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Unit - I
Chapter - 1
Introduction
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
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Section No.
1.1
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Topic Name
Page No.
Introduction to CAD
1 - 2
Product Cycle
1 - 2
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
1.7
Computer Graphics
1.8
Coordinate System
1.9
2D Transformations
1.10
3D Transformations
1.11
Line Drawing
1.12
Clipping
1.13
Viewing Transformation
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
1.2
1.3
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Introduction
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.
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·
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.
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1.2 Product Cycle
·
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.
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Introduction
The 4 life cycle stages and their marketing implications
Introduction
Growth
Decline
Maturity
Sales
Shake-out
Saturation
Take-off
· Low sales
· High cost per
customer
· Financial losses
· Innovative customers
· Few (if any) competitors
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· Increasing sales
· Cost per customer falls
· Profits rise
· Increasing No.
of customers
· More competitors
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Time
· Falling sales
· Cost per customer low
· Profits fall
· Customer base contracts
· Number of competitors fall
· Peak sales
· Cost per customer
lowest
· Profits high
· Mass market
· Stable number
of competitors
Fig. 1.2.1 Various stages in product life cycle
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Product
Lifecycle
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Fig. 1.2.2 Various steps in product life cycle
·
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.
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Introduction
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.
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ii) Manufacturing process
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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.
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Introduction
·
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.
The design process
Synthesis
Collecting relevant
design information
and feasibility study
Design definitions,
specifications and
requirements
Design
need
Analysis
Design
communication
and documentation
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Process
planning
Design
analysis
Design
optimization
Design
evaluation
The CAD process
Design
modeling and
simulation
Design
conceptualization
The manufacturing process
The CAM process
Production
planning
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Design and
procurement
of new tools
Production
Order
material
NC, CNC,
DNC
programming
Quality
control
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Packaging
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Marketing
Fig. 1.2.3 Product cycle
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1.3 Design Process
Engineering design process :
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·
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.
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Introduction
Define problem
Problem statement
Benchmarking
QFD
PDS
Project planning
Production architecture
Arrangement of physical
elements to carry out
functions
Configuration design
Preliminary selection of
material and
manfacturing
modeling/sizing of parts
Gather information
Internet
Patents
Trade
Literature
Conceptual
design
Parametric design
Robust design
Tolerances
Final dimensions
DFM
Concept generation
Brainstorming
Functional decomposition
Morphological chart
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Embodiment
design
Evaluation of
concepts
Pugh concept selection
Decision matrix
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Detail design
Detail drawing and
specifications
Fig. 1.3.1 Engineering design process
Conceptual Design
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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.
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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.
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Introduction
·
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.
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Detail Design
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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
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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.
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Introduction
Recognition of need
Definition of problem
Change the design
Synthesis
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Yes
Can the design be improved
Analysis and optimization
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Evaluation
Design impossible for the
given specification
Success
Presentation
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Fig. 1.3.2 Shigley's design process
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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
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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.
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Introduction
·
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.
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Assembly
Fig. 1.4.1 (a) Over the wall engineering
Information flow
Requirements definition
Product definition
Process definition
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Delivery and support
Errors changes and corrections
Fig. 1.4.1 (b) Sequential engineering
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Introduction
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.
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·
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.
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Life Cycle Aspects (Electrical, Mechanical
Survicing, Assembiability, Recyclability, etc.)
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Feedback loops between
different life - cycle phases
Requirements Analysis
Preliminary Design
Detailed Design
Manufacturing
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Assembly
Testing
Life - cycle phases
Installation
Time
Fig. 1.4.2 Concurrent engineering
·
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.
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Introduction
·
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.
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·
The team often consists of product design engineers, manufacturing engineers,
marketing personnel, purchasing, finance, and suppliers.
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Comparison between Concurrent and Sequential Engineering
·
Fig. 1.4.3 depicts the schematic representation of the comparison between
sequential and concurrent engineering.
Requirements definition
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Information flow
Product definition
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Process definition
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Errors, changes and corrections
CE life cycle time
(a) Sequential engineering
Delivery and support
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Requirements definition
Product definition
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Process definition
Delivery and support
CE life - cycle time
Requirements
definition
Product
definition
Time saved
Delivery and
support
Process
definition
(b) Concurrent engineering
Fig. 1.4.3 Comparison between sequential and concurrent engineering
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Introduction
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.
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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.
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1.5 Computer Aided Design (CAD)
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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)
i)
Geometric modeling
ii)
Engineering analysis
iii)
Design review and evaluation
iv)
Automated drafting.
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
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Introduction
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.
Conventional Design Process
Computer - aided design
Recognition of need
Problem definition
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Synthesis
Geometric modeling
Analysis and
optimization
Engineering analysis
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Evaluation
Presentation
Design review and evaluation
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Automated drafting
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Fig. 1.5.1 Computer aided design process
·
Geometric modeling is classified into
a) Wireframe modeling
b) Solid modeling
c) Surface modeling
ii) Engineering Analysis
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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)
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Introduction
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.
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Another advantage of animating the complete assembly of product is that whether
any component fouls the other components in its working.
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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.
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Advantages of CAD
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·
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.
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Introduction
·
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.
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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.
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Introduction
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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.
System
Major classes :
Main frame
Mini computer
Workstation
Microcomputer
Based
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Application
software
Database
(CAD model)
Application areas :
Mechanical
Architectural
Construction
Circuit design
Chip design
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Graphics
utility
Device
drivers
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Cost :
High end
Low end
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Input - output
devices
Fig. 1.6.1 CAD system architecture
1.7 Computer Graphics
User
interface
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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.
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Introduction
·
Through these input devices it is possible to provide input to the computer and
display device is an output device which shows the image.
Control signals
Link to
host computer
Control
processor
Display file
Display
processor
unit
Display
screen
Input
devices
ww
Fig. 1.7.1 A basic computer graphics layout
w.E
·
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.
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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
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Introduction
·
·
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.
ww
a) Model Coordinate System (MCS) or Database CS/ World CS / Master CS
b) Working Coordinate System (WCS) or User Coordinate System (UCS)
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c) Screen Coordinate System (SCS) or Device CS
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a) Model Coordinate System (MCS)
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Z'
Z
Y'
Z
Y
Y
nee
Y1
X'
X
(Xmax'Ymax)
X
rin
(0,0) X1
(a) MCS
(b) WCS
Fig. 1.8.1
(c) SCS
g.n
et
·
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.
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Introduction
·
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.
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c) Screen Coordinate System (SCS)
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·
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).
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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.
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Introduction
·
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.
ww
View Port
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·
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.
Window
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Original
drawing
130,100
View port 2
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View port 4
65,50
(a) Window
View port 1
(b) View port
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View port 3
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Fig. 1.8.2
+ [AU : Dec.-17]
1.9 2D Transformations
·
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.
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Introduction
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)
ww
x' = [x+Dx]
…(1.9.2)
y' = [y+Dy]
…(1.9.3)
é x ¢ù é x + D
[P'] = ê ú = ê
ë y ¢û ë y + D
w.E
xù é xù é D xù
=
+
yúû êë yúû êë D yúû
…(1.9.4)
Y
asy
E
Y'
Y
P
P
X
Z
Z'
P'
X'
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Fig. 1.9.1 Translation of a point
2D Translation of an object
X
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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.
…(1.9.5)
P ¢ = [x', y' ]=[Sx ´ X, Sy ´ Y]
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Introduction
10
9
(5, 8)
(8, 8)
8
7
Y
After translation
6
(1, 5)
5
(4, 5)
4
ww
(5, 4)
(8, 4)
3
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Original rectangle
2
1
(1, 1)
0
0
1
(4, 1)
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2
3
4
5
6
8
7
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9
10
X
Fig. 1.9.2 2D Translation of an object
éS x
[P ¢] = ê
ë0
0 ù é xù
S y úû êë yúû
…(1.9.6)
éS x
[Ts] = ê
ë0
0ù
S y úû
…(1.9.7)
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Y
[P ¢] = [Ts] × [P]
…(1.9.8)
· Fig. 1.9.3 depicts the scaling of an
object.
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P'
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
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P
SY
Y
·
X
X
SX
Fig. 1.9.3 2D Scaling of an object
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Introduction
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.
Y
P'
Y'
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P
Y
r
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0
X
X'
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X
Fig. 1.9.4 2D rotation of an objects
·
·
·
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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
nee
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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
g.n
et
= 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,
é x ¢ù é cos q - sin q ù é xù
…(1.9.9)
[P ¢] = ê ú = ê
úê ú
ë y ¢û ë sin q cos q û ë yû
The rotation matrix is given as TR .
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Introduction
é cos q
[TR ] = ê
ë sin q
- sin q ù
cos q úû
…(1.9.10)
[P ¢] = [TR ]× [P]
(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.
ww
·
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.
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10
9
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8
D1
D
7
Y 6
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After X-shear
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5
E
4
C
E1
Original part
g.n
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C1
3
2
A1
1
A
B1
B
0
0
1
2
3
4
5
6
8
7
9
10
11
12
13
X
Fig. 1.9.5 (a) X-Shear
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Introduction
·
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.
10
C1
D1
After X-shear
9
8
7
ww
Y
E1
D
6
5
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4
E
3
A1
2
1
A
0
0
C
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1
2
3
4
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5
6
X
B
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7
Fig. 1.9.5 (b) Y-Shear
·
8
9
10
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A Y-shear transforms the point (x, y) to the point (x 1, y 1) by a factor Sh 1,
x¢ = x
·
Original part
B1
…(1.9.12)
…(1.9.13)
y ¢ = Sh 1 × x + y
An X-shear transforms the point (X, Y) to (x 1, y 1), where Sh 2 is the shear factor
x ¢ = x + Sh 2 × y
…(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).
