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Engineering drawing is a language of all persons involved in engineering activity. Engineering ideas are recorded by preparing drawings and execution of work is also carried out on the basis of drawings. Communication in engineering field is done by drawings. It is called as a “Language of Engineers”.

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CHAPTER – 2 ENGINEERING CURVES

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**USES OF ENGINEERING CURVES**

Useful by their nature & characteristics. Laws of nature represented on graph. Useful in engineering in understanding laws, manufacturing of various items, designing mechanisms analysis of forces, construction of bridges, dams, water tanks etc.

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**CLASSIFICATION OF ENGG. CURVES**

1. CONICS 2. CYCLOIDAL CURVES 3. INVOLUTE 4. SPIRAL 5. HELIX 6. SINE & COSINE

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What is Cone ? It is a surface generated by moving a Straight line keeping one of its end fixed & other end makes a closed curve. The fixed point is known as vertex or apex. Vertex/Apex The closed curve is known as base. 90º If the base/closed curve is a circle, we get a cone. there is no objection If the base/closed curve is a polygon, we get a pyramid. Base

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**The line joins apex to the center of base is called axis.**

If axes is perpendicular to base, it is called as right circular cone. If axis of cone is not perpendicular to base, it is called as oblique cone. Base Vertex/Apex Cone Axis Generator The line joins vertex/ apex to the circumference of a cone is known as generator. 90º

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CONICS Definition :- The section obtained by the intersection of a right circular cone by a cutting plane in different position relative to the axis of the cone are called CONICS.

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CONICS A - TRIANGLE B - CIRCLE C - ELLIPSE D – PARABOLA E - HYPERBOLA

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TRIANGLE When the cutting plane contains the apex, we get a triangle as the section.

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CIRCLE 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

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**α > θ ELLIPSE Definition :-**

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. α θ α > θ

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PARABOLA 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. α = θ θ α

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**α = 0 α < θ HYPERBOLA Definition :-**

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. α = 0 θ α < θ

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CONICS Definition :- 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. Conic Curve M P Directrix F C V Focus Fixed straight line is called as directrix. Fixed point is called as focus.

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**The intersection of conic curve with axis is called as vertex.**

The line passing through focus & perpendicular to directrix is called as axis. The intersection of conic curve with axis is called as vertex. M C F V P Focus Conic Curve Directrix Axis Vertex

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**= Eccentricity Distance of a point from focus Ratio =**

Axis C V Focus Conic Curve Directrix Vertex M P N Q Distance of a point from focus Ratio = Distance of a point from directrix = Eccentricity = PF/PM = QF/QN = VF/VC = e

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ELLIPSE 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 P M Axis Directrix Vertex F C V Focus Eccentricity=PF/PM = QF/QN < 1. N Q

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ELLIPSE Ellipse is the locus of a point, which moves in a plane so that the sum of its distance from two fixed points, called focal points or foci, is a constant. The sum of distances is equal to the major axis of the ellipse. F1 A B P F2 O Q C D

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**PF1 + PF2 = QF1 + QF2 = CF1 +CF2 = constant = F1A + F1B = F2A + F2B**

D P CF1 +CF2 = AB O but CF1 = CF2 A B F1 F2 hence, CF1=1/2AB Q PF1 + PF2 = QF1 + QF2 = CF1 +CF2 = constant = F1A + F1B = F2A + F2B But F1A = F2B F1A + F1B = F2B + F1B = AB = Major Axis

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**Major Axis = 100 mm Minor Axis = 60 mm CF1 = ½ AB = AO**

D Major Axis = 100 mm Minor Axis = 60 mm O A B F1 F2 CF1 = ½ AB = AO C D Major Axis = 100 mm F1F2 = 60 mm O A B CF1 = ½ AB = AO F1 F2

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**Shape of tank in a tanker.**

Uses :- Shape of a man-hole. Shape of tank in a tanker. Flanges of pipes, glands and stuffing boxes. Shape used in bridges and arches. Monuments. Path of earth around the sun. Shape of trays etc.

