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”.

CHAPTER – 2

ENGINEERING
CURVES

USES OF 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.

CLASSIFICATION OF ENGG.
CURVES
1. CONICS
2. CYCLOIDAL
CURVES
3. INVOLUTE

4. SPIRAL

5. HELIX

6. SINE & COSINE

What is Cone ?
It is a surface generated by moving a
Straight line keeping one of its end fixed &
other end makes a closed curve.
The fixed point is known as vertex or apex.
The closed curve is Vertex/Apex
known as base.
If the base/closed curve
is a circle, we get a cone.
90º
If the base/closed
curve is a polygon, we
get a pyramid. Base

The line joins apex to the center of base is
called axis.
If axes is perpendicular to base, it is called as
right circular cone.

If axis of cone is not Vertex/Apex
perpendicular to base, it is Cone Axis
called as oblique cone. Generator

The line joins vertex/
90º
apex to the
circumference of a cone
is known as generator. Base

CONICS
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

A - TRIANGLE

B - CIRCLE

C - ELLIPSE

D – PARABOLA

E - HYPERBOLA

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

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

Sec Plane

Circle

Definition :- ELLIPSE
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.
θ
α>θ
α

PARABOLA
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.

θ α=θ
α

HYPERBOLA
Definition :-
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.
α=0
α<θ θθ

CONICS
Definition :- 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.
Conic Curve
M P
Directrix
F
C
V
Focus

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

The line passing through focus &
perpendicular to directrix is called as axis.

The intersection of conic curve with axis is
called as vertex.

Conic Curve
M P Axis
Directrix
F
C
V
Vertex Focus

Conic Curve
M P Axis
Directrix
F
C
V
Vertex Focus
N Q

Distance of a point from focus
Ratio =
Distance of a point from directrix
=
= PF/PM = QF/QN = VF/VC
Eccentricity
= e

ELLIPSE
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.

P Ellipse
M Axis
Directrix
Vertex F
C V
Focus Eccentricity=PF/PM
N = QF/QN
Q
< 1.

ELLIPSE
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.
C
P

O
A B
F1 F2

Q D

P C

CF1 +CF2 = AB
O
A B but CF1 = CF2
F1 F2
hence, CF1=1/2AB

Q D
PF1 + PF2 = QF1 + QF2 = CF1 +CF2 = constant
= F1A + F1B = F2A + F2B
But F1A = F2B

F1A + F1B = F2B + F1B = AB
= Major Axis

C
Major Axis = 100 mm
Minor Axis = 60 mm
O
A B
F1 F2 CF1 = ½ AB = AO

D
C
Major Axis = 100 mm
F1F2 = 60 mm
O
A B CF1 = ½ AB = AO
F1 F2

D

Uses :-

Shape of a man-hole.
Shape of tank in a tanker.

Flanges of pipes, glands and stuffing boxes.

Shape used in bridges and arches.
Monuments.

Path of earth around the sun.

Shape of trays etc.

PARABOLA
Definition :-
The 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.
Ratio (known as eccentricity) of its distances
from focus to that of directrix is constant
and equal to one (1). Parabola
M P
Directrix
Axis
Vertex F
C V
Eccentricity = PF/PM Focus
= QF/QN N Q
= 1.

Uses :-

Motor car head lamp reflector.

Sound reflector and detector.

Bridges and arches construction

Shape of cooling towers.

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

Home

HYPERBOLA
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.

P Hyperbola
M Axis
Directrix
F
C
V Eccentricity = PF/PM
Vertex Focus
N = QF/QN
Q
> 1.

Uses :-

Nature of graph of Boyle’s law

Shape of overhead water tanks

Shape of cooling towers etc.

METHODS FOR DRAWING ELLIPSE
1. Arc of Circle’s Method
2. Concentric Circle Method
3. Loop Method
4. Oblong Method
5. Ellipse in Parallelogram
6. Trammel Method
7. Parallel Ellipse
8. Directrix Focus Method

ARC OF CIRCLE’S P4 C P4
METHOD P3 P3
P2 P2
P1 P1
Rad =
= A1

B1
R

F1 O F2
A B
1 2 3 4
`R=A2

R =B2
Ta P1’ °° P 1’
ng
en
t P2’
P2’
l
ma

90° P 3’ P 3’
D
r

P4’ P4’
No

CONCENTRIC 11
10
9
CIRCLE
METHOD
12
C P10 8
P11 10 P9
N 11 9
P12 P8
T 12 8
P1 Major Axis Minor P7
Q 1
Axis
A F1 1 O 7 F2 B 7
6 P6
P2` 2
3 5
P3 4 P5
2 D P4 6
e = AF1/AQ

