Engineering Curves
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This document discusses engineering drawings and curves. It states that engineering drawings are the language used to communicate engineering ideas and execute work. Various types of curves are useful in engineering for understanding natural laws, manufacturing, design, analysis, and construction. Common engineering curves include conics, cycloids, involutes, spirals, and helices. Conics specifically include circles, ellipses, parabolas, and hyperbolas which are sections of a right circular cone cut by planes. The document provides definitions and examples of each type of conic section. It also discusses different methods for drawing ellipses like the arc of circles method.
Engineering Curves
- 1. 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”.
- 3. 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. USES OF ENGINEERING CURVES
- 4. 1. CONICS 2. CYCLOIDAL CURVES 3. INVOLUTE 4. SPIRAL 5. HELIX 6. SINE & COSINE CLASSIFICATION OF ENGG. CURVES
- 5. It is a surface generated by moving a Straight line keeping one of its end fixed & other end makes a closed curve. What is Cone ? If the base/closed curve is a polygon, we get a pyramid. If the base/closed curve is a circle, we get a cone. The closed curve is known as base. The fixed point is known as vertex or apex. Vertex/Apex 90º Base
- 6. If axis of cone is not perpendicular to base, it is called as oblique cone. The line joins vertex/ apex to the circumference of a cone is known as generator. If axes is perpendicular to base, it is called as right circular cone. Generator Cone Axis The line joins apex to the center of base is called axis. 90º Base Vertex/Apex
- 7. 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. CONICS
- 8. B - CIRCLE A - TRIANGLE CONICS C - ELLIPSE D – PARABOLA E - HYPERBOLA
- 9. When the cutting plane contains the apex, we get a triangle as the section. TRIANGLE
- 10. When the cutting plane is perpendicular to the axis or parallel to the base in a right cone we get circle the section. CIRCLE Sec Plane Circle
- 11. 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. ELLIPSE α θ α > θ
- 12. 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. PARABOLA θ α α = θ
- 13. 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. HYPERBOLA Definition :- α < θ α = 0 θ θ
- 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. Fixed point is called as focus. Fixed straight line is called as directrix. M C F V P Focus Conic Curve Directrix
- 15. The line passing through focus & perpendicular to directrix is called as axis. The intersection of conic curve with axis is called as vertex. AxisM C F V P Focus Conic Curve Directrix Vertex
- 16. N Q Ratio = Distance of a point from focus Distance of a point from directrix = Eccentricity= PF/PM = QF/QN = VF/VC = e M P F Axis C V Focus Conic Curve Directrix Vertex
- 17. Vertex 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 M N Q P C F V Axis Focus Ellipse Directrix Eccentricity=PF/PM = QF/QN < 1.
- 18. 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. ELLIPSE F1 A B P F2 O Q C D
- 19. F1 A B C D P F2 O PF1 + PF2 = QF1 + QF2 = CF1 +CF2 = constant = Major Axis Q = F1A + F1B = F2A + F2B But F1A = F2B F1A + F1B = F2B + F1B = AB CF1 +CF2 = AB but CF1 = CF2 hence, CF1=1/2AB
- 20. F1 F2 O A B C D Major Axis = 100 mm Minor Axis = 60 mm CF1 = ½ AB = AO F1 F2 O A B C D Major Axis = 100 mm F1F2 = 60 mm CF1 = ½ AB = AO
- 21. Uses :- Shape of a man-hole. Flanges of pipes, glands and stuffing boxes. Shape of tank in a tanker. Shape used in bridges and arches. Monuments. Path of earth around the sun. Shape of trays etc.
- 22. Ratio (known as eccentricity) of its distances from focus to that of directrix is constant and equal to one (1). PARABOLA 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. Definition :- Directrix Axis Vertex M C N Q F V P Focus Parabola Eccentricity = PF/PM = QF/QN = 1.
- 23. Motor car head lamp reflector. Sound reflector and detector. Shape of cooling towers. Path of particle thrown at any angle with earth, etc. Uses :- Bridges and arches construction Home
- 24. 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. Eccentricity = PF/PM Axis Directrix Hyperbola M C N Q F V P FocusVertex HYPERBOLA = QF/QN > 1.
