Conics & Engg_curves

8/12/2019 Conics & Engg_curves

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Line types


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Line types


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Conics and Engineering Curves

Conics & Engineering Curves

Indian Institute of Technology Kharagpur

Conics & Engineering Curves 1


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Divide a line in n equal segments.

Conics & Engineering 2

C i d E i i C


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Conics & Engineering Curves 3

C i d E i i C


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Drawing a perpendicularline from a point in a line

A BCE F

D

AB is the line, and C is thepoint on it

With center C and anyradius, describe equal arcsto cut AB at E and F

From E and F describe

equal arcs to intersect at D Join C and D to give the

required perpendicular


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Drawing a line parallel to a given line at agiven distance from it

AB is the given line,and c is the givendistance

From any two pointswell apart of AB, drawtwo arcs of radiusequal to c

Draw a line tangentialto the two arcs to givethe required line

c

A B


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Bisecting an angle

ABC is the given angle

From B describe an arcto cut AB and BC at Eand D respectively

With centers E and D,draw equal arcs tointersect at F

Join BF, the requiredbisector of the angle

A

B CD

EF


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conic sections



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Conic Sections

Eccentricity = distance of the point from the focusdistance of the point from directrix

e< 1Ellipse, e = 1 Parabola and e > 1Hyperbola.




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Conic Sections

1

An ellipse

Section planeAA, inclined to the axis cuts

all the generators of the

cone.

2A parabola Section plane

BB, parallel to one of the

generators cuts the cone.

A hyperbola Section

plane CC, inclined to the

axis cuts the cone on oneside of the axis.

3

4A rectangular hyperbola

Section plane DD, parallel

to the axis cuts the cone.




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Ellipse by Directrix and Focus

e = 2/3 and focus distance from directrix is known.

1Divide CF in five equal

segments. Draw VE CF

and VF = VE.

Extend CE to cover the

horizontal span of the

ellipse.

2

3 Draw a vertical at 1.

Use the compass to measure4

11

and mark P1 and P1

from F.

Any point, P2 or Pt , is C2 distance apart from the directrix and FP22or FPt apart from the focus.2




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and VF = VE.



ellipse.

2



11

and mark P1 and P1

from F.





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and VF = VE.



ellipse.

2



11

and mark P1 and P1

from F.





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and VF = VE.



ellipse.

2



11

and mark P1 and P1

from F.





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and VF = VE.



ellipse.

2



11

and mark P1 and P1

from F.





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and VF = VE.



ellipse.

2



11

and mark P1 and P1

from F.





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and VF = VE.



ellipse.

2



11

and mark P1 and P1

from F.





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and VF = VE.



ellipse.

2



11

and mark P1 and P1

from F.





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and VF = VE.



ellipse.

2



11

and mark P1 and P

1

from F.





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Tangent and Normals to Conics

1When a tangent at any point

on the curve is produced to

meet the directrix, the linejoining the focus with this

meeting point will be at

right angles to the line

joining the focus with the

point of contact.



T d N l C i


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point of contact.



T t d N l t C i


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point of contact.



T t d N l t C i


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point of contact.



Elli f M j d Mi A i


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Ellipse from Major and Minor Axis

1Draw major and minor axes

and two circles.

2Divide it for sufficient points

to draw the ellipse.

3Draw a vertical at 1.

Draw horizontal at 1

Repeat the procedure

described in the last two

steps

4

5





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and two circles.







steps

4

5





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and two circles.







steps

4

5





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and two circles.







steps

4

5





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and two circles.







steps

4

5





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and two circles.







steps

4

5



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Ellipse by Oblong method


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1Divide AO and AE in equal

segments (4)

Join 1, 2 and 3 to C

Join 1, 2 and 3 with D and

extend

2

3

4 On vertical P2P11 cut

P2x = xP11

On horizontal P2P5 cut

P2y = yP5

Find other points similarly

5

6





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segments (4)



extend

2

3


P2x = xP11


P2y = yP5


5

6





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segments (4)



extend

2

3


P2x = xP11


P2y = yP5


5

6

Conics & Engineering 31 Conics and Engineering Curves



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segments (4)



extend

2

3


P2x = xP11


P2y = yP5


5

6




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segments (4)



extend

2

3


P2x = xP11


P2y = yP5


5

6




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p y g


segments (4)



extend

2

3


P2x = xP11


P2y = yP5


5

6




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p y g


segments (4)



extend

2

3


P2x = xP11


P2y = yP5


5

6


Parabola for a given focus distance


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g

1Bisect CF for the vertex V

with proper process.

