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- 1. Engineering Curves – I1. Classification2. Conic sections - explanation3. Common Definition4. Ellipse – ( six methods of construction)5. Parabola – ( Three methods of construction)6. Hyperbola – ( Three methods of construction )7. Methods of drawing Tangents & Normals ( four cases)
- 2. Engineering Curves – II1.Classification2. Definitions3. Involutes - (five cases)4. Cycloid5. Trochoids – (Superior and Inferior)6. Epic cycloid and Hypo - cycloid7. Spiral (Two cases)8. Helix – on cylinder & on cone9. Methods of drawing Tangents and Normals (Three cases)
- 3. ENGINEERING CURVES Part- I {Conic Sections}ELLIPSE PARABOLA HYPERBOLA1.Concentric Circle Method 1.Rectangle Method 1.Rectangular Hyperbola (coordinates given)2.Rectangle Method 2 Method of Tangents ( Triangle Method) 2 Rectangular Hyperbola3.Oblong Method (P-V diagram - Equation given) 3.Basic Locus Method4.Arcs of Circle Method (Directrix – focus) 3.Basic Locus Method (Directrix – focus)5.Rhombus Metho6.Basic Locus Method Methods of Drawing (Directrix – focus) Tangents & Normals To These Curves.
- 4. 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. OBSERVE ILLUSTRATIONS GIVEN BELOW.. Ellipse Section Plane ola Section PlaneThrough Generators Hyperbola rab Parallel to Axis. Pa Section Plane Parallel to end generator.
- 5. COMMON DEFINATION OF ELLIPSE, PARABOLA & HYPERBOLA: These are the loci of points moving in a plane such that the ratio of it’s distances from a fixed point And a fixed line always remains constant. The Ratio is called ECCENTRICITY. (E) A) For Ellipse E<1 B) For Parabola E=1 C) For Hyperbola E>1 Refer Problem nos. 6. 9 & 12 SECOND DEFINATION OF AN ELLIPSE:- It is a locus of a point moving in a plane such that the SUM of it’s distances from TWO fixed points always remains constant. {And this sum equals to the length of major axis.} These TWO fixed points are FOCUS 1 & FOCUS 2 Refer Problem no.4 Ellipse by Arcs of Circles Method.
- 6. ELLIPSE BY CONCENTRIC CIRCLE METHOD Problem 1 :- Draw ellipse by concentric circle method.Take major axis 100 mm and minor axis 70 mm long. 3 2 4 Steps: 1. Draw both axes as perpendicular bisectors C of each other & name their ends as shown. 2. Taking their intersecting point as a center, 1 5 3 2 4 draw two concentric circles considering both as respective diameters. 1 5 3. Divide both circles in 12 equal parts & name as shown. A 4. From all points of outer circle draw vertical B lines downwards and upwards respectively. 5.From all points of inner circle draw 10 6 horizontal lines to intersect those vertical lines. 10 9 7 6 6. Mark all intersecting points properly as 8 those are the points on ellipse. D 7. Join all these points along with the ends of both axes in smooth possible curve. It is 9 7 required ellipse. 8
- 7. Steps: ELLIPSE BY RECTANGLE METHOD1 Draw a rectangle taking majorand minor axes as sides.2. In this rectangle draw both Problem 2axes as perpendicular bisectors Draw ellipse by Rectangle method.of each other..3. For construction, select upper Take major axis 100 mm and minor axis 70 mm long.left part of rectangle. Dividevertical small side and horizontallong side into same number of D 4 4equal parts.( here divided in fourparts) 3 34. Name those as shown..5. Now join all vertical points 2 21,2,3,4, to the upper end of minoraxis. And all horizontal points 1 1i.e.1,2,3,4 to the lower end ofminor axis. A 1 2 3 4 3 2 1 B6. Then extend C-1 line upto D-1and mark that point. Similarlyextend C-2, C-3, C-4 lines up toD-2, D-3, & D-4 lines.7. Mark all these points properlyand join all along with ends Aand D in smooth possible curve.Do similar construction in right Cside part.along with lower half ofthe rectangle.Join all points insmooth curve.It is required ellipse.