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Introduction
Y
Y
–X
X
–Y Y
P
P'
X
P'
P
X
ww
(b) Reflection about Y-axis
(a) Reflection about X-Axis
·
Fig. 1.9.6
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For reflection about x-axis the y coordinate will be negative and the following
equations should be utilized,
…(1.9.16)
P ¢ = [X ¢, Y ¢] = [X, – Y]
asy
E
é1 0 ù
[P ¢] = ê
ú
ë0 -1û
é xù
ê yú
ë û
The translation matrix is given as,
é1 0 ù
[Tm ] = ê
ú
ë0 -1û
·
ngi
…(1.9.17)
nee
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…(1.9.18)
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et
…(1.9.19)
[P ¢] = [Tm ] × [P]
For reflection about y-axis the x coordinate will be negative and the following
equations should be utilized,
…(1.9.20)
P ¢ = [X ¢, Y ¢] = [X, – Y]
é - 1 0ù é xù
[P ¢] = ê
ú ê ú
ë 0 1û ë yû
…(1.9.21)
The translation matrix is given as,
é -1 0ù
[Tm ] = ê
ú
ë 0 1û
·
…(1.9.22)
[P ¢] = [Tm ] × [P]
Thus the general form of reflection matrix could be written as,
é± 1 0 ù
[Tm ] = ê
ú
ë 0 ± 1û
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…(1.9.23)
…(1.9.24)
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Introduction
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,
…(1.9.25)
[P ¢] = [Tn ][Tn - 1 ][Tn - 2 ]...[T3 ][T2 ][T1 ]
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.
0 0ù é x ù
é x ¢ù é 1
ê
ú
ê
…(1.9.26)
[P ¢] = y ¢ = 0
1 0ú ê y ú
ê ú ê
ú ê ú
êë 1 úû êë DX DY 1úû êë 1 úû
ww
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·
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,
0 0ù
é 1
ê
[T] = 0
1 0ú
ê
ú
êë DX DY 1úû
·
·
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The Scaling matrix in homogeneous form is,
éS x 0 0ù
[S] = ê 0 S y 0ú
ê
ú
0 1úû
êë 0
The Rotation matrix in homogeneous form is,
é cos q sin q 0ù
[TR ] = ê - sin q cos q 0ú
ê
ú
0
1úû
êë 0
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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.
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Introduction
ii) Rotate the object by the given angle 'q'.
iii) Translate the point back to its original position from origin.
Y
P'
P
r
r
Y
ww
A
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O
X
X
Fig. 1.9.7 Rotation of an object about an arbitrary point
·
asy
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The following homogeneous transformation matrices should be used for the
translation operation,
i) Translate the point from point 'P' to origin 'O'
ngi
0
0ù
é 1
ê
[T1 ] = [T] = 0
1
0ú
ê
ú
êë - DX - DY 1úû
ii) Rotate the object by the given angle 'q'.
é cos q
[T2 ] = [TR ] = ê - sin q
ê
êë 0
nee
sin q 0ù
cos q 0ú
ú
0
1úû
rin
g.n
et
iii) Translate the point back to its original position from origin.
0 0ù
é 1
ê
1 0ú
[T 3 ] = [T] = 0
ê
ú
êë DX DY 1úû
iv) Final Transformation matrix after concatenation,
[T] = [T1 ] ´ [T2 ] ´ [T 3 ]
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Introduction
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 1 , y 1 )
Dx = 4;
Dy = 5
Tranformation matrix
é Dx ù
T = ê ú
ë Dy û
ww
New position of a point is,
w.E
P¢ = P + T
é x1¢ ù
é x1 ù é Dx ù é2ù é 4ù é6ù
ê y ¢ ú = ê y ú + ê Dy ú = ê3ú + ê5ú = ê8ú
ë 1û
ë 1û ë û ë û ë û ë û
Y
asy
E
ngi
y=5
P
(2, 3)
x=4
nee
P'(6, 8)
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X
Fig. 1.9.8
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(x1 , y 1 ) = (2, 4)
B(x 2 , y 2 ) = (5, 6)
T(Dx, Dy ) = (1, 3)
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Introduction
\
A¢ = A + T
é2ù é1ù é 3ù
A¢ = ê ú + ê ú = ê ú
ë 4û ë3û ë 7û
é5ù é1ù é6ù
B¢ = B + T = ê ú + ê ú = ê ú
ë6û ë3û ë9û
Similarly
Y
B' (6, 9)
A' (3, 7)
ww
w.E
y=3
B (5, 6)
asy
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A
(2, 4)
ngi
x=1
Fig. 1.9.9
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X
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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 = (x1 , y 1 ) = (1, 1)
B = (x 2 , y 2 ) = (3, 1)
C = (x 3 , y 3 ) = (1, 3)
Dx = 3;
Dy = 2 Þ T = (3, 2)
é1ù é3ù é 4ù
A¢ = A + T = ê ú + ê ú = ê ú
ë1û ë2û ë3û
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Introduction
é3ù é3ù é6ù
B¢ = B + T = ê ú + ê ú = ê ú
ë1û ë2û ë3û
é1ù é3ù é 4ù
C¢ = C + T = ê ú + ê ú = ê ú
ë3û ë2û ë5û
Y
C' (4, 5)
ww
C (1, 3)
w.E
B' (6, 3)
A' (4, 3)
y=2
B (3, 1)
asy
E
A (1, 1)
X
x=3
ngi
Fig. 1.9.10
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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) = (x1 , y 1 )
B (4, 2) = (x 2 , y 2 )
C (4, 5) = (x 3 , y 3 )
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D (2, 5) = (x 4 , y 4 )
Dx = 4;
Dy = 4 Þ T( Dx , Dy ) = (4, 4)
é2ù é 4ù é6ù
[A] = [A] + [T] = ê ú + ê ú = ê ú
ë2û ë 4û ë6û
é 4ù é 4ù é8ù
[B ¢] = [B] + [T] = ê ú + ê ú = ê ú
ë2û ë 4û ë6û
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Introduction
é 4ù é 4ù é8ù
[C ¢] = [C] + [T] = ê ú + ê ú = ê ú
ë5û ë 4û ë9û
é2ù é 4ù é6ù
[D ¢] = [D] + [T] = ê ú + ê ú = ê ú
ë5û ë 4û ë9û
Y
(6, 9) D'
ww
A' (6, 6)
w.E
(2, 5) D
C' (8, 9)
B' (8, 6)
C (4, 5)
y=4
asy
E
(2, 2) A
B (4, 2)
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x=4
Fig. 1.9.11
X
nee
rin
g.n
et
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(x1 , y 1 ) = (2, 3)
B(x 2 , y 2 ) = (6, 3)
C(x 3 , y 3 ) = (6, 6)
D(x 4 , y 4 ) = (2, 6)
Scaling factor ®
\
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S = (S x ,S y ) = (2, 2)
éS x 0 ù é2 0ù
Scaling matrix, [S] = ê
ú =ê
ú
ë 0 S y û ë0 2û
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Introduction
[A]¢ = [S] ´ [A]
é2 0ù é2ù é 4ù
= ê
ú´ê ú = ê ú
ë0 2û ë3û ë6û
[B]¢ = [S] ´ [B] Þ
é2 0ù é6ù é12ù
= ê
ú´ê ú = ê ú
ë0 2û ë3û ë 6 û
[C]¢ = [S] ´ [C]
é2 0ù é6ù é12ù
= ê
ú´ê ú = ê ú
ë0 2û ë6û ë12û
ww
w.E
Y
[D]¢ = [S] ´ [D]
é2 0ù é2ù é 4 ù
= ê
ú´ê ú = ê ú
ë0 2û ë6û ë12û
asy
E
D' (4, 12)
D (2, 6)
ngi
C (6, 6)
A' (4, 6)
A (2, 3)
C' (12, 12)
nee
rin
g.n
et
B' (12, 6)
B (6, 3)
X
Fig. 1.9.12
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Introduction
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(x1 , y 1 ) = (6, 8) ;
q = 30°
é x1¢ ù écos q
P¢ = ê ú = ê
ë y 1¢ û ë sin q
- sin q ù
cos q úû
é x1 ù écos 30 - sin 30ù
ê y ú = ê sin 30 cos 30 ú
ë 1û ë
û
é 6ù
ê 8ú
ë û
é1.196ù
[P ¢] = ê
ú
ë 9.928û
Þ
ww
Y
w.E
asy
E
P' (1.196, 9.28)
P(6, 8)
ngi
°=
30
Fig. 1.9.13
nee
X
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et
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)
B(3, 5) C(7, 5)
,
(x1 , y 1 ) (x 2 , y 2 ) (x 3 , y 3 )
i) Rotated by 45 ° in clockwise direction :
,
q = – 45°
\
é x1¢ ù écos( - 45° ) - sin ( - 45° )ù
[A]¢ = ê ú = ê
ú
ë y 1¢ û ë sin ( - 45° ) cos( - 45° ) û
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é5ù
ê2ú
ëû
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Introduction
Þ
é 4.97 ù
[A]¢ = ê
ú
ë - 2.12û
Similarly,
é x ¢2 ù écos( - 45° ) - sin ( - 45° )ù
[B]¢ = ê ú = ê
ú
ë y ¢2 û ë sin ( - 45° ) cos( - 45° ) û
Þ
é 5.65ù
[B]¢ = ê
ú
ë1.41û
Similarly,
é x ¢3 ù écos( - 45° ) - sin ( - 45° )ù
[C]¢ = ê ú = ê
ú
ë y ¢3 û ë sin ( - 45° ) cos( - 45° ) û
ww
Þ
é3ù
ê5ú
ëû
é 7ù
ê 5ú
ë û
é 8.48 ù
[C]¢ = ê
ú
ë -1.414û
w.E
ii) Rotated by 60° in anticlockwise direction (counter-clockwise) :
asy
E
q = 60°
\
é x1¢¢ ù écos 60 - sin 60ù
[A]¢¢ = ê ú = ê
ú
ë y 1¢¢û ë sin 60 cos 60 û
é5ù
ê2ú
ëû
Þ
é0.767 ù
[A]¢¢ = ê
ú
ë 5.330 û
Similarly,
é x 2¢¢ ù écos 60 - sin 60ù
[B]¢¢ = ê ú = ê
ú
ë y 2¢¢ û ë sin 60 cos 60 û
nee
Þ
é - 2.830 ù
[B]¢¢ = ê
ú
ë 5.098û
Similarly,
é x ¢¢3 ù écos 60 - sin 60ù
[C]¢¢ = ê ú = ê
ú
ë y ¢¢3 û ë sin 60 cos 60 û
Þ
é - 0.83 ù
[C]¢¢ = ê
ú
ë 8.562û
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ngi
é3ù
ê5ú
ëû
é 7ù
ê 5ú
ë û
rin
g.n
et
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Introduction
Y
(–0.830, 8.582) C"
B"
(–2.830, 5.098)
A" (0.767, 5.330)
B(3, 5)
A(5, 2)
B' (5.65, 1.41)
60°
45°
X
C(7, 5)
X
0
ww
C' (8.48, –1.41)
w.E
A' (4.97, –2.12)
asy
E
Y
ngi
Fig. 1.9.14
nee
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.
