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Engineering Drawing
UNIT- I
K. Srinivasulu Reddy
Mechanical Engineering Department
Introduction to Engineering Drawing
K.Srinivasulu Reddy, SNIST
UNIT-I
i)Introduction to Engineering Drawing: Drawing
instruments and their uses, types of lines, use of
pencils, lettering, rules of dimensioning.
ii)Construction of Polygons: Inscription &
superscription of polygons given the diameter of circle.
iii)Introduction to scales(Theory only):Scales used in
Engineering Practice and concept of Representative
Fraction.
iv)Curves used in Engineering Practice & their
Construction: Conic sections including the rectangular
Hyperbola-General Method only.K.Srinivasulu Reddy, SNIST
TEXT BOOKS:
1. Engineering Drawing by N. D. Bhatt / Charotar
Publications
2. Engineering Drawing by K.L.Narayana and
Kannaiah/Scietech Publishers
REFERENCES:
1. Engineering Drawing,K.Venugopal/G.Sreekanjana,
New Age International Publishers
2. Engineering Drawing,B.V.R Gupta,M.Roja Roy
I.K. International Publishing
K.Srinivasulu Reddy, SNIST
•Why and What is Engineering Drawing?
•Importance of Engineering Drawing
•Standards
•List of Drawing instruments
•Border lines
•Title block
•Types of lines
•Pencils recommended
•Lettering
•Dimensioning K.Srinivasulu Reddy, SNIST
What is Engineering Drawing ?
 Drawing is a graphical Language of an Engineer
which communicates ideas and information from
one mind to another.
 It communicates all needed information from the
engineer, who designed a part, to the workers,
who will make it. K.Srinivasulu Reddy, SNIST
Engineering Drawing is a two dimensional representation of
a three dimensional object. It is the graphic language, from
which a trained person can visualize the object.
 Engineering drawing is formal and precise way of
communicating information about the shape, size, features
and precision of physical objects.
 Engineering Drawing is the universal language of
Engineers
What is Engineering Drawing Contd..
K.Srinivasulu Reddy, SNIST
ENGINEERING DRAWING
The classification of Engineering Drawings include:
•Building drawing
•Machine drawing
•Production drawing
•Electrical drawing etc.
K.Srinivasulu Reddy, SNIST
A PICTURE SPEAKS A THOUSAND WORDS
In Engineering, a good
drawing is worth even
more than a thousand
words.
Importance
K.Srinivasulu Reddy, SNIST
Why instruments for drawing ?
K.Srinivasulu Reddy, SNIST
A typical study of the drawing course will lead to the
following observations.
• It develops a type of discipline in graphical techniques.
• It develops ability to analyze and communicate ideas.
• It helps in imagining the proper size, shape & form of an
object.
• It develops capacity of reasoning & judgment.
• It helps in the study of other Engineering subjects.
K.Srinivasulu Reddy, SNIST
STANDARDS
• For the convenience of exchange of information and to buy or
sell machinery from or to a foreign country and to facilitate
interchangeability of parts it is essential to have standard code
of practice
• Bureau of Indian Standards (BIS) is the National Standards
Body formed by the Government of India on 1 April 1987,
replacing Indian Standard Institution(ISI) existing earlier. BIS
– SP 46 revised as SP – 46 – 2003
• The other foreign standards are
–DIN of Germany, BS of Britain
• BIS works in association with ISO ( International Standards
Organization )
K.Srinivasulu Reddy, SNIST
List of drawing instruments
• Drawing board
• Drawing sheets & Supporting sheet
• Mini-drafter
• Instrument box
• Drawing pencils (2H,H and HB Grades)
• Eraser
• Clips and Drawing pins/Adhesive tape
• Protractor
• Paper Napkins or Handkerchief
K.Srinivasulu Reddy, SNIST
Drawing Board
o To make lines on a drawing paper smooth
and straight, a polished drawing board is
one of the top most requirements.
IS 1444:1989
K.Srinivasulu Reddy, SNIST
Sizes of drawing boards
S. No. Designation Size (mm)
1 B0 1000 X 1500
2 B1 700 X 1000
3 B2 500 X 700
4 B3 350 X 500
Drawing board
K.Srinivasulu Reddy, SNIST
Drawing sheets IS 10711:2001
• The standard sizes of the
drawing sheets recommended
by BIS (Bureau of Indian
Standards)
• A0 = 1 m2
• Drawing sheet A2.
• width to length 1:
S.NO Sheet
designation
Trimmed
size(mm)
Length x Width
1 A0 1189 x 841
2 A1 841 x 594
3 A2 594 x 420
4 A3 420 x 297
5 A4 297 x 210
2
K.Srinivasulu Reddy, SNIST
Drawing sheet formats
K.Srinivasulu Reddy, SNIST
Working with T-square
K.Srinivasulu Reddy, SNIST
Mini-Drafter
This instrument gives faster drawing as it serves the
purpose of Set-Square, Protractor and scales.
K.Srinivasulu Reddy, SNIST
Set – square
oSet-Squares are available in
450 and 300 600 angles.
From this type of set, we
can easily draw different
angle lines.
K.Srinivasulu Reddy, SNIST
Large size compass
compass
1 mm gap
K.Srinivasulu Reddy, SNIST
Use of Lengthening bar (large compass)
K.Srinivasulu Reddy, SNIST
Small bow compass Divider
K.Srinivasulu Reddy, SNIST
Mini-drafter fixed on
drawing boardK.Srinivasulu Reddy, SNIST
Drawing Clips
& Pins
K.Srinivasulu Reddy, SNIST
K.Srinivasulu Reddy, SNIST
BORDER LINES
Perfectly rectangular working space is determined by drawing
the borderlines
• 20 mm space (top-bottom & right hand edges of the paper
• On the left hand side 30mm
20
20
30
20
185
65
NOTE:
ALL
DIMENSIONS
ARE IN mmK.Srinivasulu Reddy, SNIST
SNIST
65
185
NAME:
TITLE:
ROLL NO:
SHEET NO:
SECTION
:DATE:
TITLE BLOCK
ALL DIMENSIONS ARE IN mm
1
13
K.Srinivasulu Reddy, SNIST
DRAWING PENCILS
Pencils from 9B to 9H. B means soft, H means hard; the higher the number the harder/softer
the pencil is. Use H pencils when you want lighter lines/shading, use B when you want
darker lines/shading.
K.Srinivasulu Reddy, SNIST
Pencils recommended
PENCIL LINES
2H Initial work, construction lines, center line,
hatching lines, locus lines, projectors, extension
lines etc.
H Lettering, dimensioning, figures in the assumed
position and anything given in the problem,
section lines etc.,
HB Aim of the problem (Result), outer lines
K.Srinivasulu Reddy, SNIST
TYPE LINE DESCRIPTION GENERAL APPLICATIONS
A
Continuous thick Visible outlines
B
Continuous thin
(straight or
curved)
Imaginary lines of intersection
Dimension lines
Projection lines
Leader lines
Hatching
Outlines of revolved sections in
place
Short center lines
C
Continuous thin
freehand
Limits of partial or interrupted
views and sections, if the limit is
not a chain thin line
D
Continuous thin
(straight with
zigzags)
Long break line
E
Dashed thick
Hidden outlines
Hidden edges
TYPES OF LINES:
K.Srinivasulu Reddy, SNIST
F
Dashed thin
Hidden outlines
Hidden edges
G
Chain thin
Center line
Lines of symmetry
trajectories
H Chain thin,
thick at ends
and changes
of direction
Cutting planes
J
Chain thick
Indication of lines or surfaces
to which a special requirement
applies
K
Chain thin
double dashed
Outlines of adjacent parts
Alternative and extreme
Positions of movable parts
Centroidal lines
Initial outlines prior to forming
Parts situated in front of the
cutting plane.
K.Srinivasulu Reddy, SNIST
Basic lines :
K.Srinivasulu Reddy, SNIST
LINES:
K.Srinivasulu Reddy, SNIST
K.Srinivasulu Reddy, SNIST
LETTERING
Writing titles, dimensions, notes, and other important
particulars on a drawing is called lettering.
• Single stroke letters (recommended by BIS)
i) Vertical (mostly used) ii) Inclined
• Gothic letters(main titles of ink drawing)
K.Srinivasulu Reddy, SNIST
 Inclined letters are lean to the right, the slope being
75º with the horizontal.
 lettering is generally done in capital letters
K.Srinivasulu Reddy, SNIST
LETTERING
Engineering drawings use single-stroke SANS SERIF
vertical CAPITAL letters because they are highly legible
and quick to draw.
Lower case letters are used for abbreviations like mm, cm
etc.
Left to right & top to bottom
K.Srinivasulu Reddy, SNIST
 Size of letters is measured by height ‘h’ of capital letters
as well as numerals.
 Standard heights: 1.8, 2.5, 3.5, 5, 7,10,14, 20 mm
Main title: 5 or 7 or 10 mm Sub-titles: 3.5 or 5 mm
Dimensions, notes etc: 2.5, 3.5, 5 mm
 Capital letters Size-10:7(height : width) Gap:2 or 3 mm
TOMQVAXY -10:8 ; W-10:12; I - 10:2; others 10:7
 Lower-case: 0.7 times capital letters
 Numericals(0 to 9 except 1)- 10:6 and 1 - 10:2K.Srinivasulu Reddy, SNIST
Different sizes of letters used for different purposes
• The main titles are generally written in 6mm to 8mm size.
• Sub titles in 3mm to 6mm size.
• Notes, dimension figures etc in 3mm to 5mm size.
• The drawing number in the block is written in numerals of
10 mm to 12 mm size.
• The width of the normal letter is about 0.67 times the
height.
• It is often desirable to increase or decrease the size of the
letters in order to make them neat & pleasing to the eyes.
K.Srinivasulu Reddy, SNIST
10
10
10
7 3
K.Srinivasulu Reddy, SNIST
K.Srinivasulu Reddy, SNIST
K.Srinivasulu Reddy, SNIST
Lowercase letters are rarely used in engineering
sketches except for lettering large volumes of notes.
Vertical lowercase letters are used on map drawings,
but very seldom on machine drawings.
K.Srinivasulu Reddy, SNIST
Topics Covered
1.What is Engineering drawing
2.Importance of Engineering drawing
3.Drawing Instruments
4.Lettering
K.Srinivasulu Reddy, SNIST
DIMENSIONING
Which is better ?
