This ppt contains brief details about engineering curves like parabola, ellipse, involute, helical curves and many other curves to easily draw........

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. 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.

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

5. 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
known as base.
If the base/closed curve
is a circle, we get a cone.
If the base/closed
curve is a polygon, we
get a pyramid.
Vertex/Apex
90º
Base

6. 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.
Cone Axis
Generator
90º
Base
Vertex/Apex

7. 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.

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

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

10. 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

11. ELLIPSE
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.
θ
α
α > θ

12. 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
α = θ

13. 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 < 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.
M
P
C F
V
Conic Curve
Focus
Directrix
Fixed straight line is called as directrix.
Fixed point is called as focus.

15. 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 Axis
C F
V
P
Focus
Directrix
Vertex

16. N Q
Conic Curve
Distance of a point from focus
Distance of a point from directrix
=
E= c cPFe/PnMt r= iQcFi/tQyN = VF/VC
= e
Ratio =
M P
F
Axis
C V
Focus
Directrix
Vertex

17. 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.
Vertex
ELLIPSE
M
N Q
P
C F
V
Axis
Ellipse
Focus
Directrix
Eccentricity=PF/PM
= QF/QN
< 1.

18. 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.
A B
F1
P
F2
O
Q
C
D

19. A B
F1
C
Q
D
= FA + FB = FA + FB
1122But FA = FB
12FA + FB = FB + FB = AB
1121P
F2
O
PF1 + PF2 = QF1 + QF2 = CF1 +CF2 = constant
= Major Axis
CF1 +CF2 = AB
but CF1 = CF2
hence, CF1=1/2AB

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

21. 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.

22. 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).
M
P
Parabola
Directrix Axis
Vertex
C
N Q
F
V
Focus
Eccentricity = PF/PM
= QF/QN
= 1.

23. 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

24. 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.
Axis
Eccentricity = PF/PM
Directrix
Hyperbola
M
P
C
N Q
F
V
Vertex Focus
= QF/QN
> 1.

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

26. 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

27. AARRCC OOFF CCIIRRCCLLEE’’SS
MMEETTHHOODD
R =A1
A B
Tangent
NormalP2’
1 2 3 4
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
° °

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

29. OOBBLLOONNGG MMEETTHHOODD
44
Normal
33
22
11
00
CC
MMiinnoorr AAxxiiss
11 22 33 44 11’’
44’’
33’’
22’’
00’’
44’’ 33’’ 22’’
11’’
AA BB
DD
MMaajjoorr AAxxiiss
FF11 FF22
DDiirreeccttrriixx
EE
FF
SS
PP
PP11
PP22
PP33
PP44
Tangent
PP11’’
PP22’’
PP33’’
PP44’’
Ø Ø
R=AB/2
PP00
P1’’
P2’’
P3’’
P4P ’’ 4
P3
P2
P1

30. P4
P P1 P
0 P
2 0
3
4
5
P5
A B
D
C
S4
60°
6
3
2
1
0
5 4 3 2 1 0 1 2 3 4 5 6
5
3
1
2
Q1
Q2Q3Q4
Q5
P6 O Q6
4
ELLIPSE IN PARALLELOGRAM
R4
R3 R2 R1
S1
S2
S3
G
H
I
K
J
Minor Axis
Major Axis

