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Ellipsce, parabola, hyperbola, cycloid, hypocycloid, epicycloid,rectangular method,spiral,helix

Ellipsce, parabola, hyperbola, cycloid, hypocycloid, epicycloid,rectangular method,spiral,helix

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- 1. Engineering drawing is a languageof all persons involved inengineering activity. Engineeringideas are recorded by preparingdrawings and execution of work isalso carried out on the basis ofdrawings. Communication inengineering field is done bydrawings. It is called as a“Language of Engineers”.
- 2. CHAPTER – 2ENGINEERING CURVES
- 3. 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.
- 4. CLASSIFICATION OF ENGG. CURVES1. CONICS2. CYCLOIDALCURVES3. INVOLUTE4. SPIRAL5. HELIX6. SINE & COSINE
- 5. What is Cone ?It is a surface generated by moving aStraight line keeping one of its end fixed &other end makes a closed curve.The fixed point is known as vertex or apex.The closed curve is Vertex/Apexknown as base.If the base/closed curveis a circle, we get a cone. 90ºIf the base/closedcurve is a polygon, weget a pyramid. Base
- 6. The line joins apex to the center of base iscalled axis.If axes is perpendicular to base, it is called asright circular cone.If axis of cone is not Vertex/Apexperpendicular to base, it is Cone Axiscalled as oblique cone. GeneratorThe line joins vertex/ 90ºapex to thecircumference of a coneis known as generator. Base
- 7. CONICSDefinition :- The section obtained by theintersection of a right circular cone by acutting plane in different position relativeto the axis of the cone are calledCONICS.
- 8. CONICSA - TRIANGLEB - CIRCLEC - ELLIPSED – PARABOLAE - HYPERBOLA
- 9. TRIANGLEWhen the cutting plane contains theapex, we get a triangle as thesection.
- 10. CIRCLEWhen the cutting plane is perpendicular tothe axis or parallel to the base in a rightcone we get circle the section. Sec Plane Circle
- 11. Definition :- ELLIPSE 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. θ α>θ α
- 12. PARABOLAWhen the cutting plane is inclined to the axisand parallel to one of the generators of thecone or the inclination of the plane(α) is equalto semi cone angle(θ), we get a parabola asthe section. θ α=θ α
- 13. HYPERBOLADefinition :- 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 α<θ θθ
- 14. 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 PDirectrix F C V Focus Fixed straight line is called as directrix. Fixed point is called as focus.
- 15. The line passing through focus & perpendicular to directrix is called as axis. The intersection of conic curve with axis is called as vertex. Conic Curve M P AxisDirectrix F C V Vertex Focus
- 16. Conic Curve M P AxisDirectrix F C V Vertex Focus N Q Distance of a point from focusRatio = Distance of a point from directrix = = PF/PM = QF/QN = VF/VC Eccentricity = e
- 17. 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. P Ellipse M AxisDirectrixVertex F C V Focus Eccentricity=PF/PM N = QF/QN Q < 1.
- 18. 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. C P OA B F1 F2 Q D
- 19. P C CF1 +CF2 = AB OA B but CF1 = CF2 F1 F2 hence, CF1=1/2AB Q D PF1 + PF2 = QF1 + QF2 = CF1 +CF2 = constant = F1A + F1B = F2A + F2B But F1A = F2B F1A + F1B = F2B + F1B = AB = Major Axis
- 20. C Major Axis = 100 mm Minor Axis = 60 mm OA B F1 F2 CF1 = ½ AB = AO D C Major Axis = 100 mm F1F2 = 60 mm OA B CF1 = ½ AB = AO F1 F2 D
- 21. 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.
- 22. PARABOLADefinition :- 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). Parabola M P Directrix Axis Vertex F C VEccentricity = PF/PM Focus = QF/QN N Q = 1.
- 23. 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
- 24. 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. P Hyperbola M AxisDirectrix F C V Eccentricity = PF/PM Vertex Focus N = QF/QN Q > 1.
- 25. Uses :- Nature of graph of Boyle’s law Shape of overhead water tanks Shape of cooling towers etc.
