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Assignment & Exam.

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BHARAT 2050 INFO SOLUTIONS VASUDEVAM KUTAMBAKAM

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NIT, Nagpur.Prof Yugal Video For Reference.

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NIT, Nagpur.Prof Yugal Video For Reference.

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Engineering Graphics is the language of Engineers. The concept of graphical language are used in express ideas, conveying the instruments, which are used in ...

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IIT MADRAS

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Production Drawing (IIT MADRAS)

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Ex_3

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CASE 4. SQ.PRISM STANDING & SQ.PRISM  PENETRATING

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https://youtu.be/Fad4WXyldGo

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Intersection of Solids

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Intersection of Solids.

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Development of Surface.

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Development of Surface.

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Bharat 2050 Info Solutions Vasudevam Kutambakam

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BHARAT

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Problem 3

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How to Draw an Isometric Drawing - First Year Engineering Drawing - Isometric Projections

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Isometric View

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Orthographic Projections

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An useful slideshow for all 1st year engineering students, who are afraid of the subject : ENGINEERING GRAPHICS..

learn, enjoy~

1. 1. DRAWINGS: ( A Graphical Representation) The Fact about: If compared with Verbal or Written Description,Drawings offer far better idea about the Shape, Size & Appearance of any object or situation or location, that too in quite a less time. Hence it has become the Best Media of Communication not only in Engineering but in almost all Fields.

2. 2. Drawings (Some Types) Portraits Botanical Drawings ( human faces,Nature Drawings ( plants, flowers etc.) expressions etc.)( landscape, Geographicalscenery etc.) Zoological Drawings Drawings Engineering Drawings, (creatures, animals etc.) ( maps etc.) (projections.) Building Related Drawings. Machine component Drawings Orthographic Projections Isometric ( Mech.Engg.Term.) (Fv,Tv & Sv.-Mech.Engg terms) or Perspective(Civil Engg.Term) (Plan, Elevation- Civil Engg.terms) (Working Drawings 2-D type) (Actual Object Drawing 3-D)

3. 3. ORTHOGRAPHIC PROJECTIONS:IT IS A TECHNICAL DRAWING IN WHICH DIFFERENT VIEWS OF AN OBJECT ARE PROJECTED ON DIFFERENT REFERENCE PLANES OBSERVING PERPENDICULAR TO RESPECTIVE REFERENCE PLANE Different Reference planes are Horizontal Plane (HP), Vertical Frontal Plane ( VP ) Side Or Profile Plane ( PP) And Different Views are Front View (FV), Top View (TV) and Side View (SV) FV is a view projected on VP. TV is a view projected on HP. SV is a view projected on PP. IMPORTANT TERMS OF ORTHOGRAPHIC PROJECTIONS: 1 Planes. 2 Pattern of planes & Pattern of views 3 Methods of drawing Orthographic Projections

4. 4. PLANES1 PRINCIPAL PLANES HP AND VP AUXILIARY PLANES Auxiliary Vertical Plane Auxiliary Inclined Plane Profile Plane (A.V.P.) (A.I.P.) ( P.P.) < A.I A.V.P. .P. ⊥t oV ⊥ to Hp & ∠ to Vp p ∠ & to Hp <

5. 5. PATTERN OF PLANES & VIEWS (First Angle Method) 2 THIS IS A PICTORIAL SET-UP OF ALL THREE PLANES. ARROW DIRECTION IS A NORMAL WAY OF OBSERVING THE OBJECT. BUT IN THIS DIRECTION ONLY VP AND A VIEW ON IT (FV) CAN BE SEEN. THE OTHER PLANES AND VIEWS ON THOSE CAN NOT BE SEEN. Y PROCEDURE TO SOLVE ABOVE PROBLEM:- TO MAKE THOSE PLANES ALSO VISIBLE FROM THE ARROW DIRECTION, A) HP IS ROTATED 900 DOUNWARDX B) PP, 900 IN RIGHT SIDE DIRECTION. THIS WAY BOTH PLANES ARE BROUGHT IN THE SAME PLANE CONTAINING VP. Click to view Animation On clicking the button if a warning comes please click YES to continue, this program is safe for your pc. VP PP Y FV LSV X Y X TV HP ACTUAL PATTERN OF PLANES & VIEWS HP IS ROTATED DOWNWARD 900 PP IS ROTATED IN RIGHT SIDE 90 0 OF ORTHOGRAPHIC PROJECTIONS AND AND DRAWN IN BROUGHT IN THE PLANE OF VP. BROUGHT IN THE PLANE OF VP. FIRST ANGLE METHOD OF PROJECTIONS

6. 6. 3 Methods of Drawing Orthographic ProjectionsFirst Angle Projections Method Third Angle Projections Method Here views are drawn Here views are drawn by placing object by placing object in 1st Quadrant in 3rd Quadrant.( Fv above X-y, Tv below X-y ) SYMBOLIC ( Tv above X-y, Fv below X-y ) PRESENTATION OF BOTH METHODS WITH AN OBJECT STANDING ON HP ( GROUND) TV FV ON IT'S BASE. X Y NOTE:- X Y HP term is used in 1 Angle method st & FV TV For the same G L Ground term is used in 3rd Angle method of projections

7. 7. FIRST ANGLE FOR T.V. PROJECTION IN THIS METHOD,THE OBJECT IS ASSUMED TO BESITUATED IN FIRST QUADRANT MEANS ABOVE HP & INFRONT OF VP. OBJECT IS INBETWEEN OBSERVER & PLANE. VP PP FV LSV . .V FO S RX Y R F.V FO . TV HP ACTUAL PATTERN OF PLANES & VIEWS INFIRST ANGLE METHOD OF PROJECTIONS

8. 8. THIRD ANGLE FOR T.V. PROJECTION IN THIS METHOD, THE OBJECT IS ASSUMED TO BE SITUATED IN THIRD QUADRANT ( BELOW HP & BEHIND OF VP. )PLANES BEING TRANSPERENT AND INBETWEEN OBSERVER & OBJECT. TV X Y LSV FV ACTUAL PATTERN OF PLANES & VIEWS OF . FOTHIRD ANGLE PROJECTIONS S.V R F.V FOR .

9. 9. ORTHOGRAPHIC PROJECTIONS { MACHINE ELEMENTS } OBJECT IS OBSERVED IN THREE DIRECTIONS. THE DIRECTIONS SHOULD BE NORMAL TO THE RESPECTIVE PLANES. AND NOW PROJECT THREE DIFFERENT VIEWS ON THOSE PLANES. THESE VEWS ARE FRONT VIEW , TOP VIEW AND SIDE VIEW. FRONT VIEW IS A VIEW PROJECTED ON VERTICAL PLANE ( VP ) TOP VIEW IS A VIEW PROJECTED ON HORIZONTAL PLANE ( HP ) SIDE VIEW IS A VIEW PROJECTED ON PROFILE PLANE ( PP ) FIRST STUDY THE CONCEPT OF 1 ST AND 3 RD ANGLE PROJECTION METHODSND THEN STUDY NEXT 26 ILLUSTRATED CASES CAREFULLRY TO RECOGNIZE SURFACESERPENDICULAR TO THE ARROW DIRECTIONS

10. 10. FIRST ANGLE FOR T.V. PROJECTION IN THIS METHOD,THE OBJECT IS ASSUMED TO BESITUATED IN FIRST QUADRANT MEANS ABOVE HP & INFRONT OF VP. OBJECT IS INBETWEEN OBSERVER & PLANE. VP PP FV LSV . .V FO S RX Y R F.V FO . TV HP ACTUAL PATTERN OF PLANES & VIEWS INFIRST ANGLE METHOD OF PROJECTIONS

11. 11. FOR T.V. THIRD ANGLE PROJECTION IN THIS METHOD, THE OBJECT IS ASSUMED TO BE SITUATED IN THIRD QUADRANT ( BELOW HP & BEHIND OF VP. )PLANES BEING TRANSPERENT AND INBETWEEN OBSERVER & OBJECT. TV X Y LSV FV ACTUAL PATTERN OF PLANES & VIEWS FO OF . S .V RTHIRD ANGLE PROJECTIONS F.V FOR .

12. 12. FOR T.V. 1 ORTHOGRAPHIC PROJECTIONS FRONT VIEW L.H.SIDE VIEW FO R . F. V. S.V R FO x yPICTORIAL PRESENTATION IS GIVEN TOP VIEW DRAW THREE VIEWS OF THIS OBJECT BY FIRST ANGLE PROJECTION METHOD

13. 13. FOR T.V. 2 ORTHOGRAPHIC PROJECTIONS FRONT VIEW L.H.SIDE VIEW X Y FO . R S .V F.V R .FO TOP VIEWPICTORIAL PRESENTATION IS GIVEN DRAW THREE VIEWS OF THIS OBJECT BY FIRST ANGLE PROJECTION METHOD

14. 14. FOR T.V. 3 ORTHOGRAPHIC PROJECTIONS FRONT VIEW L.H.SIDE VIEW FO . R S.V FOR F. V. X Y TOP VIEWPICTORIAL PRESENTATION IS GIVEN DRAW THREE VIEWS OF THIS OBJECT BY FIRST ANGLE PROJECTION METHOD

15. 15. 4 FOR T.V. ORTHOGRAPHIC PROJECTIONS FRONT VIEW L.H.SIDE VIEW X Y . FO S.V R F.V RFO . TOP VIEW PICTORIAL PRESENTATION IS GIVEN DRAW THREE VIEWS OF THIS OBJECT BY FIRST ANGLE PROJECTION METHOD

16. 16. FOR T.V. 5 ORTHOGRAPHIC PROJECTIONS FRONT VIEW L.H.SIDE VIEW . S.V R FO FO R F.V X Y . TOP VIEWPICTORIAL PRESENTATION IS GIVEN DRAW THREE VIEWS OF THIS OBJECT BY FIRST ANGLE PROJECTION METHOD

17. 17. 6 FOR T.V. ORTHOGRAPHIC PROJECTIONS FRONT VIEW L.H.SIDE VIEW . FO X Y S .V R R F.VFO . TOP VIEWPICTORIAL PRESENTATION IS GIVEN DRAW THREE VIEWS OF THIS OBJECT BY FIRST ANGLE PROJECTION METHOD

18. 18. 7 FOR T.V. ORTHOGRAPHIC PROJECTIONS FRONT VIEW L.H.SIDE VIEW FO R F.V . . S .V R X Y FOPICTORIAL PRESENTATION IS GIVEN TOP VIEW DRAW THREE VIEWS OF THIS OBJECT BY FIRST ANGLE PROJECTION METHOD