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**PARABOLA Definition :-**

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

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**Uses :- Motor car head lamp reflector. Sound reflector and detector.**

Bridges and arches construction Shape of cooling towers. Path of particle thrown at any angle with earth, etc. Home

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HYPERBOLA 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. Axis Directrix Hyperbola M C N Q F V P Focus Vertex Eccentricity = PF/PM = QF/QN > 1.

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**Uses :- Nature of graph of Boyle’s law Shape of overhead water tanks**

Shape of cooling towers etc.

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**METHODS FOR DRAWING ELLIPSE**

1. Arc of Circle’s Method 2. Concentric Circle Method 3. Loop Method 4. Oblong Method 5. Ellipse in Parallelogram 6. Trammel Method 7. Parallel Ellipse 8. Directrix Focus Method

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**P4 P4 P3 P3 P2 P1 P1 P1’ Tangent P2’ P2’ Normal P3’ P3’ P4’ P4’**

ARC OF CIRCLE’S METHOD P4 C D P4 P3 P3 P2 P2 P1 P1 Rad =B1 R =A1 F1 O F2 A B 1 2 3 4 `R=A2 R=B2 P1’ P1’ Tangent P2’ P2’ Normal P3’ P3’ 90° P4’ P4’

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**CONCENTRIC CIRCLE METHOD**

10 11 9 C P10 12 8 P11 P9 10 N 11 Axis Minor 9 P12 P8 T 12 8 P7 Major Axis Q 1 P1 7 A B F1 1 O 7 F2 6 P6 P2` 2 5 3 P3 4 P5 2 6 P4 D e = AF1/AQ CF1=CF2=1/2 AB 3 5 4

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**Directrix 1 2 3 4 4’ 3’ 2’ 1’ C E F Normal A F1 F2 B Tangent D**

OBLONG METHOD P4 C P1’ P2’ P3’ P4’ E F 4 4’ P3 Normal 3 3’ Directrix P2 Minor Axis 2 S R=AB/2 2’ P1 Ø 1 1’ P0 Major Axis 0’ A 1 F1 2 3 4 4’ 3’ 2’ F2 1’ B Tangent P1’’ P2’’ P3’’ P4’’ P4 P3 P2 P1 P D

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**ELLIPSE IN PARALLELOGRAM**

D C H P1 Q1 P0 1 P2 1 Q2 K 2 P3 2 Q3 3 Q4 P4 3 4 Q5 4 5 P5 Minor Axis P6 5 B A Q6 O 6 5 4 3 2 1 1 2 3 4 5 6 Major Axis S4 R4 G J S3 R3 60° S2 R2 S1 I R1

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**ELLIPSE – DIRECTRIX FOCUS METHOD**

g f < 45º e Ellipse D1 d c Eccentricity = 2/3 b P5 P6 P7 P4 a P3 P2 3 R1V1 QV1 = V1F1 2 Q P1 Directrix R=6f` R=1a 1 2 3 4 5 6 7 R1 V1 F1 Dist. Between directrix & focus = 50 mm 90° P1’ T Tangent P2’ N 1 part = 50/(2+3)=10 mm P3’ P4’ P5’ P6’ P7’ S V1F1 = 2 part = 20 mm T V1R1 = 3 part = 30 mm Normal

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PROBLEM :- The distance between two coplanar fixed points is 100 mm. Trace the complete path of a point G moving in the same plane in such a way that the sum of the distance from the fixed points is always 140 mm. Name the curve & find its eccentricity.

