Engineering Drawing
1
Introduction
Engineering Drawing is the language of engineers. It is a
graphic language, and has its own rules of grammar.
Engineers do calculations using the knowledge of various
subjects and convert them into drawings. Without deep and
sound knowledge of engineering drawing , an engineer would
not be a complete professional.
By learning how to create a drawing, students will be able to
read, understand and interpret it. Although a machine or
machine part can be described in words, generally, some
technical and hidden information cannot be explained in words
when the part or machine is of complex nature. Therefore,
Engineering Drawing is a good method of describing a machine
or machine part.
As compared to verbal or written description, this method is
brief, accurate, simple, and clear.
2
Drawing Instruments

Drawing board, T- sqaure, Drawing sheet

Instrument box containing compass, divider, etc.

Scales

Protractor

French curves

Drawing pencils

Eraser

Drawing clip/pin/adhesive tape

Sharpener

Duster
3
I . Standard Size of the Drawing Paper, Drawing Board
& T -Square , Drawing Instruments Scales Etc.
Standard Size of the Drawing Paper
4
Standard Size of the Drawing Board
Designation
Length x Width (mm)
Recommended
for use with sheet
size
D0
1500 x 1000
A0
D1
1000 x 700
A1
D2
700 x 500
A2
D3
500 x 500
A3
D 0 and D 1 for Drawing offices and D 2 for Students
5
Drawing Instruments
Objectives in
Drawing
1. Accuracy
2. Speed
3. Legibility
4. Neatness
Typical Drawing Equipment
6
Drawing Boards
Drawing Boards
The left edge and right edge
of a drawing board has a
true straight edge.
For right-handed people, the
left-hand edge of the board
is called the working edge
because the T-square head
slides against it.
For left-handed people, the
right-hand edge of the board
is called the working edge
because the T-square head
slides against it.
T-Squares
The T-square is made of a long strip called the
blade, fastened at right angles to a shorter piece
called the head.
7
Drawing Instruments
The drawing paper should be
placed close to the working
edge of the board to reduce
any error resulting from the
bending of the blade of the Tsquare.
The paper should also be
placed close enough to the
upper edge of the board to
permit space at the bottom of
the sheet for using the
T-square.
Drafting tape is used to fasten
the drawing paper to the
drawing board.
8
Drawing Pencils
High-quality drawing pencils should be used in technical drawing, never ordinary
writing pencils.
Many makes of mechanical pencils are available together with refill leads in all
grades. Choose a mechanical pencil that feels comfortable in your hand.
9
Drawing Pencils
Drawing Leads
The first consideration in the selection of a grade of lead is the type of line
work required. For light construction lines and guide lines for lettering use a
hard lead. For all other line work, the lines should be BLACK. The lead
chosen should be soft enough to produce jet black lines but hard enough not
to smudge.
HARD
MEDIUM
SOFT
9H 8H 7H 6H 5H 4H
3H 2H H F HB B
2B 3B 4B 5B 6B 7B
Hard leads are used
where extreme accuracy
is required. Generally
these leads are used for
construction lines.
Medium leads are used
for general purpose line
work in technical
drawing.
Soft leads are used for
various kinds of art work.
These leads are too soft
to be useful in
mechanical drafting.
Drawing Instruments
Drawing Lead Applications
TASK
CONSTRUCTION LINES
VISIBLE OBJECT LINES
LEAD GRADE
3H, 4H, 6H
H, F, HB
LINE WEIGHT
THIN, LIGHT
THICK, DARK
HIDDEN LINES
2H, H
THIN, DARK
CENTER LINES
2H, H
THIN, DARK
DIMENSION LINES
2H, H
THIN, DARK
EXTENSION LINES
2H, H
THIN, DARK
LEADER LINES
2H, H
THIN, DARK
H, F, HB
THICK, DARK
2H, H
THIN, DARK
H, F, HB
THIN, DARK
CUTTING PLANE LINES
PHANTOM LINES
LETTERING
11
Drawing Instruments
Drawing Horizontal and Vertical Lines
To draw a horizontal line, press the head
of the T-square against the working edge
of the board with your left hand. Lean
the pencil in the direction of the line at
an angle of approximately 60º and draw
the line from left to right. While drawing
the line, rotate the pencil to distribute
the wear uniformly on the lead to
maintain a symmetrical point.
To draw a vertical line, press the head of the T-square
against the working edge of the board with your left
hand and place a triangle against the blade of the Tsquare. Lean the pencil in the direction of the line at an
angle of approximately 60º and draw the line upward,
rotating the pencil to distribute the wear uniformly on
the lead to maintain a symmetrical point.
12
Drawing Instruments
Triangles
Most inclined lines are
drawn at standard angles
using the 45º x 45º
triangle and the 30º x 60º
triangle.
In addition to drawing
angles of 90º, 45º, 30º,
and 60º, triangles can be
combined to draw angles
of 15º increments.
13
Drawing Instruments - Scales
Scales
Scales are instruments used in
making technical drawings full size
or at a given reduction or
enlargement.
Types of scales include metric
scales, engineers’ scales, decimal
scales, mechanical engineers’
scales, and architects’ scales.
Scales are usually made of plastic
or boxwood and are either
triangular of flat in shape.
14
2. Lines - As per SLS 409
15
Lines contd…….
16
3. Lettering, Dimensioning etc.
Lettering Technique Most engineering lettering is single-stroke
Gothic font. Lettering is drawn freehand and
are drawn within light horizontal guidelines.
All lettering uses upper case letters. Lower case
letters are rarely used in technical drawings.
There are three aspects of good lettering:
.
proportions and forms of the letters,
composition and spacing, and practice.
There are six fundamental drawing strokes and
their directions in basic lettering. Horizontal
strokes are drawn from left to right, vertical
strokes are drawn from top to bottom, and
curved strokes are drawn downward.
17
Lettering Technique contd……
18
19
Units of Measurement
 International systems of units (SI) – which is based on the
meter
 Millimeter (mm) - The common SI unit of measure on
engineering drawing.
 Individual identification of linear units is not required if all
dimensions on a drawing are in the same unit (mm).
 The drawing shall however contain a note:
ALL DIMENSIONS ARE IN MM. (Bottom left corner
outside the title box)
20
Dimensioning
Indicating on a drawing, the size of the object and other details
essential for its construction and function, using lines, numerals,
symbols, notes, etc.
Dimensions indicated on a drawing should be those that are
essential for the production, inspection and functioning of the
object.
Dimensions indicated should not be mistaken as those that are
required to make the drawing of an object.
Extension line – a thin, solid line perpendicular to a dimension line,
indicating which feature is associated with the dimension.
Visible gap – there should be a visible gap of 1.5 mm between the
feature’s corners and the end of the extension line.
21
Leader line
 A thin, solid line used to indicate the feature with which a
dimension, note, or symbol is associated.
 Generally a straight line drawn at an angle that is neither
horizontal nor vertical.
 Terminated with an arrow touching the part or detail.
 On the end opposite the arrow, the leader line will have a short,
horizontal shoulder. Text is extended from this shoulder such that
the text height is centered with the shoulder line.
22
Arrows
3 mm wide and should be 1/3rd as wide as they are long - symbols placed at the
end of dimension lines to show the limits of the dimension. Arrows are uniform
in size and style, regardless of the size of the drawing.
23
Dimensioning Techniques
.D
imen
24
25
26
II - Geometric Constructions
27
TO DIVIDE A LINE
To divide a given line AB into any number of equal parts.
-Suppose the line AB is to be divided into 6 equal parts.
1. Draw a line AC of any length inclined to AB at some convenient angle (preferably
between 20° and 40°).