31. ELLIPSE – DIRECTRIX FOCUS METHOD
P6
P4
1 2 3 4 5 6 7
P3P ’ 2’
Normal
P5’ P P7’ 6’
P1
P1’

Tangent
N N
T
T
V1
P5
P4’
F1
D1 D1
R1
a b
c
d
e f
g
Q
PP 7 P 3 2
Directrix
R=6f`
90°
q < 45º
Eccentricity = 2/3
V1F1 = 2
QV1 = R1V1
R 3 1V1
Ellipse
R=1a
Dist. Between directrix
& focus = 50 mm
1 part = 50/(2+3)=10 mm
V 1F1 = 2 part = 20 mm
V1R1 = 3 part = 30 mm
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. AARRCC OOFF CCIIRRCCLLEE’’SS
MMEETTHHOODD
R =A1
R=B1
A B
Tangent
1 2 3 4
NormalG2’
G
G’
G1
G3
G2
G4 G4 G3
G2
G1
G1’
G3’
G4’ G4’
G3’
G2’
G1’
F2 F1
R=B2
`R=A2
O
° °
90°
90°
ddiirreeccttrriixx
110000
114400
GGFF11 ++ GGFF22 == MMAAJJOORR AAXXIISS == 114400
EE
AAFF11 ee
AAEE ee ==
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. D
C
75
7
45
1
A 100
O B
1
2 2
3
3
4
4
5
5
6 6
P1 7
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. 1
1
D C
P Q
F1 F2
I3
R
O
S
5500
I1 I4
AI2 B
2
2
4
4
3
3
110000

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. R = 50
M
2
A O
B P
N
2
1
1 2 60
80
Q
1
2 1
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. QQ
OO33 OO33’’
MMIINNOORR AAXXIISS
OO22 OO22’’
q q
R=AB/2
EELLLLIIPPSSEE
80mm
PP
R
40mm
89mm
33
22
11
AA
CC
33
BB
11’’ 22’’ 33’’ 33’’’’ 22’’’’ 11’’’’
DD
OO11
MMAAJJOORR AAXXIISS == PPQQ++QQRR == 112299mmmm
OO11’’
TANGENT
NORMAL
6060º
3300º
22
11
ddiirreeccttrriixx
FF11 FFMMAAJJOORR AAXXIISS 22
SS
EECCCCEENNTTRRIICCIITTYY == AAPP // AASS
?? ??

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. P 0 2
D
C
5
6
4
6
3
2
1
0
5 4 3 2 1 0 1 2 3 4 5 6
6
5
3
2
1
P3
P4
P5
Q1
Q2Q
3
Q4 Q5
A B 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. PPAARRAABBOOLLAA ––RREECCTTAANNGGLLEE MMEETTHHOODD
DD VV CC
PP11 PP11
PP22 PP22
PP33 PP33
PP44 PP44
PP55 PP55
PPAARRAABBOOLLAA
PP66 PP66
55 44 33 22
00
11
22
33
44
55
00
11
22
33
44
55
66 11 00
11 22 33 44 55 66
AA BB

48. PPAARRAABBOOLLAA –– IINN PPAARRAALLLLEELLOOGGRRAAMM
00
11’’
33’’
44’’
BB
5’ 2’ 3’ 4’ 1’ 00
11 33 44
22’’
DD
00
22
11
22
33
44
66
CC
6’
VV
55
PP’’ 55
3300°°
AA XX
55’’
55
PP11
PP22
PP33
PP44
PP55
PP’’ 44
PP’’ 33
PP’’ 22 PP’’ 11
PP’’
66
P6
P6

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

50. D
R F
D
P1
R3
90° 2 3 4
DIRECTRIX
T
N
S
N T
V 1
P2
PF
P3
P4
P1’
P2’
P3’
P4’
PF’
AXIS
RF R2
R1
R4
90°
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. 66mm
TTOOPP OOFF TTRREEEE
3m 6m FF
RROOOOTT OOFF TTRREEEE
BBUUIILLDDIINNGG
RREEQQDD..DDIISSTTAANNCCEE
AA
SSTTOONNEE FFAALLLLSS HHEERREE

53. DD CC
11
22
33
PP22
PP
PP11
PP11
PP22
6m
PP33
3m PP33
AA PPBB
44
RROOOOTT OOFF TTRREEEE
BBUUIILLDDIINNGG
PP44
RREEQQDD..DDIISSTTAANNCCEE
GGRROOUUNNDD
TTOOPP OOFF TTRREEEE
3m 6m
11
22
33
11 22 33 44 55 66
55
66
PP55
FF EE
PP66
33 22 11
00
SSTTOONNEE FFAALLLLSS HHEERREE

