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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

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Ellipsce, parabola, hyperbola, cycloid, hypocycloid, epicycloid,rectangular method,spiral,helix

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Page 1: Engg curves

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”.

Page 2: Engg curves

CHAPTER – 2

ENGINEERING

CURVES

Page 3: Engg 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

Page 4: Engg curves

1. CONICS

2. CYCLOIDAL CURVES3. INVOLUTE

4. SPIRAL

5. HELIX

6. SINE & COSINE

CLASSIFICATION OF ENGG. CURVES

Page 5: 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

Page 6: Engg curves

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

Page 7: Engg curves

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

Page 8: Engg curves

B - CIRCLE

A - TRIANGLE

CONICS

C - ELLIPSE

D – PARABOLA

E - HYPERBOLA

Page 9: Engg curves

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

TRIANGLE

Page 10: Engg curves

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

Page 11: Engg curves

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

Page 12: Engg curves

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

Page 13: Engg curves

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

Page 14: Engg curves

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

Page 15: Engg curves

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

axis is called as vertex.

AxisM

C FV

P

Focus

Conic

CurveDirectrix

Vertex

Page 16: Engg curves

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

Page 17: Engg curves

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.

Page 18: Engg curves

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

Page 19: Engg curves

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

Page 20: Engg curves

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

Page 21: Engg curves

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.

Page 22: Engg curves

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.

Page 23: Engg curves

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

Page 24: Engg curves

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

Verte

x

HYPERBOLA

= QF/QN

> 1.

Page 25: Engg curves

Nature of graph of Boyle’s

law

Shape of overhead water

tanks

Uses :-

Shape of cooling towers

etc.

Page 26: Engg curves

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

Page 27: Engg curves

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’SARC OF CIRCLE’S METHODMETHOD

Page 28: Engg curves

Axi

sM

inor

A BMajor Axis 7

8

910

11

9

8

7

6

543

2

1

12

11

P6

P5P4

P3

P2`

P1

P12

P11

P10P9

P8

P7

6

54

3

2

1

12 C10

O

CONCENTRIC CONCENTRIC CIRCLE CIRCLE

METHODMETHOD

F2F1

D

CF1=CF2=1/2 AB

T

N

Q

e = AF1/AQ

Page 29: Engg curves

Normal

Normal

00

11

22

33

44

11 22 33 44 1’1’0’0’

2’2’3’3’4’4’

1’1’

2’2’

3’3’

4’4’

AA BB

CC

DD

Major AxisMajor Axis

Min

or A

xis

Min

or A

xis

FF11 FF22

Dir

ectr

ixD

irec

trix

EE

FF

SS

PP

PP11

PP22

PP33

PP44

Tan

gent

Tan

gent

PP11’’

PP22’’

PP33’’PP44’’

ØØ

R=AB/2

R=AB/2

PP00

P1’’

P2’’

P3’’P4’’P4

P3

P2

P1

OBLONG METHODOBLONG METHOD

Page 30: Engg curves

BA

P4

P0

D

C

60°

6

5432

10

5 4 3 2 1 0 1 2 3 4 5 65

3210P1P2

P3

Q1

Q2Q3Q4

Q5

P6 Q6O

4

ELLIPSE IN PARALLELOGRAM

R4

R3 R2

R1S1

S2

S3

S4

P5

G

H

I

K

JM

inor Axis

Major Axis

Page 31: Engg curves

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

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

Page 32: Engg curves

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.

Page 33: Engg curves

ARC OF CIRCLE’S ARC OF CIRCLE’S METHODMETHOD

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

ixd

irec

trix

100100

140140

GF1 + GF2 = MAJOR AXIS = 140GF1 + GF2 = MAJOR AXIS = 140

EE

ee AF1AF1

AEAEe e = =

R=70R=70 R=70

R=70

Page 34: Engg curves

PROBLEM :-3

Two points A & B are 100 mm apart. A point C is 75 mm from A and 45 mm from B. Draw an ellipse passing through points A, B, and C so that AB is a major axis.