2. Mark off six equal divisions on AC by cutting arcs of suitable radii consecutively
starting from A. Number these divisions as 1, 2, 3, 4, 5 and 6.
3. Join 6 with B.
4. Draw lines through 5, 4, 3, 2 and 1 parallel to 6– B and cutting AB at points 5’, 4’,
3’, 2’ and 1’ respectively. Set-squares or drafter may be used for this purpose. The
divisions 1’, 2’, 3’, 4’, 5’ divide the line AB into 6 equal parts.
28
Draw a perpendicular bisector of a line
• Draw the line AB
• With A as center and radius greater
than half AB but less than AB, draw an
arc on either side of AB (shown green)
• With B as center and same radius,
draw an arc on either side of AB
(shown brown)
• Join the point of intersection of these
arcs on either side of AB
A
B
• This is the perpendicular bisector
29
Bisecting an arc
It is similar to bisecting a line
BC>BE
E
Line
Arc
30
Drawing a perpendicular to a line at a given point
• Draw the line AB
• With P as center and any
convenient radius, draw an arc
cutting AB in C (shown blue)
Q
• With the same radius cut 2 equal
divisions CD and DE (shown red)
E
D
• With same radius and centers D
and E, draw arcs (green and
brown) intersecting at Q
• PQ is the required perpendicular
A
C
P
B
31
Drawing a perpendicular to a line at a given
point (alternate method)
 Cut arcs with any radius (r1) on both sides of the point on the line AB. AB may be
extended
 With a radius greater than r1, draw arcs with centers C and D to intersect at Q
 QP is the required perpendicular
32
Drawing a perpendicular from a point to a
line
 From the external point P, draw arcs to cut the line AB at C and D. AB may be
extended
 With C and D as centers and radius greater than half CD, draw arcs to intersect at
E
 PE is the required perpendicular to AB
33
Bisecting an angle
To bisect a given angle AOB.
1. With O as centre and any convenient radius, mark arcs cutting OA and OB at C and
D respectively.
2. With C and D as centers and same or any other convenient radius, mark two arcs
intersecting each other at E.
3. Join OE.
4. OE is the bisector of AOB, i.e., AOE = 2 x EOB.
34
Geometric Constructions
Circles and Arcs
A circle is a closed curve, all points of which are the same
distance from a point called the center.
35
To divide a given circle into 8 equal parts
Draw horizontal (1, 5) and vertical (3, 7) diameters which will be at 90o to
each other.
Bisect the angles to get new diameters (2, 6) and (4, 8) at 45o to the horizontal
and vertical dimeters.
The circle is divided into 8 equal sectors
36
To divide a circle into 12 equal parts
Draw the two diameters 1–7 and 4–10, perpendicular to each other.
With 1 as a centre and radius = R (= radius of the circle), cut two arcs at
3 and 11 on the circle.
Similarly, with 4, 7 and 10 as the centres and the same radius, cut arcs
on the circle respectively at 2 and 6, 5 and 9, and 8 and 12. The points 1,
2, 3, etc., give 12 equal divisions of the circle.
37
To draw a normal and a tangent to an arc or
circle at a point P on it
With centre P and any convenient radius, mark off two arcs cutting the arc/circle
at C and D.
Obtain QR, the perpendicular bisector of arc CD. QR is the required normal.
Draw the perpendicular ST to QR for the required tangent.
38
Tangent to a given arc AB (or a circle) from a point
P outside it.
Join the centre O with P and locate the midpoint M of OP.
With M as a centre and radius = MO, mark an arc cutting the circle at Q.
Join P with Q. PQ is the required tangent.
Another tangent PQ’ can be drawn in a similar way.
39
Common external tangent to 2 circles
Given circles are with radii R1 and R2 and centers O and P respectively
Draw a circle with radius R1-R2 and center O
Draw a circle with dia. OP cutting the circle with radius R1-R2 at T
Draw a line OT extended cutting the circle with radius R1 at A
Draw a line PB parallel to OA with B lying on the circumference of circle with radius R2
Line AB is the required tangent
Circle with
radius R1
Circle with
radius R2
A
B
T
O
P
Circle with
radius R1-R2
NOTE: PT is a tangent from
point P to the circle with
radius R1-R2
See N. D. Bhatt pg. 88, 89
40
Common internal tangent to 2 circles
Given circles are with radii R1 and R2 and centers O and P respectively
Draw a circle with radius R1+R2 and center O
Draw a circle with dia. OP cutting the circle with radius R1+R2 at T
Draw a line OT cutting the circle with radius R1 at A
Draw a line PB parallel to OA with B lying on the circumference of circle with radius R2
Line AB is the required tangent
T
Circle with
radius R2
Circle with
radius R1
A
O
P
B
Circle with
radius R1+R2
41
See N. D. Bhatt pg. 89
Line parallel to another line
Example 4.24 To draw a line parallel to a given line AB and at a given distance R
from it.
Solution Refer Fig. 4.30.
1. Draw a perpendicular bisector of the line AB, cutting it at M.
2. Set off MN = R. Draw PQ perpendicular to MN at N. PQ is parallel to AB.
42
Draw an arc (radius R) touching 2 given
lines
AB and AC are the given lines
Draw a line PQ parallel to and at a distance R from AB
Draw a line EF parallel to and at a distance equal to R from AC intersecting PQ at
O
With O as center and radius R draw the arc touching to 2 lines
43
Draw an arc (radius R2) touching a given
line and another arc
CASE I
• AB is the given line
• Draw a line parallel to AB at a
distance R2
• With O ac center and radius R1R2, draw an arc EF cutting the
line at P
• With P as center and Radius R2,
draw the required arc
CASE II
• AB is the given line
• Draw a line parallel to AB at a
distance R2
• With O as center and radius
R1+R2, draw an arc EF cutting
the line at P
• With P as center and Radius R2,
draw the required arc
44
Finding the center of an arc
Draw 2 chord of the arc (CD and EF in this case)
Draw perpendicular bisectors of CD and EF intersecting each
other at O.
O is the required center.
45
Curve (given radius) joining 2 other curves
Draw arcs with radius R1 + R3
(center O) and R2 + R3 (center
P) intersecting at Q.
Draw arcs with radius R1 - R3
(center O) and R2 + R3 (center
P) intersecting at Q.
Draw arcs with radius R3 – R1
(center O) and R3 - R2 (center
P) intersecting at Q.
With center Q draw an arc with
radius R3 joining the 2 curves.
With center Q draw an arc with
radius R3 joining the 2 curves.
With center Q draw an arc
with radius R3 joining the 2
curves.
46
Geometric Constructions
Drawing a Square
Given the inscribed circle,
draw two diameters at
right angles to each other.
The intersections of these
diameters with the circle
are the vertexes of an
inscribed square.
Given the circumscribed
circle, use the T-square
and 45º triangle and draw
the four sides tangent to
the circle.
47
Geometric Constructions
Drawing a Hexagon
Given the inscribed circle, draw vertical and horizontal center lines and the diagonals (AB)
and (CD) at 30º or 60º with the horizontal.
With the 30º x 60º triangle and T-square, draw the six sides of the hexagon.
Given the circumscribed circle, draw vertical and horizontal center lines .
With the 30º x 60º triangle and T-square, draw the six sides of the hexagon tangent to the
circle.
48
To construct a regular hexagon of given side
length
With any point O as centre and radius = AB, draw a circle.
Starting from any point (say A) on the circle, mark off the five arcs of radius
= AB consecutively cutting the circle at B, C, D, E and F.
Join A, B, C, D, E and F for the required hexagon.
