Transparency CS 6965 Fall 2011 Transparency Light is an - PowerPoint PPT Presentation

Transparency CS 6965 Fall 2011

Transparency • Light is an electromagnetic wave • Some materials conduct energy at visible frequencies CS 6965 Fall 2011 2

Transmitted rays plane of incidence i n r θ 1 θ 1 η 1 boundary θ 2 t η 2 CS 6965 Fall 2011 3

Maxwell’s equations • Maxwell’s equations relate electricity and magnetism (light) as waves in space • They predict the speed of light in a vacuum: 1 c = µ 0 ε 0 µ 0 : Electric permittivity of free space ε 0 :Magnetic permeability of free space CS 6965 Fall 2011 4

Speed of light • Speed of light changes (slows) in other materials 1 v = µ ε µ : Electrical permittivity of material ε :Magnetic permeability of material v ≤ c CS 6965 Fall 2011 5

Fermat’s principle (1657) • Light takes the fastest path between two points: A t AB = t AC + t BC = C − A + B − C v a v b Simplify: B y = 0 2 + A y 2 + B y C ( ) ( ) 2 2 C x − A x B x − C x t AB = + v a v b 3.00 2.75 B 2.50 t 2.25 2.00 -1 -0.8 -0.6 -0.4 -0.2 0 0.2 0.4 0.6 0.8 1 CS 6965 Fall 2011 6

A Refraction θ A 2 + A y 2 + B y ( ) ( ) 2 2 C x − A x B x − C x C t AB = + v a v b d C x − A x C x − B x t AB = 0 = 2 + 2 + A y 2 + B y dC x ( ) ( ) 2 v a C x − A x v b B x − C x 0 = sin θ a + − sin θ b θ B B v a v b v b sin θ a = v a sin θ b sin θ a = v a sin θ b v b CS 6965 Fall 2011 7

Index of refraction Absolute index of refraction: c η abs = v where c = 2.99 x 108 m/s. Note: v ≤ c for all transparent materials, so η ≥ 1. Relative index of refraction: η = η 2 η 1 CS 6965 Fall 2011 8

Common values Medium Index of refraction Perfect vacuum 1.0 Air (1 atm, 20° C) 1.0003 Water 1.33 Acrylic 1.49 Crown glass 1.52 Diamond 2.42 CS 6965 Fall 2011 9

Snell’s Law plane of incidence sin θ 1 = v 1 = η 2 = η = η 12 sin θ 2 v 2 η 1 i n r θ 1 θ 1 η 1 boundary θ 2 t η 2 CS 6965 Fall 2011 10

Transmission N V sin θ 1 = η = S 1 = η 2 , S 2 η = S 1 sin θ 2 S 2 η 1 θ 1    cos θ 1 = C 1 = − N ⋅ V    cos θ 2 = C 2 = − N ⋅ T Square both sides of S 1 η = S 2 : V’ θ 2 T -N’ 2 η 2 = S 1 2 S 2 And use S 2 + C 2 = 1 to get cosine forms : ) η 2 = 1 − C 1 ( 2 2 1 − C 2 2 − 1 ( ) C 1 2 = 1 + C 2 η 2 2 − 1 ( ) C 1 C 2 = 1 + η 2 CS 6965 Fall 2011 11

Transmission N V                θ 1 ( ) = k ʹ″ T N + k V N V + ( k − 1) N = − ʹ″ ʹ″ + ʹ″ ʹ″   T = 1     N = C 2 N ʹ″    V’ θ 2 T          -N’ N − ʹ″ = C 2 V C 1 = − N , V , V ʹ″ ʹ″ ʹ″ = ʹ″ V’+N’ k C 1 C 1     = C 2 V V ʹ″ C 1   T k = S 2  = S 2 1 = S 2 C 1  C 2 S 1 S 1 C 2 S 1 V ʹ″ C 1 CS 6965 Fall 2011 12

               Transmission ( ) = k ʹ″ T N + k V N V + ( k − 1) N = − ʹ″ ʹ″ + ʹ″ ʹ″   T k = S 2  = S 2 1 = S 2 C 1  C 2 V N S 1 S 1 C 2 S 1 V ʹ″ C 1 θ 1 2 − 1 ( ) C 1 C 2 = 1 + η 2      = S 2 C 1 C 2 ⎛ S 2 C 1 ⎞ T V − 1 ⎟ C 2 N + ⎜ V’ C 2 S 1 C 1 C 2 S 1 T ⎝ ⎠ θ 2 -N’    = S 2 ⎛ S 2 ⎞ V’+N’ k V C 1 − C 2 ⎟ N + ⎜ S 1 S 1 ⎝ ⎠    = 1 1 ⎛ ⎞ η V η C 1 − C 2 ⎟ N + ⎜ ⎝ ⎠ 2 − 1 ⎛ ( ) ⎞    C 1 = 1 + C 1 η V ⎜ 1 + ⎟ N η − η 2 ⎜ ⎟ ⎝ ⎠ CS 6965 Fall 2011 13

Transmitted ray direction, t (cont’d.) When η 1 < η 2, t bends toward the normal direction at the hit point n i r θ 1 θ 1 η 1 η 2 θ 2 t CS 6965 Fall 2011 14

Transmitted ray direction, t (cont’d.) When η 1 > η 2, t bends away from the normal direction at the hit point n i r θ 1 θ 1 η 1 t η 2 θ 2 CS 6965 Fall 2011 15

