CS 6958 LECTURE 7 TRIANGLES, BVH January 29, 2014 Triangles 2 - PowerPoint PPT Presentation

CS 6958 LECTURE 7 TRIANGLES, BVH January 29, 2014

Triangles 2  Let’s try to derive an intersection test

Barycentric Coordinates 3 𝑄 1 𝑐 2 𝑐 3 𝑄 0 ≤ 𝑐 1 , 𝑐 2 , 𝑐 3 ≤ 1 𝑐 1 𝑐 1 + 𝑐 2 + 𝑐 3 = 1 𝑄 2 𝑄 3 𝑄 = 𝑐 1 𝑄 1 + 𝑐 2 𝑄 2 + 𝑐 3 𝑄 3 𝑄 = 𝑐 1 𝑄 1 + 𝑐 2 𝑄 2 + 1 − 𝑐 1 − 𝑐 2 𝑄 3

Barycentric Coordinates 4 𝑄 1 𝑐 2 𝑐 3 𝑄 0 ≤ 𝑐 1 , 𝑐 2 , 𝑐 3 ≤ 1 𝑐 1 𝑐 1 + 𝑐 2 + 𝑐 3 = 1 𝑄 2 𝑄 3 𝑄 = 𝑐 1 𝑄 1 + 𝑐 2 𝑄 2 + 𝑐 3 𝑄 3 𝑄 = 𝑐 1 𝑄 1 + 𝑐 2 𝑄 2 + 1 − 𝑐 1 − 𝑐 2 𝑄 3 𝑃 + 𝑢𝑊 = 𝑐 1 𝑄 1 + 𝑐 2 𝑄 2 + 1 − 𝑐 1 − 𝑐 2 𝑄 3

Barycentric Coordinates 5 𝑃 + 𝑢𝑊 = 𝑐 1 𝑄 1 + 𝑐 2 𝑄 2 + 1 − 𝑐 1 − 𝑐 2 𝑄 3 −𝑢𝑊 + 𝑐 1 𝑄 1 − 𝑄 3 + 𝑐 2 𝑄 2 − 𝑄 3 = 𝑃 − 𝑄 3 𝑄 1 𝑓 1 = 𝑄 1 − 𝑄 3 𝑓 2 = 𝑄 2 − 𝑄 3 𝑡 = 𝑃 − 𝑄 3 𝑐 2 𝑐 3 𝑄 −𝑊 𝑓 1𝑦 𝑓 2𝑦 𝑡 𝑦 𝑢 𝑦 𝑐 1 −𝑊 𝑓 1𝑧 𝑓 2𝑧 𝑐 1 𝑡 𝑧 = 𝑧 𝑄 2 𝑄 3 𝑡 𝑨 𝑐 2 −𝑊 𝑓 1𝑨 𝑓 2𝑨 𝑨

Solution – Cramer’s Rule 6 −𝑊 𝑡 𝑦 𝑓 2𝑦 −𝑊 𝑓 1𝑦 𝑡 𝑦 𝑡 𝑦 𝑓 1𝑦 𝑓 2𝑦 𝑦 𝑦 −𝑊 𝑡 𝑧 𝑓 2𝑧 −𝑊 𝑓 1𝑧 𝑡 𝑧 𝑡 𝑧 𝑓 1𝑧 𝑓 2𝑧 𝑧 𝑧 −𝑊 𝑡 𝑨 𝑓 2𝑨 −𝑊 𝑓 1𝑨 𝑡 𝑨 𝑡 𝑨 𝑓 1𝑨 𝑓 2𝑨 𝑨 𝑨 𝑢 = , 𝑐 1 = , 𝑐 2 = −𝑊 𝑓 1𝑦 𝑓 2𝑦 −𝑊 𝑓 1𝑦 𝑓 2𝑦 −𝑊 𝑓 1𝑦 𝑓 2𝑦 𝑦 𝑦 𝑦 −𝑊 𝑓 1𝑧 𝑓 2𝑧 −𝑊 𝑓 1𝑧 𝑓 2𝑧 −𝑊 𝑓 1𝑧 𝑓 2𝑧 𝑧 𝑧 𝑧 −𝑊 𝑓 1𝑨 𝑓 2𝑨 −𝑊 𝑓 1𝑨 𝑓 2𝑨 −𝑊 𝑓 1𝑨 𝑓 2𝑨 𝑨 𝑨 𝑨  In reality too slow for intersections, but we can do better!

Scalar Triple Product 7  Scalar triple product 𝐵 ∙ 𝐶 × 𝐷 = 𝐶 ∙ 𝐷 × 𝐵 = 𝐷 ∙ 𝐵 × 𝐶  Also expressed as a determinant 𝐵 𝑦 𝐵 𝑧 𝐵 𝑨 𝐶 𝑦 𝐶 𝑧 𝐶 𝑨 𝐷 𝑦 𝐷 𝑧 𝐷 𝑨  This comes in handy 𝐵 𝑈 = 𝐵

