Engineering curves-1

Hasmukh Goswami College of
Engineering
Prepared by:
Hardik Dholakiya
Guided by:
Nirav sir [E.G. Department]

Engineering Curves I
[Conic Section]

Types & Methods:
Ellipse Parabola Hyperbola
1. Concentric Circle Method 1. Rectangle Method 1. Rectangular Hyperbola
2. Rectangle Method (Coordinates given)
3. Oblong Method 2. Method of Tangents 2. Rectangular Hyperbola
4. Arcs of Circle Method (Triangle Method) ( P-V diagram)
5. Rhombus Method 3. Basic Locus Method 3. Basic Locus Method
6. Basic Locus method (Directrix – Focus) (Directrix – Focus)
(Directrix – Focus)

Common definition of ellipse , parabola & hyperbola :
These are the loci f point moving in a plane such that the ratio of it’s distances from a
fixed point and a fixed line always remain constant.
The Ratio is called ECCENTRICITY (E).
(A) For Ellipse E<1
(B) For Parabola E=1
(C) For Hyperbola E>1

Ellipse :
.
Steps:
1. Draw both axes as perpendicular bisectors of each other & name their
ends as shown.
2. Taking their intersecting point as a center, draw two concentric circles
considering both as respective diameters.
3. Divide both circles in 12 equal parts & name as shown.
4. From all points of outer circle draw vertical lines downwards and
upwards respectively.
5.From all points of inner circle draw horizontal lines to intersect those
vertical lines.
6. Mark all intersecting points properly as those are the points on ellipse.
7. Join all these points along with the ends of both axes in smooth possible
curve. It is required ellipse.
Problem 1 : Draw ellipse by concentric circle method.
Take major axis 100 mm and minor axis 70 mm long.

Problem 2 : Draw ellipse by Rectangle method. Take major axis 100 mm and minor axis
70 mm long.
Steps:
1 Draw a rectangle taking major and minor axes as sides.
2. In this rectangle draw both axes as perpendicular bisectors of each
other..
3. For construction, select upper left part of rectangle. Divide vertical
small side and horizontal long side into same number of equal parts.(
here divided in four parts)
4. Name those as shown..
5. Now join all vertical points 1,2,3,4, to the upper end of minor axis. And
all horizontal points i.e.1,2,3,4 to the lower end of minor axis.
6. Then extend C-1 line upto D-1 and mark that point. Similarly extend
C-2, C-3, C-4 lines up to D-2, D-3, & D-4 lines.
7. Mark all these points properly and join all along with ends A and D in
smooth possible curve. Do similar construction in right side part.along
with lower half of the rectangle.Join all points in smooth curve.
It is required ellipse.

Problem 3: Draw ellipse by Oblong method.
Draw a parallelogram of 100 mm and 70 mm long sides with included angle of
750.Inscribe Ellipse in it.
STEPS ARE SIMILAR TO
THE PREVIOUS CASE
(RECTANGLE METHOD)
ONLY IN PLACE OF RECTANGLE,
HERE IS A PARALLELOGRAM

PROBLEM 4.
MAJOR AXIS AB & MINORAXIS CD ARE 100AMD 70MM LONG RESPECTIVELY.
DRAW ELLIPSE BY ARCS OF CIRLES METHOD.
STEPS:
1.Draw both axes as usual.Name the ends &
intersecting point.
2.Taking AO distance I.e.half major axis, from C,
mark F1 & F2 On AB .( focus 1 and 2.)
3.On line F1- O taking any distance, mark points
1,2,3, & 4
4.Taking F1 center, with distance A-1 draw an arc
above AB and taking F2 ‘ center, with B-1 distance
cut this arc. Name the point p1.
5.Repeat this step with same centers but taking now
A-2 & B-2 distances for drawing arcs. Name the
point p2
6.Similarly get all other P points.With same steps
positions of P can be located below AB.
7.Join all points by smooth curve to get an ellipse.

STEPS:
1. Draw rhombus of given dimensions.
2. Mark mid points of all sides &name Those A,B,C,
& D.
3. Join these points to the ends of smaller diagonals.
4. Mark points 1,2,3,4 as four centers.
5. Taking 1 as center and 1-Aradius draw an arc AB.
6. Take 2 as center draw an arc CD.
7. Similarly taking 3 & 4 as centers and 3-D radius
draw arcs DA & BC.
PROBLEM 5 :
DRAW RHOMBUS OF 100 MM & 70 MM LONG DIAGONALS AND
INSCRIBE AN ELLIPSE IN IT.