- 25. Nature of graph of Boyle’s law Shape of overhead water tanks Uses :- Shape of cooling towers etc.
- 26. METHODS FOR DRAWING ELLIPSE 2. Concentric Circle Method 3. Loop Method 4. Oblong Method 5. Ellipse in Parallelogram 6. Trammel Method 7. Parallel Ellipse 8. Directrix Focus Method 1. Arc of Circle’s Method
- 27. Normal P2’ R=A1 Tangent 1 2 3 4 A B C D P1 P3 P2 P4 P4 P3 P2 P1 P1’ F2 P3’ P4’ P4’ P3’ P2’ P1’ 90° F1 Rad =B1 R=B2 `R=A2 O ° ° ARC OF CIRCLE’SARC OF CIRCLE’S METHODMETHOD
- 28. AxisMinor A B Major Axis 7 8 9 10 11 9 8 7 6 5 4 3 2 1 12 11 P6 P5 P4 P3 P2` P1 P12 P11 P10 P9 P8 P7 6 5 4 3 2 1 12 C 10 O CONCENTRICCONCENTRIC CIRCLECIRCLE METHODMETHOD F2 F1 D CF1=CF2=1/2 AB T N Q e = AF1/AQ
- 29. Normal Normal 00 11 22 33 44 11 22 33 44 1’1’ 0’0’ 2’2’3’3’4’4’ 1’1’ 2’2’ 3’3’ 4’4’ AA BB CC DD Major AxisMajor Axis MinorAxisMinorAxis FF11 FF22 DirectrixDirectrix EE FF SS PP PP11 PP22 PP33 PP44 Tangent Tangent PP11’’ PP22’’ PP33’’ PP44’’ ØØ R =A B/2 R =A B/2 PP00 P1’’ P2’’ P3’’ P4’’P4 P3 P2 P1 OBLONG METHODOBLONG METHOD
- 30. BA P4 P 0 D C 60° 6 5 4 3 2 1 0 5 4 3 2 1 0 1 2 3 4 5 6 5 3 2 1 0P1 P2 P 3 Q1 Q2 Q3Q4 Q5 P6 Q6O 4 ELLIPSE IN PARALLELOGRAM R4 R3 R2 R1 S1 S2 S3 S4 P5 G H I K J MinorAxis Major Axis
- 31. P6 Normal P5’ P7’P6’ P1 Tangent P1’ N T T V1 P5 P4’ P4 P3’ P2’ F1 D1D1 R1 b a c d e f g Q P7 P3 P2 Directrix R =6f`90° 1 2 3 4 5 6 7 Eccentricity = 2/3 3R1V1 QV1 = R1V1 V1F1 = 2 Ellipse ELLIPSE – DIRECTRIX FOCUS METHOD R=1a R=1a Dist. Between directrix & focus = 50 mm 1 part = 50/(2+3)=10 mm V1F1 = 2 part = 20 mm V1R1 = 3 part = 30 mm θ < 45º S
- 32. 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.
- 33. ARC OF CIRCLE’SARC OF CIRCLE’S METHODMETHOD Normal G2’ R=A1 Tangent 1 2 3 4 A B G G’ G1 G3 G2 G4 G4 G3 G2 G1 G1’ G3’ G4’ G4’ G3’ G2’ G1’ F2 F1 R=B1 R=B2 `R=A2 O ° ° 90° 90° directrixdirectrix 100100 140140 GF1 + GF2 = MAJOR AXIS = 140GF1 + GF2 = MAJOR AXIS = 140 EE eeAF1AF1 AEAE ee == R=70 R=70 R=70 R=70
- 34. 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.
- 36. 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.
- 37. I3 CD F1 F2 P Q R S 5050 I1 I4 A BI2 O 1 1 2 2 4 4 3 3 100100
- 38. 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.
- 39. A B P R=50 M N O 1 2 1 2 60 80 Q 1 2 12 P1 P2 Q2 Q1 R1 R2 S2 S1
- 40. 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.