Principle: The distance across opposite
corners in a regular hexagon = 2 x side
length
AD = 2 x AB
49
CONSTRUCTION OF A POLYGON
Draw side AB of specified length
Draw a perpendicular BP at B such that
BP = AB
Draw a straight line joining A and P
P
6
With B as center and radius AB draw
arc AP
5
4
Draw a perpendicular bisector of AB to
meet the line AP at 4 and arc AP at 6
Locate point 5 as the midpoint of 4-5
A square of side AB can be inscribed in
the circle with center 4 and radius A4
A
B
50
Polygons of different number of sides on
same construction
• Similarly a hexagon of side AB
can be inscribed in the circle with
center 6 and radius A6
8
• Mark points 7, 8, 9 on the
perpendicular bisector such that
5-6 = 6-7 = 7-8 = 8-9 and so on
7
5
• A heptagon of side AB can be
inscribed in the circle with center
7 and radius A7
• An octagon of side AB can be
inscribed in the circle with center
8 and radius A8…and so on
P
6
4
A
B
51
Drawing a pitch circle and marking the holes
Pitch circle: circle on which lies certain features e.g. the centers of smaller circles or
holes. Fig. shows 8 holes drawn on a pitch circle in a square plate
•Draw the pitch circle
•Since there are 8 holes, the angle between
the lines joining their centers to the center
of the pitch circle will be 45o
•Divide the pitch circle into 8 parts by
drawing lines from its center at 45o to the
adjacent one
C
•The points of intersection of these lines
and the pitch circle are the centers of the
required holes
Pitch circle
•Draw the holes with specified diameter
52
Geometric Constructions
Bisecting an angle
Given the angle (BAC) to be bisected.
Strike the large arc (R) at any convenient radius.
Strike equal arcs (r) with a radius slightly larger than half (BC) to intersect at (D).
Draw line (AD) which bisects the angle.
53
Geometric Constructions
Transferring an angle
Given the angle (BAC) to be transferred to the new position at (A’B’).
Use any convenient radius (R) and strike arcs from centers (A) and (A’).
Strike equal arcs (r) and draw side (A’C’).
54
DRAWING LINES
Find the Centre of an Arc
1.
2.
3.
Select three points A, B and C on the arc and join AB and BC
Bisect AB and BC.
Fine the intersection point of the bisecting lines/bisectors. That is the
centre of the arc.
55
DRAWING LINES
Inscribe a Circle in a Triangle
1.
2.
3.
Bisect angle ABC and angle BAC.
Fine the intersection point of the bisecting lines/bisectors. That is the
centre of the circle.
The radius of the circle is the length of a perpendicular line on any of
the sides of the triangle drawn from the centre of the circle.
56
DRAWING LINES
Circumscribe a Circle on a Triangle
1.
2.
3.
Bisect sides AC and BC.
Fine the intersection point of
the bisecting lines/bisectors.
That is the centre of the
circle.
The radius of the circle is the
length of a line joining any
one of the vertices of the
triangle to the centre of the
circle.
57
DRAWING LINES
Draw a Hexagon
• To draw a regular
hexagon given the
distance across flats
Draw a circle having a
diameter equal to the
distance across flats.
• Draw tangents to this
circle with a 60° set
square to produce the
hexagon.
58
DRAWING LINES
Draw a Hexagon
• To draw a regular
hexagon given the
distance across corners,
draw a circle having a
diameter equal to the
distance across corners
• Step off the radius round
it to give six equally
spaced points.
• Join these points to form
the hexagon.
59
Loci of Points & Application of Loci
• The Ellipse, Parabola, Hyperbola
• Curves: Roulettes or Cycloidal curves
Cycloids, Epicycloids,
Hypocycloids, Involutes,
Spirals and Helices
60
III(a)- CONIC SECTIONS
ELLIPSE, PARABOLA AND HYPERBOLA ARE CALLED CONIC SECTIONS
BECAUSE
THESE CURVES APPEAR ON THE SURFACE OF A CONE
WHEN IT IS CUT BY SOME TYPICAL CUTTING PLANES.
Hyperbola
Ellipse
Parabola
Section Plane
Through Generators
Section Plane
Parallel to Axis.
Section Plane Parallel
to end generator.
61
III (a) - Conic Sections - Engineering Curves –
Ellipse, Parabola and Hyperbola
ELLIPSE
PARABOLA
HYPERBOLA
1.Concentric Circle Method
1.Rectangle Method
1.Rectangular Hyperbola
(coordinates given)
2.Rectangle Method
2 Method of Tangents
( Triangle Method)
3.Oblong Method
4.Arcs of Circle Method
3.Basic Locus Method
(Directrix – focus)
2 Rectangular Hyperbola
(P-V diagram - Equation given)
3.Basic Locus Method
(Directrix – focus)
5.Rhombus Metho
6.Basic Locus Method
(Directrix – focus)
Methods of Drawing
Tangents & Normals
To These Curves.
62
COMMON DEFINATION OF ELLIPSE, PARABOLA & HYPERBOLA
These are the loci of points moving in a plane such that the ratio of it’s distances
from a fixed point And a fixed line always remains constant.
The Ratio is called ECCENTRICITY. (E)
A) For Ellipse
E<1
B) For Parabola E=1
C) For Hyperbola E>1
Refer Problem nos. 6. 9 & 12
SECOND DEFINATION OF AN ELLIPSE:It is a locus of a point moving in a plane
such that the SUM of it’s distances from TWO fixed points
always remains constant.
{And this sum equals to the length of major axis.}
These TWO fixed points are FOCUS 1 & FOCUS 2
Refer Problem no.4
Ellipse by Arcs of Circles Method.
63
ELLIPSE
Problem 1 :Draw ellipse by concentric circle method.
Take major axis 100 mm and minor axis 70 mm long.
BY CONCENTRIC CIRCLE METHOD
3
2
Steps:
1. Draw both axes as perpendicular
bisectors of each other & name their ends
as shown.
2. Taking their intersecting point as a
center, draw two concentric circles
considering both as respective diameters.
3. Divide both circles in 12 equal parts &
name as shown.
4. From all points of outer circle draw
vertical lines downwards and upwards
respectively.
5.From all points of inner circle draw
horizontal lines to intersect those vertical
lines.
6. Mark all intersecting points properly as
those are the points on ellipse.
7. Join all these points along with the
ends of both axes in smooth possible
curve. It is required ellipse.
4
C
1
2
3
5
4
1
5
A
B
10
10
6
9
8
D
9
7
6
7
8
64
Steps:
1 Draw a rectangle taking major
and minor axes as sides.
2. In this rectangle draw both
axes as perpendicular bisectors
of each other..
3. For construction, select upper
left part of rectangle. Divide
vertical small side and horizontal
long side into same number of
equal parts.( here divided in four
parts)
4. Name those as shown..
5. Now join all vertical points
1,2,3,4, to the upper end of
minor axis. And all horizontal
points i.e.1,2,3,4 to the lower
end of minor axis.
6. Then extend C-1 line upto D-1
and mark that point. Similarly
extend C-2, C-3, C-4 lines up to
D-2, D-3, & D-4 lines.
7. Mark all these points properly
and join all along with ends A
and D in smooth possible curve.
Do similar construction in right
side part.along with lower half of
the rectangle.Join all points in
smooth curve.
It is required ellipse.
ELLIPSE
BY RECTANGLE METHOD
Problem 2
Draw ellipse by Rectangle method.
Take major axis 100 mm and minor axis 70 mm long.
4
D
4
3
3
2
2
1
1
A
B
C
65
ELLIPSE
BY OBLONG METHOD
Problem 3:Draw ellipse by Oblong method.
Draw a parallelogram of 100 mm and 70 mm long sides with
included angle of 750.Inscribe Ellipse in it.
STEPS ARE SIMILAR TO THE PREVIOUS CASE (RECTANGLE METHOD) ONLY IN PLACE OF RECTANGLE,
HERE IS A PARALLELOGRAM.
