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- 1. VOLUME RENDERING OF UNSTRUCTURED TETRAHEDRAL GRIDS
- 2. Flow • Introduction • Problem Formulation • Tetrahedral Interpolation • Implementation • Results • Demo • Endnotes 04-06-2014 Nitesh Bhatia | CPDM 2
- 3. Introduction: Volume Rendering • Volume rendering is a technique that can be used to visualize sampled 3D scalar data as a continuous medium or extract features. Most algorithms for direct volume rendering have assumed structured data in form of rectilinear grid. • In this project we worked on a method for rendering unstructured volume; volume represented by group of tetrahedrals. 04-06-2014 Nitesh Bhatia | CPDM 3
- 4. Problem Formulation • The idea is to convert unstructured input tetrahedral grid (UG) to output structured regular grid (SG) and render it using existing ray casting system. • The data represented in UG must be interpolated to produce SG. • The UG consists of tetrahedrals bounded by four vertices numbered 1,2,3,4, the coordinates of ith vertex being (xi, yi, zi) and associated data value is denoted fi. • The data values are assumed to be the values of an unknown locally smooth trivariate function interpolate discussed in next heading. 04-06-2014 Nitesh Bhatia | CPDM 4 1/3
- 5. Problem Formulation • Let P = (x,y,z) be the point at which the value of interpolation function is to be estimated. Two different interpolation schemes are followed here: • Scheme 1: Map SG to UG • For a given point P of SG and find the tetrahedral associated with P in UG • Estimate the function values for given cell based on interpolate. 04-06-2014 Nitesh Bhatia | CPDM 5 empty SG UG SG 2/3
- 6. Problem Formulation • Scheme 2: Map UG to SG • Take a tetrahedral from UG and find points Ps lying on SG • Estimate the function value at Ps based on interpolate. 04-06-2014 Nitesh Bhatia | CPDM 6 UG empty SG SG 3/3
- 7. Tetrahedral Interpolation: interpolate • Given a tetrahedron T with vertices v1, v2, v3, v4 and function value associated with these vertices be f1, f2, f3 and f4, the problem is to find interpolated function value f for any given point P. 04-06-2014 Nitesh Bhatia | CPDM 7 .P(f) v1(f1) v2(f2)v4(f4) v3(f3) 1/4
- 8. Tetrahedral Interpolation: interpolate Geometric Solution • Take ratio of perpendicular distance of P to a face with perpendicular distance of opposite vertex to that face. • Find these ratios with all four faces and name them l1, l2, l3 and l4. • If the point P is lying inside tetrahedral these ratios will come between 0 and 1. • Sum of these ratios will always be 1. • These l1,l2, l3 and l4 are known as barycentric coordinates of point P with respect to tetrahedral T. • f = l1*f1 + l2*f2 + l3*f3 + l4*f4 04-06-2014 8 2/4 Nitesh Bhatia | CPDM
- 9. Tetrahedral Interpolation: interpolate • Mathemetical Interpretation 04-06-2014 Nitesh Bhatia | CPDM 9 3/4
- 10. Tetrahedral Interpolation: interpolate 04-06-2014 Nitesh Bhatia | CPDM 10 4/4
- 11. Implementation In implementation we are following 3 step approach 1. Load the vertices and tetrahedron information from given .ts file into CPU memory 2. Based on two schemes presented, perform computations using OpenCL (or OpenMP) to form a regular grid 3. Display the grid by Ray Casting in OpenCL using CL-GL Interoperability. 04-06-2014 Nitesh Bhatia | CPDM 11
