Engineering Graphics Whole.pdf

Contents 1. Scales 2. Engineering Curves - I 3. Engineering Curves - II 4. Loci of Points 5. Orthographic Projections - Basics 6. Conversion of Pictorial View into Orthographic Views 7. Projections of Points and Lines 8. Projection of Planes 9. Projection of Solids 10. S...

Contents 1. Scales 2. Engineering Curves - I 3. Engineering Curves - II 4. Loci of Points 5. Orthographic Projections - Basics 6. Conversion of Pictorial View into Orthographic Views 7. Projections of Points and Lines 8. Projection of Planes 9. Projection of Solids 10. Sections & Development 11. Intersection of Surfaces 12. Isometric Projections 13. Exercises 14. Solutions – Applications of Lines EXIT Scales 1. Basic Information 2. Types and important units 3. Plain Scales (3 Problems) 4. Diagonal Scales - information 5. Diagonal Scales (3 Problems) 6. Comparative Scales (3 Problems) 7. Vernier Scales - information 8. Vernier Scales (2 Problems) 9. Scales of Cords - construction 10. Scales of Cords (2 Problems) Engineering Curves – I 1. Classification 2. Conic sections - explanation 3. Common Definition 4. 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) Engineering Curves – II 1. Classification 2. Definitions 3. Involutes - (five cases) 4. Cycloid 5. Trochoids – (Superior and Inferior) 6. Epic cycloid and Hypo - cycloid 7. Spiral (Two cases) 8. Helix – on cylinder & on cone 9. Methods of drawing Tangents and Normals (Three cases) Loci of Points 1. Definitions - Classifications 2. Basic locus cases (six problems) 3. Oscillating links (two problems) 4. Rotating Links (two problems) Orthographic Projections - Basics 1. Drawing – The fact about 2. Drawings - Types 3. Orthographic (Definitions and Important terms) 4. Planes - Classifications 5. Pattern of planes & views 6. Methods of orthographic projections 7. 1st angle and 3rd angle method – two illustrations Conversion of pictorial views in to orthographic views. 1. Explanation of various terms 2. 1st angle method - illustration 3. 3rd angle method – illustration 4. To recognize colored surfaces and to draw three Views 5. Seven illustrations (no.1 to 7) draw different orthographic views 6. Total nineteen illustrations ( no.8 to 26) Projection of Points and Lines 1. Projections – Information 2. Notations 3. Quadrant Structure. 4. Object in different Quadrants – Effect on position of views. 5. Projections of a Point – in 1st quadrant. 6. Lines – Objective & Types. 7. Simple Cases of Lines. 8. Lines inclined to one plane. 9. Lines inclined to both planes. 10. Imp. Observations for solution 11. Important Diagram & Tips. 12. Group A problems 1 to 5 13. Traces of Line ( HT & VT ) 14. To locate Traces. 15. Group B problems: No. 6 to 8 16. HT-VT additional information. 17. Group B1 problems: No. 9 to 11 18. Group B1 problems: No. 9 to 1 19. Lines in profile plane 20. Group C problems: No.12 & 13 21. Applications of Lines:: Information 22. Group D: Application Problems: 14 to 23 Projections of Planes: 1. About the topic: 2. Illustration of surface & side inclination. 3. Procedure to solve problem & tips: 4. Problems:1 to 5: Direct inclinations: 5. Problems:6 to 11: Indirect inclinations: 6. Freely suspended cases: Info: 7. Problems: 12 & 13 8. Determination of True Shape: Info: 9. Problems: 14 to 17 Projections of Solids: 1. Classification of Solids: 2. Important parameters: 3. Positions with Hp & Vp: Info: 4. Pattern of Standard Solution. 5. Problem no 1,2,3,4: General cases: 6. Problem no 5 & 6 (cube & tetrahedron) 7. Problem no 7 : Freely suspended: 8. Problem no 8 : Side view case: 9. Problem no 9 : True length case: 10. Problem no 10 & 11 Composite solids: 11. Problem no 12 : Frustum & auxiliary plane: Section & Development 1. Applications of solids: 2. Sectioning a solid: Information: 3. Sectioning a solid: Illustration Terms: 4. Typical shapes of sections & planes: 5. Development: Information: 6. Development of diff. solids: 7. Development of Frustums: 8. Problems: Standing Prism & Cone: no. 1 & 2 9. Problems: Lying Prism & Cone: no.3 & 4 10. Problem: Composite Solid no. 5 11. Problem: Typical cases no.6 to 9 Intersection of Surfaces: 1. Essential Information: 2. Display of Engineering Applications: 3. Solution Steps to solve Problem: 4. Case 1: Cylinder to Cylinder: 5. Case 2: Prism to Cylinder: 6. Case 3: Cone to Cylinder 7. Case 4: Prism to Prism: Axis Intersecting. 8. Case 5: Triangular Prism to Cylinder 9. Case 6: Prism to Prism: Axis Skew 10. Case 7 Prism to Cone: from top: 11. Case 8: Cylinder to Cone: Isometric Projections 1. Definitions and explanation 2. Important Terms 3. Types. 4. Isometric of plain shapes-1. 5. Isometric of circle 6. Isometric of a part of circle 7. Isometric of plain shapes-2 8. Isometric of solids & frustums (no.5 to 16) 9. Isometric of sphere & hemi-sphere (no.17 & 18) 10. Isometric of Section of solid.(no.19) 11. Illustrated nineteen Problem (no.20 to 38)  OBJECTIVE OF THIS CD Sky is the limit for vision. Vision and memory are close relatives. Anything in the jurisdiction of vision can be memorized for a long period. We may not remember what we hear for a long time, but we can easily remember and even visualize what we have seen years ago. So vision helps visualization and both help in memorizing an event or situation. Video effects are far more effective, is now an established fact. Every effort has been done in this CD, to bring various planes, objects and situations in-front of observer, so that he/she can further visualize in proper direction and reach to the correct solution, himself. Off-course this all will assist & give good results only when one will practice all these methods and techniques by drawing on sheets with his/her own hands, other wise not! So observe each illustration carefully note proper notes given everywhere Go through the Tips given & solution steps carefully Discuss your doubts with your teacher and make practice yourself. Then success is yours !! Go ahead confidently! CREATIVE TECHNIQUES wishes you best luck !  