- 8. ELLIPSE Problem 3:- BY OBLONG METHOD 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. D 4 4 3 3 2 2 1 1A 1 2 3 4 3 2 1 B C
- 9. PROBLEM 4. ELLIPSEMAJOR AXIS AB & MINOR AXIS CD ARE BY ARCS OF CIRCLE METHOD100 AMD 70MM LONG RESPECTIVELY.DRAW ELLIPSE BY ARCS OF CIRLESMETHOD. As per the definition Ellipse is locus of point P moving in a plane such that the SUM of it’s distances from two fixedSTEPS: points (F1 & F2) remains constant and equals to the length1.Draw both axes as usual.Name the of major axis AB.(Note A .1+ B .1=A . 2 + B. 2 = AB) ends & intersecting point2.Taking AO distance I.e.half major axis, from C, mark F1 & F2 On AB . p4 C ( focus 1 and 2.) p33.On line F1- O taking any distance, p2 mark points 1,2,3, & 4 p14.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 A B O5.Repeat this step with same centers but F1 1 2 3 4 F2 taking now A-2 & B-2 distances for drawing arcs. Name the point p26.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/ D
- 10. PROBLEM 5. ELLIPSEDRAW RHOMBUS OF 100 MM & 70 MM LONG BY RHOMBUS METHODDIAGONALS AND INSCRIBE AN ELLIPSE IN IT.STEPS: 21. Draw rhombus of given dimensions.2. Mark mid points of all sides & name Those A,B,C,& D3. Join these points to the ends of A B smaller diagonals.4. Mark points 1,2,3,4 as four centers.5. Taking 1 as center and 1-A 3 4 radius 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. D C 1
- 11. PROBLEM 6:- POINT F IS 50 MM FROM A LINE AB.A POINT P IS MOVING IN A PLANESUCH THAT THE RATIO OF IT’S DISTANCES FROM F AND LINE AB REMAINS CONSTANT ELLIPSEAND EQUALS TO 2/3 DRAW LOCUS OF POINT P. { ECCENTRICITY = 2/3 } DIRECTRIX-FOCUS METHOD ELLIPSE A STEPS: 1 .Draw a vertical line AB and point F DIRECTRIX 50 mm from it. 2 .Divide 50 mm distance in 5 parts. 45mm 3 .Name 2nd part from F as V. It is 20mm and 30mm from F and AB line resp. 30mm 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. (vertex) V 5.Taking 45,60 and 75mm distances from F ( focus) 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. B
- 12. PROBLEM 7: A BALL THROWN IN AIR ATTAINS 100 M HIEGHT PARABOLA AND COVERS HORIZONTAL DISTANCE 150 M ON GROUND. RECTANGLE METHOD Draw the path of the ball (projectile)-STEPS: 6 61.Draw rectangle of above size and divide it in two equal vertical parts2.Consider left part for construction. 5 5 Divide height and length in equal number of parts and name those 1,2,3,4,5& 6 4 43.Join vertical 1,2,3,4,5 & 6 to the top center of rectangle4.Similarly draw upward vertical 3 3 lines from horizontal1,2,3,4,5 And wherever these lines intersect previously drawn inclined lines in sequence Mark those points and 2 2 further join in smooth possible curve.5.Repeat the construction on right side rectangle also.Join all in sequence. 1 1 This locus is Parabola.. 1 2 3 4 5 6 5 4 3 2 1
- 13. Problem no.8: Draw an isosceles triangle of 100 mm long base and PARABOLA110 mm long altitude.Inscribe a parabola in it by method of tangents. METHOD OF TANGENTS Solution Steps: C 1. Construct triangle as per the given 14 1 dimensions. 2. Divide it’s both sides in to same no.of 13 2 equal parts. 12 3 3. Name the parts in ascending and 11 descending manner, as shown. 4 10 4. Join 1-1, 2-2,3-3 and so on. 5 5. Draw the curve as shown i.e.tangent to 9 6 all these lines. The above all lines being 8 tangents to the curve, it is called method 7 of tangents. 7 8 6 9 5 10 4 11 3 12 2 13 1 14 A B
- 14. PROBLEM 9: Point F is 50 mm from a vertical straight line AB. PARABOLADraw locus of point P, moving in a plane such that DIRECTRIX-FOCUS METHODit always remains equidistant from point F and line AB. PARABOLASOLUTION STEPS:1.Locate center of line, perpendicular to A AB from point F. This will be initial point P and also the vertex.2.Mark 5 mm distance to its right side, name those points 1,2,3,4 and from P1those draw lines parallel to AB.3.Mark 5 mm distance to its left of P and (VERTEX) V name it 1. F O 1 2 3 44.Take O-1 distance as radius and F as center draw an arc ( focus) cutting first parallel line to AB. Name upper point P1 and lower point P2. P2(FP1=O1)5.Similarly repeat this process by taking again 5mm to right and left and locate P3 P4 . B6.Join all these points in smooth curve. It will be the locus of P equidistance from line AB and fixed point F.
