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
known as base.
Vertex/Apex
is
If the base/closed curve
is a circle, we get a cone.
If
the
base/closed
curve is a polygon, we
get a pyramid.
90º
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
perpendicular to base, it is
called as oblique cone.
The line joins vertex/
apex
to
the
circumference of a cone
is known as generator.
Vertex/Apex
Cone Axis
Generator
90º
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
ELLIPSE
Definition :When the cutting plane is inclined
axis but not parallel to generator
inclination of the cutting plane(α) is
than the semi cone angle(θ), we
ellipse as the section.
θ
α
to the
or the
greater
get an
α>θ
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.
α=θ
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
θ
α<θ
θ
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.
Directrix
M
C
Conic Curve
P
V
F
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.
Directrix
M
C
Vertex
Conic Curve
P
V
F
Focus
Axis
Directrix
M
Conic Curve
P
C
V
Vertex
N
Axis
F
Focus
Q
Distance of a point from focus
Ratio =
Distance of a point from directrix
= Eccentricity
=
=
PF/PM = QF/QN = VF/VC
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.
M
Directrix
Vertex
C
N
Ellipse
P
V
Q
Axis
F
Focus Eccentricity=PF/PM
= QF/QN
< 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.
P
A
C
O
F2
F1
Q
D
B
C
P
CF1 +CF2 = AB
A
O
F2
F1
Q
B
but CF1 = CF2
hence, CF1=1/2AB
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
A
O
F1
F2
B
D
C
A
Major Axis = 100 mm
F1F2 = 60 mm
O
F1
F2
D
CF1 = ½ AB = AO
B
CF1 = ½ AB = AO
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.
Definition :-
PARABOLA
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
Directrix
Vertex
Eccentricity = PF/PM
= QF/QN
= 1.
C
N
P
V
Q
F
Focus
Axis
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.
M
Hyperbola
Axis
P
Directrix
C
Vertex
N
V
Q
F
Eccentricity = PF/PM
Focus
= QF/QN
> 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
METHOD
P3
P4
C
P4
P3
P2
P2
P1
P1
F1
A
1
P1’
F2
O
2
3
4
P1’

P 2’
90°
B
P2’
P3’
P 4’
D
P 4’
P 3’
CONCENTRIC
CIRCLE
METHOD
12
10
11
N
C P10
P11
10
11
P12
T
Q
9
9
1
A F1
Major Axis
1
P2`
2
P7
O
2
P3
F2 B
7
6
3
P6
5
4
D P4
P5
e = AF1/AQ
CF1=CF2=1/2 AB
3
8
P8
8
12
P1
P9
4
5
6
7
4
P4 C P4’
P3
3
P2
2 S
P1
1
P2’
2’
P1’
1’
1 F1 2
4 4’
3
3’
2’ F2 1’
F
0’
B
P1’’
P1
P
3’
Major Axis
0 P0
A
4’
P3’
Minor Axis
E
Directrix
OBLONG METHOD
P2’’
P2
P3’’
P3
P4
DP4’’
ELLIPSE IN PARALLELOGRAM
0
H
P2
1
2
P
3 P 3
4
4
5P
AP6 5
6 5 4 3 2
C
P
P1
60°
Q2
0
1
0O 1
R4
R3G
S3
S2
S1
R1
D
I
R2
K
Q3
Q4 3
Q5 4
Q6 B5
5 6
2 3 4
S4
J
Q1
0
1
2
ELLIPSE – DIRECTRIX FOCUS METHOD
Directrix
D1
Ellipse
R1
D1
T
c
Q
a
b
P1 P2
d
e
f
g
 < 45º
Eccentricity = 2/3
P6 P7
P
P
5
P3 4
V1F1
QV1
2
=
=
R1V1
R1V1 3
1 2 3 4 5 6 7
V1
F1
90°
P1’
P2 ’
P3’P ’
4 P5’ P ’ P ’
6
7
S
Dist. Between directrix
& focus = 50 mm
1 part = 50/(2+3)=10 mm
V1F1 = 2 part = 20 mm
T V R = 3 part = 30 mm
1 1
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
eccentricity.