19. 19. STUDY 8 FOR T.V. ZILLUSTRATIONS ORTHOGRAPHIC PROJECTIONS FRONT VIEW 50 X Y 20 25 FO R F.V . 25 20 TOP VIEW PICTORIAL PRESENTATION IS GIVEN DRAW THREE VIEWS OF THIS OBJECT BY FIRST ANGLE PROJECTION METHOD

20. 20. 9 FOR T.V. ORTHOGRAPHIC PROJECTIONS FRONT VIEW L.H.SIDE VIEW X Y FO . R F.V S.V . RFO TOP VIEW PICTORIAL PRESENTATION IS GIVEN DRAW THREE VIEWS OF THIS OBJECT BY FIRST ANGLE PROJECTION METHOD

21. 21. 10 FOR T.V. ORTHOGRAPHIC PROJECTIONS FRONT VIEW L.H.SIDE VIEW X Y . R S.VFO FO R F. V. PICTORIAL PRESENTATION IS GIVEN TOP VIEW DRAW THREE VIEWS OF THIS OBJECT BY FIRST ANGLE PROJECTION METHOD

22. 22. FOR T.V. 11 ORTHOGRAPHIC PROJECTIONS FRONT VIEW L.H.SIDE VIEW FO R F.V X Y . . S .V RFO TOP VIEW PICTORIAL PRESENTATION IS GIVEN DRAW THREE VIEWS OF THIS OBJECT BY FIRST ANGLE PROJECTION METHOD

23. 23. 12 FOR T.V. ORTHOGRAPHIC PROJECTIONS FRONT VIEW L.H.SIDE VIEW X Y . FO S.V R F.V R .FO PICTORIAL PRESENTATION IS GIVEN TOP VIEW DRAW THREE VIEWS OF THIS OBJECT BY FIRST ANGLE PROJECTION METHOD

24. 24. STUDY 13 ZILLUSTRATIONS FOR T.V. ORTHOGRAPHIC PROJECTIONS 35 FV 35 10 x y 10 20 30 FO R 40 F.V . 70 TV O PICTORIAL PRESENTATION IS GIVEN DRAW FV AND TV OF THIS OBJECT BY FIRST ANGLE PROJECTION METHOD

25. 25. STUDY 14 Z ILLUSTRATIONS FOR T.V. ORTHOGRAPHIC PROJECTIONS FV 30 10 30 SV 30 10 30 x y . FO S.V R R F.VFO . ALL VIEWS IDENTICAL TV PICTORIAL PRESENTATION IS GIVEN DRAW THREE VIEWS OF THIS OBJECT BY FIRST ANGLE PROJECTION METHOD

26. 26. STUDY 15 Z ILLUSTRATIONS ORTHOGRAPHIC PROJECTIONS FOR T.V. ALL VIEWS IDENTICAL FV SV x y 10 40 60 . FO S.V R F.VOR 40 . TV 60 PICTORIAL PRESENTATION IS GIVEN DRAW THREE VIEWS OF THIS OBJECT BY FIRST ANGLE PROJECTION METHOD

27. 27. FOR T.V. ORTHOGRAPHIC PROJECTIONS 16 ALL VIEWS IDENTICAL FV SV x y 10 . FO S .V R R F.VFO . 40 60PICTORIAL PRESENTATION IS GIVEN TOP VIEW 40 DRAW THREE VIEWS OF THIS OBJECT BY FIRST ANGLE PROJECTION METHOD 60

28. 28. 17 ORTHOGRAPHIC PROJECTIONS FRONT VIEW L.H.SIDE VIEW 30 SQUARE 40 20 50 FO . R 20 S.V F.V X 10 Y R .FO 30 O 60 F.V. S.V. PICTORIAL PRESENTATION IS GIVEN DRAW FV AND SV OF THIS OBJECT BY FIRST ANGLE PROJECTION METHOD

29. 29. FOR T.V. ORTHOGRAPHIC PROJECTIONS 18 FV 40X 10 Y O 30 D 45 FO R F.V . O 10 50 PICTORIAL PRESENTATION IS GIVEN DRAW FV AND TV OF THIS OBJECT BY FIRST ANGLE PROJECTION METHOD 80 TV

30. 30. 19 FOR T.V. ORTHOGRAPHIC PROJECTIONS FV 40X O 10 Y 100 10 25 FO R TV 10 30 10 F.V 25 . 30 R O 20 D PICTORIAL PRESENTATION IS GIVEN DRAW FV AND TV OF THIS OBJECT BY FIRST ANGLE PROJECTION METHOD

31. 31. PICTORIAL PRESENTATION IS GIVEN ORTHOGRAPHIC PROJECTIONS 20 DRAW FV AND TV OF THIS OBJECT BY FIRST ANGLE PROJECTION METHOD 30 FOR T.V. FV RECT. SLOT 10 50 35 10 X Y 20 D FO R TV F.V . 60 D 30 D TOP VIEW O

32. 32. 21 ORTHOGRAPHIC PROJECTIONS 10 25 15 25 40 10 25 FO . R S.V F.V 25 R .FO O O 80 25 F.V. S.V. PICTORIAL PRESENTATION IS GIVEN DRAW FV AND SV OF THIS OBJECT BY FIRST ANGLE PROJECTION METHOD

33. 33. FOR T.V. 22 ORTHOGRAPHIC PROJECTIONS 450 30 FV 40 X Y 30 D FO R F.V . 40 TV OPICTORIAL PRESENTATION IS GIVEN 15 DRAW FV AND TV OF THIS OBJECT 40 BY FIRST ANGLE PROJECTION METHOD

34. 34. 23 ORTHOGRAPHIC PROJECTIONS FO HEX PART R F.V 20 30 . 20 . S.V 40 20 RFO O 50 20 15 O 30 100 60PICTORIAL PRESENTATION IS GIVEN FRONT VIEW L.H.SIDE VIEW DRAW FV ABD SV OF THIS OBJECT BY FIRST ANGLE PROJECTION METHOD

35. 35. FOR T.V. ORTHOGRAPHIC PROJECTIONS 24 20 40 FRONT VIEW F.V. 30 10 X Y O 10 30 FO R F.V 10 . 30 O 80PICTORIAL PRESENTATION IS GIVEN T.V. TOP VIEW DRAW FV AND TV OF THIS OBJECT BY FIRST ANGLE PROJECTION METHOD

36. 36. 25 ORTHOGRAPHIC PROJECTIONS 10 10 15 25 25X Y O 50 10 . FV LSV S .V FO R R FO F.V . PICTORIAL PRESENTATION IS GIVEN DRAW FV AND LSV OF THIS OBJECT BY FIRST ANGLE PROJECTION METHOD

37. 37. PICTORIAL PRESENTATION IS GIVEN26 DRAW FV AND SV OF THIS OBJECT BY FIRST ANGLE PROJECTION METHOD ORTHOGRAPHIC PROJECTIONS FO R F.V . 30 20 10 20 15 10 15 . S .V R OFO 30 15 15 X Y 50 F.V. LEFT S.V.

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Marketing & Branding Specialist Published on An useful slideshow for all 1st year engineering students, who are afraid of the subject : ENGINEERING GRAPHICS.. learn, enjoy~ 1. DRAWINGS: ( A Graphical Representation) The Fact about: If compared with Verbal or Written Description,Drawings offer far better idea about the Shape, Size & Appearance of any object or situation or location, that too in quite a less time.

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Orthographic Projections

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1. 1. Dr.R. GANESAMOORTHY.B.E.,M.E.,Ph.d.Dr.R. GANESAMOORTHY.B.E.,M.E.,Ph.d. Professor -Mechanical EngineeringProfessor -Mechanical Engineering Saveetha Engineering CollegeSaveetha Engineering College TandalamTandalam Chennai.Chennai. 09/20/18 Dr.RGM/ Prof-Mechanical/unit 1 1

2. 2. Orthographic projections In orthographic projections, the principle views - Front view, Top view, & Side views of an object are drawn by the direct observation.  These views are drawn from the pictorial view of an object. The pictorial view is a three dimensional representation. By observing pictorial view, it is very easy to visualize the shape when the object is viewed from front, top & sides. 09/20/18 Dr.RGM/ Prof-Mechanical/unit 1 2

3. 3. Orthographic Projection Multi-view drawings provide the most accurate description of 3D objects and structures for engineering, manufacturing, and construction requirements In engineering, there are two systems of projection: First Angle Projection - Europe, Asia and former European colonies in Africa and parts of Australia Third Angle Projection - Used in the USA and Canada, in Africa where American development is taking place and parts of Australia 09/20/18 Dr.RGM/ Prof-Mechanical/unit 1 3

4. 4. Conventional Projections 09/20/18 Dr.RGM/ Prof-Mechanical/unit 1 4

5. 5. ` 09/20/18 Dr.RGM/ Prof-Mechanical/unit 1 5 ORTHOGRAPHIC PROJECTIONS

6. 6. Orthographic projections 09/20/18 Dr.RGM/ Prof-Mechanical/unit 1 6

7. 7. Views What Seen Front view Length and Height-(width not consider) Top view Length and width-(Height not consider) Side view Height and width-(Length not consider) Arrangement of views 09/20/18 Dr.RGM/ Prof-Mechanical/unit 1 7

8. 8. Representation of orthographic views 09/20/18 Dr.RGM/ Prof-Mechanical/unit 1 8

9. 9. Representation of orthographic views 09/20/18 Dr.RGM/ Prof-Mechanical/unit 1 9

10. 10. Representation of orthographic views 09/20/18 Dr.RGM/ Prof-Mechanical/unit 1 10

11. 11. Representation of orthographic views 09/20/18 Dr.RGM/ Prof-Mechanical/unit 1 11

12. 12. Representation of orthographic views 09/20/18 Dr.RGM/ Prof-Mechanical/unit 1 12

13. 13. Representation of orthographic views 09/20/18 Dr.RGM/ Prof-Mechanical/unit 1 13

14. 14. Representation of orthographic views 09/20/18 Dr.RGM/ Prof-Mechanical/unit 1 14

15. 15. Representation of orthographic views 09/20/18 Dr.RGM/ Prof-Mechanical/unit 1 15

16. 16. Representation of orthographic views 09/20/18 Dr.RGM/ Prof-Mechanical/unit 1 16

17. 17. Representation of orthographic views 09/20/18 Dr.RGM/ Prof-Mechanical/unit 1 17

18. 18. Representation of orthographic views 09/20/18 Dr.RGM/ Prof-Mechanical/unit 1 18

19. 19. Representation of orthographic views 09/20/18 Dr.RGM/ Prof-Mechanical/unit 1 19

20. 20. Representation of orthographic views 09/20/18 Dr.RGM/ Prof-Mechanical/unit 1 20

21. 21. Representation of orthographic views 09/20/18 Dr.RGM/ Prof-Mechanical/unit 1 21

22. 22. Representation of orthographic views 09/20/18 Dr.RGM/ Prof-Mechanical/unit 1 22

23. 23. Representation of orthographic views 09/20/18 Dr.RGM/ Prof-Mechanical/unit 1 23

24. 24. Representation of orthographic views 09/20/18 Dr.RGM/ Prof-Mechanical/unit 1 24

25. 25. Representation of orthographic views 09/20/18 Dr.RGM/ Prof-Mechanical/unit 1 25

26. 26. Representation of orthographic views 09/20/18 Dr.RGM/ Prof-Mechanical/unit 1 26

27. 27. Representation of orthographic views 09/20/18 Dr.RGM/ Prof-Mechanical/unit 1 27

28. 28. Representation of orthographic views 09/20/18 Dr.RGM/ Prof-Mechanical/unit 1 28

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Successfully reported this slideshow. Published on 1. Dr.R. GANESAMOORTHY.B.E.,M.E.,Ph.d.Dr.R. GANESAMOORTHY.B.E.,M.E.,Ph.d. Professor -Mechanical EngineeringProfessor -Mechanical Engineering Saveetha Engineering CollegeSaveetha Engineering College TandalamTandalam Chennai.Chennai. 09/20/18 Dr.RGM/ Prof-Mechanical/unit 1 1 2. Orthographic projections In orthographic projections, the principle views - Front view, Top view, & Side views of an object are drawn by the direct observation.