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**R=B1 R =A1 F2 F1 A B 1 2 3 4 `R=A2 R=B2 Tangent Normal ARC OF CIRCLE’S**

METHOD G G’ G4 G4 AF1 AE e = e G3 G3 directrix G2 G2 G1 R=70 G1 R=70 R=B1 R =A1 F2 F1 E O A B 140 90° 100 1 2 3 4 `R=A2 R=B2 G1’ G1’ Tangent G2’ G2’ 90° Normal G3’ G3’ G4’ G4’ GF1 + GF2 = MAJOR AXIS = 140

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PROBLEM :-3 Two points A & B are 100 mm apart. A point C is 75 mm from A and 45 mm from B. Draw an ellipse passing through points A, B, and C so that AB is a major axis.

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8 D 1 P8 C 7 E P7 1 P1 45 75 6 2 P6 A B P2 100 O 5 P5 P3 3 4 P4

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PROBLEM :-5 ABCD is a rectangle of 100mm x 60mm. Draw an ellipse passing through all the four corners A, B, C and D of the rectangle considering mid – points of the smaller sides as focal points. Use “Concentric circles” method and find its eccentricity.

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1 4 R I1 I4 D C 50 P Q F1 F2 O 100 2 3 A B I2 I3 S

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PROBLEM :-1 Three points A, B & P while lying along a horizontal line in order have AB = 60 mm and AP = 80 mm, while A & B are fixed points and P starts moving such a way that AP + BP remains always constant and when they form isosceles triangle, AP = BP = 50 mm. Draw the path traced out by the point P from the commencement of its motion back to its initial position and name the path of P.

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M P2 Q2 2 2 P1 Q1 R = 50 1 1 A O B Q P 1 2 2 1 60 80 R1 S1 R2 S2 N

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PROBLEM :-2 Draw an ellipse passing through 60º corner Q of a 30º - 60º set square having smallest side PQ vertical & 40 mm long while the foci of the ellipse coincide with corners P & R of the set square. Use “OBLONG METHOD”. Find its eccentricity.

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**ELLIPSE C Q S A B P D directrix O3 O3’ 3 2 3 1 O2 60º O2’ 2 O1 O1’ 1**

TANGENT directrix NORMAL O3 O3’ Q R=AB/2 3 2 3 1 MINOR AXIS O2 60º O2’ 2 ELLIPSE 40mm O1 89mm O1’ 1 F1 F2 S 80mm MAJOR AXIS 30º A B P 1’’ 2’’ 3’’ ? R 1’ 2’ 3’ D MAJOR AXIS = PQ+QR = 129mm ECCENTRICITY = AP / AS

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PROBLEM :-4 Two points A & B are 100 mm apart. A point C is 75 mm from A and 45 mm from B. Draw an ellipse passing through points A, B, and C so that AB is not a major axis.

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**ELLIPSE H 75 45 K 100 J G I C P1 P2 1 P0 Q1 1 P3 2 Q2 2 Q3 3 P4 3 Q4 4**

H P1 45 P2 1 P0 Q1 1 P3 2 Q2 2 Q3 K 3 P4 3 Q4 4 4 P5 Q5 5 5 6 O Q6 P6 B A 6 100 6 5 4 3 2 1 1 2 3 4 5 6 J G I D

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PROBLEM :- Draw an ellipse passing through A & B of an equilateral triangle of ABC of 50 mm edges with side AB as vertical and the corner C coincides with the focus of an ellipse. Assume eccentricity of the curve as 2/3. Draw tangent & normal at point A.

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PROBLEM :- Draw an ellipse passing through all the four corners A, B, C & D of a rhombus having diagonals AC=110mm and BD=70mm. Use “Arcs of circles” Method and find its eccentricity.

