Engineering Curves

1. ENGINEERING CURVES
SH 1132 Engineering Graphics F. Y. B. Tech.
ENGINEERING
CURVES

2. ENGINEERING CURVES
SH 1132 Engineering Graphics F. Y. B. Tech.
WHY CURVES?
CIVIL ENGINEERING
Bridges, Arches, Dams, Roads, Manholes etc.
MECHANICAL ENGINEERING
Gear Teeth, Reflector Lights, Centrifugal Pumps etc
ELECTRONINICS ENGINEERING
Design of Satellites, Missiles etc, Dish Antennas, ECG & EEG Machines
CSE & IT
Computer Graphics, Networking Concepts

3. ENGINEERING CURVES
SH 1132 Engineering Graphics F. Y. B. Tech.
CURVES
CONIC SECTIONS ENGINEERING CURVES
 CIRCLE
 ELLIPSE
 PARABOLA
 HYPERBOLA
 CYCLOIDAL CURVES/ROULETTES
 Cycloid
 Epicycloid
 Hypocycloid
 Trochoid (Superior & Inferior)
 Epitrochoid (Superior & Inferior)
 Hypotrochoid (Superior & Inferior)
 INVOLUTE
 SPIRALS
 Archimedian
 Logarithmic
 Hyperbolic
 HELIX
 Cylindrical
 Conical
 SPECIAL CURVES

4. ENGINEERING CURVES
SH 1132 Engineering Graphics F. Y. B. Tech.
(A) CONICS
The curves obtained by the
intersection of a cone by cutting
plane in different positions are
called conics.
The conics are
1. CIRCLE
2. ELLIPSE
3. PARABOLA
4. HYPERBOLA

5. ENGINEERING CURVES
SH 1132 Engineering Graphics F. Y. B. Tech.
DEFINING CONICS
Curve Position of Cutting Plane
Circle Perpendicular to axis and parallel to the base
Ellipse Inclined to the axis and not parallel to any generator.
Angle of Cutting Plane > Angle of Generator
Parabola Inclined to axis, parallel to generators and passes
through the base and axis
Hyperbola Inclined to the axis and not parallel to any generator.
Angle of Cutting Plane < Angle of Generator

6. ENGINEERING CURVES
SH 1132 Engineering Graphics F. Y. B. Tech.

7. ENGINEERING CURVES
SH 1132 Engineering Graphics F. Y. B. Tech.

8. ENGINEERING CURVES
SH 1132 Engineering Graphics F. Y. B. Tech.
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
CIRCLE

9. ENGINEERING CURVES
SH 1132 Engineering Graphics F. Y. B. Tech.
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

10. ENGINEERING CURVES
SH 1132 Engineering Graphics F. Y. B. Tech.
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

11. ENGINEERING CURVES
SH 1132 Engineering Graphics F. Y. B. Tech.
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
α < θ
α = 0
θ
θ

12. ENGINEERING CURVES
SH 1132 Engineering Graphics F. Y. B. Tech.
CONICS
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

13. ENGINEERING CURVES
SH 1132 Engineering Graphics F. Y. B. Tech.
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

14. ENGINEERING CURVES
SH 1132 Engineering Graphics F. Y. B. Tech.
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

15. ENGINEERING CURVES
SH 1132 Engineering Graphics F. Y. B. Tech.
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.

16. ENGINEERING CURVES
SH 1132 Engineering Graphics F. Y. B. Tech.
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.

17. ENGINEERING CURVES
SH 1132 Engineering Graphics F. Y. B. Tech.
monuments
Bridges
Earth orbit around Sun

18. ENGINEERING CURVES
SH 1132 Engineering Graphics F. Y. B. Tech.
Arches
Elliptical tray

19. ENGINEERING CURVES
SH 1132 Engineering Graphics F. Y. B. Tech.
PROBLEM :-
Point F is 50 mm from a line AB. A point P is moving in a plane
such that the ratio of it’s distances from F and line AB remains
constant and equals to 2/3
draw locus of point P. Draw tangent and normal at 50 mm from
the focus. { Eccentricity = 2/3 }

20. ENGINEERING CURVES
SH 1132 Engineering Graphics F. Y. B. Tech.
P5
P4’ P6’P5’
P1
P1’
V
P4
P3’
P3
P2’
PF’
F
D’D
X
F’
1’
2’
3’
4’
5’
E
P6
P2
PF
Directrix
90°
1 2 3 4 5 6
Eccentricity = 2/3
3XV
VF =
2
Dist. Between directrix & focus = 50 mm
1 part = 50/(2+3)=10 mm
VF = 2 part = 20 mm
VX = 3 part = 30 mm
 < 45°