CF1=CF2=1/2 AB 3 5
4

OBLONG METHOD

P4 C P4’
E 4 P3 P3’ 4’
No
Directrix

rm 3

Minor Axis
al 3’
P2 P2’

2
B/
2 S 2’

A
P1

R=
ØØ P1’
1 1’
0 P0 Major Axis 0’
A 1 F1 2 3 4 4’ 3’ 2’ F2 1’ B
t
en

P1
ng

P1’’
Ta

P2 P2’’
P
F P3 P3’’
P4 DP4’’

ELLIPSE IN PARALLELOGRAM

0 C
H P1 P Q1 0
1 P2 1
2 P Q2
0
Q3K 2
3P 3 Q4 3

Min
4 4 Q5 4
5P
AP6 5 5
Q6 B

or
6 5 4 3 x2s 1 0O 1 2 3 4 5 6
A i
Ax i
S4 ajor
s
M R4
S3
J R3 G
60° S2 R2
S1
R1 I
D

ELLIPSE – DIRECTRIX FOCUS METHOD
g
f θ < 45º
Ellipse e
d
c Eccentricity = 2/3
b P3 P4 P5 P6 P7
a
Directrix

Q P1 P2 QV1 V1F1 2

f`
= =

=6
R=
R= R1V1 R1V1 3

R
1a
1a
R1  1 2 3 4 5 6 7
D1
V1 F1
Dist. Between directrix
90° & focus = 50 mm
P’
T Tangen 1 P ’
t 2 1 part = 50/(2+3)=10 mm
P3’P ’
4 P5’ P6’ P ’
V1F1 = 2 part = 20 mm
N
S 7
T V R = 3 part = 30 mm
Normal

1 1
N

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
directrix
G4 G G4
e =
AF1 e G3 G3
AE G2 G2
G1 70 R= G1
R= R= 70
B1
R =A1
E F1 O F2 B
A 90°
1 2 3 4 100
`R=A2

R =B2
Ta G ’
ng 1 °° G1’
en
t G2’ G2’
al
rm

90° G3’ G3’
G4’ G’ G4’
No

140
GF1 + GF2 = MAJOR AXIS = 140

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.

8
D

1 P8 C 7
8 E
1 P7
P1 7 45
75

2 2 6 6
A P 100 O P6 B
2

5 P5
P3 3
4
3 P4 5

4

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.

1 4
R
I1 I4
D 1 4 C

P F1 O Q
50
50 F2
100
2 3
AI B
I3
2

S
2 3

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.

M
P2 Q2

2 2
P1 Q1

50
1
R=
1

A O B
Q P
1 2 60 2 1
80

R1 S1

R2 S2
N

PROBLEM :-2
Draw 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.

directrix T
EN C

NO
NO
G
AN
O3 O3 ’

MINOR AXIS
T

RM
RM

2
B/
Q

AL
AL
3 3

A
R=
O2 O2 ’

θ θº
60
2

40mm
2

ELLIPSE
89m
O1 m O1 ’
1 1
F1 80mm MAJOR AXIS 30º F2
S A B
? ? 1’ P 2’ 3’ 3’’ 2’’ R 1’’

D
MAJOR AXIS = PQ+QR = 129mm ECCENTRICITY = AP / AS

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.

