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CISC 3620 Lecture 12:Rendering pipeline revisited
Topics:
Rendering overviewClipping
Polygon renderingRasterization
Display issues
Rendering overview
3
Objectives
● Examine what happens between the vertex shader and the fragment shader
● Introduce basic implementation strategies● Clipping ● Rendering
– lines– polygons
● Give a sample algorithm for each
Angel and Shreiner: Interactive Computer Graphics 7E © Addison-Wesley 2015
4
Overview
● At end of the geometric pipeline, vertices have been assembled into primitives
● Must clip out primitives that are outside the view frustum– Algorithms based on representing primitives by lists of vertices
● Must find which pixels can be affected by each primitive– Fragment generation– Rasterization or scan conversion
Angel and Shreiner: Interactive Computer Graphics 7E © Addison-Wesley 2015
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Required Tasks
● Clipping● Rasterization or scan conversion● Transformations● Some tasks deferred until fragment processing
– Hidden surface removal – Antialiasing
Angel and Shreiner: Interactive Computer Graphics 7E © Addison-Wesley 2015
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Rasterization Meta Algorithms
● Any rendering method must process every object and assign a color to every pixel
● Think of rendering algorithms as two loops– over objects– over pixels
● The order of these loops defines two strategies– Image-space approach– Object-space approach
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Object-Space Approach
● For every object, determine which pixels it covers and shade these pixels– Pipeline approach– Must keep track of depths for hidden surface removal– Cannot handle most global lighting calculations– Need entire framebuffer available at all times
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Image-Space Approach
● For every pixel, determine which object that projects on the pixel is closest to the viewer and compute the shade of this pixel– Ray tracing paradigm– Need all objects available
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Image-Space Approach:Patch renderers
● Divide framebuffer into small patches● Determine which objects affect each patch● Image-space render each patch separately using
relevant objects● Used in limited power devices such as cell phones
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Clipping
11
Objectives
● Clipping lines● First implementation of algorithms● Clipping polygons
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Clipping
● 2D against clipping window● 3D against clipping volume● Easy for line segments, polygons● Hard for curves and text
– Convert to lines and polygons first
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Clipping 2D Line Segments
● Brute force approach: compute intersections with all sides of clipping window– Inefficient: one division per intersection
● Idea: eliminate as many cases as possible without computing intersections
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Cohen-Sutherland Algorithm
● Start with four lines that determine the sides of the clipping window
x = xmaxx = xmin
y = ymax
y = ymin
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Cohen-Sutherland Algorithm
● Start with four lines that determine the sides of the clipping window
● What are the possible line-rectangle interactions?
x = xmaxx = xmin
y = ymax
y = ymin
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The Cases
● Case 1: both endpoints of line segment inside all four lines– Draw (accept) line segment as is
● Case 2: both endpoints outside all lines and on same side of a line– Discard (reject) the line segment
x = xmaxx = xmin
y = ymax
y = ymin
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Case 1Case 2
17
The Cases
● Case 3: One endpoint inside, one outside– Must do at least one intersection
● Case 4: Both outside– May have part inside (4a)– Must do at least one intersection
x = xmaxx = xmin
y = ymax
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Case 4a
Case 4b
18
Classify endpoints with Outcodes
● For each endpoint, define an outcode: b0b1b2b3
– b0 = 1 if y > ymax, 0 otherwise
– b1 = 1 if y < ymin, 0 otherwise
– b2 = 1 if x > xmax, 0 otherwise
– b3 = 1 if x < xmin, 0 otherwise
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Classify endpoints with Outcodes
● For each endpoint, define an outcode: b0b1b2b3
– b0 = 1 if y > ymax, 0 otherwise
– b1 = 1 if y < ymin, 0 otherwise
