# ENGINEERING CURVES.ppt

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ENGINEERING CURVES Part- I {Conic Sections}ELLIPSE 1.Concentric Circle Method 2.Rectangle Method 3.Oblong Method 4.Arcs of Circle Method 5.Rhombus Metho 6.Basic Locus Method (Directrix – focus)HYPERBOLA 1.Rectangular Hyperbola (coordinates given) 2 Rectangular Hyperbola (P-V diagram - Equation given) 3.Basic Locus Method (Directrix – focus)PARABOLA 1.Rectangle Method 2 Method of Tangents ( Triangle Method) 3.Basic Locus Method (Directrix – focus)

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CONIC SECTIONS ELLIPSE, PARABOLA AND HYPERBOLA ARE CALLED CONIC SECTIONS BECAUSE THESE CURVES APPEAR ON THE SURFACE OF A CONE WHEN IT IS CUT BY SOME TYPICAL CUTTING PLANES.Section Plane Through GeneratorsEllipseSection Plane Parallel to end generator.ParabolaSection Plane Parallel to Axis.Hyperbola

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COMMON DEFINITION OF ELLIPSE, PARABOLA & HYPERBOLA:

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ELLIPSE BY CONCENTRIC CIRCLE METHOD

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1234ELLIPSE BY RECTANGLE METHOD

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1234ELLIPSE BY OBLONG METHOD

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F1F21 2 3 4 ABCDp1p2p3p4ELLIPSE BY ARCS OF CIRCLE METHODO

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1423ELLIPSE BY RHOMBUS METHOD

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ELLIPSE DIRECTRIX-FOCUS METHODF ( focus)DIRECTRIXVELLIPSE(vertex)AB30mm

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123456123456PARABOLA RECTANGLE METHOD

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CABPARABOLA METHOD OF TANGENTS

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ABVPARABOLA(VERTEX)F ( focus)4 3 2 11 2 3 4PARABOLA DIRECTRIX-FOCUS METHOD

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PO1231212312HYPERBOLA THROUGH A POINT OF KNOWN CO-ORDINATESSSolution Steps:   Extend horizontal line from P to right side. 2Extend 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. 5From horizontal 1,2,3,4 draw vertical lines downwards and 6From vertical 1,2,3,4 points [from P-B] draw horizontal lines. 7Line from 1 horizontal and line from 1 vertical will meet at P1.Similarly mark P2, P3, P4 points. 8Repeat 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. Problem No.10: Point P is 40 mm and 30 mm from horizontal and vertical axes respectively.Draw Hyperbola through it.

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VOLUME:( M3 )PRESSURE ( Kg/cm2)0 HYPERBOLA P-V DIAGRAMProblem no.11: 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.

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F ( focus)V(vertex)AB45mmHYPERBOLA DIRECTRIX FOCUS METHODPROBLEM 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 }

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QTANGENTNORMALTO DRAW TANGENT & NORMAL TO THE CURVE FROM A GIVEN POINT ( Q ) JOIN POINT Q TO F1 & F2 BISECT ANGLE F1Q F2 THE ANGLE BISECTOR IS NORMAL A PERPENDICULAR LINE DRAWN TO IT IS TANGENT TO THE CURVE.ELLIPSE TANGENT & NORMALProblem 13:

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ELLIPSE TANGENT & NORMALTTNNQ900Problem 14:

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QTNNT900PARABOLA TANGENT & NORMALProblem 15:

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HYPERBOLA TANGENT & NORMALQNNTT900Problem 16

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INVOLUTE CYCLOID SPIRAL HELIX ENGINEERING CURVES Part-II (Point undergoing two types of displacements)1. Involute of a circle a)String Length = D b)String Length > D c)String Length < D 2. Pole having Composite shape. 3. Rod Rolling over a Semicircular Pole.1. General Cycloid 2. Trochoid ( superior) 3. Trochoid ( Inferior) 4. Epi-Cycloid 5. Hypo-Cycloid 1. Spiral of One Convolution. 2. Spiral of Two Convolutions.1. On Cylinder 2. On a Cone AND

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CYCLOID: IT IS A LOCUS OF A POINT ON THE PERIPHERY OF A CIRCLE WHICH ROLLS ON A STRAIGHT LINE PATH. INVOLUTE: IT IS A LOCUS OF A FREE END OF A STRING WHEN IT IS WOUND ROUND A CIRCULAR POLE SPIRAL: IT IS A CURVE GENERATED BY A POINT WHICH REVOLVES AROUND A FIXED POINT AND AT THE SAME MOVES TOWARDS IT. HELIX: IT IS A CURVE GENERATED BY A POINT WHICH MOVES AROUND THE SURFACE OF A RIGHT CIRCULAR CYLINDER / CONE AND AT THE SAME TIME ADVANCES IN AXIAL DIRECTION AT A SPEED BEARING A CONSTANT RATIO TO THE SPPED OF ROTATION. ( for more problems on helix refer topic Development of surfaces)DEFINITIONS

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INVOLUTE OF A CIRCLEPP8ASolution Steps: 1) Point or end P of string AP is exactly D distance away from A. Means if this string is wound round the circle, it will completely cover given circle. B will meet A after winding. 2) Divide D (AP) distance into 8 number of equal parts. 3)  Divide circle also into 8 number of equal parts. 4)  Name after A, 1, 2, 3, 4, etc. up to 8 on D line AP as well as on circle (in anticlockwise direction). 5)  To radius C-1, C-2, C-3 up to C-8 draw tangents (from 1,2,3,4,etc to circle). 6)  Take distance 1 to P in compass and mark it on tangent from point 1 on circle (means one division less than distance AP). 7)  Name this point P1 8)  Take 2-B distance in compass and mark it on the tangent from point 2. Name it point P2. 9)  Similarly take 3 to P, 4 to P, 5 to P up to 7 to P distance in compass and mark on respective tangents and locate P3, P4, P5 up to P8 (i.e. A) points and join them in smooth curve it is an INVOLUTE of a given circle.

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INVOLUTE OF A CIRCLE String length MORE than DPp8

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PINVOLUTE OF A CIRCLE String length LESS than D

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123456PROBLEM 20 : A POLE IS OF A SHAPE OF HALF HEXABON AND SEMICIRCLE. ASTRING IS TO BE WOUND HAVING LENGTH EQUAL TO THE POLE PERIMETER DRAW PATH OF FREE END P OF STRING WHEN WOUND COMPLETELY. (Take hex 30 mm sides and semicircle of 60 mm diameter.)SOLUTION STEPS: Draw pole shape as per dimensions. Divide semicircle in 4 parts and name those along with corners of hexagon. Calculate perimeter length. Show it as string AP. On this line mark 30mm from A Mark and name it 1 Mark D/2 distance on it from 1 And dividing it in 4 parts name 2,3,4,5. Mark point 6 on line 30 mm from 5 Now draw tangents from all points of pole and proper lengths as done in all previous involute’s problems and complete the curve.

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Last Updated: 8th March 2018