Engineering drawing is a language of all persons involved in engineering activity. Engineering ideas are recorded by preparing drawings and execution of work is also carried out on the basis of drawings. Communication in engineering field is done by drawings. It is called as a “Language of Engineers”.

CHAPTER – 2

ENGINEERING

CURVES

Useful by their nature & characteristics. Laws of nature represented on graph.

Useful in engineering in

understanding laws,

manufacturing of various items,

designing mechanisms analysis of

forces, construction of bridges,

dams, water tanks etc.

USES OF ENGINEERING CURVES

1. CONICS

2. CYCLOIDAL CURVES3. INVOLUTE

4. SPIRAL

5. HELIX

6. SINE & COSINE

CLASSIFICATION OF ENGG. CURVES

It is a surface generated by moving a Straight line keeping one of its end fixed & other end makes a closed curve.

What is Cone ?

If the base/closed curve is a polygon, we get a pyramid.

If the base/closed curve is a circle, we get a cone.

The closed curve is known as base.

The fixed point is known as vertex or apex. Vertex/Apex

90º

Base

If axis of cone is not perpendicular to base, it is called as oblique cone.

The line joins vertex/ apex to the circumference of a cone is known as generator.

If axes is perpendicular to base, it is called as right circular cone.

Generato

r

Cone Axis

The line joins apex to the center of base is called axis.

90º

Base

Vertex/Apex

Definition :- The section obtained by the intersection of a right circular cone by a cutting plane in different position relative to the axis of the cone are called CONICS.

CONICS

B - CIRCLE

A - TRIANGLE

CONICS

C - ELLIPSE

D – PARABOLA

E - HYPERBOLA

When the cutting plane contains the apex, we get a triangle as the section.

TRIANGLE

When the cutting plane is perpendicular to the axis or parallel to the base in a right cone we get circle the section.

CIRCLE

Sec Plane

Circle

Definition :-When the cutting plane is inclined to the axis but not parallel to generator or the inclination of the cutting plane(α) is greater than the semi cone angle(θ), we get an ellipse as the section.

ELLIPSE

α

θ

α > θ

When the cutting plane is inclined to the axis and parallel to one of the generators of the cone or the inclination of the plane(α) is equal to semi cone angle(θ), we get a parabola as the section.

PARABOLA

θ

α

α = θ

When the cutting plane is parallel to the axis or the inclination of the plane with cone axis(α) is less than semi cone angle(θ), we get a hyperbola as the section.

HYPERBOLADefinition :-

α < θα = 0

θθ

CONICSDefinition :- The locus of point moves in a plane such a way that the ratio of its distance from fixed point (focus) to a fixed Straight line (Directrix) is always constant.

Fixed point is called as focus.Fixed straight line is called as directrix.

M

C FV

P

Focu

s

Conic

CurveDirectrix

The line passing through focus & perpendicular to directrix is called as axis.The intersection of conic curve with

axis is called as vertex.

AxisM

CF

V

P

Focus

Conic

CurveDirectrix

Vertex

N Q

Ratio =Distance of a point from

focusDistance of a point from

directrix= Eccentricity= PF/PM = QF/QN = VF/VC = e

M P

F

Axis

CV

Focus

Conic

CurveDirectrix

Vertex

Vertex

Ellipse is the locus of a point which moves in a plane so that the ratio of its distance from a fixed point (focus) and a fixed straight line (Directrix) is a constant and less than one.

ELLIPSE

M

N Q

P

CF

V

Axi

s

Focu

s

Ellips

eDirectri

x

Eccentricity=PF/PM = QF/QN

< 1.

Ellipse is the locus of a point, which moves in a plane so that the sum of its distance from two fixed points, called focal points or foci, is a constant. The sum of distances is equal to the major axis of the ellipse.

ELLIPSE

F1

A B

P

F2

O

Q

C

D

F1

A B

C

D

P

F2

O

PF1 + PF2 = QF1 + QF2 = CF1 +CF2 =

constant

= Major Axis

Q

= F1A + F1B = F2A + F2B

But F1A = F2B

F1A + F1B = F2B + F1B = AB

CF1 +CF2 = AB

but CF1 = CF2

hence,

CF1=1/2AB

F1 F2

OA B

C

D

Major Axis = 100 mm

Minor Axis = 60 mm

CF1 = ½ AB = AO

F1 F2

OA B

C

D

Major Axis = 100 mm

F1F2 = 60 mm

CF1 = ½ AB = AO

Uses :-

Shape of a man-hole.