– b2 = 1 if x > xmax, 0 otherwise
– b3 = 1 if x < xmin, 0 otherwise
● Outcodes divide space into ____ regions● Computation of outcode requires at most _____
subtractions
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Classify endpoints with Outcodes
● For each endpoint, define an outcode: b0b1b2b3
– b0 = 1 if y > ymax, 0 otherwise
– b1 = 1 if y < ymin, 0 otherwise
– b2 = 1 if x > xmax, 0 otherwise
– b3 = 1 if x < xmin, 0 otherwise
● Outcodes divide space into 9 regions● Computation of outcode requires at most 4
subtractions
Angel and Shreiner: Interactive Computer Graphics 7E © Addison-Wesley 2015
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Classify endpoints with Outcodes
● For each endpoint, define an outcode: b0b1b2b3
– b0 = 1 if y > ymax, 0 otherwise
– b1 = 1 if y < ymin, 0 otherwise
– b2 = 1 if x > xmax, 0 otherwise
– b3 = 1 if x < xmin, 0 otherwise
● Outcodes divide space into 9 regions● Computation of outcode requires at most 4
subtractions
Angel and Shreiner: Interactive Computer Graphics 7E © Addison-Wesley 2015
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Classify endpoints with Outcodes
● For each endpoint, define an outcode: b0b1b2b3
– b0 = 1 if y > ymax, 0 otherwise
– b1 = 1 if y < ymin, 0 otherwise
– b2 = 1 if x > xmax, 0 otherwise
– b3 = 1 if x < xmin, 0 otherwise
● Outcodes divide space into 9 regions● Computation of outcode requires at most 4
subtractions
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Using Outcodes
● What are the outcodes for each of these vertices?
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Using Outcodes
● What are the outcodes for each of these vertices?
Angel and Shreiner: Interactive Computer Graphics 7E © Addison-Wesley 2015
1000
0001
1000
1000
00010000
0000 0010
0000
00100010
0010
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Using Outcodes
● AB: outcode(A) = outcode(B) = 0–
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Using Outcodes
● AB: outcode(A) = outcode(B) = 0– Accept line segment
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Using Outcodes
● CD: outcode (C) = 0, outcode(D) 0 0–
–
–
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Using Outcodes
● CD: outcode (C) = 0, outcode(D) 0 0– Compute intersection with edge– Position of 1 in outcode(D) determines which edge to intersect with– Note: if there were a segment from C to a point in a region with 2
ones in outcode, we might have to do two intersection calculations
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Using Outcodes
● EF: outcode(E) logically ANDed with outcode(F) (bitwise) 0 0–
–
–
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Using Outcodes
● EF: outcode(E) logically ANDed with outcode(F) (bitwise) 0 0– Both outcodes have a 1 bit in the same place– Line segment is outside of corresponding side of clipping window– Reject
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Using Outcodes
● GH and IJ: same outcodes, neither zero but logical AND yields zero–
–
–
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Using Outcodes
● GH and IJ: same outcodes, neither zero but logical AND yields zero– Shorten line segment by intersecting with one of sides of window– Compute outcode of intersection (new endpoint of shortened line segment)– Reexecute algorithm
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Efficiency
● In many applications, the clipping window is small relative to the size of the entire object database– Most line segments are outside one or more side of the
window and can be eliminated based on their outcodes
● Inefficiency when code has to be reexecuted for line segments that must be shortened in more than one step
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Cohen Sutherland in 3D
● Use 6-bit outcodes ● When needed, clip line segment against planes
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Clipping and Normalization
● General clipping in 3D requires intersection of line segments against arbitrary plane
● Example: oblique view
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Plane-Line Intersections
a=n⋅( po−p1 )
n⋅( p2−p1)
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Normalized Form
before normalization after normalization
top view
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● Normalization is part of viewing (pre clipping)– but after normalization, we clip against sides of right parallelepiped
● Typical intersection calculation now requires only a floating point subtraction– e.g. is x > xmax ?