Page 35: Engg curves

OA B

C

7545

1

D

100

1

2 2

3

3

4

4

5

5

66

7

7P1

P2

P3

P4

P5

P6

P7

P8

E

8

8

Page 36: Engg curves

PROBLEM :-5

ABCD is a rectangle of 100mm x 60mm. Draw an ellipse passing through all the four corners A, B, C and D of the rectangle considering mid – points of the smaller sides as focal points.

Use “Concentric circles” method and find its eccentricity.

Page 37: Engg curves

I3

CD

F1 F2P Q

R

S

5050I1 I4

A BI2

O

1

1

2

2

4

4

3

3

100100

Page 38: Engg curves

PROBLEM :-1

Three points A, B & P while lying along a horizontal line in order have AB = 60 mm and AP = 80 mm, while A & B are fixed points and P starts moving such a way that AP + BP remains always constant and when they form isosceles triangle, AP = BP = 50 mm. Draw the path traced out by the point P from the commencement of its motion back to its initial position and name the path of P.

Page 39: Engg curves

A B P

R =

50

M

N

O

1

2

1 2 6080

Q

1

2

12

P1

P2 Q2

Q1

R1

R2 S2

S1

Page 40: Engg curves

PROBLEM :-2Draw an ellipse passing through 60º corner Q of a 30º - 60º set square having smallest side PQ vertical & 40 mm long while the foci of the ellipse coincide with corners P & R of the set square.

Use “OBLONG METHOD”. Find its eccentricity.

Page 41: Engg curves

ELLIPSEELLIPSEPP

QQ

R80mm

40m

m

89mm

AA

CC

BB

DD

MAJOR AXIS = PQ+QR = 129mmMAJOR AXIS = PQ+QR = 129mm

R=A

B/2

11

22

33

1’1’ 2’2’ 3’3’ 1’’1’’2’’2’’3’’3’’

OO33 OO33’’

OO11

OO22 OO22’’

OO11’’

TANGENT

TANGENTNO

RM

AL

NO

RM

AL

6060ºº

30º30º

22

33

11

dir

ectr

ixd

irec

trix

FF11 FF22MAJOR AXISMAJOR AXIS

MIN

OR

AX

ISM

INO

R A

XIS

SS

ECCENTRICITY = AP / ASECCENTRICITY = AP / AS

?? ??

Page 42: Engg curves

PROBLEM :-4

Two points A & B are 100 mm apart. A point C is 75 mm from A and 45 mm from B. Draw an ellipse passing through points A, B, and C so that AB is not a major axis.

Page 43: Engg curves

D

C

665

43

21

0

5 4 3 2 1 0 1 2 3 4 5 66

5

32

10

P2

P3

P4

P5

Q1

Q2Q3

Q4Q5

BA O

4

ELLIPSE

100

4575

P0

P1

P6Q6

G

H

I

K

J

Page 44: Engg curves

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.

Page 45: Engg curves

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.

Page 46: Engg curves

METHODS FOR DRAWING PARABOLA

1. Rectangle Method

2. Parabola in Parallelogram

3. Tangent Method

4. Directrix Focus Method

Page 47: Engg curves

22334455

00

11

22

33

44

55

66 11 11 55443322 66

00

11

22

33

44

55

00

VVDD CC

AA BB

PP44 PP44

PP55 PP55

PP33 PP33

PP22 PP22

PP66 PP66

PP11 PP11

PARABOLA –RECTANGLE METHODPARABOLA –RECTANGLE METHOD

PARABOLA PARABOLA

Page 48: Engg curves

BB

00

2’2’

00

22

66

CC

6’6’

VV

55

P’P’55

3030°°

AA XX

DD

1’1’ 2’2’4’4’

5’5’

3’3’

11

3344

5’5’

4’4’

3’3’

1’1’

00

55

44

33

22

11

PP11

PP22

PP33

PP44

PP55

P’P’44

P’P’33

P’P’22

P’P’11

P’P’66

PARABOLA – IN PARALLELOGRAMPARABOLA – IN PARALLELOGRAM

PP66

Page 49: Engg curves

BA O

V

1

8

3

4

5

2

6

7

9

10

0

1

2

3

4

5

6

7

8

9

10

0

F

PARABOLA

TANGENT METHOD

Page 50: Engg curves

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

Page 51: Engg curves

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.