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**METHODS FOR DRAWING PARABOLA**

1. Rectangle Method 2. Parabola in Parallelogram 3. Tangent Method 4. Directrix Focus Method

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**PARABOLA –RECTANGLE METHOD**

V C P1 P1 P2 P2 PARABOLA 1 1 P3 P3 2 2 P4 P4 3 3 4 P5 P5 4 5 5 P6 P6 6 6 A B 5 4 3 2 1 1 2 3 4 5

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**PARABOLA – IN PARALLELOGRAM**

C 1’ 2’ P’ 1 P’ 2 V P’ 3 P1 3’ P’ 4 4’ P2 P’ 5 5’ D P3 P’ 6 B 6’ 5’ 1 P4 4’ 3’ 2 2’ 1’ 3 P5 1 4 2 30° 3 5 P 6 4 6 5 A X

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**PARABOLA TANGENT METHOD**

10 9 1 8 2 7 3 6 4 5 5 O V 4 6 F 3 7 2 8 1 9 10 B A

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**DIRECTRIX FOCUS METHOD**

PARABOLA DIRECTRIX FOCUS METHOD P4 P3 P2 R4 PF R3 R2 P1 RF R1 AXIS R V 1 F 2 3 4 90° T 90° N P1’ PF’ S P2’ P3’ DIRECTRIX P4’ T

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PROBLEM:- A stone is thrown from a building 6 m high. It just crosses the top of a palm tree 12 m high. Trace the path of the projectile if the horizontal distance between the building and the palm tree is 3 m. Also find the distance of the point from the building where the stone falls on the ground.

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**A F 6m BUILDING 6m 3m TOP OF TREE STONE FALLS HERE ROOT OF TREE**

REQD.DISTANCE

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TOP OF TREE D C P 6m P1 P1 P2 P3 P4 1 2 3 P2 1 2 3 P3 BUILDING F A B P4 3 2 1 6m 3m 3 2 1 4 5 6 5 6 P5 STONE FALLS HERE ROOT OF TREE E REQD.DISTANCE 3m GROUND P6

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PROBLEM:- In a rectangle of sides 150 mm and 90 mm, inscribe two parabola such that their axis bisect each other. Find out their focus points & positions of directrix.

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**A B C O 90 mm M 150 mm D 1’ 2’ 3’ 4’ 5’ P1 P2 P3 P4 P5 P1’ P2’ P3’ P4’**

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EXAMPLE A shot is discharge from the ground level at an angle 60 to the horizontal at a point 80m away from the point of discharge. Draw the path trace by the shot. Use a scale 1:100

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parabola gun shot 60º B A 80 M ground level

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**10 9 1 8 2 7 3 6 4 5 5 4 6 3 7 2 1 9 10 e = 1 gun shot ground level**

VF VE 10 e = 1 = 9 1 8 2 7 D E 3 D 6 4 5 5 O V 4 6 F 3 7 2 8 gun shot 1 9 60º 10 B A ground level

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**Connect two given points A and B by a Parabolic curve, when:-**

1.OA=OB=60mm and angle AOB=90° 2.OA=60mm,OB=80mm and angle AOB=110° 3.OA=OB=60mm and angle AOB=60°

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**1.OA=OB=60mm and angle AOB=90°**

2 Parabola 5 4 3 90 ° B O 60 1 2 3 4 5

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**2.OA=60mm,OB=80mm and angle AOB=110°**

Parabola 5 4 3 110 ° B O 80 1 2 3 4 5

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**3.OA=OB=60mm and angle AOB=60°**

1 2 Parabola 5 4 3 60 ° B O 1 2 5 4 3 60

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example Draw a parabola passing through three different points A, B and C such that AB = 100mm, BC=50mm and CA=80mm respectively.