- 26. METHODS FOR DRAWING ELLIPSE1. Arc of Circle’s Method2. Concentric Circle Method3. Loop Method4. Oblong Method5. Ellipse in Parallelogram6. Trammel Method7. Parallel Ellipse8. Directrix Focus Method
- 27. ARC OF CIRCLE’S P4 C P4METHOD P3 P3 P2 P2 P1 P1 Rad = = A1 B1 R F1 O F2A B 1 2 3 4 `R=A2 R =B2Ta P1’ °° P 1’ ng en t P2’ P2’ l ma 90° P 3’ P 3’ D r P4’ P4’ No
- 28. CONCENTRIC 11 10 9 CIRCLE METHOD 12 C P10 8 P11 10 P9 N 11 9 P12 P8 T 12 8 P1 Major Axis Minor P7 Q 1 Axis A F1 1 O 7 F2 B 7 6 P6 P2` 2 3 5 P3 4 P5 2 D P4 6 e = AF1/AQCF1=CF2=1/2 AB 3 5 4
- 29. OBLONG METHOD P4 C P4’E 4 P3 P3’ 4’ No Directrix rm 3 Minor Axis al 3’ P2 P2’ 2 B/ 2 S 2’ A P1 R= ØØ P1’ 1 1’ 0 P0 Major Axis 0’ A 1 F1 2 3 4 4’ 3’ 2’ F2 1’ B t en P1 ng P1’’ Ta P2 P2’’PF P3 P3’’ P4 DP4’’
- 30. ELLIPSE IN PARALLELOGRAM 0 C H P1 P Q1 0 1 P2 1 2 P Q2 0 Q3K 2 3P 3 Q4 3 Min 4 4 Q5 4 5PAP6 5 5 Q6 B or 6 5 4 3 x2s 1 0O 1 2 3 4 5 6 A i Ax i S4 ajor s M R4 S3J R3 G60° S2 R2 S1 R1 I D
- 31. ELLIPSE – DIRECTRIX FOCUS METHOD g f θ < 45º Ellipse e d c Eccentricity = 2/3 b P3 P4 P5 P6 P7 aDirectrix Q P1 P2 QV1 V1F1 2 f` = = =6 R= R= R1V1 R1V1 3 R 1a 1a R1 1 2 3 4 5 6 7 D1 V1 F1 Dist. Between directrix 90° & focus = 50 mm P’ T Tangen 1 P ’ t 2 1 part = 50/(2+3)=10 mm P3’P ’ 4 P5’ P6’ P ’ V1F1 = 2 part = 20 mm N S 7 T V R = 3 part = 30 mm Normal 1 1 N
- 32. PROBLEM :-The distance between two coplanarfixed points is 100 mm. Trace thecomplete path of a point G movingin the same plane in such a waythat the sum of the distance fromthe fixed points is always 140 mm.Name the curve & find itseccentricity.
- 33. ARC OF CIRCLE’SMETHOD directrix G4 G G4 e = AF1 e G3 G3 AE G2 G2 G1 70 R= G1 R= R= 70 B1 R =A1 E F1 O F2 B A 90° 1 2 3 4 100 `R=A2 R =B2 Ta G ’ ng 1 °° G1’ en t G2’ G2’ al rm 90° G3’ G3’ G4’ G’ G4’ No 140 GF1 + GF2 = MAJOR AXIS = 140
- 34. PROBLEM :-3Two points A & B are 100 mmapart. A point C is 75 mm from Aand 45 mm from B. Draw anellipse passing through points A,B, and C so that AB is a majoraxis.
- 35. 8 D 1 P8 C 7 8 E 1 P7 P1 7 45 75 2 2 6 6A P 100 O P6 B 2 5 P5 P3 3 4 3 P4 5 4
- 36. PROBLEM :-5ABCD is a rectangle of 100mm x60mm. Draw an ellipse passingthrough all the four corners A, B,C and D of the rectangleconsidering mid – points of thesmaller sides as focal points.Use “Concentric circles” methodand find its eccentricity.
- 37. 1 4 R I1 I4 D 1 4 CP F1 O Q5050 F2 100 2 3 AI B I3 2 S 2 3
- 38. PROBLEM :-1Three points A, B & P while lyingalong a horizontal line in order haveAB = 60 mm and AP = 80 mm, while A& B are fixed points and P startsmoving such a way that AP + BPremains always constant and whenthey form isosceles triangle, AP = BP =50 mm. Draw the path traced out bythe point P from the commencement ofits motion back to its initial positionand name the path of P.
- 39. M P2 Q22 2 P1 Q1 50 1 R=1 A O BQ P 1 2 60 2 1 80 R1 S1 R2 S2 N
- 40. PROBLEM :-2Draw an ellipse passing through60º corner Q of a 30º - 60º setsquare having smallest side PQvertical & 40 mm long while thefoci of the ellipse coincide withcorners P & R of the set square.Use “OBLONG METHOD”. Findits eccentricity.
