Lecture 7 - Path Tracing Welcome! , = (, ) , - PowerPoint PPT Presentation

INFOMAGR – Advanced Graphics Jacco Bikker - November 2016 - February 2017 Lecture 7 - “Path Tracing” Welcome! 𝑱 𝒚, 𝒚 ′ = 𝒉(𝒚, 𝒚 ′ ) 𝝑 𝒚, 𝒚 ′ + 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 + 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 (joules per second 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: The BRDF describes how incoming light is absorbed or scattered. More accurately: for an incoming direction 𝜕 𝑗 and an outgoing direction 𝜕 𝑝 , it defines the relation between received irradiance and reflected radiance. A physically based BRDF has some important properties: 𝝏 𝒋  𝑔 𝑠 𝜕 𝑝 , 𝜕 𝑗 ≥ 0 𝒐  𝑔 𝑠 𝜕 𝑝 , 𝜕 𝑗 = 𝑔 𝑠 𝜕 𝑗 , 𝜕 𝑝  𝛻 𝑔 𝑠 𝜕 𝑝 , 𝜕 𝑗 cos 𝜄 𝑝 𝑒𝜕 𝑝 ≤ 1 𝑏𝑚𝑐𝑓𝑒𝑝 Example: diffuse BRDF, 𝑔 𝑠 ω 𝑝 , ω 𝑗 = . 𝜌 Note that the diffuse BRDF is view independent; 𝜕 𝑝 is irrelevant. It does however take into account 𝜕 𝑗 : this affects how radiance is converted to irradiance.

Advanced Graphics – Path Tracing 7 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 tracing ng: eval aluat ating this integral usin ing a a rand andom wal alk .

Today’s Agenda:  Introduction  Path Tracing

Advanced Graphics – Path Tracing 9 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 = 𝑔 𝑠 𝑞, 𝜕 𝑝 , 𝜕 𝑗 𝑀 𝑒 𝑞, 𝜕 𝑗 cos 𝜄 𝑗 𝑒𝜕 𝑗 𝛻 𝜕 𝑗 𝑜 𝜕 𝑝 p

Advanced Graphics – Path Tracing 10 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 11 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 arriving 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 12 Path Tracing Direct Illumination 𝑀 𝑝 𝑞, 𝜕 𝑝 = 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 Here, C compensates for the fact 𝑘 𝑞, 𝜕 𝑗 cos 𝜄 𝑗 𝑒𝜕 𝑗 that we now sample the area of 𝑀 𝑝 𝑞, 𝜕 𝑗 = 𝑔 𝑠 𝑞, 𝜕 𝑝 , 𝜕 𝑗 𝑀 𝑒 the light source, instead of the 𝛻 𝑘=1 hemisphere. The probability of stumbling onto an unoccluded Obviously, sampling the entire hemisphere for each light light used to be proportional to is not necessary; we can sample the area of the light instead: solid angle; now it is always 1. C is therefore ~ (cos 𝜄 𝑝 𝐵 𝑀 𝑒 )/𝑠 2 . 𝑚𝑗𝑕ℎ𝑢𝑡 𝑘 𝑞, 𝜕 𝑗 𝐷 cos 𝜄 𝑗 𝑒𝜕 𝑗 𝑀 𝑝 𝑞, 𝜕 𝑗 = 𝑔 𝑠 𝑞, 𝜕 𝑝 , 𝜕 𝑗 𝑀 𝑒 𝐵 𝑘=1

