1. ERODE SENGUNTHAR ENGINEERING COLLEGE
(An Autonomous Institution)
(Approved by AICTE, New Delhi, Permanently Affiliated to Anna University, Chennai&
Accredited by National Board of Accreditation (NBA), New Delhi,
National Accreditation Assessment Council, Bangalore)
PERUNDURAI, ERODE - 638 057
DEPARTMENT OF MECHANICAL ENGINEERING
Presented by,
Dr.J.BALAJI., M.E., Ph.D
Associate Professor
Department of Mechanical Engineering
Erode Sengunthar Engineering College
Perundurai, Erode 638051.
2. UNIT-1 PLANE CURVES
SYNOPSIS
1. CONIC SECTIONS
I. DEFINITION AND BASICS OF CONIC SECTIONS
II. CONSTRUCTION OF ELLIPES
III. CONSTRUCTION OF PARABOLA
IV. CONSTRUCTION OF HYPERBOLA
2. CYCLOIDES
I. DEFINITION
II. CONSTRUCTION OF CYCLOID
III. CONSTRUCTION OF EPICYCLOID
IV. CONSTRUCTION OF HYPOCYCLOID
3. INVOLUTES
I. DEFINITION
II. CONSTRUCTION OF INVOLUTE OF CIRCLE
III. CONSTRUCTION OF INVOLUTE OF SQUARE
IV. CONSTRUCTION OF INVOLUTE OF SQUARE
3. CONE
DEFINITION: Cone is formed when a right angled triangle with an apex and angle θ is
rotated about its axis (altitude).
(The Length or height of the cone is equal to the altitude of the triangle and the radius of
the base of the cone is equal to the base of triangle. The apex angle of the cone is 2θ)
4. 1.CONIC SECTIONS
DEFINITION
The sections obtained by the intersection of
a right circular cone by a cutting plane in different
positions are called conic sections or conics.
CIRCLE
When the cutting plane is parallel to the base or
perpendicular to the axis, then the true shape of the
section is circle.
5. Line diagram of conic
section
CONIC SECTIONS…
Practical application of
conics curves in bridge
construction
6. ELLIPSE:
When the cutting plane is inclined to the horizontal plane and
perpendicular to the vertical plane, then the true shape of the
section is an ellipse.
Some of the practical Applications of Elliptical curve
7. PARABOLA:
When the cutting plane is inclined to the axis and is parallel
to one of the generators, then the true shape of the section
is a parabola.
CONIC SECTIONS…
Some of the practical Applications of Parabolic Curve
8. HYPERBOLA:
When the cutting plane is parallel to the axis of the cone, then the true
shape of the section is a rectangular hyperbola.
Some of the practical Applications of Hyperbolic Curve
9. Focus & Directrix
Conic may be defined as the locus of a point moving in a plane in such a way
that the ratio of its distances from a fixed point, called focus and a fixed straight line called
directrix.
Eccentricity
The is the ratio of shortest distance from the focus to the shortest distance from the
directrix is called eccentricity.
11. PROBLEM:1 Draw an ellipse when the distance between the focus and directrix is 50mm
and eccentricity is 2/3. Also draw the tangent and normal curve.
GIVEN
Given Curve is Ellipse
Distance from Focus to Directrix is 35mm
Eccentricity = ¾
I. CONSTRUCTION OF ELLIPSE
12.
13. PROBLEM : 2 Construct a parabola when the distance of the Focus from the directrix
is 50 mm. draw tangent and normal curve at any point.
GIVEN
Given Curve is Parabola
Distance from Focus to Directrix is 50mm
Eccentricity e = 1
II.CONSTRUCTION OF PARABOLA
14.
15. PROBLEM:3 Draw a hyperbola when the distance of the focus from the directrix is 50 mm
and eccentricity is 3/2. Also draw tangent and normal curve at any point.
III.CONSTRUCTION OF HYPERPARABOLA
16. 2. CYCLOIDAL CURVES
DEFINITION: Cycloidal curves are generated by a fixed point on the circumference of a
circle, which rolls without slipping along a fixed straight line or a circle.
In engineering drawing some special Curves (Cycloidal curves) are used in the profile of
teeth of gear wheels.
The rolling circle is called generating circle. The fixed straight line or circle is called
directing line or directing circle.
17. EPICYCLOID
Epicycloid is the curve generated by a
point on the circumference of a circle which
rolls without slipping along another.
HYPOCYCLOID
Hypocycloid is the curve generated by a point
on the circumference of a circle which rolls
without slipping inside another circle.
18. PROBLEM:4 A circle of 72mm diameter rolls along a straight line without slipping. Draw the
curve traced out by a point P on the circumference, for complete one revolution. Name the curve,
also draw the tangent and normal curve at a point 62 mm from the straight line.
