Lecture 6 - Path Tracing Welcome! , = (, ) , - PowerPoint PPT Presentation

INFOMAGR – Advanced Graphics Jacco Bikker - November 2019 - February 2020 Lecture 6 - “Path Tracing” Welcome! 𝑱 𝒚, 𝒚 ′ = 𝒉(𝒚, 𝒚 ′ ) 𝝑 𝒚, 𝒚 ′ + න 𝝇 𝒚, 𝒚 ′ , 𝒚 ′′ 𝑱 𝒚 ′ , 𝒚 ′′ 𝒆𝒚′′ 𝑻

Today’s Agenda: ▪ Introduction ▪ Path Tracing

Advanced Graphics – Path Tracing 3 Introduction Previously in Advanced Graphics The Rendering Equation: 𝑀 𝑝 𝑦, 𝜕 𝑝 = 𝑀 𝐹 𝑦, 𝜕 𝑝 + න 𝑔 𝑠 𝑦, 𝜕 𝑝 , 𝜕 𝑗 𝑀 𝑗 𝑦, 𝜕 𝑗 cos 𝜄 𝑗 𝑒𝜕 𝑗 𝛻 …which models light transport as it happens in the real world, by summing: ▪ Direct illumination: 𝑀 𝐹 (𝑦, 𝜕 𝑝 ) ▪ Indirect illumination, or reflected light: ׬ 𝛻 𝑔 𝑠 𝑦, 𝜕 𝑝 , 𝜕 𝑗 𝑀 𝑗 𝑦, 𝜕 𝑗 cos 𝜄 𝑗 𝑒𝜕 𝑗 We used quantities flux 𝛸 (joules per second), radiance 𝑀 (flux per 𝑛 2 per sr) and irradiance 𝐹 (flux per 𝑛 2 ).

Advanced Graphics – Path Tracing 4 Introduction Previously in Advanced Graphics Particle transport: As an alternative to discrete flux / radiance / irradiance, we can reason about light transport in terms of particle transport. ▪ Flux then becomes the number of emitted photons; ▪ Radiance the number of photons travelling through a unit area in a unit direction; ▪ Irradiance the number of photons arriving on a unit area. A BRDF tells us how many particles are absorbed, and how outgoing particles are distributed. The distribution depends on the incident and exitant direction.

Advanced Graphics – Path Tracing 5 Introduction Previously in Advanced Graphics Probabilities: We can also reason about the behavior of a single photon. In that case, the BRDF tells us the probability of a photon being absorbed, or leaving in a certain direction.

Advanced Graphics – Path Tracing 6 Introduction Previously in Advanced Graphics BRDFs:

Advanced Graphics – Path Tracing 7 Introduction Light Transport 𝑠 𝑡 ← 𝑦 ← 𝑦′ 𝑀 𝑦 ← 𝑦′ 𝐻 𝑦 ↔ 𝑦 ′ 𝑒𝐵(𝑦 ′ ) 𝑀 𝑡 ← 𝑦 = 𝑀 𝐹 𝑡 ← 𝑦 + න 𝑔 𝐵 𝑀 𝑝 𝑦, 𝜕 𝑝 = 𝑀 𝐹 𝑦, 𝜕 𝑝 + න 𝑔 𝑠 𝑦, 𝜕 𝑝 , 𝜕 𝑗 𝑀 𝑗 𝑦, 𝜕 𝑗 cos 𝜄 𝑗 𝑒𝜕 𝑗 𝛻 Radiance Radiance Radiance Irradiance BRDF

Advanced Graphics – Path Tracing 8 Introduction Bidirectional Reflectance Distribution Function BRDF: function describing the relation between radiance emitted in direction 𝜕 𝑝 and irradiance arriving from direction 𝜕 𝑗 : 𝑠 𝜕 𝑝 , 𝜕 𝑗 = 𝑀 𝑝 (𝜕 𝑝 ) 𝑀 𝑝 (𝜕 𝑝 ) = 𝑝𝑣𝑢𝑕𝑝𝑗𝑜𝑕 𝑠𝑏𝑒𝑗𝑏𝑜𝑑𝑓 𝑔 𝐹 𝑗 (𝜕 𝑗 ) = 𝑀 𝑗 𝜕 𝑗 cos θ 𝑗 𝑗𝑜𝑑𝑝𝑛𝑗𝑜𝑕 𝑗𝑠𝑠𝑏𝑒𝑗𝑏𝑜𝑑𝑓 Or, if spatially variant: 𝑠 𝑦, 𝜕 𝑝 , 𝜕 𝑗 = 𝑀 𝑝 (𝑦, 𝜕 𝑝 ) 𝑀 𝑝 (𝑦, 𝜕 𝑝 ) 𝑔 𝐹 𝑗 (𝑦, 𝜕 𝑗 ) = 𝑀 𝑗 𝑦, 𝜕 𝑗 cos θ 𝑗 Properties: ▪ Should be positive: 𝑔 𝑠 𝜕 𝑝 , 𝜕 𝑗 ≥ 0 ▪ Helmholtz reciprocity should be obeyed: 𝑔 𝑠 𝜕 𝑝 , 𝜕 𝑗 = 𝑔 𝑠 𝜕 𝑗 , 𝜕 𝑝 ▪ Energy should be conserved: ׬ 𝛻 𝑔 𝑠 𝜕 𝑝 , 𝜕 𝑗 cos 𝜄 𝑝 𝑒𝜕 𝑝 ≤ 1

