B. SC CSIT Computer Graphics Unit 2 By Tekendra Nath Yogi

Unit 2
Presented By : Tekendra Nath Yogi
Tekendranath@gmail.com
College Of Applied Business And Technology
2/9/2019 1By: Tekendra Nath Yogi

Outline
• Geometrical Transformations
• 2D Transformations:
– Basic transformations :Translation, rotation and scaling
• Matrix Representations and Homogeneous Coordinates
• Composite Transformations :
– Two successive Translations ,Rotations, Scaling
– General Pivot-Point Rotation
– General Fixed-Point Scaling
• Other 2D transformations : reflection, shear
• Two-dimensional viewing:
– Window to View Port Transformation.
• 3D Transformations:
– Basic transformations :Translation, rotation and scaling
– Other transformations: reflection, shear
• 3D Viewing
• Projections
2By: Tekendra Nath Yogi2/9/2019

Geometric Transformation
• Geometric transformation is the process of altering the
coordinate description of objects.
• Purposes:
– To move the position of objects
– To alter the shape / size of objects
– To change the orientation of objects
• Types:
– Rigid Body Transformation( Transformation without deformation of
shape)
– Non Rigid Body Transformation( Transformation with change in
shape)

Contd….
• Basic 2-D transformations:
– Translation
– Rotation
– Scaling

Contd….
• Two-Dimensional translation:
– One of rigid-body transformation, which move objects without
deformation.
– Translate an object by Adding translation distances to coordinates to
generate new coordinates positions
– Let,(x, y) be the 2-D point and tx, ty be the translation distance, we
have
– In matrix format, where T is the translation vector
xtx'x  yty'y 







y
x
P 






y
x
t
t
T






'y
'x
'P
TP'P 
Translating a point from
position P to position P'
with translation vector T.

Contd….
• Example1: Translate the given point (2,5) by translation vector
(3,3)

Contd….
• Example2: Translate a polygon with coordinates A(2,5), B(7,10) and
C(10,2) by 3 units in x direction and 4 unit in y direction.
• Solution:
Translated by T(3,4)
Fig: Before translation Fig: After translation

Contd….
• Homework: Translate the triangle with vertices A(2,4), B(1,7)
and C(4,6) by translation vector T(2,4).

Contd….
• 2-D Rotation:

Contd….
• Rotation of point position (x, y) when the pivot point (xr, yr) is
at coordinate origin(for positive value of rotation angle):

Contd….
• The original coordinates of the point in polar coordinates are:
• Transformed coordinates:
• In matrix form:
OR
Where, R is a rotation matrix and is given as:
cosrx  sinyy 


cossinsinsinsincos)sin('
sincossinsincoscos)cos('
yxrrry
yxrrrx







 



cossin
sincos
R
PR'P 

Contd….
• When coordinate positions are represented as row vectors
instead of column vectors the transformed point (x’, y’ ) is
calculated as:
• i.e.,

Contd….
• For negative value of rotation angle:

Contd….
• Example: A point (4,3) is rotated counterclockwise by an
angle 45 degree. Find the rotation matrix and the resultant
point.
• Solution:

Contd….
• Homework:
– Example2: Rotate the triangle ABC having coordinates A(1,2), B(2,3)
and C(4,5) by 60⁰ about the origin.
– Example3: A point (4,3) is rotated clockwise by an angle -45 degree.
Find the rotation matrix and the resultant point.
– Example4: Rotate the triangle with vertices A(5,8), B(12,10) and
C(10,10) about origin with angle 90 degree in anticlockwise direction.

Contd….
• 2-D scaling Transformation:
– alters the size of an object.
– carried out for polygons by multiplying the coordinate values (x, y) of
each vertex by scaling factors sx and sy to produce the transformed
coordinates (x', y'):
– scaling factors sx scales object in the x-direction and scaling factor sy
scales object in the y-direction.
– In matrix form:
xsx'x  ysyy 


















y
x
s0
0s
'y
'x
y
x
PS'P 

Contd….
• Example1: scale the polygon with coordinates A(2,5), B(7,10)
and C(10,2) by two units in x-direction and two units in y-
direction.

Contd….

Homework
• Example2: perform the scaling transformation to the triangle
with vertices A(6,9), B(10,5) and C(4,3) with scaling factors
sx=3 and sy=2.
• Example3:scale the rectangle ABCD with vertices A(2,2) ,
B(2,6), C(5,2) ad D(5,6) to double its size.
• Example4: perform the scaling transformation to the triangle
with vertices A(6,10), B(10,6) and C(4,2) with scaling factors
sx=0.5 and sy=0.5.

Matrix Representations and Homogeneous coordinates
• Many graphics applications involve sequences of geometric
transformations
– E.g., In design and picture construction applications, we perform
translations, rotations, and scaling to fit the picture components into
their proper positions.
• But this sequential transformation process is not efficient.
• Solution: Use homogeneous coordinate system!

