0% found this document useful (0 votes)
76 views14 pagesC Scaner1
Scanner
Uploaded by
Chirag UCopyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
Available Formats
Download as PDF or read online on Scribd
0% found this document useful (0 votes)
76 views14 pagesC Scaner1
Scanner
Uploaded by
Chirag UCopyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
Available Formats
Download as PDF or read online on Scribd
/ 14
Sixth Semester B.E. Degree Examination
CBCS - Model Question Paper - 1
COMPUTER GRAPHICS AND VISUALIZATION
Time: 3 hrs. Max. Marks: 100
Note : Answer any FIVE full questions, selecting ONE full question from each module.
MODULE -1
1. a, Computer graphics can be applied in all ficlds, Justify this statement. (08 Marks)
Ans, Computer-Aided Design for engineering and architectural systems.
Objects maybe displayed in a wireframe outline form, Multi-window environment is
also favored for producing various zooming scales and views. Animations are useful
for testing performance,
Presentation Graphics
To produce illustrations which summarize various kinds of data. Except 2D, 3D
graphics are good tools for reporting more complex data.
Computer Art ;
Painting packages are available. With cordless, pressure-sensitive stylus, artists can
produce electronic paintings which simulate different brush strokes, brush widths,
and colors. Photorealistic techniques, morphing and animations are very useful in
commercial art. For films, 24 frames per second are required. For video monitor, 30
frames per second are required.
Entertainment
Motion pictures, Music videos, and TV shows, Computer games
Education and Training
Training with computer-generated models of specialized systems such as the training
of ship captains and aircraft pilots.
Visualization
For analyzing scientific, engineering, medical and business data or behavior.
Converting data to visual form can help to understand mass volume of data very
efficiently.
Image Processing
Image processing is to apply techniques to modify or interpret existing pictures. It is
widely used in medical applications.
Graphical User Interface
Multiple window, icons, menus allow a computer setup to be utilized more efficiently.
b, Explain the differences between raster scan and random scan display device.
(06 Marks) Computer ureprese
VI Sem (CSE / ISE)
BCS - Model Question Paper + 1
Dave of aca Spee Handom Sean System,
yteresce
aa Tn eee 0 a | Te cc ea dete
taacron [Te Stn ee Me [he pa of ren Mee ap poramcter mode ofthe function glBepin canbe one of the following
tap to bottom. = Ee GL_POINTS
ecu GL_LINES
el aa
tem m eta pec 1428 GLILINE,sTRIP
Reaction |ifey nat we ped @ eC GLCLINE LOOP
GLZTRIANGLES
GL_TRIANGLE_STRIP
GLLTRIANGLE_FAN
GL_QUADS
OR
do you mean by interactive computer graphics? Explain (02 Marks)
itive Computer grapbics is an art of drawing pictures, lines, chars, ete.
‘comput of programming. Computer graphics image is made
up of number is the smallest addressable graphical unit represented
‘on the computer sereen.
Creation of an object + Manipulation of Object + Display of an object using Computer
bb, What are advantages of DVST over CRT? Also list some disadvantages of
ar (04 Marks)
These systems are designed for
sy| drawing snd can't display seal
peter
Drowan | Screen pomtvpnels ae wed to
eam a9 nase,
ded.
res can be displayed at very high resolution without icker
‘Compute the following
Size of 800 x 600 image at 240 pixels per inch.
image that has S12 x 512 pixels.
image 1024 x 768 to one that is 640 pixels wide with
aspect ratlo.
da of an image having height of S Inches and an aspect ra
(04 Maris)
{800/240 inch = 31 /3 inch Similarly 600 pixels
pe
io of the 1024 x 768 image is 768 / 1024 = %
‘of the image having height of 640 pixels having aspect ratio % is 640
Suns Exams Paros 5 computer Graphics And Visualisation)
VI Sen (CSE / SE)
4. Describe the mui
‘Ans. void PlotCirelePoi
ixel(centerx-yicentery-x);
lfint centers, int eentery, int radius)
point Circle Draw
Given aeiccleradius =O
CBCS - Model Question Paper - 1
¥_)=(0,10) and
responding coordinate
Bye
20
20
0
Module - 2
3. a. Explain the Boundary-Fill Algorithm w.rt computer graphics (08 Marks)
‘Ans.