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80 C 50 B A 100

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**C A B P1 P’ 1 P2 P’ 2 1 P3 P’ 3 1’ 2 P4 2’ P’ 4 3 3’ P5 4 P’ 5 4’ 5 P**

P2 P’ 2 1 P3 P’ 3 1’ 2 P4 2’ P’ 4 3 3’ P5 4 P’ 5 4’ 5 P 6 5’ A B 6 P’ 6 6’ 5 4 3 2 1 1’ 2’ 3’ 4’ 5’

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**METHODS FOR DRAWING HYPERBOLA**

1. Rectangle Method 2. Oblique Method 3. Directrix Focus Method

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**RECTANGULAR HYPERBOLA**

When the asymptotes are at right angles to each other, the hyperbola is called rectangular or equilateral hyperbola B F 6’ P6 Given Point P0 AXIS 6 1 2 3 4 5 C D P0 P1 P2 2’ P3 3’ Hyperbola P4 4’ P5 Y 5’ ASYMPTOTES X and Y A O X E 90° AXIS

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Problem:- Two fixed straight lines OA and OB are at right angle to each other. A point “P” is at a distance of 20 mm from OA and 50 mm from OB. Draw a rectangular hyperbola passing through point “P”.

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**RECTANGULAR HYPERBOLA**

F 6’ P6 Given Point P0 6 1 2 3 4 5 C D P0 P1 P2 2’ P3 3’ Hyperbola P4 4’ P5 Y = 50 5’ A O 90° X=20 E

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PROBLEM:- Two straight lines OA and OB are at 75° to each other. A point P is at a distance of 20 mm from OA and 30 mm from OB. Draw a hyperbola passing through the point “P”.

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**Oblique Hyperbola B F X = 20 P7 7’ Given Point P0 7 1 2 3 4 5 6 C D P0**

1’ P1 2’ P2 Y = 30 P3 P4 P5 P6 6’ O A E 750

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**Directrix and focus method**

4’ DIRECTRIX D P4 3’ T2 P3 2’ P2 N2 N1 1’ s P1 NORMAL TANGENT AXIS C V 1 F1 2 3 4 T1 P1’ P2’ P3’ P4’

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**CYCLOIDAL GROUP OF CURVES**

When one curve rolls over another curve without slipping or sliding, the path Of any point of the rolling curve is called as ROULETTE. When rolling curve is a circle and the curve on which it rolls is a straight line Or a circle, we get CYCLOIDAL GROUP OF CURVES. Cycloidal Curves Cycloid Epy Cycloid Hypo Cycloid Superior Hypotrochoid Inferior Trochoid Superior Trochoid Inferior Hypotrochoid Inferior Epytrochoid Superior Epytrochoid

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**Rolling Circle or Generator Directing Line or Director**

CYCLOID:- Cycloid is a locus of a point on the circumference of a rolling circle(generator), which rolls without slipping or sliding along a fixed straight line or a directing line or a director. Rolling Circle or Generator P R C C P P Directing Line or Director

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EPICYCLOID:- Epicycloid is a locus of a point(P) on the circumference of a rolling circle(generator), which rolls without slipping or sliding OUTSIDE another circle called Directing Circle. O Ø/2 P0 Rolling Circle Directing Circle r P0 P0 Rd x Ø = 2πr Rd Circumference of Generating Circle Arc P0P0 Ø = 360º x r/Rd =

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HYPOCYCLOID:- Hypocycloid is a locus of a point(P) on the circumference of a rolling circle(generator), which rolls without slipping or sliding INSIDE another circle called Directing Circle.` Directing Circle(R) P Ø /2 Ø = 360 x r R T Rolling Circle Radius (r) O Vertical Hypocycloid

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What is TROCHOID ? DEFINITION :- It is a locus of a point inside/outside the circumference of a rolling circle, which rolls without slipping or sliding along a fixed straight line or a fixed circle. If the point is inside the circumference of the circle, it is called inferior trochoid. If the point is outside the circumference of the circle, it is called superior trochoid.