Contd….
•What is homogeneous coordinates?
– A point (x, y) can be re-written in homogeneous coordinates as (xh, yh,
h)
– The homogeneous parameter h is a non- zero value such that:
– We can then write any point (x, y) as (h*x, h*y, h)
– We can conveniently choose h = 1 so that (x, y) becomes (x, y, 1)
h
x
x h

h
y
y h


Contd….
• Homogeneous coordinate system allows us to express
transformation equations as matrix multiplication.
• The homogeneous coordinates allow combined transformation,
eliminating the calculation of intermediate coordinate values
and thus save required time for transformation and memory
required to store the intermediate coordinate values.

Contd….
• Homogeneous coordinate for translation:
• Alternatively,































1100
10
01
1
'
'
y
x
t
t
y
x
y
x

Contd….
• Homogeneous coordinates for Rotation:
• Alternatively,



















 











1100
0cossin
0sincos
1
'
'
y
x
y
x



Contd….
• Homogeneous coordinates for scaling
• Alternatively,































1100
00
00
1
'
'
y
x
s
s
y
x
y
x

Inverse transformations:
• Inverse translation:
– The inverse translation matrix can be obtained by replacing the
translation distances tx and ty with their negatives: -tx and –ty.
– i.e.,
– So,













100
t10
t01
T y
x
1

































1100
10
01
1
'
'
y
x
t
t
y
x
y
x

Contd….
• Inverse Rotation:
• The inverse rotation matrix can be obtained by replacing the rotation angle with
its negative value.
• i.e.,
• So,











100
0cossin
0sincos
R 1

































1100
0cossin
0sincos
1
'
'
y
x
y
x



Contd….
• Inverse scaling:
– Inverse scaling matrix can be obtained by replacing sx and sy by 1/sx
and 1/sy respectively.
– i.e.,
– So,

















100
0
1
0
00
1
1
sy
s
S
x































1100
0/10
00/1
1
'
'
y
x
s
s
y
x
y
x

Composite transformations
• The basic purpose of composite transformations is to gain
efficiency by applying a single composed transformation to a
point, rather than applying a series of transformations, one
after the other.

Contd…
• Two successive Translations:
– If two successive translation distances (tx1, ty1) and (tx2, ty2) are applied
to coordinate position P, the final transformed location P’ is calculated
as:
– The composite transformation matrix for this sequence of translation is:
– Or
– Which demonstrates that two successive translations are additive.
PttTttT
PttTttTP
yxyx
yxyx
)}.,().,({
}).,().{,('
1122
1122



































100
10
01
100
10
01
.
100
10
01
21
21
1
1
2
2
yy
xx
y
x
y
x
tt
tt
t
t
t
t
),(),().,( 21211122 yyxxyxyx ttttTttTttT 

Contd…
• Two successive rotations:
– For Successive Rotations θ1 and θ2
– Successive Rotations are additive. i.e.,
– Now proving this statement:
PRR
PRRP
)}.().({
}).().{('
12
12




)()().( 2112   RRR

Contd…
• Which demonstrates that two successive rotations are additive.
• So the final rotated coordinates can be calculated with the composite
rotation matrix as:
PRP ).(' 21  

Contd…
• Two successive scaling:
– Successive Scaling are multiplicative
– Or,































100
0.0
00.
100
00
00
.
100
00
00
21
21
1
1
2
2
yy
xx
y
x
y
x
ss
ss
s
s
s
s
).,.(),().,( 21211122 yyxxyxyx ssssSssSssS 

Contd…
• General Pivot Rotation: can generate rotations about an
arbitrary point (xr, yr) by performing the following sequence of
translate-rotate-translate operations:
– Translate the object so that the pivot-point position is moved to the
coordinate origin.
– Rotate the object about the coordinate origin.
– Translate the object so that the pivot point is returned to its original
position.

Contd…
• This transformation sequence is illustrated in the following fig:

Contd…

Contd…
• Example: Rotate the triangle having coordinates (1,2),(2,3)
and (4,5) by 60⁰ about (2,3).
• Example: perform the counterclockwise 45O rotate of triangle
A(2,3), B(5,5), C(4,3) about point (3,4).

Contd…
• General fixed point scaling:
– Translate object so that the fixed point coincides with the coordinate
origin.
– Scale the object with respect to the coordinate origin.
– Use the inverse translation of step 1 to return the object to its original
position.

Contd…
• The above transformation sequence to achieve fixed point scaling is
illustrated in the following fig:

Homework
• Example: Find the transformation matrix that transforms the
given square ABCD to half its size about (2,2,). The
coordinates of the square are A(1,1), B(3,1), C(3,3) and
D(1,3).

Homework
• Example: Rotate ∆ABC by 45⁰ clockwise about the origin and
scale it by (2,3) about the origin.
• Example: Rotate ∆ABC by 90⁰ anti-clockwise about the (5,8)
and scale it by (2,2) about (10,10).

Other 2-D Transformations
• Reflection
• Shear

Reflection
• A reflection is a transformation that produces a mirror image
of an object.
• The mirror image for a two-dimensional reflection is
generated relative to an axis of reflection by rotating the object
180 degree about the reflection axis.