Advanced Graphics – Path Tracing 9 Introduction Bidirectional Reflectance Distribution Function The diffuse BRDF is: 𝑠 𝜕 𝑝 , 𝜕 𝑗 = 𝑏𝑚𝑐𝑓𝑒𝑝 𝑔 π So, for a total irradiance 𝐹 at surface point 𝑦 , the 𝝏 𝒋 𝑏𝑚𝑐𝑓𝑒𝑝 outgoing radiance 𝑀 𝑝 = 𝐹 𝑗 . Why the 𝜌 ? 𝒐 𝜌 Energy conservation: 𝐹 𝑝 ≤ 𝐹 𝑗 Suppose we have a directional light parallel to 𝑜 , with 𝑏𝑚𝑐𝑓𝑒𝑝 intensity 1. Then: 𝐹 𝑗 = 𝑀 𝑗 = 1. Suppose our BRDF = . 1 Then, for albedo = 1 we get: 𝐹 𝑝 = ׬ 𝛻 𝑀 𝑗 𝑔 𝑠 (𝜕 𝑝 , 𝜕 𝑗 ) cos 𝜕 𝑝 𝑒𝜕 𝑝 = ׬ 𝛻 cos 𝜕 𝑝 𝑒𝜕 𝑝 𝛻 cos 𝜕 𝑝 𝑒𝜕 𝑝 = 𝜌 ➔ 𝐹 𝑝 = 𝜌 𝐹 𝑗 . Now: ׬

Advanced Graphics – Path Tracing 10 Introduction Bidirectional Reflectance Distribution Function Mirror / Perfect specular: Reflects light in a fixed direction. 𝝏 𝒑 𝝏 𝒋 𝒐 For a given incoming direction 𝜕 𝑗 , all light is emitted in a single infinitesimal set of directions. The specular BRDF is thus 𝑠 𝑦, 𝜕 𝑝 , 𝜕 𝑗 = ቊ∞, along reflected vector 𝑔 0, otherwise. This is not practical, and therefore we will handle the pure specular case (reflection and refraction) separately.

Advanced Graphics – Path Tracing 11 Introduction Previously in Advanced Graphics Monte Carlo integration: Complex integrals can be approximated by replacing them by the expected value of a stochastic experiment. ▪ Soft shadows: randomly sample the area of a light source; ▪ Glossy reflections: randomly sample the directions in a cone; ▪ Depth of field: randomly sample the aperture; ▪ Motion blur: randomly sample frame time. In the case of the rendering equation, we are dealing with a recursive integral . Pat ath tr tracing: evaluating th this in integral us using a a ran andom walk .

Today’s Agenda: ▪ Introduction ▪ Path Tracing

Advanced Graphics – Path Tracing 13 Path Tracing Solving the Rendering Equation Let’s start with direct illumination: For a screen pixel, diffuse surface point 𝑞 with normal 𝑂 is directly visible. What is the radiance travelling via 𝑞 towards the eye? Answer: 𝑀 𝑝 𝑞, 𝜕 𝑝 = න 𝑔 𝑠 𝑞, 𝜕 𝑝 , 𝜕 𝑗 𝑀 𝑒 𝑞, 𝜕 𝑗 cos 𝜄 𝑗 𝑒𝜕 𝑗 𝛻 𝜕 𝑗 𝑂 𝜕 𝑝 p

