Unit 1 plane curves

Dr.R. GANESAMOORTHY.B.E.,M.E.,Ph.d.Dr.R. GANESAMOORTHY.B.E.,M.E.,Ph.d.
Professor –Mechanical EngineeringProfessor –Mechanical Engineering
Saveetha Engineering CollegeSaveetha Engineering College
TandalamTandalam
Chennai.Chennai.09/12/18 1Dr.RGM/Prof -Mech/UNIT 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.
09/12/18 2Dr.RGM/Prof -Mech/UNIT 1

CONIC SECTIONS
Circle:
When the cutting plane is parallel to the base or perpendicular
to the axis, then the true shape of the section is circle.
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.

CONIC SECTIONS
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.
Hyperbola:
When the cutting plane is parallel to the axis of the cone, then
the true shape of the section is a rectangular hyperbola.

Eccentricity
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 ratio of shortest distance from the focus to the shortest
distance from the directrix is called eccentricity.

Eccentricity Conti....

Draw an ellipse when the distance between the
focus and directrix is 80mm and eccentricity is ¾.
1.Draw the directrix AB and axis CC’
2.Mark F on CC’ such that CF = 80 mm.
3.Divide CF into 7 equal parts and mark V at the fourth division from C.
4.Now, e = FV/ CV = 3/4.
5.At V, erect a perpendicular VB = VF. Join CB. Through F, draw a line at
45° to meet CB produced at D. Through D, drop a perpendicular DV’on
CC’. Mark O at the midpoint of V– V’.
6.With F as a centre and radius = 1–1’, cut two arcs on the
perpendicular
through 1 to locate P1 and P1’. Similarly, with F as centre and radii =2-
2’,
3–3’, etc., cut arcs on the corresponding perpendiculars to locate P2
and P2’, P3 and P3’, etc. Also, cut similar arcs on the perpendicular
through O to locate V1 and V1’.
6. Draw a smooth closed curve passing through V, P1, P/2, P/3,., V1,
…,V’, …, V1’, … P/3’, P/2’, P1’.09/12/18 7Dr.RGM/Prof -Mech/UNIT 1

Draw an ellipse

Construction of Parabola

Construction of Parabola
Construct a parabola when the distance of the Focus
from the directrix is 60 mm. Note: Eccentricity, e = 1.
1.Draw directrix AB and axis CC’ as shown.
2.Mark F on CC’ such that CF = 60 mm.
3.Mark V at the midpoint of CF. Therefore, e = VF/ VC = 1.
At V, erect a perpendicular VB = VF. Join CB.
4.Mark a few points, say, 1, 2, 3, … on VC’ and erect
perpendiculars through them meeting CB produced at 1’,2’, 3’…
5.With F as a centre and radius = 1–1’, cut two arcs on the
perpendicular through 1 to locate P1 and P1’. Similarly, with F as
a centre and radii = 2–2’, 3–3’, etc., cut arcs on the
corresponding perpendiculars to locate P2 and P2’,P3 and P3’ etc.
6.Draw a smooth curve passing through V, P1,P2,P3 …P3’,P2’,P1’.

Construction of Parabola Conti...

Construction of Hyperbola
Hyperbolic shapes finds large number of industrial
applications like the shape of cooling towers, mirrors
Used for long distance telescopes, etc.

Draw a hyperbola when the distance of
the focus from the directrix is 55 mm and
eccentricity is 3/2.
 Draw a perpendicular line AB (directrix) and a horizontal line CC’ (axis).
 Mark the focus point F on the axis line 55 mm from the directrix.
 Divide the CF in to 5 equal parts.
 As per the eccentricity mark the vertex V, in the third division of CF
 Draw a perpendicular line from vertex V, and mark the point B, with the
distance VF.
 Join the points C& B and extend the line.
 Draw number of smooth vertical lines 1,2,3,4,5,6,etc., as shown in figure.
 Now mark the points 1′, 2′, 3′, 4′, 5′…
 Take the vertical distance of 11′ and with F as center draw an arc cutting
the vertical line 11′ above and below the axis.
 Similarly draw the arcs in all the vertical lines (22′, 33′, 44′…)
 Draw a smooth curve through the cutting points to get the required
hyperbola by free hand.

