CS 655 Advanced Computer Graphics

1
CS 655 Advanced Computer Graphics

• Ray-Object Intersections

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Ray-Object Intersections

• Many objects can be represented as implicit
  surfaces
• Sphere (with center at c and radius R)
  fsphere(p) p c2 - R2 0
• Plane (with normal n and distance to origin d)
  fplane(p) p n D 0
• To determine where a ray intersects an object
• Need to find the intersection point p of the ray
  and the object
• The ray is represented explicitly in parametric
  form
• R(t) Ro Rdt
• Plug the ray equation into the surface equation
  and solve for t
• f(R(t)) 0
• Substitute t back into ray equation to find
  intersection point p
• p R(t) Ro Rdt

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Recall Ray Representation

• A ray can be represented explicitly (in
  parametric form) as an origin (point) and a
  direction (vector)
• Origin
• Direction
• The ray consists of all points
• R(t) Ro Rdt

R(3)
R(2)
Ro Rd R(1)
Ro R(0)
R(1)
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Ray-Sphere Intersections

• To find the intersection points of a ray with a
  sphere
• Sphere is represented as
• center (point) Sc (xc, yc, zc)
• and a radius (float) r
• The surface of the sphere is the set of all
  points (x, y, z) such that
• (x-xc)2 (y - yc)2 (z - zc)2 r2
• In this form (implicit form), it is difficult to
  directly generate surface points
• However, given a point, it is easy to see if the
  point lies on the surface
• To solve the ray-surface intersection, substitute
  the ray equation into the sphere equation.

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Ray-Sphere Intersections

• First, split the ray into its component
  equations
• x xo xdt
• y yo ydt
• z zo zdt
• Next substitute the ray equation into the sphere
  equation
• (x-xc)2 (y - yc)2 (z - zc)2 r2
• Giving
• (xo xdt - xc)2 (yo ydt - yc)2 (zo zdt
  - zc)2 r2

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Ray-Sphere Intersections

• Next, simplify the equation
• (x0 xdt - xc)2 (y0 ydt - yc)2 (z0 zdt
  - zc)2 r2
• Þ (xd2 yd2 zd2)t2
• 2(xdxo - xdxc ydyo - ydyc zdzo - zdzc)t
• (xo2- 2xoxc xc2 yo2- 2yoyc yc2 zo2-2zozc
  zc2)
• r2

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Ray-Sphere Intersections

• Let
• A xd2 yd2 zd2 1
• B 2(xdxo - xdxc ydyo - ydyc zdzo - zdzc)
• C xo2- 2xoxc xc2 yo2- 2yoyc yc2 zo2-2zozc
  zc2 - r2
• Then solve using the quadratic equation

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Ray-Sphere Intersections

• If the discriminant is negative, the ray misses
  the sphere
• The smaller positive root (if one exists) is the
  closer intersection point
• We can save some computation time by computing
  the smaller root first, then only computing the
  second root if necessary

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Ray-Sphere Intersections Algorithm

• Algorithm for ray-sphere intersection
• 1. Calculate B and C of the quadratic
• 2. Calculate the discriminant D B2 4C
• 3. If D ? 0 return false (no intersection point)
• 4. Calculate smaller intersection parameter t0
• 5. If t0 ? 0 then calculate larger t-value t1
• 6. If t1 ? 0 return false (intersection point
  behind ray)
• 7. else set t t1
• 8. else set t t0
• 9. Return intersection point p r(t) ro rdt

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Ray-Sphere Intersections Normal

• The normal n at an intersection point p on a
  sphere is

n
p
r
c
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Ray-Sphere Intersections

• Computation time
• 17 adds/subtracts
• 17 multiplies
• 1 square root
• for each ray/sphere test
• Can we reduce the number of intersection
  calculations?
• use a geometric approach

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Ray-Sphere Intersections - Geometric
tca
thc
d
Sc
OC

• R0

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Ray-Sphere Intersections - Geometric

• Determine whether the rays origin is outside
  the sphere
• if R0 - Sc lt r, then the point is inside the
  sphere
• Call the vector between R0 and Sc OC
• Find the closest approach of the ray to the
  spheres center.
• let d distance from Sc to the ray
• let tca distance from R0 to closest approach of
  ray to Sc
• tca Rd OC

tca
thc
d
Sc
OC
R0
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Ray-Sphere Intersections - Geometric