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

56. EEXXAAMMPPLLEE
AA sshhoott iiss ddiisscchhaarrggee ffrroomm tthhee ggrroouunndd
lleevveell aatt aann aannggllee 6600 ttoo tthhee hhoorriizzoonnttaall
aatt aa ppooiinntt 8800mm aawwaayy ffrroomm tthhee ppooiinntt ooff
ddiisscchhaarrggee.. DDrraaww tthhee ppaatthh ttrraaccee bbyy tthhee
sshhoott.. UUssee aa ssccaallee 11::110000

57. 60º
gguunn
sshhoott
A ggrroouunndd lleevveell B
80 M
ppaarraabboollaa

58. VVFF
VVEE
DD DD
V
1
8
3
4
5
2
6
7
9
10
A ggrroouunnddO lleevveell B
0
1
2
3
4
5
6
7
8
9
10
0


F
60º
gguunn
sshhoott
== ee == 11
EE

59. CCoonnnneecctt ttwwoo ggiivveenn ppooiinnttss AA aanndd BB bbyy aa
PPaarraabboolliicc ccuurrvvee,, wwhheenn::--
11..OOAA==OOBB==6600mmmm aanndd aannggllee AAOOBB==9900°°
22..OOAA==6600mmmm,,OOBB==8800mmmm aanndd aannggllee
AAOOBB==111100°°
33..OOAA==OOBB==6600mmmm aanndd aannggllee AAOOBB==6600°°

60. Parabola
AA
1
2
60 1 2 3 4 5
66
00
3
4
5
BB
9900 °°
OO
11..OOAA==OOBB==6600mmmm aanndd aannggllee
AAOOBB==9900°°

61. 22..OOAA==6600mmmm,,OOBB==8800mmmm aanndd aannggllee
AAOOBB==AA
111100°°
BB
Parabola
8800
60
5
4
3
2
1
110
°°
1 2 3 4 5
OO

62. Parabola
1 2 3 4 5
5
4
3
2
1
AA
OO BB
66
00
6060
6600
°°
33..OOAA==OOBB==6600mmmm
aanndd aannggllee AAOOBB==6600°°

63. eexxaammppllee
DDrraaww aa ppaarraabboollaa ppaassssiinngg tthhrroouugghh tthhrreeee
ddiiffffeerreenntt ppooiinnttss AA,, BB aanndd CC ssuucchh tthhaatt AABB ==
110000mmmm,, BBCC==5500mmmm aanndd CCAA==8800mmmm
rreessppeeccttiivveellyy..

64. CC
AA BB
110000
50
8080

65. 00
1’
PP
66AA 00
BB
2’
00
22
66
P’ 33
6’
44
55
P’ 55
22
1’ 2’ 4’ 5’ 3’ 11 33 44
4’
5’
3’
55
33
11
PP11
PP22
PP33
PP44
PP55
P’ 44
P’ 22 P’ 11
P’
66
CC

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

67. RECTANGULAR HYPERBOLA
When the asymptotes are at right angles to each other, the hyperbola
is called rectangular or equilateral hyperbola
Given Point P0
D
F
1 2 3 4 5
2’
3’
4’
5’
P1
P2
P3 P4 P5
0
P6
P0
B
C
Y
O X E A
90°
6
6’
Hyperbola
AXIS
AXIS
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. RECTANGULAR HYPERBOLA
Given Point P0
D
F
1 2 3 4 5
2’
3’
4’
5’
P1
P2
P3 P4 P5
0
P6
P0
B
C
Y = 50
O X=20 E A
90°
6
6’
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. 750
P4
B F
P 7’ 7
6’
E
1’ P1
2’
1 2 3 4 5 6 D
P6
P5
P3
P2
P0
C 7
O
Y = 30
X = 20
Given Point P0
A

72. AAXXIISS
33’’
22’’
DDiirreeccttrriixx aanndd ffooccuuss mmeetthhoodd
11’’
PP11
PP22
NORMAL
DD DDIIRREECCTTRRIIXX DD
CC VV FF11
44’’
PP33
11 22 33 44
PP44
PP11’’
PP22’’
PP33’’
PP44’’
TT11
TT22
NN11
ss
TANGENT
NN22

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. 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
C
P
R
C
P P
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
ØØ == 336600º xx rr//RRdd
P0 P0
Circumference of
RRdd
RRoolllliinngg
CCiirrccllee
r
O
ØØ//
22
ØØ//
22
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)
Vertical
P P
P
ØØ /2 ØØ /2
ØØ = 360 x r
R R
T
Rolling Circle
Radius (r)
O
Hypocycloid

77. 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.