- 12. Implementation: Description Step1: • The data set given .ts file is form of list of vertices with 3D coordinates and associated function value and then a list of tetrahedrons with asociated 4 vertices. • We are first loading this information into memory. • While loading the vertices we are computing minimum and maximum values for (x,y,z) coordinates and storing it as minX, minY, minZ, maxX, maxY, maxZ. • We are then finding the difference between these minimum and maximum values and storing it as diffX, diffY, diffZ. • We are then computing diff equal to maximum of diffX, diffY and diffZ. • We are then finding dimensions of our bounding box with side equal to diff. 04-06-2014 Nitesh Bhatia | CPDM 12
- 13. Step 2: • Given • We are computing A-1 for each tetrahedron • We define a constant STEP_SIZE = 128 (or any other value) which gives dimension of our volume as 128*128*128. • We are setting the resolution (step size) of our volume as res = diff / STEP_SIZE 04-06-2014 Nitesh Bhatia | CPDM 13
- 14. Scheme 1: 04-06-2014 Nitesh Bhatia | CPDM 14 1. Finding point lying on SG 2. searching for associated T in UG and finding f value using interpolation
- 15. • This scheme is implemented in both OpenCL and OpenMP • In OpenMP implementation we are adding following two lines as compiler directive in starting of loop: • #paragma omp set_num_threads(8) • #paragma omp parallel for shared(i,j,k) • In OpenCL implementation we are setting our dimensions as 1D with • size_t global_size = {STEP_SIZE * STEP_SIZE * STEP_SIZE} • Here we are Parallelizing in terms of volume element 04-06-2014 Nitesh Bhatia | CPDM 15
- 16. Scheme2: 04-06-2014 Nitesh Bhatia | CPDM 16 1. Finding limits of points lying inside tetrahedral 2. For given limits finding f values of points associated with SG
- 17. • Scheme 2 is implemented in OpenCL. We are setting our dimensions as 1D with size_t global_size = tet_qty • Here we are Parallelizing in terms of tetrahedral quantity Step 3: • We are then giving this 1D grid of function values to existing ray tracer (provided by nVidia in their SDK) as input. 04-06-2014 Nitesh Bhatia | CPDM 17
- 18. Results • Hardware / Software for tests 04-06-2014 Nitesh Bhatia | CPDM 18 GPU Model: nVidia Quadro FX 580 Cores: 32 Core Clock: 450 MHz Memory: 512MB Memory Bandwidth: 25.6 GiB/s CPU Model: Intel Core i7 860 Cores / Threads: 4/8 Clock Speed: 2.8GHz (3.0GHz when running on full load) Memory: 8 GB Memory Bandwidth: 21GB/s OS / SDKs Microsoft Windows 7 Professional 64Bit Visual Studio 2010 32Bit Microsoft OpenMP nVidia CUDA SDK 3.2 nVidia OpenCL 1.1 Intel OpenCL 1.1 alpha Input UG Torus1.ts Torusf1.ts Torus8.ts Engine.ts
- 19. 04-06-2014 Nitesh Bhatia | CPDM 19 Scheme 1 Scheme 2 STEP_SIZE = 512 Time: 8.4 sec STEP_SIZE = 512 Time: 2.5 sec
- 20. 04-06-2014 Nitesh Bhatia | CPDM 20 Scheme 1 Scheme 2 STEP_SIZE = 512 Time: 2.5 sec STEP_SIZE = 512 Time: 3.3 sec
- 21. 04-06-2014 Nitesh Bhatia | CPDM 21
- 22. 04-06-2014 Nitesh Bhatia | CPDM 22
- 23. • Demo Video URL: https://www.youtube.com/watch?v=CaBuZ7 se7-o 04-06-2014 Nitesh Bhatia | CPDM 23
- 24. Impressions • Learning OpenCL was a challenging task but we it was interesting. • Debugging OpenCL is difficult task as stream output (“printf” function) cannot be called in openCL kernel. In Intel’s compiler is based on OpenCL 1.1 in which “printf” is supported. • Double precision computations are not supported on my card. • Graphic Driver Crash Problem 04-06-2014 Nitesh Bhatia | CPDM 24
- 25. “We now know a thousand ways not to build a light bulb” 04-06-2014 Nitesh Bhatia | CPDM 25 THANKS !

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