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et
Y
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).
C (x2, y2)
(x3, y3) D
Initially evaluate the coordinates of
the square.
Þ A(x1 , y 1 ) = (0, 0). Since place
at the origin.
Þ B(x 2 , y 2 )
B (x1, y1)
X
60°
0
A (0, 0)
30°
X
Fig. 1.9.15
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Introduction
B(x 2 , y 2 ) :
Y
Here edge length = 10 units
\
x 2 = 10 cos 30 = 8.66 units
B (x2, y2)
y2
y 2 = 10 sin 30 = 5 units
10
30°
B(x 2 , y 2 ) = (8.66, 5)
\
l=
X
x2
A (0, 0)
Fig. 1.9.16
Similarly, C(x 3 , y 3 ) :
ww
Y
w.E
(x3, y3)
asy
E
l = 10
10 sin 60
ngi
y2
60°
30°
30°
B (x2, y2)
nee
rin
10 cos 60
x2
A (0, 0)
X
g.n
et
Fig. 1.9.17
\
x 3 = x 2 - 10 cos 60
= 8.66 – 5 = 3.66 units
Y
y 3 = y 2 + 10 sin 60
= 5 + 8.66 = 13.66 units
D(x4, y4)
C(x 3 , y 3 ) = (3.66, 13.66)
\
l=1
\
10 sin 60
0
[D](x 4 , y 4 ) :
x 4 = – 10 cos 60
X
60°
– 10 cos 60
= – 5 units
Fig. 1.9.18
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Introduction
y 4 = 10 sin 60 = 8.66 units
\
D(x 4 , y 4 ) = (– 5, 8.66)
i) Rotate the square by 30° clockwise :
q = – 30°
\
\
ww
Þ
écos( -30) - sin ( - 30)ù
[A]¢ = ê
ú
ësin ( - 30) cos( - 30) û
é0ù é0ù
ê0ú = ê0ú
ëû ëû
écos( -30) - sin ( - 30)ù
[B]¢ = ê
ú
ësin ( - 30) cos( - 30) û
é 8.66ù é 9.99 ~ 10 ù
ê 5 ú = ê 0.0001 ~ 0ú
ë
û ë
û
é10 ù
[B]¢ = ê ú
ë0 û
w.E
écos( -30) - sin ( - 30)ù
[C]¢ = ê
ú
ësin ( - 30) cos( - 30) û
é 3.66 ù é10ù
ê13.66ú = ê10ú
ë
û ë û
écos( -30) - sin ( - 30)ù
[D]¢ = ê
ú
ësin ( - 30) cos( - 30) û
é -5 ù é 0ù
ê 8.66ú = ê10ú
ë
û ë û
asy
E
ngi
nee
iii) Square is rotated by 60° in counter clockwise direction :
\
q = 60°
écos 60 - sin 60ù
[A]¢¢ = ê
ú
ë sin 60 cos 60 û
é0ù é0ù
ê0ú = ê0ú
ëû ëû
écos 60 - sin 60ù
[B]¢¢ = ê
ú
ë sin 60 cos 60 û
é 8.66ù é 0 ù
ê 5 ú = ê10ú
ë
û ë û
écos 60 - sin 60ù
[C]¢¢ = ê
ú
ë sin 60 cos 60 û
é 3.66 ù é - 10ù
ê13.66ú = ê 10 ú
ë
û ë
û
écos 60 - sin 60ù
[D]¢¢ = ê
ú
ë sin 60 cos 60 û
é - 5 ù é - 10ù
ê 8.66ú = ê 0 ú
ë
û ë
û
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Introduction
(3.66, 13.66)
C
(–10, 10) C"
C' (10, 10)
D' B''
(0, 10)
B
ww
D''
(–10, 0)
w.E
60°
30°
A (0, 0)
A', A"
B' (10, 0)
Fig. 1.9.19
asy
E
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
é8ù
i) Translated by a translation matrix [T] = ê ú
ë5û
ngi
ii) Rotated by 60° in counter clockwise direction.
nee
rin
D (4, 5)
A (3, 5)
g.n
et
P
(4, 3)
B (2, 2)
C (8, 2)
Fig. 1.9.20
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Introduction
Sol. :
i) Translation :
é Dx ù é8ù
[T] = ê ú = ê ú
ë Dy û ë5û
\
Dx = 8;
Dy = 5
é x1 ù é Dx ù é3ù é8ù é11ù
[A]¢ = ê ú + ê ú = ê ú + ê ú = ê ú
ë y 1 û ë Dy û ë5û ë5û ë10 û
é x 2 ù é Dx ù é2ù é8ù é10 ù
[B]¢ = ê ú + ê ú = ê ú + ê ú = ê ú
ë y 2 û ë Dy û ë2û ë5û ë 7 û
ww
é x 3 ù é Dx ù é8ù é8ù é16ù
[C]¢ = ê ú + ê ú = ê ú + ê ú = ê ú
ë y 3 û ë Dy û ë2û ë5û ë 7 û
w.E
é x 4 ù é Dx ù é 4ù é8ù é12ù
[D]¢ = ê ú + ê ú = ê ú + ê ú = ê ú
ë y 4 û ë Dy û ë5û ë5û ë10 û
asy
E
é 4ù é Dx ù é 4ù é8ù é12ù
[P]¢ = ê ú + ê ú = ê ú + ê ú = ê ú
ë 3û ë Dy û ë3û ë5û ë 8 û
ngi
ii) Rotate through 60° in counter clockwise direction :
q = 60°
nee
écos 60 - sin 60ù
[A]¢¢ = ê
ú
ë sin 60 cos 60 û
é3ù é - 2.83ù
ê5ú = ê 5.09 ú
û
ëû ë
écos 60 - sin 60ù
[B]¢¢ = ê
ú
ë sin 60 cos 60 û
é 2ù é - 0.732ù
ê 2ú = ê 2.732 ú
ë û ë
û
écos 60 - sin 60ù
[C]¢¢ = ê
ú
ë sin 60 cos 60 û
é 8ù é 2.26ù
ê 2ú = ê7.93ú
ë û ë
û
écos 60 - sin 60ù
[D]¢¢ = ê
ú
ë sin 60 cos 60 û
é 4ù é - 2.33ù
ê 5ú = ê 5.96 ú
ë û ë
û
écos 60 - sin 60ù
[P]¢¢ = ê
ú
ë sin 60 cos 60 û
é 4ù é - 0.59ù
ê 3ú = ê 4.96 ú
ë û ë
û
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rin
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et
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12
9
B'
6
5
nee
rin
4
3
g.n
et
–1
–3
X
A''
D''
–2
B"
P"
Y
Y
0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
1
B
2
C''
A
D
ngi
7
8
C
asy
E
P
w.E
10
A'
ww
11
D'
P'
13
14
15
16
C'
X
Introduction
Fig. 1.9.21
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Introduction
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 :
Þ
é1 0 ù
[A]¢ = ê
ú
ë0 -1û
é x1 ù
êy ú
ë 1û
é1 0 ù
[A]¢ = ê
ú
ë0 -1û
é 3ù é 3 ù
ê 4ú = ê - 4ú
ë û ë û
ww
é1 0 ù é7 ù é 7 ù
[B]¢ = ê
ú ê ú =ê ú
ë0 -1û ë 4û ë - 4û
w.E
é1 0 ù é7 ù é 7 ù
[C]¢ = ê
ú ê ú =ê ú
ë0 -1û ë 6û ë - 6û
asy
E
é1 0 ù
[D]¢ = ê
ú
ë0 -1û
é 3ù é 3 ù
ê 6ú = ê - 6ú
ë û ë û
ngi
Y
nee
(3, 6)
D
A
(3, 4)
(7, 6)
C
rin
B
(7, 4)
g.n
et
X
X
(3, – 4)
A'
D'
(3, – 6)
B'(7, – 4)
C'
(7, – 6)
Y
Fig. 1.9.22
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Introduction
ii) Reflection about Y-axis :
é -1 0ù é x1 ù
[A]¢ = ê
ú ê ú
ë 0 1û ë y 1 û
é -1 0ù é 3ù é - 3ù
[A]¢ = ê
ú ê ú=ê ú
ë 0 1û ë 4û ë 4û
Þ
é -1 0ù é7 ù é - 7 ù
[B]¢ = ê
ú ê ú =ê ú
ë 0 1û ë 4û ë 4 û
é -1 0ù é7 ù é - 7 ù
[C]¢ = ê
ú ê ú =ê ú
ë 0 1û ë 6û ë 6 û
ww
é -1 0ù é 3ù é - 3ù
[D]¢ = ê
ú ê ú=ê ú
ë 0 1û ë 6û ë 6 û
w.E
(–7, 6)
(–7, 4)
C''
B''
asy
E
Y
D(3, 6)
D''(–3, 6)
ngi
A''(–3, 4)
X
nee
C(7, 6)
A(3, 4)
B(7, 4)
rin
g.n
et
X
Y
Fig. 1.9.23
iii) About origin :
é -1 0 ù
[A]¢¢¢ = ê
ú
ë 0 -1û
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é x1 ù
êy ú
ë 1û
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Introduction
Þ
é -1 0 ù
[A]¢¢¢ = ê
ú
ë 0 -1û
é 3ù é - 3 ù
ê 4ú = ê - 4ú
ë û ë
û
é -1 0 ù
[B]¢¢¢ = ê
ú
ë 0 -1û
é7 ù é - 7 ù
ê 4ú = ê - 4ú
ë û ë û
é -1 0 ù
[C]¢¢¢ = ê
ú
ë 0 -1û
é7 ù é - 7 ù
ê 6ú = ê - 6ú
ë û ë û
é -1 0 ù
[D]¢¢¢ = ê
ú
ë 0 -1û
é 3ù é - 3ù
ê 6ú = ê - 6ú
ë û ë û
Y
ww
w.E
asy
E
(3, 6) D
C (7, 6)
(3, 4) A
B (7, 4)
ngi
X
B''' (–7,–4)
A''' (–3,–4)
C''' (–7,–6)
D''' (–3,–6)
X
O
Y
Fig. 1.9.24
nee
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et
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).