Flanges of pipes, glands and stuffing

boxes.

Shape of tank in a

tanker.

Shape used in bridges and arches.

Monuments.

Path of earth around the sun.

Shape of trays etc.

Ratio (known as eccentricity) of its distances from focus to that of directrix is constant and equal to one (1).

PARABOLAThe parabola is the locus of a point,

which moves in a plane so that its distance from a fixed point (focus) and a fixed straight line (directrix) are always equal.

Definition :-

Directrix AxisVertex

M

C

N Q

FV

P

Focus

Parabol

a

Eccentricity = PF/PM = QF/QN = 1.

Motor car head lamp

reflector.Sound reflector and

detector.

Shape of cooling towers.

Path of particle thrown at any angle with earth, etc.

Uses :-

Bridges and arches construction

Home

It is the locus of a point which moves in a plane so that the ratio of its distances from a fixed point (focus) and a fixed straight line (directrix) is constant and grater than one.

Eccentricity = PF/PM

AxisDirectrix

Hyperbol

aM

C

N Q

FV

P

Focu

s

Vertex

HYPERBOLA

= QF/QN

> 1.

Nature of graph of Boyle’s

law

Shape of overhead water

tanks

Uses :-

Shape of cooling towers

etc.

METHODS FOR DRAWING ELLIPSE

2. Directrix Focus Method

1. Arc of Circle’s Method

Nor

mal

P2’

R =

A1

Tangent

1 2 3 4A B

C

D

P1

P3

P2

P4 P4 P3

P2

P1

P1’

F2

P3’ P4’ P4’P3’

P2’

P1’

90°

F1

Rad =B1

R=B2

`R=

A2

O

ARC OF CIRCLE’S METHOD

P6

Nor

mal

P5’ P7’P6’

P1

TangentP1’

N

N

T

T

V1

P5

P4’

P4

P3’P2’

F1

D1

D1

R1

ba

cd

ef

g

Q

P7P3P2

Dir

ectr

ix

R=6f`

90°

1 2 3 4 5 6 7

Eccentricity = 2/3

3R1V1

QV1 = R1V1

V1F1 = 2

Ellipse

ELLIPSE – DIRECTRIX FOCUS METHOD

R=1a

Dist. Between directrix & focus = 50 mm

1 part = 50/(2+3)=10 mm V1F1 = 2 part = 20 mmV1R1 = 3 part = 30 mm

< 45º

S

PROBLEM :-

The distance between two coplanar fixed points is 100 mm. Trace the complete path of a point G moving in the same plane in such a way that the sum of the distance from the fixed points is always 140 mm.

Name the curve & find its eccentricity.

ARC OF CIRCLE’S METHOD

Nor

mal

G2’

R =

A1

Tangent

1 2 3 4A B

G

G’

G1

G3

G2

G4 G4 G3

G2

G1

G1’

G3’ G4’ G4’G3’

G2’

G1’

F2F1

R=B1

R=B2

`R=

A2

O

90°

90°

dir

ectr

ix

100

140

GF1 + GF2 = MAJOR AXIS = 140

E

e AF1

AEe =

R=70 R=70

PROBLEM :-

Draw an ellipse passing through A & B of an equilateral triangle of ABC of 50 mm edges with side AB as vertical and the corner C coincides with the focus of an ellipse. Assume eccentricity of the curve as 2/3. Draw tangent & normal at point A.

PROBLEM :-

Draw an ellipse passing through all the four corners A, B, C & D of a rhombus having diagonals AC=110mm and BD=70mm.

Use “Arcs of circles” Method and find its eccentricity.

