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Lecture 7 - “Path Tracing” Welcome! 푰 풙, 풙′ = 품(풙, 풙′) 흐 풙, 풙′ + 흆 풙, 풙′, 풙′′ 푰 풙′, 풙′′ 풅풙′′ 푺 Today’S Agenda

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Lecture 7 - “Path Tracing” Welcome! 푰 풙, 풙′ = 품(풙, 풙′) 흐 풙, 풙′ + 흆 풙, 풙′, 풙′′ 푰 풙′, 풙′′ 풅풙′′ 푺 Today’S Agenda INFOMAGR – Advanced Graphics Jacco Bikker - November 2016 - February 2017 Lecture 7 - “Path Tracing” Welcome! 푰 풙, 풙′ = 품(풙, 풙′) 흐 풙, 풙′ + 흆 풙, 풙′, 풙′′ 푰 풙′, 풙′′ 풅풙′′ 푺 Today’s Agenda: . Introduction . Path Tracing Advanced Graphics – Path Tracing 3 Introduction Previously in Advanced Graphics The Rendering Equation: 퐿표 푥, 휔표 = 퐿퐸 푥, 휔표 + 푓푟 푥, 휔표, 휔푖 퐿푖 푥, 휔푖 cos 휃푖 푑휔푖 훺 …which models light transport as it happens in the real world, by summing: . Direct illumination: 퐿퐸(푥, 휔표) . Indirect illumination, or reflected light: 훺 푓푟 푥, 휔표, 휔푖 퐿푖 푥, 휔푖 cos 휃푖 푑휔푖 We used quantities flux 훷 (joules per second), radiance 퐿 (flux per 푚2 per sr) and irradiance 퐸 (joules per second per 푚2). 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 휔푖 and an outgoing direction 휔표, it defines the relation between received irradiance and reflected radiance. A physically based BRDF has some important properties: 흎 . 푓푟 휔표, 휔푖 ≥ 0 풊 . 푓푟 휔표, 휔푖 = 푓푟 휔푖, 휔표 풏 . 훺 푓푟 휔표, 휔푖 cos 휃표 푑휔표 ≤ 1 푎푙푏푒푑표 Example: diffuse BRDF, 푓 ω , ω = . 푟 표 푖 휋 Note that the diffuse BRDF is view independent; 휔표 is irrelevant. It does however take into account 휔푖: 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. Path tracing: evaluating this integral using a random walk. 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 푝 with normal 푛 is directly visible. What is the radiance travelling via 푝 towards the eye? Answer: 퐿표 푝, 휔표 = 푓푟 푝, 휔표, 휔푖 퐿푑 푝, 휔푖 cos 휃푖 푑휔푖 훺 휔푖 휔표 푛 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 푝 . 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 푁 2휋 퐿 푝, 휔 ≈ 푓 푝, 휔 , 휔 퐿 푝, 휔 cos 휃 표 표 푁 푟 표 푖 푑 푖 푖 푖=1 휔푖 휔표 푛 p Advanced Graphics – Path Tracing 11 Path Tracing We integrate over the hemisphere, which Direct Illumination has an area of 2휋. 푁 2휋 Do not confuse this with the 1/ 휋 factor in the BRDF, which doesn’t compensate 퐿표 푝, 휔표 ≈ 푓푟 푝, 휔표, 휔 푖 퐿푑 푝, 휔 푖 cos 휃 푁 for the surface of the hemisphere, but the 푖=1 integral of cos 휃 over the hemisphere (휋). Questions: . Why do we multiply by 2휋? . What is the radiance 퐿푑(푝, 휔푖) towards 푝 for e.g. a 100W light? . What is the irradiance 퐸 arriving at 푝 from this light? 퐿 is per sr; 퐿푑(푝, 휔푖) is proportional to the solid angle of the light as seen 2 휔푖 from p, so (cos 휃표 퐴퐿 )/푟 . 푑 휔표 푛 Note that the 100W flux is spread out over the area; irradiance is defined per 푚2. p Advanced Graphics – Path Tracing 12 Path Tracing Direct Illumination 퐿표 푝, 휔표 = 푓푟 푝, 휔표, 휔푖 퐿푑 푝, 휔푖 cos 휃푖 푑휔푖 훺 In many directions, we will not find light sources. We can improve our estimate by sampling the lights separately. 푙푖ℎ푡푠 Here, C compensates for the fact 푗 that we now sample the area of 퐿표 푝, 휔푖 = 푓푟 푝, 휔표, 휔푖 퐿 푝, 휔푖 cos 휃푖 푑휔푖 푑 the light source, instead of the 푗=1 훺 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 2 is therefore ~(cos 휃표퐴퐿푑)/푟 . 