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Engineering Curves

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1. 1. Engineering Curves - I1. Classification2. Conic sections - explanation3. Common Definition4. Ellipse - ( six methods of construction)5. Parabola - ( Three methods of construction)6. Hyperbola - ( Three methods of construction )7. Methods of drawing Tangents & Normals ( four cases)

2. 2. Engineering Curves - II1.Classification2. Definitions3. Involutes - (five cases)4. Cycloid5. Trochoids - (Superior and Inferior)6. Epic cycloid and Hypo - cycloid7. Spiral (Two cases)8. Helix - on cylinder & on cone9. Methods of drawing Tangents and Normals (Three cases)

3. 3. ENGINEERING CURVES Part- I {Conic Sections}ELLIPSE PARABOLA HYPERBOLA1.Concentric Circle Method 1.Rectangle Method 1.Rectangular Hyperbola (coordinates given)2.Rectangle Method 2 Method of Tangents ( Triangle Method) 2 Rectangular Hyperbola3.Oblong Method (P-V diagram - Equation given) 3.Basic Locus Method4.Arcs of Circle Method (Directrix - focus) 3.Basic Locus Method (Directrix - focus)5.Rhombus Metho6.Basic Locus Method Methods of Drawing (Directrix - focus) Tangents & Normals To These Curves.

4. 4. CONIC SECTIONS ELLIPSE, PARABOLA AND HYPERBOLA ARE CALLED CONIC SECTIONS BECAUSE THESE CURVES APPEAR ON THE SURFACE OF A CONE WHEN IT IS CUT BY SOME TYPICAL CUTTING PLANES. OBSERVE ILLUSTRATIONS GIVEN BELOW.. Ellipse Section Plane ola Section PlaneThrough Generators Hyperbola rab Parallel to Axis. Pa Section Plane Parallel to end generator.

5. 5. COMMON DEFINATION OF ELLIPSE, PARABOLA & HYPERBOLA: These are the loci of points moving in a plane such that the ratio of it's distances from a fixed point And a fixed line always remains constant. The Ratio is called ECCENTRICITY. (E) A) For Ellipse E<1 B) For Parabola E=1 C) For Hyperbola E>1 Refer Problem nos. 6. 9 & 12 SECOND DEFINATION OF AN ELLIPSE:- It is a locus of a point moving in a plane such that the SUM of it's distances from TWO fixed points always remains constant. {And this sum equals to the length of major axis.} These TWO fixed points are FOCUS 1 & FOCUS 2 Refer Problem no.4 Ellipse by Arcs of Circles Method.

6. 6. ELLIPSE BY CONCENTRIC CIRCLE METHOD Problem 1 :- Draw ellipse by concentric circle method.Take major axis 100 mm and minor axis 70 mm long. 3 2 4 Steps: 1. Draw both axes as perpendicular bisectors C of each other & name their ends as shown. 2. Taking their intersecting point as a center, 1 5 3 2 4 draw two concentric circles considering both as respective diameters. 1 5 3. Divide both circles in 12 equal parts & name as shown. A 4. From all points of outer circle draw vertical B lines downwards and upwards respectively. 5.From all points of inner circle draw 10 6 horizontal lines to intersect those vertical lines. 10 9 7 6 6. Mark all intersecting points properly as 8 those are the points on ellipse. D 7. Join all these points along with the ends of both axes in smooth possible curve. It is 9 7 required ellipse. 8

7. 7. Steps: ELLIPSE BY RECTANGLE METHOD1 Draw a rectangle taking majorand minor axes as sides.2. In this rectangle draw both Problem 2axes as perpendicular bisectors Draw ellipse by Rectangle method.of each other..3. For construction, select upper Take major axis 100 mm and minor axis 70 mm long.left part of rectangle. Dividevertical small side and horizontallong side into same number of D 4 4equal parts.( here divided in fourparts) 3 34. Name those as shown..5. Now join all vertical points 2 21,2,3,4, to the upper end of minoraxis. And all horizontal points 1 1i.e.1,2,3,4 to the lower end ofminor axis. A 1 2 3 4 3 2 1 B6. Then extend C-1 line upto D-1and mark that point. Similarlyextend C-2, C-3, C-4 lines up toD-2, D-3, & D-4 lines.7. Mark all these points properlyand join all along with ends Aand D in smooth possible curve.Do similar construction in right Cside part.along with lower half ofthe rectangle.Join all points insmooth curve.It is required ellipse.

8. 8. ELLIPSE Problem 3:- BY OBLONG METHOD Draw ellipse by Oblong method. Draw a parallelogram of 100 mm and 70 mm long sides with included angle of 750.Inscribe Ellipse in it. STEPS ARE SIMILAR TO THE PREVIOUS CASE (RECTANGLE METHOD) ONLY IN PLACE OF RECTANGLE, HERE IS A PARALLELOGRAM. D 4 4 3 3 2 2 1 1A 1 2 3 4 3 2 1 B C

9. 9. PROBLEM 4. ELLIPSEMAJOR AXIS AB & MINOR AXIS CD ARE BY ARCS OF CIRCLE METHOD100 AMD 70MM LONG RESPECTIVELY.DRAW ELLIPSE BY ARCS OF CIRLESMETHOD. As per the definition Ellipse is locus of point P moving in a plane such that the SUM of it's distances from two fixedSTEPS: points (F1 & F2) remains constant and equals to the length1.Draw both axes as usual.Name the of major axis AB.(Note A .1+ B .1=A . 2 + B. 2 = AB) ends & intersecting point2.Taking AO distance I.e.half major axis, from C, mark F1 & F2 On AB . p4 C ( focus 1 and 2.) p33.On line F1- O taking any distance, p2 mark points 1,2,3, & 4 p14.Taking F1 center, with distance A-1 draw an arc above AB and taking F2 center, with B-1 distance cut this arc. Name the point p1 A B O5.Repeat this step with same centers but F1 1 2 3 4 F2 taking now A-2 & B-2 distances for drawing arcs. Name the point p26.Similarly get all other P points. With same steps positions of P can be located below AB.7.Join all points by smooth curve to get an ellipse/ D

10. 10. PROBLEM 5. ELLIPSEDRAW RHOMBUS OF 100 MM & 70 MM LONG BY RHOMBUS METHODDIAGONALS AND INSCRIBE AN ELLIPSE IN IT.STEPS: 21. Draw rhombus of given dimensions.2. Mark mid points of all sides & name Those A,B,C,& D3. Join these points to the ends of A B smaller diagonals.4. Mark points 1,2,3,4 as four centers.5. Taking 1 as center and 1-A 3 4 radius draw an arc AB.6. Take 2 as center draw an arc CD.7. Similarly taking 3 & 4 as centers and 3-D radius draw arcs DA & BC. D C 1

11. 11. PROBLEM 6:- POINT F IS 50 MM FROM A LINE AB.A POINT P IS MOVING IN A PLANESUCH THAT THE RATIO OF IT'S DISTANCES FROM F AND LINE AB REMAINS CONSTANT ELLIPSEAND EQUALS TO 2/3 DRAW LOCUS OF POINT P. { ECCENTRICITY = 2/3 } DIRECTRIX-FOCUS METHOD ELLIPSE A STEPS: 1 .Draw a vertical line AB and point F DIRECTRIX 50 mm from it. 2 .Divide 50 mm distance in 5 parts. 45mm 3 .Name 2nd part from F as V. It is 20mm and 30mm from F and AB line resp. 30mm It is first point giving ratio of it's distances from F and AB 2/3 i.e 20/30 4 Form more points giving same ratio such as 30/45, 40/60, 50/75 etc. (vertex) V 5.Taking 45,60 and 75mm distances from F ( focus) line AB, draw three vertical lines to the right side of it. 6. Now with 30, 40 and 50mm distances in compass cut these lines above and below, with F as center. 7. Join these points through V in smooth curve. This is required locus of P.It is an ELLIPSE. B

12. 12. PROBLEM 7: A BALL THROWN IN AIR ATTAINS 100 M HIEGHT PARABOLA AND COVERS HORIZONTAL DISTANCE 150 M ON GROUND. RECTANGLE METHOD Draw the path of the ball (projectile)-STEPS: 6 61.Draw rectangle of above size and divide it in two equal vertical parts2.Consider left part for construction. 5 5 Divide height and length in equal number of parts and name those 1,2,3,4,5& 6 4 43.Join vertical 1,2,3,4,5 & 6 to the top center of rectangle4.Similarly draw upward vertical 3 3 lines from horizontal1,2,3,4,5 And wherever these lines intersect previously drawn inclined lines in sequence Mark those points and 2 2 further join in smooth possible curve.5.Repeat the construction on right side rectangle also.Join all in sequence. 1 1 This locus is Parabola.. 1 2 3 4 5 6 5 4 3 2 1

13. 13. Problem no.8: Draw an isosceles triangle of 100 mm long base and PARABOLA110 mm long altitude.Inscribe a parabola in it by method of tangents. METHOD OF TANGENTS Solution Steps: C 1. Construct triangle as per the given 14 1 dimensions. 2. Divide it's both sides in to same no.of 13 2 equal parts. 12 3 3. Name the parts in ascending and 11 descending manner, as shown. 4 10 4. Join 1-1, 2-2,3-3 and so on. 5 5. Draw the curve as shown i.e.tangent to 9 6 all these lines. The above all lines being 8 tangents to the curve, it is called method 7 of tangents. 7 8 6 9 5 10 4 11 3 12 2 13 1 14 A B

14. 14. PROBLEM 9: Point F is 50 mm from a vertical straight line AB. PARABOLADraw locus of point P, moving in a plane such that DIRECTRIX-FOCUS METHODit always remains equidistant from point F and line AB. PARABOLASOLUTION STEPS:1.Locate center of line, perpendicular to A AB from point F. This will be initial point P and also the vertex.2.Mark 5 mm distance to its right side, name those points 1,2,3,4 and from P1those draw lines parallel to AB.3.Mark 5 mm distance to its left of P and (VERTEX) V name it 1. F O 1 2 3 44.Take O-1 distance as radius and F as center draw an arc ( focus) cutting first parallel line to AB. Name upper point P1 and lower point P2. P2(FP1=O1)5.Similarly repeat this process by taking again 5mm to right and left and locate P3 P4 . B6.Join all these points in smooth curve. It will be the locus of P equidistance from line AB and fixed point F.