ELLIPSE

C
0 H P1 0
1 P2 P0 Q1 1
2 P3 Q2 2
Q3 K

45
3 P
75
3
4
4 Q4 4
5 P5 Q5
6 O Q6 65
A P6
6 5 4 3 2 1 100 0 1 2 3 4 56 B

J G

I
D

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. Tangent Method


PARABOLA –RECTANGLE METHOD
D V C
0 P1 P1 0
P2 P2 PARABOLA
1 1
P3 P3
2 2
P4 P4
3 3

4 P5 P5 4

5 5
P6 P6
6A 5 4 3 2 1 0 1 2 3 4 5 B6

PARABOLA – IN PARALLELOGRAM
C
0
1’
2’
P’ P’
2 P’
V 1
3’
P1 3
P’
4
P2 4’
P’
5
B 5’
D P3 P’
0 6’
6

1 5’
P4 4’
2 3’
2’
3 P 1’
5
4 1
2
0

5 3
P 4 30°
66 5
A X

PARABOLA 10 0
TANGENT METHOD 9 1
8 2
7 3
6 4
5 V
5
4 6
F
3 7
2 8θ
1 θ 9
0 10
A O B

D PARABOLA
DIRECTRIX FOCUS METHOD
P4
P3
P2

R4
PF

R3
R2
P1

RF
R1
AXIS
R V 1 F 2
90° 3 4
T 90°
N
P1’
PF’ S
DIRECTRIX

P2’
P3’
P4’

N T
D

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.

TOP OF TREE

BUILDING

A
6m
6m

6m
STONE FALLS HERE
ROOT OF TREE

F 3m

REQD.DISTANCE

TOP OF TREE
D P C
P1 P1
1 P2 P2 1

2 2
P3 P3
BUILDING 3 3
A P4 0 P4 B
6m 3 2 1 1 2 3 4 5 6
6m
5

6m 6 P5
STONE FALLS HERE
ROOT OF TREE

F 3m 3m E GROUND
P6
REQD.DISTANCE

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 mm
D 5 4 3 2 1 A
1
P1’’
P1
P1 2
P2’’
P2
3
m
M

P2
P3’’
P3 4
’’ 3’ 4’ 5’
3’ 4’ 5’ P3
5
P4 P
P4’’
P5
C 5’ 4’ 3’ O 2’ 1’ B

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

parabola

gun
shot 60º

A ground level B
80 M

VF 10 0
= e=1
VE 9 1
8 2
D 7 E 3 D

6 4
5 V
5
4 6
F
3 7
2 8

gun 1 9
shot 60º
0 10
A O B
ground level

Connect two given points A and B by a
Parabolic curve, when:-
1.OA=OB=60mm and angle AOB=90°
2.OA=60mm,OB=80mm and angle
AOB=110°
3.OA=OB=60mm and angle AOB=60°

A
1.OA=OB=60mm and angle
AOB=90°
1

2

3
60
60 Parabola

4

5

90 °
O B
1 2 63 4 5

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

1

2 Parabola
60
60
3

4

5
110
° B
O 1 2 3 4 5
80

3.OA=OB=60mm
A
and angle AOB=60°
1

2
Parabola
60

3

4

5

60
B
O ° 1 2 3 4 5
6

example

Draw a parabola passing through three
different points A, B and C such that AB =
100mm, BC=50mm and CA=80mm
respectively.

0 P2
P1 C P’
1 0
P3 P’
1 2
1’
P’
2 P4
3
2’
3 P’
4
3’
4 P5
P’ 4’
5
5 5’
P
6 6
6
P’
0 1’ 2’ 3’ 4’ 5’ 6’ 6
A 5 4 3 2 1
B

METHODS FOR DRAWING HYPERBOLA

1. Rectangle Method

2. Oblique Method


RECTANGULAR HYPERBOLA
When the asymptotes are at right angles to each other, the hyperbola
is called rectangular or equilateral hyperbola
B 6’ F
P6
Given Point P0
AXIS

C 6 0 1 2 3 4 5
D
P0 P1

2’ P2
3’ P3
P4 Hyperbola
4’
5’ P5
Y

ASYMPTOTES X and Y
O X E A
90° AXIS

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”.

RECTANGULAR HYPERBOLA
B 6’ F
P6
Given Point P0

C 6 0 1 2 3 4 5
D
P0 P1

2’ P2
3’ P3
P4 Hyperbola
Y = 50

4’
5’ P5

O X=20 E A
90°

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”.

B F
X=2
0

P7 7’

Given Point P0
C 7 1 2 3 4 5 6
P0 D
1’ P1
Y = 30

2’ P2
P3 P4 P5 P6
6’
O A
E
75 0

Directrix and focus method 4’

D
D
P4
3’
T2 P3
2’

N1
N1
P2
1’
s P1
NO
RM

T
AL

EN
NG
D
D
AXIS

TA

N2
N
C V 1 F12 3 4
T1

P1 ’
DIRECTRIX
DIRECTRIX

P2’

P3’

P4’

CYCLOIDAL GROUP OF CURVES
When 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

Inferior Superior Inferior Superior
Trochoid Trochoid Hypotrochoid Hypotrochoid

Inferior Superior
Epytrochoid Epytrochoid

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.
Rolling Circle or Generator
P
R

C C

P P
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.
P0

Rolling
Circle

r
P0 P0
Ø/ Ø/
2 2
Rd x Ø = 2πr O
Circumference of
Arc P0P0 =
d
R

Ø = 360º x r/Rd

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.`
Vertical

Rolling Circle Directing
Circle(R)
Radius (r)

P P
T

P

Ø /2 Ø /2

O 360 x r
R Hypocycloid Ø= R

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.