K.Srinivasulu Reddy, SNIST
K.Srinivasulu Reddy, SNIST
DIMESIONING
The information of size on the drawing is called “Dimensioning”. It plays
an important role as it helps in giving the correct and accurate size of the
part to be manufactured.
• Drawing without dimensions is meaningless
The elements of dimensioning are:
– Dimension line
– Extension line
– Arrowheads
– Dimension figures
– Leaders
– Notes
*Dimension lines should always be parallel to the line it dimensions.
*Extension line should extend slightly beyond the dimension line.K.Srinivasulu Reddy, SNIS
K.Srinivasulu Reddy, SNIST
K.Srinivasulu Reddy, SNIST
Lines used in Dimensioning
oDimensioning requires the use of
Dimension lines
Extension lines
Leader lines
oAll three line types are drawn thin.K.Srinivasulu Reddy, SNIST
Dimension line
oDimension line: A line terminated by arrowheads, which
indicates the direction and extension of a dimension.
oExtension line: An extension line is a thin solid line that
extends from a point on the drawing to which the
dimension refers.
oLeader Line: A straight inclined thin solid line that is
usually terminated by an arrowhead.K.Srinivasulu Reddy, SNIST
K.Srinivasulu Reddy, SNIST
Dimension line
K.Srinivasulu Reddy, SNIST
Arrowheads
• Arrowheads are drawn between the extension lines if possible.
If space is limited, they may be drawn on the outside.
K.Srinivasulu Reddy, SNIST
LEADERS
EXTENSION LINE
ARROW HEADDIMENSION LINE
K.Srinivasulu Reddy, SNIST
*FILLED IN TYPE
OPEN TYPE
3d
d
d
3d
* ISO recommends
K.Srinivasulu Reddy, SNIST
K.Srinivasulu Reddy, SNIST
Practice
K.Srinivasulu Reddy, SNIST
Practice
SYSTEM OF DIMENSIONING
1.ALIGNED SYSTEM
2.UNI-DIRECTIONAL SYSTEM
 In the aligned system, dimensions are placed
perpendicular to the dimension line in such a way
that it may be read from bottom edge or right hand
edge of the drawing sheet.
 In Unidirectional Method of Dimensioning,
dimension line should be cut at center and
dimensions should be placed in the middle of
dimension lines. K.Srinivasulu Reddy, SNIST
K.Srinivasulu Reddy, SNIST
UNI-DIRECTIONAL SYSTEM
K.Srinivasulu Reddy, SNIST
PRINCIPLES TO BE FOLLOWED IN DIMENSIONING
• The aligned system of giving dimensions should be
followed.
• As far as possible all the dimensions should be placed
outside the object placing dimension lines at least 8mm
from the outlines and from one another.
• The dimensions should never be crowded.
• The diameter can be dimensioned by giving Ф or D before
the measurement.
K.Srinivasulu Reddy, SNIST
•The angles should be dimensioned as shown below
K.Srinivasulu Reddy, SNIST
PARALELL DIMENSIONING
• Dimensions start from a common datum
• Cumulative error can be avoided.
K.Srinivasulu Reddy, SNIST
PROGRESSIVE DIMENSIONING
The datum is indicated by a dot and the dimensions are
placed in line with the projection line
K.Srinivasulu Reddy, SNIST
What are the 6 dimensioning
mistakes?
K.Srinivasulu Reddy, SNIST
Mistakes are:
K.Srinivasulu Reddy, SNIST
Correctly Dimensioned
K.Srinivasulu Reddy, SNIST
COMPUTER AS A DRAFTING TOOL
Most people who create technical drawings use
CAD. The advantages include accuracy, speed, and
the ability to present spatial and visual information in
a variety of ways. Even the most skilled CAD users
need to also be skilled in
freehand sketching, to quickly get
ideas down on paper.
One benefit of CAD is the ability to draw perfectly straight
uniform lines and other geometric elements. Making
changes to a CAD drawing takes about a tenth the time that
it takes to edit a drawing by hand.K.Srinivasulu Reddy, SNIST
Summary
Border lines are to be drawn with HB pencil.
Title block and main figures are to be drawn with HB pencil.
Arrow heads, dimensioning and lettering should be done with HB Pencil.
Construction lines, guide lines and dimension lines should be drawn with
2H pencil lightly.
Arcs and circles are to be drawn with HB pencil lead.
The main title of the drawing should be 5 – 6 mm height and all other sub
titles should be of 3 – 4 mm height.
The dimension should be given above the dimension line at the center-
Aligned dimensioning system.
K.Srinivasulu Reddy, SNIST
For vertical and inclined lines the dimensioning are given in such
a way that they can be read clearly from the right hand side of the
sheet.
All letters should be uniform in size, slope and intensity.
Leave one letter gap between words.
All dimensions are in mm.
Summary Contd..
K.Srinivasulu Reddy, SNIST
K.Srinivasulu Reddy, SNIST
Title: LETTERING AND DIMENSIONING
1. Print the letters A to Z (h=10 mm).
(Gap between letters =2mm Width of all letters: 6mm except M,
W and I. For M,W the width is 8mm and for I the width is 2mm)
2. Print letters a to z (h=10)
3. Print the Numerals 0 to 9, (h=10mm)
Exercise: Sheet No.1 (A)
K.Srinivasulu Reddy, SNIST
1.ENGINEERING DRAWING IS THE LANGUAGE OF ENGINEERS
2. SCIENTISTS STUDY THE WORLD AS IT IS!
ENGINEERS CREATE THE WORLD THAT NEVER HAS BEEN!!
3.PLAN YOUR WORK AND WORK YOUR PLAN
4.STANDARDS CONNECT THE WORLD
5.PRACTICE MAKES YOU PERFECT
6.THERE IS NOTHING PERMANENT EXCEPT CHANGE
N.D.Bhatt Exercise-1:
Fig.1-38(a to f) : 6 Figures
Fig 1-40 (a) 1-40(b): 2 Figures
Exercise Sheet 1(A) Contd..
K.Srinivasulu Reddy, SNIST
Polygons
• A Polygon is any plane figure bounded by straight sides.
• If the polygon has equal angles and equal sides, it can be inscribed in or
circumscribed around a circle and is called a regular polygon.
ii) Construction of Polygons
K.Srinivasulu Reddy, SNIST
K.Srinivasulu Reddy, SNIST
Why Geometric Construction
• Construction of primitive geometric forms (points, lines and planes etc.) that
serve as the building blocks for more complicated geometric shapes.
K.Srinivasulu Reddy, SNIST
• Draw a line MO at any convenient angle (preferably an acute angle) from point
M.
• From M and along MO, cut off with a divider equal divisions (say three) of any
convenient length.
• Draw a line joining NO.
• Draw lines parallel to NO through the remaining points on line MO. The
intersection of these lines with line MN will divide the line into 3 equal parts.K.Srinivasulu Reddy, SNIST
K.Srinivasulu Reddy, SNIST
Bisecting an angle
Angle BAC is to be bisected
 From intersection points C & B,
strike equal arcs r with radius
slightly larger than half BC, to
intersect at D
 Draw line AD, which bisect the
angle
 Strike large arc R
K.Srinivasulu Reddy, SNIST
Drawing any polygon, when side is given
1. Draw AB = given length of polygon
2. At B, Draw BP perpendicular & = AB
3. Draw Straight line AP
4. With center B and radius AB, draw arc AP.
5. The perpendicular bisector of AB meets straight line
AP and arc AP in 4 and 6 respectively.
6. Draw circles with centers as 4, 5,& 6 and radii as 4B,
5B, & 6B and inscribe a square, pentagon, &
hexagon in the respective circles.
7. Mark point 7, 8, etc with 6-7,7-8,etc. = 4-5 to get the
centers of circles of heptagon and etc.
General method-I
K.Srinivasulu Reddy, SNIST
 Draw circles by taking the points 4,5,6,7,8 as center points
and divide the respective circles with the compass
measurement equal to the distance of the line AB.
 Connect the divided points of respective circles in sequence
with straight lines to get respective polygons. Like Square,
Pentagon, Hexagon, Heptagon and Octagon.
K.Srinivasulu Reddy, SNIST
K.Srinivasulu Reddy, SNIST
1.Draw a semi circle with side of polygon as radius
2.Divide the semi circle into n equal parts.
Angle = 180/no. of sides
3. Join A and 2
4. Extend the lines from A3, A4 etc.
5. With length of side from 2 cut on the line extended A3 etc.
Heptagon
General Method-II
K.Srinivasulu Reddy, SNIST
Inscription and superscription of Polygons
When diameter is given
K.Srinivasulu Reddy, SNIST
Draw the circle with diameter AB.
Divide AB in to “n” equal parts.
Number them.
With center A & B and radius AB,
draw arcs to intersect at P.
Draw line P2 and produce it to
meet the circle at C.
AC is the length of the side of the
polygon.
K.Srinivasulu Reddy, SNIST
Side of polygon horizontal Diameter verticalK.Srinivasulu Reddy, SNIST
K.Srinivasulu Reddy, SNIST
Superscription/Describe/Circumscribe a Polygon
When Circle diameter is given
Flats are horizontal
(Vertical diameter)
Flats are vertical
(Horizontal diameter)K.Srinivasulu Reddy, SNIST
1.Construct pentagon, hexagon and octagon of side 35
mm in the same diagram adopting inscription of circle
method.
Exercise: Sheet No. 1(B)
K.Srinivasulu Reddy, SNIST
2.Inscribe a Decagon in the circle of radius 30mm.
(Inscribe a polygon in a given circle)
Pentagon
K.Srinivasulu Reddy, SNIST
DecagonK.Srinivasulu Reddy, SNIST
3.Construct a regular hexagon, given the distance
across the flats as 60mm.
Dia. of inscribed circle in the hexagon is nothing but
distance across flats. Hence draw circle of Ø 60
(Super scribe)
K.Srinivasulu Reddy, SNIST
4.Construct a regular hexagon given distance
across corners is 50mm.
Diameter of circumscribing circle is nothing but
distance across corners. Hence draw a circle of Ø 50
( Inscribe )
K.Srinivasulu Reddy, SNIST
5.Describe a hexagon about a circle of 60 mm dia.
With one side a)vertical b)horizontal
K.Srinivasulu Reddy, SNIST
6. Super scribe a regular octagon about a given
circle of 70 mm dia.
