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# Mel110 part3

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• 1. ENGINEERING CURVES Point d P i t undergoing two types of displacements i t t f di l tINVOLUTE: Locus of a free end of a string when it is gwound round a (circular) poleCYCLOID: Locus of a point on th periphery of aCYCLOID L f i t the i h fcircle which rolls on a straight line path.SPIRAL: Locus of a point which revolves around afixed point and at the same time moves towards it. pHELIX: Locus of a point which moves around thesurface of a right circular cylinder / cone and at the f f i ht i l li d d t thsame time advances in axial direction at a speedbearing a constant ratio to the speed of rotation. rotation
• 2. INVOLUTE OF A CIRCLE• Problem: P3 P1Draw Involute ofa circle. Stringlength is equal tothe P4 4 to p 4circumference of 3 5circle. i l 2 6 1 7 A 8 P5 P P8 1 2 3 4 5 6 7 8 P7 P6 πD
• 3. Problem: Draw Involute of a circle. String length is MOREthan the circumference of circle. INVOLUTE OF A CIRCLE String length MORE than πD Solution Steps: P2 In this case string length is more than Π D. But remember! Whatever may be the length of P3 P1 string, mark Π D distance horizontal i.e.along the string and divide it in 8 number of d d v de u be o equal parts, and not any other distance. Rest all steps are same as previous INVOLUTE. Draw the curve completely. 4 to p P4 4 3 5 2 6 1 P5 7 8 p8 P 1 2 3 4 5 6 7 8 P7 165 mm P6 (more than πD) πD
• 4. Problem: Draw Involute of a circle.String length is LESS than the circumference of circle. INVOLUTE OF A 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 d d v de u be oequal parts, and not any otherdistance. Rest all steps are sameas previous INVOLUTE. Drawthe curve completely. 4t p to P4 4 3 5 2 6 1 P5 7 P 8 P7 1 2 3 4 5 6 7 8 P6 150 mm (Less than πD) πD
• 5. INVOLUTE OF A PENTAGON l 5l Thread should be taut 5
• 6. Problem : A pole is of a shape of INVOLUTE OFhalf hexagon (side 30 mm) and g ( ) COMPOSIT SHAPED POLEsemicircle (diameter 60 mm). Astring is to be wound havinglength equal to the pole perimeter g q p p Calculate perimeter lengthdraw path of free end P of string P1when wound completely. P P2 1 to P P3 3 to P 3 4 2 5 1 6 A 1 2 3 4 5 6 P P4 πD/2 P6 P5
• 7. PROBLEM : Rod AB 85 mm long rolls over a semicircular pole without slipping from it’sinitially vertical position till it becomes up-side-down vertical. Draw locus of both ends A & B. B A4 4OBSERVE ILLUSTRATION CAREFULLY! B1when one end is approaching, A3other end will move away from y 3poll. πR 2 A2 B2 2 1 3 1 A1 4 A B3 B4
• 8. DEFINITIONS SUPERIOR TROCHOID: If the point in the definition ofCYCLOID: cycloid is outside the circleLOCUS OF A POINT ON THEPERIPHERY OF A CIRCLE WHICH INFERIOR TROCHOID.:ROLLS 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 the other circle,
• 9. PROBLEM: Draw locus (one cycle) of a point (P) on the periphery of a CYCLOIDcircle (diameter=50 mm) which rolls on straight line path. 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 Point C (zero radius) will not rotate and it will traverse on straight line.