METHODS FOR DRAWING PARABOLA

1. Rectangle Method

2. Parabola in Parallelogram

3. Directrix Focus Method

2345

0

1

2

3

4

5

6 1 1 5432 6

0

1

2

3

4

5

0

VD C

A B

P4 P4

P5 P5

P3 P3

P2 P2

P6 P6

P1 P1

PARABOLA –RECTANGLE METHOD

PARABOLA

B

0

2’

0

2

6

C

6’

V

5

P’5

30°

A X

D

1’2’

4’5’

3’

1

34

5’

4’

3’

1’

0

5

4

3

2

1

P1

P2

P3

P4

P5

P’4

P’3

P’2

P’1

P’6

PARABOLA – IN PARALLELOGRAM

P6

D

D

DIR

EC

TR

IX

90° 2 3 4T

TN

N

S

V 1

P1

P2

PF

P3

P4

P1’

P2’P3’

P4’

PF’

AXIS

RF

R2

R1

R3

R4

90°

R F

PARABOLA DIRECTRIX FOCUS METHOD

PROBLEM:-A stone is thrown from a building 6 m high. It just crosses the top of a palm tree 12 m high. Trace the path of the projectile if the horizontal distance between the building and the palm tree is 3 m. Also find the distance of the point from the building where the stone falls on the ground.

6m

ROOT OF TREE

BUILDING

REQD.DISTANCE

TOP OF TREE

3m

6m

F

A

STONE FALLS HERE

3m

6m

ROOT OF TREE

BUILDING

REQD.DISTANCE

GROUND

TOP OF TREE

3m

6m

1

2

3

1

2

3

321 4 5 6

5

6

EF

A B

CD

P3

P4

P2

P1

P

P1

P2

P3

P4

P5

P6

3 2 10

STONE FALLS HERE

PROBLEM:-

In a rectangle of sides 150 mm and 90 mm, inscribe two parabola such that their axis bisect each other. Find out their focus points & positions of directrix.

150 mmA

B C

D1 2 3 4 5

1

2

3

4

5

O

P1

P2

P3

P4

P5

M1’ 2’ 3’ 4’ 5’

1’

2’3’

4’5’

P1’ P

2’ P

3’ P

4’ P

5’

90 m

m

example

Draw a parabola passing through three

different points A, B and C such that AB =

100mm, BC=50mm and CA=80mm

respectively.

BA

C

100

50

80

1’

0

0

2’

0

26

6’5

P’5

1’ 2’ 4’ 5’3’134

5’

4’

3’

54

3

21

P1P

2P

3

P4

P5

P’4

P’3

P’2

P’1

P’6

P6

A B

C

METHODS FOR DRAWING HYPERBOLA

1. Rectangle Method

2. Oblique Method

3. Directrix Focus Method

D

F

1 2 3 4 5

5’4’3’

2’

P1

P2

P3P4

P5

0

P6

P0

AO EX

B

C

Y

Given Point P0

90°

6

6’

Hyperbola

RECTANGULAR HYPERBOLA

AXIS

AXIS

When the asymptotes are at right angles to each other, the hyperbola is called rectangular or equilateral hyperbola

ASYMPTOTES X and Y

Problem:-

Two fixed straight lines OA and OB are at right angle to each other. A point “P” is at a distance of 20 mm from OA and 50 mm from OB. Draw a rectangular hyperbola passing through point “P”.

D

F

1 2 3 4 5

5’4’3’

2’

P1

P2

P3P4

P5

0

P6

P0

AO EX=20

B

C

Y =

50

Given Point P0

90°

6

6’

Hyperbola

RECTANGULAR HYPERBOLA

PROBLEM:-

Two straight lines OA and OB are at 75° to each other. A point P is at a distance of 20 mm from OA and 30 mm from OB. Draw a hyperbola passing through the point “P”.

75 0

P4

E

6’

2’1’ P1

1 2 3 4 5 6 D

P6P5

P3P2

P0

7’P7

7C

B F

O

Y =

30

X = 20

Given Point P0

A

AXIS

NORMAL

C V F1

DIR

EC

TR

IXD

D

1 2 3 4

4’

3’

2’

1’P1

P2

P3

P4

P1’

P2’

P3’

P4’

T1

T2

N2

N1

TAN

GE

NT

s

Directrix and focus method

CYCLOIDAL GROUP OF CURVESWhen one curve rolls over another curve without slipping or sliding, the path Of any point of the rolling curve is called as ROULETTE.When rolling curve is a circle and the curve on which it rolls is a straight line Or a circle, we get CYCLOIDAL GROUP OF CURVES.