&
find
its
directrix
ARC OF CIRCLE’S
METHOD
E
e
AF1
=
AE
G4
e G3
G
G4
G3
G2
G2
G1
A
G1
F1
90°
G1’
1 2
3
F2
O
4 100
G1’

G2’
90°
B
G2’
G3’
G4’
G’
140
GF1 + GF2 = MAJOR AXIS = 140
G4’
G3’
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
P8
8
1
P1
7
C
E
1
P7
7
6
P6 B
6
2
2
A P
2
D
100 O
P3
5 P5
3
4
3
P4
4
5
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
50
D
P
4
R
I1
I4
4
1
O
F1
C
F2
100
AI
2
3
2
S
2
I3B
3
Q
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
Q1
P1
1
1
A
Q
1
2
O
60
80
B
2
P
1
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
C
O2
2
1
S
?
A
O1
O 3’
MINOR AXIS
Q
40mm
3
O3
O 2’
ELLIPSE
F1
80mm MAJOR AXIS
? 1’ P 2’
3’
30º F2
3’’
2’’ R 1’’
D
MAJOR AXIS = PQ+QR = 129mm
ECCENTRICITY = AP / AS
3
2
O 1’
1
B
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
1
4
2
3 P
P2
H P1
Q1
P0
P3
4
5 P5
6
A P6
6 5 4
O
3
2
1 1000
1
2
J
3
4
G
D
I
0
Q2
2
Q3 K
3
Q4 4
Q5
Q6 65
56 B
1
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
4. Directrix Focus Method
PARABOLA –RECTANGLE METHOD
0
D
V
P1
C
P1
P2
P2
PARABOLA
1
P3
1
P3
2
2
P4
P4
3
4 P5
3
P5 4
5
P6
6A 5
0
5
4
3
2
1
0
1
2
3
4
P6
5 B6
PARABOLA – IN PARALLELOGRAM
C
0
1’
P’
P’
2
1
P1
2’
P’
3
P’
5
P2
D
0
1
2
3 P
5
4
5
P6
6
A
3’
P’
4
P3
P4
30°
X
4’
5’
P’
6
B
PARABOLA
10 0
TANGENT METHOD 9
1
8
2
7
3
4
6
5
V
4
3
5
6
F
7
2

1
0
8
9
10
A
O
B
D
PARABOLA
DIRECTRIX FOCUS METHOD
P4
P3
P2
PF
P1
R
T
AXIS
V 1
90°
F 2
3 4
90°
N
P1’
P F’
P2’
D
N
S
P3’
P4’
T
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
TOP OF TREE
BUILDING
6m
A
STONE FALLS HERE
ROOT OF TREE
F
3m
REQD.DISTANCE
TOP OF TREE
D
C
P
P1 P1
6m
1
2
BUILDING
1
P2
P2
2
P3
P3
3
A
3
P4
P4 B
0
3 2 1
1 2 3 4
5 6
6m
5
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.
2’
3’O
P5
3’ 4’ 5’
P3’
5
D
90 mm1’
P2
3
2
C
P2’
P1’
4
P3
5’
P4’
P4
5
4’
P5’
1’
B
P1
1
A
1
2
3
4
150 mm
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
A
60º
ground level
80 M
B
10 0
VF
=
VE
e=1
9
1
8
7
D
2
3
E
4
6
5
5
V
4
3
D
6
F
7
2
gun 1
shot
0
A

60º
8
9
10
groundOlevel
B
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
60
2
Parabola
3
4
5
90 °
O
1
2
603
4
5
B
2.OA=60mm,OB=80mm and angle AOB=110°
A
1
Parabola
2
3
4
5
110 °
O
1
2
3
80
4
5
B
A
3.OA=OB=60mm and
angle AOB=60°
1
2
Parabola
3
4
5
60 °
O
1
2
3
60
4
5
B
example
Draw a parabola passing through three
different points A, B and C such that AB =
100mm,
BC=50mm
respectively.