- 15. Problem No.10: Point P is 40 mm and 30 mm from horizontal HYPERBOLAand vertical axes respectively.Draw Hyperbola through it. THROUGH A POINT OF KNOWN CO-ORDINATES Solution Steps: 1) Extend horizontal line from P to right side. 2 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 1 part [P-B] also at 1,2,3,4 points. 5) From horizontal 1,2,3,4 draw vertical 2 1 P 1 2 3 lines downwards and 6) From vertical 1,2,3,4 points [from P-B] draw 1 horizontal lines. 7) Line from 1 40 mm 2 horizontal and line from 1 vertical will meet at 3 P1.Similarly mark P2, P3, P4 points. O 8) Repeat the procedure by marking four points 30 mm 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
- 16. Problem no.11: A sample of gas is expanded in a cylinder HYPERBOLAfrom 10 unit pressure to 1 unit pressure.Expansion followslaw PV=Constant.If initial volume being 1 unit, draw the P-V DIAGRAMcurve of expansion. Also Name the curve. Form a table giving few more values of P & V 10 P V = C 9 +++ +++ + 10 1 = 10 5 2 = 10 8 4 2.5 = 10 2.5 4 = 10 7 2 5 = 10 1 10 = 10 6 PRESSURE Now draw a Graph of 5 ( Kg/cm2) Pressure against Volume. It is a PV Diagram and it is Hyperbola. Take pressure on vertical axis and 4 Volume on horizontal axis. 3 2 1 0 1 2 3 4 5 6 7 8 9 10 VOLUME:( M3 )
- 17. PROBLEM 12:- POINT F IS 50 MM FROM A LINE AB.A POINT P IS MOVING IN A PLANE HYPERBOLASUCH THAT THE RATIO OF IT’S DISTANCES FROM F AND LINE AB REMAINS CONSTANTAND EQUALS TO 2/3 DRAW LOCUS OF POINT P. { ECCENTRICITY = 2/3 } DIRECTRIX FOCUS METHODSTEPS: A1 .Draw a vertical line AB and point F 50 mm from it.2 .Divide 50 mm distance in 5 parts. 30mm3 .Name 2nd part from F as V. It is 20mm and 30mm from F and AB line resp. 45m It is first point giving ratio of it’s m distances from F and AB 2/3 i.e 20/304 Form more points giving same ratio such as 30/45, 40/60, 50/75 etc. (vertex) V F ( focus)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. B
- 18. ELLIPSEProblem 13: TANGENT & NORMAL TO DRAW TANGENT & NORMAL TO THE CURVE FROM A GIVEN POINT ( Q ) 1. JOIN POINT Q TO F1 & F2 2. BISECT ANGLE F1Q F2 THE ANGLE BISECTOR IS NORMAL 3. A PERPENDICULAR LINE DRAWN TO IT IS TANGENT TO THE CURVE. p4 C p3 p2 p1 A B O F1 1 2 3 4 F2 ALM NOR Q TAN GE NT D
- 19. ELLIPSEProblem 14: TANGENT & NORMALTO DRAW TANGENT & NORMAL TO THE CURVE ELLIPSE FROM A GIVEN POINT ( Q ) A DIRECTRIX1.JOIN POINT Q TO F. T2.CONSTRUCT 900 ANGLE WITH THIS LINE AT POINT F3.EXTEND THE LINE TO MEET DIRECTRIX AT T4. JOIN THIS POINT TO Q AND EXTEND. THIS IS TANGENT TO ELLIPSE FROM Q (vertex) V5.TO THIS TANGENT DRAW PERPENDICULAR F ( focus) LINE FROM Q. IT IS NORMAL TO CURVE. 900 N Q N B T
- 20. PARABOLA Problem 15: TANGENT & NORMAL TO DRAW TANGENT & NORMAL TO THE CURVE T PARABOLA FROM A GIVEN POINT ( Q ) A1.JOIN POINT Q TO F.2.CONSTRUCT 900 ANGLE WITH THIS LINE AT POINT F3.EXTEND THE LINE TO MEET DIRECTRIX AT T4. JOIN THIS POINT TO Q AND EXTEND. THIS IS VERTEX V TANGENT TO THE CURVE FROM Q 900 F5.TO THIS TANGENT DRAW PERPENDICULAR ( focus) LINE FROM Q. IT IS NORMAL TO CURVE. N Q B N T