- 41. directrix T EN C NO NO G AN O3 O3 ’ MINOR AXIS T RM RM 2 B/ Q AL AL 3 3 A R= O2 O2 ’ θ θº 60 2 40mm 2 ELLIPSE 89m O1 m O1 ’ 1 1 F1 80mm MAJOR AXIS 30º F2S A B ? ? 1’ P 2’ 3’ 3’’ 2’’ R 1’’ D MAJOR AXIS = PQ+QR = 129mm ECCENTRICITY = AP / AS
- 42. PROBLEM :-4Two points A & B are 100 mmapart. A point C is 75 mm from Aand 45 mm from B. Draw anellipse passing through points A,B, and C so that AB is not a majoraxis.
- 43. ELLIPSE C 0 H P1 0 1 P2 P0 Q1 1 2 P3 Q2 2 Q3 K 45 3 P 75 3 4 4 Q4 4 5 P5 Q5 6 O Q6 65A P6 6 5 4 3 2 1 100 0 1 2 3 4 56 BJ G I D
- 44. PROBLEM :-Draw an ellipse passing through A& B of an equilateral triangle ofABC of 50 mm edges with side ABas vertical and the corner Ccoincides with the focus of anellipse. Assume eccentricity of thecurve as 2/3. Draw tangent &normal at point A.
- 45. PROBLEM :-Draw an ellipse passing through allthe four corners A, B, C & D of arhombus having diagonalsAC=110mm and BD=70mm.Use “Arcs of circles” Method andfind its eccentricity.
- 46. METHODS FOR DRAWING PARABOLA1. Rectangle Method2. Parabola in Parallelogram3. Tangent Method4. Directrix Focus Method
- 47. PARABOLA –RECTANGLE METHOD D V C0 P1 P1 0 P2 P2 PARABOLA1 1 P3 P32 2 P4 P43 34 P5 P5 45 5P6 P66A 5 4 3 2 1 0 1 2 3 4 5 B6
- 48. PARABOLA – IN PARALLELOGRAM C 0 1’ 2’ P’ P’ 2 P’ V 1 3’ P1 3 P’ 4 P2 4’ P’ 5 B 5’ D P3 P’0 6’ 61 5’ P4 4’2 3’ 2’3 P 1’ 54 1 2 05 3P 4 30°66 5A X
- 49. PARABOLA 10 0TANGENT METHOD 9 1 8 2 7 3 6 4 5 V 5 4 6 F 3 7 2 8θ 1 θ 9 0 10 A O B
- 50. D PARABOLA DIRECTRIX FOCUS METHOD P4 P3 P2 R4 PF R3 R2 P1 RF R1 AXISR V 1 F 2 90° 3 4T 90° N P1’ PF’ S DIRECTRIX P2’ P3’ P4’ N TD
- 51. PROBLEM:-A stone is thrown from a building 6 mhigh. It just crosses the top of a palmtree 12 m high. Trace the path of theprojectile if the horizontal distancebetween the building and the palmtree is 3 m. Also find the distance ofthe point from the building where thestone falls on the ground.
- 52. TOP OF TREE BUILDING A 6m 6m6m STONE FALLS HERE ROOT OF TREE F 3m REQD.DISTANCE
- 53. TOP OF TREE D P C P1 P1 1 P2 P2 1 2 2 P3 P3 BUILDING 3 3 A P4 0 P4 B 6m 3 2 1 1 2 3 4 5 6 6m 56m 6 P5 STONE FALLS HERE ROOT OF TREE F 3m 3m E GROUND P6 REQD.DISTANCE
- 54. PROBLEM:-In a rectangle of sides 150 mm and 90mm, inscribe two parabola such thattheir axis bisect each other. Find outtheir focus points & positions of directrix.