and
CA=80mm
C
A
100
B
C
A
B
METHODS FOR DRAWING HYPERBOLA
1. Rectangle Method
2. Oblique Method
3. Directrix Focus Method
RECTANGULAR HYPERBOLA
When the asymptotes are at right angles to each other, the hyperbola is
called rectangular or equilateral hyperbola
B
P6
6’ F
AXIS
6
C
0 1 2
P0 P1
2’
3’
4’
5’
3
4
5
D
P2
P3
P4
P5
ASYMPTOTES X and Y
O
90°
X
E
A
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
P6
6
C
6’ F
0 1 2
P0 P1
2’
3’
4’
5’
O
90°
X=20
E
3
4
5
D
P2
P3
P4
P5
A
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”.
F
B
P7
7
P0
1’
2’
Y = 30
C
6’
O
E
7’
Given Point P0
1
2 3
4
P1
P2
P3
P4
5
P5
6
D
P6
A
DIRECTRIX D
Directrix and focus method
4’
T2
2’
1’
s P1
C V
T1
P4
3’
P3
P2
AXIS
1 F12
3
4
P1’
P2’
D
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
Inferior
Trochoid
Epy Cycloid
Superior
Trochoid
Inferior
Epytrochoid
Hypo Cycloid
Inferior
Hypotrochoid
Superior
Epytrochoid
Superior
Hypotrochoid
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
Rd x Ø = 2πr
Ø = 360º x r/Rd
Ø/2 Ø/2
P0
O
Circumference of
Arc P0P0 =
Generating Circle
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
Directing
Circle(R)
Rolling Circle
Radius (r)
P
P
T
P
Ø /2 Ø /2
R
O
Hypocycloid
360 x r
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 60mm.
D
Rolling
Circle
7
6
P5
5
4 P4
8
R
P6
P7
R
P8
P9
9
C0 C1 3C2 C3 C4 S C5 C6 C7 C8 C9 C10 C11 C12
1
P2
R
P10
2
10
P1
P11
12
0
1
11
P0 0 1
6
8
7
9
10 11 12P12
3
4 N5
2
P3
Directing Line
2R or D
P00
C3
8
6
5
πD/2
6
7
5
C5 8 C6
P
C4
1
1
5
7
P1
P4
P
7P
D/2
0
8
6
P2
C C 1 2 C2
6
P
πD/2
2
3
Floor
4
D/2
Wall
Take diameter of circle = 40mm
Initially distance of centre of
circle from the wall 83mm (Hale
circumference + D/2)
4
CYCLOID
P3
5
3
7
C7
C8
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.
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.
r
GIVEN:
EPICYCLOID
Rad. Of Gen. Circle (r)
P4
& Rad. Of dir. Circle (Rd)
S
P3
º P5
r
U r
Rolling
C4
Circle
C3
C5
P6
C2
C6
3 P22
N
C7
C
1
4
1
r CP
C8
01 P
5
P8 P7
0
0
6
7
Ø/2 Ø/2
O
Ø = 360º x 25/75
 = 120°
Arc P0P8 = Circumference of
Generating Circle
Rd X Ø = 2πr
Ø = 360º x r/Rd
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.
Rolling
Circle
Radias (r)
Directing
Circle
Vertical
N
1
2
C2 3
P 0 0 P1 P
2 P C1
12
3
C0
P4
11
10
C3
8
R
7
C9
4
P5
5
6
9
C5 C6 C7 C
C4
8
P6
P7
T
 
/2 /2
O
Hypocycloid
C10
P12
P
C11
P10 11 T
P
P8
9
S
C12
N
 = 360 x r
R
 =36012x 4
 = 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
2
3
C3
1
12
C
C
C4 5 6 C7 C
8
4
C2
P2
P1 P3
C10
5
C1
6
P6 O
P4 C P5
C9
C11
P7
P11
C
P8 12 P9 P10
P12
7
11
10
9
8
HYPOCYCLOID
INVOLUTE
DEFINITION :- If a straight line is rolled
round a circle or a polygon without slipping or
sliding,
points
on
INVOLUTES.
line
will
trace
out
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.