PROBLEM 6: POINT F IS 50 MM FROMA LINE AB.APOINT 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. { ECCENTRICITY = 2/3 }. [Directrix-focus
method]
STEPS:
1 .Draw a vertical line AB and point F 50 mm from it.
2 .Divide 50 mm distance in 5 parts.
3 .Name 2nd part from F as V. It is 20mm and 30mm
from F and AB line resp.It is first point giving ratio
of it’s distances from F and AB 2/3 i.e 20/30
4. Form more points giving same ratio such as 30/45,
40/60, 50/75 etc.
5. Taking 45,60 and 75mm distances from
line AB, draw three vertical lines to the
right side of it.
6. Now with 30, 40 and 50mm distances in
compass cut these lines above and below,
with F as center.
7. Join these points through V in smooth
curve.
This is required locus of P.It is an ELLIPSE.

Parabola:
PROBLEM 7: A BALL THROWN IN AIR ATTAINS 100 M HIEGHT
AND COVERS HORIZONTAL DISTANCE 150 M ON GROUND.
Draw the path of the ball (projectile). [Rectangle method]
STEPS:
1.Draw rectangle of above size and divide it in two equal vertical parts
2.Consider left part for construction . Divide height and length in equal
number of parts and name those 1,2,3,4,5& 6.
3.Join vertical 1,2,3,4,5 & 6 to the top center of rectangle
4.Similarly draw upward vertical lines from horizontal1,2,3,4,5
And wherever these lines intersect previously drawn inclined lines in
sequence Mark those points and further join in smooth possible curve.
5.Repeat the construction on right side rectangle also.Join all in sequence.
This locus is Parabola.
.

Problem 8: Draw an isosceles triangle of 100 mm long base and
110 mm long altitude . Inscribe a parabola in it by method of tangents.[Triangle method]
Steps:
1. Construct triangle as per the given dimensions.
2. Divide it’s both sides in to same no.of equal parts.
3. Name the parts in ascending and descending manner,
as shown.
4. Join 1-1, 2-2,3-3 and so on.
5. Draw the curve as shown i.e. tangent to all these lines.
The above all lines being
tangents to the curve, it is called method of tangents.

PROBLEM 9 : Point F is 50 mm from a vertical straight lineAB. Draw locus of point P, moving in
a plane such that it always remains equidistant from point F and lineAB.
[Directrix-focus method]
STEPS:
1. Locate center of line, perpendicular to AB from point F. This will be
initial point P.
2. Mark 5 mm distance to its right side, name those points 1,2,3,4 and
from those draw lines parallel to AB.
3. Mark 5 mm distance to its left of P and name it 1.
4. Take F-1 distance as radius and F as center draw an arc cutting first
parallel line to AB. Name upper point P1 and lower point P2.
5. Similarly repeat this process by taking
again 5mm to right and left and locate
P3P4.
6. Join all these points in smooth curve.
It will be the locus of P equidistance
from line AB and fixed point F.

Hyperbola:
Problem 10: Point P is 40 mm and 30 mm from horizontal
and vertical axes respectively . Draw Hyperbola through it.
[Through a point of known co-ordinates]
Steps:
1. Extend horizontal line from P to right side.
2. Extend vertical line from P upward.
3. On horizontal line from P, mark some points taking any distance
and name them after P-1, 2,3,4 etc.
4. Join 1-2-3-4 points to pole O. Let them cut part [P-B] also at 1,2,3,4
points.
5. From horizontal 1,2,3,4 draw vertical lines downwards and
6. From vertical 1,2,3,4 points [from P-B] draw horizontal lines.
7. Line from 1 horizontal and line from 1 vertical will meet at
P1.Similarly mark P2, P3, P4 points.
8. Repeat the procedure by marking four points on upward vertical
line from P and joining all those to pole O. Name this points P6, P7,
P8 etc. and join them by smooth curve.

Problem11 : A sample of gas is expanded in a cylinder from 10 unit pressure to 1 unit pressure
. Expansion follows law PV= Constant . If initial volume being 1 unit, draw the curve of
expansion. Also Name the curve. [P-V Diagram]
Form a table giving few more values of P & V
Now draw a Graph of
Pressure against Volume.
It is a PV Diagram and it is Hyperbola.
Take pressure on vertical axis and
Volume on horizontal axis.

PROBLEM 12 : 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. { ECCENTRICITY= 2/3 }. [Directrics-focus method]
STEPS:
1 .Draw a vertical line AB and point F 50 mm from it.
2 .Divide 50 mm distance in 5 parts.
3 .Name 3rd part from F as V. It is 30 mm
and 20 mm from F and AB line resp.
It is first point giving ratio of it’s
distances from F and AB = 3/2 i.e 30/20
4 Form more points giving same ratio such
as 45/30, 60/40, 75/50 etc.
5.Taking 30. 40 and 50 mm distances from
line AB, draw three vertical lines to the
right side of it.
6. Now with 45, 60 and 75 mm distances in
compass cut these lines above and
below, with F as center.
7. Join these points through V in smooth
curve.
This is required locus of P. It is Hyperbola

Bibliography
http://www.wordpress.com/engineeringcurves.pptx
http://www.imagesofengineering.com/curves
http://www.slidshere.com/engineeringcurves.ppt

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