78. 7 6 5
P2
P0
T T
2R or D
5
1
3
4
2
0
0 1 2 3 4 6 7 8 9 10 11 12
8
9
10
11 12
P1
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. PP22 CC44
CC00
PP00
77 88
PP
77
PP
66 44
PP11 11
22
33
CC22 CC33
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
88
77
00
CC55 CC66 CC77 CC88
11 22 33 44
DD//22 55 ππDD//22 66
ππDD//22 DD//22
FFlloooorr
WWaallll
CCYYCCLLOOIIDD
77
66
55
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
º
GIVEN: EEPPIICCYYCCLLOOIIDD
SRad. Of Gen. Circle (r)
& Rad. Of dir. Circle (Rd) PP44
PP22
CC11
CC00
r
CC22
CC33
U
CC44 CC55
CC66
CC77
CC88
1
0
3 2
4
5
6 7
O
RRdd
N
ØØ//22ØØ//22
PP11 PP00
PP33 PP55
PPPP 77 88
RRoolllliinngg r r
CCiirrccllee
r
ØØ == 336600º xx
rr//RRArc P0P8 = Circumference of
Rd X Ø = 2πr
Generating Circle
ØØ == 336600º xx 2255//7755
 == 112200°°

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
Normal
P P10 8
Tangent
P11
r
C1
C0
C2
C3
C4
Vertical
C5 C6 C7 C8 C9
C10
C11
C12
0
1 2
3
4
5
6
7
9 8
12
11
10
P2 P3
P4
P5 P6
PP 9 7
P12
 /
2
 /
2
 = 360 x r
R
 = 360 x 4
12
 = 120°
R
T
T
N
S
N
r
r
Rolling
Circle
Radias (r)
Directing
Circle
O
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. C2
C1
C
C4
C3
C5 C6 C7
C9
C8
C10
C11
PC12 8 O
10
5
7
9 8
P2
11
12
1
2 3 4
6
P1
P11
P3 P4 P5 P6 P7 P9 P10
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
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

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
4 5 7
0
3
02
12
P2
6
T
P01 1 2 3 4 5 6 7 8 9 10 1112
p
D P3
06
05
07
P4
P5
P6
P7
P8
P9
P10
P11
1
2
8
9
10
11
03
04
08
09
010`
011
Tangent N
N
Normal
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. RR==66ttooPP
R=7toP
PP
66 55
R21
00
1111 00
11
33
44
22
R=2toP
R=1toP
00 11 22 33 44 55 66 77 88 PP
77
88
99
1100
PP11
PP22
PP33
PP44
PP55
PP66
PP77
PP88
LL== 110000 mmmm
RR==33ttooPP
R=4toP
R=5toP
IINNVVOOLLUUTTEE
99
ø
11 mm = 30°
Then 5 mm = z
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. PPPP 1144
PP1133
3
PP1111
6 5 4
3
14
7
10 1
13
15
PP00
PP1122
O
2
8
9
11
12 1 2
PP11
PP22
PPPP 33 44
PP55
PP66
PP77
PP88 PP99 PP1100
LL==116600 mmmm
R=21mm
4 5 6 7 8 9 10 1112 131415
øø

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

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

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

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

98. PROBLEM:-
Draw an involute of a string
unwound from the given figure
from point C in anticlockwise
direction.
BB
6600°°
AA
CC
R21
3300°°