4
4
3
3
2
2
1
A
1
B
66
PROBLEM 4.
MAJOR AXIS AB & MINOR AXIS CD ARE
100 AMD 70MM LONG RESPECTIVELY
.DRAW ELLIPSE BY ARCS OF CIRLES
METHOD.
STEPS:
1.Draw both axes as usual.Name the
ends & intersecting point
2.Taking AO distance I.e.half major
axis, from C, mark F1 & F2 On AB .
( focus 1 and 2.)
3.On line F1- O taking any distance,
mark points 1,2,3, & 4
4.Taking F1 center, with distance A-1
draw an arc above AB and taking F2
center, with B-1 distance cut this arc. A
Name the point p1
5.Repeat this step with same centers
but
taking now A-2 & B-2 distances for
drawing arcs. Name the point p2
6.Similarly get all other P points.
With same steps positions of P can be
located below AB.
7.Join all points by smooth curve to get
an ellipse/
ELLIPSE
BY ARCS OF CIRCLE METHOD
As per the definition Ellipse is locus of point P moving in
a plane such that the SUM of it’s distances from two fixed
points (F1 & F2) remains constant and equals to the length
of major axis AB.(Note A .1+ B .1=A . 2 + B. 2 = AB)
p4
p3
C
p2
p1
F1
1
2
3
4
O
B
F2
D
67
ELLIPSE
PROBLEM 5.
DRAW RHOMBUS OF 100 MM & 70 MM LONG
DIAGONALS AND INSCRIBE AN ELLIPSE IN IT.
BY RHOMBUS METHOD
2
A
STEPS:
1. Draw rhombus of given
dimensions.
2. Mark mid points of all sides &
name Those A,B,C,& D
3. Join these points to the ends of
smaller diagonals.
4. Mark points 1,2,3,4 as four
centers.
5. Taking 1 as center and 1-A
radius draw an arc AB.
6. Take 2 as center draw an arc CD.
7. Similarly taking 3 & 4 as centers
and 3-D radius draw arcs DA &
BC.
B
4
3
C
D
1
68
ELLIPSE
PROBLEM 6 :-
POINT F IS 50 MM FROM A
LINE AB.A POINT P IS MOVING IN A PLANE
SUCH THAT THE RATIO OF IT’S DISTANCES
FROM F AND LINE AB REMAINS CONSTANT
AND EQUALS TO 2/3 DRAW LOCUS OF
POINT P. { ECCENTRICITY = 2/3 }
STEPS:
1 .Draw a vertical line AB and point F
50 mm from it.
2 .Divide 50 mm distance in 5 parts.
3 .Name 2nd part from F as V. It is 20mm
and 30mm from F and AB line resp.
It is first point giving ratio of it’s
distances from F and AB 2/3 i.e 20/30
4 Form more points giving same ratio such
as 30/45, 40/60, 50/75 etc.
5.Taking 45,60 and 75mm distances from
line AB, draw three vertical lines to the
right side of it.
6. Now with 30, 40 and 50mm distances in
compass cut these lines above and below,
with F as center.
7. Join these points through V in smooth
curve.
This is required locus of P. It is an ELLIPSE.
DIRECTRIX-FOCUS METHOD
A
ELLIPSE
45mm
(vertex) V
F ( focus)
B
69
PARABOLA
RECTANGLE METHOD
PROBLEM 7:
A BALL THROWN IN AIR ATTAINS 100 M HIEGHT
AND COVERS HORIZONTAL DISTANCE 150 M ON GROUND.
Draw the path of the ball (projectile)-
STEPS:
1.Draw rectangle of above size and
divide it in two equal vertical parts
2.Consider left part for construction.
Divide height and length in equal
number of parts and name those
1,2,3,4,5& 6
3.Join vertical 1,2,3,4,5 & 6 to the
top center of rectangle
4.Similarly draw upward vertical
lines from horizontal1,2,3,4,5
And wherever these lines intersect
previously drawn inclined lines in
sequence Mark those points and
further join in smooth possible curve.
5.Repeat the construction on right side
rectangle also.Join all in sequence.
This locus is Parabola.
.
6
6
5
5
4
4
3
3
2
2
1
1
1
2
3
4
5
6
5
4
3
2
1
70
Problem no.8: Draw an isosceles triangle of 100
mm long base and 110 mm long altitude. Inscribe
a parabola in it by method of tangents.
Solution Steps:
1. Construct triangle as per the
given
dimensions.
2.
Divide it’s both sides in to
same no.of equal parts.
3.
Name the parts in ascending
and descending manner, as
shown.
4.
Join 1-1, 2-2,3-3 and so on.
5.
Draw the curve as shown
i.e.tangent to all these lines.
The above all lines being
tangents to the curve, it is
called method of tangents.
A
PARABOLA
METHOD OF TANGENTS
C
B
71
PARABOLA
PROBLEM 9: Point F is 50 mm from a vertical straight line AB.
Draw locus of point P, moving in a plane such that
it always remains equidistant from point F and line AB.
DIRECTRIX-FOCUS METHOD
PARABOLA
SOLUTION STEPS:
1.Locate center of line, perpendicular to
AB from point F. This will be initial
point P and also the vertex.
2.Mark 5 mm distance to its right side,
name those points 1,2,3,4 and from
those
draw lines parallel to AB.
3.Mark 5 mm distance to its left of P and
name it 1.
4.Take O-1 distance as radius and F as
center draw an arc
cutting first parallel line to AB. Name
upper point P1 and lower point P2.
(FP1=O1)
5.Similarly repeat this process by taking
again 5mm to right and left and locate
P3P4.
6.Join all these points in smooth curve.
A
P1
O
(VERTEX) V
1 2 3 4
F
( focus)
P2
B
It will be the locus of P equidistance
from line AB and fixed point F.
72
Problem No.10: Point P is 40 mm and 30 mm from horizontal
and vertical axes respectively. Draw Hyperbola through it.
Solution Steps:
1) Extend horizontal line
from P to right side.
2) Extend vertical line from P
upward.
3) On horizontal line from P,
mark some points taking any
distance and name them after
P-1, 2,3,4 etc.
4) Join 1-2-3-4 points to pole
O. Let them cut part [P-B] also
at 1,2,3,4 points.
5) From horizontal 1,2,3,4
draw vertical lines downwards
and
6) From vertical 1,2,3,4
points [from P-B] draw
40 mm
horizontal lines.
7) Line from 1 horizontal and
line from 1 vertical will meet at
P1.Similarly mark P2, P3, P4
points.
8) Repeat the procedure by
marking four points on upward
vertical line from P and joining
all those to pole O. Name this
points P6, P7, P8 etc. and join
them by smooth curve.
HYPERBOLA
THROUGH A POINT
OF KNOWN CO-ORDINATES
2
1
2
1
P
1
2
3
1
2
3
O
30 mm
73
HYPERBOLA
P-V DIAGRAM
Problem no.11: A sample of
gas is expanded in a cylinder
from 10 unit pressure to 1 unit
pressure. Expansion follows
law PV= Constant. If initial
volume being 1 unit, draw the
curve of expansion. Also Name
the curve.
10
9
8
7
Form a table giving few more values of P & V
P
10
5
4
2.5
2
1
V = C
1 =
2 =
2.5 =
4 =
5 =
10 =
10
10
10
10
10
10
Now draw a Graph of
Pressure against Volume.
It is a PV Diagram and it is Hyperbola.
Take pressure on vertical axis and
Volume on horizontal axis.
