Chapter 6 – 2D Viewing
Co-ordinate Systems.
 Cartesian – offsets along the x and y axis from (0.0)
 Polar – (r,)
 Graphic libraries mostly using Cartesian co-ordinates
 Any polar co-ordinates must be converted to Cartesian coordinates
 Four Cartesian co-ordinates systems in computer Graphics.
1. Modeling co-ordinates
2. World co-ordinates
3. Normalized device co-ordinates
4. Device co-ordinates
2
Modeling Co-ordinates
 Also known as local coordinate.
 Ex: where individual object in a scene within separate
coordinate reference frames.
 Each object has an origin (0,0)
 So the part of the objects are placed with reference to the
object’s origin.
3
World Co-ordinates
 The world coordinate system describes the relative positions
and orientations of every generated objects.
 The scene has an origin (0,0).
 The object in the scene are placed with reference to the scenes
origin.
 World co-ordinate scale may be the same as the modeling coordinate scale or it may be different.
4
Normalized Device Co-ordinates
 Output devices have their own co-ordinates.
 Co-ordinates values: The x and y axis range from 0 to 1
 All the x and y co-ordinates are floating point numbers in the
range of 0 to 1
 This makes the system independent of the various devices
coordinates.
 This is handled internally by graphic system without user
awareness.
5
Device Co-ordinates
 Specific co-ordinates used by a device.
 Pixels on a monitor
 Points on a laser printer.
 mm on a plotter.
 The transformation based on the individual device is handled
by computer system without user concern.
6
Two-Dimensional Viewing
Window:
•A world-coordinate area selected for display.
•Window defines what is to be viewed
Viewport:
•An area on a display device to which a window is mapped.
•Viewport defines where it is to be displayed
Viewing transformation:
•The mapping of a part of a world-coordinate scene to device
coordinates.
 Most of the time, windows and viewports are usually rectangles in
standard position(i.e aligned with the x and y axes).
 In some application, others such as general polygon shape and
circles are also available
 However, other than rectangle will take longer time to process.
7
Window and viewports
8
Viewing Transformation
 In 2D (two dimensional) viewing transformation is simply
referred as the window-to-viewport transformation or the
windowing transformation.
 Mapping a window onto a viewport involves converting from
one coordinate system to another.
 If the window and viewport are in standard position, this just
involves translation and scaling.
 If the window and/or viewport are not in standard, then extra
transformation which is rotation is required.
9
Window-To-Viewport Coordinate Transformation
The sequence of transformations are:
1. Translate the window to the origin
2. Scale it to the size of the viewport
3. Translate the scaled object to the position of the viewport.
10
Window-To-Viewport Coordinate Transformation
11
Viewing Transformation
y-world
y-view
window
x-view
x-world
World and Viewport
window
1
0
1
Normalised device
12
Window-To-Viewport Coordinate Transformation
Window-to-Viewport transformation
13
Window-To-Viewport Coordinate Transformation
xv - xvmin
xw - xwmin
=
xvmax - xvmin
yv – yvmin
yvmax – yvmin
xwmax - xwmin
yw - ywmin
=
ywmax - ywmin
YWmax
From these two equations we derived
xv = xvmin + (xw – xwmin)sx
yv = yvmin + (yw – ywmin)sy
YVmax
xw,yw
YWmin
XWmin
YVmin . xv,yv
XVmin XVmax
XWmax
where the scaling factors are
sx = xvmax – xvmin
xwmax – xwmin
sy = yvmax - yvmin
ywmax - ywmin
14
Window-To-Viewport Coordinate Transformation
 Relative proportions of objects are maintained if the scaling
factors are the same (sx = sy).
 Otherwise, world objects will be stretched or contracted in
either x or y direction when displayed on output device.
15
The Viewing Pipeline
16
The Viewing Pipeline
17
Viewport-to-Normalized Device Coordinate Transformation
 From normalized coordinates, object descriptions can be
mapped to the various display devices.
 When mapping window-to-viewport transformation is done to
different devices from one normalized space, it is called
workstation transformation.
18
Clipping
• Clipping is a process of dividing an object into visible
and invisible positions and displaying the visible
portion and discarding the invisible portion.
• Generally we have clipping algorithms for the
following primitive types.
