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# Mel110 part3

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### Mel110 part3

1. 1. ENGINEERING CURVES P i t d i t t f di l tPoint undergoing two types of displacements INVOLUTE: Locus of a free end of a string when it isg wound round a (circular) pole CYCLOID L f i t th i h fCYCLOID: Locus of a point on the periphery of a circle which rolls on a straight line path. SPIRAL: Locus of a point which revolves around a fixed point and at the same time moves towards it.p HELIX: Locus of a point which moves around the f f i ht i l li d / d t thsurface of a right circular cylinder / cone and at the same time advances in axial direction at a speed bearing a constant ratio to the speed of rotationbearing a constant ratio to the speed of rotation.
2. 2. INVOLUTE OF A CIRCLEINVOLUTE OF A CIRCLE • Problem: Draw Involute of P3 P1 Draw Involute of a circle. String length is equal to the circumference of i l 3 4 5 P4 4 to p circle. P 1 2 6 7 8 P A P P8 P5 P7 P6 πD 1 2 3 4 5 6 7 8
3. 3. INVOLUTE OF A CIRCLE String length MORE than πD Problem: Draw Involute of a circle. String length is MORE than the circumference of circle. P2 Solution Steps: In this case string length is more than Π D. P3 P1 But remember! Whatever may be the length of string, mark Π D distance horizontal i.e.along the string and divide it in 8 number ofd d v de 8 u be o equal parts, and not any other distance. Rest all steps are same as previous INVOLUTE. Draw the curve completely. 2 3 4 5 P4 4 to p 1 2 3 4 5 6 7 8 P 1 6 7 8P5 p8 1 2 3 4 5 6 7 8 P7 P6 165 mm (more than πD) πD
4. 4. INVOLUTE OF A CIRCLE String length LESS than πD Problem: Draw Involute of a circle. String length is LESS than the circumference of circle. P2 String length LESS than πD Solution Steps: In this case string length is Less than Π D. P3 P1 But remember! Whatever may be the length of string, mark Π D distance horizontal i.e.along the string and divide it in 8 number of 4 t d d v de 8 u be o equal parts, and not any other distance. Rest all steps are same as previous INVOLUTE. Draw the curve completely. 2 3 4 5 6 P4 4 to p 1 2 3 4 5 6 7 8 P 1 6 7 8P5 P 1 2 3 4 5 6 7 8P7 P6 150 mm (Less than πD) πD
5. 5. INVOLUTE OF A PENTAGON ll 5l 5 Thread should be taut
6. 6. INVOLUTE OF COMPOSIT SHAPED POLE Problem : A pole is of a shape of half hexagon (side 30 mm) and COMPOSIT SHAPED POLEg ( ) semicircle (diameter 60 mm). A string is to be wound having length equal to the pole perimeter Calculate perimeter length P1 P2 P g q p p draw path of free end P of string when wound completely. 1toP P2 P 1 2 34 5 P3 3 to P 6 1 2 3 4 5 6 A P πD/2P πD/2P4 P5 P6
7. 7. PROBLEM : Rod AB 85 mm long rolls over a semicircular pole without slipping from it’s initially vertical position till it becomes up-side-down vertical. Draw locus of both ends A & B. 4 B A4 3 4 B1 A3 OBSERVE ILLUSTRATION CAREFULLY! when one end is approaching, other end will move away from πR 2 3y poll. 2 πR 2 A2 B2 1 2 3 4 1 A 4A A1 B3 BB4
8. 8. CYCLOID: DEFINITIONS SUPERIOR TROCHOID: If the point in the definition of cycloid is outside the circleCYCLOID: LOCUS OF A POINT ON THE PERIPHERY OF A CIRCLE WHICH ROLLS ON A STRAIGHT LINE PATH. cycloid is outside the circle INFERIOR TROCHOID.: If it is inside the circleROLLS ON A STRAIGHT LINE PATH. If it is inside the circle EPI-CYCLOID If the circle is rolling on another circle from outside HYPO-CYCLOID. If the circle is rolling from inside theIf the circle is rolling from inside the other circle,
9. 9. CYCLOID PROBLEM: Draw locus (one cycle) of a point (P) on the periphery of a circle (diameter=50 mm) which rolls on straight line path. p p4 p 4 C1 C2 C3 C4 C5 C6 C7 C8 p2 p3 p5 p6 2 3 5 6 C p1 p71 7 P p8 πD Point C (zero radius) will not rotate and it will traverse on straight line.