Total internal reflection There exists a critical angle, θ c, when the transmitted ray direction t is parallel to the boundary and θ 2 = π /2 θ 2 = π /2 η 1 t η 2 θ 1 θ 1 i r n CS 6965 Fall 2011 16

θ 1 approaches θ c As θ 1 increases, t bends toward the boundary t θ 2 θ 2 η 1 η 1 t η 2 η 2 θ 1 θ 1 r θ 1 θ 1 i … i r n n θ 1 << θ c θ 1 < θ c CS 6965 Fall 2011 17

θ 1 is greater than or equal to θ c When θ 1 ≥ θ c, t is parallel to the boundary and it carries no energy (total internal reflection occurs) θ 2 = π /2 η 1 η 1 t η 2 η 2 θ 1 θ 1 θ 1 θ 1 i r i r n n θ 1 = θ c θ 1 > θ c CS 6965 Fall 2011 18

Checking for TIR Recall that the angle of refraction θ 2 is given by: 1 cos θ 2 = [1 - (1 - cos2 θ 1)]1/2 η 2 CS 6965 Fall 2011 19

TIR occurs when θ 1 = θ c When θ 1 = θ c, the expression: 1 1 - (1 - cos2 θ 1) η 2 becomes zero. When θ 1 > θ c, this expression is negative, so θ 2 is imaginary and t is a complex number. CS 6965 Fall 2011 20

TIR condition Thus, TIR occurs when: 1 1 - (1 - cos2 θ 1) < 0 η 2 CS 6965 Fall 2011 21

Spawning secondary rays t 1 i r 1 c(p) = clocal(p) + kr(p)*cr(pr) + kt(p)*ct(pt) CS 6965 Fall 2011 22

Dielectric shading result = normal lambertian (or other) shading Ray rray = reflect_ray(ray, …); if (tir(ray)) { // kr = 1, kt = 0 result += traceRay(rray, depth + 1, …); } else { result += kr*traceRay(rray, depth + 1, …); Ray tray = transmit_ray(ray, …); result += kt*traceRay(tray, depth + 1, …); } CS 6965 Fall 2011 23

Pseudocode (cont’d.) bool tir(const Ray& ray) { float cosTheta = -ray.direction()*normal; float eta; if(cosTheta > 0) eta = ior_from/ior_to; else eta = ior_to/ior_from; return ((1. - (1. - cosTheta*cosTheta)/ (eta*eta)) < 0.); } CS 6965 Fall 2011 24

Pseudocode (cont’d.) bool Dielectric::transmit_ray(const Ray& ray, …) { compute eta and cosTheta the same as in tir flip normal, eta, cosTheta if cosTheta < 0 tmp = 1.f - (1. - cosTheta1*cosTheta1)/(eta*eta); cosTheta2 = sqrt(tmp); Ray tray(hit_point, ray.direction()/(cosTheta2 - cosTheta1/eta)*normal; return tray; } CS 6965 Fall 2011 25

Implementation notes • There are more efficient ways of doing it than the above pseudocode • If you want to do nested refraction, or if you want the eyepoint in something beside a vacuum, you must track current index of refraction • Directions of the normal matter! CS 6965 Fall 2011 26

Light transport • There are 4 primary ways that light interacts with a surface: • Bounces off (perfect specular reflection) • Absorbed and retransmitted in an arbitrary direction (perfect diffuse reflection) • Travels through surface (perfect specular transmission) • Absorbed and retransmitted on other side (perfect diffuse transmission) CS 6965 Fall 2011 27

Imperfect materials • Most surfaces are not perfectly smooth: CS 6965 Fall 2011 28

Material models Two choices: 1. Simulate microfacets (intractable) 2. Approximate models CS 6965 Fall 2011 29

BRDF Source: surfaceoptics.com CS 6965 Fall 2011 30

BRDF info • http://graphics.stanford.edu/~smr/brdf/bv/ • http://www.cs.utah.edu/~shirley/classes/brdf/ • http://www.cs.princeton.edu/~smr/cs348c-97/surveypaper.html CS 6965 Fall 2011 31

Questions? CS 6965 Fall 2011 32

Texturing CS 6965 Fall 2011

Texture mapping • Most real objects do not have uniform color • Texture: color and other material properties as a function of space CS 6965 Fall 2011 34

Texture mapping topics • Image textures • Texture coordinates • Linear, cylindrical, spherical mappings • Barycentric coordinates for triangles • Software architecture • Procedural textures • Simple: checkerboards, tiles, etc. • Fractal noise based: marble, granite, wood, etc. • Ken Perlin, NYU Bump mapping CS 6965 Fall 2011 35

Lambertian Shading       Compute hit position ( P = O + tV )  Call primitive to get normal (N ) (normalized)    costheta = − N ⋅ V if ( costheta < 0) normal =-normal Color light = scene.ambient*Ka foreach light source   get C L and L       = L   dist= L , L n L    cosphi = N ⋅ L n if( cosphi > 0) if(!intersect with 0 < t < dist ) light += C L *( Kd * cosphi ) result=light*surface color CS 6965 Fall 2011 36

Simple image texture Use hit position to determine image location   P P : hit position y   C : lower left corner of image  x C U : image X axis in world space   Image V : image Y axis in world space x, y: normalized image coordinates (0-1)         P = C + xU + yV    V if U ⋅ V = 0 :       ( ) ⋅ U P − C   x = U       ( ) ⋅ V P − C   y = U V World CS 6965 Fall 2011 37