• If tca lt 0 and R0 lies outside the sphere, the
  ray does not intersect the sphere
• Otherwise, compute thc, the distance from the
  closest approach to the spheres surface
• thc2 r2 - d2
• d2 OC2 - tca2, so
• thc2 r2 - OC2 tca2

tca
thc
d
Sc
OC
R0
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Ray-Sphere Intersections - Geometric

• If thc2 lt 0 the ray does not intersect the sphere
• Otherwise, calculate the intersection distance
• If R0 is outside the sphere
• t tca - thc
• If R0 is inside the sphere
• t tca thc

tca
thc
d
Sc
OC
R0
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Ray-Sphere Intersections - Geometric

• Calculate the intersection point
• (xi, yi, zi)
• (xo xdt, yo ydt, zo zdt)
• Calculate the normal at the intersection point
• This reduces the computation (worst case) by 4
  multiplies and 1 add
• Even fewer computations if the ray misses the
  sphere

tca
thc
d
Sc
OC
R0
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Example

• Find the intersection point of the ray
• and the sphere

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Example
1. Check ray origin inside/outside
sphere Compute OC 2. Find closest approach to
spheres center. 3. Check for non intersection 4.
Compute thc 5. Check thc2 lt 0 6. Calculate the
intersection distance 7. Calculate
intersection point 8. Calculate normal
First Normalize ray
1. Compute OC
Since gt r, Ro is outside the sphere
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Example
1. Check ray origin inside/outside
sphere Compute OC 2. Find closest approach to
spheres center. 3. Check for non intersection 4.
Compute thc 5. Check thc2 lt 0 6. Calculate the
intersection distance 7. Calculate
intersection point 8. Calculate normal
2. Find the closest approach of the ray to the
spheres center.
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Example
1. Check ray origin inside/outside
sphere Compute OC 2. Find closest approach to
spheres center. 3. Check for non intersection 4.
Compute thc 5. Check thc2 lt 0 6. Calculate the
intersection distance 7. Calculate
intersection point 8. Calculate normal
3. If tca lt 0 and R0 lies outside the sphere,
the ray does not intersect the sphere
R0 does lie outside the sphere, but tca gt 0,
so continue
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Example
1. Check ray origin inside/outside
sphere Compute OC 2. Find closest approach to
spheres center. 3. Check for non intersection 4.
Compute thc 5. Check thc2 lt 0 6. Calculate the
intersection distance 7. Calculate
intersection point 8. Calculate normal
4. Compute thc, the distance from the closest
approach to the spheres surface
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Example
1. Check ray origin inside/outside
sphere Compute OC 2. Find closest approach to
spheres center. 3. Check for non intersection 4.
Compute thc 5. Check thc2 lt 0 6. Calculate the
intersection distance 7. Calculate
intersection point 8. Calculate normal
5. See if thc2 lt 0
thc2 gt 0, so the ray must hit the sphere
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Example
1. Check ray origin inside/outside
sphere Compute OC 2. Find closest approach to
spheres center. 3. Check for non intersection 4.
Compute thc 5. Check thc2 lt 0 6. Calculate the
intersection distance 7. Calculate
intersection point 8. Calculate normal

• 6. Calculate the intersection distance
• Since R0 is outside the sphere

so the ray intersects the sphere at t 3.744
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Example
1. Check ray origin inside/outside
sphere Compute OC 2. Find closest approach to
spheres center. 3. Check for non intersection 4.
Compute thc 5. Check thc2 lt 0 6. Calculate the
intersection distance 7. Calculate
intersection point 8. Calculate normal
7. Calculate the intersection point
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Example
1. Check ray origin inside/outside
sphere Compute OC 2. Find closest approach to
spheres center. 3. Check for non intersection 4.
Compute thc 5. Check thc2 lt 0 6. Calculate the
intersection distance 7. Calculate
intersection point 8. Calculate normal
8. Calculate the normal at the intersection
point
Done!
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Computing the Reflection Ray

• Angle of incidence angle of reflection

I incoming ray N Surface normal R Reflected
ray (generally want to keep I and N normalized)
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Computing the Reflection Ray

• Knowing I and N, how do we compute R?