Polygon Clipping
● Not as simple as line segment clipping– Clipping a line segment yields at most one line segment– Clipping a polygon can yield multiple polygons
●
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Polygon Clipping
● Not as simple as line segment clipping– Clipping a line segment yields at most one line segment– Clipping a polygon can yield multiple polygons
● However, clipping a convex polygon can yield at most one polygon
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Tessellation and Convexity
● One strategy is to replace nonconvex (concave) polygons with a set of triangular polygons (a tessellation)
● Also makes fill easier● Tessellation through tessellation shaders
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Clipping as a Black Box
● Can consider line segment clipping as a process that takes in two vertices and produces either no vertices or the vertices of a clipped line segment
41Angel and Shreiner: Interactive Computer Graphics 7E © Addison-Wesley 2015
xi=( ymax− y1)x2−x1
y2− y1
Pipeline Clipping of Line Segments
● Clipping against each side of window is independent of other sides– Can use four independent clippers in a pipeline
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Pipeline Clipping of Line Segments
● Clipping against each side of window is independent of other sides– Can use four independent clippers in a pipeline
43Angel and Shreiner: Interactive Computer Graphics 7E © Addison-Wesley 2015
Pipeline Clipping of Line Segments
● Clipping against each side of window is independent of other sides– Can use four independent clippers in a pipeline
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Pipeline Clipping of Line Segments
● Clipping against each side of window is independent of other sides– Can use four independent clippers in a pipeline
45Angel and Shreiner: Interactive Computer Graphics 7E © Addison-Wesley 2015
Pipeline Clipping of Line Segments
● Clipping against each side of window is independent of other sides– Can use four independent clippers in a pipeline
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Pipeline Clipping of Polygons
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● Strategy used in SGI Geometry Engine● Small increase in latency● In three dimensions: add front and back clippers
Bounding Boxes
● Rather than doing clipping on a complex polygon, we can use an axis-aligned bounding box or extent– Smallest rectangle aligned with axes that encloses the polygon
– Simple to compute:
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Bounding Boxes
● Rather than doing clipping on a complex polygon, we can use an axis-aligned bounding box or extent– Smallest rectangle aligned with axes that encloses the polygon
– Simple to compute: max and min of x and y
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Bounding boxes
● Can usually determine accept/reject based only on bounding box
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Bounding boxes
● Can usually determine accept/reject based only on bounding box
reject
accept
requires detailed clipping
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Hidden surface removal
Clipping and Visibility
● Clipping has much in common with hidden-surface removal
● In both cases, we are trying to remove objects that are not visible to the camera
● Often we can use visibility or occlusion testing early in the process to eliminate as many polygons as possible before going through the entire pipeline
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Hidden Surface Removal
● Object-space approach: use pairwise testing between polygons (objects)
● Worst case complexity O(____) for n polygons
partially obscuring can draw independently
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Hidden Surface Removal
● Object-space approach: use pairwise testing between polygons (objects)
● Worst case complexity O(n2) for n polygons
partially obscuring can draw independently
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Painter’s Algorithm
● Render polygons in back-to-front order so that polygons behind others are simply painted over
● But how do we get back-to-front order?B behind A as seen by viewer Fill B then A
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Depth Sort
● Requires ordering of polygons first – O(n log n) calculation for ordering– Not every polygon is either in front or behind all
other polygons
● Order polygons and deal with easy cases first, harder later
Polygons sorted by distance from center of projection
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Depth Sort
● What are the possible depth relationships between pairs of polygons?
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Easy Cases
● A lies behind all other polygons– Can render
● Polygons overlap in z but not in either x or y– Can render independently
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Hard Cases
Overlap in all directionsbut one is fully on
one side of the other
cyclic overlap
piercing
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Back-Face Removal (Culling)
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● One way to save rendering is to only render polygons on the side of an object that faces the viewer
● In OpenGL we can simply enable culling– but may not work correctly if we
have nonconvex objects
Back-Face Removal (Culling)
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● Face is visible iff _______
Back-Face Removal (Culling)