SCALES DIMENSIONS OF LARGE OBJECTS MUST BE REDUCED TO ACCOMMODATE ON STANDARD SIZE DRAWING SHEET.THIS REDUCTION CREATES A SCALE FOR FULL SIZE SCALE OF THAT REDUCTION RATIO, WHICH IS GENERALLY A FRACTION.. R.F.=1 OR ( 1:1 ) SUCH A SCALE IS CALLED REDUCING SCALE MEANS DRAWING AND & OBJECT ARE OF THAT RATIO IS CALLED REPRESENTATIVE FACTOR. SAME SIZE. Other RFs are described SIMILARLY IN CASE OF TINY OBJECTS DIMENSIONS MUST BE INCREASED as FOR ABOVE PURPOSE. HENCE THIS SCALE IS CALLED ENLARGING SCALE. 1:10, 1:100, 1:1000, 1:1,00,000 HERE THE RATIO CALLED REPRESENTATIVE FACTOR IS MORE THAN UNITY. USE FOLLOWING FORMULAS FOR THE CALCULATIONS IN THIS TOPIC. DIMENSION OF DRAWING A REPRESENTATIVE FACTOR (R.F.) = DIMENSION OF OBJECT LENGTH OF DRAWING = ACTUAL LENGTH AREA OF DRAWING = V ACTUAL AREA VOLUME AS PER DRWG. =3 V ACTUAL VOLUME B LENGTH OF SCALE = R.F. X MAX. LENGTH TO BE MEASURED.  BE FRIENDLY WITH THESE UNITS. 1 KILOMETRE = 10 HECTOMETRES 1 HECTOMETRE = 10 DECAMETRES 1 DECAMETRE = 10 METRES 1 METRE = 10 DECIMETRES 1 DECIMETRE = 10 CENTIMETRES 1 CENTIMETRE = 10 MILIMETRES TYPES OF SCALES: 1. PLAIN SCALES ( FOR DIMENSIONS UP TO SINGLE DECIMAL) 2. DIAGONAL SCALES ( FOR DIMENSIONS UP TO TWO DECIMALS) 3. VERNIER SCALES ( FOR DIMENSIONS UP TO TWO DECIMALS) 4. COMPARATIVE SCALES ( FOR COMPARING TWO DIFFERENT UNITS) 5. SCALE OF CORDS ( FOR MEASURING/CONSTRUCTING ANGLES)  PLAIN SCALE:-This type of scale represents two units or a unit and it’s sub-division. PROBLEM NO.1:- Draw a scale 1 cm = 1m to read decimeters, to measure maximum distance of 6 m. Show on it a distance of 4 m and 6 dm. CONSTRUCTION:- DIMENSION OF DRAWING PLAIN SCALE a) Calculate R.F.= DIMENSION OF OBJECT R.F.= 1cm/ 1m = 1/100 Length of scale = R.F. X max. distance = 1/100 X 600 cm = 6 cms b) Draw a line 6 cm long and divide it in 6 equal parts. Each part will represent larger division unit. c) Sub divide the first part which will represent second unit or fraction of first unit. d) Place ( 0 ) at the end of first unit. Number the units on right side of Zero and subdivisions on left-hand side of Zero. Take height of scale 5 to 10 mm for getting a look of scale. e) After construction of scale mention it’s RF and name of scale as shown. f) Show the distance 4 m 6 dm on it as shown. 4 M 6 DM 10 0 1 2 3 4 5 METERS DECIMETERS R.F. = 1/100 PLANE SCALE SHOWING METERS AND DECIMETERS.  PROBLEM NO.2:- In a map a 36 km distance is shown by a line 45 cms long. Calculate the R.F. and construct a plain scale to read kilometers and hectometers, for max. 12 km. Show a distance of 8.3 km on it. CONSTRUCTION:- a) Calculate R.F. R.F.= 45 cm/ 36 km = 45/ 36. 1000. 100 = 1/ 80,000 PLAIN SCALE Length of scale = R.F. max. distance = 1/ 80000 12 km = 15 cm b) Draw a line 15 cm long and divide it in 12 equal parts. Each part will represent larger division unit. c) Sub divide the first part which will represent second unit or fraction of first unit. d) Place ( 0 ) at the end of first unit. Number the units on right side of Zero and subdivisions on left-hand side of Zero. Take height of scale 5 to 10 mm for getting a look of scale. e) After construction of scale mention it’s RF and name of scale as shown. f) Show the distance 8.3 km on it as shown. 8KM 3HM 10 5 0 1 2 3 4 5 6 7 8 9 10 11 KILOMETERS HECTOMETERS R.F. = 1/80,000 PLANE SCALE SHOWING KILOMETERS AND HECTOMETERS PROBLEM NO.3:- The distance between two stations is 210 km. A passenger train covers this distance in 7 hours. Construct a plain scale to measure time up to a single minute. RF is 1/200,000 Indicate the distance traveled by train in 29 minutes. CONSTRUCTION:- PLAIN SCALE a) 210 km in 7 hours. Means speed of the train is 30 km per hour ( 60 minutes) Length of scale = R.F. max. distance per hour = 1/ 2,00,000 30km = 15 cm b) 15 cm length will represent 30 km and 1 hour i.e. 60 minutes. Draw a line 15 cm long and divide it in 6 equal parts. Each part will represent 5 km and 10 minutes. c) Sub divide the first part in 10 equal parts,which will represent second unit or fraction of first unit. Each smaller part will represent distance traveled in one minute. d) Place ( 0 ) at the end of first unit. Number the units on right side of Zero and subdivisions on left-hand side of Zero. Take height of scale 5 to 10 mm for getting a proper look of scale. e) Show km on upper side and time in minutes on lower side of the scale as shown. After construction of scale mention it’s RF and name of scale as shown. f) Show the distance traveled in 29 minutes, which is 14.5 km, on it as shown. DISTANCE TRAVELED IN 29 MINUTES. 14.5 KM KM 5 2.5 0 5 10 15 20 25 KM MIN 10 0 10 20 30 40 50 MINUTES R.F. = 1/100 PLANE SCALE SHOWING METERS AND DECIMETERS. We have seen that the plain scales give only two dimensions, such as a unit and it’s subunit or it’s fraction. DIAGONAL SCALE The diagonal scales give us three successive dimensions that is a unit, a subunit and a subdivision of a subunit. The principle of construction of a diagonal scale is as follows. Let the XY in figure be a subunit. Y X From Y draw a perpendicular YZ to a suitable height. 10 Join XZ. Divide YZ in to 10 equal parts. 9 Draw parallel lines to XY from all these divisions 8 and number them as shown. 