- 41. ELLIPSEELLIPSE PP QQ R 80mm 40mm 89mm AA CC BB DD MAJOR AXIS = PQ+QR = 129mmMAJOR AXIS = PQ+QR = 129mm θ θ R=AB/2 11 22 33 1’1’ 2’2’ 3’3’ 1’’1’’2’’2’’3’’3’’ OO33 OO33’’ OO11 OO22 OO22’’ OO11’’ TANGENT TANGENT NORMAL NORMAL 6060ºº 30º30º 22 33 11 directrixdirectrix FF11 FF22MAJOR AXISMAJOR AXIS MINORAXISMINORAXIS SS ECCENTRICITY = AP / ASECCENTRICITY = AP / AS ?? ??
- 42. 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.
- 43. D C 6 6 5 4 3 2 1 0 5 4 3 2 1 0 1 2 3 4 5 6 6 5 3 2 1 0 P2 P3 P4 P5 Q1 Q2 Q3 Q4 Q5 B A O 4 ELLIPSE 100 45 75 P0 P1 P6 Q6 G H I K J
- 44. 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.
- 45. 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.
- 46. METHODS FOR DRAWING PARABOLA 1. Rectangle Method 2. Parabola in Parallelogram 3. Tangent Method 4. Directrix Focus Method
- 47. 22334455 00 11 22 33 44 55 66 11 11 55443322 66 00 11 22 33 44 55 00 VVDD CC AA BB PP44 PP44 PP55 PP55 PP33 PP33 PP22 PP22 PP66 PP66 PP11 PP11 PARABOLA –RECTANGLE METHODPARABOLA –RECTANGLE METHOD PARABOLAPARABOLA
- 48. BB 00 2’2’ 00 22 66 CC 6’6’ VV 55 P’P’55 3030°° AA XX DD 1’1’ 2’2’ 4’4’ 5’5’ 3’3’ 11 33 44 5’5’ 4’4’ 3’3’ 1’1’ 00 55 44 33 22 11 PP11 PP22 PP33 PP44 PP55 P’P’44 P’P’33 P’P’22 P’P’11 P’P’ 66 PARABOLA – IN PARALLELOGRAMPARABOLA – IN PARALLELOGRAM PP 66
- 50. D D DIRECTRIX 90° 2 3 4 T T N N S V 1 P1 P2 PF P3 P4 P1’ P2’ P3’ P4’ PF’ AXIS RF R2 R1 R3 R4 90° R F PARABOLA DIRECTRIX FOCUS METHOD
- 51. 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.
- 52. 6m6m ROOT OF TREEROOT OF TREE BUILDINGBUILDING REQD.DISTANCEREQD.DISTANCE TOP OF TREETOP OF TREE 3m3m 6m6m FF AA STONE FALLS HERESTONE FALLS HERE
- 53. 3m3m 6m6m ROOT OF TREEROOT OF TREE BUILDINGBUILDING REQD.DISTANCEREQD.DISTANCE GROUNDGROUND TOP OF TREETOP OF TREE 3m3m 6m6m 11 22 33 11 22 33 332211 44 55 66 55 66 EEFF AA BB CCDD PP33 PP44 PP22 PP11 PP PP11 PP22 PP33 PP44 PP55 PP66 33 22 11 00 STONE FALLS HERESTONE FALLS HERE
- 54. 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.