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: Given Data : Draw cycloid for one revolution of a rolling circle having diameter as 60mm. D Rolling Circle T N 6 P6 5 P5 7 P7 S R P4 P8 4 8 R P9 P3 9 C0 C1 C2 C3 C4 C5 C6 C7 C8 C9 C10 C11 C12 3 S1 R P2 2 P10 10 P1 P11 12 1 11 P0 P12 1 2 3 4 5 6 7 8 9 10 11 12 Directing Line 2R or D

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**Take diameter of circle = 40mm **

Problem 1: 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. C8 8 D/2 πD/2 C7 Take diameter of circle = 40mm Initially distance of centre of circle from the wall 83mm (Hale circumference + D/2) 7 Wall P8 C6 P7 P6 P5 6 4 P4 8 CYCLOID C5 7 5 3 P3 5 P2 2 6 C0 C1 C2 C3 C4 6 P1 7 1 5 P0 πD/2 D/2 Floor 1 2 3 4

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Problem : 2 A circle of 25 mm radius rolls on the circumference of another circle of 150 mm diameter and outside it. Draw the locus of the point P on the circumference of the rolling circle for one complete revolution of it. Name the curve & draw tangent and normal to the curve at a point 115 mm from the centre of the bigger circle.

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**First Step : Find out the included angle by using the equation**

360º x r / R = 360 x 25/75 = 120º. Second step: Draw a vertical line & draw two lines at 60º on either sides. Third step : at a distance of 75 mm from O, draw a part of the circle taking radius = 75 mm. Fourth step : From the circle, mark point C outside the circle at distance of 25 mm & draw a circle taking the centre as point C.

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**S U N Ø = 360º x 25/75 = 120° Rd X Ø = 2πr Ø = 360º x r/Rd**

EPICYCLOID GIVEN: Rad. Of Gen. Circle (r) & Rad. Of dir. Circle (Rd) P4 S P3 P5 r U r Rolling Circle C4 C3 C5 P2 C2 C6 P6 3 2 N C1 C7 4 r 1 Directing Circle P1 r C0 C8 5 P0 P8 P7 6 7 Ø/2 Ø/2 O Ø = 360º x 25/75 = 120° Arc P0P8 = Circumference of Generating Circle Rd Rd X Ø = 2πr Ø = 360º x r/Rd

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Problem :3 A circle of 80 mm diameter rolls on the circumference of another circle of 120 mm radius and inside it. Draw the locus of the point P on the circumference of the rolling circle for one complete revolution of it. Name the curve & draw tangent and normal to the curve at a point 100 mm from the centre of the bigger circle.

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**Hypocycloid r = Directing Circle Vertical Rolling Circle Radias (r)**

2 r Normal 1 r 3 P0 P12 P1 r P2 P11 4 P10 12 P3 T P8 P9 P5 P6 P7 P4 S Tangent 11 5 10 N 6 /2 7 9 8 = 360 x 4 12 360 x r R 120° O R Hypocycloid

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Problem : Show by means of drawing that when the diameter of rolling circle is half the diameter of directing circle, the hypocycloid is a straight line

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**10 5 7 8 9 11 12 1 2 3 4 6 C O Directing Circle Rolling Circle**

P2 P11 P3 P4 P7 P9 HYPOCYCLOID P1 C P5 O P8 P10 P12 P6

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INVOLUTE DEFINITION :- If a straight line is rolled round a circle or a polygon without slipping or sliding, points on line will trace out INVOLUTES. OR Involute of a circle is a curve traced out by a point on a tights string unwound or wound from or on the surface of the circle. Uses :- Gears profile

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PROB: 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. Draw tangent & normal to the curve at any point.

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**P9 T P8 P10 09 N 08 010` P7 07 P11 Normal 011 Tangent 06 P6 5 6 4 7 8**

05 3 . 9 2 P5 10 04 1 11 N P12 12 03 02 1 2 9 10 3 4 6 8 11 5 7 12 P1 P4 P2 P3 D

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PROBLEM:- Trace the path of end point of a thread when it is wound round a circle, the length of which is less than the circumference of the circle. Say Radius of a circle = 21 mm & Length of the thread = 100 mm Circumference of the circle = 2 π r = 2 x π x 21 = 132 mm So, the length of the string is less than circumference of the circle.