15. 15. Problem No.10: Point P is 40 mm and 30 mm from horizontal HYPERBOLAand vertical axes respectively.Draw Hyperbola through it. THROUGH A POINT OF KNOWN CO-ORDINATES Solution Steps: 1) Extend horizontal line from P to right side. 2 2) Extend vertical line from P upward. 3) On horizontal line from P, mark some points taking any distance and name them after P-1, 2,3,4 etc. 4) Join 1-2-3-4 points to pole O. Let them cut 1 part [P-B] also at 1,2,3,4 points. 5) From horizontal 1,2,3,4 draw vertical 2 1 P 1 2 3 lines downwards and 6) From vertical 1,2,3,4 points [from P-B] draw 1 horizontal lines. 7) Line from 1 40 mm 2 horizontal and line from 1 vertical will meet at 3 P1.Similarly mark P2, P3, P4 points. O 8) Repeat the procedure by marking four points 30 mm on upward vertical line from P and joining all those to pole O. Name this points P6, P7, P8 etc. and join them by smooth

16. 16. Problem no.11: A sample of gas is expanded in a cylinder HYPERBOLAfrom 10 unit pressure to 1 unit pressure.Expansion followslaw PV=Constant.If initial volume being 1 unit, draw the P-V DIAGRAMcurve of expansion. Also Name the curve. Form a table giving few more values of P & V 10 P V = C 9 +++ +++ + 10 1 = 10 5 2 = 10 8 4 2.5 = 10 2.5 4 = 10 7 2 5 = 10 1 10 = 10 6 PRESSURE Now draw a Graph of 5 ( Kg/cm2) Pressure against Volume. It is a PV Diagram and it is Hyperbola. Take pressure on vertical axis and 4 Volume on horizontal axis. 3 2 1 0 1 2 3 4 5 6 7 8 9 10 VOLUME:( M3 )

17. 17. PROBLEM 12:- POINT F IS 50 MM FROM A LINE AB.A POINT P IS MOVING IN A PLANE HYPERBOLASUCH THAT THE RATIO OF IT'S DISTANCES FROM F AND LINE AB REMAINS CONSTANTAND EQUALS TO 2/3 DRAW LOCUS OF POINT P. { ECCENTRICITY = 2/3 } DIRECTRIX FOCUS METHODSTEPS: A1 .Draw a vertical line AB and point F 50 mm from it.2 .Divide 50 mm distance in 5 parts. 30mm3 .Name 2nd part from F as V. It is 20mm and 30mm from F and AB line resp. 45m It is first point giving ratio of it's m distances from F and AB 2/3 i.e 20/304 Form more points giving same ratio such as 30/45, 40/60, 50/75 etc. (vertex) V F ( focus)5.Taking 45,60 and 75mm distances from line AB, draw three vertical lines to the right side of it.6. Now with 30, 40 and 50mm distances in compass cut these lines above and below, with F as center.7. Join these points through V in smooth curve.This is required locus of P.It is an ELLIPSE. B

18. 18. ELLIPSEProblem 13: TANGENT & NORMAL TO DRAW TANGENT & NORMAL TO THE CURVE FROM A GIVEN POINT ( Q ) 1. JOIN POINT Q TO F1 & F2 2. BISECT ANGLE F1Q F2 THE ANGLE BISECTOR IS NORMAL 3. A PERPENDICULAR LINE DRAWN TO IT IS TANGENT TO THE CURVE. p4 C p3 p2 p1 A B O F1 1 2 3 4 F2 ALM NOR Q TAN GE NT D

19. 19. ELLIPSEProblem 14: TANGENT & NORMALTO DRAW TANGENT & NORMAL TO THE CURVE ELLIPSE FROM A GIVEN POINT ( Q ) A DIRECTRIX1.JOIN POINT Q TO F. T2.CONSTRUCT 900 ANGLE WITH THIS LINE AT POINT F3.EXTEND THE LINE TO MEET DIRECTRIX AT T4. JOIN THIS POINT TO Q AND EXTEND. THIS IS TANGENT TO ELLIPSE FROM Q (vertex) V5.TO THIS TANGENT DRAW PERPENDICULAR F ( focus) LINE FROM Q. IT IS NORMAL TO CURVE. 900 N Q N B T

20. 20. PARABOLA Problem 15: TANGENT & NORMAL TO DRAW TANGENT & NORMAL TO THE CURVE T PARABOLA FROM A GIVEN POINT ( Q ) A1.JOIN POINT Q TO F.2.CONSTRUCT 900 ANGLE WITH THIS LINE AT POINT F3.EXTEND THE LINE TO MEET DIRECTRIX AT T4. JOIN THIS POINT TO Q AND EXTEND. THIS IS VERTEX V TANGENT TO THE CURVE FROM Q 900 F5.TO THIS TANGENT DRAW PERPENDICULAR ( focus) LINE FROM Q. IT IS NORMAL TO CURVE. N Q B N T

21. 21. HYPERBOLA Problem 16 TANGENT & NORMAL TO DRAW TANGENT & NORMAL TO THE CURVE FROM A GIVEN POINT ( Q ) A1.JOIN POINT Q TO F.2.CONSTRUCT 900 ANGLE WITH THIS LINE AT POINT F T3.EXTEND THE LINE TO MEET DIRECTRIX AT T4. JOIN THIS POINT TO Q AND EXTEND. THIS IS TANGENT TO CURVE FROM Q (vertex) V F ( focus)5.TO THIS TANGENT DRAW PERPENDICULAR 900 LINE FROM Q. IT IS NORMAL TO CURVE. N N Q B T

22. 22. ENGINEERING CURVES Part-II (Point undergoing two types of displacements) INVOLUTE CYCLOID SPIRAL HELIX1. Involute of a circle 1. General Cycloid 1. Spiral of 1. On Cylinder a)String Length = πD One Convolution. 2. Trochoid 2. On a Cone b)String Length > πD ( superior) 2. Spiral of 3. Trochoid Two Convolutions. c)String Length < πD ( Inferior) 4. Epi-Cycloid2. Pole having Composite shape. 5. Hypo-Cycloid3. Rod Rolling over a Semicircular Pole. AND Methods of Drawing Tangents & Normals To These Curves.

23. 23. DEFINITIONSCYCLOID: IS A LOCUS OF A POINT ON THE SUPERIORTROCHOID:ERIPHERY OF A CIRCLE WHICH IF THE POINT IN THE DEFINATIONOLLS ON A STRAIGHT LINE PATH. OF CYCLOID IS OUTSIDE THE CIRCLENVOLUTE: INFERIOR TROCHOID.: IF IT IS INSIDE THE CIRCLEIS A LOCUS OF A FREE END OF A STRINGHEN IT IS WOUND ROUND A CIRCULAR POLE EPI-CYCLOID IF THE CIRCLE IS ROLLING ONSPIRAL: ANOTHER CIRCLE FROM OUTSIDE IS A CURVE GENERATED BY A POINT HYPO-CYCLOID.HICH REVOLVES AROUND A FIXED POINT IF THE CIRCLE IS ROLLING FROMND AT THE SAME MOVES TOWARDS IT. INSIDE THE OTHER CIRCLE,HELIX: IS A CURVE GENERATED BY A POINT WHICHOVES AROUND THE SURFACE OF A RIGHT CIRCULARYLINDER / CONE AND AT THE SAME TIME ADVANCES IN AXIAL DIRECTIONT A SPEED BEARING A CONSTANT RATIO TO THE SPPED OF ROTATION. or problems refer topic Development of surfaces)

24. 24. Problem no 17: Draw Involute of a circle. INVOLUTE OF A CIRCLE String length is equal to the circumference of circle.Solution Steps:1) Point or end P of string AP isexactly πD distance away from A.Means if this string is wound roundthe circle, it will completely cover P2given circle. B will meet A afterwinding.2) Divide πD (AP) distance into 8 P3number of equal parts. P13) Divide circle also into 8 number 2 to pof equal parts. 34) Name after A, 1, 2, 3, 4, etc. up to pto 8 on πD line AP as well as on pcircle (in anticlockwise direction). o 1t5) To radius C-1, C-2, C-3 up to C-8draw tangents (from 1,2,3,4,etc to 4 to pcircle). P4 46) Take distance 1 to P in compass 3and mark it on tangent from point 1 5on circle (means one division less 2than distance AP). 6 p o7) Name this point P1 5t 18) Take 2-B distance in compass 7 A 8 6 to pand mark it on the tangent from 7 to Ppoint 2. Name it point P2. P5 p P8 1 2 3 4 5 6 7 89) Similarly take 3 to P, 4 to P, 5 to P7P up to 7 to P distance in compass P6 πand mark on respective tangentsand locate P3, P4, P5 up to P8 (i.e. DA) points and join them in smoothcurve it is an INVOLUTE of a givencircle.