10. 10. SUPERIOR TROCHOID PROBLEM: Draw locus of a point (P), 5 mm away from the periphery of a Circle (diameter=50 mm) which rolls on straight line path. Using 2H i p44 Using H p3 p53 5 C1 C2 C3 C4 C5 C6 C7 C8 p2 p6 2 6 C p1 p7 p8 1 7 DP p8πDP
11. 11. INFERIOR PROBLEM: Draw locus of a point , 5 mm inside the periphery of a Ci l hi h ll i h li h T k i l di t 50 TROCHOIDCircle which rolls on straight line path. Take circle diameter as 50 mm C1 C2 C3 C4 C5 C6 C7 C8 p2 p 3 p4 p5 p6 2 3 4 5 6 C P p1 p7 p8 1 2 6 7 πD
12. 12. CYCLOID SUPERIOR TROCHOID INFERIOR TROCHOID 12
13. 13. EPI CYCLOID : PROBLEM: Draw locus of a point on the periphery of a circle (dia=50mm) which rolls on a curved path (radius 75 mm). Solution Steps: Distance by smaller circle = Distance on larger circle Solution Steps: 1. When smaller circle rolls on larger circle for one revolution it covers Π D distance on arc and it will be decided by included arc angle θangle θ. 2. Calculate θ by formula θ = (r/R) x 360°. 3. Construct a sector with angle θ and radius R. 4 Divide this sector into 8 number of equal angular parts4. Divide this sector into 8 number of equal angular parts.
14. 14. EPI CYCLOID C 4 5 Generating/ Rolling Circle C2 2 3 6 7EPI-CYCLOID If the circle is rolling on Pr = CP 1 2If the circle is rolling on another circle from outside r R 3600 = Directing Circle O R
15. 15. PROBLEM : Draw locus of a point on the periphery of a circle which rolls from the inside of a curved path Take diameter of Rolling circle HYPO CYCLOID rolls from the inside of a curved path. Take diameter of Rolling circle 50 mm and radius of directing circle (curved path) 75 mm. P 7 P1 P21 6 P3 P 2 5 4 P4 P5 P6 P7 P8 3 4 O OC R ( R di f Di ti Ci l ) r R 3600 = OC = R ( Radius of Directing Circle) CP = r (Radius of Generating Circle)
16. 16. CYCLOID SUPERIOR TROCHOID INFERIOR TROCHOIDTROCHOID 16
17. 17. SPIRALProblem: Draw a spiral of one convolution. Take distance PO 40 mm. IMPORTANT APPROACH FOR CONSTRUCTION! FIND TOTAL ANGULAR AND TOTAL LINEAR DISPLACEMENT AND DIVIDE BOTH IN TO SAME NUMBER OF EQUAL PARTS. 2 P 13 P2 P1 P3 P4 3 P4 O 7 6 5 4 3 2 1 P4 P6P5 P7 O 5 7 6
18. 18. SPIRAL of two convolutions Problem: Point P is 80 mm from point O. It starts moving towards O and reaches it in two revolutions around. It Draw locus of point P (To draw a Spiral of TWO convolutions). 2,10 P2 1,93,11 P1 P3 P 16 13 10 8 7 6 5 4 3 2 1 P 4 12 8,16P4 P8 P9 P10 P11 4,12 , P12 P13 P14 P15 5,13 7,15 P5 P6 P7 6,14
19. 19. Problem: A link 60 mm long, swings on a point Oob e : 60 o g, sw gs o a po O from its vertical position of the rest to the left through 60° and returns to its initial position at uniformp velocity. During that period a point P moves at uniform speed along the center line of the link from Op g at reaches the end of link. Draw the locus of P. O, P N M
20. 20. P8 HELIX (UPON A CYLINDER) PROBLEM: Draw a helix of one convolution, upon a cylinder. Given 80 mm 6 7 8 P6 P7 8 pitch and 50 mm diameter of a cylinder. 4 5 6 P4 P5 6 Pitch: Axial advance during one 2 3 P2 P3 4 complete revolution . 1 P1 P 2 6 F T 57 P 1 3 4 2 3
21. 21. P8 HELIX (UPON A CONE) PROBLEM: Draw a helix of one convolution, upon a cone, diameter of base 70 mm, axis P5 P6 P7 ( )p , , 90 mm and 90 mm pitch. P4 P2 P3 6 P P1 X Y 57 P5P6 P P 4 P1 P3 P4 P7 P8 1 2 3 P2
22. 22. I i f S lid /Interpenetration of Solids / Intersection of Surfaces /Intersection of Surfaces / Lines & Curves of Intersection More points common to both the solids Basic required knowledge: ~ Projections of Solid ~ Section of Solid 22 ~ Section of Solid
23. 23. Problem: A cone 40 mm diameter and 50 mm axis is . Projections of Solid Problem: A cone 40 mm diameter and 50 mm axis is resting on one generator on Hp which makes 300 inclination with Vp. Draw it’s projections. More number of generators Better approximation. o’ a’1 h’ b’1 g’1 h’1 F g X Ya’ b’ d’ e’c’ g ’ f’h’ o’ g1 g c’1 d’1e’1 f’1 o1 o1 30 h a e f a1 h1f1 e1 a1 be f1 g1 h1 o1 o1 T o 23b c d d1 c1 b1 c1 b1 d1 e1 1st. Angle
24. 24. For TVFor TVSection of Solid SECTIONSECTION PLANEPLANE I t t ti f S lidLi & Cxx yy Apparent ShapeApparent Shape Interpenetration of Solids Intersection of Surfaces Lines & Curves of Intersection SECTION LINESSECTION LINES Apparent ShapeApparent Shape of sectionof section (45(4500 to XY)to XY) SECTIONAL T.V.SECTIONAL T.V.
