CHAPTER – 1

ENGINEERING

CURVES

Useful by their nature & characteristics.

Laws of nature represented on graph.

Useful in engineering in understandinglaws, manufacturing of various items,designing mechanisms analysis of forces,construction of bridges, dams, watertanks etc.

USES OF ENGINEERING CURVES

Be it an arc

Be it an arch in construction in civil engineering

Be it an any spring in mechanical engineering

Be it any component of electronics and computer

engineering component

Measuring the distance

On the map

For navigation

In space technology

1. CONICS

2. CYCLOIDAL CURVES

3. INVOLUTE

4. SPIRAL

5. HELIX

6. SINE & COSINE

CLASSIFICATION OF ENGG. CURVES

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

What is Cone ?

If the base/closedcurve is a polygon, weget a pyramid.

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

The closed curve isknown 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.

Generator

Cone Axis

The line joins apex to the center of base is called

axis.

90º

Base

Vertex/Apex

Definition :- The section obtained by theintersection of a right circular cone by acutting plane in different position relativeto 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 theapex, we get a triangle as the section onthe surface of the cone.

TRIANGLE

When the cutting plane is perpendicular tothe axis or parallel to the base in a right conewe get circle as the section on the surface ofthe cone.

CIRCLE

Sec Plane

Circle

Definition :-When the cutting plane is inclined to the axisbut not parallel to generator or theinclination of the cutting plane(α) is greaterthan the semi cone angle(θ), we get an ellipseas the section on the surface of the cone.

ELLIPSE

α

θ

α > θ

When the cutting plane is inclined to the axisand parallel to one of the generators of the coneor the inclination of the plane(α) is equal to semicone angle(θ), we get a parabola as the section.

PARABOLA

θ

α

α = θ

When the cutting plane is parallel to theaxis or the inclination of the plane withcone 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 aplane such a way that the ratio of itsdistance from fixed point (focus) to a fixedStraight line (Directrix) is always constant.

Fixed point is called as focus.

Fixed straight line is called as directrix.

M

CF

V

P

Focus

Conic Curve

Directrix

The line passing through focus &

perpendicular to directrix is called as axis.

The intersection of conic curve with axis iscalled as vertex.

AxisM

CF

V

P

Focus

Conic Curve

Directrix

Vertex

N Q

Ratio =Distance of a point from focus

Distance of a point from directrix

= Eccentricity

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

M P

F

Axis

CV

Focus

Conic Curve

Directrix

Vertex

Vertex

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

ELLIPSE

M

N Q

P

CF

V

Axis

Focus

Ellipse

Directrix

Eccentricity=PF/PM

= QF/QN

< 1.

Ellipse is the locus of a point, which moves in aplane so that the sum of its distance from twofixed points, called focal points or foci, is aconstant. The sum of distances is equal to themajor 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

APPLICATION :-

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 distancesfrom focus to that of directrix is constant andequal to one (1).

PARABOLA

The parabola is the locus of a point, which movesin a plane so that its distance from a fixed point(focus) and a fixed straight line (directrix) arealways equal.

Definition :-

Directrix AxisVertex

M

C

N Q

FV

P

Focus

Parabola

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 aplane so that the ratio of its distances froma fixed point (focus) and a fixed straightline (directrix) is constant and grater thanone.

Eccentricity = PF/PM

AxisDirectrix

HyperbolaM

C

NQ

FV

P

FocusVertex

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. Concentric Circle Method

3. Loop Method

4. Oblong Method

5. Ellipse in Parallelogram

6. Trammel Method

7. Parallel Ellipse

8. Directrix Focus Method

1. Arc of Circle’s Method

P2’

1 2 3 4A B

C

D

P1

P3

P2

P4 P4P3

P2

P1

P1’

F2

P3’P4’ P4’

P3’

P2’

P1’

90°

F1 O

ARC OF CIRCLE’S METHOD

A BMajor Axis 7

8

910

11

9

8

7

6

54

3

2

1

12

11

P6

P5P4

P3

P2`

P1

P12

P11

P10P9

P8

P7

6

54

3

2

1

12C10

O

CONCENTRIC

CIRCLE

METHOD

F2F1

D

CF1=CF2=1/2 AB

T

N

Q

e = AF1/AQ

0

1

2

3

4

1 2 3 4 1’0’

2’3’4’

1’

2’

3’

4’

A B

C

D

Major Axis

Min

or

Axis

F1 F2

Dir

ectr

ix

E

F

S

P

P1

P2

P3

P4

P1’

P2’

P3’P4’

P0

P1’’

P2’’

P3’’

P4’’P4

P3

P2

P1

OBLONG METHOD

BA

P4

P0

D

C

60°

6

543

21

0

5 4 3 2 1 0 1 2 3 4 5 6

5

3210P1P2

P3

Q1

Q2Q3Q4

Q5

P6 Q6O

4

ELLIPSE IN PARALLELOGRAM

R4

R3

R2R1

S1

S2

S3

S4

P5

G

H

I

K

J

P6

P5’ P7’P6’

P1

P1’

T

T

V1

P5

P4’

P4

P3’P2’

F1

D1

D1

R1

ba

cd

ef

g

Q

P7P3P2

Directr

ix

90°

1 2 3 4 5 6 7

Eccentricity = 2/3

3R1V1

QV1=

R1V1

V1F1=

2

Ellipse

ELLIPSE – DIRECTRIX FOCUS METHOD

Dist. Between directrix

& focus = 50 mm

1 part = 50/(2+3)=10

mm V1F1 = 2 part = 20 mm

V1R1 = 3 part = 30 mm

< 45º

S

METHODS FOR DRAWING PARABOLA

1. Rectangle Method

2. Parabola in Parallelogram

3. Tangent Method

4. 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

2’

