This document provides information about a textbook on Computer Aided Design and Manufacturing. It begins with the title, authors, subject code and other publishing details. The textbook covers key topics on CAD and CAM across 5 chapters. It provides fundamental concepts and step-by-step explanations of important topics with illustrations and examples. The textbook is intended to help students understand the basic concepts of CAD/CAM. It will be useful for both students and subject teachers.

1. (i)
PUBLICATIONS
TECHNICAL
An Up-Thrust for Knowledge
®
SINCE 1993
SUBJECT CODE : ME8691
Strictly as per Revised Syllabus of
Anna University
Choice Based Credit System (CBCS)
Semester - VI (MECH)
Computer Aided Design
& Manufacturing
A. Jacob Moses
M.E. (Ph.D.)
Assistant Professor,
Department of Mechanical Engineering,
Loyola-ICAM College of Engineering & Technology (LICET),
Chennai
Anup Goel
B.E. Mechanical
Post Graduation in Tool Design with CAD/CAM
Managing Director of AG Engineering Study Centre,
Akurdi, Pune
13 Years Teaching Experience
Renjin J. Bright
M.E. (Ph.D.)
Assistant Professor,
Department of Mechanical Engineering,
National Engineering College, Kovilpatti
Ruchi Agarwal
B.E. (MECH), GATE Qualified

2. (ii)
AU 17
9 [1]
788194382515
ã Copyright with Authors
All publishing rights reserved with . No part of this book
should be reproduced in any form, Electronic, Mechanical, Photocopy or any information storage and
retrieval system without prior permission in writing, from Technical Publications, Pune.
(printed and ebook version) Technical Publications
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Published by :
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Email : sales@technicalpublications.org Website : www.technicalpublications.org
PUBLICATIONS
TECHNICAL
An Up-Thrust for Knowledge
®
SINCE 1993
ISBN 978-81-943825-1-5
Price : 395/-
`
9 7 8 8 1 9 4 3 8 2 5 1 5
Semester - VI (Mechanical Engineering)
Subject Code : ME8691
Computer Aided Design
& Manufacturing
First Edition : January 2020

3. Preface
The importance of Computer Aided Design and Manufacturing is well known in
various engineering fields. Overwhelming response to our books on various subjects
inspired us to write this book. The book is structured to cover the key aspects of the
subject Computer Aided Design and Manufacturing.
The book uses plain, lucid language to explain fundamentals of this subject. The
book provides logical method of explaining various complicated concepts and stepwise
methods to explain the important topics. Each chapter is well supported with necessary
illustrations, practical examples and solved problems. All the chapters in the book are
arranged in a proper sequence that permits each topic to build upon earlier studies. All
care has been taken to make students comfortable in understanding the basic concepts
of the subject.
Representative questions have been added at the end of each Chapter to help the
students in picking important points from that Chapter.
The book not only covers the entire scope of the subject but explains the philosophy
of the subject. This makes the understanding of this subject more clear and makes it
more interesting. The book will be very useful not only to the students but also to the
subject teachers. The students have to omit nothing and possibly have to cover nothing
more.
We wish to express our profound thanks to all those who helped in making this
book a reality. Much needed moral support and encouragement is provided on
numerous occasions by our whole family. We wish to thank the Publisher and the
entire team of Technical Publications who have taken immense pain to get this book
in time with quality printing.
Any suggestion for the improvement of the book will be acknowledged and well
appreciated.
Authors
Anup Goel
A. Jacob Moses
Renjin J. Bright
Ruchi Agarwal
Dedicated to Family, Friends & Dear Students
(iii)

4. Syllabus
Computer Aided Design and Manufacturing
[ME8691]
Unit I Introduction
Product cycle- Design process- sequential and concurrent engineering - Computer aided design
- CAD system architecture- Computer graphics - co-ordinate systems - 2D and 3D
transformations- homogeneous coordinates - Line drawing - Clipping - viewing
transformation-Brief introduction to CAD and CAM - Manufacturing Planning, Manufacturing
control- Introduction to CAD/CAM - CAD/CAM concepts - Types of production - Manufacturing
models and Metrics - Mathematical models of Production Performance. (Chapter - 1)
Unit II Geometric Modeling
Representation of curves - Hermite curve- Bezier curve - B-spline curves-rational
curves-Techniques for surface modeling - surface patch- Coons and bicubic patches - Bezier
and B-spline surfaces. Solid modeling techniques - CSG and B-rep. (Chapter - 2)
Unit III CAD Standards
Standards for computer graphics - Graphical Kernel System (GKS) - standards for exchange
images - Open Graphics Library (OpenGL) - Data exchange standards - IGES, STEP, CALS etc. -
communication standards. (Chapter - 3)
Unit IV Fundamental of CNC and Part Programing
Introduction to NC systems and CNC - Machine axis and Co-ordinate system - CNC machine
tools- Principle of operation CNC- Construction features including structure- Drives and CNC
controllers- 2D and 3D machining on CNC- Introduction of Part Programming, types - Detailed
Manual part programming on Lathe & Milling machines using G codes and M codes - Cutting
Cycles, Loops, Sub program and Macros- Introduction of CAM package. (Chapter - 4)
Unit V Cellular Manufacturing and Flexible
Manufacturing System (FMS)
Group Technology(GT),Part Families - Parts Classification and coding - Simple Problems in
Opitz Part Coding system - Production flow Analysis - Cellular Manufacturing - Composite part
concept - Types of Flexibility - FMS - FMS Components - FMS Application & Benefits - FMS
Planning and Control - Quantitative analysis in FMS. (Chapter - 5)
(iv)

5. Table of Contents
Unit - I
Chapter - 1 Introduction (1 - 1) to (1 - 116)
1.1 Introduction to CAD...........................................................................................1 - 2
1.2 Product Cycle.....................................................................................................1 - 2
1.3 Design Process...................................................................................................1 - 5
1.4 Sequential and Concurrent Engineering............................................................1 - 8
1.5 Computer Aided Design (CAD).........................................................................1 - 12
1.6 CAD System Architecture.................................................................................1 - 15
1.7 Computer Graphics..........................................................................................1 - 16
1.8 Coordinate System ..........................................................................................1 - 18
1.9 2D Transformations.........................................................................................1 - 20
1.9.1 Homogeneous Coordinates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1 - 27
1.9.2 Solved Examples on 2D Transformation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1 - 29
1.10 3D Transformations.......................................................................................1 - 51
1.11 Line Drawing..................................................................................................1 - 52
1.11.1 DDA Algorithm. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1 - 53
1.11.2 Bresenham's Line Drawing Algorithm . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1 - 55
1.11.3 Solved Examples on DDA Algorithm and Bresenham's Algorithm. . . . . . . . . . 1 - 57
1.12 Clipping..........................................................................................................1 - 62
1.13 Viewing Transformation ................................................................................1 - 69
1.14 Brief Introduction to CAD and CAM..............................................................1 - 71
1.14.1 Computer Aided Design (CAD) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1 - 71
1.15 Computer Aided Manufacturing (CAM).........................................................1 - 77
1.15.1 CAD-CAM Integration . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1 - 77
1.15.2 Manufacturing Planning . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1 - 78
(v)

6. 1.15.3 Manufacturing Control . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1 - 79
1.16 Types of Production Systems.........................................................................1 - 80
1.17 Manufacturing Models and Metrics..............................................................1 - 84
1.17.1 Production Performance Measures . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1 - 85
1.17.2 Manufacturing Costs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1 - 96
1.18 Break Even Analysis - A Tool for Manufacturing Control.............................1 - 100
Review Questions ............................................................................................... 1 - 103
Part A : Two Marks Question with Answers .................................................. 1 - 104
Part B : University Questions ........................................................................ 1 - 114
Unit - II
Chapter - 2 Geometric Modeling (2 - 1) to (2 - 48)
2.1 Introduction...................................................................................................... 2 - 3
2.2 Methods of Geometric Modeling ..................................................................... 2 - 3
2.2.1 Wire Frame Modeling. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2 - 4
2.2.2.1 Advantages and Disadvantages of Wire Frame Modeling . . . . . . . . . 2 - 7
2.3 Representation of Curves ................................................................................ 2 - 8
2.4 Parametric and Non-parametric Curves........................................................... 2 - 9
2.5 Order of Continuity......................................................................................... 2 - 11
2.6 Interpolation and Approximation of Curve..................................................... 2 - 12
2.6.1 Difference between Interpolation Curve and Approximation Curve . . . . . . . . 2 - 12
2.7 Hermite Cubic Curve....................................................................................... 2 - 13
2.7.1 Solved Examples on Hermite Cubic Curve . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2 - 15
2.8 Bezier Curve.................................................................................................... 2 - 18
2.8.1 Solved Examples on Bezier Curve . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2 - 20
2.9 B-Spline Curve ................................................................................................ 2 - 22
2.9.1 Difference between Hermite Cubic Spline, Bezier Wave and B-Spline Curve . 2 - 23
2.10 Rational Curve............................................................................................... 2 - 23
2.11 Surface Modeling ......................................................................................... 2 - 24
(vi)

7. 2.11.1 Classification of Surfaces in Geometric Modeling . . . . . . . . . . . . . . . . . . . . . . 2 - 24
2.11.2 Blending Function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2 - 26
2.12 Parametrization of Surface Patch ................................................................. 2 - 27
2.12.1 Bicubic Patches . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2 - 28
2.13 Bezier Surface ............................................................................................... 2 - 28
2.14 B-spline Surface ........................................................................................... 2 - 29
2.15 Boolean Operation........................................................................................ 2 - 30
2.16 Solid Modeling ............................................................................................. 2 - 31
2.17 Constructive Solid Geometry ....................................................................... 2 - 33
2.18 Boundary Representation ............................................................................ 2 - 35
2.18.1 Different between C-rep Modeling and B-rep Modeling . . . . . . . . . . . . . . . . 2 - 35
2.19 Cell Consumption ......................................................................................... 2 - 36
2.20 Spatial Occupancy Enumeration .................................................................. 2 - 36
2.21 Sweep Representation.................................................................................. 2 - 37
Part A : Two Marks Questions with Answers .................................................. 2 - 38
Part B : University Questions with Answers .................................................... 2 - 43
Unit - III
Chapter - 3 CAD Standards (3 - 1) to (3 - 40)
3.1 Introduction...................................................................................................... 3 - 2
3.2 Standards for Computer Graphics .................................................................... 3 - 4
3.2.1 Graphics Kernel System (GKS) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3 - 5
3.3 Standards for Exchange of Images.................................................................. 3 - 10
3.3.1 Open Graphics Library (OpenGL). . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3 - 10
3.4 Data Exchange Standards ............................................................................... 3 - 13
3.4.1 IGES - Initial Graphics Exchange Specification . . . . . . . . . . . . . . . . . . . . . . . . . . 3 - 13
3.4.2 STEP - Standard for the Exchange of Product Data . . . . . . . . . . . . . . . . . . . . . . 3 - 18
3.4.3 CALS - Continuous Acquisition and Life-Cycle Support . . . . . . . . . . . . . . . . . . . 3 - 20
(vii)

8. 3.4.4 PDES - Product Data Exchange Specification . . . . . . . . . . . . . . . . . . . . . . . . . . . 3 - 23
3.4.5 DXF (Data Exchange Format) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3 - 24
3.5 Communication Standards ............................................................................. 3 - 26
3.5.1 Local Area Networks. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3 - 27
3.5.2 Wide Area Networks . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3 - 28
3.5.3 Levels of Communication Standards. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3 - 30
Review Questions ................................................................................................ 3 - 31
Part A : Two Marks Questions with Answers................................................... 3 - 32
Part B : University Questions with Answers .................................................... 3 - 38
Unit - IV
Chapter - 4 Fundamental of CNC and Part Programming
(4 - 1) to (4 - 118)
4.1 Introduction...................................................................................................... 4 - 3
4.2 Numerical Control............................................................................................. 4 - 3
4.2.1 Basic Elements of NC System. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4 - 4
4.3 Classification of NC System .............................................................................. 4 - 7
4.3.1 According to Tool Positioning . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4 - 7
4.3.1.1 Comparison of Absolute and Incremental System. . . . . . . . . . . . . 4 - 8
4.3.2 According to Motion Control System . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4 - 8
4.3.3 According to Servo Control System. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4 - 10
4.3.3.1 Comparison of Open Loop and Closed Loop System . . . . . . . . . . . 4 - 11
4.3.4 According to Feedback Devices . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4 - 11
4.4 Advantages of NC System............................................................................... 4 - 12
4.5 Disadvantages of NC System .......................................................................... 4 - 12
4.6 Applications of NC System.............................................................................. 4 - 12
4.7 Types of Numerical Control System................................................................ 4 - 13
4.8 Conventional Numerical Control (NC) ............................................................ 4 - 13
4.9 Direct Numerical Control (DNC)...................................................................... 4 - 13
(viii)

9. 4.10 Computerized Numerical Control (CNC)....................................................... 4 - 14
4.11 Constructional Features of CNC Machines ................................................... 4 - 15
4.11.1 Machine Structure . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4 - 16
4.11.2 Drives. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4 - 17
4.11.3 Actuation System . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4 - 18
4.11.4 Slideways for Machines . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4 - 21
4.11.5 Automatic Tool Changer (ATC) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4 - 24
4.11.6 Automatic Pallet Changer (APC) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4 - 25
4.11.7 Transducers/Control Elements . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4 - 27
4.11.8 Feedback Devices . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4 - 27
4.12 Advantages and Disadvantages of CNC Machines ....................................... 4 - 28
4.13 Comparison between NC, CNC and DNC System ......................................... 4 - 29
4.14 Adaptive Control System (ACS) .................................................................... 4 - 30
4.15 Machining Centre ......................................................................................... 4 - 31
4.15.1 Horizontal Machining Centre (HMC) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4 - 32
4.15.2 Vertical Machining Centre (VMC) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4 - 32
4.16 Program Reader ........................................................................................... 4 - 34
4.17 New Trends in Tool Materials ..................................................................... 4 - 34
4.18 Tool Inserts .................................................................................................. 4 - 35
4.19 Work Holding in CNC Machines .................................................................. 4 - 36
4.20 Axis Nomenclature for CNC Machines ......................................................... 4 - 36
4.21 Part Programming ........................................................................................ 4 - 39
4.21.1 Manual Part Programming. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4 - 39
4.21.2 Preparatory Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4 - 43
4.22 Procedure to Write a Part Program ............................................................. 4 - 49
4.23 Part Programming for Lathe ........................................................................ 4 - 50
4.24 Part Programming for Milling and Drilling ................................................... 4 - 67
4.25 Subroutine ................................................................................................... 4 - 90
(ix)

10. 4.26 Canned Cycle ................................................................................................ 4 - 93
4.26.1 Comparison between Subroutine and Canned Cycle . . . . . . . . . . . . . . . . . . . 4 - 93
4.26.2 Slot Milling (G74) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4 - 93
4.26.3 Rectangular Pocket Milling (G75) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4 - 95
4.27 Automatically Programmed Tools (APT) ...................................................... 4 - 96
4.27.1 Structure of APT . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4 - 96
4.28 Micromachining ........................................................................................... 4 - 99
4.28.1 Wafer Machining . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4 - 99
4.29 Part Programming using APT ..................................................................... 4 - 100
4.30 Introduction of CAM Package .................................................................... 4 - 107
Part A : Two Marks Questions with Answers ................................................ 4 - 111
Unit - V
Chapter - 5 Cellular Manufacturing and Flexible
Manufacturing System (FMS) (5 - 1) to (5 - 54)
5.1 Group Technology ............................................................................................ 5 - 3
5.1.1 Benefits of Group Technology . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5 - 3
5.2 Part Families .................................................................................................... 5 - 3
5.2.1 Identification of Part Families . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5 - 4
5.2.1.1 Visual Inspection Method . . . . . . . . . . . . . . . . . . . . . . . 5 - 4
5.2.1.2 Parts Classification and Coding . . . . . . . . . . . . . . . . . . . . 5 - 5
5.2.1.3 Production Flow Analysis . . . . . . . . . . . . . . . . . . . . . . . 5 - 5
5.3 Parts Classification and Coding......................................................................... 5 - 5
5.3.1 Part Design Attributes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5 - 5
5.3.2 Part Manufacturing Attributes. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5 - 5
5.4 Structures used for Classifying and Coding the Parts ..................................... 5 - 5
5.4.1 Hierarchical Structure . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5 - 5
5.4.2 Chain-type Structure . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5 - 6
5.4.3 Mixed Mode Symbol . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5 - 6
(x)

11. 5.5 OPTIZ Coding System ....................................................................................... 5 - 6
5.5.1 Solved Examples of Optiz Part Coding System . . . . . . . . . . . . . . . . . . . . . . . . . . 5 - 8
5.6 Production Flow Analysis ............................................................................... 5 - 10
5.6.1 Production Flow Analysis (PFA) Procedure . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5 - 10
5.6.1.1 Data Collection . . . . . . . . . . . . . . . . . . . . . . . . . . . 5 - 11
5.6.1.2 Sortation of Process Routings . . . . . . . . . . . . . . . . . . . . 5 - 12
5.6.1.3 PFA Chart . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5 - 12
5.6.1.4 Cluster Analysis . . . . . . . . . . . . . . . . . . . . . . . . . . 5 - 13
5.7 Rank Order Clustering..................................................................................... 5 - 13
5.7.1 Advantage of PFA . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5 - 19
5.7.2 Disadvantage of PFA. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5 - 19
5.8 Cellular Manufacturing .................................................................................. 5 - 19
5.8.1 Objective of Cellular Manufacturing . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5 - 19
5.9 Machine Cell Design ....................................................................................... 5 - 20
5.9.1 Types of Machine Cell. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5 - 20
5.9.1.1 Single Machines . . . . . . . . . . . . . . . . . . . . . . . . . . 5 - 20
5.9.1.2 Group Machine Cell with Manual Handling . . . . . . . . . . . . . . . 5 - 20
5.9.1.3 Group Machine Cell with Semi-integrated Handling . . . . . . . . . . . 5 - 20
5.9.1.4 Flexible Manufacturing Cell or Flexible Manufacturing System (FMS) . . . . 5 - 20
5.9.2 Types of Layout. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5 - 21
5.9.2.1 Inline Layout . . . . . . . . . . . . . . . . . . . . . . . . . . . 5 - 21
5.9.2.2 Loop Layout . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5 - 22
5.9.2.3 Rectangular Layout . . . . . . . . . . . . . . . . . . . . . . . . . 5 - 22
5.10 Part Movement between the Cell ................................................................ 5 - 23
5.11 Key Machine ................................................................................................ 5 - 24
5.12 Composite Part Concept............................................................................... 5 - 24
5.13 Flexible Manufacturing System (FMS).......................................................... 5 - 25
5.13.1 Flexibility Tests . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5 - 25
(xi)

12. 5.14 Types of Flexibility ....................................................................................... 5 - 26
5.14.1 Single Machine Cell. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5 - 27
5.14.2 Flexible Manufacturing Cell (FMC) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5 - 28
5.14.3 Flexible Manufacturing System (FMS) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5 - 28
5.14.4 Difference between FMC and FMS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5 - 29
5.15 Level of Flexibility ........................................................................................ 5 - 30
5.15.1 Dedicated FMS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5 - 30
5.15.2 Random Order FMS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5 - 30
5.16 Components of FMS ..................................................................................... 5 - 31
5.16.1 Work Stations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5 - 31
5.16.1.1 Load / Unload Work Stations . . . . . . . . . . . . . . . . . . . . 5 - 31
5.16.1.2 Machining Work Stations . . . . . . . . . . . . . . . . . . . . . . 5 - 32
5.16.1.3 Assembly Work Stations . . . . . . . . . . . . . . . . . . . . . . 5 - 32
5.16.1.4 Supporting Work Stations. . . . . . . . . . . . . . . . . . . . . . 5 - 32
5.16.1.5 Other work stations . . . . . . . . . . . . . . . . . . . . . . . . 5 - 32
5.16.2 Material Handling and Storage System. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5 - 33
5.16.2.1 FMS Layout Configurations . . . . . . . . . . . . . . . . . . . . . 5 - 33
5.16.3 Computer Control System . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5 - 38
5.16.4 Human Resources. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5 - 39
5.17 Applications of FMS ...................................................................................... 5 - 39
5.18 Advanages of FMS ........................................................................................ 5 - 40
5.19 FMS Planning and Control ........................................................................... 5 - 41
5.19.1 FMS Planning Issues . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5 - 41
5.19.2 FMS Design Issues . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5 - 41
5.19.3 FMS Operational Issues . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5 - 42
5.20 Quantitative Analysis in FMS ........................................................................ 5 - 43
5.20.1 Bottle Neck Model . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5 - 43
Part A : Two Marks Questions with Answers................................................... 5 - 49
Part B : University Questions with Answers .................................................... 5 - 52
Solved Model Question Paper ...................................................... (M - 1) to (M - 4)
(xii)

13. Syllabus : Product cycle- Design process- sequential and concurrent engineering-
Computer aided design – CAD system architecture- Computer graphics –
co-ordinate systems- 2D and 3D transformations- homogeneous coordinates -
Line drawing -Clipping- viewing transformation-Brief introduction to CAD
and CAM – Manufacturing Planning, Manufacturing control- Introduction to
CAD/CAM –CAD/CAM concepts ––Types of production - Manufacturing
models and Metrics – Mathematical models of Production Performance
Section No. Topic Name Page No.
1.1 Introduction to CAD 1 - 2
1.2 Product Cycle 1 - 2
1.3 Design Process 1 - 5
1.4 Sequential and Concurrent Engineering 1 - 8
1.5 Computer Aided Design (CAD) 1 - 12
1.6 CAD System Architecture 1 - 15
1.7 Computer Graphics 1 - 16
1.8 Coordinate System 1 - 18
1.9 2D Transformations 1 - 20
1.10 3D Transformations 1 - 51
1.11 Line Drawing 1 - 52
1.12 Clipping 1 - 62
1.13 Viewing Transformation 1 - 69
1.14 Brief Introduction to CAD and CAM 1 - 71
1.15 Computer Aided Manufacturing (CAM) 1 - 77
1.16 Types of Production Systems 1 - 80
1.17 Manufacturing Models and Metrics 1 - 84
1.18 Break Even Analysis - A Tool for Manufacturing Control 1 - 100
Part A : Two Marks Question with Answers 1 - 104
Part B : University Questions with Answers 1 - 114
1 - 1 Computer Aided Design and Manufacturing
Chapter - 1
Introduction
Unit - I

14. 1.1 Introduction to CAD
· CAD (Computer Aided Design) is the use of computer software to design and
document a product's design process.
· Engineering drawing entails the use of graphical symbols such as points, lines,
curves, planes and shapes.
· Essentially, it gives detailed description about any component in a graphical form.
· The use of orthographic projections was formally introduced by the French
mathematician Gaspard Monge in the eighteenth century.
· Since visual objects transcend languages, engineering drawings have evolved and
become popular over the years.
· While earlier engineering drawings were handmade, studies have shown that
engineering designs are quite complicated.
· A solution to many engineering problems requires a combination of organization,
analysis, problem solving principles and a graphical representation of the
problem.
· Objects in engineering are represented by a technical drawing (also called as
drafting) that represents designs and specifications of the physical object and data
relationships.
· CAD is used to design, develop and optimize products.
· While it is very versatile, CAD is extensively used in the design of tools and
equipment required in the manufacturing process as well as in the construction
domain.
· CAD enables design engineers to layout and to develop their work on a computer
screen, print and save it for future editing.
· When it was introduced first, CAD was not exactly an economic proposition
because the machines at those times were very costly.
· The increasing computer power in the later part of the twentieth century, with the
arrival of minicomputer and subsequently the microprocessor, has allowed
engineers to use CAD files that are an accurate representation of the
dimensions / properties of the object.
1.2 Product Cycle + [AU : Dec.-18]
· In the design and manufacture of a product various activities and functions must
be accomplished. These activities and functions are referred to as the
"Product Cycle".
· The product cycle includes all the activities starting from identification for
product to deliver the finished product to the customer. Fig. 1.2.1 explains
various stages in product life cycle. Fig. 1.2.2 depicts various steps in the product
cycle and Fig. 1.2.3 explains product cycle in in a detailed manner.
1 - 2 Computer Aided Design and Manufacturing
Introduction

15. · Two main processes in the product cycle are :
i) Design process
ii) Manufacturing process.
i) Design process
· The activities involved in the design process can be classified into :
· Synthesis
· Analysis.
1 - 3 Computer Aided Design and Manufacturing
Introduction
Sales
Introduction Growth
The 4 life cycle stages and their marketing implications
Maturity Decline
· Low sales · Increasing sales · Peak sales · Falling sales
· High cost per
customer
· Cost per customer falls · Cost per customer
lowest
· Cost per customer low
· Financial losses
· Profits rise
· Profits high
· Profits fall
· Innovative customers
· Increasing No.
of customers · Mass market
· Customer base contracts
· Few (if any) competitors · More competitors · Stable number
of competitors
· Number of competitors fall
Time
Take-off
Shake-out
Saturation
Fig. 1.2.1 Various stages in product life cycle
Design
E
n
d
o
f
L
i
f
e
D
i
s
t
r
i
b
u
t
i
o
n
C
u
s
to
m
e
r
Manufactu
r
i
n
g
Product
Lifecycle
Fig. 1.2.2 Various steps in product life cycle

16. Synthesis of design :
· The philosophy, functionality, and uniqueness of the product are all determined
during synthesis.
· During synthesis, a design takes the form of sketches and layout drawings that
show the relationship among the various product parts.
· Most of the information generated and handled in the synthesis sub process is
qualitative and consequently it is hard to capture in a computer system.
Analysis of design :
· The analysis begins with an attempt to put the conceptual design into the context
of engineering sciences to evaluate the performance of the expected product.
· This requires design modeling and simulation. An important aspect of analysis is
the questions that helps to eliminate multiple design choices and find the best
solution to each design problem.
· Bodies with symmetries in their geometry and loading are usually analyzed by
considering a portion of the model. Example : Stress analysis pressure vessels,
couplings etc.
· The quality of the results obtained from these activities is directly related to and
limited by the quality of the analysis model chosen.
· Prototypes may be built for the design evaluation. Prototypes can be constructed
for the given design by using software packages (CAM).
· The outcome of analysis is the design documentation in the form of engineering
drawings.
ii) Manufacturing process
· Manufacturing process begins with process planning, using the drawings from the
design process, and it ends with the actual products.
· Process planning is a function that establishes which processes and the proper
parameters for the processes are to be used.
· It also selects the most efficient sequence for the production of the product.
· The outcome of the process planning is a production plan, tools procurement,
materials order, and machine programming.
· Other special requirements, such as design of jigs and fixtures, are also planned.
The relationship of process planning to the manufacturing process is analogous to
that of synthesis to the design process. It involves considerable human experience
and qualitative decisions.
· This description implies that it would be difficult to computerize process
planning.
· Once process planning has been completed, the actual product is produced and
inspected against quality requirements.
1 - 4 Computer Aided Design and Manufacturing
Introduction

17. · Parts that pass the quality control inspection are assembled, functionally tested,
packaged, labeled, and shipped to customers.
· Market feedback is usually incorporated into the design process.
· This feedback give birth to a closed-loop product cycle.
1.3 Design Process + [AU : Dec.-16]
Engineering design process :
· The engineering design process is the formulation of a plan to help an engineer
build a product with a specified performance goal. It is a decision making process
in which the basic sciences, mathematics, and engineering sciences are applied to
convert resources optimally to meet a stated objective. Fig. 1.3.1 explains
engineering design process in a detailed manner.
· The fundamental elements of the design process are the establishment of
objectives and criteria, synthesis, analysis, construction, testing and evaluation.
· The engineering design process is a multi-step process including the research,
conceptualization, feasibility assessment, establishing design requirements,
preliminary design, detailed design, production planning and tool design and
finally production.
1 - 5 Computer Aided Design and Manufacturing
Introduction
Design
need
Design definitions,
specifications and
requirements
Collecting relevant
design information
and feasibility study
Analysis
Design
communication
and documentation
Design
evaluation
Design
optimization
Design
analysis
Design
modeling and
simulation
Design
conceptualization
The CAD process
Synthesis
The design process
The manufacturing process
The CAM process
Process
planning
Production
planning
Design and
procurement
of new tools
Order
material
NC, CNC,
DNC
programming
Production
Quality
control
Packaging Shipping
Marketing
Fig. 1.2.3 Product cycle

18. Conceptual Design
It is a process in which we initiate the design and come up with a number of design
concepts and then narrow down to the single best concept. This involved the following
steps.
· Identification of customer needs : To identify the customers' needs and to
communicate them to the design team.
· Problem definition : The main goal of this activity is to create a statement that
describes what are the needs to be accomplished to meet the needs of the
customers' requirements.
· Gathering information : In this step, all the information that can be helpful for
developing and translating the customers' needs into engineering design are
collected.
· Conceptualization : In this step, broad sets of concepts are generated that can
potentially satisfy the problem statement.
· Concept selection : The main objective of this step is to evaluate the various
design concepts, modifying and evolving into a single preferred concept.
Embodiment Design
· It is a process where the structured development of the design concepts takes
place.
· It is in this phase that decisions are made on strength, material selection, size
shape and spatial compatibility.
1 - 6 Computer Aided Design and Manufacturing
Introduction
Define problem
Problem statement
Benchmarking
QFD
PDS
Project planning
Gather information
Internet
Patents
Trade
Literature
Concept generation
Brainstorming
Functional decomposition
Morphological chart
Evaluation of
concepts
Pugh concept selection
Decision matrix
Conceptual
design
Production architecture
Configuration design
Parametric design
Detail design
Arrangement of physical
Preliminary selection of
Robust design
Detail drawing and
specifications
elements to carry out
material and
Tolerances
functions
manfacturing
Final dimensions
modeling/sizing of parts
DFM
Embodiment
design
Fig. 1.3.1 Engineering design process

19. · Embodiment design is concerned with three major tasks - Product architecture,
configuration design, and parametric design.
· Product architecture : It is concerned with dividing the overall design system
into small subsystems and modules. It is in this step we decide how the physical
components of the design is to be arranged in order to combine them to carry out
the functional duties of the design.
· Configuration design: In this process we determine what all features are required
in the various parts / components and how these features are to be arranged in
space relative to each other.
· Parametric design : It starts with information from the configuration design
process and aims to establish the exact dimensions and tolerances of the product.
Also, final decisions on the material and manufacturing processes are done if it
has not been fixed in the previous process. One of the important aspects of
parametric designs is to examine if the design is robust or not.
Detail Design
· It is in this phase the design is brought to a state where it has the complete
engineering description of a tested and a producible product. Any missing
information about the arrangement, form, material, manufacturing process,
dimensions, tolerances etc. of each part is added and detailed engineering drawing
suitable for manufacturing are prepared.
Shigley's Design Process
· Fig. 1.3.2 explain the step by step procedure of Shigley's design process model.
(See Fig. 1.3.2 on next page)
· Recognition of need : The problems in the existing products (or) potential for
new products in market has to be identified.
· Definition of problem : The problem in the existing product or specification of
the new product is specified as design brief to the designers. It includes the
specification of physical and functional characteristics, cost, quality, performance
requirements etc. and requirement of design brief.
· Analysis and optimization : Each design from the synthesis stages are analysed
and optimum one is selected. It should be noted that synthesis and analysis are
highly iterative. A certain component or subsystem of the overall system
conceived by the designer in the synthesis stage is subjected to analysis. Based
on the analysis, improvements are made and redesigned. The process is repeated
until the design optimized within all the constraints imposed by designer.
· Evaluation : In this stage optimized design from the previous stage is checked
for all the specification mentioned in the design brief. A prototype of the product
is developed and experimentally checked for its performance, quality, reliability
and other aspects of product. If any discrepancies/problems are faced, it should be
fed back to the designer in the synthesis stage.
1 - 7 Computer Aided Design and Manufacturing
Introduction

20. · Presentation : After the product design passing through the evaluation stage,
drawings, diagrams, material specification, assembly lists, bill of materials etc.
which are required for product manufacturing are prepared and given to process
planning department and production department.
1.4 Sequential and Concurrent Engineering + [AU : May-17, Dec.-18]
Sequential Engineering (Over The Wall Engineering)
· In sequential engineering design has been carried out as a sequential set of
activities with distinct non-overlapping phases as shown in Fig. 1.4.1.
· Sequential engineering is the term used to describe the method of production in a
linear format. The different steps are done one after another, with all attention
and resources focused on that one task. After it is completed it is left alone and
everything is concentrated on the next task.
· In such an approach, the life-cycle of a product starts with the identification of
the need for that product. These needs are converted into product requirements
which are passed on to the design department.
1 - 8 Computer Aided Design and Manufacturing
Introduction
Recognition of need
Definition of problem
Synthesis
Analysis and optimization
Evaluation
Presentation
Success
Change the design
Can the design be improved
Design impossible for the
given specification
Fails No
Yes
Fig. 1.3.2 Shigley's design process

21. · The designers design the product's form, fit, and function to meet all the
requirements, and pass on the design to the manufacturing department.
· After the product is manufactured it goes through the phases of assembly, testing
and installation. This type of approach to life-cycle development is also known as
`over the wall' approach, because the different life-cycle phases are hidden or
isolated from each other.
· Each phase receives the output of the preceding phase as if the output had been
thrown over the wall. In such an approach, the manufacturing department, for
example, does not know what it will actually be manufacturing until the detailed
design of the product is over.
· There are a lot of disadvantages of the sequential engineering process. The
designers are responsible for creating a design that meets all the specified
requirements. They are usually not concerned with how the product will be
manufactured or assembled.
· Problems and inconsistencies in the designs are therefore, detected when the
product reaches into the later phases of its life-cycle.
· At this stage, the only possible option is to send the product back for a re-design.
The whole process becomes iterative and it not until after a lot of re-designs has
taken place that the product is finally manufactured.
1 - 9 Computer Aided Design and Manufacturing
Introduction
Design Manufacturing
· · · · · · · ·
Assembly
Fig. 1.4.1 (a) Over the wall engineering
Requirements definition Product definition Process definition Delivery and support
Errors changes and corrections
Information flow
Fig. 1.4.1 (b) Sequential engineering

22. Concurrent Engineering
· Due to the large number of changes, and hence iterations, the product's
introduction to market gets delayed. In addition, each re-design, re-work,
re-assembly etc. incurs cost, and therefore the resulting product is costlier than
what it was originally thought to be. The market share is lost because of the
delay in product's introduction to market, and customer faith is lost.
· Concurrent engineering is a dramatically different approach to product
development in which various life-cycle aspects are considered simultaneously
right from the early stages of design as shown in Fig. 1.4.2.
· These life-cycle aspects include product's functionality, manufacturability,
testability, assimilability, maintainability, and everything else that could be
affected by the design. In addition, various life-cycle phases overlap each other,
and there in no "wall" between these phases.
· The completion of a previous life-cycle phase is not a pre-requisite for the start
of the next life-cycle phase. In addition, there is a continuous feedback between
these life-cycle phases so that the conflicts are detected as soon as possible.
· The concurrent approach results in less number of changes during the later phases
of product life-cycle, because of the fact that the life-cycle aspects are being
considered all through the design.
· The benefits achieved are reduced lead times to market, reduced cost, higher
quality, greater customer satisfaction, increased market share etc.
· In concurrent engineering, different tasks are tackled at the same time, and not
necessarily in the usual order. This means that info found out later in the process
can be added to earlier parts, improving them, and also saving a lot of time.
1 - 10 Computer Aided Design and Manufacturing
Introduction
Life - cycle phases
Requirements Analysis
Detailed Design
Preliminary Design
Manufacturing
Assembly
Testing
Installation
Life Cycle Aspects (Electrical, Mechanical
Survicing, Assembiability, Recyclability, etc.)
Time
Feedback loops between
different life - cycle phases
Fig. 1.4.2 Concurrent engineering

23. · Concurrent engineering is a method by which several teams within an
organization work simultaneously to develop new products and services and
allows a more streamlined approach.
· The concurrent engineering is a non-linear product or project design approach
during which all phases of manufacturing operate at the same time
-simultaneously.
· Both product and process design run in parallel and occur in the same time
frame.
· Product and process are closely coordinated to achieve optimal matching of
requirements for effective cost, quality, and delivery. Decision making involves
full team participation and involvement.
· The team often consists of product design engineers, manufacturing engineers,
marketing personnel, purchasing, finance, and suppliers.
Comparison between Concurrent and Sequential Engineering
· Fig. 1.4.3 depicts the schematic representation of the comparison between
sequential and concurrent engineering.
1 - 11 Computer Aided Design and Manufacturing
Introduction
Requirements definition Product definition Process definition Delivery and support
Errors, changes and corrections
Information flow
Requirements definition
Product definition
Process definition
Delivery and support
CE life - cycle time Time saved
Requirements
definition
Product
definition
Process
definition
Delivery and
support
CE life cycle time
(a) Sequential engineering
(b) Concurrent engineering
Fig. 1.4.3 Comparison between sequential and concurrent engineering

24. Sr. No. Sequential engineering Concurrent engineering
1. Sequential engineering is the term
used to explain the method of
production in a linear system. The
various steps are done one after
another, with all attention and
resources focused on that single
task.
In concurrent engineering, various tasks
are handled at the same time, and not
essentially in the standard order. This
means that info found out later in the
course can be added to earlier parts,
improving them, and also saving time.
2. Sequential engineering is a system
by which a group within an
organization works sequentially to
create new products and services.
Concurrent engineering is a method by
which several groups within an
organization work simultaneously to create
new products and services.
3. The sequential engineering is a
linear product design process during
which all stages of manufacturing
operate in serial.
The concurrent engineering is a non-linear
product design process during which all
stages of manufacturing operate at the
same time.
4. Both process and product design
run in serial and take place in the
different time.
Both product and process design run in
parallel and take place in the same time.
5. Process and product are not
matched to attain optimal matching.
Process and product are coordinated to
attain optimal matching of requirements for
effective quality and delivery.
6. Decision making done by only
group of experts.
Decision making involves full team
involvement.
1.5 Computer Aided Design (CAD) + [AU : May-17]
The conventional design process has been accomplished on drawing boards with
design being documented in the form of a detailed engineering drawing. This process is
iterative in nature and is time consuming. The computer can be beneficially used in the
design process. The various tasks performed by a modern computer aided design system
can be grouped into four functional areas.
i) Geometric modeling
ii) Engineering analysis
iii) Design review and evaluation
iv) Automated drafting.
i) Geometric Modeling
· The geometric modeling is concerned with computer compatible mathematical
description of geometry of an object.
· The mathematical description should be such that the image of the object can be
displayed and manipulated in the computer terminal, modification on the
1 - 12 Computer Aided Design and Manufacturing
Introduction

25. geometry of the object can be done easily, it can be stored in the computer
memory, and can be retrieve back on the computer screen for review analysis or
alteration.
· Geometric modeling is classified into
a) Wireframe modeling
b) Solid modeling
c) Surface modeling
ii) Engineering Analysis
· The computer can be used to aid the analysis work such as stress-strain analysis,
heat transfer analysis, etc. The analysis can be done by using specific program
generated for it or by using general purpose software commercially available in
the market.
· The geometric models generated can be used for the analysis by properly
interfacing the modeling software with the analysis software.
· Two types of engineering analysis are
a) Analysis for mass properties
b) Finite Element Analysis (FEA)
1 - 13 Computer Aided Design and Manufacturing
Introduction
Recognition of need
Conventional Design Process Computer - aided design
Problem definition
Synthesis
Evaluation
Presentation
Analysis and
optimization
Geometric modeling
Engineering analysis
Design review and evaluation
Automated drafting
Fig. 1.5.1 Computer aided design process

26. iii) Design Review and Evaluation
· The accuracy of the design can be checked and rectified if required in the screen
itself.
· Layering feature available in software are very useful for design review purpose.
· Similarly, using the layer procedure, every stage of production can be checked.
· Suppose a new mechanism is to be designed, the same mechanism can be
simulated in the computer.
· By animation, the working of the mechanism can be checked.
· These will relieve the designer from tedious conventional method of mechanism
checking.
· Another advantage of animating the complete assembly of product is that whether
any component fouls the other components in its working.
iv) Automated Drafting
· Automated drafting is the process of creating hard copies of design drawing.
· The important features of drafting software's are automated dimensioning, scaling
of the drawing and capable of generating sectional views.
· The enlargement of minute part details and ability to generate different views of
the object like orthographic, oblique, isometric and perspective views are possible.
· Thus, CAD systems can increase productivity on drafting.
Advantages of CAD
· Efficiency, effectiveness and creativity of the designer are drastically improved.
· Faster, consistent and more accurate.
· Easy modification (copy) and improvement (edit).
· Repeating the design drawing is not needed when modifying.
· Manipulation of various dimensions, attributes is easy.
· Parametric and possess parent-child relationship.
· Inspecting tolerance and interface is easy.
· Use of standard components from part library makes fast modeling.
· Excellent graphical representation.
· Co-ordination among the groups and sharing the design data is possible.
· Exchange of e-drawing and storage of several data are easily possible.
· Graphical Simulation and animation studies the real-time behavior.
· 3D visualization of model in several orientations eliminates prototype.
· Documentation at various design phases is efficient, easier, flexible and economical.
1 - 14 Computer Aided Design and Manufacturing
Introduction

27. · Linkage to Manufacturing to carry out the production (NC/CNC programming).
· Engineering applications of CAD.
Applications of CAD :
· Structural design of Aircraft
· Aircraft simulation
· Real time simulation
· Automobile industries
· Architectural design
· Pipe routing and plan layout design
· Electronic industries
· Dynamic analysis of mechanical systems
· Kinematic analysis
· Mesh data preparation for finite element analysis.
1.6 CAD System Architecture
· In CAD, computer architecture is a set of disciplines that explains the
functionality, the organization and the introduction of computer systems; that is, it
describes the capabilities of a computer and its programming method in a
summary way, and how the internal organization of the system is designed and
executed to meet the specified facilities.
· Computer architecture engages different aspects, including instruction set
architecture design, logic design, and implementation.
· The implementation includes integrated circuit design, power, and cooling.
Optimization of the design needs expertise with compilers, operating systems and
packaging.
· Its use in designing electronic systems is known as Electronic Design
Automation, or EDA. In mechanical design it is known as Mechanical Design
Automation (MDA) or Computer-Aided Drafting (CAD), which includes the
process of creating a technical drawing with the use of computer software.
· CAD software for mechanical design uses either vector-based graphics to depict
the objects of traditional drafting, or may also produce raster graphics showing
the overall appearance of designed objects.
· However, it involves more than just shapes. As in the manual drafting of
technical and engineering drawings, the output of CAD must convey information,
such as materials, processes, dimensions, and tolerances, according to
application-specific conventions.
1 - 15 Computer Aided Design and Manufacturing
Introduction

28. · CAD may be used to design curves and figures in two-dimensional (2D) space;
or curves, surfaces, and solids in three-dimensional (3D) space.
· CAD is an important industrial art extensively used in many applications,
including automotive, shipbuilding, and aerospace industries, industrial and
architectural design, prosthetics, and many more. CAD is also widely used to
produce computer animation for special effects in movies, advertising and
technical manuals, often called DCC (Digital Content Creation).
· Fig. 1.6.1 explains CAD system architecture.
1.7 Computer Graphics
· Computer graphics involves creation, display, manipulation and storage of
pictures and experimental data for proper visualization using a computer.
· Typically, a graphics system comprises of a host computer which must have a
support of a fast processor, a large memory and frame buffer along with a few
other crucial components.
· The first of them is the display devices. Colour monitors are one example of such
display device.
· There are other examples of output devices like LCD panels, laser printers, colour
printers, plotters etc.
· Set of input devices are also needed. Typical examples are the mouse, keyboard,
joystick, touch screen, trackball etc.
1 - 16 Computer Aided Design and Manufacturing
Introduction
Database
(CAD model)
Application
software
Graphics
utility
Device
drivers
Input - output
devices
User
interface
System
Major classes :
Main frame
Mini computer
Workstation
Microcomputer
Based
Application areas :
Mechanical
Architectural
Construction
Circuit design
Chip design
Cost :
High end
Low end
Fig. 1.6.1 CAD system architecture

29. · Through these input devices it is possible to provide input to the computer and
display device is an output device which shows the image.
· The first and most important of them is the GUI as it is called. It has various
components.
· A graphical interface is basically a piece of interface or a program which exists
between the user and the graphics application program.
· It helps the graphics system to interact with the user both in terms of input and
output.
· Typical components which are used in a graphical user interface are menus,
icons, cursors, dialog boxes and scrollbars.
· Grids are used in two dimensional graphics packages to align the objects along a
set of specific coordinates or positions. It can be switched on and off and
displayed on the screen.
· Sketching is an example which is used to draw lines, arcs, poly lines and various
other objects.
· The most difficult part of the GUI is three dimensional interfaces which is
normally available at the bottom of screen.
· It is easy to interact and handle with two dimensional objects but for interacting
with the three dimensional objects three dimensional interface is needed to pick
up one of the 3D objects from a two dimensional screen.
· Essentially the computer monitor is just a two dimensional ray of pixels where
the entire picture is projected and the picture could represent a three dimensional
scene. Special facilities for 3D interface to handle or manipulate three
dimensional objects are needed.
Classification of computer graphics
· Based on the control the user has over the image
a) Passive computer graphics - The user has no control
b) Interactive graphics - The user may interact with the graphics
1 - 17 Computer Aided Design and Manufacturing
Introduction
Control
processor
Input
devices
Display file
Display
processor
unit
Display
screen
Link to
host computer
Control signals
Fig. 1.7.1 A basic computer graphics layout

30. · Based on the way the image is generated
a) Vector graphics - The image comprises of number of lines.
b) Raster graphics - Manipulation of the colour and intensity of points, pixels.
· Based on the space
a) Image-space graphics - Image itself is directly manipulated to create a picture.
b) Object-space graphics - Separate model is manipulated.
1.8 Coordinate System
Three types of coordinate systems are generally used in CAD/CAM operations as
shown in Fig. 1.8.1.
a) Model Coordinate System (MCS) or Database CS/ World CS / Master CS
b) Working Coordinate System (WCS) or User Coordinate System (UCS)
c) Screen Coordinate System (SCS) or Device CS
a) Model Coordinate System (MCS)
· It is the reference space of the model with respect to which all the model
geometrical data is stored.
· It is a Cartesian system which forms the default coordinate system used by a
particular software program.
· The X, Y, and Z axes of the MCS can be displayed on the computer screen.
· The choice of origin is arbitrary.
· The three default sketch planes of a CAD/CAM system define the three planes of
MCS, and their intersection point is the MCS origin.
1 - 18 Computer Aided Design and Manufacturing
Introduction
Z Z
Y Y
X X
Z'
Y'
X'
(0,0) X1
Y1
(X Y )
max' max
(a) MCS (b) WCS (c) SCS
Fig. 1.8.1

31. · When a CAD designer begins sketching, the origin becomes a corner point of the
profile being sketched. The sketch plane defines the orientation of the profile in
the model 3D space.
· Existing CAD/CAM software uses the MCS as the default WCS.
· The MCS is the only coordinate system that the software recognizes when storing
or retrieving graphical information from a model database. Many existing
software package allow the user to input Cartesian and cylindrical coordinates.
This input information is transformed to (x, y, z) coordinates relative to the MCS
before being stored in the database.
b) Working Coordinate System (WCS) or User Coordinate System (UCS)
· This is basically an auxiliary coordinate system used in place of MCS. For
convenience while we develop the geometry by data input this kind of coordinate
system is useful.
· It is very useful when a plane (face) in MCS is not aligned (easily defined) along
any orthogonal planes.
· It can be established at any position and orientation in space that the user desires.
· The user can define a Cartesian coordinate system whose XY plane is coincident
with the desired plane of construction. That new system is called as WCS.
· It is a user defined system that facilitates the geometrical construction. While user
inputs data in WCS the software transforms it to MCS before storing the data.
· There is only one active WCS at any one time. If the user defines multiple
WCSs in one session, the software recognizes only the last one.
c) Screen Coordinate System (SCS)
· In contrast to MCS and WCS, Screen Coordinate System is a two-dimensional
device-independent system whose origin is usually located at the lower left corner
of the graphic display (display screen).
· The physical dimensions of the device screen and the type of device determine
the range of the SCS. A 1024 ´ 1024 display has an SCS with a range of (0, 0)
to (1024,1024).
· The SCS is important for display, screen input and digitizing tasks.
· A transformation operation from MCS coordinates to SCS coordinates is
performed by the software before displaying the model views and graphics.
· For a geometric model, there is a data structure to store its geometric data
(relative to MCS), and a display file to store its display data (relative to SCS).
Window and View Port
Window
· When a design package is initiated, the display will have a set of co-ordinate
values. These are called default co-ordinates.
1 - 19 Computer Aided Design and Manufacturing
Introduction

32. · A user co-ordinate system is one in which the designer can specify his own
coordinates for a specific design application.
· These screen independent coordinates can have large or small numeric range, or
even negative values, so that the model can be represented in a natural way.
· It may, however, happen that the picture is too crowded with several features to
be viewed clearly on the display screen.
· Therefore, the designer may want to view only a portion of the image, enclosed
in a rectangular region called a window.
· Different parts of the drawing can thus be selected for viewing by placing the
windows.
· Portions inside the window can be enlarged, reduced or edited depending upon
the requirements. Fig. 1.8.2 (a) depicts the use of windowing to enlarge an image.
View Port
· It may be sometimes desirable to display different portions or views of the
drawing in different regions of the screen.
· A portion of the screen where the contents of the window are displayed is called
a view port. Fig. 1.8.2 (b) explains a view port.
1.9 2D Transformations + [AU : Dec.-17]
· Geometric transformations provide a means by which an image can be enlarged
in size, or reduced, rotated, or moved.
· These changes are brought about by changing the co-ordinates of the picture to a
new set of values depending upon the requirements.
· The basic transformations are translation, scaling, rotation, reflection or mirror
and shear.
1 - 20 Computer Aided Design and Manufacturing
Introduction
Window
Original
drawing
65,50
130,100 View port 1
View port 4 View port 3
View port 2
(a) Window (b) View port
Fig. 1.8.2

33. a) 2D Translation
· This moves a geometric entity in space in such a way that the new entity is
parallel at all points to the old entity. Translation of a point is shown in
Fig. 1.9.1.
· Let's consider a point on the object, represented by P which is translated along X
and Y axes by DX and DY respectively to a new position P '.
· The new coordinates after transformation are given by following equations.
P' = [x', y' ] …(1.9.1)
x' = [x+Dx] …(1.9.2)
y' = [y+Dy] …(1.9.3)
[P'] =
¢
¢
é
ë
ê
ù
û
ú
x
y
=
x x
y y
+
+
é
ë
ê
ù
û
ú
D
D
=
x
y
é
ë
ê
ù
û
ú +
D
D
x
y
é
ë
ê
ù
û
ú …(1.9.4)
2D Translation of an object
Fig. 1.9.2 explains the transformation of a rectangle. Consider a rectangle of
coordinates (1,1), (4,1), (1,5) and (4,5). The rectangle is translated by 3 units along
x-direction (Dx) and 3 units along y-direction (Dy). (See Fig. 1.9.2 on next page)
b) 2D Scaling
· Scaling is the transformation applied to change the scale of an entity.
· To achieve scaling, the original coordinates would be multiplied uniformly by the
scaling factors.
Sx = Scaling factor along x-direction
Sy = Scaling factor along y-direction
Ts = Scaling matrix
· The scaling operations could be explained by the equations stated below.
¢
P = [x', y' ]=[Sx ´ X, Sy ´ Y] …(1.9.5)
1 - 21 Computer Aided Design and Manufacturing
Introduction
P
P Z
Y
X
X
P'
X'
Y'
Z'
Y
Fig. 1.9.1 Translation of a point

34. [ ¢
P ] =
S 0
0 S
x
y
x
y
é
ë
ê
ù
û
ú
é
ë
ê
ù
û
ú …(1.9.6)
[Ts] =
S 0
0 S
x
y
é
ë
ê
ù
û
ú …(1.9.7)
[ ¢
P ] = [Ts] × [P] …(1.9.8)
· Fig. 1.9.3 depicts the scaling of an
object.
c) 2D Rotation
· Rotation is another important
geometric transformation. The
final position and orientation of a
geometric entity is decided by the
angle of rotation (q) and the base
point about which the rotation, is
to be done.
· If rotation is made in clockwise
direction 'q' is considered as
1 - 22 Computer Aided Design and Manufacturing
Introduction
0 1 2 3 4 5 6 7 8 9 10
X
(1, 1) (4, 1)
(1, 5) (4, 5)
(5, 4) (8, 4)
(5, 8) (8, 8)
Y
0
1
2
3
4
5
6
7
8
9
10
Original rectangle
After translation
Fig. 1.9.2 2D Translation of an object
X
SX
SY
Y
P
P'
Y
X
Fig. 1.9.3 2D Scaling of an object

35. negative and if rotation is made in counter clockwise (anti-clockwise) direction
' q' is considered as positive.
· Fig. 1.9.4 depicts rotation of an object.
· To develop the transformation matrix for transformation, consider a point P
located in XY-plane, being rotated in the counter clockwise direction to the new
position, ¢
P by an angle 'q' as shown in Fig. 1.9.4. The new position ¢
P is given
by
¢
P = [ ¢
x , ¢
y ]
· From the figure the original position is specified by
x = r cos a
y = r sin a
· The new position ¢
P is specified by
¢
x = r cos ( +
a q) = r cos q cos a – r sin q sin a
= x cos q – y sin q
Also, ¢
y = r sin ( +
a q) = r sin q cos a + r cos q sin a
= x sin q + y cos q
· Thus the transformation matrix for a rotation operation could be derived as
follows,
[P ]
¢ =
¢
¢
é
ë
ê
ù
û
ú
x
y
=
cos sin
sin cos
x
y
q q
q q
-
é
ë
ê
ù
û
ú
é
ë
ê
ù
û
ú …(1.9.9)
· The rotation matrix is given as TR .
1 - 23 Computer Aided Design and Manufacturing
Introduction
X'
X
P'
r
P
X
Y
Y'
Y
0
Fig. 1.9.4 2D rotation of an objects

36. [TR ] =
cos sin
sin cos
q q
q q
-
é
ë
ê
ù
û
ú …(1.9.10)
[P ]
¢ = [T [P]
R ]× (1.9.11)
d) 2D Shearing
· A shearing transformation produces distortion of an object or an entire image.
There are two types of shears : X-shear and Y-shear.
· A transformation that slants the shape of an object is called the shear
transformation.
· One shifts X coordinates values and other shifts Y coordinate values. However;
in both the cases only one coordinate changes its coordinates and other preserves
its values.
· Shearing is also termed as skewing.
· The X-shear as shown in the Fig. 1.9.5 (a), preserves the Y coordinate and
changes are made to X coordinates, which causes the vertical lines to tilt right or
left.
1 - 24 Computer Aided Design and Manufacturing
Introduction
11 12 13
After X-shear
Original part
D1
A1
B1
A B
E1
C1
C
D
0 1 2 3 4 5 6 7 8 9 10
X
Y
0
1
2
3
4
5
6
7
8
9
10
E
Fig. 1.9.5 (a) X-Shear

37. · The Y-shear as shown in the Fig. 1.9.5 (b) preserves the X coordinates and
changes the Y coordinates which causes the horizontal lines to transform into
lines which slopes up or down.
· A Y-shear transforms the point (x, y) to the point (x1, y1) by a factor Sh1,
¢
x = x …(1.9.12)
¢
y = Sh x y
1 × + …(1.9.13)
· An X-shear transforms the point (X, Y) to (x1, y1), where Sh2 is the shear factor
¢
x = x + Sh y
2 × …(1.9.14)
¢
y = y …(1.9.15)
e) 2D Reflection/Mirror
· Reflection is the mirror image of original object.
· Mirroring is a convenient method used for copying an object while preserving its
features.
· In reflection transformation, the size of the object does not change.
· Reflection could be done along both x and y directions as shown in the
Fig.1.9.6(a) and 1.9.6(b).
1 - 25 Computer Aided Design and Manufacturing
Introduction
After X-shear
Original part
D1
C1
B1
E1
0 1 2 3 4 5 6 7 8 9 10
X
Y
0
1
2
3
4
5
6
7
8
9
10
E
D
C
B
A
A1
Fig. 1.9.5 (b) Y-Shear

38. · For reflection about x-axis the y coordinate will be negative and the following
equations should be utilized,
¢
P = [X , Y ]
¢ ¢ = [X, – Y] …(1.9.16)
[P ]
¢ =
1 0
0 1
-
é
ë
ê
ù
û
ú
é
ë
ê
ù
û
ú
x
y
…(1.9.17)
The translation matrix is given as,
[T ]
m =
1 0
0 1
-
é
ë
ê
ù
û
ú …(1.9.18)
[P ]
¢ = [T ] [P]
m × …(1.9.19)
· For reflection about y-axis the x coordinate will be negative and the following
equations should be utilized,
¢
P = [X , Y ]
¢ ¢ = [X, – Y] …(1.9.20)
[P ]
¢ =
-
é
ë
ê
ù
û
ú
é
ë
ê
ù
û
ú
1 0
0 1
x
y
…(1.9.21)
The translation matrix is given as,
[T ]
m =
-
é
ë
ê
ù
û
ú
1 0
0 1
…(1.9.22)
[P ]
¢ = [T ] [P]
m × …(1.9.23)
· Thus the general form of reflection matrix could be written as,
[T ]
m =
±
±
é
ë
ê
ù
û
ú
1 0
0 1
…(1.9.24)
1 - 26 Computer Aided Design and Manufacturing
Introduction
(a) Reflection about X-Axis
P
Y
X
Y
–
Y
P'
– X X
Y
X
P
P'
(b) Reflection about Y-axis
Fig. 1.9.6

39. 1.9.1 Homogeneous Coordinates
Concatenation of Transformations
· Sometimes it becomes necessary to combine the individual transformations in
order to achieve the required results. In such cases the combined transformation
matrix can be obtained by multiplying the respective transformation matrices as
shown below,
[P ]
¢ = [T ][T ][T ]...[T ][T ][T ]
n n 1 n 2 3 2 1
- - …(1.9.25)
· In order to concatenate the transformation, all the transformation matrices should
be multiplicative type. The following form known as homogeneous form should
be used to convert the translation matrix into a multiplication type.
[P ]
¢ =
¢
¢
é
ë
ê
ê
ê
ù
û
ú
ú
ú
x
y
1
=
1 0 0
0 1 0
D D
X Y 1
x
y
1
é
ë
ê
ê
ê
ù
û
ú
ú
ú
é
ë
ê
ê
ê
ù
û
ú
ú
ú
…(1.9.26)
· The three dimensional representation of a two dimensional plane is called
homogeneous coordinates and the transformation using the homogeneous
co-ordinates is called homogeneous transformation.
· The translation matrix in homogeneous form is,
[T] =
1 0 0
0 1 0
D D
X Y 1
é
ë
ê
ê
ê
ù
û
ú
ú
ú
· The Scaling matrix in homogeneous form is,
[S] =
S 0 0
0 S 0
0 0 1
x
y
é
ë
ê
ê
ê
ù
û
ú
ú
ú
· The Rotation matrix in homogeneous form is,
[T ]
R =
cos sin
sin cos
q q
q q
0
0
0 0 1
-
é
ë
ê
ê
ê
ù
û
ú
ú
ú
Need for homogeneous transformation
· Consider the need for rotating an object about an arbitrary point as shown in
Fig. 1.9.7.
· The transformation given earlier for rotation is about the origin of the axes
system.
· To derive the necessary transformation matrix, the following complex procedure
would be required.
i) Translate the point 'P' to 'O', the origin of the axes system.
1 - 27 Computer Aided Design and Manufacturing
Introduction

40. ii) Rotate the object by the given angle 'q'.
iii) Translate the point back to its original position from origin.
· The following homogeneous transformation matrices should be used for the
translation operation,
i) Translate the point from point 'P' to origin 'O'
[T ]
1 = [T] =
1 0 0
0 1 0
1
- -
é
ë
ê
ê
ê
ù
û
ú
ú
ú
D D
X Y
ii) Rotate the object by the given angle 'q'.
[T ]
2 = [T ]
R =
cos sin
sin cos
q q
q q
0
0
0 0 1
-
é
ë
ê
ê
ê
ù
û
ú
ú
ú
iii) Translate the point back to its original position from origin.
[T ]
3 = [T] =
1 0 0
0 1 0
1
D D
X Y
é
ë
ê
ê
ê
ù
û
ú
ú
ú
iv) Final Transformation matrix after concatenation,
[T] = [T ] [T ] [T ]
1 2 3
´ ´
1 - 28 Computer Aided Design and Manufacturing
Introduction
A
X
Y
r
r
P'
P
O
Y
X
Fig. 1.9.7 Rotation of an object about an arbitrary point

41. 1.9.2 Solved Examples on 2D Transformation
Example 1.9.1 : Translate a point P(2, 3) by four units in x-direction and 5 units in
y-direction.
Solution : Given : P(2, 3) Þ (x , y )
1 1
Dx = 4; Dy = 5
Tranformation matrix
T =
D
D
x
y
é
ë
ê
ù
û
ú
New position of a point is,
¢
P = P + T
¢
¢
é
ë
ê
ù
û
ú
x
y
1
1
=
x
y
x
y
1
1
é
ë
ê
ù
û
ú +
é
ë
ê
ù
û
ú
D
D
=
2
3
4
5
é
ë
ê
ù
û
ú +
é
ë
ê
ù
û
ú =
6
8
é
ë
ê
ù
û
ú
Example 1.9.2 : A line AB, A(2, 4) and B(5, 6) is to be translated 1 unit along +ve
direction of 'x' and 3 units along +ve direction of y. Find the translated coordinates.
Solution : Given :
For line AB ® A(x , y )
1 1 = (2, 4)
B(x , y )
2 2 = (5, 6)
T x y
(D , D ) = (1, 3)
1 - 29 Computer Aided Design and Manufacturing
Introduction
y = 5
P
(2, 3)
P'(6, 8)
X
Y
x = 4
Fig. 1.9.8

42. ¢
A = A + T
¢
A =
2
4
1
3
é
ë
ê
ù
û
ú +
é
ë
ê
ù
û
ú =
3
7
é
ë
ê
ù
û
ú
Similarly ¢
B = B + T =
5
6
1
3
é
ë
ê
ù
û
ú +
é
ë
ê
ù
û
ú =
6
9
é
ë
ê
ù
û
ú
Example 1.9.3 : Translate a triangle ABC having coordinates A(1, 1), B(3, 1) and
C(1, 3) about the origin by 3 - units in x - direction and 2 - units in y - direction.
Solution : Given : Triangle ABC,
A = (x , y )
1 1 = (1, 1)
B = (x , y )
2 2 = (3, 1)
C = (x , y )
3 3 = (1, 3)
Dx = 3; Dy = 2 Þ T = (3, 2)
¢
A = A + T =
1
1
3
2
é
ë
ê
ù
û
ú +
é
ë
ê
ù
û
ú =
4
3
é
ë
ê
ù
û
ú
1 - 30 Computer Aided Design and Manufacturing
Introduction
x = 1
A
(2, 4)
B (5, 6)
B' (6, 9)
A' (3, 7)
X
Y
y = 3
Fig. 1.9.9

43. ¢
B = B + T =
3
1
3
2
é
ë
ê
ù
û
ú +
é
ë
ê
ù
û
ú =
6
3
é
ë
ê
ù
û
ú
¢
C = C + T =
1
3
3
2
é
ë
ê
ù
û
ú +
é
ë
ê
ù
û
ú =
4
5
é
ë
ê
ù
û
ú
Example 1.9.4 : A rectangular lamina ABCD having co-ordinates A(2, 2), B(4, 2),
C(4, 5) and D(2, 5) is translated by 4 units in x - direction and 4 units in y - direction.
Find out the translated coordinates and plot the rectangle before and after translation.
Solution : Given : Rectangle ABCD,
A (2, 2) = (x , y )
1 1
B (4, 2) = (x , y )
2 2
C (4, 5) = (x , y )
3 3
D (2, 5) = (x , y )
4 4
Dx = 4; Dy = 4 Þ T( , )
D D
x y = (4, 4)
[A] = [A] + [T] =
2
2
4
4
é
ë
ê
ù
û
ú +
é
ë
ê
ù
û
ú =
6
6
é
ë
ê
ù
û
ú
[B ]
¢ = [B] + [T] =
4
2
4
4
é
ë
ê
ù
û
ú +
é
ë
ê
ù
û
ú =
8
6
é
ë
ê
ù
û
ú
1 - 31 Computer Aided Design and Manufacturing
Introduction
x = 3
y = 2
C (1, 3)
A' (4, 3)
B (3, 1)
A (1, 1)
B' (6, 3)
C' (4, 5)
X
Y
Fig. 1.9.10