1. qI q R 2. I, N, and R are coplanar 3. R aI
bN 4. -I ? N cos(qI) 5. N ? R cos(q
R) 6. Using 1, 4, and 5, we get 7. -I ? N N ?
R, so 8. -I ? N N ? (aI bN) 9. -I ? N (N ?
aI) (N ? bN) 10. -I ? N a(N ? I) b (N ?
N) 11. -I ? N a(N ? I) b, since N is
normalized 12. Without loss of generality, let a
1, then 13. b -2 (N ? I), so 14. R I - 2(N
? I)N
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Refraction (transparency)

• When an object is transparent, it transmits light
• Light travels through different materials at
  different speeds.
• Thus, we get light bending.
• E.g., pole in water

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Refraction (transparency)

• The manner in which light travels through a
  material is accounted for by the objects index
  of refraction
• the ratio of the speed of light through the
  material to the speed of light in a vacuum
• Light bend computed by Snells law

I
n
Medium 1
T
Medium 2
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Refraction Indices

• Index of refraction for various materials
• Material Index
• Vacuum 1.0
• Air 1.0003
• Water 1.33
• Alcohol 1.36
• Fused quartz 1.46
• Crown glass 1.52
• Flint glass 1.65
• Sapphire 1.77
• Heavy flint glass 1.89
• Diamond 2.42

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Refraction

• How do we compute T?

1. Let nIT n1/n2 sin(f)/sin(q) 2. then nIT2
sin2(f)/sin2(q) 3. sin2(q) nIT2 sin2(f)
4. since sin2(q) cos2(q) 1, we get 5. (1 -
cos2(q)) nIT2 1 - cos2(f), so 6. (1 - cos2(q))
nIT2 - 1 - cos2(f) 7. (1 - cos2(q)) nIT2 - 1
- (-N ? T)2 8. (1 - cos2(q)) nIT2 - 1 - (-N ?
(aI bN))2 9. (1 - cos2(q)) nIT2 - 1 - (-a(N
? I) b(-N ? N))2 10. (1 - cos2(q)) nIT2 - 1
-(a(cos(f)) - b)2 11. We want T normalized, so T
? T 1, so 12. 1 (aI bN) ? (aI bN) 13. 1
a2(I ? I) 2ab(I ? N) ? b2(N ? N) 14. 1 a2
2abcos(q) ? b2
I
n
Medium 1
T
Medium 2
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Refraction

• How do we compute T?

I
n
15. Solve equations 10 and 14 simultaneously 16.
(1 - cos2(q)) nIT2 - 1 -(a(cos(f)) - b)2
1 a2 2abcos(q) ? b2 17. After solving, we
get 18. So,
Medium 1
T
Medium 2
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Refraction Example

• Compute T for

Water
Glass
T
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Refraction Example
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Shadows

• The intensity of what we see at at pixel will
  depend on whether or not a light actually hits
  the object
• The point of intersection between ray 1 and the
  object is in direct illumination
• The point of intersection between ray 2 and the
  object is blocked from the light source (in
  shadow)
• The final pixel intensity must be computed taking
  this into account

Light
ray 1
ray 2
Eye
Image plane
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Shadows

• Approach for computing shadows (hard shadows)
• For each ray generated, find the first object
  intersected
• From that intersection point, send a ray directly
  to each light source
• If the ray intersects an object en route to the
  light source, the object is occluded from that
  light source
• For shadow rays, we dont care which object the
  ray intersects, just the fact that it does
  intersect

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Ray-Plane Intersections

• To find the intersection points of a ray with an
  infinite plane
• Ray equation
• Origin ro (xo, yo, zo)
• Direction rd (xd, yd, zd)
• Plane equation
• ax by cz d 0
• with a2 b2 c2 1
• normal vector pn (a, b, c)
• distance from (0, 0, 0) to plane is d
• Substitute the ray equation into the plane
  equation
• a(x0 xdt) b(y0 ydt) c(z0 zdt) d
  0
• Þ ax0 axdt by0 bydt cz0 czdt d
  0
• Solving for t we get
• t -(ax0 by0 cz0 d) / (axd byd czd)