- 21. HYPERBOLA Problem 16 TANGENT & NORMAL TO DRAW TANGENT & NORMAL TO THE CURVE FROM A GIVEN POINT ( Q ) A1.JOIN POINT Q TO F.2.CONSTRUCT 900 ANGLE WITH THIS LINE AT POINT F T3.EXTEND THE LINE TO MEET DIRECTRIX AT T4. JOIN THIS POINT TO Q AND EXTEND. THIS IS TANGENT TO CURVE FROM Q (vertex) V F ( focus)5.TO THIS TANGENT DRAW PERPENDICULAR 900 LINE FROM Q. IT IS NORMAL TO CURVE. N N Q B T
- 22. ENGINEERING CURVES Part-II (Point undergoing two types of displacements) INVOLUTE CYCLOID SPIRAL HELIX1. Involute of a circle 1. General Cycloid 1. Spiral of 1. On Cylinder a)String Length = πD One Convolution. 2. Trochoid 2. On a Cone b)String Length > πD ( superior) 2. Spiral of 3. Trochoid Two Convolutions. c)String Length < πD ( Inferior) 4. Epi-Cycloid2. Pole having Composite shape. 5. Hypo-Cycloid3. Rod Rolling over a Semicircular Pole. AND Methods of Drawing Tangents & Normals To These Curves.
- 23. DEFINITIONSCYCLOID: IS A LOCUS OF A POINT ON THE SUPERIORTROCHOID:ERIPHERY OF A CIRCLE WHICH IF THE POINT IN THE DEFINATIONOLLS ON A STRAIGHT LINE PATH. OF CYCLOID IS OUTSIDE THE CIRCLENVOLUTE: INFERIOR TROCHOID.: IF IT IS INSIDE THE CIRCLEIS A LOCUS OF A FREE END OF A STRINGHEN IT IS WOUND ROUND A CIRCULAR POLE EPI-CYCLOID IF THE CIRCLE IS ROLLING ONSPIRAL: ANOTHER CIRCLE FROM OUTSIDE IS A CURVE GENERATED BY A POINT HYPO-CYCLOID.HICH REVOLVES AROUND A FIXED POINT IF THE CIRCLE IS ROLLING FROMND AT THE SAME MOVES TOWARDS IT. INSIDE THE OTHER CIRCLE,HELIX: IS A CURVE GENERATED BY A POINT WHICHOVES AROUND THE SURFACE OF A RIGHT CIRCULARYLINDER / CONE AND AT THE SAME TIME ADVANCES IN AXIAL DIRECTIONT A SPEED BEARING A CONSTANT RATIO TO THE SPPED OF ROTATION. or problems refer topic Development of surfaces)
- 24. Problem no 17: Draw Involute of a circle. INVOLUTE OF A CIRCLE String length is equal to the circumference of circle.Solution Steps:1) Point or end P of string AP isexactly πD distance away from A.Means if this string is wound roundthe circle, it will completely cover P2given circle. B will meet A afterwinding.2) Divide πD (AP) distance into 8 P3number of equal parts. P13) Divide circle also into 8 number 2 to pof equal parts. 34) Name after A, 1, 2, 3, 4, etc. up to pto 8 on πD line AP as well as on pcircle (in anticlockwise direction). o 1t5) To radius C-1, C-2, C-3 up to C-8draw tangents (from 1,2,3,4,etc to 4 to pcircle). P4 46) Take distance 1 to P in compass 3and mark it on tangent from point 1 5on circle (means one division less 2than distance AP). 6 p o7) Name this point P1 5t 18) Take 2-B distance in compass 7 A 8 6 to pand mark it on the tangent from 7 to Ppoint 2. Name it point P2. P5 p P8 1 2 3 4 5 6 7 89) Similarly take 3 to P, 4 to P, 5 to P7P up to 7 to P distance in compass P6 πand mark on respective tangentsand locate P3, P4, P5 up to P8 (i.e. DA) points and join them in smoothcurve it is an INVOLUTE of a givencircle.