44. [C ]
¢ = [C] + [T] =
4
5
4
4
é
ë
ê
ù
û
ú +
é
ë
ê
ù
û
ú =
8
9
é
ë
ê
ù
û
ú
[D ]
¢ = [D] + [T] =
2
5
4
4
é
ë
ê
ù
û
ú +
é
ë
ê
ù
û
ú =
6
9
é
ë
ê
ù
û
ú
Example 1.9.5 : A polygon ABCD is having the coordinates A(2, 3), B(6, 3), C(6, 6),
D(2, 6). Scale the polygon by 2 units along x-axis and y-axis.
Solution : Given :
Polygon ABCD ® A(x , y )
1 1 = (2, 3)
B(x , y )
2 2 = (6, 3)
C(x , y )
3 3 = (6, 6)
D(x , y )
4 4 = (2, 6)
Scaling factor ® S = (S ,S )
x y = (2, 2)
Scaling matrix, [S] =
S 0
0 S
x
y
é
ë
ê
ù
û
ú =
2 0
0 2
é
ë
ê
ù
û
ú
1 - 32 Computer Aided Design and Manufacturing
Introduction
(6, 9)
(6, 6)
C' (8, 9)
x = 4
(2, 5)
(2, 2)
C (4, 5)
B (4, 2)
A
D
(8, 6)
B'
D'
y = 4
A'
X
Y
Fig. 1.9.11

45. [A]¢ = [S] ´ [A]
=
2 0
0 2
2
3
é
ë
ê
ù
û
ú ´
é
ë
ê
ù
û
ú =
4
6
é
ë
ê
ù
û
ú
[B]¢ = [S] ´ [B] Þ
=
2 0
0 2
6
3
é
ë
ê
ù
û
ú ´
é
ë
ê
ù
û
ú =
12
6
é
ë
ê
ù
û
ú
[C]¢ = [S] ´ [C]
=
2 0
0 2
6
6
é
ë
ê
ù
û
ú ´
é
ë
ê
ù
û
ú =
12
12
é
ë
ê
ù
û
ú
[D]¢ = [S] ´ [D]
=
2 0
0 2
2
6
é
ë
ê
ù
û
ú ´
é
ë
ê
ù
û
ú =
4
12
é
ë
ê
ù
û
ú
1 - 33 Computer Aided Design and Manufacturing
Introduction
X
Y
A (2, 3)
D (2, 6)
A' (4, 6)
C (6, 6)
B' (12, 6)
C' (12, 12)
D' (4, 12)
B (6, 3)
Fig. 1.9.12

46. Example 1.9.6 : Rotate the point P(6, 8) about the origin at an angle 30 ° in
anti-clock wise direction and obtain the new position of the point.
Solution : Given
P(x , y )
1 1 = (6, 8) ; q = 30°
¢
P =
¢
¢
é
ë
ê
ù
û
ú
x
y
1
1
=
cos sin
sin cos
q q
q q
-
é
ë
ê
ù
û
ú
é
ë
ê
ù
û
ú
x
y
1
1
=
cos sin
sin cos
30 30
30 30
-
é
ë
ê
ù
û
ú
é
ë
ê
ù
û
ú
6
8
Þ [P ]
¢ =
1.196
9.928
é
ë
ê
ù
û
ú
Example 1.9.7 : A triangle ABC, A(5, 2), B(3, 5), C(7, 5). Find the transformed
position if,
i) The triangle is rotated by 45 ° in clockwise direction.
ii) The triangle is rotated by 60 ° in anti-clockwise direction.
Solution : Given : DABC Þ A(5, 2)
(x , y )
,
B(3, 5)
(x , y )
,
C(7, 5)
(x , y )
1 1 2 2 3 3
i) Rotated by 45 ° in clockwise direction :
q = – 45°
[A]¢ =
¢
¢
é
ë
ê
ù
û
ú
x
y
1
1
=
cos( ) sin( )
sin( ) cos( )
- ° - - °
- ° - °
é
ë
ê
ù
û
ú
é
ë
ê
ù
û
45 45
45 45
5
2ú
1 - 34 Computer Aided Design and Manufacturing
Introduction
Y
X
30° =
P' (1.196, 9.28)
P(6, 8)
Fig. 1.9.13

47. Þ [A]¢ =
4.97
2.12
-
é
ë
ê
ù
û
ú
Similarly, [B]¢ =
¢
¢
é
ë
ê
ù
û
ú
x
y
2
2
=
cos( ) sin( )
sin( ) cos( )
- ° - - °
- ° - °
é
ë
ê
ù
û
ú
é
ë
ê
ù
û
45 45
45 45
3
5ú
Þ [B]¢ =
5.65
1.41
é
ë
ê
ù
û
ú
Similarly, [C]¢ =
¢
¢
é
ë
ê
ù
û
ú
x
y
3
3
=
cos( ) sin( )
sin( ) cos( )
- ° - - °
- ° - °
é
ë
ê
ù
û
ú
é
ë
ê
ù
û
45 45
45 45
7
5ú
Þ [C]¢ =
8.48
1.414
-
é
ë
ê
ù
û
ú
ii) Rotated by 60° in anticlockwise direction (counter-clockwise) :
q = 60°
[A]¢¢ =
¢¢
¢¢
é
ë
ê
ù
û
ú
x
y
1
1
=
cos sin
sin cos
60 60
60 60
5
2
-
é
ë
ê
ù
û
ú
é
ë
ê
ù
û
ú
Þ [A]¢¢ =
0.767
5.330
é
ë
ê
ù
û
ú
Similarly, [B]¢¢ =
¢¢
¢¢
é
ë
ê
ù
û
ú
x
y
2
2
=
cos sin
sin cos
60 60
60 60
3
5
-
é
ë
ê
ù
û
ú
é
ë
ê
ù
û
ú
Þ [B]¢¢ =
-
é
ë
ê
ù
û
ú
2.830
5.098
Similarly, [C]¢¢ =
¢¢
¢¢
é
ë
ê
ù
û
ú
x
y
3
3
=
cos sin
sin cos
60 60
60 60
7
5
-
é
ë
ê
ù
û
ú
é
ë
ê
ù
û
ú
Þ [C]¢¢ =
-
é
ë
ê
ù
û
ú
0.83
8.562
1 - 35 Computer Aided Design and Manufacturing
Introduction

48. Example 1.9.8 : A square with an edge length of 10 units is located in the origin with
one of the edges inclined at an angle 30 ° with X-axis. Calculate the new position of the
square, i) If it is rotated by an angle 30 ° in clock-wise direction. ii) If it is rotated by
60° in counter-clockwise direction.
Solution : Given : Square of edge
length 10 units.
® Positioned in such a way that, it
is located in origin ® with one of the
edges inclined by 30 ° to 'X'
(Refer Fig.1.9.15).
Initially evaluate the coordinates of
the square.
Þ A(x , y )
1 1 = (0, 0). Since place
at the origin.
Þ B(x , y )
2 2
1 - 36 Computer Aided Design and Manufacturing
Introduction
0
A" (0.767, 5.330)
(–0.830, 8.582)
(–2.830, 5.098)
60°
45°
B(3, 5)
C(7, 5)
A(5, 2)
A' (4.97, –2.12)
C' (8.48, –1.41)
B' (5.65, 1.41)
X
Y
Y
X
B"
C"
Fig. 1.9.14
60° 30°
(x , y )
3 3
(x , y )
1 1
(x , y )
2 2
D
B
C
Y
X X
A (0, 0)
0
Fig. 1.9.15

49. B(x , y )
2 2 :
Here edge length = 10 units
x2 = 10 cos 30 = 8.66 units
y 2 = 10 sin 30 = 5 units
B(x , y )
2 2 = (8.66, 5)
Similarly, C(x , y )
3 3 :
x 3 = x2 -10 60
cos
= 8.66 – 5 = 3.66 units
y 3 = y 2 +10 60
sin
= 5 + 8.66 = 13.66 units
C(x , y )
3 3 = (3.66, 13.66)
[D](x , y )
4 4 :
x4 = – 10 cos 60
= – 5 units
1 - 37 Computer Aided Design and Manufacturing
Introduction
Y
A (0, 0)
X
y2
x2
l = 10
30°
(x , y )
2 2
B
Fig. 1.9.16
Y
A (0, 0)
X
y2
x2
(x , y )
2 2
(x , y )
3 3
10 sin 60
B
60°
30°
30°
l = 10
10 cos 60
Fig. 1.9.17
Y
X
l=10 10 sin 60
– 10 cos 60
D(x , y )
4 4
60°
Fig. 1.9.18

50. y 4 = 10 sin 60 = 8.66 units
D(x , y )
4 4 = (– 5, 8.66)
i) Rotate the square by 30° clockwise :
q = – 30°
[A]¢ =
cos( ) sin( )
sin( ) cos( )
- - -
- -
é
ë
ê
ù
û
ú
é
ë
ê
ù
û
ú
30 30
30 30
0
0
=
0
0
é
ë
ê
ù
û
ú
[B]¢ =
cos( ) sin( )
sin( ) cos( )
- - -
- -
é
ë
ê
ù
û
ú
é
ë
ê
ù
û
ú
30 30
30 30 5
8.66
=
9.99 ~ 10
0.0001~ 0
é
ë
ê
ù
û
ú
Þ [B]¢ =
10
0
é
ë
ê
ù
û
ú
[C]¢ =
cos( ) sin( )
sin( ) cos( )
- - -
- -
é
ë
ê
ù
û
ú
é
ë
30 30
30 30
3.66
13.66
ê
ù
û
ú =
10
10
é
ë
ê
ù
û
ú
[D]¢ =
cos( ) sin( )
sin( ) cos( )
- - -
- -
é
ë
ê
ù
û
ú
-
é
ë
ê
ù
û
30 30
30 30
5
8.66ú =
0
10
é
ë
ê
ù
û
ú
iii) Square is rotated by 60° in counter clockwise direction :
q = 60°
[A]¢¢ =
cos sin
sin cos
60 60
60 60
0
0
-
é
ë
ê
ù
û
ú
é
ë
ê
ù
û
ú =
0
0
é
ë
ê
ù
û
ú
[B]¢¢ =
cos sin
sin cos
60 60
60 60
-
é
ë
ê
ù
û
ú
é
ë
ê
ù
û
ú
8.66
5
=
0
10
é
ë
ê
ù
û
ú
[C]¢¢ =
cos sin
sin cos
60 60
60 60
-
é
ë
ê
ù
û
ú
é
ë
ê
ù
û
ú
3.66
13.66
=
-
é
ë
ê
ù
û
ú
10
10
[D]¢¢ =
cos sin
sin cos
60 60
60 60
5
-
é
ë
ê
ù
û
ú
-
é
ë
ê
ù
û
ú
8.66
=
-
é
ë
ê
ù
û
ú
10
0
1 - 38 Computer Aided Design and Manufacturing
Introduction

51. Example 1.9.9 : For a planar lamina ABCD with A(3, 5), B(2, 2), C(8, 2) and D(4, 5)
in XY plane with point P(4, 3) in the inetrior is to be
i) Translated by a translation matrix [T] =
8
5
é
ë
ê
ù
û
ú
ii) Rotated by 60° in counter clockwise direction.
1 - 39 Computer Aided Design and Manufacturing
Introduction
60° 30°
D''
(–10, 0)
B' (10, 0)
C' (10, 10)
C
(3.66, 13.66)
C"
(–10, 10) D' B''
(0, 10)
B
A (0, 0)
A', A"
Fig. 1.9.19
A (3, 5) D(4, 5)
P
(4, 3)
B (2, 2) C (8, 2)
Fig. 1.9.20

52. Sol. : i) Translation :
[T] =
D
D
x
y
é
ë
ê
ù
û
ú =
8
5
é
ë
ê
ù
û
ú
Dx = 8; Dy = 5
[A]¢ =
x
y
x
y
1
1
é
ë
ê
ù
û
ú +
é
ë
ê
ù
û
ú
D
D
=
3
5
8
5
é
ë
ê
ù
û
ú +
é
ë
ê
ù
û
ú =
11
10
é
ë
ê
ù
û
ú
[B]¢ =
x
y
x
y
2
2
é
ë
ê
ù
û
ú +
é
ë
ê
ù
û
ú
D
D
=
2
2
8
5
é
ë
ê
ù
û
ú +
é
ë
ê
ù
û
ú =
10
7
é
ë
ê
ù
û
ú
[C]¢ =
x
y
x
y
3
3
é
ë
ê
ù
û
ú +
é
ë
ê
ù
û
ú
D
D
=
8
2
8
5
é
ë
ê
ù
û
ú +
é
ë
ê
ù
û
ú =
16
7
é
ë
ê
ù
û
ú
[D]¢ =
x
y
x
y
4
4
é
ë
ê
ù
û
ú +
é
ë
ê
ù
û
ú
D
D
=
4
5
8
5
é
ë
ê
ù
û
ú +
é
ë
ê
ù
û
ú =
12
10
é
ë
ê
ù
û
ú
[P]¢ =
4
3
x
y
é
ë
ê
ù
û
ú +
é
ë
ê
ù
û
ú
D
D
=
4
3
8
5
é
ë
ê
ù
û
ú +
é
ë
ê
ù
û
ú =
12
8
é
ë
ê
ù
û
ú
ii) Rotate through 60° in counter clockwise direction :
q = 60°
[A]¢¢ =
cos sin
sin cos
60 60
60 60
3
5
-
é
ë
ê
ù
û
ú
é
ë
ê
ù
û
ú =
-
é
ë
ê
ù
û
ú
2.83
5.09
[B]¢¢ =
cos sin
sin cos
60 60
60 60
-
é
ë
ê
ù
û
ú
é
ë
ê
ù
û
ú
2
2
=
-
é
ë
ê
ù
û
ú
0.732
2.732
[C]¢¢ =
cos sin
sin cos
60 60
60 60
-
é
ë
ê
ù
û
ú
é
ë
ê
ù
û
ú
8
2
=
2.26
7.93
é
ë
ê
ù
û
ú
[D]¢¢ =
cos sin
sin cos
60 60
60 60
4
-
é
ë
ê
ù
û
ú
é
ë
ê
ù
û
ú
5
=
-
é
ë
ê
ù
û
ú
2.33
5.96
[P]¢¢ =
cos sin
sin cos
60 60
60 60
4
-
é
ë
ê
ù
û
ú
é
ë
ê
ù
û
ú
3
=
-
é
ë
ê
ù
û
ú
0.59
4.96
1 - 40 Computer Aided Design and Manufacturing
Introduction

53. 1 - 41 Computer Aided Design and Manufacturing
Introduction
1
2
3
4
5
6
7
8
9
10
11
12
13
14
Y
–3
–2
–1
0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
D''
A''
C''
A
D
B
C
B'
A'
D'
C'
P'
X
X
Y
P"
B"
P
Fig. 1.9.21

54. Example 1.9.10 : Derive an appropriate 2D transformation method to reflect the
rectangle ABCD, A(3, 4), B(7, 4), C(7, 6), D(3, 6). Reflect about i) X-axis, ii) Y-axis.
Solution : i) For reflection about X-axis :
[A]¢ =
1 0
0 1
-
é
ë
ê
ù
û
ú
é
ë
ê
ù
û
ú
x
y
1
1
Þ [A]¢ =
1 0
0 1
-
é
ë
ê
ù
û
ú
é
ë
ê
ù
û
ú
3
4
=
3
4
-
é
ë
ê
ù
û
ú
[B]¢ =
1 0
0 1
-
é
ë
ê
ù
û
ú
é
ë
ê
ù
û
ú
7
4
=
7
4
-
é
ë
ê
ù
û
ú
[C]¢ =
1 0
0 1
-
é
ë
ê
ù
û
ú
é
ë
ê
ù
û
ú
7
6
=
7
6
-
é
ë
ê
ù
û
ú
[D]¢ =
1 0
0 1
-
é
ë
ê
ù
û
ú
é
ë
ê
ù
û
ú
3
6
=
3
6
-
é
ë
ê
ù
û
ú
1 - 42 Computer Aided Design and Manufacturing
Introduction
A
(3, 4)
B
D' C'
(7, 4)
(3, – 6) (7, – 6)
D
(3, 6)
C
(7, 6)
A'
(3, – 4) B'(7, – 4)
X
Y
Y
X
Fig. 1.9.22

55. ii) Reflection about Y-axis :
[A]¢ =
-
é
ë
ê
ù
û
ú
é
ë
ê
ù
û
ú
1 0
0 1
x
y
1
1
Þ [A]¢ =
-
é
ë
ê
ù
û
ú
é
ë
ê
ù
û
ú
1 0
0 1
3
4
=
-
é
ë
ê
ù
û
ú
3
4
[B]¢ =
-
é
ë
ê
ù
û
ú
é
ë
ê
ù
û
ú
1 0
0 1
7
4
=
-
é
ë
ê
ù
û
ú
7
4
[C]¢ =
-
é
ë
ê
ù
û
ú
é
ë
ê
ù
û
ú
1 0
0 1
7
6
=
-
é
ë
ê
ù
û
ú
7
6
[D]¢ =
-
é
ë
ê
ù
û
ú
é
ë
ê
ù
û
ú
1 0
0 1
3
6
=
-
é
ë
ê
ù
û
ú
3
6
iii) About origin :
[A]¢¢¢ =
-
-
é
ë
ê
ù
û
ú
é
ë
ê
ù
û
ú
1 0
0 1
x
y
1
1
1 - 43 Computer Aided Design and Manufacturing
Introduction
(–7, 6)
D''(–3, 6) D(3, 6)
A''(–3, 4)
(–7, 4)
A(3, 4) B(7, 4)
C(7, 6)
X
Y
Y
X
C''
B''
Fig. 1.9.23

56. Þ [A]¢¢¢ =
-
-
é
ë
ê
ù
û
ú
é
ë
ê
ù
û
ú
1 0
0 1
3
4
=
-
-
é
ë
ê
ù
û
ú
3
4
[B]¢¢¢ =
-
-
é
ë
ê
ù
û
ú
é
ë
ê
ù
û
ú
1 0
0 1
7
4
=
-
-
é
ë
ê
ù
û
ú
7
4
[C]¢¢¢ =
-
-
é
ë
ê
ù
û
ú
é
ë
ê
ù
û
ú
1 0
0 1
7
6
=
-
-
é
ë
ê
ù
û
ú
7
6
[D]¢¢¢ =
-
-
é
ë
ê
ù
û
ú
é
ë
ê
ù
û
ú
1 0
0 1
3
6
=
-
-
é
ë
ê
ù
û
ú
3
6
2-D Transformation Problems based on Homogeneous Coordinate System
(Concatenation)
Example 1.9.11 : A rectangle ABCD has coordinates A(2, 3), B(6, 3), C(6, 6) and
D(2, 6). Calculate the combined transformation matrix (concatenation) for the following
operations. Also find the resultant coordinates.
i) Translation by 2 units in x - direction and 3 units in y - direction.
ii) Scaling by 4 - units in x - direction and 2 - units in y - direction.
iii) Rotation by 30 ° in counter clockwise direction about z - axis, passing through a
point (3, 3).
1 - 44 Computer Aided Design and Manufacturing
Introduction
C''' (–7,–6) D''' (–3,–6)
B''' (–7,–4) A''' (–3,–4)
A
(3, 4) B (7, 4)
D
(3, 6) C (7, 6)
X
Y
O
Y
X
Fig. 1.9.24

57. Solution : Given :
Rectangle ABCD ® A(2, 3) Þ (x , y )
1 1
B(6, 3) Þ (x , y )
2 2
C(6, 6) Þ (x , y )
3 3
D(2, 6) Þ (x , y )
4 4
i) Translation matrix in homogeneous form :
Given, Dx = 2
Dy = 3
Homogeneous Translation Matrix
[T]( , )
2 3 =
1 0 0
0 1 0
2 3 0
é
ë
ê
ê
ê
ù
û
ú
ú
ú
ii) Scaling matrix in homoeneous form,
[ ]
' S '
( , )
4 2 Given, Sx = 4
Sy = 2
[S]( , )
4 2 =
4 0 0
0 2 0
0 0 1
é
ë
ê
ê
ê
ù
û
ú
ú
ú
iii) Rotation matrix in homoeneous form,
[R]Þ at point (3, 3)
Note : In normal cases rotation is done with suspect to origin.
· But in this problem rotation has to be made at point (3, 3), which is not possible
by normal method.
· Therefore, initially the rectangle will be tanslated to origin, it will be rotated at
origin.
· After rotating at origin, the retangle will be translated back to point (3, 3).
Procedure for rotation at (3, 3)
Step 1 : Translate rectangle ABCD at origin [T ]
1 .
Step 2 : Rotate rectangle ABCD at origin [T ]
11 .
Step 3 : Translate rectangle ABCD from origin [T ]
111 to point [3, 3].
Step 4 : Final rotation matrix [R] = [ ] [ ] [ ]
T T T
1 11 111
´ ´
1 - 45 Computer Aided Design and Manufacturing
Introduction

58. Step 1 : Translation matrix for translating ABCD to origin form point (3, 3) [T ]
I
Here, DxI = – 3, DyI = – 3
[TI ] =
1 0 0
0 1 0
1
1 0 0
0 1 0
3 3 1
D D
x y
I I
é
ë
ê
ê
ê
ù
û
ú
ú
ú
é
ë
ê
ê
ê
ù
û
ú
ú
ú
– –
Step 2 : Rotate rectangle ABCD at origin [T ]
II .
Here, q = 30° (counter - clockwise)
[TII ] =
cos sin
– sin cos
cos sin
– sin
q q
q q
0
0
0 1
30 30 0
30
0
é
ë
ê
ê
ê
ù
û
ú
ú
ú
= cos30 0
0 0 1
é
ë
ê
ê
ê
ù
û
ú
ú
ú
Þ [T] =
0866 0 5 0
0 5 0866 0
0 0 1
. .
– . .
é
ë
ê
ê
ê
ù
û
ú
ú
ú
Step 3 : Translate ABCD from origin to (3, 3) [T]III
Here DxIII = 3 ; DyIII = 3
[T]III =
1 0 0
0 1 0
1
D D
x y
III III
é
ë
ê
ê
ê
ù
û
ú
ú
ú
é
ë
ê
ê
ê
ù
û
ú
ú
ú
1 0 0
0 1 0
3 3 1
Rotation matrix at (3, 3) in homogeneous form Þ [R]
[R] = [T] [T]
I II III
´ ´
[ ]
T
=
1 0 0
0 1 0
3 3 1
0866 0 5 0
0 5 0866 0
0 0 1
– –
. .
– . .
é
ë
ê
ê
ê
ù
û
ú
ú
ú
´
é
ë
ê
ê
ê
ù
û
ú
ú
ú
´
é
ë
ê
ê
ê
ù
û
ú
ú
ú
1 0 0
0 1 0
3 3 1
Þ [R] =
0.866 0.5 0
–0.5 0.866 0
1. 902 – 1.908 1
é
ë
ê
ê
ê
ù
û
ú
ú
ú
So we obtained all the three matrixes for evaluating the combined matrix [ ]
m
[T](2,3) =
1 0 0
0 1 0
2 3 0
é
ë
ê
ê
ê
ù
û
ú
ú
ú
1 - 46 Computer Aided Design and Manufacturing
Introduction

59. [S](4,2) =
4 0 0
0 2 0
0 0 1
é
ë
ê
ê
ê
ù
û
ú
ú
ú
[R]( = 30 )
q ° =
0866 0 5 0
0 5 0866 0
1902 1098 1
. .
– . .
. – .
é
ë
ê
ê
ê
ù
û
ú
ú
ú
Combined transformation matrix [ ]
Tc
[ ]
Tc = [T] [S] [R]
(2,3) (4,2) ( = 30 )
´ ´ °
q
=
1 0 0
0 1 0
2 3 0
4 0 0
0 2 0
0 0 1
0866 0 5 0
é
ë
ê
ê
ê
ù
û
ú
ú
ú
´
é
ë
ê
ê
ê
ù
û
ú
ú
ú
´
. .
–0 5 0866 0
1902 1098 1
. .
. – .
é
ë
ê
ê
ê
ù
û
ú
ú
ú
Þ [T ]
c =
3.464 2 0
–1 1.732 0
5.93 8.098 1
é
ë
ê
ê
ê
ù
û
ú
ú
ú
To find the resultant co-ordinates after combined transformations operations.
Given, A B C D Þ A (x , y )
1 1 = (2, 3)
B (x , y )
2 2 = (6, 3)
C (x , y )
3 3 = (6, 6)
D (x , y )
4 4 = (2, 6)
Coordinates an homogeneous form,
A
B
C
D
é
ë
ê
ê
ê
ê
ù
û
ú
ú
ú
ú
=
x y
x y
x y
x y
1 1
2 2
3 3
4 4
1
1
1
1
2 3 1
6 3 1
6 6 1
2 6 1
é
ë
ê
ê
ê
ê
ù
û
ú
ú
ú
ú
=
é
ë
ê
ê
ê
ê
ù
û
ú
ú
ú
ú
Resultant co-ordinates (After transformation)
¢
¢
¢
¢
é
ë
ê
ê
ê
ê
ù
û
ú
ú
ú
ú
A
B
C
D
=
A
B
C
D
[Tc
é
ë
ê
ê
ê
ê
ù
û
ú
ú
ú
ú
´ ] =
2 3 1
6 3 1
6 6 1
2 6 1
3464 2 0
1 1732 0
593 8098 1
é
ë
ê
ê
ê
ê
ù
û
ú
ú
ú
ú
´
.
– .
. .
é
ë
ê
ê
ê
ù
û
ú
ú
ú
1 - 47 Computer Aided Design and Manufacturing
Introduction
4 × 3 × 3 × 3

60. ¢
¢
¢
¢
é
ë
ê
ê
ê
ê
ù
û
ú
ú
ú
ú
A
B
C
D
=
9.858 17.294 1
23.714 25.294 1
20.714 30.49 1
6.888 22.49 1
é
ë
ê
ê
ê
ê
ù
û
ú
ú
ú
ú
¢
A = (x , y )
1 1
¢ ¢ = (9.858, 17.294)
¢
B = (x , y )
2 2
¢ ¢ = (23.714, 25.294)
¢
C = C (x , y )
3 3
¢ ¢ = (20.714, 30.49)
¢
D = (x , y )
4 4
¢ ¢ = (6.888, 22.49)
Example 1.9.12 : Find the transformartion matrix to transform polygon ABCD to 3/4
of its original size, with uts centre (4, 4.5) remians at the same position. Coordinates of
ABCD are A(2, 3), B(6, 3), C(6, 6), D(2, 6).
Solution : Given : Polygon ABCD ®
A (x , y )
1 1 = (2, 3)
B (x , y )
2 2 = (6, 3)
C (x , y )
3 3 = (6, 6)
D (x , y )
4 4 = (2, 6)
1 - 48 Computer Aided Design and Manufacturing
Introduction
D(2,6)
A(2,3)
C(6,6)
B(6,3)
(20.714, 30.49)
(23.714, 25.94)
A'
(9.858, 17.294)
(6.88, 22.49)
Y
X
B'
C'
D'
Fig. 1.9.25

61. · In this problem rectangle ABCD has to scaled to 3/4 of its size.
Sx = 0.75 and Sy = 0.75
· Scaling has to be done such that centre point (4, 4.5) remains at same position.
· This is not possible by normal means of scaling operation.
· Here scaling can be performed only with repect to origin.
· So initially translate ABCD to orgin, perform scaling at origin and translate back
to point (4, 4.5).
Procedure for performing scaling at (4, 4.5)
Step 1 : Translate ABCD from (4, 4.5) to origin (0, 0) [TI ]
Step 2 : Scale ABCD at orgin (0, 0) [TII ]
Step 3 : Translate ABCD from orgin (0, 0) and point (4, 4.5) [T ]
III
Step 4 : Evaluate scaling matrix [S]
[ ]( . , . )
S 0 75 0 75 = [T ] [T ] [T ]
I II III
´ ´
Step 5 : Find out the resultant co-ordinates after scaling.
Step 1 : Translate ABCD from (4, 4.5) to (0, 0)
Here, DxI = –4 and DyI = – 4.5
[ ]
TI in homogeneous form,
[TI ] =
1 0 0
0 1 0
0
D D
x y
I I
é
ë
ê
ê
ê
ù
û
ú
ú
ú - -
é
ë
ê
ê
ê
ù
û
ú
ú
ú
1 0 0
0 1 0
4.5 4.5 1
Step 2 : Scaling to
3
4
of its size at origin (0,0)
Here, Sx = 0.75; Sy = 0.75
[S] in homogeneous form,
[TII ] =
S
S
0 0
0.75 0 0
0 0.75 0
0 0 1
x
y
0 0
0 0
1
é
ë
ê
ê
ê
ù
û
ú
ú
ú
=
é
ë
ê
ê
ê
ù
û
ú
ú
ú
Step 3 : Translation of ABCD to (4, 4.5) from (0, 0) [T]III
Here Dx = 4 ; Dy = 4.5
1 - 49 Computer Aided Design and Manufacturing
Introduction

62. [T]III =
1 0 0
0 1 0
1
D D
x y
é
ë
ê
ê
ê
ù
û
ú
ú
ú
=
é
ë
ê
ê
ê
ù
û
ú
ú
ú
1 0 0
0 1 0
4 4.5 1
Final scaling matrix [S] 3
4
,
3
4
æ
è
ç
ö
ø
÷
[S] = [T ] T [T ]
I II III
´ ´
[ ]
=
1 0 0
0 1 0
–4 –4.5 1
0.75 0 0
–0.5 0.75 0
0 0 1
é
ë
ê
ê
ê
ù
û
ú
ú
ú
´
é
ë
ê
ê
ê
ù
û
ú
ú
ú
´
é
ë
ê
ê
ê
ù
û
ú
ú
ú
1 0 0
0 1 0
4 4.5 1
Þ [
,
S] 3
4
3
4
æ
è
ç
ö
ø
÷
=
0.75 0 0
0 0.75 0
1 1.125 1
é
ë
ê
ê
ê
ù
û
ú
ú
ú
Resultant Co-ordinates
Given, A (x , y )
1 1 = (2, 3)
B (x , y )
2 2 = (6, 3)
C (x , y )
3 3 = (6, 6)
D (x , y )
4 4 = (2, 6)
ABCD in homogeneous form,
A
B
C
D
é
ë
ê
ê
ê
ê
ù
û
ú
ú
ú
ú
=
x y
x y
x y
x y
1 1
2 2
3 3
4 4
1
1
1
1
é
ë
ê
ê
ê
ê
ù
û
ú
ú
ú
ú
=
é
ë
ê
ê
ê
2 3 1
6 3 1
6 6 1
2 6 1
ê
ù
û
ú
ú
ú
ú
Resultant Co-ordinates
¢
¢
¢
¢
é
ë
ê
ê
ê
ê
ù
û
ú
ú
ú
ú
A
B
C
D
=
A
B
C
D
[S
ì
í
ï
ï
î
ï
ï
ü
ý
ï
ï
þ
ï
ï
´ ]( / , / )
3 4 3 4
Þ
¢
¢
¢
¢
é
ë
ê
ê
ê
ê
ù
û
ú
ú
ú
ú
A
B
C
D
=
2 3 1
6 3 1
6 6 1
2 6 1
075 0 0
0 075 0
1 1125 1
é
ë
ê
ê
ê
ê
ù
û
ú
ú
ú
ú
´
é
ë
ê
ê
ê
ù
.
.
. û
ú
ú
ú
=
2.5 3.375 1
5.5 3.375 1
5.5 5.625 1
2.5 5.625 1
é
ë
ê
ê
ê
ê
ù
û
ú
ú
ú
ú
1 - 50 Computer Aided Design and Manufacturing
Introduction

63. ¢
A = (2.5, 3.375)
¢
B = (5.5, 3.375)
¢
C = (5.5, 5.625)
¢
D = (2.5, 5.625)
1.10 3D Transformations
· It is often necessary to display objects in
3-D on the graphics screen.
· The transformation matrices developed
for 2-dimensions can be extended to 3-D.
· Fig. 1.10.1 represents 3D translation of a
donut.
3D Translation
· 3D translation matrix is explained by,
[T] =
1 0 0 0
0 1 0 0
0 0 1 0
1
D D D
x y z
é
ë
ê
ê
ê
ê
ù
û
ú
ú
ú
ú
· D D
x y
, and Dy explains distance of translation along x, y and z direction
respectively.
1 - 51 Computer Aided Design and Manufacturing
Introduction
Y
X
D(2, 6)
D' (2.5, 5.625) C' (5.5, 5.625)
A' (2.5, 3.375) B' (5.5, 3.375)
C(6, 6)
A(2, 3) B(6, 3)
P (4, 4.5)
Fig. 1.9.26
Fig. 1.10.1.3D Translation of a donut

64. 3D Scaling
· 3D scaling matrix is explained by,
[ ]
Ts =
S
S
S
x
y
y
0 0 0
0 0 0
0 0 0
0 0 0 1
é
ë
ê
ê
ê
ê
ù
û
ú
ú
ú
ú
· S ,S and S
x y z represents the scaling factors along x, y and z direction
respectively.
3D Rotation
· 3D rotation matrices are given by,
i) Rotation along Z-axis by angle 'q'
[ ]
Rz =
cos
cos
1
q q
q q
sin
– sin
0 0
0 0
0 0 0
0 0 0 1
é
ë
ê
ê
ê
ê
ù
û
ú
ú
ú
ú
ii) Rotation along X-axis by angle 'f'
[ ]
Rx =
1
cos
cos
0 0 0
0 0
0 0
0 0 0 1
f f
f f
– sin
sin
é
ë
ê
ê
ê
ê
ù
û
ú
ú
ú
ú
iii) Rotation along X-axis by angle 'f'
[ ]
Ry =
cos
cos
f f
f f
0 0
0 1 0 0
0 0
0 0 0 1
sin
sin
é
ë
ê
ê
ê
ê
ù
û
ú
ú
ú
ú
1.11 Line Drawing + [AU : Dec.-16, May-18]
· Straight line segments are used a great deal in computer generated pictures.
· The following criteria have been stipulated for line drawing displays :
i) Lines should appear straight
ii) Lines should terminate accurately
iii) Lines should have constant density
iv) Line density should be independent of length and angle
v) Line should be drawn rapidly
1 - 52 Computer Aided Design and Manufacturing
Introduction

65. · The process of turning on the pixels for a line segment is called vector
generation. If the end points of the line segment are known, there are several
schemes for selecting the pixels between the end pixels. One method of
generating a line segment is a symmetrical Digital Differential Analyzer (DDA).
1.11.1 DDA Algorithm
· The digital differential analyzer algorithm generates lines from their differential
equations.
· The DDA works on the principle that X and Y are simultaneously incremented
by small steps proportional to the first derivatives of X and Y.
· In the case of a straight line the first derivatives are constant and are proportional
to dX and dY, where 'd' is a small quantity.
· In the real world of limited precision displays, addressable pixels only must be
generated. This can be done by rounding to the next integer after each
incremental step.
· After rounding, a pixel is displayed at the resultant X and Y locations. An
alternative to rounding is the use of arithmetic overflow. X and Y are kept in
registers that have integer and fractional parts.
· The incrementing values which are less than unity are repeatedly added to the
fractional part and whenever the result overflows the corresponding integer part is
incremented. The integer parts of X and Y are used to plot the line.
· This would normally have the effect of truncating. The DDA is therefore
initialized by adding 0.5 in each of the fractional parts to achieve true rounding.
· The symmetrical DDA generates reasonably accurate lines since a displayed pixel
is never away from a true line by half the pixel unit.
Procedure for line drawing using DDA algorithm
Consider a line segment with coordinates (x , y )and(x , y )
1 1 2 2 with slope 'm' as
shown in the Fig. 1.11.1.
Step 1 : Identify (x , y )and(x , y )
1 1 2 2
Step 2 : Calculate number of steps.
If D D
x y
> , No. of steps = Dy
Else If D D
x y
> , No. of steps = Dx
Step 3 : Find the slope 'm'
m =
D
D
y
x
2 1
2 1
y y
x x
=
( – )
( – )
· If m £ 1, Assume Dx= 1
1 - 53 Computer Aided Design and Manufacturing
Introduction
(x , y )
1 1
(x , y )
2 2
Y
= mx+c
X
Y
Fig. 1.11.1 Line segment

66. ¡ Then, Dy = m x
D = ( – )
y y m(x – x )
2 1 2 1
=
¡ This could be written as,
y y
i 1 i
+ - = m(x x ), where(i 1, 2, 3, )
i+1 i
– = ¼
and since, Dx = 1, y y
i+1 i
– = m
· Therefore, If m £ 1
x x 1
i 1 i
+ - - =
y y m
i 1 i
+ - =
· From this it could be observed that x value will be incremented by 1 and y value
will be incremented by slope 'm'.
· If m >1, Assume Dy = 1
¡ Then, m =
D
D
y
x
Dx =
Dy
m
( )
x x
2 1
- =
1
m
y – y
2 1
( )
¡ This could be written as,
( – )
x x
i+1 i =
1
m
y y
i+1 i
( – ), where (i = 1, 2, 3, …)
· Therefore, If m > 1
x x
i+1 i
– - = 1/m
y y
i+1 i
– = 1
· From this it could be observed that x value will be incremented by 1/m and y
value will be incremented by 1.
Step 4 : Find x and y increment values
xincrement =
Dx
No. of steps
y increment =
Dy
No. of steps
Note :
1. If m £ 1, increment 'x' by 1 and increment
'y' by 'y-increment' value and round off to
nearest value.
2. If m > 1, increment 'y' by 1 and increment
'x' by 'x-increment value and round off to
nearest value.
Step 5 : Plot the x and y points in raster scan
display as shown in Fig. 1.11.2.
1 - 54 Computer Aided Design and Manufacturing
Introduction
Fig. 1.11.2

67. · Fig. 1.11.3 represents flow chart for the DDA Algorithm.
1.11.2 Bresenham's Line Drawing Algorithm
· Bresenham's algorithm selects optimum raster locations with minimum computation.
· This algorithm always increments by one unit in either X or Y depending upon
the slope of the line.
· The increment in the other variable either zero or one is determined by
examining the distance (error) between the actual line location and the nearest
grid location. This distance is called decision variable or error.
· Only the sign of this error needs be examined.
· This is accurate and efficient method of raster generation algorithms to display
lines, circles, ellipse and other curves incorporating only incremental integer
calculation.
· Bresenham's line drawing algorithm rectifies the disadvantage of DDA algorithm.
1 - 55 Computer Aided Design and Manufacturing
Introduction
Set values for x , y , x and y
1 1 2 2
Calculate No. of steps x and y
Calculate slope 'm'
Yes
No
Evaluate and plot x and y points
Stop
Start
If m 1
y – y = 1
i+1 i
x – x = 1/m
i+1 i–
x – x = 1
i+1 i
–
y – y = m
i+1 i
Fig. 1.11.3 Flowchart for DDA algorithm

68. Procedure for Bresenham's Algorithm
Consider the line segment shown in Fig. 1.11.2.
Step 1 : Identify (x , y )and(x , y )
1 1 2 2
Step 2 : Calculate number of steps.
If D D
y x
> No. of steps = Dy
Else If D D
x y
> , No. of steps = Dx
Step 3 : Evaluate error or deviation 'e', e 2 y x
i = D D
–
Step 4 : If e 0, e
i i
³ +1 = e 2 y 2 y
i + D D
– , increment 'x' and 'y' by 1.
If e 0, e
i i
< +1 = e 2 y
i + D increment 'x' by 1.
Step 5 : Plot the x and y points in raster scan display.
Fig. 1.11.4 represents flowchart for the Bresenham's algorithm
1 - 56 Computer Aided Design and Manufacturing
Introduction
Set values for x , y , x and y
1 1 2 2
Calculate No. of steps x and y
Evaluate error 'e'
Yes
No
Evaluate and plot x and y points
Stop
Start
If e 0 e – e + 2 y – 2 x
i+1 i
e = + 2 y
i+1 ei
Fig. 1.11.4

69. 1.11.3 Solved Examples on DDA Algorithm and Bresenham's Algorithm
Example 1.11.1 : Explain rasterization by performing DDA algorithm for a line AB,
A(2, 3) and B(12, 8).
Solution :
Given, Line AB Þ A(x , y )
1 1 = (2, 3)
A(x , y )
2 2 = (12, 8)
Step 1 : Identity (x , y )
1 1 and (x , y )
2 2
x1 = 2; y1 = 3
x2 = 12; y 2 = 8
Step 2 : Calculate the number of steps
Dy = y y
2 1
- = 5
Dx = x x
2 1
- = 10
Q Dx > Dy Þ No. of steps = Dx = 10
Step 3 : Find the slope (m)
Slope, (m) =
D
D
y
x
=
5
10
= 0.5
m = 0.5 < 1
£
- =
- =
ü
ý
ï
þ
ï
+
+
when m 1
x x 1
y y m
'x'
i 1 i
i 1 i
willbe intremented by 1
' y' willbe intremented by m
Step 4 : Find the increment value
xincrement =
Dx
No. of. steps
=
10
10
= 1
y increment =
Dy
No. of. steps
=
5
10
= 0.5 = m
Sample solution
xi = 2; y i = 3
First increment :xi 1
+ = 2 + 1 = 3
y i 1
+ = 3 + 0.5 = 3.5
Second step : xi 2
+ = 3 + 1 = 4
y i 2
+ = 3.5 + 0.5 = 4
1 - 57 Computer Aided Design and Manufacturing
Introduction

70. Similarly Step 3 : xi 3
+ = 4 + 1 = 5
y i 3
+ = 4 + 0.5 = 4.5
Follow the iterations until 10 steps are completed as shown in table.
Step 5 : Identify initial and final values
No. of steps x initial yinitial x final
(Round off)
yfinal
(Round off)
0 2 3 2 3
1 3 3.5 3 4
2 4 4 4 4
3 5 4.5 5 5
4 6 5 6 5
5 7 5.5 7 6
6 8 6 8 6
7 9 6.5 9 7
8 10 7 10 7
9 11 7.5 11 8
10 12 8 12 8
Step 6 : Plot
1 - 58 Computer Aided Design and Manufacturing
Introduction
0 1 2 3 4 5 6 7 8 9 10 11 12
1
2
3
4
5
6
7
8
yvalues
xvalues
Step 6 : Plot
Fig. 1.11.5

71. Example 1.11.2 : Using DDA algorithm rasterize line AB, A(0, 0), B(4,6).
Solution :
Given, AB Þ A(0, 0) ; B(4, 6)
A(x , y )
1 1 ; B(x , y )
2 2
Step 1 : Identity (x , y )
1 1 and (x , y )
2 2
x1 = 0; y1 = 0
x2 = 4; y 2 = 6
Step 2 : Calculate the number of steps
Dy = y y
2 1
- = (6 – 0) = 6
Dx = x x
2 1
- = (4 – 0) = 4
Q Dy > Dx Þ No. of steps = Dy = 6
Step 3 : Find the slope (m)
Slope (m) =
D
D
y
x
=
6
4
= 1.5
Q m 1
x x
1
m
y y = 1
i 1 i
i 1 i
>
- =
-
ü
ý
ï
þ
+
+
ï
'x' willbe incremented by
1
m
and
' y' willbe intremented by ' '
1
Step 4 : Find the increment value :
xincrement =
Dx
No. of steps
=
4
6
= 0.66
Also, 0.66 =
1
m
=
1
1.5
y increment =
Dy
No. of steps
=
6
6
= 1
Initial values : xi = 0 ; y i = 0
1 - 59 Computer Aided Design and Manufacturing
Introduction

72. First step : xi 1
+ = 0 + 0.66 = 0.66; y i 1
+ = 0 + 1 = 1
Second step : xi 2
+ = 0.66 + 0.66 = 1.32; y i 1
+ = 1 + 1 = 2
Follow this until 6-steps are completed.
Step 5 : Find initial and final values of 'x' and 'y'
No. of steps x initial yinitial x final
(Round off)
yfinal
(Round off)
0 0 0 0 0
1 0.66 1 1 1
2 1.32 2 1 2
3 1.98 3 2 3
4 2.64 4 3 4
5 3.30 5 3 5
6 3.96 6 4 6
Step 6 : Plot
1 - 60 Computer Aided Design and Manufacturing
Introduction
0 1 2 3
1
2
3
4
5
6
4
Fig. 1.11.6

73. Bresenham's Line Algorithm
Example 1.11.3 : Rasterize line A,B, A(3,3) and B(11, 7) using Bresenham's line
drawing algorithm.
Solution :
Given, AB Þ A(x , y )
1 1 = (3, 3)
B(x , y )
2 2 = (11, 7)
Step 1 : Identity (x , y )
1 1 and (x , y )
2 2
x1 = 3; y1 = 3
x2 = 11; y 2 = 7
Step 2 : Calculate the number of steps
Dx = x x
2 1
- = 11 – 3 = 8
Dy = y y
2 1
- = 7 – 3 = 4
Here Dx > Dy; No. of steps = Dx = 8
Step 3 : Evaluate error and deviation.
Initial values of error and deviation,
ei = 2D D
y x
-
= 2 4 8
´ - = 0
ei = 0
Step 4 :
1) e1 = 0 = e0 (if ei ³ 0; ei 1
+ = ei + -
2 2
D D
y x)
e1 = e0 + -
2 2
D D
y x x = x + 1 = 3 + 1 = 4
Þ e1 = 0 + 8 – 16 = –8 y = y + 1 = 3 + 1 = 4
Since, e1 = – 8 (if ei < 0, ei + 1 = ei + 2Dy)
e2 = e7 + 2Dy x = x + 1 = 4 + 1 = 5
Þ = – 8 + 8 = 0 y = 4
Note : From procedure it is clear that,
· If ei ³ 0 ® Increment x and y by 1
1 - 61 Computer Aided Design and Manufacturing
Introduction

74. · If ei < 0 ® Increment x by 1 and keep 'y' as it is.
· The results could be tabulated as below.
No. of Steps x y Error
0 3 3 0
1 4 4 –8
2 5 4 0
3 6 5 –8
4 7 5 0
5 8 6 – 8
6 9 6 0
7 10 7 – 8
8 11 7 0
Step 5 : Plot the points
1.12 Clipping + [AU : Dec.-16, 17]
· Various projections of an object's geometry can be defining views.
· A view requires a view window.
1 - 62 Computer Aided Design and Manufacturing
Introduction
0 1 2 3 4 5 6 7 8 9 10 11
1
2
3
4
5
6
7
Fig. 1.11.7

75. · If any part of the geometry is not inside the window, it is made invisible by the
CAD software through a process known as clipping.
· Clipping is the process of determining the visible portions of a drawing lying
within a window.
· Clipping, in the context of computer graphics, is a method to selectively enable
or disable rendering operations within a defined region of interest.
· In clipping each graphic element of the display is examined to determine whether
or not it is completely inside the window, completely outside the window or
crosses a window boundary.
· Portions outside the boundary are not drawn. Any geometry lying wholly outside
the view boundary is not mapped to the screen, and any geometry lying partially
inside and partially outside is cut off at the boundary before being mapped.
· Typical clipping algorithm : Cohen-Sutherland clipping algorithm.
· If clipping is not done properly a CAD system will produce incorrect pictures
due to an overflow of internal coordinate registers. This effect is known as wrap
round.
Applications of Clipping :
· Extracting part of defined scene for viewing
· Identifying visible surfaces in 3D Views
· Anti-aliasing line segments or object boundaries
· Creating objects using solid modelling procedures
· Drawing and painting operations
Types of Clipping :
i. Point clipping ii. Line clipping iii. Area clipping (Polygon)
iv. Curve clipping v. Text clipping
i. Point Clipping
· The clip window is a rectangle
as shown in the Fig. 1.12.1.
· The point P = (x,y) will be
displayed if the following
inequalities are satisfied :
x x x
wmin wmax
£ <
y y y
wmin wmax
£ <
1 - 63 Computer Aided Design and Manufacturing
Introduction
ywmax
ywmin
xwmin xwmax
P (x, y)
Fig. 1.12.1 Point clipping

76. (x , x )
wmin wmax and (y , y )
wmin wmax are the edges of the clip window.
· If any one of these four inequalities is not satisfied, the point is clipped.
ii. Line Clipping
· Line that do not intersect the clipping window are either completely inside the
window or completely outside the window.
· In the case of line clipping, four different cases are possible.
Different Cases for line Clipping
· Case1 : Both endpoints of the line lie within the clipping area as shown in
Fig. 1.12.2. This means that the line is included completely in the clipping area,
so that the whole line must be drawn.
· Case 2 : One end point of the line lies within the other outside the clipping area
as shown in Fig. 1.12.3. It is necessary to determine the intersection point of the
line with the bounding rectangle of the clipping area. Only a part of the line
should be drawn.
1 - 64 Computer Aided Design and Manufacturing
Introduction
A
B
A
B
Clip
rectangle
Fig. 1.12.2 Line clipping case 1
A
B
A
B
Clip
rectangle
C
D'
D
C
D'
Fig. 1.12.3 Line clipping case 2

77. · Case 3 : Both end points are
located outside the clipping area
and the line do not intersect the
clipping area as shown in
Fig. 1.12.4. In the case, the line
lies completely outside the clipping
area and can be neglected for the
scene.
· Case 4 : Both endpoints are
located outside the clipping area
and the line intersect the clipping
area as shown in Fig. 1.12.5. The two intersection points of the line with the
clipping are must be determined. Only the part of the line between these two
intersection points should be drawn.
Cohen-Sutherland Line Clipping Algorithm
· This algorithm divides a 2D space into 9 parts, of which only the middle part is
visible.
· Cohen-Sutherland subdivision line clipping algorithm was developed by Dan
Cohen and lvan Sutherland.
· This method is used to :
· To save a line segment.
· To discard line segment or,
· To divide the line according to window co-ordinate.
· Cohen Sutherland performs line clipping in two phases :
· Phase 1 : Find visibility of line.
· Phase 2 : Clip the line falling in category 3 (candidate for clipping).
1 - 65 Computer Aided Design and Manufacturing
Introduction
F
E
Clip
rectangle
Fig. 1.12.4 Line Clipping Case 3
Clip
rectangle
C
D'
D
Fig. 1.12.5 Line Clipping Case 4

78. · Every line end point in a
picture is assigned a four-bit
binary code (TBRL), called a
region code, that identify the
location of the point relative
to the boundaries of the
clipping rectangle.
· Region are set up as top (T),
bottom (B), right (R) and left
(L) as shown in Fig. 1.12.6.
· Every bit position in the
region code is used to
indicate one of the four
relative coordinate position of
the point with respect to the
clip window to the left, right,
top and bottom.
· This rule is also called TBRL code (Top-Bottom-Right-Left).
· Cohen Sutherland clipping could be explained by the Fig. 1.12.7.
· Following rules are used for clipping by Cohen-Sutherland line clipping
algorithm :
· Visible : Any lines that are completely contained within the window boundaries
have a region code of '0000' for both endpoints, and we trivially accept these
lines. For example, Line segment P1P2 is visible in the figure.
1 - 66 Computer Aided Design and Manufacturing
Introduction
0010
1010
0110
1000
0100
Left Right
xwmin xwmax
Top
Bottom
Window
0000
1001
0001
0101
ywmin
ywmax
y
x
Fig. 1.12.6 Bit code (TBRL code) for
cohen- sutherland clipping
0010
1010
0110
1000
0100
Left Right
xwmin xwmax
Top
Bottom
Window
0000
1001
0001
0101
ywmin
ywmax
y
x
P8
P7
P3
P1
P2
P4
P5
P6
Fig. 1.12.7 Cohen, Sutherland line clipping technique

79. · Invisible : Any lines that have a 1 in the same bit position in the region code for
each endpoint are completely outside the clipping rectangle and the line segment
is invisible, and these lines will be trivially rejected. The line that has a region
code of '0001' will be discarded. For one endpoint and a code of '0101' for the
other endpoint. Both end points of the line are at left of the clipping rectangle, as
indicated by the '1' in the first position of each region code.
· Clipping candidate or indeterminate : A line segment is said to be indeterminate
if the bitwise logical AND of the region codes of the end points is equal to
'0000'.
· For example : Line segment P3P4 having endpoint codes '0100' and '0010' and
P7P8 having endpoints codes '0010' and '1000' in the figure. These line segments
may or may not process the window boundaries as line segment P7P8 is invisible
but line segment P3P4 is partially visible and must be clipped.
iii. Polygon Clipping
· The simplest curve is a line segment or simply a line.
· A sequence of line where the following line starts where the previous one ends is
called a polyline.
· If the last line segment of the polyline ends where the first line segment started,
the polyline is called a polygon.
· A polygon is defined by 'n' number of sides in the polygon.
· We can divide polygon into two classes.
a) Convex polygon
b) Concave polygon
· A convex polygon as shown in Fig. 1.12.8 (a), is
a polygon such that for any two points inside
the polygon, all the point of the line segment
connecting them are also inside the polygon. A
triangle is always a convex one.
· A polygon is said to be a concave, if the line
joining any two interior points of the polygon
does not lie completely inside the polygon, as
shown in Fig. 1.12.8 (b)
· There are four possible cases when processes
vertices in sequence around the parameter of the
polygon. As each pair of adjacent polygon
vertices is passed to a window boundary clipper,
the following test could be made,
· Case 1 : If the first vertex is outside the window boundary and second vertex is
inside the window boundary then both the intersection point of a polygon edge
with the window boundary and second vertex are added to the output vertex list.
1 - 67 Computer Aided Design and Manufacturing
Introduction
P
Q
Fig. 1.12.8 (a) Convex polygon
Fig. 1.12.8 (b) Concave polygon

80. · Case 2 : If both input vertices are inside the window boundary. Only the
second vertex is added to the vertex list.
· Case 3 : If the first vertex is inside the window boundary and the second vertex
is outside the window boundary then only the edge intersection with the
window boundary is added to the output vertex list.
· Case 4 : If both the input vertices are outside the window boundary then
nothing is save to the output list.
· Fig. 1.12.9 explains an example for polygon clipping.
iv. Curve Clipping
· The boundary rectangle for a circle or other curved object can be used first to
test for overlap with a rectangular boundary 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.
· Fig. 1.12.10 explains curve
clipping.
1 - 68 Computer Aided Design and Manufacturing
Introduction
1
2
3
4
5 6 7 8
9
10
11 12
13
(a) Clip polygon
(b) Clipped polygon
2
3
4
5
6 7
8
x
Fig. 1.12.9 Example for polygon clipping
Fig. 1.12.10 Curve clipping

81. v. Text Clipping
Text clipping can be of two types :
a) All or none string-clipping : In this
clipping as shown in figure, if all the
string is inside a clip window, it will
be kept. Otherwise, the string is
discarded.
b) All or none character-clipping : In this clipping as shown in figure, only those
characters that are not completely inside the window will be discarded.
1.13 Viewing Transformation + [AU : Dec.-16, May-18]
· The process that converts object coordinates in WCS to normalized device
coordinates is called window-to-view port mapping or normalization
transformation.
· The process that maps normalized device coordinates to discrete device / image
coordinates is called workstation transformation, which is essentially a second
window-to-view port mapping, with a workstation window in the normalized
device coordinate system and a workstation view port in the device coordinate
system.
· Collectively, these two coordinate-mapping operations are referred to as viewing
transformation.
· The step by step procedure for viewing transformation is shown in Fig. 1.13.1.
1 - 69 Computer Aided Design and Manufacturing
Introduction
STRING 1
STRING 2
Widow
Fig. 1.12.11 All or none string-clipping
STRING 1
STRING
2
Widow
ING 1
STRING 4
Widow
STRING 1 STRING 4 TRING 1
STR
Fig. 1.12.12 All or none character-clipping
2D Object
data
Window
transformation
Clipping
Computer
display
Frame
buffer
Scan
conversion
Visaport
transformation
Object
transformation
a b c
d
f e
Fig. 1.13.1 Viewing transformation

82. · A window is specified by four world coordinates, xwmin , xwmax, y wmin , and
y wmax as shown in Fig. 1.13.2 (a).
· Similarly, a view port is described by four normalized device coordinates :
xvmin , xvmax, y vmin , and y vmax as shown in Fig. 1.13.2 (b).
· The objective of window-to-view port mapping is to convert the world
coordinates (xw , y w ) of an arbitrary point to its corresponding normalized
device coordinates (xv , y v ).
· In order to maintain the same relative placement of the point in the view port as
in the window, we require :
x – x
x – x
w w min
w max w min
=
x – x
x – x
v v min
v max v min
and
y – y
y – y
w w min
w max w min
=
y – y
y – y
v v min
v max v min
· Solving the above two equations we get,
x = x +(x – x )S
y = y +(y – y )S
v v min w w min x
v v min w w min y
ì
í
î
· Sx and Sy are scaling factors,
Sx =
x – x
x – x
v max v min
w max w min
Sy =
y – y
y – y
v max v min
w max w min
· Since the eight coordinate values that define the window the view port are just
constants, these two formulas for computing (xv , y v ) from (xw , y w ) can be
expressed in terms of a translate-scale-translate transformation 'N',
x
y
1
v
v
æ
è
ç
ç
ç
ö
ø
÷
÷
÷
= N
x
y
1
w
w
×
æ
è
ç
ç
ç
ö
ø
÷
÷
÷
1 - 70 Computer Aided Design and Manufacturing
Introduction
ywmax
ywmin
xwmin xwmax
(x , y )
w w
Window
Fig. 1.13.2 (a) Window
yvmax
yvmin
xvmin xvmax
(x , y )
v v
View port
Fig. 1.13.2 (b) View Port

83. N =
1 0 x
0 1 y
0 0 1
S 0 0
0 S 0
0 0 1
v min
v min
x
y
æ
è
ç
ç
ç
ö
ø
÷
÷
÷
×
æ
è
ç
ç
ç
ö
ø
÷
÷
÷
×
1 0 – x
0 1 – y
0 0 1
w min
w min
æ
è
ç
ç
ç
ö
ø
÷
÷
÷
· An example for viewing transformation is shown in Fig. 1.13.3.
1.14 Brief Introduction to CAD and CAM
· CAD and CAM are important tools in designing and manufacturing.
· Before the advent of computers and especially PC in the eighties, draftsmen
performed an important role in designing in companies.
· Today hand drafting for designing has become outdated and the days of
compasses and protractors are virtually over.
· CAD and CAM are important terms in the field of design and manufacture and
refer to Computer Aided Design and Computer Aided Manufacture respectively.
1.14.1 Computer Aided Design (CAD)
· CAD is the intersection of computer graphics, geometric modeling, and design
tools as explained in Fig. 1.14.1. (Refer Fig. on next page)
· CAD is defined as any design activity that involves the effective use of a
computer to create, modify analyze, optimize and document an engineering
design.
· CAD software for design uses either vector-based graphics to explain the objects
of traditional drafting or may also develop raster graphics showing the overall
look of designed objects.
· During the manual drafting of engineering drawings, the output of CAD must
convey information, like dimensions, materials, processes, and tolerances.
· CAD is a significant industrial art used for many purposes, including industrial
and architectural design, shipbuilding, automotive, and aerospace industries etc.
1 - 71 Computer Aided Design and Manufacturing
Introduction
(x , y )
w w
Window
ywmax
ywmin
xwmin xwmax
ywmax
ywmin
xvmin xvmax
0
0
Fig. 1.13.3 Viewing transformation

84. · CAD software packages
provide the designer with a
multi window environment
with animation.
· The animations using wire
frame modeling helps the
designer to see into the
interior of the object and to
observe the behaviors of
the inner components of the
assembly during the motion.
Typical Components CAD/CAM System
CAD hardware :
CAD hardware consists of a central processing unit, storage devices, one or more
graphic display terminals and other input and output devices as shown in Fig. 1.14.2.
CAD hardware components :
· Central Processing Unit (CPU)
· Memory
· Hard disk, Floppy disk, CD-ROM
· External storage devices
· The monitor
1 - 72 Computer Aided Design and Manufacturing
Introduction
Computer
graphics
Design
engineering
Geometric
modelling
CAD
Fig. 1.14.1
Computing machine
Hardware Software
Graphics device
Display processing unit
Display device
Input device
Output device
CAD/CAM
system
Fig. 1.14.2 Components of a CAD/CAM system

85. · Printers and Plotters
· Digitizer, Puck and Mouse
CAD software :
· CAD software allows the designer to create and manipulate a shape interactively
and store it. CAD software consists of (a) System software and (b) Application
software.
a) System software : System software control the operations of a computer. It is
responsible for making the hardware components to work and interact with
each other and the end user. Example : Operating system, compiler, interpreter
etc.
b) Application software : Application software or application programs are used
for general or customized CAD problems. Example : Auto CAD, CATIA,
CREO, Solidworks, ADAMS, ANSYS, ABAQUS, NASTRAN etc.
· A CAD software or program contains hundreds of functions that enable you to
accomplish specific drawing tasks.
· The drawing tasks may involve drawing an object, editing an existing drawing,
displaying a view of the drawing, printing or saving it, or controlling any other
operation of the computer.
· The functions are organized into modules that provide easy access to all the
commands.
· The CAD modules are :
i. Draw : The draw module enables to draw lines, arcs, circles, ellipses, text,
dimensions, symbols, borders and many other drawing components.
ii. Edit : The edit module lets to change existing drawing elements and
manipulate them in a number of ways such as move, copy or erase drawing
components.
iii. Data output : The data output module enables to display drawings on the
screen and then print them on paper. There are two separate sets of functions
that help accomplish this :
a) View-display functions
b) Print/plot functions
iv. System control : The system control module enables to control how the CAD
software works such as setting default units, dimension style, precision, line
type, colour etc.
v. Data storage and management : The data storage and management module
enable the storage you can store drawings as files on the hard disk.
vi. Special features : CAD provides certain special features which makes working
with CAD easier, such as rendering, animation, spreadsheets creation etc.
1 - 73 Computer Aided Design and Manufacturing
Introduction

86. Basic Elements of a CAD System
The basic elements of a CAD system are
shown in Fig. 1.14.3.
a) Geometric Modeling :
· Geometric modeling is a branch of
applied mathematics and computational
geometry that studies methods and
algorithms for the mathematical
description of shapes.
· The shapes studied in geometric
modeling are mostly two or
three-dimensional, although many of
its tools and principles can be applied
to sets of any finite dimension.
· Today most geometric modeling is done with computers and for computer-based
applications.
· Two-dimensional models are important in computer typography and technical
drawing. Three-dimensional models are central to computer aided design and
manufacturing (CAD/CAM).
· Basic geometric modeling techniques are,
· Wireframe modeling
· Surface modeling
· Solid modeling
b) Engineering analysis
· Checking the designed object for its functionality is called as engineering
analysis.
· In almost all the engineering design related projects some or the other analysis is
required. It can be stress-strain calculations, heat transfer measurements, or using
differential equations to find the dynamic behavior of the system, which is being
designed.
· One of the most commonly used and powerful feature of the CAD software to
carry out engineering analysis is Finite Element Analysis.
· To carry out the analysis of object by using FEA, the object is divided into finite
number of small elements of shapes like rectangular or triangular. These objects
form the interconnected network of the concentrated nodes.
· To carry out the analysis of whole object each and every node of the network is
analyzed and their results are synthesized to get the complete analysis of the
object. Each and every node can be analyzed for various properties like
1 - 74 Computer Aided Design and Manufacturing
Introduction
Geometric modeling
Engineering analysis
Design review and evaluation
Automated drafting
Fig. 1.14.3 Basic elements of a CAD system

87. stress-strain, heat transfer or any other characteristics depending upon the type of
application. The interrelating behavior of all the nodes gives the behavior of the
whole object.
· The CAD software has the option of defining the nodes and network structure as
per the designer's requirement.
· The output of the finite element analysis can be observed through the graphical
user interface. If the user finds that the output results are undesirable, they can
change the shape and dimensions of the object and carry out FEA again.
· Some of the common FEA packages in CAD for carrying out engineering
analysis are ANSYS, ABAQUS, NASTRAN etc.
· CAD also has provision for kinematic analysis of the design. ADAMS is the
commonly used CAD package for kinematic analysis.
c) Design Review and Evaluation
· Review and evaluation is checking whether the designed part has been designed
properly or not and if they will fail in practical situations.
· CAD software makes the process of design review and evaluation has become
much faster and convenient.
· Design review and evaluation features offered by CAD software are zoom in and
out, layering, checking interference, animation capability.
d) Automated drafting
· Drafting is the process of making the drawings of the designed parts.
· After designing of the object its assembly and detail parts drawings have to be
made which includes specifications of various materials also called as bill of
materials used for the manufacturing the components of the object.
· Automated drafting is one of the most important applications of the CAD
software.
Reasons for Implementing CAD or Advantages or Benefits of CAD
i. Increase in the productivity of the designer : The CAD software helps designer
in visualizing the final product that is to be made, it subassemblies and the
constituent parts. The product can also be given animation and see how the actual
product will work, thus helping the designer to immediately make the modifications
if required. CAD software helps designer in synthesizing, analyzing, and
documenting the design. All these factors help in drastically improving the
productivity of the designer that translates into fast designing, lower designing cost
and shorter project completion times.
ii. Improve the quality of the design : With the CAD software the designing
professionals are offered large number of tools that help in carrying out thorough
engineering analysis of the proposed design. The tools also help designers to
consider large number of investigations. Since the CAD systems offer greater
1 - 75 Computer Aided Design and Manufacturing
Introduction

