Ray Representation. A ray can be represented explicitly (in parametric form) as an origin (point) and a direction (vector):

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1 Ray Tracing Part II

2 Ray Representation A ray can be represented explicitly (in parametric form) as an origin (point) and a direction (vector): Origin: r Direction: o The ray consists of all points: r(t) = r o + r d t x y z r d o o o x y z d d d r o = r(0) r(2) r o + r d = r(1) r(3) r( 1) 2

3 Ray-Object Intersections Many objects can be represented as implicit surfaces: Sphere (with center at c and radius R): f s (p) = p c 2 - R 2 = 0 Plane (with normal n and distance to origin d): f p (p) = p n + D = 0 To determine where a ray intersects an object: Plug the ray equation into the surface equation and solve for t: f(r o + r d t) = 0 Substitute t back into ray equation to find intersection point p: p = r(t) = r o + r d t 3

4 Ray-Sphere Intersections To find the intersection points of a ray with a sphere: Sphere is represented as: center: c = (x c, y c, z c ) radius: R The sphere is the set of all points (x, y, z) such that: (x-x c ) 2 + (y - y c ) 2 + (z - z c ) 2 = R 2 4

5 Ray-Sphere Intersections First, split the ray into its component equations: x = x o + x d t y = y o + y d t z = z o + z d t Then substitute the ray equation into the sphere equation: Giving: (x-x c ) 2 + (y - y c ) 2 + (z - z c ) 2 = R 2 (x o + x d t - x c ) 2 + (y o + y d t - y c ) 2 + (z o + z d t - z c ) 2 = R 2 5

6 Ray-Sphere Intersections Next multiply out the squared terms: (x 0 + x d t - x c ) 2 + (y 0 + y d t - y c ) 2 + (z 0 + z d t - z c ) 2 = R 2 (x d2 + y d2 + z d2 )t 2 + 2(x d x o - x d x c + y d y o - y d y c + z d z o - z d z c )t + (x o2-2x o x c + x c2 + y o2-2y o y c + y c2 + z o2-2z o z c + z c2 ) = R 2 How do we solve for t? Use the Quadratic Equation 6

7 Ray-Sphere Intersections Let A = x d2 + y d2 + z d2 = 1 B = 2(x d x o - x d x c + y d y o - y d y c + z d z o - z d z c ) C = x o2-2x o x c + x c2 + y o2-2y o y c + y c2 + z o2-2z o z c + z c2 - r 2 So At 2 + Bt + C = 0 and we can solve this using the quadratic equation: t B B 2 2 4C 7

8 Ray-Sphere Intersections t B B 2 2 4C 3 possibilities: Case 1: B 2 4C < 0 Zero real roots. No intersection. Case 2: B 2 4C = 0 One real root t 0 = t 1 = -B/2 Case 3: B 2 4C > 0 Two real roots Subcase a: t 0 < 0, t 1 > 0 t 1 is the correct answer Subcase b: 0 < t 0 < t 1 t 0 is the correct answer t 0 t 0 t 1 t 1 case 1 case 2 case 3a case 3b 8

9 Ray-Sphere Intersections If the discriminant B 2 4C < 0, the ray misses the sphere. The smaller positive root (if one exists) is the closest intersection point. We can save time by computing the small root first and only computing the large root if necessary. t 0 B B 2 2 4C t 1 B B 2 2 4C 9

10 Ray-Sphere Intersections: Algorithm Algorithm for ray-sphere intersection: Calculate B and C of the quadratic Calculate the discriminant: D = B 2 4AC If D < 0 return false (no intersection point) Calculate smaller t-value t 0 : If t 0 0 then B D calculate larger t-value t 1 : t1 2 If t 1 0 return false (intersection is behind ray) else set t = t 1 else set t = t 0 Return intersection point: p = r(t) = r o + r d t t 0 B 2 D 10

11 11 The normal n at an intersection point p on a sphere is: Ray-Sphere Intersections: Normal R z z R y y R x x R c i c i c i c p n n p c R

12 Ray-Sphere Intersections Computation time per ray-sphere test: 17 additions / subtractions 17 multiplies 1 square root Can we reduce the number of intersection calculations? Yes, use a geometric approach. 12

13 Ray-Sphere Intersections - Geometric t ca t hc r 0 OC d c 13

14 Ray-Sphere Intersections - Geometric 1. What is OC? t ca t hc r d OC c r o r 0 OC d c 2. What is t ca? tca OC cos, t ca OC cos OC r d cos t or, equivalently: OC ca r d 14

15 Ray-Sphere Intersections - Geometric 1. Determine if r 0 is inside the sphere: If c - r 0 < R it s inside the sphere. r 0 t ca OC t hc d c r d Let OC = c - r 0 2. Find the closest point on the ray to c. Let d = closest distance to c. Let t ca = distance from r 0 to the point closest to c. t ca = r d OC 15