Contd…
• Reflection about x-axis:
– Also known as reflection about line y=0
– Change the sign of y coordinate but x-coordinate remains same
• X’= X
• Y’=- Y
– In matrix form
– Equivalent to 180O rotation
about x-axis

Contd…
• Reflection about y-axis:
– Also known as reflection about line x=0.
• Change the sign of x-coordinate but y-coordinate remains same
• X’= -X
• Y’= Y
– In matrix form:
about y-axis

Contd…
• Reflection about origin:
– Also known as reflection about axis that is perpendicular to xy-plane
and that passes through the coordinate origin.
– Change the sign of both x-coordinate and y-coordinate.
» x’=-x
» y’=-y
– In matrix form:
about origin

Contd…
• Reflection about arbitrary fixed point:
– same as a 180o rotation in the xy plane using the reflection point as the
pivot point.

Contd…
• Reflection about a diagonal line y=x: can be obtained by
performing following sequence of transformations:
– Clock wise Rotation with 450
– reflection with x-axis
– Inverse rotation with 450










100
001
010

Contd…
1. first perform a clockwise rotation through a 45"
angle
2.Next, we perform a reflection with respect to the x
axis.
3. back to its original position with a counterclockwise
rotation through 45".

Contd…
• Reflection about a diagonal line y=-x:
– Can be obtained by performing the following sequence of
transformations:
• clockwise rotation
• reflection about the y axis, and
• counterclockwise rotation .












100
001
010

Contd…
• Reflections about any line y = mx+ c in the xy plane:
– can be accomplished with a combination of :
1. translate the Line so that it passes through the origin i.e., T(0, -c)
2. rotate the line onto one of the coordinate axes and reflect about
that axis. i.e.,
1. R(-theta)
2. Reflection with coordinate axis-axis
3. restore the line to its original position with the inverse rotation
and translation transformations i.e.,
1. R(theta)
2. T(0.c)

Contd…
• Example: Find out final transformation matrix, when point
p(x, y) is to be reflected about a line y= mx+c.
• Solution:

Contd…

Contd…
• Question: Determine the transformation for reflection about
y=x+3

Contd…
• Question: Translate the given triangle A(1,3), B(-2,-1) and
C(6,-2) with translation vector T(-2,-6) and then reflect about
y=x axis.

Shearing
59By: Tekendra Nath Yogi
A transformation that distorts the shape of an object.
2/9/2019

Cont…

Homework
1. Scale the rectangle ABCD with vertices A(2,2),B(2,6), C(5,2) and D(5,6) to double its
size about fixed point (2,3) then translate it using translation vector (1,3).
2. Rotate the triangle ABC with vertices A(3,5), B(7,2) and C(4,9) in counterclockwise
direction by 90 degree and then enlarge it to double its size.
3. A unit square is located at origin shear it along horizontal direction with shearing
coefficient 2.
4. Find out the final coordinates of a figure bounded by the coordinate (1,1), (3,4), (5,7),
(10,3) when rotated about a point (8,8) by 30 degree in clockwise direction and scaled
by two units in x-direction and three units y-direction.
5. Show that 2D reflection through x-axis followed by 2-D reflection through the line y=-
x is equivalent to a pure rotation about the origin.
6. Show that transformation matrix for reflection about a line y=x is equivalent to
reflection to x-axis followed by counter clockwise rotation of 90degree.
7. Reflect rectangle PQRS, P(1,2), Q(1,6), R(9,6) and S(9,2) on y=5 axis, then scale about (2,3) with scale
factor (4,2) and finally rotate with rotation angle +45o about origin. Use homogenous coordinate system
to calculate transformed rectangle coordinates.

3D Transformations
Presented By : Tekendra Nath Yogi
College Of Applied Business And Technology

3D Transformations
• Extend from two-dimensional transformation by including
considerations for the z coordinates.
• Translation and scaling are similar to two-dimension, include the three
Cartesian coordinates
• Rotation method is less straight forward
• The most common 3D transformations are:
– Translation
– Rotation
– Scaling
– Reflection and
– shears

Contd…
• Three-dimensional translation:
• A point P (x, y, z) in three-dimensional space translate to new
location p’(x’, y’, z’) with the translation distance T (tx, ty, tz) as :
• In matrix form:
xtx'x  yty'y 






































11000
100
010
001
1
'
'
'
z
y
x
t
t
t
z
y
x
z
y
x
PT'P 
ztz'z 

Contd…
• Questions: Translate the 3D square with vertices A(4,5,7), B(-
3,-2,1) , C(2,4,7) and D(-5,3,2) using translation vector (1,3,4).