fand paint the interior outward
is recursive method which may
COLOR current; current=GetPixel(
if (current boundary) and (current fll) then
boundary); Boundary}
‘boundary); Boundary ill(x.y-1 6 CBCS - Model Question Paper - 1
8
- OR
4, a. Write explanatory note on i) Flood Fill ji) Text generation Methods in OpenGL
(08 Marks)
‘BO
‘these associative grouping. Fy
he matrix B, then we
(9 Boundary-Fill. The difference is that Flood-Fill isto fill at
ii) Text Generation Methods in OpenGL
‘There are 3 basic methods of generating characters, they are namely stroke method,
ist method and BitMap Method.
supports the scaling of a character.
"= In this method a fix pattern of line segments are used to
character generator:
the sereen by copying pixel values from the character array into some portion of the
‘screens frame buffer. The opengl function is void glut Bitmap Character(void *font,
int character).
b. Explain the different types of polygons supported by openGL. (07 Marks)
1 Fay from the iin
specifies the angle of lation "PMS Pit (sy. 2). The angle parameter Ans Polygong{GL_FOLYGON) a 5 ecimea bal
in degrees. ’ "Successive vertices define line segments and a line Segment con nal
Sealing ;
by vrextyi terior is filled according tothe state of relevant attributes.
eee 3 fost graphics systems allow us to fill the polygon with color or pattern orto draw
erent mati by a
es = that stretches, shrinks.or ref the lines around the edges but not to do both.
ects © We use the function g!PolygonMode to tell what we want.
sli ts Sicina' Bis Senscen Esme Poastars
9 CBCS - Model Question Paper - 1
eee
transformation as a matrix transformation, we 1
tye det xe ple bo oI
veers oR EG
bell
a
transformation
{In matrix form
sone of the positive x,
. The canonical rotations are defined
‘Then the construction ofthe
Rotation about the z axis,
We know
displaced a given distance anc
oo eal pene ofthe translat Rye:
: Rotation about the y Axis
‘An analogous derivation leads to
F -
acs -ModebquestionPaper"2
OR
(08 Marks)
re by coating
subtractive
presents black because all components are subtracts and origin
ite light.
© Gray can be produce among main diagonal by using all three color in equal
«© Printing process often use CMY model generates a color points wit
‘of four ink dots, one for each primary color C, M, and Y and one
Fig. «= Diverging ray paths from a po
Teasonable approximation for source whose size is
pared to the size of objector may beat sufficient distance so that we
san see it as point source. For example sun can be taken as point source on earth.
1s Anearby souree such as the long fluorescent light is more accurately modelled 3s
a distributed light source.
Sumstan Seame Parioner 13 y VI Sen (CSE / ISE) Computer Graphicy anw Visualiz
In this case the illumination effects cannot be approximated with poin
because the area of the source is not small compare to the size of object.
© When light is falls on the surface the part of, the light is reflected and part of
the light is absorbed. Amount of reflected and absorbed light is depends ON the 4
property of the object surface. For example shiny surface reflect more light while
dull surface reflect less light.
c. Perform a 45 degree rotation of a triangle A(0,0),B(1,1),C(52) _—idabout the
Source
origin (06 Maris)
* Ans. We represent the triangle by a matrix formed from the homogeneous co-ordinates of
the vertices:
o1s
on 2
i) The matrix rotation is
R45° = cos 45° -sin4s° 0
Sin45° cos 45° 0
0 0 1
Since the coordinated A’B’C’ of the rotated triangle ABC can be found as
[AB’C’]=R45[ABC]=0 0 32/2
0 v2 7N2/2
.
Module - 4
7. a, What is viewing transformation? Explain (04 Marks) —
Ans. Conversion of objection descriptions from world to viewing coordinates is equivalent
to a transformation that superimposes the viewing reference frame onto the world
frame using the basic geometric translate-rotate operations:
1, Translate the view reference point to the origin of the world-coordinate system.
2. Apply rotations to align the x,, y,, and z, axes (viewing coordinate system) with
the world x,, y,,Z, axes, respectively. 4
@
b, Explain, why there is a need for visible surface detection? Differentiate between
object precision and image precision methods for detecting visible surface.