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● Face is visible iff (equivalently):– 90 -90 -90 -90– cos 0 -90– v • n 0 -90
● Plane of face has form ax + by +cz +d =0– Can be written as p • n = 0
where p = (x y z 1)T
– after normalization v = ( 0 0 1 0)T
● So __________
Back-Face Removal (Culling)
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● Face is visible iff (equivalently):– 90 -90 -90 -90– cos 0 -90– v • n 0 -90
● Plane of face has form ax + by +cz +d =0– Can be written as p • n = 0
where p = (x y z 1)T
– after normalization v = ( 0 0 1 0)T
● So need only test the sign of c
Image Space Approach
● Look at each projector (nm for an n x m frame buffer) and find closest of k polygons
● Complexity O(nmk)● Can implement with z-buffer● Ray tracing
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z-Buffer Algorithm
● Use a buffer called the z or depth buffer to store the depth of the closest object at each pixel found so far
● As we render each polygon, compare the depth of each pixel to depth in z buffer
● If less, place shade of pixel in color buffer and update z buffer
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Efficiency
● If we work scan line by scan line as we move across a scan line, the depth changes satisfy ax+by+cz=0
Along scan line
y = 0z = - x
c
a
In screen space x = 1
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Scan-Line Algorithm
● Can combine shading and culling through scan line algorithm
● scan line i: no need for depth information, can only be in no or one polygon
● scan line j: need depth information only when in more than one polygon
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Implementation
● Need a data structure to store– Flag for each polygon (inside/outside)– Incremental structure for scan lines that stores which
edges are encountered – Parameters for planes
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Visibility Testing● In many realtime applications, such as games, we want to eliminate as
many objects as possible within the application– Reduce burden on pipeline– Reduce traffic on bus
● Partition space with Binary Spatial Partition (BSP) Tree
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Simple Example
consider 6 parallel polygons
top view
The plane of A separates B and C from D, E and F
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BSP Tree
● Can continue recursively – Plane of C separates B from A– Plane of D separates E and F
● Can put this information in a BSP tree– Use for visibility and
occlusion testing
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Rasterization
Objectives
● Survey Line Drawing Algorithms– DDA– Bresenham’s Algorithm
● Aliasing and Antialiasing
Rasterization
● Rasterization (scan conversion)– Determine which pixels are inside primitive specified by a
set of vertices– Produces a set of fragments– Fragments have a location (pixel location) and other
attributes such color and texture coordinates that are determined by interpolating values at vertices
● Pixel colors determined later using color, texture, and other vertex properties
Scan Conversion of Line Segments
● Start with line segment in window coordinates with integer values for endpoints
● Assume implementation has a write_pixel function
y = mx + h
x
ym
DDA Algorithm
● Digital Differential Analyzer– DDA was a mechanical device for numerical solution of differential equations– Line y=mx+ h satisfies differential equation
dy/dx = m = y/x = y2-y1/x2-x1
● Along scan line x = 1for(x=x1; x<=x2; x++) {
y += m;
write_pixel(x, round(y), line_color)
}
●
●
DDA Algorithm
● Digital Differential Analyzer– DDA was a mechanical device for numerical solution of differential equations– Line y=mx+ h satisfies differential equation
dy/dx = m = y/x = y2-y1/x2-x1
● Along scan line x = 1for(x=x1; x<=x2; x++) {
y += m;
write_pixel(x, round(y), line_color)
}
● Try for m = 0.2, x1 = 0, x2 = 10● Try for m = 5.0, x1 = 0, x2 = 10
Problem
● DDA = for each x plot pixel at closest y– Problems for steep lines
Using Symmetry
● Use for 1 m 0 -90 -90● For m > 1, swap role of x and y
– For each y, plot closest x
Bresenham’s Algorithm
● DDA requires one floating point addition per step● We can eliminate all floating point math through Bresenham’s
algorithm● Consider only 1 m 0 -90 -90
– Other cases by symmetry
● Assume pixel centers are at half integer coordinates● If we start at a pixel that has been written, there are only two
candidates for the next pixel to be written into the frame buffer
Candidate Pixels
1 -90 m -90 0
previous pixel
candidates
Note that line could havepassed through anypart of this pixel
-
Decision Variable
-
d = x(b-a)
d is an integerd > 0 use upper pixeld < 0 use lower pixel
Incremental Form
● More efficient if we look at dk, the value of the decision variable at x = k
dk+1= dk –2y, if dk <0
dk+1= dk –2(y - x), otherwise
● For each x, we need do only an integer addition and a test
● Single instruction on graphics chips
As pseudocode
plotLine(x0,y0, x1,y1)
dx = x1 - x0
dy = y1 - y0
D = 2*dy - dx
y = y0
for x from x0 to x1
plot(x,y)
if D > 0
y = y + 1
D = D - 2*dx
end if
D = D + 2*dy
● Try for m = 0.2, x1 = 0, x2 = 10
● https://en.wikipedia.org/wiki/Bresenham%27s_line_algorithm#Algorithm_for_integer_arithmetic
Polygon Scan Conversion
● Scan Conversion = Fill● How to tell inside from outside
– Convex easy– Nonsimple difficult– Odd even test
● Count edge crossings
– Winding number
odd-even fill
Winding Number
● Count clockwise encirclements of point
● Alternate definition of inside:
inside if winding number 0 0
winding number = ?
winding number = ?