GIVEN
Given Curve is cycloid
Rolling circle diameter 72mm
Straight line = π × 72 = 226 mm
I.CONSTRUCTION OF CYCLOID
19.
20. PROBLEM 5 : Construct a epi-cycloid rolling circle 60mm diameter and directing circle 122mm
radius. Draw also the tangent and normal at M, which on the curve.
GIVEN
Given Curve is Epicycloid
Rolling circle diameter 60mm
Directing Diameter 122mm
Angle = (r/R) × 360°
Angle = 177°
II.CONSTRUCTION OF EPICYCLOID
21. 1.With O as Centre and radius OP (base circle radius), draw an arc PQ.
2.The included angle θ = (r/R) x 360°.
3.With O as Centre and OC as radius, draw an arc to represent locus of Centre.
4.Divide arc PQ in to 12 equal parts and name them as 1’, 2’, …., 12’.
5.Join O1’, O2’, … and produce them to cut the locus of centers at C1, C2, ….C12. Taking C1 as
Centre, and radius equal to r, draw an arc cutting the arc through 1 at P1.
6.Taking C2 as Centre and with the same radius, draw an arc cutting the arc through 2 at P2
Similarly obtain points P3, P3, …., P12.
7.Draw a smooth curve passing through P1, P2….. , P12, which is the required epicycloid.
22. PROBLEM 6 : Draw a hypocycloid of a circle of 40 mm diameter, which rolls inside of another
circles of 160mm diameter for one revolution counter clock wise. Draw a tangent and a normal to it
at a point 65m from the center of the directing circle.
II.CONSTRUCTION OF HYPOCYCLOID
23. 1. With O as canter and 80mm as radius, draw the directing arc. draw an arc BA.
2. Draw the radial line OA and produce it to D such that BD = 20mm.
3. With D as Centre and 20 mm as radius, draw the generating circle. Let point A be the generating
point.
4. Mark point B on the directing arc such that the arc length AB is equal to the perimeter of the circular
disk. θ = (r/R) x 360°.
5. Divide the circle and the line AB into the same number of equal parts, say 12 and name the points as
shown in the figure.
6. From point D, draw an arc DC with O as canter and OD as radius such that <BOA = <COD. Divide
the arc DC into 12 equal parts and mark the points as O1, O2, O3…..
7. With O1 as canter radius equal to 20mm,draw an arc cutting the arc passing through 1 at 1.
8. Similarly get the points 21,31,41…. And join these points by drawing a smooth curve obtained the
required hypocycloid.
24. 3.INVOLUTES
An involutes is a curve traced by a point on a perfectly flexible string, while unwinding
from around a circle or polygon the string being kept taut (Tight). It is also a curve traced by a
point on a straight line while the line is rolling around a circle or polygon without slipping.
Application of involute curves
25. I.CONSTRUCTION OF INVOLUTES OF CIRCLE
1. Draw the circle with O as centre and OA as radius.(20mm).
2. Draw line P-P12 = 2π D, tangent to the circle at P
3. Divide the circle into 12 equal parts. Number them as1, 2…
4. Divide the line PQ into 12 equal parts and number
as1΄,2΄…
5. Draw tangents to the circle at 1, 2,3….
6. Locate points P1, P2 such that 1-P1 = P1΄,
2-P2 = P2΄ Join P
, P1,P2...
7. The tangent to the circle at any point on
it is always normal to the its involutes.
PROBLEM 7 : Draw the involute of the circle of diameter 40mm.Also draw a tangent and normal at
any point on the curve.
26. II.CONSTRUCTION OF INVOLUTES OF SQUARE
1. Draw the given squareABCD of side a.(a=30mm)
2. Taking D as the starting point, with CentreA and radius
DA=a, draw an arc to intersect the line BA produced at P
3. With Centre B and radius BP1 = 2 a, draw on arc to
intersect the line CB produced at P2.
4. Similarly, locate the points P3 and P4.
5.The curve through D, PI' P2, P3 and P4 is the
required involutes.
6. DP 4 is equal to the perimeter of the square.
PROBLEM 8 : Draw the involute of the square of side 30mm. Also draw a tangent and normal at
any point on the curve.
27. III.CONSTRUCTION OF INVOLUTES OF EQUILATERAL TRIANGLE
PROBLEM 9 : Draw the involute of the equilateral triangle of side 40mm. Also draw a
tangent and normal at any point on the curve.
1. Draw the given triangle ABC of side
a.(a=40mm)
2. With B as Centre and AB as radius draw
an arc from A to P1 on the extension line
of BC.
3. With C as canter and CP1 as radius draw
an arc from P1 and P2 on the extension
line of AC.
4. Repeat the same step with A as center
and AP2 as radius to get the point P3.
The curve AP11P2P3 is the required
involute for the given triangle with
40mm side.