Contd…
• Three-dimensional scaling:
– Relative to the coordinate origin, just include the parameter
for z coordinate scaling in the transformation matrix
– Relative to a fixed point (xf, yf zf)
• Perform a translate-scaling-translate composite transformation

















1000
)1(00
)1(00
)1(00
),,(),,(),,(
fzz
fyy
fxx
fffzyxfff
zss
yss
xss
zyxTsssSzyxT
PS'P 






































1
z
y
x
1000
0s00
00s0
000s
1
'z
'y
'x
z
y
x

Contd…
• The sequence of transformations for fixed point scaling is demonstrated in
Fig below.

Contd…
• Question1: A 3D object with vertices A(-2,-4,4), B(3,-6, -8), C(-
6,1,0) and D(3.-6.2) is required to scale with scale factor S(2,4,6)
about origin. Find the final coordinates.
• Question2: Given a solid PQT , p( 2,-5,0), Q(4,-2,3) and T(6,-7,-1)
is translated with T(9,0,-2) and then scale about origin with scaling
factor (2,4,5). Find the final coordinates.
• Question3: suppose we have a quadrilateral ABCD with vertices
A(3,1,4), B(-7,-4,2) C(5, -8, -3) and D(3,0,1). What is the
coordinates of quadrilateral if it is to be doubled in size about origin.
Then translate obtained coordinates with (-4,3,1)
• Question 4: scale the triangle with vertices A(-3,-2,1), B(4,5,6),
C(5,7,9) to double its size about fixed point (1,2,3).

3D Rotation
• To generate a rotation transformation for an object, we
must designate an axis of rotation (about which the object
is to be rotated) and the amount of angular rotation.
• Axis of rotation can be:
– Coordinate axis
– Parallel axis, or
– Arbitrary axis.
• Angle of rotation( ) can be:
– Positive : indicate counter clock wise rotation, or
– Negative: indicate the clock wise rotation.

Coordinate-Axes Rotations:
• A) Rotation about Z-axis

Contd..
• Any point p(x, y, z) can be rotated about z-axis with angle as
follows:
• In matrix form:
























 













1
z
y
x
1000
0100
00cossin
00sincos
1
'z
'y
'x


zz
yxy
yxx



'
cossin'
sincos'


PRP z ).(' 












 

1000
0100
00cossin
00sincos
)(


zR
2/9/2019

Contd..
• B) Rotation About X-axis

Contd..
• Any point p(x, y, z) can be rotated about x-axis with angle as
follows:
• In matrix form:
 














1000
0cossin0
0sincos0
0001


xx RR
x'x
coszsiny'z
sinzcosy'y











































1
.
1000
0cossin0
0sincos0
0001
1
'
'
'
z
y
x
z
y
x


PRP x ).(' 

2/9/2019

Contd..
• C) Rotation About Y-axis

Contd..
• Any point p(x, y, z) can be rotated about y-axis with angle as
follows:
• In matrix form:
y'y
cosxsinz'x
sinxcosz'z





 














1000
0cos0sin
0010
0sin0cos


yy RR






































1
.
1000
0cos0sin
0010
0sin0cos
1
'
'
'
z
y
x
z
y
x


PRP y ).(' 

2/9/2019

Contd…
• Question1: A coordinate point p(3,2,1) is translated in x, y and z direction
by -2, -2 and -2 unit respectively followed by rotation with 60o about x-
axis. Find the final position of a point
• Question 2: A cube of length 10 units having one of its corner at origin (0,
0, 0) and three edges along three principle axis. If the cube is to be rotated
about z-axis by an angle of 45 degree in counterclockwise direction, then
calculate the new position of the cube.
• Question 3: The pyramid defined by the coordinates A(0,0,0) B(1,0,0),
C(0,1,0) and D(0,0,1) is scaled to double its size about point (0,1,0), then
translated to new position by translation vector (4,5,6) and rotated by angle
45 degree in clockwise direction about origin. Use homogeneous
transformation matrix to find the new vertices of pyramid.

Contd..
• Rotation about axis parallel to coordinate axis
– when an object is to be rotated about an axis that is
parallel to one of the coordinate axes, we can attain the
desired rotation with the following transformation
sequence.
• Translate the object so that the rotation axis coincides with the
parallel coordinate axis.
• Perform the specified rotation about that axis.
• Translate the object so that the rotation axis is moved back to its
original position.

Contd..
• The sequence of transformations for rotation about axis parallel to
coordinate axis(e.g.., x-axis) is demonstrated in figure below:
step1
step2
step3

Contd..
• Rotation about arbitrary axis that is not parallel to
one of the coordinate axes :
– when an object is to be rotated about an axis that is not parallel to one of
the coordinate axes, we can attain the desired rotation with the following
transformation sequence.

Contd..
– Translate the object so that the rotation axis passes through the coordinate
origin.
– Rotate the object so that the axis of rotation coincides with one of the coordinate
axes.
– perform the specified rotation about that coordinate axis.
– Apply inverse rotation to bring the rotation axis back to its original orientation.
– Apply the inverse translation to bring the rotation axis back to its original
position.