Advanced Graphics – Path Tracing 14 Path Tracing 𝑀 𝑝 𝑞, 𝜕 𝑝 = න 𝑔 𝑠 𝑞, 𝜕 𝑝 , 𝜕 𝑗 𝑀 𝑒 𝑞, 𝜕 𝑗 cos 𝜄 𝑗 𝑒𝜕 𝑗 𝛻 Solving the Rendering Equation = 𝑏𝑚𝑐𝑓𝑒𝑝 න 𝑀 𝑒 𝑞, 𝜕 𝑗 cos 𝜄 𝑗 𝑒𝜕 𝑗 𝜌 𝛻 Q: What about distance attenuation? Let’s start with direct illumination: A: A far-away light is found by fewer directions 𝜕 𝑗 : it’s solid ang angle on the hemisphere is smaller. For a screen pixel, diffuse surface point 𝑞 with normal 𝑂 is directly visible. What is the radiance travelling via 𝑞 towards the eye? Q: What about the diffuse surface? Answer: A: The BRDF is independent of 𝜕 𝑝 (it doesn’t appear in the equation), but as 𝜕 𝑗 approaches 𝑀 𝑝 𝑞, 𝜕 𝑝 = න 𝑔 𝑠 𝑞, 𝜕 𝑝 , 𝜕 𝑗 𝑀 𝑒 𝑞, 𝜕 𝑗 cos 𝜄 𝑗 𝑒𝜕 𝑗 the horizon, cos 𝜄 𝑗 approaches zero. 𝛻 𝜕 𝑗 𝑜 𝜕 𝑝 p

Advanced Graphics – Path Tracing 15 Path Tracing Direct Illumination We can solve this integral using Monte-Carlo integration: ▪ Chose N random directions over the hemisphere for 𝑞 ▪ Find the first surface in each direction by tracing a ray ▪ Sum the luminance of the encountered surfaces ▪ Divide the sum by N and multiply by 2π 𝑂 𝑀 𝑝 𝑞, 𝜕 𝑝 ≈ 2𝜌 𝑂 ෍ 𝑔 𝑠 𝑞, 𝜕 𝑝 , 𝜕 𝑗 𝑀 𝑒 𝑞, 𝜕 𝑗 cos 𝜄 𝑗 𝑗=1 𝜕 𝑗 𝑜 𝜕 𝑝 p

Advanced Graphics – Path Tracing 16 Path Tracing We integrate over the hemisphere, which Direct Illumination has an area of 2 𝜌 . 𝑂 Do not confuse this with the 1/ 𝜌 factor 𝑀 𝑝 𝑞, 𝜕 𝑝 ≈ 2𝜌 in the BRDF, which doesn’t compensate 𝑂 ෍ 𝑔 𝑠 𝑞, 𝜕 𝑝 , 𝜕 𝑗 𝑀 𝑒 𝑞, 𝜕 𝑗 cos 𝜄 for the surface of the hemisphere, but the 𝑗=1 integral of cos 𝜄 over the hemisphere ( 𝜌 ). Questions: ▪ Why do we multiply by 2𝜌 ? ▪ What is the radiance 𝑀 𝑒 (𝑞, 𝜕 𝑗 ) towards 𝑞 for e.g. a 100W light? ▪ What is the irradiance 𝐹 at 𝑞 from this light? 𝑀 is per sr; 𝑀 𝑒 (𝑞, 𝜕 𝑗 ) is proportional to the solid angle of the light as seen 𝜕 𝑗 from p , so: ~ (cos 𝜄 𝑝 𝐵 𝑀 𝑒 )/𝑠 2 . 𝑜 𝜕 𝑝 Note that the 100W flux is spread out over the area; irradiance is defined per 𝑛 2 . p

Advanced Graphics – Path Tracing 17 Path Tracing Direct Illumination 𝑀 𝑝 𝑞, 𝜕 𝑝 = න 𝑔 𝑠 𝑞, 𝜕 𝑝 , 𝜕 𝑗 𝑀 𝑒 𝑞, 𝜕 𝑗 cos 𝜄 𝑗 𝑒𝜕 𝑗 𝛻 In many directions, we will not find light sources. We can improve our estimate by sampling the lights separately. 𝑚𝑗𝑕ℎ𝑢𝑡 𝑘 𝑞, 𝜕 𝑗 cos 𝜄 𝑗 𝑒𝜕 𝑗 𝑀 𝑝 𝑞, 𝜕 𝑗 = ෍ න 𝑔 𝑠 𝑞, 𝜕 𝑝 , 𝜕 𝑗 𝑀 𝑒 𝛻 𝑘=1 Obviously, sampling the entire hemisphere for each light is not necessary; we can sample the area of the light instead: 𝑚𝑗𝑕ℎ𝑢𝑡 𝑘 𝑞, 𝜕 𝑗 cos 𝜄 𝑗 𝑒𝜕 𝑗 𝑀 𝑝 𝑞, 𝜕 𝑗 = ෍ න 𝑔 𝑠 𝑞, 𝜕 𝑝 , 𝜕 𝑗 𝑀 𝑒 𝐵 𝑘=1