25. 25. 25
26. 26. 26
27. 27. 27
28. 28. How to make these shapes
29. 29. CURVES OF INTERSECTIONS are shown by WHITE ARROWS. Machine component having Intersection of a Cylindrical two intersecting cylindrical surfaces with the axis at acute angle to each other. y main and Branch Pipe.Industrial Dust collector. Intersection of two cylinders. WHEN TWO OBJECTS ARE TO BE JOINED TOGATHER, MAXIMUM Pump lid having shape of a h l P i d SURFACE CONTACT BETWEEN BOTH BECOMES A BASIC REQUIREMENT FOR STRONGEST & LEAK PROOF hexagonal Prism and Hemi-sphere intersecting each other. Feeding Hopper. STRONGEST & LEAK-PROOF JOINT.
30. 30. Minimum Surface Contact. ( Point Contact) (Maximum Surface Contact) Li f I i C f I iLines of Intersections. Curves of Intersections. Square Pipes. Circular Pipes. Square Pipes. Circular Pipes. MAXIMUM SURFACE CONTACT BETWEEN BOTH BECOMES A BASIC REQUIREMENT FOR STRONGEST & LEAK-PROOF JOINT. Two plane surfaces (e-g. faces of prisms and pyramids) intersect in a straight line Q intersect in a straight line. The line of intersection between two curved surfaces (e-g. of cylinders and cones) or between a plane surface and aof cylinders and cones) or between a plane surface and a curved surface is a curve. 30 More points common to both the solids
31. 31. How to find Lines/Curves of Intersection Generator line Method Cutting Plane Method T F T FF 75 31 75
32. 32. Problem: Find intersection curve . Draw convenient number of lines on the surface of one of the solids. Transfer point of intersection to their corresponding positions in other When one solid completely penetrates another, there will be two curves of intersection. 32 p p g p views.
33. 33. Problem: A cylinder 50mm dia. & 70mm axis is completely penetrated by another of 40 mm dia. & 70 mm axis horizontally. Both axes intersect & bisect each other. Draw projections showing curves of intersections 4” 1”3” 2”1’ 2’4’ 3’ Draw projections showing curves of intersections. a” b”h” a’ b ’h’ g” c” f” d” c’g’ d’f’ X Y e” f” dd f a’ 4 1 3 2 33
34. 34. Problem: CYLINDER (50mm dia.and 70mm axis ) STANDING & SQ.PRISM (25 mm sides and 70 mm axis) PENETRATING. Both axes Intersect & bisect each other. All faces of i ll i li d t H D j ti h i f i t ti 4” 1”3” 2”1’ 2’4’ 3’ prism are equally inclined to Hp. Draw projections showing curves of intersections. a” d”d” b” Projections ofProjections ofFind lines of X Y c” j Solid 1. j Solid2. Find lines of intersectionMissing critical points at which curve changes direction 4 direction. 1 3 2 34
35. 35. Problem. SQ.PRISM (30 mm base sides and 70mm axis ) STANDING & SQ.PRISM (25 mm sides and 70 mm axis) PENETRATING. Both axes intersects & bisect each other. All faces of prisms are equally inclined to Vp Draw projections showing curves of 1’ 2’4’ 3’ faces of prisms are equally inclined to Vp. Draw projections showing curves of intersections. a”a’ a’ 1’ 2’4’ 3’ 4” 1”3” 2” d” b” d’ b’ d’ b’ X Y c” c’ c’ 4 Preference of object line over dash line 1 3 dash line… 2 35
36. 36. Problem. SQ.PRISM (30 mm base sides and 70mm axis ) STANDING & SQ.PRISM (25 mm sides and 70 mm axis) PENETRATING. Both axes Intersect & bisect each other. Two faces of penetrating prism are 300 inclined to Hp. Draw projections showing curves of 1’ 2’4’ 3’ 4” 1”3” 2” p g p p p j g intersections in I angle projection system. f” a’ f’ e’ c” c’ d’ b’ X Y 4 300 d 1 3 2 od" 36 Other possible arrangements ???? b"
37. 37. Problem: A vertical cylinder 50mm dia. and 70mm axis is completely penetrated by a triangular prism of 45 mm sides and 70 mm axis, horizontally. One flat face of prism is parallel to Vp and Contains axis of cylinder Draw projections showingprism is parallel to Vp and Contains axis of cylinder. Draw projections showing curves of intersections in I angle projection system. 4” 1”3” 2”1’ 2’4’ 3’ aaa b c d bb c d X Y e d f e e f f 4 1 3 2 37
38. 38. Problem: A vertical cone, base diameter 75 mm and axis 100 mm long, is completely penetrated by a cylinder of 45 mm diameter and axis 100 mm long. The axis of the cylinder is parallel to Hp and Vp and intersects axis of the cone at a point 28 mm above the base. Draw o”o’ projections showing curves of intersection in FV & TV. in I angle projection system 11 1 2 3 46 7 8 33 7, 8,22 4 g a’ b’h’ c’g’ d’f’ e’ g” f”h” a”e” b”d” c” 4 5 6 X Y 5 5 64 4 28 h g f a e b c d 38
39. 39. Problem: Vertical cylinder (80 mm diameter & 100 mm height) is completely penetrated by a horizontal cone (80 mm diameter and 120 mm height). Both axes intersect & bisect each other. Draw projections showing curve of intersections in I angle projection systemDraw projections showing curve of intersections in I angle projection system. 7’7 6’ 8’ 1’ 5’ 2’ 4’ X Y 3’ 1 2 8 Intersection of a curve with 3 7 curve with another !!!! Generator lines.. 4 6 5 39