– Point clipping
– Line clipping
– Area clipping (Polygon)
Clipping Window
World Coordinates
ywmax
The clipping window is
mapped into a viewport.
ywmin
xwmin
xwmax
Viewport
yvmax
yvmin
Viewport Coordinates
xvmin
xvmax
20
Point Clipping
clip
rectangle
y = ymax
(xmax, ymax)
(x1, y1)
x = xmin
(xmin , ymin )
x = xmax
y = ymin
For a point (x,y) to be inside the clip rectangle:
xmin  x  xmax
ymin  y  ymax
Line Clipping
clip
rectangle
Cases for clipping lines
Line Clipping
B
B
A
A
clip
rectangle
Cases for clipping lines
Line Clipping
D
D'
D'
C
C
B
B
A
A
clip
rectangle
Cases for clipping lines
Line Clipping
F
D
D'
D'
C
C
B
B
A
A
E
clip
rectangle
Cases for clipping lines
Line Clipping
F
D
D'
D'
C
C
B
H
B
E
H'
A
G'
clip
rectangle
H'
A
G
Cases for clipping lines
G'
Line Clipping
F
D
D'
D'
C
C
B
H
B
E
H'
J
A
G'
G'
J'
clip
rectangle
H'
A
G
I'
I
Cases for clipping lines
Cohen-Sutherland line clipping algorithm
• This is one of the most popular line clipping
algorithms.
• This was introduced by Dan Cohen and Ivan
Sutherland.
• It was designed not only to find the endpoints very
rapidly but also to reject even more rapidly any line
that is clearly invisible.
• This makes it a very good algorithm.
Cohen-Sutherland line clipping algorithm
• Assign a four bit code to all regions.
• Every line end point in a picture is
assigned a 4-bit binary code known as
region code.
• That identifies the location of the point
relative to the boundaries of the clipping
rectangle.
• Each bit position in the region code is
used to indicate one of the 4 rectangle
coordinates positions of the point with
respected to clipping window to the left,
right, below and top.
Cohen-Sutherland Clipping-Region Outcodes
1001
1000
1010
0001
0000
0010
0101
0100
0110
ymax
ymin
x min
x max
Cohen-Sutherland line clipping algorithm
–
–
–
–
Bit 1 – left
Bit 2 – right
Bit 3 – below
Bit 4 – top
• A value of 1 in any bit position indicates that the
point is in the relative position otherwise the bit
position is set to zero.
• If a point is within the clipping the region code is
0000.
Cohen-Sutherland line clipping algorithm
•
If the code of both the endpoints are 0000
then the line is totally visible and hence
draw the line.
•
Bit values in the region codes are determined
by comparing endpoint co-ordinate values
(x, y) to the clip boundaries.
Cohen-Sutherland line clipping algorithm
• Bit 1 is set to 1 if x < Xwmin
• The other three bit values can be determined
using similar comparison or calculating
differences between endpoint co-ordinates and
clipping boundaries.
• Use the resultant sign of each differences
calculation to set the corresponding value in
the region code.
Cohen-Sutherland line clipping algorithm
–
–
–
–
•
Bit 1 is the sign bit of x-xwmin
Bit 2 is the sign bit of xwmax-x
Bit 3 is the sign bit of y-ywmin
Bit 4 is the sign bit of ywmax-y
If any line have 1 in the same bit position in
the region codes for each endpoint are
completely outside the clipping rectangle, so
we discard the line
Cohen-Sutherland line clipping algorithm
• For a line with end points co-ordinates (x1, y1)
and (x2, y2) then the y coordinate of the
intersection point with a vertical boundary can
be obtained with the calculation
• y=y1 + m(x-x1) (1)
• Where the x value is set either to xwmin or
xwmax.
Cohen-Sutherland line clipping algorithm
•
•
•
Similarly if we are looking for the
intersection with a horizontal boundary, the x
co-ordinate can be calculated as
x = x1 + 1/m (y-y1)  (2)
where y set either ywmin or ywmax.
Cohen-Sutherland Clipping Trivial Acceptance:
O(P0) = O(P1) = 0
1001
1000
ymax
1010
P1
0001
0000
0010
0100
0110
P0
ymin
0101
x min
x max
Cohen-Sutherland Clipping: Trivial Rejection:
O(P0) & O(P1)  0
P0
1001
1000
1010
P1
ymax
P0
0001
ymin
0000
0010
P0
0101
x min
P1
P1
0100
0110
x max
Cohen-Sutherland Clipping:
1001
O(P0) =0 , O(P1)  0
1000
1010
ymax
P0
0001
0000
P1
0010
P0
P0
ymin
0100
0101
0110
P1
x min
x max
P1
Cohen-Sutherland Clipping: The Algorithm
1. Compute the outcodes for the two vertices
2. Test for trivial acceptance or rejection
3. Select a vertex for which outcode is not zero
– There will always be one
4. Select the first nonzero bit in the outcode to define
the boundary against which the line segment will be
clipped.