- 25. INVOLUTE OF A CIRCLEProblem 18: Draw Involute of a circle. String length MORE than πDString length is MORE than the circumference of circle.Solution Steps: P2In this case string length is morethan Π D. But remember!Whatever may be the length of P3 P1string, mark Π D distance 2 to phorizontal i.e.along the stringand divide it in 8 number of 3 toequal parts, and not any other p pdistance. Rest all steps are same o 1tas previous INVOLUTE. Drawthe curve completely. 4 to p P4 4 3 5 2 p o 5t 6 1 P5 7 8 7 p8 1 P 6 to p to p 2 3 4 5 6 7 8 P7 165 mm P6 (more than πD) πD
- 26. Problem 19: Draw Involute of a circle. INVOLUTE OF A CIRCLEString length is LESS than the circumference of circle. String length LESS than πDSolution Steps: P2In this case string length is Lessthan Π D. But remember!Whatever may be the length of P3 P1string, mark Π D distancehorizontal i.e.along the stringand divide it in 8 number of 2 to p 3 toequal parts, and not any other pdistance. Rest all steps are sameas previous INVOLUTE. Draw op 1tthe curve completely. 4 to p P4 4 3 5 2 p o 6 5t 1 6 to p P5 7 to 7 P p 8 P7 1 2 3 4 5 6 7 8 P6 150 mm (Less than πD) πD
- 27. PROBLEM 20 : A POLE IS OF A SHAPE OF HALF HEXABON AND SEMICIRCLE. ASTRING IS TO BE WOUND HAVING LENGTH EQUAL TO THE POLE PERIMETER INVOLUTE DRAW PATH OF FREE END P OF STRING WHEN WOUND COMPLETELY. OF (Take hex 30 mm sides and semicircle of 60 mm diameter.) COMPOSIT SHAPED POLESOLUTION STEPS:Draw pole shape as perdimensions. P1Divide semicircle in 4parts and name those Palong with corners of P2hexagon.Calculate perimeterlength. 1 to PShow it as string AP.On this line mark 30mm 2 tofrom A P oPMark and name it 1 AtMark πD/2 distance on itfrom 1And dividing it in 4 parts P3name 2,3,4,5. 3 to P 3Mark point 6 on line 30 4 2mm from 5Now draw tangents from 5 1all points of pole oPand proper lengths as A 4tdone in all previous 6 5 to P involute’s problems and 1 2 3 4 5 6 P 6t oPcomplete the curve. πD/2 P4 P6 P5
- 28. PROBLEM 21 : Rod AB 85 mm long rollsover a semicircular pole without slippingfrom it’s initially vertical position till itbecomes up-side-down vertical. BDraw locus of both ends A & B. A4 Solution Steps? 4 If you have studied previous problems B1 properly, you can surely solve this also. Simply remember that this being a rod, A3 it will roll over the surface of pole. 3 Means when one end is approaching, other end will move away from poll.OBSERVE ILLUSTRATION CAREFULLY! πD 2 A2 B2 2 1 3 1 A1 4 A B3 B4
- 29. PROBLEM 22: DRAW LOCUS OF A POINT ON THE PERIPHERY OF A CIRCLE CYCLOIDWHICH ROLLS ON STRAIGHT LINE PATH. Take Circle diameter as 50 mm p4 4 p3 p5 3 5 C p2 C1 C2 C3 C4 C5 C6 C7 p6 C 8 2 6 p1 1 p7 7 P p8 πD Solution Steps: 1) From center C draw a horizontal line equal to πD distance. 2) Divide πD distance into 8 number of equal parts and name them C1, C2, C3__ etc. 3) Divide the circle also into 8 number of equal parts and in clock wise direction, after P name 1, 2, 3 up to 8. 4) From all these points on circle draw horizontal lines. (parallel to locus of C) 5) With a fixed distance C-P in compass, C1 as center, mark a point on horizontal line from 1. Name it P. 6) Repeat this procedure from C2, C3, C4 upto C8 as centers. Mark points P2, P3, P4, P5 up to P8 on the horizontal lines drawn from 2, 3, 4, 5, 6, 7 respectively. 7) Join all these points by curve. It is Cycloid.