K.Srinivasulu Reddy, SNIST
7. Divide a of a line of 90 into 8 equal parts
K.Srinivasulu Reddy, SNIST
8. Divide a circle into 12 and 8 parts.
K.Srinivasulu Reddy, SNIST
Practical Methods
K.Srinivasulu Reddy, SNIST
Inscribe a circle inside a regular polygon
• Bisect any two adjacent internal angles
of the polygon (angle bisection).
• From the intersection of these lines,
draw a perpendicular to any one side of
the polygon (say OP). With OP as radius,
draw the circle with O as center.
K.Srinivasulu Reddy, SNIST
Title: PLANE FIGURES
Sheet No.1 (C)
Q. Construct the following:
i) Equilateral triangle of side 50mm by using compass (Equilateral triangle)
ii) Isosceles triangle of base 50mm and altitude 70mm.
iii) Rectangle of length 60mm and width 40mm
iv) Square of side 50mm
v) Rhombus of diagonal 80mm & 50mm.
vi) Regular pentagon side 30mm
vii) Regular hexagon of side 30mm.
viii) Regular octagon of side 20mm
Exercise- Sheet No.1(C)
K.Srinivasulu Reddy, SNIST
Title: INSCRIPTION OF POLYGONS
1. Inscribe the following polygon in a circle of diameter 70mm.
i) Pentagon 720
ii) Hexagon 600
iii) Heptagon 520
iv) Octagon 450
Exercise- Sheet No.2(A)
K.Srinivasulu Reddy, SNIST
Title: DESCRIPTION OF POLYGONS
Sheet No.2(B)
1. Describe the following polygons about a circle of diameter
60mm.
i) Square
ii) Pentagon
iii) Hexagon
iv) Octagon
K.Srinivasulu Reddy, SNIST
 Similarly in case of tiny objects dimensions must be increased for
above purpose. Hence this scale is called ENLARGING SCALE.
Here the ratio called representative factor(RF) is more than unity.
R.F. = LENGTH OF DRAWING in cm
ACTUAL LENGTH in cm
 Dimensions of large objects must be reduced to accommodate on
standard size drawing sheet. This reduction creates a scale of that
reduction ratio, which is generally a fraction. Such a scale is called
REDUCING SCALE and that ratio is called REPRESENTATIVE
FRACTION (RF) is less than unity.
iii) Scales
K.Srinivasulu Reddy, SNIST
Scale recommended for use in ED
Full
scale
Reduced
scale
Enlarged
scale
1 : 1 1 : 2 1 : 20 10 : 1
1 : 2.5 1 : 50 5 : 1
1 : 5 1 : 100 2 : 1
1 : 10 1 : 200
FOR FULL SIZE SCALE
R.F.=1 OR ( 1:1 )
MEANS DRAWING
& OBJECT ARE OF
SAME SIZE.
K.Srinivasulu Reddy, SNIST
iv) Curves used in Engineering Practice and
their construction
CONIC SECTIONS
K.Srinivasulu Reddy, SNIST
1.Conic Sections
2.Cycloidal Curves
3.Involutes
4.Evolutes
5.Spirals
TYPES OF CURVES
K.Srinivasulu Reddy, SNIST
1.CONIC SECTIONS
These are the intersections of a right regular cone,
by a cutting plane in different positions relative to
the axis of the cone.
K.Srinivasulu Reddy, SNIST
ELLIPSE -Definition
A regular oval shape, traced by a
point moving in a plane so that
the sum of its distances from two
fixed points (the foci) is constant,
or resulting when a cone is cut by
an oblique plane which does not
intersect the base.
It cuts all generators so we get
closed Ellipse.
α > θ
α
Base
B
B
K.Srinivasulu Reddy, SNIST
It is a curve traced by a
point moving such that at any
position, its distance from a
fixed point (focus) is always
less to its distance from fixed
straight line (directrix).
e<1
Ellipse in Nature K.Srinivasulu Reddy, SNIST
PARABOLA -Definition
It is a curve traced by a point moving such
that at any position its distance from a fixed
point (focus) is always equal to its distance
from fixed straight line ( directrix).
e=1
Cutting plane angle is say α , Cone Apex
half angle is say Ѳ then for Parabola α = Ѳ.
Section plane A-A parallel to a generator
It is not closed curve.
The size of parabola depends upon the
distance of the section plane from the
generator.
Base
K.Srinivasulu Reddy, SNIST
Parabola Applications
K.Srinivasulu Reddy, SNIST
K.Srinivasulu Reddy, SNIST
HYPERBOLA -Definition
 It is a curve traced by a point moving such that
at any position, its distance from a fixed point
(focus) is always greater to its distance from
fixed straight line i. e directrix. e>1
 Cutting plane angle is say α , Cone Apex half
angle is say Ѳ then for Parabola α < Ѳ.
 Even α = 0,provided section plane parallel to axis &
not passing through the apex of the cone is also
Hyper Parabola.
 If section plane passes through Apex the
section produced is Isosceles Triangle.
 If double cone is cut by section plane we get
symmetric Hyperbola.
Axis
Base
Ѳ
α
K.Srinivasulu Reddy, SNIST
Cooling Towers of Nuclear
Reactors and Coal-fired Power
Plants
The shadow of a lampshade or a flashlig
Hyperbola Applications
K.Srinivasulu Reddy, SNIST
K.Srinivasulu Reddy, SNIST
CONSTRUCTION OF ELLIPSE
BY
DIFFERENT METHODS
K.Srinivasulu Reddy, SNIST
Conics: The sections obtained by the
intersection of a right circular cone by a
plane in different positions relative to
the axis of the cone are called conics.
Ellipse: Inclined to axis, cuts all the generators on one side of
apex, not passing through base
Parabola: Inclined to axis, parallel to one of the generators,
passing through base
Hyperbola: Not parallel to generator, passing through base
Rectangular hyperbola: cutting plane is parallel to the axis of
cone(but not pass through Apex)K.Srinivasulu Reddy, SNIST
• Conic : The locus of a point moving in a plane in such a way
that the ratio of it distances from a fixed straight line is
always constant .
• The fixed point is called FOCUS.
• The fixed line is DIRECTRIX.
ECCENTRICITY (e) = The ratio of Distance of the point
from the Focus to the Distance of the point from the
Directrix.
Ellipse : e < 1
Parabola: e = 1
Hyperbola: e > 1 K.Srinivasulu Reddy, SNIST
K.Srinivasulu Reddy, SNIST
ELLIPSE
K.Srinivasulu Reddy, SNIST
B
D
I
R
E
C
T
R
I
X
AXIS
Focus from Directrix is 50 mm
F
E = 2/3
C
V
VE = VF
E
1
1’
2
2’
3
3’
4
4’
F = Centre , 1- 1’ = Radius, cut 1-1’ at
P1
F = Centre , 2- 2’ = Radius, cut 2-2’ at
P2P1
P2
P3
P4
F = Centre , 3- 3’ = Radius, cut 3-3’ at
P3F = Centre , 4- 4’ = Radius, cut 4-4’ at
P4
A
K.Srinivasulu Reddy, SNIST
Problem: Construct an Ellipse, with distance of the focus from the directrix as
50 and eccentricity as 2/3. Also draw normal and tangent to the curve at a point
40 from the directrix.
Directrix
K.Srinivasulu Reddy, SNIST
K.Srinivasulu Reddy, SNIST
Eccentricity method-Construction procedure
1. Draw the axis AB and directrix CD at right angle to each other.
2. Mark focus F1 on the axis such that AF1=50
3. Divide AF1 into 5 equal parts.
4. Locate the vortex V on the third division from point A.
5. Draw a line VE perpendicular to AB such that VE=Vf1
6. Join A,E and extend. By construction VE/VA = VF/VA=2/3 the eccentricity.
7. Mark a number of points 1,2,3 …. To the right of V on the axis which need not be
equidistant.
K.Srinivasulu Reddy, SNIST
8. Through the points 1,2,3… draw lines perpendicular to the axis and to meet the line
AE extended at 1’ , 2’ , 3’ etc.
9. With centre F1 and radius 1-1’ draw arcs intersecting the line through 1 at P1 and P1’.
P1 and p1’ are the points on the ellipse, because the distance of P1 from f1 is 1-1’
and from CD, it is A-1 and 1-1’/A-1 =VE/VA=2/3 the eccentricity.
10. Similarly locate the points P2,P2’: P3,P3’etc., on either side of the axis.
11. Join the points by a smooth curve, forming the required ellipse.
K.Srinivasulu Reddy, SNIST
A B
C
D
F1 F21 2 3
B1
A1
A1
B1
O
P1
P1
P1
P1
½ AB
P2
P2
P2
P2
K.Srinivasulu Reddy, SNIST
Problem: The major and minor axes of an Ellipse are 120 and 80. Draw an Ellipse.
Construction of an Ellipse –Foci/Arcs of circles method
F1, F2 centers; A1,B1 radius. CF1= CF2=½ AB=AO
PF1 + PF2=AB=Major axis
CF1 + CF2= ABor
K.Srinivasulu Reddy, SNIST
Construction Procedure
1. Draw the major and minor axes and locate the centre O.
2. With centre C or D and radius OA(=OB) draw arcs intersecting the major axis at F1 and F2 the
foci.
3. Mark a number of points 1,2,3 etc., between F1 and O which need not be equidistant.
4. With centers F1 and F2 and radii A-1 and B-1 respectively, draw arcs intersecting at points P1
and P1’.
5. With centers F1 and F2 and radii B-1 and A-1 respectively, draw arcs intersecting at points Q1
and Q1’.
6. Repeat the steps 4 and 5 with the remaining points 2,3,4 etc., and obtain additional points on
the curve.
7. Join all the points we get ellipse.
K.Srinivasulu Reddy, SNIST
A B
C
D
O
P1
P2
P3
P4
K.Srinivasulu Reddy, SNIST
Problem: The major and minor axes of an Ellipse are 120 and 80. Draw an Ellipse.
Construction of Ellipse-Concentric circles method
K.Srinivasulu Reddy, SNIST
Construction procedure
1.Draw the major and minor axes and locate the centre O.
2.With centre O and major and minor axes as diameters draw
two concentric circles.
3.Divide both the circles into the same number of parts say 12 by
radial lines.