Cycloidal Curves

Cycloid Epy Cycloid Hypo Cycloid

Rolling Circle or Generator

CYCLOID:-

Cycloid is a locus of a point on the circumference of a rolling circle(generator), which rolls without slipping or sliding along a fixed straight line or a directing line or a director.

C

P P

P

R

C

Directing Line or Director

EPICYCLOID:-Epicycloid is a locus of a point(P) on the circumference of a rolling circle(generator), which rolls without slipping or sliding OUTSIDE another circle called Directing Circle.

2πr

Ø = 360º x r/Rd

Circumference of Generating Circle

R d

Rolling Circle

r

O

Ø/2

Ø/2

P0 P0

Arc P0P0 =

Rd x Ø =

P0

HYPOCYCLOID:-Hypocycloid is a locus of a point(P) on the circumference of a rolling circle(generator), which rolls without slipping or sliding INSIDE another circle called Directing Circle.`

DirectingCircle(R)

P

Ø /2 Ø /2

Ø =360 x r

RR

T

Rolling CircleRadius (r)

O

Vertical

Hypocycloid

P P

P0

2R or D 5

T

T

1

2

1 2 3 4 6 7 8 9 10 110 120

3

4

567

8

9

1011 12

P1

P2

P3

P4

P5 P7

P8

P9

P11

P12

C0 C1 C2 C3 C4 C5 C6 C7 C8 C9 C10C11

Directing Line

C12

N

N

S

S1

R

P6

R

P10

R

: Given Data :

Draw cycloid for one revolution of a rolling circle having diameter as 60mm.

Rolling Circle

D

C0

P0

78

P64

P1

1

2

3

C2 C3

P2 C4

Problem 1:A circle of diameter D rolls without

slip on a horizontal surface (floor) by Half revolution and then it rolls up a vertical surface (wall) by another half revolution. Initially the point P is at the Bottom of circle touching the floor. Draw the path of the point P.

5

6 C1

P3

P4

P5

P7

P8

70

C 5

C 6

C 7

C 8

1 2 3 4

56

D/2

πD

/2

πD/2 D/2Floor

Wal

l

CYCLOID

56

7

8

Take diameter of circle = 40mmInitially distance of centre of

circle from the wall 83mm (Hale circumference + D/2)

Problem : 2

A circle of 25 mm radius rolls on the

circumference of another circle of 150

mm diameter and outside it. Draw the

locus of the point P on the

circumference of the rolling circle for

one complete revolution of it. Name

the curve & draw tangent and normal

to the curve at a point 115 mm from

the centre of the bigger circle.

First Step : Find out the included angle by using the equation

360º x r / R = 360 x 25/75 = 120º.

Second step: Draw a vertical line & draw two lines at 60º on either sides.

Third step : at a distance of 75 mm from O, draw a part of the circle taking radius = 75 mm.

Fourth step : From the circle, mark point C outside the circle at distance of 25 mm & draw a circle taking the centre as point C.

P6

P4

rP2

C1

C0

C2

C3C4 C5

C6

C7

C8

1

0

23

4

5

6 7

O

R d

Ø/2 Ø/2

P1P0

P3 P5

P7P8

r rRolling Circle

r

Rd X Ø = 2πrØ = 360º x r/Rd

Arc P0P8 = Circumference of Generating Circle

EPICYCLOIDGIVEN:Rad. Of Gen. Circle (r)& Rad. Of dir. Circle (Rd) S

ºU

N

Ø = 360º x 25/75

= 120°

Problem :3

A circle of 80 mm diameter rolls on

the circumference of another circle of

120 mm radius and inside it. Draw the

locus of the point P on the

circumference of the rolling circle for

one complete revolution of it. Name

the curve & draw tangent and normal

to the curve at a point 100 mm from

the centre of the bigger circle.