16 Ray-Sphere Intersections - Geometric 3. If t ca < 0 and r 0 is outside the sphere, the ray does not intersect the sphere. 4. Otherwise, compute t hc, the distance from the closest point to the sphere s surface t hc2 = R 2 - d 2 d 2 = OC 2 - t ca 2 r 0 t ca OC t hc d c r d t hc2 = R 2 - OC 2 + t ca 2 16

17 Ray-Sphere Intersections - Geometric 5. If t hc2 < 0 the ray does not intersect the sphere 6. Otherwise, calculate the intersection distance r 0 t ca OC t hc d c r d If r 0 is outside the sphere: t = t ca - t hc If r 0 is inside the sphere: t = t ca + t hc 17

18 Ray-Sphere Intersections - Geometric 7. Calculate the intersection point: t ca t hc r d (x i, y i, z i ) = (x o + x d t, y o + y d t, z o + z d t) r 0 OC d c 8. Calculate the normal at the intersection point: n x i x r c y i r y c z i z r c This reduces the computation by 4 multiplies and 1 add in the worst case. Even fewer computations if the ray misses the sphere 18

19 Ray-Plane Intersections To find the intersection point of a ray with a plane: Plane equation: ax +by +cz +d = 0 with a 2 + b 2 +c 2 = 1 normal n = (a, b, c) distance from (0, 0, 0) to plane is d Ray equation: Origin: r o = (x o, y o, z o ) Direction: r d = (x d, y d, z d ) Substitute the ray equation into the plane equation: a(x 0 + x d t) + b(y 0 + y d t) + c(z 0 + z d t) + d = 0 ax 0 + ax d t + by 0 + by d t + cz 0 + cz d t + d = 0 Solving for t we get: t = -(ax 0 + by 0 + cz 0 + d) / (ax d + by d + cz d ) 19

20 Ray-Plane Intersections In vector form: If nr d 0 the ray is parallel to the plane and does not intersect n r d 0 t If the plane s normal points away from the ray and thus the plane is culled. If t 0 then intersection point is behind the ray, so no real intersection occurs Otherwise, compute intersection: p = r o + r d t nro d nr 20 d

21 Ray/Plane Intersection Basic Algorithm: v d = n r d if v d 0 and plane has one side then return if v d 0 then return (ray is parallel to plane) t = -(n r o + d) / v d if t < 0 then return return r = (x o + x d t, y o + y d t, z o + z d t) 21

22 Ray/Polygon Intersection Assume planar polygons First, intersect the plane the polygon is on with the ray. Next, determine if the intersection is within the polygon. How do we quickly tell if the intersect is within the polygon? 22

23 Ray/Polygon Intersection Idea: shoot a ray from the intersection point in an arbitrary direction if the ray crosses an even number of polygon edges, the point is outside the polygon if the ray crosses an odd number of polygon edges, the point is inside the polygon 11 crossings - point inside polygon 23 4 crossings - point outside polygon

24 Ray/Polygon Intersection Polygon: a set of N points P j = (x j, y j, z j ), j = 0, 1, N-1 Plane: ax + by + cz + d = 0, with normal n = (a, b, c) Intersection point: p i = (x i, y i, z i ), p i on the plane 24

25 Ray/Polygon Intersection Algorithm Step 1: Project the polygon onto a coordinate plane Simply discard one of the coordinates. This will project the polygon onto the plane defined by the other two coordinates. This doesn t preserve area, but does preserve topology. Discard the coordinate whose magnitude in the plane normal is the largest. If n = (3, -1, -5) we would throw away the z coordinates. 25

26 Ray/Polygon Intersection Algorithm Step 2: Translate the polygon so that the intersection point is at the origin. Translate all its vertices. Step 3: Send a ray down a coordinate axis and count the number of times it intersects the polygon. This can be done simply by counting the number of edges that cross the coordinate axis. 26

27 Ray/Polygon Intersection Algorithm Problems: Vertices that lie exactly on the ray Edges that lie exactly on the ray 27

28 Ray/Polygon Intersection Algorithm How do we handle these cases? Essentially, shift the vertex or edge up by e so that it will be just above the axis. 28

29 Ray/Polygon Intersection Algorithm The effect of doing this e shift is: These cases all work correctly now. 29

30 The Complete Algorithm - part 1 1. Find the coordinate of n with the largest magnitude. 2. For each vertex (x j, y j, z j ), j = 0, 1,, N-1 of the polygon, project the vertex onto the coordinate plane of the other two coordinates. This gives us a new coordinates system (u j, v j ). 3. Project the intersection point (x i, y i, z i ) onto the same coordinate plane as the vertices. 4. Translate all polygon vertices by (-u i, -v i ). (u j, v j ) = (u j, v j ) - (u i, v i ). 5. Set numcrossings = 0 30