4.Considering radial line O-1’-1 draw a horizontal line from 1’
and vertical line from 1 intersecting at P1.
5.Repeat the construction through all the points and obtain
P2,P3, etc.,
6. Join all the points we get ellipse.
K.Srinivasulu Reddy, SNIST
A B
C
D
O
1’
2’
3’
4’
1 2 3 4
P1
P22
P3
P4 01
02
03
04
P
5 P
6
P7
P8
P12
K.Srinivasulu Reddy, SNIST
Problem: The major and minor axes of an Ellipse are 120 and 80. Draw an Ellipse.
K.Srinivasulu Reddy, SNIST
Construction Procedure-Oblong method
1.Draw the major and minor axes and locate the centre O.
2.Draw the rectangle KLMN passing through D,B,C and A.
3.Divide AO and AN into the same equal number of points.
4.Join C with the points 1’,2’, and 3’.
5.Join D with 1,2 and 3 and extend till they meet the above lines C-1’, C-2’ and C-3’
respectively. At points P1 ,P2 and P3.
6.Repeat the steps 3 to 5 and obtain the points in the remaining quadrants.
7.Join all the points we get ellipse.
K.Srinivasulu Reddy, SNIST
Problem:
A parallelogram has sides 100 and 80,at an angle of 700 . Inscribe an Ellipse
in the parallelogram. Find the major and minor axes of the curve.
K.Srinivasulu Reddy, SNIST
Construction Procedure
1.Draw the parallelogram KLMN of given sides and inclined. Two axes EF and GH
are called the conjugate axes (diameters).
2.Divide EO and EN into the same number of equal parts and number the
division points.
3.Join G with 1’,2’ and 3’.
4.Join H with 1,2,3 and extend till these meet the lines G1’,G2’ and G3’ at P1,P2
and P3 respectively.
5.Repeat the steps 2 to 4 and obtain the points in the remaining quadrants.
K.Srinivasulu Reddy, SNIST
6.With O as centre and OG as radius, draw the semi-circle cutting the ellipse at a
point I
7.Draw the line GI
8.Through O draw a line parallel to GI and cutting the ellipse at points C and D. CD is
the minor axis
9.Through O, draw a line perpendicular to CD and cutting the ellipse at points A and
B. AB is the major axis
K.Srinivasulu Reddy, SNIST
Construct an ellipse when the major axis is 120 and the
distance between the foci is 108. Determine the length of the
minor axis.
Tip: OF1=OF2= AB/2= 108/2=56
OA as radius, F1 or F2 as centers, draw an arc to cut the
ellipse, which is minor axis.
Exercise
K.Srinivasulu Reddy, SNIST
CONSTRUCTION OF PARABOLA
BY
DIFFERENT METHODS
K.Srinivasulu Reddy, SNIST
Problem: Construct a Parabola , with the distance of the focus from the directrix as 50. Also draw normal and
Tangent to the Curve, at a point 40 from the directrix.
Parabola-Eccentricity Method
K.Srinivasulu Reddy, SNIST
Parabola- General Method
K.Srinivasulu Reddy, SNIST
Construction Procedure
1. Draw the axis AB and the directrix CD at right angles to each other.
2. Mark the focus F on the axis such that AF=50.
3. Locate vortex V on AB such that AV=VF=25.
4. Draw a line VE , perpendicular to AB such that VE=VF.
5. Join A,E and extend. By construction VE/VA = VF/VA = 1.
6. Locate a number of points 1,2,3,etc., to the right of V on the axis, which need
not be equidistant.
7. Through the points 1,2,3, etc., draw lines perpendicular to the axis and to meet
the line AE extended at 1’ , 2’ , 3’, etc.
8. With center F and radius 1-1’ draw arcs intersecting the line through 1 at P1 and
p1’. These p1 and p1’ are the points on the Parabola because the distance of p1
(p1’) from F is 1-1’ and from CD it is A-1 and 1-1’/A-1 = VE/VA = VF/VA = 1.
9. Similarly locate the points P2,P2’ : P3,P3’: etc., on either side of the axis.
10.Join the points by a smooth curve , forming the required parabola.
K.Srinivasulu Reddy, SNIST
Construction Procedure –Tangent and Normal
1.Locate the point M which is at 40 from the directrix.
2.Join M to F .
3.Draw a line through F perpendicular to MF meeting the directrix at T.
4.Join T and M and extend ,T-T is the tangent to the Parabola.
5.Draw a line perpendicular to tangent T-T name as N-N is the normal to the
Parabola.
K.Srinivasulu Reddy, SNIST
Problem: Construct a parabola ,with the length of base as 60 and axis 30 long. Also , draw a
tangent to the curve at a point 25 from the base.
K.Srinivasulu Reddy, SNIST
A BC
D
P1
P2
1
2
1’ 2’
P4
P3
FE
K.Srinivasulu Reddy, SNIST
A B
O
E
F
1
2
3
4
5
6 1’
2’
3’
4’
5’
6’
K.Srinivasulu Reddy, SNIST
Construction Procedure
1. Draw the base AB=60 and the axis CD=30 such that CD is perpendicular
bisector to AB.
2. Produce CD to E such that DE=CD.
3. Join E ,A and E,B. These are the tangents to the Parabola at A and B.
4. Divide AE and BE into the same number of equal parts and number the
points.
5. Join 1-1’ , 2-2’ , 3-3’, etc., forming the tangents to the required parabola.
A smooth curve passing through A,D and B and tangential to the above line is
the required Parabola.
K.Srinivasulu Reddy, SNIST
Construction Procedure –Tangent and Normal
1. Locate the point M which is 25 from the base.
2. Draw a horizontal line through M meeting the axis at F.
3. Mark G on the extension of the axis such that DG=FD.
4. Join G and M and extend ,forming the tangent to the
Parabola at M.
K.Srinivasulu Reddy, SNIST
Problem: Construct a parabola with the base 60 and length of the axis 40. Draw a tangent to
the curve at a point 20 from the base. Also, locate the focus and directrix to the Parabola.
K.Srinivasulu Reddy, SNIST
Construction Procedure
1. Draw the base AB=60 and the axis CD=40 such that CD is perpendicular
bisector to AB.
2. Construct the rectangle ABKL passing through D.
3. Divide AC and AL into the same number of equal parts and number the
points.
4. Join 1,2,3 to D.
5. Through 1’,2’,3’ draw lines parallel to axis intersecting the lines
1-D ,2-D,3-D at P1,P2 and P3 respectively.
6. Parabola is symmetric so draw the remaining half portion.
Join all the points we get Parabola K.Srinivasulu Reddy, SNIST
Problem:
Construct a Parabola in a parallelogram of sides 100 and 60, and with an included angle of 750 .
K.Srinivasulu Reddy, SNIST
Construction Procedure
1. Draw the base AB=60 and the axis CD=40 such that CD is perpendicular
bisector to AB.
2. Construct the rectangle ABKL passing through D.
3. Divide AC and AL into the same number of equal parts and number the
points.
4. Join 1,2,3 to D.
5. Through 1’,2’,3’ draw lines parallel to axis intersecting the the lines
1-D ,2-D,3-D at P1,P2 and P3 respectively.
6. Parabola is symmetric so draw the remaining half portion.
Join all the points we get Parabola
K.Srinivasulu Reddy, SNIST
CONSTRUCTION OF HYPERBOLA
BY
DIFFERENT METHODS
K.Srinivasulu Reddy, SNIST
Problem:
Construct a Hyperbola ,
with the distance
between the focus and
the directrix as 50 and
eccentricity as 3/2. Also
draw normal and
tangent to the curve at a
point 30 from the
directrix.
Construction of Hyperbola-
Eccentricity method
K.Srinivasulu Reddy, SNIST
1.Draw the axis AB and the directrix CD at right angle to each other.
2.Mark focus F on the axis such that AF=50.
3.Divide AF in 5 equal parts.
4.Locate the vertex V on the second division point from A.
5.Draw a line VE perpendicular to AB such that VE=VF.
6.Join A,E and extend. By construction , VE/VA = VF/VA = 3/2 ,the eccentricity.
7. Locate a number of points 1,2,3,etc., to the right of V on the axis, which need not be
equidistant.
7. Through the points 1,2,3, etc., draw lines perpendicular to the axis and to meet the line
AE extended at 1’ , 2’ , 3’, etc.
8. With center F and radius 1-1’ draw arcs intersecting the line through 1 at P1 and p1’.
These p1 and p1’ are the points on the Hyperbola.
9. Similarly locate the points P2,P2’ : P3,P3’: etc., on either side of the axis.
10.Join the points by a smooth curve , forming the required Hyperbola.
Construction Procedure
K.Srinivasulu Reddy, SNIST
Problem:
Construct a Hyperbola with
its foci 70 apart and the
major axis 45. Draw a
tangent to the curve at a
point 20 from the focus.
Also determine the
eccentricity of the curve.
K.Srinivasulu Reddy, SNIST
Construction Procedure
1.Draw the axis AB and locate a point O on it.
2.Locate the foci F1,F2 such that F1F2=70 and locate vertices V1,V2 such that V1V2 =45 on AB.
3.Mark a number of points 1,2,3 etc., on AB to the right of F2 which need not be equidistant .
4.With center F1 and radius v1-1 draw arcs on either side if the transverse axis.
5. With center F2 and radius v2-1 draw arcs intersecting the above arcs at P1 and P1’.
6. With center F2 and radius v1-1 draw arcs on either side if the transverse axis.
7. With center F1 and radius v2-1 draw arcs intersecting the above arcs at Q1 and Q1’.
8.Repeat the steps 4 to 7 and obtain the two branches of the hyperbola. K.Srinivasulu Reddy, SNIST
Construction Procedure –Tangent and Normal
1. Locate the point M which is 20 from the focus say F2.
2. Then join M to foci F1 and F2..
3.Draw a line T-T bisecting <F1MF2 forming the required
tangent.
K.Srinivasulu Reddy, SNIST
Eccentricity calculation
Eccentricity e = OF1/OV1
= OF2/OV2
= 35/22.5
e = 1.56
K.Srinivasulu Reddy, SNIST
Rectangular Hyperbola
 A hyperbola for which the asymptotes
are perpendicular, also called an equilateral
hyperbola or right hyperbola.
Ex: Boyle’s law PV=C
As the pressure
increases, its volume
decreases.