- 55. 150 mm D 5 4 3 2 1 A 1 P1’’ P1 P1 2 P2’’ P2 3 m M P2 P3’’ P3 4’’ 3’ 4’ 5’ 3’ 4’ 5’ P3 5 P4 P P4’’ P5 C 5’ 4’ 3’ O 2’ 1’ B
- 56. EXAMPLEA shot is discharge from the groundlevel at an angle 60 to the horizontalat a point 80m away from the point ofdischarge. Draw the path trace by theshot. Use a scale 1:100
- 57. parabolagunshot 60º A ground level B 80 M
- 58. VF 10 0 = e=1 VE 9 1 8 2 D 7 E 3 D 6 4 5 V 5 4 6 F 3 7 2 8 gun 1 9shot 60º 0 10 A O B ground level
- 59. Connect two given points A and B by aParabolic curve, when:-1.OA=OB=60mm and angle AOB=90°2.OA=60mm,OB=80mm and angleAOB=110°3.OA=OB=60mm and angle AOB=60°
- 60. A 1.OA=OB=60mm and angle AOB=90° 1 2 36060 Parabola 4 5 90 ° O B 1 2 63 4 5
- 61. 2.OA=60mm,OB=80mm and angleAOB=110°A1 2 Parabola 60 60 3 4 5 110 ° B O 1 2 3 4 5 80
- 62. 3.OA=OB=60mm A and angle AOB=60° 1 2 Parabola 60 3 4 5 60 BO ° 1 2 3 4 5 6
- 63. exampleDraw a parabola passing through threedifferent points A, B and C such that AB =100mm, BC=50mm and CA=80mmrespectively.
- 64. C 80 50 50A B 100
- 65. 0 P2 P1 C P’ 1 0 P3 P’ 1 2 1’ P’ 2 P4 3 2’ 3 P’ 4 3’ 4 P5 P’ 4’ 5 5 5’ P6 6 6 P’ 0 1’ 2’ 3’ 4’ 5’ 6’ 6A 5 4 3 2 1 B
- 66. METHODS FOR DRAWING HYPERBOLA1. Rectangle Method2. Oblique Method3. Directrix Focus Method
- 67. RECTANGULAR HYPERBOLAWhen the asymptotes are at right angles to each other, the hyperbolais called rectangular or equilateral hyperbola B 6’ F P6 Given Point P0AXIS C 6 0 1 2 3 4 5 D P0 P1 2’ P2 3’ P3 P4 Hyperbola 4’ 5’ P5 Y ASYMPTOTES X and Y O X E A 90° AXIS
- 68. Problem:-Two fixed straight lines OA and OB areat right angle to each other. A point “P”is at a distance of 20 mm from OA and50 mm from OB. Draw a rectangularhyperbola passing through point “P”.
- 69. RECTANGULAR HYPERBOLA B 6’ F P6 Given Point P0 C 6 0 1 2 3 4 5 D P0 P1 2’ P2 3’ P3 P4 Hyperbola Y = 50 4’ 5’ P5 O X=20 E A90°
- 70. PROBLEM:-Two straight lines OA and OB are at75° to each other. A point P is at adistance of 20 mm from OA and 30mm from OB. Draw a hyperbolapassing through the point “P”.
- 71. B F X=2 0 P7 7’ Given Point P0 C 7 1 2 3 4 5 6 P0 D 1’ P1 Y = 30 2’ P2 P3 P4 P5 P6 6’ O A E75 0
- 72. Directrix and focus method 4’ D D P4 3’ T2 P3 2’ N1 N1 P2 1’ s P1 NO RM T AL EN NG D D AXIS TA N2 N C V 1 F12 3 4 T1 P1 ’ DIRECTRIX DIRECTRIX P2’ P3’ P4’
- 73. 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 CycloidInferior Superior Inferior SuperiorTrochoid Trochoid Hypotrochoid Hypotrochoid Inferior Superior Epytrochoid Epytrochoid
- 74. CYCLOID:-Cycloid is a locus of a point on thecircumference of a rolling circle(generator),which rolls without slipping or sliding along afixed straight line or a directing line or adirector. Rolling Circle or Generator P R C C P P Directing Line or Director
- 75. 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. P0 Rolling Circle r P0 P0 Ø/ Ø/ 2 2Rd x Ø = 2πr O Circumference of Arc P0P0 = d R Ø = 360º x r/Rd
- 76. HYPOCYCLOID:-Hypocycloid is a locus of a point(P) on the circumference ofa rolling circle(generator), which rolls without slipping or slidingINSIDE another circle called Directing Circle.` VerticalRolling Circle Directing Circle(R) Radius (r) P P T P Ø /2 Ø /2 O 360 x r R Hypocycloid Ø= R
- 77. What is TROCHOID ?DEFINITION :- It is a locus of a pointinside/outside the circumference of a rollingcircle, which rolls without slipping or slidingalong a fixed straight line or a fixed circle.If the point is inside the circumference of thecircle, it is called inferior trochoid.If the point is outside the circumference of thecircle, it is called superior trochoid.