Draw a vertical at 1 and cut2

FP and FP1 = C1.1

3 Repeat for more points.




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g




FP and FP1 = C1.1





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FP and FP1 = C1.1





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FP and FP1 = C1.1



Parabola for a given base and axis


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1Draw the base AB and axis

EF from mid point E.

2Construct rectangle ABCD

and divide AB and CD in

equal parts.

3Draw lines F1, F2, F3 and

perpendiculars 1P1, 2P2

and 3P3.

4 Cut points like P ....1




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equal parts.



and 3P3.





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equal parts.



and 3P3.





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equal parts.



and 3P3.





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equal parts.



and 3P3.



Parabola in Parrallelogram


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Follow similar process.


Hyperbola for a given eccentricity


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Follow similar process.


Rectangular Hyperbola through a given point


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( e =

2 and equation may be assumed as xy = 1 not x2

y

2

= 1)

1Draw the axes OA, OB and

mark point P.

2 Draw CD I OA, EF I OB .

Mark points 1, 2, 3 ... (may

not be equidistant)

3

4Join O1 and 1P1 I OB and

1P1 I OA to find P1

Find the other points

similarly.

5




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( e =


y2

= 1)


mark point P.



not be equidistant)

3


1P1 I OA to find P1


similarly.

5




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( e =


y

2

= 1)


mark point P.



not be equidistant)

3


1P1 I OA to find P1


similarly.

5




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( e =


y

2

= 1)


mark point P.



not be equidistant)

3


1P1 I OA to find P1


similarly.

5




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( e =


y

2

= 1)


mark point P.



not be equidistant)

3


1P1 I OA to find P1


similarly.

5

Conics & Engineering51 Conics and Engineering Curves



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( e =

2 and equation may be assumed as xy = 1 not x

2y

2

= 1)


mark point P.



not be equidistant)

3


1P1 I OA to find P1


similarly.

5


Length of an Arc


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Length of an arc subtending an angle less than 600 and length of

circumference.


Cycloidal Curves


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The curve generated by a fixed point on the circumference of a circle, which rolls

without slipping along a fixed straight line or a circle. Circle = generating circle. The

line or circle = directing line or circle. Used in tooth profile of gears of a dial gauge.1

4

Let P is the generating point and PA is the circumference of the circle.

Divide PA and the circle in equal parts say 12.

Draw CBI PA and CB = PA and mark 1, 2, 3 .. and C1, C2, .. .

Draw lines through 1, 2, ... I PA.

With centerC1 and radius R mark P1 on line through 1.

2

3

5


Cycloidal Curves


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4






2

3

5


Cycloidal Curves


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4






2

3

5


Cycloidal Curves


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4






2

3

5


Tangent to a Cycloid Curve


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The normal at any point on a cycloidal curve will pass through the correspondingpoint of contact between the generating circle and the directing line or circle.


Trochoid Curves


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curve generated by a point fixed to a circle, within or outside its circumference, as the

circle rolls along a straight line. Point within the circle = inferior trochoid, Point

outside the circle = superior trochoid,1

Process is similar to the previous. We need to extend the lines C1P1, C2P2, ..

and cut the appropriate lengths like radius R1 or R2 from the lines.


Involute


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The curve traced out by an end of a piece of thread unwound from a circle or a

plygon, the thread being kept tight.

1Divide PQ (= perimeter)

and circle in equal parts

(say 12).

2

Draw tangents at 1 and cutP1 on the tangent.

3 Repeat for further points.

To draw tangent and normal

at any point N draw CN,

With diameter CN describe

a semicircle to find tangent

point M.

4

5


Involute


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(say 12).

2







point M.

4

5


Involute


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(say 12).

2







point M.

4

5


Involute


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(say 12).

2







point M.

4

5


Involute


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plygon, the thread being kept tight.1

Divide PQ (= perimeter)


(say 12).

2







point M.

4

5

Conics & Engineering64

Conics & Engg_curves

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