Page 52: Engg curves

6m

6m

ROOT OF TREEROOT OF TREE

BUILDINGBUILDING

REQD.DISTANCEREQD.DISTANCE

TOP OF TREETOP OF TREE

3m3m

6m

6m

FF

AA

STONE FALLS HERESTONE FALLS HERE

Page 53: Engg curves

3m3m

6m

6m

ROOT OF TREEROOT OF TREE

BUILDINGBUILDING

REQD.DISTANCEREQD.DISTANCE

GROUNDGROUND

TOP OF TREETOP OF TREE

3m3m

6m

6m

11

22

33

11

22

33

332211 44 55 66

55

66

EEFF

AA BB

CCDD

PP33

PP44

PP22

PP11

PP

PP11

PP22

PP33

PP44

PP55

PP66

33 22 1100

STONE FALLS HERESTONE FALLS HERE

Page 54: Engg curves

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.

Page 55: Engg curves

150 mmAA

BB CC

DD11 22 33 44 55

11

22

33

44

55

OO

PP11

PP22

PP33

PP44

PP55

MM1’1’ 2’2’ 3’3’ 4’4’ 5’5’

1’1’

2’2’

3’3’

4’4’

5’5’

PP11’’

PP22’’

PP33’’ PP

44’’

PP55’’

90 m

m

Page 56: Engg curves

EXAMPLEEXAMPLE

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

Page 57: Engg curves

ground levelground level BA

60ºgun gun shotshot

80 M

parabolaparabola

Page 58: Engg curves

ground levelground level BA O

V

1

8

3

4

5

2

6

7

9

10

0

1

2

3

4

5

6

7

8

9

10

0

F

60ºgun gun shotshot

DD DD

VFVF

VEVE== e = 1e = 1

EE

Page 59: Engg curves

Connect two given points A and B by a Connect two given points A and B by a Parabolic curve, when:-Parabolic curve, when:-

1.OA=OB=60mm and angle AOB=90°1.OA=OB=60mm and angle AOB=90°

2.OA=60mm,OB=80mm and angle 2.OA=60mm,OB=80mm and angle AOB=110°AOB=110°

3.OA=OB=60mm and angle AOB=60°3.OA=OB=60mm and angle AOB=60°

Page 60: Engg curves

6600

66 00

1 2 3 4 5

Parabola

5

4

3

2

1

AA

BB90 90 °°

OO

1.OA=OB=60mm and angle 1.OA=OB=60mm and angle AOB=90AOB=90°°

Page 61: Engg curves

AA

BB

8800

6600

5

4

3

2

1

1 2 3 4 5

110 °°

Parabola

OO

2.OA=60mm,OB=80mm and angle 2.OA=60mm,OB=80mm and angle AOB=110°AOB=110°

Page 62: Engg curves

54321

5

4

3

2

1

AA

BBOO

Parabola

6600

6060

60 60 °°

3.OA=OB=60mm 3.OA=OB=60mm and angle and angle AOB=60°AOB=60°

Page 63: Engg curves

exampleexample

Draw a parabola passing through three Draw a parabola passing through three

different points A, B and C such that AB = different points A, B and C such that AB =

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

respectively. respectively.

Page 64: Engg curves

BBAA

CC

100100

5050

8080

Page 65: Engg curves

1’1’

00

00

2’2’

00

2266

6’6’55

P’P’55

1’1’ 2’2’ 4’4’ 5’5’3’3’113344

5’5’

4’4’

3’3’

5544

33

2211

PP11PP

22

PP33

PP44

PP55

P’P’44

P’P’33

P’P’22

P’P’11

P’P’66

PP66

AA BB

CC

Page 66: Engg curves

METHODS FOR DRAWING HYPERBOLA

1. Rectangle Method

2. Oblique Method

3. Directrix Focus Method

Page 67: Engg curves

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

Page 68: Engg curves

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”.