- 78. : Given Data : Draw cycloid for one revolution of a rolling circle having diameter as T 60mm. Rolling D N Circle 6 P6 7 5 P5 P7 S8 4 P4 R P8 R P9 P39 C0 C1 3 2 C3 C4 S C5 C6 C7 C8 C9 CT C10 C11 C12 1 P2 R 2 P1010 P1 P11 11 12 0 1 P0 0 1 2 3 4 N 5 6 7 8 9 10 11 12P12 Directing Line 2R or D
- 79. Problem 1: A circle of diameter D rolls withoutslip on a horizontal surface (floor) byHalf revolution and then it rolls up avertical surface (wall) by another halfrevolution. Initially the point P is at C8 C8 8the Bottom of circle touching the floor.Draw the path of the point P. C7 C7 WallTake diameter of circle = 40mm 7 Initially distance of centre of Pcircle from the wall 83mm (Hale 8 C6 C P P6circumference + D/2) P 7 6 8 4 P4 πD/2 πD/2 CYCLOID C5 C 5 7 5 3 P3 5 P2 6 C C1 2 C2 C3 C4 6 0 P1 7 1 5 P00 1 πD/2 2 3 4 D/2 Floor
- 80. Problem : 2A circle of 25 mm radius rolls on thecircumference of another circle of 150 mmdiameter and outside it. Draw the locus ofthe point P on the circumference of therolling circle for one complete revolution ofit. Name the curve & draw tangent andnormal to the curve at a point 115 mm fromthe centre of the bigger circle.
- 81. First Step : Find out the included angle by using theequation 360º x r / R = 360 x 25/75 = 120º.Second step: Draw a vertical line & draw two lines at60º on either sides.Third step : at a distance of 75 mm from O, draw apart of the circle taking radius = 75 mm.Fourth step : From the circle, mark point C outside thecircle at distance of 25 mm & draw a circle taking thecentre as point C.
- 82. GIVEN: EPICYCLOIDRad. Of Gen. Circle (r) P4& Rad. Of dir. Circle (Rd) S P3 º P5Rolling r U rCircle C3 C4 C5 3 P2 2 C2 C6 P6 C1 N C7 4 r 1 r CP1 P 0 C8 5 0 P8 P7 0 6 7 Ø/2 Ø/2 O Ø = 360º x 25/75 Arc P0P8 = Circumference of Rd = 120° Rd X Ø = 2πr Generating Circle Ø = 360º x
- 83. Problem :3A circle of 80 mm diameter rolls on thecircumference of another circle of 120 mmradius and inside it. Draw the locus of thepoint P on the circumference of the rollingcircle for one complete revolution of it.Name the curve & draw tangent and normalto the curve at a point 100 mm from thecentre of the bigger circle.
- 84. Directing Circle Vertical Rolling N CircleRadias (r) C5 C6 C7 C Norm C4 8 2 C3 C9 1 C2 3 C10 al P11 P12 r P0 0 P1 Pr r C C11 12 2 P3 1 4 P8 P9 P10 T C0 P4 P5 P6 P7 S e nt C12 11 5 Tang T 10 / / N 6 9 7 2 2 8 R O = 360 x r R Hypocycloid = 360 x 4 12 = 120°
- 85. Problem :Show by means of drawing thatwhen the diameter of rolling circle ishalf the diameter of directing circle,the hypocycloid is a straight line
- 86. Directing CircleRolling Circle C4 C5 C6 C7 2 3 C8 4 C9 C3 1 C2 C10 5 C1 C11 P2 P1112 6 P1 P3 P4 C P5 P6 O C12 P12 P7 P8 P9 P10 7 11 HYPOCYCLOID 10 8 9
- 87. INVOLUTEDEFINITION :- If a straight line is rolledround a circle or a polygon without slipping orsliding, points on line will trace outINVOLUTES. ORInvolute of a circle is a curve traced out by apoint on a tights string unwound or wound fromor on the surface of the circle.Uses :- Gears profile
- 88. PROB: A string is unwound from acircle of 20 mm diameter. Draw thelocus of string P for unwounding thestring’s one turn. String is kept tightduring unwound. Draw tangent &normal to the curve at any point.
