1. CS G140 Graduate Computer Graphics
Prof. Harriet Fell
Spring 2006
Lecture 2 - January 23, 2006
April 27, 2019

2. Today’s Topics Ray Tracing More Math Ray-Sphere Intersection
Light: Diffuse Reflection
Shadows
Phong Shading
More Math
Matrices
Transformations
Homogeneous Coordinates
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3. Ray Tracing a World of Spheres
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4. What is a Sphere Vector3D center; // 3 doubles double radius;
double R, G, B; // for RGB colors between 0 and 1
double kd; // diffuse coeficient
double ks; // specular coefficient
int specExp; // specular exponent 0 if ks = 0
(double ka; // ambient light coefficient)
double kgr; // global reflection coefficient
double kt; // transmitting coefficient
int pic; // > 0 if picture texture is used
April 27, 2019

5. -.01 .01 500 800 // transform theta phi mu distance 1 // antialias
1 // numlights
// Lx, Ly, Lz
9 // numspheres
//cx cy cz radius R G B ka kd ks specExp kgr kt pic
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6. World of Spheres Vector3D VP; // the viewpoint int numLights;
Vector3D theLights[5]; // up to 5 white lights
double ka; // ambient light coefficient
int numSpheres;
Sphere theSpheres[20]; // 20 sphere max
int ppmT[3]; // ppm texture files
View sceneView; // transform data
double distance; // view plane to VP
bool antialias; // if true antialias
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7. Simple Ray Tracing for Detecting Visible Surfaces
select window on viewplane and center of projection
for (each scanline in image) {
for (each pixel in the scanline) {
determine ray from center of projection
through pixel;
for (each object in scene) {
if (object is intersected and
is closest considered thus far)
record intersection and object name; }
set pixel’s color to that of closest object intersected;
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8. Ray Trace 1 Finding Visible Surfaces
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9. Ray-Sphere Intersection
Given
Sphere
Center (cx, cy, cz)
Radius, R
Ray from P0 to P1
P0 = (x0, y0, z0) and P1 = (x1, y1, z1)
View Point
(Vx, Vy, Vz)
Project to window from (0,0,0) to (w,h,0)
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10. Sphere Equation y Center C = (cx, cy, cz) Radius R (x, y, z) x
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11. Ray Equation P0 = (x0, y0, z0) and P1 = (x1, y1, z1)
The ray from P0 to P1 is given by:
P(t) = (1 - t)P0 + tP1 0 <= t <= 1
= P0 + t(P1 - P0) P0 P1
t 
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12. Intersection Equation
P(t) = P0 + t(P1 - P0) 0 <= t <= 1
is really three equations
x(t) = x0 + t(x1 - x0)
y(t) = y0 + t(y1 - y0)
z(t) = z0 + t(z1 - z0) 0 <= t <= 1
Substitute x(t), y(t), and z(t) for x, y, z, respectively in
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13. Solving the Intersection Equation
is a quadratic equation in variable t.
For a fixed pixel, VP, and sphere,
x0, y0, z0, x1, y1, z1, cx, cy, cz, and R
eye pixel sphere
are all constants.
We solve for t using the quadratic formula.
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14. The Quadratic Coefficients
Set dx = x1 - x0
dy = y1 - y0
dz = z1 - z0
Now find the the coefficients:
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15. Computing Coefficients
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16. The Coefficients
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17. Solving the Equation
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18. The intersection nearest P0 is given by:
To find the coordinates of the intersection point:
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19. First Lighting Model Ambient light is a global constant.
Ambient Light = ka (AR, AG, AB)
ka is in the “World of Spheres”
0  ka  1
(AR, AG, AB) = average of the light sources
(AR, AG, AB) = (1, 1, 1) for white light
Color of object S = (SR, SG, SB)
Visible Color of an object S with only ambient light
CS= ka (AR SR, AG SG, AB SB)
For white light
CS= ka (SR, SG, SB)
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20. Visible Surfaces Ambient Light
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21. Second Lighting Model
Point source light L = (LR, LG, LB) at (Lx, Ly, Lz)
Ambient light is also present.
Color at point p on an object S with ambient & diffuse reflection
Cp= ka (AR SR, AG SG, AB SB)+ kd kp(LR SR, LG SG, LB SB)
For white light, L = (1, 1, 1)
Cp= ka (SR, SG, SB) + kd kp(SR, SG, SB)
kp depends on the point p on the object and (Lx, Ly, Lz)
kd depends on the object (sphere)
ka is global
ka + kd  1
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22. Diffuse Light
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23. Lambertian Reflection Model Diffuse Shading
For matte (non-shiny) objects
Examples
Matte paper, newsprint
Unpolished wood
Unpolished stones
Color at a point on a matte object does not change with viewpoint.
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24. Physics of Lambertian Reflection
Incoming light is partially absorbed and partially transmitted equally in all directions
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25. Geometry of Lambert’s LawN L  L
90 -  dA dA 
dAcos()
90 - 
Surface 1
Surface 2
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26. cos()=NL Cp= ka (SR, SG, SB) + kd NL (SR, SG, SB) Surface 2 NdA
90 -  
dAcos() L N
Cp= ka (SR, SG, SB) + kd NL (SR, SG, SB)
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27. Finding N N = (x-cx, y-cy, z-cz) |(x-cx, y-cy, z-cz)| normal (x, y, z)
radius
(cx, cy, cz)
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28. Diffuse Light 2
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29. Time for a Break
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30. Shadows on Spheres
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31. More Shadows
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32. Finding Shadows
Shadow Ray P
Pixel gets shadow color
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33. Shadow Color
Given
Ray from P (point on sphere S) to L (light)
P= P0 = (x0, y0, z0) and L = P1 = (x1, y1, z1)
Find out whether the ray intersects any other object (sphere).
If it does, P is in shadow.
Use only ambient light for pixel.
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34. Shape of Shadows
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35. Different Views
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36. Planets
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37. Starry Skies
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38. Shadows on the Plane
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39. Finding Shadows on the Back Plane
Shadow Ray P
Pixel in Shadow
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40. Close up
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41. On the Table
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42. Phong Highlight
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43. Phong Lighting Model Light The viewer only sees the light when  is 0.
Normal
Reflected
View
The viewer only sees the light when  is 0. N L R V   
We make the highlight maximal when  is 0, but have it fade off gradually.
Surface
April 27, 2019