K.Srinivasulu Reddy, SNIST
Construct a Rectangular Hyperbola ,when a point P on it is at a
distance of 18 and 34 from two asymptotes. Also draw a tangent to
the curve at a point 20 from an asymptote .
K.Srinivasulu Reddy, SNIST
Construction Procedure
1.Draw the asymptotes OA and OB at right angle to each other and locate the
given point P.
2.Draw the lines CD and EF passing through point P and parallel to OA and OB
respectively.
3.Locate a number of points 1,2,3,etc., along the line CD , which need not be
equi -distant.
4.Join 1,2,3,etc., to O and extend if necessary , till these lines meet the line EF
at points 1’ , 2’ , 3’, etc.
5.Draw lines through 1,2,3,etc.,parallel to EF and through 1’ , 2’ , 3’, etc. parallel
to CD to intersect at P1,P2,P3,etc.
K.Srinivasulu Reddy, SNIST
Tangent to rectangular Hyperbola
1.Locate Point M on the curve by drawing a line GM parallel to OA and at a distance of 20 from it.
2.Locate the point H on OB such that GH = OG.
3.The line HT passing through M is the required tangent to the curve.
K.Srinivasulu Reddy, SNIST
Projection Methods:
Notation
K.Srinivasulu Reddy, SNIST
Projection Symbol
Projection Symbol
First Angle European
Third Angle American
K.Srinivasulu Reddy, SNIST
k.srinivasulureddy@gmail.com

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Engineering drawing unit 1-snist

  • 1. Engineering Drawing UNIT- I K. Srinivasulu Reddy Mechanical Engineering Department Introduction to Engineering Drawing K.Srinivasulu Reddy, SNIST
  • 2. UNIT-I i)Introduction to Engineering Drawing: Drawing instruments and their uses, types of lines, use of pencils, lettering, rules of dimensioning. ii)Construction of Polygons: Inscription & superscription of polygons given the diameter of circle. iii)Introduction to scales(Theory only):Scales used in Engineering Practice and concept of Representative Fraction. iv)Curves used in Engineering Practice & their Construction: Conic sections including the rectangular Hyperbola-General Method only.K.Srinivasulu Reddy, SNIST
  • 3. TEXT BOOKS: 1. Engineering Drawing by N. D. Bhatt / Charotar Publications 2. Engineering Drawing by K.L.Narayana and Kannaiah/Scietech Publishers REFERENCES: 1. Engineering Drawing,K.Venugopal/G.Sreekanjana, New Age International Publishers 2. Engineering Drawing,B.V.R Gupta,M.Roja Roy I.K. International Publishing K.Srinivasulu Reddy, SNIST
  • 4. •Why and What is Engineering Drawing? •Importance of Engineering Drawing •Standards •List of Drawing instruments •Border lines •Title block •Types of lines •Pencils recommended •Lettering •Dimensioning K.Srinivasulu Reddy, SNIST
  • 5. What is Engineering Drawing ?  Drawing is a graphical Language of an Engineer which communicates ideas and information from one mind to another.  It communicates all needed information from the engineer, who designed a part, to the workers, who will make it. K.Srinivasulu Reddy, SNIST
  • 6. Engineering Drawing is a two dimensional representation of a three dimensional object. It is the graphic language, from which a trained person can visualize the object.  Engineering drawing is formal and precise way of communicating information about the shape, size, features and precision of physical objects.  Engineering Drawing is the universal language of Engineers What is Engineering Drawing Contd.. K.Srinivasulu Reddy, SNIST
  • 7. ENGINEERING DRAWING The classification of Engineering Drawings include: •Building drawing •Machine drawing •Production drawing •Electrical drawing etc. K.Srinivasulu Reddy, SNIST
  • 8. A PICTURE SPEAKS A THOUSAND WORDS In Engineering, a good drawing is worth even more than a thousand words. Importance K.Srinivasulu Reddy, SNIST
  • 9. Why instruments for drawing ? K.Srinivasulu Reddy, SNIST
  • 10. A typical study of the drawing course will lead to the following observations. • It develops a type of discipline in graphical techniques. • It develops ability to analyze and communicate ideas. • It helps in imagining the proper size, shape & form of an object. • It develops capacity of reasoning & judgment. • It helps in the study of other Engineering subjects. K.Srinivasulu Reddy, SNIST
  • 11. STANDARDS • For the convenience of exchange of information and to buy or sell machinery from or to a foreign country and to facilitate interchangeability of parts it is essential to have standard code of practice • Bureau of Indian Standards (BIS) is the National Standards Body formed by the Government of India on 1 April 1987, replacing Indian Standard Institution(ISI) existing earlier. BIS – SP 46 revised as SP – 46 – 2003 • The other foreign standards are –DIN of Germany, BS of Britain • BIS works in association with ISO ( International Standards Organization ) K.Srinivasulu Reddy, SNIST
  • 12. List of drawing instruments • Drawing board • Drawing sheets & Supporting sheet • Mini-drafter • Instrument box • Drawing pencils (2H,H and HB Grades) • Eraser • Clips and Drawing pins/Adhesive tape • Protractor • Paper Napkins or Handkerchief K.Srinivasulu Reddy, SNIST
  • 13. Drawing Board o To make lines on a drawing paper smooth and straight, a polished drawing board is one of the top most requirements. IS 1444:1989 K.Srinivasulu Reddy, SNIST
  • 14. Sizes of drawing boards S. No. Designation Size (mm) 1 B0 1000 X 1500 2 B1 700 X 1000 3 B2 500 X 700 4 B3 350 X 500 Drawing board K.Srinivasulu Reddy, SNIST
  • 15. Drawing sheets IS 10711:2001 • The standard sizes of the drawing sheets recommended by BIS (Bureau of Indian Standards) • A0 = 1 m2 • Drawing sheet A2. • width to length 1: S.NO Sheet designation Trimmed size(mm) Length x Width 1 A0 1189 x 841 2 A1 841 x 594 3 A2 594 x 420 4 A3 420 x 297 5 A4 297 x 210 2 K.Srinivasulu Reddy, SNIST
  • 18. Mini-Drafter This instrument gives faster drawing as it serves the purpose of Set-Square, Protractor and scales. K.Srinivasulu Reddy, SNIST
  • 19. Set – square oSet-Squares are available in 450 and 300 600 angles. From this type of set, we can easily draw different angle lines. K.Srinivasulu Reddy, SNIST
  • 20. Large size compass compass 1 mm gap K.Srinivasulu Reddy, SNIST
  • 21. Use of Lengthening bar (large compass) K.Srinivasulu Reddy, SNIST
  • 22. Small bow compass Divider K.Srinivasulu Reddy, SNIST
  • 23. Mini-drafter fixed on drawing boardK.Srinivasulu Reddy, SNIST
  • 26. BORDER LINES Perfectly rectangular working space is determined by drawing the borderlines • 20 mm space (top-bottom & right hand edges of the paper • On the left hand side 30mm 20 20 30 20 185 65 NOTE: ALL DIMENSIONS ARE IN mmK.Srinivasulu Reddy, SNIST
  • 27. SNIST 65 185 NAME: TITLE: ROLL NO: SHEET NO: SECTION :DATE: TITLE BLOCK ALL DIMENSIONS ARE IN mm 1 13 K.Srinivasulu Reddy, SNIST
  • 28. DRAWING PENCILS Pencils from 9B to 9H. B means soft, H means hard; the higher the number the harder/softer the pencil is. Use H pencils when you want lighter lines/shading, use B when you want darker lines/shading. K.Srinivasulu Reddy, SNIST
  • 29. Pencils recommended PENCIL LINES 2H Initial work, construction lines, center line, hatching lines, locus lines, projectors, extension lines etc. H Lettering, dimensioning, figures in the assumed position and anything given in the problem, section lines etc., HB Aim of the problem (Result), outer lines K.Srinivasulu Reddy, SNIST
  • 30. TYPE LINE DESCRIPTION GENERAL APPLICATIONS A Continuous thick Visible outlines B Continuous thin (straight or curved) Imaginary lines of intersection Dimension lines Projection lines Leader lines Hatching Outlines of revolved sections in place Short center lines C Continuous thin freehand Limits of partial or interrupted views and sections, if the limit is not a chain thin line D Continuous thin (straight with zigzags) Long break line E Dashed thick Hidden outlines Hidden edges TYPES OF LINES: K.Srinivasulu Reddy, SNIST
  • 31. F Dashed thin Hidden outlines Hidden edges G Chain thin Center line Lines of symmetry trajectories H Chain thin, thick at ends and changes of direction Cutting planes J Chain thick Indication of lines or surfaces to which a special requirement applies K Chain thin double dashed Outlines of adjacent parts Alternative and extreme Positions of movable parts Centroidal lines Initial outlines prior to forming Parts situated in front of the cutting plane. K.Srinivasulu Reddy, SNIST
  • 35. LETTERING Writing titles, dimensions, notes, and other important particulars on a drawing is called lettering. • Single stroke letters (recommended by BIS) i) Vertical (mostly used) ii) Inclined • Gothic letters(main titles of ink drawing) K.Srinivasulu Reddy, SNIST
  • 36.  Inclined letters are lean to the right, the slope being 75º with the horizontal.  lettering is generally done in capital letters K.Srinivasulu Reddy, SNIST
  • 37. LETTERING Engineering drawings use single-stroke SANS SERIF vertical CAPITAL letters because they are highly legible and quick to draw. Lower case letters are used for abbreviations like mm, cm etc. Left to right & top to bottom K.Srinivasulu Reddy, SNIST
  • 38.  Size of letters is measured by height ‘h’ of capital letters as well as numerals.  Standard heights: 1.8, 2.5, 3.5, 5, 7,10,14, 20 mm Main title: 5 or 7 or 10 mm Sub-titles: 3.5 or 5 mm Dimensions, notes etc: 2.5, 3.5, 5 mm  Capital letters Size-10:7(height : width) Gap:2 or 3 mm TOMQVAXY -10:8 ; W-10:12; I - 10:2; others 10:7  Lower-case: 0.7 times capital letters  Numericals(0 to 9 except 1)- 10:6 and 1 - 10:2K.Srinivasulu Reddy, SNIST