99. BB
6600°°
R2
11
AA
CC
3300°°
XX
XX
X+A
11
X+A2
44
33
XX++AA
33
R
=X+AB
X+A5
X+A4
XX++AA
BB
X+66+BC
11
22
55
C0
C1
C2 C3
C4
C5
C6
C7
C8

100. PROBLEM:-
A stick of length equal to tthhee cciirrccuummffeerreennccee ooff aa
sseemmiicciirrccllee,, iiss iinniittiiaallllyy ttaannggeenntt ttoo tthhee sseemmiicciirrccllee
oonn tthhee rriigghhtt ooff iitt.. TThhiiss ssttiicckk nnooww rroollllss oovveerr tthhee
cciirrccuummffeerreennccee ooff aa sseemmiicciirrccllee wwiitthhoouutt sslliiddiinngg ttiillll
iitt bbeeccoommeess ttaannggeenntt oonn tthhee lleefftt ssiiddee ooff tthhee
sseemmiicciirrccllee.. DDrraaww tthhee llooccii ooff ttwwoo eenndd ppooiinntt ooff tthhiiss
ssttiicckk.. NNaammee tthhee ccuurrvvee.. TTaakkee RR== 4422mmmm..

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

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. 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)

106. o
12 11 9 8 7 6 5 4 3 2 1 0
PP1111
PP1100
1
2
3
4
5
6
7
8
9
10
11
0
12
PP11
PP22 PP33
PP44
PP55
PP66
PP77 PP88
PP99
PP1122

107. 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

108. Diff. in length of any two radius vectors
Angle between them in radians
8 4
9 7
6 2
5
3 1
2
10
11
1
3
4
T 5
T
6
N
N 7
8
9
10
12
11
P1
P2
P3
P4
P5
P6
P7
P8 P9
P10
P11
P12 O
S
R min
R max
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.
40 25
B A O

110. P1
P2
P3 P4
11
BB111098 76 54 32 1 AA OO
P5
P6
P7
P8
P9
P10
P11
P12
21
31
41
51
61
71
81
91
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 PP Initial PPoossiittiioonn ooff ppooiinntt PP OO
PP11
PP22
PP33
PP44 PP55
11
33
44
55
66
77
OO
PP88
PP66
PP77
22
11
22
33
44
55
66
77
2/3 X 360°
= 240°
88
112200º

113. 00
EXAMPLE: A link AABB,
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.
Linear Travel of point PP on AABB
= 96 =16x (6 div.)
AAnngguullaarr SSwwiinngg
ooff lliinnkk AABB == 118800°° ++ 9900°°
PP11’’
AA
PP66
PP55
PP44
PP33
PP22
AAPP 66 00
BB
AA11
AA22
AA33
AA44
AA55
PP11
PP22’’
NORMAL
PP33’’
PP44’’ PP55’’
PP66’’
9966
LLiinnkk AABB == 9966
CC
Tangent
== 227700 °°
==4455 °°XX 66 ddiivv..
AARRCCHHIIMMEEDDIIAANN
SSPPIIRRAALL
DD
MM
NN

114. Arch.Spiral Curve Constant BBCC
= 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. 12
3
4
5
6
7
8
9
10
11
12
14
15
θ
o
13
19
20
21
1 0 2
3
16
18
4 5 6
7
8
9
10
11
14 13 12
15
16 17 18 19
20
21
24 23 22
23
22
24
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. RRoolllliinngg cciirrccllee
PP11
SS
normal
CC00 CC11 CC22 CC33 CC44
PP22
PP33
PP44
PP00
PP66
PP77
PP88
PP55
PP99
PP1100
11
22
33
44 5
66
77 88
99
1100
CC55 CC66 CC77 CC88 CC99 CC1100
DDiirreeccttiinngg lliinnee
00 11 22 33 44 55 66 77 88 99 1100
LLeennggtthh ooff ddiirreeccttiinngg lliinnee == 33P
554400 ° == 336600° ++ 118800°
554400 ° == PDD ++
TTPooDDttaa//ll22 ll ee nn gg tt h ffoorr 554400 ° rroottaattiioonn == 33PDD//22