Here are the steps to solve this problem: 1. The diameter of the directing circle is 150 mm. So its radius (R) is 150/2 = 75 mm. 2. The radius of the rolling circle (r) is 25 mm. 3. We are required to draw the locus of the point P which lies on the circumference of the rolling circle. 4. Since the rolling circle rolls outside the directing circle, the curve traced is an epicycloid. 5. To draw an epicycloid, we use the formula: Arc of epicycloid = Rd * Θ Where, Rd is the radius of the directing circle and Θ is the central angle swept in one

1. CHAPTER – 1
ENGINEERING
CURVES

2. 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

3. Be it an arc
Be it an arch in construction in civil engineering
Be it an any spring in mechanical engineering
Be it any component of electronics and computer
engineering component
Measuring the distance
On the map
For navigation
In space technology

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 on
the surface of the cone.
TRIANGLE

10. When the cutting plane is perpendicular to
the axis or parallel to the base in a right cone
we get circle as the section on the surface of
the cone.
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 on the surface of the cone.
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. APPLICATION :-
Shape of a man-hole.
Flanges of pipes, glands and stuffing boxes.
Shape of tank in a tanker.
Shape used in bridges and arches.
Monuments.
Path of earth around the sun.
Shape of trays etc.

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. P2’
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 O
 
ARC OF CIRCLE’S
METHOD

28. 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
CONCENTRIC
CIRCLE
METHOD
F2
F1
D
CF1=CF2=1/2 AB
T
N
Q
e = AF1/AQ

29. 0
1
2
3
4
1 2 3 4 1’
0’
2’3’4’
1’
2’
3’
4’
A B
C
D
Major Axis
MinorAxis
F1 F2
Directrix
E
F
S
P
P1
P2
P3
P4
P1’
P2’
P3’
P4’
P0
P1’’
P2’’
P3’’
P4’’P4
P3
P2
P1
OBLONG 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

31. P6
P5’ P7’P6’
P1
P1’
T
T
V1
P5
P4’
P4
P3’
P2’
F1
D1D1
R1
b
a
c
d
e
f
g
Q
P7
P3
P2
Directrix
90°
1 2 3 4 5 6 7
Eccentricity = 2/3
3R1V1
QV1
=
R1V1
V1F1
=
2
Ellipse
ELLIPSE – DIRECTRIX FOCUS METHOD
Dist. Between directrix
& focus = 50 mm
1 part = 50/(2+3)=10
mm
V1F1 = 2 part = 20 mm
V1R1 = 3 part = 30 mm
 < 45º
S

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

33. 2345
0
1
2
3
4
5
6 1 1 5432 6
0
1
2
3
4
5
0
VD C
A B
P
4
P
4
P
5
P
5
P
3
P
3
P
2
P
2
P
6
P
6
P
1
P
1
PARABOLA –RECTANGLE METHOD
PARABOLA

34. B
2’
0
6
C
P’5
30°
A X
D
5’
4’
3’
1’
0
5
4
3
2
1
P1
P2
P3
P4
P5
P’4
P’3
P’2P’1
P’6
PARABOLA – IN PARALLELOGRAM
P6

35. BA O
V
1
8
3
4
5
2
6
7
9
10
0
1
2
3
4
5
6
7
8
9
1
0
0


F
PARABOLA
TANGENT METHOD

36. D
D
90° 2 3 4
T
T
N
N
S
V 1
P1
P2
PF
P
3
P4
P1’
P2’
P3’
P4’
PF’
AXIS
90°
R F
PARABOLA
DIRECTRIX FOCUS METHOD

37. 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.

38. 6m
ROOT OF TREE
BUILDING
REQD.DISTANCE
TOP OF TREE
3m
6m
F
A
STONE FALLS HERE

39. 3m
6m
ROOT OF TREE
BUILDING
REQD.DISTANCE
GROUND
TOP OF TREE
3m
6m
1
2
3
1
2
3
321 4 5 6
5
6
EF
A B
CD
P3
P4
P2
P1
P
P1
P2
P3
P4
P5
P6
3 2 1
0
STONE FALLS HERE

40. 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

41. ground level BA
60º
gunshot
80 M
parabola

42. ground level BA O
V
1
8
3
4
5
2
6
7
9
10
0
1
2
3
4
5
6
7
8
9
1
0
0


F
60º
gunshot
D D
VF
VE = e = 1
E

43. CYCLOIDAL GROUP OF CURVES
Superior
Hypotrochoid
Cycloidal Curves
Cycloid Epy Cycloid Hypo Cycloid
Superior
Trochoid
Inferior
Trochoid
Superior
Epytrochoi
d
Inferior
Epytrochoi
d
Inferior
Hypotrochoi
d

44. 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

45. EPICYCLOID:-
Epicycloid is a locus of a point(P) on the circumference
of a rolling circle(generator), which rolls without slipping or
sliding OUTSIDE another circle called Directing Circle.
2πr
Ø = 360º x r/Rd
Circumference of
Generating Circle
Rolling
Circle
r
O
Ø/2 Ø/2
P0 P0
Arc P0P0 =
Rd x Ø =
P0

46. 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
Ø /2 Ø /2
Ø =
360 x r
RR
T
Rolling Circle
Radius (r)
O
Vertical
Hypocycloi
P P

47. 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.