Winding Number
● Count clockwise encirclements of point
● Alternate definition of inside:
inside if winding number 0 0
winding number = 2
winding number = 1
Filling in the Frame Buffer
● Fill at end of pipeline– Convex Polygons only– Nonconvex polygons assumed to have been tessellated– Shades (colors) have been computed for vertices
(Gouraud shading)– Combine with z-buffer algorithm
● March across scan lines interpolating shades● Incremental work easy
Using Interpolation
span
C1
C3
C2
C5
C4
scan line
● C1 C2 C3 specified by glColor or by vertex shading● C4 determined by interpolating between C1 and C2● C5 determined by interpolating between C2 and C3● interpolate between C4 and C5 along span
Flood Fill● Fill can be done recursively if we know a seed point located inside (WHITE)● Scan convert edges into buffer in edge/inside color (BLACK)
flood_fill(int x, int y) { if(read_pixel(x,y) == WHITE) { write_pixel(x,y,BLACK); flood_fill(x-1, y); flood_fill(x+1, y); flood_fill(x, y+1); flood_fill(x, y-1); } }
By André Karwath aka Aka - Own work, CC BY-SA 2.5, https://commons.wikimedia.org/w/index.php?curid=481651
Scan Line Fill
● Can also fill by maintaining a data structure of all intersections of polygons with scan lines– Sort by scan line– Fill each span
vertex order generated by vertex list desired order
Scan Line Fill
● Can also fill by maintaining a data structure of all intersections of polygons with scan lines– Sort by scan line– Fill each span
vertex order generated by vertex list desired order
Data Structure
Aliasing
● Ideal rasterized line should be 1 pixel wide
● Choosing best y for each x (or visa versa) produces aliased raster lines
Antialiasing by Area Averaging
● Color multiple pixels for each x depending on coverage by ideal line
original antialiased
magnified
Polygon Aliasing
● Aliasing problems can be serious for polygons– Jaggedness of edges– Small polygons
neglected– Need compositing so
color of one polygon does not totally determine color of pixel
All three polygons should contribute to color
Display issues
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Objectives
● Consider perceptual issues related to displays● Introduce chromaticity space
Color systems Color transformations
● Standard Color Systems
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No Display Can Be Perfect
● An analog display device such as a CRT takes digital input (pixels) and outputs a small spot of color
● A Digital display such as a LCD display outputs discrete spots
● The eye merges (filters) these spots● Sampling theory shows this process cannot be
done perfectly
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Perception Review
● Light is the part of the electromagnetic spectrum between ~350-750 nm
● A color C() is a distribution of energies within this range
● The human visual system has three types of cones on the retina, each with its own spectral sensitivity
● Consequently, only three values, the tristimulus values, are “seen” by the brain
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Tristimulus Values
● The human visual center has three cones with sensitivity curves S1(), S2(), and S3()
● For a color C(), the cones output the tristimulus values
dCST )()(11 dCST )()(22 dCST )()(33
C()
T1, T2, T3cones
optic nerve
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Three Color Theory
● Any two colors with the same tristimulus values are perceived to be identical
● Thus a display (CRT, LCD, film) must only produce the correct tristimulus values to match a color
● Is this possible? Not always Different primaries (different sensitivity curves) in
different systems
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The Problem
● The sensitivity curves of the human are not the same as those of physical devices
● Human: curves centered in blue, green, and green-yellow
● CRT: RGB● Print media: CMY or CMYK● Which colors can we match and, if we cannot
match, how close can we come?