- 55. 150 mm AA BB CC DD11 22 33 44 55 11 22 33 44 55 OO PP11 PP22 PP33 PP44 PP55 MM 1’1’ 2’2’ 3’3’ 4’4’ 5’5’ 1’1’2’2’3’3’4’4’5’5’ PP11’’ PP22’’ PP33’’ PP44’’ PP55’’ 90mm
- 56. EXAMPLEEXAMPLE A shot is discharge from the groundA shot is discharge from the ground level at an angle 60 to the horizontallevel at an angle 60 to the horizontal at a point 80m away from the point ofat a point 80m away from the point of discharge. Draw the path trace by thedischarge. Draw the path trace by the shot. Use a scale 1:100shot. Use a scale 1:100
- 57. ground levelground level BA 60º gungun shotshot 80 M parabolaparabola
- 58. ground level BA O V 1 8 3 4 5 2 6 7 9 10 0 1 2 3 4 5 6 7 8 9 10 0 F 60º gungun shotshot DD DD VFVF VEVE == e = 1e = 1 EE
- 59. Connect two given points A and B by aConnect two given points A and B by a Parabolic curve, when:-Parabolic curve, when:- 1.OA=OB=60mm and angle AOB=90°1.OA=OB=60mm and angle AOB=90° 2.OA=60mm,OB=80mm and angle2.OA=60mm,OB=80mm and angle AOB=110°AOB=110° 3.OA=OB=60mm and angle AOB=60°3.OA=OB=60mm and angle AOB=60°
- 60. 66 6060 1 2 3 4 5 Parabola 5 4 3 2 1 AA BB 9090 °° OO 1.OA=OB=60mm and angle1.OA=OB=60mm and angle AOB=90AOB=90°°
- 61. AA BB 8080 6060 5 4 3 2 1 1 2 3 4 5 110 °° Parabola OO 2.OA=60mm,OB=80mm and angle2.OA=60mm,OB=80mm and angle AOB=110°AOB=110°
- 62. 54321 5 4 3 2 1 AA BB OO Parabola 66 6060 6060 °° 3.OA=OB=60mm3.OA=OB=60mm and angle AOB=60°and angle AOB=60°
- 63. exampleexample Draw a parabola passing through threeDraw a parabola passing through three different points A, B and C such that AB =different points A, B and C such that AB = 100mm, BC=50mm and CA=80mm100mm, BC=50mm and CA=80mm respectively.respectively.
- 65. 1’1’ 00 00 2’2’ 00 22 66 6’6’55 P’P’55 1’1’ 2’2’ 4’4’ 5’5’3’3’113344 5’5’ 4’4’ 3’3’ 55 44 33 22 11 PP 11 PP 22 PP 33 PP 44 PP 55 P’P’44 P’P’33 P’P’22 P’P’11 P’P’66 PP 66 AA BB CC
- 66. METHODS FOR DRAWING HYPERBOLA 1. Rectangle Method 2. Oblique Method 3. Directrix Focus Method
- 67. D F 1 2 3 4 5 5’ 4’ 3’ 2’ P1 P2 P3 P4 P5 0 P6 P0 AO EX B C Y Given Point P0 90° 6 6’ Hyperbola RECTANGULAR HYPERBOLA AXIS AXIS When the asymptotes are at right angles to each other, the hyperbola is called rectangular or equilateral hyperbola ASYMPTOTES X and Y
- 68. 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”.
- 69. D F 1 2 3 4 5 5’ 4’ 3’ 2’ P1 P2 P3 P4 P5 0 P6 P0 AO EX=20 B C Y=50 Given Point P0 90° 6 6’ Hyperbola RECTANGULAR HYPERBOLA
- 70. 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”.
- 71. 75 0 P4 E 6’ 2’ 1’ P1 1 2 3 4 5 6 D P6 P5 P3 P2 P0 7’P7 7C B F O Y=30 X = 20 Given Point P0 A
- 72. AXISAXIS NORMAL NORMAL CC VV FF11 DIRECTRIXDIRECTRIXDDDD 11 22 33 44 4’4’ 3’3’ 2’2’ 1’1’ PP11 PP22 PP33 PP44 PP11’’ PP22’’ PP33’’ PP44’’ TT11 TT22 NN 22 NN 11 TANGENT TANGENT ss Directrix and focus methodDirectrix and focus method
- 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. Superior Hypotrochoid Cycloidal Curves Cycloid Epy Cycloid Hypo Cycloid Superior Trochoid Inferior Trochoid Superior Epytrochoid Inferior Epytrochoid Inferior Hypotrochoid
- 74. Rolling Circle or Generator 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. C P P P R C 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. 2πr Ø = 360Ø = 360ºº x r/Rx r/Rdd Circumference ofRRdd RollingRolling CircleCircle r O Ø/Ø/ 22 Ø/Ø/ 22 P0 P0 Arc P0P0 = Rd x Ø = P0
- 76. 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 ØØ /2 ØØ /2 ØØ = 360 x r RR T Rolling Circle Radius (r) O Vertical Hypocycloid P P
- 77. 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. 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.