- 89. T P9 P8 P10 09 ` N 0 P7 08 01 07 P11 al Ta rm 011 No ngP6 06 en 5 6 7 t 4 3 8 05 . 2 9 P5 1 10 T 04 01211N P12 02 03 P4 P1 1 2 3 4 5 6 7 8 9 10 1112 0 P3 P2 π D
- 90. PROBLEM:-Trace the path of end point of a threadwhen it is wound round a circle, thelength of which is less than thecircumference of the circle.Say Radius of a circle = 21 mm &Length of the thread = 100 mmCircumference of the circle = 2 π r = 2 x π x 21 = 132 mmSo, the length of the string is less thancircumference of the circle.
- 91. Ø = 30° x 5 /11 = 13.64 ° 11 mm = 30° P3 Then 5 mm = ζ P4 P2 INVOLUTE P5 R= R= R= 4tto R= 4o R=3toP R=3toP 5tto 5o PP to P P1 P P6 R=6toP P 2 R= P 7to 7 6 5 P7 R= 4 P P8 8ø 1 to P 3 R= 9 0 1 2 R2 10 11 0 1 0 1 2 3 4 5 6 7 8 P 9S = 2 x π x r /12 L= 100 mm
- 92. PROBLEM:-Trace the path of end point of a threadwhen it is wound round a circle, thelength of which is more than thecircumference of the circle.Say Radius of a circle = 21 mm &Length of the thread = 160 mmCircumference of the circle = 2 π r = 2 x π x 21 = 132 mmSo, the length of the string is more thancircumference of the circle.
- 93. P4 P3 P2P5 P1P6 6 5 7 4 8 3 15 9 O ø PP 2 14 R=21mm 10 11 113 P13 P0 14P7 12 1 2P 3 4 5 6 7 8 9 10 11 12 13 1415 12 P11 L=160 mm P8 P9 P
- 94. PROBLEM:-Draw an involute of a pantagon having sideas 20 mm.
- 95. INVOLUTE P3 OF A POLYGON ∗0 1 P2 R=3 R=4∗ R=2∗01Given : 2 01 3 1Side of a polygon P4 4 50 R=01 P1 P0 T 01 =5 ∗ N R S P5
- 96. PROBLEM:-Draw an involute of a squarehaving side as 20 mm.
- 97. INVOLUTE OF A SQUARE P1 01 P0 R= 0 R P4 =2 1 4 ∗ 01 2 3 R=P2 4∗ 01 01 3∗ N R= S N P3
- 98. PROBLEM:-Draw an involute of a stringunwound from the given figurefrom point C in anticlockwisedirection. B C 60° R 21 30° A
- 99. C8 C7 CC 6 +B +B +666 X+ R B A =X+ B X+A 5 C6 B C 60° 4 R 3 30° 2 1 X X X+ 2 X+ C0 X 1 A5 A5 A 3 3 C5 X X X+A X+A +A A +A 2 X+ +A 44 1 C1 X C2 C4 C
- 100. PROBLEM:-A stick of length equal to the circumference of asemicircle, is initially tangent to the semicircleon the right of it. This stick now rolls over thecircumference of a semicircle without sliding tillit becomes tangent on the left side of thesemicircle. Draw the loci of two end point of thisstick. Name the curve. Take R= 42mm.
- 101. INVOLUTE B A6 6 A5 B1 5 A4 4 B2 3A3 2 B3 2 3 4 A2 1 1 5 B4 A1 A C O B6 B5
- 102. SPIRALSIf a line rotates in a plane about one of itsends and if at the same time, a point movesalong the line continuously in onedirection, the curves traced out by themoving 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.
- 103. The angle between the radius vector and theline in its initial position is called theVECTORIAL ANGLE.Each complete revolution of the curve istermed as CONVOLUTION. SpiralArche Median Spiral for Clock Semicircle Quarter Logarithmic Circle
- 104. ARCHEMEDIAN SPIRALIt is a curve traced out by a pointmoving in such a way that itsmovement towards or away from thepole is uniform with the increase ofvectorial angle from the starting line.USES :-Teeth profile of Helical gears.Profiles of cams etc.