푙푖ℎ푡푠 푗 퐿표 푝, 휔푖 = 푓푟 푝, 휔표, 휔푖 퐿푑 푝, 휔푖 퐶 cos 휃푖 푑휔푖 푗=1 퐴 Advanced Graphics – Path Tracing 13 Path Tracing Direct Illumination 푙푖ℎ푡푠 푗 퐿표 푝, 휔푖 = 푓푟 푝, 휔표, 휔푖 퐿푑 푝, 휔푖 퐶 cos 휃푖 푑휔푖 푗=1 퐴 Using Monte-Carlo: 푁 퐴 cos 휃 cos 휃 1 퐿 푖 표 퐿 푝, 휔 ≈ 푙푔ℎ푡푠 ∗ 푓 푝, 휔 , 푝 ′ 퐿푗 푝, 푝 ′ 푉 푝 ↔ 푝 ′ 푑 표 푖 푁 푟 표 푑 ∥ 푝 − 푝 ′ ∥2 푖=1 where 푗 ′ . 퐿푑 is the direct light to p from random point 푝 on random light 푗 . 퐴 is the area of this light source 퐿푑 . 푉 푝 ↔ 푝 ′ is the mutual visibility between p and p’. 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: 푁 2휋 퐿 푝, 휔 ≈ 푓 푝, 휔 , 휔 퐿 푝, 휔 cos 휃 표 표 푁 푟 표 푖 푑 푖 푖 푖=1 2. By sampling the lights directly: 푁 퐴 cos 휃 cos 휃 1 퐿 푖 표 퐿 푝, 휔 ≈ 푙푔ℎ푡푠 ∗ 푓 푝, 휔 , 푝 ′ 퐿푗 푝, 푝 ′ 푉 푝 ↔ 푝 ′ 푑 표 푖 푁 푟 표 푑 ∥ 푝 − 푝 ′ ∥2 푖=1 For 푁 = ∞, 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: 푁 2휋 퐿 푝, 휔 ≈ 푓 푝, 휔 , 휔 퐿 푝, 휔 cos 휃 표 표 푁 푟 표 푖 푑 푖 푖 푖=1 2. By sampling the lights directly (three point notation): 푁 퐴 cos 휃 cos 휃 1 퐿 푖 표 퐿 푠 ← 푝 ≈ 푙푔ℎ푡푠 ∗ 푓 푠 ← 푝 ← 푞 퐿푗 푝 ← 푞 푉 푝 ↔ 푞 푑 표 푁 푟 푑 ∥ 푝 − 푞 ∥2 푖=1 For 푁 = ∞, these yield the same result. Advanced Graphics – Path Tracing 16 Path Tracing 푁 2휋 퐿 푝, 휔 ≈ 푓 푝, 휔 , 휔 퐿 푝, 휔 cos 휃 표 표 푁 푟 표 푖 푑 푖 푖 Verification 푖=1 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; } Advanced Graphics – Path Tracing 17 푁 Path Tracing 퐴 cos 휃 cos 휃 1 퐿 푖 표 퐿 푝, 휔 ≈ 푙푔ℎ푡푠 ∗ 푓 푝, 휔 , 푝′ 퐿푗 푝, 푝 ′ 푉 푝 ↔ 푝 ′ 푑 표 푖 푁 푟 표 푑 ∥ 푝 − 푝 ′ ∥2 Verification 푖=1 Method 2 in a small C# ray tracing framework: // construct vector to random point on light Vector3 L = Scene.RandomPointOnLight() - I; float dist = L.Length(); L /= dist; float cos_o = Vector3.Dot( -L, lightNormal ); float cos_i = Vector3.Dot( L, ray.N ); if ((cos_o <= 0) || (cos_i <= 0)) return BLACK; // light is not behind surface point, trace shadow ray Ray r = new Ray( I + EPSILON * L, L, dist - 2 * EPSILON ); Scene.Intersect( r ); if (r.objIdx != -1) return Vector3.Zero; // light is visible (V(p,p’)=1); calculate transport Vector3 BRDF = material.diffuse * INVPI; float solidAngle = (cos_o * Scene.LIGHTAREA) / (dist * dist); return BRDF * lightCount * Scene.lightColor * solidAngle * cos_i; Advanced Graphics – Path Tracing 18 Path Tracing 0.1s Advanced Graphics – Path Tracing 19 Path Tracing 0.5s Advanced Graphics – Path Tracing 20 Path Tracing 2.0s Advanced Graphics – Path Tracing 21 Path Tracing 30.0s Advanced Graphics – Path Tracing 22 Path Tracing Rendering using Monte Carlo Integration In the demonstration, we sampled each light using only 1 sample. The (very noisy) result is directly visualized. To get a better estimate, we average the result of several frames (and thus: several samples). Observations: 1. The light sampling estimator is much better than the hemisphere estimator; 2. Relatively few samples are sufficient for a recognizable image; 3. Noise reduces over time, but we quickly get diminishing returns. Advanced Graphics – Path Tracing 23 Path Tracing Indirect Light Returning to the full rendering equation: We know how to evaluate direct lighting: What remains is indirect light. This is the light that is not emitted by the surface in direction 휔푖, but reflected.
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