25. 25. INVOLUTE OF A CIRCLEProblem 18: Draw Involute of a circle. String length MORE than πDString length is MORE than the circumference of circle.Solution Steps: P2In this case string length is morethan Π D. But remember!Whatever may be the length of P3 P1string, mark Π D distance 2 to phorizontal i.e.along the stringand divide it in 8 number of 3 toequal parts, and not any other p pdistance. Rest all steps are same o 1tas previous INVOLUTE. Drawthe curve completely. 4 to p P4 4 3 5 2 p o 5t 6 1 P5 7 8 7 p8 1 P 6 to p to p 2 3 4 5 6 7 8 P7 165 mm P6 (more than πD) πD

26. 26. Problem 19: Draw Involute of a circle. INVOLUTE OF A CIRCLEString length is LESS than the circumference of circle. String length LESS than πDSolution Steps: P2In this case string length is Lessthan Π D. But remember!Whatever may be the length of P3 P1string, mark Π D distancehorizontal i.e.along the stringand divide it in 8 number of 2 to p 3 toequal parts, and not any other pdistance. Rest all steps are sameas previous INVOLUTE. Draw op 1tthe curve completely. 4 to p P4 4 3 5 2 p o 6 5t 1 6 to p P5 7 to 7 P p 8 P7 1 2 3 4 5 6 7 8 P6 150 mm (Less than πD) πD

27. 27. PROBLEM 20 : A POLE IS OF A SHAPE OF HALF HEXABON AND SEMICIRCLE. ASTRING IS TO BE WOUND HAVING LENGTH EQUAL TO THE POLE PERIMETER INVOLUTE DRAW PATH OF FREE END P OF STRING WHEN WOUND COMPLETELY. OF (Take hex 30 mm sides and semicircle of 60 mm diameter.) COMPOSIT SHAPED POLESOLUTION STEPS:Draw pole shape as perdimensions. P1Divide semicircle in 4parts and name those Palong with corners of P2hexagon.Calculate perimeterlength. 1 to PShow it as string AP.On this line mark 30mm 2 tofrom A P oPMark and name it 1 AtMark πD/2 distance on itfrom 1And dividing it in 4 parts P3name 2,3,4,5. 3 to P 3Mark point 6 on line 30 4 2mm from 5Now draw tangents from 5 1all points of pole oPand proper lengths as A 4tdone in all previous 6 5 to P involute's problems and 1 2 3 4 5 6 P 6t oPcomplete the curve. πD/2 P4 P6 P5

28. 28. PROBLEM 21 : Rod AB 85 mm long rollsover a semicircular pole without slippingfrom it's initially vertical position till itbecomes up-side-down vertical. BDraw locus of both ends A & B. A4 Solution Steps? 4 If you have studied previous problems B1 properly, you can surely solve this also. Simply remember that this being a rod, A3 it will roll over the surface of pole. 3 Means when one end is approaching, other end will move away from poll.OBSERVE ILLUSTRATION CAREFULLY! πD 2 A2 B2 2 1 3 1 A1 4 A B3 B4

29. 29. PROBLEM 22: DRAW LOCUS OF A POINT ON THE PERIPHERY OF A CIRCLE CYCLOIDWHICH ROLLS ON STRAIGHT LINE PATH. Take Circle diameter as 50 mm p4 4 p3 p5 3 5 C p2 C1 C2 C3 C4 C5 C6 C7 p6 C 8 2 6 p1 1 p7 7 P p8 πD Solution Steps: 1) From center C draw a horizontal line equal to πD distance. 2) Divide πD distance into 8 number of equal parts and name them C1, C2, C3__ etc. 3) Divide the circle also into 8 number of equal parts and in clock wise direction, after P name 1, 2, 3 up to 8. 4) From all these points on circle draw horizontal lines. (parallel to locus of C) 5) With a fixed distance C-P in compass, C1 as center, mark a point on horizontal line from 1. Name it P. 6) Repeat this procedure from C2, C3, C4 upto C8 as centers. Mark points P2, P3, P4, P5 up to P8 on the horizontal lines drawn from 2, 3, 4, 5, 6, 7 respectively. 7) Join all these points by curve. It is Cycloid.

30. 30. PROBLEM 23: DRAW LOCUS OF A POINT , 5 MM AWAY FROM THE PERIPHERY OF A SUPERIOR TROCHOID CIRCLE WHICH ROLLS ON STRAIGHT LINE PATH. Take Circle diameter as 50 mm 4 p4 p3 p5 3 5 p2 C C1 C C3 C4 C5 C6 C7 C8 p 6 2 6 2 p7 1 p1 7 P πD p8Solution Steps:1) Draw circle of given diameter and draw a horizontal line from it's center C of length Π D and divide it in 8 number of equal parts and name them C1, C2, C3, up to C8.2) Draw circle by CP radius, as in this case CP is larger than radius of circle.3) Now repeat steps as per the previous problem of cycloid, by dividing this new circle into 8 number of equal parts and drawing lines from all these points parallel to locus of C and taking CP radius wit different positions of C as centers, cut these lines and get different positions of P and join4) This curve is called Superior Trochoid.

31. 31. PROBLEM 24: DRAW LOCUS OF A POINT , 5 MM INSIDE THE PERIPHERY OF A INFERIOR TROCHOID CIRCLE WHICH ROLLS ON STRAIGHT LINE PATH. Take Circle diameter as 50 mm p4 4 p3 p5 3 5 p2 C C1 C2 C3 C4 C5 C6 C7 p6 C8 2 6 p1 p7 1 7 P p8 πDSolution Steps:1) Draw circle of given diameter and draw a horizontal line from it's center C of length Π D and divide it in 8 number of equal parts and name them C1, C2, C3, up to C8.2) Draw circle by CP radius, as in this case CP is SHORTER than radius of circle.3) Now repeat steps as per the previous problem of cycloid, by dividing this new circle into 8 number of equal parts and drawing lines from all these points parallel to locus of C and taking CP radius with different positions of C as centers, cut these lines and get different positions of P and join those in curvature.4) This curve is called Inferior Trochoid.

32. 32. PROBLEM 25: DRAW LOCUS OF A POINT ON THE PERIPHERY OF A CIRCLEWHICH ROLLS ON A CURVED PATH. Take diameter of rolling Circle 50 mm EPI CYCLOID :And radius of directing circle i.e. curved path, 75 mm.Solution Steps:1) When smaller circle will roll onlarger circle for one revolution it willcover Π D distance on arc and it will bedecided by included arc angle θ.2) Calculate θ by formula θ = (r/R) x Generating/3600. Rolling Circle3) Construct angle θ with radius OC 4 5and draw an arc by taking O as center C2 C3 C1 C4OC as radius and form sector of angle 3 6θ. C C 54) Divide this sector into 8 number of 7equal angular parts. And from C 2 C6onward name them C1, C2, C3 up toC8. 1 P r = CP C75) Divide smaller circle (Generatingcircle) also in 8 number of equal parts.And next to P in clockwise direction Directing Circlename those 1, 2, 3, up to 8. R C 86) With O as center, O-1 as radiusdraw an arc in the sector. Take O-2, O- = r 3600 + R3, O-4, O-5 up to O-8 distances withcenter O, draw all concentric arcs in Osector. Take fixed distance C-P incompass, C1 center, cut arc of 1 at P1.Repeat procedure and locate P2, P3,P4, P5 unto P8 (as in cycloid) and jointhem by smooth curve. This is EPI -CYCLOID.

33. 33. PROBLEM 26: DRAW LOCUS OF A POINT ON THE PERIPHERY OF A CIRCLEWHICH ROLLS FROM THE INSIDE OF A CURVED PATH. Take diameter of HYPO CYCLOIDrolling circle 50 mm and radius of directing circle (curved path) 75 mm. Solution Steps: 1) Smaller circle is rolling here, inside the larger circle. It has to rotate anticlockwise to move P 7 ahead. 2) Same steps should be P1 6 taken as in case of EPI - CYCLOID. Only change is 1 P2 C2 C1 C3 in numbering direction of 8 C4 number of equal parts on C C P3 5 5 the smaller circle. 2 C 3) From next to P in 6 anticlockwise direction, 4 P4 C name 1,2,3,4,5,6,7,8. 3 7 4) Further all steps are P5 P8 that of epi - cycloid. This P6 P7 is called C8 HYPO - CYCLOID. r 3600 = R + O OC = R ( Radius of Directing Circle) CP = r (Radius of Generating Circle)

34. 34. HELIX (UPON A CYLINDER)PROBLEM: Draw a helix of one convolution, upon a cylinder. P8Given 80 mm pitch and 50 mm diameter of a cylinder. 8(The axial advance during one complete revolution is called P7The pitch of the helix) 7 P6 6 P5SOLUTION: 5Draw projections of a cylinder.Divide circle and axis in to same no. of equal parts. ( 8 ) 4 P4Name those as shown. 3Mark initial position of point 'P' P3Mark various positions of P as shown in animation. 2 P2Join all points by smooth possible curve.Make upper half dotted, as it is going behind the solid 1 P1and hence will not be seen from front side. P 6 7 5 P 4 1 3 2

35. 35. Problem 9: A particle which is initially on base circle of a cone, standingon Hp, moves upwards and reaches apex in one complete turn around the cone.Draw it's path on projections of cone as well as on it's development.Take base circle diameter 50 mm and axis 70 mm long. It's a construction of curve o' Helix of one turn on cone : 7' DEVELOPMENT Draw Fv & Tv & dev.as usual HELIX CURVE 6' On all form generators & name. A Construction of curve Helix:: 5' 4' B Show 8 generators on both views 3' Divide axis also in same parts. 1 2' Draw horizontal lines from those 1' C points on both end generators. X Y 2 1' is a point where first horizontal a' h'b' c' g g' f' d' e' D Line & gen. b'o' intersect. 3 2' is a point where second horiz. h f O 4 Line & gen. c'o' intersect. E 7 6 6 5 In this way locate all points on Fv. 5 7 Project all on Tv.Join in curvature. F a e O 4 For Development: Then taking each points true 3 G Distance From resp.generator b 1 d 2 from apex, Mark on development c H & join. A

36. 36. STEPS: InvoluteDRAW INVOLUTE AS USUAL. Method of DrawingMARK POINT Q ON IT AS DIRECTED. Tangent & NormalJOIN Q TO THE CENTER OF CIRCLE C.CONSIDERING CQ DIAMETER, DRAWA SEMICIRCLE AS SHOWN. INVOLUTE OF A CIRCLE l maMARK POINT OF INTERSECTION OF r NoTHIS SEMICIRCLE AND POLE CIRCLEAND JOIN IT TO Q. QTHIS WILL BE NORMAL TO INVOLUTE. Ta ngDRAW A LINE AT RIGHT ANGLE TO enTHIS LINE FROM Q. tIT WILL BE TANGENT TO INVOLUTE. 4 3 5 C 2 6 1 7 8 P P8 1 2 3 4 5 6 7 8 π D

37. 37. STEPS:DRAW CYCLOID AS USUAL. CYCLOIDMARK POINT Q ON IT AS DIRECTED. Method of DrawingWITH CP DISTANCE, FROM Q. CUT THE Tangent & NormalPOINT ON LOCUS OF C AND JOIN IT TO Q.FROM THIS POINT DROP A PERPENDICULARON GROUND LINE AND NAME IT NJOIN N WITH Q.THIS WILL BE NORMAL TOCYCLOID.DRAW A LINE AT RIGHT ANGLE TOTHIS LINE FROM Q. al No r mIT WILL BE TANGENT TO CYCLOID. CYCLOID Q Tang e nt CP C C1 C2 C3 C4 C5 C6 C7 C8 P N πD

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Published on 1. Engineering Curves - I1. Classification2. Conic sections - explanation3. Common Definition4. Ellipse - ( six methods of construction)5. Parabola - ( Three methods of construction)6. Hyperbola - ( Three methods of construction )7. Methods of drawing Tangents & Normals ( four cases) 2. Engineering Curves - II1.Classification2.