Construction of Hyperbola

Cycloidal curves
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.

CYCLOIDS
Cycloid is a curve generated by a point on the circumference
of a circle which rolls along a straight line.
Draw a circle with diameter 50mm and mark the center O.
Divide the circle in to12 equal parts as1,2,3…12.
Draw horizontal line from the bottom points of the circle, with the
distance equal to the circumference of the circle (ПD) and mark
the other end point B.
Divide the line AB in to 12 equal parts. (1′, 2′, 3′…12′)
Draw a horizontal line from O toand mark the equal
distance point O1, O2,
O3…O12.
Draw smooth horizontal lines from the points 1,2,3…12.
When the circle starts rolling towards right hand side, the point 1
coincides with 1′ at the same time the center O moves to O1.

Take OA as radius, O1 as center draw an arc to cut the
horizontal line 1 to mark the point a1.
Similarly O2 as center and with same radius OA draw an arc
to cut the horizontal line 2 to mark the point a2.
Similarly mark a3, a4…a11.
Draw a smooth curve through the points a1, a2, a3,…. a11, B by
free hand.
The obtained curve is a cycloid.

Epicycloids
Epicycloid is the curve generated by a point on the
circumference of a circle which rolls without slipping
along another circle outside it.
 With O as centre and radius OP (base circle radius), draw an arc
PQ.
 The included angle θ = (r/R) x 360°.
 With O as centre and OC as radius, draw an arc to represent locus
of centre.
 Divide arc PQ in to 12 equal parts and name them as 1’, 2’, …., 12’.
 Join O1’, O2’, … and produce them to cut the locus of centres at
C1, C2, ….C12. Taking C1 as centre, and radius equal to r, draw
an arc cutting the arc through 1 at P1.
 Taking C2 as centre and with the same radius, draw an arc cutting
the arc through 2 at P2Similarly obtain points P3, P3, …., P12.
 Draw a smooth curve passing through P1, P2….. , P12, which is
the required epiclycloid.

Epicycloids

Hypocycloid
Hypocycloid is the curve generated by a point on the
circumference of a circle which rolls without slipping
inside another circle.
1.With O as centre and radius OB (base circle radius), draw an arc BA.
2.The included angle θ = (r/R) x 360°. With O as centre and OB as
radius, draw an arc to represent locus of centre.
3.Divide arc AB in to 12 equal parts and name them as 1’, 2’, …., 12’.
Join O1’, O2’, …, O12’ so as to cut the locus of centres at C1, C2,….C12.
4.Taking C1 as centre, and radius equal to r, draw an arc cutting the arc
through 1 at P1. Taking C2 as centre and with the same radius, draw an arc
cutting the arc through 2 at P2.
5.Similarly obtain points P3, P3, …., P12.
6.Draw a smooth curve passing through P1, P2….. , P12, which is the
required hypocycloid.

Hypocycloid

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.

Draw an involute of a given square.
1. Draw the given square ABCD of side a.
2. Taking D as the starting point, with centre A and radius
DA=a, draw an arc to intersect the line BA produced at P1.
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.

involutes of a square

Construction of Involutes of circle
1. Draw the circle with O as centre and OA as radius.
2. Draw line P-P12 = 2π D, tangent to the circle at P
3. Divide the circle into 12 equal parts. Number them as
1, 2…
4. Divide the line PQ into 12 equal parts and number as
1΄, 2΄…..
5. Draw tangents to the circle at 1, 2,3….
6. Locate points P1, P2 such that 1-P1 = P1΄, 2-P2 = P2΄
7. Join P, P1,P2...
8. The tangent to the circle at any point on it is always
normal to the its involutes.

Involutes of circle

Unit 1 plane curves

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Unit 1 plane curves