Visible Surface Detection
Shmuel Wimer
Bar Ilan Univ., Eng. Faculty
Technion, EE Faculty
May 2012
1
Back-Face Detection
A point  x, y, z  is behind a polygon surface if Ax  By  Cz  D  0,
where A, B, C and D are the plane parameters of the polygon. Test is
simplified by considering normal of face and viewing direction.
N   A, B, C 
Vview
A polygon is back face if Vview  N  0. In a right-handed
viewing system viewing direction is along negative z axis,
Vview   0, 0, 1 . Then, a polygon is a back face if C  0.
May 2012
2
Depth-Buffer Methods
S3
S2
S1
view plane
yv
xv
 x, y 
zv
Three surfaces overlapping pixel position (x,y) on the view
plane. The visible surface, S1, has the smallest depth value.
May 2012
3
Z - Buffer Algorithm
Initialize each pixel ( x, y ) of depth buffer and frame buffer (color):
depthBuf  x, y   1.0; // Z is normalized to [0,1.0]
frameBuf  x, y   background color; // initialize to background
for each polygon { // traverse all polygons
for each pixel  x, y   polygon { // rasterization
if (z  x, y   depthBuf  x, y  ) { // check closer pixel
get (compute) color  x, y  ; // compute color of pixel
depthBuf  x, y   z  x, y  ; frameBuf  x, y   color  x, y  ;
}}}
May 2012
4
Efficient Depth Calculation
Given the depth value in a vertex of a polygon, the depth
of any other point in the plane containing the polygon can
be calculated efficiently (additions only).
Depth is calculated from plane equatuion z    Ax  By  D  C .
At rasterization x along scan-line changes by  1. Therefore,
z  x  1    A  x  1  By  D  C  z  A C , and the ratio  A C
is constant for the entire polygon.
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5
Scanning can start at the top vertex of polygon and
obtain the first depth value. Progressing from scan line
to next one y is decreased by 1, which eases the depth
calculation of first pixel along the scan line.
top scan line
y scan line
y+1 scan line
x
May 2012
x`
6
If m is the slope of the edge then x  x  1 m . Therefore the
depth at first pizel in new scan line is z   z   A m  B  C .
All values of 1 m can be stored in a lookup table (memory).
 A m  B
C is calculated once for the entire edge, and
A C and B C calculated once for the entire polygon.
If the edge is vertical the slope is infinite and z  z  B C.
The starting point of scan line can use mid-point method
or Bernsham-type algorithm.
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7
A-Buffer Method
background
opaque surface
foreground
transparent
surface
It is named so since z-buffer is used for depth semantics and
in some sense a-buffer has the nature of the other end of
alphabet. It is an extension using antialiasing, area-averaging
visibility detection method, supporting transparency.
May 2012
8
Ray Tracing
opaque
opaque
transparent
Projection
reference
point
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9
Highly realistic. It detects visibility, transparency, shadows,
illumination effects, and generates perspective views.
Computational intensive.
May 2012
10
Binary Ray-Tracing Tree
R4
S3 transparent
S4 opaque
R3
T3
R1
R2
S2 opaque
T1
S1
R1
T1
S3
S1 transparent
R3
Projection
reference
point
S2
T3
R2
S4
R4
May 2012
11
Left branches represent reflection paths, right
branches represent transmission paths.
A path is terminated if any of the following occurs:
The ray intersects no surface.
The ray intersects a light source that is not a reflecting
surface.
The tree reached a depth limit.
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12
R
L
Ν
u – incoming ray
u
H
R – unit reflected ray
N – unit surface normal
L – unit vector pointing to light source
H – unit vector half way between –u and L
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13
The paths along the
direction of L is referred
as the shadow ray.
source.
If any object intersects the shadow ray between the
surface and the point light source, the surface position
is in shadow with respect to that source.
For reflection path there exists R  u  2  u   N  N,
hence, R  u   2u  N  N.
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14
In transparent surface light is transmitted through
material. The unit transmission vector T is obtained
from u and N and refraction indices.


i
i
T  u   cos r  cos i  N
r
r


r
T
N
u
May 2012
i

cosr  1  i r  1  cos2 i
2

15
Calculating Pixel Intensity
It works bottom up, where the intensity at a child node
is added to parent with an attenuation relative to the
distance of the child’s surface from the parent’s
surface. The intensity assigned to the pixel is the sum
of attenuated intensities at the root.
If the primary ray for a pixel doesn’t intersect with an
object of the scene, the ray-tracing tree is empty and
the pixel is assigned the background intensity.
May 2012
16
Ray-Surface Intersection Calculations
projection
reference point
pixel point
y
P0
u
x
z
A point along the ray at distance s from P0 is given
by P  P0  su.