Faster Solution 8 −𝑊 −𝑊 −𝑊 𝑦 𝑦 𝑦 𝑓 1𝑦 𝑓 1𝑧 𝑓 1𝑨 𝑒𝑓𝑜𝑝𝑛 = 𝑓 2𝑦 𝑓 2𝑧 𝑓 2𝑨 = −𝑊 ∙ 𝑓 1 × 𝑓 2 = −𝑓 1 ∙ 𝑓 2 × 𝑊 = 𝑓 1 ∙ 𝑊 × 𝑓 2 𝑒𝑓𝑜𝑝𝑛 = 𝑓 1 ∙ 𝑊 × 𝑓 2 𝑢 = 𝑓 2 ∙ 𝑡 × 𝑓 1 𝑒𝑓𝑜𝑝𝑛 𝑐 1 = 𝑡 ∙ 𝑊 × 𝑓 2 𝑒𝑓𝑜𝑝𝑛 𝑐 1 = 𝑊 ∙ 𝑡 × 𝑓 1 𝑒𝑓𝑜𝑝𝑛

Ray-Triangle Intersection 9 𝑓 1 = 𝑄 1 − 𝑄 3 𝑓 2 = 𝑄 2 − 𝑄 3 𝑠 1 = 𝑊 × 𝑓 2 𝑒𝑓𝑜𝑝𝑛 = 𝑓 1 ∙ 𝑠 1 if( 𝑏𝑐𝑡 𝑒𝑓𝑜𝑝𝑛 < 𝜗 ) miss, return;

Ray-Triangle Intersection 10 𝑓 1 = 𝑄 1 − 𝑄 3 𝑓 2 = 𝑄 2 − 𝑄 3 𝑠 1 = 𝑊 × 𝑓 2 𝑒𝑓𝑜𝑝𝑛 = 𝑓 1 ∙ 𝑠 1 if( 𝑏𝑐𝑡 𝑒𝑓𝑜𝑝𝑛 < 𝜗 ) miss, return; 𝑗𝑜𝑤𝐸𝑓𝑜𝑝𝑛 = 1 𝑒𝑓𝑜𝑝𝑛 s = 𝑃 − 𝑄 3 𝑐 1 = 𝑡 ∙ 𝑠 1 𝑗𝑜𝑤𝐸𝑓𝑜𝑝𝑛 if( 𝑐 1 < 0 ∥ 𝑐 1 > 1 ) miss, return;

Ray-Triangle Intersection 11 𝑓 1 = 𝑄 1 − 𝑄 3 𝑓 2 = 𝑄 2 − 𝑄 3 𝑠 1 = 𝑊 × 𝑓 2 𝑒𝑓𝑜𝑝𝑛 = 𝑓 1 ∙ 𝑠 1 if( 𝑏𝑐𝑡 𝑒𝑓𝑜𝑝𝑛 < 𝜗 ) miss, return; 𝑗𝑜𝑤𝐸𝑓𝑜𝑝𝑛 = 1 𝑒𝑓𝑜𝑝𝑛 s = 𝑃 − 𝑄 3 𝑐 1 = 𝑡 ∙ 𝑠 1 𝑗𝑜𝑤𝐸𝑓𝑜𝑝𝑛 if( 𝑐 1 < 0 ∥ 𝑐 1 > 1 ) miss, return; 𝑠 2 = 𝑡 × 𝑓 1 𝑐 2 = 𝑊 ∙ 𝑠 2 𝑗𝑜𝑤𝐸𝑓𝑜𝑝𝑛 if( 𝑐 2 < 0 ∥ 𝑐 1 + 𝑐 2 > 1 ) miss, return; t = 𝑓 2 ∙ 𝑠 2 𝑗𝑜𝑤𝐸𝑓𝑜𝑝𝑛 hit! save 𝑐 1 and 𝑐 2 for interpolation

Operation add / sub / mult compare divide 𝑓 1 = 𝑄 1 − 𝑄 3 3 12 𝑓 2 = 𝑄 2 − 𝑄 3 3 𝑠 1 = 𝑊 × 𝑓 2 9 𝑒𝑓𝑜𝑝𝑛 = 𝑓 1 ∙ 𝑠 1 5 if( 𝑏𝑐𝑡 𝑒𝑓𝑜𝑝𝑛 < 𝜗 ) miss, return; 2 𝑗𝑜𝑤𝐸𝑓𝑜𝑝𝑛 = 1 𝑒𝑓𝑜𝑝𝑛 1 s = 𝑃 − 𝑄 3 3 𝑐 1 = 𝑡 ∙ 𝑠 1 𝑗𝑜𝑤𝐸𝑓𝑜𝑝𝑛 6 if( 𝑐 1 < 0 ∥ 𝑐 1 > 1 ) miss, return; 2 𝑠 2 = 𝑡 × 𝑓 1 9 𝑐 2 = 𝑊 ∙ 𝑠 2 𝑗𝑜𝑤𝐸𝑓𝑜𝑝𝑛 6 if( 𝑐 2 < 0 ∥ 𝑐 1 + 𝑐 2 > 1 ) miss, return; 1 2 t = 𝑓 2 ∙ 𝑠 2 𝑗𝑜𝑤𝐸𝑓𝑜𝑝𝑛 6 hit! save 𝑐 1 and 𝑐 2 for interpolation 2 total 20 / 29 / 45 / 51 2 / 4 / 6 / 8 0 / 1 / 1 / 1

Normals 13  Flat shaded triangles 𝑂 = 𝑓 1 × 𝑓 2  Smooth shaded (per-vertex normals) 𝑂 = 𝑐 1 𝑂 1 + 𝑐 2 𝑂 2 + 1 − 𝑐 1 − 𝑐 2 𝑂 3

Updates to Hitrecord 14  include barycentric coordinates  include computed normal  alternatively, save per-vertex normals and interpolate before shading

Which Operation Most Costly? 15 foreach frame foreach pixel foreach sample generate ray intersect ray with objects shade intersection point

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