# Ray Tracing Ray Casting

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```					Ray Tracing

G. Drew Kessler   1
An Ideal Ray Tracer
    Cast light rays from all light sources.
    Follow them to the eye point.
 Reflect off of solid objects.

 Pass through transparent objects (refraction).

 Obtain color by light absorption.

 Diffuse reflection/refraction multiplies rays.

    For rays that “hit” the eye point, intersect with window
on view plane, which corresponds to a screen pixel.
    Add the light ray’s color intensities to the pixel.
    Problem: It could take forever to draw an image!
G. Drew Kessler, Larry F. Hodges                                      2
Ray Tracing That Works
    Observation: We are only interested in rays that cross
the eye point.
 Idea: Start at eye point and trace backwards.

Light Source

Eye
Object
Eye Ray

Object
Light Source
G. Drew Kessler, Larry F. Hodges                                           3
Visible Surface Determination
    Ray tracing can be used as a visible surface
determination algorithm
 Often called ray casting.

 (A ray is a uni-directional line segment).

    The algorithm:
 For each pixel

   Determine the world coordinates of the pixel’s center.
   Cast a ray from the eye through the pixel’s center.
   Check for intersection of this ray with any object.
   For closest intersection, determine color of object at that point.

G. Drew Kessler, Larry F. Hodges                                                           4
Color Determination
    To determine the color at the intersection of an eye ray
with an object, we can:
 Simply assign the color of the object (no shading).

 Determine the shading of the object.

 Calculate the contribution of a texture at that point.

 Some combination of the above

 Cast ray from intersection point to each light source.

not intersect an object in the scene.
G. Drew Kessler, Larry F. Hodges                                           5
Intersections
    Ray tracing algorithms are dominated by intersection
testing and determination
 Need efficient methods to test for intersection

   Does the eye ray or shadow ray intersect an object?
   Need efficient method to calculate intersection points
and normal vectors (for lighting calculations).

    Ray/Sphere
    Ray/Plane
    Ray/Polygon

G. Drew Kessler, Larry F. Hodges                                            6
Ray/Sphere Intersection
(Algebraic Solution)
Ray is defined by R(t) = Ro + Rd*t where t > 0.

Ro = Origin of ray at (xo, yo, zo)
Rd = Direction of ray [xd, yd, zd] (unit vector)

Sphere's surface is defined by the set of points {(xs, ys, zs)}
satisfying the equation:

(xs - xc)2 + (ys - yc)2 + (zs - zc)2 - rs2 = 0

Center of sphere: (xc, yc, zc)
G. Drew Kessler, Larry F. Hodges                                        7
Possible Cases of Ray/Sphere
Intersection
1. Ray intersects
1                                sphere twice with t>0
2
2. Ray tangent to
sphere
3. Ray intersects
sphere with t<0

3                                    4. Ray originates
4                   inside sphere
5   5. Ray does not
intersect sphere
G. Drew Kessler, Larry F. Hodges                                   8
Solving For t
Substitute the basic ray equation:
x = xo + xd*t
y = yo + yd*t
z = zo + zd*t
into the equation of the sphere:
(x0 + xdt - xc)2 + (y0 + ydt - yc)2 + (z0 + zdt - zc)2 - rs2 = 0
This is a quadratic equation in t: At2 + Bt + C = 0, where
A = xd2 + yd2 + zd2
B = 2[xd(x0 - xc) + yd(y0 - yc) + zd(z0 - zc)]
C = (x0 - xc)2 + (y0 - yc)2 + (z0 - zc)2 - rs2
Note: A=1
G. Drew Kessler, Larry F. Hodges                                                     9
Relation of t to Intersection
We want the smallest positive t - call it ti

t0
t1                       Discriminant = 0
t0

t0

t1

t1

Discriminant < 0

G. Drew Kessler, Larry F. Hodges                                                10
Actual Intersection
Intersection point, (xi, yi, zi) = (xo+xd*ti, yo+yd*ti, zo+zd*ti)
Unit vector normal to the surface at this point is
N = [(xi - xc) / rs, (yi - yc) / rs, (zi - zc) / rs]
If the ray originates inside the sphere, N should be negated
so that it points back toward the center.

N
N

G. Drew Kessler, Larry F. Hodges                                           11
Summary (Algebraic Solution)
1.    Calculate A, B and C of the quadratic
2.    Calculate discriminant (If < 0, then no intersection)
3.    Calculate t0
4.    If t0 < 0, then calculate t1 (If t1 < 0, no intersection point on ray)
5.    Calculate intersection point
6.    Calculate normal vector at point

   Precompute rs2
   Precompute 1/rs
   If computed t is very small then, due to rounding error, you may
not have a valid intersection

G. Drew Kessler, Larry F. Hodges                                                     12
Ray/Sphere Intersection
(Geometric Solution)
In the case of the sphere there are a number of tests that we can make
to determine that there is not an intersection. This approach allows us
to avoid calculations and result in a faster algorithm.

1. Find if the ray's origin is outside the sphere.

Ro
OC             rs
Origin to center vector, OC = C - Ro
C
Length squared of OC = ||OC||2

If ||OC||2 < rs2 then the ray origin is inside the sphere (go to step 4)
Else if ||OC||2 > rs2 then the ray origin is outside the sphere

G. Drew Kessler, Larry F. Hodges                                                  13
R/S Geometric Solution (cont.)
2. Find the length of the ray to the point along the ray
closest to the sphere's center.

