1. 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”.

2. CHAPTER – 2
ENGINEERING
CURVES

3. 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

4. 1. CONICS
2. CYCLOIDAL
CURVES
3. INVOLUTE
4. SPIRAL
5. HELIX
6. SINE & COSINE
CLASSIFICATION OF ENGG.
CURVES

5. 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

6. 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

7. 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

8. B - CIRCLE
A - TRIANGLE
CONICS
C - ELLIPSE
D – PARABOLA
E - HYPERBOLA

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

10. 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

11. 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

12. 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

13. 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.
HYPERBOLA
Definition :-
α < θ
α = 0
θ
θ

14. 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.
Fixed point is called as focus.
Fixed straight line is called as directrix.
M
C
F
V
P
Focus
Conic Curve
Directrix

15. 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
F
V
P
Focus
Conic Curve
Directrix
Vertex

16. 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
C
V
Focus
Conic Curve
Directrix
Vertex

17. 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
C
F
V
Axis
Focus
Ellipse
Directrix
Eccentricity=PF/PM
= QF/QN
< 1.

18. 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

19. 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

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

21. 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.

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

23. 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

24. 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
Axis
Directrix
Hyperbola
M
C
N
Q
F
V
P
FocusVertex
HYPERBOLA
= QF/QN
> 1.

25. Nature of graph of Boyle’s law
Shape of overhead water tanks
Uses :-
Shape of cooling towers etc.

26. 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

27. Normal
P2’
R=A1
Tangent
1 2 3 4
A 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

28. AxisMinor
A B
Major Axis 7
8
9
10
11
9
8
7
6
5
4
3
2
1
12
11
P6
P5
P4
P3
P2`
P1
P12
P11
P10
P9
P8
P7
6
5
4
3
2
1
12
C
10
O
CONCENTRICCONCENTRIC
CIRCLECIRCLE
METHODMETHOD
F2
F1
D
CF1=CF2=1/2 AB
T
N
Q
e = AF1/AQ

29. 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
MinorAxisMinorAxis
FF11 FF22
DirectrixDirectrix
EE
FF
SS
PP
PP11
PP22
PP33
PP44
Tangent
Tangent
PP11’’
PP22’’
PP33’’
PP44’’
ØØ
R
=A
B/2
R
=A
B/2
PP00
P1’’
P2’’
P3’’
P4’’P4
P3
P2
P1
OBLONG METHODOBLONG METHOD

30. BA
P4
P
0
D
C
60°
6
5
4
3
2
1
0
5 4 3 2 1 0 1 2 3 4 5 6
5
3
2
1
0P1
P2
P
3
Q1
Q2
Q3Q4
Q5
P6 Q6O
4
ELLIPSE IN PARALLELOGRAM
R4
R3
R2
R1
S1
S2
S3
S4
P5
G
H
I
K
J
MinorAxis
Major Axis

31. P6
Normal
P5’ P7’P6’
P1
Tangent
P1’

N
T
T
V1
P5
P4’
P4
P3’
P2’
F1
D1D1
R1
b
a
c
d
e
f
g
Q
P7
P3
P2
Directrix
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 mm
V1R1 = 3 part = 30 mm
θ < 45º
S

32. 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.

33. ARC OF CIRCLE’SARC OF CIRCLE’S
METHODMETHOD
Normal
G2’
R=A1
Tangent
1 2 3 4
A B
G
G’
G1
G3
G2
G4 G4
G3
G2
G1
G1’
G3’
G4’ G4’
G3’
G2’
G1’
F2
F1
R=B1
R=B2
`R=A2
O
° °
90°
90°
directrixdirectrix
100100
140140
GF1 + GF2 = MAJOR AXIS = 140GF1 + GF2 = MAJOR AXIS = 140
EE
eeAF1AF1
AEAE
ee ==
R=70
R=70 R=70
R=70

34. 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.

35. OA B
C
75
45
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

36. 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.

37. I3
CD
F1 F2
P Q
R
S
5050 I1 I4
A BI2
O
1
1
2
2
4
4
3
3
100100

38. 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.

39. A B
P
R=50
M
N
O
1
2
1 2 60
80
Q
1
2
12
P1
P2 Q2
Q1
R1
R2 S2
S1

40. 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.

41. ELLIPSEELLIPSE
PP
QQ
R
80mm
40mm
89mm
AA
CC
BB
DD
MAJOR AXIS = PQ+QR = 129mmMAJOR AXIS = PQ+QR = 129mm
θ
θ
R=AB/2
11
22
33
1’1’ 2’2’ 3’3’ 1’’1’’2’’2’’3’’3’’
OO33 OO33’’
OO11
OO22 OO22’’
OO11’’
TANGENT
TANGENT
NORMAL
NORMAL
6060ºº
30º30º
22
33
11
directrixdirectrix
FF11 FF22MAJOR AXISMAJOR AXIS
MINORAXISMINORAXIS
SS
ECCENTRICITY = AP / ASECCENTRICITY = AP / AS
?? ??

