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 RayObject 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 RaySphere 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: (xx c ) 2 + (y  y c ) 2 + (z  z c ) 2 = R 2 4
5 RaySphere 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: (xx 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 RaySphere 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 o22x o x c + x c2 + y o22y o y c + y c2 + z o22z o z c + z c2 ) = R 2 How do we solve for t? Use the Quadratic Equation 6
7 RaySphere 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 o22x o x c + x c2 + y o22y o y c + y c2 + z o22z 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 RaySphere 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 RaySphere 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 RaySphere Intersections: Algorithm Algorithm for raysphere 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 tvalue t 0 : If t 0 0 then B D calculate larger tvalue 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: RaySphere 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 RaySphere Intersections Computation time per raysphere test: 17 additions / subtractions 17 multiplies 1 square root Can we reduce the number of intersection calculations? Yes, use a geometric approach. 12
13 RaySphere Intersections  Geometric t ca t hc r 0 OC d c 13
14 RaySphere 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 RaySphere 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 RaySphere 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 RaySphere 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 RaySphere 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 RayPlane 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 RayPlane 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, N1 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,, N1 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 N1 (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 N1 \\ (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 upsidedown. Sorry. The positive v axis points down. P 0 =(1,3) 113*(5(1))/(13) = 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 tvalues is t [0, t l ] Due to numerical (floatingpoint) 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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