5. Compute the intersection and replace the vertex with
the intersection point
6. Compute the outcode for the new point and iterate
Cohen-Sutherland Clipping
A
1001
1000
1010
B
ymax
C
0001
0000
0010
0101
0100
0110
ymin
x min
x max
Cohen-Sutherland Clipping
A
1001
1000
1010
B
ymax
C
0001
0000
0010
0101
0100
0110
ymin
x min
x max
Cohen-Sutherland Clipping: Example 1
1001
1000
1010
B
ymax
C
0001
0000
0010
0101
0100
0110
ymin
x min
x max
Cohen-Sutherland Clipping
1001
1000
1010
E
ymax
D
0001
0000
ymin
C
0010
B
A
0100
0101
x min
0110
x max
Cohen-Sutherland Clipping
1001
1000
1010
E
ymax
D
0001
0000
ymin
C
0010
B
0100
0101
x min
0110
x max
Cohen-Sutherland Clipping
1001
1000
1010
ymax
D
0001
0000
ymin
C
0010
B
0100
0101
x min
0110
x max
Cohen-Sutherland Clipping
1001
1000
0001
0000
1010
ymax
ymin
C
0010
B
0100
0101
x min
0110
x max
Example
Clip the line with the boundaries (0, 0) and (15, 15)
and the points are (2, 3) and (9, 10).
Solution: Given (x1, y1) = (2, 3) & (x2, y2) = (9, 10)
Here
xwmin = 0, ywmin = 0
Xwmax = 15, ywmax = 15
Example
•
•
•
•
Bit 1 is the sign bit of x-xwmin
Bit 2 is the sign bit of xwmax-x
Bit 3 is the sign bit of y-ywmin
Bit 4 is the sign bit of ywmax-y
• For (2, 3) the region code is 0000.
• For (9, 10) the region code is 0000.
• So the line should be drawn.
Example
• Clip the line with the boundaries (0, 0) and (15, 15) and
the points are (2, -5) and (2, 18).
• Solution: Region code for (2, -5) is
2-0=0,
15-2=0,
-5-0=1,
15+5=0.
So it is in 0100 region.
So (2, -5) is not in the clipping region.
Example
• Region code for (2, 18) is
2-0=0, 15-2=0, 18-0=0, 15-18=1.
The code is 1000.
Clearly this point is out of clipping window.
• Now (0100) & (1000) = 0000
• So we have to find the horizontal intercept
point
Example
• Horizontal intercept:
• point (x=1/m (ybottom-y1) + x1, ybottom);
• m=y2-y1 / x2-x1 = 18-(-5) / 2-2 = 23/0 = ∞
• Then x= 1/∞ (0-(-5)) + 2 = 2 and y=ybottom=0.
• The point is (2, 0).
• So we have to draw the line from (2, 0).
Example
• Point is coordinate (x=1/m (ytop-y1) + x1, ytop)
• m=y2-y1 / x2-x1 = ∞.
• y = ytop = 15.
• x= 1/∞ (15-2) + 2 =2.
• The point is (2, 15).
• Then draw the line between (2, 15) and (2, 0).
Liang-Barsky Line Clipping Algorithm
• Using Parametric equations of line, Cyrus and
Beck developed an algorithm.
• Later Liang and Barsky modified and make
even faster parametric line-clipping algorithm.
54
Liang-Barsky Line Clipping Algorithm
• Consider the parametric definition of a line:
– x = x1 + u∆x
– y = y1 + u∆y
Where ∆x = (x2 - x1), ∆y = (y2 - y1), 0 ≤ u ≤ 1
• Mathematically, this means
– xmin ≤ x1 + u∆x ≤ xmax
– ymin ≤ y1 + u∆y ≤ ymax
• Rearranging, we get
–
–
–
–
-u∆x ≤ (x1 - xmin)
u∆x ≤ (xmax - x1)
-u∆y ≤ (y1 - ymin)
u∆y ≤ (ymax - y1)
• In general: u * pk ≤ qk
p1=-∆x
p2= ∆x
p3= -∆y
p4= ∆y
q1= x1 - xmin
q2= xmax - x1
q3= y1 - ymin
q4= ymax - y1
55
Where K=1,2,3,4.