• 10. PROBLEM: Draw locus of a point (P), 5 mm away from theperiphery of a Circle (diameter=50 mm) which rolls on straight SUPERIOR TROCHOIDline path. Using 2H Using H i 4 p4 p3 p5 3 5 p2 C C1 C C3 C4 C5 C6 C7 C8 p62 6 2 p7 1 p1 7 P πD D p8
• 11. PROBLEM: Draw locus of a point , 5 mm inside the periphery of a INFERIORCi l which rolls on straight line path. T k circle diameter as 50Circle hi h ll i h li h Take i l di t TROCHOIDmm p4 4 p p5 3 5 p2 3 C C1 C2 C3 C4 C5 C6 C7p6 C8 2 6 p1 p7 1 7 P p8 πD
• 12. CYCLOIDSUPERIOR TROCHOID INFERIOR TROCHOID 12
• 13. EPI CYCLOID :PROBLEM: Drawlocus of a point on the periphery of a circle(dia=50mm) which rolls on a curved path (radius 75 mm). Distance by smaller circle = Distance on larger circleSolution 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 θ θ.2. Calculate θ by formula θ = (r/R) x 360°.3. Construct a sector with angle θ and radius R.4.4 Divide this sector into 8 number of equal angular parts. parts
• 14. EPI CYCLOID Generating/ Rolling Circle 4 5 C2 3 6EPI-CYCLOID 7If the circle is rolling on 2another circle from outside 1 P r = CP Directing Circle = r 3600 R O
• 15. PROBLEM : Draw locus of a point on the periphery of a circle whichrolls from the inside of a curved path Take diameter of Rolling circle path. HYPO CYCLOID50 mm and radius of directing circle (curved path) 75 mm. P 7 P1 6 1 P2 P3 5 2 4 P4 3 P8 P5 P6 P7 r = 3600 R O OC = R ( Radius of Directing Ci l ) R di f Di ti Circle) CP = r (Radius of Generating Circle)
• 16. CYCLOID SUPERIOR TROCHOIDINFERIORTROCHOID 16
• 17. Problem: Draw a spiral of one convolution. Take distance PO 40 mm. SPIRAL IMPORTANT APPROACH FOR CONSTRUCTION!FIND TOTAL ANGULAR AND TOTAL LINEAR DISPLACEMENTAND DIVIDE BOTH IN TO SAME NUMBER OF EQUAL PARTS. 2 P2 3 1 P1 P3 4 P4 O P 7 6 5 4 3 2 1 P7 P5 P6 5 7 6
• 18. SPIRAL of two convolutionsProblem: Point P is 80 mm from point O. It starts movingtowards O and reaches it in two revolutions around. It Draw locusof point P (To draw a Spiral of TWO convolutions). 2,10 P2 3,11 P1 1,9 P3 P10 P9 P11 P4 16 13 10 8 P7 6 5 4 3 2 1 P 8 8,16 , 4,12 4 12 P12 P15 P13 P14 P7 P5 P6 5,13 7,15 6,14
• 19. Problem: A link 60 mm long, swings on a point O ob e : o g, sw gs o pofrom its vertical position of the rest to the left through60° and returns to its initial position at uniform pvelocity. During that period a point P moves atuniform speed along the center line of the link from O p gat reaches the end of link. Draw the locus of P. O, P N M
• 20. PROBLEM: Draw a helix of one HELIXconvolution, upon a cylinder. Given 80 mm (UPON A CYLINDER) P8pitch and 50 mm diameter of a cylinder. 8 P7 7 P6 6 P5 5 Pitch: Axial advance during one 4 P4 complete revolution . 3 P3 2 P2 1 P1 P F 6 T 7 5 P 4 1 3 2
• 21. PROBLEM: Draw a helix of one convolution, P8 HELIXupon a cone, diameter of base 70 mm, axis p , , ) (UPON A CONE) ( P790 mm and 90 mm pitch. P6 P5 P4 P3 P2 P1 X P Y 6 7 5 P6 P5 P7 P4 P 4 P8 P1 P3 1 3 P2 2
• 22. Interpenetration of Solids / I i f S lid Intersection of Surfaces / Lines & Curves of IntersectionMore points common to both the solidsBasic required knowledge:~ Projections of Solid~ Section of Solid 22
• 23. Projections of Solid .Problem: A cone 40 mm diameter and 50 mm axis isresting on one generator on Hp which makes 300inclination with Vp. Draw it’s projections. More number of generators Better approximation. o’ a’1 h’1 b’1 F g’1 f’1 c’ X a’ h’b’ c’ g f’ d’ e’ o’ e’1 d’1 1 o1 Y 30 ’ g g1 g1 o1 h f1 h1 h1 T f f1 a1 a o e e1 a1 o1 e1 b1 1st. Angle b d d1 b1 d1 23 c c1 c1
• 24. Section of Solid For TV SECTION PLANE I Li Lines & CurvesSolids t C Interpenetration of S lid t ti f x y of Intersection Intersection of Surfaces Apparent Shape of section SECTION LINES (450 to XY) SECTIONAL T.V.