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Introduction
Solution : Given :
Rectangle ABCD ®
A(2, 3) Þ (x1 , y 1 )
B(6, 3) Þ (x 2 , y 2 )
C(6, 6) Þ (x 3 , y 3 )
D(2, 6) Þ (x 4 , y 4 )
i) Translation matrix in homogeneous form :
Dx = 2
Given,
Dy = 3
ww
Homogeneous Translation Matrix
é1 0 0ù
= ê0 1 0ú
ê
ú
êë2 3 0úû
w.E
[T](2,3)
asy
E
ii) Scaling matrix in homoeneous form,
'[ S ] (4 ,2)'
Given, S x = 4
Sy = 2
\
[S](4 ,2)
é 4 0 0ù
= ê0 2 0ú
ê
ú
êë0 0 1úû
ngi
nee
iii) Rotation matrix in homoeneous form,
[R] Þ at point (3, 3)
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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 [T1 ].
Step 2 : Rotate rectangle ABCD at origin [T11 ].
Step 3 : Translate rectangle ABCD from origin [T111 ] to point [3, 3].
Step 4 : Final rotation matrix [R] = [T1 ] ´ [T11 ] ´ [T111 ]
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Introduction
Step 1 : Translation matrix for translating ABCD to origin form point (3, 3) [TI ]
Here, DxI = – 3, Dy I = – 3
é 1
[TI ] = ê 0
ê
êë DxI
\
0
1
Dy I
0ù
0ú
ú
1úû
é 1 0 0ù
ê 0 1 0ú
ê
ú
êë –3 –3 1úû
Step 2 : Rotate rectangle ABCD at origin [TII ].
q = 30° (counter - clockwise)
Here,
\
ww
Þ
é cos q sin q 0ù
é cos 30 sin 30 0ù
ê
ú
[TII ] = – sin q cos q 0 = ê – sin 30 cos30 0ú
ê
ú
ê
ú
1úû
0
0
1úû
êë 0
êë 0
w.E
.
0.5 0ù
é0866
ê
[T] = – 0.5 0866
.
0ú
ê
ú
0
1úû
êë 0
asy
E
Step 3 : Translate ABCD from origin to (3, 3) [T] III
Here DxIII = 3 ; Dy III = 3
[T]III
é 1
= ê 0
ê
êë DxIII
0
1
Dy III
ngi
0ù
0ú
ú
1úû
nee
é1 0 0 ù
ê0 1 0 ú
ê
ú
êë3 3 1úû
\ Rotation matrix at (3, 3) in homogeneous form Þ [R]
[R] = [T]I ´ [T ]II ´ [T]III
rin
0.5 0ù é1 0 0ù
.
é 1 0 0ù é0866
ê
ú
ê
= 0 1 0 ´ – 0.5 0866
0ú ´ ê0 1 0ú
.
ê
ú ê
ú ê
ú
0
1úû êë3 3 1úû
êë –3 –3 1úû êë 0
Þ
g.n
et
0.5
0ù
é 0.866
ê
[R] = –0.5
0.866 0 ú
ê
ú
êë1 . 902 – 1.908 1úû
So we obtained all the three matrixes for evaluating the combined matrix [m ]
[T](2,3)
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é1 0 0ù
= ê0 1 0ú
ê
ú
êë2 3 0úû
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Introduction
[S](4,2)
[R]( q= 30°)
é 4 0 0ù
= ê0 2 0ú
ê
ú
êë0 0 1úû
0.5
0ù
.
é0866
ê
= – 0.5 0866
0ú
.
ê
ú
1úû
.
–1098
.
êë1902
\ Combined transformation matrix [Tc ]
[Tc ] = [T](2,3) ´ [S](4,2) ´ [R]( q= 30°)
ww
Þ
=
w.E
0.5
0ù
.
é1 0 0ù é 4 0 0ù é0866
ê0 1 0ú ´ ê0 2 0ú ´ ê –0.5 0866
.
0ú
ê
ú ê
ú ê
ú
.
–1098
.
1úû
êë2 3 0úû êë0 0 1úû êë1902
2
0ù
é3.464
ê
–1
1.732 0 ú
[Tc ] =
ê
ú
êë 5.93 8.098 1úû
asy
E
ngi
To find the resultant co-ordinates after combined transformations operations.
Given,
A B C D Þ A (x1 , y 1 ) = (2, 3)
B (x 2 , y 2 ) = (6, 3)
C (x 3 , y 3 ) = (6, 6)
D (x 4 , y 4 ) = (2, 6)
nee
Coordinates an homogeneous form,
éA ù
é x1
ê Bú
êx
ê ú = ê 2
êCú
êx 3
ê Dú
êx
ë û
ë 4
y1
y2
y3
y4
1ù é2 3
1ú ê6 3
ú = ê
1ú ê6 6
1úû êë2 6
rin
1ù
1ú
ú
1ú
1úû
g.n
et
\ Resultant co-ordinates (After transformation)
éA ù
é A ¢ù
ê Bú
ê B¢ ú
ê ú = ê ú ´ [Tc ] =
êCú
êC ¢ú
ê Dú
ê D ¢ú
ë û
ë û
é2
ê6
ê
ê6
ê2
ë
3
3
6
6
1ù
2
0ù
é3.464
1ú ê
ú ´ –1 1732
0ú
.
ú
ê
1ú
593
8098
1
.
.
úû
ê
ë
1úû
4×3×3×3
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Introduction
é A ¢ù
é 9.858 17.294 1ù
ê B¢ ú
ê 23.714 25.294 1ú
ê ú = ê
ú
êC ¢ú
ê 20.714 30.49 1ú
ê D ¢ú
ê 6.888 22.49 1ú
ë û
ë
û
A ¢ = (x1¢ , y 1¢ ) = (9.858, 17.294)
B ¢ = (x ¢2 , y ¢2 ) = (23.714, 25.294)
C ¢ = C (x ¢3 , y ¢3 ) = (20.714, 30.49)
D ¢ = (x ¢4 , y ¢4 ) = (6.888, 22.49)
(20.714, 30.49)
C'
ww
Y
w.E
B'
(23.714, 25.94)
D'
(6.88, 22.49)
asy
E
ngi
nee
A'
(9.858, 17.294)
D(2,6)
C(6,6)
A(2,3)
B(6,3)
rin
g.n
et
X
Fig. 1.9.25
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 (x1 , y 1 ) = (2, 3)
B (x 2 , y 2 ) = (6, 3)
C (x 3 , y 3 ) = (6, 6)
D (x 4 , y 4 ) = (2, 6)
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Introduction
·
In this problem rectangle ABCD has to scaled to 3/4 of its size.
\
S x = 0.75 and S y = 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 ]
ww
Step 3 : Translate ABCD from orgin (0, 0) and point (4, 4.5) [TIII ]
w.E
Step 4 : Evaluate scaling matrix [S]
[S ](0.75, 0.75) = [TI ] ´ [TII ] ´ [TIII ]
asy
E
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 Dy I = – 4.5
\
é 1
[TI ] = ê 0
ê
êë DxI
Step 2 :
ngi
[TI ] in homogeneous form,
Scaling to
0
1
Dy I
0ù
0ú
ú
0úû
nee
0
0ù
é 1
ê 0
1
0ú
ê
ú
êë - 4.5 - 4.5 1úû
3
of its size at origin (0,0)
4
Here, S x = 0.75; S y = 0.75
\
rin
g.n
et
[S] in homogeneous form,
éS x 0
[TII ] = ê 0 S y
ê
0
êë 0
0ù é 0.75
0
0ù
ú
ê
0 = 0
0.75 0ú
ú ê
ú
1úû êë 0
0
1úû
Step 3 : Translation of ABCD to (4, 4.5) from (0, 0) [T] III
Here D x = 4 ; D y = 4.5
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Introduction
\
\
[T]III
0 0ù
é1
é1 0 0 ù
ê
ú
= 0
1 0 = ê0 1 0 ú
ê
ú
ê
ú
êë Dx Dy 1úû
êë 4 4.5 1úû
Final scaling matrix [S]æ 3
3ö
ç , ÷
è4 4ø
[S] = [TI ] ´ [TII ] ´ [TIII ]
ww
0 0ù é0.75 0 0ù é1 0 0ù
é1
ê
= 0
1 0ú ´ ê –0.5 0.75 0ú ´ ê0 1 0ú
ê
ú ê
ú ê
ú
0 1úû êë 4 4.5 1úû
êë –4 –4.5 1úû êë 0
Þ
0
0ù
é0.75
ê
=
0
0.75 0 ú
ê
ú
1.125 1úû
êë 1
w.E
[S]æ 3
3ö
ç , ÷
è 4 4ø
Resultant Co-ordinates
asy
E
Given, A (x1 , y 1 ) = (2, 3)
ngi
B (x 2 , y 2 ) = (6, 3)
C (x 3 , y 3 ) = (6, 6)
D (x 4 , y 4 ) = (2, 6)
ABCD in homogeneous form,
éA ù
é x1
ê Bú
êx
ê ú = ê 2
êCú
êx 3
ê Dú
êx
ë û
ë 4
y1
y2
y3
y4
1ù é 2 3
1ú ê6 3
ú = ê
1ú ê6 6
1úû êë 2 6
nee
1ù
1ú
ú
1ú
1úû
rin
g.n
et
Resultant Co-ordinates
é A ¢ù
ìA ü
ê B¢ ú
ïï B ïï
ê ú = í ý ´ [S ](3/ 4 , 3/ 4 )
êC ¢ú
ïC ï
ê D ¢ú
ïî D ïþ
ë û
Þ
é A ¢ù
é2
ê B¢ ú
ê6
ê ú = ê
êC ¢ú
ê6
ê D ¢ú
ê2
ë û
ë
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3
3
6
6
1ù
é 2.5
0
0ù ê
é0.75
ú
1 ê
5.5
ú´ 0
0.75 0ú = ê
ú ê 5.5
1ú ê
1úû ê
.
