/m² to the surface of the Earth.

So a 10 square meter solar panel facing the sun would receive 7000W of power.
So a 1.0e-10 square meter CCD pixel open for 1.0e-3 seconds would receive 7.0e-11 joules of light energy, or a couple hundred million photons (visible light photons have energy of about 4e-19 Joules).

Irradiance as perceived by the human eye can be measured in "lux" (lumens per square meter).

Radiance: "brightness" as seen by a camera or your eye.

Measured in (get this) watts per square meter [of receiver area] per steradian [of source coverage, as measured from the receiver], or just W/m²/sr. (See below for details on steradians)
E.g., from the Earth the Sun is about 1 degree across, or about 1/60 radian, so assuming it's a tiny square it would cover 1/3600 steradians. The radiance of the sun is thus about 700W/m² per 1/3600 steradians, or 2.3 megawatts per square meter per steradian (2.3 MW/m²/sr).

This means if a lens sets it up so from one spot you see the sun at a coverage of 1 steradian, that spot receives an irradiance of 2.3MW/m² of power!

Suprisingly, radiance is unchanged along a straight line through free space.

That is, radiance does *not* depend directly on distance: if the source coverage is unchanged, the radiance is unchanged. That is, the pixel doesn't get brighter just because the geometry is right in front of it!
Suprisingly, radiance is even unchanged when passing through a *lens*. Sunlight focused through a lens has the same radiance, but because the lens changes the Sun's *angle* (and hence solid angle coverage) of the light, the received irradiance is bigger.
Yes, radiance changes when light hits any non-transparent object, like a brick.

Solid angle: "bigness" of light source as seen by a receiver.

Measured in "Steradians" (abbreviated "sr").
A light source's solid angle in steradians is defined as the ratio of the area of the light source when projected onto the surface of a sphere centered on the receiver. This area is then divided by the sphere's radius squared, or equivalently, is always performed with a unit-radius sphere.

See entries in Mathworld or the figure in the Lighting Design Glossary.

Examples:

The maximum coverage is the whole sphere, which has area 4 pi * radius squared, so the maximum coverage is 4 pi steradians.
A hemisphere is half the sphere, so the coverage is 2 pi steradians.
A small square that measures T radians across covers approximately T² steradians. This expression is exact for an infinitesimal square.

A flat receiver surface only receives part of the light from a light source low on its horizon--this is Lambert's cosine illumination law. Hence for a flat receiver, we've got to weight the incoming solid angle by a factor of "cosine(angle to normal)".

For a source that lies directly along the receiver's normal, this doesn't affect the solid angle--it's a scaling by 1.0.
For a source that lies right at the receiver's horizon, this factor totally eliminates the source's contribution--it's a scaling by 0.0.
For a hemisphere, this cosine factor varies across the surface, but it integrates out to a weighted solid angle of just 1 pi steradians (the unweighted solid angle of a hemisphere is 2 pi steradians).

You only want the solid angle for computing illumination from a polygon to a point. For illumination between two polygons, you actually want to compute the "form factor" between the polygons, which you can either approximate using the solid angle, or compute exactly using Schröder and Hanrahan's 1990 paper.

See the Light Measurement Handbook for many more units and nice figures.

To do global illumination, we start with the radiance of each light source (in W/m²/sr). For a given receiver patch, we compute the cosine-weighted solid angle of that light source, in steradians. The incoming radiance times the solid angle gives the incoming irradiance (W/m²). Most surfaces reflect some of the light that falls on them, so some fraction of the incoming irradiance leaves. If the outgoing light is divided equally among all possible directions (i.e., the surface is a diffuse or Lambertian reflector), for a flat surface it's divided over a solid angle of pi (cosine weighted) steradians. This means the outgoing radiance for a flat diffuse surface is just M/pi (W/m²/sr), if the outgoing irradiance (or radiant exitance) is M (W/m²).

To do global illumination for diffuse surfaces, then, for each "patch" of geometry we:

Compute how much light comes in. For each light source:

Compute the cosine-weighted solid angle for that light source.

This depends on the light source's shape and position.
It also depends on what's in the way: what shapes occlude/shadow the light.

Compute the radiance of the light source in the patch's direction.

This might just be a fixed brightness, or might depend on view angle or location.

Multiply the two: incoming irradiance = radiance times weighted solid angle.

