Light Field = Radiance(Ray)
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The Light Field Light field = radiance function on rays Surface and field radiance Conservation of radiance Measurement Irradiance from area sources Measuring rays Form factors and throughput Conservation of throughput From London and Upton CS348B Lecture 5 Pat Hanrahan, 2004 Light Field = Radiance(Ray) Page 1 Incident Surface Radiance Definition: The incoming surface radiance (luminance) is the power per unit solid angle per unit projected area arriving at a receiving surface dω dA dx2Φ (,ω ) Lx(,ω )≡ i i ddAωi CS348B Lecture 5 Pat Hanrahan, 2004 Exitant Surface Radiance Definition: The outgoing surface radiance (luminance) is the power per unit solid angle per unit projected area leaving at surface dω dA dx2Φ (,ω ) Lx(,ω )≡ o o ddAωi Alternatively: the intensity per unit projected area leaving a surface CS348B Lecture 5 Pat Hanrahan, 2004 Page 2 Field Radiance Definition: The field radiance (luminance) at a point in space in a given direction is the power per unit solid angle per unit area perpendicular to the direction dω dA Lx(,ω ) CS348B Lecture 5 Pat Hanrahan, 2004 Environment Maps L(,θ ϕ ) Miller and Hoffman, 1984 CS348B Lecture 5 Pat Hanrahan, 2004 Page 3 Spherical Gantry ⇒ Light Field Lxy(,,,θ ϕ ) (,θ ϕ ) Capture all the light leaving an object - like a hologram CS348B Lecture 5 Pat Hanrahan, 2004 Two-Plane Light Field 2D Array of Cameras 2D Array of Images L(,,,)uvst CS348B Lecture 5 Pat Hanrahan, 2004 Page 4 Multi-Camera Array ⇒ Light Field CS348B Lecture 5 Pat Hanrahan, 2004 Properties of Radiance Page 5 Properties of Radiance 1. Fundamental field quantity that characterizes the distribution of light in an environment. ∴ Radiance is a function on rays ∴ All other field quantities are derived from it 2. Radiance invariant along a ray. ∴ 5D ray space reduces to 4D 3. Response of a sensor proportional to radiance. CS348B Lecture 5 Pat Hanrahan, 2004 1st Law: Conversation of Radiance The radiance in the direction of a light ray remains constant as the ray propagates 2 2 22 d Φ1 d Φ2 ddΦ12=Φ dA1 dω1 2 L1 dLddAΦ=1111ω dω2 dA2 2 L2 dLddAΦ=2222ω r dA dA ddAωω==12 ddA 11r 2 2 2 ∴L12= L CS348B Lecture 5 Pat Hanrahan, 2004 Page 6 Radiance: 2nd Law The response of a sensor is proportional to the radiance of the surface visible to the sensor. Ω Aperture A Sensor RLddALT==∫∫ ω TddA= ∫∫ ω A Ω A Ω L is what should be computed and displayed. T quantifies the gathering power of the device. CS348B Lecture 5 Pat Hanrahan, 2004 Quiz Does the brightness that a wall appears to the sensor depend on the distance? No!! CS348B Lecture 5 Pat Hanrahan, 2004 Page 7 Irradiance from a Uniform Surface Source Irradiance from the Environment 2 dxΦ=ii(,ω ) Lx (,ωθ )cos dAd ω Lxi (,ω ) dE( x ,ω )= Li ( x ,ωθω )cos d dω θ dA E()xLxd= (,ω )cosθω ∫ i H2 CS348B Lecture 5 Pat Hanrahan, 2004 Page 8 Irradiance from an Area Source A E()xLd= ∫ cosθ ω H2 Ω = Ld∫cosθ ω Ω =ΩL Ω CS348B Lecture 5 Pat Hanrahan, 2004 Projected Solid Angle dω θ cosθ dω ∫ cosθ dωπ= H2 CS348B Lecture 5 Pat Hanrahan, 2004 Page 9 Uniform Disk Source Geometric Derivation Algebraic Derivation cosαπ 2 r Ω= ∫∫cosθφdd cosθ 10 cosα α cos2 θ = 2π 2 h 1 = παsin2 sinα r 2 = π rh22+ Ω= π sin2 α CS348B Lecture 5 Pat Hanrahan, 2004 Spherical Source Geometric Derivation Algebraic Derivation r Ω= ∫ cosθ dω R = π sin2 α α r 2 = π R2 Ω= π sin2 α CS348B Lecture 5 Pat Hanrahan, 2004 Page 10 The Sun Solar constant (normal incidence at zenith) Irradiance 1353 W/m2 Illuminance 127,500 lm/m2 = 127.5 kilolux Solar angle α = .25 degrees = .004 radians (half angle) 22 Ω= π sin απα ≈ = 6 x 10-5 steradians Solar radiance EWm1.353× 1032 / W L == =2.25 × 107 Ω× 610−52sr m ⋅ sr CS348B Lecture 5 Pat Hanrahan, 2004 Polygonal Source CS348B Lecture 5 Pat Hanrahan, 2004 Page 11 Polygonal Source CS348B Lecture 5 Pat Hanrahan, 2004 Polygonal Source CS348B Lecture 5 Pat Hanrahan, 2004 Page 12 Lambert’s Formula γ iiN γ i 3 Ai ∑ A1 i=1 −A2 −A3 nn Aiii=•γ NN ∑∑Aiii= γ NN• ii==11 CS348B Lecture 5 Pat Hanrahan, 2004 Penumbras and Umbras CS348B Lecture 5 Pat Hanrahan, 2004 Page 13 Measuring Rays = Throughput Throughput Counts Rays Define an infinitesimal beam as the set of rays intersecting two infinitesimal surface elements ru(,,112 v u , v 2 ) dA111(,) u v dA222(,) u v 2 dA12 dA dT= 2 x12− x Measure the number of rays in the beam This quantity is called the throughput CS348B Lecture 5 Pat Hanrahan, 2004 Page 14 Parameterizing Rays Parameterize rays wrt to receiver ru(,,,)2222 v θ φ dω222(,)θφ dA222(,) u v 2 dA1 dT==2 dA222 dω dA xx12− CS348B Lecture 5 Pat Hanrahan, 2004 Parameterizing Rays Parameterize rays wrt to source ru(,,,)1111 v θ φ dA111(,) u v dω111(,)θφ 2 dA2 d T== dA1112 dA dω xx12− CS348B Lecture 5 Pat Hanrahan, 2004 Page 15 Parameterizing Rays Tilting the surfaces reparameterizes the rays ru(,, v u , v ) 112 2 dA111(,) u v dA222(,) u v 2 cosθθ12 cos d T= 2 dA12 dA xx12− = ddAω11i = ddAω22i All these throughputs must be equal. CS348B Lecture 5 Pat Hanrahan, 2004 Parameterizing Rays: S2 × R2 Parameterize rays by rxy(,,,)θ φ ω Projected area A()ω Measuring the number or rays that hit a shape Td= ∫∫ωθϕ(, ) dAxy (,) SR22 = Ad(,θϕ ) ωθϕ (, ) Sphere: ∫ 22 S2 TA==44ππ R = 4π A CS348B Lecture 5 Pat Hanrahan, 2004 Page 16 Parameterizing Rays: M2 × S2 Parameterize rays by ruv(,,θφ , ) (,)θφ (,)uv TdAuv= (,) cosθ dωθϕ (, ) ∫∫ MH22 ()N N S π 22 Sphere: TS==ππ4 R S Crofton’s Theorem: 4ππASA= ⇒= 4 CS348B Lecture 5 Pat Hanrahan, 2004 Form Factors Page 17 Types of Throughput 1. Infinitesimal beam of rays cosθ cosθ′ d2 T(, dA dA′′ )≡ dA ()() x dA x xx− ′ 2 2. Infinitesimal-finite beam cosθθ cos ′ dT(,) dA A′′ dA≡ dA () x dA () x ∫ 2 A′ xx− ′ 3. Finite-finite beam cosθ cosθ′ TAA(,′′ )≡ dAxdAx () () ∫∫ 2 AA′ xx− ′ CS348B Lecture 5 Pat Hanrahan, 2004 Probability of Ray Intersection Probability of a ray hitting A’ given