40. 40. Problem: CONE STANDING & SQ.PRISM PENETRATING (BOTH AXES VERTICAL) 2’ 1’ 5’ 3’ X Ya’ b’h’ c’g’ d’f’ e’ 4’ 6’ Y h g f 8 a b h c g d f e a e f 1 6 10 9 7 Problem: A cone70 mm base diameter and 90 mm axis is completely penetrated by a square prism from top b d 2 3 4 5 with it’s axis // to cone’s axis and 5 mm away from it. A vertical plane containing both axes is parallel to Vp. Take all faces of sq.prism equally inclined to Vp. Base Side of prism is 30 mm and axis is 100 mm long c 5 mm OFF-SET Base Side of prism is 30 mm and axis is 100 mm long. Draw projections showing curves of intersections. 40
41. 41. Intersection of two cylindersIntersection of two cylinders oblique to each other –Use PAVoblique to each other Use PAV III angle projection 41
42. 42. Intersection of two cylindersIntersection of two cylinders oblique to each other –Use AVoblique to each other Use AV 42
43. 43. Intersection of Cone and ObliqueIntersection of Cone and Oblique cylinder using PAVcylinder using PAV 43
44. 44. Intersection of irregular Prism &g offset Cylinder III angle projection 44
45. 45. Intersection of irregular Prism &g offset Cylinder InvisibleInvisible Visible 45
46. 46. Intersection of irregular Prism &g offset Cylinder InvisibleInvisible Visible 46
47. 47. Intersection of irregular Prism &g offset Cylinder 47
48. 48. DEVELOPMENT OF SURFACES OF SOLIDSDEVELOPMENT OF SURFACES OF SOLIDSDEVELOPMENT OF SURFACES OF SOLIDSDEVELOPMENT OF SURFACES OF SOLIDS Solids are bounded by geometric surfaces:Solids are bounded by geometric surfaces: LATERLAL SURFACELATERLAL SURFACE IS SURFACE EXCLUDING SOLID’S TOP & BASE.IS SURFACE EXCLUDING SOLID’S TOP & BASE. Development ~ obtaining the area of the surfaces of a solid. Development of the solid when folded or rolled gives the solid - Plane Prism, Pyramid Single curved Cone Cylinder Development of the solid, when folded or rolled, gives the solid. - Single curved Cone, Cylinder - Double curved sphere 48
49. 49. Examples Prism – Made up of same number of rectangles as sides of the base One side: Height of the prismg p Other side: Side of the base Cylinder – Rectangle One side: Height of the cylinder Other side: Circumference of the baseOther side: Circumference of the base Pyramid – Number of triangles in contact h The base may be included if present πd T. L. 49
50. 50. Methods to Develop Surfacesp 1. Parallel-line development: Used for prisms (full or truncated), c linders (f ll or tr ncated) Parallel lines are dra n along thecylinders (full or truncated). Parallel lines are drawn along the surface and transferred to the development Cylinder: A Rectangle H Cylinder: A Rectangle πD D H= Height D= base diameterg Prisms: No.of Rectangles H 50SS H= Height S = Edge of base
51. 51. Ex:Ex: DOTTED LINES are neverDOTTED LINES are never shown on developmentshown on development 51 shown on developmentshown on development
52. 52. 4, d 2, b Complete de elopment of c be c t b c tting plane (inclined to HP at 52 Complete development of cube cut by cutting plane (inclined to HP at 30 degrees and perpendicular to VP)
53. 53. 4, d 2, b 53
54. 54. Cylinder cut by h i j k φ50 Cylinder cut by three planes a g hl three planes G T F b c d e f Cc’ d' e' f' g' D E F G 45o H I J K IA B C a' b' c 15o 100 AI 3 4 5 6 7 8 9 1041 51 61 71 54 πx50 11 2 3 4 10 11 12 11 21 31 41
55. 55. Problem: Development of a solid is a parabolaProblem: Development of a solid is a parabola with a 180 mm base and a 90 mm height. Draw the projections of solidprojections of solid. 180=π D D=57.3 mm 55
56. 56. Methods to Develop Surfacesp 1. Parallel-line development 2. Radial-line development: Used for pyramids, cones etc. in which the true length of the slant edge or generator is used as radius Cone: (Sector of circle) Pyramids: (No.of triangles) θ R=Base circle radius. L= Slant edge θ = R L 3600 L=Slant height. L Slant edge. S = Edge of base 56
57. 57. Parallel vs Radial line method Parallel line method Radial line method 57
58. 58. DEVELOPMENT OF DEVELOPMENT OF FRUSTUMSFRUSTUMS Base side Top side DEVELOPMENT OF FRUSTUM OF CONE DEVELOPMENT OF FRUSTUM OF SQUARE PYRAMID Top side θ θ = R L 3600 R= Base circle radius of cone L= Slant height of cone L= Slant edge of pyramid L1 = Slant edge of cut part. L1 = Slant height of cut part. 1.1. DevelopmentDevelopment isis a shape showing AREA, means it’s a 2a shape showing AREA, means it’s a 2--D plain drawing.D plain drawing. 22 AllAll dimensions of it must be TRUE dimensionsdimensions of it must be TRUE dimensions ImportantImportant pointspoints.. 2.2. AllAll dimensions of it must be TRUE dimensions.dimensions of it must be TRUE dimensions. 3. As it is representing shape of an un3. As it is representing shape of an un--folded sheet, no edges can remain hiddenfolded sheet, no edges can remain hidden and hence DOTTED LINES are never shown on development.and hence DOTTED LINES are never shown on development.