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Ray-Plane Intersections

• In vector form we have
• If the ray is parallel to the plane and does
  not intersect
• If the normal of the plane is pointing away
  from the ray, and thus the plane is culled
• If t lt 0 then intersection point is behind the
  ray, so no real intersection occurs
• Otherwise, compute intersection point p ro
  rdt

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Ray/Plane Intersection

• Basic Algorithm
• Compute vd pn rd
• If vd gt 0 and assuming 1 sided planes then return
• If vd 0 then return (ray parallel to plane)
• Compute vo -(pn rd d)
• Compute t vo / vd
• If t lt 0 return
• If vd gt 0 reverse the planes normal
• Return r (xo xdt, yo ydt, zo zdt)

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Ray/Polygon Intersection

• Assume planar polygons
• First perform the ray/plane intersection with the
  plane in which the polygon lies
• Next, determine whether the intersection point
  lies within the polygon

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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
4 crossings - point outside polygon
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Ray/Polygon Intersection

• Polygon
• a set of N points Gn (xn, yn, zn), n 0, 1,
  N-1
• Plane
• Ax By Cz D 0, with normal Pn (A, B, C)
• Intersection point
• Ri (xi, yi, zi), Ri on the plane

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Ray/Polygon Intersection Algorithm

• Step 1
• Project the polygon onto a 2D plane
• one of the coordinate planes
• simply discard one of the coordinates
• this will project the polygon onto the plane
  defined by the other two coordinates
• topology preserving but not area preserving
• throw away the coordinate whose magnitude in the
  plane normal is greatest
• e.g., if Pn (3, -1, -5) we would throw away all
  z coordinates

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Ray/Polygon Intersection Algorithm

• Step 2
• Translate the polygon so that the intersection
  point is at the origin
• apply this translation to all points in the
  polygon
• Step 3
• Send a ray down one of the coordinate axes and
  count the number of intersections of that ray
  with the polygon
• This is done in practice by looking at each
  polygon edge, one at a time, to see if it crosses
  the coordinate axis

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Ray/Polygon Intersection Algorithm

• Problems
• Vertices that lie exactly on the ray
• Edges that lie exactly on the ray

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Ray/Polygon Intersection Algorithm

• How do we handle these cases?
• In essence, shift the vertex or edge up by e so
  that it will lie in the plane just above the axis
• This is done by checking for a y value gt 0
• What about if the intersection point is exactly a
  vertex?
• Perform a similar e shift
• What if the intersection point lies exactly on an
  edge?
• Specify it as either in or out - as long as we do
  the specification consistently, it will work okay
• If the edge is shared by two polygons, the point
  will be in one polygon and out of the other.

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Ray/Polygon Intersection Algorithm

• The effect of doing this e shift is
• These cases all work correctly now.

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The Actual Algorithm - part 1

• Determine the dominant coordinate as the largest
  magnitude component of Pn
• For each vertex (xn, yn, zn), n 0, 1, , N-1 of
  the polygon, project the vertex onto the dominant
  coordinate axis, giving (un, vn) vertices.
• Project the intersection point (xint, yint, zint)
  onto the same coordinate plane as the vertices.
• Translate all polygon vertices by (-uint, -vint),
  giving (un, vn) vertices.
• Set numCrossings 0

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The Actual Algorithm - part 2

• If v0 lt 0, set signHolder -1, otherwise set
  signHolder 1
• For i 0 to N-1 (note - when i N-1, i1 should
  be 0)
• if vi1 lt 0 set nextSignHolder -1 else set
  nextSignHolder 1
• if (signHolder ltgt nextSignHolder)
• if (ui gt 0 and ui1 gt 0) this edge crosses u
  so increment numCrossings
• else if (ui gt 0 or ui1gt0) the edge might cross
  u, so compute the intersection with the u axis
  
• ucross ui -vi (ui1 -ui)/(vi1 -vi)
• if ucross gt 0 the edge crosses u so increment
  numCrossings
• set signHolder nextSignHolder
• If numCrossings is odd, the point is inside the
  polygon

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Example

• Given a polygon
• G0 (-3, -3, 7)
• G1 (3, -4, 3)
• G2 (4, -5, 4)
• and intersection point
• Ri (-2, -2, 4)
• Does the intersection point lie within the
  polygon?