31 The Complete Algorithm - part 2 6. If v 0 < 0, set sign = -1, otherwise set sign = 1 7. For i = 0 to N-1 (let u N = u 0, v N = v 0 ) if v i+1 < 0 set nextsign = -1 else set nextsign = 1 if (sign nextsign) if (u i > 0 and u i+1 > 0) numcrossings++ else if (u i > 0 or u i+1 >0) \\ the edge might cross +u \\ compute the intersection with the u axis u c = u i - v i * (u i+1 -u i )/(v i+1 -v i ) if u c > 0 numcrossings++ sign = nextsign 8. If numcrossings is odd, the intersection is inside the polygon. 31

32 Example Given a polygon: P 0 = (-3, -3, 7) P 1 = (3, -4, 3) P 2 = (4, -5, 4) P 0 P1 and intersection point R i = (-2, -2, 4) P 2 Does the intersection point lie within the polygon? 32

33 Example Step 1: Get the plane normal, determine dominant coordinate n can be computed from the cross product of two vectors in the plane The vertices of the polygon can be used to compute vectors in the plane P 0 P 2 P1 P 0 = (-3, -3, 7) P 1 = (3, -4, 3) P 2 = (4, -5, 4) R i = (-2, -2, 4) 33

34 Example Compute the Normal: First, compute edge vectors from the vertices v 1 = P 0 - P 1 = (-3, -3, 7) - (3, -4, 3) = (-6, 1, 4) v 2 = P 2 - P 1 = (4, -5, 4) - (3, -4, 3) = (1, -1, 1) P 0 P 2 P1 P 0 = (-3, -3, 7) P 1 = (3, -4, 3) P 2 = (4, -5, 4) R i = (-2, -2, 4) 34

35 Example The plane normal is then v 2 x v 1 n = (-6, 1, 4) x (1, -1, 1) = (5, 10, 5) P 0 n P1 So the dominant coordinate is y P 2 P 0 = (-3, -3, 7) P 1 = (3, -4, 3) P 2 = (4, -5, 4) R i = (-2, -2, 4) 35

36 Example Step 2: Project the vertices project P 0 = (-3, -3, 7) 3,7 project P 1 = (3, -4, 3) 3,3 project P 2 = (4, -5, 4) 4,4 Step 3: Project the intersection point project R i = (-2, -2, 4) 2,4 R i P 0 v P 1 P 2 u 36

37 Example Step 4: Translate the vertices P 0 = (-3, 7) - (-2, 4) 1,3 P 1 = (3, 3) - (-2, 4) 5,1 P 2 = (4, 4) - (-2, 4) 6, R i = (-2, 4) - (-2, 4), P 1 R i u P 2 P 0 v 37

38 Example Step 5: Set numcrossings = 0 Step 6: v 0 = 3, so sign = 1 P 0 P 1 R i v P 2 u 38

39 Example Step 7: For i = 0 to N-1 \\ (let un = u0, vn = v0 ) if vi+1 < 0 set nextsign = -1 else set nextsign = 1 if (sign nextsign) if (ui > 0 and ui+1 > 0) numcrossings++ else if (ui > 0 or ui+1 >0) \\ the edge might cross +u \\ compute the intersection with the u axis uc = ui - vi * (ui+1 -ui )/(vi+1 -vi ) if uc > 0 numcrossings++ sign = nextsign i sign nextsign numcrossings intersection point (Figure is drawn upside-down. Sorry. The positive v axis points down. P 0 =(-1,3) 1-1-3*(5-(-1))/(-1-3) = 3.5 P 1 =(5,-1) R i u v P 2 = (6,0) Since numcrossings is even, the point is outside the polygon 39

40 Shadows Send a shadow ray from intersection point to the light: Compute the following shadow ray properties: Shadow ray direction: s d = (l p) / l p Distance to the light from the intersection point: t l = l p Test if shadow ray intersects an object before reaching the light: In theory, the possible range of t-values is t [0, t l ] Due to numerical (floating-point) error, test in the range t [e, t 1 ] where e is some small value (such as 2 16 ) 40

41 Ray Tracing (images by Nan Schaller) Ambient Light only 47

42 Ray Tracing With shadows 48

43 Ray Tracing Primary rays only 49

44 Ray Tracing With Phong shading 50

45 Ray Tracing With reflections 51

46 Ray Tracing With transmissions 52

47 Ray Tracing Movies 53

48 Other Ray Traced Images 54

49 Other Ray Traced Images 55

50 Other Ray Traced Images 56

51 Other Ray Traced Images 57

52 Other Ray Traced Images 58

53 Other Ray Traced Images 59

54 Other Ray Traced Images 60

55 Other Ray Traced Images 61

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