Page 69: Engg curves

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

Page 70: Engg curves

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”.

Page 71: Engg curves

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

Page 72: Engg curves

AXISAXIS

NORMAL

NORMAL

CC VV FF11

DIR

EC

TR

IXD

IRE

CT

RIX

DDDD

11 22 33 44

4’4’

3’3’

2’2’

1’1’PP11

PP22

PP33

PP44

PP11’’

PP22’’

PP33’’

PP44’’

TT11

TT22

NN22

NN11

TAN

GE

NT

TAN

GE

NT

ss

Directrix and focus methodDirectrix and focus method

Page 73: Engg curves

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.

SuperiorHypotrochoid

Cycloidal Curves

Cycloid Epy Cycloid Hypo Cycloid

SuperiorTrochoid

Inferior

Trochoid

SuperiorEpytrochoid

InferiorEpytrochoid

InferiorHypotrochoid

Page 74: Engg curves

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

Page 75: Engg curves

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 = 360Ø = 360ºº x r/R x r/Rdd

Circumference of Generating Circle

RR dd

Rolling Rolling CircleCircle

r

O

Ø/Ø/22

Ø/Ø/22

P0 P0

Arc P0P0 =

Rd x Ø =

P0

Page 76: Engg curves

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

ØØ /2 ØØ /2

ØØ =360 x r

RR

T

Rolling CircleRadius (r)

O

Vertical

Hypocycloid

P P

Page 77: Engg curves

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.

Page 78: Engg curves

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

Page 79: Engg curves

CC00

PP00

7788

PP

6644

PP11

11

22

33

CC22 CC33

PP22 CC44

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.

55

66 CC11

PP33

PP44

PP

55

PP

77

PP

88

7700

CC55

CC66

CC77

CC88

11 22 33 44

5566

DD/2/2

ππDD

/2/2

ππDD/2/2 DD/2/2FloorFloor

Wal

lW

all

CYCLOIDCYCLOID

5566

77

88

Take diameter of circle = 40mmInitially distance of centre of

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

Page 80: Engg curves

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.

Page 81: Engg curves

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.

Page 82: Engg curves

PP66

PP44

rPP22

CC11

CC00

CC22

CC33CC44 CC55

CC66

CC77

CC88

1

0

23

4

5

6 7

O

RR dd

Ø/2Ø/2 Ø/2Ø/2

PP11PP00

PP33 PP55

PP77PP88

r rRolling Rolling CircleCircle

r

Rd X Ø = 2πrØ = 360Ø = 360ºº x x r/Rr/Rdd

Arc P0P8 = Circumference of Generating Circle

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

ºU

N

Ø = 360Ø = 360ºº x 25/75 x 25/75

= 120°= 120°

Page 83: Engg curves

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.

Page 84: Engg curves

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

Page 85: Engg curves

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

Page 86: Engg curves

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

Page 87: Engg curves

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.

Page 88: Engg curves

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.

Page 89: Engg curves

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.

Page 90: Engg curves

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.

Page 91: Engg curves

PP

R=7toPR=7toPR=6toPR=6toP

R21

R21

00

00 11 22 33 44 55 66 77 88 PP1111 00

11

22

33

445566

77

88

99

1010

PP11

PP22

PP33

PP44

PP55

PP66

PP77

PP88

L= 100 mmL= 100 mm

R=1toPR=1toP

R=2

toP

R=2

toPR

=3

toP

R=

3to

PR=4toP

R=4toPR=5toP

R=5toP

INVOLUTEINVOLUTE

99

ø

11 mm = 30°

Then 5 mm = Ø = 30° x 5 /11 = 13.64 °

S = 2 x π x r /12

Page 92: Engg curves

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.

Page 93: Engg curves

PP1313

PP1111

313

14

15

PP00

PP1212

O

7

10 123

45689

1112 1 2

PP11

PP22

PP33PP44

PP55

PP66

PP77

PP88 PP99 PP1010

PP1414PP

L=160 L=160

mmmm

R=21mm

64 5 7 8 9 101112131415

øø

Page 94: Engg curves

PROBLEM:-

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

Page 95: Engg curves

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

Page 96: Engg curves

PROBLEM:-

Draw an involute of a square

having side as 20 mm.