  • 39. Different sizes of letters used for different purposes • The main titles are generally written in 6mm to 8mm size. • Sub titles in 3mm to 6mm size. • Notes, dimension figures etc in 3mm to 5mm size. • The drawing number in the block is written in numerals of 10 mm to 12 mm size. • The width of the normal letter is about 0.67 times the height. • It is often desirable to increase or decrease the size of the letters in order to make them neat & pleasing to the eyes. K.Srinivasulu Reddy, SNIST
  • 43. Lowercase letters are rarely used in engineering sketches except for lettering large volumes of notes. Vertical lowercase letters are used on map drawings, but very seldom on machine drawings. K.Srinivasulu Reddy, SNIST
  • 44. Topics Covered 1.What is Engineering drawing 2.Importance of Engineering drawing 3.Drawing Instruments 4.Lettering K.Srinivasulu Reddy, SNIST
  • 45. DIMENSIONING Which is better ? K.Srinivasulu Reddy, SNIST
  • 47. DIMESIONING The information of size on the drawing is called “Dimensioning”. It plays an important role as it helps in giving the correct and accurate size of the part to be manufactured. • Drawing without dimensions is meaningless The elements of dimensioning are: – Dimension line – Extension line – Arrowheads – Dimension figures – Leaders – Notes *Dimension lines should always be parallel to the line it dimensions. *Extension line should extend slightly beyond the dimension line.K.Srinivasulu Reddy, SNIS
  • 50. Lines used in Dimensioning oDimensioning requires the use of Dimension lines Extension lines Leader lines oAll three line types are drawn thin.K.Srinivasulu Reddy, SNIST
  • 51. Dimension line oDimension line: A line terminated by arrowheads, which indicates the direction and extension of a dimension. oExtension line: An extension line is a thin solid line that extends from a point on the drawing to which the dimension refers. oLeader Line: A straight inclined thin solid line that is usually terminated by an arrowhead.K.Srinivasulu Reddy, SNIST
  • 54. Arrowheads • Arrowheads are drawn between the extension lines if possible. If space is limited, they may be drawn on the outside. K.Srinivasulu Reddy, SNIST
  • 55. LEADERS EXTENSION LINE ARROW HEADDIMENSION LINE K.Srinivasulu Reddy, SNIST
  • 56. *FILLED IN TYPE OPEN TYPE 3d d d 3d * ISO recommends K.Srinivasulu Reddy, SNIST
  • 59. SYSTEM OF DIMENSIONING 1.ALIGNED SYSTEM 2.UNI-DIRECTIONAL SYSTEM  In the aligned system, dimensions are placed perpendicular to the dimension line in such a way that it may be read from bottom edge or right hand edge of the drawing sheet.  In Unidirectional Method of Dimensioning, dimension line should be cut at center and dimensions should be placed in the middle of dimension lines. K.Srinivasulu Reddy, SNIST
  • 62. PRINCIPLES TO BE FOLLOWED IN DIMENSIONING • The aligned system of giving dimensions should be followed. • As far as possible all the dimensions should be placed outside the object placing dimension lines at least 8mm from the outlines and from one another. • The dimensions should never be crowded. • The diameter can be dimensioned by giving Ф or D before the measurement. K.Srinivasulu Reddy, SNIST
  • 63. •The angles should be dimensioned as shown below K.Srinivasulu Reddy, SNIST
  • 64. PARALELL DIMENSIONING • Dimensions start from a common datum • Cumulative error can be avoided. K.Srinivasulu Reddy, SNIST
  • 65. PROGRESSIVE DIMENSIONING The datum is indicated by a dot and the dimensions are placed in line with the projection line K.Srinivasulu Reddy, SNIST
  • 66. What are the 6 dimensioning mistakes? K.Srinivasulu Reddy, SNIST
  • 69. COMPUTER AS A DRAFTING TOOL Most people who create technical drawings use CAD. The advantages include accuracy, speed, and the ability to present spatial and visual information in a variety of ways. Even the most skilled CAD users need to also be skilled in freehand sketching, to quickly get ideas down on paper. One benefit of CAD is the ability to draw perfectly straight uniform lines and other geometric elements. Making changes to a CAD drawing takes about a tenth the time that it takes to edit a drawing by hand.K.Srinivasulu Reddy, SNIST
  • 70. Summary Border lines are to be drawn with HB pencil. Title block and main figures are to be drawn with HB pencil. Arrow heads, dimensioning and lettering should be done with HB Pencil. Construction lines, guide lines and dimension lines should be drawn with 2H pencil lightly. Arcs and circles are to be drawn with HB pencil lead. The main title of the drawing should be 5 – 6 mm height and all other sub titles should be of 3 – 4 mm height. The dimension should be given above the dimension line at the center- Aligned dimensioning system. K.Srinivasulu Reddy, SNIST
  • 71. For vertical and inclined lines the dimensioning are given in such a way that they can be read clearly from the right hand side of the sheet. All letters should be uniform in size, slope and intensity. Leave one letter gap between words. All dimensions are in mm. Summary Contd.. K.Srinivasulu Reddy, SNIST
  • 73. Title: LETTERING AND DIMENSIONING 1. Print the letters A to Z (h=10 mm). (Gap between letters =2mm Width of all letters: 6mm except M, W and I. For M,W the width is 8mm and for I the width is 2mm) 2. Print letters a to z (h=10) 3. Print the Numerals 0 to 9, (h=10mm) Exercise: Sheet No.1 (A) K.Srinivasulu Reddy, SNIST
  • 74. 1.ENGINEERING DRAWING IS THE LANGUAGE OF ENGINEERS 2. SCIENTISTS STUDY THE WORLD AS IT IS! ENGINEERS CREATE THE WORLD THAT NEVER HAS BEEN!! 3.PLAN YOUR WORK AND WORK YOUR PLAN 4.STANDARDS CONNECT THE WORLD 5.PRACTICE MAKES YOU PERFECT 6.THERE IS NOTHING PERMANENT EXCEPT CHANGE N.D.Bhatt Exercise-1: Fig.1-38(a to f) : 6 Figures Fig 1-40 (a) 1-40(b): 2 Figures Exercise Sheet 1(A) Contd.. K.Srinivasulu Reddy, SNIST
  • 75. Polygons • A Polygon is any plane figure bounded by straight sides. • If the polygon has equal angles and equal sides, it can be inscribed in or circumscribed around a circle and is called a regular polygon. ii) Construction of Polygons K.Srinivasulu Reddy, SNIST
  • 77. Why Geometric Construction • Construction of primitive geometric forms (points, lines and planes etc.) that serve as the building blocks for more complicated geometric shapes. K.Srinivasulu Reddy, SNIST
  • 78. • Draw a line MO at any convenient angle (preferably an acute angle) from point M. • From M and along MO, cut off with a divider equal divisions (say three) of any convenient length. • Draw a line joining NO. • Draw lines parallel to NO through the remaining points on line MO. The intersection of these lines with line MN will divide the line into 3 equal parts.K.Srinivasulu Reddy, SNIST
  • 80. Bisecting an angle Angle BAC is to be bisected  From intersection points C & B, strike equal arcs r with radius slightly larger than half BC, to intersect at D  Draw line AD, which bisect the angle  Strike large arc R K.Srinivasulu Reddy, SNIST
  • 81. Drawing any polygon, when side is given 1. Draw AB = given length of polygon 2. At B, Draw BP perpendicular & = AB 3. Draw Straight line AP 4. With center B and radius AB, draw arc AP. 5. The perpendicular bisector of AB meets straight line AP and arc AP in 4 and 6 respectively. 6. Draw circles with centers as 4, 5,& 6 and radii as 4B, 5B, & 6B and inscribe a square, pentagon, & hexagon in the respective circles. 7. Mark point 7, 8, etc with 6-7,7-8,etc. = 4-5 to get the centers of circles of heptagon and etc. General method-I K.Srinivasulu Reddy, SNIST
  • 82.  Draw circles by taking the points 4,5,6,7,8 as center points and divide the respective circles with the compass measurement equal to the distance of the line AB.  Connect the divided points of respective circles in sequence with straight lines to get respective polygons. Like Square, Pentagon, Hexagon, Heptagon and Octagon. K.Srinivasulu Reddy, SNIST
  • 84. 1.Draw a semi circle with side of polygon as radius 2.Divide the semi circle into n equal parts. Angle = 180/no. of sides 3. Join A and 2 4. Extend the lines from A3, A4 etc. 5. With length of side from 2 cut on the line extended A3 etc. Heptagon General Method-II K.Srinivasulu Reddy, SNIST
  • 85. Inscription and superscription of Polygons When diameter is given K.Srinivasulu Reddy, SNIST
  • 86. Draw the circle with diameter AB. Divide AB in to “n” equal parts. Number them. With center A & B and radius AB, draw arcs to intersect at P. Draw line P2 and produce it to meet the circle at C. AC is the length of the side of the polygon. K.Srinivasulu Reddy, SNIST
  • 87. Side of polygon horizontal Diameter verticalK.Srinivasulu Reddy, SNIST
  • 89. Superscription/Describe/Circumscribe a Polygon When Circle diameter is given Flats are horizontal (Vertical diameter) Flats are vertical (Horizontal diameter)K.Srinivasulu Reddy, SNIST
  • 90. 1.Construct pentagon, hexagon and octagon of side 35 mm in the same diagram adopting inscription of circle method. Exercise: Sheet No. 1(B) K.Srinivasulu Reddy, SNIST
  • 91. 2.Inscribe a Decagon in the circle of radius 30mm. (Inscribe a polygon in a given circle) Pentagon K.Srinivasulu Reddy, SNIST
  • 93. 3.Construct a regular hexagon, given the distance across the flats as 60mm. Dia. of inscribed circle in the hexagon is nothing but distance across flats. Hence draw circle of Ø 60 (Super scribe) K.Srinivasulu Reddy, SNIST
  • 94. 4.Construct a regular hexagon given distance across corners is 50mm. Diameter of circumscribing circle is nothing but distance across corners. Hence draw a circle of Ø 50 ( Inscribe ) K.Srinivasulu Reddy, SNIST