DIRECTRIX-FOCUS METHOD
X
T’
T
ELLIPSE
AXIS
6’
P0

21. ENGINEERING CURVES
SH 1132 Engineering Graphics F. Y. B. Tech.
Ratio (known as eccentricity) of its distances from
focus to that of directrix is constant and equal to one.
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.
Directrix Axis
Vertex
M
C
N Q
F
V
P
Focus
Parabola
Eccentricity = PF/PM
= QF/QN
= 1.

22. ENGINEERING CURVES
SH 1132 Engineering Graphics F. Y. B. Tech.
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

23. ENGINEERING CURVES
SH 1132 Engineering Graphics F. Y. B. Tech.
Headlamp
Gateway Arch
monument

24. ENGINEERING CURVES
SH 1132 Engineering Graphics F. Y. B. Tech.
PROBLEM :
Point F is 40 mm from a vertical Straight line
AB. Draw locus of point P, moving in a plane such That
it always remains equidistant from point F and line AB.
Draw tangent and normal at a point S lying on a curve
below the axis at a distance of 55 mm from the line.
(Fixed line)

25. ENGINEERING CURVES
SH 1132 Engineering Graphics F. Y. B. Tech.
D
D’
2 3 4
T
T’N
N’
S
V 1
P1
P2
PF
P3
P4
P1’
P2’
P3’
P4’
PF’
AXIS
90°
X
F
PARABOLA
 = 45°
Eccentricity = 1/1
2XV
VF
=
2
Dist. Between directrix & focus = 40 mm
1 part = 40/(2+2)=10 mm
VF = 2 part = 20 mm
VX = 2 part = 20 mm
X
E
1’
F’
2’
3’
4’

P0

26. ENGINEERING CURVES
SH 1132 Engineering Graphics F. Y. B. Tech.
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
Focus
Vertex
HYPERBOLA
= QF/QN
> 1.

27. ENGINEERING CURVES
SH 1132 Engineering Graphics F. Y. B. Tech.
Nature of graph of Boyle’s law
Shape of overhead water tanks
Uses :-
Shape of cooling towers etc.

28. ENGINEERING CURVES
SH 1132 Engineering Graphics F. Y. B. Tech.
Cooling tower
Nature of Boyle’s law

29. ENGINEERING CURVES
SH 1132 Engineering Graphics F. Y. B. Tech.
PROBLEM :-
POINT F IS 50 MM FROM A LINE
AB.A POINT P IS MOVING IN A PLANE SUCH
THAT THE RATIO OF IT’S DISTANCES FROM
F AND LINE AB REMAINS CONSTANT AND
EQUALS TO 3/2. DRAW LOCUS OF POINT P.
{ ECCENTRICITY = 3/2 }. DRAW THE TANGENT
AND NORMAL AT ANY CONVENIENT POINT.

30. ENGINEERING CURVES
SH 1132 Engineering Graphics F. Y. B. Tech.
AXISC
V
F1
DIRECTRIXDD
1 2 3 4
4’
3’
2’
1’
P1
P2
P3
P4
P1’
P2’
P3’
P4’
T1
T2
s
Eccentricity = 3/2
2XV
VF =
3
Dist. Between directrix & focus = 50 mm
1 part = 50/(3+2)=10 mm
VF = 3 part = 30 mm
VX = 2 part = 20 mm
 > 45°

31. ENGINEERING CURVES
SH 1132 Engineering Graphics F. Y. B. Tech.
CYCLOID
A cycloid is a curve generated by a point on the circumference of a
circle as the circle rolls along a straight line without slipping.
The rolling circle is called generating circle and the line along which it
rolls is called base line or directing line.