7 From geometry we know that similar triangles have their like sides proportional. 6 5 Consider two similar triangles XYZ and 7’ 7Z, 4 we have 7Z / YZ = 7’7 / XY (each part being one unit) 3 Means 7’ 7 = 7 / 10. x X Y = 0.7 XY :. 2 Similarly 1 1’ – 1 = 0.1 XY 2’ – 2 = 0.2 XY Z Thus, it is very clear that, the sides of small triangles, which are parallel to divided lines, become progressively shorter in length by 0.1 XY. The solved examples ON NEXT PAGES will make the principles of diagonal scales clear.  PROBLEM NO. 4 : The distance between Delhi and Agra is 200 km. In a railway map it is represented by a line 5 cm long. Find it’s R.F. Draw a diagonal scale to show single km. And maximum 600 km. DIAGONAL Indicate on it following distances. 1) 222 km 2) 336 km 3) 459 km 4) 569 km SCALE SOLUTION STEPS: RF = 5 cm / 200 km = 1 / 40, 00, 000 Length of scale = 1 / 40, 00, 000 X 600 X 105 = 15 cm Draw a line 15 cm long. It will represent 600 km.Divide it in six equal parts.( each will represent 100 km.) Divide first division in ten equal parts.Each will represent 10 km.Draw a line upward from left end and mark 10 parts on it of any distance. Name those parts 0 to 10 as shown.Join 9th sub-division of horizontal scale with 10th division of the vertical divisions. Then draw parallel lines to this line from remaining sub divisions and complete diagonal scale. 569 km 459 km 336 km 222 km 10 9 8 7 6 KM 5 4 3 2 1 0 KM 100 50 0 100 200 300 400 500 KM R.F. = 1 / 40,00,000 DIAGONAL SCALE SHOWING KILOMETERS.  PROBLEM NO.5: A rectangular plot of land measuring 1.28 hectors is represented on a map by a similar rectangle of 8 sq. cm. Calculate RF of the scale. Draw a diagonal scale to read single meter. Show a distance of 438 m on it. SOLUTION : DIAGONAL 1 hector = 10, 000 sq. meters SCALE 1.28 hectors = 1.28 X 10, 000 sq. meters Draw a line 15 cm long. = 1.28 X 104 X 104 sq. cm 8 sq. cm area on map represents It will represent 600 m.Divide it in six equal parts. = 1.28 X 104 X 104 sq. cm on land ( each will represent 100 m.) 1 cm sq. on map represents Divide first division in ten equal parts.Each will = 1.28 X 10 4 X 104 / 8 sq cm on land represent 10 m. 1 cm on map represent Draw a line upward from left end and mark 10 parts on it of any distance. = 1.28 X 10 4 X 104 / 8 cm Name those parts 0 to 10 as shown.Join 9th sub-division = 4, 000 cm of horizontal scale with 10th division of the vertical divisions. 1 cm on drawing represent 4, 000 cm, Means RF = 1 / 4000 Then draw parallel lines to this line from remaining sub divisions Assuming length of scale 15 cm, it will represent 600 m. and complete diagonal scale. 438 meters 10 9 8 7 6 5 M 4 3 2 1 0 M 100 50 0 100 200 300 400 500 M R.F. = 1 / 4000 DIAGONAL SCALE SHOWING METERS. PROBLEM NO.6:. Draw a diagonal scale of R.F. 1: 2.5, showing centimeters and millimeters and long enough to measure up to 20 centimeters. SOLUTION STEPS: DIAGONAL R.F. = 1 / 2.5 SCALE Length of scale = 1 / 2.5 X 20 cm. = 8 cm. 1.Draw a line 8 cm long and divide it in to 4 equal parts. (Each part will represent a length of 5 cm.) 2.Divide the first part into 5 equal divisions. (Each will show 1 cm.) 3.At the left hand end of the line, draw a vertical line and on it step-off 10 equal divisions of any length. 4.Complete the scale as explained in previous problems. Show the distance 13.4 cm on it. 13.4 CM 10 9 8 7 6 MM 5 4 3 2 1 0 CM 5 4 3 2 1 0 5 10 15 CENTIMETRES R.F. = 1 / 2.5 DIAGONAL SCALE SHOWING CENTIMETERS. COMPARATIVE SCALES: EXAMPLE NO. 7 : These are the Scales having same R.F. A distance of 40 miles is represented by a line but graduated to read different units. 8 cm long. Construct a plain scale to read 80 miles. These scales may be Plain scales or Diagonal scales Also construct a comparative scale to read kilometers and may be constructed separately or one above the other. upto 120 km ( 1 m = 1.609 km ) SOLUTION STEPS: CONSTRUCTION: Scale of Miles: Take a line 16 cm long and divide it into 8 parts. Each will represent 10 miles. 40 miles are represented = 8 cm Subdivide the first part and each sub-division will measure single mile. : 80 miles = 16 cm R.F. = 8 / 40 X 1609 X 1000 X 100 = 1 / 8, 04, 500 CONSTRUCTION: Scale of Km: On the top line of the scale of miles cut off a distance of 14.90 cm and divide Length of scale it into 12 equal parts. Each part will represent 10 km. = 1 / 8,04,500 X 120 X 1000 X 100 Subdivide the first part into 10 equal parts. Each subdivision will show single km. = 14. 90 cm 10 0 10 20 30 40 50 60 70 80 90 100 110 KM 5 10 5 0 10 20 30 40 50 60 70 MILES R.F. = 1 / 804500 COMPARATIVE SCALE SHOWING MILES AND KILOMETERS  SOLUTION STEPS: Scale of km. COMPARATIVE SCALE: length of scale = RF X 60 km = 1 / 4,00,000 X 60 X 105 = 15 cm. EXAMPLE NO. 8 : CONSTRUCTION: A motor car is running at a speed of 60 kph. Draw a line 15 cm long and divide it in 6 equal parts. On a scale of RF = 1 / 4,00,000 show the distance ( each part will represent 10 km.) traveled by car in 47 minutes. Subdivide 1st part in `0 equal subdivisions. ( each will represent 1 km.) Time Scale: Same 15 cm line will represent 60 minutes. Construct the scale similar to distance scale. It will show minimum 1 minute & max. 60min. 47 MINUTES 10 5 0 10 20 30 40 50 MINUTES MIN. KM 5 40 10 0 10 20 30 50 KM 47 KM R.F. = 1 / 4,00,000 COMPARATIVE SCALE SHOWING MINUTES AND KILOMETERS EXAMPLE NO. 9 : A car is traveling at a speed of 60 km per hour. A 4 cm long line represents the distance traveled by the car in two hours. Construct a suitable comparative scale up to 10 hours. The scale should be able to read the distance traveled in one minute. Show the time required to cover 476 km and also distance in 4 hours and 24 minutes. SOLUTION: COMPARATIVE 4 cm line represents distance in two hours , means for 10 hours scale, 20 cm long line is required, as length SCALE: of scale.This length of scale will also represent 600 kms. ( as it is a distance traveled in 10 hours) CONSTRUCTION: Distance Scale ( km) Draw a line 20 cm long. Divide it in TEN equal parts.