48. P0
2R or D
5
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
S1
R
P6
R
P10
R
: Given Data :
Draw cycloid for one revolution of a rolling circle having
diameter as 60mm.
Rolling
Circle
D

49. 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.

50. 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.

51. P6
P4
r
P2
C1
C0
C2
C3
C4 C5
C6
C7
C8
1
0
23
4
5
6 7
O
Ø/2 Ø/2
P1
P0
P3 P5
P7P8
r rRolling
Circle
r
Rd X Ø = 2πr
Ø = 360º x r/Rd
Arc P0P8 = Circumference of
Generating Circle
EPICYCLOIDGIVEN:
Rad. Of Gen. Circle (r)
& Rad. Of dir. Circle (Rd)
S
º
U
N
Ø = 360º x 25/75
 = 120°

52. 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.

53. P0 P1 P11
C0
C1
C2
C3
C4
C5
C6 C7 C8
C9
C10
C1
1
C12
P1
0
P8
0
1 2
3
4
5
6
7
89
10
1
1
1
2
P
2
P3
P4
P5 P6
P9
P7
P12

/2

/2
 =360 x 4
12
 = 360 x r
R
 =120°
R
T
T
N
S
N
Rolling
Circle
Radias (r)
Directing
Circle
O
Vertical
Hypocycloi

54. INVOLUTE
DEFINITION :- If a straight line is rolles
round a circle or a polygon without slipping or
sliding, the locus of points on the straight line
is an INVOLUTES to the curve.
OR
Uses :- Gears profile
Involute of a circle is a curve traced out by a
point on a tights string unwound or wound from
or on the surface of the circle.

55. PROBLEM:
A string is unwound from a
circle of 20 mm diameter. Draw the
locus of string P for unwinding the
string’s one turn. String is kept tight
during unwound. Draw tangent &
normal to the curve at any point.

56. P12
P2
012
6
P1
1 20 9 103 4 6 8 115 7 12
D
P3
P4
P5
P6
P7
P8
P9
P10
P11
1
2
3
4
5
7
8
9
10
11N
.

57. 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.

58. P
R=6toP
0
0 1 2 3 4 5 6 7 8 P
11 0
1
2
3
4
56
7
8
9
10
P1
P2
P3
P4
P5
P6
P7
P8
L= 100 mm
R=3toP
INVOLUTE
9
ø
11 mm = 30°
Then 5 mm = Ø = 30° x 5 /11 = 13.64 °
S = 2 x π x r /12

59. 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.

60. P13
P11
3
13
14
15
P0
P12
O
7
10 1
2
3
456
8
9
11
12 1 2
P1
P2
P3P4
P5
P6
P7
P8 P9 P10
P14P
L=160 mm
R=21mm
64 5 7 8 9 10 11 12 131415
ø

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

62. P
5
P
0
P
1
P
2
P
3
P
4
2
3
4 5
1
INVOLUTE
OF A POLYGON
Given :
Side of a polygon 0

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

64. P2
1
2 3
0
4
P0
P1
P3
P4
INVOLUTE OF A SQUARE

65. PROBLEM:-
Draw an involute of a string unwound
from the given figure from point C in
anticlockwise direction.
60°
A
B
C
30°

66. 60°
A
B
C
30°
X
X+A3
X+AB
1
2
3
4
5
C0
C1
C2
C3
C4
C5
C6
C7
C8

67. 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.

68. 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

69. 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.

70. 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:

71. P10
1
2
3
4
5
6
7
8
9
10
11
12
0
8 7 012345691112
P1
P2
P3
P4
P5
P6
P7
P8
P9
P11
P12
o

72. 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

73. 3 1
26
5
8 4
79
1
011
2
1
3
4
5
6
7
8
9
10
11
12
P1
P2
P3
P4
P5
P6
P7
P8 P9
P1
0
P1
1
P12
O
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

74. 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.

75. A
Initial Position of point PPO
P1
P2
P3
P4
P5
P6
P7
P8
2
1
3
4
5
6
7
O
1
2
3
4
5
6
7
8
2/3 X 360°
= 240°
120º

76. 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.

77. θ
o
01
2
3
4
5 6
7
8
9
10
11
121314
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

78. http://www.engineering108.com/pages/Engineeri
ng_graphics/Engineering_graphics_tutorials_free
_download.html
A text book of engineering graphics- Prof. P.J
SHAH
Engineering Drawing-N.D.Bhatt
Engineering Drawing-P.S.Gill

79. Thank You