Unit vector along initial ray path is u  Ppixel  P0 | Ppixel  P0 |.
At each surface intersection P0 and u are updated acoording
to R and T.
May 2012
17
Ray-tracing is computations intensive, requiring root
finding to get the value of parameter s at ray-surface
incidence point. Efficient techniques exists for spherical,
planar and also spline surfaces.
P
y
P0
r
u
Pc
Ray-Sphere
Intersections
x
z
| P  Pc |2 r 2  0  | P0  su  Pc |2 r 2  0.
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

Set Pc  P0  ΔP, then s 2  2  u  ΔP  s  | ΔP |2 r 2  0.
At intersection there exists
s  u  ΔP 
2
2
u

ΔP

|
ΔP
|

r
.


2
If the discriminant is negative then either the ray
doesn’t intersect the sphere or the sphere is behind P0.
In either case the sphere is eliminated from further
consideration.
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s  u  ΔP 
2
2
u

ΔP

|
ΔP
|

r
is susceptible to


2
round-off error if the sphere is small and very far,
since r
2
2
| ΔP | .
This can be avoided in most cases by the
rearrangement s  u  ΔP  r  | ΔP   u  ΔP  u | ,
2
2
yielding two terms of same order of magnitude.
May 2012
20
Ray-Polyhedron Intersections
Computing intersection with polyhedron is expensive. It is
useful to include polyhedron in the smallest enclosing
sphere and exclude any intersection calculation if the ray
doesn’t intersect the sphere.
Front faces of the polyhedron are identified first by u  N  0.
For those we find first the intersection point of the ray with
D  N  P0
face's plane N  P   D  N   P0  su    D  s  
N u
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To find whether the point is within the face an oddeven test is in order. Among all the faces intersected
by the ray, the one with the smallest s is chosen. If
there’s no intersection with any face, the polyhedron is
excluded from further consideration.
May 2012
22
Reducing Intersection Calculations
Ray tracer consumes 95% time on intersection
calculations. It is possible to enclose groups of adjacent
objects in spheres, calculate intersection with enclosing
sphere first, and proceed to real objects only if intersection
exists.
Another approach is space-subdivision, where the entire
scene is enclosed with a cube. The cube is successively
divided into smaller cubes until the number of spaces
contained in a cube exceeds a predefined limit or the
size of cube reaches some threshold.
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23
pixel ray
Cells are traced through the individual cells. intersection
tests are performed only for cells containing surfaces.
The first intersected surface is the visible for that ray.
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24
N2
u
Pin
N1
Pout
N1   1, 0, 0 
N 2   0, 1, 0 
N3   0, 0, 1
N3
Given a unit vector u of the ray and the entry point Pin , the
candidate faces of exit point are those satisfying u  Nk  0.
The exit point satisfies Pout,k  Pin  sk u, sk is the distance along
the ray from Pin to Pout,k . Exit point satisfies Nk  Pout,k   Dk .
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Solving for sk yields sk    Dk  N k  Pin   N k  u  .
The smallest value of sk defines the exit face.
Notice that other 2 intersection must occure outside
the cube since cube is convex. It is a good practice
to look at the largest component of u and start with
the face on the perpendicular plane.
May 2012
26
Consider the cube plane attempted
6
for exit point calculation. The cube
1
face divides the plane into nine
sectors. If the exit point is falling in
5
2
sector 0 we are done. If it is falling
in any of 1, 3, 2 or 4 we know the
0
4
7
face containing the exit point and
3
then can find it. In case of 5, 6, 7
and 8 further test is required.
May 2012
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27