Advanced Graphics – Path Tracing 13 Path Tracing Direct Illumination 𝑚𝑗𝑕ℎ𝑢𝑡 𝑘 𝑞, 𝜕 𝑗 𝐷 cos 𝜄 𝑗 𝑒𝜕 𝑗 𝑀 𝑝 𝑞, 𝜕 𝑗 = 𝑔 𝑠 𝑞, 𝜕 𝑝 , 𝜕 𝑗 𝑀 𝑒 𝐵 𝑘=1 Using Monte-Carlo: 𝑂 𝑞 ′ 𝐵 𝑀 𝑒 𝑘 cos 𝜄 𝑗 cos 𝜄 𝑝 𝑀 𝑝 𝑞, 𝜕 𝑗 ≈ 𝑚𝑗𝑕ℎ𝑢𝑡 ∗ 1 𝑘 𝑞, 𝑞 ′ 𝑀 𝑒 𝑞 ′ 𝑊 𝑞 ↔ 𝑂 𝑔 𝑠 𝑞, 𝜕 𝑝 , 𝑞′ ∥ 2 ∥ 𝑞 − 𝑗=1 where 𝑘 is the direct light to p from random point 𝑞 ′ on random light  𝑀 𝑒 𝑘  𝐵 𝑀 𝑒 𝑘 is the area of this light source 𝑞 ′ is the mutual visibility between p and p’.  𝑊 𝑞 ↔

Advanced Graphics – Path Tracing 14 Path Tracing Direct Illumination We now have two methods to estimate direct illumination using Monte Carlo integration: 1. By random sampling the hemisphere: 𝑂 𝑀 𝑝 𝑞, 𝜕 𝑝 ≈ 2𝜌 𝑂 𝑔 𝑠 𝑞, 𝜕 𝑝 , 𝜕 𝑗 𝑀 𝑒 𝑞, 𝜕 𝑗 cos 𝜄 𝑗 𝑗=1 2. By sampling the lights directly: 𝑂 𝐵 𝑀 𝑒 𝑘 cos 𝜄 𝑗 cos 𝜄 𝑝 𝑀 𝑝 𝑞, 𝜕 𝑗 ≈ 𝑚𝑗𝑕ℎ𝑢𝑡 ∗ 1 𝑘 𝑞, 𝑞 ′ 𝑀 𝑒 𝑞 ′ 𝑊 𝑞 ↔ 𝑞 ′ 𝑂 𝑔 𝑠 𝑞, 𝜕 𝑝 , 𝑞′ ∥ 2 ∥ 𝑞 − 𝑗=1 For 𝑂 = ∞ , these yield the same result.

Advanced Graphics – Path Tracing 15 Path Tracing Direct Illumination We now have two methods to estimate direct illumination using Monte Carlo integration: 1. By random sampling the hemisphere: 𝑂 𝑀 𝑝 𝑞, 𝜕 𝑝 ≈ 2𝜌 𝑂 𝑔 𝑠 𝑞, 𝜕 𝑝 , 𝜕 𝑗 𝑀 𝑒 𝑞, 𝜕 𝑗 cos 𝜄 𝑗 𝑗=1 2. By sampling the lights directly (three point notation): 𝑂 𝐵 𝑀 𝑒 𝑘 cos 𝜄 𝑗 cos 𝜄 𝑝 𝑀 𝑝 𝑡 ← 𝑞 ≈ 𝑚𝑗𝑕ℎ𝑢𝑡 ∗ 1 𝑘 𝑞 ← 𝑂 𝑔 𝑠 𝑡 ← 𝑞 ← 𝑟 𝑀 𝑒 𝑟 𝑊 𝑞 ↔ 𝑟 𝑟 ∥ 2 ∥ 𝑞 − 𝑗=1 For 𝑂 = ∞ , these yield the same result.

Advanced Graphics – Path Tracing 16 Path Tracing 𝑂 𝑀 𝑝 𝑞, 𝜕 𝑝 ≈ 2𝜌 𝑂 𝑔 𝑠 𝑞, 𝜕 𝑝 , 𝜕 𝑗 𝑀 𝑒 𝑞, 𝜕 𝑗 cos 𝜄 𝑗 𝑗=1 Verification Method 1 in a small C# ray tracing framework: In: Ray ray, with members O, D, N, t. Already calculated: intersection point I = O + t * D. Vector3 R = RTTools.DiffuseReflection( ray.N ); Ray rayToHemisphere = new Ray( I + R * EPSILON, R, 1e34f ); Scene.Intersect( rayToHemisphere ); if (rayToHemisphere.objIdx == LIGHT) { Vector3 BRDF = material.diffuse * INVPI; float cos_i = Vector3.Dot( R, ray.N ); return 2.0f * PI * BRDF * Scene.lightColor * cos_i; }