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Example

• Step 1 Get the plane normal, determine dominant
  coordinate
• Pn 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

v1 G0 - G1 (-3, -3, 7) - (3, -4, 3)
(-6, 1, 4)
v2 G2 - G1 (4, -5, 4) - (3, -4, 3) (1,
-1, 1)
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Example

• The plane normal is then v1 x v2 (assuming
  clockwise vertex specification)

Pn (-6, 1, 4) x (1, -1, 1) (5, 10, 5)
So the dominant coordinate is y
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Example

• Step 2 Project the vertices
• Step 3 Project the intersection point

G0 proj of (-3, -3, 7) Þ (-3, 7) G1 proj of
(3, -4, 3) Þ (3, 3) G2 proj of (4, -5, 4) Þ (4,
4)
G1
Ri
G2
G0
Ri proj of (-2, -2, 4) Þ (-2, 4)
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Example

• Step 4 Translate the vertices

G0 (-3, 7) - (-2, 4) Þ (-1, 3) G1 (3, 3) -
(-2, 4) Þ (5, -1) G2 (4, 4) - (-2, 4) Þ (6,
0) Ri (-2, 4) - (-2, 4) Þ (0, 0)
G1
Ri
G2
G0
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Example

• Step 5 Set numCrossings 0
• Step 6 v0 3, so
• signHolder 1

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Example

• Step 7
• i signHolder nextSignHolder numCrossings intersect
  ion point

u
v
1
0
0
-1
-1
1
-1-3(5-(-1))/(-1-3) 3.5
1
1
1
2
2
1
Since numCrossings is even, the point is outside
the polygon
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Ray/Box Intersection

• i.e., intersecting with bounding boxes
• We will deal with the case of boxes with parallel
  sides with normals parallel to the coordinate
  axes.
• Box
• Minimum extent Bl (xl, yl, zl)
• Maximum extent Bh (xh, yh, zh)
• Ray
• R(t) Ro Rdt
• Ro (xo, yo, zo)
• Rd (xd, yd, zd)

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Ray/Box Intersection (2)

• Algorithm
• Set tnear -8, tfar 8
• For the pair of X planes
• If xd 0, the ray is parallel to the planes
• If xo lt xl or xo gt xh then return FALSE
  (origin not between planes)
• Else the ray is not parallel to the planes, so
  calculate intersection distances of planes
• t1 (xl - xo) / xd (time at which the ray
  intersects the minimum x plane)
• t2 (xh - xo) / xd (time at which the ray
  intersects the maximum x plane)
• If t1 gt t2 swap t1 and t2
• If t1 gt tnear , set tnear t1
• If t2 lt tfar , set tfar t2
• If tnear gt tfar box is missed so return FALSE
• If tfar lt 0 box is behind ray so return FALSE
• Repeat step 2 for Y then Z
• All tests survived, so return TRUE

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Ray/Box Intersection example 1
tfar
tnear
yh
yl
Ro
xl
xh
60
Ray/Box Intersection example 2
tfar
tnear
yh
yl
Ro
xl
xh
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Ray/Box Intersection Example 3

• Given the ray
• Ro (0, 4, 2)
• Rd (0.213, -0.436, 0.873)
• And the box
• Bl (-1, 2, 1)
• Bh (3, 3, 3)
• Does the ray hit the box?

62
Ray/Box Intersection Example 3

• t1x (-1 - 0) / 0.218 -4.59
• t2x (3 - 0) / 0.218 13.8
• So tnear -4.59, tfar 13.8
• t1y (2 - 4) / (-.436) 4.59
• t2y (3 - 4) / (-.436) 2.29
• t1y gt t2y so swap
• Then tnear 2.29, tfar 4.59
• t1z (1 - 2) / (0.873) -1.15
• t2z (3 - 2) / (0.873) 1.15
• tnear stays the same
• tfar 1.15
• So, tnear gt tfar so the ray misses the box