P9
P8
P10
P7
P11
P6
4
8
9
10
1
P12
01211N
P01 1 2 3 4 5 6 7 8 9 10 1112
D
P3 P2
3
2
P5
P4
5 6 7
.
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.
11 mm = 30°
Then 5 mm = 
Ø = 30° x 5 /11 = 13.64 °
P3
P4
P5
P6
INVOLUTE
R=3toP
P2
P1
R=6toP
P7
P8 8ø
P
9
7
6
5
4
3
0
10
11 0
0
S = 2 x π x r /12
2
1
1
2
3
4
5
6
L= 100 mm
7
8 P 9
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.
P3
P4
P2
P5
P1
P6
6 5
4
8
3 15
O
ø2PP14
9
14
R=21mm 10
P0
11 113 P13
12 1 2P 3 4 5 6 7 8 9 10 11 12 13 1415
12
7
P7
P11
P8
P9
P10
L=160 mm
PROBLEM:Draw an involute of a pantagon having side
as 20 mm.
INVOLUTE
OF A POLYGON
Given :
Side of a polygon
P4
P3
P2
3
4
P5
2
1
50
P0
P1
PROBLEM:Draw an involute of a square
having side as 20 mm.
INVOLUTE OF A SQUARE
P1
P0
0
P2
1
4
2
3
P3
P4
PROBLEM:-
Draw an involute of a string
unwound from the given figure from
point C in anticlockwise direction.
B
C
60°
30°
A
C8
C7
B
C
X+AB
5
60°
4
3
30°
2
X
A
1
C5
X+A3
C0
C6
C1
C2
C3
C4
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
6
5
A4
A6
A5
B1
4
B2
3
A3
A2
2
1
A1 A
2
3
1
O
B3
4
5
B4
C
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
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.
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
Shortest radius =
mm
&
00 mm ( at centre or at pole)
3
4
2
P3
P2
P4
5
P1 1
P5
6
o
P6
P1212
P7
0
11
P11 9 8 7 6 5 4 3 2 1 012
P10
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 =
3
4
P3
P4
Angle between them in radians
2
P2
OP – OP3
=
Π/2
P11
5
P5
P12
6 P
6
O
108 6 4 2
11
9 7 5 3 1 12
P11
7
P7
P8
P10
P9
8
9
100 – 90
=
Π/2
10
11
=
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
25 P.
40
B
A
O
31
21
41
11
P2
40
P3
P5
P1 25
B 11109 8 76 5 4 3 2 1 A
P12
51
P4
61
P6
O
P7
P11
111
P10
71
P8
P9
101
9
81
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.
Initial Position of point P
PO
A
2/3 X 360°
1
1
2
3
4
5
6
120ºP7
= 240°
P1
2
P2
3
8
O
P3
P7
P6
P5
P4
4
8
5
7
6
M
Link AB = 96
Linear Travel of point P on AB
= 96 =16x (6 div.)
A0
P6
P5
Angular Swing
of link AB = 180° + 90°
= 270 °
=45 °X 6 div.
A1
P4
96
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.
P3
ARCHIMEDIAN
SPIRAL
P2
P1 P1’
C
P0
A6
A2
P6’
P2’
B
P3’
N
P4’
A4
D
P5’
A3
A5
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
2
P3
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
1
0
θ
P15
3
15
2
4
14
16
5
6
1317 18 1224
P9
8
7
10
11
20
1923
22
9
21
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
7
P0
8
C0
6
P8
P1
91
10 2C1
C2
SC
3P2
4
0
C4
P6C5
C6
C7
C8
C9
C10
7
8
9
10
P3 P5
3
5
P9
P7
1
2
3
P4 Directing line
4
6
5
Length of directing line = 3D/2
540 = 360 + 180
540 = D + D/2
Total length for 540 rotation = 3D/2
P10