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.

: Given Data :
Draw cycloid for one revolution of a rolling circle having
diameter as T
60mm.
Rolling
D N
Circle
6 P6
7 5 P5 P7
S
8 4 P4 R
P8
R P9
P3
9 C0 C1 3 2 C3 C4 S C5 C6 C7 C8 C9
CT C10 C11 C12
1
P2 R
2 P10
10 P1 P11
11 12 0 1
P0 0 1 2 3 4 N 5 6 7 8 9 10 11 12P12
Directing Line
2R or D

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 C8
C8
8
the Bottom of circle touching the floor.
Draw the path of the point P. C7
C7

Wall
Take diameter of circle = 40mm 7
Initially distance of centre of P
circle from the wall 83mm (Hale 8
C6
C P P6
circumference + D/2) P
7
6

8
4 P4
πD/2
πD/2
CYCLOID C5
C 5

7
5 3 P3 5

P2
6
C C1 2 C2 C3 C4

6
0

P1
7 1

5
P00 1 πD/2 2 3 4 D/2
Floor

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.

GIVEN: EPICYCLOID
Rad. Of Gen. Circle (r)
P4
& Rad. Of dir. Circle (Rd)
S
P3 º P5
Rolling r U r
Circle C3 C4 C5
3 P2 2
C2 C6 P6
C1 N C7
4
r

1
r CP1 P
0 C8
5 0 P8 P7
0
6 7 Ø/2 Ø/2

O
Ø = 360º x 25/75 Arc P0P8 = Circumference of
Rd
 = 120° Rd X Ø = 2πr
Generating Circle
Ø = 360º x

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.

Directing
Circle Vertical
Rolling N
Circle
Radias (r)
C5 C6 C7 C

Norm
C4 8
2 C3 C9
1 C2 3 C10

al
P11 P12
r
P0 0 P1 Pr

r
C C11
12 2
P3 1 4 P8 P9 P10 T
C0 P4 P5 P6 P7 S
e nt C12
11 5 Tang
T
10 / / N
6
9 7 2 2
8
R O  = 360 x r
R
Hypocycloid  = 360 x 4
12
 = 120°

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

Directing Circle

Rolling Circle
C4 C5 C6 C7
2 3 C8
4 C9
C3
1 C2 C10
5
C1 C11
P2 P11
12 6
P1 P3 P4 C P5 P6 O C12 P12
P7 P8 P9 P10

7
11 HYPOCYCLOID
10 8
9

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
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.

Uses :- Gears profile

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.

T
P9
P8 P10

09

`

N
0
P7
08

01
07 P11
al

Ta
rm 011
No

ng
P6 06

en
5 6 7

t
4
3 8
05 .
2 9
P5 1 10

T
04

01211N P12
02
03

P4 P1 1 2 3 4 5 6 7 8 9 10 1112
0
P3 P2 π
D

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

Ø = 30° x 5 /11 = 13.64 ° 11 mm = 30°
P3 Then 5 mm = ζ
P4 P2 INVOLUTE

P5

R=
R=
R=

4tto
R=

4o
R=3toP
R=3toP
5tto
5o

PP

to P
P1
P

P6 R=6toP
P

2
R=
P
7to 7
6 5
P7 R=
4

P
P8 8ø

1 to
P 3

R=
9 0
1

2
R2

10
11 0 1
0 1 2 3 4 5 6 7 8 P 9
S = 2 x π x r /12
L= 100 mm

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

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

P4 P3
P2

P5
P1

P6
6 5
7 4
8 3 15
9 O ø PP
2 14
R=21mm 10
11 113 P13 P0
14

P7 12 1 2P 3 4 5 6 7 8 9 10 11 12 13 1415
12
P11 L=160 mm
P8 P9 P

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

INVOLUTE P3

OF A POLYGON ∗0
1
P2
R=3

R=4∗
R=2∗01
Given : 2

01
3 1
Side of a polygon P4 4 50
R=01 P1
P0

T

01
=5 ∗
N R
S

P5

PROBLEM:-

Draw an involute of a square

having side as 20 mm.