32. ENGINEERING CURVES
SH 1132 Engineering Graphics F. Y. B. Tech.
P0
2R or D
5
1
2
1 2 3 4 6 7 8 9 10 110 120
3
4
5
6
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
P6
R
P10
R
PROBLEM: DRAW CYCLOID FOR ONE REVOLUTION OF A ROLLING CIRCLE HAVING DIAMETER AS 50MM.
Rolling Circle
D
N
R
S

33. ENGINEERING CURVES
SH 1132 Engineering Graphics F. Y. B. Tech.
P
C1 C2 C3 C4 C5 C6 C7 C8
p1
p2
p3
p4
p5
p6
p7
p8
1
2
3
4
5
6
7
C
D
PROBLEM 4: DRAW LOCUS OF A POINT ON THE PERIPHERY OF A CIRCLE WHICH ROLLS ON STRAIGHT LINE
PATH. Take Circle diameter as 40 mm
Solution Steps:
1) From center C draw a horizontal line equal to D distance.
2) Divide D distance into 8 number of equal parts and name them C1, C2, C3__ etc.
3) Divide the circle also into 8 number of equal parts and in clock wise direction, after P name 1, 2, 3 up to 8.
4) From all these points on circle draw horizontal lines. (parallel to locus of C)
5) With a fixed distance C-P in compass, C1 as center, mark a point on horizontal line from 1. Name it P.
6) Repeat this procedure from C2, C3, C4 up to C8 as centers. Mark points P2, P3, P4, P5 up to P8 on the
horizontal lines drawn from 2, 3, 4, 5, 6, 7 respectively.
7) Join all these points by curve. It is Cycloid.
8

34. ENGINEERING CURVES
SH 1132 Engineering Graphics F. Y. B. Tech.
Rolling circle
P1
P2
P3
P4
P0
P6
P7
P8
P5
P9
P10
1
2
3
4
5
6
7 8
9
10
1 2 3 4 5 6 7 8 9 100
C0 C1 C2 C3
C4
Directing line
Length of directing line = 5D/4
450 = 360 + 90
450 = D + D/4
Total length for 450 rotation = 5D/4
C5 C6 C7
C8 C9 C10
Problem :
Draw a cycloid for a rolling circle, 60 mm diameter rolling along a straight line without slipping for 450°
revolution. Take initial position of the tracing point at the highest point on the rolling circle.

35. ENGINEERING CURVES
SH 1132 Engineering Graphics F. Y. B. Tech.
C0
P0
786
4
P1
1
2
3
C2
C3
P2
C4
Problem:
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.
5
6
C1
P3
P4
P5
P7
P8
7
0
C5C6C7C8
1 2 3 4
5
P6
D/2πD/2
πD/2 D/2
Floor
Wall
CYCLOID
5
6
7
8
Take diameter of circle = 40mm
Initially distance of centre of circle from the wall
83mm (Half circumference + D/2)
4

36. ENGINEERING CURVES
SH 1132 Engineering Graphics F. Y. B. Tech.
An epicycloid is a curve generated by a point on the circumference of
a circle which rolls on the outside of another circle without sliding or
slipping.
The rolling circle is called generating circle and the outside circle on
which it rolls is called the directing circle or the base circle.
EPICYCLOID

37. ENGINEERING CURVES
SH 1132 Engineering Graphics F. Y. B. Tech.
PROBLEM 5: DRAW LOCUS OF A POINT ON THE PERIPHERY OF A CIRCLE WHICH ROLLS ON A CURVED
PATH. Take diameter of rolling Circle 50 mm And radius of directing circle i.e. curved path, 75
OP
OP=Radius of directing circle=75mm
C
PC=Radius of generating circle=25mm
θ
1
2
3
4
5
6
7
8 9 1
0
1
1 1
2
1’
2’
3’
4’
5’
6’
7’
8’
9’
10’
11’
12’
c1
c2
c3
c4
c5
c6
c7
c8
c9 c10
c11
c12
Directing
circle
Rolling circle
or generating
circle
θ=r/R X 360º
= 25/75 X 360º
=120º

38. ENGINEERING CURVES
SH 1132 Engineering Graphics F. Y. B. Tech.
A hypocycloid is a curve generated by a point on the circumference of
a circle which rolls on the inside of another circle without sliding or
slipping.
The rolling circle is called generating circle/ hypo circle and the inside
circle on which it rolls is called the directing circle or the base circle.
HYPOCYCLOID