(08 Marks)
Ans, The surfaces that are blocked or hidden from view must be “removed” in order to
construct a realistic view of the 3D scene. The identification and removal of these «
surfaces is called the visible surface detection or hidden — surface removal problem. 3
14 Senstar Exam Parton Sixth Semester B.E, Degree Examination
CS - Model Question Paper - 2
COMPUTER GRAPHICS AND VISUALIZATION
lecting ONE ful question from each
Idof statistics uses graphics system(computer plotting packay
plots that aid the viewer in understanding the information in
ography), MRI(Magnetic Resonance
1g), Ultrasound and PET(pasition-emission Tomography) uses computer
fie Visualization provides graphical tools that hel
interpret the vast quantity of data that they generate.
ive graphical tools in computer aided design(CAD)
including as architecture, mec!
simulators led to the use of computer graphics for an
nt picture and advertising industries.
BCS - Model Question Paper - 2
‘manage multp
‘working with computer grap
gram, explain the architecture of raster scan display «
iy 2 advantages and disadvantages. (08 Marks)
‘Ans. Raster sean display
‘roceeno
ronnie (5
ncnccanont
Fetnaneater
Fig. Architecture of a raster display.
cture of Raster display. It consists of display controller,
fresh buffer, keyboard, mouse and CRT.
stored inthe form of I's and 0's in the refresh buffer.
reads this refresh buffer and produces the actual image on
me Tine ata time from top to bottom & then back to the top.
Fig. Raster scan CRT.
Samstar Eran Partcs 3 Computer Graphicy and Visualization cs - Model Question Paper - 2
e
BO
VI Senv (CSE / ISE)
————————— ae
5 ba signa are zeneratedtompag
A, Here beam is gy
tis ON. When beam is moved from ri
}FF and process of moving beam fiom right to left after comp
is known as Horizontal Retrace.
al the bottom of the screen. It is made O °F and rapidly
ack 10 the (op left o start again and process of moving back to top ig
e. What is an OpenGL? Explain the different OpenGL Primitives with example
for cach. (04 Marks)
OpenGL is a software inte Si is i *
Open rare interface to graphics hardware. This interface consists of
needed to produce interactive three dimensional ap
a streamlined, hardware-independent interfa
‘hardware platforms.
165 are specified using a sequence of
Sunstan Exam Partacr
“The most predominant type of display has been the Cathode Ray Tube (CRT).
vertices
2)GL.1
3) GL_LINE_STRIP
4) GL_LINE_LOOP
5) GL_TRIANGLES
6) GL_QUADS
7)GL_POLYGON
8) GL_TRIANGLE_FAN
L_QUAD_FAN
siBegin(Gl
glVertex2f(x
plVerten2Mx2, ¥
glENDO:
‘Definea pair of pointsas: glBegin(GL_POINTS); glVertex2Axt, yl); glVertex2Mx2,
glENDO;
OR
Describe the working of an output device with an example, (06 Marks)
Various parts of a CRT:
@ Electron Gun — emits electron beam which
‘on Plates — controls the direction of beam. The output of the computer
js converted by digita-to-analog converters 0 voliges across x & y deflection
+ free image, the image on the sereen has
strikes the phosphor coating to emit
plates. ;
« Refresh Rate In order to view a flicker
tobe :
‘© reiraced by the beam ata high rate (modem
© 2iypes of refresh:
systems operate at 85112)
27
Sunscas Exame Partacr y
VI Senv (CSE / ISt) Computer Graphicy and Visuadi,
eV_ ion
ced display: Pixels are displayed cow by row atthe refresh rate
coos -odelquetionPeper=2
‘Odd rows and even rows are refreshed alternately, x x Phot points (Rounded to Insp
b.Wh tewecesaryal rel sale 0 foros
for scan converting oe a
(8 Marta 1.3240 66=1.98 (23)
L
2
3 | 1.98+0.66-2.64 Bay
12.60.6633 [eS 13.5)
UI
tween the endpoint positions are
late dx=xb-xa and dy=yb-ya).
sr magnitude determines the value of parameter
izes of DDA Algorithm
wmetic in DDA algorithm
ion to get command line arguments and initializes system.
