Graphics 2009/2010, period 1. Lecture 8: ray tracing

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1 Graphics 2009/2010, period 1 Lecture 8 Ray tracing

2 Original plan Finish the basic stuff (book ch 1-9) today: We will not look into signal processing (chapter 4) We will look at triangle rasterization (cf. chapter 3, but differently covered here) Also at shading (cf. chapter 9, also some differences here) And again at Z-buffering (not covered in the book) And then cover some advanced topics : Ray tracing (chapter 10) Texture mapping (chapter 11) Radiosity and shadows (not in the book) Graphics pipeline, part II (chapter 12) But you need ray tracing & texture mapping for P2

3 New plan Finish the basic stuff (book ch 1-9) today next Wednesday: We will not look into signal processing (chapter 4) We will look at triangle rasterization (cf. chapter 3, but differently covered here) Also at shading (cf. chapter 9, also some differences here) And again at Z-buffering (not covered in the book) And then cover some advanced topics : Ray tracing (chapter 10) today Texture mapping (chapter 11) next Monday Radiosity and shadows (not in the book) Graphics pipeline, part II (chapter 12) But you need ray tracing & texture mapping for P2 new DL 6.11.

4 Outline

5 Ray tracing / ray casting Idea: for every pixel Compute ray from viewpoint through pixel center Determine first object hit by ray (including intersection point) Calculate shading for the pixel (possibly with recursion)

6 Triangles Rays Intersections Triangles: barycentric coordinates Recall that the plane V through the points a, b, and c can be written as p = a + β(b a) + γ(c a). a p c b Q: When does a point p in V lie in the triangle formed by a, b, and c?

7 Triangles Rays Intersections Triangles: barycentric coordinates Recall that the plane V through the points a, b, and c can be written as p = a + β(b a) + γ(c a). a p c b Q: When does a point p in V lie in the triangle formed by a, b, and c?

8 Triangles Rays Intersections Triangles: barycentric coordinates Recall that the plane V through the points a, b, and c can be written as p = a + β(b a) + γ(c a). a p c b Q: When does a point p in V lie in the triangle formed by a, b, and c?

9 Rays: parametric representation Triangles Rays Intersections A ray starting in the point e with direction d can be written as p = e + td. p Q: are there any conditions on t? d e

10 Intersecting a ray and a triangle Triangles Rays Intersections c If there is a unique intersection between a ray and a triangle, then the intersection point p satisfies both the plane equation and the ray equation, as well as the conditions on β, γ and t. a p b d e

11 Intersecting a ray and a triangle So we can write: Triangles Rays Intersections x e + tx d = x a + β(x b x a ) + γ(x c x a ) y e + ty d = y a + β(y b y a ) + γ(y c y a ) z e + tz d = z a + β(z b z a ) + γ(z c z a ) which can be rewritten as or as (x a x b )β + (x a x c )γ + x d t = x a x e (y a y b )β + (y a y c )γ + y d t = y a y e (z a z b )β + (z a z c )γ + z d t = z a z e x a x b x a x c x d β x a x e y a y b y a y c y d γ = y a y e z a z b z a z c z d t z a z e

12 Intersecting a ray and a triangle Triangles Rays Intersections Now, if we write x a x b x a x c x d β x a x e y a y b y a y c y d γ = y a y e z a z b z a z c z d t z a z e as then we see that β x a x e A γ = y a y e t z a z e β x a x e γ = A 1 y a y e t z a z e

13 Outline Triangles Rays Intersections

14 Snell s law Getting rid of sines Finding the refraction vector Light traveling from one transparent medium into another one is refracted.

15 Snell s law Snell s law Getting rid of sines Finding the refraction vector Angles before and after refraction are related as follows: λ s sin θ = λ t sin φ. where λ s and λ t are the refractive indices of the source and target media, respectively, and θ and φ the angles indicated in the image. n d θ φ t

16 Getting rid of sines Snell s law Getting rid of sines Finding the refraction vector An equation that relates sines of the angles θ and φ is not as convenient as an equation that relates the cosines of the angles. With the identity sin 2 φ + cos 2 φ = 1 we derive the following equation from Snell s law: n d θ φ t cos 2 φ = 1 λ2 s(1 cos 2 θ) λ 2 t

17 Getting rid of sines Snell s law: λ s sin θ = λ t sin φ sin φ = λs λ t sin θ Snell s law Getting rid of sines Finding the refraction vector Trigonometric identity: sin 2 φ + cos 2 φ = 1 sin 2 φ = 1 cos 2 φ and cos 2 φ = 1 sin 2 φ cos 2 φ = 1 λ2 s(1 cos 2 θ) λ 2 t

18 Constructing an orthonormal basis Snell s law Getting rid of sines Finding the refraction vector How do we find the refracted vector t? Assume the incoming vector d and the normal n are normalized. First, t lies in the plane spanned by d and n. n d θ b φ t Next, we can set up an orthonormal basis in this plane by picking an appropriate vector b.