88. accuracy, the errors are reduced drastically in the designed product leading to better
design. Eventually, better design helps carrying out manufacturing faster and
reducing the wastages that could have occurred because of the faulty design.
iii.Flexibility in design : It is easy to change and alter the design in CAD as per the
user's requirement. For example, a building plan might contain separate overlays for
its structural, electrical, and plumbing components. In CAD layers can be used for
this purpose. The user could display, edit, and print layers separately or in
combination. Also, the layers could be named to track content, and layers could be
locked so that they can't be altered. Assigning settings such as color, linetype, or
lineweight to layers helps to comply with industry standards. Assigning a plot style
to a layer makes all the objects drawn on that layer plot in a similar manner.
iv.Improved design analysis : CAD helps in performing advanced engineering
analysis of the design with the aid of CAD packages such as ANSYS, ABAQUS
for FEA and ADAMS for kinematic analysis.
v. Better communications : The next important part after designing is making the
drawings. With CAD software better and standardized drawings can be made easily.
The CAD software helps in better documentation of the design, fewer drawing
errors, and greater legibility.
vi.Creating documentation of the designing : Creating the documentation of
designing is one of the most important parts of designing and this can be made
very conveniently by the CAD software. The documentation of designing includes
geometries and dimensions of the product, its subassemblies and its components,
material specifications for the components, bill of materials for the components etc.
vii. Creating the database for manufacturing : When the creating the data for the
documentation of the designing most of the data for manufacturing is also created
like products and component drawings, material required for the components, their
dimensions, shape etc.
viii. Saving of design data and drawings for future reference : All the data used for
designing can easily be saved and used for the future reference, thus certain
components don't have to be designed again and again. Similarly, the drawings can
also be saved and any number of copies can be printed whenever required. Some of
the component drawings can be standardized and be used whenever required in any
future drawings.
ix. Better design accuracy
x. Better visualization of drawing : CAD has provision for rendering and 3D
visualization. Rendering helps in the visualization of design in required
environment.
1 - 76 Computer Aided Design and Manufacturing
Introduction

89. 1.15 Computer Aided Manufacturing (CAM)
· Computer-Aided Manufacturing (CAM) is an application technology that uses
computer software and machinery to facilitate and automate manufacturing
processes.
· CAM reduces waste and energy for enhanced manufacturing and production
efficiency via increased production speeds, raw material consistency and tool
accuracy.
· CAM uses computer-driven manufacturing processes for additional automation of
management, material tracking, planning and transportation. CAM also
implements advanced productivity tools like simulation and optimization to
leverage professional skills.
· CAM is often linked with CAD for more enhanced and streamlined
manufacturing, efficient design and superior machinery automation.
1.15.1 CAD-CAM Integration
· CAD/CAM integration allows the transfer of information from the design stage
into the planning stage of the manufacture of a product.
· The design is stored CAD database and is further processed by CAM into the
necessary data and instructions for operating and controlling the production
machinery, material handling equipment and automated testing and inspection.
· Fig. 1.15.1 explains basic elements of the CAD/CAM integration.
1 - 77 Computer Aided Design and Manufacturing
Introduction
Material
selection
Determination
of dimensions
Part
drawing
Part
arrangement
Conceptual
design
Thickness
determination
Simulation
Material
handling Jigs
and
fixtures
Automatic
assembly
Automatic
cutting
Assembly
method
Database
CAD
CAM
Fig. 1.15.1 CAD/ CAM integration

90. 1.15.2 Manufacturing Planning
The information processing activities in manufacturing planning includes,
i. Process planning
ii. Master scheduling
iii.Materials requirement planning
iv.Capacity planning
i. Process Planning
· Purpose of process planning is to translate design requirement to manufacturing
process detail.
· Process planning acts as a bridge between design and manufacturing.
· The process starts with selection of raw material and ends with the completion of
the part.
· Process planning could be achieved by means of manual as well as by means of
computer aided approach
· Computer Aided Process Planning (CAPP) : Two basic approaches of CAPP
are retrieval CAPP approach and generative CAPP approach.
ii. Master Scheduling
· Scheduling is the process of arranging, controlling and optimizing work and
workloads in a production process or manufacturing process.
· Scheduling is used to allocate plant and machinery resources, plan human
resources, plan production processes and purchase materials.
· The master production schedule is a detailed plan of production. It drives the
MRP system by referencing inventory, requirements and bill of materials.
· For the purpose of materials requirements planning, the time periods must be
identical with those used in MRP system.
· Master production schedule represents the plan for manufacturing products. It
consists quantities, dates and configurations. Typical MPS is a table containing
the following information :
· Demand forecast
· Allocated, reserved and unplanned slots
· Planned order - Planned and firm
· Projected Available Balance (PAB)
· Available To Promise (ATP)
1 - 78 Computer Aided Design and Manufacturing
Introduction

91. iii. Materials Requirement Planning
· MRP is a production planning, scheduling, and inventory control system used to
manage manufacturing processes. Most MRP systems are software-based, but it is
possible to conduct MRP by hand as well.
· An MRP system is intended to simultaneously meet three objectives :
· Ensure materials are available for production and products are available for
delivery to customers.
· Maintain the lowest possible material and product levels in store
· Plan manufacturing activities, delivery schedules and purchasing activities.
iv. Capacity Planning
· Capacity planning is the process of determining the production capacity needed
by an organization to meet changing demands for its products.
· In the context of capacity planning, design capacity is the maximum amount of
work that an organization is capable of completing in a given period.
· Effective capacity is the maximum amount of work that an organization is
capable of completing in a given period due to constraints such as quality
problems, delays, material handling, etc.
1.15.3 Manufacturing Control
· Manufacturing control is concerned with managing and controlling the physical
operations in the factory to implement the manufacturing plans.
· Factory operations included in the manufacturing control are,
(i) Shop floor control
(ii) Inventory control
(iii)Quality control
(i) Shop Floor Control
· Shop Floor Control is a system of methods and tools that are used to track,
schedule and report on the progress of work in a manufacturing plant.
· Shop Floor Control systems generally evaluate the portion of an order or
operation that has been completed.
· That percentage of work in process is useful for resource planning, inventory
evaluations and supervisor and operator productivity on a shop floor.
(ii) Inventory Control
· Inventory control, also known as stock control, involves regulating and
maximizing company's inventory.
· The goal of inventory control is to maximize profits with minimum inventory
investment, without impacting customer satisfaction levels.
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92. · Inventory control is an important aspect for the growth of company.
· Stores inventory is the heart of an industry.
· Inventory control or stock control can be broadly defined as "the activity of
checking a shop's stock".
· Major field of application of inventory control :
· In operations management, logistics and supply chain management, the
technological system and the programmed software necessary for managing
inventory.
· In economics and operations management, the inventory control problem, which
aims to reduce overhead cost without hurting sales. It answers the 3 basic
questions of any supply chain : When ? Where ? How much ?
· In the field of loss prevention, systems designed to introduce technical barriers
to shoplifting.
(iii) Quality Control
· Quality control is a process by which entities review the quality of all factors
involved in production.
· Visual inspection is a major component of quality control, where a physical
product is examined visually and the inspectors will provide list of unacceptable
products with defects such as cracks or surface blemishes.
· ISO 9000 defines quality control as "A part of quality management focused on
fulfilling quality requirements". This approach places an emphasis on three
aspects,
i. Elements such as controls, job management, defined and well managed
processes, performance and integrity criteria, and identification of records.
ii. Competence, such as knowledge, skills, experience, and qualifications.
iii. Soft elements, such as personnel, integrity, confidence, organizational culture,
motivation, team spirit, and quality relationships.
1.16 Types of Production Systems + [AU : May-17]
The production system of an organization is that part, which produces products of an
organization. It is that activity whereby resources, flowing within a defined system are
combined and transformed in a controlled manner to add value in accordance with the
policies communicated by management.
The production system has the following characteristics :
· Production is an organized activity, so every production system has an objective.
· The system transforms the various inputs to useful outputs.
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93. · It does not operate in isolation from the other organization system.
· There exists a feedback about the activities, which is essential to control and
improve system performance.
Classification of Production System
Fig. 1.16.1 represents the classification
of production system.
(i) Job Production
· Under this method peculiar,
special or non-standardized
products are produced in
accordance with the orders
received from the customers.
· As each product is non-
standardized varying in size and
nature, it requires separate job for
production.
· The machines and equipment's are
adjusted in such a manner so as to suit the requirements of a particular job.
· Job production involves intermittent process as the work is carried as and when
the order is received.
· It consists of bringing together of material, parts and components in order to
assemble and commission a single piece of equipment or product.
· Ship building, dam construction, bridge building, book printing are some of the
examples of job production. Third method of plant layout viz., stationery material
layout is suitable for job production.
Characteristics
The job production possesses the following characteristics :
· A large number of general purpose machines are required.
· A large number of workers conversant with different jobs will have to be
employed.
· There can be some variations in production.
· Some flexibility in financing is required because of variations in work load.
· A large inventory of materials, parts and tools will be required.
· The machines and equipment setting will have to be adjusted and readjusted to
the manufacturing requirements.
· The movement of materials through the process is intermittent.
1 - 81 Computer Aided Design and Manufacturing
Introduction
Production/Operations
volume
Mass production
Batch production
Job-shop
production
Output/Product variety
Fig. 1.16.1 Classification of production system

94. Limitations
Job production has the following limitations :
· The economies of large scale production may not be attained because production
is done in short-runs.
· The demand is irregular for some products.
· The use of labour and equipment may be inefficient.
· The scientific assessment of costs is difficult.
(ii) Batch Production
· Batch production pertains to repetitive production.
· It refers to the production of goods, the quantity of which is known in advance.
· It is that form of production where identical products are produced in batches on
the basis of demand of customers' or of expected demand for products.
· This method is generally similar to job production except the quantity of
production.
· Instead of making one single product as in case of job production, a batch or
group of products are produced at one time.
· It should be remembered here that one batch of products may not resemble with
the next batch.
· Under batch system of production, the work is divided into operations and one
operation is done at a time.
· After completing the work on one operation it is passed on to the second
operation and so on till the product is completed.
· Batch production can be explained with the help of an example.
™ An enterprise wants to manufacture 20 electric motors. The work will be
divided into different operations. The first operation on all the motors will be
completed in the first batch and then it will pass on to the next operation. The
second group of operators will complete the second operation before the next
and so on. Under job production the same operators will manufacture full
machine and not one operation only.
· Batch production can fetch the benefits of repetitive production to a large extent,
if the batch is of a sufficient quantity.
· Thus, batch production may be defined as the manufacture of a product in small
or large batches or lots by series of operations, each operation being carried on
the whole batch before any subsequent operation is operated.
· This method is generally adopted in case of biscuit and confectionery and motor
manufacturing, medicines, tinned food and hardware's like nuts and bolts etc.
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95. Characteristics
The batch production method possesses the following characteristics,
· The work is of repetitive nature.
· There is a functional layout of various manufacturing processes.
· One operation is carried out on whole batch and then is passed on to the next
operation and so on.
· Same type of machines is arranged at one place.
· It is generally chosen where trade is seasonal or there is a need to produce great
variety of goods.
(iii) Mass or Flow Production
· This method involves a continuous production of standardized products on a large
scale.
· Under this method, production remains continuous in anticipation of future
demand. Standardization is the basis of mass production.
· Standardized products are produced under this method by using standardized
materials and equipment.
· There is a continuous or uninterrupted flow of production obtained by arranging
the machines in a proper sequence of operations.
· Process layout is best suited method for mass production units.
· Flow production is the manufacture of a product by a series of operations, each
article going on to a succeeding operation as soon as possible.
· The manufacturing process is broken into separate operations.
· The product completed at one operation is automatically passed on to the next till
its completion.
· There is no time gap between the work done at one process and the starting at
the next.
· The flow of production is continuous and progressive.
Characteristics
The mass or flow production possesses the following characteristics :
· The unit's flow from one operation point to another throughout the whole process.
· There will be one type of machine for each process.
· The products, tools, materials and methods are standardized.
· Production is done in anticipation of demand.
· Production volume is usually high.
· Machine set ups remain unchanged for a considerable long period.
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96. · Any fault in flow of production is immediately corrected otherwise it will stop
the whole production process.
Suitability of Flow/Mass Production :
· There must be continuity in demand for the product.
· The products, materials and equipment must be standardized because the flow of
line is inflexible.
· The operations should be well defined.
· It should be possible to maintain certain quality standards.
· It should be possible to find time taken at each operation so that flow of work is
standardized.
· The process of stages of production should be continuous.
Advantages of Mass Production
A properly planned flow production method, results in the following advantages :
· The product is standardized and any deviation in quality etc. is detected at the
spot.
· There will be accuracy in product design and quality.
· It will help in reducing direct labour cost.
· There will be no need of work-in-progress because products will automatically
pass on from operation to operation.
· Since flow of work is simplified there will be lesser need for control.
· A weakness in any operation comes to the notice immediately.
· There may not be any need of keeping work-in-progress, hence storage cost is
reduced.
1.17 Manufacturing Models and Metrics + [AU : Dec.-16, May-17]
· Manufacturing metrics are effectively utilized to quantitatively measure the
performance of a manufacturing company.
· It is used to track the performance of a company in successive periods
(eg. Months and years)
· It provides the facility to try new technologies and new systems to determine the
merits, identify problems with performance, compare alternate methods and make
good decisions.
· Manufacturing metrics can be divided into two basic categories :
a) Production performance measures
b) Manufacturing costs
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97. 1.17.1 Production Performance Measures
Metrics that indicate production performance include
i) Production rate
ii) Plant capacity
iii) Proportion uptime on equipment (Reliability)
iv) Manufacturing head time (v) work in process
i) Production rate
· Production rate for an individual production operation is expressed as the work
units completed per hour (P hr)
c
· The starting point to evaluate production rate is cycle time (T )
c .
Cycle time (T )
c
· Cycle time is defined as the time that one work unit spends being processed or
assembled.
· It is the time between when one work unit begins processing and when next unit
begins.
· Cycle time is expressed as the sum of (1) actual processing (or) assembly
operation time (T )
o (min P )
c
2) Handling time (loading and unloading time) min Pc (T )
n
3) Tool handling time (tool changing time) min Pc (T )
th
Te = T T T
o n th
+ + min Pc
Production rate evaluation for 3 types of production (R )
p
1) Batch production :
· In batch production, the time to process one batch consisting of 'Q' work units is
the sum of set up time and processing time.
Batch processing time = T (min)
b
Set up time = T (min batch)
su
Q = Batch quantity (P )
c
Tc = Cycle time (min P )
c
Tb = T Q T
su c
+ ´
'Q Tc
´ ' Þ Representing processing time (min)
· List average production time per min be 'Tp' (min P )
c
Tp =
T
Q
b
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98. · The average production rate for the machine is simply the reciprocal of average
production time expressed in hourly rate.
Rp =
60
Tp
P hr
c
2) Job shop production :
· For job shop production, usually the value of quantity, Q = 1
· Then average production time per work unit, Tp = T 1 T
su c
+ ´
Tp = T T
su c
+ min Pc
The production rate, Rp =
60
Tp
P hr
c
· If 'Q' is greater than 'i', Tp is evaluated as explained for batch production
3) Mass production :
· For mass production setup time could be neglected (T ~ 0)
su and Q = 1
· Then average production time per work unit,
Tp = 0 Tc
+
· Therefore Tp = Tc
· Here production rate could be expressed as,
Rp =
60
Tp
=
60
Tc
P hr
c
ii) Plant capacity or Production capacity
· Production capacity is defined as the maximum rate of output that a production
facility is able to produce under a given set of assumed production operating
conditions.
· The production facility is usually referred to a plant (or) factory, and so the term
plant capacity is often used.
· The assumed operating conditions are,
· No. of shifts per day (1, 2 or 3)
· No. of days in the week or month that the plant operates
· Employment levels
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99. · Production capacity for a plant is given by,
Pc =
nS H R
n
w sh p
o
×
Where, Pc = weekly production capacity of the facility (units/week)
Sw = no. of shifts per period (shift/week)
Hsh = hr/shift (hr)
Rp = Hourly production rate of each work center (units/hr)
no = Number of distinct operations through which work
units are routed
Problems On Production Capacity
Example 1.17.1 : The Turret lathe section has six machines, all devoted to the
production of the same part. The section operates 10 shifts/week. The number of hours
per shift averages 8.0. Average production rate of each machine is 17 units/hr.
Determine the weekly production capacity of the turret lathe section.
Solution :
n = Number of work centers
= Number of machines in turret lathe section = 6
Sw = Number of shifts in (shifts/week) = 10 shift/week
Hsh = Number of hours/shift = 8 hrs/shift
RP = Hourly production rate of each work center = 17 units/hr
Production capacity, Pc =
n S H R
n
w sh P
0
Here,
n0 = 1, number of distinct operations is one since this problem deals with turret lathe
section with production of same part.
Pc =
6 10 8 17
1
´ ´ ´
= 8160 Units/week
Pc = 8160 unit/week
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100. Example 1.17.2 : A production facility has 5 works centers, all devoted to the
production of the same part. The facility operates 8 hr/shift, 2 shifts/day and
5 days/week. Average production rate for each machine is 15 units/hr. Compute the
weekly production capacity of the production facility.
Solution : Given : Number of work centers, n = 5,
Hsh = Number of hours per shift = 8 hr/shift,
Sw = Number of shifts per week.
Here,
Shifts per day - 2 shifts /day
Working days in a week = 5 days/week
Sw = 2 shifts /day ´ 5 days/week
Sw = 10 Shifts/week
RP = Hourly production rate = 15 units/hr
Production capacity =
Pc =
n S H R
n
w sh P
o
Here, no = 1, since production facility deals with production of same part.
Pc =
5 10 8 5
1
´ ´ ´
= 6000 units/week
Pc = 6000 units/week
iii) Proportion Uptime On Equipment (A reliability measure)
The common reliability measures for an equipment are,
a) Utilization
b) Availability
A) Utilization (u)
· Utilization is defined as the amount of output of a production facility relative to
its capacity.
· Utilization is expressed as,
u =
Q
Pc
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Introduction

101. Where,
Q = Actual quantity produced by the facility during a given time period (P / week
c ).
Pc = Production capacity for same period (P / week
c ).
Note :
i) Utilization can be assessed for an entire plant, a single machine in the plant or
any other productive resources such as labour.
ii) Utilization is often defined as the proportion of time that the facility is operating
relative to the time available.
iii) Utilization is usually expressed as percentage.
iv) If utilization is high, that means the facility is being operated to its full capacity.
B) Availability (A)
· Availability is defined using two other reliability measure terms.
· MTBF = Mean Time Between Failures
· MTTR = Mean Time To Repair.
A =
MTBF MTTR
MTBF
-
· MTBF = Average length of time the piece of equipment runs between
breakdowns.
· MTTR = Average time required to service the equipment and put it back into
operation.
· Availability is also expressed as percentage.
· Availability provides a measure of how well the equipment on the plant are
service and maintained.
1 - 89 Computer Aided Design and Manufacturing
Introduction
Breakdown
Repairs completed
Equipment
operating
MTBF
TIme
Fig. 1.17.1 Time scale showing MTBF and MTTR used to define availability

102. · Effect of utilization find availability on plant or production capacity.
Pc = U A
(n S H R )
n
w sh P
o
´ ´
Problems based of utilization and availability
Example 1.17.3 : A production machine operates 80 hrs/week (2 shifts, 5 days) at full
capacity. During a certain week the machine produced 1000 parts and was idle in the
remaining part. a) Determine the production capacity of the machine. b) What was the
utilization of the machine during the week under consideration. c) Compute the expected
plant capacity if the availability of the machine is, A = 90 % and with the effect of
computed utilization ‘U’.
Solution : Given : n = No. of work centers = 1,
Sw = No. of shift /week, Hsh = number of hrs/shift,
RP = Production rate = 20 units/hr
Here,
Production machine operation at full capacity = 80 hrs/week
H S
sh w
´ = 80 hrs/week
a) Production capacity
Pc =
n S H R
n
w sh P
o
no = 1, since same parts are being produced
Pc =
1 80 20
1
´ ´
= 1600 units/week
Pc = 1600 units/week
b) Utilization (u)
u =
Q
Pc
Q = Actual quantity produced = 1000 Pc/week
u =
1000
1600
= 0.625
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103. ‘U’ is usually represented in percentage,
U = 62.5 %
c) Expected plant capacity if availability is 90 % and utilization 62.5 %
Here, A = 90 %
U = 62.5 %
Pc = U A
n S H R
n
w sh P
o
´ ´ = 0.9 0.625 1600
´ ´ = 900
Pc = 900 units/week
Example 1.17.4 : The mean time between failures for a certain production machine is
200 hours and the mean time to repair is 5 hours. Determine the availability of the
machine.
Solution : Given : MTBF = 200 hr, MTTR = 5 hr
Availability, A =
MTBF MTTR
MTBF
-
=
200 5
200
-
= 0.975
Availability is usually represented in percentage, A = 97.5 %
ii) Manufacturing Lead Time (MLT)
· Manufacturing Lead Time (MLT) is defined as the total time required to process
a given part or product through the plant including any lost time due to delays,
time spent in storage and reliability problems.
· MLT represents both operation and non operation elements.
· An operation is performed on a work unit when it is in the production machine.
· Non operation elements include handling, temporary storage, inspections and
other sources of delay when work unit is not in the machine.
· MLT for 3 types of production.
a) MLT for batch production :
The manufacturing lead time for the batch production is given by,
MLT = n (T QT T )
o su c n o
+ +
no = No. of distinct operations through which work units are
routed.
Tsu = Set up time (min/batch)
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104. Q = Batch quantity (P )
c
Tc = Cycle time (min / P )
c
Tno
= Non-operation time operated with machine (min)
b) MLT for job shop production
· For job shop production, the batch quantity, Q = 1
MLT = n (T T T )
o su c n o
+ +
c) MLT for mass production
· For mass production
no = Number of distinct operations through which work units
are routed
= 1
Tsu = Set up time is neglected ~ 0
Tno
= Non operation time is neglected ~ 0
MLT = Tc
Problems On Manufacturing Lead Time
Example 1.17.5 : A certain part is produced in a batch size of 100 units. The batch
must be routed through 5 operations to complete the processing of the parts. Average
set up time is 3 hr/operations and average operation time is 6 min (0.1 hour). Average
non operation time due to handling, delays, inspection etc. Is 7 hours for each
operation. Determine how many days it will take to complete the batch, assuming that
the plant runs for 8 hr shift / day.
Solution : Given, no = No. of distinct operations = 5,
Tsu = Average setup time = 3 hr/ operation,
Tc = Average operation time = Average cycle time = 0.1 hr,
Tn o = Average non operation time = 7 hours,
Q = Batch size = 100 Pc
· For batch Production,
MLT = n (T QT T )
o su c n o
+ +
MLT = 5(3 100 0.1 7)
+ ´ + = 100 hours
MLT = 100 hours
· If the plant runs for 8 hr shifts/day
MLT =
100
8
= 12.5 days
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105. Example 1.17.6 : A certain part is routed through six machines in a batch production
plant. The set up and operation times for each machines are given in the table below.
The batch size is 100 and the average non operation time per machine is 12 hours.
Determine.
i) Manufacturing Lead Time (MLT)
ii) Production Rate for operation 3
Solution : Given, Q = Batch size = 100,
Tn o = Avg. non operation time = 12 hr = 720 min
i) MLT
MLT = n (T Q.T T )
o su c n o
+ +
Here, Tsu = Average Set up time =
4 2 8 3 3 4
6
+ + + + +
= 4 hr
Tsu = 4 hours = 240 min
Also, Tc = Average operation time (Cycle time)
=
5 3 1 1 4 2
6
+ + + + +
. . . .
5 0 9 1 5
= 4.5 min
Tc = 4.5 min
MLT = n (T Q.T T )
o su c n o
+ + = 6 (240 (100 4.5) 720)
+ ´ +
= 8460 min
MLT = 141 hr
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Introduction
Machine Set up time ( )
hr Operation time (min)
1 4 5
2 2 3.5
3 8 1.0
4 3 1.9
5 3 4.1
6 4 2.5

106. ii) Production rate ‘RP’ for operation 3
For batch processing, Batch processing Time,
Tb = T Q.T
su c
+
For operation 3 Þ Tsu = 8 hrs = 480 min
Tc = 10 min
Tc = 1480 min
· The average production time,
TP =
T
Q
b
=
1480
1000
= 14.8 min
· The production rate RP (in hours)
RP =
60
Tb
=
60
14.8
= 4.05 hr
RP = 4.05 hr
v) Work In Process (WIP)
· WIP is defined as the quantity of parts or products currently located in the
factory that are either being processed or are between processing operation.
· WIP can be expressed as,
WIP =
A U Pc MLT
Sw Hsh
´ ´ ´
Where,
A = Availability
U = Utilization
Pc = Plant (or) production capacity
MLT = Manufacturing Lead Time
Sw = Number of shifts/ week
Hsh = Hours/shift
· Work In Process (WIP) is the inventory that is in the state of being transformed
from raw material to finished product.
· WIP represents investment by a firm that cannot be converted to revenue until all
processing is finished.
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107. Problems Based On Work In Process
Example 1.17.7 : The average part produced in a batch manufacturing plant must be
processed subsequentially through six machines on average. Twenty new batches of
parts are launched each week. Average operation time is 6 min, average setup time
= 5 hours, average batch size = 36 parts and average non operation time per batch
= 10 hrs/ machine. There are 18 machines in the plant working in parallel. Each of the
machine in the plant working in parallel. Each of the machine can be set up any type of
job processed in the plant. The plant operates at an average of 70 production hours per
week. Scrap rate is negligible. Determine manufacturing lead time for an average part,
plant capacity, plant utilization and work in process.
Solution : Given, n0 = 6, Tc = Average operation time = 6 min,
Tsu = 5 hr = 300 min, Q = Average batch size = 25 units,
Tno
= Average non operation time = 10 hr = 600 min,
n = Number of work centers = 18, Plant operation time = 70 hr/week.
Here, Sw = No. of shifts/week
Hsh = No. of hours/week
S H
w sh
´ = 70 hr/week
· MLT for batch production
MLT = n (T Q.T T )
o su c n o
+ + = 6(300 2 6 6 )
+ ´ +
5 00 = 6300 min
= 105 hr
If plant runs for 70 hr/week
MLT =
105
70
= 1.5 week
· Production rate, RP
Batch processing time,
TP = T QT
su c
+ = 300 25 6
+ ´ = 450 min = 450 min
Average production Time/week, (TP)
TP =
T
Q
b
=
450
25
= 18 min
Production rate, RP =
60
TP
=
60
18
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108. RP = 3.33 P / hr
c
· Plant capacity or production capacity
Pc =
n S H R
n
w sh P
o
Here, no = 6 since operations done through six different machines
Pc =
18 70 3.33
6
´ ´
= 699.3 = 699 units/week
· Plant utilization (u)
U =
Q
Po
Q = Actual output quantity
= No. of batches ´ Batch size = 20 25
´ = 500 Pc
U =
500
699
= 0.715
Utilization is usually represented in percentage
u = 71.5 %
· Work In Process (WIP)
WIP is given as
WIP =
AUP MLT
S H
c
w sh
Assume 100 % availability
WIP =
1 0.715 700 1.5
70
´ ´ ´
= 10.78
WIP = 11 parts
1.17.2 Manufacturing Costs
· Manufacturing cost is the second defining element of manufacturing metrics.
· Decision on automation and production system are usually based on the relative
cost of alternatives.
· Manufacturing cost can be classified into two major categories
1) Fixed cost
2) Variable cost
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109. 1) Fixed Cost (Fc) :
A fixed cost is the one which remains constant for any level of production output.
Example : Cost of equipment and erection, insurance and property tax etc.
· Fixed costs are generally expressed as annual amounts since insurance and
property taxes are annual costs.
· Capital costs such as equipments and erection costs are converted to annual
amounts based on interest rates.
2) Variable costs (Vc) :
· Variable cost is the one that varies in proportion to the level of production
output.
Example : Direct labour, raw materials, electric power to operate equipments etc.
· As output increases, variable cost also increases.
· Total cost (Tc) could be obtained by adding fixed cost (Fc) and variable cost
(Vc).
Tc = F V
c c
+
Direct Labour, Material and Overhead Costs
· An alternate classification divide manufacturing cost into,
1) Direct labour cost
2) Material cost
3) Overhead cost
1) Direct labour cost :
It is the sum of wages and benefits paid to the worker who operates the equipment
and performs processing tasks.
2) Material cost :
It is the cost of all raw materials required to make the product.
Example : For a rolling mill which works on steel sheet stocks, iron or and scrap iron
raw materials, out of which sheet is rolled.
3) Overhead cost
· Overhead costs the expenses other than direct labour cost and material cost
associated with running a firm.
· They are divided into two major categories
a) Factory overheads
b) Corporate overheads
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110. a) Factory overheads
Some of the typical factory overheads are,
· Plant supervision
· Applicable taxes
· Factory depreciation
· Material handling
· Power for machinery
· Security personnel
· Insurance
· Tool crib attendant
· Clerical support
· Heat and Air conditioning
· Light
· Payroll services
b) Corporate overheads
Some of the typical corporate overheads are,
· Sales and Marketing
· Accounting department
· Research and development
· Office space
· Finance department
· Legal counsel
· Corporate executives
Various components of manufacturing costs,
Various components of manufacturing costs are,
1) Prime cost
2) Factory or Work cost
3) Manufacturing or Production cost
4) Total cost or Ultimate cost
5) Selling prices
6) Market price
1) Prime cost :
It is the direct cost associated with production
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111. Prime cost = Direct labour cost + Direct material cost + Direct expenses
2) Factory (or) work costs :
Factory cost = Prime cost + Factory expenses (Factory on cost)
3) Production (or) Manufacturing cost :
Production cost = Factory + Administrative expenses
4) Total cost (or) ultimate cost :
Total cost = Manufacturing or production cost + Selling expenses
+ Distribution expenses
5) Selling price :
Selling price = Total cost + Profit
6) Market Price :
Market price = Selling price + Discount
Problems based on components of cost
Example 1.17.8 : A certain piece of work is purchased by a firm in batches of 100.
The direct material cost of 100 pieces is ` 200 and direct labour cost is ` 300.
Overhead cost is 30 % of factory cost and factory on cost is 20 % of the total material
and labour cost. If management want to make a profit of 15 % on the gross cost.
Determine selling price of each product.
Solution : Given, no = 100, Direct material cost = ` 200,
Direct labour cost = ` 300, Direct expenses = 0 (not specified),
Overhead charges = 30 % of factory cost,
Factory on cost = 20 % of total material and labour cost.
· Prime cost or direct cost
= Direct material cost + Direct labour cost + Direct expenses
= 200 300 0
+ + = ` 500
Factory cost = Prime cost + Factory expenses (Factory on cost)
Factory expenses = Factory on cost
=
20
100
200 300
( )
+ = `100
Q Factory cost = 500 100
+ = ` 600
· Selling price for each article for a profit of 15 % of gross cost
Selling price = Total cost Profit
+
1 - 99 Computer Aided Design and Manufacturing
Introduction

112. Total cost = Factory cost Overhead cost
+
Overhead cost = 30% of factory cost
=
30
100
600
´ = ` 180
Total cost = 600 + 180 = ` 780
Selling price = 780
15
100
Selling price
+ ´ =
780
0.85
= ` 917.65
Selling price per product = ` 917.65
1.18 Break Even Analysis - A Tool for Manufacturing Control
· Break even analysis is defined as the study of inter-relationships among a joins
sales, cost and operating profit at various levels of output.
· It analyses the relationship between fixed cost, variable cost, business volume etc.
· It is a technique widely used by production management and management
accountants.
· It is also known as cost-volume profit analysis.
Break Even Point
· Break Even point is defined as the level of sale volume, sales value (or)
production at which the business makes neither a profit nor a loss.
· It is also known as no profit - no loss point.
Break Even Chart
· Break Even chart is a graphical representation of costs at various levels of
activity as variation on income (Sales or revenue).
· The Fig. 1.18.1 below explains Break even analysis.
1 - 100 Computer Aided Design and Manufacturing
Introduction
Total revenue
line
Break even point
Fixed cost
Variable cost
Total cost line
Volume (units)
Loss
Profit
Cost
( )
`
Fig. 1.18.1 Break even analysis

113. Mathematical Model for Break Even Analysis
· Break Even Quantity (Q BEP)
· Break even quantity defines the sales volume at Break even point.
Q BEP =
F
S V
c
p c
-
Where,
Fc = Fixed cost
Vc = Variable cost
Sp = Selling price/unit
Also, Total cost = F V Q
c c
+ ´
Q = Quantity solid (units)
Total revenue = S Q
P ´
At Break even point Q = Q BEP
Q BEP =
F
S V
c
p c
- units
Þ Break even sales (SBEP)
SBEP =
F
1
V
S
c
c
P
-
æ
è
ç
ö
ø
÷
in rupees
Problems On Break Even Analysis
Example 1.18.1 : The fixed cost for the year 2015-2016 are ` 700000, variable cost
per unit is ` 45.Each unit is sold at ` 200. Determine.
i) Break even point in terms of sales volume.
ii) Break even point in terms of rupees.
iii) If sales volume of 6000 units has been expected, what will be the profit earned ?
iv) If a profit target of ` 150000 has be budgeted, compute the number of units to be
sold.
1 - 101 Computer Aided Design and Manufacturing
Introduction

114. Solution : Given, Fc = 700000 `, Vc = 45 `, Sp = ` 200
i) BEP in terms of sales volume (Q )
BEP
Q BEP =
F
S V
c
p c
-
=
700000
200 45
-
= 4516.12
Q BEP = 4516 units
ii) BEP in terms of Rupees (SBEP)
SBEP =
F
1
V
S
c
c
P
-
=
700000
1
45
200
-
æ
è
ç
ö
ø
÷
SBEP = ` 903225.8
iii) If Q BEP = 6000 units
Profit = ? (If profit is there, we can write)
Also, Q BEP =
F Profit
S V
c
p c
+
-
= Q
6000 =
700000 Profit
200 45
+
-
= ` 23000
iv) Number of units to be sold if profit target = ` 150000
Here, Q BEP = Q, since there is profit
Q =
F Profit
S V
c
p c
+
-
=
700000 150000
200 45
+
-
= 5483.8
Q = 5484 units
1 - 102 Computer Aided Design and Manufacturing
Introduction

115. Review Questions
1. What is meant by product cycle ? Explain it with a neat sketch.
2. Explain the Shigley's design model with a neat diagram.
3. Explain concurrent and sequential engineering with neat diagram and also
mention its advantages and disadvantages.
4. Explain the CAD process with a neat sketch and also state its applications ?
5. Explain the 2-D transformation matrix for the various transformation processes.
6. Explain the 3-D transformation matrix for the various transformation processes.
7. What is the need of homogeneous coordinates ? Mention the homogeneous
coordinates for translation, rotation and scaling.
8. Explain the DDA hidden line algorithm with an example.
9. Explain the Bresenham's line algorithm with an example.
10. Explain Cohen Sutherland clipping algorithm with an example.
11. Elaborate on the basic requirements that a CAD software has to satisfy.
12. Distinguish between modes of the design process and models of designs.
13. Describe the various database models which are generally used.
14. What are the differences between the sequential approach to the product
development process and the concurrent engineering approach ? Why should the
latter be adopted ?
15. A scaling factor of 2 is applied in the Y direction while no scaling is applied in
the X direction to the line whose two end points are at coordinates (1, 3) and
(3,6). The line is to be rotated subsequently through 300, in the
counter-clockwise direction. Determine the necessary transformation matrix for
the operation and the new coordinates of the end points.
16. What are the reasons for implementing a computer aided design system.
17. The vertices of a triangle are situated at points (15, 30), (25, 35) and (5, 45).
Find the coordinates of the vertices if the triangle is first rotated 100' counter
clockwise direction about the origin and then scaled to twice its size.
18. Describe the basic types of coordinate transformation in CAD, and then show
how these may all be calculated using matrix operations through the
1 - 103 Computer Aided Design and Manufacturing
Introduction

116. homogeneous coordinate with an example of matrix. How may a general rotation
transformation be expressed in terms of a combination of other transformation.
19. What is meant by interactive computer graphics ? Explain its various elements
20. Briefly explain the clipping and line drawing with an example.
21. Explain and bring out their differences between CAD and CAM.
22. Explain break-even analysis.
23. Explain briefly the types of production systems.
24. Explain the manufacturing models and metrics in detail.
Part A : Two Marks Question with Answers
Q.1 State any two benefits of CAD. + [ AU : May 2017 ]
Ans. : · Efficiency, effectiveness and creativity of the designer are drastically improved.
· Faster, Consistent and More accurate.
· Easy modification (copy) and Improvement (Edit).
· Inspecting tolerance and interface is easy.
· Use of standard components from part library makes fast modeling.
· 3D visualization of model in several orientations eliminates prototype.
Q.2 What is concurrent engineering. + [ AU : Dec. 2016, May 2017 ]
Ans. : In concurrent engineering, various tasks are handled at the same time, and not
essentially in the standard order. This means that info found out later in the course can be
added to earlier parts, improving them, and also saving time. Concurrent engineering is a
method by which several groups within an organization work simultaneously to create new
products and services.
Q.3 What are the advantages of concurrent engineering ? + [ AU : May 2018 ]
Ans. : · Both product and process design run in parallel and take place in the same time.
· Process and Product are coordinated to attain optimal matching of requirements for
effective quality and delivery.
· Decision making involves full team involvement.
· Reduced lead times to market
· Reduced cost
· Higher quality
· Greater customer satisfaction
· Increased market share
1 - 104 Computer Aided Design and Manufacturing
Introduction

117. Q.4 What is meant by concatenation transformation ?
+ [ AU : Dec. 2018, May 2018 ]
Ans. : Sometimes it becomes necessary to combine the individual transformations in order to
achieve the required results. In such cases the combined transformation matrix can be obtained
by multiplying the respective transformation matrices as shown below,
[P ]
¢ = [T ] [T ] [T ]........[T ] [T ] [T ]
n n 1 n 2 3 2 1
- -
Q.5 List the various activities involved in product development.
+ [ AU : Dec. 2018 ]
Ans. : i. Design process
· Synthesis
· Analysis
ii. Manufacturing process.
· Process planning
· Process control
Q.6 What is meant by homogeneous coordinates? + [AU : Dec. 2016 ]
Ans. : · The three dimensional representation of a two dimensional plane is called
homogeneous coordinates and the transformation using the homogeneous co-ordinates is called
homogeneous transformation.
· In order to concatenate the transformation, all the transformation matrices should be
multiplicative type. The following form known as homogeneous form which should
be used to convert the translation matrix into a multiplicative type.
[P ]
¢ =
x
y
1
¢
¢
é
ë
ê
ê
ê
ù
û
ú
ú
ú
=
1 0 0
0 1 0
1
D D
X Y
x
y
1
é
ë
ê
ê
ê
ù
û
ú
ú
ú
é
ë
ê
ê
ê
ù
û
ú
ú
ú
Q.7 What do you mean by synthesis of design ? + [AU : Dec. 2016 ]
Ans. : · The philosophy, functionality, and uniqueness of the product are all determined during
synthesis.
· During synthesis, a design takes the form of sketches and layout drawings that show
the relationship among the various product parts.
Q.8 What is meant by view port and windowing ? + [ AU : Dec. 2017 ]
Ans. : · View Port
· It may be sometimes desirable to display different portions or views of the drawing
in different regions of the screen.
· A portion of the screen where the contents of the window are displayed is called a
view port.
1 - 105 Computer Aided Design and Manufacturing
Introduction

118. · Window
· When a design package is initiated, the display will have a set of co-ordinate values.
These are called default co-ordinates.
· A user co-ordinate system is one in which the designer can specify his own
coordinates for a specific design application.
· Therefore, the designer may want to view only a portion of the image, enclosed in a
rectangular region called a window.
Q.9 List out the fundamental reason for implementing a CAD system.
+ [ AU : May 2015, Dec. 2013, Dec. 2011 ]
Ans. : · To Increase in the productivity of the designer
· To Improve the quality of the design
1 - 106 Computer Aided Design and Manufacturing
Introduction
65,50
130, 100
View port 1
View port 2
View port 3
View port 4
Fig. 1.1
Window
Original
drawing
Fig. 1.2

119. · To provide flexibility in design
· To improve design and analysis
· For creating documentation of the design
· For better communications
· For creating the database for manufacturing:
· For saving of design data and drawings for future reference
· To obtain better design accuracy
· For better visualization of drawing
Q.10 What are the components of a CAD system ? + [ AU : Dec. 2011 ]
Ans. : CAD hardware components :
· Central Processing Unit (CPU)
· Memory
· Hard Disk, Floppy Disk, CD-ROM
· External storage devices
· The monitor
· Printers and Plotters
· Digitizer, Puck and Mouse
CAD software :
CAD software allows the designer to create and manipulate a shape interactively and store it.
CAD software consists of (a) System software and (b) Application software.
· System Software : System software control the operations of a computer. It is
responsible for making the hardware components to work and interact with each
other and the end user. Example: Operating system, Compiler, Interpreter etc.
· Application Software : Application software or application programs are used for
general or customized CAD problems. Example : Auto CAD, CATIA, CREO, Solid
works, ADAMS, ANSYS, ABAQUS, NASTRAN etc.
Q.11 Define product cycle.
Ans. : Product cycle is the process of managing the entire lifecycle of a product from starting,
through design and manufacture, to repair and removal of manufactured products.
Q.12 List out fundamentals of product life cycle management.
Ans. : · Customer Relationship Management (CRM)
· Supply Chain Management (SCM)
· Enterprise Resource Planning (ERP)
· Product Planning and Development (PPD).
1 - 107 Computer Aided Design and Manufacturing
Introduction

120. Q.13 What is conceptualization in design process ?
Ans. : A concept study is the stage of project planning that includes developing ideas and
taking into account the all features of executing those ideas. This stage of a project is done to
reduce the likelihood of assess risks, error and evaluate the potential success of the planned
project.
Q.14 What is meant by design process ? Mention the steps involved in Shigley's
model for the design process.
Ans. : · Design process is an approach for breaking down a large project into
m Manageable portions.
m Recognition of need
m Definition of Problem
m Synthesis
m Analysis and Optimization
m Evaluation
m Presentation
Q.15 Mention any four applications of computer aided design in mechanical
engineering.
Ans. : · Computer-Aided Engineering (CAE) and Finite Element Analysis (FEA)
· Computer-Aided Manufacturing (CAM) including instructions to Computer
Numerical Control (CNC) machines
· Photorealistic rendering and motion simulation.
· Document management and revision control using product data management.
Q.16 List and differentiate the types of 2D geometric transformations.
Ans. : · Translation - Moves an object to a different position on the screen.
· Rotation - Rotate the object at particular angle q (theta) from its origin.
· Scaling - Change the size of an object
· Reflection - Mirror image of original object
· Shear - Slants the shape of an object
Q.17 List the various stages in the life cycle of a product
Ans. : · Developing the product concept
· Evolving the design
· Engineering the Product
· Manufacturing the product
1 - 108 Computer Aided Design and Manufacturing
Introduction

121. · Marketing
· Servicing
Q.18 Define clipping.
Ans. : Any procedure that identifies those portions of a picture that are either inside or outside
of a specified region or space is known as clipping.
· Types of clipping
· Point clipping
· Line clipping
· Area clipping
· Text clipping
Q.19 What is viewing transformations ?
Ans. : · The mapping of a part of a world-coordinate scene to device coordinates is referred to
as a viewing transformation. Sometimes the two-dimensional viewing transformation is simply
referred to as the window-to-viewport transformation or the windowing transformation.
· A world-coordinate area selected for display is called a window.
· An area on a display device to which a window is mapped is called a viewport.
· The window defines what is to be viewed; the viewport defines where it is to be
displayed.
Q.20 Describe Computer Aided Design.
Ans. : CAD is the function of computer systems to support in the creation, modification,
analysis, or optimization of a design. CAD software is used to raise the productivity of the
designer, progress the quality of design, progress communications through documentation, and to
generate a database for manufacturing.
Q.21 State the importance of Computer Architecture in CAD.
Ans. : In CAD, Computer architecture is a set of disciplines that explains the functionality, the
organization and the introduction of computer systems; that is, it describes the capabilities of a
computer and its programming method in a summary way, and how the internal organization of
the system is designed and executed to meet the specified facilities.
Q.22 What are the steps involved in architecture implementation ?
Ans. : Computer architecture engages different aspects, including instruction set architecture
design, logic design, and implementation. The implementation includes Integrated Circuit
Design, Power, and Cooling. Optimization of the design needs expertise with Compilers,
Operating Systems and Packaging.
1 - 109 Computer Aided Design and Manufacturing
Introduction

122. Q.23 Differentiate clockwise and counter clockwise rotation matrix.
Ans. : The direction of vector rotation is counter-clockwise if q is positive and clockwise if q
is negative.
R( )
q =
cos sin
sin cos
q q
q q
-
é
ë
ê
ù
û
ú
R( )
-q =
cos sin
sin cos
q q
q q
-
é
ë
ê
ù
û
ú
Q.24 What is design process ?
Ans. : The Engineering Design process is a multi-step process including the research,
conceptualization, feasibility assessment, establishing design requirements, preliminary design,
detailed design, production planning and tool design, and finally production.
Q.25 What is meant by analysis ?
Ans. : · The analysis begins with an attempt to put the conceptual design into the context of
engineering sciences to evaluate the performance of the expected product.
· This requires design modeling and simulation. An important aspect of analysis is the
questions that helps to eliminate multiple design choices and find the best solution to
each design problem.
· Bodies with symmetries in their geometry and loading are usually analyzed by
considering a portion of the model. Example : Stress analysis pressure vessels,
couplings etc.
Q.26 What are the applications of CAD ?
Ans. : · Structural design of aircraft
· Aircraft simulation
· Real time simulation
· Automobile industries
· Architectural design
· Pipe routing and plan layout design
· Electronic industries
· Dynamic analysis of mechanical systems
· Kinematic analysis
· Mesh data preparation for finite element analysis.
1 - 110 Computer Aided Design and Manufacturing
Introduction

123. Q.27 Differentiate between sequential and concurrent engineering.
Ans. :
Sequential engineering Concurrent engineering
Sequential engineering is the term used to
explain the method of production in a linear
system. The various steps are done one
after another, with all attention and resources
focused on that single task.
In concurrent engineering, various tasks are
handled at the same time, and not
essentially in the standard order. This means
that info found out later in the course can be
added to earlier parts, improving them, and
also saving time.
Sequential engineering is a system by which
a group within an organization works
sequentially to create new products and
services.
Concurrent engineering is a method by which
several groups within an organization work
simultaneously to create new products and
services.
Q.28 Define computer graphics.
Ans. : · Computer Graphics involves creation, display, manipulation and storage of pictures and
experimental data for proper visualization using a computer.
· Typically, a graphics system comprises of a host computer which must have a
support of a fast processor, a large memory and frame buffer along with a few other
crucial components.
· The first of them is the display devices. Colour monitors are one example of such
display device.
· There are other examples of output devices like LCD panels, laser printers, colour
printers, plotters etc.
Q.29 What is transformation ? List its types.
Ans. : · Geometric transformations provide a means by which an image can be enlarged in
size, or reduced, rotated, or moved.
· These changes are brought about by changing the co-ordinates of the picture to a
new set of values depending upon the requirements.
· The basic transformations are translation, scaling, rotation, reflection or mirror and
shear.
Q.30 Define translation.
Ans. : This moves a geometric entity in space in such a way that the new entity is parallel at
all points to the old entity. Translation of a point is shown below,
1 - 111 Computer Aided Design and Manufacturing
Introduction

124. Q.31 Write the features needed to be satisfied for line drawing algorithm.
Ans. : · Lines should appear straight
· Lines should terminate accurately
· Lines should have constant density
· Line density should be independent of length and angle
· Line should be drawn rapidly
Q.32 Differentiate preliminary design and detailed design.
Ans. :
Preliminary design Detailed design
The preliminary design fills the gap between the
design concept and the detailed design phase. The
system configuration is defined, and schematics,
diagrams, and layouts of the project will offer early
project configuration. In detailed design and
optimization, the parameters of the part being
produced will change, but the preliminary design
focuses on creating the common framework to
construct the project.
The next phase of preliminary design
is the Detailed Design which may
include procurement also. This phase
builds on the already developed
preliminary design, aiming to further
develop each phase of the project by
total description through drawings,
modeling as well as specifications.
Q.33 What are the types of production systems ?
Ans. : · Mass production
· Batch production
· Job-Shop production
Q.34 Define - production capacity
Ans. : Production capacity is defined as the maximum rate of output that a production facility
is able to produce under a given set of assumed production operation conditions such as,
· No of shifts/day
1 - 112 Computer Aided Design and Manufacturing
Introduction
Z P'
P
Z
P
Y'
X'
X
X
Y
Y
Fig. 1.3

125. · No. of days in the week (or) month that the plant operates.
· Employment levels.
Q.35 Define - Utilization and Availability
Ans. : Utilization (U) is defined as the amount of output of a production facility relative to its
capacity.
· Utilization is expressed as,
U =
Q
Pc
Q = Actual quantity produced by the facility during a given
time period. (pieces/week)
Pc = Production capacity for the same period. (pieces/week)
Availability (A) is defined using the two reliability measure terms.
MTBF = Mean Time Between Failures
MTTR = Mean Time to Repair
A =
MTBF MTTR
MTBF
-
Q.36 Define - Manufacturing Lead Time.
Ans. : Manufacturing lead time (MLT) is defined as the total time required to process a given
part (or) product through the plant including any lost time due to delays, time spent in storage
and reliability problems.
Q.37 Define - Fixed Cost and Variable Cost.
Ans. : · Fixed cost : A fixed cost is the one which remains constant for any level of
production output. Eg. Cost of equipment, erection, insurance, property tax etc.
· Variable cost : Variable cost is the one that varies in proportion to the level of
production output. Eg. Direct labour, Raw materials, Electric power to operate
equipment etc.
Q.38 Define - Direct Labour, Material and Overhead Cost
Ans. : · Direct Labour Cost : It is the sum of the wages and benefits paid to the worker
who operates the equipment and performs processing tasks.
· Material cost : It is the cost of all raw materials required to make a product.
· Overhead cost : Overhead costs are the expenses other than direct labour cost and
material cost associated with a running firm.
· Factory overhead expenses
· Corporate overhead expenses
1 - 113 Computer Aided Design and Manufacturing
Introduction

126. Q.39 Name five typical factory overhead expenses.
Ans. : · Plant supervision
· Applicable taxes
· Factory depreciation
· Material handling
· Power for machinery
Q.40 Name five typical corporate overhead expenses.
Ans. : · Sales and Marketing
· Accounting department
· Research and Development
· Office space
· Finance department
· Legal counsel
Part B : University Questions with Answers
Dec.-2016
1. Describe various stages of design process with an example. (Refer section 1.3) [8]
2. Explain a line drawing algorithm. (Refer section 1.11) [8]
3. Define clipping. Also explain the working of a simple line clipping algorithm.
(Refer section 1.12) [8]
4. Deduce windowing and viewing transformation parametrically.
(Refer section 1.13) [8]
5. Write in detail about the production performance metrics.
(Refer section 1.17) [8]
6. The average part produced in a batch manufacturing plant must be processed
sequentially through six machines on average. Twenty new batches of parts are
launched each week. Average operation time = 6 min., average set up time
= 5 hours, average batch size = 36 parts, and average non-operation time per
batch=10 hrs/machine. There are 18 machines in the plant working in parallel.
Each of the machines can be set up for any type of job processed in the plant.
The plant operates an average of 70 production hours per week. Scrap rate is
negligible. Determine manufacturing lead time for an average part, plant
capacity and plant utilization.) (Refer example 1.17.7) [16]
1 - 114 Computer Aided Design and Manufacturing
Introduction

127. May-2017
7. Compare and Contrast sequential and concurrent engineering with suitable
examples. (Refer section 1.4) [16]
8. Explain with block diagram, CAD process with suitable examples.
(Refer section 1.5) [16]
9. The average part produced in a batch manufacturing plant must be processed
sequentially through six machines on average. Twenty new batches of parts are
launched each week. Average operation time = 6 min., average set up time
= 5 hours, average batch size = 36 parts, and average non-operation time per
batch = 10 hrs/ machine. There are 18 machines in the plant working in
parallel. Each of the machines can be set up for any type of job processed in the
plant. The plant operates an average of 70 production hours per week. Scrap
rate is negligible. Determine manufacturing lead time for an average part, plant
capacity and plant utilization. (Refer example 1.17.7) [16]
10. Explain in detail job shop production and mass production.
(Refer section 1.16) [16]
Dec.-2017
11. Explain the different types of 2D transformations with examples.
(Refer section 1.9) [13]
12. Explain the Cohen-Sutherland line-clipping approach with proper sketches.
(Refer section 1.12) [13]
May-2018
13. Explain the various graphic transformations required for manipulating the
geometric transformation. (Refer section 1.13) [13]
14. Describe and Demonstrate DDA line drawing algorithm.
(Refer section 1.11.1 and example 1.11.1) [13]
Dec.-2018
15. Describe the stages in product life cycle and importance of each stage.
(Refer section 1.2) [13]
16. Discuss the significance of concurrent engineering approach in limiting design
changes. (Refer section 1.4) [13]
Introduction ends ...
1 - 115 Computer Aided Design and Manufacturing
Introduction

128. Notes
1 - 116 Computer Aided Design and Manufacturing
Introduction

129. Syllabus : Representation of curves- Hermite curve- Bezier curve- B-spline
curves-rational curves-Techniques for surface modeling – surface patch-
Coons and bicubic patches- Bezier and B-spline surfaces. Solid modeling
techniques- CSG and B-rep
Section No. Topic Name Page No.
2.1 Introduction 2 - 3
2.2 Methods of Geometric Modeling 2 - 3
2.3 Representation of Curves 2 - 8
2.4 Parametric and non-parametric Curves 2 - 9
2.5 Order of Continuity 2 - 11
2.6 Interpolation and Approximation of Curve 2 - 12
2.7 Hermite Cubic Curve 2 - 13
2.8 Bezier Curve 2 - 18
2.9 B-Spline Curve 2 - 22
2.10 Rational Curve 2 - 23
2.11 Surface Modelling 2 - 24
2.12 Parametrization of Surface Patch 2 - 27
2.13 Bezier Surface 2 - 28
2.14 B-spline Surface 2 - 29
2.15 Boolean Operation 2 - 30
2 - 1 Computer Aided Design and Manufacturing
Chapter - 2
Geometric Modeling
Unit - II

130. 2.16 Solid Modeling 2 - 31
2.17 Constructive Solid Geometry 2 - 33
2.18 Boundary Representation 2 - 35
2.19 Cell Consumption 2 - 36
2.20 Spatial Occupancy Enumeration 2 - 36
2.21 Sweep Representation 2 - 37
Part A : Two Marks Question Marks Answers 2 - 38
Part B : University Questions with Answers 2 - 43
2 - 2 Computer Aided Design and Manufacturing
Geometric Modeling

131. 2.1 Introduction
It is a branch of computational geometry and applied mathematics under which we
study different algorithm for the description of different shapes. Now a days-geometric
modeling is done with the computers and for computer based applications. 3D models and
2D models are widely used in many technical fields such as mechanical and civil
engineering, architecture, medical image processing.
Following are the requirements of geometric modeling.
· At the time of parts inspection, inter changeable manufacturing tolerance is
required.
· There should be an automatic assembly of the model for checking interference,
modeling, etc.
· Kinematic analysis and finite element analysis is required.
· Properties and geometrical evaluation of area, volume, weight, density, etc are
required.
· There should be a graphical visualization of cross-section, line that are hidden in
the geometry.
· The boundary of the solid should be uniquely identified.
2.2 Methods of Geometric Modeling + [AU : May-16, 17, Dec.-17]
There are basically three different methods to represents the geometric modeling.
1. Wire frame modeling
2. Surface modeling
3. Solid modeling
1. Wire frame modeling : It is one of the methods used in geometric modeling
system. Wireframe represents a solid shape in the form of lines, edges and points. It is
used to represent mathematically in the computer.
2 - 3 Computer Aided Design and Manufacturing
Geometric Modeling
Fig. 2.2.1 Wire frame model

132. 2. Surface modeling : It is used to represent the complex object that cannot be
represented by the wireframe modeling. It provides more and less ambiguous
representation.
Surface representation can be done in both parametric and non-parametric form.
Surfaces available in the CAD/CAM systems are Bezier surface, B-spline surface, plane
surface, coons path, surface of revolution, etc.
3. Solid modeling : It provides the complete information of the object as compared
with the surface modeling. It stores the geometric data and topological information of the
object. (Refer Fig. 2.2.3 on next page)
2.2.1 Wire Frame Modeling
Wire frame modeling has been used in computer aided engineering which helps in the
visualization of a design. It consist of finite sets of points which together form various
pairs of edges which makes the visualization of the object simple. If allows to calculate
the positions of different points quickly and accurately.
Wireframe models are classified into three types :
1) 2D model : It represents the flat object that is
it has only two dimensions, such as length and
width without having thickness.
2 - 4 Computer Aided Design and Manufacturing
Geometric Modeling
Control
polygon
Control point
Fig. 2.2.2 Surface modeling
2D Model
Fig. 2.2.4

133. 2. 2 1/2 model : It represents
beyond the 2D view that
represents the 3D object
which does not have any
side wall details. It
simplifies the data
representation which
increases the efficiency of
the software.
2 - 5 Computer Aided Design and Manufacturing
Geometric Modeling
Torus Elipsoid
Sphere Box
Paraboloid Tube Cone
Truncated
cone
Cylinder
Fig. 2.2.3 Solid modeling
2 D Model
1
2
Fig. 2.2.5

134. 3. 3D model : It allows the
modeling of the complex
geometry which gives the
complete information of the
object.
A wireframe representation is a
3D line drawing of an object which
shows only the edges without any
side surface in between.
· 3D object consist of a finite set of points and the connecting edges which define
the object adequately.
· An edge can be a circle, any arc, a defined space curves or a line segment.
Wireframes offers a 3D model with relatively small computing resources.
· It contains two types of data i.e. geometric data which include the positions of
node in the 3D geometry and it contain topological data which includes the
relation between the pair of the points.
· Consider a simple wire
frame model of the cuboid
as shown in Fig. 2.2.7,
which consist of 8 vertex
points and 12 edges which
join these joints. The
geometric and topological
data can be enlisted as
shows in the following
table.
Vertex Edge Edge Type
V1 E1[ – ]
1 2 Linear
V2 E2[ – ]
2 3 Linear
V3 E3[ – ]
3 4 Linear
V4 E4[ – ]
1 4 Linear
V5 E5[ – ]
2 6 Linear
V6 E6[ – ]
5 6 Linear
E7[ – ]
1 5 Linear
2 - 6 Computer Aided Design and Manufacturing
Geometric Modeling
3D Model
Fig. 2.2.6
z
2
E1
1 E4 4
x
E 5
E7
E2
E6
E 11
E8
6
5
E3
E12
E10
7
y
3
E9
8
Fig. 2.2.7 Wire frame of cuboid

135. E8[ – ]
6 7 Linear
E9[ – ]
7 8 Linear
E10[ – ]
5 8 Linear
E11[ – ]
3 7 Linear
E12[ – ]
4 8 Linear
· Considering the fire frame
model of cone which consist
of an apex and a circular
base as shown in Fig. 2.2.8.
Vertex Edge Edge Type
V1 E1[ – ]
1 2 Linear
V2 E2[ – ]
1 3 Linear
V3 E3[ – ]
2 3 Semi
circle
E4[ – ]
3 2 Semi
circle
2.2.2.1 Advantages and Disadvantages of Wire Frame Modeling
Advantages :
· Only lines can be seen at the intersections of surfaces.
· It conveys the information of 3D object more efficiently and quickly.
· It is used as input for CNC machines to generate the parts.
· Used for finite element analysis.
· It provides the information to create higher order models.
Disadvantages :
· Volumes and surfaces of the object are not defined.
· Hidden lines cannot be removed from the object.
· More complex images causes confusion.
· Not able to determine computational information.
· It contains limited ability for checking interference between mating parts.
· It does not represent the actual solid object.
2.3 Representation of Curves + [AU : Dec.-18]
2 - 7 Computer Aided Design and Manufacturing
Geometric Modeling
y E1 E2
E4
2
3
x
E3
z
1
Fig. 2.2.8

136. 1. Analytical curves : The curves which are defined as those that can be described by
analytic equation such as lines, circle and conics.
a) Point : A point is an exact position or location on a plane surface.
b) Circle : It is a set of all points
in a place that are at a given
distance from a given point are
equidistance from the curve.
The parametric equation of the
circle is given as
x y
2 2
+ = r2
where x = a cos q
y = a sin q
c) Ellipse : It is a plane curve
which have two focal points such
that all points on the curve and
the sum of the two distances to
focal points are constant.
d) Parabola : It is a curve where
any point is at an equal distance
from the fixed point i.e. focus
and from the straight line i.e.
from directrix. (Refer Fig. 2.3.3)
e) Hyperbola : It is a smooth curve
which is defined by its geometric
properties or by the solution set
of the equation. It is an open
curve with two branches, the intersection of a plane with both halves of a double
cone. (Refer Fig. 2.3.4)
2 - 8 Computer Aided Design and Manufacturing
Geometric Modeling
Focus
F2
Center
F1
Semi - major
axis
Vertex
Semi minor axis
Fig. 2.3.2 Ellipse
Parabola
Focus
F
Axis of symmetry
Vertex
Directrix
Fig. 2.3.3 Parabola
P(x, y)
O
a
x
y
Q
Fig. 2.3.1

137. 2. Synthetic curves : The curves which are described by a set of data points or the
control points such as splines, Bezier curve, B-spline curve, etc.
2.4 Parametric and Non-parametric Curves
The data storage of the curve can be represented by the array of coordinates or in the
form of the analytical equation. But the storage of data points in the form of array occupy
large space and does not provide the exact shape of the curve as graphical manipulation is
not exact, whereas the analytical equation provides all the information about the curve
behaviour, continuity between the curve etc.
Curves are categorized into two forms :
i) Non-parametric form ii) Parametric form
i) Non-parametric form : Non-parametric form represents the curve in form of
coordinate with the reference frame. It can be classifieds as :
· Explicit non-parametric form
· Implicit non-parametric form
· Explicit form : In explicit form any two points of the coordinates can be
expressed in the form of the third variable.
For e.g. y can be expressed in the form of x to form 2D curve.
P = [x y]
P = [x f(x)]
similarly for 3D curve the coordinate of y and z will be expressed in the forms of x.
P = [x y z]
P = [x f(x) g(x)]
· Implicit form : It is represented by the intersection of two surface which can be
given as
f(x, y, z) = 0
g(x, y, z) = 0
where the value of y and z can be computed for some value of x.
2 - 9 Computer Aided Design and Manufacturing
Geometric Modeling
a
Vertex
Focus
b
Center
F1
F2
a : Semi major axis
c : Linear eccentricity
c/a : Eccentricity
Fig. 2.3.4 Hyperbola

138. Disadvantages of non-parametric representation
1. To obtain a smooth curve we need to check the slope at point to ensure that it
should not tend to infinity which is difficult to deal with computer.
2. Non-parametric curves requires large calculation part.
3. These curves are independent of coordinate system.
ii) Parametric form : It can overcome the difficulties faced in non-parametric form. In
parametric form all the three dimensions are expressed in one single parameter
which acts as a global coordinate for the points on the curve.
Thus expressing the cartesian coordinates as the parameter of u.
Considering point P(x, y, z). In the parametric from it can be written as
P(u) = [x y z] = [x(u) y(u) z(u)]
Umin £ U Umax
£
It establishes one to one mapping between the parametric and the cartesian space
which is shown in the Fig. 2.4.1 below.
2 - 10 Computer Aided Design and Manufacturing
Geometric Modeling
y
u
0
y'(u)
u
z
u
0
z'(u)
u
x
u
0 umin umax
x'(u)
y
z
P(x,y,z) umax
umin P(u)
P'(u)
x
Cartesian form
A
Fig. 2.4.1