  • 95. 5.Describe a hexagon about a circle of 60 mm dia. With one side a)vertical b)horizontal K.Srinivasulu Reddy, SNIST
  • 96. 6. Super scribe a regular octagon about a given circle of 70 mm dia. K.Srinivasulu Reddy, SNIST
  • 97. 7. Divide a of a line of 90 into 8 equal parts K.Srinivasulu Reddy, SNIST
  • 98. 8. Divide a circle into 12 and 8 parts. K.Srinivasulu Reddy, SNIST
  • 100. Inscribe a circle inside a regular polygon • Bisect any two adjacent internal angles of the polygon (angle bisection). • From the intersection of these lines, draw a perpendicular to any one side of the polygon (say OP). With OP as radius, draw the circle with O as center. K.Srinivasulu Reddy, SNIST
  • 101. Title: PLANE FIGURES Sheet No.1 (C) Q. Construct the following: i) Equilateral triangle of side 50mm by using compass (Equilateral triangle) ii) Isosceles triangle of base 50mm and altitude 70mm. iii) Rectangle of length 60mm and width 40mm iv) Square of side 50mm v) Rhombus of diagonal 80mm & 50mm. vi) Regular pentagon side 30mm vii) Regular hexagon of side 30mm. viii) Regular octagon of side 20mm Exercise- Sheet No.1(C) K.Srinivasulu Reddy, SNIST
  • 102. Title: INSCRIPTION OF POLYGONS 1. Inscribe the following polygon in a circle of diameter 70mm. i) Pentagon 720 ii) Hexagon 600 iii) Heptagon 520 iv) Octagon 450 Exercise- Sheet No.2(A) K.Srinivasulu Reddy, SNIST
  • 103. Title: DESCRIPTION OF POLYGONS Sheet No.2(B) 1. Describe the following polygons about a circle of diameter 60mm. i) Square ii) Pentagon iii) Hexagon iv) Octagon K.Srinivasulu Reddy, SNIST
  • 104.  Similarly in case of tiny objects dimensions must be increased for above purpose. Hence this scale is called ENLARGING SCALE. Here the ratio called representative factor(RF) is more than unity. R.F. = LENGTH OF DRAWING in cm ACTUAL LENGTH in cm  Dimensions of large objects must be reduced to accommodate on standard size drawing sheet. This reduction creates a scale of that reduction ratio, which is generally a fraction. Such a scale is called REDUCING SCALE and that ratio is called REPRESENTATIVE FRACTION (RF) is less than unity. iii) Scales K.Srinivasulu Reddy, SNIST
  • 105. Scale recommended for use in ED Full scale Reduced scale Enlarged scale 1 : 1 1 : 2 1 : 20 10 : 1 1 : 2.5 1 : 50 5 : 1 1 : 5 1 : 100 2 : 1 1 : 10 1 : 200 FOR FULL SIZE SCALE R.F.=1 OR ( 1:1 ) MEANS DRAWING & OBJECT ARE OF SAME SIZE. K.Srinivasulu Reddy, SNIST
  • 106. iv) Curves used in Engineering Practice and their construction CONIC SECTIONS K.Srinivasulu Reddy, SNIST
  • 108. 1.CONIC SECTIONS These are the intersections of a right regular cone, by a cutting plane in different positions relative to the axis of the cone. K.Srinivasulu Reddy, SNIST
  • 109. ELLIPSE -Definition A regular oval shape, traced by a point moving in a plane so that the sum of its distances from two fixed points (the foci) is constant, or resulting when a cone is cut by an oblique plane which does not intersect the base. It cuts all generators so we get closed Ellipse. α > θ α Base B B K.Srinivasulu Reddy, SNIST
  • 110. It is a curve traced by a point moving such that at any position, its distance from a fixed point (focus) is always less to its distance from fixed straight line (directrix). e<1 Ellipse in Nature K.Srinivasulu Reddy, SNIST
  • 111. PARABOLA -Definition It is a curve traced by a point moving such that at any position its distance from a fixed point (focus) is always equal to its distance from fixed straight line ( directrix). e=1 Cutting plane angle is say α , Cone Apex half angle is say Ѳ then for Parabola α = Ѳ. Section plane A-A parallel to a generator It is not closed curve. The size of parabola depends upon the distance of the section plane from the generator. Base K.Srinivasulu Reddy, SNIST
  • 114. HYPERBOLA -Definition  It is a curve traced by a point moving such that at any position, its distance from a fixed point (focus) is always greater to its distance from fixed straight line i. e directrix. e>1  Cutting plane angle is say α , Cone Apex half angle is say Ѳ then for Parabola α < Ѳ.  Even α = 0,provided section plane parallel to axis & not passing through the apex of the cone is also Hyper Parabola.  If section plane passes through Apex the section produced is Isosceles Triangle.  If double cone is cut by section plane we get symmetric Hyperbola. Axis Base Ѳ α K.Srinivasulu Reddy, SNIST
  • 115. Cooling Towers of Nuclear Reactors and Coal-fired Power Plants The shadow of a lampshade or a flashlig Hyperbola Applications K.Srinivasulu Reddy, SNIST
  • 117. CONSTRUCTION OF ELLIPSE BY DIFFERENT METHODS K.Srinivasulu Reddy, SNIST
  • 118. Conics: The sections obtained by the intersection of a right circular cone by a plane in different positions relative to the axis of the cone are called conics. Ellipse: Inclined to axis, cuts all the generators on one side of apex, not passing through base Parabola: Inclined to axis, parallel to one of the generators, passing through base Hyperbola: Not parallel to generator, passing through base Rectangular hyperbola: cutting plane is parallel to the axis of cone(but not pass through Apex)K.Srinivasulu Reddy, SNIST
  • 119. • Conic : The locus of a point moving in a plane in such a way that the ratio of it distances from a fixed straight line is always constant . • The fixed point is called FOCUS. • The fixed line is DIRECTRIX. ECCENTRICITY (e) = The ratio of Distance of the point from the Focus to the Distance of the point from the Directrix. Ellipse : e < 1 Parabola: e = 1 Hyperbola: e > 1 K.Srinivasulu Reddy, SNIST
  • 122. B D I R E C T R I X AXIS Focus from Directrix is 50 mm F E = 2/3 C V VE = VF E 1 1’ 2 2’ 3 3’ 4 4’ F = Centre , 1- 1’ = Radius, cut 1-1’ at P1 F = Centre , 2- 2’ = Radius, cut 2-2’ at P2P1 P2 P3 P4 F = Centre , 3- 3’ = Radius, cut 3-3’ at P3F = Centre , 4- 4’ = Radius, cut 4-4’ at P4 A K.Srinivasulu Reddy, SNIST
  • 123. Problem: Construct an Ellipse, with distance of the focus from the directrix as 50 and eccentricity as 2/3. Also draw normal and tangent to the curve at a point 40 from the directrix. Directrix K.Srinivasulu Reddy, SNIST
  • 125. Eccentricity method-Construction procedure 1. Draw the axis AB and directrix CD at right angle to each other. 2. Mark focus F1 on the axis such that AF1=50 3. Divide AF1 into 5 equal parts. 4. Locate the vortex V on the third division from point A. 5. Draw a line VE perpendicular to AB such that VE=Vf1 6. Join A,E and extend. By construction VE/VA = VF/VA=2/3 the eccentricity. 7. Mark a number of points 1,2,3 …. To the right of V on the axis which need not be equidistant. K.Srinivasulu Reddy, SNIST
  • 126. 8. Through the points 1,2,3… draw lines perpendicular to the axis and to meet the line AE extended at 1’ , 2’ , 3’ etc. 9. With centre F1 and radius 1-1’ draw arcs intersecting the line through 1 at P1 and P1’. P1 and p1’ are the points on the ellipse, because the distance of P1 from f1 is 1-1’ and from CD, it is A-1 and 1-1’/A-1 =VE/VA=2/3 the eccentricity. 10. Similarly locate the points P2,P2’: P3,P3’etc., on either side of the axis. 11. Join the points by a smooth curve, forming the required ellipse. K.Srinivasulu Reddy, SNIST
  • 127. A B C D F1 F21 2 3 B1 A1 A1 B1 O P1 P1 P1 P1 ½ AB P2 P2 P2 P2 K.Srinivasulu Reddy, SNIST
  • 128. Problem: The major and minor axes of an Ellipse are 120 and 80. Draw an Ellipse. Construction of an Ellipse –Foci/Arcs of circles method F1, F2 centers; A1,B1 radius. CF1= CF2=½ AB=AO PF1 + PF2=AB=Major axis CF1 + CF2= ABor K.Srinivasulu Reddy, SNIST
  • 129. Construction Procedure 1. Draw the major and minor axes and locate the centre O. 2. With centre C or D and radius OA(=OB) draw arcs intersecting the major axis at F1 and F2 the foci. 3. Mark a number of points 1,2,3 etc., between F1 and O which need not be equidistant. 4. With centers F1 and F2 and radii A-1 and B-1 respectively, draw arcs intersecting at points P1 and P1’. 5. With centers F1 and F2 and radii B-1 and A-1 respectively, draw arcs intersecting at points Q1 and Q1’. 6. Repeat the steps 4 and 5 with the remaining points 2,3,4 etc., and obtain additional points on the curve. 7. Join all the points we get ellipse. K.Srinivasulu Reddy, SNIST
  • 131. Problem: The major and minor axes of an Ellipse are 120 and 80. Draw an Ellipse. Construction of Ellipse-Concentric circles method K.Srinivasulu Reddy, SNIST
  • 132. Construction procedure 1.Draw the major and minor axes and locate the centre O. 2.With centre O and major and minor axes as diameters draw two concentric circles. 3.Divide both the circles into the same number of parts say 12 by radial lines. 4.Considering radial line O-1’-1 draw a horizontal line from 1’ and vertical line from 1 intersecting at P1. 5.Repeat the construction through all the points and obtain P2,P3, etc., 6. Join all the points we get ellipse. K.Srinivasulu Reddy, SNIST
  • 133. A B C D O 1’ 2’ 3’ 4’ 1 2 3 4 P1 P22 P3 P4 01 02 03 04 P 5 P 6 P7 P8 P12 K.Srinivasulu Reddy, SNIST
  • 134. Problem: The major and minor axes of an Ellipse are 120 and 80. Draw an Ellipse. K.Srinivasulu Reddy, SNIST
  • 135. Construction Procedure-Oblong method 1.Draw the major and minor axes and locate the centre O. 2.Draw the rectangle KLMN passing through D,B,C and A. 3.Divide AO and AN into the same equal number of points. 4.Join C with the points 1’,2’, and 3’. 5.Join D with 1,2 and 3 and extend till they meet the above lines C-1’, C-2’ and C-3’ respectively. At points P1 ,P2 and P3. 6.Repeat the steps 3 to 5 and obtain the points in the remaining quadrants. 7.Join all the points we get ellipse. K.Srinivasulu Reddy, SNIST