0

6

C

P’5

30°

A X

D5’

4’

3’

1’

0

5

4

3

2

1

P1

P2

P3

P4

P5

P’4

P’3P’2P’1

P’6

PARABOLA – IN PARALLELOGRAM

P6

BA O

V

1

8

3

4

5

2

6

7

9

10

0

1

2

3

4

5

6

7

8

9

1

0

0

F

PARABOLA

TANGENT METHOD

D

D

90° 2 3 4T

TN

N

S

V 1

P1

P2

PF

P3

P4

P1’

P2’P3’

P4’

PF’

AXIS

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 1

0

STONE FALLS HERE

EXAMPLE

A shot is discharge from the ground level at an angle 60 to the horizontal at a point 80m away from the point of discharge. Draw the path trace by the shot. Use a scale 1:100

ground level BA

60º

gun shot

80 M

parabola

ground level BA O

V

1

8

3

4

5

2

6

7

9

10

0

1

2

3

4

5

6

7

8

9

1

0

0

F

60º

gun shot

D D

VF

VE = e = 1

E

CYCLOIDAL GROUP OF CURVES

SuperiorHypotrochoid

Cycloidal Curves

Cycloid Epy Cycloid Hypo Cycloid

SuperiorTrochoid

InferiorTrochoid

SuperiorEpytrochoi

d

InferiorEpytrochoi

d

InferiorHypotrochoi

d

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

Rolling

Circle

r

O

Ø/2 Ø/2P0 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

Hypocycloi

d

P P

If the point is inside the circumference of the circle,

it is called inferior trochoid.

If the point is outside the circumference of the

circle, it is called superior trochoid.

What is TROCHOID ?

DEFINITION :- It is a locus of a point

inside/outside the circumference of a rolling

circle, which rolls without slipping or sliding

along a fixed straight line or a fixed circle.

P0

2R or D

51

2

1 2 3 4 6 7 8 9 10 110 120

3

4

56

7

8

9

10

11 12P1

P2

P3

P4

P5 P7

P8

P9

P11

P12

C0 C1 C2 C3 C4 C5 C6 C7 C8 C9 C10C11

Directing Line

C12

N

S1

R

P6

R

P10

R

: Given Data :

Draw cycloid for one revolution of a rolling circle having

diameter as 60mm.

Rolling

Circle

D

Problem : 2A 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

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) Sº

U

N

Ø = 360º x 25/75

= 120°

Problem :3A 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 P11

C0

C1

C2

C3

C4

C5C6 C7 C8

C9

C10

C1

1

C12

P1

0P8

0

1 23

4

5

67

89

10

1

1

1

2

P2

P3

P4P5 P6

P9P7

P12

/2

/2

=360 x 412

= 360 x rR

=120°

R

T

T

N

S

N

Rolling Circle

Radias (r)

DirectingCircle

O

Vertical

Hypocycloi

d

INVOLUTEDEFINITION :- If a straight line is rollesround a circle or a polygon without slipping or sliding, the locus of points on the straight line is an INVOLUTES to the curve.

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.

PROBLEM:A string is unwound from a

circle of 20 mm diameter. Draw the

locus of string P for unwinding the

string’s one turn. String is kept tight

during unwound. Draw tangent &

normal to the curve at any point.

P12

P2

012

6

P11 20 9 103 4 6 8 115 7 12

DP3

P4

P5

P6

P7

P8

P9

P10

P11

123

45

78

910

11N

.

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=6toP

0

0 1 2 3 4 5 6 7 8 P

11 01

2

3

456

7

8

9

10

P1

P2

P3

P4

P5

P6

P7

P8

L= 100 mm

R=

3to

P

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

456

89

1112 1 2

P1

P2

P3P4

P5

P6

P7

P8 P9 P10

P14P

L=160 mm

R=21mm

64 5 7 8 9 10 11 12 131415

ø

PROBLEM:-

Draw an involute of a pantagon having

side as 20 mm.

P5

P0

P1

P2

P3

P4

23

4 5

1

INVOLUTE

OF 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

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

30°

60°

A

B

C

30°

X

X+

A3

X+AB

12

3

4

5

C0

C1

C2

C3

C4

C5

C6

C7

C8

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

CircleLogarithmic

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

2

3

4

5

6

7

89

10

11

120

8 7 012345691112

P1

P2P3

P4

P5

P6

P7P8

P9

P11

P12

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

1

011

2

1

34

5

6

7

89

10

11

12

P1

P2P3

P4

P5

P6

P7

P8 P9

P1

0

P1

1

P12

O

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 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

PROBLEM :

A monkey at 20 m slides down from a

rope. It swings 30° either sides of

rope initially at vertical position. The

monkey initially at top reaches at

bottom, when the rope swings about

two complete oscillations. Draw the

path of the monkey sliding down

assuming motion of the monkey and

the rope as uniform.

θ

o

012

3

45 6

78

9

1011

121314

15

16 17 18 1920

21

222324

23

13

22

24

12

34

56

78

910

1112

1415

16

1819

2021

17

P3

P9

P15

http://www.engineering108.com/pages/Engineeri

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_download.html

A text book of engineering graphics- Prof. P.J

SHAH

Engineering Drawing-N.D.Bhatt

Engineering Drawing-P.S.Gill

Thank You