• 25. 25
• 26. 26
• 27. 27
• 28. How to make theseshapes
• 29. CURVES OF INTERSECTIONS are shown by WHITE ARROWS.Machine component having Intersection of a Cylindrical ytwo intersecting cylindrical Industrial Dust collector. main and Branch Pipe.surfaces with the axis at acute angle to each other. Intersection of two cylinders. WHEN TWO OBJECTS ARE TO BE JOINED TOGATHER, MAXIMUM SURFACE CONTACT BETWEEN BOTH BECOMES A BASIC REQUIREMENT FOR Pump lid having shape of a STRONGEST & LEAK-PROOF h LEAK PROOF hexagonal Prism and lP i dFeeding Hopper. JOINT. Hemi-sphere intersecting each other.
• 30. Minimum Surface Contact. ( Point Contact) (Maximum Surface Contact) Lines of Intersections. Li fI i Curves of Intersections. C fI iSquare Pipes. Circular Pipes. Square Pipes. Circular Pipes. MAXIMUM SURFACE CONTACT BETWEEN BOTH BECOMES A BASIC REQUIREMENT FOR STRONGEST & LEAK-PROOF JOINT. Q Two plane surfaces (e-g. faces of prisms and pyramids) intersect in a straight line. line The line of intersection between two curved surfaces (e-g. of cylinders and cones) or between a plane surface and a curved surface is a curve. More points common to both the solids 30
• 31. How to find Lines/Curves of IntersectionGenerator line Method Cutting Plane Method T T F F 75 31 75
• 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 p p p g positions in otherviews. When one solid completely penetrates another, there will be two curves of intersection. 32
• 33. Problem: A cylinder 50mm dia. & 70mm axis is completely penetrated by anotherof 40 mm dia. & 70 mm axis horizontally. Both axes intersect & bisect each other.Draw projections showing curves of intersections intersections. 1’ 2’ 4’ 3’ 4” 1”3” 2” a’ a” b ’h’ h” b” c’g’ g” c” df d’f’ f” d d” a’ e”X Y 4 1 3 33 2
• 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 prism are equally i li d t H D i ll inclined to Hp. Draw projections showing curves of i t j ti h i f intersections. ti 1’ 2’ 4’ 3’ 4” 1”3” 2” a” d” b” Projections of j Find lines ofof Solid 1. critical Solid2. Missing intersection c” points at whichX Y curve changes direction. direction 4 1 3 34 2
• 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 Vp. intersections. 1’ 2’4’ 3’ 4” 1”3” 2” a’ a’ a” b’ b’ d” b” d’ d’ c’ c’ c”X Y 4 Preference of object line over dash line… line 1 3 2 35
• 36. Problem. SQ.PRISM (30 mm base sides and 70mm axis ) STANDING & SQ.PRISM (25mm sides and 70 mm axis) PENETRATING. Both axes Intersect & bisect each other. Twofaces of penetrating prism are 300 inclined to Hp. Draw p j p gp p projections showing curves of gintersections in I angle projection system. 1’ 2’4’ 3’ 4” 1”3” 2” a’ f’ f” e’ b’ c’ c” d d’ X 300 Y 4 1 3 d" o 2 Other possible arrangements ???? b" 36
• 37. Problem: A vertical cylinder 50mm dia. and 70mm axis is completely penetrated bya triangular prism of 45 mm sides and 70 mm axis, horizontally. One flat face ofprism is parallel to Vp and Contains axis of cylinder Draw projections showing cylinder.curves of intersections in I angle projection system. 1’ 2’4’ 3’ 4” 1”3” 2” a a a b b b c c d e e d e f f f X Y 4 1 3 2 37
• 38. Problem: A vertical cone, base diameter 75 mm and axis 100 mm long, is completelypenetrated by a cylinder of 45 mm diameter and axis 100 mm long. The axis of the cylinder isparallel to Hp and Vp and intersects axis of the cone at a point 28 mm above the base. Drawprojections showing curves of intersection in FV & TV. in I angle projection system o’ o” 1 1 2 8,2 8 1 2 7 3 3 3 7, 64 6 4 4 5 28 5 5X a’ b’h’ c’g’ d’f’ e’ g” f”h” a”e” b”d” c” Y g h f a e b d 38 c