P0 P1

Tangent

P11

r

C0

C1

C2

C3

C4C5

C6 C7 C8 C9

C10

C11

C12

P10P8

0

1 2 3

4

5

67

89

10

11

12P2 P3

P4P5 P6

P9P7

P12

/2

/2

=360 x 412

= 360 x rR

= 120°

R

T

T

N

S

N

Norm

alr

r

Rolling Circle

Radias (r)

DirectingCircle

O

Vertical

Hypocycloid

Problem :

Show by means of drawing that when the diameter of rolling circle is half the diameter of directing circle, the hypocycloid is a straight line

C

C1

C2

C3

C4C5

C6 C7

C9

C8

C10

C11

C12P8O

10

5

7

89

11

12

1

2 34

6P1

P2

P3 P4P5 P6 P7 P9

P10

P11

P12

Directing Circle

Rolling Circle

HYPOCYCLOID

INVOLUTE

DEFINITION :- If a straight line is rolled

round a circle or a polygon without

slipping or sliding, points on line will

trace out INVOLUTES.OR

Uses :- Gears profile

Involute of a circle is a curve traced out

by a point on a tights string unwound or

wound from or on the surface of the

circle.

PROB:

A string is unwound from a circle of 20 mm diameter. Draw the locus of string P for unwounding the string’s one turn. String is kept tight during unwound. Draw tangent & normal to the curve at any point.

P12

P2

002 12

6

P11 20 9 103 4 6 8 115 7 12

DP3

P4

P5

P6

P7

P8

P9

P10

P11

123

45 7

8910

11

03

04

05

06

07

08

09

010`

011

Tangent

N

N

Norm

alT

T.

PROBLEM:-

Trace the path of end point of a thread when it is wound round a circle, the length of which is less than the circumference of the circle.

Say Radius of a circle = 21 mm & Length of the thread = 100 mm

Circumference of the circle = 2 π r = 2 x π x 21 = 132 mm

So, the length of the string is less than circumference of the circle.

P

R=7toPR=6toP

R21

0

0 1 2 3 4 5 6 7 8 P11 0

1

2

3

456

7

8

9

10

P1

P2

P3

P4

P5

P6

P7

P8

L= 100 mm

R=1toP

R=2

toPR

=3

toPR

=4toPR=5toP

INVOLUTE

9

ø

11 mm = 30°Then 5 mm =

Ø = 30° x 5 /11 = 13.64 °

S = 2 x π x r /12

PROBLEM:-

Trace the path of end point of a thread when it is wound round a circle, the length of which is more than the circumference of the circle.

Say Radius of a circle = 21 mm & Length of the thread = 160 mm

Circumference of the circle = 2 π r = 2 x π x 21 = 132 mm

So, the length of the string is more than circumference of the circle.

P13

P11

313

14

15

P0

P12

O

7

10 123

45689

1112 1 2

P1

P2

P3P4

P5

P6

P7

P8 P9 P10

P14P

L=160

mm

R=21mm

64 5 7 8 9 101112131415

ø

PROBLEM:-

Draw an involute of a pantagon having side as 20 mm.

P5

R=01

R=201

P0

P1

P2

P3

P4

R=3

01

R=401

R=501

23

4 51

T

TN

NS

INVOLUTEOF A POLYGON

Given : Side of a polygon 0

PROBLEM:-

Draw an involute of a square

having side as 20 mm.

P2

1

2 3

04

P0

P1

P3

P4

N

NS

R=3

01

R=401

R=201

R=0

1

INVOLUTE OF A SQUARE

PROBLEM:-

Draw an involute of a string unwound from the given figure from point C in anticlockwise direction.

60°

A

B

C

R2130°

R21

60°

A

B

C

30°X

X+A1

XX+

A2

X+

A3

X+A5

X+

A4

X+AB

R

=X+AB

X+

66+

BC

12

3

45

C0

C1

C2C3

C4

C5

C6

C7

C8

A stick of length equal to the circumference of a

semicircle, is initially tangent to the semicircle

on the right of it. This stick now rolls over the

circumference of a semicircle without sliding till

it becomes tangent on the left side of the

semicircle. Draw the loci of two end point of this

stick. Name the curve. Take R= 42mm.

PROBLEM:-

A6

B6

5

A

B

C

B1

A1

B2

A2

B3A3

B4

A4

B5

A5

12 3

4

5

O

1

2

3

4

6

INVOLUTE

SPIRALS

If a line rotates in a plane about one of its

ends and if at the same time, a point moves

along the line continuously in one

direction, the curves traced out by the

moving point is called a SPIRAL.

The point about which the line rotates is called a POLE.

The line joining any point on the curve with the pole is called the RADIUS VECTOR.

The angle between the radius vector and the line in its initial position is called the VECTORIAL ANGLE.