- 30. PROBLEM 23: DRAW LOCUS OF A POINT , 5 MM AWAY FROM THE PERIPHERY OF A SUPERIOR TROCHOID CIRCLE WHICH ROLLS ON STRAIGHT LINE PATH. Take Circle diameter as 50 mm 4 p4 p3 p5 3 5 p2 C C1 C C3 C4 C5 C6 C7 C8 p 6 2 6 2 p7 1 p1 7 P πD p8Solution Steps:1) Draw circle of given diameter and draw a horizontal line from it’s center C of length Π D and divide it in 8 number of equal parts and name them C1, C2, C3, up to C8.2) Draw circle by CP radius, as in this case CP is larger than radius of circle.3) Now repeat steps as per the previous problem of cycloid, by dividing this new circle into 8 number of equal parts and drawing lines from all these points parallel to locus of C and taking CP radius wit different positions of C as centers, cut these lines and get different positions of P and join4) This curve is called Superior Trochoid.
- 31. PROBLEM 24: DRAW LOCUS OF A POINT , 5 MM INSIDE THE PERIPHERY OF A INFERIOR TROCHOID CIRCLE WHICH ROLLS ON STRAIGHT LINE PATH. Take Circle diameter as 50 mm p4 4 p3 p5 3 5 p2 C C1 C2 C3 C4 C5 C6 C7 p6 C8 2 6 p1 p7 1 7 P p8 πDSolution Steps:1) Draw circle of given diameter and draw a horizontal line from it’s center C of length Π D and divide it in 8 number of equal parts and name them C1, C2, C3, up to C8.2) Draw circle by CP radius, as in this case CP is SHORTER than radius of circle.3) Now repeat steps as per the previous problem of cycloid, by dividing this new circle into 8 number of equal parts and drawing lines from all these points parallel to locus of C and taking CP radius with different positions of C as centers, cut these lines and get different positions of P and join those in curvature.4) This curve is called Inferior Trochoid.
- 32. PROBLEM 25: DRAW LOCUS OF A POINT ON THE PERIPHERY OF A CIRCLEWHICH ROLLS ON A CURVED PATH. Take diameter of rolling Circle 50 mm EPI CYCLOID :And radius of directing circle i.e. curved path, 75 mm.Solution Steps:1) When smaller circle will roll onlarger circle for one revolution it willcover Π D distance on arc and it will bedecided by included arc angle θ.2) Calculate θ by formula θ = (r/R) x Generating/3600. Rolling Circle3) Construct angle θ with radius OC 4 5and draw an arc by taking O as center C2 C3 C1 C4OC as radius and form sector of angle 3 6θ. C C 54) Divide this sector into 8 number of 7equal angular parts. And from C 2 C6onward name them C1, C2, C3 up toC8. 1 P r = CP C75) Divide smaller circle (Generatingcircle) also in 8 number of equal parts.And next to P in clockwise direction Directing Circlename those 1, 2, 3, up to 8. R C 86) With O as center, O-1 as radiusdraw an arc in the sector. Take O-2, O- = r 3600 + R3, O-4, O-5 up to O-8 distances withcenter O, draw all concentric arcs in Osector. Take fixed distance C-P incompass, C1 center, cut arc of 1 at P1.Repeat procedure and locate P2, P3,P4, P5 unto P8 (as in cycloid) and jointhem by smooth curve. This is EPI –CYCLOID.