( Each will show 60 km) Sub-divide 1st part in SIX subdivisions.( Each will represent 10 km) At the left hand end of the line, draw a vertical line and on it step-off 10 equal divisions of any length. And complete the diagonal scale to read minimum ONE km. Time scale: Draw a line 20 cm long. Divide it in TEN equal parts.( Each will show 1 hour) Sub-divide 1st part in SIX subdivisions.( Each will represent 10 minutes) At the left hand end of the line, draw a vertical line and on it step-off 10 equal divisions of any length. And complete the diagonal scale to read minimum ONE minute. TIME SCALE TO MEASURE MIN 1 MINUTE. 10 5 MIN. 0 60 0 1 2 3 4 5 6 7 8 9 4 hrs 24 min. ( 264 kms ) HOURS 476 kms ( 7 hrs 56 min.) 10 kM 5 0 kM 60 0 60 120 180 240 300 360 420 480 540 DISTANCE SCALE TO MEASURE MIN 1 KM KILOMETERS Vernier Scales: These scales, like diagonal scales , are used to read to a very small unit with great accuracy. It consists of two parts – a primary scale and a vernier. The primary scale is a plain scale fully divided into minor divisions. As it would be difficult to sub-divide the minor divisions in ordinary way, it is done with the help of the vernier. The graduations on vernier are derived from those on the primary scale. Figure to the right shows a part of a plain scale in which length A-O represents 10 cm. If we divide A-O B 9.9 7.7 5.5 3.3 1.1 0 into ten equal parts, each will be of 1 cm. Now it would not be easy to divide each of these parts into ten equal divisions to get measurements in millimeters. A 9 8 7 6 5 4 3 2 1 0 Now if we take a length BO equal to 10 + 1 = 11 such equal parts, thus representing 11 cm, and divide it into ten equal divisions, each of these divisions will represent 11 / 10 – 1.1 cm. The difference between one part of AO and one division of BO will be equal 1.1 – 1.0 = 0.1 cm or 1 mm. This difference is called Least Count of the scale. Minimum this distance can be measured by this scale. The upper scale BO is the vernier.The combination of plain scale and the vernier is vernier scale. Example 10: Draw a vernier scale of RF = 1 / 25 to read centimeters upto Vernier Scale 4 meters and on it, show lengths 2.39 m and 0.91 m SOLUTION: CONSTRUCTION: ( vernier) Length of scale = RF X max. Distance Take 11 parts of Dm length and divide it in 10 equal parts. = 1 / 25 X 4 X 100 Each will show 0.11 m or 1.1 dm or 11 cm and construct a rectangle = 16 cm Covering these parts of vernier. CONSTRUCTION: ( Main scale) Draw a line 16 cm long. TO MEASURE GIVEN LENGTHS: Divide it in 4 equal parts. (1) For 2.39 m : Subtract 0.99 from 2.39 i.e. 2.39 -.99 = 1.4 m ( each will represent meter ) The distance between 0.99 ( left of Zero) and 1.4 (right of Zero) is 2.39 m Sub-divide each part in 10 equal parts. (2) For 0.91 m : Subtract 0.11 from 0.91 i.e. 0.91 – 0.11 =0.80 m ( each will represent decimeter ) The distance between 0.11 and 0.80 (both left side of Zero) is 0.91 m Name those properly. 2.39 m 0.91 m 1.1.99.77.55.33.11 0 1.0.9.8.7.6.5.4.3.2.1 0 1 1.4 2 3 METERS METERS Example 11: A map of size 500cm X 50cm wide represents an area of 6250 sq.Kms. Construct a vernier scaleto measure kilometers, hectometers and decameters Vernier Scale and long enough to measure upto 7 km. Indicate on it a) 5.33 km b) 59 decameters. SOLUTION: CONSTRUCTION: ( Main scale) TO MEASURE GIVEN LENGTHS: AREA OF DRAWING Draw a line 14 cm long. a) For 5.33 km : RF = Divide it in 7 equal parts. Subtract 0.33 from 5.33 V ACTUAL AREA i.e. 5.33 - 0.33 = 5.00 ( each will represent km ) = 500 X 50 cm sq. Sub-divide each part in 10 equal parts. The distance between 33 dm V 6250 km sq. ( left of Zero) and ( each will represent hectometer ) 5.00 (right of Zero) is 5.33 k m = 2 / 105 Name those properly. (b) For 59 dm : Subtract 0.99 from 0.59 Length of CONSTRUCTION: ( vernier) i.e. 0.59 – 0.99 = - 0.4 km scale = RF X max. Distance ( - ve sign means left of Zero) Take 11 parts of hectometer part length = 2 / 105 X 7 kms The distance between 99 dm and and divide it in 10 equal parts. -.4 km is 59 dm = 14 cm Each will show 1.1 hm m or 11 dm and (both left side of Zero) Covering in a rectangle complete scale. 59 dm 5.33 km Decameters 99 77 55 33 11 90 70 50 30 10 10 0 1 2 3 4 5 6 HECTOMETERS KILOMETERS  800 900 700 600 SCALE OF CORDS 500 400 300 200 100 00 A O 0 10 20 30 40 50 60 70 80 90 CONSTRUCTION: 1. DRAW SECTOR OF A CIRCLE OF 900 WITH ‘OA’ RADIUS. ( ‘OA’ ANY CONVINIENT DISTANCE ) 2. DIVIDE THIS ANGLE IN NINE EQUAL PARTS OF 10 0 EACH. 3. NAME AS SHOWN FROM END ‘A’ UPWARDS. 4. FROM ‘A’ AS CENTER, WITH CORDS OF EACH ANGLE AS RADIUS DRAW ARCS DOWNWARDS UP TO ‘AO’ LINE OR IT’S EXTENSION AND FORM A SCALE WITH PROPER LABELING AS SHOWN. AS CORD LENGTHS ARE USED TO MEASURE & CONSTRUCT DIFERENT ANGLES IT IS CALLED SCALE OF CORDS. PROBLEM 12: Construct any triangle and measure it’s angles by using scale of cords. CONSTRUCTION: SCALE OF CORDS First prepare Scale of Cords for the problem. 