139. Thus the derivative of the three coordinate system represent the tangent vector which
is given as
¢
P (u) =
dP(u)
du
= [x y z ]
¢ ¢ ¢
¢
P (u) = [x (u) y (u) z (u)]
¢ ¢ ¢
¢
P (u) =
dx
du
dy
du
dz
du
é
ë
ê
ù
û
ú
Umin £ u £ Umax
2.5 Order of Continuity
To achieve the smoothness of a function is measured by the number of derivative it
has that are continuous and for that certain continuity conditions has to be imposed.
Depending on the condition following curves can be defined as -
1) Zero order continuity C0
: It
ensures that the two curves meet at a
point where the values remain same
and such a curve is called as zero
order continuity curve or C0
curve.
x (u )
C1 max = x (u )
C2 min
y (u )
C1 max = y (u )
C2 min
z (u )
C1 max = z (u )
C2 min
2) First order continuity C1 :
It ensures that the slope at
the end of the curve C1 is
same as the slope at the
starting of the curve C2 and
thus we obtained smoother
curve.
The slope can be found by
differentiating the parametric
equation.
¢
x (u )
C1 max = ¢
x (u )
C2 min
¢
y (u )
C1 max = ¢
y (u )
C2 min
¢
z (u )
C1 max = ¢
z (u )
C2 min
2 - 11 Computer Aided Design and Manufacturing
Geometric Modeling
u of C
max 1 u of C
min 2
u of
min C1
u of C
max 2
P1
P3
C1
C2
P2
Fig. 2.5.1 Zero order continuity curve
Slope of C1
C1 Curve
C2 Curve
Slope of C2
u of C
max 1 u of C
max 2
u of C
max 2
P2
P1
P3
u of C
min 1
Fig. 2.5.2 First order continuity curve

140. 3) Second order continuity C2 :
When the first order equation
are differentiated further then
the condition of second order
continuity is obtained i.e. they
satisfy both slope as well as
curvature continuity.
¢¢
x (u )
C1 max = ¢¢
x (u )
C2 min
¢¢
y (u )
C1 max = ¢¢
y (u )
C2 min
¢¢
z (u )
C1 max = ¢¢
z (u )
C2 min
2.6 Interpolation and Approximation of Curve
When the curve passes through all the control points then such a curve is known as
inter polated curve.
Generally we use Lagrangian polynomial for interpolated curve but it is unsuitable as
they tend to oscillate about control point and thus it become inconvinent for storing in the
system.
Fig. 2.6.1
Whereas when it is not necessary to pass through all the control points then the
resulting curve is known as approximated curve.
2.6.1 Difference between Interpolation Curve and Approximation Curve
Sr. No Interpolation curve Approximation curve
1. When all the data points are located on
the created figure is called interpolation
curve segment.
When all the data points are not located
on the created figure is called as
approximation curve segment.
2. If interpolation is applied on n-tuple of
points then the resulting curve is of the
degree n - 1
Modeling can be applied to the whole
ordered set, n-tuple of points or it can
be modeled by parts.
3. Not possible to change the shape of the
curve locally.
Possible to change the shape of a curve
locally.
2 - 12 Computer Aided Design and Manufacturing
Geometric Modeling
C1 Curve
C2 Curve
u of C
max 1
u of C
min 1
u of C
min 2
u of C
max 2
P2
P3
P1
Tangent
Center of
curvature
Fig. 2.5.3 Second order continuity curve
Interpolation curve Approximation curve

141. 4. Modeling is not flexible. Modeling is flexible.
5. Ferguson cubic curve come under this
category.
Bezier cubic, B-spline come under this
category.
2.7 Hermite Cubic Curve + [AU : Dec.-16]
When the curve is defined by the
two end points and their slope are
termed as Hermite cubic curve. These
types of curve are generally used to
interpolate a curve for a given data
points. It is commonly known as
splines.
Cubic equation of the spline is
given as :
P(u) = C u
i
i
i=0
3
å where 0 u 1
£ £
P(u) = C u C u + C C
3
3
2
2
1 0
+ +
u
…(2.7.1)
· In matrix form it can be written as
P(u) = [C C C C ]
u
u
u
1
3 2 1 0
3
2
é
ë
ê
ê
ê
ê
ù
û
ú
ú
ú
ú
· In scalar from the equation can be represented as :
x u
y u
( )
( )
= + + +
= + + +
C u C u C u C
C u C u C u C
3x
3
2x
2
1x 0x
3y
3
2y
2
1y 0y
z u
( ) = + + +
ü
ý
ï
þ
ï
C u C u C u C
3z
3
2z
2
1z 0z
…(2.7.2)
To define the Hermite cubic spline we need the tangent vectors at the points which
can be found by differentiating equation (2.7.1) w.r.t u
¢
P (u) = iC u
i
i–1
i=0
3
å
¢
P (u) = 3C u C u C
3
2
2 1
+ +
2 …(2.7.3)
2 - 13 Computer Aided Design and Manufacturing
Geometric Modeling
y
P (u = 0)
0
P'0
P (u = 1)
1
P'
1
0
y
x
Fig. 2.7.1 Hermite cubic curve

142. where 0 £ u £ 1
Thus to find the coefficients C3, C2, C1 and C0 we need to take two end points
(P ,P )
0 1 and their tangent vector ( ¢
P0 and ¢
Pi).
Taking u = 0, 1 in the equation (2.7.1) and (2.7.3) we get
P0 = C0 …(i)
¢
P0 = C1 …(ii)
P1 = C C C C
3 2 1 0
+ + + …(iii)
¢
P1 = 3C C C
3 2 1
+ +
2 …(iv)
Substituting (i) and (ii) in (iii) and (iv) we get
P1 = C C P + P
3 2 0 0
+ + ¢ …(v)
¢
P1 = 3C C P
3 2 0
+ + ¢
2 …(vi)
Performing 3(v) – (vi) We get
3P – P
1 1
¢ = C 2P 3P
2 0 0
+ ¢ +
C2 = 3P – P – 2P – 3P
1 1 0 0
¢ ¢
C2 = 3(P – P )– (2P P )
1 0 0 1
¢ + ¢ …(vii)
Performing (vi) – 2(v) we get
¢
P – 2P
1 1 = C P – 2P – 2P
3 0 0 0
+ ¢ ¢
C3 = ¢ ¢
P – 2P + P + 2P
1 1 0 0
C3 = 2(P – P )+ P + P
0 1 0 1
¢ ¢ …(viii)
Substituting (i), (ii), (vii) and (viii) in equation (2.7.1) we get,
P(u) = (2P – 2P + P + P )u
0 1 0 1
3
¢ ¢ + ¢ ¢
( – – – )
3 2
P P P P u
1 0 0 1
2 + ¢
P u + P
0 0
P(u) = (2u – 3u +1)P (–2u 3u )P
3 2
0
3 2
1
+ +
+ ¢ + ¢
(u – 2u + u)P (u – u )P
3 2
0
3 2
1 …(2.7.4)
where 0 £ u £ 1
In matrix form it can be represented as
P(u) = [P P P P ]
2u – 3u 1
– 2u 3u
u – 2u u
u – u
0 1 0 1
3 2
3 2
3 2
3 2
¢ ¢
+
+
+
é
ë
ê
ê
ê
ê
ù
û
ú
ú
ú
ú
2 - 14 Computer Aided Design and Manufacturing
Geometric Modeling

143. P(u) = [P P P P ]
u
u
u
1
0 1 0 1
3
2
¢ ¢
é
ë
ê
ê
ê
ê
ù
û
ú
ú
ú
ú
2 3 0 1
2 3 0 0
1 2 1 0
1 1 0 0
–
–
–
–
é
ë
ê
ê
ê
ê
ù
û
ú
ú
ú
ú
Hermite matrix =
2 3 0 1
2 3 0 0
1 2 1 0
1 1 0 0
–
–
–
–
é
ë
ê
ê
ê
ê
ù
û
ú
ú
ú
ú
Similarly, the equation for tangent vector is found by differentiating equation (2.7.4)
w.r.t
¢
P (u) = (6u – 6u)P (– 6u 6u)P
2
0
2
1
+ +
+ ¢ + ¢
(3u – 4u + 1)P (3u – 2u)P
2
0
2
1 …(2.7.5)
where 0 £ u £ 1
In matrix form it can be written as
¢
P (u) = [P P P P ]
6u – 6u
– 6u 6u
3u 4u+1
3u – 2u
0 1 0 1
2
2
2
2
¢ ¢
+
é
ë
ê
ê
ê
ê
ù
û
ú
ú
– ú
ú
¢
P (u) = [P P P P ]
u
u
u
1
0 1 0 1
3
2
¢ ¢
é
ë
ê
ê
ê
ê
ù
û
ú
ú
ú
ú
0 6 6 0
0 6 6 0
0 3 4 1
0 3 2 0
–
–
–
–
é
ë
ê
ê
ê
ê
ù
û
ú
ú
ú
ú
Hermite tangent matrix =
0 6 6 0
0 6 6 0
0 3 4 1
0 3 2 0
–
–
–
–
é
ë
ê
ê
ê
ê
ù
û
ú
ú
ú
ú
2.7.1 Solved Examples on Hermite Cubic Curve
Example 2.7.1 : Find the parametric equation of Hermite cubic spline with the end
point P (1,1)
0 and P (7,4)
1 whose tangent vector for end are given as P (5,6)
2 and
P (10,7)
3 .
2 - 15 Computer Aided Design and Manufacturing
Geometric Modeling

144. Solution : Given : P (1,1)
0 ; P (7,4)
1 ; P (5,6)
2 ; P (10,7)
3
To find P (u)
x ; P (u)
y
P0x = 1
¢
P0x = 5 – 1 = 4
¢
P1x = 10 – 7 = 3
P1x = 7
Parametric equation for x - coordinates are given as
P (u)
x = [P P P P ]
u
u
u
1
0 1 0 1
3
2
¢ ¢
é
ë
ê
ê
ê
ê
ù
û
ú
ú
ú
ú
2 3 0 1
2 3 0 0
1 2 1 0
1 1 0 0
–
–
–
–
é
ë
ê
ê
ê
ê
ù
û
ú
ú
ú
ú
= [1 7 4 ]
u
u
u
1
3
2
3
2 3 0 1
2 3 0 0
1 2 1 0
1 1 0 0
–
–
–
–
é
ë
ê
ê
ê
ê
ù
û
ú
ú
ú
ú
é
ë
ê
ê
ê
ê
ù
û
ú
ú
ú
ú
P (u)
x = [– 5 7 4 ]
u
u
u
1
3
2
1
é
ë
ê
ê
ê
ê
ù
û
ú
ú
ú
ú
P (u)
x = – 5u 7u 4u 1
3 2
+ + + …Ans.
Parametric equation for y-coordinate
P0y = 1
¢
P0y = 6 – 1 = 5
¢
P1y = 7 – 4 = 3
P1y = 4
P (u)
y = [1 4 5 ]
u
u
u
1
3
2
3
2 3 0 1
2 3 0 0
1 2 1 0
1 1 0 0
–
–
–
–
é
ë
ê
ê
ê
ê
ù
û
ú
ú
ú
ú
é
ë
ê
ê
ê
ê
ù
û
ú
ú
ú
ú
2 - 16 Computer Aided Design and Manufacturing
Geometric Modeling

145. P (u)
y = [2 – 4 5 ]
u
u
u
1
3
2
1
é
ë
ê
ê
ê
ê
ù
û
ú
ú
ú
ú
P (u)
y = 2u – 4u 5u 1
3 2 + + …Ans.
Example 2.7.2 : The Hermite cubic spline curve has the end points P (1,1)
0 and
P (7,4)
1 . The tangent vector for end P0 is defined by the line between P0 and another
point P (8,7)
2 whereas the tangent vector for end P1 is defined by the line between P1
and point P (8,7)
2 . Evaluate the value of u = 0, 0.2, 0.4, 0.6, 0.8 and 1.0
Solution : Given P0 = (1, 1)
P1 = (7, 4)
P2 = (8, 7)
To find : Parametric equation at P (u)
x , P (u)
y
P0x = 1
¢
P0x = 8 – 1 = 7
¢
P1x = 7 – 8 = 1
P1x = 7
· Parametric equation for x - coordinates are given as
P (u)
x = [P P P P ]
u
u
u
1
0 1 0 1
3
2
¢ ¢
é
ë
ê
ê
ê
ê
ù
û
ú
ú
ú
ú
2 3 0 1
2 3 0 0
1 2 1 0
1 1 0 0
–
–
–
–
é
ë
ê
ê
ê
ê
ù
û
ú
ú
ú
ú
= [1 7 7 ]
u
u
u
1
3
2
–
–
–
–
–
1
2 3 0 1
2 3 0 0
1 2 1 0
1 1 0 0
é
ë
ê
ê
ê
ê
ù
û
ú
ú
ú
ú
é
ë
ê
ê
ê
ê
ù
û
ú
ú
ú
ú
= [– 6 5 7 ]
u
u
u
1
3
2
1
é
ë
ê
ê
ê
ê
ù
û
ú
ú
ú
ú
2 - 17 Computer Aided Design and Manufacturing
Geometric Modeling

146. P (u)
x = – 6u 5u 7u 1
3 2
+ + + …Ans.
· Parametric equation for y-coordinate
P0y = 1
¢
P0y = 7 – 1 = 6
¢
P1y = 4 – 7 = – 3
P1y = 4
P (u)
y = [1 4 6 ]
u
u
u
1
3
2
–
–
–
–
–
3
2 3 0 1
2 3 0 0
1 2 1 0
1 1 0 0
é
ë
ê
ê
ê
ê
ù
û
ú
ú
ú
ú
é
ë
ê
ê
ê
ê
ù
û
ú
ú
ú
ú
P (u)
y = [– 3 0 6 ]
u
u
u
1
3
2
1
é
ë
ê
ê
ê
ê
ù
û
ú
ú
ú
ú
P (u)
y = – 3u + 6u 1
3 + …Ans.
· Point on the Hermite cubic spline
u 0 0.2 0.4 0.6 0.8 1
P (u)
x 1 2.552 4.216 5.704 6.728 7
P (u)
y 1 2.176 3.208 3.952 4.264 4
2.8 Bezier Curve + [AU : May-16, 17, 18, Dec.-16, 17]
These are approximation curve as in Bezier curve it is not necessary that the curve
should pass through all the data points, but the shape of the curve is influenced by the
control points.
· Bezier curve does not use first order differential as used in case of cubic spline
curve.
· The order of the curve depend on the number of control points. (Refer 2.8.1)
· Characteristics of Bezier curve are :
¡ The Bezier curve passes through start point and the end point.
¡ The control points define the order, derivative and the shape of the curve. As
the position of control point changes the shape of the curve would change.
2 - 18 Computer Aided Design and Manufacturing
Geometric Modeling

147. ¡ Bezier curve is always tangential to the first and the last control point as
shown in Fig. 2.8.2.
Bezier curve is based on Bernstein polynomial which is given by
P(u) = P C ui(1 – u)
i (n, i)
n–i … (2.8.1)
2 - 19 Computer Aided Design and Manufacturing
Geometric Modeling
P1 P2
u = 0
P0
Start
point
u = 1
P3 End
point
P and P = control points
1 2
Fig. 2.8.1
P0
P3
P1 P2
(i)
P0
P1
P2
P3
(ii)
P0
P3
P2
P1
(iii)
P0
P1
P2
P3
(iv)
Fig. 2.8.2

148. The expanded form of the equation is given as
P(u) = P C u (1 – u)
0 (n,0)
0 n–0 +P C u (1 – u)
1 (n,1)
1 n–1
+P C u (1 – u)
2 (n,2)
2 n–2+ +
L P C u (1 – u)
n (n,n)
n n–n
where 0 £ u £ 1
The general equation of Bezier curve is
P(u) = P (1 u) P C u(1 u) P C u (1 u) ...
0
n
1 (n,1)
n 1
2 (n,2)
2 n 2
- + - + - + +
- - P u
n
n … (2.8.2)
where 0 £ u £ 1
where C(n,i) =
n!
i(n i)!
-
For the four control points
Pn = P3
C(3,0) =
3!
0(3 0)!
-
= 1
C(3,1) =
3!
1(3 1)!
-
= 3
C(3,2) =
3!
2 (3 2)!
-
= 3
C(3,3) =
3!
3(3 3)!
-
= 1
Thus substituting in the equation (2.8.2) we will get,
P(u) = P (1 u) 3P u(1 u) 3P u (1 u) P u
0
3
1
2
2
2
3
3
- + - + - + … (2.8.3)
where 0 £ u £ 1
2.8.1 Solved Examples on Bezier Curve
Example 2.8.1 : Find the parametric equation of the Bezier curve whose end points
are P (0,0)
0 and P (7,0)
3 . The other control points are P (7,0)
1 and P (7,6)
2 .
Solution : Given :
P0 = (0, 0)
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149. P1 = (7, 0)
P2 = (7, 6)
P3 = (7, 0)
To find : P (u)
x , P (u)
y
By using the equation (2.8.3) for the four control points.
P (u)
x = P (1 – u) 3P u(1 – u) 3P u (1 – u) P u
0x
3
1x
2
2x
2
3x
3
+ + +
P0x = 0 ; P1x = 7 ; P2x = 7 ; P3x = 7
P (u)
x = 0(1 – u) 3.7(1 – u) u 3.7u (1 – u) 7u
3 2 2 3
+ + +
P (u)
x = 21u(1 – u) 21u (1 – u) 7u
2 2 3
+ +
P (u)
x = 21u(1 + u – 2u) 21u (1 – u) 7u
2 2 3
+ +
P (u)
x = 21u + 21u – 42u 21u u 7u
3 2 2 3
+ +
– 21 3
P (u)
x = 7u – 21u 21u
3 2 + … Ans.
Parametric equation along y-coordinate.
P (u)
y = P (1 – u) 3 P u (1 – u) 3 P u (1 – u) P
0y
3
1y
2
xy
2
3y
+ + + u3
P0y = 0 ; P1y = 0 ; P2y = 6 ; P3y = 0
P (u)
y = 0 + 0 + 3.6 u (1 – u) 0
2 +
P (u)
y = 18u (1 – u)
2
P (u)
y = 18u – 18u
2 3 … Ans.
Example 2.8.2 : Find the equation of a Bezier curve which is defined by four control
points as (80, 30, 0), (100, 100, 0), (200, 100, 0) and (250, 30, 0).
Solution : Given : P0 = (80, 30, 0)
P1 = (100, 100, 0)
P2 = (200, 100, 0)
P3 = (250, 30, 0)
To find : Equation of a Bezier curve.
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150. Since all the coordinates along z are equal to zero therefore we have a parametric
curve along the x and y coordinates.
Parametric curve along X - coordinate.
P (u)
x = P (1 u) 3P u(1 u) + 3P u (1 u) P u
0x
3
1x
2
2x
2
3x
3
- + - - +
P0x = 80, P1x = 100, P2x = 200 P3x = 250
P (u)
x = 80(1 u) 3.100 u(1 u) 3.200 u (1 u)+ 250 u
3 2 2 3
- + - + -
= 80(1 3u 3u u ) 300 u(1 u 2u)+ 600 u (1 u) 250 u
2 3 2 2 3
- + - + + - - +
= 80 240 u 240 u 80 u 300 u 300 u 600 u 600 u 600 u 25
2 3 3 2 2 3
- + - + + - + - + 0 u3
P (u)
x = - + + +
130 u 240 u 60 u 80
3 2 …Ans.
Parametric curve along y-coordinate
P (u)
y = P (1 u) 2P u(1 u) 3P u (1 u) P u
0y
3
1y
2
2y
2
3y
3
- + - + - +
P0y = 30; P1y = 100; P2y = 100; P3y = 30
P (u)
y = 30(1 u) 3.100 u(1 u) 3.100 u (1 u) 30 u
3 2 2 3
- + - + - +
= 30(1 3u 3u u ) 300 u(1 u 2u)+ 300 u (1 u) 30 u
2 3 2 2 3
- + - + + - - +
P (u)
y = 30 90 u 90 u 30 u 300 u 300 u 600 u 300 u 300 u 30 u
2 3 3 2 2 3 3
- + - + + - + - +
P (u)
y = - + +
210 u 210 u 30
2 …Ans.
2.9 B-Spline Curve
It is a generelised form of the Bezier curve. It is similar to the Bezier curve which is
been defined by the number of control points. The main difference between the B-spline
curve and the Bezier curve is that they have an ability to control the shape of the curve
locally then the global control in the Bezier curve.
Characteristics of B-spline curve :
· The degree of the curve is independent of the number of control points. It can be
linear, quadratic or cubic.
· Local control of the curve is possible in B-spline curve.
· As the control point changes the shape of the curve changes without changing the
degree of the polynomial.
· It is widely used as compared to Bezier curve for the more complex objects.
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151. 2.9.1 Difference between Hermite Cubic Spline, Bezier Wave and B-Spline
Curve
Sr.
No.
Hermite cubic spline Bezier curve B-spline curve
1. It is represented by the
polynomial of degree 3.
Curve with
(n + 1) data points are
represented by the
polynomial of nth degree.
Curve with
(n + 1) data points are
represented by the
polynomial of nth degree.
2. To draw the curve it needs
two data points and two
tangent vector.
To draw Bezier curve it
require two data points and
one or more control points
in between is required.
To draw a curve it require
two data points and one or
more control points in
between is required.
3. Degree of polynomial is
independent of data points.
Degree of polynomial
depends on the number of
data points.
Degree of polynomial is
independent of the number
of data points.
4. The shape of the curve
depends on the tangent
vectors at the end.
The shape is controlled by
the control points.
The shape is controlled by
the control points.
5. It is not convenient to
control the shape of the
curve.
The curve is affected by
the movement of the
control points.
It affects the curve locally
by the movement of the
control points.
2.10 Rational Curve
The non-uniform rational B-spline curve is also known as NURB which include both
the Bezier and B-spline curve. Hence it is a standard curve defintion for data exchange.
Rational curve is a function in which one polynomial curve is divided by another
polynomial curve, which is given as :
2 - 23 Computer Aided Design and Manufacturing
Geometric Modeling
Original curve
Control point
Control point
Fig. 2.9.1 Change due to modified control point

152. P(u) =
P W C u)
W C u)
i i i,n
i= 0
n
i i,n
i= 0
n
å
å
(
(
where u Î [0, 1]
where Wi = Weighting factor
When Wi = 1
then the above expression changes into conventional Bezier curve.
2.11 Surface Modeling
· It is a mathematical method used by the computer-aided design applications for
displaying solid appearing objects.
· Surface modeling is a popular technique for architectural design and rendering.
· Surface models are preferred for the representation of complex objects such as
car, ship, aircraft and casting.
· Surface models helps the designers to obtain good visualization of the entire
surface.
· The advanced surface models can be used for generating NC tool path.
· It only stores the geometry of the object and not its topology. The surface is
generated by connecting the points of the *wireframe.
· CAD uses the two basic method for the creation of surfaces. The first begins
with the construction curves from which 3D surface is swept.
· The second method is direct creation of the surface with manipulation of the
surface poles or control points.
2.11.1 Classification of Surfaces in Geometric Modeling
· Data is required for the creation of different surfaces which depends on the
application.
· To create a ruled surface we need two boundaries while to create the surface of
revolution we need one entity.
· The surfaces can be either analytical or synthetic some of the analytical surfaces
are :
1. Plane surface : It is one of the simplest form of the analytical surface which
require three non-coincident points to define the plane as shown in the Fig. 2.11.1.
2. Loafed surface : It is a linear surface which is formed by interpolating between the
boundaries. It needs two boundaries to define ruled or a loafed surface as shown in
the Fig. 2.11.2.
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153. 3. Edge surfaces : It is an extension to the ruled surface as in this the surface is been
patched between the boundaries as shown in the Fig. 2.11.3.
4. Surface of revolution : It is used for axisymmetric object which can be revolved
around the axis to form the surface. The revolution can be controlled by controlling
the angle of revolution as shown in the Fig. 2.11.4.
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Fig. 2.11.1 Plane surface
Boundary 1
Boundary 2
Fig. 2.11.2 Ruled surface
Boundary 1
Boundary 3
Boundary 2
Boundary 4
Fig. 2.11.3 Edge surfaces
Axis of
revolution
Generator
curve
Generated
part
Fig. 2.11.4 Surface of revolution

154. 5. Tabulated surface : It is used to generate the surface by extending the planar curve
in either direction as required in the object. This method is suitable for identical
curved cross-section as shown in the Fig. 2.11.5.
6. Coons patch : This surface
is formed by curves which
form the closed boundaries.
Hence, we required different
forms of curve to obtain
Coons path.
2.11.2 Blending Function
Blending functions are also known as
Basis function which is an element from a
particular basis for a function space. As every
vector in a vector space can be represented as
a linear combination of basic vector.
Blending function determine how the
control points influence the shape of the given
curve for values of the parameters u over the
range from 0 to 1.
For e.g. : In Bezier spline curve we have
two points and two tangents which are used to
plot the graph whose parametric range varies
2 - 26 Computer Aided Design and Manufacturing
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Fig. 2.11.5 Tabulated surface
Boundaries
Coon patch
Fig. 2.11.6 Coons patch
P(u)
P0 P3
u = 0 u = 1 u
Fig. 2.11.7

155. from 0 to 1. The equation of Bezier curve is given by
P(u) = P (1 u) 3 P u(1 u) 3 P u (1 u) P u
0
3
1
2
2
2
3
3
- + - + - +
Put, u = 0
P(0) = P0
Put, u = 1
P(1) = P3 Whose graphs can be
plotted as shown in Fig. 2.11.7.
Now by differentiating the
above equation we get the tangent
vector which can be given as
shown in Fig. 2.11.8.
2.12 Parametrization of Surface Patch + [AU : May-18, Dec.-18]
The parametrization of the
surface is viewed as a
one-to-one mapping from the
surface to a suitable domain.
The parameter domain itself will
be a surface and constructing a
parametrization means mapping
one surface into another.
Parametrization has many
application in the field of
science and engineering, CAD
model, computer graphics to
enhance the visual quality, 3D
scanning, surface approximation.
CAD/CAM system prefer
parametric form of surface representation. A vector r(u,v) is described in two variables i.e.
u and v. It can be represented as follows in the Fig. 2.12.1.
r(u,v) = [x y z]
= [x(u,v) y(u,v) z(u,v)]
When umin £ u umax
£
vmin £ v vmax
£
2 - 27 Computer Aided Design and Manufacturing
Geometric Modeling
P(u)
P1 P2
u = 0 u = 1 u
Fig. 2.11.8
u = umax
v = vmax
ru
rv
r (x, y, z)
u=umin
v = vmin
v
u
v = const
u = const
z
y
x
Fig. 2.12.1 Parametric representation

156. 2.12.1 Bicubic Patches
· It is generated by the four boundary curves which are of bicubic polynomial.
· The patch is defined by the 16 control points i.e. 4 control points of each curve.
· The bicubic patch can be written as,
P(u, v) = ( )
(
(
V V V 1 M
q (u)
q u)
q (u)
q u)
3
1
2
3
4
2
é
ë
ê
ê
ê
ê
ù
û
ú
ú
ú
ú
where M = Matrix that describe the cubic curve
q u), q u), q u),q u)
1 3 4
( ( ( (
2 = Control points which passes through the curves.
The value of V varies between 0 to 1.
2.13 Bezier Surface + [AU : May-18]
Bezier surfaces are guaranted by using Bezier curve as shown in the Fig. 2.13.1. The
surface is not necessarily pass from all the control points as it was there in the Bezier
curve. The surface can be twisted by using different control points. The surface can be
controlled globally not locally.
Characteristics of Bezier surface :
1. The surface generally follows the shape of the defining polygon net.
2. The surface is contained in the convex hull of the polygon net.
3. Each of the boundary curve in this case is a Bezier curve.
4. The degree of the surface in each polynomial direction is one less than the number
of defining polygon vertical in that direction.
5. The continuity of the surface in each parametric direction is too less than the
number of defining polygon vertices.
2 - 28 Computer Aided Design and Manufacturing
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u = 0
v = 1
u = 1
v = 0
Increasing v Increasing u
u = 0, v = 0
Fig. 2.12.2 Bicubic surface patch

157. 2.14 B-spline Surface
The surface which is formed by
using B-spline curve are known as
B-spline surface. Its not necessary
to pass the surface from all control
points. The surface obtained have
the local control and these
modification of the surface can be
carried out locally by adjusting
local area control points.
Characteristics of B-spline surface :
1. The surface is invariant to an affine transformation.
2. If the number of polygon vertices is equal to the order of basis in that direction and
if there are no interior Knot values, then B-spline surface reduces to a Bezier
surface.
3. The highest order in parametric direction is limited to the number of defining
polygon vertices in that direction.
2 - 29 Computer Aided Design and Manufacturing
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(a) (b)
Fig. 2.13.1 Bezier surfaces
Fig. 2.14.1 B-spline surface

158. 2.15 Boolean Operation
It is important for advanced 3D modeling in mechanical and architectural design.
There are three Boolean operations which are basically used to form the complex geometry
i.e.
i) Union,
ii) Subtract,
iii) Intersect.
i) Union operation : This operation fuses solids or region together into one object. It
is used when we need to joint mechanical assemblies.
For eg. :
After performing union operation the resulting object will be
ii) Subtract operation : In this process one object get subtracted out from the other to
form the resultant object. It is most frequently used to form holes or remove any
material.
In the above example if subtract operation was done then resulting object will be
2 - 30 Computer Aided Design and Manufacturing
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159. iii) Intersect operation : It creates
the new shape. Out of the overlapping parts
between the given two primitives to get the
desired object.
In the above example if intersect operation
was done then the resulting object will be as
shown in Figure.
2.16 Solid Modeling
For designing 3-Dimensional solid geometry we used solid modelling technique which
are often known as Primitive Instancing. Some of the primitives which is been utilized by
the solid modules are shown below.
2 - 31 Computer Aided Design and Manufacturing
Geometric Modeling
y
P
x
z W D
Block
H
R
P
x
y
z
Cylinder
H
z
R H
P
y
x
Cone
y
P
R
x
z
Sphere
P
H
z
W
D x
y
Wedge
R2
z
y
P
R1
R0
x
Torus
R1
Fig. 2.16.1 Solid modelling primitives

160. By using these primitives we can obtain different geometric objects by adding or
subtracting the primitives. To form the complex solid geometry we can use the Boolean
operator such as union, intersection, difference can be performed to obtain the different
object. But to form the solid model we should have geometric data i.e. the coordinate
position of the object in space and the connectivity or topological data that relates the
object with each other.
For e.g. : Performing different Boolean operation on the parts A and B.
i) Union between A and B (A È B) :
ii) Intersection between A and B (A Ç B) :
iii) Difference of A from B (A – B) :
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A
B

161. iv) Difference of B from A (B – A) :
2.17 Constructive Solid Geometry + [AU : May-17, Dec.-18]
CSG method is also known as building block approach. In this method we need to
select the set of solid primitives to build the model such as sphere, cylinder, cone,
rectangular block, etc.
· It is the popular method for constructing complex solids by dividing the complex
object into the set of primitive and combining these primitives by using the sets
of boolean operation to form the required object.
· Consider two solids A and B as shown in the Fig. 2.17.1 on previous page and
the resultant solids after performing the boolean operation.
· By using CSG method it is easy to construct the complex model just by adding,
subtracting, performing intersection operation.
· CSG model is been represented in the form of binary tree which gives the
complete information about the model and the number of boolean operation
required to construct the binary tree as shown the Fig. 2.17.2 on previous page.
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A
B
A + B A – B B – A A n B
Union Subtraction Intersection
Resulting solids
Fig. 2.17.1

162. · The above binary tree is an unbalanced tree which require more computation
hence it is feasible to obtain the balanced binary tree whose left and right
subtrees have almost equal number of nodes.
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C
A B
D
A B
C
D
S = A – B
1
S = S + C
2 1
S = S + D
3 2
Fig. 2.17.2 Unbalanced tree
A B C D
S = A – B
1
S = S + S
3 1 2
S = C + D
2
Fig. 2.17.3 Balanced tree

163. 2.18 Boundary Representation + [AU : Dec.-16, 18]
· It is one of the popular method of solid representation. It consist of set of faces
in the solid which requires its topological and geometric information.
· Topological information about the model : It provides the relation between the
vertices, edges and the faces which is been used in wire frame model. It also
included the orientation of edges and faces.
· Geometric information : It provides the equation of the edges and the faces.
· In such a representation the orientation
of faces are important which is given in
anticlockwise fashion and normal
vectors are represented by the right
hand thumb rule as shown in the
Fig. 2.18.1.
· It stores information only about the
surfaces of the solid which can provide
with the geometric and mass properties
of the given object and with the help of Gauss Divergence theorem it is possible
to relate the surface integral with the volume integral.
· It is useful for the complex geometries which could not be represented by CSG
method such as geometry of automobile body, wing shapes, etc.
· B-rep requires the more storage space but less computation compared with the
CSG model.
· In B-rep it is very simple to convert back and forth from solid model to the wire
frame model.
2.18.1 Different between C-rep Modeling and B-rep Modeling
C-rep modeling B-rep modeling
Amount of data Small Large
Surface
representation
Difficult Relatively easy
Speed Slow Fast
Local modification Difficult Easy
Structure of an
object
Simple Complicated
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4
3
2
7
6
5
1
Fig. 2.18.1

164. 2.19 Cell Consumption
It is a special type of
decomposition model which
subdivides the 3D space into simple
solids. It provides convenient ways
for computing certain topological
properties of a solid such as
number of pieces and the number
of holes.
It is represented by the list of
cell it occupies, but the cells may
not be necessarily cubes, nor they
must be identical.
2.20 Spatial Occupancy Enumeration
· It has a spatial cells occupied by the solid.
· The cells are called voxel which are of fixed size and arranged in the spatial
grid.
· The 3D object are divided into the cubical cells to obtain a particular resolution.
Smaller the size of the cell, more accurate the representation will be.
· Each cell is been represented by the co-ordinate of a single point.
· It requires a large amount of storage to obtain the accurate resolution of the
object.
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7
8 6
5
1
2 3 4
9 11
10
Fig. 2.19.1 Cellular decomposition (11 cell)
(a) First level

165. 2.21 Sweep Representation
· Sweep is a twist angle which specifies the amount of rotation along the path. It
creates a 3D solid or 3D surface by sweeping a 2D object or sub object along an
open or closed path.
· Open-ended objects create the 3D surfaces, which the object that encloses the
area inside it forms either 3D solids or 3D surfaces.
· Sweep is based on a motion of a point, curves or a surface along the path. There
are different types of sweep representation which are as follows -
1) Extruded solid : This is also known as linear sweep. In this the path is linear
or circular vector. It is been further divided into transnational sweep and
rotational sweep. In transnational sweep, a planar two dimensional boundary
can be moved a given distance in space in a perpendicular direction as shown
in the Fig. 2.21.1.
2 - 37 Computer Aided Design and Manufacturing
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(b) Second level
Fig. 2.20.1 Spatial occupancy enumeration
Directrix
Boundary of point
set to translate
Translational sweep
Rotational axis Boundary of point
set to rotate
Rotational sweep
Fig. 2.21.1 Linear sweep

166. 2) Non-linear sweep : It follows the path of a curve which is described by a
higher order equation i.e. quadratic, cubic or higher. Non-linear sweep is
similar to the linear sweep instead of vector as a curve it use direction to
sweep the object as shown in the Fig. 2.21.2.
3) Hybrid sweep : It combines both linear and non-linear sweep by using
different Boolean operation as shown in Fig. 2.21.3.
Part A : Two Marks Questions with Answers
Q.1 Define curve ?
Ans. : A curve is an infinitely large sets of points whose each point has two neighbors
except endpoints.
Q.2 How the curves are classified ?
Ans. : Curves can be classified into three ways :
i) Explicit curves ii) Implicit curves iii) Parametric curves.
Q.3 What are the different methods of geometric modeling ?
Ans. : Basically there are three different methods i.e.
i) Wireframe modeling ii) Surface modeling iii) solid modeling
Q.4 Define wire frame modeling.
Ans. : Wire frame modeling represents a solid shape in the form of lines, edges and
points which is used to represent mathematically in the computer.
2 - 38 Computer Aided Design and Manufacturing
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Boundary of point set to move
Directrix
Fig. 2.21.2 Non-linear sweep
Gluing area
Directrix
Directrix
Fig. 2.21.3 Hybrid sweep

167. Q.5 Define surface modeling.
Ans. : It is used to represent the complex object that cannot be represented by the wire
frame modeling. It can be done in both parametric and non-parametric form.
Q.6 Define solid modeling.
Ans. : It provides the complete information of the object as compared with the surface
modeling. It stores the geometric data and topological information of the object.
Q.7 Define analytical curves.
Ans. : Analytical curves : The curves which are defined as those that can be
described by analytic equation such as lines, circle and conics.
Q.8 Define synthetic curves.
Ans. : Synthetic curves : The curves which are described by a set of data points or
the control points such as splines, Bezier curve, B-spline curve, etc.
Q.9 Enlist different entities of wire frame modeling under analytical curve.
Ans. : Different entities of wire frame modeling are :
i) Point ii) Circle iii) Ellipse iv) Hyperbola v) Parabola
Q.10 Define interpolation curves.
Ans. : When the curve passes through all the control points then such a curve is
known as interpolated curve.
Q.11 Define approximation curve ?
Ans. : When it is not necessary to pass the curve through all the control points then
such a curve is called approximated curve.
Q.12 What are the types of continuity curves ?
Ans. : Different types of continuity curves
i) Zero order continuty ii) First order continuity iii) Second order continuity
Q.13 Define zero order
continuity.
Ans. : Zero order continuity C0
:
It ensures that the two curves meet
at a point where the values remain
same and such a curve is called as
zero order continuity curve or C0
curve.
2 - 39 Computer Aided Design and Manufacturing
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u of C
max 1 u of C
min 2
u of
min C1
u of C
max 2
P1
P3
C1
C2
P2
Fig. 2.1 Zero order continuity curve

168. x (u )
C1 max = x (u )
C2 min
y (u )
C1 max = y (u )
C2 min
z (u )
C1 max = z (u )
C2 min
Q.14 Define first order continuity.
Ans. : First order continuity C1 :
It ensures that the slope at the end
of the curve C1 is same as the
slope at the starting of the curve
C2 and thus we obtained smoother
curve.
The slope can be found by
differentiating the parametric
equation.
¢
x (u )
C1 max = ¢
x (u )
C2 min
¢
y (u )
C1 max = ¢
y (u )
C2 min
¢
z (u )
C1 max = ¢
z (u )
C2 min
Q.15 Define second under continuity.
Ans. : Second order continuity
C2 : When the first order
equation are differentiated further
then the condition of second order
continuity is obtained i.e. they
satisfy both slope as well as
curvature continuity.
¢¢
x (u )
C1 max = ¢¢
x (u )
C2 min
¢¢
y (u )
C1 max = ¢¢
y (u )
C2 min
¢¢
z (u )
C1 max = ¢¢
z (u )
C2 min
Q.16 Define Hermite cubic curve.
Ans. : When the curve is defined by the two end points and their slope are termed as
Hermite cubic curve. These types of curve are generally used to interpolate a curve for
a given data points. It is commonly known as splines.
2 - 40 Computer Aided Design and Manufacturing
Geometric Modeling
Slope of C1
C1 Curve
C2 Curve
Slope of C2
u of C
max 1 u of C
max 2
u of C
max 2
P2
P1
P3
u of C
min 1
Fig. 2.2 First order continuity curve
C1 Curve
C2 Curve
u of C
max 1
u of C
min 1
u of C
min 2
u of C
max 2
P2
P3
P1
Tangent
Center of
curvature
Fig. 2.3 Second order continuity curve

169. Q.17 Define Bezier curve.
Ans. : · Bezier curve does not use first order differential as used in case of cubic
spline curve.
· The order of the curve depend on the number of control points. (Refer Q.16.1)
Q.18 Define B-spline curve.
Ans. : It is a generelised form of the Bezier curve. It is similar to the Bezier curve
which is been defined by the number of control points. The main difference between
the B-spline curve and the Bezier curve is that they have an ability to control the
shape of the curve locally then the global control in the Bezier curve.
Q.19 How the surfaces are classified in geometric modeling application ?
Ans. : The surface can be either analytical or synthetic. The analytical surfaces are
2 - 41 Computer Aided Design and Manufacturing
Geometric Modeling
y
P (u = 0)
0
P'0
P (u = 1)
1
P'
1
0
y
x
Fig. 2.4 Hermite cubic curve
P1 P2
u = 0
P0
Start
point
u = 1
P3 End
point
P and P = control points
1 2
Fig. 2.5

170. 1) Plane surface
2) Loafed surface
3) Edge surface
4) Surface of revolution
5) Tabulated surface
6) Coons path
Q.20 What is blending function ?
Ans. : Wire frame modeling has been used in computer aided engineering which helps
in the visualization of a design. It consist of finite sets of points which together form
various pairs of edges which makes the visualization of the object simple. If allows to
calculate the positions of different points quickly and accurately.
Q.21 Enlist the different Boolean operations is solid modeling.
Ans. : There are three Boolean operations which are basically used to form the
complex gemetrics : i) Union ii) Subtract 3) Intersect.
Q.22 Define union operation.
Ans. : Union operation : This operation fuses solids or region together into one object.
It is used when we need to joint mechanical assemblies.
For eg. :
After performing union operation the resulting object will be
2 - 42 Computer Aided Design and Manufacturing
Geometric Modeling

171. Q.23 Define subtract operation.
Ans. : Subtract operation : In this process
one object get subtracted out from the other
to form the resultant object. It is most
frequently used to form holes or remove any
material. In the above example if subtract
operation was done then resulting object will
be
Q.24 Define intersect operation.
Ans. : Intersect operation : It creates the
new shape. Out of the overlapping parts
between the given two primitives to get the
desired object.
In the above example if intersect operation
was done then the resulting object will be
Q.25 Define constructive solid geometry.
Ans. : CSG method is also known as building block approach. In this method we need
to select the set of solid primitives to build the model such as sphere, cylinder, cone,
rectangular block, etc.
Q.26 What are the limitations of hermite curves ?
Ans. : Limitations of hermite curves is as follows :
1. To draw the hermite curve it heads two data points and two tangent vector.
2. Shape of the curve depends on the tangent vector.
3. It is not convenient to control the shape of the curve.
Q.27 What are the advantages and disadvantages of wireframe modelling ?
(Refer section 2.2.2.1) + {AU : May 16)
Q.28 State advantages of bezier curves. (Refer section 2.8) + {AU : Dec. 16)
Q.29 Why B-rep modeling approach are widely followed than CSG approach ?
(Refer section 2.18) + {AU : Dec. 16)
Q.30 Define quadratic bezier curve. (Refer section 2.8) + {AU : May 17)
Q.31 What is the significance of CSG. (Refer section 2.17) + {AU : May 17)
Q.32 Write the equation of a circle in parametric form.
(Refer section 2.3) + {AU : Dec. 17)
Q.33 Mention the various limitations of using wire frame models.
(Refer section 2.2.2.1) + {AU : Dec. 17)
2 - 43 Computer Aided Design and Manufacturing
Geometric Modeling

172. Q.34 List out the various bezier curves based on control points. (Refer section 2.8)
+ {AU : May 18)
Q.35 What is the use of surface patch ? (Refer section 2.12) + {AU : May 18)
Q.36 Distinguish between analytic curve and synthetic curve.
(Refer section 2.3) + {AU : Dec. 18)
Part B : University Questions with Answers
May - 2016
1. What are bezier curves ? Discuss its important properties. (Refer section 2.8) [16]
Dec.-2016
2. Explain different features of a bezier curve with construction details.
(Refer section 2.8) [8]
3. Derive the transformation matrix for a hermite curve. (Refer section 2.7) [8]
May - 2017
4. Explain different types of geometric modeling with suitable examples.
(Refer section 2.2) [16]
Dec.-2017
5. A set of control points is given by P0 4 4 4
= ( , , ), P1 6 8 6
=( , , ) and P2 10 3 4
( , , ),
compute bezier curve with two intermediate points. (Refer example 2.8.2) [6]
May - 2018
6. Briefly discuss about the bezier surface and composite surface. [13]
Ans. : Refer section 2.13.
· It is a collection of connected surfaces i.e. A surface that contains atleast one
composite chain as a boundary or internal curve is a composite surface. The
composite chain consist of different types of curves which satisfies tangent and
curvature continuity. Composite surface are composed of different sets of surfaces
but treated as a single entity.
· The orientation of the composite surface is determined by the orientation of the
first surface. Since all the member of the composite surface are oriented in the
same way with their neighbors.
· For example if there are three or more adjacent surfaces need to be composited,
all the surfaces may not be composited into a single surface then in this case
2 - 44 Computer Aided Design and Manufacturing
Geometric Modeling

173. different subsets of the surface may be composite as shown in the figure below.
As we have three surface A, B and C adjacent to each other. The common curve
between A and B is AB, the common curve between B and C is BC and
common curve between A and C is CA. Hence, there can be two subsets in
which the composite surface can be forced as shown in the Fig. 2.6.
7. Sketch the CSG tree for each of the two solids shown below. [15]
2 - 45 Computer Aided Design and Manufacturing
Geometric Modeling
AB
A
B
C
D
CA
BC
BC
CA
Composite
Composite
AB BC
B
C
D
D
Fig. 2.6 Composite surface
C3 C1
C3 C2
B3 B4
B1 B2
C1
C5 C6
C2
B1 B2
B3 B4
(a) Solid S1 (b) Solid S2
Not part of S2
Fig. 2.7

174. Ans. :
2 - 46 Computer Aided Design and Manufacturing
Geometric Modeling
S = S + S + S
4 1 2 3
S = A + B + C
1
A
D E
B
C
S = F + G + H
3
S = D + E
2
G
H
F
Fig. 2.8 solid S1

175. Dec.-2018
8. Write short notes on parametric representation of synthetic surface.
(Refer section 2.12) [13]
9. Discuss the following for B-rep and CSG scheme : i) how to represent surface
normals and neighborhoods ii) how to develop a classification algorithm.
(Refer sections 2.17 and 2.18) [13]
Geometric Modeling ends ...
2 - 47 Computer Aided Design and Manufacturing
Geometric Modeling
S = S + S
3 1 2
S = A + B – C
1 S = D + E – F
2
A
B
C
Solid S2
F
E
D
Fig. 2.9 solid S2

176. Notes
2 - 48 Computer Aided Design and Manufacturing
Geometric Modeling

177. Syllabus : Standards for computer graphics-Graphical Kernel System (GKS)- Standards
for exchange of images-Open Graphics Library (Open GL)- Data Exchange
standards-IGES, STEP, CALS etc.- Communication standards
Section No. Topic Name Page No.
3.1 Introduction 3 - 2
3.2 Standards for Computer Graphics 3 - 4
3.3 Standards for Exchange of Images 3 - 10
3.4 Data Exchange Standards 3 - 13
3.5 Communication Standards 3 - 26
Part A : Two Marks Question Marks Answers 3 - 32
Part B : University Questions with Answers 3 - 38
3 - 1 Computer Aided Design and Manufacturing
Chapter - 3
Cad standards
Unit - III

178. 3.1 Introduction + [AU : May-18, Dec.-18]
The heart of any CAD model is the component database. It includes the graphics
entities like points, lines, arcs, circles etc. and the coordinate points, which define the
location of these entities. In designing a data structure for CAD database, the following
factors are to be considered :
· Data must be neutral
· Data structure must be user-friendly
· Data must be portable.
In order to achieve the above requirements, some type of standardization has to be
followed by the CAD software designers. The basic elements associated with a CAD
system are :
· Operator (user)
· Graphics support system
· Other user interface support system
· Application functions
· Database
The reasons for evolving a graphic standard thus include,
(Needs for Standardization)
· To exchange graphic data between different computer systems.
· To exchange the graphic data between various software packages
· To provide a clear distinction between modeling and reviewing aspects.
Fundamental incompatibilities among entity representations greatly complicate
exchanging modeling data among CAD/CAM systems. Even simple geometric entities such
as circular arcs are represented by incompatible forms in many systems. Some systems use
NURBS (Non-Uniform Rational Basis Spline) to represent them, while others use the usual
parametric representation. The transfer of data between dissimilar CAD/CAM systems must
embrace the complete description of a product stored in its database. Four types of model
data are,
· Shape Data - consists of both geometrical and topological information.
· Non-shape Data - includes shaded images and measuring units
· Design Data - includes finite element analysis (FEA) data.
· Manufacturing Data - includes tolerancing and bill of materials
Exchange of data would not suffer any difficulty when similar CAD/CAM systems are
operated by both parties (eg. both parties use Auto CAD software). However, when
dissimilar CAD/CAM systems are in existence communication problems arise
(eg. Execution of same geometry faces difficulty when two parties uses different software
3 - 2 Computer Aided Design and Manufacturing
CAD Standards

179. such as Auto CAD and CATIA). The two methods of exchanging the data among the
different CAD/CAM systems are,
Direct Translators
· Direct translators convert data directly in one step.
· The direct translator entails translating the modeling data directly from one
CAD/CAM system format to another usually in one step.
· This solution converts the data (database) format from one native format to
another.
· It requires knowledge of both the native formats.
· Direct translators provide a satisfactory solution when only a small number of
systems are involved, but as this number increases the number of translator
programs that need to be written becomes prohibitive.
Indirect Translators
· It adopts the philosophy of creating a neutral database structure (also called a
neutral file) which is independent of any existing or future CAD/CAM system.
· This structure acts as an intermediary and a focal point of communication among
the dissimilar database structures of CAD/CAM systems.
· This solution converts native formats to a neutral format that all CAD/CAM
systems can interpret and understand.
· Indirect translators utilize some neutral file format, which reflects the neutral
database structure, with each system having its own pair of processors to transfer
data to and from this neutral format.
Comparison between direct and indirect translators
3 - 3 Computer Aided Design and Manufacturing
CAD Standards
System
5
System
4
System
1
System
2
System
3
(a) Direct translators
Neutral
file
System
5
(b) Indirect translators
System
4
System
3
System
1
System
2
Fig. 3.1.1 Translators

180. Direct translators run more quickly than the indirect ones, and the data files they
produce are smaller than the neutral files created by indirect translators. Indirect translator
philosophy provides stable communication between CAD/CAM systems, protects against
system obsolescence, and unlike direct translators eliminates dependence on single system
supplier. A side benefit of neutral files is that they can potentially be archived. Some
companies in the aerospace industry for example need to keep CAD/CAM databases for
20 to 50 years. Indirect translators based on a standard neutral file format are now the
common practice, while direct translators are seldom used.
CAD system with and without graphic standard
· Fig. 3.1.2 explains CAD system with and without graphics standards
3.2 Standards for Computer Graphics + [AU : Dec.-16, 17]
Evolution of Graphic Standards
A Graphic Standards Planning Committee (GSPC) was formed in 1974 by
ACM-SIGGRAPH (Association of Computing Machinery's Special Interest Group on
Graphics and Interactive Techniques). A committee for the development of computer
graphics standard was formed by DIN in 1975. IFIP organized a workshop on
3 - 4 Computer Aided Design and Manufacturing
CAD Standards
Graphics
data base
Graphics
function
Input/Output
devices
Application
programme
CAD without graphics standard
Input/Output
devices
Graphics
data base
Device
driver
Application
programme
Kernel
system
X
Y
CAD with graphics standard
Fig. 3.1.2 CAD system with and without graphics standard

181. Methodology in Computer Graphics in 1976. A significant development in CAD standards
is the publication of Graphical Kernel System (GKS) in 1982.
3.2.1 Graphics Kernel System (GKS)
GKS (Graphical Kernel System) is the first ISO standard. GKS standardizes
two-dimensional graphics functionality at a relatively low level.
The primary objectives of the GKS standard are :
1. To provide the complete range of graphical facilities in 2D, including the interactive
capabilities.
2. To control all types of graphic devices such as plotters and display devices in a
consistent manner.
3. To be small enough for a variety of programs.
The major contribution of GKS for the graphics programming is in terms of the layer
model, as shown in following figure. An environment for user to work is termed as work
station in GKS. This could be VDU, plotter or printer. For a programmer, all work
stations are identical. The characteristics of these workstations are built into GKS. It is
also possible to work simultaneously on more than one work station. The layer diagram of
GKS is shown in Fig. 3.2.1.
The GKS consists of three basic parts :
i. An informal exposition of contents of the standard, which includes such things as
positioning of text, filling of polygons etc.
ii. A formalization of the expository material outlined in (i) by way of abstracting the
ideas into functional descriptions (input/output parameters), effect of each function
etc.
iii. Language bindings, which are the implementations of the abstract functions,
described in (ii) in a specific computer language like FORTRAN, Ada or C.
3 - 5 Computer Aided Design and Manufacturing
CAD Standards
Application program
Application oriented layer
Language-independent layer
Graphic kernel system
Operating system
Other resources Graphical resources
Fig. 3.2.1 Layer diagram of GKS

182. The features of GKS include :
· Device independence : The standard does not assume that the input or output
devices have any particular features or restrictions.
· Text/Annotations : All text or annotations are in a natural language like English.
· Display management : A complete suite of display management functions,
cursor control and other features are provided.
· Graphics functions : Graphics functions are defined in 2D or 3D. The drivers in
GKS also include metafile drivers. Metafiles are devices with no graphic
capability like a disc unit.
· The GKS always works in a rectangular window or world coordinate system.
· The window also defines a scaling factor used to map the created picture into the
internal co-ordinate system of GKS called normalized device co-ordinates.
· Windows and view ports can then work in this co-ordinate system.
· GKS offers two routines to define the user created pictures, primitive functions
and attribute functions.
Basic Graphic Primitives of GKS
i. POLYLINE
· The GKS function for drawing
line segments is called polyline.
· The polyline function takes an
array of X-Y coordinates and
draws line segments connecting
them.
· An example of polyline primitive is shown in the Fig. 3.2.2.
ii. POLYMARKER
· It is a marker type of polyline primitive.
· It is used for drawing certain set of markers or shapes as shown in Fig.3.2.3.
3 - 6 Computer Aided Design and Manufacturing
CAD Standards
Fig. 3.2.2 Polyline (N, XPTS, YPTS)
Fig. 3.2.3 Polymarker

183. iii. FILL AREA
· Used for filling and hatching desired areas in a picture.
· Certain examples of fill area is shown in Fig. 3.2.4.
iv. GKS TEXT
· This primitive is mainly utilized for providing annotation of the drawings such as
name of the drawing, bill of materials, dimensions etc.
· An example of GKS text primitive is shown in the Fig. 3.2.5.
v. GKS CELL ARRAY
· This primitive is used to plot the raster image corresponding to pixels.
· An example for GKS cell array is shown in the Fig. 3.2.6.
· An example of utilization of GKS primitives for the drawing of a duck is shown
in the Fig. 3.2.7.
3 - 7 Computer Aided Design and Manufacturing
CAD Standards
Fig. 3.2.4 Fill area (N, XPTS, YPTS)
B
Character
height
Top
Cap
Half
Base
Bottom
Left Right
Center
Fig. 3.2.5 GKS text

184. 3 - 8 Computer Aided Design and Manufacturing
CAD Standards
Fig. 3.2.6 GKS cell array
Fig. 3.2.7 Drawing a duck using GKS primitive

185. Attribute functions of GKS
· The attribute functions define the appearance of the image e.g. color, line type
etc. Current level of GKS is GKS-3D, which is an extension to GKS that allows
the creation of 3-D objects.
· Objectives of GKS 3-D
n To define and display of 3D graphical primitives.
n To control the appearance of primitives including optional support for hidden
line and/or hidden surface removal but excluding light source, shading, and
shadow computation
· GKS 3-D drawing primitives
n Polyline
3DCALL GPL3(N, PXA, PYA, PZA)
n Polymarker
3DCALL GPM3(N, PXA, PYA, PZA)
n Fill Area
3DCALL GFA3(N, PXA, PYA, PZA)
· Graphical Kernal System draws graphical elements into a window defined using a
real-valued user coordinate system and transformed into a viewport defined rising
Normalized device coordinates (NDCs) in which coordinates values are defined to
lie within the range 0_<x_<1 and 0_<y_<1.
· Display characteristics for primitives, such as line-style and thickness, colour, text
font and text angle, are defined by attributes, the values of which are set using
the SET command.
· GKS also allows attributes to be bundled, that is grouped together and modified
as a single entity.
· The other features of the GKS standard include the handling of user interaction
and a wide range of levels of operation for input and output.
· The other graphics standard system which was widely used in Europe is Graphics
Kernal System (GKS).
· Graphics Kernal System implementations have been made by many hardware
manufacturers, for many languages.
· On the other side, GKS is not satisfactory for dynamic graphics, nor as a tool for
programming large graphics applications, and hence a variety of alternative
approaches have been developed.
3 - 9 Computer Aided Design and Manufacturing
CAD Standards

186. 3.3 Standards for Exchange of Images + [AU : Dec.-15, 18, May-16]
· There are set of important graphics standards concerning the storage and
exchange of images produced using computer graphics.
· These standards may be divided into those concerned with images that are a
collection of graphics primitives.
· Those images are concerned and stored as bitmaps.
· The former includes the Computer Graphics Metafile (CGM), which establishes a
format for device independent definition, capture, storage and transfer of vector
graphics, images and the companion Computer Graphics Interface (CGI) provides
a interface for the CGM primitives
· Large number of bitmap storage formats including the Graphics Interchange
Format (GIF), Tag Image File Format (TIFF or TIF), the windows Bitmap
Format (BMP) and many others.
Bitmaps
· At the lowest level in the computer graphics hierarchy is the pixel raster
displayed on a graphics device.
· One bit per pixel allows only 'black-and-white' images. Colour bitmaps commonly
assign 4, 8 or 24 bits/pixel to give 16 or 256 colours or 256 levels of each of the
red, green and blue colour guns respectively.
· The simplest way of storing a bitmap is simply to write the numbers that the
pixels represent to the file, together with a header giving information about the
file.
· A large number of bitmap storage formats have been developed over the years
for a variety of purposes.
· For this reason, Silicon Graphics Inc. (SGI) developed the OpenGL
Application-Programming Interface (API) for the development of 2D and 3D
graphics applications.
3.3.1 Open Graphics Library (OpenGL)
· OpenGL provides a set of commands that allow the specification of geometric
objects in two or three dimensions, using the provided primitives, together with
commands that control how these objects are rendered into the frame buffer. It is
often referred to as the assembler language of computer graphics.
· OpenGL is a low-level graphics library specification. OpenGL makes available to
the programmer a small set of geometric primitives - points, lines, polygons,
images, and bitmaps.
· It provides a means of drawing and rendering geometric objects like points, line
segments, polygons for which specifying how they should be coloured, and how
they should be mapped from the model space to the screen.
3 - 10 Computer Aided Design and Manufacturing
CAD Standards

187. · OpenGL does not require high performance display hardware to be present, but it
does require a frame buffer - memory that stores the raster display bitmap.
· OpenGL draws directly into the frame buffer but also allows the use of multiple
buffers where, for example one buffer is displayed while second is being updated.
· Fig. 3.3.1 shows a schematic diagram of OpenGL.
· Commands go into OpenGL on the left. The majority commands may be
collected in a 'display list' for executing at a later time. If not, commands are
successfully sent through a pipeline for processing.
· The first stage gives an effective means for resembling curve and surface
geometry by estimating polynomial functions of input data.
· The next stage works on geometric primitives explained by vertices. In this stage
vertices are converted, and primitives are clipped to a seeing volume in creation
for the next stage.
· All 'fragment' created is supplied to the next stage that executes processes on
personal fragments before they lastly change the structural buffer.
· These operations contain restricted updates into the structural buffer based on
incoming and formerly saved depth values, combination of incoming colors with
stored colors, as well as covering and other logical operations on fragment values.
· To end with pixels and bitmaps bypass the vertex processing part of the pipeline
to move a group of fragments in a straight line to the individual fragment actions,
finally rooting a block of pixels to be written to the frame buffer.
· Values can also be read back from the frame buffer or duplicated from one part
of the frame buffer to another. These transfers may contain several type of
encoding or decoding.
3 - 11 Computer Aided Design and Manufacturing
CAD Standards
Vertex
data
Per-vertex
operations
Primitive
assembly
Display
list
Evaluator
Pixel
operations
Rasteriz-
ation
Per-
fragment
operations
Frame buffer
Texture
memory
Pixel
data
Fig. 3.3.1 Schematic representation of OpenGL

188. Features of OpenGL
i) Based on IRIS GL : OpenGL is supported on Silicon Graphics' Integrated Rater
Imaging System Graphics Library (IRIS GL). Though it would have been potential
to have designed a totally new Application Programmer's Interface (API), practice
with IRIS GL offered insight into what programmers need and don't need in a
Three-Dimensional graphics API. Additional, creation of OpenGL similar to
Integrated Rater Imaging System Graphics Library where feasible builds OpenGL
most likely to be admitted; there are various successful IRIS GL applications, and
programmers of IRIS GL will have a simple time switching to OpenGL.
ii) Low-Level : A critical target of OpenGL is to offer device independence while
still permitting total contact to hardware. Therefore, the API gives permission to
graphics operations at the lowest level that still gives device independence. Hence,
OpenGL does not give a suggestion for modeling complex geometric objects.
iii) Fine-Grained Control : Due to minimize the needs on how an application
utilizing the Application Programmer's Interface must save and present its
information, the API must give a suggestion to state entity parts of geometric
entities and operations on them. This fine-grained control is necessary so that these
mechanism and operations may be defined in any order and so that control of
rendering operations is comfortable to contain the needs of various applications.
iv) Modal : A modal Application Programmer's Interface arises in executions in
which processes function in parallel on different primitives. In that cases, a mode
modify must be transmit to all processors so that all collects the new parameters
before it processes its next primitive. A mode change is thus developed serially,
stopping primitive processing until all processors have collected the modifications,
and decreasing performance accordingly.
v) Frame Buffer : Most of OpenGL needs that the graphics hardware has a frame
buffer. This is a realistic condition since almost all interactive graphics run on
systems with frame buffers. Some actions in OpenGL are attained only during
exposing their execution using a frame buffer. While OpenGL may be applied to
give data for driving such devices as vector displays, such use is minor.
vi) Not Programmable : OpenGL does not give a programming language. Its
function may be organized by turning actions on or off or specifying factors to
operations, but the rendering algorithms are basically fixed. One basis for this
decision is that, for performance basis, graphics hardware is generally designed to
apply particular operations in a defined order; changing these operations with
random algorithms is generally infeasible. Programmability would variance with
maintenance of the API close to the hardware and thus with the objective of
maximum performance.
vii) Geometry and Images : OpenGL gives support for managing both 3D and 2D
geometry. An Application Programmer's Interface for utilize with geometry should
also give guidance for reading, writing, and copying images, because geometry and
3 - 12 Computer Aided Design and Manufacturing
CAD Standards

189. images are regularly joint, as when a Three-Dimensional view is laid over a
background image. Various per-fragment processes that are applied to fragments
beginning from geometric primitives apply uniformly well to fragments
corresponding to pixels in an image, making it simple to mix images with
geometry.
3.4 Data Exchange Standards + [AU : Dec.-15, 16, 17, 18, May-16, 17, 18]
· The increase in CAD applications in many parts of the engineering industry has
been accomplished by growth in product variety and broadening of the range of
companies involved in the design of a particular product.
· The easiest way for two companies to exchange data is to use the same CAD
software, operating at the same revision level.
· The transfer of data between the systems has been made possible by the neutral
format of data exchange.
· The software packages will have pre-processors to convert drawing data to neutral
file format and post processors to convert neutral file data to drawing file, as
explained in Fig. 3.4.1.
Most commonly used types of neutral files formats are,
i. IGES files ii. STEP files
iii. CALS iv. PDES
3.4.1 IGES - Initial Graphics Exchange Specification
· IGES was established in the year 1979.
· The CAD/CAM Integrated Information Network (CIIN) of Boeing served as the
preliminary basis of IGES.IGES version 1.0 was released in 1980. IGES
continues to undergo revisions.
3 - 13 Computer Aided Design and Manufacturing
CAD Standards
Pre-processor
Post-processor
Neutral format
Neutral format
Post-processor
Pre-processor
CAD
software A
CAD
software B
Fig. 3.4.1 CAD data exchange using neutral file format