39. ENGINEERING CURVES
SH 1132 Engineering Graphics F. Y. B. Tech.
OP
OP=Radius of directing circle=75mm
C
PC=Radius of generating circle=25mm
θ
θ=r/R X360º
= 25/75 X360º
=120º
1
2
3
4
5
6
7
8 9 10
11
12
c2
c1
c3
c4
c5
c6
c7
c8 c9 c1
0
c1
1
c1
2
1’
2’
3’
4’
5’
6’
7’
8’
9’
10’
11’
12’
PROBLEM 6: DRAW LOCUS OF A POINT ON THE PERIPHERY OF A CIRCLE
WHICH ROLLS FROM THE INSIDE OF A CURVED PATH. Take diameter of
rolling circle 50 mm and radius of directing circle (curved path) 75 mm. Directing circle
Rolling circle or
generating circle

40. ENGINEERING CURVES
SH 1132 Engineering Graphics F. Y. B. Tech.
An involute is a curve traced by a point as it unwinds from around a
circle or polygon.
The concerned circle or polygon is called as evolutes.
INVOLUTE

41. ENGINEERING CURVES
SH 1132 Engineering Graphics F. Y. B. Tech.
PROBLEM: 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.
P12
P2
0 12
6
P1
1 20 9 103 4 6 8 115 7 12
DP3
P4
P5
P6
P7
P8
P9
P10
P11
1
2
3
4
5
7
8
9
10
11
.

42. ENGINEERING CURVES
SH 1132 Engineering Graphics F. Y. B. Tech.
PROBLEM: DRAW AN INVOLUTE OF A PENTAGON HAVING SIDE AS 20 MM.
P5
P0
P1
P2
P3
P4
2
3
4 5
1
0

43. ENGINEERING CURVES
SH 1132 Engineering Graphics F. Y. B. Tech.
1
2
34
5
6
1 2 3 4 5 6
A
P
D/2
P1
1toP
P2
P3
3 to P
P4
P
P5
P6
PROBLEM : A POLE IS OF A SHAPE OF HALF HEXABON AND SEMICIRCLE.
ASTRING IS TO BE WOUND HAVING LENGTH EQUAL TO THE POLE PERIMETER
DRAW PATH OF FREE END P OF STRING WHEN WOUND COMPLETELY.
(Take hex 15 mm sides and semicircle of 30 mm diameter.)

44. ENGINEERING CURVES
SH 1132 Engineering Graphics F. Y. B. Tech.
PROBLEM NO 7: DRAW INVOLUTE OF A CIRCLE. STRING LENGTH IS EQUAL TO
THE CIRCUMFERENCE OF CIRCLE. DIAMETER OF CIRCLEIS 50MM.
1 2 3 4 5 6 7 8
P
P8
1
2
3
4
5
6
7
8
P3
P4
4 to p
P5
P7
P6
P2
P1
D
A

45. ENGINEERING CURVES
SH 1132 Engineering Graphics F. Y. B. Tech.
1 2 3 4 5 6 7 8
P
1
2
3
4
5
6
7
8
P3
P4
4 to p
P5
P7
P6
P2
P1
165 mm
(more than
D)D
p8
PROBLEM 8: DRAW INVOLUTE OF A CIRCLE.
STRING LENGTH IS MORE THAN THE CIRCUMFERENCE OF CIRCLE.
DIAMETER OF CIRCLE IS 50 MM & STRING LENGTH IS 165 MM

46. ENGINEERING CURVES
SH 1132 Engineering Graphics F. Y. B. Tech.
1 2 3 4 5 6 7 8
P
1
2
3
4
5
6
7
8
P3
P4
4 to p
P5
P7
P6
P2
P1
150 mm
(Less than
D)
D
PROBLEM : DRAW INVOLUTE OF A CIRCLE.
STRING LENGTH IS LESS THAN THE CIRCUMFERENCE OF CIRCLE.
DIAMETER OF CIRCLE IS 50 MM & STRING LENGTH IS 150 MM.

47. ENGINEERING CURVES
SH 1132 Engineering Graphics F. Y. B. Tech.
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= 21MM.
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

48. ENGINEERING CURVES
SH 1132 Engineering Graphics F. Y. B. Tech.
1
2
3
4
D
1
2
3
4
A
B
A1
B1
A2
B2
A3
B3
A4
B4
PROBLEM : ROD AB 85 MM LONG ROLLS
OVER A SEMICIRCULAR POLE WITHOUT
SLIPPING FROM IT’S INITIALLY VERTICAL
POSITION TILL IT BECOMES UP-SIDE-
DOWN VERTICAL.
DRAW LOCUS OF BOTH ENDS A & B.
TAKE R= 25MM.

49. ENGINEERING CURVES
SH 1132 Engineering Graphics F. Y. B. Tech.