êë 1 1125
ú
1û
ë 2.5
1 - 50
3.375
3.375
5.625
5.625
1ù
1ú
ú
1ú
1úû
Computer Aided Design and Manufacturing
Downloaded From: www.EasyEngineering.net
Introduction
\
A ¢ = (2.5, 3.375)
B ¢ = (5.5, 3.375)
C ¢ = (5.5, 5.625)
D ¢ = (2.5, 5.625)
Y
C(6, 6)
D(2, 6)
D' (2.5, 5.625)
ww
w.E
C' (5.5, 5.625)
P (4, 4.5)
asy
E
A' (2.5, 3.375)
A(2, 3)
B' (5.5, 3.375)
ngi
Fig. 1.9.26
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.
B(6, 3)
nee
X
rin
g.n
et
3D Translation
·
·
3D translation matrix is explained by,
0
0 0ù
é 1
ê 0
1
0 0ú
ê
ú
[T] =
0
1 0ú
ê 0
êD
ú
ë x D y D z 1û
Fig. 1.10.1.3D Translation of a donut
D x , D y and D y explains distance of translation along x, y and z direction
respectively.
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Introduction
3D Scaling
·
3D scaling matrix is explained
éS x 0
ê0 S
y
[ Ts ] = ê
0
ê0
ê0
0
ë
·
S x ,S y and S z represents the scaling factors along x, y and z direction
by,
0 0ù
0 0ú
ú
S y 0ú
0 1úû
respectively.
3D Rotation
·
3D rotation matrices are given by,
i) Rotation along Z-axis by angle 'q'
ww
é cos q
ê – sin q
[R z ] = ê
ê 0
ê 0
ë
w.E
sin q
0 0ù
cos q 0 0ú
ú
0
1 0ú
0
0 1úû
asy
E
ii) Rotation along X-axis by angle 'f'
ngi
0
0
é1
ê0 cos f – sin f
[R x ] = ê
ê0 sin f cosf
ê0
0
0
ë
0ù
0ú
ú
0ú
1úû
iii) Rotation along X-axis by angle 'f'
é cos f
ê 0
[R y ] = ê
ê sin f
ê 0
ë
0 sin f 0ù
1
0
0ú
ú
0 cos f 0ú
0
0
1úû
nee
rin
g.n
et
+ [AU : Dec.-16, May-18]
1.11 Line Drawing
·
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
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Introduction
·
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.
ww
w.E
asy
E
ngi
Procedure for line drawing using DDA algorithm
nee
rin
g.n
et
Consider a line segment with coordinates (x 1 , y 1 ) and (x 2 , y 2 ) with slope 'm' as
shown in the Fig. 1.11.1.
Y
Step 1 : Identify (x 1 , y 1 ) and(x 2 , y 2 )
(x2, y2)
Step 2 : Calculate number of steps.
If D x > D y , No. of steps = D y
Y
Else If D x > D y , No. of steps = D x
Step 3 : Find the slope 'm'
m =
·
Dy
Dx
c
x+
=m
(x1, y1)
=
(y 2 – y1)
( x 2 – x1 )
X
Fig. 1.11.1 Line segment
If m £ 1, Assume D x = 1
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Introduction
¡
Then, D y = mDx = ( y 2 – y 1 ) = m (x 2 – x1 )
¡
This could be written as,
y i + 1 - y i = m(x i+1 – x i ), where(i = 1, 2, 3,¼ )
and since, D x = 1, y i+1 – y i = m
· Therefore, If m £ 1
x i+1 - x i - = 1
·
·
y i+1 - y i = m
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 D y = 1
Dy
¡ Then, m =
Dx
Dy
Dx =
m
1
(y – y1)
( x 2 - x1 ) =
m 2
¡ This could be written as,
1
(y
– y i ), where (i = 1, 2, 3, …)
( x i+1 – x i ) =
m i+1
Therefore, If m > 1
x i+1 – x i - = 1/m
ww
w.E
·
·
asy
E
ngi
nee
rin
y i+1 – y 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
Dx
No. of steps
Dy
=
No. of steps
x increment =
y increment
g.n
et
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.
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Fig. 1.11.2
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Introduction
·
Fig. 1.11.3 represents flow chart for the DDA Algorithm.
Start
Set values for x1, y1, x2 and y2
Calculate No. of steps x and y
Calculate slope 'm'
ww
w.E
If m
No
1
xi+1 – xi– = 1/m
yi+1 – yi = 1
Yes
asy
E
xi+1– xi– = 1
yi+1– yi = m
ngi
Evaluate and plot x and y points
Stop
nee
rin
Fig. 1.11.3 Flowchart for DDA algorithm
g.n
et
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.
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Introduction
Procedure for Bresenham's Algorithm
Consider the line segment shown in Fig. 1.11.2.
Step 1 : Identify (x 1 , y 1 ) and (x 2 , y 2 )
Step 2 : Calculate number of steps.
If D y > D x No. of steps = D y
Else If D x > D y , No. of steps = D x
Step 3 : Evaluate error or deviation 'e', e i = 2 Dy – Dx
Step 4 : If e i ³ 0, e i + 1 = e i + 2 Dy – 2 Dy , increment 'x' and 'y' by 1.
ww
If e i < 0, e i +1 = e i + 2 Dy increment 'x' by 1.
w.E
Step 5 : Plot the x and y points in raster scan display.
Fig. 1.11.4 represents flowchart for the Bresenham's algorithm
asy
E
Start
ngi
Set values for x1, y1, x2 and y2
Calculate No. of steps x and y
Evaluate error 'e'
If e
0
No
nee
rin
ei+1 – ei + 2 y – 2 x
g.n
et
Yes
ei+1 = ei + 2 y
Evaluate and plot x and y points
Stop
Fig. 1.11.4
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Introduction
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(x1 , y 1 ) = (2, 3)
A(x 2 , y 2 ) = (12, 8)
Step 1 : Identity (x1 , y 1 ) and (x 2 , y 2 )
ww
x1 = 2;
y1 = 3
x 2 = 12;
y2 = 8
Step 2 : Calculate the number of steps
w.E
Dy = y 2 - y 1 = 5
Dx = x 2 - x1 = 10
Q
Dx
asy
E
> Dy Þ No. of steps = Dx = 10
Step 3 : Find the slope (m)
Dy
5
=
= 0.5
Dx 10
Slope, (m) =
m = 0.5 < 1
\ when m £ 1
xi + 1 - xi = 1
y i+ 1 - y i = m
ngi
nee
ü
ï ' x' will be intremented by 1
ý
ï ' y' will be intremented by m
þ
Step 4 : Find the increment value
x increment =
Dx
10
=
=1
No. of. steps 10
y increment =
Dy
5
=
= 0.5 = m
No. of. steps 10
rin
g.n
et
Sample solution
x i = 2;
First increment : x i +
yi = 3
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
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Introduction
Similarly Step 3 : x i +3 = 4 + 1 = 5
yi
+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
y initial
x final
(Round off)
y final
(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
7
5.5
7
6
8
6
8
6
ww
7
w.E
9
6.5
9
7
8
10
7
10
7
9
11
7.5
11
8
10
12
8
12
8
5
6
asy
E
ngi
Step 6 : Plot
yvalues 8
7
nee
rin
g.n
et
6
5
4
3
2
1
0
1
2
3
4
5
6
7
8
9
10
11
12 xvalues
Step 6 : Plot
Fig. 1.11.5
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Introduction
Example 1.11.2 : Using DDA algorithm rasterize line AB, A(0, 0), B(4,6).
Solution :
AB Þ A(0, 0) ; B(4, 6)
Given,
A(x1 , y 1 ) ; B(x 2 , y 2 )
Step 1 : Identity (x 1 , y 1 ) and (x 2 , y 2 )
ww
x1 = 0;
y1 = 0
x 2 = 4;
y2 = 6
Step 2 : Calculate the number of steps
w.E
Dy = y 2 - y 1 = (6 – 0) = 6
asy
E
Dx = x 2 - x1 = (4 – 0) = 4
Q
Dy > Dx Þ No. of steps = Dy = 6
Step 3 : Find the slope (m)
Dy 6
Slope (m) =
= = 1.5
Dx 4
Q
m >1
xi +
1
yi +
1
1
- xi =
m
- yi =1
ngi
ü
ï ' x' will be incremented by 1 and
ý
m
ï ' y' will be intremented by '1'
þ
Step 4 : Find the increment value :
x increment =
Also,
0.66 =
y increment =
Initial values :
Dx
4
= = 0.66
No. of steps 6
rin
g.n
et
1
1
=
m 1.5
Dy
6
= =1
No. of steps 6
xi = 0 ;
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Introduction
First step :
x i + 1 = 0 + 0.66 = 0.66;
y i +1 = 0 + 1 = 1
Second step :
x i + 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
y initial
x final
(Round off)
y final
(Round off)
0
0
0
0
0
1
ww
0.66
1
1
1
2
1.32
2
1
2
1.98
3
2
3
4
2.64
4
3
4
5
3.30
5
3
5
6
3.96
4
6
3
w.E
asy
E
ngi
Step 6 : Plot
6
6
nee
rin
5
4
3
g.n
et
2
1
0
1
2
3
4
Fig. 1.11.6
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Introduction
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 :
AB Þ A(x1 , y 1 ) = (3, 3)
Given,
B(x 2 , y 2 ) = (11, 7)
Step 1 : Identity (x 1 , y 1 ) and (x 2 , y 2 )
ww
x1 = 3;
y1 = 3
x 2 = 11;
y2 = 7
w.E
Step 2 : Calculate the number of steps
Dx = x 2 - x1 = 11 – 3 = 8
asy
E
Dy = y 2 - y 1 = 7 – 3 = 4
Here
Dx > Dy; \ No. of steps = Dx = 8
Step 3 : Evaluate error and deviation.