Contd..
Fig: Five transformation steps for obtaining a composite matrix for rotation about arbitrary axis,
with the rotation axis projected onto the z-axis2/9/2019

Contd…

Contd…
• Example: Find the new co-ordinates of a unit cube 90 degree
rotated about an axis defined by its end points A(2,1,0) and
B(3,3,1).
A unit cube

Contd…

Contd…
• Question1: Rotate triangle with vertices A(), B(0,2,1),(2,3,0)
and (1,2,1) about the line joining the point (0,0,0) and (1,1,1)
with the angle of rotation 450 in counter clockwise direction.
• Question1: Rotate triangle with vertices A(0,2,1),B(2,3,0)
and C(1,2,1) about the line joining the point (2,2,2) and (1,1,1)
with the angle of rotation 450 in counter clockwise direction.

Reflections
• A three-dimensional reflection can be performed relative to a
selected reflection axis or with respect to a selected refection plane.
• Reflections relative to a given axis are equivalent to 18o0 rotations
about that axis.
• Reflections with respect to a plane are equivalent to 1800 rotations
in four-dimensional space.
• Reflection plane can be a coordinate plane (either xy, xz, or yz)

Contd…
• Reflection in xy plane:
• This transformation changes the sign of the z coordinates, Leaving the x
and y-coordinate values unchanged.
– x’ = x
– y’ = y
– z’ = -z
• The matrix representation for this reflection of points relative to the x-y
plane is:

Contd…
• Reflection in yz plane:
– This transformation changes the sign of the x coordinates, Leaving the
z and y-coordinate values unchanged.
• x’ = -x
– y’=y
– z’=z
– The matrix representation for this reflection of points relative to the x-y
plane is:

Contd…
Reflection in XZ plane:
– This transformation changes the sign of the y coordinates, Leaving the
z and x-coordinate values unchanged.
– y’= -y
– x’=x
– z’=z
– The matrix representation for this reflection of points relative to the x-y
plane is:

Shears
• Shearing transformations produce distortions in the shapes of
objects.
• The transformation equations for a Z-axis shear is :
– x’ = x + a.z
– y’ = y + b.z
– z’ = z
The matrix representation for a z-axis shear is:





































1
.
1000
0100
010
001
1
'
'
'
z
y
x
b
a
z
y
x
2/9/2019

Contd…
• Similarly, The matrix representation for a x-axis shear is:
• The matrix representation for a y-axis shear is:





































1
.
1000
010
001
0001
1
'
'
'
z
y
x
b
a
z
y
x





































1
.
1000
010
0010
001
1
'
'
'
z
y
x
b
a
z
y
x

Contd…
• Question1: shear the triangle with vertices A(1,2,1), B(5,4,7)
and C(3,6,9) about x-axis using shearing coefficient shy=3 and
shz=5.

2D viewing
• Pictures are stored in memory using Cartesian coordinate
system referred to as world coordinate system.
• But, pictures are displayed in display device using physical
device coordinate system.
• So, to display picture requires the transformation from world
coordinate to device coordinate.
• This mapping of coordinate can be achieved with the use of
viewing transformation.

Contd…
• window :
– A world-coordinate area selected for display is called a window.
– defines what is to be viewed
• viewport :
– An area on a display device to which a window is mapped is called a
viewport.
– defines where it is to be displayed.

Contd..
• viewing transformation:
– The mapping of a part of a world-coordinate scene to device
coordinates is referred to as a viewing transformation.
– 2D viewing transformation referred to as the window-to-viewport
transformation or the windowing transformation.
Fig: A viewing transformation using standard rectangles for the window and viewport.

Contd…
• viewing pipeline:
– Takes the object coordinates through several intermediate coordinates
systems before finishing with device coordinates as shown in figure
below:
Fig: The 2D viewing-transformation pipeline

Window to Viewport Transformation
• A point at position (xw, yw) in a designated window is mapped to
viewport coordinates (xv, yv) so that relative positions in the two
areas are the same.
• i.e., If a coordinate position is at the center of the viewing window, for
instance, it will be displayed at the center of the viewport.

Contd…
• To maintain the same relative placement, the following ratios
must be equal.
minmax
min
minmax
min
minmax
min
minmax
min
ywyw
ywyw
yvyv
yvyv
xwxw
xwxw
xvxv
xvxv











Contd…
• By solving these equations for the unknown viewport position
(xv, yv):
• Where the scaling factors are:
syywywyvyv
sxxwxwxvxv
)(
)(
m inm in
m inm in


m inm ax
m inm ax
m inm ax
m inm ax
ywyw
yvyv
xwxw
xvxv
sy
sx







Contd…
• Given a window and viewport, what is the transformation
matrix that maps the window from world coordinates into the
viewport in screen coordinates?
• This matrix can be developed by three step transformation
composition as:
– The object together with its widow is translated until the lower left
corner of the window is at the origin.
– Object and window are scaled until the window has the dimension of
the viewport.
– Translate the viewport to its correct position on the screen.

Contd…
• Above three steps are illustrated in the following figure:
step0.
Step 3.step2.
step1.