PRESSURE
( Kg/cm2)
6
5
4
3
2
1
0
1
2
3
4
5
6
7
8
10
9
VOLUME:( M3 )
74
HYPERBOLA
DIRECTRIX
FOCUS METHOD
PROBLEM 12:- POINT F IS 50 MM FROM A LINE AB.A
POINT P IS MOVING IN A PLANE SUCH THAT THE
RATIO OF IT’S DISTANCES FROM F AND LINE AB
REMAINS CONSTANT AND EQUALS TO 2/3 DRAW
LOCUS OF POINT P. { ECCENTRICITY = 2/3 }
STEPS:
1 .Draw a vertical line AB and point F
50 mm from it.
2 .Divide 50 mm distance in 5 parts.
3 .Name 2nd part from F as V. It is 20mm
and 30mm from F and AB line resp.
It is first point giving ratio of it’s
distances from F and AB 2/3 i.e 20/30
4 Form more points giving same ratio such
as 30/45, 40/60, 50/75 etc.
5.Taking 45,60 and 75mm distances from
line AB, draw three vertical lines to the
right side of it.
6. Now with 30, 40 and 50mm distances in
compass cut these lines above and below,
with F as center.
7. Join these points through V in smooth
curve.
This is required locus of P.It is an ELLIPSE.
A
30mm
(vertex)
V
F ( focus)
B
75
ELLIPSE
TANGENT & NORMAL
Problem 13:
TO DRAW TANGENT & NORMAL
TO THE CURVE FROM A GIVEN POINT ( Q )
1.
2.
3.
JOIN POINT Q TO F1 & F2
BISECT ANGLE F1Q F2 THE ANGLE BISECTOR IS NORMAL
A PERPENDICULAR LINE DRAWN TO IT IS TANGENT TO THE
CURVE.
p4
p3
C
p2
p1
A
F1
1
2
3
4
O
B
F2
Q
D
76
Ellipse Construction
1. Draw the axes AB and CD and draw circles (called auxiliary circles) on
them as diameters.
2. Divide the circles into a number of equal parts, by radial lines through O.
Each of the radial lines intersect the major and minor auxiliary circle.
3. Through the points where radial lines cut the major auxiliary circles drop
vertical perpendiculars, and through the points where the radial lines cut
the minor auxiliary circle draw horizontals to cut the verticals. These
77
intersections are points on the ellipse.
ELLIPSE
TANGENT & NORMAL
Problem 14:
TO DRAW TANGENT & NORMAL
TO THE CURVE FROM A GIVEN POINT ( Q )
A
1. JOIN POINT Q TO F.
2. CONSTRUCT 900 ANGLE WITH
THIS LINE AT POINT F
3. EXTEND THE LINE TO MEET
DIRECTRIX AT T
4. JOIN THIS POINT TO Q AND EXTEND.
THIS IS TANGENT TO ELLIPSE FROM Q
5.T O THIS TANGENT DRAW
PERPENDICULAR LINE FROM Q. IT IS
NORMAL TO CURVE.
ELLIPSE
T
(vertex) V
F ( focus)
900
N
N
Q
B
78
T
PARABOLA
TANGENT & NORMAL
Problem 15:
TO DRAW TANGENT & NORMAL
TO THE CURVE
FROM A GIVEN POINT ( Q )
1. JOIN POINT Q TO F.
2. CONSTRUCT 900 ANGLE WITH
THIS LINE AT POINT F
3. EXTEND THE LINE TO MEET
DIRECTRIX
AT T
4. JOIN THIS POINT TO Q AND
EXTEND.
THIS IS TANGENT TO THE
CURVE
FROM Q
5. TO THIS TANGENT DRAW
ERPENDICULAR
LINE FROM Q. IT IS NORMAL TO
T
PARABOLA
A
VERTEX
V
900
F
( focus)
N
Q
N
B
T
79
HYPERBOLA
TANGENT & NORMAL
Problem 16
TO DRAW TANGENT & NORMAL
TO THE CURVE
FROM A GIVEN POINT ( Q )
A
T
1.JOIN POINT Q TO F.
2.CONSTRUCT 900 ANGLE WITH THIS LINE AT
POINT F
3.EXTEND THE LINE TO MEET DIRECTRIX AT T
4. JOIN THIS POINT TO Q AND EXTEND. THIS IS
TANGENT TO CURVE FROM Q
5.TO THIS TANGENT DRAW PERPENDICULAR
LINE FROM Q. IT IS NORMAL TO CURVE.
(vertex)
F ( focus)
V
900
N
N
Q
B
T
80
III (b) - ENGINEERING CURVES : LOCI
Point undergoing two types of displacements)
INVOLUTE
1. Involute of a circle
a)String Length = D
b)String Length > D
c)String Length < D
2. Pole having Composite
shape.
3. Rod Rolling over
a Semicircular Pole.
CYCLOID
1. General Cycloid
2. Trochoid
( superior)
3. Trochoid
( Inferior)
4. Epi-Cycloid
SPIRAL
HELIX
1. Spiral of
One Convolution.
1. On Cylinder
2. On a Cone
2. Spiral of
Two Convolutions.
5. Hypo-Cycloid
AND
Methods of Drawing
Tangents & Normals
To These Curves.
81
DEFINITIONS
CYCLOID:
SUPERIORTROCHOID
IF THE POINT IN THE DEFINATION
OF CYCLOID IS OUTSIDE THE
CIRCLE
IT IS A LOCUS OF A POINT ON THE
PERIPHERY OF A CIRCLE WHICH
ROLLS ON A STRAIGHT LINE PATH.
INFERIOR TROCHOID
INVOLUTE:
EPI-CYCLOID
IT IS A LOCUS OF A FREE END OF A STRING
WHEN IT IS WOUND ROUND A CIRCLE OR POLYGON
SPIRAL:
IT IS A CURVE GENERATED BY A POINT
WHICH REVOLVES AROUND A FIXED POINT
AND AT THE SAME MOVES TOWARDS IT.
IF IT IS INSIDE THE CIRCLE
IF THE CIRCLE IS ROLLING ON
ANOTHER CIRCLE FROM OUTSIDE
HYPO-CYCLOID.
IF THE CIRCLE IS ROLLING FROM
INSIDE THE OTHER CIRCLE,
HELIX:
IT IS A CURVE GENERATED BY A POINT WHICH
MOVES AROUND THE SURFACE OF A RIGHT CIRCULAR
CYLINDER / CONE AND AT THE SAME TIME ADVANCES IN AXIAL DIRECTION
AT A SPEED BEARING A CONSTANT RATIO TO THE SPPED OF ROTATION.
( for problems refer topic Development of surfaces)
82
CYCLOID
• The cycloid is the locus of a point on the
rim of a circle rolling along a straight line.
83
CYCLOID
Problem 22: Draw locus of a point on the periphery of a circle which rolls on
straight line path. Take circle diameter as 50 mm. Draw normal and tangent
on the curve at a point 40 mm above the directing line.
6
7
p6
p5
5
p8
4 p4
9
C1
C
p3
C2
3
C3
C4
C5
p2
10
2
11
Solution Steps:
7)
p1
12 P
C6
C7
C8
C9
p9
C10
C11
40mm
8
1)
2)
3)
4)
5)
6)
p7
p10
p11
p12
1
1’
2’
C12
3’
4’
5’
6’
7’
D
8’
9’
10’
11’
12’
Q
From center C draw a circle of 50mm dia. and from point P draw a horizontal line PQ equal to D length.
Divide the circle in 12 equal parts and in anticlockwise direction, after P name 1, 2, 3 up to 12.
Also divide the straight line PQ into 12 number of equal parts and after P name them 1’,2’,3’__ etc.
From all these points on circle draw horizontal lines. (parallel to locus of C)
With a fixed distance C-P in compass, C1 as center, mark a point on horizontal line from 1. Name it P 1.