42. 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.

43. D
C
6
6
5
4
3
2
1
0
5 4 3 2 1 0 1 2 3 4 5 6
6
5
3
2
1
0
P2
P3
P4
P5
Q1
Q2
Q3
Q4
Q5
B
A O
4
ELLIPSE
100
45
75
P0
P1
P6
Q6
G
H
I
K
J

44. 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.

45. 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.

46. METHODS FOR DRAWING PARABOLA
1. Rectangle Method
2. Parabola in Parallelogram
3. Tangent Method
4. Directrix Focus Method

47. 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
PARABOLAPARABOLA

48. 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
33
44
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
PP
66

49. 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
θ
θ
F
PARABOLA
TANGENT METHOD

50. D
D
DIRECTRIX
90° 2 3 4
T
T
N
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

51. 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.

52. 6m6m
ROOT OF TREEROOT OF TREE
BUILDINGBUILDING
REQD.DISTANCEREQD.DISTANCE
TOP OF TREETOP OF TREE
3m3m
6m6m
FF
AA
STONE FALLS HERESTONE FALLS HERE

53. 3m3m
6m6m
ROOT OF TREEROOT OF TREE
BUILDINGBUILDING
REQD.DISTANCEREQD.DISTANCE
GROUNDGROUND
TOP OF TREETOP OF TREE
3m3m
6m6m
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 11
00
STONE FALLS HERESTONE FALLS HERE

54. 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.

55. 150 mm
AA
BB CC
DD11 22 33 44 55
11
22
33
44
55
OO
PP11
PP22
PP33
PP44
PP55
MM
1’1’ 2’2’ 3’3’ 4’4’ 5’5’
1’1’2’2’3’3’4’4’5’5’
PP11’’
PP22’’
PP33’’
PP44’’
PP55’’
90mm

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

57. ground levelground level BA
60º
gungun
shotshot
80 M
parabolaparabola

58. 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
10
0


F
60º
gungun
shotshot
DD DD
VFVF
VEVE
== e = 1e = 1
EE

59. Connect two given points A and B by aConnect 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 angle2.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°

60. 66
6060
1 2 3 4 5
Parabola
5
4
3
2
1
AA
BB
9090 °°
OO
1.OA=OB=60mm and angle1.OA=OB=60mm and angle
AOB=90AOB=90°°

61. AA
BB
8080
6060
5
4
3
2
1
1 2 3 4 5
110
°°
Parabola
OO
2.OA=60mm,OB=80mm and angle2.OA=60mm,OB=80mm and angle
AOB=110°AOB=110°

62. 54321
5
4
3
2
1
AA
BB
OO
Parabola
66
6060
6060
°°
3.OA=OB=60mm3.OA=OB=60mm
and angle AOB=60°and angle AOB=60°

63. exampleexample
Draw a parabola passing through threeDraw 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=80mm100mm, BC=50mm and CA=80mm
respectively.respectively.

64. BBAA
CC
100100
5050
8080

65. 1’1’
00
00
2’2’
00
22
66
6’6’55
P’P’55
1’1’ 2’2’ 4’4’ 5’5’3’3’113344
5’5’
4’4’
3’3’
55
44
33
22
11
PP
11
PP
22
PP
33
PP
44
PP
55
P’P’44
P’P’33
P’P’22
P’P’11
P’P’66
PP
66
AA BB
CC

66. METHODS FOR DRAWING HYPERBOLA
1. Rectangle Method
2. Oblique Method
3. Directrix Focus Method

67. D
F
1 2 3 4 5
5’
4’
3’
2’
P1
P2
P3
P4
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

68. 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”.

69. D
F
1 2 3 4 5
5’
4’
3’
2’
P1
P2
P3
P4
P5
0
P6
P0
AO EX=20
B
C
Y=50
Given Point P0
90°
6
6’
Hyperbola
RECTANGULAR HYPERBOLA

70. 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”.