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Engineering Curves

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1. 1. ENGINEERING CURVES SH 1132 Engineering Graphics F. Y. B. Tech. ENGINEERING CURVES

2. 2. ENGINEERING CURVES SH 1132 Engineering Graphics F. Y. B. Tech. WHY CURVES? CIVIL ENGINEERING Bridges, Arches, Dams, Roads, Manholes etc. MECHANICAL ENGINEERING Gear Teeth, Reflector Lights, Centrifugal Pumps etc ELECTRONINICS ENGINEERING Design of Satellites, Missiles etc, Dish Antennas, ECG & EEG Machines CSE & IT Computer Graphics, Networking Concepts

3. 3. ENGINEERING CURVES SH 1132 Engineering Graphics F. Y. B. Tech. CURVES CONIC SECTIONS ENGINEERING CURVES  CIRCLE  ELLIPSE  PARABOLA  HYPERBOLA  CYCLOIDAL CURVES/ROULETTES  Cycloid  Epicycloid  Hypocycloid  Trochoid (Superior & Inferior)  Epitrochoid (Superior & Inferior)  Hypotrochoid (Superior & Inferior)  INVOLUTE  SPIRALS  Archimedian  Logarithmic  Hyperbolic  HELIX  Cylindrical  Conical  SPECIAL CURVES

4. 4. ENGINEERING CURVES SH 1132 Engineering Graphics F. Y. B. Tech. (A) CONICS The curves obtained by the intersection of a cone by cutting plane in different positions are called conics. The conics are 1. CIRCLE 2. ELLIPSE 3. PARABOLA 4. HYPERBOLA

5. 5. ENGINEERING CURVES SH 1132 Engineering Graphics F. Y. B. Tech. DEFINING CONICS Curve Position of Cutting Plane Circle Perpendicular to axis and parallel to the base Ellipse Inclined to the axis and not parallel to any generator. Angle of Cutting Plane > Angle of Generator Parabola Inclined to axis, parallel to generators and passes through the base and axis Hyperbola Inclined to the axis and not parallel to any generator. Angle of Cutting Plane < Angle of Generator

6. 6. ENGINEERING CURVES SH 1132 Engineering Graphics F. Y. B. Tech.

7. 7. ENGINEERING CURVES SH 1132 Engineering Graphics F. Y. B. Tech.

8. 8. ENGINEERING CURVES SH 1132 Engineering Graphics F. Y. B. Tech. When the cutting plane is perpendicular to the axis or parallel to the base in a right cone we get circle the section. Sec Plane Circle CIRCLE

9. 9. ENGINEERING CURVES SH 1132 Engineering Graphics F. Y. B. Tech. 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

10. 10. ENGINEERING CURVES SH 1132 Engineering Graphics F. Y. B. Tech. 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

11. 11. ENGINEERING CURVES SH 1132 Engineering Graphics F. Y. B. Tech. 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 α < θ α = 0 θ θ

12. 12. ENGINEERING CURVES SH 1132 Engineering Graphics F. Y. B. Tech. CONICS 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

13. 13. ENGINEERING CURVES SH 1132 Engineering Graphics F. Y. B. Tech. 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

14. 14. ENGINEERING CURVES SH 1132 Engineering Graphics F. Y. B. Tech. 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

15. 15. ENGINEERING CURVES SH 1132 Engineering Graphics F. Y. B. Tech. 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.

16. 16. ENGINEERING CURVES SH 1132 Engineering Graphics F. Y. B. Tech. 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.

17. 17. ENGINEERING CURVES SH 1132 Engineering Graphics F. Y. B. Tech. monuments Bridges Earth orbit around Sun

18. 18. ENGINEERING CURVES SH 1132 Engineering Graphics F. Y. B. Tech. Arches Elliptical tray

19. 19. ENGINEERING CURVES SH 1132 Engineering Graphics F. Y. B. Tech. PROBLEM :- Point F is 50 mm from a line AB. A point P is moving in a plane such that the ratio of it's distances from F and line AB remains constant and equals to 2/3 draw locus of point P. Draw tangent and normal at 50 mm from the focus. { Eccentricity = 2/3 }

20. 20. ENGINEERING CURVES SH 1132 Engineering Graphics F. Y. B. Tech. P5 P4' P6'P5' P1 P1' V P4 P3' P3 P2' PF' F D'D X F' 1' 2' 3' 4' 5' E P6 P2 PF Directrix 90° 1 2 3 4 5 6 Eccentricity = 2/3 3XV VF = 2 Dist. Between directrix & focus = 50 mm 1 part = 50/(2+3)=10 mm VF = 2 part = 20 mm VX = 3 part = 30 mm  < 45°  DIRECTRIX-FOCUS METHOD X T' T ELLIPSE AXIS 6' P0

21. 21. ENGINEERING CURVES SH 1132 Engineering Graphics F. Y. B. Tech. Ratio (known as eccentricity) of its distances from focus to that of directrix is constant and equal to one. 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. Directrix Axis Vertex M C N Q F V P Focus Parabola Eccentricity = PF/PM = QF/QN = 1.

22. 22. ENGINEERING CURVES SH 1132 Engineering Graphics F. Y. B. Tech. 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

23. 23. ENGINEERING CURVES SH 1132 Engineering Graphics F. Y. B. Tech. Headlamp Gateway Arch monument

24. 24. ENGINEERING CURVES SH 1132 Engineering Graphics F. Y. B. Tech. PROBLEM : Point F is 40 mm from a vertical Straight line AB. Draw locus of point P, moving in a plane such That it always remains equidistant from point F and line AB. Draw tangent and normal at a point S lying on a curve below the axis at a distance of 55 mm from the line. (Fixed line)

25. 25. ENGINEERING CURVES SH 1132 Engineering Graphics F. Y. B. Tech. D D' 2 3 4 T T'N N' S V 1 P1 P2 PF P3 P4 P1' P2' P3' P4' PF' AXIS 90° X F PARABOLA  = 45° Eccentricity = 1/1 2XV VF = 2 Dist. Between directrix & focus = 40 mm 1 part = 40/(2+2)=10 mm VF = 2 part = 20 mm VX = 2 part = 20 mm X E 1' F' 2' 3' 4'  P0

26. 26. ENGINEERING CURVES SH 1132 Engineering Graphics F. Y. B. Tech. 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 Focus Vertex HYPERBOLA = QF/QN > 1.

27. 27. ENGINEERING CURVES SH 1132 Engineering Graphics F. Y. B. Tech. Nature of graph of Boyle's law Shape of overhead water tanks Uses :- Shape of cooling towers etc.

28. 28. ENGINEERING CURVES SH 1132 Engineering Graphics F. Y. B. Tech. Cooling tower Nature of Boyle's law

29. 29. ENGINEERING CURVES SH 1132 Engineering Graphics F. Y. B. Tech. PROBLEM :- POINT F IS 50 MM FROM A LINE AB.A POINT P IS MOVING IN A PLANE SUCH THAT THE RATIO OF IT'S DISTANCES FROM F AND LINE AB REMAINS CONSTANT AND EQUALS TO 3/2. DRAW LOCUS OF POINT P. { ECCENTRICITY = 3/2 }. DRAW THE TANGENT AND NORMAL AT ANY CONVENIENT POINT.

30. 30. ENGINEERING CURVES SH 1132 Engineering Graphics F. Y. B. Tech. AXISC V F1 DIRECTRIXDD 1 2 3 4 4' 3' 2' 1' P1 P2 P3 P4 P1' P2' P3' P4' T1 T2 s Eccentricity = 3/2 2XV VF = 3 Dist. Between directrix & focus = 50 mm 1 part = 50/(3+2)=10 mm VF = 3 part = 30 mm VX = 2 part = 20 mm  > 45°

31. 31. ENGINEERING CURVES SH 1132 Engineering Graphics F. Y. B. Tech. CYCLOID A cycloid is a curve generated by a point on the circumference of a circle as the circle rolls along a straight line without slipping. The rolling circle is called generating circle and the line along which it rolls is called base line or directing line.

32. 32. ENGINEERING CURVES SH 1132 Engineering Graphics F. Y. B. Tech. P0 2R or D 5 1 2 1 2 3 4 6 7 8 9 10 110 120 3 4 5 6 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 P6 R P10 R PROBLEM: DRAW CYCLOID FOR ONE REVOLUTION OF A ROLLING CIRCLE HAVING DIAMETER AS 50MM. Rolling Circle D N R S

33. 33. ENGINEERING CURVES SH 1132 Engineering Graphics F. Y. B. Tech. P C1 C2 C3 C4 C5 C6 C7 C8 p1 p2 p3 p4 p5 p6 p7 p8 1 2 3 4 5 6 7 C D PROBLEM 4: DRAW LOCUS OF A POINT ON THE PERIPHERY OF A CIRCLE WHICH ROLLS ON STRAIGHT LINE PATH. Take Circle diameter as 40 mm Solution Steps: 1) From center C draw a horizontal line equal to D distance. 2) Divide D distance into 8 number of equal parts and name them C1, C2, C3__ etc. 3) Divide the circle also into 8 number of equal parts and in clock wise direction, after P name 1, 2, 3 up to 8. 4) From all these points on circle draw horizontal lines. (parallel to locus of C) 5) With a fixed distance C-P in compass, C1 as center, mark a point on horizontal line from 1. Name it P. 6) Repeat this procedure from C2, C3, C4 up to C8 as centers. Mark points P2, P3, P4, P5 up to P8 on the horizontal lines drawn from 2, 3, 4, 5, 6, 7 respectively. 7) Join all these points by curve. It is Cycloid. 8

34. 34. ENGINEERING CURVES SH 1132 Engineering Graphics F. Y. B. Tech. Rolling circle P1 P2 P3 P4 P0 P6 P7 P8 P5 P9 P10 1 2 3 4 5 6 7 8 9 10 1 2 3 4 5 6 7 8 9 100 C0 C1 C2 C3 C4 Directing line Length of directing line = 5D/4 450 = 360 + 90 450 = D + D/4 Total length for 450 rotation = 5D/4 C5 C6 C7 C8 C9 C10 Problem : Draw a cycloid for a rolling circle, 60 mm diameter rolling along a straight line without slipping for 450° revolution. Take initial position of the tracing point at the highest point on the rolling circle.