Liang-Barsky Line Clipping: Observation
1. pk = 0
• Line is parallel to boundaries.
• If for the same k, qk < 0, reject
• Else, accept
u>1
p2
u[0,1]
p1
u<0
Liang-Barsky Line Clipping: Observation
2. pk < 0
• Line proceeds from outside to inside
• rk = qk / pk
u>1
• u1 = max(0, rk, u1)
p2
u[0,1]
iL
u2
u1
p1
u<0
potential entrance
potential exit
Liang-Barsky Line Clipping: Observation
3. pk > 0
• Line proceeds from inside to out side
• rk = qk / pk
• u2 = min(1, rk, u2)
u[0,1]
u<0
p2
u>1
iT
u2
p1
u1
potential entrance
potential exit
Liang-Barsky: Algorithm
p2
wxmin
u>1
iT
wymax
u[0,1]
u2
u
iL 1
p1
u<0
wymin
potential entrance
potential exit
Liang-Barsky Line Clipping: Observation
p2
u>1
u[0,1]
If u1 > u2, the line is completely outside
u<0
p1
iR
u1= 0
iT u2= -1/6
Opps! u1 > u2
potential entrance
potential exit
Liang-Barsky Line Clipping: Observation
If u1 > u2, the line is completely outside
iR
iT
u>1
u[0,1]
iB
p2
iL
u1=1.25
u2=1
Opps! u1 > u2
p1
potential entrance
u<0
potential exit
Advantage of Liang-Barsky Line Clipping
This is more efficient than Cohen-Sutherland Alg,
which computes intersection with clipping window
borders for each undecided line, as a part of the
feasibility tests.
62
Example 1
• Clipping window (1,2) and (9,8)
Line end points A(11,6) and B(11,10)
Solution:
P1=-dx =0
q1=x1-xmin=10
P2= dx = 0
q2=xmax-x1=-2
P3=-dy =-4
q3=y1-ymin=4
P4= dy =4
q4=ymax-y1=2
P2=0 and q2<0 so discard the line
63
Liang-Barsky Line Clipping: Observation
u>1
p2
u[0,1]
p1
u<0
Example 2
• Clipping window (1,2) and (9,8)
Line end points A(3,7) and B(3,10)
Solution:
P1=-dx =0
q1=x1-xmin=2
P2= dx = 0
q2=xmax-x1=6
P3=-dy =-3
q3=y1-ymin=5
P4= dy =3
q4=ymax-y1=1
u1=max(0,-5/3)=0
u2=min(1/3,1)=1/3
r3=-5/3
r4=1/3
65
Liang-Barsky Line Clipping: Observation
u>1
u2=1/3
p2
u[0,1]
p1
u<0
• Here u1<u2 so there is a visible section
New endpoints of line:
• x1’=x1+u1*dx=3
• y1’=y1+u1*dy=7
• x2’=x1+u2*dx=3
• y2’=y1+u2*dy=8.
• Hence visible line will be from (3,7) to (3,8)
67
Example 3
• Clipping window (1,2) and (9,8)
Line end points A(2,3) and B(8,4)
Solution:
P1=-dx =-6
q1=x1-xmin=1
P2= dx = 6
q2=xmax-x1=7
P3=-dy =-1
q3=y1-ymin=1
P4= dy =1
q4=ymax-y1=5
u1=max(0,-1/6,-1)=0
u2=min(7/6,5,1)=1
r1=-1/6
r2=7/6
r3=-1
r4=5
68
Liang-Barsky Line Clipping: Observation
u>1
iR
p2
u[0,1]
p1
u<0
iT
u2=1
u1=0
iL
iB
potential entrance
potential exit
• Here u1<u2 so there is a visible section
• u1=0 and u2=1
• So line itself inside the clipping window.
• Hence visible whole line.
70
Example 4
• Clipping window (1,2) and (9,8)
Line end points A(-1,7) and B(11,1)
Solution:
P1=-dx =-12
q1=x1-xmin=-2
P2= dx = 12
q2=xmax-x1=10
P3=-dy =6
q3=y1-ymin=5
P4= dy =-6
q4=ymax-y1=1
u1=max(0,1/6,-1/6)=1/6
u2=min(5/6,5/6,1)=5/6
r1=1/6
r2=5/6
r3=5/6
r4=-1/6
71
• Here u1<u2 so there is a visible section
New endpoints of line:
• x1’=x1+u1*dx=1
• y1’=y1+u1*dy=6
• x2’=x1+u2*dx=9
• y2’=y1+u2*dy=2.
