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

Transparency

CS 6965 Fall 2011

Fall 2011 CS 6965

• Light is an electromagnetic wave
• Some materials conduct energy

at visible frequencies

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Transparency

Fall 2011 CS 6965

Transmitted rays

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n r i t boundary η1 η2 plane of incidence θ2 θ1 θ1

Fall 2011 CS 6965

Maxwell’s equations

• Maxwell’s equations relate electricity and magnetism (light) as

waves in space

• They predict the speed of light in a vacuum:

c = 1 µ0ε0 µ0 : Electric permittivity of free space ε0:Magnetic permeability of free space

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Fall 2011 CS 6965

Speed of light

• Speed of light changes (slows) in other materials

v = 1 µε µ : Electrical permittivity of material ε:Magnetic permeability of material v ≤ c

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Fall 2011 CS 6965

Fermat’s principle (1657)

• Light takes the fastest path between two points:

A B C

tAB = tAC + tBC = C − A va + B − C vb Simplify: By = 0 tAB = Cx − Ax

( )

2 + Ay 2

va + Bx − Cx

( )

2 + By 2

vb

6 2.00 2.25 2.50 2.75 3.00

• 1
• 0.8 -0.6 -0.4 -0.2

0.2 0.4 0.6 0.8 1 t

Fall 2011 CS 6965

Refraction

tAB = Cx − Ax

( )

2 + Ay 2

va + Bx − Cx

( )

2 + By 2

vb d dCx tAB = 0 = Cx − Ax va Cx − Ax

( )

2 + Ay 2 +

Cx − Bx vb Bx − Cx

( )

2 + By 2

0 = sinθa va + −sinθb vb vb sinθa = va sinθb sinθa sinθb = va vb

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A B C θA θB

Fall 2011 CS 6965

Index of refraction

ηabs = where c = 2.99 x 108 m/s. Note: v ≤ c for all transparent materials, so η ≥ 1. Relative index of refraction: Absolute index of refraction: η = η2 η1 c v

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Fall 2011 CS 6965

Common values

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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

Fall 2011 CS 6965

Snell’s Law

sinθ1 sinθ2 = v1 v2 = η2 η1 = η = η12

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n r i t boundary η1 η2 plane of incidence θ2 θ1 θ1

Fall 2011 CS 6965

Transmission

sinθ1 sinθ2 = η2 η1 = η = S1 S2 ,S2η = S1 cosθ1 = C1 = −N  ⋅V   cosθ2 = C2 = −N  ⋅T   Square both sides of S1η = S2 : S2

2η2 = S1 2

And use S2 + C 2 = 1 to get cosine forms : 1− C2

2

( )η2 = 1− C1

2

C2

2 = 1+

C1

2 −1

( )

η2 C2 = 1+ C1

2 −1

( )

η2

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N V

• N’

V’ T θ2 θ1

Fall 2011 CS 6965

Transmission

T   = − ʹ″ N    + k ʹ″ V   + ʹ″ N   

( ) = k ʹ″

V   + (k −1) ʹ″ N    T   = 1 ʹ″ N    = C2 N  ʹ″ V   C1 = − ʹ″ N    , ʹ″ V   = − ʹ″ N    C1 , ʹ″ V   = C2 C1 ʹ″ V   = C2 C1 V   k = S2 S1 T   ʹ″ V   = S2 S1 1 C2 C1 = S2C1 C2S1

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N V

• N’

V’ T θ2 θ1 V’+N’ k

Fall 2011 CS 6965

Transmission

T   = − ʹ″ N    + k ʹ″ V   + ʹ″ N   

( ) = k ʹ″

V   + (k −1) ʹ″ N    k = S2 S1 T   ʹ″ V   = S2 S1 1 C2 C1 = S2C1 C2S1 C2 = 1+ C1

2 −1

( )

η2 T   = S2C1 C2S1 C2 C1 V   + S2C1 C2S1 −1 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ C2 N  = S2 S1 V   + S2 S1 C1 − C2 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ N  = 1 ηV   + 1 η C1 − C2 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ N  = 1 ηV   + C1 η − 1+ C1

2 −1

( )

η2 ⎛ ⎝ ⎜ ⎜ ⎞ ⎠ ⎟ ⎟ N 

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N V

• N’

V’ T θ2 θ1 V’+N’ k

Fall 2011 CS 6965

Transmitted ray direction, t (cont’d.)