- 78. P0 2R or D 5 T T 1 2 1 2 3 4 6 7 8 9 10 110 12 0 3 4 56 7 8 9 10 11 12 P1 P2 P3 P4 P5 P7 P8 P9 P11 P12 C0 C1 C2 C3 C4 C5 C6 C7 C8 C9 C10 C11 Directing Line C12 N N S S1 R P6 R P10 R : Given Data : Draw cycloid for one revolution of a rolling circle having diameter as 60mm. Rolling Circle D
- 79. CC00 PP00 7788 PP 66 44 PP11 11 22 33 CC22 CC33 PP22 CC44 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. 55 66 CC11 PP33 PP44 PP 55 PP 77 PP 88 77 00 CC55CC66CC77CC88 11 22 33 44 5566DD/2/2ππDD/2/2 ππDD/2/2 DD/2/2 FloorFloor WallWall CYCLOIDCYCLOID 55 66 77 88 Take diameter of circle = 40mm Initially distance of centre of circle from the wall 83mm (Hale circumference + D/2)
- 80. 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.
- 81. 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.
- 82. PP66 PP44 r PP22 CC11 CC00 CC22 CC33 CC44 CC55 CC66 CC77 CC88 1 0 23 4 5 6 7 O RRdd Ø/2Ø/2 Ø/2Ø/2 PP11 PP00 PP33 PP55 PP77PP88 r rRollingRolling CircleCircle r Rd X Ø = 2πr Ø = 360Ø = 360ºº xx Arc P0P8 = Circumference of Generating Circle EPICYCLOIDEPICYCLOIDGIVEN: Rad. Of Gen. Circle (r) & Rad. Of dir. Circle (Rd) S º U N Ø = 360Ø = 360ºº x 25/75x 25/75 = 120°= 120°
- 83. 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.
- 84. P0 P1 Tangent P11 r C0 C1 C2 C3 C4 C5 C6 C7 C8 C9 C10 C11 C12 P10P8 0 1 2 3 4 5 6 7 89 10 11 12 P2 P3 P4 P5 P6 P9 P7 P12 / 2 / 2 = 360 x 4 12 = 360 x r R = 120° R T T N S N Normal r r Rolling Circle Radias (r) Directing Circle O Vertical Hypocycloid
- 85. 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
- 86. C C1 C2 C3 C4 C5 C6 C7 C9 C8 C10 C11 C12 P8 O 10 5 7 8 9 11 12 1 2 3 4 6 P1 P2 P3 P4 P5 P6 P7 P9 P10 P11 P12 Directing Circle Rolling Circle HYPOCYCLOID
- 87. 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 Uses :- Gears profile 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.
- 88. 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.
- 89. P12 P2 0 02 12 6 P1 1 20 9 103 4 6 8 115 7 12 π D P3 P4 P5 P6 P7 P8 P9 P10 P11 1 2 3 4 5 7 8 9 10 11 03 04 05 06 07 08 09 010` 011 Tangent N N Norm al T T .
- 90. 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.
- 91. PP R=7toP R=7toP R=6toPR=6toP R21 R21 00 00 11 22 33 44 55 66 77 88 PP 1111 00 11 22 33 44 5566 77 88 99 1010 PP11 PP22 PP33 PP44 PP55 PP66 PP77 PP88 L= 100 mmL= 100 mm R=1toP R=1toP R=2toP R=2toP R=3toPR=3toP R=4toP R=4toP R=5toP R=5toP INVOLUTEINVOLUTE 99 ø 11 mm = 30° Then 5 mm = ζ Ø = 30° x 5 /11 = 13.64 ° S = 2 x π x r /12
- 92. 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.
- 93. PP1313 PP1111 3 13 14 15 PP00 PP1212 O 7 10 1 2 3 456 8 9 11 12 1 2 PP11 PP22 PP33PP44 PP55 PP66 PP77 PP88 PP99 PP PP1414PP L=160 mmL=160 mm R=21mm 64 5 7 8 9 10 1112 131415 øø
- 94. PROBLEM:- Draw an involute of a pantagon having side as 20 mm.
- 95. P5 R=01 R=2∗01 P0 P1 P2 P3 P4 R=3∗01 R=4∗01 R=5∗01 2 3 4 5 1 T T N N S INVOLUTE OF A POLYGON Given : Side of a polygon 0
- 96. PROBLEM:- Draw an involute of a square having side as 20 mm.