190. · IGES (Initial Graphics Exchange Specification) is the first standard exchange
format developed to address the concept of communicating product data among
dissimilar CAD/CAM systems.
· IGES supports solid modeling, including both B-rep and CSG schemes.
IGES Entities or Data Types
· IGES entities are classified into three, they are;
i) Geometric Entities
ii) Annotation Entities
iii) Structure Entities
Table 3.4.1 IGES geometric entities
Entity Number Entity Description Entity Number Entity Description
100 Circular Arc 136 Finite Element
102 Composite Curve 140 Offset Surface
104 Conic Arc 138 Nodal Display and Rotation
106 Copious Data 140 Offset Surface
108 Plane 142 Curve on A Parametric
Surface
110 Line 144 Trimmed Parametric Surface
112 Parametric Spline Curve 146 Nodal Results
114 Parametric Spline Surface 148 Element Results
116 Point 150 Block
118 Ruled Surface 152 Right Angular Wedge
120 Surface of Revolution 154 Right Circular Cylinder
122 Tabulated Cylinder 156 Right Circular Cone
124 Transformation Matrix 158 Sphere
126 Rational B-Spline Curve 160 Torus
128 Rational B-Spline Surface 162 Solid of Revolution
130 Offset Curve 164 Solid of Linear Extrusion
132 Connect Point 186 Ellipsoid
i. Geometric Entities
· Defines the product shape and include curves, surfaces and solids.
· IGES reserves entity numbers 100 to 199 inclusive for its geometric entities.
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191. · Sample entity type numbers used by IGES are shown in the table.
· Specifications and descriptions or entities, including geometric entities, in IGES
follow one pattern.
· Each entity has two main types of data :
¡ Directory data - Defines the entity type number
¡ Parameter data - Defines the parameter required to define the entity
completely.
ii. Annotation Entities
· Defines various types of dimensions (linear, angular and ordinate), centerlines,
notes, general labels, symbols and cross-hatching.
· Many IGES annotation entities are constructed by using other basic entities that
IGES defines, such as copious data (centerline, section etc), leader (arrow) and a
general note.
· An annotation entity may be defined in the modeling space (WCS) or in the
drawing space (a given drawing).
· If a dimension is inserted by the user in model mode, then it requires a
transformation matrix pointer when it is translated into IGES.
· Annotation entities of IGES are shown in the Table 3.4.2
Table 3.4.2 IGES annotation entities
Entity Number Entity Description
202 Angular Dimension
206 Diameter Dimension
208 Flag Note
210 General Label
212 General Note
214 Leader Arrow
216 Linear Dimension
218 Ordinate Dimension
220 Point Dimension
222 Radius Dimension
228 General Symbol
230 Sectional Area
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192. iii. Structure Entities
· The previous two sections slow how geometric and drafting data can be
represented in IGES.
· Product definition includes much more information. IGES permits a valuable set
of product data to be represented via its structure entities.
· These entities include associativity, drawing, view, external reference, property,
subfigure, macro and attribute entities.
· Attributes include line fonts, text fonts and color definition.
· Table 3.4.3 define the IGES structure entities.
Table 3.4.3 IGES structure entities
Entity Number Entity Description Entity Number Entity Description
302 Associativity Definition 406 Property
304 Line Font Definition 408 Singular Subfigure Instance
306 Macro Definition 410 View
308 Subfigure Definition 412 Rectangle Array
310 Text Font Definition 414 Circular Array
312 Text Display Template 416 External Reference
314 Colour Definition 418 Nodal Load
320 Network Subfigure
Definition
420 Network Subfigure Instance
402 Associativity Instance 600 699
Macro Instance (user Defined)
404 Drawing 10000 9999
IGES File Structure
· IGES file consists of six subsections as shown in the Fig. 3.4.2.
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Combined in
compressed
ASCII format
Flag section
Start section
Global section
Directory entry section
Parameter data section
Terminate section
Fig. 3.4.2 IGES file structure

193. i. Flag Section
· The flag section is used only with compressed ASCII and binary format.
· It is a single line that precedes the start section in a IGES file.
· In binary file format, the flag section is called as binary file information section
ii. Start Section
· This section is setup manually by the person initiating the IGES file.
· This contains the information such as the name of the sending and receiving
CAD/CAM systems and a brief description of the product.
iii. Global Section
· This section provides the 24 field parameters necessary to translate the file.
· It includes the delimiter characters (1 and 2), sender's identifier (3), file name
(4), ID of the software which generate file (5), version of IGES processor
(6), precision of integer (7 to 11), receiver's identifier (12), model space (13),
units (14), name of the units (15), maximum number of line thickness (16, 17),
file generated time (18), smallest distance (19), largest coordinate value (20),
person and organization creating file (21 and 22), IGES version (23), drafting
standards (24).
iv. Directory Section (DE)
· Contains attribute information such as color, line type, etc.
· This section is generated by IGES pre-processor.
· This section also contains the entry for each entity in the file comprising a code
representing the entity type and subtype.
v. Parameter Data Section (PD)
· This section contains the entity-specific data such as coordinate values, annotation
text, number of spline data points and etc.
· This section includes the geometric, annotation and structure entities to explain
the specific code for a drawing (Explained in section above).
· The Parameter Data (PD) section contains the data that defines the entity.
· The Directory section organizes and gives structure to the information in the
Parameter Data section.
· There can be only one directory entry for each Parameter Data section entity.
· Directory section entries may reference other Directory section entries. This
happens when we specify a transformation matrix and when we represent
structures.
· The supported parameter data is the data being communicated.
vi. Termination Section
· This section marks the end of the data file.
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194. · This section contains subtotals of records for data transmission check purposes.
· The terminate section consists of one physical card image record.
Error Sources in IGES File Processing
· Program errors in the processor
· Misinterpretation of the IGES standard
Limitations of IGES
· IGES files are not information rich - IGES files were developed to exchange
product definition data instead of product data such as mass property information,
product life cycle information etc.
· IGES files cannot carry Model Based Definition (MBD), Product and
Manufacturing Information (PMI) etc.
· IGES files often needed to be repaired.
· Last updated version of IGES file format was released in 1996, which makes this
file format old.
3.4.2 STEP - Standard for the Exchange of Product Data
· The development of STEP started in 1984 as a worldwide collaboration. The goal
was to define a standard to cover all aspects of a product (i.e. geometry,
topology, tolerances, materials, etc.), during its lifetime.
· STEP is a collection of standards to represent and exchange product information.
· The Standard for the Exchange of Product Data (STEP) is the enabler for such
seamless data exchange.
· It provides a worldwide standard for storing, sharing and exchanging product
information among different CAD systems.
· It includes methods of representing all critical product specifications such as
shape information, materials, tolerances, finishes and product structure.
· Information is modeled using the EXPRESS language
· EXPRESS has elements of Pascal, C and other languages
· EXPRESS describes geometry and other information in a standard, unambiguous
way
Comparison between STEP and IGES
· Both are "neutral file formats".
· They were developed to be compatible with different 3D packages
· The oldest is IGES. It was developed in the mid '70s by the defense industry to
solve compatibility issues between different software packages. STEP was created
in the '80s by ISO as an improvement on IGES
· The most widespread format is IGES but it can only contain basic 2D or 3D data
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195. · STEP is more versatile and contains additional information such as material
information and tolerances
· STEP is often viewed as a replacement for IGES, though IGES is still expected
to be in active use for some more time in the future.
· Although the current focus of STEP is on mechanical parts, STEP is a data
exchange standard that would apply to a wide range of product areas, including
electronics, architectural, engineering and construction, apparel and ship building.
Three Layer Architecture of STEP
· STEP has a three-layer architecture as shown the Fig. 3.4.3.
Layer 1 : Implementation Methods :
It comprises of the techniques for the implementation of STEP, in which the models
are related to the EXPRESS language, and through this physical file.
Layer 2 : Resource Information Models :
· Provides context - independent information such as the description of the
geometry, topology or product structure.
· Resource models are so called because they provide resource to the 3rd layer.
Layer 3 : Application Protocols :
· The third layer contains information related to a particular application domain
such as draughting or electrical product modeling.
· This layer describes constrained subsets of the STEP standard which should
ensure that the implementations by different vendors are very much more
compatible than IGES implementations.
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Layer-3
Application protocols
Layer-1
Implementation methods
(Express design language)
Layer-2
Resource information
models
Physical files
Conformance
testing
+
Test
suites
Fig. 3.4.3 STEP three - layer architecture

196. · This standard itself is so large that it is being developed incrementally as a series
of separate standards called parts.
Classes of STEP Parts :
i. Introductory : Consists of overview and general principles.
ii. Description methods : Consist of parts related to the express language.
iii. Implementation method : Describes how EXPRESS is mapped to physical file
and other storage mechanisms.
iv. Conformance testing methodology and frame work : This provides the
methods for testing implementations and test suites to be used during
conformance testing.
v. Integrated resource : This part includes generic resources such as geometry and
structure representation.
vi. Application protocols : This part describes implementations of STEP specific to
particular industrial applications.
vii. Abstract test suites : This part provides test suites for each of the application
protocols.
viii. Application interpreted construct : This part describes various model entity
constructs and specific modeling approaches.
3.4.3 CALS - Continuous Acquisition and Life-Cycle Support
· CALS is a United States Department of Defense (DoD) initiative for
electronically capturing military documentation and linking related information.
· In the past, technical data such as engineering drawings, illustrations and textual
data for a weapon system was delivered to the Government in paper form.
· This made it necessary for DoD activities involved in managing the acquisition of
a weapon system to orient their processes around handling paper-based
documentation.
· These processes, however, were slow, error-prone and manpower intensive.
· In the mid 1980's the DoD sought to capitalize on advances in computer
hardware and in the areas of computer-aided design, computer-aided engineering,
and concurrent engineering.
· DoD structured a series of military specifications and standards that facilitated the
handling of weapon system technical data in open, digital formats.
· This initiative grew into a joint DoD-industry Continuous Acquisition and
Lifecycle Support (CALS) initiative and led to acquisition processes between
defense contractors and DoD acquisition managers being conducted with technical
data in digital formats. With this change, there came a need for data management
systems that could receive, store and manipulate technical data in its various
formats.
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197. · Additionally, many of the acquisition management processes required
"re-engineering" using concepts such as Business Process Reengineering in order
to truly reap the benefits of receiving, handling and managing technical data in
digital formats.
· This CALS initiative has developed a number of standard specifications
(protocols) for the exchange of electronic data with commercial suppliers.
· It was thought that the CALS initiative could help the DoD reduce its costs for
acquiring technical documentation while making it more accurate, current and
timely.
· In its beginnings, CALS primarily dealt with the logistics of support
documentation. It was originally called Computer-Aided Logistics Support.
· As the benefits of the CALS initiative became better known, DoD acquisition
managers sought to incorporate CALS concepts into weapon systems
procurements. In 1988, CALS was renamed the Computer-aided Acquisition
and Logistics Support initiative.
· This better reflected its use in managing the technical information associated with
weapon system acquisition.
· Close integration among buyers and vendors or the different units of an
enterprise, created and sustained through application of standard technologies
(such as electronic data interchange or EDI), streamlining of business processes
(business process engineering), and effective use of business and technical
information.
· Developments in the field of Concurrent Engineering (CE) eventually led the
CALS initiative to encompass all aspects of weapon system acquisition : design,
production and logistics support processes.
· Similarly, advances in telecommunications such as enterprise networking and
digital information exchange protocols led to more technical documentation to be
exchanged between businesses.
· Terms such as electronic commerce and electronic data interchange soon became
associated with CALS.
· Now renamed Continuous Acquisition and Life-cycle Support, the CALS
initiative has been expanded from its roots in technical documentation and
logistics support to CE and integrated business processes.
· These standards are often referred to as simply "CALS". CALS standards have
been adopted by several other allied nations.
· It has gained acceptance outside the DoD and defense industries to become a
joint DoD-industry managed initiative.
· The CALS initiative has also been accepted and implemented within international
defense departments in Canada, Europe, Asia and Australia.
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198. · This data exchange standard was produced with an aim of applying computer
technology to the process of specifying, ordering, operating, supporting and
maintaining the weapons systems used by the US armed forces, although it can
be adopted by any industry, not just defense industry.
· The CALS initiative has endorsed IGES and STEP as formats for digital data
CALS includes standards for electronic data interchange, electronic technical
documentation, and guidelines for process improvement.
· The Vision is for all or part of a single enterprise (e.g., an original equipment
manufacturer and its suppliers, or a consortium of public and private groups and
academia), to be able to work from a common digital data base, in real time, on
the design, development, manufacturing, distribution and servicing of products.
· The direct benefits would come through substantial reductions in product-to
market time and costs, along with significant enhancements in quality and
performance.
· CALS has developed a further standard which defines subsets of IGES to be used
for specific applications including technical illustrations, engineering drawings,
electronic engineering data, and geometry for manufacture by numerical control
machines. It is expected that CALS will use the STEP standards for product data,
and will also extend into such areas as electronic hardware description and office
document exchange.
· The Spacecraft Industries CALS is to introduce a exchanging of interface data,
which are applied in the collaborative manufacturing of spacecraft by the space
industry.
· The data is exchanged by way of CALS standard such as Word, Excel, PDF,
TIFF and 3D-CAD etc. through the network.
· Generally, several companies conduct manufacturing of spacecraft simultaneously,
for which control documents named ICD (Interface Control Documentation and
Drawing) with 3D-CAD are utilized to provide information on interface work.
Two CALS systems :
i. Joint Computer-Aided Acquisition and Logistics Support (JCALS)
· JCALS concept originated from the US Army's Technical Information
Management System (TIMS)
· JCALS is an information management system that will support acquisition,
logistics support, engineering, manufacturing, configuration control and materiel
management processes throughout the life-cycle of a weapon system.
· It uses multi weapon system IWSDBs and Global Data Dictionary and Directory
(GDD/D) Services that are connected by a wide area computer network.
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199. · The interface for users is the JCALS Workbench that provides an environment to
access all of JCAL's functionality transparently to the user.
ii. Joint Engineering Data Management Information and Control System (JEDMICS).
3.4.4 PDES - Product Data Exchange Specification
· PDES was developed in the year 1984 by IGES organization in the year 1984 in
order to overcome the difficulties of IGES.
· PDES is used to support any industrial application such as mechanical, electric,
plant design, and architecture and engineering construction.
PDES includes all four types of data which is relevant to the entire life-cycle of a
product : design, analysis, manufacturing, quality assurance, testing, support, etc.
The basic configuration of PDES is explained in Fig. 3.4.4.
· Product data is exchanged through discipline models or mental models, which
should be known by both sender and receiver.
· These discipline models are standardized by PDES architecture which contains
three- layer architecture.
· The three-layer architecture includes Application layer, Logical layer and the
Physical layer.
· The application layer is the interface between user and PDES. It contains all the
information and descriptions of the required application areas.
· The logical layer provides a consistent computer independent description of the
product.
· The physical layer consists of data structures and file syntax of PDES.
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Discipline
model Pre-processor Post-processor
Three layer
architecture
Data exchange
unit
Product
data
Product
data
Discipline
model
Archival
product
data
Fig. 3.4.4 PDES file structure

200. · Preprocessor enable the user to specify the processor actions from entities of their
own system and entities of PDES.
· The pre-processor acts on the basis of the action of application layer followed by
the logical and physical layer.
3.4.5 DXF (Data Exchange Format)
· DXF is an AutoCAD format. Auto Desk Inc., the maker of AutoCAD, publishes,
supports and maintains it.
· DXF 3D is a format that translates CAD models (part files), while DXF/DWG is
a format that translates drawing files.
· DXF/DWG does not and cannot translate part files.
· DXF files come in two formats : ASCII and binary.
· The ASCII version is the most widely used in industry.
· Important sections of DXF file : Header, Tables, Blocks and Entities.
· AutoCAD DXF (Drawing Interchange Format or Drawing Exchange Format) is a
CAD data file format developed by Autodesk for enabling data interoperability
between AutoCAD and other programs.
· DXF was originally introduced in December 1982 as part of AutoCAD 1.0, and
was intended to provide an exact representation of the data in the AutoCAD
native file format, DWG (Drawing), for which Autodesk for many years did not
publish specifications and because of this, correct imports of DXF files have been
difficult.
· Autodesk now publishes the DXF specifications as a PDF on its website.
· Versions of AutoCAD from Release 10 (October 1988) and up support both
ASCII and binary forms of DXF. Earlier versions support only ASCII.
· As AutoCAD has become more powerful, supporting more complex object types,
DXF has become less useful.
· Certain object types, including ACIS solids and regions, are not documented.
· Other object types, including AutoCAD 2006's dynamic blocks, and all of the
objects specific to the vertical market versions of AutoCAD, are partially
documented, but not well enough to allow other developers to support them.
· For these reasons many CAD applications use the DWG format which can be
licensed from Autodesk or non-natively from the Open Design Alliance.
· DXF coordinates are always without dimensions so that the reader or user needs
to know the drawing unit or has to extract it from the textual comments in the
sheets.
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201. Sections of DXF Format
HEADER SECTION :
· Contains general information about the drawing.
· The HEADER section of a DXF file contains the settings of variables associated
with the drawing.
· System settings such as dimension style and layers.
· Each variable is specified by a 9 group code giving the variable's name, followed
by groups that supply the variable's value.
CLASSES SECTION
· Holds the information for application-defined classes whose instances appear in
BLOCKS, ENTITIES and OBJECTS sections of the database.
· Generally, does not provide sufficient information to allow interoperability with
other programs.
· It is assumed that a class definition is permanently fixed in the class hierarchy.
TABLES SECTION
· Contains definitions of named items.
· The TABLES section contains several tables, each of which can contain a
variable number of entries.
· These codes are also used by AutoLISP and ObjectARX applications in entity
definition lists.
· It includes line styles and user- coordinate systems.
· More primitives in tables section :
™ Dimension Style (DIMSTYPE) table
™ Layer (LAYER) table
™ Line type (LTYPE) table
™ Text style (STYLE) table
™ User Coordinate System (UCS) table
™ View (VIEW) table
BLOCKS SECTION
· This section contains an entry for each block reference in the drawing.
· The BLOCKS section of the DXF file contains all the block definitions, including
anonymous blocks generated by the HATCH command and by associative
dimensioning.
· Each block definition contains the entities that make up that block as it is used in
the drawing.
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202. · The format of the entities in this section is identical to those in the ENTITIES
section.
ENTITIES SECTION
· This section presents the group codes that apply to graphical objects.
· It includes entity definition and data.
· eg.: Circle, Ellipse, Leader, Light
OBJECTS SECTION
· This section presents the group codes that apply to nongraphical objects.
· Also used by Auto LISP and Object ARX applications.
THUMB NAIL IMAGE SECTION
· Contains the preview image for the DXF file.
· This section exists only if a preview image has been saved with the DXF file.
· Virtually all user-specified information in a drawing file can be represented in
DXFformat.
Advantages
· The DXF file format is the most compatible vector file type
· DXF files are used to exchange data between different CAD programs
· The DXF file format is easy to parse
· The DXF file specification is publicly available
Limitations
· DXF does not support application specific CAD elements
· Complex DXF files can become large in size
· Some applications cannot deal with line widths in DXF Files
3.5 Communication Standards + [AU : Dec.-15, May-17]
· Data exchange depends not only on the compatibility of the applications data
formats between the communicating systems, but also on compatibility of the
physical means of communication Computers are arranged to communicate with
each other.
· Local connections are known as Local Area Networks (LANs) and involves the
connection of digital devices over distance from a few meters up to a few
kilometers.
· Wide Area Networks (WANs) are used to connect the computers or machines of
a number of university campuses, even if these sites are in different countries.
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203. 3.5.1 Local Area Networks
· A Local Area Network (LAN) is a network that is restricted to smaller physical
areas e.g. a local office, school or house.
· Approximately all current LANs whether wired or wireless are based on Ethernet.
On a 'Local Area Network' data transfer speeds are higher than WAN and MAN
that can extend to a 10.0 Mbps (Ethernet network) and 1.0 Gbps (Gigabit
Ethernet).
· Computers and servers (provide services to other computers like printing, file
storage and sharing) can connect to each other via cables or wirelessly in a same
LAN.
· Wireless access in conjunction with wired network is made possible by Wireless
Access Point (WAP).
· A WAP is able to connect hundreds or even more of wireless users to a network.
· Servers in a LAN are mostly connected by a wire since it is still the fastest
medium for network communication.
· But for workstations (Desktop, laptops, etc.) wireless medium is a more suitable
choice, since at some point it is difficult and expensive to add new work stations
into an existing system already having complex network wiring.
· The most obvious area of difference between a WAN and a LAN is in the
topology of network itself.
· LAN topology is generally rather simpler than the mesh arrangement.
· The method of controlling access to the network is also achieved in a number of
different ways, of which perhaps the most important is the use of a control token,
and Carrier Sense Multiple Access with Collision Detection (CSMA/CD).
· A wireless LAN or WLAN is a wireless Local Area Network, which is the
linking of two or more computers without using wires.
· It uses radio communication to accomplish the same functionality that a wired
LAN has.
· WLAN utilizes spread-spectrum technology based on radio waves to enable
communication between devices in a limited area, also known as the basic service
set.
· This gives users the mobility to move around within a broad coverage area and
still be connected to the network.
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204. · LAN topology is depicted in Fig 3.4.5.
3.5.2 Wide Area Networks
· Wide Area Network is a computer network that covers relatively larger
geographical area such as a state, province or country.
· Contrast with Personal Area Networks (PAN's), Local Area Networks (LAN's) or
Metropolitan Area Networks (MAN's) that are usually limited to a room, building
or campus.
· WAN's are used to connect Local Area Networks (LAN's) together, so that users
and computers in one location can communicate with users and computers in
other locations.
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(a) Star
(c) Bus/Tree
(b) Ring
Fig. 3.4.5 LAN topology

205. · It provides a solution to companies or organizations operating from distant
geographical locations who want to communicate with each other for sharing and
managing central data or for general communication.
· WAN is made up of two or more Local Area Networks (LANs) or Metropolitan
Area Networks (MANs) that are interconnected with each other, thus users and
computers in one location can communicate with users and computers in other
locations.
· In 'Wide Area Network', Computers are connected through public networks, such
as the telephone systems, fiber-optic cables and satellite links or leased lines.
· The largest and most well-known example of a WAN is the Internet.
· WANs are mostly private and are building for a particular organization by
'Internet Service Providers (ISPs)' which connects the LAN of the organization to
the internet.
· WANs are frequently built using expensive leased lines where with each end of
the leased line a router is connected to extend the network capability across sites.
· For low cost solutions, WAP is also built using a 'circuit switching' or 'packet
switching' methods.
· WAN topology is depicted in Fig. 3.4.6.
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PSE
PSE
PSE
PSE
Computer
site B
Computer
site D
Computer
site C
Computer
site A
PSE Packet
Switching
Exchange
Fig. 3.4.6 WAN topology

206. 3.5.3 Levels of Communication Standards
· The levels of communication standard is depicted in Fig. 3.4.7
LEVEL 1 :
· Here the communication of data is between application software and output
devices such as printers, plotters etc.
· Virtual Device Interface (VDI) or Computer Graphics Interface (CGI) is the most
important standard in this category.
· These standards specify the format for transfer of data to application software and
output devices
LEVEL 2 :
· Here communication of data takes place between application software and
graphics utility
· GKS is mainly proposed for this purpose
LEVEL 3 :
· Here the communication of data takes place between CAD systems and also
among CAD systems and CAD database.
· IGES is an example for standard used for such communication.
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CAD systems
CAD database
Graphics utility
Output devices
VDI/CGI
GKS
IGES
Application software
Level - 3
Level - 2
Level - 1
Fig. 3.4.7 Levels of communication standards

207. Review Questions
1. (i) Define graphics Keneral system. Explain briefly with suitable examples.
(ii) Examine IGES data exchange format.
2. Describe the data exchange standard and development in data exchange
format.
3. (i) Write short notes on data base management.
(ii) Identiy the thrust involved in developing CAD standards.
4. (i) Summarize the standards for exchanging images.
(ii) Discuss about open graphics library.
5. (i) Describe (1) Local area network (2) Wide area network.
(ii) Describe Standard for the Exchange of product model Data (STEP) in
detail.
6. (i) Describe the types of graphics standards.
(ii) Discuss the two basic items of GKS such as primitives and attributes.
7. (i) Examine the features of open GL.
(ii) Illustrate (1) Direct CAD system export (2) Direct translation software
(3) Neutral data exchange format.
8. Classify the neutral file formats and explain in detail.
9. Analyze Continuous acquisition and life cycle support (CALS) and elaborate
their types.
10. Explain product data exchange standard and draw the three-layer architecture
of PDES.
11. Infer the different levels of graphics standard communication and elaborate the
each.
12. Compare IGES and STEP
13. Elaborate (i) HTML (ii) VRML (iii) CGM (iv) BITMAPS (v) Computer aided
design interface.
14. Compare various testing methods of IGES processors.
15. Explain STEP architecture with neat sketch.
16. List and explain the basic requirements and principles of communication
protocol.
17. Explain IGES entities.
18. Explain GKS primitives with an example.
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208. Part A : Two Marks Questions with Answers
Q.1 What is an annotation entity ? (Dec 2018)
Ans. : Defines various types of dimensions (linear, angular and ordinate), centerlines, notes,
general labels, symbols, and cross-hatching.
Example : Entity No. 202- Angular Dimension, Entity No. 206-Diameter Dimension
Q.2 Mention the need of standardization in computer graphics. (Dec 2018)
Ans. : · To exchange graphic data between different computer systems.
· To provide a clear distinction between modeling and reviewing aspects.
Q.3 List out the international organizations involved in developing the graphics
standards. (May 2018).
Ans. : · ACM (Association for Computer Machinery)
· ANSI (American National Standards Institute)
· ISO (International Standards Organization)
· GIN (German Standards Institute)
Q.4 What is the objective of GKS- 3D standard ? (May 2018)
Ans. : · To define and display of 3D graphical primitives
· To control the appearance of primitives including optional support for
hidden line and/or hidden surface removal but excluding light source,
shading and shadow computation
Q.5 What is open graphics library (Open GL) ? (Dec 2017, May 2017)
Ans. : Open GL is a low-level graphics library specification that comprises a set of several
hundred procedures and functions that allow a programmer to specify the objects and operations
involved in the production of colour graphical images of three-dimensional objects. OpenGL
makes available to the programmer a small set of geometric primitives - points, lines, polygons,
images and bitmaps.
Q.6 What is meant by IGES ? (Dec 2017)
Ans. : IGES (Initial Graphics Exchange Specification) enables the exchange of model data
basis among CAD system.
Q.7 Define Graphics Kernel System. (May 2017)
Ans. : GKS (Graphics Kernel System) provides a set of drawing features for two-dimensional
vector graphics.
Q.8 State the needs for data exchange standards. (Dec 2016, Dec 2015)
Ans. : Data exchange standards are required to translate data between different CAD systems
since all CAD systems possess their own database formats. Data conversions between different
CAD systems can be attained by data exchange formats such as IGES, STEP, DXF etc.
3 - 32 Computer Aided Design and Manufacturing
CAD Standards

209. Q.9 What is GKS cell array ? (Dec 2016)
Ans. : GKS cell array is used to plot the raster image corresponding to pixels. An example of
GKS cell array is shown below.
Q.10 Write any three CAD standards of exchange of modelling data. (May 2016)
Ans. : IGES, STEP, CALS.
Q.11 Compare the IGES and STEP. (May 2016)
OR Compare the shape based and product data exchange-based standards. Shape
Based Data Exchange Model- IGES Product Data Exchange Model-STEP
(Dec 2015)
Ans. :
IGES - INITIAL GRAPHICS EXCHANGE
SPECIFICATION
STEP - STANDARD FOR THE EXCHANGE
OF PRODUCT DATA
l IGES files tends to be surface models. l Step files tends to be solid models.
l IGES is an older file format since the final
version of IGES file was released during
1996.
l STEP is a newer technology with
periodical updating.
l Defines the product shape and include
curves, surfaces and solids.
l Used to support any industrial application
such as mechanical, electric, plant design
and architecture and engineering
construction.
3 - 33 Computer Aided Design and Manufacturing
CAD Standards
Fig. 3.1

210. Q.12 List down the various elements of CAD/CAM structure with and without
graphics system.
Ans. :
Q.13 Give the types of graphics standard.
Ans. : · GKS- Graphics Kernel System
· Open GL- Open Graphics Library
Q.14 Predict the features of GKS.
Ans. : The features of GKS are,
· Device independence : The standard does not assume that the input or output
devices have any particular features or restrictions.
· Text/Annotations : All text or annotations are in a natural language like English.
· Display management : A complete suite of display management functions,
cursor control and other features are provided.
· Graphics functions : Graphics functions are defined in 2D or 3D. The drivers in
GKS also include metafile drivers. Metafiles are devices with no graphic
capability like a disc unit.
3 - 34 Computer Aided Design and Manufacturing
CAD Standards
Graphics
data base
Graphics
function
Input/Output
devices
Application
programme
CAD without graphics standard
Input/Output
devices
Graphics
data base
Device
driver
Application
programme
Kernel
system
X
Y
CAD with graphics standard
Fig. 3.2

211. Q.15 Sketch the layer model of GKS.
Ans. : GKS layer model is shown below,
Q.16 Draw a neat sketch of image exchange using OpenGL.
Ans. : The image exchange using Open GL is demonstrated below,
Q.17 Explain the limitations of IGES.
Ans. : · IGES files are not information rich- IGES files were developed to exchange product
definition data instead of product data such as mass property information, product life cycle
information etc.
· IGES files cannot carry Model Based Definition (MBD), Product and
Manufacturing Information (PMI) etc.
· IGES files often needed to be repaired.
· Last updated version of IGES file format was released in 1996, which makes
this file format old.
3 - 35 Computer Aided Design and Manufacturing
CAD Standards
Application program
Application oriented layer
Language-independent layer
Graphic kernel system
Operating system
Other resources Graphical resources
Fig. 3.3
Vertex
data
Per-vertex
operations
Primitive
assembly
Display
list
Evaluator
Pixel
operations
Rasteriz-
ation
Per-
Fragment
operations
Framebuffer
Texture
memory
Pixel
data
Fig. 3.4

212. Q.18 Obtain a sketch of the file structure for product data exchange.
Ans. :
Q.19 Assess the various file section in IGES.
Ans. : · Flag Section
· Start Section
· Global Section
· Directory Entry Section
· Parameter Data Section
· Terminate Data Section
Q.20 Summarize the eight major areas in step documentation.
Ans. : · Introductory section :
™ Consists of overview and general principles.
· Description methods
™ Consist of parts related to the express language.
· Implementation method
™ Describes how EXPRESS is mapped to physical file and other storage
mechanisms.
· Conformance testing methodology and frame work
™ This provides the methods for testing implementations and test suites to be
used during conformance testing.
· Integrated resource
™ This part includes generic resources such as geometry and structure
representation.
3 - 36 Computer Aided Design and Manufacturing
CAD Standards
Discipline
model Pre-processor Post-processor
Three layer
architecture
Data exchange
unit
Product
data
Product
data
Product
data
Archival
product
data
Fig. 3.5

213. · Application protocols
™ This part describes implementations of STEP specific to particular industrial
applications.
· Abstract test suites
™ This part provides test suites for each of the application protocols.
· Application interpreted construct
™ This part describes various model entity constructs and specific modeling
approaches.
Q.21 Create the flow diagram to communicate between two CAD systems using
IGES.
Ans. : Flow diagram explaining communication between two CAD system using IGES is
shown below,
Q.22 Discuss the advantage of open graphics library.
Ans. : · OpenGL provides a set of commands that allow the specification of geometric
objects in two or three dimensions
· OpenGL makes available to the programmer a small set of geometric
primitives - points, lines, polygons, images and bitmaps.
· It provides a means of drawing and rendering geometric objects like points,
line segments, polygons
· Open GL does not require high performance display hardware to be present.
· Open GL draws directly into the frame buffer but also allows the use of
multiple buffers where, for example one buffer is displayed while second is
being updated.
Q.23 Discuss about Local Area Network (LAN) and Wide Area Network (WAN).
Ans. : · A Local Area Network (LAN) is a network that is restricted to smaller physical
areas e.g. a local office, school, or house.
· Wide area network is a computer network that covers relatively larger
geographical area such as a state, province or country.
3 - 37 Computer Aided Design and Manufacturing
CAD Standards
Native database Native database
Pre-processor Post-processor
System 1 System 2
IGES
Archival
database
Fig. 3.6

214. Q.24 Sketch LAN and WAN topologies.
Ans. :
For WAN topology : Refer Fig. 3.4.6
Q.25 List the various levels of communication standard.
Ans. : Refer Fig. 3.4.7.
Part B : University Questions with Answers
Dec.-2015
1. Explain Initial Graphics Exchange Specification (IGES) methodology.
(Refer section 3.4.1) [16]
2. Write short notes on OpenGL. (Refer section 3.3.1) [8]
3. Write Short notes on communication standards. (Refer section 3.5) [16]
May-2016
4. Briefly explain any one of the known graphics standards. (Refer section 3.3) [16]
3 - 38 Computer Aided Design and Manufacturing
CAD Standards
(a) Star
(c) Bus/Tree
(b) Ring
Fig. 3.7 LAN topologies

215. 5. Write short notes on Drawing Exchange Format (DXF). (Refer section 3.4.5) [16]
Dec.-2016
6. Explain Initial Graphics Exchange Specification (IGES) methodology.
(Refer section 3.4.1) [16]
7. Explain Graphics Kernel System (GKS). (Refer section 3.2.1) [16]
May-2017
8. Explain Initial Graphics Exchange Specification (IGES) methodology.
(Refer section 3.4.1) [16]
9. Write short notes on data exchange standards. (Refer section 3.4.5) [8]
10. Write short notes on communication standards. (Refer section 3.5) [8]
Dec.-2017
11. Explain the concept of product data exchange using STEP.
(Refer section 3.4.2) [13]
12. Explain Graphics Kernel System (GKS). (Refer section 3.2.1) [7]
13. Explain CALS. (Refer section 3.4.3) [6]
May-2018
14. Explain Initial Graphics Exchange Specification (IGES) methodology.
(Refer section 3.4.1) [13]
15. List and discuss the major available modules in CAD software packages.
(Refer section 3.1 and Fig. 3.1.2) [7]
16. Explain the concept of product data exchange using STEP.
(Refer section 3.4.2) [6]
Dec.-2018
17. Explain Initial Graphics Exchange Specification (IGES) methodology.
(Refer section 3.4.1) [13]
18. Write short notes on computer graphics. (Refer section 3.1) [7]
19. Write short notes on OpenGL. (Refer section 3.3.1) [6]
CAD Standards ends.....
3 - 39 Computer Aided Design and Manufacturing
CAD Standards

216. Notes
3 - 40 Computer Aided Design and Manufacturing
CAD Standards

217. Syllabus : Introduction to NC systems and CNC - Machine axis and Co-ordinate
system- CNC machine tools- Principle of operation CNC- Construction
features including structure- Drives and CNC controllers- 2D and 3D
machining on CNC- Introduction of Part Programming, types - Detailed
Manual part programming on Lathe & Milling machines using G codes and
M codes- Cutting Cycles, Loops, Sub program and Macros- Introduction of
CAM package.
Section No. Topic Name Page No.
4.1 Introduction 4 - 3
4.2 Numerical Control 4 - 3
4.3 Classification of NC System 4 - 7
4.4 Advantages of NC System 4 - 12
4.5 Disadvantages of NC System 4 - 12
4.6 Applications of NC System 4 - 12
4.7 Types of Numerical Control System 4 - 13
4.8 Conventional Numerical Control (NC) 4 - 13
4.9 Direct Numerical Control (DNC) 4 - 13
4.10 Computerized Numerical Control (CNC) 4 - 14
4.11 Constructional Features of CNC Machines 4 - 15
4.12 Advantages and Disadvantages of CNC Machines 4 - 28
4.13 Comparison between NC, CNC and DNC System 4 - 29
4.14 Adaptive Control System (ACS) 4 - 30
4 - 1 Computer Aided Design and Manufacturing
Chapter - 4
Fundamental of cnc and part
programming
Unit - IV

218. 4.15 Machining Centre 4 - 31
4.16 Program Reader 4 - 34
4.17 New Trends in Tool Materials 4 - 34
4.18 Tool Inserts 4 - 35
4.19 Work Holding in CNC Machines 4 - 36
4.20 Axis Nomenclature for CNC Machines 4 - 36
4.21 Part Programming 4 - 39
4.22 Procedure to Write a Part Program 4 - 49
4.23 Part Programming for Lathe 4 - 50
4.24 Part Programming for Milling and Drilling 4 - 67
4.25 Subroutine 4 - 90
4.26 Canned Cycle 4 - 93
4.27 Automatically Programmed Tools (APT) 4 - 96
4.28 Micromachining 4 - 99
4.29 Part Programming Using APT 4 - 100
4.30 Introduction of CAM Package 4 - 107
Two Marks Questions with Answers 4 - 111
4 - 2 Computer Aided Design and Manufacturing
Fundamental of CNC and Part Programming

219. 4.1 Introduction :
· When any machine tool is manually operated, the operator controls the relative
movements of the workpiece and tool.
· The accuracy of these movements is controlled by reference to some form of
measuring device fitted on the machine slide or lead screw.
· The operator has to perform functions like starting and stopping the machine,
turning the coolant on and off, etc.
· With manual control accuracy of final workpiece, quality and time required to
manufacture depends on the skill, concentration and experience of the operator.
· When many batches of identical parts are required, it is preferable to use jigs,
fixtures and templates.
· Automatic machine tools are also used in order to minimize errors and variable
quality of manual operation.
· So, to avoid human errors, minimize production cost and due to many other
reasons, NC i.e. Numerical Control machines comes into the picture.
· On a numerically controlled machine tool the decisions which govern the
operation of the machine are made by a series of numbers in binary code which
are interpreted by an electrical system.
· The electronic system converts these numerical commands into the physical
movement of the machine elements.
· Now a days, these NC machines are modified into different machines as follows :
m Special purpose CNC machine tool with vertical and horizontal machining
centre.
m Flexible Manufacturing System (FMS).
m Gear cutting machines.
m Electro-discharge machines with CNC.
m Co-ordinate Measuring Machines (CMM).
4.2 Numerical Control :
· "Numerical control is a programmable automation in which actions are controlled
by means of coded numbers, letters and other symbols."
· The numerical data which is required for producing a part is maintained on
punched tape.
· This data is arranged in the form of blocks of information.
· The block contains cutting speed, feed, dimensional information and contour
form.
4 - 3 Computer Aided Design and Manufacturing
Fundamental of CNC and Part Programming

220. · For preparing this data, part programmer is required which should have the
knowledge of tools, cutting fluids, use of machinability data and process
engineering.
· Fig. 4.2.1 (a) shows the block diagram for the procedure of production through NC.
· Fig. 4.2.1 (b) shows the block diagram for NC machine tool system.
4.2.1 Basic Elements of NC System
A Numerical Control machine consists of following elements :
1. Machine Control Unit (MCU),
2. Machine tool and NC tooling,
3. Part program and drawings. (Refer Fig. 4.2.1(b)).
1. Machine Control Unit (MCU) :
· It is the heart of NC machine tool system and consists of many sub-units inside
it.
· The first sub-unit is tape reader which receives the coded data from punched
tape.
4 - 4 Computer Aided Design and Manufacturing
Fundamental of CNC and Part Programming
Fig. 4.2.1 (a) : The procedure of production through numerical control
Fig. 4.2.1 (b) : Elements of a NC system

221. · The tape reader reads this data and passes data to the buffer storage through the
decoding circuits.
· The buffer storage stores the received information, till it is required and transfers
it to the required area.
· This unit is also called as Data Processing Unit (DPU).
· MCU also consists of sub-units like control unit, decoding circuits, feed control
units, etc.
· Almost all the operations like tool movements, tool change, speed and feed
change and many others can be controlled by MCU.
2. Machine tool and NC tooling :
· It is the manufacturing arm of NC machine tool system.
· It receives the raw material and performs different operations like turning,
milling, drilling, grinding, etc.
· For performing these operations, it should receive the information from the MCU.
· As per the information, the desired shape and size is modified.
3. Part program and drawings :
· NC machine operates as per coded information, which is Input Data for the
machine.
· The feeding of this data may be manually or automatic.
· The manual feeding of input data includes operator and hence chances of error
increases.
· Hence, data is fed by automatic means and for this purpose punched tape is
mostly used.
· As punched tape is most widely used hence, it becomes a standard and due to
standardisation, similar tape punchers and tape readers are used in all the systems.
· Punched tape uses a binary coded decimal system for containing operating
information of NC tool.
· Punched tape have eight vertical columns (channels) numbering from 1 to 8 and
one feeding holes column between them.
· Channel 1 to 3 is on one side and 4 to 8 on another side of the punched tape.
· Also, it carries horizontal rows, which represent a code number, letter code or a
word.
· The instructions are marked on the tape in the form of holes in binary codes
format, which is decoded in MCU and electric pulse is generated.
· These pulses are fed further to the servo systems and mechanisms.
· Generally, these tapes are manufactured by paper, which may be oiled or unoiled.
4 - 5 Computer Aided Design and Manufacturing
Fundamental of CNC and Part Programming

222. · Punched tapes are cheap but they tear easily. Hence, laminated tapes are also
used.
· Fig. 4.2.2 shows standard EIA (Electronics Industries Association) punched tape.
· The other input media instead of punched tapes are :
m Punched cards m Magnetic tapes m Diskettes, etc.
· Now a days, to enter the program, instead of these media, magnetic cassettes,
floppy discs, compact disc (CD) are used.
· In many machines, MCU carries a keyboard also. This is used directly to
manipulate and feed the part program.
· Due to this method, there is saving in machining time, hence now a days it is the
most popular type of input media to feed part program.
4 - 6 Computer Aided Design and Manufacturing
Fundamental of CNC and Part Programming
Fig. 4.2.2 : Main features of a standard EIA Tape

223. 4.3 Classification of NC System :
The classification of NC systems can be done in the following ways :
1. According to tool positioning or modes of programming :
a) Absolute system b) Incremental system
2. According to motion control system :
a) Point to point system b) Straight line or straight cut system
c) Continuous or contouring path system
3. According to servo control system :
a) Open loop system b) Closed loop system
4. According to the types of feedback devices :
a) Analog transducer b) Digital transducer
4.3.1 According to Tool Positioning
a) Absolute system :
· In this system, all the positions are indicated from a reference point, which is a
fixed zero point or set point.
· Fig. 4.3.1 (a) shows that all positions are marked with a set point.
· In Fig. 4.3.1 (a) point 'A' is a set point.
b) Incremental system :
· In this method, the tool positions are indicated with respect to previous point.
· Fig. 4.3.1 (b) shows an example of this system.
4 - 7 Computer Aided Design and Manufacturing
Fundamental of CNC and Part Programming
Fig. 4.3.1 (a) : Absolute system

224. · The main disadvantage of this system is that if an error occurs into the
dimensions of any location, all the locations marked after that will carry the same
error.
4.3.1.1 Comparison of Absolute and Incremental System
Sr. No. Absolute positioning system Incremental positioning system
1. In this system, all the positions are
indicated from a reference point,
which is a fixed zero point or set
point.
In this method, the tool positions are
indicated with respect to previous
point.
2. The coordinates of each point are
independent of each other.
The coordinates of each point are
dependent on each other.
3. If an error occurs into the dimensions
of any location, then the error will be
restricted to that location only.
The main disadvantage of this system
is that if an error occurs into the
dimensions of any location, all the
locations marked after that will carry
the same error.
4. A reference point is a must here. Reference point is not needed here.
4.3.2 According to Motion Control System
a) Point to Point (PTP) system :
· In this system, tool is accurately located at some specified position.
· Fig. 4.3.2 (a) shows path of tool movement for drilling number of holes.
· The spindle is first brought to the starting point, then moved to the next location
i.e. hole 1 along the marked path.
· On that location, drilling operation is performed and then tool moves to next
location.
4 - 8 Computer Aided Design and Manufacturing
Fundamental of CNC and Part Programming
Fig. 4.3.1 (b) : Incremental system

225. b) Straight line or straight cut system :
· In this system, the cutting tool can be moved along a straight line only, which is
parallel to the principal axes of motion.
· It is not possible to combine the motion of axes. Hence, the tool motion is only
along the X-axis, Y-axis and Z-axis.
· Due to this, angular cuts cannot be produced. (Refer Fig. 4.3.2 (b)).
c) Continuous or Contouring path system :
· In this, there is relative motion between the tool and workpiece, during the whole
operation.
· Due to this relative motion, different curves and profiles can be cut.
· Actually, it is a combination of PTP and straight cut system.
· Fig. 4.3.2 (c) shows an example of continuous path system for a component on
NC-milling machine.
4 - 9 Computer Aided Design and Manufacturing
Fundamental of CNC and Part Programming
Fig. 4.3.2 (b) : Straight cut system
Fig. 4.3.2 (a) : Point to point system

226. 4.3.3 According to Servo Control System
Servo control is a group of electrical, mechanical, hydraulic and pneumatic devices,
which are used to control the slide position of NC machine tool.
a) Open loop system :
· It is a simpler and cheaper system.
· It involves feeding of tape, interpretation of information by tape reader, storing
the data in buffer storage.
· After storing, it converts into electrical signal and send this signal to the control
unit.
· The control unit is connected to servo control which controls the slide movement.
Refer Fig. 4.3.3 (a) which shows the block diagram of open loop NC system.
· In open loop system, there is no feedback, to ensure whether the obtained slide
movement is same as desired or not and if not, what error is present.
b) Closed loop system :
· This is almost similar to open loop system, only carries an additional feed back
device.
· This device is nothing but a transducer and accompanied by a comparator.
· As this is similar to open loop system, the motion is same upto servo control.
· The transducer fed back the slide displacement corresponding to the applied
signals, as shown in Fig. 4.3.3 (b).
· The comparator compares the obtained slide motion with applied slide motion and
error, if any, is fed back to control unit through an amplifier.
4 - 10 Computer Aided Design and Manufacturing
Fundamental of CNC and Part Programming
Amplifier
Fig. 4.3.3 (a) : Open loop NC system
Fig. 4.3.2 (c) : Continuous path system

227. · Then control unit sends correct commands to servo motor (control) and cycle
continues.
4.3.3.1 Comparison of Open Loop and Closed Loop System
Sr. No. Open loop system Closed loop system
1. It is a system that involves feeding of
tape, interpretation of information by
tape reader, storing the data in buffer
storage, converting it into electrical
signal and sending this signal to the
control unit.
It is a system that carries an
additional feedback device along with
a transducer, accompanied by a
comparator.
2. It is a simpler and cheaper system. It is more complicated and costly than
the open loop system.
3. Feedback device is absent. Feedback device is present.
4. With no feedback device, chance of
error is always present.
As the comparator compares the
obtained slide motion, chance of error
is greatly reduced.
4.3.4 According to Feedback Devices
A comparing mechanism is always needed to compare actual slide position with
applied slide position to ensure accuracy. Feedback devices are units which convey the
actual slide position to the control unit, so that comparison is easily done.
a) Analog transducers :
· It produces a variable electrical voltage, which varies with rotational speed of the
shaft.
· This voltage can be easily measured and converted into linear distances to
indicate machine tool position.
4 - 11 Computer Aided Design and Manufacturing
Fundamental of CNC and Part Programming
Fig. 4.3.3 (b) : Closed loop NC system

228. b) Digital transducers :
· It converts the rotary motion of machine screw into countable electrical pulses.
· The number of electrical pulses indicates linear distance moved by the machine
table corresponding to the lead screw rotation.
4.4 Advantages of NC System :
· High productivity : Due to less set up and lead time, productivity is higher.
· Less scrap : As human errors are eliminated, accurate components are machined
hence, scrap is reduced.
· Reduced jigs and fixtures : Work and tool positioning is done by NC tape,
hence less requirement of jigs and fixtures.
· High quality : Due to higher accuracy of NC systems, the quality of products is
easily controlled.
· Flexibility in design : In NC system complicated profiles can be easily
produced at faster rate.
· Utilization of manpower : In NC system, there is greater utilization of
manpower because, after setting a component, operator can perform other
operations.
· Reduction in the inventory.
· Safety to the operator and machine tool.
· There is a greater flexibility in the manufacturing.
· Less floor space is required.
· As no jigs and fixtures are required, hence tooling cost is low.
· Skilled operator is not required.
4.5 Disadvantages of NC System :
· High initial cost : Initial investment is high.
· High maintenance cost : Maintenance is costly and complicated.
· Costly control system : Control systems are also costly.
· Skilled operator : For part programming well trained and highly skilled operator
is required.
· Unemployment : As only one operator is required, there is increase in
unemployment.
4.6 Applications of NC System :
· NC system is used where 100 % inspection is required.
4 - 12 Computer Aided Design and Manufacturing
Fundamental of CNC and Part Programming

229. · NC system is suitable for machining of parts where frequent changes in design
occurs.
· Repetitive production of precise parts in small and medium size production can
be done by NC system.
· When accuracy requirement is high, NC system is suitable.
· When high amount of material is to be removed, NC system is preferred.
· For complex machining operations also, NC system is required.
4.7 Types of Numerical Control System :
With the same basic elements and same principle, NC machines can operate on
different systems of numerical control. The common types of NC systems used in machine
tools are :
1. Conventional Numerical Control (NC)
2. Direct Numerical Control (DNC)
3. Computerized Numerical Control (CNC)
4.8 Conventional Numerical Control (NC) :
· It is a hard wired NC system having IC's (Integrated Circuits) which are
permanently wired and arranged on circuit boards.
· It is a hardware based system, and it is difficult to change features of MCU.
· In conventional NC system, there are chances of mistakes while punching of tape.
· There are no provisions for speed change, feed change hence, not suitable for
large production.
· It is simple and cheaper than other NC systems.
4.9 Direct Numerical Control (DNC) :
· It is a manufacturing system in which a number of machines are controlled by a
central computer through a direct connection of telecommunication lines and in
real time.
· Instead of using a tape reader as in NC machines, the part program is transmitted
to the machine directly from computer memory. One computer can control more
than 200 separate machines.
· The computer used for DNC system is designed in such a way that, on demand it
will provide instructions to each machine tool.
· The Direct Numeric Control system consists of four components :
1) Central computer
4 - 13 Computer Aided Design and Manufacturing
Fundamental of CNC and Part Programming

230. 2) Bulk memory, which stores the NC part programs
3) Telecommunication lines
4) Machine tools. (Refer Fig. 4.9.1)
· There are two types of Direct Numeric Control system, these are :
m Behind the Tape Reader system (BTR)
m Special machine control unit
Advantages :
· Control of more than one NC machine.
· Elimination of punched tape and tape reader.
· Convenient storage of NC part programs in computer files.
· Greater computational capability and flexibility.
· The data for tools and cutters can be centrally maintained and updated.
· The data related to manufacturing can be effectively collected and hence,
inventory can be better controlled.
Disadvantages :
· The crucial disadvantage of Direct Numerical Control system is that, if the
central computer goes down, all machines become inactive.
· Initial cost is too high.
4.10 Computerized Numerical Control (CNC) :
· In CNC, there is absence of hard-wired logic systems.
· The functions of hard-wired are performed by the software program of the
computer. Hence called as software based system.
4 - 14 Computer Aided Design and Manufacturing
Fundamental of CNC and Part Programming
Fig. 4.9.1 : Direct Numerical Control system (DNC)

231. · A separate computer is attached with each machine tool, with stored
programmable logic, hence termed as self contained NC system.
· The computer which is used, is known as mini-computer.
· The main change in CNC is the hardware of NC is replaced with the software to
the maximum possible extent.
· The program is entered into the computer early through the tape, but now a days
through a keyboard.
· The program is stored in computer memory, which can be recalled whenever
required.
· Program can be easily edited and modified as per the requirement.
· An extra feature in this system is the diagnostic software which enables easy
trouble shooting, if CNC system fails.
· The cause can be easily detected and rectified through this software. Refer block
diagram of CNC system as shown in Fig. 4.10.1.
4.11 Constructional Features of CNC Machines :
A CNC machine, which is now a days very popular, consists of following features :
1. Machine structure 2. Drives
3. Actuation system 4. Slideways for machines
5. Automatic Tool Changer (ATC) 6. Automatic pallet changer
7. Transducers/Control system 8. Feedback devices
4 - 15 Computer Aided Design and Manufacturing
Fundamental of CNC and Part Programming
Fig. 4.10.1 : CNC system

232. 4.11.1 Machine Structure
The CNC machine tool structure consists of following main parts :
a) Bed b) Table c) Column
a) Bed :
· The bed of CNC machine is generally made of high quality cast iron with heavy
ribbing to provide high stiffness and low weight.
· The cast iron structure provides the necessary damping, to reduce the vibrations
produced due to high speed, large material removal rates and heavy duty
machining.
· Another area of consideration is the design from chip disposal point of view.
Fig. 4.11.1 shows a slant bed structure used in turning centres
· These allows the chips to fall off from the cutting zone. It also provides the
operator easier and better access to the workpiece and tooling.
b) Table :
· The table is mounted on the bed which provides the machining centre with the
z-axis (linear movement).
4 - 16 Computer Aided Design and Manufacturing
Fundamental of CNC and Part Programming
–
+
Z
–
+
X
Slant bed
Fig. 4.11.1 : Slant bed construction

233. c) Column :
· Column is mounted on the saddle and designed with high torsional strength, to
prevent distortion and deflection while machining.
4.11.2 Drives
· There are two basic applications where drives are used in CNC machines.
1) Spindle drives 2) Feed drives
1) Spindle drives :
· Spindle drives are used to provide the main spindle power for cutting.
· As large material removal rates are used in CNC, large power motors are used
for spindle drives.
· Also, the speed required during operations is infinitely variable.
· Hence to provide such a speed control for infinitely variable speed DC motors
are used.
· The speed control for DC motors can be achieved by varying the voltage
infinitely.
· The use of AC motors are preffered in the generation of currents in CNC
machine tools. This is achieved by developments in the frequency converter.
2) Feed drives :
· CNC machines are provided with independent axis drive to provide the feed
movements for the slides.
· In order to obtain fast response and positional accuracy a special type of motor
called servomotor is used to power the slides.
· Following are the feed drives that are used in the CNC machine tools :
i) DC servomotors ii) Brushless DC servomotors
iii) AC servomotor iv) Stepper motor
v) Linear motor
i) DC servomotor :
· DC servomotors are characterized by high overload capacity, excellent dynamic
response and low moment of inertia.
· These are made up of permanent magnet type and have high acceleration torque.
· The speed control in DC servomotors is achieved by flux control method, voltage
control method and rheostatic control.
ii) Brushless DC servomotor :
· In brushless DC servomotor, the motoring action achieved by electrical
commutation rather than by mechanical commutation.
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234. · The flux created by current carrying conductor should rotate around the inside of
the stator.
· The speed in this motor is proportional to the frequency of the applied voltage
and number of poles.
iii) AC servomotor :
· AC servomotors are used for low power servomechanisms and have constant
acceleration to maximum speed.
· These motors are highly reliable and have high frequency response.
· The speed control is achieved by controlling the applied controlled voltage.
· Following are some methods to control the speed :
m Supply frequency control
m Supply voltage control
m Controlling number of poles
m Adding external resistance in rotor circuit
m Cascade control
iv) Stepper motor :
· Stepper motors convert the input pulses into a precisely defined increment in the
shaft position.
· Each size or step angle is determined by construction of motor and type of drive.
· These motors are suitable for position control systems in plottors, disk drives,
machine tools, robotics, etc.
· The speed and position is controlled by input pulses.
v) Linear motor :
· Linear motors are widely used in high performance CNC machine tools.
· These motors give higher positional accuracy at higher feeds and speeds.
· Also, linear motors have higher acceleration and deceleration rate.
· The maximum speed of linear motor is limited by the bus voltage and speed of
control electronics.
4.11.3 Actuation System
An important element of actuation system is recirculating ball screw.
Recirculating ball screw :
· In this, the sliding friction is replaced by rolling friction.
· It consists of screw with circular form threads and nut assembly with internal
helical ball groove, to allow a continuous flow of steel balls.
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235. · Either the nut or the screw rotates and causes the rolling of balls through a
helical path, as shown in Fig. 4.11.2.
· A return tube deflects the balls and recirculates them.
· The rigidity of the drive system can be improved by preloading the ball screw
and nut assembly.
· Also, precise positioning can be achieved by the application of preloading. Here
axial displacement is eliminated and hence reduce the backlash.
· Preloading can be achieved by fitting the balls in the component tightly while
assembly.
· One of the method used for preloading the ball screws is keeping the spacer
between two nuts as shown in Fig. 4.11.4.
· Spacer provided between nut A and nut B avoid the axial movement of nut and
hence the balls are tightly fitted in the gap.
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Fine pitch worm gear
Sector gear
Ball bearings
Recirculation channel
Nut (shown cutaway)
Fig. 4.11.2 : Re-circulating ball screw
Circular form
threads
Nut
Return tube
Ball screw
Balls
Fig. 4.11.3 : Recirculating ball screws

236. Comparison of Conventional Screw and Re-circulating Ball Screw
Table 4.11.1 : Conventional screw and Re-circulating ball screw
Sr.
No.
Parameter Conventional power screw
(Square, trapezoidal)
Re-circulating ball screw
Advantages of recirculating ball screw
1. Efficiency There is sliding friction,
hence high input torque is
required to overcome
friction. Thus efficiency of
screw is as low as 40 %.
There is rolling friction,
hence low input torque is
required to overcome
friction. Thus efficiency of
screw is as high as 90 %.
2. Load carrying capacity It has lower load carrying
capacity as compared to
recirculating ball screw.
It has high load carrying
capacity as compared to
conventional power screw.
For the same load carrying
capacity, recirculating ball
screw is more compact and
light weight.
3. 'Stick-slip' phenomenon In conventional power
screw, 'stick-slip'
phenomenon is observed.
This is due to difference
between the value of
coefficient of static friction
and coefficient of sliding
friction.
The operation of
recirculating ball screw is
smooth and free from any
'stick-slip' phenomenon.
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Nut B Spacer Nut A
Fa
FB
FA
Fig. 4.11.4 : Preloading of the recirculating ball screw and nut arrangement

237. 4. Compensation for wear
and tear
It requires periodic
adjustment to compensate
for wear on the surfaces of
the screw and nut.
It is virtually wear-free due
to presence of lubricating
film between the contacting
surfaces.
Limitations of recirculating ball screw
5. Cost It is low at initial cost. The initial cost of
recirculating ball screw is
very high.
6. Special operating
environment
It can be operated in any
environment with
satisfactory life.
It requires high degree of
cleanliness and restricted
entry of foreign particles.
7. Lubrication It can be easily lubricated
by grease.
It requires a continuous thin
film of lubricant between
the balls and grooves in the
nut and screw.
8. Self-locking and
over-hauling
Due to high friction between
thread surfaces, these
screws are self-locking.
Due to negligible friction
between balls and thread
surfaces, these screws are
over-hauling. Hence special
brake is required to hold
the load in its place.
Applications of Recirculating Ball Screw
Re-circulating ball screw is used in high speed applications such as :
(i) Automobile steering gears
(ii) Power actuators
(iii) Hospital bed adjusters
(iv) Machine tool controls
(v) X-Y recorders of CNC machines.
4.11.4 Slideways for Machines
· Precise positioning and repeatability of machine tool slides are the major
functional requirements of CNC machine.
· To eliminate stick slip, there are different slideway systems such as rolling
friction slideway and slideways with low friction PTFE (Poly Tetra Fluoro
Ethylene).
· These slides have low wear, good vibration damping, easy machinability, low
coefficient of friction and low price.
· The plastic coated slideway have static coefficient of friction, which is less than
dynamic coefficient of friction.
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238. · With increase in speed, dynamic coefficient of friction increases upto a certain
value and then remains constant.
· Following are the techniques used to meet requirements in CNC machine tool
slideways :
i) Hydrostatic slideways ii) Linear bearings with balls
iii) Rollers or Needles iv) Surface coatings
i) Hydrostatic slideways :
· Fig. 4.11.5 shows the working of hydrostatic slideways.
· The carriage or slides in contact with the slideway are provided with the small
pockets or cavities.
· Air or oil is pumped into small pockets and then flows out from pockets between
slide and slideways.
· These slideways provides almost frictionless condition for slide movement. The
slideways should be kept clean for the efficient operation.
· The only disadvantage is that, it requires very large surface area to provide
proper support.
ii) Linear bearings with balls :
· In this technique, the sliding friction is replaced with rolling friction.
· Fig. 4.11.6 shows the linear ball bearing with recirculating balls.
· These ball bearings are designed to give frictionless movement for shaft rotation
as well as over varying strokes of length with high linear precision.
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Oil
inlet
Oil pocket
Oil inlet
Fig. 4.11.5 : Hydrostatic slideway

239. iii) Rollers or Needle bearing :
· Rollers or needle bearings are also called as tachoway and are provided for the
movement along a flat plane.
· Linear roller bearings provides unlimited linear movement.
· Rollers in the roller bearings are guided between shoulders of the supporting
element with very close tolerance. Refer Fig. 4.11.7.
· The guiding element prevents the falling of rollers from the shoulders and also
recirculate them with ease.
· While using the rollers the bed should be machined accurately and surface in
contact should be hardened.
iv) Surface coatings :
· In this technique, the guiding surface is coated with low friction material such as
polytetrafluroethylene (PTFE).
· Sometimes, replaceable strips of low friction material are used on guide ways.
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Rollers
Fig. 4.11.7 : Linear roller bearing
Shaft
Ball
Cage
Fig. 4.11.6 : Linear ball bearing

240. 4.11.5 Automatic Tool Changer (ATC)
· The complicated jobs can be machined on NC and CNC machines and for that
they require different types of tools.
· For changing and resetting the tool, more time is required.
· For this purpose, tools can be automatically changed with the help of Automatic
Tool Changer i.e. ATC.
· By using ATC, a complete job can be machined in one pass, on one machine,
with one program only.
· Hence, productivity and repeatability of manufacturing increases.
· As per part program, the machining centre select a desired tool and machining is
done.
· Generally tools are stored in drum type or chain type magazine as shown in
Fig. 4.11.8 (a) and (b).
· Fig. 4.11.8 (c) shows a 180º type of rotation tool changer mechanism.
4 - 24 Computer Aided Design and Manufacturing
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(c)
Fig. 4.11.8 : Automatic tool changer

241. · Generally, machining centres with 16 to 24 tools are used, but now a days
machining centres with 160-200 tools are used.
· MCU receives tool change command and send spindle to its fixed tool change
co-ordinates.
· At that time, tool magazine is indexed to the proper position and tool changer
rotates.
· This tool changer engages the tool in the spindle and tool in the magazine at that
time.
· Both the tools are removed by tool changer from their places and turns by 180º
to swap both tools.
· Thus old tool is returned to magazine and new in the spindle. Hence, completes
the tool changing operation.
4.11.6 Automatic Pallet Changer (APC)
· Machine downtime because of loading-unloading, clamping-releasing, etc. can be
minimized with the help of automatic workpiece loader/unloader system.
· In this system, the workpieces are mounted on the pallet and the pallets are
moved around the machine in a logical manner. This system is called as pallet
changing system.
· According to logical movement of pallet, the system can be linear or rotary.
Linear pallet changer system :
· A typical linear pallet changer system is as shown in Fig. 4.11.9.
· In this system, the table moves in a linear motion, hence called as linear pallet
changer system.
· The workpiece can move in two ways.
i) Linear motion ii) Inverted U-path.
i) Linear motion :
· In Fig. 4.11.9 (a) the workpiece is on left side track waiting for completion of
machining operation of earlier workpiece.
· In Fig. 4.11.9 (b) after completion of earlier workpiece, it moves onto the
unloading table and the next component is ready to move onto the machining
table.
· In Fig. 4.11.9 (c) the next component moves onto the machining table and the
system continuous.
ii) Inverted U-path :
· In Fig. 4.11.9 (d) the table is in linear motion but the component is rotated in an
inverted U-path to move onto the machining table and then moves linearly.
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242. Rotary pallet changer system :
· Rotary pallet changer system, is shown in Fig. 4.11.10 which is same as linear
pallet changer system except that the table is rotated for the movement of the
workpieces.
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Table 1
Table 2 Setting
station 1
(a) (b) (c)
ATC
Machine
spindle
Pallets
(d) Inverted U-path
Fig. 4.11.9 : Linear pallet changer system
Machine
spindle
Pallets
Fig. 4.11.10 : Rotary pallet changer system

243. · For the maching purpose of workpieces, the table is moved in rotary motion with
the help of indexing mechanism.
4.11.7 Transducers/Control Elements
· The control unit should indicate the current status and position of various
machine tool elements.
· The control unit part, for allowing manual control and programming of machine,
may be housed on machine structure itself.
· To monitor the position of slides, linear and rotary transducers are used.
4.11.8 Feedback Devices
· In closed loop control system, feedback devices are required for proper position
control of slides or drives, for holding workpieces and for controlling tool
motions.
· The feedback devices can be either of rotary or linear form.
· There are two types of rotary transducers which are resolvers and encoders that
can be connected directly to the ball screw.
· Linear transducers have a portion attached to the structure and their other part is
fixed to the slide which moves over the stationary part.
Encoder (Rotary transducer)
· Encoders are numerical devices which indicates output in digital form directly
and widely used as position and motion sensors.
· It consists of a glass disc with accurately etched lines at regular intervals. Refer
Fig. 4.11.11.
· The glass disc rotates between the light source and photodiodes.
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Light source
Photo diode
Glass disc
Fig. 4.11.11 : Encoder