Each complete revolution of the curve is termed as CONVOLUTION.

Spiral

Arche Median Spiral for Clock

Semicircle Quarter Circle Logarithmic

ARCHEMEDIAN SPIRAL

It is a curve traced out by a point moving in such a way that its movement towards or away from the pole is uniform with the increase of vectorial angle from the starting line.

USES :-

Teeth profile of Helical gears.

Profiles of cams etc.

To construct an Archemedian Spiral of one convolutions, given the radial movement of the point P during one convolution as 60 mm and the initial position of P is the farthest point on the line or free end of the line.

Greatest radius = 60 mm &

Shortest radius = 00 mm ( at centre or at pole)

PROBLEM:

P10

1

23

4

5

6

7

89

10

11

120

8 7 012345691112

P1

P2P3

P4

P5

P6

P7P8

P9

P11P12

o

To construct an Archemedian Spiral of one convolutions, given the greatest & shortest(least) radii.

Say Greatest radius = 100 mm &

Shortest radius = 60 mm

To construct an Archemedian Spiral of one convolutions, given the largest radius vector & smallest radius vector.

OR

3 1

26

5

8 4

79

1011

2

1

34

5

6

7

89

10

11

12

P1

P2P3

P4

P5

P6

P7

P8 P9

P10

P11

P12

O

N

N

T

T

S

R m

in

R max

Diff. in length of any two radius vectors

Angle between them in radiansConstant of the curve =

=OP – OP3

Π/2

100 – 90=

Π/2

= 6.37 mm

PROBLEM:-

A slotted link, shown in fig rotates in the horizontal plane about a fixed point O, while a block is free to slide in the slot. If the center point P, of the block moves from A to B during one revolution of the link, draw the locus of point P. OAB

40 25

B A O1234567891011

P1

P2

P3 P4

P5

P6

P7

P8

P9P10

P11

P12

11

21

31

41

51

61

71

81

91

101

111

40 25

PROBLEM:-

A link OA, 100 mm long rotates about O in clockwise direction. A point P on the link, initially at A, moves and reaches the other end O, while the link has rotated thorough 2/3 rd of the revolution. Assuming the movement of the link and the point to be uniform, trace the path of the point P.

AInitial Position of point PPO

P1

P2

P3

P4P5

P6

P7

P8

21

34567O

1

2

3

4

56

7

8

2/3 X 360°

= 240°

120º

A0

Linear Travel of point P on AB

= 96 =16x (6 div.)

EXAMPLE: A link AB, 96mm long initially is vertically upward w.r.t. its pinned end B, swings in clockwise direction for 180° and returns back in anticlockwise direction for 90°, during which a point P, slides from pole B to end A. Draw the locus of point P and name it. Draw tangent and normal at any point on the path of P.

P1’

A

B

A1

A2

A3

A4

A5

A6P0

P1

P2

P3

P4

P5

P6

P2’

P3’

P4’P5’

P6’

9 6

Link AB = 96

C

Tangent

Angular Swing of link AB = 180° + 90°

= 270 ° =45 °X 6 div.

ARCHIMEDIAN SPIRAL

D

NO

RM

AL

M

N

Arch.Spiral Curve Constant BC

= Linear Travel ÷Angular Swing in Radians

= 96 ÷ (270º×π /180º)

=20.363636 mm / radian

Problem : 2

Draw a cycloid for a rolling circle, 60

mm diameter rolling along a straight

line without slipping for 540°

revolution. Take initial position of the

tracing point at the highest point on

the rolling circle. Draw tangent &

normal to the curve at a point 35 mm

above the directing line.

First Step : Draw a circle having diameter of 60 mm.

Second step: Draw a straight line tangential to the circle from bottom horizontally equal to

(540 x ) x 60 mm= 282.6 mm i.e. 1.5 x x 60 mm 360

Third step : take the point P at the top of the circle.

Rolling circle

P1

P2

P3

P4

P0

P6

P7

P8

P5

P9

P10

1

2

34

5

6

7 89

10

1 2 3 4 5 6 7 8 9 100

C0 C1 C2C3 C4

Directing line

Length of directing line = 3D/2

540 = 360 + 180

540 = D + D/2 Total length for 540 rotation = 3D/2

C5 C6 C7C8 C9 C10

S

norm

al