19 Finding the refraction vector Snell s law Getting rid of sines Finding the refraction vector We have t = b sin φ n cos φ d = b sin θ n cos θ n d θ b φ t

20 Finding the refraction vector Snell s law Getting rid of sines Finding the refraction vector We have t = b sin φ n cos φ d = b sin θ n cos θ So we can solve for b: b = d+n cos θ sin θ and for t: t = = sin φ(d+n cos θ) sin θ λs(d+n cos θ) λ t = λs(d n(d n)) λ t n cos φ n cos φ n 1 λ2 s (1 (d n)2 ) λ 2 t d θ n b φ t

21 Outline Snell s law Getting rid of sines Finding the refraction vector

22 Copying and transforming objects Copying and transforming objects Ray/instance intersection is an elegant technique to place various transformed copies of an object in a scene. Expl.: circle elipse O M 1 O M 3 O M 2 O M 4 O

23 Copying and transforming objects Copying and transforming objects Ray/instance intersection Instead of making actual copies, we simply store a reference to a base object, together with a transformation matrix. O M 1 O M 3 O M 2 O M 4 O

24 Ray/instance intersection Copying and transforming objects Ray/instance intersection O p M 1 r MO Mp r

25 Ray/instance intersection Copying and transforming objects Ray/instance intersection To determine the intersection q of a ray r with an instance MO, we first compute the intersection p of the inverse transformed ray M 1 r and the original object O. O p M 1 r The point q is then simply Mp. MO Mp This way, complicated intersection tests (e.g. ray/ellipsoid) can often be replaced by simpler tests (ray/sphere). r

26 Outline Copying and transforming objects Ray/instance intersection

27 Combining shapes Computing intersections C S For ray tracing, we can basically use any object that allows us to calculate its intersection with a 3D line. Using Constructive Solid Geometry (CSG) we can build complex objects from simple ones with set operations. C S S C C S C S

28 Intersections and CSG Combining shapes Computing intersections Big advantage: instead of actually constructing the objects, we can calculate ray-object intersections with the original objects and perform set operations on the resulting intervals. S C C S C S

29 Intersections and CSG Combining shapes Computing intersections S For every base object, we maintain an interval (or set of intervals) representing the part of the ray inside the object. The intervals for combined objects are computed with the same set operations that are applied to the base objects. C C S C S

30 Outline Combining shapes Computing intersections

31 The bottleneck in ray tracing Bounding boxes Uniform spatial subdivision Octrees BSP trees Our algorithm for testing ray/object intersections is more or less the same as that of [Whitted, 1980]: test all objects, and report the nearest one. It is estimated that 75% to 95% of the time in ray/tracing is spent on ray/object intersections [Chang, 2001].

32 Bounding boxes Bounding boxes Uniform spatial subdivision Octrees BSP trees A common technique to improve ray/object intersection query times is the use of bounding boxes. One advantage: We don t need the actual intersection point but just a yes or no answer to the intersection test.

33 Bounding boxes Bounding boxes Uniform spatial subdivision Octrees BSP trees How do we get that?

34 Hierarchical bounding boxes Bounding boxes Uniform spatial subdivision Octrees BSP trees But why stop with bounding objects? We can also bound groups of bounding boxes, and build a hierarchy.

35 Hierarchical bounding boxes Bounding boxes Uniform spatial subdivision Octrees BSP trees But why stop with bounding objects? We can also bound groups of bounding boxes, and build a hierarchy. In practice, the choice of what items to group is a hard problem.

36 Current research Bounding boxes Uniform spatial subdivision Octrees BSP trees Packing an object in more than one box makes the ray/object test more expensive if there is a hit, but may drastically reduce false positives.

