Clipping
CS 445/645
Introduction to Computer Graphics
David Luebke, Spring 2003
Admin
● Call roll
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Recap: Homogeneous Coords
● Intuitively:
■ The w coordinate of a homogeneous point is typically 1
■ Decreasing w makes the point “bigger”, meaning further
from the origin
■ Homogeneous points with w = 0 are thus “points at
infinity”, meaning infinitely far away in some direction.
(What direction?)
■ To help illustrate this, imagine subtracting two
homogeneous points: the result is (as expected) a vector
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Recap: Perspective Projection
● When we do 3-D graphics, we think of the
screen as a 2-D window onto the 3-D world:
How tall should
this bunny be?
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Recap: Perspective Projection
● The geometry of the situation:
View
plane
X
P (x, y, z)
x’ = ?
(0,0,0)
Z
d
● Desired
result:
David Luebke
dx
x
x' 

,
z
z d
5
dy
y
y' 

, zd
z
z d
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Recap: Simple Perspective
Projection Matrix
● Example:
 x  1
 y  0


 z  0

 
 z d  0
0
1
0
0
0
1
0 1d
0  x 



0  y 
0  z 
 
0  1 
● Or, in 3-D coordinates:
 x

,
z d
David Luebke

y
, d 
zd

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Recap: OpenGL Perspective
Projection Matrix
● OpenGL’s gluPerspective() command
generates a slightly more complicated matrix:
 f
 aspect

 0

 0

 0
where
0
0
f
0
 Ζ far  Z near 


Z Z 
far 
 near
1
0
0
0
0
 2  Z far  Z near

 Z Z
near
far

0









 fov y 

f  cot 
 2 
■ Can you figure out what this matrix does?
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Next Topic: Clipping
● We’ve been assuming that all primitives (lines,
triangles, polygons) lie entirely within the viewport
● In general, this assumption will not hold:
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Clipping
● Analytically calculating the portions of primitives
within the viewport
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Why Clip?
● Bad idea to rasterize outside of framebuffer bounds
● Also, don’t waste time scan converting pixels outside
window
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Clipping
● The naïve approach to clipping lines:
for each line segment
for each edge of viewport
find intersection points
pick “nearest” point
if anything is left, draw it
● What do we mean by “nearest”?
● How can we optimize this?
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Trivial Accepts
● Big optimization: trivial accept/rejects
● How can we quickly determine whether a line
segment is entirely inside the viewport?
● A: test both endpoints.
xmin
xmax
ymax
ymin
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Trivial Rejects
● How can we know a line is outside viewport?
● A: if both endpoints on wrong side of same edge, can
trivially reject line
xmin
xmax
ymax
ymin
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Cohen-Sutherland
Line Clipping
● Divide viewplane into regions defined by viewport
edges
● Assign each region a 4-bit outcode:
xmin
1001
xmax
1000
1010
0000
0010
0100
0110
ymax
0001
ymin
0101
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Cohen-Sutherland
Line Clipping
● To what do we assign outcodes?
● How do we set the bits in the outcode?
● How do you suppose we use them?
xmin
1001
xmax
1000
1010
0000
0010
0100
0110
ymax
0001
ymin
0101
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Cohen-Sutherland
Line Clipping
● Set bits with simple tests
x > xmax
y < ymin
etc.
● Assign an outcode to each vertex of line
■ If both outcodes = 0, trivial accept
■ bitwise AND vertex outcodes together
■ If result  0, trivial reject
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Cohen-Sutherland
Line Clipping
● If line cannot be trivially accepted or rejected,
●
●
●
●
subdivide so that one or both segments can be
discarded
Pick an edge that the line crosses (how?)
Intersect line with edge (how?)
Discard portion on wrong side of edge and assign
outcode to new vertex
Apply trivial accept/reject tests; repeat if necessary
■ Q: How did the programmer get stuck in the shower?
■ A: Lather, Rinse, Repeat
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Cohen-Sutherland
Line Clipping
● Outcode tests and line-edge intersects are quite fast
(how fast?)
● But some lines require multiple iterations:
■ Clip top
■ Clip left
■ Clip bottom
■ Clip right
● Fundamentally more efficient algorithms:
■ Cyrus-Beck uses parametric lines
■ Liang-Barsky optimizes this for upright volumes
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Coming Up…
● We know how to clip a single line segment
■ How about a polygon in 2D?
■ How about in 3-D?
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Clipping Polygons
● Clipping polygons is more complex than clipping the
individual lines
■ Input: polygon
■ Output: polygon, or nothing
● When can we trivially accept/reject a polygon as
opposed to the line segments that make up the
polygon?
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Why Is Clipping Hard?
● What happens to a triangle during clipping?
● Possible outcomes:
triangletriangle
trianglequad
triangle5-gon
● How many sides can a clipped triangle have?
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Why Is Clipping Hard?
● A really tough case:
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Why Is Clipping Hard?
● A really tough case:
concave polygonmultiple polygons
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Sutherland-Hodgman Clipping
● Basic idea:
■ Consider each edge of the viewport individually
■ Clip the polygon against the edge equation
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7/27/2016
Sutherland-Hodgman Clipping
● Basic idea:
■ Consider each edge of the viewport individually
■ Clip the polygon against the edge equation
■ After doing all planes, the polygon is fully clipped
David Luebke
25
7/27/2016
Sutherland-Hodgman Clipping
● Basic idea:
■ Consider each edge of the viewport individually
■ Clip the polygon against the edge equation
■ After doing all planes, the polygon is fully clipped
David Luebke
26
7/27/2016
Sutherland-Hodgman Clipping
● Basic idea:
■ Consider each edge of the viewport individually
■ Clip the polygon against the edge equation
■ After doing all planes, the polygon is fully clipped
David Luebke
27
7/27/2016
Sutherland-Hodgman Clipping
● Basic idea:
■ Consider each edge of the viewport individually
■ Clip the polygon against the edge equation
■ After doing all planes, the polygon is fully clipped
David Luebke
28
7/27/2016
Sutherland-Hodgman Clipping
● Basic idea:
■ Consider each edge of the viewport individually
■ Clip the polygon against the edge equation
■ After doing all planes, the polygon is fully clipped
David Luebke
29
7/27/2016
Sutherland-Hodgman Clipping
● Basic idea:
■ Consider each edge of the viewport individually
■ Clip the polygon against the edge equation
■ After doing all planes, the polygon is fully clipped
David Luebke
30
7/27/2016
Sutherland-Hodgman Clipping
● Basic idea:
■ Consider each edge of the viewport individually
■ Clip the polygon against the edge equation
■ After doing all planes, the polygon is fully clipped
David Luebke
31
7/27/2016
Sutherland-Hodgman Clipping
● Basic idea:
■ Consider each edge of the viewport individually
■ Clip the polygon against the edge equation
■ After doing all planes, the polygon is fully clipped
David Luebke
32
7/27/2016
Sutherland-Hodgman Clipping:
The Algorithm
● Basic idea:
■ Consider each edge of the viewport individually
■ Clip the polygon against the edge equation
■ After doing all planes, the polygon is fully clipped
● Will this work for non-rectangular clip regions?
● What would 3-D
clipping involve?
●
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