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Representing Colors
● Consider a color C()● It generates tristimulus values T1, T2, T3
Write C = (T1, T2, T3 )
Conventionally,we assume 1 -90 T1, T2, T3 -90 0 because there is a maximum brightness we can produce and energy is nonnegative
C is a point in color solid
C1
11
T1
T2
T3
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Producing Colors
● Consider a device such as a CRT with RGB primaries and sensitivity curves
● Tristimulus values
dCRT )()(1 dCGT )()(2 dCBT )()(3
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Matching
● This T1, T2, T3 is dependent on the particular device
● If we use another device, we will get different values and these values will not match those of the human cone curves
● Need a way of matching and a way of normalizing
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Color Systems
● Various color systems are used Based on real primaries:
• NTSC RGB• UVW• CMYK• HLS
Theoretical• XYZ
● Prefer to separate brightness (luminance) from color (chromatic) information Reduce to two dimensions
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Tristimulus Coordinates
TTTTt
321
11
TTTTt
321
22
For any set of primaries, define
TTTTt
321
33
1ttt 321 0,,1 ttt 321
-90 -90Note
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Maxwell Triangle
Project onto 2D: chromaticity space
1
1T1 + T2+T3 =1
1
color solid
t1
t2
1
1
t1 +
t2 =1 possible colors
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NTSC RGB
1
1
r
g
r+g+b=1
r+g=1
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Producing Other Colors
● However colors producible on one system (its color gamut) is not necessarily producible on any other
● Not that if we produce all the pure spectral colors in the 350-750 nm range, we can produce all others by adding spectral colors
● With real systems (CRT, film), we cannot produce the pure spectral colors
● We can project the color solid of each system into chromaticity space (of some system) to see how close we can get
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Color Gamuts
spectral colorsprinter colors
CRT colors
350 nm
750 nm
600 nm
producible color on CRT but not on printer
producible color on both CRT and printer
unproducible color
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XYZ
● Reference system in which all visible pure spectral colors can be produced
● Theoretical systems as
there are no corresponding
physical primaries● Standard reference system
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Color Systems
● Most correspond to real primaries
National Television Systems Committee (NTSC) RGB matches phosphors in CRTs
● Film both additive (RGB) and subtractive (CMY) for positive and negative film
● Print industry CMYK (K = black) K used to produce sharp crisp blacks Example: ink jet printers
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Color Transformations
● Each additive color system is a linear transformation of another
R
R’
GG’
B
B’
C = (T1, T2, T3) = (T’1, T’2, T’3)
in RGB system
in R’G’B’system
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RGB, CMY, CMYK
● Assuming 1 is max of a primary
C = 1 – R
M = 1 – G
Y = 1 – B● Convert CMY to CMYK by
K = min(C, M, Y)
C’ = C – K
M’ = M – K
Y’ = Y - K
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Color Matrix
● Exists a 3 x 3 matrix to convert from representation in one system to representation in another
● Example: XYZ to NTSC RGB find in colorimetry references
● Can take a color in XYZ and find out if it is producible by transforming and then checking if resulting tristimulus values lie in (0,1)
TTT
T'T'T'
3
2
1
3
2
1
M
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YIQ
● NTSC Transmission Colors● Here Y is the luminance
Arose from need to separate brightness from chromatic information in TV broadcasting
● Note luminance shows high green sensitivity
B
G
R
0.3110.523-0.212
0.321-0.275-0.596
0.1140.5870.299
Q
I
Y
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Other Color Systems
● UVW: equal numerical errors are closer to equal perceptual errors
● HLS: perceptual color (hue, saturation, lightness)
Polar representation of color space Single and double cone versions
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Gamma
● Intensity vs CRT voltage is nonlinear
I = cV
● Can use a lookup table to correct● Human brightness response is logarithmic
Equal steps in gray levels are not perceived equally
Can use lookup table● CRTs cannot produce a full black
Limits contrast ratio
Angel and Shreiner: Interactive Computer Graphics 7E © Addison-Wesley 2015
sRGB
● Standard for Internet● Adjust colors to match standard gamma of
panels match gamma over most of the range enhance less bright colors
● OpenGL (soon WebGL?) can input sRGB and convert to RGB for processing and then back to sRGB
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Summary
● Rendering overview● Clipping● Polygon rendering● Rasterization● Display issues