71. 75 0
P4
E
6’
2’
1’ P1
1 2 3 4 5 6 D
P6
P5
P3
P2
P0
7’P7
7C
B F
O
Y=30
X = 20
Given Point P0
A

72. AXISAXIS
NORMAL
NORMAL
CC VV
FF11
DIRECTRIXDIRECTRIXDDDD
11 22 33 44
4’4’
3’3’
2’2’
1’1’
PP11
PP22
PP33
PP44
PP11’’
PP22’’
PP33’’
PP44’’
TT11
TT22
NN
22
NN
11
TANGENT
TANGENT
ss
Directrix and focus methodDirectrix and focus method

73. 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.
Superior
Hypotrochoid
Cycloidal Curves
Cycloid Epy Cycloid Hypo Cycloid
Superior
Trochoid
Inferior
Trochoid
Superior
Epytrochoid
Inferior
Epytrochoid
Inferior
Hypotrochoid

74. 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

75. 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/Rx r/Rdd
Circumference ofRRdd
RollingRolling
CircleCircle
r
O
Ø/Ø/
22
Ø/Ø/
22
P0 P0
Arc P0P0 =
Rd x Ø =
P0

76. 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.`
Directing
Circle(R)
P
ØØ /2 ØØ /2
ØØ =
360 x r
RR
T
Rolling Circle
Radius (r)
O
Vertical
Hypocycloid
P P

77. 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.

78. P0
2R or D
5
T
T
1
2
1 2 3 4 6 7 8 9 10 110 12
0
3
4
56
7
8
9
10
11 12
P1
P2
P3
P4
P5 P7
P8
P9
P11
P12
C0 C1 C2 C3
C4 C5 C6 C7
C8 C9 C10
C11
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

79. CC00
PP00
7788
PP
66
44
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
77
00
CC55CC66CC77CC88
11 22 33 44
5566DD/2/2ππDD/2/2
ππDD/2/2 DD/2/2
FloorFloor
WallWall
CYCLOIDCYCLOID
55
66
77
88
Take diameter of circle = 40mm
Initially distance of centre of
circle from the wall 83mm (Hale
circumference + D/2)

80. 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.

81. 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.

82. PP66
PP44
r
PP22
CC11
CC00
CC22
CC33
CC44 CC55
CC66
CC77
CC88
1
0
23
4
5
6 7
O
RRdd
Ø/2Ø/2 Ø/2Ø/2
PP11
PP00
PP33 PP55
PP77PP88
r rRollingRolling
CircleCircle
r
Rd X Ø = 2πr
Ø = 360Ø = 360ºº xx
Arc P0P8 = Circumference of
Generating Circle
EPICYCLOIDEPICYCLOIDGIVEN:
Rad. Of Gen. Circle (r)
& Rad. Of dir. Circle (Rd)
S
º
U
N
Ø = 360Ø = 360ºº x 25/75x 25/75
 = 120°= 120°

83. 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.

84. P0 P1
Tangent
P11
r
C0
C1
C2
C3
C4
C5
C6 C7 C8
C9
C10
C11
C12
P10P8
0
1 2
3
4
5
6
7
89
10
11
12
P2
P3
P4
P5 P6
P9
P7
P12
 /
2
 /
2
 = 360 x 4
12
 = 360 x r
R
 = 120°
R
T
T
N
S
N
Normal
r
r
Rolling
Circle
Radias (r)
Directing
Circle
O
Vertical
Hypocycloid

85. 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

86. C
C1
C2
C3
C4
C5
C6 C7
C9
C8
C10
C11
C12
P8
O
10
5
7
8
9
11
12
1
2 3
4
6
P1
P2
P3 P4
P5 P6 P7 P9
P10
P11
P12
Directing Circle
Rolling Circle
HYPOCYCLOID

87. 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.

88. 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.

89. P12
P2
0
02
12
6
P1
1 20 9 103 4 6 8 115 7 12
π
D
P3
P4
P5
P6
P7
P8
P9
P10
P11
1
2
3
4
5
7
8
9
10
11
03
04
05
06
07
08
09
010`
011
Tangent
N
N
Norm
al
T
T
.

90. 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.

91. PP
R=7toP
R=7toP
R=6toPR=6toP
R21
R21
00
00 11 22 33 44 55 66 77 88 PP
1111 00
11
22
33
44
5566
77
88
99
1010
PP11
PP22
PP33
PP44
PP55
PP66
PP77
PP88
L= 100 mmL= 100 mm
R=1toP
R=1toP
R=2toP
R=2toP
R=3toPR=3toP
R=4toP
R=4toP
R=5toP
R=5toP
INVOLUTEINVOLUTE
99
ø
11 mm = 30°
Then 5 mm = ζ
Ø = 30° x 5 /11 = 13.64 °
S = 2 x π x r /12

92. 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.