• 39. Problem: Vertical cylinder (80 mm diameter & 100 mm height) is completely penetrated by ahorizontal cone (80 mm diameter and 120 mm height). Both axes intersect & bisect each other.Draw projections showing curve of intersections in I angle projection system. system 7 7’ 6’ 8’ 1’ 5’ 2’ 4’ 3’X Y 1 28 Intersection of a curve with 37 another !!!! Generator lines.. 46 5 39
• 40. Problem: CONE STANDING & SQ.PRISM PENETRATING (BOTH AXES VERTICAL) 2’ 1’ 3’ 5’ 4’ 6’X a bh a’ b’h’ cg c’g’ df d’f’ e e’ Y g h 8 f 9 7 10 6 Problem: A cone70 mm base diameter and 90 mm axis a 1 e is completely penetrated by a square prism from top 2 with it’s axis // to cone’s axis and 5 mm away from it. 3 5 A vertical plane containing both axes is parallel to Vp. 4 Take all faces of sq.prism equally inclined to Vp. b d Base Side of prism is 30 mm and axis is 100 mm long.long c Draw projections showing curves of intersections. 40 5 mm OFF-SET
• 41. Intersection of two cylindersoblique to each other –Use PAV Use III angle projection 41
• 42. Intersection of two cylindersoblique to each other –Use AV Use 42
• 43. Intersection of Cone and Oblique cylinder using PAV 43
• 44. Intersection of irregular Prism & g offset Cylinder III angle projection 44
• 45. Intersection of irregular Prism & g offset Cylinder Invisible Visible 45
• 46. Intersection of irregular Prism & g offset Cylinder Invisible Visible 46
• 47. Intersection of irregular Prism & g offset Cylinder 47
• 48. DEVELOPMENT OF SURFACES OF SOLIDS Solids are bounded by geometric surfaces:LATERLAL SURFACE 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. solid, rolled, solid - Plane Prism, Pyramid - Single curved Cone, Cone Cylinder - Double curved sphere 48
• 49. ExamplesPrism – Made up of same number of rectangles as sides of the baseOne side: Height of the p g prismOther side: Side of the baseCylinder – RectangleOne side: Height of the cylinderOther side: Circumference of the base hPyramid – Number of triangles in contact πdThe base may be included T. L.if present 49
• 50. Methods to Develop Surfaces p1. Parallel-line development: Used for prisms (full or truncated), cylinders (full c linders (f ll or tr ncated) Parallel lines are dra n along the truncated). drawn surface and transferred to the development Cylinder: A Rectangle H πD D H= Height D= base diameter g Prisms: No.of Rectangles H S S H= Height S = Edge of base 50
• 51. Ex: DOTTED LINES are never shown on development 51
• 52. 4, d 2, bComplete development of c be c t b c tting plane (inclined to HP at de elopment cube cut by cutting30 degrees and perpendicular to VP) 52
• 53. 4, d2, b 53
• 54. φ50 j Cylinder cut by k l i h three planes a g b f T c e d F G g F H 45o f E I e d D J c c’ C K I b A B A 15o a100 1 71 6 7 8 51 6 5 9 41 3 4 10 31 11 12 21 2 11 1 1 54 πx50
• 55. Problem: Development of a solid is a parabolawith a 180 mm base and a 90 mm height. Draw theprojections of solid solid. 180=π D D=57.3 mm 55
• 56. Methods to Develop Surfaces p1. Parallel-line development2. 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 L= Slant edge. edge L=Slant height. S = Edge of base θ = R 3600 L 56
• 57. Parallel vs Radial line method Parallel line method Radial line method 57
• 58. FRUSTUMS DEVELOPMENT OF DEVELOPMENT OF FRUSTUM OF CONE FRUSTUM OF SQUARE PYRAMID Base side Top side θ θ = R 3600 L L= Slant edge of pyramid R= Base circle radius of cone L1 = Slant edge of cut part. L= Slant height of cone L1 = Slant height of cut part. Important points. points.1. Development is a shape showing AREA, means it’s a 2-D plain drawing. 2-2.2 All dimensions of it must be TRUE dimensions. dimensions3. As it is representing shape of an un-folded sheet, no edges can remain hidden un- and hence DOTTED LINES are never shown on development.