- 98. PROBLEM:- Draw an involute of a string unwound from the given figure from point C in anticlockwise direction. 60°60° AA BB CC R 21 R 2130°30°
- 100. A stick of length equal to the circumference of aA stick of length equal to the circumference of a semicircle, is initially tangent to the semicirclesemicircle, is initially tangent to the semicircle on the right of it. This stick now rolls over theon the right of it. This stick now rolls over the circumference of a semicircle without sliding tillcircumference of a semicircle without sliding till it becomes tangent on the left side of theit becomes tangent on the left side of the semicircle. Draw the loci of two end point of thissemicircle. Draw the loci of two end point of this stick. Name the curve. Take R= 42mm.stick. Name the curve. Take R= 42mm. PROBLEM:-
- 102. 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.
- 103. 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 CircleLogarithmic
- 104. 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.
- 105. 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 = 60 mm & Shortest radius = 00 mm ( at centre or at pole) PROBLEM:
- 107. To construct an Archemedian Spiral of one convolutions, given the greatest & shortest(least) radii. Say Greatest radius = 100 mm & Shortest radius = 60 mm To construct an Archemedian Spiral of one convolutions, given the largest radius vector & smallest radius vector. OR
- 108. 3 1 26 5 8 4 79 10 11 2 1 3 4 5 6 7 8 9 10 11 12 P1 P2 P3 P4 P5 P6 P7 P8 P9 P10 P11 P12 O N N T T S Rmin R m ax Diff. in length of any two radius vectors Angle between them in radians Constant of the curve = = OP – OP3 Π/2 100 – 90 = Π/2 = 6.37 mm
- 109. 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. OAB 40 25
- 110. BB AA OO1234567891011 P1 P2 P3 P4 P5 P6 P7 P8 P9 P10 P11 P12 11 21 31 41 51 61 71 81 9 101 111 40 25
- 111. 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.
- 112. AA Initial Position of point PInitial Position of point PPPOO PP11 PP22 PP33 PP44 PP55 PP66 PP77 PP88 22 11 33 44 55 66 77 OO 11 22 33 44 55 66 77 88 2/3 X 360° = 240° 120º120º
- 113. AA00 Linear Travel of point PP on ABAB = 96 =16x (6 div.) EXAMPLE: A link ABAB, 96mm long initially is vertically upward w.r.t. its pinned end BB, swings in clockwise direction for 180° and returns back in anticlockwise direction for 90°, during which a point PP, slides from pole BB to end AA. Draw the locus of point PP and name it. Draw tangent and normal at any point on the path of PP. PP11’’ AA BB AA11 AA22 AA33 AA44 AA55 AA66 PP00 PP11 PP22 PP33 PP44 PP55 PP66 PP22’’ PP33’’ PP44’’ PP55’’ PP66’’ 9696 LinkLink ABAB = 96= 96 CC Tangent Tangent AngularAngular SwingSwing ofof linklink AB =AB = 180° + 90°180° + 90° == 270270 °° =45 °X 6 div.=45 °X 6 div. ARCHIMEDIANARCHIMEDIAN SPIRALSPIRAL DDNORMAL NORMAL MM NN
- 114. Arch.Spiral Curve Constant BCBC = Linear Travel ÷Angular Swing in Radians = 96 ÷ (270º×π /180º) =20.363636 mm / radian
- 115. 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.
- 116. θ o 01 2 3 4 5 6 7 8 9 10 11 1213 14 15 16 17 18 19 20 21 222324 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 P9 P15
- 117. 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.
- 118. 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= 282.6 mm i.e. 1.5 x x 60 mm 360 Third step : take the point P at the top of the circle.
- 119. Rolling circleRolling circle PP11 PP22 PP33 PP44 PP00 PP66 PP77 PP88 PP55 PP99 PP1010 11 22 33 44 55 66 77 88 99 1010 11 22 33 44 55 66 77 88 99 101000 CC00 CC11 CC22 CC33 CC44 Directing lineDirecting line Length of directing line = 3Length of directing line = 3ΠΠ 540540 °° = 360= 360°° + 180+ 180°° 540540 °° == ΠΠD +D + ΠΠD/2D/2 Total length for 540Total length for 540 °° rotation = 3rotation = 3ΠΠD/2D/2 CC55 CC66 CC77 CC88 CC99 CC1010 SS norm al
Editor's Notes
- #6: there is no objection