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**11 mm = 30° Then 5 mm = Ø = 30° x 5 /11 = 13.64 ° P3 P4 P2 INVOLUTE**

R=3toP P2 INVOLUTE R=4toP R=2toP P5 R=5toP P1 P6 R=6toP 11 1 2 3 4 5 6 7 8 9 10 R=1toP R=7toP P7 ø P8 P R21 1 2 3 4 5 6 7 8 P L= 100 mm 9 S = 2 x π x r /12

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PROBLEM:- Trace the path of end point of a thread when it is wound round a circle, the length of which is more than the circumference of the circle. Say Radius of a circle = 21 mm & Length of the thread = 160 mm Circumference of the circle = 2 π r = 2 x π x 21 = 132 mm So, the length of the string is more than circumference of the circle.

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P3 P4 P2 P5 P1 P6 7 10 1 2 3 4 5 6 8 9 11 12 P14 O 13 14 15 P ø R=21mm P13 P0 P7 1 L=160 mm 2 P12 3 6 4 5 7 8 9 10 11 12 13 14 15 P11 P8 P9 P10

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PROBLEM:- Draw an involute of a pantagon having side as 20 mm.

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**INVOLUTE OF A POLYGON Given : Side of a polygon P3 R=301 P2 R=201**

4 5 P1 P4 P0 N R=501 T S P5

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PROBLEM:- Draw an involute of a square having side as 20 mm.

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**INVOLUTE OF A SQUARE R=301 P1 R=01 P0 R=201 P4 1 4 2 3 P2 R=401 S N**

R=201 P4 1 4 2 3 P2 R=401 R=301 N S P3

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PROBLEM:- Draw an involute of a string unwound from the given figure from point C in anticlockwise direction. 60° A B C R21 30°

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**C8 C7 C6 C0 C5 C1 C2 C4 C3 B 5 C 4 3 2 X 1 A X+66+BC R =X+AB X+AB 60°**

30° X+AB 1 2 3 4 5 C6 X+A5 X X+A4 X+A3 C0 X X+A2 X+A1 C5 C1 C2 C4 C3

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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= 42mm.

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**INVOLUTE B A6 6 A5 B1 5 4 A4 B2 3 A3 B3 2 3 2 4 1 A2 1 5 B4 A C A1 B5**

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SPIRALS If a line rotates in a plane about one of its ends and if at the same time, a point moves along the line continuously in one direction, the curves traced out by the moving point is called a SPIRAL. The point about which the line rotates is called a POLE. The line joining any point on the curve with the pole is called the RADIUS VECTOR.

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**Spiral for Clock Semicircle Quarter Circle**

The angle between the radius vector and the line in its initial position is called the VECTORIAL ANGLE. Each complete revolution of the curve is termed as CONVOLUTION. Spiral Arche Median Spiral for Clock Semicircle Quarter Circle Logarithmic

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ARCHEMEDIAN SPIRAL It is a curve traced out by a point moving in such a way that its movement towards or away from the pole is uniform with the increase of vectorial angle from the starting line. USES :- Teeth profile of Helical gears. Profiles of cams etc.

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PROBLEM: To construct an Archemedian Spiral of one convolutions, given the radial movement of the point P during one convolution as 60 mm and the initial position of P is the farthest point on the line or free end of the line. Greatest radius = mm & Shortest radius = mm ( at centre or at pole)

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3 4 2 P3 P2 P4 5 1 P1 P5 o 6 P6 P12 8 7 1 2 3 4 5 6 9 11 12 12 P11 P10 P7 P9 P8 7 11 8 10 9

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**To construct an Archemedian Spiral of one convolutions, given the greatest & shortest(least) radii.**

To construct an Archemedian Spiral of one convolutions, given the largest radius vector & smallest radius vector. Say Greatest radius = 100 mm & Shortest radius = mm

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**Diff. in length of any two radius vectors**