Contd…
• Step1: translation













100
10
01
min
min
w
W
Y
X
T

Contd…
• Step2: scaling about origin
• Where,
minmax
minmax
minmax
minmax
ywyw
yvyv
xwxw
xvxv
sy
sx

















100
00
00
y
x
s
s
S

Contd…
• Step 3: inverse translation











100
10
01
)( min
min
v
v
Y
X
Tinverse

Contd…
• The overall transformation matrix for window to viewport
transformation is:













100
).(0
).(0
minmin
minmin
syywyvs
sxxwxvs
y
x












100
10
01
min
min
w
W
Y
X










100
00
00
y
x
s
s











100
10
01
min
min
v
v
Y
X

Contd…
• Example:

Contd…
• Step2: calculating the scaling factor and then scaling
transformation matrix
• Sx=(1-0.5)/(20-10)=0.5/10= 0.05
• Sy=(1-0.5)/(20-10)= 0.5/10= 0.05

Contd…
• Step3: Inverse translation matrix to maintain the relative
position in view port is:

Contd…
• So, the final composite view transformation matrix is:
0 0

Contd…
• Example1: Translate the triangle with vertices A(10,10), B(20, 30), and
C(15,20) from window to viewport where window is defined by xwmin=50,
xwmax=100, ywmin=20 and ywmax=100 and viewport is defined as xvmin=20,
xvmax=45, yvmin=40 and yvmax=80
• Example2: Translate the triangle with vertices A(2,2), B(3, 5), and C(5,8)
from window to viewport where window is defined by xwmin=1, xwmax=6,
ywmin=2 and ywmax=10 and viewport is defined as xvmin=0, xvmax=2
yvmin=0 and yvmax=4
• Example3: consider the window is located from 0 to 100 and a point is
located in (30, 30). Identify the point position in the viewport located from
0 to 50.

Clipping
Presented by: Tekendra Nath Yogi
College of applied business and technology

Clipping
• Any procedure that identifies those portions of a picture that are either
inside or outside of a specified region of a space is referred to as
clipping.
• The region against which an object is to be clipped is called a clip
window. Depending on the application a clip window can be polygon or
curved boundaries.
• World- coordinate clipping removes the primitives outside the window
from further consideration; thus eliminating the processing necessary to
transform these primitives to device space.
• Clipping type- point ,line, polygon, curved areas and etc.

Contd…
• Applications of clipping include:
– Extracting part of a defined scene for viewing
– Identifying visible surfaces in three-dimensional views
– Antialiasing line segments or object boundaries
– Creating objects using solid-modeling procedures
– Displaying a multi-window environment
– Drawing and painting operations that allow parts of a picture to be
selected for copying, moving, erasing, or duplicating.

Point Clipping
• In a rectangular clip window save a point P = (x, y) for display(i.e., not clipped)
if the following inequalities are satisfied:
• Xwmin ≤ x ≤ xwmax
• ywmin ≤ y ≤ ywmax
• If any one of these four inequalities is not satisfied, the point is clipped
(not saved for display).
• The equal sign indicates that point on the window boundary are included within
the window.

Contd…

Line clipping
• The visible segment of a straight line can be determined by
inside – outside test:
– A line with both endpoints inside clipping boundary, such as the line
from p1 to p2, is saved.

Contd…
• A line with both endpoints outside the clip boundary, such as
the line from p3 to p4, is not saved.

Contd…
• If the line is not completely inside or completely outside (e.g., p7 to p8 ),
then perform intersection calculations with one or more clipping
boundaries.

Cohen-Sutherland Line Clipping
• This algorithm clips the line by performing following steps:
• Step1: Region code assignment
– divide the whole picture region into nine regions by extending the window
boundaries as shown in figure below and then assign a 4-bit region code to
each region as follows:
1001 1000 1010
0001
0000
Window
0010
0101 0100 0110
Rules For assigning region code:
T=1 , if the region is above the window,
= 0 , otherwise.
B= 1, if the region is below the window,
= 0, otherwise.
R=1, if the region is right of window,
= 0, otherwise.
L= 1, if the region is left of window,
= 0, otherwise.

Contd…
• Every end-point is labelled with the appropriate region code
• For example:
wymax
wymin
wxmin wxmax
Window
P3 [0001]
P6 [0000]
P5 [0000]
P7 [0001]
P10 [0100]
P9 [0000]
P4 [1000]
P8 [0010]
P12 [0010]
P11 [1010]
P13 [0101] P14 [0110]

Contd…
• Step2: Trivial acceptance of line segment
– Lines completely contained within the window boundaries have region code
[0000] for both end-points so are not clipped. i.e., accept these lines
trivially.
– E.g., line p5p6
wymax
wymin
wxmin wxmax
Window
P3 [0001]
P6 [0000]
P5 [0000]
P7 [0001]
P10 [0100]
P9 [0000]
P4 [1000]
P8 [0010]
P12 [0010]
P11 [1010]
P13 [0101] P14 [0110]