- 105. PROBLEM:To construct an Archemedian Spiralof one convolutions, given the radialmovement of the point P during oneconvolution as 60 mm and the initialposition of P is the farthest point onthe line or free end of the line.Greatest radius = 60 mm &Shortest radius = 00 mm ( at centre or at pole)
- 106. 3 4 2 P3 P2 P4 5 P1 1 P56 o 0 P6 P12 11 9 8 7 6 5 4 3 2 1 0 12 P11 12 P10 P7 P9 P8 7 11 8 10 9
- 107. To construct an ArchemedianSpiral of one convolutions,given the greatest &shortest(least) radii. ORTo construct an ArchemedianSpiral of one convolutions,given the largest radius vector& smallest radius vector.Say Greatest radius = 100 mm & Shortest radius = 60 mm
- 108. Diff. in length of any two radius vectorsConstant of the curve = Angle between them in radians 3 2 OP – OP3 4 P3 = P2 Π/2 P4 T 5 P1 1 100 – 90 P5 = n R mi Π/2 N P12 10 8 6 4 2 6 P O 11 9 7 5 3 1 12 = 6.37 mm 6 S P11 R P7 P10max 11 7 P8 P9 8N 10T 9
- 109. PROBLEM:-A slotted link, shown in fig rotates in thehorizontal plane about a fixed point O,while a block is free to slide in the slot. Ifthe center point P, of the block movesfrom A to B during one revolution of thelink, draw the locus of point P. 40 25 B A O
- 110. 31 21 41 11 P3 P4 51 P2 P1 25 P5 40B 11109 8 76 5 4 3 2 1 A O P6 61P12 P7 P11 111 71 P8 P10 P9 101 81 9
- 111. PROBLEM:-A link OA, 100 mm long rotates about O inclockwise direction. A point P on the link,initially at A, moves and reaches the other endO, while the link has rotated thorough 2/3 rd ofthe revolution. Assuming the movement of thelink and the point to be uniform, trace the pathof the point P.
- 112. PO Initial Position of point P2/3 X 360° A 1 1= 240° 2 P1 3 4 2 5 P2 6 120ºP7 O 8 P3 3 P7 P6 P4 P5 8 4 7 5 6
- 113. EXAMPLE: A link AB, AB Angular Swing A0 of link AB = 180° + 90°96mm long initially is P6vertically upward w.r.t. its = 270 ° P5 =45 °X 6 div.pinned end B, swings in A1 P4clockwise direction for 96 P3 96180° and returns back in ARCHIMEDIANanticlockwise direction for P2 SPIRAL90°, during which a point P1 P1’ C A6P, slides from pole B to P0 P 2’ A2 B NORM NORMAend A. Draw the locus of P 6’point P and name it. Draw P 3’ ALtangent and normal at any L Npoint on the path of P. P 5’ P 4’ A3Link AB = 96 M D A5 tLinear Travel of point P on AB Ta ngen= 96 =16x (6 div.) A4
- 114. Arch.Spiral Curve Constant BC = Linear Travel ÷Angular Swing in Radians = 96 ÷ (270º×π /180º) =20.363636 mm / radian
- 115. PROBLEM :A monkey at 20 m slides downfrom a rope. It swings 30° eithersides of rope initially at verticalposition. The monkey initially attop reaches at bottom, when therope swings about two completeoscillations. Draw the path of themonkey sliding down assumingmotion of the monkey and the ropeas uniform.
- 116. o 1 P3 2 θ 3 4 5 6 7 P9 8 9 10 11 P15 12 13 14 15 16 17 18 19 20 3 21 9 22 2115 23 2 24 8 4 1 0 7 10 14 5 6 11 20 16 1317 18 1224 1923 22
- 117. Problem : 2Draw a cycloid for a rolling circle, 60 mmdiameter rolling along a straight line withoutslipping for 540° revolution. Take initialposition of the tracing point at the highestpoint on the rolling circle. Draw tangent &normal to the curve at a point 35 mm abovethe directing line.
- 118. First Step : Draw a circle having diameter of 60 mm.Second step: Draw a straight line tangential to the circlefrom bottom horizontally equal to(540 x ) x 60 mm= 282.6 mm i.e. 1.5 x x 60 mm 360Third step : take the point P at the top of the circle.
- 119. al rm Rolling circle no P0 P8 8 P1 P7 P9 7 91 C0 SC C4 C8 C9 P106 10 2C1 C2 P P 3 2 P6 C5 C6 C7 C10 3 P3 P5 5 4 P4 Directing line 0 1 2 3 4 5 6 7 8 9 10 Length of directing line = 3Π 540 ° = 360° + 180° 540 ° = Π D + Π D/2 Total length for 540 ° rotation = 3Π D/2

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