• 59. Development of Sphere usingFrustum of Cones: Zone Method Zone 1: Cone Zone 2: Frustum of cone Zone 3: Frustum of cone Zone 4: Frustum of cone Z 4 F t f θ = R 3600 L θ 59
• 60. Development by Radial MethodPyramids (full or Truncated) &Cones (full or Truncated) Truncated). If 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. Ex: 61
• 62. Development of Oblique Objects• Right regular objects – Axis of object perpendicular to base.• Axis of any regular object (prism (prism, pyramid, cylinder, cone, etc.) inclined at angle other than right angle – Oblique OBJECT. Use ARC method. 62
• 63. Ob que p s Oblique prism e dg f j c h a i b Parallel f’ a b’ c a b f fg h’ i j g h i g 63
• 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 φ30 g G A • Draw projection lines from top edge of cylinder a F B such that they are perpendicular to end generator E D C T • Mark distances AB, BC etc. from one projector line to the next to complete the profile F G’ A’ A • Do the similar process for the bottom edge B C50 A G 75o g’ a a A1 A1 g a 64 A1
• 65. Oblique Cone 65
• 66. Intersection of Plane & Pyramid. y 4 1 D Development of C Develop B 0 D 1-D-A-2-1 resulting lateral g 1 2-A-B-3-2 B 3-B-C-4-3 A truncated Pyramid2 3 1-D-C-4-1 2 1 X 0 Y B’B B 4 A D’ o 1’ 3 23 66 2
• 67. Methods to Develop Surfaces p1. Parallel-line development:2. Radial-line development:3. Triangulation development: Complex shapes are divided into a number of triangles and transferred into the development EXAMPLES:- EXAMPLES:- Tetrahedron: Four Equilateral Triangles Boiler Shells & chimneys, Pressure Vessels, Shovels, Trays Vessels Shovels Trays, Boxes & Cartons, Feeding All sides Hoppers, Large Pipe sections, equal in length Body P t f t B d & Parts of automotives, ti Ships, Aero planes. 67
• 68. Connect two hollow objects having different base. Transition Three surfaces Pieces Triangulation Method: Dividing a surface into a number of triangles and transfer them to the development. development 68
• 69. Ex: In air conditioning system, a square duct of 50mm by 50mm is connected toanother square duct of 25mm by 25 mm by using a connector (transition piece) ofheight 25mm Draw development of lateral surface of the connector (Neglect 25mm.thickness of connector). Pyramids: (No.of triangles) O O’ a’ A L= Slant edge. S = Edge of base g b’ B b a O 69 c
• 70. Development of Transition Piece for D l t fT iti Pi f Difference Shapes and Sizes p• Connect a Square pipe with circular pipe.• Ex: Imagine a transition piece (height = 25) to connect a chimney of square cross section 50mm * 50 mm to circular pipe of 30mm diameter Draw the projections and diameter. develop lateral surface of the transition piece. 70
• 71. 2’, 8’ 1’ a’, b’ 1/8 ofc b circumference i f 3 4 1 2 1 5 6 8 A B 7d a 8 1 I angle projection p ojection 71
• 72. Development ofSphere/Hemisphereusing Lune Method 25% circ- i umference ence rcumfere 50% cir 72