• Hence visible line will be from (1,6) to (9,2)
72
Example 5
• Clip the line p1(-15,-30) , p2(30,60) against
the window having diagonally opposite
corners as (0,0) and (15,15)
Solution:
P1=-dx =-45
q1=x1-xmin=-15
r1=1/3
P2= dx = 45
q2=xmax-x1=30
r2=2/3
P3=-dy =-90
q3=y1-ymin=-30
r3=1/3
P4= dy =90
q4=ymax-y1=45
r4=1/2
u1=max(0,1/3,1/3)=1/3
u2=min(2/3,1/2,1)=1/2
73
Liang-Barsky: Algorithm
p2
wxmin
u>1
iR
iT
wymax
u[0,1]
u2=0.5
u =1/3
iL 1
iB
p1
u<0
wymin
potential entrance
potential exit
• Here u1<u2 so there is a visible section
New endpoints of line:
• x1’=x1+u1*dx=0
• y1’=y1+u1*dy=0
• x2’=x1+u2*dx=7.5
• y2’=y1+u2*dy=15.
• Hence visible line will be from (0,0) to (7.5,15)
75
Nicholl-Lee-Nicholl Line Clipping Algorithm(NLN)
• Generate region codes (Cohen-Suther.) & use
trivial accept and reject
• When trivial case fails , creates more regions
around clipping window to avoid multiple line
intersection calculations.
• Performs fewer comparisons and divisions than
Cohen-Sutherland and Liang-Barsky, but cannot
be extended to 3D.
76
P0
P0
P0
• Examine first where the starting point p0 is located.
• We will consider only 3 regions.
• If point is in any of the other six regions then
perform transformation such that that point will
convert in any of this 3 regions.
77
Case 1: Inside case for P0
The location of Pend in each
T
L
P0
B
R
region defines what edge the
line  P0 , Pend  is intersecting.
78
Case 2: edge case for P0
Detecting whether Pend is in
LT
L
P0
L
LR
any of regions L is immediate.
L
LB
Else, Pend is detected for being positioned in any of
LB, LR or LT, case where  P0 , Pend  is clipped with
left border and bottom, right or top border, resp.
79
The slope of  P0 , Pend  is
Pend LT
compared to  P0 , Pcorner 
L
P0
L
LR
L
LB
for each corner to find the
region of Pend .
• Pend is in region LT if
slop P0Ptr < slop P0Pend < slop P0Ptl
Yt-Y1
Xr-X1
< Y2-Y1 < Yt-Y1
X2-X1
XL-X1
Once one of LT, LR or LB regions is found,
intersection point with appropriate border is calculated.80
The slope of  P0 , Pend  is
Pend
LT
compared to  P0 , Pcorner 
L
P0
L
LR
L
LB
for each corner to find the
region of Pend .
 P0 , Pend  is entierely clipped if Pend is positioned
outside the regions.
81
Case 3: corner case for P0
P0
P0
T
TR
L
T
T
TR
L
LB
L
LR
LB
TB
There are two cases, depending on whether P0 is
closer to left or top borders.
82
Polygon Clipping
using line clipping not good enough
Not a polygon
Polygon Clipping
• Need to maintain closed polyline
boundary!
Note, clipping yields 2 polygons
Trivial Accept & Reject
Accept
Reject
Need Clipping
bounding box easy to compute
Sutherland-Hodgman Clipping
•
Ivan Sutherland, Gary W. Hodgman: Reentrant Polygon
Clipping. Communications of the ACM, vol. 17, pp. 32-42,
1974
• Basic idea:
– Consider each edge of the viewport individually
– Clip the polygon against the edge equation
Sequentially Clip polygon across each side
1
3
2
clip
right
clip
left
5
4
clip
bottom
clip
top
Sutherland-Hodgman Polygon Clipping
• Input each edge (vertex pair) successively.
• Output is a new list of vertices.
• Each edge goes through 4 clippers.