it hits A TAA(,)′ Pr(AA′ | ) = TA() A′ TAA(,)′ = ′ πA dA() x cosθ cosθ′ TAA(,)′′= dAxdAx () () ∫∫ 2 AA′ xx− ′ A TA()=π A dA() x CS348B Lecture 5 Pat Hanrahan, 2004 Page 18 Probability of Ray Intersection cosθ cosθ′ TAA(,)′′= dAxdAx () () ∫∫ 2 AA′ xx− ′ =π Gxx(,)()()′′ dAx dAx A′ ∫∫ AA′ dA() x′ cosθθ cos ′ Gxx(,′ )≡ 2 Vxx (,′ ) π xx− ′ 0 ¬visible A Vxx(,′ )= 1 visible dA() x CS348B Lecture 5 Pat Hanrahan, 2004 Form Factor Probability of a ray hitting A’ given it hits A Pr(A′′′′ |ATA ) ( )== Pr( AA | ) TA ( ) TAA ( , ) A′ Form factor definition FAA(,)Pr(|)′′= A A dA() x′ FAA(,′′ )= Pr(| AA ) Form factor reciprocity FAAA(,)′′′= FAAA (,) A TA()=π A dA() x CS348B Lecture 5 Pat Hanrahan, 2004 Page 19 Form Factor Power transfer from a constant source Φ=(,AA′′ ) LTAA (, ) TAA(,′ ) = LT() A′ A′ TA()′ dA() x′ =Φ()(,)A′′FAA A TA()=π A dA() x CS348B Lecture 5 Pat Hanrahan, 2004 Differential Form Factor Probability of a ray leaving dA(x) hitting A’ TdAAdA(,)′ TdAAdA (,)′′ TdAA (,) Pr(AdA′ | ) === TdA() ππ dA = ∫Gxx(,′′ ) dAx ( ) A′ A′ dA() x′ dA() x CS348B Lecture 5 Pat Hanrahan, 2004 Page 20 Form Factor Power transfer from a constant source ddAALTdAAΦ=(,)′′ (,) TdAA(,)′ = LT() A′ A′ TA()′ dA() x′ =Φ()(,)A′′FdAA A dA() x CS348B Lecture 5 Pat Hanrahan, 2004 Radiant Exitance Definition: The radiant (luminous) exitance is the energy per unit area leaving a surface. dΦ Mx()≡ o dA Wlm = lux mm22 In computer graphics, this quantity is often referred to as the radiosity (B) CS348B Lecture 5 Pat Hanrahan, 2004 Page 21 Uniform Diffuse Emitter M = Ldcosθ ω ∫ o H2 Lo(,x ω ) = L cosθ dω o ∫ H2 dω = πLo θ M L = o π dA CS348B Lecture 5 Pat Hanrahan, 2004 Form Factor Irradiance from a constant source ddAAΦ=Φ(,)′′ ()(,) AFdAA ′ Φ()A′ = A′FdAA(,)′ A′ A′ = M()(, A′′ F A dA ) dA dA() x′ EdA()= MA ()(,)′′ FAdA A dA() x CS348B Lecture 5 Pat Hanrahan, 2004 Page 22 Form Factors and Throughput Throughput measures the number of rays in a set of rays Form factors represent the probability of ray leaving a surface intersecting another surface Only a function of surface geometry Differential form factor Irradiance calculations Form factors Radiosity calculations (energy balance) CS348B Lecture 5 Pat Hanrahan, 2004 Conservation of Throughput Throughput conserved during propagation Number of rays conserved Assuming no attenuation or scattering n2 (index of refraction) times throughput invariant under the laws of geometric optics Reflection at an interface Refraction at an interface Causes rays to bend (kink) Continuously varying index of refraction Causes rays to curve; mirages CS348B Lecture 5 Pat Hanrahan, 2004 Page 23 Conservation of Radiance Radiance is the ratio of two quantities: 1. Power 2. Throughput ∆Φ∆()Td Φ Lr()== lim ∆→T 0 ∆TdT Since power and throughput are conserved, ∴ Radiance conserved CS348B Lecture 5 Pat Hanrahan, 2004 Quiz Does radiance increase under a magnifying glass? No!! CS348B Lecture 5 Pat Hanrahan, 2004 Page 24.