35. 35. ENGINEERING CURVES SH 1132 Engineering Graphics F. Y. B. Tech. C0 P0 786 4 P1 1 2 3 C2 C3 P2 C4 Problem: 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. 5 6 C1 P3 P4 P5 P7 P8 7 0 C5C6C7C8 1 2 3 4 5 P6 D/2πD/2 πD/2 D/2 Floor Wall CYCLOID 5 6 7 8 Take diameter of circle = 40mm Initially distance of centre of circle from the wall 83mm (Half circumference + D/2) 4

36. 36. ENGINEERING CURVES SH 1132 Engineering Graphics F. Y. B. Tech. An epicycloid is a curve generated by a point on the circumference of a circle which rolls on the outside of another circle without sliding or slipping. The rolling circle is called generating circle and the outside circle on which it rolls is called the directing circle or the base circle. EPICYCLOID

37. 37. ENGINEERING CURVES SH 1132 Engineering Graphics F. Y. B. Tech. PROBLEM 5: DRAW LOCUS OF A POINT ON THE PERIPHERY OF A CIRCLE WHICH ROLLS ON A CURVED PATH. Take diameter of rolling Circle 50 mm And radius of directing circle i.e. curved path, 75 OP OP=Radius of directing circle=75mm C PC=Radius of generating circle=25mm θ 1 2 3 4 5 6 7 8 9 1 0 1 1 1 2 1' 2' 3' 4' 5' 6' 7' 8' 9' 10' 11' 12' c1 c2 c3 c4 c5 c6 c7 c8 c9 c10 c11 c12 Directing circle Rolling circle or generating circle θ=r/R X 360º = 25/75 X 360º =120º

38. 38. ENGINEERING CURVES SH 1132 Engineering Graphics F. Y. B. Tech. A hypocycloid is a curve generated by a point on the circumference of a circle which rolls on the inside of another circle without sliding or slipping. The rolling circle is called generating circle/ hypo circle and the inside circle on which it rolls is called the directing circle or the base circle. HYPOCYCLOID

39. 39. ENGINEERING CURVES SH 1132 Engineering Graphics F. Y. B. Tech. OP OP=Radius of directing circle=75mm C PC=Radius of generating circle=25mm θ θ=r/R X360º = 25/75 X360º =120º 1 2 3 4 5 6 7 8 9 10 11 12 c2 c1 c3 c4 c5 c6 c7 c8 c9 c1 0 c1 1 c1 2 1' 2' 3' 4' 5' 6' 7' 8' 9' 10' 11' 12' PROBLEM 6: 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 50 mm and radius of directing circle (curved path) 75 mm. Directing circle Rolling circle or generating circle

40. 40. ENGINEERING CURVES SH 1132 Engineering Graphics F. Y. B. Tech. An involute is a curve traced by a point as it unwinds from around a circle or polygon. The concerned circle or polygon is called as evolutes. INVOLUTE

41. 41. ENGINEERING CURVES SH 1132 Engineering Graphics F. Y. B. Tech. PROBLEM: 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. P12 P2 0 12 6 P1 1 20 9 103 4 6 8 115 7 12 DP3 P4 P5 P6 P7 P8 P9 P10 P11 1 2 3 4 5 7 8 9 10 11 .

42. 42. ENGINEERING CURVES SH 1132 Engineering Graphics F. Y. B. Tech. PROBLEM: DRAW AN INVOLUTE OF A PENTAGON HAVING SIDE AS 20 MM. P5 P0 P1 P2 P3 P4 2 3 4 5 1 0

43. 43. ENGINEERING CURVES SH 1132 Engineering Graphics F. Y. B. Tech. 1 2 34 5 6 1 2 3 4 5 6 A P D/2 P1 1toP P2 P3 3 to P P4 P P5 P6 PROBLEM : A POLE IS OF A SHAPE OF HALF HEXABON AND SEMICIRCLE. ASTRING IS TO BE WOUND HAVING LENGTH EQUAL TO THE POLE PERIMETER DRAW PATH OF FREE END P OF STRING WHEN WOUND COMPLETELY. (Take hex 15 mm sides and semicircle of 30 mm diameter.)

44. 44. ENGINEERING CURVES SH 1132 Engineering Graphics F. Y. B. Tech. PROBLEM NO 7: DRAW INVOLUTE OF A CIRCLE. STRING LENGTH IS EQUAL TO THE CIRCUMFERENCE OF CIRCLE. DIAMETER OF CIRCLEIS 50MM. 1 2 3 4 5 6 7 8 P P8 1 2 3 4 5 6 7 8 P3 P4 4 to p P5 P7 P6 P2 P1 D A

45. 45. ENGINEERING CURVES SH 1132 Engineering Graphics F. Y. B. Tech. 1 2 3 4 5 6 7 8 P 1 2 3 4 5 6 7 8 P3 P4 4 to p P5 P7 P6 P2 P1 165 mm (more than D)D p8 PROBLEM 8: DRAW INVOLUTE OF A CIRCLE. STRING LENGTH IS MORE THAN THE CIRCUMFERENCE OF CIRCLE. DIAMETER OF CIRCLE IS 50 MM & STRING LENGTH IS 165 MM

46. 46. ENGINEERING CURVES SH 1132 Engineering Graphics F. Y. B. Tech. 1 2 3 4 5 6 7 8 P 1 2 3 4 5 6 7 8 P3 P4 4 to p P5 P7 P6 P2 P1 150 mm (Less than D) D PROBLEM : DRAW INVOLUTE OF A CIRCLE. STRING LENGTH IS LESS THAN THE CIRCUMFERENCE OF CIRCLE. DIAMETER OF CIRCLE IS 50 MM & STRING LENGTH IS 150 MM.

47. 47. ENGINEERING CURVES SH 1132 Engineering Graphics F. Y. B. Tech. PROBLEM: A STICK OF LENGTH EQUAL TO THE CIRCUMFERENCE OF A SEMICIRCLE, IS INITIALLY TANGENT TO THE SEMICIRCLE ON THE RIGHT OF IT. THIS STICK NOW ROLLS OVER THE CIRCUMFERENCE OF A SEMICIRCLE WITHOUT SLIDING TILL IT BECOMES TANGENT ON THE LEFT SIDE OF THE SEMICIRCLE. DRAW THE LOCI OF TWO END POINT OF THIS STICK. NAME THE CURVE. TAKE R= 21MM. A6 B6 5 A B C B1 A1 B2 A2 B3 A3 B4 A4 B5 A5 1 2 3 4 5 O 1 2 3 4 6 INVOLUTE

48. 48. ENGINEERING CURVES SH 1132 Engineering Graphics F. Y. B. Tech. 1 2 3 4 D 1 2 3 4 A B A1 B1 A2 B2 A3 B3 A4 B4 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. TAKE R= 25MM.

49. 49. ENGINEERING CURVES SH 1132 Engineering Graphics F. Y. B. Tech.

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STUDENT at V.J.T.I. MUMBAI Published on 1. ENGINEERING CURVES SH 1132 Engineering Graphics F. Y. B. Tech. ENGINEERING CURVES 2. ENGINEERING CURVES SH 1132 Engineering Graphics F. Y. B. Tech. WHY CURVES? CIVIL ENGINEERING Bridges, Arches, Dams, Roads, Manholes etc.

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19/1/21

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, M.Tech in Mechanical Design at Indian Institute of Technology, Delhi

Published on

Engineering Graphics/ Drawing, Drawing Instruments, Dimension, Scales

1. 1. What is an Engineering Drawing ? "An Engineering Drawing is a technical (not artistic) drawing which clearly defines and communicates a design to other interested parties. Other parties may have an interest in design collaboration, procurement / purchasing, costing, manufacturing, quality control, marketing, handling / packaging." 13/01/2017 Introduction to Engg. Graphics 1

2. 2. What is Engineering Drawing 13/01/2017 Introduction to Engg. Graphics 2

3. 3. Role of Graphics * Visualization * Communication * Documentation 13/01/2017 Introduction to Engg. Graphics 3

4. 4. Role of Graphics 1200 CE, 73 meter high minar with 14.32 meter diameter at the base and 2.75 meter diameter at the peak. It has inside stair with 379 steps. 13/01/2017 Introduction to Engg. Graphics 4

5. 5. Ships Applications of Engineering Drawing 13/01/2017 Introduction to Engg. Graphics 5

6. 6. Manufacturing of Automobiles Applications of Engineering Drawing 13/01/2017 Introduction to Engg. Graphics 6

7. 7. Construction Applications of Engineering Drawing 13/01/2017 Introduction to Engg. Graphics 7

8. 8. Introduction Drawing: The way of conveying the ideas through the systematic lines on the paper. The art of representation of an object by systematic lines on a paper. Classification: 1. Artistic Drawing 2. Engineering Drawing 13/01/2017 Introduction to Engg. Graphics 8

9. 9. 1. Artistic Drawing The art of representation of an object by the artist by his imagination or by keeping the object before him. e.g. paintings, cinema posters, advertisement boards, etc. 13/01/2017 Introduction to Engg. Graphics 9

10. 10. 2. Engineering Drawing The art of representation of engineering objects. e.g. buildings, roads, machines, etc. 13/01/2017 Introduction to Engg. Graphics 10

11. 11. Drawing Instruments 1. Drawing Board 2. Drawing Sheet 3. Drawing Sheet Holder 4. Set-squares - 45o and 30o - 60o 5. Large size Compass 6. Small bow Compass 7. Large size Divider 8. Small bow Divider 9. Scales - 6" and 12" 10. Protractor 11. French Curve 12. Drawing Pencils - H, 2H, HB 13. Sand Paper 14. Eraser (Rubber) 15. Drawing Pins and Clips 16. Cello Tape 17. Duster or Handkerchief 18. Drafting Machine / Mini Drafter 19. Sketch Book (Medium size) 20. Roller Scale 21. Pencil Sharpener 22. Sheet Folder 13/01/2017 Introduction to Engg. Graphics 11

12. 12. Drawing Instruments 13/01/2017 Introduction to Engg. Graphics 12

13. 13. Use of Drawing Instruments 13/01/2017 Introduction to Engg. Graphics 13

14. 14. Layout of Drawing Sheet Title Block 30 20 2020 All the dimensions are in millimeters. 150 50 19/01/2017 Type of Lines, Dimensioning and Lettering 14

15. 15. SHEET NO.: 01 CHECKED BY: NAME: ROLL NO.: SEM. & SEC.: SUB. DATE: GRADE: LINES & LETTERING NOTES: All the dimensions are in millimeters. Name and Roll No. should be written by ink-pen. 50 50 50 10 20101010 10101010 Title Block (Sample) 19/01/2017 Type of Lines, Dimensioning and Lettering 15