ised for initializing the glut library and also to initiate»
session dows system. The f not take any arguments end
should be the frst function to be called within the main promis, mm
TaitDisplayModeO requests properties forthe window (the renlering conten
Bea ected [UT_RGB. It specifies that a 3 color mode
RGB color- specified by the argument GL
needs tobe used. P
Single bufering - GLUT_SINGLE: specifies thatthe images are static and col @
single frame bulfer is require to store the pixels
Sates Esa Pastas 29 Computer Graphics and Visa
VI Semv (CSE / ISE)
IBLE: specifies that the images are ani
snd back are required for rendering a smooth
y
lutWindowSiz in pixels
lutWindow Position from top-left comer of display
flutCreateWindow create window with a particular title
ns
‘ ae (08
ods involved in filling polygons are:
each edge of the polyon from Ymin to Ymax.Name
.on.From the diagram itis named a
thm makes use of four-connected and
- According to this technique instead of
ize the value of the seed point(seedx seedy), feolor and deol.
fine he boundary alas fhe yg
if the current seed point is ofthe default colorthen re
the boundary pixels reached ne
thange the default color withthe fil colo atthe seed point.
Recursively follow the procedure with four neighbourhood points.
‘According to this 8 connected region we have
‘the seed point(seedx. seedy) feolot and dol.
color atthe seed point.
the procedure with 4 neighbourhood points.
ied oan object by repositioning it along asricht
location to another xl"xeTx yI=y*Ty (TT) ~
‘A two dimensional rotation is applied to an object by repositioning it
ag a circular path in the xy plane.
along circular pat inthe xy pl
Re [se sie 7}
Hiadcose (0: rotation angle
ing: A scaling transformation alters the size of an object
cy 1=y.SySx and Sy are scaling factors. Thee ae two types of scaling they
ii) Rot
xr Sealing: Here Sx and Sy are assigned the same value,
tial Sealing: Unequal values fr Sx and Sy.
Mlection: A reflection isa transformation tht produces a mitror imoge of 20
object.
By line y=O(s-axi
Transformation Matrixis, 1 OO
o 4 0
oo 8
By the X acs, the transformation mauixis <1 6
oot
sear Atnsfomasos tations ie snp emo
shape appears asf theoject composed of intema) E> NS caused
slide over each other is called shear
bry
31
iste Cone Dane OR
-xamples explain some
carious (ypes of views that are =
i ‘empoyed
(06 May
Janar geometric projections
eel
perspective
pave
|
|
I | Ih Gee 2psint spline
aultbiew axonametie oblave
orthographic
Isometric dimetieg —trimettis
Fig Types of Views
and parallel proj i
viewing. Perspective projection has a COP where a
jrojection has parallel projectors. Here
‘So here we have a “Direction of projection (DOP)” instead.
ies-hollow with a color border,
bas
cant by comp
le examples.
transformation
‘by the other, We can set up a matrix
fransformation matrix by calculating the matrix
{lowed by a second plane transformation
transformation Tw!
Te TIT.
of transformation
12,
the
formation ean be achieved by concatena
=
rom SS one sees
sagen:
mpd = x) (7H £72) (73) (74) .... Ta}
is Preecon plone
(b) Parallel Projection
ifferent results as in general
1B} . [A] and the order of
to gain efficiency
rather than applying
the order of transformation
nis not cumulative. tha
‘object looks like from its multiview orthographie projection
somposed transformation 10 9 poi
one after another.
sbitrary point (X,, ¥,)s We have t0 ¢arry
origin.
tae tbe center of rotation back where i belonged.
33
32 Be 34
Sour Ene Paco i Samer Esa Parnet
Comper urupracs crue Vistalé
Vr Senv (CSE / ISE)
cacs « Model Question Paper -2
taken as largest of set consisting of 0 and various values of. The
the boundaries for which lines proceeds
wes gliPerspective(GLdouble fovy.
double far) ac
inthe xz plane and must be between
for Pand q are calculated and used by function Ci
1c ean be rejected or whether the intersection parameter can be adjusted. When
rameter © is used 10
p=0 and q<0,
thas not been
of clipped
are determined from values of ul and w2.
thm is more efficient than the Cohen-Sutherland algorithm.
Module - 3
a ¢ Liang-Darksy Line Clipping algorithm (08 Marks),
ns. line clippers have
= xl) and dy = 2-y1)
ipping condition in parametric be
‘Ans.
Te
arnt sate sie rs pal
eee a clipping, we require an ‘that will gene
lin we teases a Theoapeein cr
should be a sequence of vertices that defines the clipped.
1p Suterland-Hodgeman concept polygon