Repeat this procedure from C2, C3, C4 up to C12 as centers. Mark points P2, P3, P4, P5 up to P12 on the
horizontal lines drawn from 1,2, 3, 4, 5, 6, 7 respectively.
Join all these points by curve. It is Cycloid.
84
STEPS:
DRAW CYCLOID AS USUAL.
MARK POINT Q ON IT AS DIRECTED.
CYCLOID
Method of Drawing
Tangent & Normal
WITH CP DISTANCE, FROM Q. CUT THE
POINT ON LOCUS OF C AND JOIN IT TO Q.
FROM THIS POINT DROP A PERPENDICULAR
ON GROUND LINE AND NAME IT N
JOIN N WITH Q.THIS WILL BE NORMAL TO
CYCLOID.
DRAW A LINE AT RIGHT ANGLE TO
THIS LINE FROM Q.
IT WILL BE TANGENT TO CYCLOID.
CYCLOID
Q
C
C1
C2
C3
C4
P
C5
C6
C7
C8
N
D
85
CYCLOID
Problem 22: Draw locus of a point on the periphery of a circle which rolls on
straight line path. Take circle diameter as 50 mm. Draw normal and tangent
on the curve at a point 40 mm above the directing line.
6
7
p6
p5
5
p8
4 p4
9
C1
C
p3
C2
3
C3
C4
C5
C6
p2
10
2
11
Solution Steps:
7)
p1
12 P
C7
C8
C9
C10
p9
C11
40mm
8
1)
2)
3)
4)
5)
6)
p7
C12
p10
p11
p12
1
1’
2’
3’
4’
5’
6’
D
7’
8’
9’
10’
11’
12’
Q
From center C draw a circle of 50mm dia. and from point P draw a horizontal line PQ equal to D length.
Divide the circle in 12 equal parts and in anticlockwise direction, after P name 1, 2, 3 up to 12.
Also divide the straight line PQ into 12 number of equal parts and after P name them 1’,2’,3’__ etc.
From all these points on circle draw horizontal lines. (parallel to locus of C)
With a fixed distance C-P in compass, C1 as center, mark a point on horizontal line from 1. Name it P 1.
Repeat this procedure from C2, C3, C4 up to C12 as centers. Mark points P2, P3, P4, P5 up to P12 on the
horizontal lines drawn from 1,2, 3, 4, 5, 6, 7 respectively.
Join all these points by curve. It is Cycloid.
86
STEPS:
DRAW CYCLOID AS USUAL.
MARK POINT Q ON IT AS DIRECTED.
CYCLOID
Method of Drawing
Tangent & Normal
WITH CP DISTANCE, FROM Q. CUT THE
POINT ON LOCUS OF C AND JOIN IT TO Q.
FROM THIS POINT DROP A PERPENDICULAR
ON GROUND LINE AND NAME IT N
JOIN N WITH Q.THIS WILL BE NORMAL TO
CYCLOID.
DRAW A LINE AT RIGHT ANGLE TO
THIS LINE FROM Q.
IT WILL BE TANGENT TO CYCLOID.
CYCLOID
Q
C
C1
C2
C3
C4
P
C5
C6
C7
C8
N
D
87
HYPOCYCLOID
The curve produced by fixed point P
on the circumference of a small circle
of radius a rolling around the inside
of a large circle of radius b.
88
EPICYCLOID
The path traced out by a point P on the edge of
a circle of radius a rolling on the outside of a
89
circle of radius b.
What is an involute
?
• Attach a string to a point on a curve.
• Make the string a tangent to the curve at the
point of attachment.
• Then wind the string up, keeping it always taut.
The locus of points traced out by the end of the
string is called the involute of the original curve.
• An involute can be described as the path traced
by the end point of a string as it is unwound from
a line, a polygon, or circumference of a circle
90
Involute of a line (AB):
A
B
C
91
Problem: Draw involute of an equilateral triangle of 35 mm sides.
B
C
A
35
3X35
92
Problem: Draw involute of a square of 25 mm sides
C
B
D
A
25
100
93
INVOLUTE OF A CIRCLE
Problem no 23: Draw Involute of a circle of 40 mm diameter.
Also draw normal and tangent to it at a point 100 mm from the
centre of the circle.
Solution Steps:
1) Point or end P of string AP is
exactly D distance away from A.
Means if this string is wound round
the circle, it will completely cover
given circle. B will meet A after
winding.
2) Divide D (AP) distance into 12
number of equal parts.
3) Divide circle also into 12 number
of equal parts.
P5
4) Name after A, 1, 2, 3, 4, etc. up
to 12 on D line AP as well as on
circle (in anticlockwise direction).
5) To radius C-1’, C-2’, C-3’ up to C12’ draw tangents (from 1’,2’,3’,4’,
etc to circle).
6) Take distance 1 to P in compass
P6 from point 1’
and mark it on tangent
on circle (means one division less
than distance AP).
7) Name this point P1
8) Take 2-P distance in compass
and mark it on the tangent from
point 2’. Name it point PP2.7
9) Similarly take 3 to P, 4 to P, 5 to
P up to 11 to P distance in compass
and mark on respective tangents
and locate P3, P4, P5 up to P12 (i.e. P8
A) points and join them in smooth
curve it is an INVOLUTE of a given
circle.
P3
P4
P2
P1
6’
7’
8’
5’
4’
9’
c
3’
10’
2’
11’
12’
A
P11
P9
P10
1’
1
P
2
3
4
5
6
7
8
9
10
11
D
94
12
STEPS:
DRAW INVOLUTE AS USUAL.
Involute
Method of Drawing
Tangent & Normal
INVOLUTE OF A CIRCLE
MARK POINT Q ON IT AS DIRECTED.
JOIN Q TO THE CENTER OF CIRCLE C.
CONSIDERING CQ DIAMETER, DRAW
A SEMICIRCLE AS SHOWN.
Q
MARK POINT OF INTERSECTION OF
THIS SEMICIRCLE AND POLE CIRCLE
AND JOIN IT TO Q.
THIS WILL BE NORMAL TO INVOLUTE.
DRAW A LINE AT RIGHT ANGLE TO
THIS LINE FROM Q.
IT WILL BE TANGENT TO INVOLUTE.
4
3
5
2
C
6
7
1
8
P8
1
2
3
4
5
6
P
8
7

D
95
Archimedean Spiral
• Spiral of Archimedes is a spiral , a curve generated by a
point that moves at a uniform speed along a straight line
while the straight line rotates with uniform angular
velocity about a fixed point.
96
Geometric Constructions
Solids
Solids bounded
by plane
surfaces are
called
polyhedra.
The surfaces
are called
faces.
If the faces are
equal regular
polygons the
solids are called
regular
polyhedra.
97
PROBLEM 25: DRAW LOCUS OF A POINT ON THE PERIPHERY OF A CIRCLE WHICH ROLLS ON A CURVED PATH. Take diameter
of rolling Circle 50 mm And radius of directing circle i.e. curved path, 75 mm.Also draw normal and tangent on the
curve at 110mm from the centre of directing circle.
Solution Steps:
1) When smaller circle will roll on
larger circle for one revolution it will
cover D distance on arc and it will
be decided by included arc angle .
2) Calculate  by formula  = (r/R)
x 360º.
c9
c8
3) Construct angle  with radius
c7
OC and draw an arc by taking O as
c6
center OC as radius and form sector
of angle .
c5
9
4) Divide this sector into 12
8
7
number of equal angular parts. And Rolling circle or
6
generating circle
from C onward name them C1, C2,
c4
C3 up to C12.
5
5) Divide smaller circle (Generating
c3
circle) also in 12 number of equal
4
parts. And next to P in anticlockwDirecting circle
ise direction name those 1, 2, 3, up
3
c2
to 12.