244. · The lines make and break this photoelectric beam which generates a pulse signal
and this signal is amplified to give a square wave output.
· Number of signals generated per revolution depends on the number of lines on
the disc.
Linear transducer :
· The principle of linear transducer is similar to rotary transducer except that, the
signal directly translates into linear displacements of the slide.
· In linear transducer instead of disc, glass scale with line grating is used, which
have line graduations. Refer Fig. 4.11.12.
· The relative movement between glass scale (fixed to slide) and photocells (fixed
to guide) generates an electric pulse.
· The amount of pulses produced for a given resolution of the gratings decide the
magnitude of the travel.
4.12 Advantages and Disadvantages of CNC Machines :
Advantages
Advantages of CNC machines are similar to NC machine. Some additional advantages
due to additional feature in CNC machine over conventional machines are as follows :
· Program storage : As computer is available, hence multiple programs can be
stored in the machine.
· Reliability of system : As the data is directly entered with the help of computer,
no need to use punched tape. It also improves reliability of the system.
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Light source
Slide motion
Photocells
Grated
glass
scale
Fig. 4.11.12 : Linear transducer

245. · Online part programming : The part program can be done online with editing,
if required.
· Flexibility of system : The system is too flexible, as new systems can be added
at low costs.
· Metric conversions : Part program, which is written in inches, can be easily
converted into millimeters i.e. metric conversion is easy.
· Interpolations : In NC system, there is interpolation for straight and circular
path, but in CNC it is available for helical, parabolic and cubic curves also.
· Expanded tool compensations : For the purpose of tool offset and tool wear,
tool compensation is provided.
Disadvantages
Disadvantages of CNC machines are similar to NC machines.
· High initial cost : Initial investment is high.
· High maintenance cost : Maintenance is costly and complicated.
· Costly control system : Control systems are also costly.
· Skilled operator : For part programming well trained and highly skilled operator
is required.
· Unemployment : As only one operator is required, there is increase in
unemployment.
· Computer problem : If there is any problem with computer, then the whole
machine will get stop.
· Costly software : The software required for the operation of CNC machines is
also costly.
4.13 Comparison between NC, CNC and DNC System :
Sr. No. Parameters NC CNC DNC
1. Flexibility Less High High
2. Tape editing Not possible
at site
Is possible Is possible
3. Productivity Less High Highest
4. Number of programs
stored
Only one at
a time
Multiple programs
can be stored
Multiple programs
can be stored.
5. Number of operations
done at a time
One One Multiple
6. Initial cost Low High Highest
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246. 4.14 Adaptive Control System (ACS) :
· Adaptive control system is the logical extension or improvement over the NC and
CNC systems.
· In conventional systems, the total production time is increased due to the
non-productive time such as time required for workpiece handling, setup, tool
changing time, operators delay, lead time between order processing and
production, etc.
· Hence NC, CNC systems find major application in the areas where reduction in
non productive time is the primary requirement.
· The adaptive control system automatically determines the process variables such
as cutting speed, feed, depth of cut during the machining process and make
changes in the prescribed limit as per the requirement. Refer Fig. 4.14.1.
· Also, it makes optimal use of machine capability to reduce the non-productive
time.
· Following are the two common systems of adaptive control :
1. Adaptive Control with Optimisation (ACO)
2. Adaptive Control with Constraints (ACC)
1. Adaptive Control with Optimisation (ACO) :
· In this system, the overall performance of the process is indicated by performance
index (PI) or merit figure and it is given by,
PI =
Material Removal Rate(MRR)
Tool Wear Rate(TWR)
· Sensors mounted on the machine tool measures the various parameters such as
tool wear, cutting temperature and torque, machine vibration, etc.
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Input
Program
NC system
NC machine
tool
Sensors
Adaptive
controller
Speed and feed
corrections
Process constraints
Strategy
Performance
index
Measured
data
Command
signals
Position
feedback
Fig. 4.14.1 : The adaptive control system

247. · These sensors fed this data to adaptive controller along with the process
constraints such as speed, feed rate, etc.
· This performance index is then compared with the set value and maintain it by
continuously changing the process variables.
2. Adaptive Control with Constraints (ACC)
· In this system the various constraints such as torque, cutting forces, motor power,
tool wear, cutting temperature, etc. are specified.
· When the process is in progress, the ACC system manipulate spindle speed, feed
rates to maintain the specified constraints in prescribed limit.
· These constraints are measured with the help of sensors and transducers and
compare with the set value.
4.15 Machining Centre :
· Generally, each and every machine tool is designed to perform basically one type
of operation. But, during the manufacturing process most of the components
require various operations on their different surfaces.
· For example, consider any workpiece on which various machining operations like
milling, drilling, boring, threading, etc are to be performed.
· The conventional method to perform these operations is move the workpiece from
one machine tool to another machine tool until all the operations are completed.
But, this method will take more time and may be some errors.
· For this purpose Computerised Numerical Control (CNC) machine tool is used
which is designed to perform various cutting operations on different surfaces of
workpiece.
· CNC machines are available in the form of Lathe (CNC-Lathe), Milling
(CNC-Machining center), EDM (CNC-EDM), etc.
· Machining centre or CNC milling machine is capable of performing milling,
drilling, boring, counter-boring, threading and so many operations.
· In these machines, the workpiece does not have to be moved to another machine
tool for other operations.
· Some of the machining centres are provided with two work tables called as
pallets.
· When the workpiece on one pallet is being machined, the operator set the
workpiece on the free pallet. After machining, the pallet changer moves the pallet
of finished workpiece away from the operator and the other pallet comes with the
new workpiece for machining.
· In machining centres, generally the workpiece is stationary and the tool is moving
(rotating).
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248. · Machining centres are classified as follows :
i) Horizontal machining centre ii) Vertical machining centre.
4.15.1 Horizontal Machining Centre (HMC)
· HMC carries horizontal machining
spindle (head) which can slide along
the horizontal guideways. Refer
Fig. 4.15.1.
· In these machines, software is used to
move the tool or workpiece.
· These machines are used for machining
of large workpieces on its all the
surfaces.
· HMC's are very heavy in construction.
Fig. 4.15.1 shows the principal parts of
HMC, which are described as follows :
i) Bed : It is a heavy structure which
supports the complete machine and
carries guide-ways over its top surface.
It is generally made of cast iron.
ii) Saddle : It is mounted over the
guide-ways on the bed and also carries
column over it. Generally, it provides
X-axis movement to the machining
centre.
iii) Table : It is mounted over the guide-ways provided on the saddle. It is made of
cast iron. For mounting the work holding devices, T-slots are provided on the
table. It provides Z-axis movement to the machining centre.
vi) Column : It is mounted over the saddle. It provides Y movement to the machining
centre. The column can be of fixed type or travelling type.
v) Automatic Tool Changer (ATC) : It is used to change the tool from the machine
spindle. It is placed closed to the spindle and enables the tool change rapidly.
vi) Spindle and servo system : Spindle is mounted on the headstock and it provides
Z-axis movement to the machining centre. Servo system consists of servo motors
and feedback system. It provides accurate and rapid movement along all the axes.
4.15.2 Vertical Machining Centre (VMC)
· It carries a vertical machining spindle (head) which can slide along the vertical
guide-ways provided on the column. Refer Fig. 4.15.2.
4 - 32 Computer Aided Design and Manufacturing
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Table
Bed
Saddle
Spindle
Column
+
–
+
–
+
–
X
Z
Y
Fig. 4.15.1 : Horizontal machining centre

249. · The vertical head can be tilted
(swivelled) in either direction.
· These machines are suitable for
machining flat surfaces with deep
cavities like the manufacturing of
moulds, dies, etc.
· These machines are also very heavy
in construction. Fig. 4.15.2 shows
the principal parts of VMC which
are almost similar to HMC.
i) Bed : It is a heavy structure which
supports the complete machine and
carries guide-ways over its top
surface. It is generally made of cast
iron.
ii) Saddle : It is mounted over the
guide-ways on the bed and also
carries column over it. It provides
Y-axis movement to the machining
centre.
iii) Table : It is mounted over the
guide-ways provided on the saddle.
It is made of cast iron. For
mounting the work holding devices,
T-slots are provided on the table. It provides X-axis movement to the machining
centre.
iv) Column : It is mounted over the saddle. It provides Z-axis movement to the
machining centre.
v) Automatic Tool Changer (ATC) : It is used to change the tool from the machine
spindle, rapidly.
vi) Spindle and servo system : Spindle is mounted on the headstock and it provides
Z-axis movement to the machining centre. Servo system consists of servo motors
and feedback system. It provides accurate and rapid movement along all the axes.
Advantages of machining centre :
· Machining centres have high metal removal rate (MRR) capability.
· Machining centres are highly versatile.
· It increases productivity.
· It consists of automatic tool changer (ATC) and automatic pallet changer system
(APC), hence it is more flexible and economical then the conventional machines.
4 - 33 Computer Aided Design and Manufacturing
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Table
Bed
Saddle
Spindle
Column
+
–
+
–
+
– X
Z
Y
Fig. 4.15.2 Vertical machining centre

250. · As various operations can be performed at one place, handling of the workpiece
and time of machining is minimum.
· Due to less handling of the workpiece, errors are minimum.
· It can machine the components with closed tolerances.
· Complicated components are also machined very easily.
· Faster cutting speed, heavier cutting depths and feeds can be obtained.
4.16 Program Reader :
It is a device used to read the coded instructions from the program of instructions.
They are classified on the basis of programming input medium.
1) Punched tapereader :
· When a punched tape is passed through a tapereader, the electric connections are
either close or open depending on whether there is a hole punched at a particular
track or not.
· The coded instructions on tape are transformed and utilized for various machine
tool functions.
· The commonly used tapereaders are :
i) Pneumatic ii) Photo electrical iii) Mechanical.
2) Card readers :
· It reads the information punched into a card, converting the presence or absence
of hole into an electric signal representing a binary 0 or 1.
· It operates at speeds ranging from 12 to 1000 cards per minute.
4.17 New Trends in Tool Materials :
· Trends in the manufacturing industries leads to the new trends in the tool
material.
· Now-a-days changes in the various workpiece materials and manufacturing
processes affects the material to be used for tooling.
· As the industries continually looking for new manufacturing materials that are
lighter and stronger, the tool makers must develop the tools that can easily
machine these materials keeping the highest possible rate of productivity.
· New trends in the tool material involves the various combinations of tool material
compositions, coatings and tool geometries.
New tool materials :
· Refer section 6.4 for various tool materials used for tooling.
· Along with these materials, polycrystalline diamond (PCD) cutting tools and
polycrystalline boron nitrides (PCBN) cutting tools are presently dominate in
turning, milling, drilling of various alloys.
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251. · Cermets cutting tools, comprised of titanium carbonitride (TiCN), are hard and
chemically stable, leading to high wear resistance.
· Cermet tools are also effective in dry machining.
· Development in ceramic tool technology enables these tools to move into new
areas of applications. For example, silicon nitride tools offer improved fracture
resistance as compare to other ceramic materials.
Coatings :
· Coatings for tool inserts can be classified as Chemical Vapour Deposition (CVD)
coatings and Physical Vapour Deposition (PVD) coatings.
· These coatings includes titanium carbonited, titanium aluminium nitride which
offers high hardness, increased toughness and wear resistance.
· Recent development in coatings includes soft coatings that are used in dry
machining.
· Diamond coated tools are used for machining hard materials.
4.18 Tool Inserts :
· Inserts are removable cutting tips used in CNC tooling for high cutting speeds
and feeds.
· Tool inserts are usually indexable. They can be rotated, flipped or changed
without disturbing the overall geometry of the tooling system.
· These inserts are available in different shapes with varying geometry.
· Inserts may be in the form of triangle, square, round, rhombus, diamond shapes
with different angles.
· Coated and uncoated cemented carbides are most commonly used cutting inserts
in the market.
Coated Inserts :
· Coated inserts are used while working with ferrous materials such as steel, cast
iron, iron, stainless steel, etc.
· For machining super alloys, it is best to use coated inserts.
· Also it is beneficial to use coated inserts for titanium alloys.
· Both the coating material and coating processes are considered while selecting
inserts for a particular operation.
· Physical or chemical vapour deposition coatings are used for carbide tool
materials to improve productivity and tool life.
· Coating materials : TiN, TiC, Al2O3 , TiCN, TiAlN, etc.
Uncoated Inserts :
· Uncoated inserts are used while machining soft materials due to its sharp and
uncoated cutting edge.
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252. · It is ideal for non-ferrous materials such as aluminium, brasses, bronzes and
many composites and wood.
· Uncoated inserts avoid formation of built up edge.
4.19 Work Holding in CNC Machines :
· CNC machines performed number of operations using variety of tools, on
different faces of the workpiece to accomplish finished component in single
setting.
· This requires that the workpiece should be operated from different sides without
repositioning it.
· Hence, the work holding devices in the CNC machine tool has to bear
multidirectional cutting forces.
· For very simple components conventional work holding devices such as chucks,
vices are used but for complex shapes of workpiece it becomes necessary to use
modular fixturing for work holding purpose.
· Grid plate modular fixturing facilitates precise and exact positioning of the
component.
· Also, this fixture in conjunction with a rotory table will allow to be used as an
indexing fixture.
· It provides clamping of more workpieces in a single fixture.
· This type of work holding reduces the clamping and unclamping time.
Requirements of work holding devices
· Work holding devices should restrict the linear and rotory movement of the
workpiece.
· It should allow quick loading and unloading of the workpiece.
· Holding force as well as cutting force should not distort or deflect the workpiece.
· It should ensure the proper loading of the workpiece.
· It should allow the workpiece to be operated on different faces in single setting.
· It should have provision for easy chip removal.
4.20 Axis Nomenclature for CNC Machines :
· A program in CNC system, specify the various axis about which motion is
required.
· For this purpose, a standard axis system is considered due to which relative tool
position with respect to work must be obtained.
4 - 36 Computer Aided Design and Manufacturing
Fundamental of CNC and Part Programming

253. · A right hand rule for machine tool axis is as shown in Fig. 4.20.2 (a).
· Machine tool co-ordinate axis is defined for providing a means of locating a tool
in relation to the workpiece.
· According to the machine, a point can be located by several methods. Generally,
that point is at the origin.
· In NC machine, the origin is defined in two ways which are fixed zero and
floating zero.
· In a fixed zero method, the origin is always predefined. It is generally at the
lowermost left hand corner of the worktable. Refer Fig. 4.20.1 (a).
· All other points on the worktable are defined from this point.
· But in modern NC machines, floating zero concept is provided which allows the
operator to define his origin.
· This makes it convenient to develop programs of symmetrical components by
providing the origin at the point of symmetry in the workpiece.
Refer Fig. 4.20.1 (b).
· This setting of the zero is done manually by
the operator, by positioning the tool about
the point at which origin is to be defined
and by pressing the zero button at that
point.
· Fig. 4.20.2 (b) shows the axis nomenclature
for different machines.
· The motion of three axes i.e. X, Y and Z
are specified as follows :
m Z-axis is always spindle axis or parallel to spindle axis.
m X-axis is always horizontal axis and parallel to the surface of the work.
m Y-axis is perpendicular to both X and Z-axis.
4 - 37 Computer Aided Design and Manufacturing
Fundamental of CNC and Part Programming
(0, 0)
(0, 0)
(a) Fixed zero (b) Floating zero
Fig. 4.20.1 : Fixed zero and floating zero
Fig. 4.20.2 (a) : Machine tool
co-ordinate axis

254. Example 4.20.1 : Explain with neat sketch 2, 21 2
/ and 3 axes of CNC machines.
Solution. : i) 2 axes of CNC Machines :
If the machine tool controls 2 axes at the same time, then it is called as a 2-axes CNC
machine.
4 - 38 Computer Aided Design and Manufacturing
Fundamental of CNC and Part Programming
+X
+Y
+Z
+Y
+Z
+X
+X
+Y
+Z
+X +Z
+X
+Y
+Z
+Y
+X
+Z
(Drilling) (Turning) (Milling)
Fig. 4.20.2 (b) : X, Y, Z motions
Z Axis control plane
Tool
path
+ Z
+ Y
+ X
XY Plane
Fig. 4.20.3 2-Axes Machine tool

255. ii) 2 1
2 axes of CNC Machines :
If the tool can be controlled to
follow an inclined Z-axis control
plane, the machine tool is called as
2 1
2 axes CNC machine.
iii) 3 Axes machine tool :
If the machine tool can control
X, Y and Z axes simultaneously at
the same time, then the machine tool
is called as 3 axes machine tool.
4.21 Part Programming :
· It is a set of instructions which instruct machine tool about processing steps to be
performed to manufacture the component.
· The various techniques used for generating CNC instructions are as follows :
m Manual CNC part programming.
m Computer assisted part programming.
m CAD-CAM based programming.
m Modelling based programming.
m Automatically programmed tools (APT).
· Instructions given by part program carry dimensional and non-dimensional data,
which is written in specific format.
4.21.1 Manual Part Programming
· The program for machining any type of workpiece not only varies from person to
person but also varies from machine to machine.
4 - 39 Computer Aided Design and Manufacturing
Fundamental of CNC and Part Programming
Tool
path
+ Z
+ Y
+ X
Fig. 4.20.5 Axes machine tool
Z Axis control plane
Tool
path
+ Z
+ Y
+ X
XY Plane
Fig. 4.20.4 2 1
2 axes of machine tool

256. · Various controllers use different syntax during instructing the machine tools.
· But, there are similarities between the codes and by understanding the basics, one
can easily adapt to other controllers with minor changes.
· It is necessary that, the programmer understands the different processes involved
by carefully studying the drawings, fixtures and machine tools.
· A typical block diagram of this process is shown in Fig. 4.21.1.
· It is a simplest method of part programming, in which program is written
manually on a paper.
· The program is nothing but an instruction set for machine tool which defines the
tool position relative to the workpiece.
· After verifying the program, corresponding punch tape is prepared.
· Each line of program is called as block, which consists of an operation number
word, data word, etc.
· The format of each block is as follows :
N…. G…. X…. Y…. Z…. U…. V….…. W…. F…. S…. T…. M….;
Above each letter signifies a particular operation, which is as follows :
a) N is a Sequence Number and used for giving the number to the lines. It is
generally, minimum three digit number. It is written as N001, N002, N003,
etc.
b) G is a Preparatory Function which changes the control mode of the
machine and called as G-codes. Generally, G-codes are followed by two
digit number. It is written as G01, G02, etc. Some common G-codes are
tabulated in the Table 4.21.1.
c) 'X, Y, Z and U, V, W' represents co-ordinate positions of tools. For two axes
system, only two letters are specified. In case of multiple axes (Milling
machine) other additional letters i.e. 'U, V, W' are specified. 'X, Y, Z' can be
positive or negative according to dimension. Generally 'X, Y, Z' are called as
dimensional data.
d) F is the Feed rate function, which defines feed rate of operation. For
example, F100, it means feed rate is 100 mm/min. If it is specified once,
then no need to specify again. It continues unless and until another value is
specified.
4 - 40 Computer Aided Design and Manufacturing
Fundamental of CNC and Part Programming
Tape Program Drawing
Punch Programmer
Machine tool
MCU
Fig. 4.21.1 : Manual part programming

257. e) S is a Cutting Speed Function which specifies spindle speed. For example,
S2000 means spindle is rotating at a speed of 2000 rpm.
f) T is a Tool Change Function. Generally, all the CNC machines are having
'ATC'. For programming, each tool is associated with an index number. For
example, T-04 it means tool number 4 is in ready position.
g) M is a Miscellaneous Function which is generally called as M-codes. By
specifying M-codes, other auxiliary operations are performed. Some common
M-codes are given in the Table 4.21.2.
h) ; is called as End of Block (EOB) which is written after each and every
line.
G code Function G code Function
G00 Positioning (rapid
traverse)
G42 Tool nose
compensation right
G01 Linear interpolation
(feed)
G63 Tapping mode
G02 Circular interpolation
clockwise
G65 Macro calling
G03 Circular interpolation
anti-clockwise
G66 Macro modal call
G04 Dwell G67 Macro modal call
cancel
G10 Data setting G68 Mirror image for
double turret ON
G17 XPYP plane selection G69 Mirror image for
double turret OFF
G18 ZP XP plane selection G70 Inch data input
G19 YP ZP plane selection G71 Metric data input
G20 Outer diameter /
internal diameter
cutting cycle
G72 Finishing cycle
G21 Thread cutting cycle G73 Stock removal in
turning
G22 Stored stroke limit
function ON
G74 Stock removal in
facing
G23 Stored stroke limit
function OFF
G75 Pattern repeating
G24 End face turning cycle G76 Peck drilling on Z-axis
G25 Spindle speed
fluctuation detect OFF
G77 Grooving on X-axis
4 - 41 Computer Aided Design and Manufacturing
Fundamental of CNC and Part Programming

258. G26 Spindle speed
fluctuation detect ON
G78 Multiple threading
cycle
G27 Reference point return
check
G81 Drilling cycle on milling
G28 Return to reference
point
G84 Tapping cycle
G30 2
nd
, 3
rd
, 4
th
reference
point return
G88 Boring manual dwell
G31 Skip cutting G90 Absolute programming
G33 Thread cutting G91 Incremental
programming
G34 Variable-lead thread
cutting
G92 Co-ordinate system
setting, maximum
spindle speed setting
G36 Automatic tool
compensation X
G94 Per minute feed
G37 Automatic tool
compensation Z
G95 Per revolution feed
G40 Tool nose radius
compensation cancel
G96 Constant surface
speed control
G41 Tool nose radius
compensation left
G97 Constant surface
speed control cancel
Table 4.21.1 : List of G-codes
M Code Function M Code Function
M00 Program stop M18 Turret reverse rotation
M01 Optional stop M22 Chip conveyor forward
M02 Program stop-reset M23 Chip conveyor reverse
M03 Spindle normal rotation M24 Chip conveyor stop
M04 Spindle reverse rotation M30 Program stop-Reset and
rewind
M05 Spindle stop M31 Tailstock base unclamp
M06 Tool change M32 Tailstock base clamp
M08 Coolant ON M40 Spindle neutral gear
4 - 42 Computer Aided Design and Manufacturing
Fundamental of CNC and Part Programming

259. M09 Coolant OFF M41 Spindle low gear
M10 Chuck clamp M42 Spindle high gear
M11 Chuck unclamp M51 Air blow ON
M12 Tail stock quill OUT M52 Air blow OFF
M13 Tailstock quill IN M70 Tool presetter arm down
M17 Turret forward rotation
Table 4.21.2 : List of M-codes
4.21.2 Preparatory Functions
· Preparatory function changes the control mode of the machine.
· G-codes are generally followed by two digit number. Refer Table 4.21.1.
· Out of Table 4.21.1, few G-codes are required in almost all programs and hence
discussed below :
i) G00 (Rapid traverse function) :
· It is the positioning function and enables rapid movement for tool positioning.
· This code is used during a typical situation in a machining operation where, the
tool is to be positioned near the cutting surface in smallest possible time, without
any machining.
· This code remains valid till it is cancelled by another G code.
ii) G01 (Linear interpolation function) :
· Any machining during straight or taper lines is done using this function G01.
· The feed rate at which the cutting tool is required to move is also specified by
using G01.
· The use of G00 and G01 is explained with the help of Fig. 4.21.2 and
corresponding instruction blocks :
N001 G00 X10 Y40;
(From origin (0,0) the tool moves at rapid feed rate to position X = 10 and Y = 40
i.e. point P)
N002 G01 X30 Y10 F200;
(In linear interpolation the tool moves to point Q i.e. X = 30 and Y = 10 with a
feed of 200 mm/min).
4 - 43 Computer Aided Design and Manufacturing
Fundamental of CNC and Part Programming

260. iii) G02/G03 (Circular interpolation function) :
· If the cutting tool is required to move along an arc then circular interpolation,
functions are used. It permits the cutting tool to move along an arc of circle in
clockwise or counter-clockwise direction.
· When the circular interpolation is to be used, it is necessary to determine the
plane in which the arc is positioned. For this purpose plane selection codes G17,
G18 or G19 are used.
· To define the movement of cutting tool along the circle, find the end point
coordinates and the respective radius vectors in a given plane (write the values of
I J K i.e. coordinates of centre of arc with respect to starting point).
· Sometimes, I J represents the coordinates of the centre of the arc. But generally
this method is not used.
G02 (Clockwise circular interpolation)
· For clockwise circular interpolation G02 code is used. The use of G02 is
explained with the help of Fig. 4.21.3 and corresponding instruction blocks :
N001 G02 X50 Y30 I–10 J–30;
(Clockwise circular interpolation from A to B, where I J represents the value centre
of arc with respect to A). Refer Fig. 4.21.3 (a).
· If direct radius method is used then,
N001 G02 X50 Y30 R31.62;
(Clockwise circular interpolation from A to B, where R= 30 10 31.62)
2 2
+ = .
· For Fig. 4.21.3 (b) we can write,
N002 G02 X30 Y20 I–40 J20;
(clockwise circular interpolation from A to B).
· If direct radius method is used then,
N002 G02 X30 Y20 R44.72;
4 - 44 Computer Aided Design and Manufacturing
Fundamental of CNC and Part Programming
10 20 30 40
10
20
30
40
0,0
P
Q
Y
X
Fig. 4.21.2 : Linear interpolation

261. (Clockwise circular interpolation from A to B, where R = 40 20 44.72)
2 2
+ = .
G03 (Anticlockwise circular interpolation)
· For anticlockwise circular interpolation G03 code is used. The use of G03 is
explained with the help of Fig. 4.21.4 and corresponding instruction blocks :
N001 G03 X30 Y50 I–30 J–10;
(Anticlockwise circular interpolation from B to A). Refer Fig. 4.21.4 (a).
· If direct radius method is used then
N001 G03 X30 Y50 R31.62;
(Anticlockwise circular interpolation from B to A, where R = 30 10 31.62)
2 2
+ = .
4 - 45 Computer Aided Design and Manufacturing
Fundamental of CNC and Part Programming
10
20
30
40
50
60
10
(0, 0) (0, 0)
20 30 40 50 60
J = 30
I = – 10
A
B
R
10
20
30
40
50
60
10 20 30 40 50 60
J = 20
I = – 40
A
B
R
X
Y
70
70
X
Y
(a) (b)
Fig. 4.21.3 : Clockwise circular interpolation
10
20
30
40
50
60
10
(0, 0) (0, 0)
20 30 40 50 60
J = 10
I = – 30
A
B
R
10
20
30
40
50
60
10 20 30 40 50 60
J = 50
I = – 10
A
B
R
X
Y
70
70
X
Y
(a) (b)
Fig. 4.21.4 : Anticlockwise circular interpolation

262. · For Fig. 4.21.4 (b) we can write,
N002 G03 X60 Y50 I–10 J50;
(Anticlockwise circular interpolation from B to A).
· If direct radius method is used then,
N002 G03 X60 Y50 R 50.99;
(Anticlockwise circular interpolation from B to A, where R = 50 10 50.99)
2 2
+ = .
Note : For XY plane I J values are specified, similarly for YZ plane J K values are
specified.
iv) G43/G49 (Tool length compensation)
· We know that, tools used for machining can vary in length.
· This method must be employed to compensate for these varied lengths.
· There are two methods to account of tool lengths.
i) Premeasuring the tools.
ii) Tool length compensation using CNC controller's.
· In the first method, the tool length is measured and known length can then be
added in the program's Z-axis dimensions to account for the tool. This is known
as presetting the tool.
· In a CNC machine setup, a programmable tool register is provided which is a
memory location in the computer where the tool length may be stored.
· When a particular tool is called, the required information for the tool offset is
called from the tool register.
4 - 46 Computer Aided Design and Manufacturing
Fundamental of CNC and Part Programming
Tool
offset
Tool
offset
Tool 1 Tool 2 Tool 3
Tool
offset
Fig. 4.21.5 : Tool length offset

263. · The MCU then shifts the Z-axis by the amount stored in the shift register.
Generally, the values of the offset are entered by the operator at the time of
programming. Refer Fig. 4.21.5 for tool length offset.
· Cutter length compensation is given by G43 and cancelled by G49.
v) G40/G41/G42 (Cutter radius compensation)
· During profile cutting operations, an allowance is provided to the cutter radius in
the programmed co-ordinates.
· Consider a component as shown in Fig. 4.21.6 in which the tool path centre line
is decided by the spindle axis centre-line, whereas the workpiece edge is offset
from it by the cutter radius.
· As per the cutter radius, the programmer will have to generate new co-ordinate
positions.
· Usually, the CNC machines have a built-in feature called as cutter diameter
compensation which allows the user to input the cutter diameter compensation for
each tool.
· Cutter compensation is accomplished by using following G codes :
(a) Cutter diameter compensation Left (G41) : When G41 command is given, the
tool will compensate to the left of the programmed surface when seen in the
direction of the tool movement. Refer Fig. 4.21.7.
4 - 47 Computer Aided Design and Manufacturing
Fundamental of CNC and Part Programming
Tool
path
Component
profile
Fig. 4.21.6 : Cutter diameter compensation
G 42
G 41
Fig. 4.21.7 : Cutter diameter compensation to the LEFT and RIGHT

264. (b) Cutter diameter compensation right (G42) : When G42 command is given, the
tool will compensate to the right of the programmed surface when seen in the
direction of the tool movement. Refer Fig. 4.21.7.
(c) Cutter diameter compensation cancel (G40) : G40 command cancels any
compensation as applied previously. The tool will change from a compensated
position to an uncompensated position.
vi) G90/G91 (Absolute and Incremental programming)
G90 (Absolute programming system)
· In this system, all the positions are indicated from a reference point, which is a
fixed zero point or set point.
· Fig. 4.21.8 (a) shows that all positions are marked with a set point.
· In Fig. 4.21.8 (a) point 'A' is a set point.
N001 G90 G01 X10 Y5 F200; (Linear interpolation from A to B in absolute mode).
N002 X20 Y10; (Linear interpolation and tool moves from B to C).
N003 X30 Y15; (Linear interpolation and tool moves from C to D).
G91 (Incremental programming system)
· In this method, the tool positions are indicated with respect to previous point.
· Fig. 4.1.82 (b) shows an example of this system.
· The main disadvantage of this system is that if an error occurs into the
dimensions of any location, all the locations marked after that will carry the same
error.
4 - 48 Computer Aided Design and Manufacturing
Fundamental of CNC and Part Programming
B
C
D
X
Y
Fig. 4.21.8 (a) : Absolute system

265. N001 G91 G01 X10 Y5 F200; (Linear interpolation from A to B in
incremental mode).
N002 X10 Y5; (Linear interpolation from B to C).
[Coordinates of C are given with respect to B].
N003 X10 Y5; (Linear interpolation from C to D).
[Coordinates of D are given with respect to C].
4.22 Procedure to Write a Part Program :
The general procedure to write a part program for any component is as follows :
· Study the given part or component carefully.
· Assume required data like feed, speed, cutter diameter, etc. and decide the path to
be followed by the cutter.
· Mark the tool path on a separate dimensionless drawing and give the number to
various points along the path.
· Write the coordinates of all the points on a separate table or on the drawn
drawing.
· The first four blocks and last two blocks are almost same in each program.
Note :
i) G codes, M codes and other data remains active or continue, unless and until it is
cancelled by other codes and data. Hence there is no need to write this data again
and again on each line. Even if it is written, it is not wrong.
ii) Number of blocks can be written as N001, N01, N1, N10, N100 in any of the way.
iii)We can write more than one G codes or M codes in one block.
iv)We can write any number as a program number.
v) The initial position (position 1 which is not reference position) can be assumed any
where near the workpiece surface.
4 - 49 Computer Aided Design and Manufacturing
Fundamental of CNC and Part Programming
A
B
C
D
X
Y
Fig. 4.21.8 (b) : Incremental system

266. 4.23 Part Programming for Lathe :
Example 4.23.1 : Write a manual part program to finish the stepped shaft in f 40 mm
section as shown in Fig. 4.23.1. Assume spindle speed as 350 rpm and feed rate
0.4 mm/rev.
Solution : Given data :
S = 350 rpm and F = 0.4 mm/rev.
Assume that the machine by default is in diameter programming mode.
Refer Fig. 4.23.1 (a).
Program Description
N001 G90 G71; (Absolute programming mode and data input is in metric
mode)
N002 G95 G28 U0 W0; (Feed in mm/rev. and go to home position)
N003 M06 T0101; (Tool change, tool no. 01 with offset value 01)
N004 S350 M03; (Spindle speed 350 rpm and in clockwise direction)
N005 G00 X0 Z0 M08; (Rapid tool positioning at point 1 and coolant ON)
N006 G01 X40 F0.4; (Finish the face till point 2 at 0.4 mm/rev.)
4 - 50 Computer Aided Design and Manufacturing
Fundamental of CNC and Part Programming
50
0
40
Fig. 4.23.1
50
0
40
4
3 2
1
+ X
+ Z
(Home position)
Fig. 4.23.1 (a)

267. N007 Z – 40; (Finish the turn upto point 3)
N008 X52; (Finish cut face till point 4)
N009 G00 G28 U0 W0 M09; (Rapid traverse to home position and coolant OFF)
N0010 M05 M30; (Spindle stop and program stop).
Example 4.23.2 : Write a manual part program for turning and facing the barstock to
the required dimensions as shown in Fig. 4.23.2. Assume one roughing and one finish
face cut; two roughing and two finish turning cut. Assume spindle speed 450 rpm and
feed 0.5 mm/rev.
Solution : Given data :
S = 450 rpm and F = 0.5 mm/rev.
Assume that the machine by default is in diameter programming mode. Refer
Fig. 4.23.2 (a).
4 - 51 Computer Aided Design and Manufacturing
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65
50
50
Fig. 4.23.2
11
8
5
1296
7
1
4
10
50
69
72
65
32
2
1 52
50
+X
+Z
Fig. 4.23.2 (a)

268. Program Description
N001 G90 G71; (Absolute programming mode and data input is in metric
mode)
N002 G95 G28 U0 W0; (Feed in mm/rev. and go to home position)
N003 M06 T0101; (Tool change, tool no. 01 with offset value 01)
N004 S450 M03; (Spindle speed is 450 rpm and in clockwise direction)
N005 G00 X75 Z2 M08; (Rapid traverse to point 1 and coolant ON)
N006 G01 X0 F0.5; (Rough cut face till point 2 at a feed of 0.5 mm/rev.)
N007 Z0; (Advance or depth of cut upto point 3)
N008 X60; (Finish cut face till point 4)
N009 Z – 48; (Rough turning upto point 5)
N0010 X70; (Rough turning upto point 6)
N0011 G00 X60 Z0; (Rapid traverse to point 4)
N0012 G01 X55; (Rough turning upto point 7)
N0013 Z – 49; (Rough turning upto point 8)
N0014 X70; (Rough turning upto point 9)
N0015 G00 X60 Z0; (Rapid traverse to point 4)
N0016 G01 X50; (Rough turning upto point 10)
N0017 Z – 50; (Rough turning upto point 11)
N0018 X70; (Rough turning till point 12)
N0019 G00 G28 U0 W0 M09; (Rapid traverse to home position and coolant OFF)
N0020 M05 M30; (Spindle stop and program stop)
Explanation :
The tool cycles are as follows :
1-2 (Rough facing) 6-4 (Rapid traverse)
2-3 (Feed for finishing) 4-7-8-9 (Again rough turning of shaft and
shoulder)
3-4 (Finish facing) 9-4 (Rapid traverse)
4-5-6 (Rough turning of shaft and
shoulder)
4-10-11-12 (Finish turning of shaft and
shoulder)
4 - 52 Computer Aided Design and Manufacturing
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269. Example 4.23.3 : Write a manual part program for finishing a forged component as
shown in Fig. 4.23.3. Assume speed 300 rpm and feed 0.4 mm/rev. Assume 1 mm
material is to be removed radially from external diameter.
Solution : Given data :
S = 300 rpm and F = 0.4 mm/rev.
Assume that the machine by default is in diameter programming mode.
Refer Fig. 4.23.3 (a).
Program Description
N001 G90 G71; (Absolute programming mode and data input is in metric
mode)
N002 G95 G28 U0 W0; (Feed in mm/rev. and go to home position)
N003 M06 T0101; (Tool change, tool no. 01 with offset value 01)
4 - 53 Computer Aided Design and Manufacturing
Fundamental of CNC and Part Programming

50

100
R5
20 30 30
Fig. 4.23.3
5
6
4 3
2 1
100
60
50
R5
25
30
60
80
+X
+Z
(Home position)
Fig. 4.23.3 (a)

270. N004 S300 M03; (Spindle speed is 300 rpm and in clockwise direction)
N005 G00 X50 Z2 M08; (Rapid traverse to point 1 and coolant ON)
N006 G01 Z – 25 F0.4; (Finish turning upto point 2 with a feed of 0.4 mm/rev.)
N007 G02 X60 Z – 30 R5; (Clockwise interpolation upto point 3 with a radius of 5)
N008 G01 Z – 60; (Finish turning upto point 4)
N009 X100 Z – 80; (Finish turning upto point 5)
N0010 G00 X102; (Rapid traverse to point 6)
N0011 G28 U0 W0 M09; (Return to home position and coolant OFF)
N0012 M05 M30; (Spindle stop and program stop)
Example 4.23.4 : Write a part program for the component as shown Fig. 4.23.4.
Assume that spindle speed 500 rpm and feed is 0.3 mm/rev.
Solution : Given data :
S = 500 rpm and Feed = 0.3 mm/rev.
Assume that the machine by default is in diameter programming mode.
Refer Fig. 4.23.4 (a).
4 - 54 Computer Aided Design and Manufacturing
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60
20
40
20 30 20
R10
Fig. 4.23.4
7
8
6
5 4
3
2 1
60
R10
20
40
50
70
(Home position)
20
40
+X
+Z
Fig. 4.23.4 (a)

271. Program Description
N001 G90 G71; (Absolute programming mode and data input is in metric
mode)
N002 G95 G28 U0 W0; (Feed in mm/rev. and go to home position)
N003 M06 T0101; (Tool change, tool no. 01 with offset value 01)
N004 S500 M03; (Spindle speed 500 rpm and in clockwise direction)
N005 G00 X0 Z2 M08; (Rapid traverse to point 1 and coolant ON)
N006 G01 Z0 F0.3; (Linear interpolation upto point 2 with a feed of
0.3 mm/rev.)
N007 X20; (Finish turn upto point 3)
N008 X40; Z – 20; (Finish turning upto point 4)
N009 Z – 40; (Finish turning upto point 5)
N0010 G02 X60 Z – 50 R10; (Circular interpolation upto point 6 with a radius of 10)
N0011 G01 Z – 70; (Finish turning upto point 7)
N0012 G00 X65; (Rapid traverse to point 8)
N0013 G28 U0 W0 M09; (Return to home position and coolant OFF)
N0014 M05 M30; (Spindle stop and program stop)
Example 4.23.5 : Write a manual part program to finish the following component
(Refer Fig. 4.23.5). Assume spindle speed 600 rpm and feed 0.45 mm/rev.
Solution : Given data :
S = 600 rpm and F = 0.45 mm/rev.
4 - 55 Computer Aided Design and Manufacturing
Fundamental of CNC and Part Programming
R15
Ø
30
Fig. 4.23.5

272. Assume that the machine by default is in diameter programming mode.
Refer Fig. 4.23.5 (a).
Program Description
N001 G90 G71; (Absolute programming mode and data input is in metric
mode)
N002 G95 G28 U0 W0; (Feed in mm/rev. and go to home position)
N003 M06 T0101; (Tool change, Tool no. 01 with offset value 01)
N004 S600 M03; (Spindle speed 600 rpm and in clockwise direction)
N005 G00 X0 Z2 M08; (Rapid traverse to point 1 and coolant ON)
N006 G01 Z0 F0.45; (Linear interpolation upto point 2 with a feed of
0.45 mm/rev.)
N007 G03 X30 Z – 15 R15; (Anticlockwise interpolation upto point 3 with a radius of
15)
N008 G00 X32; (Rapid traverse upto point 4)
N009 G28 U0 W0 M09; (Return to home position and coolant OFF)
N0010 M05 M30; (Spindle stop and program stop)
4 - 56 Computer Aided Design and Manufacturing
Fundamental of CNC and Part Programming
R15
Ø
30
3
4
2 1
(Home
position)
+X
+Z
Fig. 4.23.5 (a)

273. Example 4.23.6 : Fig. 4.23.6 shows the finished size of a round bar. The original
diameter of the bar was 28 mm. Make a part program for facing, parting and
reduction of diameter. Take feed = 200 mm/min, spindle speed = 640 rpm and
depth of cut = 2 mm per cut.
Solution : Given data :
Feed F = 200 mm/min, Spindle speed S = 640 rpm, Depth of cut D = 2 mm
It is assumed that the machine by default is in diameter programming mode. Refer
Fig. 4.23.6 (a) for turning and Fig. 4.23.6 (b) for parting.
4 - 57 Computer Aided Design and Manufacturing
Fundamental of CNC and Part Programming
Ø
20
Z
X
Ø
24
Ø
28
40
60
Fig. 4.23.6
Z
X
40
60
Home position
5
7
1
4 3
2
8
9
10
11
6
12
Fig. 4.23.6 (a)

274. Program Description
N001 G90 G71; (Absolute programming mode and data input is in metric
mode)
N002 M06 T0101; (Tool change, tool No.1 with offset value 01)
N003 M03 S640 G94; (Spindle speed 640 rpm and in clockwise direction, Feed in
mm/min)
N004 G00 X0 Z5 M08; (Rapid traverse to point 1 and coolant ON)
N005 G73 P6 Q11 U2 W1 D2
F200;
(See explanation at the end)
N006 G01 X0 Z0; (Linear interpolation upto point 2)
N007 X20; (Linear interpolation upto point 3)
N008 Z – 40; (Linear interpolation upto point 4)
N009 X24; (Linear interpolation upto point 5)
N0010 Z – 60; (Linear interpolation upto point 6)
N0011 X30; (Linear interpolation upto point 7)
N0012 G72 P6 Q11; (See explanation at the end)
N0013 G28 U0 W0; (Return to home position)
N0014 M06 T0202; (Tool change for parting with offset value 02)
N0015 G00 X – 30 Z – 62; (Rapid traverse to parting position)
N0016 G01 X0 F50; (Linear interpolation for parting with Feed of 50 mm/min)
N0017 G28 U0 W0; (Return to home position)
N0018 M03 M09 M30; (Coolant OFF, spindle stop and program stop).
Line N005 : G73 P6 Q11 U2 W1 D2 F200;
G73 – Rough facing and rough turning
P6 – Indicates sequence number at which the machining starts
Q11 – Indicates sequence number at which the machining ends
U2 – Leave 2 mm stock on X axis for machining
W1 – Leave 1 mm stock on Z axis for finishing
4 - 58 Computer Aided Design and Manufacturing
Fundamental of CNC and Part Programming
Ø
20
Ø
24
Ø
28
Parting Tool
Fig. 4.23.6 (b)

275. D2 – Depth of cut for each roughing pass is 2 mm
F200 – Feed is 200 mm/min
Line N0012 : G72 P6 Q11;
G72 – Finish facing and turning
P6 – Indicates sequence number at which finish pass starts
Q11 – Indicates sequence number at which finish pass ends
Example 4.23.7 : Write a part program for that part shown in Fig. 4.23.7.
Solution : Given data :
Assume, S = 500 rpm and F = 0.3 mm/rev
Program Description
N001 G90 G71; (Absolute programming mode and data input is in metric
mode)
N002 G95 G28 U0 W0; (Feed in mm/rev. and go to home position)
N003 M06 T0101; (Tool change, tool no. 01 with offset value 01)
N004 S500 M03; (Spindle speed 500 rpm and in clockwise direction)
N005 G00 X0 Z0 M08; (Rapid traverse to point A and coolant ON)
N006 G01 X5 Z – 25 F0.3; (Linear interpolation upto point D with a feed of
0.3 mm/rev.)
N009 Z – 60; (Finish turn upto point E)
N008 G00 X15; (Rapid traverse to point C)
N009 X15 Z – 20; (Rapid traverse to point B)
4 - 59 Computer Aided Design and Manufacturing
Fundamental of CNC and Part Programming
 10
40 20
A
D
E
B
(0, 0)
10 mm
5 mm
5
 20
C
–x
+x
–z +z
Fig. 4.23.7

276. N010 G28 U0 W0 M09; (Return to home position and coolant OFF)
N011 M05 M30; (Spindle stop and program stop)
Example 4.23.8 : A 110 mm long cylindrical rod of f 75 mm is to be turned into a
component as shown in Fig. 4.23.8, using a CNC lathe. Write a CNC program for
manufacturing this component.
Solution : Given data :
Assume spindle speed 500 rpm and feed 0.15 mm/rev. Refer Fig. 4.23.8 (a).
4 - 60 Computer Aided Design and Manufacturing
Fundamental of CNC and Part Programming
Ø
70
R12
31
Ø
55
Ø
55
Ø
44
Ø
44
Ø
30
20
45
67
95
15º
Fig. 4.23.8
Ø
70
R12
31
Ø
55
Ø
55
Ø
44
Ø
44
Ø
30
20
45
67
95
15º
2
1
3
4
5
6
7
8
9
Fig. 4.23.8 (a)

277. Program Description
N001 G90 G71; (Absolute programming mode and data input is in metric
mode)
N002 G95 G28 U0 W0; (Feed in mm/rev and go to home position)
N003 M06 T0101; (Tool change, tool no. 01 with offset value 01)
N004 S500 M03; (Spindle speed 500 rpm and in clockwise direction)
N005 G74 P006 Q015 D2 R1; (Canned cycle starts from line 006 to 015 with depth of
cut 2 mm and relief 1 mm)
N006 G00 X0 Z0 F0.15 M08; (Rapid traverse to origin, coolant ON and feed is
0.15 mm/rev)
N007 G01 X28 Z0; (Linear interpolation upto point 1)
N008 X30 Z – 1; (Linear interpolation upto point 2)
N009 X30 Z – 20; (Linear interpolation upto point 3)
N010 G03 X44 Z – 31 R12; (Circular interpolation upto point 4)
N011 G01 X44 Z – 45; (Linear interpolation upto point 5)
N011 X55 Z – 45; (Linear interpolation upto point 6)
N012 X55 Z – 67; (Linear interpolation upto point 7)
N013 X70 Z – 69; (Linear interpolation upto point 8)
N014 X70 Z – 95; (Linear interpolation upto point 9)
N015 G72 P006 Q014; (Finish facing and turning from line 006 to 014)
N016 G28 U0 W0 M09; (Return to home position and coolant OFF)
N017 M05 M30; (Spindle stop and program stop)
Example 4.23.9 : Write a manual part program to turn the component shown on a
CNC Lathe from 75 mm bar stock. The following data may be assumed :
i) There will be two rough turnings and one finish turning. The first cut is with a depth
of 3 mm for a length of 58 mm, the second with a depth of 3 mm for a length of
59 mm and the third with a depth of 1.5 mm for the full length of 60 mm.
ii) The shoulder of the work-piece is also machined during each cut.
iii) The spindle speed is 400 rpm and the feed rate is 0.5 mm/rev.
Make a free-hand sketch showing relevant points of tool positions for each of the three
turning operations and then write the manual part program. State also what each line
of the program does.
Note : If the exact G-codes and M-codes are not known, the student can use his/her
own code-numbers, but the function of such codes must be clearly stated.
4 - 61 Computer Aided Design and Manufacturing
Fundamental of CNC and Part Programming

278. Solution : Given data :
S = 400 rpm, F = 0.5 mm/rev
Assume that the machine is in diameter programming mode.
Also, assuming one rough and one finish facing cut.
Program Description
N001 G90 G71; (Absolute programming mode and data input is in metric
mode)
N002 G95 G28 U0 W0; (Feed in mm/rev. and go to home position)
N003 M06 T0101; (Tool change, tool no. 01 with offset value 01)
N004 S400 M03; (Spindle speed is 400 rpm and in clockwise direction)
4 - 62 Computer Aided Design and Manufacturing
Fundamental of CNC and Part Programming

75

60
60
+X
+Z
Fig. 4.23.9
11
8
5
1296
7
1
4
10 60
63
69
75
32
2
1 62
60
+X
+Z
Fig. 4.23.9 (a)

279. N005 G00 X80 Z2 M08; (Rapid traverse to point 1 and coolant ON)
N006 G01 X0 F0.5; (Rough cut face till point 2 at a feed of 0.5 mm/rev.)
N007 Z0; (Advance or depth of cut upto point 3)
N008 X69; (Finish cut face till point 4)
N009 Z – 58; (Rough turning upto point 5)
N0010 X80; (Rough turning of shoulder upto point 6)
N0011 G00 X69 Z0; (Rapid traverse to point 4)
N0012 G01 X63; (Rough turning upto point 7)
N0013 Z – 59; (Rough turning upto point 8)
N0014 X80; (Rough turning upto point 9)
N0015 G00 X69 Z0; (Rapid traverse to point 4)
N0016 G01 X60; (Rapid traverse to point 10)
N0017 Z – 60; (Finish turning upto point 11)
N0018 X80; (Finish turning of shoulder till point 12)
N0019 G00 G28 U0 W0 M09; (Rapid traverse to home position and coolant OFF)
N0020 M05 M30; (Spindle stop and program stop)
Explanation :
The tool cycles are as follows :
1-2 (Rough facing) 6-4 (Rapid traverse)
2-3 (Feed for finishing) 4-7-8-9 (Again rough turning of shaft and
shoulder)
3-4 (Finish facing) 9-4 (Rapid traverse)
4-5-6 (Rough turning of shaft and
shoulder)
4-10-11-12 (Finish turning of shaft and
shoulder)
Example 4.23.10 : Write a complete part program using G and M codes for the job
shown in Fig. 4.23.10. Assume suitable speed and feed for machining.
Billet size - Diameter : 60 mm and Length : 90 mm.
Thread : Major Diameter, D0 = 20 mm, Minor Diameter,
Dc = 17 mm and Pitch : 2.5 mm, Groove : Width = 5 mm and depth = 2.5 mm.
4 - 63 Computer Aided Design and Manufacturing
Fundamental of CNC and Part Programming

280. Solution :
Program Description
N01 G90 G71 ; Absolute programming mode and data input is in metric mode.
N02 G95 G21 ; Feed in mm/rev ; Metric mode
Facing
N03 M42 T01 ; Tool change, tool number 01
4 - 64 Computer Aided Design and Manufacturing
Fundamental of CNC and Part Programming
20 10 5
20 20

20

40

60
R10
Thread
Groove
Fig. 4.23.10
1
2
3
4
5
6
7
8
9
10 X
Z
Fig. 4.23.10 (a)

281. N04 S200 M03 M07 ; Spindle speed 200 rpm, clockwise ON, coolant ON
N05 G00 X0 Z3 ; Rapid to point 1
N06 G01 Z0 F0.5 ; Facing operation begins on the stock
N07 X62 ; Facing ends at point 2
Turning
N08 G71 U2 R1 ; See explanation below
N09 P010 Q015 W0
F0.5 ;
See explanation below
N010 G00 X20 Z0 ; Rapid to point 2
N011 G01 Z–25 ; Turning begins till point 6
N012 G02 X40 Z–35 ; Circular clockwise interpolation upto point 7
N013 G01 X40 Z–45 ; Linear interpolation upto point 8
N014 G01 X60 Z–65 ; Linear interpolation upto point 9
N015 G01 X60 Z–85 ; Linear interpolation upto point 10
Finishing
N016 M42 T02 ; High gear, tool No. 02
N017 G70 P010 Q015 ; See explanation below
N018 G28 U0 V0 W0
M09 ;
Return to home position, coolant off
N019 M05 ; Spindle off
Grooving
N020 M42 T03 ; High gear, tool no. 03
N021 G95 G21 G96 ; Feed per revolution, metric mode, constant surface speed
N022 S400 M03 M07 ; Spindle speed 400 rpm, clockwise ON coolant ON
N023 G00 X30 Z0 ; Rapid to point 2
N024 Z–20 ; Rapid to point 3
N025 X30 ; Rapid to point 3
N026 G75 R0.5 ; See explanation
N027 X15 Z–25 P1000
Q500 F0.15 ;
See explanation
N028 X30 ; Rapid to point 6
4 - 65 Computer Aided Design and Manufacturing
Fundamental of CNC and Part Programming

282. N029 Z0 X30 ; Rapid to point 2
Threading Tool
N030 M42 T04 ; High gear, tool no. 0.4
N031 S400 M03 M07 ; Spindle speed 400 rpm, clockwise ON, coolant ON
N032 G00 X20 Z3 ; Rapid to point 2
N033 G76 P020060
Q300 R100 ;
See explanation below
N034 X17 Z–22 P2000
Q300 F2.5 ;
See explanation below
N035 G28 U0 V0 W0
M09 ;
Return to home position, coolant off
N036 M05 M30 ; Spindle off and program stop
Explanation :
N08 G71 U2 R1
Sequence
Number
Stock removal
in turning
Depth of each roughing
pass in 2 mm
Tool escape of 1 mm
N09 P010 Q015 W0 F0.5
Sequence
Number
Sequence
Number at
beginning
Sequence
number at end
Facing operation is done
and there is no stock left
on 2 axis
Feed rate is
0.5 mm/ rev
N017 G70 P010 Q015
Sequence
Number
G code for finish
face and turn
Sequence number
at beginning
Sequence number at end
N026 G75 R0.5
Sequence
Number
G code for
Grove cutting
Tool escape
of 0.5 mm
N027 X14 Z–23 P1000 Q500 F2.5
Sequence
Number
Finished
Grooved
diameter is
15 mm
End point of
the groove
Incremental
depth of cut on
X axis
1000/1000 = 1
mm
Tool
advance
on Z axis
500/1000 =
0.5 mm
Feed rate of
0.15 mm/ rev
4 - 66 Computer Aided Design and Manufacturing
Fundamental of CNC and Part Programming

283. N033 G76 P020060 Q300 R100
Sequence
Number
Multiple
threading
cycle
Double finish pass
indicated by 02, No
chamfer (00) and
angle of tool 60º
Minimum cutting
depth,
300/1000 = 0.3 mm
Finishing allowance,
100/1000 = 0.1 mm
N034 X17 Z–22 P2000 Q300 F2.5
Sequence
Number
Minor
diameter is
17 mm
Thread
length
Height of
thread =
2000/1000 =
2 mm
Depth of
initial cut is
300/1000 =
0.3 mm
Pitch of
threads is
2.5 mm
Example 4.23.11 : Write CNC part program for the component shown in Fig. 4.23.11
Mention the assumption made.
Solution : Refer example 4.23.10.
4.24 Part Programming for Milling and Drilling :
Example 4.24.1 : Write a part program to machine a workpiece as shown in
Fig. 4.24.1 (a). Assume cutter diameter as 10 mm end mill type, depth of workpiece
10 mm and feed rate 200 mm/min. Take spindle speed 500 r.p.m.
4 - 67 Computer Aided Design and Manufacturing
Fundamental of CNC and Part Programming
100.00
80.00
R10
29.00
32.00

54.56
32.00
45.00
65.00
120.00
130.00
150.00
R8
Grove 3 DEEP
Chramfer 3 45º

M20 = 2.5 Thread
root diameter 16.75
Billet diameter 110 mm
Billet length 200 mm
All dimensions in mm
Fig. 4.23.11

284. Solution : For the tool path refer Fig. 4.24.1 (b).
Programs Description
101; Program number.
N01 G28 U0 V0 W0; Return to machine reference position.
N02 G90 G71 G94; Absolute programming mode, metric (mm) data input and
feed in mm/min.
N03 G17 M06 T01; Selection of XY plane, tool change and select tool no.01.
N04 G41 M03 S500; Cutter radius compensation to left, spindle start in clockwise
direction and spindle speed is 500 r.p.m.
N05 G00 X–10 Y–10 Z5
M08;
Rapid travel to position (–10, – 10) and cutter is 5 mm
above the workpiece surface, coolant ON.
N06 G01 X0 Y0 Z–10 F200; Linear interpolation and move tool 10 mm downward along
Z-axis with feed 200 mm/min to position (0, 0).
N07 X50 Y0; Linear interpolation and tool move to position (50, 0)
N08 X50 Y20; Linear interpolation and tool move to position (50, 20)
4 - 68 Computer Aided Design and Manufacturing
Fundamental of CNC and Part Programming
X
Y
(a) Workpiece
20
20
30
50
X
Y
10
X
Z
– 10
X
Z
P0(–10, –10)
P1(0, 0)
P2(50, 0)
P3(50, 20)
P4(20, 20)
P5(20, 50)
P6(0, 50)
(b) Tool path
Fig. 4.24.1

285. N09 X20 Y20; Linear interpolation and tool move to position (20, 20)
N10 X20 Y50; Linear interpolation and tool move to position (20, 50)
N11 X0 Y50; Linear interpolation and tool move to position (0, 50)
N12 X0 Y0; Linear interpolation and tool move to position (0, 0)
N13 G00 Z5; Rapid travel 5 mm above the workpiece surface.
N14 G28 U0 V0 W0; Return to machine reference position.
N15 G40 M05; Cutter radius compensation cancel and spindle stop.
N16 M09 M30; Coolant off, program end and tape rewind.
Example 4.24.2 : Write a part program to machine a workpiece as shown in
Fig. 4.24.2 (a). Assume cutter diameter as 10 mm end mill type, depth of workpiece
10 mm and feed rate 150 mm/min. Take spindle speed 600 r.p.m.
Solution : For the tool path refer Fig. 4.24.2 (b).
Program Description
102; Program number
N01 G28 U0 V0 W0; Return to machine reference position.
N02 G90 G71 G94; Absolute programming mode, metric (mm) data input and
feed in mm/min.
4 - 69 Computer Aided Design and Manufacturing
Fundamental of CNC and Part Programming
P1(0, 0)
(b) Tool path
X X
Y
(a) Workpiece
R20
40
50
10
X
Z
Y
I = 0
and J = 20
– 10
X
Z
P0(–10, –10)
P2(50, 0)
P3(50, 40)
P6(0, 40)
Fig. 4.24.2

286. N03 G17 M06 T01; Selection of XY plane, tool change and select tool no. 01.
N04 G41 M03 S600; Cutter radius compensation to left, spindle ON with speed
600 r.p.m.
N05 G00 X–10 Y–10 Z5 M08; Rapid travel to position (–10, –10) and cutter is 5 mm
above the workpiece surface, coolant ON.
N06 G01 X0 Y0 Z–10 F150; Linear interpolation and move tool 10 mm downward along
Z-axis with feed 150 mm/min.
N07 X50 Y0; Linear interpolation and tool move to position (50, 0).
N08 G03 X50 Y40 I0 J20; Anticlockwise circular interpolation and tool move to position
(50, 40).
N09 G01 X0 Y40; Linear interpolation and tool move to position (0, 40)
N10 X0 Y0; Linear interpolation and tool move to position (0, 0).
N11 G00 Z5; Rapid travel 5 mm above the workpiece surface.
N12 G28 U0 V0 W0; Return to machine reference position.
N13 G40 M05; Cutter radius compensation cancel and spindle stop.
N16 M09 M30; Coolant off, program end and tape rewind.
Example 4.24.3 : Write a part program to machine a workpiece as shown in
Fig. 4.24.3 (a). Assume cutter diameter as 10 mm end mill type, depth of workpiece
10 mm and feed rate 100 mm/min. Take spindle speed 800 r.p.m.
4 - 70 Computer Aided Design and Manufacturing
Fundamental of CNC and Part Programming
R15
20 20
R15
R15
20
20
X
Y
X
Y
P0(–10, –10)
P1(0, 0) P2(20, 0) P3(50, 0) P4(70, 0)
P5(70, 20)
P6(70, 50)
P7(70, 70)
P8(50, 70)
P9(20, 70)
P10(0, 70)
(b) Tool path
(a) Workpiece
Fig. 4.24.3

287. Solution : For tool path refer Fig. 4.24.3 (b).
Program Description
103; Program number.
N01 G28 U0 V0 W0; Return to machine reference position.
N02 G90 G71 G94; Absolute programming mode, metric (mm) data input and
feed in mm/min.
N03 G17 M06 T01; Selection of XY plane, tool change and select tool no. 01.
N04 G41 M03 S800; Cutter radius compensation to left, spandle ON with speed
800 r.p.m.
N05 G00 X–10 Y–10 Z5 M08; Rapid travel to position (–10, –10) and cutter is 5 mm
above the workpiece surface, coolant ON.
N06 G01 X0 Y0 Z–10 F100; Linear interpolation and move tool 10 mm downward along
Z-axis with feed 100 mm/min to position (0, 0).
N07 X20 Y0; Linear interpolation and tool move to position (20, 0).
N08 G02 X50 Y0 I15 J0; Clockwise circular interpolation and tool move to position
(50, 0).
N09 G01 X70 Y0; Linear interpolation and tool move to position (70, 0).
N10 X70 Y20; Linear interpolation and tool move to position (70, 20).
N11 G02 X70 Y50 I0 J15; Clockwise circular interpolation and tool move to position
(70, 50).
N12 G01 X70 Y70; Linear interpolation and tool move to position (70, 70).
N13 X50 Y70; Linear interpolation and tool move to position (50, 70).
N14 G02 X20 Y70 I–15 J0; Clockwise circular interpolation and tool move to position
(20, 70).
N15 G01 X0 Y70; Linear interpolation and tool move to position (0, 70).
N16 X0 Y0; Linear interpolation and tool move to position (0, 0).
N17 G00 Z5; Rapid travel to 5 mm above the workpiece surface.
N18 G28 U0 V0 W0; Return to machine reference position.
N19 G40 M05; Cutter radius compensation cancel and spindle stop.
N20 M09 M30; Coolant off, program end and tape rewind.
4 - 71 Computer Aided Design and Manufacturing
Fundamental of CNC and Part Programming

288. Example 4.24.4 : Write the program for the job as shown in Fig. 4.24.4 (a). Use the
following machining data :
Speed = 800 r.p.m., feed = 10 mm/min,
Depth of cut = 3 mm, Thickness of job = 3 mm.
Solution : For tool path refer Fig. 4.24.4 (b).
Program Description
104; Program number.
N01 G28 U0 V0 W0; Return to machine reference position.
N02 G90 G71 G94; Absolute programming mode, metric (mm) data input
and feed in mm/min.
N03 G17 M06 T01; Selection of XY plane, tool change and select tool
no.01.
N04 G41 M03 S800; Cutter radius compensation to left, spindle ON with
speed 800 r.p.m.
N05 G00 X–10, Y–10 Z5 M08; Rapid travel to position (–10, –10) and cutter is
5 mm above the workpiece surface, coolant ON.
N06 G01 X0 Y0 Z–3 F10; Linear interpolation and move tool 3 mm downward
along Z-axis with feed 10 mm/min to position (0, 0).
N07 X110 Y0; Linear interpolation and tool move to position (110,0).
4 - 72 Computer Aided Design and Manufacturing
Fundamental of CNC and Part Programming
P0(–10, –10)
110
100
R20
45
R12
20
160
75
(0, 0)
R12
P1(0, 0)
P2(110, 0)
X
P3(110, 80)
P4(130, 100)
P5(160, 100)
P6(160, 133)
P7(148, 145)
P8(75, 145)
P9(75, 165)
P10(12, 165)
P11(0, 153)
Y
(a) Workpiece (b) Tool path
Fig. 4.24.4

289. N08 X110 Y80; Linear interpolation and tool move to position
(110, 80).
N09 G02 X130 Y100 I20 J0; Clockwise circular interpolation and tool move to
position (130, 100).
N10 G01 X160 Y100; Linear interpolation and tool move to position
(160, 100).
N11 X160 Y133; Linear interpolation and tool move to position
(160, 133).
N12 G03 X148 Y145 I–12 J0; Anticlockwise circular interpolation and tool move to
position (148, 145).
N13 G01 X75 Y145; Linear interpolation and tool move to position
(75, 145).
N14 X75 Y165; Linear interpolation and tool move to position
(75, 165).
N15 X12 Y165; Linear interpolation and tool move to position
(12, 145).
N16 G03 X0 Y153 I0 J–12; Anticlockwise circular interpolation and tool move to
position (0, 153).
N17 G01 X0 Y0; Linear interpolation and tool move to position (0, 0).
N18 G00 Z5; Rapid travel to 5 mm above the workpiece surface.
N19 G28 U0 V0 W0; Return to machine reference position.
N20 G40 M05; Cutter radius compensation cancel and spindle stop.
N21 M09 M30; Coolant off, program end and tape rewind.
Example 4.24.5 : Write the program for the job as shown in Fig. 4.24.5 (a).
4 - 73 Computer Aided Design and Manufacturing
Fundamental of CNC and Part Programming
A B
C
(0, 0) 80 120
100
80
P0(0, 0) P1(80, 0)
P6(80, 20)
P2(200, 0)
P3(200, 100)
P4(0, 100)
P5(0, 20)
Fig. 4.24.5