Compute how much light goes out. For a diffuse patch, outgoing radiance is just the total incoming irradiance (from all light sources) divided by pi.

For a general surface, the outgoing radiance in a particular direction is the integral, over all incoming directions, of the product of the incoming radiance and the surface's BRDF (Bidirectional Reflectance Distribution Function). For diffuse surfaces, where the surface's radiance-to-radiance BRDF is just the cosine-weighted solid angle over pi, this is computed exactly as above.

Computing Solid Angles

So one important trick in doing this is being able to compute cosine-weighted solid angles of light sources. Luckily, there are several funny tricks for this.

The oldest, best-known, and least accurate way to compute a solid angle is to assume the light source is small and/or far away. If either of these are true, then the light source projects to a little dot on the hemisphere, and we just need to figure out how the area (and hence solid angle) of that dot scales with distance. Well, the height of the dot scales like 1/z, and the width of the dot scales like 1/z, so together the area (width times height) scales like 1/z². That's the old, familiar inverse-square "law" you've heard since middle-school physics class and which, unfortunately, is actually a lie.

First, inverse-square falloff is only valid for isotropic light sources (same in all directions). Non-isotropic sources (think a laser beam, or a flashlight, that shines only in one direction) don't generally fall off this way.

Second, say the light source is an isotropic 1-meter radius disk, sitting z meters away and directly facing you. What's the falloff with z? It turns out to be 1/(z²+1), not 1/z². These are similar for large z, but at small z the disk isn't as bright as an equivalent point source. That is, inverse-square breaks down when you get too close--and it's a good thing too, because you'd reach infinite brightness at z=0!

Solid Angle from Point To Polygon Light Source

Luckily, there's a fairly simple, if bizarre, algorithm for computing the solid angle of a polygon transmitter from any receiver point. This algorithm was presented by Hottel and Sarofin in 1967 (for heat transfer engineering, not computer graphics!), and again in a 1992 Graphics Gem by Filippo Tampieri.

vec3 light_vector=vec3(0.0);
for (int i=0;i<light.size();i++) { // Loop over vertices of light source
    vec3 Ri=normalize(receiver-light[i]); // Points from light source vertex to receiver
    vec3 Rp=normalize(receiver-light[i+1]); // Points from light source's *next* vertex to receiver
    light_vector += acos(dot(Ri,Rp)) * normalize(cross(Ri,Rp));
}
float solid_angle = dot(receiver_normal,light_vector);

You can actually just dump this whole thing directly into GLSL, and it will run. Two years ago, the "acos" kicked it out into software rendering, which was 1000x slower. Today, on the *same* hardware, the *same* GLSL program runs in graphics hardware, and is pretty dang fast too--a sincere thank you goes out to the ATI GLSL driver writers!

You can speed it up a bit (and mess it up too at close range!) by first noting that

    light_vector += acos(dot(Ri,Rp)) * normalize(cross(Ri,Rp))

is an angle-weighted, normalized cross product.

In theory we could compute the angle weighting just as well with

    light_vector += asin(length(cross(Ri,Rp))) * normalize(cross(Ri,Rp))

This isn't actually useful in practice, because it breaks down for nearby receivers--the angle between Ri and Rp exceeds 90 degrees, and asin starts giving the wrong answer, while acos keeps working all the way out to 180.

However, for distant receivers, which have small angles between Ri and Rp, we're close to having asin x == x. Thus the above is very nearly equal to:

    light_vector += length(cross(Ri,Rp))) * normalize(cross(Ri,Rp))

Now, the length of a vector times the direction of the vector gives the original vector, so we can actually just sum up the cross products:

    light_vector+=cross(Ri,Rp);

In practice, this is very difficult to distinguish visually from the real equation, except very close to the light source, where this approximation comes out a bit darker than the real thing.

Solid Angle's Achilles Heel--Occlusion

The above equation works great if the receiver can see the whole light source. If something gets in the way of that light source, you still compute solid angle, but only where the light source actually shines on the receiver--a much more complicated computation to perform.

It is actually possible to clip away the occluded parts of the light source, and then compute the solid angle of the non-occluded pieces. I've done this in software. It's actually tolerably fast as long as you're willing to limit yourself to a few dozen polygons per scene. But with any realistic number of polygons, computing the exact solid angle becomes very painfully slow. This is where approximations come in, which we'll cover next class.