Sutherland-Hodgman Polygon Clipping:
Four possible scenarios at each clipper
1. If first input vertex is outside, and second is inside, output
the intersection and the second vertex
2. If both input vertices are inside, then just output second
vertex
3. If first input vertex is inside, and second is outside, output
is the intersection
4. If both vertices are outside, output is nothing
outside
inside
v2
outside
inside
v2
v1’
outside
inside
v2
inside
v2
v1’
v1
v1
v1
Outside to inside:
Output: v1’ and v2
outside
Inside to inside:
Output: v2
Inside to outside:
Output: v1’
v1
Outside to outside:
Output: nothing
Complete Example
2
2
2
2'
2'
2'
3
2' '
I – In
3'
O - Out
3’
1
Left Clip
(1,2) (I/I) → {2}
(2,3) (I/O) → {2'}
(3,1) (O/I) → {3',1}
1
Right Clip
3’
Bottom Clip
1’
1
Top Clip
(2,2'): (I/I) → {2'}
(2',3'): (I/I) → {3'} (2',3'): (I/O) → {2''}
(3',1): (I/I) → {1} (3',1): (O/O) → {}
(2'',1'): (I/I) → {1'}
(1,2): (I/I) → {2} (1,2): (O/I) → {1',2}
(1',2): (I/I) → {2}
(2,2'): (I/I) → {2'} (2,2'): (I/I) → {2'}
(2',2''): (I/I) → {2''}
Disadvantage of Sutherland - Hodgeman
• Convex polygons are correctly clipped.
• Concave polygons may not be correctly clipped so we will use
Weiler-Atherton.
• Properties of Convex polygon.
1. Every internal angle is less than or equal to180 degrees.
2. Every line segment between two vertices remains inside or on
the boundary of the polygon.
91
Weiler-Atherton Polygon Clipping
• Developed as a method for identifying visible surfaces
• It can be applied with arbitrary polygon-clipping
•
region.
• Not always proceeding around polygon edges
• Sometimes follows the window boundaries
For clockwise processing of polygon vertices
• For an outside-to-inside pair of vertices, follow the
polygon boundary.
• For an inside-to-outside pair of vertices, follow the
window boundary in clockwise direction.
92
Weiler-Atherton Polygon Clipping
93
Weiler-Atherton Polygon Clipping
• For clockwise processing of polygon vertices
• For an outside-to-inside pair of vertices, follow the polygon
boundary.
• For an inside-to-outside pair of vertices, follow the window
boundary in clockwise direction.
1
1’
2
1
1’
2
1
3
1’
3’
2
In -> In
Add end vertex
1’
3
4
Out -> In
Add clip vertex
Add end vertex
1
In -> Out
Add clip vertex
Cache old direction
2
3
4
3’
Follow clip edge until
(a) new crossing found
(b) reach vertex already
added
Weiler-Atherton Polygon Clipping
• For clockwise processing of polygon vertices
• For an outside-to-inside pair of vertices, follow the polygon
boundary.
• For an inside-to-outside pair of vertices, follow the window
boundary in clockwise direction.
1
4
1’
3’
2
1
3
1’
3’
4
2
1’
3
3’
4’
4’
2
1
3
Out -> In
Add clip vertex
Add end vertex
1’
6’
6’
6
2
3’
4’
5
5
Continue from
cached vertex and
direction
1
3
5
6
Follow clip edge until
In -> Out
(a) new crossing found
Add clip vertex
(b) reach vertex already
Cache old direction
added
Weiler-Atherton Polygon Clipping
• For clockwise processing of polygon vertices
• For an outside-to-inside pair of vertices, follow the polygon
boundary.
• For an inside-to-outside pair of vertices, follow the window
boundary in clockwise direction.
1’
2
1
1’
3’
2
1
3
3’
4’
4’
6
6’
5
Continue from
cached vertex and
direction
1’
6
6’
2
3
5
Nothing added
Finished
3’
4’
6’
3
5
Final Result:
2 unconnected
polygons
Other Clipping
• Curve clipping
• Use bounding rectangle to test for overlap with a
rectangular clip window.
97
Text Clipping
• There are several techniques that can be used to
provide text clipping in a graphics packages.
• The choice of clipping method depends on how
characters
are
generated
and
what
requirements we have for displaying character
strings.
98
Text Clipping
All-or-none string-clipping
• If all of the string is inside a clip window, we
keep it.
• Otherwise the string is discarded.
99
Text Clipping
All-or-none character-clipping
Here we discard only those characters that are not
completely inside the window
100
Text Clipping
Clip the components of individual characters
• We treat characters in much the same way that
we treated lines.
• If an individual character overlaps a clip window
boundary, we clip off the parts of the character
that are outside the window
101
Done with Clipping
• Point Clipping
– Easy, just do inequalities
• Line Clipping
– Cohen-Sutherland
Any
– Liang-Barsky
– Nicholl-Lee-Nicholl
• Polygon Clipping
– Sutherland-Hodgeman
– Weiler-Atherton
• Text Clipping
Question…?