INVOLUTE OF A SQUARE
P1

01
P0

R=
0
R
P4
=2
1 4
∗
01 2 3 R=
P2 4∗
01
01
3∗
N

R=

S
N

P3

PROBLEM:-

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

C 60°

R
21
30°

A

C8

C7
CC
6 +B
+B
+666
X+

R B
A
=X+
B X+A
5 C6
B
C 60° 4
R 3
30° 2
1
X
X

X+
2

X+
C0 X 1

A5
A5
A
3
3
C5

X
X
X+A
X+A

+A
A

+A
2
X+

+A

44
1

C1
X

C2 C4
C

PROBLEM:-

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.

INVOLUTE
B A6
6
A5 B1
5

A4 4
B2
3

A3 2 B3
2 3
4
A2 1
1 5 B4
A1 A C
O B6 B5

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
Logarithmic Circle

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.

PROBLEM:

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)

3
4 2
P3 P2

P4
5 P1 1

P5

6
o 0
P6 P12 11 9 8 7 6 5 4 3 2 1 0
12
P11 12
P10
P7 P9
P8

7 11

8 10
9

To construct an Archemedian
Spiral of one convolutions,
given the greatest &
shortest(least) radii.
OR
To construct an Archemedian
Spiral of one convolutions,
given the largest radius vector
& smallest radius vector.
Say Greatest radius = 100 mm &
Shortest radius = 60 mm

Diff. in length of any two radius vectors
Constant of the curve =
Angle between them in radians
3
2 OP – OP3
4 P3 =
P2 Π/2
P4

T
5 P1 1 100 – 90
P5 =

n
R mi
Π/2
N

P12 10 8 6 4 2
6 P O 11
9 7 5 3 1 12
= 6.37 mm
6
S P11
R
P7 P10max 11
7
P8 P9
8
N

10
T

9

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.
40 25
B A O

31
21 41

11 P3 P4 51
P2
P1 25 P5
40

B 11109 8 76 5 4 3 2 1 A O P6 61
P12

P7
P11
111 71
P8
P10
P9
101 81
9

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.

PO Initial Position of point P
2/3 X 360° A
1 1
= 240° 2 P1
3
4 2
5 P2
6
120ºP7
O
8
P3 3
P7

P6 P4
P5

8 4

7 5
6

EXAMPLE: A link AB, AB Angular Swing
A0 of link AB = 180° + 90°
96mm long initially is P6
vertically upward w.r.t. its = 270 °
P5 =45 °X 6 div.
pinned end B, swings in A1
P4
clockwise direction for 96
P3
96
180° and returns back in ARCHIMEDIAN
anticlockwise direction for P2 SPIRAL
90°, during which a point P1 P1’
C A6
P, slides from pole B to P0 P 2’ A2
B

NORM
NORMA
end A. Draw the locus of P 6’
point P and name it. Draw
P 3’

AL
tangent and normal at any

L
N
point on the path of P. P 5’
P 4’ A3
Link AB = 96 M D A5
t
Linear Travel of point P on AB
Ta ngen
= 96 =16x (6 div.) A4

Arch.Spiral Curve Constant BC

= Linear Travel ÷Angular Swing in Radians
= 96 ÷ (270º×π /180º)
=20.363636 mm / radian

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.

o
1
P3 2
θ 3
4
5
6
7 P9
8
9
10
11
P15 12
13
14
15
16
17
18
19
20
3 21 9
22 21
15 23
2 24 8
4 1 0 7 10
14 5 6 11 20
16 1317 18 1224 1923 22

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.

al
rm
Rolling circle

no
P0 P8
8 P1 P7 P9
7
91
C0
SC C4 C8 C9 P10
6 10 2C1 C2 P
P
3 2 P6 C5 C6 C7 C10

3 P3 P5
5
4
P4 Directing line
0 1 2 3 4 5 6 7 8 9 10
Length of directing line = 3Π 

540 ° = 360° + 180°
540 ° = Π D +
Π D/2
Total length for 540 ° rotation = 3Π D/2

Engg curves

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