Initial values of error and deviation,
e i = 2Dy - Dx
ngi
= 2´ 4-8 = 0
\
nee
rin
ei = 0
Step 4 :
e1 = 0 = e 0 (if e i ³ 0; e i
1)
+1
= e i + 2Dy - 2Dx)
g.n
et
\
e1 = e 0 + 2Dy - 2Dx
x=x+1=3+1=4
Þ
e1 = 0 + 8 – 16 = –8
y=y+1=3+1=4
Since,
e1 = – 8 (if e i < 0, e i + 1 = e i + 2Dy)
e 2 = e7 + 2Dy
\
Þ
x=x+1=4+1=5
= –8+8=0
y=4
Note : From procedure it is clear that,
·
If e i ³ 0 ® Increment x and y by 1
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Introduction
·
If e i < 0 ® Increment x by 1 and keep 'y' as it is.
·
The results could be tabulated as below.
No. of Steps
x
y
0
3
3
0
1
4
4
–8
2
5
4
0
3
6
5
–8
4
7
5
0
5
8
6
– 8
9
6
0
10
7
– 8
11
7
0
ww
w.E
6
7
8
asy
E
ngi
Step 5 : Plot the points
7
6
5
Error
nee
rin
4
g.n
et
3
2
1
0
1
2
3
4
5
6
7
8
9
10
11
Fig. 1.11.7
+ [AU : Dec.-16, 17]
1.12 Clipping
·
Various projections of an object's geometry can be defining views.
·
A view requires a view window.
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Introduction
·
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.
ww
·
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.
w.E
Applications of Clipping :
asy
E
·
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
ngi
Types of Clipping :
i. Point clipping
ii.
Line clipping
iv. Curve clipping
v.
Text clipping
i. Point Clipping
·
The clip window is a rectangle
as shown in the Fig. 1.12.1.
y wmin £ y < y wmax
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iii. Area clipping (Polygon)
ywmax
·
The point P = (x,y) will be
displayed if the following
inequalities are satisfied :
x wmin £ x < x wmax
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P (x, y)
ywmin
xwmin
xwmax
Fig. 1.12.1 Point clipping
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Introduction
(x wmin , x wmax ) and (y wmin , y 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.
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B
B
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A
A
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Clip
rectangle
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Fig. 1.12.2 Line clipping case 1
·
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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.
D
D'
D'
B
A
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B
C
C
A
Clip
rectangle
Fig. 1.12.3 Line clipping case 2
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Introduction
·
·
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.
E
F
Clip
rectangle
Case 4 : Both endpoints are
located outside the clipping area
Fig. 1.12.4 Line Clipping Case 3
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.
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D
D'
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Clip
rectangle
C
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Fig. 1.12.5 Line Clipping Case 4
Cohen-Sutherland Line Clipping Algorithm
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et
·
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).
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Introduction
·
·
·
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.
y
1000
1010
Window
ywmin
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1001
ywmax
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.
Right
Left
0001
0000
0010
0101
0100
0110
xwmax
xwmin
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This rule is also called TBRL code (Top-Bottom-Right-Left).
·
Cohen Sutherland clipping could be explained by the Fig. 1.12.7.
Left
1001
ywmax
ywmin
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P8
1000
Window
P6
0001
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1010
P7
0010
P4
rin
Top
P1
0100
0110
P5
xwmin
x
Right
P2
0000
0101
Bottom
Fig. 1.12.6 Bit code (TBRL code) for
cohen- sutherland clipping
·
y
Top
P3
xwmax
Bottom
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et
x
Fig. 1.12.7 Cohen, Sutherland line clipping technique
·
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.
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Introduction
·
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.
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iii. Polygon Clipping
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·
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
·
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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.
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P
Q
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Fig. 1.12.8 (a) Convex polygon
·
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 Fig. 1.12.8 (b) Concave 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.
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Introduction
·
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.
13
10
11
9
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6
5
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12
1
8
7
2
4
3
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(a) Clip polygon
5
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7
6
4
3
x
8
(b) Clipped polygon
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2
Fig. 1.12.9 Example for polygon clipping
iv. Curve Clipping
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·
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.
Fig. 1.12.10 Curve clipping
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Introduction
v. Text Clipping
Text clipping can be of two types :
Widow
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.
STRING 1
STRING 2
Fig. 1.12.11 All or none string-clipping
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.
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Widow
STRING 1
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STRING 1
G
IN
TR
S
STRING 4
Widow
2
ING 1
R
ST
STRING 4
TRING 1
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Fig. 1.12.12 All or none character-clipping
1.13 Viewing Transformation
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+ [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.
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·
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.
2D Object
data
Computer
display
a
b
c
Object
transformation
Window
transformation
Clipping
f
e
Frame
buffer
Scan
conversion
d
Visaport
transformation
Fig. 1.13.1 Viewing transformation
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Introduction
·
A window is specified by four world coordinates, x wmin , x wmax , y wmin , and
y wmax as shown in Fig. 1.13.2 (a).
·
Similarly, a view port is described by four normalized device coordinates :
x vmin , x vmax , y vmin , and y vmax as shown in Fig. 1.13.2 (b).
yvmax
ywmax
(xv, yv)
(xw, yw)
yvmin
ywmin
ww
xwmin
w.E
xvmin
xwmax
Window
Fig. 1.13.2 (a) Window
xvmax
View port
Fig. 1.13.2 (b) View Port
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·
The objective of window-to-view port mapping is to convert the world
coordinates (x w , y w ) of an arbitrary point to its corresponding normalized
device coordinates (x v , y v ).
·
In order to maintain the same relative placement of the point in the view port as
in the window, we require :
x w – x w min
x v – x v min
y w – y w min
y v – y v min
=
and
=
x w max – x w min
x v max – x v min
y w max – y w min
y v max – y v min
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nee
·
Solving the above two equations we get,
ì x v = x v min + (x w – x w min )S x
íy = y
v min + (y w – y w min )S y
î v
·
S x and S y are scaling factors,
·
Sx =
x v max – x v min
x w max – x w min
Sy =
y v max – y v min
y w max – y w min
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et
Since the eight coordinate values that define the window the view port are just
constants, these two formulas for computing (x v , y v ) from (x w , y w ) can be
expressed in terms of a translate-scale-translate transformation 'N',
æ xw ö
æ xv ö
÷
÷
ç
ç
ç y v ÷ = N× ç y w ÷
ç 1 ÷
ç 1 ÷
ø
ø
è
è
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Introduction
æ1 0
ç
N = ç0 1
ç0 0
è
·
0 ö æ 1 0 – x w min ö
÷ ç
÷
0 ÷ × ç 0 1 – y w min ÷
÷
1
1 ÷ø çè 0 0
ø
x v min ö æ S x 0
÷ ç
y v min ÷ × ç 0 S y
1 ÷ø çè 0
0
An example for viewing transformation is shown in Fig. 1.13.3.
Window
ywmax
(xw, yw)
ywmax
ywmin
ywmin
ww
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xwmin
0
xwmax
0
xvmin
xvmax
asy
E
Fig. 1.13.3 Viewing transformation
1.14 Brief Introduction to CAD and CAM
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·
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.
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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.
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Introduction
·
·
CAD software packages
provide the designer with a
multi window environment
with animation.
Computer
graphics
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.
CAD
Design
engineering
Geometric
modelling
Fig. 1.14.1
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Typical Components CAD/CAM System
CAD/CAM
system
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Hardware
Computing machine
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Software
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Graphics device
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Display processing unit
Display device
Input device
Output device
Fig. 1.14.2 Components of a CAD/CAM system
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et
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
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Introduction
·
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.
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·
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·
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.
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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.
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Introduction
Basic Elements of a CAD System
The basic elements of a CAD system are
shown in Fig. 1.14.3.
Geometric modeling
a) Geometric Modeling :
Engineering analysis
·
·
Geometric modeling is
applied mathematics and
geometry that studies
algorithms for the
description of shapes.
a branch of
computational
methods and
mathematical
Design review and evaluation
Automated drafting
The shapes studied in geometric
modeling
are
mostly
two
or
three-dimensional, although many of Fig. 1.14.3 Basic elements of a CAD system
its tools and principles can be applied
to sets of any finite dimension.
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·
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
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b) Engineering analysis
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·
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
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Introduction
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.
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c) Design Review and Evaluation
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·
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
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·
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.
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Reasons for Implementing CAD or Advantages or Benefits of CAD
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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
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Introduction
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.
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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.
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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.
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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.
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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.
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Introduction
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.
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·
1.15.1
·
CAM is often linked with CAD for more enhanced and streamlined
manufacturing, efficient design and superior machinery automation.
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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.
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·
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.
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CAD
Determination
of dimensions
Material
selection
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Assembly
method
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Part
drawing
Conceptual
design
Thickness
determination
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Part
arrangement
Database
Automatic
cutting
Simulation
Material
handling
Jigs
and
fixtures
Automatic
assembly
CAM
Fig. 1.15.1 CAD/ CAM integration
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Introduction
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.
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ii. Master Scheduling
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·
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)
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Introduction
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
ww
·
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.
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1.15.3 Manufacturing Control
· Manufacturing control is concerned with managing and controlling the physical
operations in the factory to implement the manufacturing plans.
·
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Factory operations included in the manufacturing control are,
(i) Shop floor control
(ii) Inventory control
(iii)Quality control
(i) Shop Floor Control
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·
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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Introduction
·
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 ?
ww
·
In the field of loss prevention, systems designed to introduce technical barriers
to shoplifting.
w.E
(iii) Quality Control
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·
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.
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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.
+ [AU : May-17]
1.16 Types of Production Systems
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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Introduction
·
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
(i) Job Production
Production/Operations volume
Fig. 1.16.1 represents the classification
of production system.
·
Under this method peculiar,
special
or
non-standardized
products
are
produced
in
accordance with the orders
received from the customers.
ww
·
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As each product is nonstandardized varying in size and
nature, it requires separate job for
production.
Mass production
Batch production
Job-shop
production
asy
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Output/Product variety
Fig. 1.16.1 Classification of production system
·
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.
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Characteristics
The job production possesses the following characteristics :
· A large number of general purpose machines are required.
g.n
et
·
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.
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Introduction
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.
ww
·
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g.n
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·
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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Introduction
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
ww
·
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.
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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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Introduction
·
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.
ww
·
The process of stages of production should be continuous.