6) With O as center, O-1 as radius
2
3’
draw an arc in the sector. Take O-2,
4’
2’
O-3, O-4, O-5 up to O-12 distances
c1
1
5’
with center O, draw all concentric
1’
arcs in sector. Take fixed distance CP in compass, C1 center, cut arc of 1
6’
12’
at P1.
C
P
Repeat procedure and locate P2, P3,
O
P4, P5 unto P12 (as in cycloid) and
11’
OP=Radius of directing circle=75mm
7’
join them by smooth curve. This is
PC=Radius of generating circle=25mm
EPI – CYCLOID.
8’
10’
θ=r/R X360º= 25/75 X360º=120º
9’
c1
c1
0
c1
1
2
10
11
12
θ
98
Problem 17: A circle of 50 mm diameter rolls on another circle of 175 mm diameter and outside it.
Draw the curve traced by a point P on its circumference for one complete revolution.Also draw normal
and tangent on the curve at 125 mm from the centre of directing circle.
Draw a horizontal line OP of 87.5 mm and draw an arc with O as centre and PO as radius
Draw a horizontal line CP of 25 mm and draw a circle with C as centre and CP as radius.
θ=(OP/PC) X 360º = (25/87.5) X 360º = 102.8º ≈103º
Divide the rolling circle in 8 equal parts
Also divide the angle in 8 equal parts using angle bisectors
Directing
circle
Rolling circle or
generating circle
C
P
θ=103º O
99
PROBLEM 26: DRAW LOCUS OF A POINT ON THE PERIPHERY OF A CIRCLE
WHICH ROLLS FROM THE INSIDE OF A CURVED PATH. Take diameter of
rolling circle 50 mm and radius of directing circle (curved path) 75 mm. Also draw normal and
tangent on the curve at a point 40mm from the centre of directing circle
Directing circle
Solution Steps:
1) Smaller circle is rolling here,
inside the larger circle. It has to
rotate anticlockwise to move
ahead.
2) Same steps should be taken
as in case of EPI – CYCLOID.
Only change is in numbering
direction of 12 number of equal
parts on the smaller circle.
3) From next to P in clockwise
direction, name
1,2,3,4,5,6,7,8,9,10,11,12
4) Further all steps are that of
epi – cycloid. This is called
HYPO – CYCLOID.
10
11
12
6
5
4
c6
3
c7
c8
c9
c10
c11
c12
c5
c4
2
3’
c3
2’
4’
c2
1
1’
5’
c1
θ
12’
6’
C
P
11’
Rolling circle or
generating circle
9
8
7
7’
O
8’
10’
9’
OP=Radius of directing circle=75mm
PC=Radius of generating circle=25mm
θ=r/R X360º= 25/75 X360º=120º
100
Problem 28: A point P moves towards another point O, 75 mm from it and
reaches it while moving around it once. Its movement towards O being uniform
with its movement around it. Draw the curve traced out by point P.
SPIRAL
Important approach for construction
Find total angular and total linear displacement and divide both in to same number of equal parts. Total
linear movement 75 mm. Total angular movement 360º
2’
With OP radius & O as center draw a
circle and divide it in EIGHT
parts. Name those 1’,2’,3’,4’, etc.
up to 8’
Similarly
divided line PO also in
EIGHT parts and name those
1,2,3, starting from P.
Take O-1 distance from OP line and
draw an arc up to O1’ radius
vector. Name the point P1
Similarly mark points P2, P3, P4 up to
P8
And join those in a smooth curve. It is
a SPIRAL of one convolution.
P2
3’
P1
1’
P3
P4
4’
O
P5
7 6 5 4 3 2 1
P7
P
P6
7’
5’
6’
101
Draw an Archemedian spiral of one convolution, greatest and least radii being 115mm and 15 mm
respectively. Draw a normal and tangent to the spiral at a point 65 mm from the pole.
Important approach for construction!
Find total angular and total linear displacement and divide both in to same number of equal parts.
Angular displacement =360º, Linear displacement = 100mm
3’
Solution Steps
1. With PO & QO radii draw two
circles and divide them in
twelve equal parts. Name those
1’,2’,3’,4’, etc. up to 12’
2 .Similarly divided line PQ also in
twelve parts and name those
1,2,3,-- as shown.
3. Take O-1 distance from OP line
and draw an arc up to O1’ radius
vector. Name the point P1
4. Similarly mark points P2, P3, P4
up to P12
And join those in a smooth curve.
It is a SPIRAL of one convolution.
2’
4’
P3
P2
P4
5’
P1
P5
N
c Q
P6
6’
O
12 1110 9 8 7 6 5 4 3 2 1
P11
P7
P10
P8
P9
7’
C=(Rmax-Rmin)/No. of
convolutions in radians
= (115-15)/3.14 X 2 =15.92
1’
11’
8’
9’
10’
102
P
12’
Draw an Archemedian spiral of one and half convolution, greatest and least radii being 115mm and 15 mm
respectively. Draw a normal and tangent to the spiral at a point 70 mm from the pole.
Important approach for construction
Find total angular and total linear displacement and divide both in to same number of equal parts. Total
Angular displacement 540º. Total Linear displacement 100 mm
3’ 15’
1 Draw a 115 mm long line OP.
16’ 4’
2 Mark Q at 15 mm from O
3 with O as centre draw two circles with OP
and OQ radius
4 Divide the circle in 12 equal divisions and
17’ 5’
mark the divisions as 1’,2’ and so on up to 18’ P
5
5 Divide the line PQ in 18 equal divisions as 1,2,3 and so on upto 18
6.Take O-1 distance from OP line and
draw an arc up to O1’ radius vector.
P6
18’ 6’
Name the point P1
7.Similarly mark points P2, P3, P4 up
to P18.
8. And join those in a smooth curve.
It is a SPIRAL of one and half
convolution.
2’ 14’
P3
P2
P4
P1
P15
1’13’
P14
P16
P13
P17
P18
P12
Q
18
O
16
14
12
10
8
6
4
P
12’
2
P11
P7
P10
P8
7’
P9
C=(Rmax-Rmin)/No. of
convolutions in radians
= (115-15)/3.14 X3 =10.61
8’
9’
11’
10’
103
Spiral.
Method of Drawing
Tangent & Normal
SPIRAL (ONE CONVOLUSION.)
2
P2
3
P1
Q
Difference in length of any radius vectors
1
Constant of the Curve =
Angle between the corresponding
radius vector in radian.
P3
=
P4
4
O
P5
7 6 5 4 3 2 1
P7
P6
7
5
6
P
OP – OP2
/2
=
OP – OP2
1.57
= 3.185 m.m.
STEPS:
*DRAW SPIRAL AS USUAL.
DRAW A SMALL CIRCLE OF RADIUS EQUAL TO THE
CONSTANT OF CURVE CALCULATED ABOVE.
* LOCATE POINT Q AS DISCRIBED IN PROBLEM AND
THROUGH IT DRAW A TANGENTTO THIS SMALLER
CIRCLE.THIS IS A NORMAL TO THE SPIRAL.
*DRAW A LINE AT RIGHT ANGLE
*TO THIS LINE FROM Q.
IT WILL BE TANGENT TO CYCLOID.
104
SPIRAL
of
two convolutions
Problem 28
Point P is 80 mm from point O. It starts moving towards O and reaches it in two
revolutions around.it Draw locus of point P (To draw a Spiral of TWO convolutions).
IMPORTANT APPROACH FOR CONSTRUCTION!
FIND TOTAL ANGULAR AND TOTAL LINEAR DISPLACEMENT
AND DIVIDE BOTH IN TO SAME NUMBER OF EQUAL PARTS.