• 73. Summarizing DEVELOPMENT OF SPHERE • All three views -> CIRCLE. • Approximate development by dividing sphere i t a series of zones. h into i fCone,Cone Frustumof cone,Symmetry 73
• 74. 12 such developments.Draw the development of a hemispherical bowl of radius 3 cm by any method 10 6 8 10 8 2 4 6 4 1 2 a 10 e 9 10 e 9 10 e 9 10 e 9 10 e 9 10 E 9 10 e 9 10 e 9 10 E b c de O o 1 3 2 3 c 65 7 57 9 8 This “Lune 4 d b 4 3 b a2 1 a o πR method” is also 0 OE = 5 8 known as 7 c 6 2 d T “Polycylindric o a2 1 a2 1 a2 1 b 4 3 b 4 3 b 4 3 c 65 c 65 c 65 method method” or 8 7 d F “Gore method”. O o 8 a 7 b d c c o 8 d 7 e For a reasonable d o accuracy circle 1 a2 1 a2 1 a2 1 3 b 43 b 4 3 b 43 65 c 65 c 65 c 65 8 7 needs to be d o divided in 16 or 8 7 d more parts. o 8 7 74 d o 8 7
• 75. Draw the development of a hemispherical bowl of radius 3 cm by any method o4 o4 o a b c d e oa θ0 = × 360o oa ob θ1 = × 360o o3 o1b T oc o3 F o2 o θ2 = × 360o o2 c o2 a b od o1 c c θ3 = × 360o o o o1 d o3 d e a o oe b θ4 = × 360o c o4 e d e
• 76. PROBLEM: A room is of size L=6.5m, D=5m, H=3.5m. An electricbulb hangs 1m below the center of ceiling. A switch is p g g placed in oneof the corners of the room, 1.5m above the flooring. Draw theprojections and determine real distance between the bulb & switch. Ceiling TV Bulb Switch D 76
• 77. PROBLEM:- A picture frame 2 m wide and 1 m tall is resting on horizontalwall railing makes 350 inclination with wall. It is attached to a hook in thewall by two strings. The hook is 1.5 m above wall railing. Determine length of each strings 15 railingchain and true angle between them 350 Wall railing 77
• 78. PROBLEM: Line AB is 75 mm long and it is 300 & 400 Inclined to HP & VPrespectively. End A is 12mm above Hp and 10 mm in front of VP. Drawprojections. Line is in 1st quadrant. b’ b’1 FV TL θ a’ X Y a LFV Ø 1 TV TL b b1
• 79. PROBLEM: Line AB 75mm long makes 450 inclination with VP while it’s FVmakes 550. End A is 10 mm above HP and 15 mm in front of VP. If line is in 1stquadrant draw it’s projections and find it’s inclination with HP. b’ b’1 LOCUS OF b1’ 550 a’ X Y a LFV 1 LOCUS OF b b b1
• 80. PROBLEM: FV of line AB is 500 inclined to XY and measures 55 mm long whileit’s TV is 600 inclined to XY line. If end A is 10 mm above HP and 15 mm infront of VP, draw it’s projections,find TL, inclinations of line with HP & VP. VP projections find TL VP b’ b’1 500 θ a’ X Y a Φ 600 b1 b
• 81. PROBLEM :- Line AB is 75 mm long . It’s FV and TV measure 50 mm & 60 mmlong respectively. End A is 10 mm above Hp and 15 mm in front of Vp. Draw g p y p pprojections of line AB if end B is in first quadrant. Find angle with HP and VP. b’ b’1 θ LTV 1’ a’ X Y a LFV 1 Φ b1 b
• 82. PROBLEM.Length (L) depth (D) and (L), (D),height (H) of a room are6.5m, 5m and 3.5mrespectively. An elect ic bulb especti el electric b lbhangs 1m below the center Ceilingof ceiling. A switch is placed TVin one of the corners of the Bulbroom, 1.5m above theflooring. gDraw the projections anddetermine real distance Switchbetween the bulb and Dswitch.