Advantages of Mass Production
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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.
asy
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·
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.
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1.17 Manufacturing Models and Metrics
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+ [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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Introduction
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 (Pc hr)
·
The starting point to evaluate production rate is cycle time (Tc ).
ww
Cycle time (Tc )
·
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.
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·
asy
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Cycle time is expressed as the sum of (1) actual processing (or) assembly
operation time (To ) (min Pc )
2) Handling time (loading and unloading time) min Pc (Tn )
ngi
3) Tool handling time (tool changing time) min Pc (Tth )
Te = To + Tn + Tth min Pc
nee
Production rate evaluation for 3 types of production (R p )
1) Batch production :
·
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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 = Tb (min)
Set up time = Tsu (min batch)
Q = Batch quantity (Pc )
Tc = Cycle time (min Pc )
Tb = Tsu + Q ´ Tc
\
·
'Q ´ Tc ' Þ Representing processing time (min)
List average production time per min be 'Tp ' (min Pc )
\
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Tp =
Tb
Q
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Introduction
·
The average production rate for the machine is simply the reciprocal of average
production time expressed in hourly rate.
Rp =
60
Tp
Pc hr
2) Job shop production :
·
For job shop production, usually the value of quantity, Q = 1
·
Then average production time per work unit, Tp = Tsu + 1 ´ Tc
Tp = Tsu + Tc
ww
The production rate,
·
Rp =
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60
Tp
min Pc
Pc hr
If 'Q' is greater than 'i', Tp is evaluated as explained for batch production
3) Mass production :
asy
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·
For mass production setup time could be neglected (Tsu ~ 0) and Q = 1
·
Then average production time per work unit,
Tp = 0 + Tc
ngi
·
Therefore
·
Here production rate could be expressed as,
Tp = Tc
Rp
60
60
=
=
Tp
Tc
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Pc hr
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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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Introduction
·
Production capacity for a plant is given by,
nS w H sh × R p
Pc =
no
Pc = weekly production capacity of the facility (units/week)
Where,
S w = no. of shifts per period (shift/week)
H sh = hr/shift (hr)
R p = Hourly production rate of each work center (units/hr)
ww
n o = Number of distinct operations through which work
units are routed
Problems On Production Capacity
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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.
asy
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ngi
Solution :
n = Number of work centers
nee
= Number of machines in turret lathe section = 6
rin
S w = Number of shifts in (shifts/week) = 10 shift/week
H sh = Number of hours/shift = 8 hrs/shift
g.n
et
R P = Hourly production rate of each work center = 17 units/hr
\ Production capacity, Pc =
n S w H sh R P
n0
Here,
n 0 = 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
= 8160 Units/week
1
Pc = 8160 unit/week
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Introduction
Example 1.17.2 : A production
production of the same part.
5 days/week. Average production
weekly production capacity of the
facility has 5 works centers, all devoted to the
The facility operates 8 hr/shift, 2 shifts/day and
rate for each machine is 15 units/hr. Compute the
production facility.
Solution : Given : Number of work centers, n = 5,
H sh = Number of hours per shift = 8 hr/shift,
S w = Number of shifts per week.
Here,
Shifts per day - 2 shifts /day
ww
Working days in a week = 5 days/week
\ S w = 2 shifts /day ´ 5 days/week
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S w = 10 Shifts/week
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R P = Hourly production rate = 15 units/hr
Production capacity =
Pc =
n S w H sh R P
no
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Here, no = 1, since production facility deals with production of same part.
\
\
Pc =
5 ´ 10 ´ 8 ´ 5
= 6000 units/week
1
Pc = 6000 units/week
iii) Proportion Uptime On Equipment (A reliability measure)
The common reliability measures for an equipment are,
a) Utilization
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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
Where,
Q = Actual quantity produced by the facility during a given time period (Pc / week).
Pc = Production capacity for same period (Pc / week).
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.
ww
iv)
If utilization is high, that means the facility is being operated to its full capacity.
B) Availability (A)
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·
Availability is defined using two other reliability measure terms.
·
MTBF = Mean Time Between Failures
·
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MTTR = Mean Time To Repair.
MTBF - MTTR
A =
MTBF
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nee
·
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.
Breakdown
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Repairs completed
Equipment
operating
MTBF
TIme
Fig. 1.17.1 Time scale showing MTBF and MTTR used to define availability
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Introduction
·
Effect of utilization find availability on plant or production capacity.
Pc = U ´ A ´
(n S w H sh R P )
no
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’.
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Solution : Given : n = No. of work centers = 1,
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S w = No. of shift /week,
H sh = number of hrs/shift,
R P = Production rate = 20 units/hr
Here,
ngi
Production machine operation at full capacity = 80 hrs/week
\
H sh ´ S w = 80 hrs/week
a) Production capacity
Pc =
n S w H sh R P
no
nee
rin
n o = 1, since same parts are being produced
\
Pc =
1 ´ 80 ´ 20
= 1600 units/week
1
g.n
et
Pc = 1600 units/week
b) Utilization (u)
u =
Q
Pc
Q = Actual quantity produced = 1000 Pc /week
\
u =
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1000
= 0.625
1600
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Introduction
‘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 %
n S w H sh R P
= 0.9 ´ 0.625 ´ 1600 = 900
\
Pc = U ´ A ´
no
Pc = 900 units/week
ww
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.
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Solution : Given : MTBF = 200 hr, MTTR = 5 hr
\ Availability,
A =
ngi
MTBF - MTTR 200 - 5
=
= 0.975
MTBF
200
nee
\ Availability is usually represented in percentage, A = 97.5 %
ii) Manufacturing Lead Time (MLT)
rin
·
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.
g.n
et
a) MLT for batch production :
The manufacturing lead time for the batch production is given by,
MLT = n o (Tsu + QTc + Tn o )
n o = No. of distinct operations through which work units are
routed.
Tsu = Set up time (min/batch)
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Introduction
Q = Batch quantity (Pc )
Tc = Cycle time (min / Pc )
Tn o = Non-operation time operated with machine (min)
b) MLT for job shop production
· For job shop production, the batch quantity, Q = 1
MLT = n o (Tsu + Tc + Tn o )
\
c) MLT for mass production
· For mass production
n o = Number of distinct operations through which work units
are routed
= 1
Tsu = Set up time is neglected ~ 0
Tn o = Non operation time is neglected ~ 0
ww
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asy
E
MLT = Tc
\
Problems On Manufacturing Lead Time
ngi
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.
nee
Solution : Given, n o
·
\
\
·
\
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g.n
et
= 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 o (Tsu + QTc + Tn o )
MLT = 5(3 + 100 ´ 0.1 +7) = 100 hours
MLT = 100 hours
If the plant runs for 8 hr shifts/day
MLT =
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100
= 12.5 days
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Introduction
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
Machine
ww
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
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6
asy
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4
Solution : Given, Q = Batch size = 100,
2.5
ngi
Tn o = Avg. non operation time = 12 hr = 720 min
i) MLT
MLT = n o (Tsu + Q.Tc + Tn o )
Here,
Tsu = Average Set up time =
nee
rin
4+ 2+ 8+ 3+ 3+ 4
= 4 hr
6
Tsu = 4 hours = 240 min
Also,
Tc = Average operation time (Cycle time)
=
g.n
et
5 + 3.5 + 10 + 1 .9 + 4 .1 + 2 .5
= 4.5 min
6
Tc = 4.5 min
\
MLT = n o (Tsu + Q.Tc + Tn o ) = 6 (240 + (100 ´ 4.5) +720)
= 8460 min
MLT = 141 hr
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Introduction
ii) Production rate ‘R P ’ for operation 3
For batch processing, Batch processing Time,
Tb = Tsu + Q.Tc
For operation 3 Þ Tsu = 8 hrs = 480 min
Tc = 10 min
·
Tc = 1480 min
The average production time,
T
1480
= 14.8 min
TP = b =
Q
1000
ww
·
The production rate R P (in hours)
w.E
\
RP =
60
60
=
= 4.05 hr
Tb
14.8
asy
E
R P = 4.05 hr
v) Work In Process (WIP)
ngi
·
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
nee
rin
g.n
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MLT = Manufacturing Lead Time
S w = Number of shifts/ week
H sh = 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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Introduction
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.
ww
Solution : Given, n 0 = 6, Tc = Average operation time = 6 min,
w.E
Tsu = 5 hr = 300 min, Q = Average batch size = 25 units,
Tn o = Average non operation time = 10 hr = 600 min,
asy
E
n = Number of work centers = 18, Plant operation time = 70 hr/week.
Here,
S w = No. of shifts/week
ngi
H sh = No. of hours/week
nee
\
S w ´ H sh = 70 hr/week
· MLT for batch production
MLT = n o (Tsu + Q.Tc + Tn o ) = 6(300 + 25 ´ 6 + 600) = 6300 min
= 105 hr
If plant runs for 70 hr/week
MLT =
·
105
= 1.5 week
70
rin
g.n
et
Production rate, R P
Batch processing time,
TP = Tsu + QTc = 300 + 25 ´ 6 = 450 min = 450 min
Average production Time/week, (TP )
TP =
Production rate, R P =
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Tb
450
=
= 18 min
Q
25
60
60
=
TP
18
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Introduction
R P = 3.33 Pc / hr
·
Plant capacity or production capacity
n S w H sh R P
Pc =
no
Here, n o = 6 since operations done through six different machines
18 ´ 70 ´ 3.33
= 699.3 = 699 units/week
6
Pc =
·
Plant utilization (u)
ww
U =
\
Q
Po
Q = Actual output quantity
w.E
= No. of batches ´ Batch size = 20 ´ 25 = 500 Pc
U =
asy
E
500
= 0.715
699
Utilization is usually represented in percentage
u = 71.5 %
\
· Work In Process (WIP)
WIP is given as
WIP =
AUPc MLT
S w H sh
ngi
nee
Assume 100 % availability
\
WIP =
rin
1 ´ 0.715 ´ 700 ´ 1.5
= 10.78
70
g.n
et
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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Introduction
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.
ww
·
As output increases, variable cost also increases.