Page 97: Engg curves

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

Page 98: Engg curves

PROBLEM:-

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

60°60°

AA

BB

CC

R21R2130°30°

Page 99: Engg curves

R2R2

11

60°60°

AA

BB

CC

30°30°XX

X+AX+A

11

XXX+

A2

X+

A2

X+

AX

+A

33

X+A5X+A5

X+

A4

X+

A4

X+AX+ABB

R R

=X+AB=X+AB

X+

66+

BX

+66+

BCC

1122

33

4455

C0

C1

C2C3

C4

C5

C6

C7

C8

Page 100: Engg curves

A stick of length equal to the circumference of a A stick of length equal to the circumference of a

semicircle, is initially tangent to the semicircle semicircle, is initially tangent to the semicircle

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

circumference of a semicircle without sliding till circumference of a semicircle without sliding till

it becomes tangent on the left side of the it becomes tangent on the left side of the

semicircle. Draw the loci of two end point of this semicircle. Draw the loci of two end point of this

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

PROBLEM:-

Page 101: Engg curves

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

Page 102: Engg curves

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.

Page 103: Engg curves

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

Page 104: Engg curves

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.

Page 105: Engg curves

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:

Page 106: Engg curves

PP1010

1

23

4

5

6

7

89

10

11

120

8 7 012345691112

PP11

PP22PP33

PP44

PP55

PP66

PP77PP88

PP99

PP1111PP1212

o

Page 107: Engg curves

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

Page 108: Engg curves

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

Page 109: Engg curves

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

Page 110: Engg curves

BB AA OO1234567891011

P1

P2

P3 P4

P5

P6

P7

P8

P9P10

P11

P12

11

21

31

41

51

61

71

81

91

101

111

40 25

Page 111: Engg curves

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.

Page 112: Engg curves

AAInitial Position of point PInitial Position of point PPPOO

PP11

PP22

PP33

PP44PP55

PP66

PP77

PP88

2211

3344556677OO

11

22

33

44

5566

77

88

2/3 X 360°

= 240°

120120ºº

Page 113: Engg curves

AA00

Linear Travel of point PP on ABAB

= 96 =16x (6 div.)

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

PP11’’

AA

BB

AA11

AA22

AA33

AA44

AA55

AA66PP00

PP11

PP22

PP33

PP44

PP55

PP66

PP22’’

PP33’’

PP44’’PP55’’

PP66’’

99 66

Link Link ABAB = 96 = 96

CC

TangentTangent

AngularAngular SwingSwing ofof linklink AB = AB = 180° + 90°180° + 90°

= = 270270 ° ° =45 °X 6 div. =45 °X 6 div.

ARCHIMEDIAN ARCHIMEDIAN SPIRAL SPIRAL

DD

NO

RM

AL

NO

RM

AL

MM

NN

Page 114: Engg curves

Arch.Spiral Curve Constant BCBC

= Linear Travel ÷Angular Swing in Radians

= 96 ÷ (270º×π /180º)

=20.363636 mm / radian

Page 115: Engg curves

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.

Page 116: Engg curves

θ

o

012

3

4 5 67

8

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

Page 117: Engg curves

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.

Page 118: Engg curves

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.

Page 119: Engg curves

Rolling circleRolling circle

PP11

PP22

PP33

PP44

PP00

PP66

PP77

PP88

PP55

PP99

PP1010

11

22

3344

55

66

77 8899

1010

11 22 33 44 55 66 77 88 99 101000

CC00 CC11 CC22CC33 CC44

Directing lineDirecting line

Length of directing line = 3Length of directing line = 3

540540 = 360 = 360 + 180 + 180

540540 = = D + D + D/2 D/2 Total length for 540Total length for 540 rotation = 3 rotation = 3D/2D/2

CC55 CC66 CC77CC88 CC99 CC1010

SS

norm

al