93. PP1313
PP1111
3
13
14
15
PP00
PP1212
O
7
10 1
2
3
456
8
9
11
12 1 2
PP11
PP22
PP33PP44
PP55
PP66
PP77
PP88 PP99 PP
PP1414PP
L=160 mmL=160 mm
R=21mm
64 5 7 8 9 10 1112 131415
øø

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

95. P5
R=01
R=2∗01
P0
P1
P2
P3
P4
R=3∗01
R=4∗01
R=5∗01
2
3
4 5
1
T
T
N
N
S
INVOLUTE
OF A POLYGON
Given :
Side of a polygon 0

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

97. P2
1
2 3
0
4
P0
P1
P3
P4
N
N
S
R=3∗01
R=4∗01
R
=2∗01
R=01
INVOLUTE OF A SQUARE

98. PROBLEM:-
Draw an involute of a string
unwound from the given figure
from point C in anticlockwise
direction.
60°60°
AA
BB
CC
R
21
R
2130°30°

99. R
2
R
2
11
60°60°
AA
BB
CC
30°30°XX
X+A
X+A
11
XX X+A2
X+A2
X+AX+A
33
X+A5
X+A5
X+A4
X+A4
X+AX+A
BB
RR
=X+AB
=X+AB
X+66+BC
X+66+BC
11
22
33
44
55
C0
C1
C2
C
C4
C5
C6
C7
C8

100. A stick of length equal to the circumference of aA stick of length equal to the circumference of a
semicircle, is initially tangent to the semicirclesemicircle, is initially tangent to the semicircle
on the right of it. This stick now rolls over theon the right of it. This stick now rolls over the
circumference of a semicircle without sliding tillcircumference of a semicircle without sliding till
it becomes tangent on the left side of theit becomes tangent on the left side of the
semicircle. Draw the loci of two end point of thissemicircle. Draw the loci of two end point of this
stick. Name the curve. Take R= 42mm.stick. Name the curve. Take R= 42mm.
PROBLEM:-

101. A6
B6
5
A
B
C
B1
A1
B2
A2
B3
A3
B4
A4
B5
A5
1
2 3
4
5
O
1
2
3
4
6
INVOLUTE

102. 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.

103. 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

104. 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.

105. 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:

106. PP1010
1
2
3
4
5
6
7
8
9
10
11
12
0
8 7 012345691112
PP11
PP22
PP33
PP44
PP55
PP66
PP77
PP88
PP99
PP1111
PP1212
o

107. 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

108. 3 1
26
5
8 4
79
10
11
2
1
3
4
5
6
7
8
9
10
11
12
P1
P2
P3
P4
P5
P6
P7
P8 P9
P10
P11
P12
O
N
N
T
T
S
Rmin
R
m
ax
Diff. in length of any two radius vectors
Angle between them in radians
Constant of the curve =
=
OP – OP3
Π/2
100 – 90
=
Π/2
= 6.37 mm

109. 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

110. BB AA OO1234567891011
P1
P2
P3 P4
P5
P6
P7
P8
P9
P10
P11
P12
11
21
31
41
51
61
71
81
9
101
111
40 25

111. 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.

112. AA
Initial Position of point PInitial Position of point PPPOO
PP11
PP22
PP33
PP44
PP55
PP66
PP77
PP88
22
11
33
44
55
66
77
OO
11
22
33
44
55
66
77
88
2/3 X 360°
= 240°
120º120º

113. 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
AA66
PP00
PP11
PP22
PP33
PP44
PP55
PP66
PP22’’
PP33’’
PP44’’ PP55’’
PP66’’
9696
LinkLink ABAB = 96= 96
CC
Tangent
Tangent
AngularAngular SwingSwing
ofof linklink AB =AB = 180° + 90°180° + 90°
== 270270 °°
=45 °X 6 div.=45 °X 6 div.
ARCHIMEDIANARCHIMEDIAN
SPIRALSPIRAL
DDNORMAL
NORMAL
MM
NN

114. Arch.Spiral Curve Constant BCBC
= Linear Travel ÷Angular Swing in Radians
= 96 ÷ (270º×π /180º)
=20.363636 mm / radian

115. 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.

116. θ
o
01
2
3
4
5 6
7
8
9
10
11
1213
14
15
16 17 18 19
20
21
222324
23
13
22
24
1
2
3
4
5
6
7
8
9
10
11
12
14
15
16
18
19
20
21
17
P3
P9
P15

117. 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.

118. 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.

119. Rolling circleRolling circle
PP11
PP22
PP33
PP44
PP00
PP66
PP77
PP88
PP55
PP99
PP1010
11
22
33
44
55
66
77 88
99
1010
11 22 33 44 55 66 77 88 99 101000
CC00 CC11 CC22
CC33 CC44
Directing lineDirecting line
Length of directing line = 3Length of directing line = 3ΠΠ
540540 °° = 360= 360°° + 180+ 180°°
540540 °° == ΠΠD +D +
ΠΠD/2D/2
Total length for 540Total length for 540 °° rotation = 3rotation = 3ΠΠD/2D/2
CC55 CC66 CC77
CC88 CC99 CC1010
SS
norm
al