  • 136. Problem: A parallelogram has sides 100 and 80,at an angle of 700 . Inscribe an Ellipse in the parallelogram. Find the major and minor axes of the curve. K.Srinivasulu Reddy, SNIST
  • 137. Construction Procedure 1.Draw the parallelogram KLMN of given sides and inclined. Two axes EF and GH are called the conjugate axes (diameters). 2.Divide EO and EN into the same number of equal parts and number the division points. 3.Join G with 1’,2’ and 3’. 4.Join H with 1,2,3 and extend till these meet the lines G1’,G2’ and G3’ at P1,P2 and P3 respectively. 5.Repeat the steps 2 to 4 and obtain the points in the remaining quadrants. K.Srinivasulu Reddy, SNIST
  • 138. 6.With O as centre and OG as radius, draw the semi-circle cutting the ellipse at a point I 7.Draw the line GI 8.Through O draw a line parallel to GI and cutting the ellipse at points C and D. CD is the minor axis 9.Through O, draw a line perpendicular to CD and cutting the ellipse at points A and B. AB is the major axis K.Srinivasulu Reddy, SNIST
  • 139. Construct an ellipse when the major axis is 120 and the distance between the foci is 108. Determine the length of the minor axis. Tip: OF1=OF2= AB/2= 108/2=56 OA as radius, F1 or F2 as centers, draw an arc to cut the ellipse, which is minor axis. Exercise K.Srinivasulu Reddy, SNIST
  • 140. CONSTRUCTION OF PARABOLA BY DIFFERENT METHODS K.Srinivasulu Reddy, SNIST
  • 141. Problem: Construct a Parabola , with the distance of the focus from the directrix as 50. Also draw normal and Tangent to the Curve, at a point 40 from the directrix. Parabola-Eccentricity Method K.Srinivasulu Reddy, SNIST
  • 143. Construction Procedure 1. Draw the axis AB and the directrix CD at right angles to each other. 2. Mark the focus F on the axis such that AF=50. 3. Locate vortex V on AB such that AV=VF=25. 4. Draw a line VE , perpendicular to AB such that VE=VF. 5. Join A,E and extend. By construction VE/VA = VF/VA = 1. 6. Locate a number of points 1,2,3,etc., to the right of V on the axis, which need not be equidistant. 7. Through the points 1,2,3, etc., draw lines perpendicular to the axis and to meet the line AE extended at 1’ , 2’ , 3’, etc. 8. With center F and radius 1-1’ draw arcs intersecting the line through 1 at P1 and p1’. These p1 and p1’ are the points on the Parabola because the distance of p1 (p1’) from F is 1-1’ and from CD it is A-1 and 1-1’/A-1 = VE/VA = VF/VA = 1. 9. Similarly locate the points P2,P2’ : P3,P3’: etc., on either side of the axis. 10.Join the points by a smooth curve , forming the required parabola. K.Srinivasulu Reddy, SNIST
  • 144. Construction Procedure –Tangent and Normal 1.Locate the point M which is at 40 from the directrix. 2.Join M to F . 3.Draw a line through F perpendicular to MF meeting the directrix at T. 4.Join T and M and extend ,T-T is the tangent to the Parabola. 5.Draw a line perpendicular to tangent T-T name as N-N is the normal to the Parabola. K.Srinivasulu Reddy, SNIST
  • 145. Problem: Construct a parabola ,with the length of base as 60 and axis 30 long. Also , draw a tangent to the curve at a point 25 from the base. K.Srinivasulu Reddy, SNIST
  • 148. Construction Procedure 1. Draw the base AB=60 and the axis CD=30 such that CD is perpendicular bisector to AB. 2. Produce CD to E such that DE=CD. 3. Join E ,A and E,B. These are the tangents to the Parabola at A and B. 4. Divide AE and BE into the same number of equal parts and number the points. 5. Join 1-1’ , 2-2’ , 3-3’, etc., forming the tangents to the required parabola. A smooth curve passing through A,D and B and tangential to the above line is the required Parabola. K.Srinivasulu Reddy, SNIST
  • 149. Construction Procedure –Tangent and Normal 1. Locate the point M which is 25 from the base. 2. Draw a horizontal line through M meeting the axis at F. 3. Mark G on the extension of the axis such that DG=FD. 4. Join G and M and extend ,forming the tangent to the Parabola at M. K.Srinivasulu Reddy, SNIST
  • 150. Problem: Construct a parabola with the base 60 and length of the axis 40. Draw a tangent to the curve at a point 20 from the base. Also, locate the focus and directrix to the Parabola. K.Srinivasulu Reddy, SNIST
  • 151. Construction Procedure 1. Draw the base AB=60 and the axis CD=40 such that CD is perpendicular bisector to AB. 2. Construct the rectangle ABKL passing through D. 3. Divide AC and AL into the same number of equal parts and number the points. 4. Join 1,2,3 to D. 5. Through 1’,2’,3’ draw lines parallel to axis intersecting the lines 1-D ,2-D,3-D at P1,P2 and P3 respectively. 6. Parabola is symmetric so draw the remaining half portion. Join all the points we get Parabola K.Srinivasulu Reddy, SNIST
  • 152. Problem: Construct a Parabola in a parallelogram of sides 100 and 60, and with an included angle of 750 . K.Srinivasulu Reddy, SNIST
  • 153. Construction Procedure 1. Draw the base AB=60 and the axis CD=40 such that CD is perpendicular bisector to AB. 2. Construct the rectangle ABKL passing through D. 3. Divide AC and AL into the same number of equal parts and number the points. 4. Join 1,2,3 to D. 5. Through 1’,2’,3’ draw lines parallel to axis intersecting the the lines 1-D ,2-D,3-D at P1,P2 and P3 respectively. 6. Parabola is symmetric so draw the remaining half portion. Join all the points we get Parabola K.Srinivasulu Reddy, SNIST
  • 154. CONSTRUCTION OF HYPERBOLA BY DIFFERENT METHODS K.Srinivasulu Reddy, SNIST
  • 155. Problem: Construct a Hyperbola , with the distance between the focus and the directrix as 50 and eccentricity as 3/2. Also draw normal and tangent to the curve at a point 30 from the directrix. Construction of Hyperbola- Eccentricity method K.Srinivasulu Reddy, SNIST
  • 156. 1.Draw the axis AB and the directrix CD at right angle to each other. 2.Mark focus F on the axis such that AF=50. 3.Divide AF in 5 equal parts. 4.Locate the vertex V on the second division point from A. 5.Draw a line VE perpendicular to AB such that VE=VF. 6.Join A,E and extend. By construction , VE/VA = VF/VA = 3/2 ,the eccentricity. 7. Locate a number of points 1,2,3,etc., to the right of V on the axis, which need not be equidistant. 7. Through the points 1,2,3, etc., draw lines perpendicular to the axis and to meet the line AE extended at 1’ , 2’ , 3’, etc. 8. With center F and radius 1-1’ draw arcs intersecting the line through 1 at P1 and p1’. These p1 and p1’ are the points on the Hyperbola. 9. Similarly locate the points P2,P2’ : P3,P3’: etc., on either side of the axis. 10.Join the points by a smooth curve , forming the required Hyperbola. Construction Procedure K.Srinivasulu Reddy, SNIST
  • 157. Problem: Construct a Hyperbola with its foci 70 apart and the major axis 45. Draw a tangent to the curve at a point 20 from the focus. Also determine the eccentricity of the curve. K.Srinivasulu Reddy, SNIST
  • 158. Construction Procedure 1.Draw the axis AB and locate a point O on it. 2.Locate the foci F1,F2 such that F1F2=70 and locate vertices V1,V2 such that V1V2 =45 on AB. 3.Mark a number of points 1,2,3 etc., on AB to the right of F2 which need not be equidistant . 4.With center F1 and radius v1-1 draw arcs on either side if the transverse axis. 5. With center F2 and radius v2-1 draw arcs intersecting the above arcs at P1 and P1’. 6. With center F2 and radius v1-1 draw arcs on either side if the transverse axis. 7. With center F1 and radius v2-1 draw arcs intersecting the above arcs at Q1 and Q1’. 8.Repeat the steps 4 to 7 and obtain the two branches of the hyperbola. K.Srinivasulu Reddy, SNIST
  • 159. Construction Procedure –Tangent and Normal 1. Locate the point M which is 20 from the focus say F2. 2. Then join M to foci F1 and F2.. 3.Draw a line T-T bisecting <F1MF2 forming the required tangent. K.Srinivasulu Reddy, SNIST
  • 160. Eccentricity calculation Eccentricity e = OF1/OV1 = OF2/OV2 = 35/22.5 e = 1.56 K.Srinivasulu Reddy, SNIST
  • 161. Rectangular Hyperbola  A hyperbola for which the asymptotes are perpendicular, also called an equilateral hyperbola or right hyperbola. Ex: Boyle’s law PV=C As the pressure increases, its volume decreases. K.Srinivasulu Reddy, SNIST
  • 162. Construct a Rectangular Hyperbola ,when a point P on it is at a distance of 18 and 34 from two asymptotes. Also draw a tangent to the curve at a point 20 from an asymptote . K.Srinivasulu Reddy, SNIST
  • 163. Construction Procedure 1.Draw the asymptotes OA and OB at right angle to each other and locate the given point P. 2.Draw the lines CD and EF passing through point P and parallel to OA and OB respectively. 3.Locate a number of points 1,2,3,etc., along the line CD , which need not be equi -distant. 4.Join 1,2,3,etc., to O and extend if necessary , till these lines meet the line EF at points 1’ , 2’ , 3’, etc. 5.Draw lines through 1,2,3,etc.,parallel to EF and through 1’ , 2’ , 3’, etc. parallel to CD to intersect at P1,P2,P3,etc. K.Srinivasulu Reddy, SNIST
  • 164. Tangent to rectangular Hyperbola 1.Locate Point M on the curve by drawing a line GM parallel to OA and at a distance of 20 from it. 2.Locate the point H on OB such that GH = OG. 3.The line HT passing through M is the required tangent to the curve. K.Srinivasulu Reddy, SNIST
  • 166. Projection Symbol Projection Symbol First Angle European Third Angle American K.Srinivasulu Reddy, SNIST