• Step3: Trivial rejection and clipping
– 3.1 : Any lines that have a 1 in the same bit position in the
region-codes for each endpoint are completely outside and we
reject these lines.
– The AND operation can efficiently check this: If the logical AND of both
region codes result is not 0000, the line is completely outside the
clipping region so clipped.
Contd…
wymax
wymin
wxmin wxmax
Window
P3 [0001]
P6 [0000]
P5 [0000]
P7 [0001]
P10 [0100]
P9 [0000]
P4 [1000]
P8 [0010]
P12 [0010]
P11 [1010]
P13 [0101] P14 [0110]

3.2:If the logical AND operation results in 0000 then
a) Choose an endpoint of the line that is outside the window.
b) Find the intersection point at the window boundary by using the
following formula:
c) Replace endpoint with the intersection point and update the region code.
Above process is repeated until we find a clipped line either trivially accepted
or trivially rejected.
Contd…

Contd…
• Algorithm:

Contd..
• Example: Use the Cohen-Sutherland line clipping algorithm to clip line
defined with end points p1(0,120) and (130,5) against a window defined by the
following four points : (10,10), (10,100), (150,10) and (150,100)

Contd…
• Solution:

Contd…

• Example1: Let ABCD be the rectangular window with A(20, 20), B(90,
120), C(90, 70), and D(20,70). Find the region codes for end points and
use Cohen- Sutherland algorithm to clip the line p q with p( 10,30) and
q(80, 90).
• Example2: Use the Cohen-Sutherland algorithm to clip line defined with
end points p1(40,15) and p2(75,45) against a window A(50,10),
B(80,10), C(80,40) and D(50,40).
• Example3: Use the Cohen-Sutherland algorithm to clip line defined with
end points p1(70, 20) and p2(100,10) against a window A(50,10),
B(80,10), C(80,40) and D(50,40).
By: Tekendra Nath Yogi 139
Homework
2/9/2019

Polygon Clipping
• Polygon clipping algorithm generates one or more closed bounded areas as
shown in figure below:
By: Tekendra Nath Yogi 1402/9/2019

Contd…
• Sutherland-Hodgman Polygon Clipping:
– Begin with the initial set of polygon vertices, first clip the polygon against
the left rectangle boundary to produce a new sequence of vertices.
• The new set of vertices could be successively passed to a right boundary
clipper, a bottom boundary clipper, and a top boundary clipper as shown in
figure below.

142
Contd…
Left
Clipper
Right
Clipper
Bottom
Clipper
Top Clipper
At each step, a new sequence of output vertices is generated and passed to the next
window boundary clipper.
2/9/2019
By: Tekendra Nath Yogi

As each pair of adjacent polygon vertices is passed to a next window
boundary clipper, we make the following tests:
1. If the first vertex is outside the window boundary and the second vertex
is inside Then , both the intersection point of the polygon edge with the
window boundary and the second vertex are added to the output vertex
list.
For example:
v1- outside and
v2- inside
Movement is out in
Output vertex – intersection point v1’ and
destination point v2

Contd…
2. If both input vertices are inside the window boundary. Then,
only the second vertex is added to the output vertex list.
• For example:
– v1- inside
– V2- inside
– Movement in in
– Output vertex = destination vertex
– Therefore save V2

Contd…
3. If the first vertex is inside the window boundary and the second
vertex is outside. Then, only the edge intersection with the
window boundary is added to the output vertex list.
• For Example:
– V1-inside
– V2- outside
– Movement = in out
– Therefore, save v1’

Contd….
4. If both input vertices are outside the window boundary.
Then, nothing is added to the output vertex list.
• For Example:
– V1- outside
– V2- outside
– Movement = out out
– Therefore, save nothing
By: Tekendra Nath Yogi 146
2/9/2019

Contd…
• Example1 :We illustrate this algorithm by processing the area in figure against
the left window boundary.
• Vertices 1 and 2 are outside of the boundary.
• Vertex 3, which is inside, 1' and vertex 3 are saved.
• Vertex 4 and 5 are inside, and they also saved.
• Vertex 6 is outside, 5' is saved.
• Using the five saved points, we would repeat the process for the next window
boundary.

Contd…
• Example2: clip polygon ABCDE against window PQRS. The
coordinate of the polygon are A(80, 200), B(220, 120), C( 150,
100), D(100, 30), E(10, 120). Coordinates of the window are
P(200,50), Q(50, 150), R(200, 150), S(50, 50).

Homework
• Example 3: You are provided with the clipping rectangle with co-
ordinates A(10, 10), B(10, 20), C(20, 10) and D(20, 20). Clip the
given polygon PQRS with coordinates P(5, 15), Q(15, 25), R(15,
15) and S(10, 10) using Sutherland - Hodgeman polygon clipping
algorithm.

Curve Clipping
• The bounding rectangle for a circle or other curved object can
be used first to test for overlap with a rectangular clip window.
• If the bounding rectangle for the object is completely inside
the window, we save the object.
• If the rectangle is determined to be completely outside the
window, we discard the object. In either case, there is no
further computation necessary.
• But if the bounding rectangle test fails, we can look for other
computation-saving approaches.