16. 16. Sheet Size 19/01/2017 Type of Lines, Dimensioning and Lettering 16

17. 17. LINES Line Thickness: Thickness varied according to the use of pen or pencil and the size & type of the drawing. For pencil, the lines can be divided into two line-groups: Line-group (mm) Thickness Lines 0.2 Medium Out lines, dotted lines, cutting plane lines 0.1 Thin Centre lines, section lines, dimension lines, extension lines, construction lines, leader lines, short-break lines and long-break lines. Important Notes: In the finished drawing, all lines except construction lines should be dense, clean and uniform. Construction lines should be drawn very thin and faint and should be hardly visible. 19/01/2017 Type of Lines, Dimensioning and Lettering 17

18. 18. Types of Lines Lines Description General Applications A Continuous thick A1 A2 Visible outlines Visible edges B Continuous thin (straight / curve) B1 B2 B3 B4 B5 B6 B7 Imaginary lines of intersection Dimension lines Projection lines Leader lines Hatching or section lines Outlines of revolved sections in plane Short centre lines C Continuous thin (free-hand) C1 C2 Limits of partial or interrupted views and sections Short-break lines D Continuous thin (straight with zigzags) D1 Long-break lines 19/01/2017 Type of Lines, Dimensioning and Lettering 18

19. 19. Lines Description General Applications E Dashed thick E1 E2 Hidden outlines Hidden edges F Dashed thin F1 F2 Hidden outlines Hidden edges G Chain thin G1 G2 G3 Centre lines Lines of symmetry Trajectories H Chain thin, thick at ends and changes of direction H1 Cutting planes J Chain thick J1 Indication of lines or surfaces to which a special treatment applies K Chain thin double-dashed K1 K2 K3 K4 Outlines of adjacent parts Alternative and extreme positions of movable parts Centroidal lines Parts situated in front of the cutting plane19/01/2017 Type of Lines, Dimensioning and Lettering 19

20. 20. Application of various types of lines according to B.I.S. PP 80o G1 B5 B2 A1 J1 C1 D1 K2 G3 H1 E1 90 Ø10 B4 19/01/2017 Type of Lines, Dimensioning and Lettering 20

21. 21. 19/01/2017 Type of Lines, Dimensioning and Lettering 21

22. 22. 19/01/2017 Type of Lines, Dimensioning and Lettering 22

23. 23. Lettering Writing of titles, dimensions, notes and other important particulars on a drawing is lettering Classification: 1. Single-stroke Letters: The thickness of the line of the letter is obtained in one stroke of the pencil. Recommended by B.I.S. It has two types: i. Vertical ii. Inclined (slope 75o with the horizontal) * The ratio of height to width varies but in most of the cases it is 6:5. * Lettering is generally done in capital letters. * The lower-case letters are generally used in architectural drawings. 19/01/2017 Type of Lines, Dimensioning and Lettering 23

24. 24. * The spacing between two letters should not be necessarily equal. * The letters should be so placed that they do not appear too close together too much apart. * The distance between the words must be uniform and at least equal to the height of the letters. * Lettering, except the dimension figures, should be underlined to make them more prominent. Size of Alphabets for Drawing: Main titles -----------------------------------------6-8 mm Sub titles ------------------------------------------3-6 mm Notes, dimension figures, etc. ---------------3-5 mm Drawing no. --------------------------------------10-12 mm 19/01/2017 Type of Lines, Dimensioning and Lettering 24

25. 25. ABCDEFGHIJKLMN OPQRSTUVWXYZ 1234567890 abcdefghijklmnopqrstuvwxy z dd/2 19/01/2017 Type of Lines, Dimensioning and Lettering 25

26. 26. Dimensioning The art of writing the various sizes or measurement on the finished drawing of an object. Types of Dimensioning: i. Size or Functional Dimensions (S): It indicates sizes. e.g. length, breadth, height, diameter, etc. ii. Location or Datum Dimensions (L): It shows location or exact position of various constructional details within the object. L SL S L 19/01/2017 Type of Lines, Dimensioning and Lettering 26

27. 27. Notations of Dimensioning Symbol Extension line Dimension line 2 Holes, Ø12 90 Arrow-head Dimension figure Leader (at 30o,45o,60o) Note 3X X Arrow-head19/01/2017 Type of Lines, Dimensioning and Lettering 27

28. 28. 1. Dimension line: Thin continuous line used to indicate the measurement. 2. Extension line: Thin continuous line extending beyond the outline of the object. 3. Arrow-head: Used to terminate the dimension line. Length : width ratio is 3:1. Space filled up. 4. Note: Gives information regarding specific operation relating to a feature. 5. Leader: Thin continuous line connecting a note or a dimension figure with the feature to which it is applied. Terminated by arrow-head or dot. 6. Symbol: The representation of any object by some mark on the drawing. It saves time and labour.19/01/2017 Type of Lines, Dimensioning and Lettering 28

29. 29. Units of Dimensioning As for as possible all dimensions should be given in millimeters omitting the abbreviation mm. If another unit is used, only the dimension figures should be written. But a foot note such as 'All the dimensions are in centimeters' is inserted in a prominent place near the title box. e.g. 15.50 0.75 (Zero must precede the decimal point.) 15.50 ± .75 ( Zero is omitted.) 19/01/2017 Type of Lines, Dimensioning and Lettering 29

30. 30. The ways of Placing the Dimensions in a Series 20 15 25 60 35 20 Chain Parallel 19/01/2017 Type of Lines, Dimensioning and Lettering 30

31. 31. The ways of Placing the Dimensions in a Series 90 15 30 30 15 0 1560 45 30 30 Combined Progressive 19/01/2017 Type of Lines, Dimensioning and Lettering 31

32. 32. 1. Chain Dimensioning: Dimensions are arranged in a straight line. 2. Parallel Dimensioning: All the dimensions are shown from a common base line. The smaller dimension is placed nearer the view. 3. Combined Dimensioning: Chain and parallel dimensioning used simultaneously. 4. Progressive Dimensioning: One datum or surface is selected which reads as zero. All the dimensions are referred to that point or surface. The ways of Placing the Dimensions in a Series 19/01/2017 Type of Lines, Dimensioning and Lettering 32

33. 33. Types of Dimensioning 1. Aligned System In the aligned system the dimensions are placed perpendicular to the dimension line in such a way that it may be read from bottom edge or right hand edge of the drawing sheet. 2. Unidirectional System In the unidirectional system, the dimensions are so oriented such that they can be read from the bottom of the drawing. 19/01/2017 Type of Lines, Dimensioning and Lettering 33

34. 34. Some Important Rules for Dimensioning 1. All the dimensions necessary for the correct functioning of the part should be expressed directly on the drawing. 2. Every dimension should be given, but none should be given more than once. 3. A dimension should be placed on the view where its use is shown more clearly. 4. Dimensions should be placed outside the view, as for as possible. 5. Mutual crossing of dimension lines and dimensioning between hidden lines should be avoided. Also it should not cross any other line of the drawing. 6. An outline or a centre line should never be used as a dimension line. A centre line may be extended to serve as an extension line. 7. Aligned system of dimensioning is recommended. 8. Dimension lines should be drawn at least 8 mm away from the outlines and from each other. 9. The extension line should be extended by about 3 mm beyond the dimension line. 19/01/2017 Type of Lines, Dimensioning and Lettering 34

35. 35. 15 10. When the space is too narrow, the arrow-head may be placed outside. Also a dot may be used to replace an arrow-head. 11. The various methods of dimensioning different sizes of circles are as follows: 10 10 15 15 10 Ø10 Ø20 Ø20 12. Arcs of circles should be dimensioned by their respective radii. R3 R8 R6 19/01/2017 Type of Lines, Dimensioning and Lettering 35

36. 36. 13. Radii of a spherical surface and square cross section of a rod is shown as below: Sphere R20 SQ24 14. Angular dimension may be given as follows: 25o 40 45 Length of Chord Length of Arc19/01/2017 Type of Lines, Dimensioning and Lettering 36

37. 37. 15. Method of dimensioning of Chamfer: 2×45o 16. Dimensioning of Tapered Surface: L h H Slope or Taper = (H-h)/L 19/01/2017 Type of Lines, Dimensioning and Lettering 37

38. 38. FOR FULL SIZE SCALE R.F.=1 OR ( 1:1 ) MEANS DRAWING & OBJECT ARE OF SAME SIZE. Other RFs are described as 1:10, 1:100, 1:1000, 1:1,00,000 SCALES DIMENSIONS OF LARGE OBJECTS MUST BE REDUCED TO ACCOMMODATE ON STANDARD SIZE DRAWING SHEET.THIS REDUCTION CREATES A SCALE OF THAT REDUCTION RATIO, WHICH IS GENERALLY A FRACTION.. SUCH A SCALE IS CALLED REDUCING SCALE AND THAT RATIO IS CALLED REPRESENTATIVE FACTOR. SIMILARLY IN CASE OF TINY OBJECTS DIMENSIONS MUST BE INCREASED FOR ABOVE PURPOSE. HENCE THIS SCALE IS CALLED ENLARGING SCALE. HERE THE RATIO CALLED REPRESENTATIVE FACTOR IS MORE THAN UNITY. REPRESENTATIVE FACTOR (R.F.) = = = = A USE FOLLOWING FORMULAS FOR THE CALCULATIONS IN THIS TOPIC. B LENGTH OF SCALE = R.F. MAX. LENGTH TO BE MEASURED. DIMENSION OF DRAWING DIMENSION OF OBJECT LENGTH OF DRAWING ACTUAL LENGTH AREA OF DRAWING ACTUAL AREA VOLUME AS PER DRWG. ACTUAL VOLUME V V 3 19/01/2017 Type of Lines, Dimensioning and Lettering 38

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M.Tech in Mechanical Design at Indian Institute of Technology, Delhi Published on Engineering Graphics/ Drawing, Drawing Instruments, Dimension, Scales 1. What is an Engineering Drawing ? "An Engineering Drawing is a technical (not artistic) drawing which clearly defines and communicates a design to other interested parties.

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Problem

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Example for Circle, Cylinder

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(Hindi) How to draw Isometric CYLINDER #Engineering Drawing

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Problem 3

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How to Draw an Isometric Drawing - First Year Engineering Drawing - Isometric Projections

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Problem 2

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HOW TO DRAW ISOMETRIC VIEW (QUE NO.6) AND ISOMETRIC PROJECTION IN ENGINEERING DRAWING

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Problem 1

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HOW TO DRAW ISOMETRIC VIEW (QUE NO.1) AND ISOMETRIC PROJECTION IN ENGINEERING DRAWING

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Other Examples

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Syllabus of Engineering Graphics.

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Conic Section: Learn to draw Ellipse, Parabola & Hyperbola - YouTube

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Construct an Ellipse - Rectangle Method - Engineering Curves - Engineering Drawing

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ED ELlipse by arc of circle method #Sheet 4 Example - 5 #MSU-Polytechnic

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2  How to draw an ellipse by Concentric Circles Method