·
Total cost (Tc ) could be obtained by adding fixed cost (Fc ) and variable cost
(V c ).
w.E
Tc = Fc + V c
asy
E
Direct Labour, Material and Overhead Costs
·
ngi
An alternate classification divide manufacturing cost into,
1) Direct labour cost
2) Material cost
3) Overhead cost
1) Direct labour cost :
nee
rin
g.n
et
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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Introduction
a) Factory overheads
Some of the typical factory overheads are,
· Plant supervision
·
Applicable taxes
·
Factory depreciation
·
Material handling
·
Power for machinery
·
Security personnel
·
Insurance
ww
·
Tool crib attendant
·
Clerical support
·
Heat and Air conditioning
·
Light
w.E
asy
E
· Payroll services
b) Corporate overheads
ngi
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,
nee
rin
g.n
et
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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Introduction
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
ww
5) Selling price :
Selling price = Total cost + Profit
w.E
6) Market Price :
asy
E
Market price = Selling price + Discount
Problems based on components of cost
ngi
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.
nee
Solution : Given, n o = 100, Direct material cost = ` 200,
Direct labour cost = ` 300, Direct expenses = 0 (not specified),
rin
g.n
et
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
=
(200 + 300) = `100
100
Factory cost = 500 + 100 = ` 600
Q
· Selling price for each article for a profit of 15 % of gross cost
Selling price = Total cost + Profit
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Introduction
Total cost = Factory cost + Overhead cost
Overhead cost = 30% of factory cost
=
30
´ 600 = ` 180
100
Total cost = 600 + 180 = ` 780
Selling price = 780 +
780
15
= ` 917.65
´ Selling price =
0.85
100
Selling price per product = ` 917.65
1.18 Break Even Analysis - A Tool for Manufacturing Control
ww
·
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.
w.E
asy
E
ngi
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
nee
rin
·
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.
g.n
et
e
ue
en
ev
lr
ta
To
lin
it
rof
l
Tota
P
Break even
Cost
(`)
cost
line
ost
le c
iab
Var
point
Fixed cost
ss
Lo
Volume (units)
Fig. 1.18.1 Break even analysis
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Introduction
Mathematical Model for Break Even Analysis
·
Break Even Quantity (Q BEP )
·
Break even quantity defines the sales volume at Break even point.
Q BEP =
Fc
S p - Vc
Where,
Fc = Fixed cost
ww
V c = Variable cost
Also,
S p = Selling price/unit
w.E
Total cost = Fc + V c ´ Q
Q = Quantity solid (units)
asy
E
Total revenue = S P ´ Q
At Break even point Q = Q BEP
Q BEP =
ngi
Fc
S p - V c units
Þ Break even sales (S BEP )
S BEP
Fc
=
æV ö
1- ç c ÷
è SP ø
in rupees
nee
rin
g.n
et
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.
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Introduction
Solution : Given, Fc = 700000 `, V c = 45 `, S p = ` 200
i) BEP in terms of sales volume (Q BEP )
Fc
700000
=
= 4516.12
S p - Vc
200 - 45
Q BEP =
Q BEP = 4516 units
ii) BEP in terms of Rupees (S BEP )
Fc
700000
=
V
æ 45 ö
1- c
1 -ç
÷
SP
è 200 ø
ww
S BEP =
w.E
S BEP = ` 903225.8
asy
E
iii) If Q BEP = 6000 units
Profit = ? (If profit is there, we can write)
Also,
Q BEP =
6000 =
Fc + Profit
=Q
S p - Vc
ngi
nee
700000 + Profit
= ` 23000
200 - 45
iv) Number of units to be sold if profit target = ` 150000
Here, Q BEP = Q, since there is profit
\
Q =
=
Fc + Profit
S p - Vc
rin
g.n
et
700000 + 150000
= 5483.8
200 - 45
Q = 5484 units
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Introduction
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.
ww
7. What is the need of homogeneous coordinates ? Mention the homogeneous
coordinates for translation, rotation and scaling.
w.E
8. Explain the DDA hidden line algorithm with an example.
9. Explain the Bresenham's line algorithm with an example.
asy
E
10. Explain Cohen Sutherland clipping algorithm with an example.
11. Elaborate on the basic requirements that a CAD software has to satisfy.
ngi
12. Distinguish between modes of the design process and models of designs.
13. Describe the various database models which are generally used.
nee
14. What are the differences between the sequential approach to the product
rin
development process and the concurrent engineering approach ? Why should the
latter be adopted ?
g.n
et
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
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Introduction
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.
ww
Part A : Two Marks Question with Answers
Q.1
+ [ AU : May 2017 ]
State any two benefits of CAD.
w.E
Ans. : · Efficiency, effectiveness and creativity of the designer are drastically improved.
· Faster, Consistent and More accurate.
asy
E
· Easy modification (copy) and Improvement (Edit).
· Inspecting tolerance and interface is easy.
ngi
· 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.
nee
+ [ AU : Dec. 2016, May 2017 ]
rin
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 ?
g.n
et
+ [ 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
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Introduction
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¢ ] = [Tn ] [Tn -1 ] [Tn - 2 ]........[T 3 ] [T2 ] [T1 ]
Q.5
List the various activities involved in product development.
+ [ AU : Dec. 2018 ]
Ans. : i. Design process
· Synthesis
ww
· Analysis
ii. Manufacturing process.
w.E
· Process planning
· Process control
Q.6
asy
E
+
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.
ngi
· 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.
0 0ù é x ù
é x¢ ù é 1
ê
ú
ê
[P¢ ] = y ¢ = 0
1 0ú ê y ú
ê ú ê
úê ú
êë 1 úû êë DX DY 1úû êë 1 úû
Q.7
nee
What do you mean by synthesis of design ?
rin
g.n
et
+ [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 ?
Ans. : · View Port
+ [ AU : Dec. 2017 ]
· 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.
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Introduction
130, 100
View port 2
View port 1
View port 4
View port 3
65,50
Fig. 1.1
ww
· Window
· When a design package is initiated, the display will have a set of co-ordinate values.
w.E
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.
asy
E
· Therefore, the designer may want to view only a portion of the image, enclosed in a
rectangular region called a window.
ngi
Window
nee
Original
drawing
rin
g.n
et
Fig. 1.2
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
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Introduction
· 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
+ [ AU : Dec. 2011 ]
What are the components of a CAD system ?
ww
Ans. : CAD hardware components :
· Central Processing Unit (CPU)
w.E
· Memory
· Hard Disk, Floppy Disk, CD-ROM
asy
E
· External storage devices
· The monitor
· Printers and Plotters
· Digitizer, Puck and Mouse
CAD software :
ngi
nee
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.
rin
· 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.
g.n
et
· 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).
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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
ww
Q.15
w.E
asy
E
ngi
Mention any four applications of computer aided design in mechanical
engineering.
nee
Ans. : · Computer-Aided Engineering (CAE) and Finite Element Analysis (FEA)
· Computer-Aided Manufacturing (CAM)
Numerical Control (CNC) machines
including
· Photorealistic rendering and motion simulation.
instructions
rin
to
Computer
g.n
et
· 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
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Introduction
· 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
ww
· Text clipping
Q.19
What is viewing transformations ?
w.E
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.
asy
E
· 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.
ngi
· The window defines what is to be viewed; the viewport defines where it is to be
displayed.
Q.20
Describe Computer Aided Design.
nee
rin
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.
g.n
et
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.
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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.
écos q
R(q ) = ê
ë sin q
- sin q ù
cos q úû
é cos q sin q ù
R( -q ) = ê
ú
ë - sin q cos q û
Q.24
What is design process ?
Ans. : The Engineering Design process is a multi-step process including the research,
ww
conceptualization, feasibility assessment, establishing design requirements, preliminary design,
detailed design, production planning and tool design, and finally production.
Q.25
w.E
What is meant by analysis ?
Ans. : · The analysis begins with an attempt to put the conceptual design into the context of
asy
E
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.
ngi
nee
· 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
rin
g.n
et
· Architectural design
· Pipe routing and plan layout design
· Electronic industries
· Dynamic analysis of mechanical systems
· Kinematic analysis
· Mesh data preparation for finite element analysis.
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Introduction
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.
ww
Q.28
w.E
Define computer graphics.
Ans. : · Computer Graphics involves creation, display, manipulation and storage of pictures and
asy
E
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.
ngi
· The first of them is the display devices. Colour monitors are one example of such
display device.
nee
rin
· There are other examples of output devices like LCD panels, laser printers, colour
printers, plotters etc.
Q.29
What is transformation ? List its types.
g.n
et
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,
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Introduction
Y'
Y
Z
P'
X'
P
Y
P
Z
X
X
Fig. 1.3
ww
Q.31
Write the features needed to be satisfied for line drawing algorithm.
Ans. : · Lines should appear straight
w.E
· Lines should terminate accurately
· Lines should have constant density
asy
E
· Line density should be independent of length and angle
· Line should be drawn rapidly
Q.32
ngi
Differentiate preliminary design and detailed design.
Ans. :
Preliminary 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.
Q.33
nee
Detailed design
rin
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.
g.n
et
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
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Introduction
· 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,
Q
U =
Pc
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Q = Actual quantity produced by the facility during a given
time period. (pieces/week)
Pc = Production capacity for the same period. (pieces/week)
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Availability (A) is defined using the two reliability measure terms.
MTBF = Mean Time Between Failures
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MTTR = Mean Time to Repair
A =
Q.36
MTBF - MTTR
MTBF
Define - Manufacturing Lead Time.
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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. : ·
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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
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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
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· Office space
· Finance department
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· Legal counsel
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Part B : University Questions with Answers
Dec.-2016
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1. Describe various stages of design process with an example. (Refer section 1.3) [8]
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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)
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4. Deduce windowing and viewing transformation parametrically.
(Refer section 1.13)
5. Write in detail about the production performance metrics.
(Refer section 1.17)
[8]
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[8]
[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)
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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]
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10. Explain in detail job shop production and mass production.
(Refer section 1.16)
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[16]
Dec.-2017
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11. Explain the different types of 2D transformations with examples.
(Refer section 1.9)
[13]
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12. Explain the Cohen-Sutherland line-clipping approach with proper sketches.
(Refer section 1.12)
May-2018
[13]
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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 ...
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Notes
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