59. 59. Development of Sphere using Frustum of Cones: Zone MethodFrustum of Cones: Zone Method Zone 1: Cone Zone 2: Frustum of cone Zone 3: Frustum of cone Z 4 F t fZone 4: Frustum of cone θ = R L 3600 θ 59
60. 60. Development by Radial Method Pyramids (full or Truncated) & Cones (full or Truncated)Cones (full or Truncated). If the slant height of a cone is equal to its diameter of base then its development isIf the slant height of a cone is equal to its diameter of base then its development is a semicircle of radius equal to the slant height. 60
61. 61. Ex: 61
62. 62. Development of Oblique Objects • Right regular objects – Axis of object perpendicular to base. • Axis of any regular object (prismAxis of any regular object (prism, pyramid, cylinder, cone, etc.) inclined at angle other than right angle Obliqueangle other than right angle – Oblique OBJECT. Use ARC method. 62
63. 63. Oblique prismOb que p s c de fg j a b cf h i g Parallel j a b f a' b’f’ c' f h’ i'g' j' h ig gg 63
64. 64. Draw the development of an oblique circular cylinder with base diameter 30 mm and axis inclined at 75o with the base. Height of the cylinder is 50 mm • Divide the surface of the cylinder into equal parts as shown, with the generator lines parallel to the end generators • Draw projection lines from top edge of cylinder φ30 AGg a T • Draw projection lines from top edge of cylinder such that they are perpendicular to end generator • Mark distances AB, BC etc. from one projector A E B CD F g a F A line to the next to complete the profile • Do the similar process for the bottom edgeA’G’ B C A 75o G 50 A1 75 g’ a' a A1 64 A1 g a
65. 65. Oblique Cone 65
66. 66. Intersection of Plane & Pyramid. 14y Development of resulting lateral 0 C D Develop 1-D-A-2-1 2-A-B-3-21 B D g truncated Pyramid 23 A B 3-B-C-4-3 1-D-C-4-12 1 B YX B’ 0 A B D’ 4 23 1’ o 3 2 66
67. 67. Methods to Develop Surfacesp 1. Parallel-line development: 2. Radial-line development: 3. Triangulation development: Complex shapes are divided into a number of triangles and transferred into the development Tetrahedron: Four Equilateral Triangles EXAMPLES:EXAMPLES:-- Boiler Shells & chimneys,Boiler Shells & chimneys, Pressure Vessels Shovels TraysPressure Vessels Shovels Trays All sides equal in length Pressure Vessels, Shovels, Trays,Pressure Vessels, Shovels, Trays, Boxes & Cartons, FeedingBoxes & Cartons, Feeding Hoppers, Large Pipe sections,Hoppers, Large Pipe sections, B d & P t f t tiB d & P t f t ti equal in length Body & Parts of automotives,Body & Parts of automotives, Ships, Aero planes.Ships, Aero planes. 67
68. 68. Transition Connect two hollow objects having different base. Pieces Three surfaces Triangulation Method: Dividing a surface into aDividing a surface into a number of triangles and transfer them to the developmentdevelopment. 68
69. 69. Ex: In air conditioning system, a square duct of 50mm by 50mm is connected to another square duct of 25mm by 25 mm by using a connector (transition piece) of height 25mm Draw development of lateral surface of the connector (Neglectheight 25mm. Draw development of lateral surface of the connector (Neglect thickness of connector). Pyramids: (No.of triangles) O’O L= Slant edge. S = Edge of base a’ A g b’ B a b O 69 c
70. 70. D l t f T iti Pi fDevelopment of Transition Piece for Difference Shapes and Sizesp • Connect a Square pipe with circular pipe. • Ex: Imagine a transition piece (height = 25) to connect a chimney of square crossto connect a chimney of square cross section 50mm * 50 mm to circular pipe of 30mm diameter Draw the projections and30mm diameter. Draw the projections and develop lateral surface of the transition piece. 70
71. 71. 1’2’, 8’ 1/8 of i fc a’, b’ circumferencebc 2 3 4 1 1 2 5 8 I angle p ojectionad 6 7 8 18 A B 71 I angle projectionad 18
