Engineering Drawing 1 Introduction Engineering Drawing is the language of engineers. It is a graphic language, and has its own rules of grammar. Engineers do calculations using the knowledge of various subjects and convert them into drawings. Without deep and sound knowledge of engineering drawing , an engineer would not be a complete professional. By learning how to create a drawing, students will be able to read, understand and interpret it. Although a machine or machine part can be described in words, generally, some technical and hidden information cannot be explained in words when the part or machine is of complex nature. Therefore, Engineering Drawing is a good method of describing a machine or machine part. As compared to verbal or written description, this method is brief, accurate, simple, and clear. 2 Drawing Instruments Drawing board, T- sqaure, Drawing sheet Instrument box containing compass, divider, etc. Scales Protractor French curves Drawing pencils Eraser Drawing clip/pin/adhesive tape Sharpener Duster 3 I . Standard Size of the Drawing Paper, Drawing Board & T -Square , Drawing Instruments Scales Etc. Standard Size of the Drawing Paper 4 Standard Size of the Drawing Board Designation Length x Width (mm) Recommended for use with sheet size D0 1500 x 1000 A0 D1 1000 x 700 A1 D2 700 x 500 A2 D3 500 x 500 A3 D 0 and D 1 for Drawing offices and D 2 for Students 5 Drawing Instruments Objectives in Drawing 1. Accuracy 2. Speed 3. Legibility 4. Neatness Typical Drawing Equipment 6 Drawing Boards Drawing Boards The left edge and right edge of a drawing board has a true straight edge. For right-handed people, the left-hand edge of the board is called the working edge because the T-square head slides against it. For left-handed people, the right-hand edge of the board is called the working edge because the T-square head slides against it. T-Squares The T-square is made of a long strip called the blade, fastened at right angles to a shorter piece called the head. 7 Drawing Instruments The drawing paper should be placed close to the working edge of the board to reduce any error resulting from the bending of the blade of the Tsquare. The paper should also be placed close enough to the upper edge of the board to permit space at the bottom of the sheet for using the T-square. Drafting tape is used to fasten the drawing paper to the drawing board. 8 Drawing Pencils High-quality drawing pencils should be used in technical drawing, never ordinary writing pencils. Many makes of mechanical pencils are available together with refill leads in all grades. Choose a mechanical pencil that feels comfortable in your hand. 9 Drawing Pencils Drawing Leads The first consideration in the selection of a grade of lead is the type of line work required. For light construction lines and guide lines for lettering use a hard lead. For all other line work, the lines should be BLACK. The lead chosen should be soft enough to produce jet black lines but hard enough not to smudge. HARD MEDIUM SOFT 9H 8H 7H 6H 5H 4H 3H 2H H F HB B 2B 3B 4B 5B 6B 7B Hard leads are used where extreme accuracy is required. Generally these leads are used for construction lines. Medium leads are used for general purpose line work in technical drawing. Soft leads are used for various kinds of art work. These leads are too soft to be useful in mechanical drafting. Drawing Instruments Drawing Lead Applications TASK CONSTRUCTION LINES VISIBLE OBJECT LINES LEAD GRADE 3H, 4H, 6H H, F, HB LINE WEIGHT THIN, LIGHT THICK, DARK HIDDEN LINES 2H, H THIN, DARK CENTER LINES 2H, H THIN, DARK DIMENSION LINES 2H, H THIN, DARK EXTENSION LINES 2H, H THIN, DARK LEADER LINES 2H, H THIN, DARK H, F, HB THICK, DARK 2H, H THIN, DARK H, F, HB THIN, DARK CUTTING PLANE LINES PHANTOM LINES LETTERING 11 Drawing Instruments Drawing Horizontal and Vertical Lines To draw a horizontal line, press the head of the T-square against the working edge of the board with your left hand. Lean the pencil in the direction of the line at an angle of approximately 60º and draw the line from left to right. While drawing the line, rotate the pencil to distribute the wear uniformly on the lead to maintain a symmetrical point. To draw a vertical line, press the head of the T-square against the working edge of the board with your left hand and place a triangle against the blade of the Tsquare. Lean the pencil in the direction of the line at an angle of approximately 60º and draw the line upward, rotating the pencil to distribute the wear uniformly on the lead to maintain a symmetrical point. 12 Drawing Instruments Triangles Most inclined lines are drawn at standard angles using the 45º x 45º triangle and the 30º x 60º triangle. In addition to drawing angles of 90º, 45º, 30º, and 60º, triangles can be combined to draw angles of 15º increments. 13 Drawing Instruments - Scales Scales Scales are instruments used in making technical drawings full size or at a given reduction or enlargement. Types of scales include metric scales, engineers’ scales, decimal scales, mechanical engineers’ scales, and architects’ scales. Scales are usually made of plastic or boxwood and are either triangular of flat in shape. 