0 Then construct a triangle of given sides. ( You are supposed to measure angles x, y and z) 800 90 700 To measure angle at x: 600 Take O-A distance in compass from cords scale and mark it on lower side of triangle 500 as shown from corner x. Name O & A as shown. Then O as center, O-A radius 400 draw an arc upto upper adjacent side.Name the point B. Take A-B cord in compass and place on scale of cords from Zero. 300 It will give value of angle at x To measure angle at y: 200 Repeat same process from O1. Draw arc with radius O1A1. 100 Place Cord A1B1 on scale and get angle at y. To measure angle at z: A O 00 Subtract the SUM of these two angles from 1800 to get angle at z. 0 10 20 30 40 50 60 70 80 90 B1 z B 550 300 y x O1 A1 A O Angle at z = 180 – ( 55 + 30 ) = 950 PROBLEM 12: Construct 250 and 1150 angles with a horizontal line , by using scale of cords. CONSTRUCTION: SCALE OF CORDS First prepare Scale of Cords for the problem. 0 Then Draw a horizontal line. Mark point O on it. 0 800 90 70 To construct 250 angle at O. 600 Take O-A distance in compass from cords scale and mark it on on the line drawn, from O 500 Name O & A as shown. Then O as center, O-A radius draw an arc upward.. 400 Take cord length of 250 angle from scale of cords in compass and from A cut the arc at point B.Join B with O. The angle AOB is thus 25 0 300 To construct 1150 angle at O. This scale can measure or construct angles upto 900 only directly. 200 Hence Subtract 1150 from 1800.We get 750 angle , 100 which can be constructed with this scale. A O Extend previous arc of OA radius and taking cord length of 75 0 in compass cut this arc 00 at B1 with A as center. Join B1 with O. Now angle AOB1 is 750 and angle COB1 is 1150. 0 10 20 30 40 50 60 70 80 90 B B1 250 A O 750 1150 C A O To construct 250 angle at O. To construct 1150 angle at O.  ENGINEERING CURVES Part- I {Conic Sections} ELLIPSE PARABOLA HYPERBOLA 1.Concentric Circle Method 1.Rectangle Method 1.Rectangular Hyperbola (coordinates given) 2.Rectangle Method 2 Method of Tangents ( Triangle Method) 2 Rectangular Hyperbola 3.Oblong Method (P-V diagram - Equation given) 3.Basic Locus Method 4.Arcs of Circle Method (Directrix – focus) 3.Basic Locus Method (Directrix – focus) 5.Rhombus Metho 6.Basic Locus Method Methods of Drawing (Directrix – focus) Tangents & Normals To These Curves.  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 Section Plane Through Generators Hyperbola Parallel to Axis. Section Plane Parallel to end generator.  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 E1 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.  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 B 4. From all points of outer circle draw vertical 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 Steps: ELLIPSE BY RECTANGLE METHOD 1 Draw a rectangle taking major and minor axes as sides. 2. In this rectangle draw both Problem 2 axes as perpendicular bisectors of Draw ellipse by Rectangle method. each other.. 3. For construction, select upper Take major axis 100 mm and minor axis 70 mm long. left part of rectangle. Divide vertical small side and horizontal long side into same number of D 4 4 equal parts.( here divided in four parts) 3 3 4. Name those as shown.. 5. Now join all vertical points 2 2 1,2,3,4, to the upper end of minor 1 1 axis. And all horizontal points i.e.1,2,3,4 to the lower end of minor axis. A B 6. Then extend C-1 line upto D-1 and mark that point. Similarly extend C-2, C-3, C-4 lines up to D-2, D-3, & D-4 lines. 7. Mark all these points properly and join all along with ends A and D in smooth possible curve. Do similar construction in right C side part.along with lower half of the rectangle.Join all points in smooth curve. It is required ellipse.  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 75 0.Inscribe Ellipse in it. STEPS ARE SIMILAR TO THE PREVIOUS CASE (RECTANGLE METHOD) ONLY IN PLACE OF RECTANGLE, HERE IS A PARALLELOGRAM. 4 4 3 3 2 2 1 1 A B PROBLEM 4. ELLIPSE MAJOR AXIS AB & MINOR AXIS CD ARE BY ARCS OF CIRCLE METHOD 100 AMD 70MM LONG RESPECTIVELY.DRAW ELLIPSE BY ARCS OF CIRLES METHOD. As per the definition Ellipse is locus of point P moving in a plane such that the SUM of it’s distances from two fixed STEPS: points (F1 & F2) remains constant and equals to the length 1.Draw both axes as usual.Name the of major axis AB.(Note A.1+ B.1=A. 2 + B. 2 = AB) ends & intersecting point 2.Taking AO distance I.e.half major axis, from C, mark F1 & F2 On AB. p4 C ( focus 1 and 2.) p3 3.On line F1- O taking any distance, p2 mark points 1,2,3, & 4 p1 4.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 B 5.Repeat this step with same centers but A O F1 1 2 3 4 F2 taking now A-2 & B-2 distances for drawing arcs. Name the point p2 6.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 PROBLEM 5. ELLIPSE DRAW RHOMBUS OF 100 MM & 70 MM LONG BY RHOMBUS METHOD DIAGONALS AND INSCRIBE AN ELLIPSE IN IT. STEPS: 2 1. Draw rhombus of given dimensions. 2. Mark mid points of all sides & name Those A,B,C,& D 3. 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 PROBLEM 6:- POINT F IS 50 MM FROM A LINE AB.A POINT P IS MOVING IN A PLANE ELLIPSE 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. { ECCENTRICITY = 2/3 } DIRECTRIX-FOCUS METHOD ELLIPSE A STEPS: 1.Draw a vertical line AB and point F 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. 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 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 6 1.Draw rectangle of above size and divide it in two equal vertical parts 2.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 4 3.Join vertical 1,2,3,4,5 & 6 to the top center of rectangle 4.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 Problem no.8: Draw an isosceles triangle of 100 mm long base and PARABOLA 110 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 dimensions. 