72. 72. Development of Sphere/HemisphereSphere/Hemisphere using Lune Method i25% circ- umference encercumfere50%cir 72
73. 73. Summarizing DEVELOPMENT OF SPHERE • All three views -> CIRCLE.All three views CIRCLE. • Approximate development by dividing h i t i fsphere into a series of zones. Cone FrustumCone, Frustum of cone, Symmetry 73
74. 74. Draw the development of a hemispherical bowl of radius 3 cm by any method12 such developments. 42 6 8 10 4 8 10 6 o a cb d e 1 3 5 42 7 9 O E 1 3 5 7 9 4 2 d a c b o e 4 2 8 10 6 0 This “Lune method” is also T 1 3 5 7 9 d a c b o e 1 3 5 7 9 4 2 8 10 6 4 8 10 6 method” is also known as “Polycylindric method” or 2 OE Rπ = o'F 9 d a c b O E 1 3 5 7 9 4 2 6 d a c b o e 4 2 8 10 6 method or “Gore method”. a' b' c' 1 3 5 7 9 d a c b o e 1 3 5 7 9 4 2 8 10 6 4 8 10 6 For a reasonable accuracy circle c d' e' 7 9 d a c b o e 1 3 5 7 9 4 2 6 d a c b o e 4 2 8 10 6 needs to be divided in 16 or more parts. o 1 3 5 7 9 d a c b o e 1 3 5 7 4 2 8 10 6 8 10 6 74
75. 75. Draw the development of a hemispherical bowl of radius 3 cm by any method o4 o4' o a b c d e o 360 ' 0 ×= oa oa θ T o3 o 360 '1 1 ×= oc bo ob θ o ' o'F a' b' c' o2 o o 360 360 '2 2 × ×= od co oc θ θ o' o1' o2' o3 o 'c d' e' o 360 360 ' 4 3 3 ×= ×= oe do θ θ o a' b' ' o' o1' '4 4 eo c' d' e'
76. 76. PROBLEM: A room is of size L=6.5m, D=5m, H=3.5m. An electric bulb hangs 1m below the center of ceiling. A switch is placed in oneg g p of the corners of the room, 1.5m above the flooring. Draw the projections and determine real distance between the bulb & switch. Ceiling TV Bulb TV Switch DD 76
77. 77. PROBLEM:- A picture frame 2 m wide and 1 m tall is resting on horizontal wall railing makes 350 inclination with wall. It is attached to a hook in the wall by two strings The hook is 1 5 m above wall railing Determine length of eachwall by two strings. The hook is 1.5 m above wall railing. Determine length of each chain and true angle between them 350 Wall railingWall railing 77
78. 78. PROBLEM: Line AB is 75 mm long and it is 300 & 400 Inclined to HP & VP respectively. End A is 12mm above Hp and 10 mm in front of VP. Draw b’ b’1 projections. Line is in 1st quadrant. TL FV a’ θ a X Y Ø 1 LFV TLTV b b1
79. 79. PROBLEM: Line AB 75mm long makes 450 inclination with VP while it’s FV makes 550. End A is 10 mm above HP and 15 mm in front of VP. If line is in 1st b’1b’ quadrant draw it’s projections and find it’s inclination with HP. LOCUS OF b1’ 0 Y a’ 550 X Y a 1 LFV b1 b LOCUS OF b
80. 80. PROBLEM: FV of line AB is 500 inclined to XY and measures 55 mm long while it’s TV is 600 inclined to XY line. If end A is 10 mm above HP and 15 mm in front of VP draw it’s projections find TL inclinations of line with HP & VP b’ b’1 front of VP, draw it’s projections,find TL, inclinations of line with HP & VP. X a’ Y 500 θ X Y a Φ 600 Φ b b1
81. 81. PROBLEM :- Line AB is 75 mm long . It’s FV and TV measure 50 mm & 60 mm long respectively. End A is 10 mm above Hp and 15 mm in front of Vp. Draw b’1 b’ g p y p p projections of line AB if end B is in first quadrant. Find angle with HP and VP. X Y a’ 1’LTVθ X Y a 1 LFV Φ b1 b
82. 82. PROBLEM. Length (L) depth (D) andLength (L), depth (D), and height (H) of a room are 6.5m, 5m and 3.5m especti el An elect ic b lbrespectively. An electric bulb hangs 1m below the center of ceiling. A switch is placed Ceiling TV in one of the corners of the room, 1.5m above the flooring. Bulb g Draw the projections and determine real distance between the bulb and Switch Dbetween the bulb and switch. D