When η1 < η2, t bends toward the normal direction at the hit point

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η2 η1 n r i t θ2 θ1 θ1

Fall 2011 CS 6965

Transmitted ray direction, t (cont’d.)

When η1 > η2, t bends away from the normal direction at the hit point

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η2 η1 n r i t θ2 θ1 θ1

Fall 2011 CS 6965

Total internal reflection

There exists a critical angle, θc, when the transmitted ray direction t is parallel to the boundary and θ2 = π/2

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η2 η1 n r i t θ1 θ2 = π/2 θ1

Fall 2011 CS 6965

θ1 approaches θc

As θ1 increases, t bends toward the boundary

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η2 η1 n r i t θ1 θ2 θ1 η2 η1 n r i t θ1 θ2 θ1

θ1 << θc θ1 < θc

…

Fall 2011 CS 6965

θ1 is greater than or equal to θc

When θ1 ≥ θc, t is parallel to the boundary and it carries no energy (total internal reflection occurs)

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η2 η1 n r i t θ1 θ2 = π/2 θ1

θ1 = θc

η2 η1 n r i θ1 θ1

θ1 > θc

Fall 2011 CS 6965

Checking for TIR

cos θ2 = [1 - 1 η2 (1 - cos2 θ1)]1/2 Recall that the angle of refraction θ2 is given by:

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Fall 2011 CS 6965

TIR occurs when θ1 = θc

1 - 1 η2 (1 - cos2 θ1) When θ1 = θc, the expression: becomes zero. When θ1 > θc, this expression is negative, so θ2 is imaginary and t is a complex number.

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Fall 2011 CS 6965

TIR condition

1 - 1 η2 (1 - cos2 θ1) < 0 Thus, TIR occurs when:

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Fall 2011 CS 6965

Spawning secondary rays

c(p) = clocal(p) + kr(p)*cr(pr) + kt(p)*ct(pt)

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i r1 t1

Fall 2011 CS 6965

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, …); }

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Fall 2011 CS 6965

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.); }

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Fall 2011 CS 6965

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; }

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Fall 2011 CS 6965

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!

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Fall 2011 CS 6965

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)

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Fall 2011 CS 6965

Imperfect materials

• Most surfaces are not perfectly smooth:

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Fall 2011 CS 6965

Material models

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Two choices:

• 1. Simulate microfacets (intractable)
• 2. Approximate models

Fall 2011 CS 6965

BRDF

Source: surfaceoptics.com

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Fall 2011 CS 6965

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

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Fall 2011 CS 6965

Questions?

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Texturing

CS 6965 Fall 2011

Fall 2011 CS 6965

Texture mapping

• Most real objects do not have

uniform color

• Texture: color and other material

properties as a function of space

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Fall 2011 CS 6965

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.

• Bump mapping

Ken Perlin, NYU

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Fall 2011 CS 6965

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 CL and L   dist= L   ,Ln   = L   L   cosphi = N  ⋅ Ln   if(cosphi > 0) if(!intersect with 0 < t < dist) light += CL *(Kd *cosphi) result=light*surface color

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Fall 2011 CS 6965

Simple image texture

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

( )⋅U

  U   y = P   − C  

( )⋅V

  V  

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World

P C x y U V Image

Fall 2011 CS 6965

Simple image texture

x = P   − C  

( )⋅U

  U   y = P   − C  

( )⋅V

  V   ix = (int) x xres −1

( )

( ), fx = ix − x xres −1

( )

iy = (int) y yres −1

( )

( ), fy = iy − y yres −1

( )

Bilinear interpolation of image(ix,iy) - image(ix+1,iy+1) using fx,fy as interpolation factors Use interpolated color in lambertian shading

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World

P C x y U V Image