2. Divide it’s both sides in to same no.of equal parts. 3. Name the parts in ascending and descending manner, as shown. 4. Join 1-1, 2-2,3-3 and so on. 5. Draw the curve as shown i.e.tangent to all these lines. The above all lines being tangents to the curve, it is called method of tangents. A B PROBLEM 9: Point F is 50 mm from a vertical straight line AB. PARABOLA Draw locus of point P, moving in a plane such that DIRECTRIX-FOCUS METHOD it always remains equidistant from point F and line AB. PARABOLA SOLUTION 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 P1 those draw lines parallel to AB. 3.Mark 5 mm distance to its left of P and (VERTEX) V name it 1. O 1 2 3 4 F 4.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 P3P4. B 6.Join all these points in smooth curve. It will be the locus of P equidistance from line AB and fixed point F. Problem No.10: Point P is 40 mm and 30 mm from horizontal HYPERBOLA and 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 curve. Problem no.11: A sample of gas is expanded in a cylinder HYPERBOLA from 10 unit pressure to 1 unit pressure.Expansion follows law PV=Constant.If initial volume being 1 unit, draw the P-V DIAGRAM curve 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 ) PROBLEM 12:- POINT F IS 50 MM FROM A LINE AB.A POINT P IS MOVING IN A PLANE HYPERBOLA 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. { ECCENTRICITY = 2/3 } DIRECTRIX FOCUS METHOD STEPS: A 1.Draw a vertical line AB and point F 50 mm from it. 2.Divide 50 mm distance in 5 parts. 30mm 3.Name 2nd part from F as V. It is 20mm and 30mm from F and AB line resp. 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 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  ELLIPSE Problem 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 Q D  ELLIPSE Problem 14: TANGENT & NORMAL TO DRAW TANGENT & NORMAL TO THE CURVE ELLIPSE FROM A GIVEN POINT ( Q ) A 1.JOIN POINT Q TO F. T 2.CONSTRUCT 900 ANGLE WITH THIS LINE AT POINT F 3.EXTEND THE LINE TO MEET DIRECTRIX AT T 4. JOIN THIS POINT TO Q AND EXTEND. THIS IS TANGENT TO ELLIPSE FROM Q (vertex) V 5.TO THIS TANGENT DRAW PERPENDICULAR F ( focus) LINE FROM Q. IT IS NORMAL TO CURVE. 900 N Q N B T  PARABOLA Problem 15: TANGENT & NORMAL TO DRAW TANGENT & NORMAL TO THE CURVE T PARABOLA FROM A GIVEN POINT ( Q ) A 1.JOIN POINT Q TO F. 2.CONSTRUCT 900 ANGLE WITH THIS LINE AT POINT F 3.EXTEND THE LINE TO MEET DIRECTRIX AT T 4. JOIN THIS POINT TO Q AND EXTEND. THIS IS VERTEX V TANGENT TO THE CURVE FROM Q 900 F 5.TO THIS TANGENT DRAW PERPENDICULAR ( focus) LINE FROM Q. IT IS NORMAL TO CURVE. N Q B N T  HYPERBOLA Problem 16 TANGENT & NORMAL TO DRAW TANGENT & NORMAL TO THE CURVE FROM A GIVEN POINT ( Q ) A 1.JOIN POINT Q TO F. 2.CONSTRUCT 900 ANGLE WITH THIS LINE AT POINT F T 3.EXTEND THE LINE TO MEET DIRECTRIX AT T 4. 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  ENGINEERING CURVES Part-II (Point undergoing two types of displacements) INVOLUTE CYCLOID SPIRAL HELIX 1. Involute of a circle 1. General Cycloid 1. Spiral of 1. On Cylinder a)String Length = D One Convolution. 2. Trochoid 2. On a Cone b)String Length > D ( superior) 2. Spiral of 3. Trochoid Two Convolutions. c)String Length < D ( Inferior) 4. Epi-Cycloid 2. Pole having Composite shape. 5. Hypo-Cycloid 3. Rod Rolling over a Semicircular Pole. AND Methods of Drawing Tangents & Normals To These Curves.  DEFINITIONS CYCLOID: IT IS A LOCUS OF A POINT ON THE SUPERIORTROCHOID: PERIPHERY OF A CIRCLE WHICH IF THE POINT IN THE DEFINATION ROLLS ON A STRAIGHT LINE PATH. OF CYCLOID IS OUTSIDE THE CIRCLE INFERIOR TROCHOID.: INVOLUTE: IF IT IS INSIDE THE CIRCLE IT IS A LOCUS OF A FREE END OF A STRING WHEN IT IS WOUND ROUND A CIRCULAR POLE EPI-CYCLOID IF THE CIRCLE IS ROLLING ON ANOTHER CIRCLE FROM OUTSIDE SPIRAL: HYPO-CYCLOID. IT IS A CURVE GENERATED BY A POINT IF THE CIRCLE IS ROLLING FROM WHICH REVOLVES AROUND A FIXED POINT INSIDE THE OTHER CIRCLE, AND AT THE SAME MOVES TOWARDS IT. HELIX: IT IS A CURVE GENERATED BY A POINT WHICH MOVES AROUND THE SURFACE OF A RIGHT CIRCULAR CYLINDER / CONE AND AT THE SAME TIME ADVANCES IN AXIAL DIRECTION AT A SPEED BEARING A CONSTANT RATIO TO THE SPPED OF ROTATION. ( for problems refer topic Development of surfaces)  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 is exactly D distance away from A. Means if this string is wound round the circle, it will completely cover P2 given circle. B will meet A after winding. 2) Divide D (AP) distance into 8 P3 number of equal parts. P1 3) Divide circle also into 8 number of equal parts. 4) Name after A, 1, 2, 3, 4, etc. up to 8 on D line AP as well as on circle (in anticlockwise direction). 5) To radius C-1, C-2, C-3 up to C-8 draw tangents (from 1,2,3,4,etc to 4 to p circle). P4 4 6) Take distance 1 to P in compass 5 and mark it on tangent from point 1 3 on circle (means one division less 6 than distance AP). 2 7) Name this point P1 7 8) Take 2-B distance in compass 1 p8 and mark it on the tangent from P P5 P8 point 2. Name it point P2. 1 2 3 4 5 6 7 8 9) Similarly take 3 to P, 4 to P, 5 to P7 P up to 7 to P distance in compass P6  and mark on respective tangents and locate P3, P4, P5 up to P8 (i.e. D A) points and join them in smooth curve it is an INVOLUTE of a given circle.  