Page 2 : 1st angle projection and third angle projection

Page 3 : Construction of polygons, • Bisector

Page 4 : • Divide a line, •, •, •, •, •, , Draw AB in a given length, Draw AC at any angle to AB, Make equal division in AC, Join B5, Draw lies parallel to B5

Page 6 : • Hexagonal drawing, •, •, •, •, , Draw circle of 2x side of hexagon, Using compass, cut side length continuously, Or, Take a side , and using protractor plot another side of 120o along the, circumference of the circle, • Or, • Draw circle with 2 x side of the hexagon, • With A and D as centre of same diameter mark point B,C,E,F points, , • Referance: Google

Page 7 : • Pentagon, • Total interior angle = (n-2) x 180= (5-2) x 180 =540o, • Interior angle =(n-2) x 180/n=720o/6 =108o, , • Hexagon, • Total interior angle = (n-2) x 180= (6-2) x 180 =720o, • Interior angle =(n-2) x 180/n=720o/6 =120o

Page 8 : Draw any polygon when side is given, • Draw AB , given length of the polygon, • At B draw BP = AB , perpendicular to AB, • Draw straight line AP, • Draw arc AP with center B and radius AB, • The perpendicular bisector of the AB meets straight line AP and arc AP n 4 and, 6 respectively, • Draw circle with radius 4B,5B,6B,7B,8B with centre 4,5,6,7 and 8 to get to, inscribe square , pentagon , hexagon , heptagon and octagon inside the circle

Page 9 : Engineering curves (Conic section), Engineering, curve, ellipse, , Parabola, , hyperbola

Page 10 : Contd..

Page 11 : What is engineering curves, • 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), • It is the ratio of the distances from any point of the conic section to, its focus to the same point to its corresponding directrix. The, eccentricity of a conic section tells the measure of how much the, curve deviates from being circular., A)For Ellipse, e<1, B)For Parabola e=1, C)For Hyperbola e>1

Page 12 : 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.

Page 14 : Ellipse- concentric circle method, • Steps:, • 1. Draw both axes as perpendicular bisectors, of each other & name their ends as shown., • 2. Taking their intersecting point as a center,, draw two concentric circles considering both, as respective diameters., • 3. Divide both circles in 12 equal parts &, name as shown., • 4. From all points of outer circle draw vertical, lines downwards and upwards respectively., • 5.From all points of inner circle draw, horizontal lines to intersect those vertical lines., • 6. Mark all intersecting points properly as, those are the points on ellipse., • 7. Join all these points along with the ends of, both axes in smooth possible curve. It is, required ellipse.

Page 15 : Ellipse- Rectangular method, • Steps:, • 1 Draw a rectangle taking major and minor axes as sides., • 2. In this rectangle draw both axes as perpendicular bisectors of each, other.., , • 3. For construction, select upper left part of rectangle. Divide vertical small, side and horizontal long side into same number of equal parts.(here divided, in four parts), • 4. Name those as shown.., • 5. Now join all vertical points 1,2,3,4, to the upper end of minor axis. And, all horizontal points i.e.1,2,3,4 to the lower end of minor axis., • 6. Then extend C-1 line upto D-1 and mark that point. Similarly extend C2, 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 side part.along, with lower half of the rectangle.Join all points in smooth curve., • It is required ellipse.

Page 16 : Ellipse – eccentricity method

Page 17 : Parabola- eccentricity method

Page 18 : Helix and involute, • Involute, • IT IS A LOCUS OF A FREE END OF A STRING, • WHEN IT IS WOUND ROUND A CIRCULAR POLE, , • 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

Page 19 : Draw Involute of a circle., String length is equal to the circumference of circle, , 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 given circle. B will meet A after winding., 2) Divide D (AP) distance into 8 number of, equal parts., 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 circle)., 6) Take distance 1 to P in compass and mark it, on tangent from point 1 on circle (means one, division less than distance AP)., 7) Name this point P1, 8) Take 2-B distance in compass and mark it, on the tangent from point 2. Name it point P2., 9) Similarly take 3 to P, 4 to P, 5 to P up to 7 to, P distance in compass and mark on respective, tangents and locate P3, P4, P5 up to P8 (i.e. A), points and join them in smooth curve it is an, INVOLUTE of a given circle.