• 83. PROBLEM: L=6.5m, D=5m, H=3.5m. H=3 5m An electric bulb hangs 1m below the center of ceiling. A switch is 6.5m placed in one of the corners of the room, 1.5m above 1m b’ b’1 the flooring. Draw the3.5m projections and determine a’ real distance between the 1.5x y bulb and switch. a 5m b Answer: a’ b’1
• 84. PROBLEM: APICTURE FRAME 2 m FRAME,WIDE & 1 m TALL,RESTING ONHORIZONTAL WALLRAILINGMAKES 350INCLINATION WITH 350WALL. IT IS AATTAACHED TO A BHOOK IN THE WALLBY TWO STRINGS. DTHE HOOK IS 1.5 mABOVE WALL RAILING.DETERMINE LENGTH OF C Wall railingEACH CHAIN AND TRUEANGLE BETWEEN THEM
• 85. PROBLEM- A picture frame 2 m wide & 1 m tall is resting on horizontal wall railing makes 350 h’ inclination with wall. It is attached to h k i th t a hook in the wall by two strings. ll b t t i (chains) The hook is 1.5 m above the wall ab a’b’ railing. 1.5 m DETERMINE LENGTH OF EACH 1m CHAIN AND TRUE ANGLE BETWEEN THEM c’d’ (wall railing)X Y a1 ad (frame) h (chains) Answers: b1 bc Length of each chain= hb1 True angle between chains =
• 86. PROBLEM:- Two mangos on a tree A & B are 1.5 m and 3.0 m above ground andthose are 1.2 m & 1.5 m from a 0.3 m thick wall but on opposite sides of it. If thedistance 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 0.3M THICK A 86
• 87. PROBLEM:- Two mangos on a tree A & B are 1.5 m and 3.00 15 3 00 b’ m above ground and those are 1.2 m & 1.5 m from a 0.3 m thick wall but on opposite pp sides of it. If the distance a’ measured between them 3.0 along the ground and parallel to wall is 2.6 m, Then find 26m 1.5 real distance between them by drawing their projections. 2.6 bB 1.51.212 a
• 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’ Ground 1.5 mx y Wall Q B 1.5 m a 3.5 m 1m b 5.5 m Wall P 88
• 89. PROBLEM :- A top view of object ( p j (three rods OA, OB and OC whose ends ,A,B & C are on ground and end O is 100mm above ground) contains threelines oa, ob & oc having length equal to 25mm, 45mm and 65mmrespectively. These three lines are equally inclined and the shortest line isvertical. Draw their projections and find length of each rod. Tv O C A Fv B
• 90. PROBLEM :- A top view of object (three rods OA, OB and OC whose ends A,B & C are onground and end O is 100mm above ground) contains three lines oa, ob & oc having lengthequal to 25mm, 45 l t 25 45mm and 65d 65mm respectively. Th ti l These th three li lines are equally inclined and the ll i li d d thshortest line is vertical. Draw their projections and find length of each rod. o’ TL2 TL1 b1’ b’ a’ a1’ c’ c1’ x y a o Answers: TL1 TL2 & TL3 b c
• 91. PROBLEM:- A pipeline from point A has a downward gradient 1:5 and it runsdue South - East. Another Point B is 12 m from A and due East of A and in samelevel of A. Pipe line from B runs 150 Due East of South and meets pipeline fl l f i li f f h d i li fromA at point C. Draw projections and find length of pipe line from B and it’sinclination with ground. 5 1 Bearing of a LINE: Horizontal angle between line & A 12 M meridian ( idi (north south th th B E line)----Measured in DEGREES (0 to 90◦). Measured in Top View. S 45° E. C S 15° E.