INVOLUTE OF A CIRCLE Problem 18: Draw Involute of a circle. String length MORE than D String length is MORE than the circumference of circle. Solution Steps: P2 In this case string length is more than  D. But remember! Whatever may be the length of P3 P1 string, mark  D distance horizontal i.e.along the string and divide it in 8 number of 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 7 8 p8 1 P P5 2 3 4 5 6 7 8 P7 165 mm P6 (more than D) D Problem 19: Draw Involute of a circle. INVOLUTE OF A CIRCLE String length is LESS than the circumference of circle. String length LESS than D Solution Steps: P2 In this case string length is Less than  D. But remember! Whatever may be the length of P3 P1 string, mark  D distance horizontal i.e.along the string and divide it in 8 number of 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 P 8 P7 1 2 3 4 5 6 7 8 P6 150 mm (Less than D) D  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 POLE SOLUTION STEPS: Draw pole shape as per dimensions. P1 Divide semicircle in 4 parts and name those P along with corners of P2 hexagon. Calculate perimeter length. 1 to P Show it as string AP. On this line mark 30mm from A Mark and name it 1 Mark D/2 distance on it from 1 And dividing it in 4 parts P3 name 2,3,4,5. 3 to P 3 Mark point 6 on line 30 4 2 mm from 5 5 1 Now draw tangents from all points of pole and proper lengths as done in all previous 6 A involute’s problems and 1 2 3 4 5 6 P complete the curve. D/2 P4 P6 P5 PROBLEM 21 : 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. B Draw 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! D 2 A2 B2 2 1 3 1 A1 4 A B3 B4 PROBLEM 22: DRAW LOCUS OF A POINT ON THE PERIPHERY OF A CIRCLE CYCLOID WHICH 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 D 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 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.  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 p6 2 6 2 p7 1 p1 7 P D p8 Solution Steps: 1) Draw circle of given diameter and draw a horizontal line from it’s center C of length  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 join 4) This curve is called Superior Trochoid.  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 p p5 3 5 p2 3 C C1 C2 C3 C4 C5 C6 C7p6 C8 2 6 p1 p7 1 7 P p8 D Solution Steps: 1) Draw circle of given diameter and draw a horizontal line from it’s center C of length  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. PROBLEM 25: DRAW LOCUS OF A POINT ON THE PERIPHERY OF A CIRCLE WHICH 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 on larger circle for one revolution it will cover  D distance on arc and it will be decided by included arc angle . 2) Calculate  by formula  = (r/R) x Generating/ 3600. Rolling Circle 3) Construct angle  with radius OC 4 5 and draw an arc by taking O as center C2 OC as radius and form sector of angle 3 6 . 4) Divide this sector into 8 number of 7 equal angular parts. And from C 2 onward name them C1, C2, C3 up to 1 C8. r = CP P 5) Divide smaller circle (Generating circle) also in 8 number of equal parts. And next to P in clockwise direction Directing Circle name those 1, 2, 3, up to 8. 6) With O as center, O-1 as radius r 3600 draw an arc in the sector. Take O-2, O- = R 3, O-4, O-5 up to O-8 distances with center O, draw all concentric arcs in O sector. Take fixed distance C-P in compass, C1 center, cut arc of 1 at P1. Repeat procedure and locate P2, P3, P4, P5 unto P8 (as in cycloid) and join them by smooth curve. This is EPI – CYCLOID. PROBLEM 26: DRAW LOCUS OF A POINT ON THE PERIPHERY OF A CIRCLE WHICH ROLLS FROM THE INSIDE OF A CURVED PATH. Take diameter of HYPO CYCLOID rolling 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 in numbering direction of 8 number of equal parts 5 on the smaller circle. P3 3) From next to P in 2 anticlockwise direction, 4 P4 name 1,2,3,4,5,6,7,8. 3 4) Further all steps are P5 P8 that of epi – cycloid. This P6 P7 is called HYPO – CYCLOID. r = 3600 R O OC = R ( Radius of Directing Circle) CP = r (Radius of Generating Circle) Problem 27: Draw a spiral of one convolution. Take distance PO 40 mm. SPIRAL IMPORTANT APPROACH FOR CONSTRUCTION! FIND TOTAL ANGULAR AND TOTAL LINEAR DISPLACEMENT AND DIVIDE BOTH IN TO SAME NUMBER OF EQUAL PARTS. 2 P2 Solution Steps 3 1 P1 1. With PO radius draw a circle and divide it in EIGHT parts. P3 Name those 1,2,3,4, etc. up to 8 2.Similarly divided line PO also in EIGHT parts and name those 4 P4 O P 1,2,3,-- as shown. 7 6 5 4 3 2 1 3. Take o-1 distance from op line P7 and draw an arc up to O1 radius P5 P6 vector. Name the point P1 4. Similarly mark points P2, P3, P4 up to P8 And join those in a smooth curve. 5 7 It is a SPIRAL of one convolution. 6 Problem 28 SPIRAL Point P is 80 mm from point O. It starts moving towards O and reaches it in two of revolutions around.it Draw locus of point P (To draw a Spiral of TWO convolutions). two convolutions IMPORTANT APPROACH FOR CONSTRUCTION! FIND TOTAL ANGULAR AND TOTAL LINEAR DISPLACEMENT AND DIVIDE BOTH IN TO SAME NUMBER OF EQUAL PARTS. 2,10 P2 3,11 P1 1,9 SOLUTION STEPS: P3 Total angular displacement here P10 is two revolutions And P9 Total Linear displacement here P11 is distance PO. 16 13 10 8 7 6 5 4 3 2 1 P Just divide both in same parts i.e. 4,12 P4 P8 8,16 P12 Circle in EIGHT parts. P15 ( means total angular displacement P13 P14 in SIXTEEN parts) Divide PO also in SIXTEEN parts. P7 Rest steps are similar to the previous P5 problem. P6 5,13 7,15 6,14  HELIX (UPON A CYLINDER) PROBLEM: Draw a helix of one convolution, upon a cylinder. P8 Given 80 mm pitch and 50 mm diameter of a cylinder. 8 (The axial advance during one complete revolution is called P7 The pitch of the helix) 7 P6 6 P5 SOLUTION: 5 Draw projections of a cylinder. Divide circle and axis in to same no. of equal parts. ( 8 ) 4 P4 Name those as shown. 3 Mark initial position of point ‘P’ P3 Mark various positions of P as shown in animation. 