83. 83. PROBLEM: L=6.5m, D=5m, H=3 5m An electric bulbH=3.5m. An electric bulb hangs 1m below the center of ceiling. A switch isof ceiling. A switch is placed in one of the corners of the room, 1.5m above 6.5m the flooring. Draw the projections and determine real distance between the a’ b’ b’1 3.5m 1m real distance between the bulb and switch. a x y 1.5 b5m b5m Answer: a’ b’1
84. 84. PROBLEM: A PICTURE FRAME 2 mPICTURE FRAME, 2 m WIDE & 1 m TALL, RESTING ON HORIZONTAL WALLHORIZONTAL WALL RAILING MAKES 350 INCLINATION WITH WALL. IT IS ATTAACHED TO A 350 A B HOOK IN THE WALL BY TWO STRINGS. THE HOOK IS 1.5 m D THE HOOK IS 1.5 m ABOVE WALL RAILING. DETERMINE LENGTH OF EACH CHAIN AND TRUE Wall railingC ANGLE BETWEEN THEM
85. 85. PROBLEM- A picture frame 2 m wide & 1 m tall is resting onwide & 1 m tall is resting on horizontal wall railing makes 350 inclination with wall. It is attached t h k i th ll b t t i h’ to a hook in the wall by two strings. The hook is 1.5 m above the wall railing.a’b’ (chains) DETERMINE LENGTH OF EACH CHAIN AND TRUE ANGLE BETWEEN a b 1.5 m 1m THEM c’d’ (wall railing) X Y ad h a1 (frame) h (chains) Answers: bcb1 Answers: Length of each chain= hb1 True angle between chains =
86. 86. PROBLEM:- Two mangos on a tree A & B are 1.5 m and 3.0 m above ground and those are 1.2 m & 1.5 m from a 0.3 m thick wall but on opposite sides of it. If thethose are 1.2 m & 1.5 m from a 0.3 m thick wall but on opposite sides of it. If the distance measured between them along the ground and parallel to wall is 2.6 m, Then find real distance between them by drawing their projections. TV B A 0.3M THICK 86
87. 87. PROBLEM:- Two mangos on a tree A & B are 1 5 m and 3 00 b’ tree A & B are 1.5 m and 3.00 m above ground and those are 1.2 m & 1.5 m from a 0.3 m thick wall but on opposite 3.0 a’ pp sides of it. If the distance measured between them along the ground and parallel to wall is 2 6 m Then find 1.5 to wall is 2.6 m, Then find real distance between them by drawing their projections. b 2.6 B 1.5 1 2 a 1.2
88. 88. PROBLEM:-Flower A is 1.5 m & 1 m from walls Q (parallel to reference line) & P (perpendicular to reference line) respectively. Flower is 1.5 m above the ground. Orange B is 3.5m & 5.5m from walls Q & P respectively. Drawing projection, find distance between them If orange is 3.5 m above ground. b’ b’b 1 3.5 m a’ x y 1.5 m Ground a 1.5 m Wall Q B 1 m 3.5 m b 5.5 m Wall P 88
89. 89. PROBLEM :- A top view of object (three rods OA, OB and OC whose endsp j ( , A,B & C are on ground and end O is 100mm above ground) contains three lines oa, ob & oc having length equal to 25mm, 45mm and 65mm respectively. These three lines are equally inclined and the shortest line isrespectively. These three lines are equally inclined and the shortest line is vertical. Draw their projections and find length of each rod. Tv O A C Fv B
90. 90. PROBLEM :- A top view of object (three rods OA, OB and OC whose ends A,B & C are on ground and end O is 100mm above ground) contains three lines oa, ob & oc having length l t 25 45 d 65 ti l Th th li ll i li d d thequal to 25mm, 45mm and 65mm respectively. These three lines are equally inclined and the shortest line is vertical. Draw their projections and find length of each rod. o’ TL1 TL2 a’b’ c’ c1’b1’ a1’ x y a o Answers: b c Answers: TL1 TL2 & TL3
91. 91. PROBLEM:- A pipeline from point A has a downward gradient 1:5 and it runs due South - East. Another Point B is 12 m from A and due East of A and in same l l f i li f 0 f h d i li flevel of A. Pipe line from B runs 150 Due East of South and meets pipeline from A at point C. Draw projections and find length of pipe line from B and it’s inclination with ground. 1 5 Bearing of a LINE: A 1 Bearing of a LINE: Horizontal angle between line & idi ( th thA B12 M E meridian (north south line)----Measured in DEGREES (0 to 90◦). C Measured in Top View. S 45° E. S 15° E.