2,10
SOLUTION STEPS:
Total angular displacement here
is two revolutions And
Total Linear displacement here
is distance PO.
Just divide both in same parts
i.e.
Circle in EIGHT parts.
( means total angular
displacement
in SIXTEEN parts)
Divide PO also in SIXTEEN
parts.
Rest steps are similar to the
previous
problem.
P2
P1
3,11
1,9
P3
P10
P9
P11
4,12
16
13
10
P4
P12
8 7 6 5 4 3 2 1
P
P8
8,16
P15
P13
P14
P7
P5
5,13
P6
7,15
105
6,14
Problem No.7:
A Link OA, 80 mm long oscillates around O,
600 to right side and returns to it’s initial vertical
Position with uniform velocity.Mean while point
P initially on O starts sliding downwards and
reaches end A with uniform velocity.
Draw locus of point P
Solution Steps:
Point P- Reaches
End A (Downwards)
1) Divide OA in EIGHT equal parts and from O to
A after O name 1, 2, 3, 4 up to 8. (i.e. up to point
A).
2) Divide 600 angle into four parts (150 each) and
mark each point by A1, A2, A3, A4 and for return A5,
A6, A7 andA8. (Initial A point).
3) Take center O, distance in compass O-1 draw an
arc upto OA1. Name this point as P1.
1) Similarly O center O-2 distance mark P2 on line
O-A2.
2) This way locate P3, P4, P5, P6, P7 and P8 and
join them.
( It will be thw desired locus of P )
OSCILLATING LINK
O
p
1
p1
p2
p3
p4
2
3
p5
A4
4
5
p6
A3
6
7
A8
p8
p7
A5
A2
A6
A1
A7
A8
106
OSCILLATING LINK
Problem No 8:
A Link OA, 80 mm long oscillates around O,
600 to right side, 1200 to left and returns to it’s initial
vertical Position with uniform velocity.Mean while point
P initially on O starts sliding downwards, reaches end A
and returns to O again with uniform velocity.
Draw locus of point P
Op
16
15
1
14
Solution Steps:
( P reaches A i.e. moving downwards.
& returns to O again i.e.moves upwards )
1.Here distance traveled by point P is PA.plus A
12
AP.Hence divide it into eight equal parts.( so
total linear displacement gets divided in 16
parts) Name those as shown.
2.Link OA goes 600 to right, comes back to
A
A13 11
original (Vertical) position, goes 600 to left
and returns to original vertical position. Hence
total angular displacement is 2400.
Divide this also in 16 parts. (150 each.)
Name as per previous problem.(A, A1 A2 etc)
3.Mark different positions of P as per the
procedure adopted in previous case.
and complete the problem.
p1
p2
p3
p4
2
13
p
3
12
A4
5
4
11
p6
5
10
A10
A14
6
9 7
A9
A15
A3
8
A p8
A8
A16
p7
A1
A7
A5
A2
A6
107
ROTATING LINK
Problem 9:
Rod AB, 100 mm long, revolves in clockwise direction for one revolution.
Meanwhile point P, initially on A starts moving towards B and reaches B.
Draw locus of point P.
1) AB Rod revolves around
center O for one revolution and
point P slides along AB rod and
reaches end B in one
revolution.
2) Divide circle in 8 number of
equal parts and name in arrow
direction after A-A1, A2, A3, up
to A8.
3) Distance traveled by point P
is AB mm. Divide this also into 8
number of equal parts.
4) Initially P is on end A. When
A moves to A1, point P goes
one linear division (part) away
from A1. Mark it from A1 and
name the point P1.
5) When A moves to A2, P will
be two parts away from A2
(Name it P2 ). Mark it as above
from A2.
6) From A3 mark P3 three
parts away from P3.
7) Similarly locate P4, P5, P6,
P7 and P8 which will be eight
parts away from A8. [Means P
has reached B].
8) Join all P points by smooth
curve. It will be locus of P
A2
A1
A3
p1
p2
p6
p5
A
P
1
2
3
p7
p3
p4
A7
4
5
p8
B A4
6
7
A5
A6
108
Problem 10 :
Rod AB, 100 mm long, revolves in clockwise direction for one revolution.
Meanwhile point P, initially on A starts moving towards B, reaches B
And returns to A in one revolution of rod.
Draw locus of point P.
ROTATING LINK
A2
Solution Steps
1) AB Rod revolves around center O
for one revolution and point P slides
along rod AB reaches end B and
returns to A.
2) Divide circle in 8 number of equal
parts and name in arrow direction
after A-A1, A2, A3, up to A8.
3) Distance traveled by point P is AB
plus AB mm. Divide AB in 4 parts so
those will be 8 equal parts on return.
4) Initially P is on end A. When A
moves to A1, point P goes one
linear division (part) away from A1.
Mark it from A1 and name the point
P1.
5) When A moves to A2, P will be
two parts away from A2 (Name it P2
). Mark it as above from A2.
6) From A3 mark P3 three parts
away from P3.
7) Similarly locate P4, P5, P6, P7
and P8 which will be eight parts away
from A8. [Means P has reached B].
8) Join all P points by smooth curve.
It will be locus of P
The Locus will follow the loop
path two times in one revolution.
A1
A3
p5
p1
p4
A
P
p8
p2
1+7
2+6 p
6
+3 5
4
+B
p7 p3
A7
A5
A6
109
A4
Problem 28: A link OA, 100 mm long rotates about O in
anti-clockwise direction. A point P on the link, 15 mm
away from O, moves and reaches the end A, while the link
has rotated through 2/5 of a revolution. Assuming that the
movements of the link to be uniform trace the path of
point P.
θ= 2/5 X 360º = 144º
Total angular movement = 144º
Total linear movement = 85 mm
To divide both of them in equal
no. of parts ( say 8)
5’
6’
4’
7’
3’
P6
P7
8’
P5
2’
P4
P8
P3
1’
P2
144º
P1
O
15
P
1
2
3
100
4
5
6
7
8
110
A
Logarithmic Spiral:
If a point moves around a pole in such a way that
The value of vectorial angle are in arithmatic progression and
The corresponding values of radius vectors are in geometric progression, then the curve
traced by the point is known as logarithmic spiral.
A3
A2
P3
A1
P2
θ
θ
P1
θ
O
A
P
Let OA be a straight line and P be a point on it at radius vector OP from O.
Now let the line moves at uniform angular speed to a new position OA1 ,at vectorial angle θ from
OA and the point moves to a new position P1 , at radius vector OP1 from O.
The line now gradually moves to the new position OA2, OA3 at vectorial angle θ and the point to P2
and P3 , at radius vectors OP2 and OP3 respectively.
In Logarithmic spiral OP3/OP2 =OP2/OP1=OP1/OP
111
Problem37: In a logarithmic spiral, the shortest radius is 40mm. The length of adjacent radius vectors
enclosing 30º are in the ratio of 9:8 Construct one revolution of the spiral. Draw tangent to the spiral at a
point 70 mm from it.
First step is to draw logarithmic scale.
Draw two straight lines OA & OB at angle of 30º.
B
P12
Mark a point P on OA at 40 mm from O.
Calculate OP1 such that OP1/OP = 9/8. => OP1 = 45 mm
Mark a OP1 on OB at 45 mm from O.
Join P with P1.
P4
P3
Draw an arc of radius OP1 from OB to OA.
P2
O
P5
Draw a line parallel to PP1 from P1 on OA to intersect OB at P2.
P
Repeat the steps to get the points P3,P4 and so1 on up to P12.
P1
30º
40
P
P2P3
P4
P
P5 6
P7
P8
P9
P10
P11
P1P2 P3 P4 P5 P6 P7 P8 P9 P10 P11 P12
P6
P
P12
P7
P11
P8
P9
P10
112
A