290. Solution : For tool path refer Fig. 4.24.5 (b).
Assume speed 500 r.p.m., thickness of job 5 mm and feed rate as 20 mm/min.
Program Description
105; Program number.
N01 G28 U0 V0 W0; Return to machine reference position.
N02 G90 G71 G94; Absolute programming mode, metric (mm) data input
and feed in mm/min.
N03 G17 M06 T01; Selection of XY plane, tool change and select tool
no.01.
N04 G41 M03 S500; Cutter radius compensation to left, spindle ON with
speed 500 r.p.m.
N05 G00 X0, Y0 Z5 M08; Rapid travel to position (0, 0) and cutter is 5 mm
above the workpiece surface, coolant ON.
N06 X80 Y0 Z–5 F20; Rapid travel to position (80, 0) and move tool 5 mm
downward along Z-axis with feed 20 mm/min.
N07 G01 X200 Y0; Linear interpolation and tool move to position (200,0).
N08 X200 Y100; Linear interpolation and tool move to position
(200, 100).
N09 X0 Y100; Linear interpolation and tool move to position (0, 100).
N10 X0 Y20; Linear interpolation and tool move to position (0, 20).
N11 X80 Y20; Linear interpolation and tool move to position (80, 20).
N12 X80 Y0; Linear interpolation and tool move to position (80, 0).
N13 G00 Z5; Rapid travel to 5 mm above the workpiece surface.
N14 G28 U0 V0 W0; Return to machine reference position.
N15 G40 M05; Cutter radius compensation cancel and spindle stop.
N16 M09 M30; Coolant off, program end and tape rewind.
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291. Example 4.24.6 : Write a part program for the workpiece as shown in Fig. 4.24.6.
Assume cutter diameter 10 mm, depth of workpiece 10 mm, spindle speed 600 r.p.m.
and feed 15 mm/min.
Solution : For tool path refer Fig. 4.24.6 (b).
Program Description
107; Program number.
N01 G28 U0 V0 W0; Return to machine reference position.
N02 G90 G71 G94; Absolute programming mode, metric (mm) data
input and feed in mm/min.
N03 G17 M06 T01; Selection of XY plane, tool change and select
tool no.01.
N04 G41 M03 S600; Cutter radius compensation to left, spindle ON
with speed 600 r.p.m.
N05 G00 X–10, Y–10 Z5 M08; Rapid travel to position (–10, –10) and cutter is
5 mm above the workpiece surface, coolant ON.
N06 G01 X0 Y0 Z–10 F15; Linear interpolation and move tool 10 mm
downward along Z-axis with feed 15 mm/min to
position (0, 0).
N07 X20 Y0; Linear interpolation and tool move to position
(20, 0).
N08 X35 Y30; Linear interpolation and tool move to position
(35, 30).
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P0(–10, –10)
P1(0, 0)
(b) Tool path
20 15 25
30
20
35
R15
R10
P2(20, 0)
P3(35, 30) P4(60, 30)
P5(60, 50)
P6(45, 50)
P7(35, 60)
P8(0, 60)
P9(0, 30)
(a) Workpiece
Fig. 4.24.6

292. N09 X60 Y30; Linear interpolation and tool move to position
(60, 30).
N10 X60 Y50; Linear interpolation and tool move to position
(60, 50).
N11 X45 Y50; Linear interpolation and tool move to position
(45, 50).
N12 G03 X35 Y60 I–10 J0; Anticlockwise circular interpolation and tool move
to position (35, 60).
N13 G01 X0 Y60; Linear interpolation and tool move to position
(0, 60).
N14 G03 X0 Y30 I0 J–15; Anticlockwise circular interpolation and tool move
to position (0, 30).
N15 G01 X0 Y0; Linear interpolation and tool move to position
(0, 0).
N16 G00 Z5; Rapid travel 5 mm above the workpiece surface.
N17 G28 U0 V0 W0; Return to machine reference position.
N18 G40 M05; Cutter radius compensation cancel and spindle
stop.
N19 M09 M30; Coolant off, program end and tape rewind.
Example 4.24.7 : Write a part program for the component as shown in Fig. 4.24.7 (a).
Take feed 30 mm/min; spindle speed 1000 r.p.m. and thickness of component as 10 mm.
Use end mill type cutter of diameter 10 mm.
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P0(0, 0)
(a) Workpiece (b) Tool path
15 40
40
55
20
55
15
20
Y
R15
X
Y
X
P1(35, 20) P2(75, 20)
P3(90, 35)
P4(90, 75)
P5(75, 90)
P6(20, 90)
P7(20, 35)
Fig. 4.24.7

293. Solution : For tool path refer Fig. 4.24.7 (b).
Program Description
108; Program number.
N01 G28 U0 V0 W0; Return to machine reference position.
N02 G90 G71 G94; Absolute programming mode, metric (mm) data input
and feed in mm/min.
N03 G17 M06 T01; Selection of XY plane, tool change and select tool
no.01.
N04 G41 M03 S1000; Cutter radius compensation to left, spindle ON with
speed 1000 r.p.m.
N05 G00 X0, Y0 Z5 M08; Rapid travel to position (0, 0) and cutter is 5 mm
above the workpiece surface, coolant ON.
N06 X35 Y20 Z–10 F30; Rapid travel to position (35, 20) and cutter is 10 mm
downward along Z-axis with feed 30 mm/min.
N07 G01 X75 Y20; Linear interpolation and tool move to position
(75, 20).
N08 G02 X90 Y35 I15 J0; Clockwise circular interpolation and tool move to
position (90, 35).
N09 G01 X90 Y75; Linear interpolation and tool move to position
(90, 75).
N10 X75 Y90; Linear interpolation and tool move to position
(75, 90).
N11 X20 Y90; Linear interpolation and tool move to position
(20, 90).
N12 X20 Y35; Linear interpolation and tool move to position
(20, 35).
N13 X35 Y20; Linear interpolation and tool move to position
(35, 20).
N14 G00 X0 Y0 Z5; Rapid travel to position (0, 0) and cutter is 5 mm
above the surface of workpiece.
N15 G28 U0 V0 W0; Return to machine reference position.
N16 G40 M05; Cutter radius compensation cancel and spindle stop.
N17 M09 M30; Coolant off, program end and tape rewind.
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294. Example 4.24.8 : Prepare a part program for the given job as shown in
Fig. 4.24.8 (a). By using following data :
Speed = 1000 r.p.m., Feed = 10 mm/min, Depth of cut = 3 mm
Tool position from the surface of the workpiece is 10 mm above. Thickness of
job = 3 mm.
Solution : For tool path refer Fig. 4.24.8 (b).
Program Description
109; Program number.
N01 G28 U0 V0 W0; Return to machine reference position.
N02 G90 G71 G94; Absolute programming mode, metric (mm) data input and
feed in mm/min.
N03 G17 M06 T01; Selection of XY plane, tool change and select tool no.01.
N04 G41 M03 S1000; Cutter radius compensation to left, spindle ON with speed
1000 r.p.m.
N05 G00 X–10 Y–10 Z10
M08;
Rapid travel to position (–10, –10) and cutter is 10 mm
above the workpiece surface, coolant ON.
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Fundamental of CNC and Part Programming
P0(–10, –10)
(a) Workpiece (b) Tool path
55
10
10 55
55
R10
P1(0, 0)
(0, 0)
P2(55, 0)
P3(65, 10)
P4(65, 65)
P5(10, 65)
P6(0, 55)
R10
Fig. 4.24.8

295. N06 G01 X0 Y0 Z–13 F10; Linear interpolation and move tool 13 mm downward along
Z-axis with feed 10 mm/min to position (0, 0).
N07 X55 Y0; Linear interpolation and tool move to position (55, 0).
N08 G03 X65 Y10 I0 J10; Anticlockwise circular interpolation and tool move to position
(65, 10).
N09 G01 X65 Y65; Linear interpolation and tool move to position (65, 65).
N10 X10 Y65; Linear interpolation and tool move to position (10, 65).
N11 X0 Y55; Linear interpolation and tool move to position (0, 55).
N12 X0 Y0; Linear interpolation and tool move to position (0, 0).
N13 G00 Z10; Rapid travel 10 mm above the surface of workpiece.
N14 G28 U0 V0 W0; Return to machine reference position.
N15 G40 M05; Cutter radius compensation cancel and spindle stop.
N16 M09 M30; Coolant off, program end and tape rewind.
Example 4.24.9 : Write a manual program for drilling holes as shown in
Fig. 4.24.9. The spindle is initially at the lower left corner of the workpiece. Take
spindle speed 1200 rpm.
Solution : Given data :
S = 1200 rpm, Refer Fig. 4.24.9 (a).
4 - 79 Computer Aided Design and Manufacturing
Fundamental of CNC and Part Programming
25
35
55
12
30
60
# 1
# 2
# 3
65
75
10 Drill (3 Holes)
Fig. 4.24.9

296. Program Description
N001 G90 G71; (Absolute programming mode and data input is in metric
mode)
N002 M06 T0101; (Tool change, Tool no.1 with offset value 01)
N003 S1200 M03; (Spindle speed 1200 rpm and in clockwise direction)
N004 G00 X12 Y55 M08; (Rapid traverse to point 1 and coolant ON)
N005 M00; (Auto-stop for drilling)
N006 X30 Y35; (Rapid traverse to point 2)
N007 M00; (Auto-stop for drilling)
N008 X60 Y25; (Rapid traverse to point 3)
N009 M00; (Auto-stop for drilling)
N0010 X – 40 Y – 30; (Rapid traverse to any arbitrary point)
N0011 M09 M05 M30; (Coolant OFF, spindle stop and program stop).
Example 4.24.10 : Write a manual part program for drilling two holes of 8 mm
diameter and two holes of 6 mm diameter. Take spindle speed as 1200 rpm for 8 mm
diameter hole and 1600 rpm for 6 mm diameter hole. Refer Fig. 4.24.10.
4 - 80 Computer Aided Design and Manufacturing
Fundamental of CNC and Part Programming
# 1
# 2
# 3
(Home position)
+X
+Y
Fig. 4.24.9 (a)

297. Solution : Given data :
S = 1200 rpm (for 8 mm diameter)
S = 1600 rpm (for 6 mm diameter). Refer Fig. 4.24.10 (a).
Program Description
N001 G90 G71; (Absolute programming mode and data input is in metric
mode)
N002 M06 T0101; (Tool change, tool no.1 with offset value 01)
N003 S1200 M03; (Spindle speed 1200 rpm and in clockwise direction)
N004 G00 X20 Y60 M08; (Rapid traverse to point 1 and coolant ON)
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30
60
75
20
40
R2.5
90
100
10
20
40
8, Drill (2 Holes)
6, Drill (2 Holes)
# 1
# 2
# 3
# 4
Fig. 4.24.10
# 1
# 2 # 3
# 4
(Home position)
+X
+Y
Fig. 4.24.10 (a)

298. N005 M00; (Auto-stop for drilling)
N006 Y30; (Rapid traverse to point 2)
N007 M00; (Auto-stop for drilling)
N008 X – 10 Y0; (Rapid traverse to tool change position)
N009 M00; (Change of drill tool from 8 mm diameter to
6 mm diameter)
N0010 G00 G90 G71; (Rapid traverse, absolute and metric mode)
N0011 S1600 M03; (Spindle speed 1600 rpm and in clockwise direction)
N0012 X90 Y30; (Rapid traverse to point 3)
N0013 M00; (Auto-stop for drilling)
N0014 Y10; (Rapid traverse to point 4)
N0015 M00; (Auto-stop for drilling)
N0016 X – 10 Y0; (Rapid traverse to tool change position)
N0017 M09 M05 M30; (Coolant OFF, spindle stop and program stop).
Example 4.24.11 : Write a manual part program for drilling and milling the
machine component as shown in Fig. 4.24.11. Assume 10 mm diameter milling
cutter. Take spindle speed 2000 rpm (for milling) and 3000 rpm (for drilling). Feed
1400 mm/min.
Solution : Given data :
D = 10 mm, S = 2000 rpm (for milling), S = 3000 rpm (for drilling),
4 - 82 Computer Aided Design and Manufacturing
Fundamental of CNC and Part Programming
60
30
15
15
60
80
90
50
15
30
8 Typ. R5 Typ.
Fig. 4.24.11

299. F = 1400 mm/min. Refer Fig. 4.24.11 (a).
It is assumed that the machine (0, 0) is at the left lower most corner of the workpiece.
Program Description
N001 G90 G71 G94; (Absolute programming mode, data input is in mertic
mode and feed in mm/min)
N002 G00 X – 5, Y – 5 S2000
M03;
(Rapid traverse to point 1 and spindle speed is
2000 rpm in clockwise direction)
N003 M00; (Allows the operator to clamp the spindle)
N004 G01 X65 F1400 M08; (Machining upto point 2 with a feed of 1400 mm/min
and coolant ON)
N005 Y25; (Machining upto point 3)
N006 X95; (Machining upto point 4)
N007 Y65; (Machining upto point 5)
N008 X25; (Machining upto point 6)
N009 Y35; (Machining upto point 7)
N0010 X – 5; (Machining upto point 8)
N0011 Y – 5; (Machining upto point 1)
N0012 M00; (Auto-stop to raise the spindle)
N0013 G00 X – 10 Y0; (Rapid traverse to tool change position)
N0014 M00; (Change to drill tool of 8 mm diameter)
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#1 #2
#3 #4
#5
#6
#7
#8
#9
Home
Position
#10
Fig. 4.24.11 (a)

300. N0015 G00 G90 G71; (Rapid traverse, absoulte and metric mode)
N0016 X15 Y15 S3000 M03; (Spindle speed 3000 rpm in clockwise direction and
locate point 9)
N0017 M00; (Auto stop for drilling)
N0018 X80 Y45; (Locate point 10)
N0019 M00; (Auto-stop for drilling)
N0020 X – 10 Y0; (Rapid traverse to tool change position)
N0021 M09 M05 M30; (Coolant OFF, spindle stop and program stop)
Example 4.24.12 : Write the part program for the work piece shown in Fig. 4.24.12.
Material : Aluminium, Work piece size : 100 mm ´ 80 mm ´ 15 mm.
Solution : Given data :
Workpiece size = 100 mm ´ 80 mm ´ 15 mm.
Workpiece material = Aluminium
Assuming the machining zero at the origin and milling cutter is selected of f 20 mm.
Refer Fig. 4.24.12 (a).
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20
60
5
R20
80
x
y
R20
20
5
Fig. 4.24.12

301. As, cutter diameter D = 20 mm
Generally for aluminium, cutting speed V = 60 m/min
But, V =
pDN
1000
m/min
Spindle speed, N =
1000 1000 60
20
V
D
p p
=
´
´
= 954.929 » 1000 rpm
Assuming Feed F = 1400 mm/min
Program Description
N001 G90 G71 G94; (Absolute programming mode,data input is in metric mode
and feed in mm/min)
N002 M06 T0101; (Tool change, Tool no. 1 with offset value 01)
N003 S1000 M03; (Spindle speed 1000 rpm and in clockwise direction)
N004 G00 X10 Y–10 Z–15
M08;
(Rapid traverse to point 1 and coolant ON)
N005 G41 D20; (Tool compensation on RIGHT and cutter diameter is
20 mm)
N006 G01 Y70; (Linear interpolation upto point 2)
N007 X90; (Linear interpolation upto point 3)
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Home position
X
Y
1
2 3
4
5
6
7 8
9
10
11
Fig. 4.24.12 (a)

302. N008 Y10; (Linear interpolation upto point 4)
N009 X10; (Linear interpolation upto point 5)
N0010 Y50; (Linear interpolation upto point 6)
N0011 X30 Y70; (Linear interpolation upto point 7)
N0012 X70; (Linear interpolation upto point 8)
N0013 G02 X90 Y50 R20; (Clockwise interpolation upto point 9 and radius is 20 mm)
N0014 G01 Y30; (Linear interpolation upto point 10)
N0015 G03 X70 Y10 R20; (Anticlockwise interpolation upto point 11 and radius is
20 mm)
N0016 G00 Z15; (Rapidly move spindle upwards)
N0017 G28 U0 W0 Z0 (Return to home position)
N0018 M03 M09 M30; (Coolant OFF, spindle stop and program stop)
Example 4.24.13 : Write a program (manual part program) to drill five holes in the
locations shown in Fig. 4.24.13 and pause at each location where a hole should be
drilled.
Solution : Given data :
S = 300 rpm, F = 50 mm/min, Thickness of plate = 10 mm
Refer Fig. 4.24.13 (a).
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Start
point
1
2
3 4
5
6
Drill 5 holes-12.5 DIA
25
32
19
44
25
100
25
25
100
Fig. 4.24.13

303. Program Description
N001 G90 G71 G94; (Absolute programming mode and data input is in metric
mode and feed in mm/min)
N002 M06 T0101; (Tool change, tool no. 01 with offset value 01)
N003 M03 S300; (Spindle speed is 300 rpm and in clockwise direction)
N004 G00 X25 Y25 Z10; (Rapid traverse to start point and tool is above the plate)
N005 M08 X25 Y25; (Coolant ON and Rapid traverse to point 1)
N006 G04 X2; (Stoppage of axis motion i.e. pause for 2 sec.)
N007 G01 Z–10 F50; (Linear interpolation, drill the hole with a feed of 50 mm/min.)
N008 G01 Z10; (Move spindle upwards)
N009 G00 X31 Y57 Z10; (Rapid traverse of point 2 and tool is above the plate)
N0010 G04 X2; (Stoppage of axis motion i.e. pause for 2 sec.)
N0011 G01 Z – 10 F50; (Linear interpolation, drill the hole with a feed of 50 mm/min)
N0012 G01 Z10; (Move spindle upwards)
N0013 G00 X31 Y76 Z10; (Rapid traverse to point 3 and tool is above the plate)
4 - 87 Computer Aided Design and Manufacturing
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Start
point
1
2
3 4
5
6
25
32
19
44
25
100
25
Home position
25
100
X
Y
Fig. 4.24.13 (a)

304. N0014 G04 X2; (Pause for 2 sec)
N0015 G01 Z – 10 F50; (Linear interpolation, drill the hole with a feed of 50 mm/min)
N0016 G01 Z10; (Move spindle upwards)
N0017 G00 X75 Y76 Z10; (Rapid traverse to point 4 and tool is above the plate)
N0018 G04 X2; (Pause for 2 sec)
N0019 G01 Z – 10 F50; (Linear interpolation, drill the hole with a feed of 50 mm/min)
N0020 G01 Z10; (Move spindle upwards)
N0021 G00 X75 Y25 Z10; (Rapid traverse to point 5 and tool is above the plate)
N0022 G04 X2; (Pause for 2 sec)
N0023 G01 Z–10 F50; (Linear interpolation, drill the hole with a feed of 50 mm/min)
N0024 G01 Z10; (Move spindle upwards)
N0025 G28 U0 W0; (Return to home position)
N0026 M03 M09 M30; (Coolant OFF, spindle stop and program stop)
Example 4.24.14 : Write a part program for drilling holes in the part shown in
Fig. 4.24.14. The plate thickness is 20 mm.
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15
25
60
90
25
50
80
A
1
3
2
4
95
+Y
+X
–z
+z
20
Fig. 4.24.14

305. Solution : Given data :
Assume, S = 300 rpm and F = 50 mm/min, Thickness of plate = 20 mm,
Also, assume size of hole = 10 mm
Refer Fig. 4.24.14 (a).
Program Description
N001 G90 G71 G94; (Absolute programming mode and data input is in metric
mode and feed in mm / min)
N002 M06 T0101; (Tool change, tool no. 1 with offset value 01)
N003 M03 S300; (Spindle speed 300 rpm and in clockwise direction)
N004 G00 X25 Y25 Z2 M08; (Rapid traverse to point 1 and coolant ON and tool is
2 mm above the plate)
N005 G01 Z – 20; (Tool moves 20 mm inside the workpiece)
N006 G00 Z2; (Rapid traverse of tool 2 mm above the plate)
N007 G00 X50 Y60; (Rapid traverse to point 2)
N008 G01 Z – 20; (Tool moves 20 mm inside the workpiece)
N009 G00 Z2; (Rapid traverse of tool 2 mm above the plate)
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15
25
60
90
25
50
80
A
1
3
2
4
95
+Y
+X
Home
position
Fig. 4.24.14 (a)

306. N010 G00 X80 Y90; (Rapid traverse of tool to point 3)
N011 G01 Z – 20; (Tool moves 20 mm inside the workpiece)
N012 G00 Z2; (Rapid traverse of tool 2 mm above the plate)
N013 G00 X95 Y15; (Rapid traverse of tool to point 4)
N014 G01 Z – 20; (Tool moves 20 mm inside the workpiece)
N015 G00 Z2; (Rapid traverse of tool 2 mm above the plate)
N016 G00 X – 10 Y – 10; (Rapid traverse of tool to home position)
N017 M09 M05 M30; (Coolant OFF, spindle stop and program stop)
4.25 Subroutine :
· Subroutine is also called as subprograms.
· When a similar machining operation is to be performed repeatedly then the
general programming method will not be used because,
¡ It is very lengthy.
¡ It is tedious.
¡ It is more time consuming.
¡ It consumes more space in the computer memory.
· In such case, subroutine method is used. It is a time saving technique.
· It is an independent program similar to general program and stored in the
computer memory under separate program number.
· It can be called anywhere in the main program and for any number of times.
· To call the subroutine in the main program, the miscellaneous code M98 is used.
The instruction block for subroutine can be written as follows :
N10 M98 P50 L1;
where,
M98 indicates a call to subroutine,
P50 indicates the program number (here 50),
L1 indicates to call subroutine only one time.
· After execution of subroutine return back to the main program and continue it.
· To end the subroutine and return back to the main program M99 code is used.
· It is important to note that, the main program is written in absolute programming
mode (G90) and subroutine is written in incremental programming mode (G91).
· Hence, use code G91 at the start of subroutine and G90 before use of M99.
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307. Example 4.25.1 : Write a part program using subroutine for milling a square pocket
of 30 30
´ mm and 5 mm depth. Refer Fig.4.25.1. Take diameter of cutter as 5 mm,
speed 500 r.p.m. and feed 100 mm/min.
Solution : Fig. 4.25.1 (b) shows the enlarged view of pocket and corresponding tool path.
From Fig. 4.25.1 (a) coordinates of point P1 = (15, 30) and of point P5 = (60, 30). In this
case we will first write a subroutine program which is then called in the main program.
Subroutine program
Program Description
50; Subroutine program number.
N01 G91; Set to incremental mode.
N02 G01 Z–10 F100; Linear interpolation and cutter moves 5 mm
downward along Z-axis with feed 100 mm/min.
(Total depth = 5 5 10
+ = )
N03 Y20; Cutter moves 20 mm in Y direction i.e.
position A.
N04 X20; Cutter moves 20 mm in X direction i.e.
position B.
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1 2
10
5
15 10
30
15
30
25
P1 P5
P (0, 20)
A
P (20, 0)
B
P (0, –20)
C
P (–20, 0)
D
A B
C
D
(a) Workpiece
(b) Enlarged view
of pocket
Incremental
mode
Fig. 4.25.1

308. N05 Y–20; Cutter moves 20 mm in Y direction i.e.
position C.
N06 X–20; Cutter moves 20 mm in X direction i.e.
position D.
N07 G00 Z10; Rapidly move the cutter 10 mm in
Z-direction.
N08 G90 M99; Set absolute programming mode and end of
subroutine program.
Main program
Program Description
51; Main program number.
N10 G28 U0 V0 W0; Return to machine reference position.
N15 G90 G71 G94; Absolute programming mode, metric (mm)
data input and feed in mm/min.
N20 G17 M06 T01; Selection of XY plane, tool change and tool
no. 01 is selected.
N25 G41 M03 S500; Cutter radius compensation to left, spindle
ON with speed 500 r.p.m.
N30 G00 X15 Y30 Z5 M08; Rapid travel to position P1 (15, 30) and
cutter is 5 mm above the surface of
workpiece, coolant ON.
N35 M98 P50 L1; Call for subroutine, program number 50 and
for only one time.
N40 G00 X60 Y30; Rapid travel to position P5 (60, 30).
N45 M98 P50 L1; Call for subroutine, program number 50 and
only once.
N50 G28 U0 V0 W0; Return to machine reference position.
N55 G40 M05; Cutter radius compensation cancel and
spindle stop.
N60 M09 M30; Coolant off, program end and tape rewind.
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309. 4.26 Canned Cycle :
· Canned cycle is also called as multiple-repetitive cycle.
· It is a set of instructions stored in the computer memory which is used to
perform a fixed sequence of operation.
· It is commonly used for repetitive operation where material is to be cut in
number of passes.
· The main advantage of canned cycle is that, it reduces length of the program
hence memory space and the complexity of the program.
· In this cycle, the final position is mentioned in the instruction block and the
cutter path is automatically plotted by the controller itself.
· Canned cycle can be called and cancelled by using G-codes (preparatory codes).
· According to the shape of workpiece various G-codes are used for canned cycle.
For example : G74 for slot or rectangular pocket milling and G77 for circular
pocket milling.
· Canned cycle can be cancelled by using G80 code.
4.26.1 Comparison between Subroutine and Canned Cycle
Give the comparison between subroutine and canned cycle.
Sr. No. Subroutine Canned cycle
1. It is a separately written program which
is called in the main program.
It is not a separately written program. It
is the part of the program.
2. It is used when multiple passes are
required at different places.
It is used when multiple passes are
required at the same place.
3. It is called and executed by using
miscellaneous function (M-codes).
It is written by using preparatory
function (G-codes).
4. It is a separate program hence separate
program number is given.
It is not a separate program hence no
need to give any program number.
5. It is given in every block of instruction
till the operation is completed.
In this cycle, directly final point is given
in the instruction block.
6. The path of cutter for every point is
mentioned by the programmer.
The path of cutter for every pass is
automatically generated.
4.26.2 Slot Milling (G74)
· Slot milling is commonly used for producing keyway in shafts as shown in
Fig. 4.26.1.
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310. · For this purpose G74 code is used which is written
in the following format :
N05 XI YI ZI ;
N06 G74 XF YF ZF K S F;
where,
X Y Z
I I I be the initial position of the tool
centre point of the slot.
X Y Z
F F F be the final position of the tool
centre point of the slot.
K be the peck depth.
S be the peck feed.
F be the feed.
· For example : Write a program to produce a 10 mm deep slot by using a cutter
of 10 mm diameter. Take peck feed 50 mm/min, feed 100 mm/min and peck
depth 2 mm. Refer Fig. 4.26.2 (a).
Solution : Fig. 4.26.2 (b) shows the tool path or peck depth. The starting blocks and end
blocks are similar to all programs hence only the block of slot milling is explained as
follows :
Program Description
N10 G00 X10 Y20 Z5; Rapid travel to initial position i.e. P1 (10, 20)
and cutter is 5 mm above the workpiece
surface.
N15 G74 X60 Y20 Z–10 K2 S50 F100; Slot milling cycle and cutter moves to
P2 (60, 20) and 10 mm downward along
Z-axis with peck feed 50 mm/min and feed
100 mm/min.
N20 G80; Canned cycle cancel.
4 - 94 Computer Aided Design and Manufacturing
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P1 P2
10 50
40
(a) Workpiece
Peck depth K = 2 mm
Start
End
(b) Tool path
Fig. 4.26.2
End mill
cutter
Keyway
(closed)
Shaft
Fig. 4.26.1 : Keyway milling

311. 4.26.3 Rectangular Pocket Milling (G75)
· It is used for producing rectangular or square pockets in the component.
· Initially the tool will move by a distance equal to crossover towards the outer
surface and then it will move in clockwise direction.
· At the end of one cycle the tool will again move outward by the distance equal
to crossover and take a cut.
· For this cycle instruction block is written in the following format :
N10 G75 X Y Z I K S F ;
where,
XYZ be the width, thickness and depth of pocket respectively.
I be the crossover (generally
2
3
times the cutter diameter).
· For example : Write a part program to mill 30 20
´ mm pocket for 10 mm depth.
Take cutter diameter 10 mm, peck feed 30 mm/min and feed 100 mm/min. Refer
Fig. 4.26.3 (a).
Solution : Fig. 4.26.3 (b) shows the tool path for pocket. The starting blocks and end
blocks are similar to all programs, hence only the block of pocket milling is explained as
follows :
Program Description
N10 G00 X30 Y20 Z5; Rapid travel to initial position i.e. centre of
pocket and cutter is 5 mm above the
workpiece surface.
N15 G75 X20 Y10 Z–10 I6 K2 S30 F100; Rectangular pocket milling cycle with all
values.
N20 G80; Canned cycle cancel.
4 - 95 Computer Aided Design and Manufacturing
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60
40
(a) Workpiece (b) Tool path for pocket
20
10
R5
Start
End
Crossover
Fig. 4.26.3

312. 4.27 Automatically Programmed Tools (APT) :
· APT is a language of computer assisted part programming for CNC machine
tools. In this case, the programmer gives instruction to the computer in the form
of programming language.
· These instructions are stored in source file. Language processor converts these
statements into a center line data file. This file has cutter centerline locations
along with spindle, cutter diameter comensation ON and OFF.
· Postprocessor is used over here to convert CL (cutter location) file into tape
commands needed by NC control.
· APT is widely used language as it can generate complex geometries in all axis
simultaneously.
· Elements involved in computer assisted part programming are shown in Fig. 4.27.1.
4.27.1 Structure of APT
· APT consists of different types of statements made of letters, numbers and
punctuation marks.
· Punctuation marks used in APT are given in the following table :
Symbol Name Description
/ Slash It divides a statement. Major words are at the left
of slash while minor words, symbols modifying the
words are on the right side. example : Go/To, L6
, Comma It is used as a separator between various
elements. Normally, it is on right side of slash.
4 - 96 Computer Aided Design and Manufacturing
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Drawing Source file Translation
CL Data
Post processor
NC Tape commands
NCtapecommonds Machine tool
MCU
APT
Fig. 4.27.1 : Elements of computer assisted part programming

313. = Equal to It is used to assign an entity to a symbolic name.
Example : SP = POINT / 80, 30, 50
( ) Parenthesis These are used to enclose the nested definitions.
$ Dollar It is placed at the end of line. This indicates that
the statement continues in next line.
$ $ Double Dollar Any statement after this sign is comment. It is not
a part of program
· Words in the statement contain maximum six letters. It can contain alphabets or
numbers but no special characters.
· Keywords are the commands in APT to perform a certain function only.
Keywords are of following two types :
i) Major word : This defines type of statement.
ii) Minor word : This defines various parameters.
· Symbols are the words used for geometrical definitions and numerical values.
They must be defined before using in the program.
· Lables are the words used to refer a statement so that control can be passed to it.
They can be defined with all characters.
· Mathematical operations are represented in APT as follows :
Symbol Operation Symbol Operation
+ Addition – Subtraction
* Multiplication / Division
* * Exponential ABS Absolute value
SQRT Square root SIN Sin of angle
COS Cosine of angle TAN Tangent of angle
EXP Value of e to the power LOG Natural log
Table 4.27.1 : Mathematical Operations in APT
· An APT program consists of four major statements :
i) Geometry statement
· Geometry of a given component can be defined by the number of elements.
A geometry statement in APT is written as follows :
<Geometry name> = Major word/ <definition>
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i) Geometry statement ii) Motion statement
iii) Post processor statement iv) Auxiliary statement

314. · Geometry name is a major word such as POINT, LINE, CIRCLE, PLANE,
etc.
· <definition> is the statement defining geometry of the major word.
· Following are some examples of geometry statements.
a) Point : In 3D system point is defined by X, Y and Z co-ordinates. It is given
as,
P1 = POINT/5, 70, 0
b) Line : Lines are considered to have infinite length and do not have any
specific direction. It is represented as follows,
<symbol> = LINE / point 1, point 2
L1 = LINE / 50, 60, 90, 94
C) Circle : Circle is considered as an infinitely long cylinder kept perpendicular
to XY plane. It's radius value can not be negative. It is represented as,
<symbol> = CIRCLE / X1, Y1 radius
C1 = CIRCLE / 61, 62, 30
ii) Motion statement :
· This statement represents motion control of the machine. This defines tool
path as per geometry. This is written as follows :
· Major word/ Minor word, scalar
· In 3D system, path of cutting tool is defined by three intersecting surfaces,
they are Drive Surface (DS), Part Surface (PS) and Check Surface (CS).
· Tool moves along intersection of part and drive surface. Tool is stopped by
check surface.
· They are divided in the following three subgroups :
a) Setup command : In this case, end point of a motion is starting point of
subsequent motion. FROM command is used to show starting point of cutter
for first motion.
FROM/ X1, Y1
b) Point to point motion command : In this type of command motion between
two points are given in operations like drilling, milling etc. The following
statements are used :
· GODLA : This specifies relative movement along the specified axis.
GODLA/dx, dy, dz.
· GOTO : This is an absolute movement statement. It moves spindle to a
specific point from a current position.
GOTO/ x, y
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a) Setup commands
b) Point to point motion command
c) Continuous path motion command

315. c) Continuous path motion commands : In milling or turning, these are used to
specify the continuous path to generate different surfaces.
iii) Post processor statement :
· In this command, machine tool functions and their settings are specified.
This statement converts CL data to machine tool co-ordinate. This can be
stated as follows for cooling fluid. This command tells us about when to
start or stop cutting fluid.
COOLNT /
ON
OFF
MIST
FLOOD
ì
í
ï
ï
î
ï
ï
ü
ý
ï
ï
þ
ï
ï
iv) Auxiliary statement
· This statement control output of the program. This also makes computer to
accept a part program and make it readable. For example,
PARTNO / <literal string> : PARTNO (This statment is used to identify part
program).
4.28 Micromachining :
· Micromachining is a advanced machining process which is used to fabricate the
components having dimensions in micrometers.
· It usually involves chemical etching process on a very fine scale.
· The following are the examples of components manufactured by micromachining :
m Probes and sensors m Measuring devices
m Microactuators m Microgimbals
· Micromachining has important role in fabrication of microelectronic devices such
as Printed Circuit Boards (PCB).
· PCB technology is now-a-days foundation for information systems,
telecommunications, automotive controls, robotics, aerospace, military weapons,
etc.
· The most commonly semiconductor material used is silicon for micromachined
parts.
4.28.1 Wafer Machining
· Wafer machining is the primary process used in manufacturing of
microelectronics devices.
· After purifying the silicon used for fabrication, a single crystal slilicon is
obtained through the process known as Czochralski process.
· This process utilizes a seed crystal that is dipped into a silicon melt and then
slowly pulling out while being rotated.
· This results in silicon crystal of 150 mm - 300 mm in diameter and over 1 m in
length.
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316. · This crystal is then sliced into individual wafers by using a inner diameter blade
known as wafer machining. (Refer Fig. 4.28.1)
· In this method, rotating blade with inner diameter is used for cutting and then the
wafers are cut to a thickness of required microns (upto 05 103
. ´ mm).
· This thickness provides necessary physical and mechanical support for
temperature absorption and fabrication.
· Finally, these wafers are cleaned and polished to get surface without damage.
The fabrication of whole microelectronic device takes place over this wafer surface.
4.29 Part Programming using APT :
Example 4.29.1 : Write an APT
program for drilling holes on a
component as shown in Fig. 4.29.1 (a)
The component is 10 mm thick.
The post processor statement is
MACHIN/MM. Assume spindle speed
as 950 rpm and feed as 0.5 mm/rev.
4 - 100 Computer Aided Design and Manufacturing
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SP (70, 35)
6,4 Holes
(15, 10)
(25, 30)
(50, 40)
(45, 20)
10
Z
X
through
Fig. 4.29.1 (a)
Step 1 :
Silicon crystal boule
Step 2 :
Slice off ends
Step 3 :
Drill center bore
Step 4 :
Grind outer diameter
between live centers
Step 5 :
Grind orientation flat
Step 6 :
Slice with diamond-coated
wire saw
Fig. 4.28.1 : Wafer Machining

317. Solution : Tool path for the given part is shown in Fig. 4.29.1 (b). By using point to
point programming method, the program is written as follows :
PARTNO / EXAMPLE 1
MACHIN/MM, 1
PRINT/ON $ $ PRINT GEOMETRY
CLPRINT/ON $$ PRINT CLDATA
SP = POINT/70, 35, 40 $$ (Starting point co-ordinates 70, 35, 40)
P1 = POINT/15, 10, 5 $$ (Co-ordinates of point 1 (P1) 15, 10, 5)
P2 = POINT/45, 20, 5 $$ (Co-ordinates of point 2 (P2) 45, 20, 5)
P3 = POINT/25, 30, 5 $$ (Co-ordinates of point 3 (P3) 25, 30, 5)
P4 = POINT/50, 40, 5 $$ (Co-ordinates of point 4 (P4) 50, 40, 5)
$$ DRILL DIA 6 MM
SPINDL/950, CLW $$ (Spindle speed is 950 rpm clockwise)
SPINDL/ON $$ (Spindle ON)
COOLNT/MIST $$ (Coolant ON and its mist type)
FROM/SP
RAPID, GOTO/P1 $$ (From starting point go to point P1)
$$ DRILLING WITH FEED RATE 0.5 MM/REV
CYCLE/DRILL, – 15, MMPR, 0.5, 2 $$ (Drill 15 mm below origin with feed
of 0.5 mm/rev and 2 mm offset)
GOTO/P2
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P1
P2
P3
P4
SP (70, 35)
(15, 10)
(25, 30)
(50, 40)
(45, 20)
Fig. 4.29.1 (b)

318. CYCLE/DRILL, – 15, MMPR, 0.5, 2 $$ (Drill 15 mm below origin with feed of
0.5 mm/rev and 2 mm offset)
GOTO/P3
CYCLE/DRILL, – 15, MMPR, 0.5,2 $$ (Drill 15 mm below origin with feed of
0.5 mm/rev and 2 mm offset)
GOTO/P4
CYCLE/DRILL, – 15, MMPR, 0.5, 2 $$ (Drill 15 mm below origin with feed of
0.5 mm/rev and 2 mm offset)
CYCLE / OFF
COOLNT / OFF $$ (cycle and coolant OFF)
RAPID, GODLTA/15
RAPID, GOTO / SP $$ (Go to starting point)
REWIND
FINI $$ (Program finish)
Example 4.29.2 : Write an APT program for a component shown in Fig.4.29.2 (a). It
is 20 mm thick. Post processor statement is MACHIN/MMPOST, 3. The mill diameter is
10 mm.
Solution : The tool path for the above geometry is shown in Fig. 4.29.2(b)
PARTNO/EXAMPLE 2
MACHIN / MMPOST, 3
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R5
50 70
95
60
50
Fig. 4.29.3 (a)

319. PRINT / ON
CLPRINT / ON
SP = POINT / 0, 0, 40 $$ (Starting point co–ordinates 0, 0, 40)
L1 = LINE/XAXIS $$ (Line L1 is along X - axis)
L2 = LINE / Y AXIS $$ (Line L2 is along Y - axis)
L3 = LINE/PARLEL, L2, X LARGE, 50 $$ (Line L3 parallel to L2 at 50 mm)
L4 = LINE/PARLEL, L3, X LARGE 70 $$ (Line L4 parallel to L3 at 70 mm)
L5 = LINE/PARLEL, L4, X LARGE, 50 $$ (Line L5 parallel to L4 at 50 mm)
L6 = LINE/PARLEL, L1, Y LARGE, 95 $$ (Line L6 parallel to L1 at 95 mm)
L7 = LINE/PARLEL, L6, Y LARGE, 60 $$ (Line L7 parallel to L6 at 60 mm)
$$ DEFINE PLANE 2 MM ABOVE THE PART
PL1 = PLANE/0, 0, 1, 2 $$ (Select plane)
$$ = SETUP STATEMENTS
CUTTER/10 $$ (Cutter diameter is 10 mm)
SPINDL/800, CLW $$ (Spindle speed 800 rpm, clockwise)
SPINDL/ON $$ (Spindle ON)
FEDRAT/MMPM, 240 $$ (Feed rate is 240 mm/min)
COOLNT/ON $$ (Coolant ON)
$$ MOTION STATEMENTS
RAPID, GOTO, SP $$ (Goto starting point rapidly)
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3 4
2 5
6
1
7
8
L5
Y
SP X
50 70
95
60
50
L7
L4
L3
L2
L6
L1
Fig. 4.29.2 (b)

320. GO/TO, L6, TO, PL1, TO, L3 $$ (Go from L6 to PL1 and then from PL1 to L3)
GODLTA/ – 18
GOLFT/L6, PAST, L2 $$ (Tool moves from 1 to 2)
GORGT/L2, PAST, L7 $$ (Tool moves from 2 to 3)
GORGT/L7, PAST, L5 $$ (Tool moves from 3 to 4)
GORGT/L5, PAST, L6 $$ (Tool moves from 4 to 5)
GORGT/L6, TO, L4 $$ (Tool moves from 5 to 6)
GOLFT/L4, PAST, L1 $$ (Tool moves from 6 to 7)
GORGT/L1, PAST, L3 $$ (Tool moves from 7 to 8)
GORGT/L3, TO, L6 $$ (Tool moves from 8 to 1)
RAPID, GODLTA/18
RAPID, GOTO/SP $$ (Go to starting point rapidly)
COOLNT/OFF $$ (Coolant OFF)
SPINDL/OFF $$ (Spindle OFF)
REWIND
FINI $$ (Program finish)
Example 4.29.3 : Write an APT program for a part as shown in Fig. 4.29.3(a).
Thickness of plate is 10 mm. End mill diameter is 10 mm. Assume feed rate as 0.3
mm/rev and spindle speed as 1000 rpm.
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100
150
50
R20
R20
Fig. 4.29.3 (a)

321. Solution : The tool path for given part is shown in Fig. 4.29.3 (b)
PART NO/ EXAMPLE 3
MACHINE/MMPOST, 3
PRINT/ON
CLPRINT/ON
$$ PRINT GEOMETRY
SP = POINT / –10, –10, 50 $$ (Starting point co-ordinates –10, – 10, 50)
L1 = LINE/XAXIS $$ (Line L1 along X-axis)
C1 = CIRCLE/150, 0, 20 $$ (Co-ordinates of center of circle C1)
L4 = LINE/YAXIS $$ (Line L2 along Y-axis)
L2 = LINE/PARLEL, L4, XLARGE, 100 $$ (Line L2 parallel to L4 at 100 mm)
C2 = CIRCLE/20, 100, 20 $$ (Co-ordinates of center of circle C2)
P0 = POINT/150, 50 $$ (Co-ordinates of point P0)
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100
150
50
R20
R20
C2
L4
L3
L1
C1
P0
L2
X
Z
X
Y
2
3
4
5
6
SP
10
1
Fig. 4.29.3 (b)

322. L3 = LINE/P0, RIGHT, TANTO, C2 $$ (Line L3 tangent to C2)
$$ DEFINE PLANE 15 MM Below THE PART SURFACE
PL1 = PLANE/0, 0, 1, – 15 $$ (Select plane)
$$ SETUP STATEMENTS
CUTTER/10 $$ (Diameter of cutter is 10 mm)
SPINDL/1000, CLW $$ (Spindle speed 1000 rpm, clockwise)
FEDRAT/MMPR, 0, 3 $$ (Feedrate 0.3 mm/rev)
SPINDL/ON $$ (Spindle ON)
COOLNT/ON $$ (Coolant ON)
$$ MOTION STATEMENTS
RAPID, GOTO/SP $$ (Rapid Go to starting point)
GO/TO, L1, TO, PL1, TO, L4 $$ (Go from L1 to PL1 and then from PL1 to L4)
$$ CUTTER POSITION 2
GOFWD/L1, PAST, 1, INTOF, C1 $$ (Cutter moves from 1 to 2)
$$ (CUTTER POSITION 3)
GOLFT/C1, PAST, 1, INT OF, L2 $$ (Cutter moves from 2 to 3)
GOLFT/L2, PAST, L3 $$ (Cutter moves from 3 to 4)
GOLFT/L3, TANTO, C2 $$ (Cutter moves from 4 to 5)
GOFWD/C2, TANTO, L4 $$ (Cutter moves from 5 to 6)
GOFWD/L4, PAST, L1 $$ (Cutter moves from 6 to 1)
RAPID, GODLTA/50
COOLNT/OFF $$ (Coolant off)
SPINDL/OFF $$ (Spindle off)
RAPID, GOTO/SP
FINI $$ (Program finish)
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323. 4.30 Introduction of CAM Package :
Computer Aided Manufacturing (CAM) is an application technology that uses
computer software and machinery to facilitate and automate the processes. It often use the
computer aided design, real time controls and robotics.
Thus it helps in reducing waste and energy for enchanced manufacturing and
production efficiency via increased speed, raw material and precise tooling accuracy.
The CAM systems can be linked with the CAD softwares such as Auto CAD, Solid
works, Catia and many more to produce the given product by using different CAM
software such as
i) Master cam
ii) Virtual Gibbs
iii) Surf cam
iv) Edge cam
v) Smart cam
vi) Alpha cam
By using any of the CAM software it helps us to develop the CNC program for the
given product. The most commonly used software in industries as well as in education
institutes is master cam. It provides an easy assess with the popular CAD system which
helps take care of the investments in the industries.
The software incorperates a comprehension cutting tool database for cutting process
parameter. It used the actual tools from the various tool manufactures, which helps us to
manufacture as per the industrial requirements. Master cam also contains many inbuilt tool
path modules for different tool path generation thus it helps to manufacture the parts
easily.
Thus by using the CAM software one can manufacture the given part in a specific
time with more accuracy.
Review Questions
1. What is mean by 'Numerical control' ?
2. What are the types of NC system ?
3. Write note on punch tape.
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324. 4. What are the advantages, limitations and applications of NC systems over
conventional system ?
5. What is machining centre ? Briefly describe the horizontal and vertical
machining centre.
6. What are the elements of NC systems ?
7. Explain NC motion control system.
8. Differentiate between open loop and closed loop system.
9. Write note on CNC.
10. What are the constructional features of CNC machines ?
11. Explain automatic tool changer.
12. Write a short note on DNC.
13. Explain briefly NC part programming.
14. Write note on G codes and M codes.
15. Explain following codes:
a) G03 b) G90 c) G96
d) M08 e) M30 f) T05
16. Explain axis nomenclature of NC system.
17. Write a manual part program for the following components. Assume spindle
speed 500 rpm and feed 0.5 mm/rev.
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100 50
R5
30
30
20
30 22 10
R
6
15 6 10 5
Fig. 4.1 (a) Fig. 4.1 (b)

325. 4 - 109 Computer Aided Design and Manufacturing
Fundamental of CNC and Part Programming
80
70
19
21
49
61
+X
+Z
R9
26
18
24
20
R4.5
51
Fig. 4.1 (c)
+X
+Z
R20
50
10
Fig. 4.1 (d)
R5
2×45º 2×45º
R40
50
36
40
7
15
22
42
47
Fig. 4.1 (e)

326. 18. What is NC machine ?
19. Name the advantages of NC machines.
20. Mention the different components of DNC.
21. Draw the simple configuration of CNC.
22. Write the main difference between CNC and DNC.
23. Write the advantages of CNC.
24. Mention the requirements of spindles in CNC.
25. Write the requirements of sideways.
26. What are capabilities of MCU ?
27. Classify machining centres.
28. List the data needed essentially to make a part program.
29. Name the methods of creatings part programming.
30. What is absolute and incremental programming ?
31. Enlist the important steps followed in preparing a part program.
32. Explain the meaning of following G codes :
i) G94 iii) G90
ii) G03 iv) G70
33. Explain the meaning of following M codes :
i) M03 iii) M09
ii) M05 iv) M30
34. What is recirculating ball screw ?
35. Explain any two applications of NC system.
36. What is the difference between open loop and closed loop system ?
37. What is mean by continuous path system ?
38. What do you mean by straight line system ?
39. Classify NC system.
40. What is pallet changer system ?
41. Explain Automatic Tool Changer.
42. Explain axis nomenclature for milling machine.
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327. Part A : Two Marks Questions with Answers :
Q.1 What is numerical control ?
Ans. : Numerical control is a programmable automation in which actions are controlled by
means of coded numbers, letters and other symbols.
Q.2 Name the basic elements of NC system.
Ans. : A numerical control machine consists of following elements :
i) Machine Control Unit (MCU)
ii) Machine tool and NC tooling
iii) Part program and drawings.
Q.3 How NC system is classified ?
Ans. : NC system is classified as,
i) According to tool positioning or modes of programming
a. Absolute system b. Incremental system
ii) According to motion control system
a. Point to point system
b. Straight line or straight cut system
c. Continous or contouring path system
iii)According to servo control
a. Open loop b. Closed loop
iv)According to type of feedback device
a. Ananlog transducer b. Digital transducer.
Q.4 Explain absolute and incremental system.
Ans. : Absolute system : In this system, all the positions are indicated from a reference point,
which is a fixed zero point or set point.
Incremental system : In this system, the tool positions are indicated with repsect to
previous point.
Q.5 State any four advantages of NC system.
Ans. : i) High productivity.
ii) Less scrap.
iii) Flexibility in design.
iv) Reduction in inventory.
v) Less floor space required.
vi) Skilled operator not required.
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328. Q.6 Mention some disadvantages of NC system.
Ans. : i) High initial cost.
ii) High maintainance cost. iii) Costly control system.
iv) Uneployment.
Q.7 What are the applications of NC system ?
Ans. : i) NC system is used where 100 % inspection is required.
ii) Where frequent changes in design occur.
iii) Where repetitive production is required.
iv) For complex machining operations.
Q.8 Mention the types of NC systems.
Ans. : The common types of NC systems used in machine tools are
i) Coventional Numerical Control (NC)
ii) Computerized Numerical Control (CNC)
iii) Direct Numerical Control (DNC)
Q.9 What is the purpose of Automatic Tool Changer (ATC).
Ans. :
i) For complicated job on NC and CNC machines different types of tools are
required; for changing and resetting the tool, more time is required.
ii) For this purpose, tools can automatically be changed with the help of automatic
tool changer (ATC), hence productivity and repeatability of manufacturing
increases.
Q.10 What is Direct Numerical Control (DNC).
Ans. : DNC is a manufacturing system in which a number of machines are controlled by a
central computer through a direct connection of telecommunication lines and in real time.
Q.11 What are G codes and M codes ?
Ans. : i) G is a preparatory function which changes the control mode of the machine.
e.g. G01 - Linear interpolation
G04 - Dwell
ii) M is a miscellaneous function which is generally called as M codes.
e.g. - M00 - Program stop
M06 - Tool change.
Q.12 Explain fixed zero and floating zero method.
Ans. :
i) Fixed zero : In this method, the origin is always predefined. It is generally at
the lowermost left hand corner of the worktable.
ii) Floating zero : Floating zero concept is provided which allows the operator to define his
origin.
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329. Q.13 List the commonly used co-ordinate systems of CNC machine tools.
Ans. : i) Absolute co-ordinate system.
ii) Incremental co-ordinate system.
Q.14 What is point-to-point (PTP) system ?
Ans. : In this system, tool is accurately located at some specified position. The spindle is brought
to the starting point, then moved to the next location i.e. from point 1 to point then point 3 etc.
On that location, operation is performed and then tool moves to next location.
Q.15 What are G-codes and M-codes ? Give examples.
Ans. : G-codes or preparatory function, changes the control mode of the machine. G-codes are
followed by two digit number. It is written as G 01, G 02, etc.
M-codes or miscellaneous function, controls other auxiliary operations. It is also followed by two
digit number. It is written as M 03, M 04, etc.
Q.16 What is the difference between incremental and absolute system ?
Ans. : In absolute system, all the co-ordinates are indicated from a referance point which is a
fixed zero or set point. In incremental system, all the co-ordinates or tool positions are indicated
with respect to previous point.
Q.17 What is the role of computer for CNC machine tool ?
Ans. : The program is entered into the computer directly through keyboard. The program is
stored in computer memory, which can be recalled whenever required. Programs can be easily
edited and modified as per the requirement. These features makes the system flexible.
Q.18 Differentiate between fixed zero and floating zero in CNC terminology.)
Ans. : In a fixed zero method, the origin is always predefined. It is generally at the lowermost
left hand corner of the worktable. All other points are defined from this point.
In a floating zero concept, operator can define his origin, which makes it convenient to develope
programs of symmetrical components by providing the origin at the point of symmetry in the
workpiece.
Q.19 Name the various elements of CNC machines.
Ans. : i) Mini computer
ii) Machine tool and CNC tooling
iii) Part program and drawings.
4 - 113 Computer Aided Design and Manufacturing
Fundamental of CNC and Part Programming

330. Q.20 What are the classifications of NC machines ? What are the types of motion
control system used in NC machines ?
Ans. :
1. According to the tool positioning
a) Absolute system b) Incremental system
2. According to the motion control system
a) Point to point system b) Straight line system
c) Continuous path system.
3. According to servo control system
a) Open loop system b) Closed loop system
Q.21 Define NC.
Ans. : NC is a programmable automation in which various actions are controlled by means of
coded numbers, letters and other symbols.
Q.22 Mention the major elements of NC machines.
Ans. : i) Machine Control Unit (MCU)
ii) Machine tool and NC tooling
iii) Part program and drawings.
Q.23 Compare closed loop NC system with open loop NC system.
Ans. : In open loop system, there is no feedback, to ensure whether the obtained slide movement
is same as desired or not and if not, what error is present.
In closed loop system, there is a feedback device to compare the slide movement. It is nothing
but a transducer.
Q.24 Show the axes of a CNC horizontal boring machine ?
Ans. :
Q.25 What are the basic assumptions made while programming in APT language?
4 - 114 Computer Aided Design and Manufacturing
Fundamental of CNC and Part Programming
+X
+Y
+Z

331. Ans. : In APT programming, it is assumed that the workpiece remains stationary and cutting tool
does all the movements.
Q.26 What is mean by APT language ?
Ans. : Automatic Programming of Tools (APT) is the oldest and one of the most powerful NC
processor languges. It is generally used on large capability computers and can perform the
mathematics required for complex curves using four or five axis contouring techniques.
Q.27 Compare a closed loop NC system with open loop NC system.
Ans. : Open loop system involves feeding of tape, interpretation of information by tape reader,
storing the data in buffer storage and it converts into electrical signal and send this signal to the
control unit.
Closed loop system is almost similar to open loop system only it carries an additional feed back
device which is nothing but a transducer and accompanied by a comparator.
Q.28 What is a preparatory function ? How is it important in CNC Programming ?
Ans. : G is the preparatory function which changes the control mode of the machine and it is
called as G-codes. Generally, they are followed by two digit number. It is written as G01, G02
etc.
Q.29 Distinguish between point to point and continuous path systems.
Ans. : Point to point system : In this system, tool is accurately located at some specified
position. Fig. 4.2 (a) shows path of tool movement for drilling number of holes.
Continuous path system : In this, there is relative motion between the tool and workpiece,
during the whole operation. Due to this relative motion, different curves and profiles can be
cut. Actually, it is a combination of PTP and straight cut system.
4 - 115 Computer Aided Design and Manufacturing
Fundamental of CNC and Part Programming
Fig. 4.2 (a) : Point to point system (PTP)

332. Q.30 What do you mean by machining centre with respect to NC machines ?
Ans. : It is a versatile NC machine tool, which can perform different operations in a single
setting and can automatically change tool under the programmable control. Several operations
like milling, boring, drilling, reaming, tapping, counter-boring, etc. can be performed in a single
set up, in any sequence.
Q.31 What is meant by 'tool magazine' in a CNC machine ?
Ans. :
i) Tool magazine is used in the automatic tool changer for storing the different tools.
ii) Tool magazines are of two types such as drum type magazine and chain type magazine
which can store 60 or more tools.
Q.32 What is the function of subroutine in NC part programming ?
Ans. : i) A subroutine is a sequence of operations which is seperately defined and stored by
the user.
ii) Subroutine can be called at any point in the main part programes.
Q.33 With reference to CNC manual part programming, state what is linear
interpolation.
Ans. : Any machining during straight or taper lines is done using linear interpolation function
G01. The feed rate at which the cutting tool is required to move is also specified by linear
interpolation.
Q.34 Mention the advantages of stepping motor.
Ans. : Advantages :
i) Stepper motors offer precise rototation control.
ii) Stepper motor exhibit excellent positional accuracy and errors are non cummulative.
iii) Stepper motor has long maintenance free life and hence it is cost effective.
iv) These are directly compatible with digital control technique.
4 - 116 Computer Aided Design and Manufacturing
Fundamental of CNC and Part Programming
Fig. 4.2 (b) : Continuous path system

333. Q.35 State the differences between CNC and DNC.
Ans. :
Sr. No. Parameter CNC DNC
1 Productivity High Highest
2 Number of operations done at a time One Multiple
3 Initial cost High Highest
Q.36 What do you understand by 'canned cycle' in manual part programming ?
Ans. : · Canned cycle is a fixed sequence of particular operations.
· This set of instructions are permanently stored in the control system and can be
called and used by a single command in the part programming.
For example : Canned cycle for drilling, tapping, boring, etc.
Q.37 Define CNC and DNC.
Ans. : DNC (Direct Numerical Control) (Refer Two Marks Q.10 of Chapter - 4)
CNC (Computerised Numerical Control) : CNC is a manufacturing system in
which a seperate compuer is attached to each machine tool, with stored programmable
logic, with the absence of hard-wired logic systems.
Q.38 What is adaptive control ?
Ans. : Adaptive control is the control technique which automatically determines the process
variables like feed, cutting speed etc. during machining and makes changes in the prescribed
limit as per requirement.
Q.39 How are various functions timed in NC machines ?
Ans. : The various functions in NC machines are timed by using punched tape. It uses a binary
coded decimal system for containing operating information of NC tool.
Q.40 Distinguish a fixed zero and floating zero.
Ans. :
Fixed Zero Floating Zero
1) In this method, the origin is
always predefined.
1) In this method, the setting
of zero is done manually.
2) 2)
4 - 117 Computer Aided Design and Manufacturing
Fundamental of CNC and Part Programming
(0, 0)
(a) Fixed zero
(0, 0)
(b) Floating zero

334. Q.41 State the functions of the following G and M codes :
G00 G03 M06 M03
Ans. :
G00 = Positioning (Rapid traverse)
G03 = Circular interpolation anticlockwise
M06 = Tool change
M03 = Spindle normal rotation
Q.42 Define "micromachining" with the help of an example.
Ans. :
· Micromachining is a advanced machining process which is used to fabricate the
components having dimensions in micrometers.
· It usually involves chemical etching process on a very fine scale.
· The following are the examples of components manufactured by micromachining :
m Probes and sensors m Measuring devices
m Microactuators m Microgimbals
Fundamental of CNC and Part Programming ends ...
4 - 118 Computer Aided Design and Manufacturing
Fundamental of CNC and Part Programming

335. Syllabus : Group Technology(GT),Part Families–Parts Classification and
coding–Simple Problems in Opitz Part Coding system–Production flow
Analysis–Cellular Manufacturing–Composite part concept–Types of
Flexibility - FMS – FMS Components – FMS Application & Benefits –
FMS Planning and Control– Quantitative analysis in FMS.
Section No. Topic Name Page No.
5.1 Group Technology 5 - 3
5.2 Part Families 5 - 3
5.3 Parts Classification and Coding 5 - 5
5.4 Structures used for Classifying and Coding the Parts 5 - 5
5.5 OPTIZ Coding System 5 - 6
5.6 Production Flow Analysis 5 - 10
5.7 Rank Order Clustering 5 - 13
5.8 Cellular Manufacturing 5 - 19
5.9 Machine Cell Design 5 - 20
5.10 Part Movement between the Cell 5 - 23
5.11 Key Machine 5 - 24
5.12 Composite Part Concept 5 - 24
5.13 Flexible Manufacturing System (FMS) 5 - 25
5.14 Types of Flexibility 5 - 26
5 - 1 Computer Aided Design and Manufacturing
Chapter - 5
Cellular Manufacturing and
Flexible Manufacturing system (fms)
Unit - V

336. 5.15 Level of Flexibility 5 - 30
5.16 Components of FMS 5 - 31
5.17 Applications of FMS 5 - 39
5.18 Advanages of FMS 5 - 40
5.19 FMS Planning and Control 5 - 41
5.20 Quantitative Analysis in FMS 5 - 43
Part A : Two Marks Questions with Answers 5 - 49
Part B : University Questions with Answers 5 - 52
5 - 2 Computer Aided Design and Manufacturing
Cellular Manufacturing and Flexible Manufacturing System (FMS)

337. 5.1 Group Technology
Group technology is a manufacturing values in which similar configured parts are
identified and grouped together to take the benefit of their similarities in design and
production.
Similar parts are arranged into part families.
Each part family has similar design and/or manufacturing characteristics.
For example,
If a plant producing 20,000 different components, may be able to group these
components based on the design/manufacturing into 30-40 different families. Processing of
each part family is similar.
The efficiency of this group technology can be achieved by grouping the machine cell
according to the part families. Arranging the Machines based on part family is called
Cellular Manufacturing.
5.1.1 Benefits of Group Technology
When the company want implements group technology, they must do identify the part
families and re-arranging the production machines into cell.
Group technology produces more benefits, such as
· It promotes standardization of tooling, fixturing and setup.
· Process planning and production scheduling are simplified.
· Material handing is reduced.
· Setup time and work in process are reduced.
· Workers satisfaction improves.
· Higher quality of work is accomplished.
5.2 Part Families + [AU : Dec.-17]
Part family is a collection of parts in which similar parts are grouped based on the
manufacturing and design considerations. The parts within the family may differ but they
are close enough to include in the particular part family.
5 - 3 Computer Aided Design and Manufacturing
Cellular Manufacturing and Flexible Manufacturing System (FMS)

338. Fig. 5.2.1 shows two parts of similar shape and size but different manufacturing
requirements.
a) 15000 pieces / year , tolerance limit ± 0.0001 mm, material is CR steel
b) 7500 pieces / year , tolerance limit ± 0.01 mm, material is SS
Fig. 5.2.2 shows the various components that are different in shape but similar in
terms of design and manufacturing.
5.2.1 Identification of Part Families
There are three methods are used to identify the part families.
1. Visual inspection method.
2. Parts classification and coding.
3. Production flow analysis.
5.2.1.1 Visual Inspection Method
It is the simple and economic process. In which parts are identified by looking the
physical parts or photograph and grouped together to form a part family. But this method
is considered as a least accurate among three.
5 - 4 Computer Aided Design and Manufacturing
Cellular Manufacturing and Flexible Manufacturing System (FMS)
(a) (b)
Fig 5.2.1
Fig 5.2.2

339. 5.2.1.2 Parts Classification and Coding
Similarities or differences between the parts are identified using coding scheme. We
will discuss this in next section.
5.2.1.3 Production Flow Analysis
In this approach parts are grouped based on the product route sheet.
5.3 Parts Classification and Coding
Design and manufacturing areas gets benefited on part classification and coding. Most
classification and coding systems are one of the following :
1. Based on part design attributes
2. Based on part manufacturing attributes
3. Based on both design and manufacturing attributes.
5.3.1 Part Design Attributes
Basic external shape, basic internal shape, cylindrical or rectangular shape, L/D ratio,
aspect ratio, material types, part function, major and minor dimensions, tolerances and
surface finish.
5.3.2 Part Manufacturing Attributes
Major and minor processes, operation sequence, major dimensions, surface finish,
machine tool, production cycle time, batch size, annual production, fixtures required and
cutting tools used for manufacturing.
5.4 Structures used for Classifying and Coding the Parts
There are three coding structure used according to the symbols in the code.
1. Hierarchical structure
2. Chain-type structure
3. Mixed-mode structure
5.4.1 Hierarchical Structure
· It is also known as monocode.
· Interpretation of the each successive symbol depends on the value of the
preceding symbols.
· More Information can be included.
Example : Two digit part code is 15 or 25
5 - 5 Computer Aided Design and Manufacturing
Cellular Manufacturing and Flexible Manufacturing System (FMS)