Page 20 : Draw a helix of one convolution, upon a cylinder. Given 80 mm pitch and 50 mm diameter of a, cylinder. (The axial advance during one complete revolution is called The pitch of the helix), , Draw projections of a cylinder., Divide circle and axis in to same no. of equal parts., Name those as shown., Mark initial position of point ‘P’, Mark various positions of P as shown in figure., Join all points by smooth possible curve., Make upper half dotted, as it is going behind the solid, and hence will not be seen from front side.

Page 21 : Draw a helix of one convolution, upon a cone, diameter of base 70 mm, axis 90 mm and 90, mm pitch. (The axial advance during one complete revolution is called The pitch of the helix), , SOLUTION:, Draw projections of a cone, Divide circle and axis in to same no. of equal parts. ( 8 ), Name those as shown., Mark initial position of point ‘P’, Mark various positions of P as shown in animation., Join all points by smooth possible curve., Make upper half dotted, as it is going behind the solid, and hence will not be seen from front side.

Page 22 : 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)

Page 23 : 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, a) Calculate R.F.=, , PLAIN SCALE, , 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, DECIMETERS, , 0, , 1, , 2, , 3, , R.F. = 1/100, PLANE SCALE SHOWING METERS AND DECIMETERS., , 4, , 5 METERS

Page 24 : 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., From Y draw a perpendicular YZ to a suitable height., Join XZ. Divide YZ in to 10 equal parts., Draw parallel lines to XY from all these divisions, and number them as shown., From geometry we know that similar triangles have, their like sides proportional., , X, , Y, 10, 9, 8, 7, , 6, 5, , Consider two similar triangles XYZ and 7’ 7Z,, we have 7Z / YZ = 7’7 / XY (each part being one unit), Means 7’ 7 = 7 / 10. x X Y = 0.7 XY, :., Similarly, 1’ – 1 = 0.1 XY, 2’ – 2 = 0.2 XY, 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., , 4, 3, 2, , 1, Z

Page 25 : 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., Indicate on it following distances. 1) 222 km 2) 336 km 3) 459 km 4) 569 km, SOLUTION STEPS:, , DIAGONAL, SCALE, , 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, , KM, , 222 km, , KM, , 10, 9, 8, 7, 6, 5, 4, 3, 2, 1, 0, , 100, , 50, , 0, , 100, , 200, , 300, , R.F. = 1 / 40,00,000, DIAGONAL SCALE SHOWING KILOMETERS., , 400, , 500 KM

Page 26 : 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, 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., 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., , B, , 9.9, , 7.7, , 5.5, , A 9, , 8 7, , 6 5 4 3 2 1, , 3.3, , 1.1 0, , 0

Page 27 : Example 10:, Draw a vernier scale of RF = 1 / 25 to read centimeters upto, 4 meters and on it, show lengths 2.39 m and 0.91 m, SOLUTION:, Length of scale = RF X max. Distance, = 1 / 25 X 4 X 100, = 16 cm, CONSTRUCTION: ( Main scale), Draw a line 16 cm long., Divide it in 4 equal parts., ( each will represent meter ), Sub-divide each part in 10 equal parts., ( each will represent decimeter ), Name those properly., , Vernier Scale, , CONSTRUCTION: ( vernier), Take 11 parts of Dm length and divide it in 10 equal parts., Each will show 0.11 m or 1.1 dm or 11 cm and construct a rectangle, Covering these parts of vernier., , TO MEASURE GIVEN LENGTHS:, (1) For 2.39 m : Subtract 0.99 from 2.39 i.e. 2.39 - .99 = 1.4 m, The distance between 0.99 ( left of Zero) and 1.4 (right of Zero) is 2.39 m, (2) For 0.91 m : Subtract 0.11 from 0.91 i.e. 0.91 – 0.11 =0.80 m, The distance between 0.11 and 0.80 (both left side of Zero) is 0.91 m, , 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, METERS, , 1, , 1.4, , 2, , 3 METERS