L             R(t)

ß            rs
Ro       OC
C

L = ||OC|| cos ß = ||OC|| ||Rd|| cos ß = OC•Rd

If the ray is outside and points away from the sphere
(L < 0), the ray must miss the sphere.

G. Drew Kessler, Larry F. Hodges                                 14
R/S Geometric Solution (cont.)
3. Else find the squared distance from the closest approach
along the ray to the sphere's center to the sphere's
surface (HC2 = half chord distance squared).
HC

R(t)
L

D
rs
D2   =   ||OC||2    -   L2        Ro     OC
C

HC2 = rs2 - D2 = rs2 - ||OC||2 + L2
If HC2 < 0, the ray misses the sphere.

G. Drew Kessler, Larry F. Hodges                                    15
R/S Geometric Solution (cont.)
4. Else find t for the intersection
ti = L - HC for rays originating outside the sphere
ti = L + HC for rays originating inside or on the sphere

5. Calculate intersection point (xi, yi, zi) by plugging ti into
Ro + tRd

6. Calculate normal vector
[(xi - xc)/rs (yi - yc)/rs, (zi - zc)/rs]
(negate if ray originates inside sphere)
G. Drew Kessler, Larry F. Hodges                                         16
Summary (Geometric Solution)
1.    Find distance squared between ray origin and center of sphere,
test for ray outside sphere.
2.    Calculate ray distance which is closest to the center of the sphere.
Test if the ray points away from the sphere.
3.    Find square of half chord intersection distance.
Test if square is negative.
4.    Calculate intersection distance.
5.    Find intersection point.
6.    Calculate normal vector at point.

These techniques are no slower than the algebraic solution and
often allow you to determine that there is no intersection using
fewer calculations.
G. Drew Kessler, Larry F. Hodges                                               17
Ray/Plane Intersection
Ray is defined by R(t) = Ro + Rd*t where t > 0
Ro = Origin of ray at (xo, yo, zo)
Rd = Direction of ray [xd, yd, zd] (unit vector)

Plane is defined by [A, B, C, D]
Ax + By + Cz + D = 0 for a point in the plane
Normal Vector, N = [A, B, C] (unit vector)
A2 + B2 + C2 = 1

G. Drew Kessler, Larry F. Hodges                           18
Ray/Plane (cont.)
Substitute the ray equation into the plane equation:
A(xo + xdt) + B(yo + ydt) + C(zo +zdt) + D = 0

Solve for t:
t = -(Axo + Byo + Czo + D) / (Axd + Byd + Czd)
t = -(N • Ro + D) / (N • Rd)

Note: The normal vector of the plane should usually (except for
backface removal) be for the surface facing the ray. If it isn't then it
should be reversed before illumination is calculated.
If N • Rd > 0 then use N = -[A, B, C]

G. Drew Kessler, Larry F. Hodges                                                 19
What Can Happen?
N • Rd = 0
N • Rd > 0

Ro                Ro

t<0

G. Drew Kessler, Larry F. Hodges              t>0         20
Ray/Plane Summary
Intersection point:
(xi, yi, zi) = (xo + xdti, yo + ydti, zo + zdti)

1.    Calculate N • Rd and compare it to zero.
2.    Calculate ti and compare it to zero.
3.    Compute intersection point.
4.    Compare N • Rd to zero and reverse normal if
appropriate

G. Drew Kessler, Larry F. Hodges                                     21
Ray/Polygon Intersection
A polygon is defined by a set of p points: Gn = {(xn, yn, zn) | n = 0, 1, ..., (p-1)}
The polygon is in a plane, Ax + By + Cz + D = 0, Normal N = [A, B, C]
1.  Find the intersection of the ray with the plane of the polygon.
2.  Throw away the coordinate of each point for which the corresponding
normal component is of the greatest magnitude (dominant coordinate).
3.  You now have a 2-D polygon defined by the set of points
{(un, vn) | n = 0, 1, .., p-1}
4.  Translate the polygon so that the intersection point of the ray with the
plane of the polygon goes to the origin of the u,v system. Call these
translated points
{(u´n, v´n) | n = 0, 1, .., p-1}
5.  Determine whether the origin is within the 2D polygon.

G. Drew Kessler, Larry F. Hodges                                                   22
Ray/Polygon (cont.)
Y
V

V´
X
U

Z

U´

G. Drew Kessler, Larry F. Hodges                     23
Ray/Polygon Algorithm
To determine whether the origin is within a 2D polygon, we only need to
count the number of times the polygon’s edges cross the positive u´-axis
as we walk around the polygon. Odd number crossings: origin is within.
num_crossings = 0;
sign_holder = sign(v´0);
for(a=0; a < p; a++){
b = (a+1)% p;
next_sign_holder = sign(v´b)
if (sign_holder != next_sign_holder) {
if (u´a > 0 and u´b > 0)
num_crossings = num_crossings + 1;
else if (u´a > 0 or u´b > 0){
int = u´a - v´a(u´b - u´a) / (v´b - v´a)
if(int > 0)
num_crossings = num_crossings + 1;
}
sign_holder = next_sign_holder;
}
}
if(num_crossings % 2 == 1) ray intersects polygon

G. Drew Kessler, Larry F. Hodges                                              24

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 views: 8 posted: 8/16/2011 language: English pages: 24