340. m First digit represents general shape of the part.
1 = Cylindrical object
2 = Rectangular object
m In hierarchical structure, the value of the second digit depends on the first
digit. So,
· If the first digit is 1 , then second digit (5) represents L/D ratio for cylindrical
object.
· If the first digit is 2, then second digit (5) represents aspect ratio for rectangular
object.
5.4.2 Chain-type Structure
· It is also known as polycode.
· Interpretation of the each symbol in the sequence is always the same. It doesnot
depend on the value of the preceding symbol.
5.4.3 Mixed Mode Symbol
· It is a hybrid type.
· Uses a combination of both hierarchical and chain type structure.
· Most commonly used structure.
The number of digits in the code between 6 and 12. But now a days includes both
design and manufacturing data so the length of the code is larger which contains more
than 30. But in modern technology the computer will do data processing.
Some of the important part classification and coding systems are,
· Opitz classification system - The University of Aachen in Germany,
nonproprietary, Chain type.
· Brisch System - (Brisch-Birn Inc.)
· CODE (Manufacturing Data System, Inc.)
· CUTPLAN (Metcut Associates)
· DCLASS (Brigham Young University)
· MultiClass (OIR : Organization for Industrial Research), hierarchical or
decision-tree coding structure
· Part Analog System (Lovelace, Lawrence & Co., Inc.)
5.5 OPTIZ Coding System + [AU : Dec.-16, 17, May-17]
· It is one of the first published coding scheme for mechanical parts.
· It was developed by H.Optiz (University of Aachen in Germany)
· It is proposed for machined parts.
5 - 6 Computer Aided Design and Manufacturing
Cellular Manufacturing and Flexible Manufacturing System (FMS)

341. · Optiz coding scheme consists of 9 digits can be extended for adding 4 more
digits.
Primary Design
Attributes
Eg: External Shape,
Machined Features
Manufacturing
Attributes
Eg: Dimensions, Work
Materials, Accuracy,
Surface finish, etc.,
Depends on
Manufacturer, Varies
from Each companies
The first 5 digits (12345) are called form code which describes design attributes of the
part such as external shape and machined features. The next four digits (6789) are called
supplementary code which indicates the code related with manufacturing such as
dimensions, work materials, accuracy, surface finish etc., The additional four digits
(ABCD) depends on the manufacturer.
Basic structure of the Optize system of parts classifiction and coding
5 - 7 Computer Aided Design and Manufacturing
Cellular Manufacturing and Flexible Manufacturing System (FMS)
1 2 3 4 5 6 7 8 9 A B C D
Form code Supplementary code Secondary code
0
1
2
3
4
5
6
7
8
9
Nonrotational
Rotational
L/D 0.5
0.5 < L/D < 3
L/D 3
With deviation
L/D 2
With deviation
L/D 2
Special
Special
A/B 3
A/C 4
A/B > 3
A/B 3
A/C < 4
External
shape
element
Main
shape
Main
shape
Main
shape
Main
shape
Main bore and
rotational
machining
Rotational
machining
Internal
shape
element
Machining
of plane
surfaces
Machining
of plane
surfaces
Machining
of plane
surfaces
Other
holes and
teeth
Other holes,
teeth and
forning
Other holes,
teeth and
forning
Dimensions
Material
Original
shape
of
raw
materials
Accuracy
Digit 1
Part class Main shape
Form code
Digit 2 Digit 3 Digit 4 Digit 5
Digit
Supplementary
code
Rotational
machining
Plane surface
machining
Additional holes
teeth and forming 6 7 8 9
Fig. 5.5.1

342. Form code (digits 1-5) for rotational parts in the Optiz coding system
5.5.1 Solved Examples of Optiz Part Coding System
Example 5.5.1 : Interpret the form code if the part code is 2080.
Solution :
· 2 - Parts has L/D ratio >= 3
· 0 - No shape element (external shape elements)
· 8 - Operating thread
· 0 - No surface machining
· 1 - Part is axial
5 - 8 Computer Aided Design and Manufacturing
Cellular Manufacturing and Flexible Manufacturing System (FMS)
Part class
Auxiliary holes
and gear teeth
L/D 0.5 No auxiliary hole
L/D 3
Axial on pitch
circle diameter
Radial, not on
pitch circle
diameter
0.5 < L/D < 3
Axial, not on pitch
circle diameter
0 0
1 1
2 2
3 3
4 4
5 5
6 6
7 7
8 8
9 9
Nonrotational
parts
With
gear
teeth
Rotational
parts
No
gear
teeth
Digit 1 Digit 5
0
1
2
3
4
5
6
7
8
9
External shape,
external shape elements
Stepped
to
one
end
Digit 2
Smooth, no shape
elements
No shape
elements
No shape
elements
No shape
elements
No shape
elements
Or
smooth
Thread
Functional
groove
Functional
groove
Thread
Stepped
to
both
ends
Functional cone
Operating thread
All others
0
1
2
3
4
5
6
7
8
9
Internal shape,
internal shape elements
Smooth
or
stepped
to
one
end
Digit 3
No hole,
no breakthrough
Thread
Functional
groove
Functional
groove
Thread
Stepped
to
both
ends
Functional cone
Operating thread
All others
0
1
2
3
4
5
6
7
8
9
Plane surface
machining
Digit 4
No surface
machining
Surface plane and/or
curved in one
direction, external
External plane surface
related by graduation
around the circle
External groove
and/or slot
External plane surface
and/or slot,
external spline
Internal plane surface
and/or slot
Internal spline
(polygon)
Internal and external
polygon, groove
and/or slot
External spline
(polygon)
All others All others
Axial and/or radial
and/or other
direction
Axial and/or radial
on PCD and/or
other directions
Spur gear teeth
Bevel gear teeth
Other gear teeth

343. Example 5.5.2 : Prepare the OPTIZ form code for the given part.
Prepare the code using OPTIZ coding system table. (Fig. No 5.5.1)
Solution : · The total length of the part is 1.75, overall diameter 1.25, L/D = 1.4 so the
code is 1.
· External shape - a rotational part that is stepped on both with one thread. So the
code is 5.
· Internal shape - a through hole (code 1)
· By examining the drawing of the part (code 0)
· No auxiliary holes and gear teeth (code 0)
The form code is
1 5 1 0 0
Example 5.5.3 : Develop the OPTIZ form code for the given part.
5 - 9 Computer Aided Design and Manufacturing
Cellular Manufacturing and Flexible Manufacturing System (FMS)
1.000
0.500
0.875
1.500
0.250 0.750
1
2
13 UNC
Fig. 5.5.2
26.0 20.0
(Both ends)
19.5
26.6
42.0
95.0
121.0
19.5 19.5
3.0 wide groove
2.0 deep (both ends)
Spur gear
36.0 pitch diameter
Fig. 5.5.3

344. Solution :
L/D = 121/36 = 3.361 1st Digit = 2
External shape : Stepped both ends with functional groove 2nd Digit = 6
Internal shape : No hole 3rd Digit = 0
Plane surface machining : None 4th Digit = 0
Auxiliary holes and gear teeth : Spur gear 5th Digit = 6
Optiz form code is
2 6 0 0 6
5.6 Production Flow Analysis
Production Flow Analysis (PFA) is a technique in which part families are identified
using route sheet rather than part drawings. Similar part producing machines are grouped
together to form a machine cell.
PFA overcomes two possible anomalies that occur in part classification coding system
1. Basic geometry is different but similar in process routings.
2. Basic geometry is quite similar but the process routing is different.
5.6.1 Production Flow Analysis (PFA) Procedure
PFA starts with deciding the population of parts to be analyzed that means number of
parts to be included for the study. It may contains entire parts in the shop or the sample
of the parts. Then the PFA procedure contains following steps.
5 - 10 Computer Aided Design and Manufacturing
Cellular Manufacturing and Flexible Manufacturing System (FMS)
Data collection
Sortation of process
routings
PFA chart
Cluster analysis
Fig. 5.6.1

345. 5.6.1.1 Data Collection
· The data needed for the analysis are part number and operation sequence.
· Both are contained in the product route sheets.
· Each operation is usually associated with particular machine so determining the
operation sequence also determines the machine sequence.
· The following image shows the example of process route sheets.
Pcs. per pur size
Weight
Assy. No
TA 1279
Sub. Assy. No.
Date supplied
Issued by
Oper
No.
Operational descirption Dept. Machine Setup
Hr.
Rate
Pc. Hr.
Tools
20 Drill hole 0.32 + 0.015
– 0.005
Drill Mach
513
Drill
1.5 254 Drill fixture
L-76
Jig # 10393
30 Deburr 0.312 + 0.015
– 0.005 dia hole
Drill Mach
510
Drill
0.1 424 Multitooth
burring tool
40 Chamfer 0.009/0.875, bore
0.878/0.875 dia
(2 passes), bore
0.7600/0.7625 (1 pass)
Lathe Mach
D 109
lathe
1.0 44 Ramet-I TPG
221, chamfer
tool
50 Tap hole as designated 1/4
min full thread
Tap Mach
517
drill tap
2.0 180 Fixture # CR
353 tap 4 flute
sp.
60 Bore hole 1.33 to 1.138 dia Lathe H & H
E107
3.0 158 1.44 turret
fixture Hartord
Superspacer pl
# 45
holder # L46
FDTW - 100
insert #21 Chk.
fixture
70 Deburr 0.005 to 0.010 both
sides, hand feed to hand stop
Lathe E162
lathe
0.3 175 Collect CR
#179
1327 RPM
80 Broach keyway to remove
thread burrs
Drill Mach.
507 drill
0.4 91 B87 fixture L59
broach tap
0.875120G-H6
90 Hone thread I.D. 0.822/0.828 Grind Grinder 1.5 120
95 Home 0.7600/0.7625 Grind Grinder 1.5 120
5 - 11 Computer Aided Design and Manufacturing
Cellular Manufacturing and Flexible Manufacturing System (FMS)

346. 5.6.1.2 Sortation of Process Routings
· In this steps, parts are arranged into groups according to the similarity of their
process routings.
· All machines which are producing the part are allocated with code numbers
according to the sequence followed.
Example
Operation or machine Code
Lathe 01
Milling machine 02
Drilling machine 03
CNC - Milling 04
Grinding machine 05
· A sortation procedure is then used to arrange parts into "packs".
· "Packs" are group of parts with identical routings.
· Pack contains many parts, and these will form a part family.
· If the pack contains only one part number it shows the uniqueness of the
processing of the part.
5.6.1.3 PFA Chart
· PFA chart also known as "Part Machine Incidence Matrix".
· Parts are identified using alphabets along Row (i), Machines are identified in
numbers along column (j).
· In this matrix entries have value xij =1 or 0.
· Value 1 indicates the corresponding part i requires processing on machine j.
· If the value is 0 , no processing of components i is accomplished on machine j.
· Usually "0"s are not indicated in the part incidence matrix for better clarity.
5 - 12 Computer Aided Design and Manufacturing
Cellular Manufacturing and Flexible Manufacturing System (FMS)

347. Example :
5.6.1.4 Cluster Analysis
From the previous step matrix, related groupings are identified and rearranged to a
new pattern with similar machine sequences. One possible rearrangement of previous
example shown in Fig. 5.6.2.
Different machine groupings are identified with in blocks. That block considered as a
machine cell.
If the some packs do not fit into logical groupings, these parts must be analyzed using
revised process sequence. If not, these parts must continue with conventional process
layout.
5.7 Rank Order Clustering + [AU : Dec.-16, May-17]
Rank order clustering technique is used for performing cluster analysis to form a
machine cell.
5 - 13 Computer Aided Design and Manufacturing
Cellular Manufacturing and Flexible Manufacturing System (FMS)
A B C D E F G H I
1
2
3
4
5
6
7
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
Machines
Parts
A B
C D
E F G
H
I
1
2
3
4
5
6
7
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
Machines
Parts
Fig. 5.6.2

348. The procedure for doing rank order clustering technique as follows.
1. Develop part incidence matrix according to the route sheets.
In this matrix "1" Indicates the parts to be machined corresponding machines. "0"s not
indicated and left as a blank for better clarity.
Machines Parts
A B C D E F G H I
1 1 1 1
2 1 1
3 1 1 1
4 1 1
5 1 1
6 1 1
7 1 1 1
2. Find decimal equivalent and ranking the matrix in row wise
In each row of the matrix , from left to right allocate the binary number and find the
decimal equivalent as below and rank the rows in the decreasing value. If tie appears rank
the rows in the same order as they appear in the current matrix.
Eg :
To find decimal equivalent for row1,
= (1 2 ) (1 2 ) (1 2 )
8 5 1
´ + ´ + ´
= 256 + 32 + 2
= 290
Binary
Values
28 27 26 25 24 23 22 21 20 Decimal
Equivalent
Rank
Parts
Machines A B C D E F G H I
1 1 1 1 290 1
2 1 1 17 7
3 1 1 1 81 5
5 - 14 Computer Aided Design and Manufacturing
Cellular Manufacturing and Flexible Manufacturing System (FMS)

349. 4 1 1 136 4
5 1 1 258 2
6 1 1 65 6
7 1 1 1 140 3
3. Re-order the rows in the part machine incidence matrix
Reorder the rows in the part machine incidence matrix by listing them in decreasing
order starting from top.
Binary
Values
28 27 26 25 24 23 22 21 20 Decimal
Equivalent
Rank
Parts
Machines A B C D E F G H I
1 1 1 1 290 1
5 1 1 258 2
7 1 1 1 140 3
4 1 1 136 4
3 1 1 1 81 5
6 1 1 65 6
2 1 1 17 7
4. Find decimal equivalent and ranking the matrix in column wise
Parts
Machines A B C D E F G H I Binary
Values
1 1 1 1 26
5 1 1 25
7 1 1 1 24
4 1 1 23
3 1 1 1 22
5 - 15 Computer Aided Design and Manufacturing
Cellular Manufacturing and Flexible Manufacturing System (FMS)

350. 6 1 1 21
2 1 1 20
Decimal
Equivalent
96 24 6 64 5 24 16 96 7
Rank 1 4 8 3 9 5 6 2 7
5. Re-order the columns in the part machine incidence matrix
Reorder the columns in the part machine incidence matrix by listing them in
decreasing order starting with left column.
Parts
Machines A H D B F G I C E
1 1 1 1
5 1 1
7 1 1 1
4 1 1
3 1 1 1
6 1 1
2 1 1
The final solution is above in the table. Block represents machine cell. So the first
machine cell consists of machine number 1 and 5 which can produce parts A, H an D.
Part Family 1 : (A, H, D) and (1 & 5)
Part Family 2 : (B, F, G) and (7 & 4)
Part Family 3 : (I, C, E) and (3,6 & 2)
Example 5.7.1 : Apply the rank order clustering technique to the part-machine
incidence matrix in the following table to identify logical part families and machine
groups. Parts area identified by letters and machines are identified numerically.
5 - 16 Computer Aided Design and Manufacturing
Cellular Manufacturing and Flexible Manufacturing System (FMS)

351. Machines A B C D E F
1 1 1
2 1 1
3 1 1
4 1 1
5 1 1
6 1 1 1
Solution :
Step 1 :
Binary
Values
25 24 23 22 21 20
Machines A B C D E F Decimal
Equivalent
Rank
1 1 1 34 2
2 1 1 4 6
3 1 1 48 1
4 1 1 12 4
5 1 1 18 3
6 1 1 1 12 5
Step 2 :
Binary
Values
25 24 23 22 21 20
Machines A B C D E F Decimal
Equivalent
Rank
1 1 1 34 2
2 1 1 4 6
5 - 17 Computer Aided Design and Manufacturing
Cellular Manufacturing and Flexible Manufacturing System (FMS)

352. 3 1 1 48 1
4 1 1 12 5
5 1 1 18 3
6 1 1 1 12 4
Step 3 :
Binary
Values
25 24 23 22 21 20
Machines A B C D E F Decimal
Equivalent
Rank
3 1 1 48 1
1 1 1 34 2
5 1 1 18 3
6 1 1 1 12 4
4 1 1 12 5
2 1 1 4 6
Step 4 :
Machines 25 24 23 22 21 Binary
Values
A B C D E F
3 1 1 25
1 1 1 24
5 1 1 23
6 1 1 1 22
4 1 1 21
2 1 1 20
Decimal
Equivalent
48 40 6 7 24 5
Rank 1 2 5 4 3 6
5 - 18 Computer Aided Design and Manufacturing
Cellular Manufacturing and Flexible Manufacturing System (FMS)

353. Step 5 :
Machines A B E D C F
3 1 1
1 1 1
5 1 1
6 1 1 1
4 1 1
2 1 1
Decimal
Equivalent
48 40 24 7 6 5
Rank 1 2 3 4 5 6
Part Family 1 : (A, B, E) and (3, 1 & 5)
Part Family 2 : (D, C, F) and (6, 4 & 2)
5.7.1 Advantage of PFA
· It requires less time for parts classification and coding procedure.
· This is attractive for many companies those who want to adapt group technology.
5.7.2 Disadvantage of PFA
· The raw data used for this analysis is existing production sheet. These sheets
prepared by different process planners based on their skill.
· The routings may contain operations that are non-optimal, illogical on
unnecessary. So final machine groupings may be suboptimal.
5.8 Cellular Manufacturing
When part families are determined using any of the three methods (Visual Inspection,
Part classification an coding or PFA) there is advantage in producing those parts using GT
cells rather than traditional process-type machine layout.
Cellular manufacturing is an application of group technology in which machines are
grouped into cells for producing dedicated part family / parts / limited group of families.
5.8.1 Objective of Cellular Manufacturing
· To reduce the lead time by reducing set up time, work part handling time and
waiting time.
5 - 19 Computer Aided Design and Manufacturing
Cellular Manufacturing and Flexible Manufacturing System (FMS)

354. · To reduce work-in process inventory - smaller batch sizes and shorter lead times
reduce work-in process.
· To improve quality.
· To simplify production scheduling.
· To reduce setup times.
5.9 Machine Cell Design + [AU : Dec.-18]
Cell design determines high performance of the cell. Group technology manufacturing
cells can be classified according to the number of machines and the material flow between
the machines.
5.9.1 Types of Machine Cell
1. Single machine cell.
2. Group machine cell with manual handling.
3. Group machine cell with semi-integrated handling.
4. Flexible manufacturing cell or flexible manufacturing system.
5.9.1.1 Single Machines
As name implies it consists of one machine along with supporting fixtures and tooling.
This type of cell suitable for the parts having only one basic operations such as
turning or milling.
5.9.1.2 Group Machine Cell with Manual Handling
It have more than one machine to produce one or more part families.
There is no automated part handling instead human operators have to run the cell for
performing material handling functions.
Sometimes it can be achieved in conventional process layout without rearranging the
equipment.
5.9.1.3 Group Machine Cell with Semi-integrated Handling
It have more than one machine along with mechanized handling system such as
conveyor to move parts between the cell.
5.9.1.4 Flexible Manufacturing Cell or Flexible Manufacturing System (FMS)
It combines a fully integrated material handling system with automated processing
stations.
5 - 20 Computer Aided Design and Manufacturing
Cellular Manufacturing and Flexible Manufacturing System (FMS)

355. The FMS is the most highly automated of the Group Technology (GT) machine cells.
(The following chapter contains more details about FMS).
5.9.2 Types of Layout
1. U - Shaped machine layout
2. Inline layout
3. Loop layout
4. Rectangular layout
U - Shape layout
It is most appropriate when there is variation in the work flow among the parts made
in the cell.
It allows workers in the cell moves easily between the machines.
It also has below advantages
1. Easier changeover to one model to the next.
2. Improved quality.
3. Visual control of work in process.
4. Lower initial investment.
5. Greater worker satisfaction.
6. More flexibility.
5.9.2.1 Inline Layout
Inline layout using mechanized operations between machines.
Machines are arranged in sequences according to the part family.
5 - 21 Computer Aided Design and Manufacturing
Cellular Manufacturing and Flexible Manufacturing System (FMS)
Proc
Man
Proc
Man
Proc
Man
Proc
Man
Proc
Man
Work in
Work out
Manual handling
between machines
Fig. 5.9.1 U-shaped macine cell with manual part handling between machines

356. It occupies more space so it is suitable for when there is adequate space is exist.
Job travelling time is more. Its suitable for assembly sections.
5.9.2.2 Loop Layout
Loop layout allows variations in part routing between machines.
It occupies less space compare than in-line layout.
Better co-ordination between workers since work in and out are in the same location.
5.9.2.3 Rectangular Layout
Rectangular layout allows variations in part routing and allows for return for work
carriers if they are used.
5 - 22 Computer Aided Design and Manufacturing
Cellular Manufacturing and Flexible Manufacturing System (FMS)
Proc
Man
Work in Work out
Proc
Man
Proc
Man
Proc
Man
Mechanized work handling
Fig. 5.9.2 In-line layout using mechanised work handling between mechines
Work in
Work out
Proc
Man
Proc
Man
Proc
Man
Proc
Man
Proc
Man
Mechanized
work handling
Fig. 5.9.3 Loop layout allows variations in part routing between machines
Proc
Man
Proc
Man
Proc
Man
Work in
Work out
Fig. 5.9.4 Rectangular layout also allows variations in part routing and allows of return of
work carriers if they are used

357. 5.10 Part Movement between the Cell
Determining the most appropriate cell layout depends on the routing of parts produced
in cell. There are four types of part movement can be distinguished in a mixed model part
production system.
1. Repeat operations 2. In-sequence moves
3. By-passing move 4. Back tracking move
"proc" = Processing operations, "man" = Manual
Repeat operations : In which a consecutive operation is carried out on the same
machine so that the parts does not actually move.
In sequence moves : In which the part moves forward from the current machine to an
immediate neighbor machine.
By-passing move : In which the part moves forward to another machine that is two or
more machines ahead.
Backtracking move : In which the part moves backward from the current machine to
another machine.
Example : Compute a) the percentage of in-sequence moves b) the percentage eof
bypass moves c) the percentage of back tracking moves for the below
5 - 23 Computer Aided Design and Manufacturing
Cellular Manufacturing and Flexible Manufacturing System (FMS)
Proc
Man
Proc
Man
Proc
Man
Proc
Man
(1) Repeat
operation
(2)
In-sequence
move
(3) By-passing move
(4) Backtracking
move
Fig. 5.10.1
1
2
3 4
10 15
40 30 25
5
50 in 30 out
20 out
10
Fig. 5.10.2

358. Number of in sequence moves = 40 + 30 + 25 = 95
Number of by-passing moves = 10 + 15 = 25
Number of back tracking moves = 5 +10 = 15
Percentage of moves = Number of moves / Total number of moves
Percentage of in sequence moves = 95 / 135 = 70.4 %
Percentage of by-passing moves = 25 / 135 = 18.5 %
Percentage of back tracking moves = 15 / 135 = 11.1 %
5.11 Key Machine
A machine that is more expensive to operate than the other machines or that performs
certain critical operations in the plant is referred as key machine. Other machines in the
cell referred as supporting machines and they should be organized in the cell to keep the
key machine busy.
5.12 Composite Part Concept
The composite part is a hypothetical part that includes all of the design and
manufacturing attributes of the family. In general, an individual part in the family will
have some of the features of the family, but not all of them. A production cell for the part
family would consist of those machines required to make the composite part. Such a cell
would be able to produce any family member, by omitting operations corresponding to
features not possessed by that part.
5 - 24 Computer Aided Design and Manufacturing
Cellular Manufacturing and Flexible Manufacturing System (FMS)
1 2
3
4
5 6
7
(a) (b)
XX
Fig. 5.12.1

359. The following example shows the composite part consisting of all 7 design and
processing attributes representing a family of rotational parts features defined in the part B.
Associated with each feature is a certain machining operations as summarized in table
below. A machine cell to produce this part family would be designed with the capability
to accomplish all seven operations required to produce the composite part. The number of
design and manufacturing attributes is greater than seven, allowances must be made for
variations in over all size and shape of the parts in the family.
Design feature Corresponding Operation
1. External cylinder Turing
2. Face of cylinder Facing
3. Cylindrical step Turning
4. Smooth surface External cylindrical grinding
5. Axical hole Drilling
6. Counterbore Counterboring
7. Internal threads Tapping
5.13 Flexible Manufacturing System (FMS) + [AU : Dec.-16]
FMS is one of the Machine cell type used to implement cellular manufacturing. It is
the most automated and technologically sophisticated of the group technology cells.
"A Flexible Manufacturing System (FMS) is a highly automated Group Technology
(GT) Machine cell, consisting of a group of processing workstations, interconnected by
an automated material handling and storage system and controlled by a distributed
computer system".
Three capabilities that a manufacturing system must possess in order to be flexible,
i) The ability to identify and distinguish among the different incoming part or product
styles processed by the system.
ii) Quick changeover of operating instructions.
iii) Quick changeover of physical setup.
5.13.1 Flexibility Tests
There are four reasonable tests of flexibility in an automated manufacturing systems. A
manufacturing system should satisfy these four test being flexible.
5 - 25 Computer Aided Design and Manufacturing
Cellular Manufacturing and Flexible Manufacturing System (FMS)

360. Part variety test Can the system process different part styles ?
Schedule change test Can system readily accept changes in the production
schedule ?
Error recovery test Can the system recover equipment malfunctions and
breakdown ?
New part test Can new part designs be introduced into the existing product
mix easily ?
If the answer to be yes for all the above questions, then the system can be considered
as flexible.
5.14 Types of Flexibility + [AU : May-18]
Flexibility type Definition Depending factor
Machine flexibility Capability to adapt a given
machine to a wide range of
production operations and
part styles. The greater the
range of operations and part
styles the greater the
machine flexibility.
Setup or changeover time
Ease of machine
reprogramming .
Tool storage capacity of
machines.
Skill and versatility of
workers in the system.
Production flexibility The range or universe of
part styles that can be
produced on the system.
Machine flexibility of
individual stations. Range of
machine flexibility of all
stations in the system.
Mix flexibility Ability to change the product
mix while maintaining the
same total production
quantity.
Similarity of parts in the mix
machine flexibility .
Product Flexibility Which is related with
adaptability of design
change.
How close with new part
design matches with exited
part family.
Routing flexibility Capacity to produce parts
through alternative
workstation sequences in
response to equipment break
downs, tool failure and other
interruptions.
Similarity of workstations
Similarity of parts in the mix
machine flexibility.
5 - 26 Computer Aided Design and Manufacturing
Cellular Manufacturing and Flexible Manufacturing System (FMS)

361. Volume flexibility Ability to produce parts
economically.
Level of manual labour
performing production
investment amount.
Expansion flexibility Ease with which system can
be expanded to increase the
total production quantities.
Expense of adding
workstations easiness of
expansion.
Type of part handling system
used.
5.14.1 Single Machine Cell
5 - 27 Computer Aided Design and Manufacturing
Cellular Manufacturing and Flexible Manufacturing System (FMS)
Based on number
of machines
Based on level
of flexibility
· Single machine cell (SMC)
·
·
Flexible manufacturing cell
Flexible manufacturing system
· Dedicated FMS
· Random order FMS
FMS
Fig. 5.14.1
Tool
storage
Shuttle cart
(empty)
Shuttle track
Pallet holder
(empty)
Pallet
(with part)
Pallet rack
Pallet (empty)
Spindle
Shuttle cart
(with pallet and part)
CNC machining
center
Fig. 5.14.2

362. · It consists of only one CNC machining center combine with automated material
handling and storage system.
· Raw materials are loaded into the machine and unloaded from the machining
center.
· The cell can operated in batch mode, a flexible mode and even both.
5.14.2 Flexible Manufacturing Cell (FMC)
A flexible manufacturing cell consists of two or three processing workstations
(CNC machining Centers) with material handling system.
The material handling system is connected with Load/unload station.
5.14.3 Flexible Manufacturing System (FMS)
It has four or more processing stations connected each other by common material hand
ling system and electronically by a distributed computer system.
5 - 28 Computer Aided Design and Manufacturing
Cellular Manufacturing and Flexible Manufacturing System (FMS)
Computer control
system
Workstation Workstation Workstation
Material handling
system
Loading
station
Unloading
station
Fig. 5.14.3

363. 5.14.4 Difference between FMC and FMS
FMC FMS
It has two or three machines. It has four or more.
It does not have any supporting machines. It has supporting machines which do not
directly participate on it.
Needed simple computer control system. Needed larger and more sophisticated
Computer control system.
Limited error recovery. Minimized effect of machine breakdown.
Simple than FMS. More complex.
5 - 29 Computer Aided Design and Manufacturing
Cellular Manufacturing and Flexible Manufacturing System (FMS)
Computer control
system
Workstation Workstation Workstation Workstation
Material handling system
Loading
station
Unloading
station
Fig. 5.14.4
Single
machine
cell
Flexible
mfg.
cell
Flexible
mfg.
system
1 2 or 3 4 or more Number of
machines
Investment,
production
rate,
valume
annual
Fig. 5.14.5 Features of three categories of flexible cells and system

364. The figure depicts three categories of FMS related with production rate and annual
volume. Though FMS investment is high production rate and annual volume and
production rate is also high. Single machine cell suitable only for low production.
5.15 Level of Flexibility
Another way to classify the flexible manufacturing system si according to the level of
flexibility designed into the system.
The two category of flexibility are,
1) Dedicated FMS 2) Random - Order FMS
5.15.1 Dedicated FMS
· A dedicated FMS is designed to produce limited variety of parts styles.
· The part family is likely to be based on product rather than geometric similarities.
· The machine designed for producing specific process in a part family.
· It is suitable when the part family is small.
5.15.2 Random Order FMS
· It is suitable for larger part family.
· It is well suitable for accommodate variations in part design and part
configurations.
· Random order FMS must be more flexible than dedicated FMS.
· More sophisticated computer control system is required for this FMS.
· It is equipped with general purpose machine to deal with variations in the
product.
· The above image shows dedicated FMS has less flexibility but produce rate and
annual volume is more.
5 - 30 Computer Aided Design and Manufacturing
Cellular Manufacturing and Flexible Manufacturing System (FMS)
Random-
order FMS
Production rate annual volume
Flexibility,
part
variety
Dedicated
FMS
Q
Fig. 5.15.1

365. System type Part Variety Schedule
change
Error recovery New part
Dedicated FMS Limited Limited changes Limited but
sequential
process.
Not possible,
Introducing new
part is difficult.
Random order
FMS
Yes, part
variations are
possible.
Frequent and
significant
changes are
possible.
Machine
redundancy
minimizes the
effect of
machine break
down.
Yes, can be
introduced new
part.
5.16 Components of FMS + [AU : Dec.-16, 17, 18]
The basic components of an FMS are : Workstations, material handling and storage
systems, computer control system and the personnel that manage and operate the system.
These components are discussed in greater detail in the sub-sections below.
5.16.1 Work Stations
The processing and assembly equipment used in FMS depends on the type of work
accomplished by the system. The following work stations are most common work stations
in FMS.
5.16.1.1 Load / Unload Work Stations
Physical interface between the FMS and the rest of the factory, it is where raw parts
enter the system, and completely-processed parts exit the system. Loading and unloading
can be performed manually by personnel or it can be automated as part of the material
5 - 31 Computer Aided Design and Manufacturing
Cellular Manufacturing and Flexible Manufacturing System (FMS)
Work stations
Material
handling
Computer
control system
Human
resources
FMS
Fig. 5.16.1

366. handling system. Should be designed to permit the safe movement of parts and may be
supported by various mechanical devices (e.g. cranes, forklifts). The station includes a data
entry unit and monitor for communication between the operator and computer system,
regarding parts to enter the system, and parts to exit the system. In some FMSs, various
pallet fixtures to accommodate different pallet sizes may have to be put in place at
load/unload stations.
5.16.1.2 Machining Work Stations
The most common FMS application occurs at machining stations. These are usually
CNC machine tools with appropriate automatic tool changing and tool storage features to
facilitate quick physical changeover, as necessary. Machining centres can be ordered with
automatic pallet changers that can be readily interfaced with the FMS part handling
system. Machining centres used for non-rotational parts; for rotational parts turning centres
are used. Milling centres may also be used where there are requirements for multi-tooth
rotational cutters.
5.16.1.3 Assembly Work Stations
Associated with assembly FMSs, the assembly operation usually consists of a number
of workstations with industrial robots that sequentially assemble components to the base
part to create the overall assembly. They can be programmed to perform tasks with
variations in sequence and motion pattern to accommodate the different product styles
assembled in the system.
5.16.1.4 Supporting Work Stations
Supporting stations may include inspection stations where various inspection tasks may
be carried out. Co-ordinate measuring machines, special inspection probes, and machine
vision may all be used here. Other supporting stations may include pallet and part washing
stations for particularly dirty or oily FMSs, and temporary storage stations for both parts
and pallets.
5.16.1.5 Other work stations
Other possible stations may be found in specific industries, such as-for example-sheet
metal fabrication, which has stations for press-working operations, such as punching,
shearing, and certain bending and forming processes. Forging is another labour intensive
operation which may be broken into specific station categories, such as heating furnace,
forging press and trimming station.
5 - 32 Computer Aided Design and Manufacturing
Cellular Manufacturing and Flexible Manufacturing System (FMS)

367. 5.16.2 Material Handling and Storage System
Functions of the material handling and storage system in FMSs include :
· Allows random, independent movement of workparts between stations;
· Enables handling of a variety of workpart configurations;
· Provision of temporary storage;
· Provision of convenient load and unload stations; and
· Creation of compatibility with computer control.
FMS material handling equipment uses a variety of conventional material transport
equipment, in-line transfer mechanisms and industrial robotics.
FMS material handling system classified into two types :
1) Primary material handling system.
2) Secondary material handling system.
Primary system establishes the basic layout and is responsible for moving parts
between stations in the system.
Secondary system consists of transfer devices, automatic pallet changers etc., The
function of secondary system is to transfer the parts from the primary system to the
workstations.
Other purposes of the secondary handling system include : (1) re-orientation of the
part if necessary to present the surface that is to be processed; and (2) to act as buffer
storage as the workstation, should this be needed.
5.16.2.1 FMS Layout Configurations
The material handling system establishes the FMS layout. The common layouts are
· In-line layout
· Loop layout
· Ladder layout
· Open-field layout
· Robot-centered layout
In-line layout
· The machines and handling system are arranged in a straight line.
· Parts progress from one workstation to the next in a well-defined sequence.
· Work always moving in one direction and with no back-flow.
5 - 33 Computer Aided Design and Manufacturing
Cellular Manufacturing and Flexible Manufacturing System (FMS)

368. · Routing flexibility can be increased by installing a linear transfer system with
bi-directional flow. (Fig. 5.16.2 (b))
Loop layout
· Workstations are organized in a loop and it is served by a looped parts handling
system.
· In Fig. 5.16.3 parts usually flow in one direction around the loop with the
capability to stop and be transferred to any station.
· Each station has secondary handling equipment so that part can be brought-to and
transferred from the station work head to the material handling loop.
· Load/unload stations are usually located at one end of the loop.
· Alternative for of loop layout is rectangular layout. In this layout, the returning
pallets reaches the starting point.
5 - 34 Computer Aided Design and Manufacturing
Cellular Manufacturing and Flexible Manufacturing System (FMS)
Load
Man
Unld
Man
Mach
Aut
Mach
Aut
Mach
Aut
Mach
Aut
Starting
workparts
Completed
parts
Part transport system
Work flow
Partially completed
work parts
Load
Unld
Man
Mach
Aut
Mach
Aut
Mach
Aut
Mach
Aut
Starting
workparts
Shuttle cart
Work flow
Primary line
Secondary
handling
system
Completed parts
(a)
(b)
Fig. 5.16.2

369. Ladder layout
· This consists of a loop with rungs like ladder upon which workstations are
located.
· The rungs increase the number of possible ways of getting from one machine to
the next and obviates the need for a secondary material handling system.
· It reduces average travel distance and minimizes congestion in the handling
system, thereby reducing transport time between stations.
5 - 35 Computer Aided Design and Manufacturing
Cellular Manufacturing and Flexible Manufacturing System (FMS)
Parts transport
loop
Load
Unld
Man
Mach
Aut
Insp.
Aut
Mach
Aut
Mach
Aut
Mach
Aut
Mach
Aut
Completed
parts
Starting
workparts
Direction of
work flow
Mach
Aut
Mach
Aut
Mach
Aut
Mach
Aut
Unld
Man
Load
Man
Starting
workparts
Completed
parts
Returning pallets
Forward loop
Return loop
Fig. 5.16.3

370. Open field layout
· Consists of multiple loops and ladders and may include sidings also.
· This layout is generally used to process a large family of parts, although the
number of different machine types may be limited and parts are usually routed to
different workstations-depending on which one becomes available first.
5 - 36 Computer Aided Design and Manufacturing
Cellular Manufacturing and Flexible Manufacturing System (FMS)
Mach
Aut
Mach
Aut
Starting
worksparts
Completed
parts
Direction of
workflow
Mach
Aut
Mach
Aut
Mach
Aut
Fig. 5.16.4

371. Robot-centered layout
5 - 37 Computer Aided Design and Manufacturing
Cellular Manufacturing and Flexible Manufacturing System (FMS)
Insp
Aut
Mach
Aut
Insp
Aut Insp
Aut
Insp
Aut
Insp
Aut
clng
Aut
Load
unlad
Man
Completed
parts
Starting
work parts
AGV AGV guidpath
Rechg Rechg
Open field FMS layout. Key : Load = Parts loading,
UnLd = Parts unloading, Mach = Machining,
Clng = Cleaning,Insp = Inspection. Man = Manual,
Aut = Automated, AGV = Automated guided vehicle.
Rechg = Battery recharging station for AGVs.
Fig. 5.16.5
Machine worktable
Machine tool
Robot
Parts carousel
Fig. 5.16.6

372. · This layout uses one or more robots as the material handling system.
· Robot centered layouts used to process cylindrical or disk-shaped parts.
5.16.3 Computer Control System
To operate, the FMS uses a distributed computer system that is interfaced with all
workstations in the system, as well as with the material handling system and other
hardware components.
It consists of a central computer and a series of micro-computers that control
individual machines in the FMS.
The central computer co-ordinates all the activities of the components to achieve
smooth operational control of the system.
The following control functions may be noted :
Workstation control - Fully automated FMSs use some form of workstation control at
each station, often in the form of CNC control.
Distribution of control - Instructions to workstations-a central computer is required to
handle the processing occurring at different workstations; this involves the distribution of
part programmes to individual workstations, based upon an overall schedule held by the
central computer.
Production control - Management of the mix and rate at which various parts are
launched into the system is important; alongside data input of a number of essential
5 - 38 Computer Aided Design and Manufacturing
Cellular Manufacturing and Flexible Manufacturing System (FMS)
Parts
Finished
goods
Terminal
Pallet
Tools
Load Unload
Computer
control
room
Machine
Machine
Fig. 5.16.7

373. metrics, such as : Daily desired production rates, number of raw workparts available,
work-in-progress etc.
Traffic control - Management of the primary handling system is essential so that parts
arrive at the right location at the right time and in the right condition.
Shuttle control - Management of the secondary handling system is also important, to
ensure the correct delivery of the workpart to the station's workhead Workpiece monitoring -
The computer must monitor the status of each cart or pallet in the primary and secondary
handling systems, to ensure that we know the location of every element in the system.
Tool control - This is concerned with managing tool location (keeping track of the
different tools used at different workstations, which can be a determinant on where a part can
be processed), and tool life (keeping track on how much usage the tool has gone through, so
as to determine when it should be replaced). Performance monitoring and reporting -
The computer must collected data on the various operations on-going in the FMS and present
performance findings based on this.
Diagnostics - The computer must be able to diagnose, to a high degree of accuracy,
where a problem may be occurring in the FMS.
5.16.4 Human Resources
Human personnel manage the overall operations of the system. Humans are also
required in the FMS to perform a variety of supporting operations in the system; these
include :
· Loading raw work parts into the system;
· Unloading finished parts or assemblies from the system;
· Changing and setting tools;
· Performing equipment maintenance and repair;
· Performing NC part programming;
· Programming and operating the computer system and
· Over all managing the system.
5.17 Applications of FMS + [AU : May-18, Dec.18]
Flexible manufacturing systems are typically used for mid-volume, mid-variety
production. If the part or product is made in high quantities with no style variations, then
a transfer line or similar dedicated production system is most appropriate. If the parts are
low volume with high variety, then numerical control or even manual methods would be
more appropriate.
5 - 39 Computer Aided Design and Manufacturing
Cellular Manufacturing and Flexible Manufacturing System (FMS)

374. 5.18 Advanages of FMS + [AU : May-18, Dec.-18]
· Increased machine utilization-owing to 24 hr per day operation, automatic tool
changing of machine tools, automatic pallet changing at workstations, queues of
parts at stations and dynamic scheduling of production that compensates for
irregularities.
· Fewer machines required-because existing machines are highly flexible, and
because of higher machine utilization.
· Reduction in the amount of factory floor space required.
· Greater responsiveness to change-owing to the inherent flexibility of the system.
· Reduced inventory requirements-work-in-process is reduced because different
parts are processed together, and not in batches.
· Lower manufacturing lead times.
· Reduced direct labour requirements and higher labour productivity-higher
production rates and lower reliance on direct labour mean greater productivity per
labour hour with an FMS than with conventional production methods.
· Opportunity for unattended production-the system can operate for long periods
without human attention.
5 - 40 Computer Aided Design and Manufacturing
Cellular Manufacturing and Flexible Manufacturing System (FMS)
Stand alone NC
machines
Flexible
manufacturing
system
Transfer
lines
High
Medium
Low
Low
Medium
Product
variety
High
Production volume
Fig. 5.17.1 Application characteristics of FMS

375. 5.19 FMS Planning and Control
Implementation of FMS is a major investment. It is important that the installation must
be proceed thorough planning and design.
5.19.1 FMS Planning Issues
Major issues of planning for the creation of FMSs include : Part family considerations;
processing requirements; physical characteristics of the workparts; and production volume.
These are listed in the below table.
Issue Description
Part family considerations. A choice has to be made regarding group
technology and the part family to be
produced on the FMS.
There must be some consideration on the
creation of a composite part, with all
possible physical attributes of the parts that
may be processed in the FMS.
Processing requirements. Once the entire range of possible parts to
be processed are known, we must use this
information to choose associated processing
requirements for each part.
Physical characteristics of the workparts. Size and weight of workparts determine size
of the machines required to process the
parts. It also determines the size of the
material handling system needed.
Production volume. Production quantities must be determined, as
these tell us how many machines of each
type will be required.
5.19.2 FMS Design Issues
Once planning issues sort out, the following design issues may considered.
Issue Description
Types of workstations. Workstation choices have to be made,
depending on part processing requirements.
We must consider position and use of load
and unload stations also.
5 - 41 Computer Aided Design and Manufacturing
Cellular Manufacturing and Flexible Manufacturing System (FMS)

376. Variations in process routings and FMS
layout.
If part processing variations are minimal, we
may decide to use an in-line flow; if part
processing variations are high, we may
instead opt for a loop flow, or higher still, for
an open field layout.
Material handling system. We must select an appropriate primary and
secondary material handling system to suit
the layout chosen.
Work-in-process (WIP) and storage capacity. Determining an appropriate level of WIP
allowed is important, as it affects the level of
utilization and efficiency of the FMS. Storage
capacity must be compatible with the level of
WIP chosen.
Tooling We must determine the number and type of
tools required at each workstation. How
much duplication of tooling should occur at
workstations ? Duplication allows for efficient
re-routing in the system should breakdowns
occur.
Pallet fixtures For non-rotational parts, selection of a few
types of pallet fixtures is important. Factors
that influence the decision include : Levels
of WIP chosen and differences in part style
and size.
5.19.3 FMS Operational Issues
Once FMS installed, then resources must be optimized to meet production
requirements and achieve operational objectives related to profit, quality and customer
satisfaction.
Issue Description
Scheduling and dispatching. Scheduling must be considered for the FMS, based upon
the master production schedule. Dispatching is concerning
with launching the parts into the system at the appropriate
times.
Machine loading We must choose how to allocate specific parts to specific
machines in the system, based upon their tooling resources
and routing considerations of the FMS.
Part routing Routing decisions involve the choice of a route that should
be followed by a part in the system. Consideration should
be given to other parts travelling in the system, traffic
management etc.
5 - 42 Computer Aided Design and Manufacturing
Cellular Manufacturing and Flexible Manufacturing System (FMS)

377. Part grouping Part types must be grouped for simultaneous production,
given limitations on available tooling and other resources at
workstations.
Tool management We must determine when best to change tools, how long
each tool can last before requiring maintenance, and how to
allocate tooling to workstations in the system.
Pallet and fixture allocation How do we allocate specific pallets and fixtures to certain
parts to be launched into the system ? Different parts
require different pallet fixtures, and before a given part style
can be launched into the system, a fixture for that part must
be made available.
5.20 Quantitative Analysis in FMS + [AU : Dec.-18]
Design and operation issues which are identified should be addressed through
quantitative analysis. These are classified into
1) Deterministic model 2) Queuing model
3) Discrete event simulation 4) Other approaches including heuristics.
5.20.1 Bottle Neck Model
· Performance of FMS can be mathematically described by a deterministic model
called bottle neck model.
· This model is simple and intuitive.
· This model is suitable for any production system not limited to FMS.
· The term bottle neck refers to the fact that the output of production system has
an upper limit given that the product mix an upper limit, given that the product
mix flowing through the system is FIXED.
Terminology and symbol used
i) Part mix
The mix of the various part or product style produced by the system is defined by pj.
pj
j 1
p
=
å = 1.0
pj = Fraction of total system output of style j.
P = Total number of different parts styles made in total number
of different parts styles made in the FMS during period
of interest
5 - 43 Computer Aided Design and Manufacturing
Cellular Manufacturing and Flexible Manufacturing System (FMS)

378. ii) Workstation and servers :
It is possible to have two or more machines capable of performing the same
operations.
Number of servers at workstation i, where i = 1,2……n.
iii) Process routings
The process routing defines sequence of operations, the work stations at which they
are performed and associated processing times.
t ijk = Processing time which is total time that a production unit occupies a given work
station server i = Station; j = Part or product; k = Sequence of operation.
iv) Work handling system
Work handling system is designated as n + 1 and Sn + 1 number of carriers in the FMS
handling system.
v) Transport time
t n 1
+ = The mean transport time required to move a part from one work station to next
work station in the process routing.
vi) Operation frequency
It is defined as the expected number of times a given operation in the process routing
is performed for each work unit.
fijk = Operation frequency for operation k in the process plan j at station i.
FMS Operational Parameters
The average work load for the given station is defined as the mean total time spent
at the station per part. It is calculated as :
WLi = t f p
ijk ijk j
k
j
å
å
WLi = Average work load for station i(min),
t ijk = Processing time for operation k in the process j at station i (min),
fijk = Operation freq for oper k in part j at stn i.
pj = Part mix fraction for part j.
The average number of transports is equal to the mean number of operations in the
process routing minus one.
5 - 44 Computer Aided Design and Manufacturing
Cellular Manufacturing and Flexible Manufacturing System (FMS)

379. nt = f p
ijk j
k
j
-
å
å
å 1
i
nt = Mean number of transports.
Computing the work load of the handling
System :
WLn 1
+ = n t
t n 1
+
WLn 1
+ = Work load of handling system (min),
nt = Mean number of transports
t n+ 1 = Mean transport time per move (min)
2. System Performance measures
Assumptions : 1. FMS producing at max possible rate; 2. Rate is constrained by
bottleneck station in the system (highest workload per server).
Work load per server is
WLi / si
The bottleneck is identified by finding max value of the ratio among all stations.
Let WL s t
* * *
, , equal to WL, No. of servers, processing time for the bottleneck station
resp.
FMS max production rate of all parts is
Rp
* =
s
WL
*
*
The above equation is valid if product mix is constant.
Individual part production rates can be obtained by multiplying Rp
* by the
respective part mix rations.
Rpj
* = p (R )
j p
* = p
s
WL
j
*
*
Rpj
* = Max prod rate of part style j (pc/min)
and pj = Part mix fraction for style j
The mean utilization of each workstation is the proportion of time that the servers
at the station are working and not idle.
5 - 45 Computer Aided Design and Manufacturing
Cellular Manufacturing and Flexible Manufacturing System (FMS)

380. Ui =
WL
s
(R )
i
i
p
* =
WL
s
,
s
WL
i
i
*
*
Ui = Utilization of station i,
WLi = Workload of station i(min/pc),
Si = Number of servers at workstation i,
and Rp
* = Overall production rate (pc/min).
The utilization of the bottleneck station is 100 % at Rp
*
The average station utilization including transport system as
U =
i 1
n 1
i
U
n 1
=
+
å
+
U = Is an unweighted average of all workstations utilization.
Useful measure is overall FMS utilization which is based on number of servers at
each station.
Us =
i 1
n
i i
i 1
n
i
s U
s
=
=
å
å
Us = Overall FMS utilization.
Number of busy servers at each station is
BSi = WL (R )
i p
* = WL
s
WL
i
*
*
BSi = Number of busy servers on average at station i
and WLi = Workload at station i.
Example
A flexible manufacturing cell consists of two maching workstations plus a load/unload
station. The load/unload station 1. Station 2 performs milling operations and consists of
one server (one CNC milling machine). Station 3 has one server that performs drilling
(one CNC drill press). The three stations are connected by a part handling system that has
one work carrier. The mean transport time is 2.5 min. The FMC produces three parts,
5 - 46 Computer Aided Design and Manufacturing
Cellular Manufacturing and Flexible Manufacturing System (FMS)

381. A, B and C. The part mixs fractions and process routing for the three parts ae presented in
the table below. The operation frequency fijk = 1.0 for all operations.
Determine :
a) Maximum production rate of the FMC,
b) Corresponding production rates of each product.
c) Utilization of each machine in the system,
d) Number of busy servers at each station.
Part j Part mix pj Operation
k
Description Station i Process time
tijk
A 0.2 1 Load 1 3 min
2 Mill 2 20 min
3 Drill 3 12 min
4 Unload 1 2 min
B 0.3 1 Load 1 3 min
2 Mill 2 15 min
3 Drill 3 30 min
4 Unload 1 2 min
C 0.5 1 Load 1 3 min
2 Mill 3 14 min
3 Drill 2 22 min
4 Unload 1 2 min
WL2 = 20 (0.2)(1.0) 15(0.3)(1.0) 22(0.5)(1.0)
+ + = 19.5 min
WL3 = 12 (0.2)(1.0) 30(0.3)(1.0) 14(0.5)(1.0)
+ + = 18.4 min
5 - 47 Computer Aided Design and Manufacturing
Cellular Manufacturing and Flexible Manufacturing System (FMS)
t f p
ijk ijk j
WL =
i
j k
WL = (3+2)(0.2)(1.0) + (3+2)(0.3)(1.0) + (3+2)(0.5)(1.0) = 5.0 min
i
a)

382. nt =
i j k
ijk j
f P 1
å å å -
nt = 3(0.2)(1.0) 3(0.3)(1.0) 3(0.5)(1.0)
+ + = 3
Bottleneck station has largest WL / s
i i ratio
Station WLi /si
1 (load/unload) 5.0/1 = 5.0 min
2 (mill) 19.5/1 = 19.5 min ¬ Bottleneck
3 (drill) 18.4/1 = 18.4 min
4 (material handling) 7.5/1 = 7.5 min
Rp
* =
s
WL
*
*
Bottleneck is station 2 : Rp
* = 1/19.5 = 0.05128 pc/min = 3.077 pc/hr. Individual part
production rates can be obtained by multiplying.
5 - 48 Computer Aided Design and Manufacturing
Cellular Manufacturing and Flexible Manufacturing System (FMS)
WL =
n+1 n t
t n+1
WL = 3(2.5) = 7.5 min
4
R = p (R* ) = p
*
pj j p j
S*
WL*
R = 0.05128(0.2) = 0.01026 pc/min =
pA 0.6154 pc/hr
b)
R = 0.05128(0.3) = 0.01538 pc/min =
pB 0.9231 pc/hr
R = 0.05128(0.5) = 0.02564 pc/min =
pC 1.5385 pc/hr

383. Part A : Two Marks Questions with Answers
Q.1 What is cellular manufacturing ? + (AU : Dec. 16)
Ans. : Cellular manufacturing is an application of group technology in which machines are
grouped into cells for producing dedicated part family / parts / limited group of families.
Q.2 Explain composite part concept. + (AU : Dec. 16)
Ans. : The composite part is a hypothetical part that includes all of the design and
manufacturing attributes of the family. In general, an individual part in the family will have
some of the features of the family, but not all of them. A production cell for the part family
would consist of those machines required to make the composite part.
Q.3 State any four benefits of FMS. + (AU : Dec. 16)
Ans. : Increased machine utilization
Fewer machines required
Lower manufacturing lead times
Opportunity for unattended production
Q.4 What is part family ? + (AU : Dec. 17)
Ans. : Part family is a collection of parts in which similar parts are grouped based on the
manufacturing and design considerations. The parts within the family may differ but they are
close enough to include in the particular part family.
5 - 49 Computer Aided Design and Manufacturing
Cellular Manufacturing and Flexible Manufacturing System (FMS)
BS = WL (R* ) = WL
i i p i
S*
WL*
BS = (5.0)(0.05128) = 0.256 servers
1
BS = (19.5)(0.05128) = 1.0 servers
2
BS = (18.4)(0.05128) = 0.944 servers
3
BS = (7.5)(0.05128) = 0.385 servers
4
U =
i
U = (5.0/1)(0.05128) = 0.256 =
1 25.6 %
WLi WLi
WL*
Si Si
S*
(R* ) =
p
U = (19.5/1)(0.05128) = 1.0 =
2 100 %
U = (18.4/1)(0.05128) = 0.944 =
3 94.4 %
U = (7.5/1)(0.05128) = 0.385 =
4 38.5 %

384. Q.5 How the machine cells are classified. + (AU : Dec. 17)
Ans. : · Single machine cell
· Group machine cell with manual handling
· Group machine cell with semi-integrated handing
· Flexible manufacturing cell or flexible manufacturing system.
Q.6 What are the components of FMS ? + (AU : Dec. 17)
Ans. : Work stations
· Material handling and storage system
· Computer control system
· Man power
Q.7 How the part families are identified ? + (AU : May 16)
List out the methods of part family formation. + (AU : Dec. 18)
Ans. : Visual inspection method
· Parts classification and coding
· Production flow analysis
Q.8 What are the problems in implementing group technology ? + (AU : May 16)
Ans. : · Creating part family.
· Rearranging the machines into machine cell.
· Resistance to change (from conventional to new).
· Changing roles of operators new responsibilities to supervisor.
· Extensive education and Hands on training required.
Q.9 List out the four test for flexibility in FMS research ? + (AU : May 16)
Ans. : Part variety test
· Schedule change test
· Error recovery test
· New part test
Q.10 What are the production conditions under which group technology and cellular
manufacturing are most applicable ? + (AU : May 18)
Ans. : · The plant currently uses traditional batch production and a process type layout, and
this results in much material handling effort, high in-process inventory and long manufacturing
lead times; and
· The parts can be grouped into part families.
Q.11 What is the application of rank order clustering ? + (AU : May 18)
Ans. : · Rank order clustering technique is used for performing cluster analysis to form a
machine cell.
· Using this technique machines are grouped into machine cell.
5 - 50 Computer Aided Design and Manufacturing
Cellular Manufacturing and Flexible Manufacturing System (FMS)

385. Q.12 What are the three capabilities that a manufacturing system must possess in
order to be flexible ? + (AU : May 18)
Ans. : · Can machine can run different part configurations ?
· It allows changes in the production schedule.
· It capable of continuing to operate event though one machine is break down.
Q.13 List the applications of FMS. + (AU : May 18)
Ans. : · Flexible manufacturing systems are typically used for mid-volume, mid-variety
production. If the part or product is made in high quantities with no style variations, then a
transfer line or similar dedicated production system is most appropriate. If the parts are low
volume with high variety, then numerical control or even manual methods would be more
appropriate
Q.14 What are the applications of group technology ? + (AU : Dec. 18)
Ans. : · Informal scheduling and routing of similar parts through selected machines.
· Virtual machine cells
· Formal machine cells
· Process planning, family tooling and numerical control programs etc.,
Q.15 What are the objective of FMS ? + (AU : Dec. 18)
Ans. : · The objective of FMS is to automate the manufacturing system and produce the
component within a short span of time.
Q.16 List any two advantages and disadvantages of FMS implementation.
+ (AU : Dec. 18)
Ans. : Advantages
· Reduced lead time of components
· Improved quality
Disadvantages
· Investment cost
· Part family considerations
Q.17 Mention the factors to be considered in selection of coding system.
+ (AU : May 19)
Ans. : 1. Designing attributes
2. Manufacturing attributes
Q.18 Define the term key machine in the cellular manufacturing. + (AU : May 19)
Ans. : A machine that is more expensive to operate than the other machines or that performs
certain critical operations in the plant is referred as key machine. Other machines in the cell
referred as supporting machines and they should be organized in the cell to keep the key
machine busy.
5 - 51 Computer Aided Design and Manufacturing
Cellular Manufacturing and Flexible Manufacturing System (FMS)

386. Q.19 What is the difference between FMC and FMS systems. + (AU : May 19)
Ans. :
FMC FMS
It has two or three machines It has four or more
It does not have any supporting machines It has supporting machines which do not
directly participate on it.
Needed simple computer control system Needed larger and more sophisticated
Computer control system
Limited error recovery Minimized effect of machine breakdown
Simple than FMS More complex
Part B : University Questions with Answers
Dec. 2016
1. Apply rank order clustering technique to the part-machine incidence matrix in the
following table to identify logical part families and machine groups. N parts are
identified by letters and machines are identified numerically.
(Refer example 5.7.1) [16]
Machines
A B C D E
1 1
2 1 1
3 1 1
4 1 1
5 1
2. Define FMS and explain in detail about the FMS components.
(Refer sections 5.13 and 5.16) [8]
3. Explain OPTIZ parts classification and coding system. (Refer section 5.5) [8]
5 - 52 Computer Aided Design and Manufacturing
Cellular Manufacturing and Flexible Manufacturing System (FMS)

387. May 2017
4. Apply rank order clustering technique to the part-machine incidence matrix in
the following table to identify logical part families and machine groups. Parts
are identified by letters and machines are identified numerically.
(Refer example 5.7.1) [16]
Machines A B C D E F G
1 1 1
2 1 1
3 1 1 1 1
4 1 1 1
5 1 1 1 1
5. Explain OPTIZ parts classification and coding system. (Refer section 5.5) [16]
Dec. 2017
6. List out the methods of part family formation. (Refer section 5.2.1) [8]
7. List out the various machines used in FMS. (Refer section 5.16.1) [8]
8. Explain OPTIZ parts classification and coding system. (Refer section 5.5) [16]
May 2018
9. Sketch and explain the layout of typical FMS. (Refer section 5.14.1) [13]
10. Explain about machine cell design and layout. (Refer sections 5.17 and 5.18) [13]
Dec. 2018
11. Discuss the functions, application, advantage and disadvantage of FMS.
(Refer section 5.9) [13]
12. Explain about machine cell design and layout. (Refer sections 5.17 and 5.18) [7]
13. Summarize the following FMS layouts with the neat sketches.
i) Open field layout
ii) Ladder layout
iii) Robot centered layout (Refer section 5.16.2.1) [6]
5 - 53 Computer Aided Design and Manufacturing
Cellular Manufacturing and Flexible Manufacturing System (FMS)

388. 14. Flexible manufacturing cell has just been created. After considering a number of
designs, the system engineer chose a layout that consists of two machining
workstations plus a load/unload station. In detail, the layout consists of : The
load/unload station is station 1. Station 2 Performs milling operations and
consists of one server (one CNC milling machine) Station 3 has one server that
performs drilling (one CNC drill press). The three stations are connected by a
part handling system, that has one work carrier. The mean transport time in the
system is 4 min. The FMC produces three parts, A, B and C. The part mix
fraction and process routings for the three parts are presented in the table
below. The operation frequency fijk = 1.0 for all operations.
Determine (i) Maximum production rate of the FMC, (ii) Corresponding
production rates of each product, (iii) Utilization of each machine in the system
and (iv) Number of busy servers at each station. (Refer section 5.20) [16]
Part j Part Mix pj Operation k Description Station i Process time tijk
A 0.4 1 Load 1 3
2 Mill 2 20
3 Drill 3 12
4 Unload 1 2
Cellular Manufacturing and Flexible Manufacturing System (FMS) ends …
5 - 54 Computer Aided Design and Manufacturing
Cellular Manufacturing and Flexible Manufacturing System (FMS)

389. Time : Three Hours] [Maximum Marks : 100
Answer All Questions
PART A - (10 ´ 2 = 20 Marks)
Q.1 What is concurrent engineering. (Refer Two Marks Q.2 of chapter 1)
Q.2 What is meant by concatenation transformation ?
(Refer Two Marks Q.4 of chapter 1)
Q.3 How the curves are classified ? (Refer Two Marks Q.2 of chapter 2)
Q.4 Define zero order continuity. (Refer Two Marks Q.13 of chapter 2)
Q.5 What is the objective of GKS- 3D standard ?
(Refer Two Marks Q.4 of chapter 3)
Q.6 State the needs for data exchange standards.
(Refer Two Marks Q.8 of chapter 3)
Q.7 Explain absolute and incremental system.
(Refer Two Marks Q.4 of chapter 4)
Q.8 What are G codes and M codes ? (Refer Two Marks Q.11 of chapter 4)
Q.9 What is cellular manufacturing ? (Refer Two Marks Q.1 of chapter 5)
Q.10 How the part families are identified ? (Refer Two Marks Q.7 of chapter 5)
PART B - (5 ´ 13 = 65 Marks)
Q.11 a) i) Describe various stages of design process with an example.
(Refer section 1.3) [8]
ii) Explain a line drawing algorithm. (Refer section 1.11) [5]
OR
b) Write short notes on parametric representation of synthetic surface.
(Refer section 2.12) [13]
Q.12 a) i) Write short notes on OpenGL. (Refer section 3.3.1) [8]
ii) Explain the concept of product data exchange using STEP.
(Refer section 3.4.2) [5]
M -omputer Aided Design and Maufacturing
(M - 1)
SOLVED MODEL QUESTION PAPER
[As per New Syllabus]
Computer Aided Design and Manufacturing
Semester - VI (Mechanical Engineering)

390. OR
b) Write a manual part program to finish the stepped shaft in f 40 mm section as
shown in Fig. 1. Assume spindle speed as 350 rpm and feed rate 0.4 mm/rev.
(Refer example 4.23.1) [13]
Q.13 a) i) Explain about machine cell design and layout.
(Refer sections 5.17 and 5.18) [8]
ii) Write short notes on OpenGL. (Refer section 3.3.1) [5]
OR
b) Flexible manufacturing cell has just been created. After considering a number
of designs, the system engineer chose a layout that consists of two machining
workstations plus a load/unload station. In detail, the layout consists of : The
load/unload station is station 1. Station 2 Performs milling operations and
consists of one server (one CNC milling machine) Station 3 has one server that
performs drilling (one CNC drill press). The three stations are connected by a
part handling system, that has one work carrier. The mean transport time in
the system is 4 min. The FMC produces three parts, A, B and C. The part mix
fraction and process routings for the three parts are presented in the table
below. The operation frequency fijk = 1.0 for all operations.
Determine (i) Maximum production rate of the FMC, (ii) Corresponding
production rates of each product, (iii) Utilization of each machine in the system
and (iv) Number of busy servers at each station. (Refer section 5.20) [13]
Part j Part Mix pj Operation k Description Station i Process time tijk
A 0.4 1 Load 1 3
2 Mill 2 20
3 Drill 3 12
4 Unload 1 2
M - 2 Computer Aided Design and Manufacturing
Solved Model Question Paper

50

0
40
Fig. 1

391. Q.14 a) i) A production machine operates 80 hrs/week (2 shifts, 5 days) at full capacity.
During a certain week the machine produced 1000 parts and was idle in the
remaining part.
a) Determine the production capacity of the machine.
b) What was the utilization of the machine during the week under
consideration.
c) Compute the expected plant capacity if the availability of the machine is,
A = 90 % and with the effect of computed utilization ‘U’.
(Refer example 1.17.3) [5]
ii) Write short notes on communication standards. (Refer section 3.5) [8]
OR
b) Discuss the functions, application, advantage and disadvantage of FMS.
(Refer section 5.9) [13]
Q.15 a) i) What are bezier curves ? Discuss its important properties.
(Refer section 2.8) [5]
ii) Write short notes on data exchange standards. (Refer section 3.4.5) [8]
OR
b) Write a part program for the component as shown Fig. 2. Assume that spindle
speed 500 rpm and feed is 0.3 mm/rev. (Refer example 4.23.4) [13]
PART C - (1 ´ 15 = 15 Marks)
Q.16 a) Explain Graphics Kernal System (GKS). (Refer section 3.2.1) [15]
OR
b) Explain OPTIZ parts classification and coding system. (Refer section 5.5)[15]
Solved Model Question Paper ends ...
M - 3 Computer Aided Design and Manufacturing
Solved Model Question Paper
60
20
40
20 30 20
R10
Fig. 2