• 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 a’ 5 b’ 1FV c’ c’1 c’2 x y NW b a 450 EAST 150TV c = Inclination of pipe line BC SOUTH
• 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 dueNorth-West direction of observer and object B is due West direction. Drawprojections of situation and find distance of objects from observer and fromtower also. O 300 450 A S B W
• 94. PROBLEM: A person observes two objects, A & B, on the ground, from a tower, 15 M high, o’ at the angles of depression 300 & 450 300 450. Object A is due North-West 15M direction of observer and object a’1 B is due West direction. Draw a’ ’ b’ projections of situation and find distance of objects from a N observer and from tower also.W E b o Answers: Distances of objects from observe o’a’1 & o’b’ From tower S oa & ob
• 95. PROBLEM:- A tank of 4 m height is to be strengthened by four rods from each corner byfixing their other ends to the flooring, at a point 1.2 m and 0.7 m from two adjacent wallsrespectively, as shown. Determine graphically length and angle of each rod with flooring. TV A 4M B
• 96. PROBLEM:- A tank of 4 m height is to be strengthened by four rods fromeach corner b fi i th i other ends t th fl i h by fixing their th d to the flooring, at a point 1 2 m and t i t 1.2 d0.7 m from two adjacent walls respectively, as shown. Determinegraphically length and angle of each rod with flooring. FV a’ True Length Answers: Length of each rod = a’b’1 Angle with Hp. = X b’1 Y b’ a b TV
• 97. SHORTEST DISTANCE BETWEEN POINT and LINE Measure distance between Point P and Line AB for given TV & FV arrangement. a b p T x y F p p’ a A b D B a’ pD is ⊥ to q ab b b’ 97
• 98. Point view of line a2, b2 b1 Shortest distance a2 p2 a1 p2 Point p1x1 a b y1 pxT y F p’ a’ ’ b’
• 99. Shortest Distance between 2 skew Lines (AB & CD) Skew (oblique) lines Lines that are not parallel & do not intersect •Find T.L. of one of the lines Distance measured alongc Line ⊥ to both. both and project its p p j point view using auxiliary plane b methoda d •Project the other line also T F in each view b •The perpendicular distancea between the point view of pc one line and the other line is the required shortest d distance b t di t between the two th t lines 99
• 100. c Primary auxiliary view TL Secondary auxiliary view d d, c Required distancec a ba d T d b Fa b a B P Ac dP is ⊥ to q ab d In mines, this method might be used to locate a connecting tunnel.
• 101. True Angle between 2 Skew Lines (AB & CD) Measure angle in view that Draw P.A. V. such that one line (AB)Shows both lines in true length shows its True Length b Draw S.A.V. view with reference line perpendicular to the True Length of the a line (AB) to get the point view of the line d T c Draw a tertiary auxiliary view withx y reference line parallel to the other line in order to get its True Length F d d’ Since the secondary auxiliary view had a’ the point view of the first line, the tertiary auxiliary view will have the True Length of th first line also. L th f the fi t li l c’ b’ 101
• 102. c3 a3 TERTIARY AUXILIARY VIEWANGLE BETWEEN TWOLINES TRUE LENGTH OF Parallel BOTH LINES b3 d3 True Angle PRIMARY 2 c2 between the two b hAUXILIARY VIEW lines a2 ,b2 b1 d2 True length SECONDARY AUXILIARY VIEWc1 Point view of one line a1 d1 b Parallel a d Angle between two T c nonintersecting lines x y is measurable in a F d’ view that shows both a’ lines in true shape. shape c’ b’
• 103. Angle between 2 planes Line of intersection of the 2 planes (here it is True Length) e f d c ax T b • Obtain an auxiliary view such that the y reference line is perpendicular to the True F a a’ b’ Length of the line of intersection of the planes l • In this case, the intersection line is parallel to both principle planes and hence is in True c’ Length i both front and top views L th in b th f t dt i d’ • Both planes will be seen as edge views in the auxiliary view. e e’ f’ • The angle between the edge views is the angle between the planes
• 104. Line of intersection of the 2 planes (here it is True Length) x1 e f f1, e1 , PRIMARY AUXILIARY VIEW d c c1, d1 a b1, a1x T b •Obtain an auxiliary view such that the y reference line is perpendicular to the True F a a’ b’ y1 1 Length of the line of intersection of the planes l •In this case, the intersection line is parallel to both principle planes and hence is in True c’ Length i both front and top views L th in b th f t dt i d’ •Both planes will be seen as edge views in the auxiliary view. e e’ f’ •The angle between the edge views is the angle between the planes
• 105. Piercing point of a line with a plane Edge view of the plane T A1 True length of g principal line p1 In a view showing the g p plane as an edge, theLine piercing point appears T where the line F intersects the edge p’ Plane view. Draw auxiliary view to Part of the line get EDGE VIEW. Principal line hidden by the plane should b shown h ld be h dotted
• 106. Find the shortest distance of point P from the body diagonal AB of the cube of side 50 mm as shown b1 a1 a1‘, b1’ , p1 1 10 p1’ dg d’g’ b’, b’ e’Requireddistance 10 p’ Draw an auxiliary view to get the 50 true length of the line Draw an auxiliary view to get the point view of the diagonal F, A c’, d’ Project the point P in these views to f,d c,b get the required distance p a,g d,e