Constant of the curve = Angle between them in radians 3 2 OP – OP3 4 P3 P2 = T P4 Π/2 1 5 P1 100 – 90 = Π/2 P5 R min N 10 8 6 4 2 6 P12 12 = 6.37 mm P6 O 11 9 7 5 3 1 S P11 P7 P10 11 7 R max P8 P9 8 10 9

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PROBLEM:- A slotted link, shown in fig rotates in the horizontal plane about a fixed point O, while a block is free to slide in the slot. If the center point P, of the block moves from A to B during one revolution of the link, draw the locus of point P. 25 40 B A O

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11 21 31 41 51 61 71 81 91 101 111 P3 P4 P2 P5 P1 40 25 P6 B A O 1 2 3 4 5 6 7 8 9 10 11 P12 P7 P11 P8 P10 P9

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PROBLEM:- A link OA, 100 mm long rotates about O in clockwise direction. A point P on the link, initially at A, moves and reaches the other end O, while the link has rotated thorough 2/3 rd of the revolution. Assuming the movement of the link and the point to be uniform, trace the path of the point P.

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**Initial Position of point P**

2/3 X 360° 1 2 3 4 5 6 7 8 2 1 3 4 5 6 7 = 240° P1 P2 120º P8 O P3 P7 P6 P4 P5

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**Angular Swing of link AB =**

EXAMPLE: A link AB, 96mm long initially is vertically upward w.r.t. its pinned end B, swings in clockwise direction for 180° and returns back in anticlockwise direction for 90°, during which a point P, slides from pole B to end A. Draw the locus of point P and name it. Draw tangent and normal at any point on the path of P. Angular Swing of link AB = A A0 180° + 90° P6 96 = 270 ° P5 =45 °X 6 div. A1 P4 P3 ARCHIMEDIAN SPIRAL P2 P1 P1’ A6 C P0 P2’ A2 B P6’ NORMAL P3’ M N P5’ P4’ A3 Link AB = 96 D A5 Linear Travel of point P on AB = 96 =16x (6 div.) Tangent A4

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**Arch.Spiral Curve Constant BC**

= Linear Travel ÷Angular Swing in Radians = 96 ÷ (270º×π /180º) = mm / radian

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PROBLEM : A monkey at 20 m slides down from a rope. It swings 30° either sides of rope initially at vertical position. The monkey initially at top reaches at bottom, when the rope swings about two complete oscillations. Draw the path of the monkey sliding down assuming motion of the monkey and the rope as uniform.

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o 23 13 22 24 1 2 3 4 5 6 7 8 9 10 11 12 14 15 16 18 19 20 21 17 P3 θ P9 P15 9 3 21 15 2 8 1 7 10 4 5 6 11 14 20 16 13 17 18 12 24 19 23 22

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Problem : 2 Draw a cycloid for a rolling circle, 60 mm diameter rolling along a straight line without slipping for 540° revolution. Take initial position of the tracing point at the highest point on the rolling circle. Draw tangent & normal to the curve at a point 35 mm above the directing line.

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**First Step : Draw a circle having diameter of 60 mm.**

Second step: Draw a straight line tangential to the circle from bottom horizontally equal to (540 x ) x 60 mm= mm i.e. 1.5 x x 60 mm 360 Third step : take the point P at the top of the circle.

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**S normal Length of directing line = 3D/2 540 = 360 + 180**

Rolling circle P0 P8 P1 8 P7 P9 7 1 9 S C0 C3 C4 C8 C9 P10 6 10 2 C1 C2 P2 P6 C5 C6 C7 C10 P3 3 P5 5 4 P4 Directing line 1 2 3 4 5 6 7 8 9 10 Length of directing line = 3D/2 540 = 360 + 180 540 = D + D/2 Total length for 540 rotation = 3D/2

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(Point undergoing two types of displacements)

(Point undergoing two types of displacements)

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