CS488 - Introduction to Computer Graphics - Lecture 19

Ray Tracing

The idea is simple:

for each pixel
  calculate the eye/pixel ray
  project the ray into the scene
  find the first intersection
  apply illumination and shading calculations
  colour the pixel with the result

Concept

Until now rendering has

started with the graphics primitive as it is found in the Scene DAG, then
looked back towards illuminants
looked forward towards the eye

Ray tracing is different. It

starts at the eye,
looks backward to the graphics primitive
then look further backward to the source of illumination

Benefit

Until now we looped over graphics primitives

ray tracing loops over the pixels
which is bounded above, unlike scene complexity.

Drawback

Loop over primitives is inside the loop over pixels

Ingredients of Ray Tracing

Eye-pixel ray

E + t*(P - E)

Object in scene

Set of points {Q} satisfying

f(Q) = 0
- Usually f(Q) is a distance function from the surface.
- f(Q) > 0 means outside,
- f(Q) < 0 means inside
This is the surface of an object. (How do you know?)
Why don't we care about inside points?

Solve the equation

f(E + t*(P-E)) = 0 for t.
For example, all points, Q, on the surface of a sphere, centred at C, of radius r, satisfies |Q-C| = r
Thus, we solve | E - C + t*(P-E) | = r
- which amounts to (E - C + t*(P-E)).(E - C + t*(P-E)) = r^2
- This is the quadratic equation
  - t^2 | P-E |^2 + 2t (E-C).(P-E) + | E-C |^2 - r^2 = 0
  which has zero, one or two solutions.
Most of the time there is no solution
- Do you remember the disciminant, D?
- D = ((E-C) \dot (E-P))^2 - |P-E|^2 (r^2 - |E-C|^2)
- Exercise for the reader. Work it out.

The surface might be defined by an algorithm, then use a root-finding method

For example, float f(E + t*(P - E)) returns distance of point from surface
Submit the function to your favourite root-finder

Object is planar, e.g. a triangle

Equation of the plane is \sum_i ai*xi = b. Where do the parameters come from?

Normal vector is n = (a0, a1, a2, 0)
Take the cross product of any two edges n = (Vn - Vn+1) x (Vm - Vm+1). Now we know the ai.
Substitute and vertex Vn = (v0, v1, v2)
b = \sum ai*vi

Substitute eye ray into the plane equation and solve for t, which is just finding the solution of a linear equation

This point is on the plane: is it inside the polygon?
Triangle: calculate barycentric coordinates; if < 0 outside.
Convex polygon: do n half-plane tests
Either: determine coordinates of the intersection point(s) in the plane of the polygon
- Clip the point against the polygon edges

Another idea: intersect in object coordinates

In object coordinates each point is given by the object definition.
In world coordinates each point in object coordinates is Q = MQ', where Q' is the point in object coordinates
- M is an affine transformation, therefore invertible
  f'(Q') is the object definition in object coordinates
Use invM to transform E,P to E',P', then solve the equation for t
Find intersection point in world coordinates as E + t*(P-E)

Shading for Ray Tracing

When we have found the intersection point we need to calculate a colour.

We know

The eye ray, V
Surface properties
The normal vector
- from a planar primitive
- by interpolation from vertex normals
- by calculation from equation of surface

For each light

Calculate illumination: RGB, l
Accumulate outgoing light

How about shadows?

In principle,

Do an intersection test from the light
- in the direction of the eye ray intersection
If it hits the eye ray intersection, then
- intersection illuminated by that light
else
- intersection in shadow from that light