Contd…
• For a circle, we can use the coordinate extents of individual
quadrants and then octants for preliminary testing before
calculating curve-window intersections.
• For an ellipse, we can test the coordinate extents of individual
quadrants.

Contd…
• Figure below illustrates circle clipping against a rectangular
window.

Contd…
• Similar procedures can be applied when clipping a curved
object against a general polygon clip region.
• On the first pass, we can clip the bounding rectangle of the
object against the bounding rectangle of the clip region.
• If the two regions overlap, we will need to solve the
simultaneous line-curve equations to obtain the clipping
intersection points.

Homework
1. What is clipping? Explain the polygon clipping algorithm with suitable
example.
2. Where do you require ellipse clipping algorithm? Explain in detail about
ellipse clipping algorithm.
3. Why clipping is required in computer graphic? Explain the circle clipping
algorithm.

3D viewing

Introduction
• 2D viewing transfers position from the world coordinate plane
to plane of the output device.
• But in 3D viewing such direct transformation is not possible
due to the different views of the object (e.g., top view, front
view, back view, side view and etc).
• So requires projection transformation.

Contd…
• Viewing pipeline:
Fig: General 3D transformation pipeline, from modeling coordinates to final device coordinates

Contd…
• Construct the shape of individual objects in a scene within
modeling coordinate, and place the objects into appropriate
positions within the scene (world coordinate).

Contd…
• World coordinate positions are converted to viewing
coordinates

Contd…
• Convert the viewing coordinate description of the scene to
coordinate positions on the projection plane.

Contd…
• Positions on the projection plane, will then mapped to the
Normalized coordinate and output device.

Projection

Projections
• In general, projection transform points in a coordinate system of dimension n
into points in a coordinate system of dimension less than n.
• Projection is the mapping or transformation of a three-dimensional (3D) object
into a two-dimensional (2D) object.

Contd…
• How projections are constructed?
– projections are constructed by linearly mapping points in 3D space
to points on a 2D projection plane.
– The projected point on the plane is chosen such that it
is collinear with the corresponding 3D point and the centre of
projection.
– The lines connecting these points are referred to as projectors.
– The centre of projection is the location of the observer
– The plane of projection (view plane)is the surface on which the 2D
projected image of the scene is recorded/viewed.

Contd…
• Properties of projection:
– Projections map points from one space to another coordinate
space of lower dimension, and hence involves loss of
information.
– Projections are not invertible.
– All points on a projector map to the same point on the plane
of projection.

Types of projection
• Projections are of the following types:

Contd..
• Parallel Projection:
– A parallel projection is a projection of an object in three-
dimensional space onto a projection plane(2D), where the projection
lines are parallel to each other.

Contd…
• How ?
– Parallel projection discards z-coordinate and parallel lines from each vertex
on the object are extended until they intersect the view plane.
– The point of intersection is the projection of the vertex.
– we connect the projected vertices by line segments which correspond to
connections on the original object.
• A parallel projection preserves relative proportion of objects, but
dose not give us a realistic representation of the appearance of
object.

Contd…
• Various types of parallel projections are :

Contd…
• Oblique projection:
– The projectors in oblique projection intersect the projection plane at an
oblique angle( not perpendicular) to produce the projected image.

Contd…
• Orthographic Projection:
– Orthographic projection used to produce the front, side, and
top views of an object.
– Projectors are perpendicular to the projection plane.

Contd…
• Perspective Projection:
– In this projection projectors all converge at a single point
called center of projection or projection reference point.

Contd…
• How?
– The object positions are transformed to the view plane along converged
projectors and the projected view of an object is determined by calculating
the intersection of the converged projectors with the view plane as shown in
figure.

Contd…
• Advantages and Disadvantages of perspective projection:
– Objects further from viewer are projected smaller than the same sized
objects closer to the viewer.
• Looks realistic
– Equal distances along lines are not projected into equal distances (non-
uniform foreshortening)
– Angles preserved only in planes parallel to the projection plane
– More difficult to construct by hand than parallel projections (but not
more difficult by computer)

Comparison between Parallel and Perspective
• Parallel Projections:
– The center of projection
is at infinity.
– The projectors are parallel
to each other.
• Perspective Projections:
– The center of projection is
a finite point.
– The projectors intersect at
the center of projection.

Contd…
• Types of perspective projection:
Assignment!

Homework
1. What is projection? Differentiate between parallel and perspective projection.
2. What is perspective projection? Explain the different types of perspective
projection.
3. Write a procedure to perform a two-point perspective projection of an object.

Thank You !

B. SC CSIT Computer Graphics Unit 2 By Tekendra Nath Yogi

In this document

More Related Content

What's hot

Similar to B. SC CSIT Computer Graphics Unit 2 By Tekendra Nath Yogi

More from Tekendra Nath Yogi

Recently uploaded

B. SC CSIT Computer Graphics Unit 2 By Tekendra Nath Yogi