37 Uniform spatial subdivision Bounding boxes Uniform spatial subdivision Octrees BSP trees An alternative method for intersection test speed-up is to put a regular grid over the object space, and to traverse from cell to cell. Q: What s the best grid size? And do we need a regular grid?

38 Octrees Bounding boxes Uniform spatial subdivision Octrees BSP trees One way to get a grid that somehow resembles the distribution of objects: Octrees An octree is the 3D version of the Quadtree.

39 Octrees Bounding boxes Uniform spatial subdivision Octrees BSP trees The idea is as follows: given a set of objects, we first compute an axis-parallel bounding box that contains all of them.

40 Octrees Bounding boxes Uniform spatial subdivision Octrees BSP trees Next, if the box contains more than a predetermined number of objects, we split it evenly along all dimensions. We continue until the condition on the maximal number of objects in a node is satisfied

41 Octrees Bounding boxes Uniform spatial subdivision Octrees BSP trees Next, if the box contains more than a predetermined number of objects, we split it evenly along all dimensions. We continue until the condition on the maximal number of objects in a node is satisfied

42 Octrees Bounding boxes Uniform spatial subdivision Octrees BSP trees Next, if the box contains more than a predetermined number of objects, we split it evenly along all dimensions. We continue until the condition on the maximal number of objects in a node is satisfied

43 Octrees Bounding boxes Uniform spatial subdivision Octrees BSP trees Next, if the box contains more than a predetermined number of objects, we split it evenly along all dimensions. We continue until the condition on the maximal number of objects in a node is satisfied

44 Octrees Bounding boxes Uniform spatial subdivision Octrees BSP trees Traversal of the nodes is similar to the traversal in uniform spatial subdivision but somewhat more complicated.

45 Octrees Bounding boxes Uniform spatial subdivision Octrees BSP trees Instead of splitting evenly, we could also do balanced splits, based on the object distribution.

46 BSP trees Bounding boxes Uniform spatial subdivision Octrees BSP trees We have seen BSP trees before. Apart from speeding up projective rendering, they can also be used for ray tracing However, in ray tracing, we do not only deal with triangles, so finding splitting planes is a bit more complicated

47 BSP trees Bounding boxes Uniform spatial subdivision Octrees BSP trees We try to find a splitting plane that splits the objects into two groups of more or less equal size In practice, we limit ourselves to axis-parallel splitting planes.

48 BSP trees Bounding boxes Uniform spatial subdivision Octrees BSP trees We try to find a splitting plane that splits the objects into two groups of more or less equal size In practice, we limit ourselves to axis-parallel splitting planes.

49 BSP trees Bounding boxes Uniform spatial subdivision Octrees BSP trees We go into recursion on the two groups, continuing until every group has at most a predetermined number of objects. Usually, we split along alternating dimensions.

50 BSP trees Bounding boxes Uniform spatial subdivision Octrees BSP trees We go into recursion on the two groups, continuing until every group has at most a predetermined number of objects. Usually, we split along alternating dimensions.

51 BSP trees Bounding boxes Uniform spatial subdivision Octrees BSP trees We go into recursion on the two groups, continuing until every group has at most a predetermined number of objects. Usually, we split along alternating dimensions.

52 BSP trees Bounding boxes Uniform spatial subdivision Octrees BSP trees We go into recursion on the two groups, continuing until every group has at most a predetermined number of objects. Usually, we split along alternating dimensions.

53 BSP trees Bounding boxes Uniform spatial subdivision Octrees BSP trees Finding a good splitting strategy is hard, and topic of ongoing research.

54 Outline Bounding boxes Uniform spatial subdivision Octrees BSP trees

55 More ray tracing... Bounding boxes Uniform spatial subdivision Octrees BSP trees... in the 2nd programming assignment Triangles and lots of texture mapping (covered in the next lecture) Related changes: New deadline for P2: Nov 6 (instead of Oct 30) Two add. practicals in the exam week (Nov 2 and Nov 4) Only one practical/ta today and next Monday (go to room 402) Don t use this as a motivation to start late!!! Keep in mind that P2 is larger than P1

56 Programming assignments Bounding boxes Uniform spatial subdivision Octrees BSP trees A few words on P1: We hope to finish (preliminary) grading by Mon, Oct 12 Final code checks for randomly selected groups on Wed, Oct 14 Selected teams will be contacted by before Monday

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