- 33. PROBLEM 26: DRAW LOCUS OF A POINT ON THE PERIPHERY OF A CIRCLEWHICH ROLLS FROM THE INSIDE OF A CURVED PATH. Take diameter of HYPO CYCLOIDrolling circle 50 mm and radius of directing circle (curved path) 75 mm. Solution Steps: 1) Smaller circle is rolling here, inside the larger circle. It has to rotate anticlockwise to move P 7 ahead. 2) Same steps should be P1 6 taken as in case of EPI – CYCLOID. Only change is 1 P2 C2 C1 C3 in numbering direction of 8 C4 number of equal parts on C C P3 5 5 the smaller circle. 2 C 3) From next to P in 6 anticlockwise direction, 4 P4 C name 1,2,3,4,5,6,7,8. 3 7 4) Further all steps are P5 P8 that of epi – cycloid. This P6 P7 is called C8 HYPO – CYCLOID. r 3600 = R + O OC = R ( Radius of Directing Circle) CP = r (Radius of Generating Circle)
- 34. HELIX (UPON A CYLINDER)PROBLEM: Draw a helix of one convolution, upon a cylinder. P8Given 80 mm pitch and 50 mm diameter of a cylinder. 8(The axial advance during one complete revolution is called P7The pitch of the helix) 7 P6 6 P5SOLUTION: 5Draw projections of a cylinder.Divide circle and axis in to same no. of equal parts. ( 8 ) 4 P4Name those as shown. 3Mark initial position of point ‘P’ P3Mark various positions of P as shown in animation. 2 P2Join all points by smooth possible curve.Make upper half dotted, as it is going behind the solid 1 P1and hence will not be seen from front side. P 6 7 5 P 4 1 3 2
- 35. Problem 9: A particle which is initially on base circle of a cone, standingon Hp, moves upwards and reaches apex in one complete turn around the cone.Draw it’s path on projections of cone as well as on it’s development.Take base circle diameter 50 mm and axis 70 mm long. It’s a construction of curve o’ Helix of one turn on cone : 7’ DEVELOPMENT Draw Fv & Tv & dev.as usual HELIX CURVE 6’ On all form generators & name. A Construction of curve Helix:: 5’ 4’ B Show 8 generators on both views 3’ Divide axis also in same parts. 1 2’ Draw horizontal lines from those 1’ C points on both end generators. X Y 2 1’ is a point where first horizontal a’ h’b’ c’ g g’ f’ d’ e’ D Line & gen. b’o’ intersect. 3 2’ is a point where second horiz. h f O 4 Line & gen. c’o’ intersect. E 7 6 6 5 In this way locate all points on Fv. 5 7 Project all on Tv.Join in curvature. F a e O 4 For Development: Then taking each points true 3 G Distance From resp.generator b 1 d 2 from apex, Mark on development c H & join. A
- 36. STEPS: InvoluteDRAW INVOLUTE AS USUAL. Method of DrawingMARK POINT Q ON IT AS DIRECTED. Tangent & NormalJOIN Q TO THE CENTER OF CIRCLE C.CONSIDERING CQ DIAMETER, DRAWA SEMICIRCLE AS SHOWN. INVOLUTE OF A CIRCLE l maMARK POINT OF INTERSECTION OF r NoTHIS SEMICIRCLE AND POLE CIRCLEAND JOIN IT TO Q. QTHIS WILL BE NORMAL TO INVOLUTE. Ta ngDRAW A LINE AT RIGHT ANGLE TO enTHIS LINE FROM Q. tIT WILL BE TANGENT TO INVOLUTE. 4 3 5 C 2 6 1 7 8 P P8 1 2 3 4 5 6 7 8 π D
- 37. STEPS:DRAW CYCLOID AS USUAL. CYCLOIDMARK POINT Q ON IT AS DIRECTED. Method of DrawingWITH CP DISTANCE, FROM Q. CUT THE Tangent & NormalPOINT ON LOCUS OF C AND JOIN IT TO Q.FROM THIS POINT DROP A PERPENDICULARON GROUND LINE AND NAME IT NJOIN N WITH Q.THIS WILL BE NORMAL TOCYCLOID.DRAW A LINE AT RIGHT ANGLE TOTHIS LINE FROM Q. al No r mIT WILL BE TANGENT TO CYCLOID. CYCLOID Q Tang e nt CP C C1 C2 C3 C4 C5 C6 C7 C8 P N πD

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1 day left for my xam hope to quickly understand these topic through ur ppt