2 P2 Join all points by smooth possible curve. Make upper half dotted, as it is going behind the solid 1 P1 and hence will not be seen from front side. P 6 7 5 P 4 1 3 2  HELIX PROBLEM: Draw a helix of one convolution, upon a cone, P8 (UPON A CONE) diameter of base 70 mm, axis 90 mm and 90 mm pitch. (The axial advance during one complete revolution is called P7 The pitch of the helix) P6 P5 SOLUTION: Draw projections of a cone Divide circle and axis in to same no. of equal parts. ( 8 ) P4 Name those as shown. Mark initial position of point ‘P’ P3 Mark various positions of P as shown in animation. Join all points by smooth possible curve. P2 Make upper half dotted, as it is going behind the solid and hence will not be seen from front side. P1 X P Y 6 7 5 P6 P5 P7 P4 P 4 P8 P1 P3 1 3 P2 2 STEPS: Involute DRAW INVOLUTE AS USUAL. Method of Drawing MARK POINT Q ON IT AS DIRECTED. Tangent & Normal JOIN Q TO THE CENTER OF CIRCLE C. CONSIDERING CQ DIAMETER, DRAW A SEMICIRCLE AS SHOWN. INVOLUTE OF A CIRCLE MARK POINT OF INTERSECTION OF THIS SEMICIRCLE AND POLE CIRCLE AND JOIN IT TO Q. Q THIS WILL BE NORMAL TO INVOLUTE. DRAW A LINE AT RIGHT ANGLE TO THIS LINE FROM Q. IT WILL BE TANGENT TO INVOLUTE. 4 3 5 C 2 6 1 7 8 P P8 1 2 3 4 5 6 7 8  D STEPS: DRAW CYCLOID AS USUAL. CYCLOID MARK POINT Q ON IT AS DIRECTED. Method of Drawing WITH CP DISTANCE, FROM Q. CUT THE Tangent & Normal POINT ON LOCUS OF C AND JOIN IT TO Q. FROM THIS POINT DROP A PERPENDICULAR ON GROUND LINE AND NAME IT N JOIN N WITH Q.THIS WILL BE NORMAL TO CYCLOID. DRAW A LINE AT RIGHT ANGLE TO THIS LINE FROM Q. IT WILL BE TANGENT TO CYCLOID. CYCLOID Q C C1 C2 C3 C4 C5 C6 C7 C8 P N D  Spiral. Method of Drawing Tangent & Normal SPIRAL (ONE CONVOLUSION.) 2 P2 3 1 Difference in length of any radius vectors Q P1 Constant of the Curve = Angle between the corresponding radius vector in radian. P3 OP – OP2 OP – OP2 = = /2 1.57 4 P4 O P = 3.185 m.m. 7 6 5 4 3 2 1 P7 STEPS: *DRAW SPIRAL AS USUAL. P5 P6 DRAW A SMALL CIRCLE OF RADIUS EQUAL TO THE CONSTANT OF CURVE CALCULATED ABOVE. * LOCATE POINT Q AS DISCRIBED IN PROBLEM AND 5 7 THROUGH IT DRAW A TANGENTTO THIS SMALLER CIRCLE.THIS IS A NORMAL TO THE SPIRAL. *DRAW A LINE AT RIGHT ANGLE 6 *TO THIS LINE FROM Q. IT WILL BE TANGENT TO CYCLOID.  LOCUS It is a path traced out by a point moving in a plane, in a particular manner, for one cycle of operation. The cases are classified in THREE categories for easy understanding. A} Basic Locus Cases. B} Oscillating Link…… C} Rotating Link……… Basic Locus Cases: Here some geometrical objects like point, line, circle will be described with there relative Positions. Then one point will be allowed to move in a plane maintaining specific relation with above objects. And studying situation carefully you will be asked to draw it’s locus. Oscillating & Rotating Link: Here a link oscillating from one end or rotating around it’s center will be described. Then a point will be allowed to slide along the link in specific manner. And now studying the situation carefully you will be asked to draw it’s locus. STUDY TEN CASES GIVEN ON NEXT PAGES  Basic Locus Cases: PROBLEM 1.: Point F is 50 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. P7 A P5 SOLUTION STEPS: 1.Locate center of line, perpendicular to P3 AB from point F. This will be initial point P. P1 2.Mark 5 mm distance to its right side, name those points 1,2,3,4 and from those draw lines parallel to AB. 3.Mark 5 mm distance to its left of P and p name it 1. 1 2 3 4 F 4 3 2 1 4.Take F-1 distance as radius and F as center draw an arc cutting first parallel line to AB. Name upper point P1 and lower point P2. P2 5.Similarly repeat this process by taking again 5mm to right and left and locate P4 P3P4. 6.Join all these points in smooth curve. P6 B P8 It will be the locus of P equidistance from line AB and fixed point F.  Basic Locus Cases: PROBLEM 2 : A circle of 50 mm diameter has it’s center 75 mm from a vertical line AB.. Draw locus of point P, moving in a plane such that it always remains equidistant from given circle and line AB. P7 P5 A SOLUTION STEPS: P3 1.Locate center of line, perpendicular to 50 D AB from the periphery of circle. This will be initial point P. P1 2.Mark 5 mm distance to its right side, name those points 1,2,3,4 and from those draw lines parallel to AB. 3.Mark 5 mm distance to its left of P and p name it 1,2,3,4. C 4 3 2 1 1 2 3 4 4.Take C-1 distance as radius and C as center draw an arc cutting first parallel line to AB. Name upper point P1 and lower point P2. P2 5.Similarly repeat this process by taking again 5mm to right and left and locate P3P4. P4 6.Join all these points in smooth curve. B P6 It will be the locus of P equidistance P8 from line AB and given circle. 75 mm  PROBLEM 3 : Basic Locus Cases: Center of a circle of 30 mm diameter is 90 mm away from center of another circle of 60 mm diameter. Draw locus of point P, moving in a plane such that it always remains equidistant from given two circles. SOLUTION STEPS: 1.Locate center of line,joining two 60 D centers but part in between periphery P7 of two circles.Name it P. This will be P5 30 D initial point P. P3 2.Mark 5 mm distance to its right side, name those points 1,2,3,4 and P1 from those draw arcs from C1 As center. 3. Mark 5 mm distance to its right p side, name those points 1,2,3,4 and C1 C2 4 3 2 1 1 2 3 4 from those draw arcs from C2 As center. 4.Mark various positions of P as per P2 previous problems and name those P4 similarly. P6 5.Join all these points in smooth curve. P8 It will be the locus of P equidistance from given two circles. 95 mm