44. Cp= ka (SR, SG, SB) + kd NL (SR, SG, SB) + ks (RV)n(1, 1, 1)
Phong Lighting Model
cos() = RV
We use cosn().
The higher n is, the faster the drop off. N L R V
For white light   
Surface
Cp= ka (SR, SG, SB) + kd NL (SR, SG, SB) + ks (RV)n(1, 1, 1)
April 27, 2019

45. Powers of cos()
cos10()
cos20()
cos40()
cos80()
April 27, 2019

46. Computing R L + R = (2 LN) N R = (2 LN) N - L LN LN R L N L+R R L 
April 27, 2019

47. Cp= ka (SR, SG, SB) + kd NL (SR, SG, SB) + ks (HN)n (1, 1, 1)
The Halfway Vector
From the picture
+  =  -  + 
So  = /2.
H = L+ V
|L+ V|
Use HN
instead of RV.
H is less expensive to compute than R. N H
This is not generally true. Why? L R  V   
Surface
Cp= ka (SR, SG, SB) + kd NL (SR, SG, SB) + ks (HN)n (1, 1, 1)
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48. Varied Phong Highlights
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49. Varying Reflectivity
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50. More Math Matrices Transformations Homogeneous Coordinates
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51. Matrices We use 2x2, 3x3, and 4x4 matrices in computer graphics.
We’ll start with a review of 2D matrices and transformations.
April 27, 2019

52. Basic 2D Linear Transforms
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53. Scale by .5
(1, 0)
(0.5, 0)
(0, 1)
(0, 0.5)
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54. Scaling by .5 y x y x
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55. General Scaling y x y x
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56. General Scaling y x y x sx 1 sy
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57. Rotation 
sin()
cos()
-sin() 
cos()
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58. Rotation y x y x 
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59. Reflection in y-axis
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60. Reflection in y-axis y x y x
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61. Reflection in x-axis
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62. Reflection in x-axis y y x x
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63. Shear-x s
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64. Shear x y x y x
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65. Shear-y s
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66. Shear y y x y x
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67. Linear Transformations
Scale, Reflection, Rotation, and Shear are all linear transformations
They satisfy: T(au + bv) = aT(u) + bT(v)
u and v are vectors
a and b are scalars
If T is a linear transformation
T((0, 0)) = (0, 0)
April 27, 2019

68. Composing Linear Transformations
If T1 and T2 are transformations
T2 T1(v) =def T2( T1(v))
If T1 and T2 are linear and are represented by matrices M1 and M2
T2 T1 is represented by M2 M1
T2 T1(v) = T2( T1(v)) = (M2 M1)(v)
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69. Reflection About an Arbitrary Line (through the origin)x y x
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70. Reflection as a Compositiony x
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71. Decomposing Linear Transformations
Any 2D Linear Transformation can be decomposed into the product of a rotation, a scale, and a rotation if the scale can have negative numbers.
M = R1SR2
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72. Rotation about an Arbitrary Pointx  y x 
This is not a linear transformation. The origin moves.
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73. Translation (x, y)(x+a,y+b) (a, b)
This is not a linear transformation. The origin moves.
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74. Homogeneous Coordinatesy y
Embed the xy-plane in R3 at z = 1.
(x, y)  (x, y, 1) x x z
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75. 2D Linear Transformations as 3D Matrices
Any 2D linear transformation can be represented by a 2x2 matrix
or a 3x3 matrix
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76. 2D Linear Translations as 3D Matrices
Any 2D translation can be represented by a 3x3 matrix.
This is a 3D shear that acts as a translation on the plane z = 1.
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77. Translation as a Shear y y x x z
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