92. 92. PROBLEM:- A pipe line from point A has a downward gradient 1:5 and it runs due South - East. Another Point B is 12 m from A and due East of A and in same level of A. Pipe line from B runs 150 Due East of South and meets pipe line from A at point C. Draw projections and find length of pipe line from B and it’s inclination with ground. 12m 1 5a’ b’ FV x y N c’2c’ c’1 a b y 450 N EAST W 150 TV c = Inclination of pipe line BC SOUTH = Inclination of pipe line BC
93. 93. PROBLEM: A person observes two objects, A & B, on the ground, from a tower, 15 M high, at the angles of depression 300 & 450 respectively. Object A is due North-West direction of observer and object B is due West direction. Draw projections of situation and find distance of objects from observer and from tower also. OO 300 450 A W S B
94. 94. PROBLEM: A person observes two objects, A & B, on thetwo objects, A & B, on the ground, from a tower, 15 M high, at the angles of depression 300 & 450 Object A is due North-West o’ 300 450 45 . Object A is due North-West direction of observer and object B is due West direction. Draw projections of situation and find a’1 ’ 30 15M projections of situation and find distance of objects from observer and from tower also.Na b’a’ W E ob Answers: S Answers: Distances of objects from observe o’a’1 & o’b’ From tower SFrom tower oa & ob
95. 95. PROBLEM:- A tank of 4 m height is to be strengthened by four rods from each corner by fixing their other ends to the flooring, at a point 1.2 m and 0.7 m from two adjacent walls respectively, as shown. Determine graphically length and angle of each rod with flooring. TV A 4 M B
96. 96. PROBLEM:- A tank of 4 m height is to be strengthened by four rods from h b fi i th i th d t th fl i t i t 1 2 deach corner by fixing their other ends to the flooring, at a point 1.2 m and 0.7 m from two adjacent walls respectively, as shown. Determine graphically length and angle of each rod with flooring. a’ True Length FV True Length Answers: Length of each rod = a’b’1 Angle with Hp. a b’b’1 = X Y a b TVTV
97. 97. SHORTEST DISTANCE BETWEEN POINT and LINE Measure distance between Point P and Line AB for given TV & FV arrangement. aa p b y T F x p p’ ba p A BD a’ b’ ba A B q D pD is ⊥ to ab 97 bq ab
98. 98. a2, b2 Point view of line a2, b2 Shortest distance a2 p2 a1 b1 p2 Point x1 p1 a 1 y1b y p T F x ’ p’ a’ b’
99. 99. Shortest Distance between 2 skew Lines (AB & CD)Shortest Distance between 2 skew Lines (AB & CD) Skew (oblique) lines Lines that are not parallel & do not intersect c •Find T.L. of one of the lines and project its point viewDistance measured along Line ⊥ to both b c p j p using auxiliary plane method Line ⊥ to both. a b d •Project the other line also in each view T F a b •The perpendicular distance between the point view of c p one line and the other line is the required shortest di t b t th t 99 d distance between the two lines
100. 100. Primary auxiliary viewc TL d d, c Secondary auxiliary view c Required distance b c a a b d T F d a b ab B AP c q dP is ⊥ to ab d In mines, this method might be used to locate a connecting tunnel.
101. 101. True Angle between 2 Skew Lines (AB & CD) Draw P.A. V. such that one line (AB) shows its True Length Measure angle in view that Shows both lines in true length Draw S.A.V. view with reference line perpendicular to the True Length of the line (AB) to get the point view of the line a b d T F Draw a tertiary auxiliary view with reference line parallel to the other line in order to get its True Length x y d’ c d Since the secondary auxiliary view had the point view of the first line, the tertiary auxiliary view will have the True L th f th fi t li l a’ d Length of the first line also. c’ b’ 101
102. 102. TRUE LENGTH OF TERTIARY AUXILIARY VIEW ANGLE BETWEEN TWO LINES a3 c3 TRUE LENGTH OF BOTH LINES True Angle b h Parallel LINES b3 2 d3 PRIMARY AUXILIARY VIEW between the two lines a2 ,b2 c2 Point view of one line SECONDARY AUXILIARY VIEWTrue length b1 c1 d2 one line Parallel a1 b d1 T Parallel a c d Angle between two nonintersecting linesT F x y a’ d’ nonintersecting lines is measurable in a view that shows both lines in true shapea’ c’ b’ lines in true shape.
103. 103. Angle between 2 planesAngle between 2 planes Line of intersection of the 2 planes (here it is True Length) e fe f a c d x T F y a’ b’ a b • Obtain an auxiliary view such that the reference line is perpendicular to the True Length of the line of intersection of the l a planes • In this case, the intersection line is parallel to both principle planes and hence is in True L th i b th f t d t i e’ d’ c’ f’ Length in both front and top views • Both planes will be seen as edge views in the auxiliary view. e f’ • The angle between the edge views is the angle between the planes
104. 104. Line of intersection of the 2 planes (here it is True Length) x1 e f f1, e1 PRIMARY AUXILIARY VIEW e f , a c d b1, a1 c1, d1 x •Obtain an auxiliary view such that the reference line is perpendicular to the True Length of the line of intersection of the l T F y y1 a’ b’ a b b1, a1 planes •In this case, the intersection line is parallel to both principle planes and hence is in True L th i b th f t d t i 1 a Length in both front and top views •Both planes will be seen as edge views in the auxiliary view. e’ d’ c’ f’ •The angle between the edge views is the angle between the planes e f’
105. 105. Piercing point of a line with a plane Edge view of the plane True length of T A1 p1 g principal line In a view showing the Line p T g plane as an edge, the piercing point appears where the line Planep’ F where the line intersects the edge view. Principal line Part of the line hidden by the plane h ld b h Draw auxiliary view to get EDGE VIEW. should be shown dotted
106. 106. Find the shortest distance of point P from the body diagonal AB of the cube of side 50 1 mm as shown a1 b1 a1‘, b1’ d’g’ b’ e’ p1 p ’ 10 1 , 1 p’ d g 10 50 b’, e’p1’Required distance Draw an auxiliary view to get the 50Draw an auxiliary view to get the true length of the line Draw an auxiliary view to get the point view of the diagonal c’, d’F, A c,bf,d p point view of the diagonal Project the point P in these views to get the required distance d,ea,g