14 2. Lines - As per SLS 409 15 Lines contd……. 16 3. Lettering, Dimensioning etc. Lettering Technique Most engineering lettering is single-stroke Gothic font. Lettering is drawn freehand and are drawn within light horizontal guidelines. All lettering uses upper case letters. Lower case letters are rarely used in technical drawings. There are three aspects of good lettering: . proportions and forms of the letters, composition and spacing, and practice. There are six fundamental drawing strokes and their directions in basic lettering. Horizontal strokes are drawn from left to right, vertical strokes are drawn from top to bottom, and curved strokes are drawn downward. 17 Lettering Technique contd…… 18 19 Units of Measurement International systems of units (SI) – which is based on the meter Millimeter (mm) - The common SI unit of measure on engineering drawing. Individual identification of linear units is not required if all dimensions on a drawing are in the same unit (mm). The drawing shall however contain a note: ALL DIMENSIONS ARE IN MM. (Bottom left corner outside the title box) 20 Dimensioning Indicating on a drawing, the size of the object and other details essential for its construction and function, using lines, numerals, symbols, notes, etc. Dimensions indicated on a drawing should be those that are essential for the production, inspection and functioning of the object. Dimensions indicated should not be mistaken as those that are required to make the drawing of an object. Extension line – a thin, solid line perpendicular to a dimension line, indicating which feature is associated with the dimension. Visible gap – there should be a visible gap of 1.5 mm between the feature’s corners and the end of the extension line. 21 Leader line A thin, solid line used to indicate the feature with which a dimension, note, or symbol is associated. Generally a straight line drawn at an angle that is neither horizontal nor vertical. Terminated with an arrow touching the part or detail. On the end opposite the arrow, the leader line will have a short, horizontal shoulder. Text is extended from this shoulder such that the text height is centered with the shoulder line. 22 Arrows 3 mm wide and should be 1/3rd as wide as they are long - symbols placed at the end of dimension lines to show the limits of the dimension. Arrows are uniform in size and style, regardless of the size of the drawing. 23 Dimensioning Techniques .D imen 24 25 26 II - Geometric Constructions 27 TO DIVIDE A LINE To divide a given line AB into any number of equal parts. -Suppose the line AB is to be divided into 6 equal parts. 1. Draw a line AC of any length inclined to AB at some convenient angle (preferably between 20° and 40°). 2. Mark off six equal divisions on AC by cutting arcs of suitable radii consecutively starting from A. Number these divisions as 1, 2, 3, 4, 5 and 6. 3. Join 6 with B. 4. Draw lines through 5, 4, 3, 2 and 1 parallel to 6– B and cutting AB at points 5’, 4’, 3’, 2’ and 1’ respectively. Set-squares or drafter may be used for this purpose. The divisions 1’, 2’, 3’, 4’, 5’ divide the line AB into 6 equal parts. 28 Draw a perpendicular bisector of a line • Draw the line AB • With A as center and radius greater than half AB but less than AB, draw an arc on either side of AB (shown green) • With B as center and same radius, draw an arc on either side of AB (shown brown) • Join the point of intersection of these arcs on either side of AB A B • This is the perpendicular bisector 29 Bisecting an arc It is similar to bisecting a line BC>BE E Line Arc 30 Drawing a perpendicular to a line at a given point • Draw the line AB • With P as center and any convenient radius, draw an arc cutting AB in C (shown blue) Q • With the same radius cut 2 equal divisions CD and DE (shown red) E D • With same radius and centers D and E, draw arcs (green and brown) intersecting at Q • PQ is the required perpendicular A C P B 31 Drawing a perpendicular to a line at a given point (alternate method) Cut arcs with any radius (r1) on both sides of the point on the line AB. AB may be extended With a radius greater than r1, draw arcs with centers C and D to intersect at Q QP is the required perpendicular 32 Drawing a perpendicular from a point to a line From the external point P, draw arcs to cut the line AB at C and D. AB may be extended With C and D as centers and radius greater than half CD, draw arcs to intersect at E PE is the required perpendicular to AB 33 Bisecting an angle To bisect a given angle AOB. 1. With O as centre and any convenient radius, mark arcs cutting OA and OB at C and D respectively. 2. With C and D as centers and same or any other convenient radius, mark two arcs intersecting each other at E. 3. Join OE. 4. OE is the bisector of AOB, i.e., AOE = 2 x EOB. 34 Geometric Constructions Circles and Arcs A circle is a closed curve, all points of which are the same distance from a point called the center. 35 To divide a given circle into 8 equal parts Draw horizontal (1, 5) and vertical (3, 7) diameters which will be at 90o to each other. Bisect the angles to get new diameters (2, 6) and (4, 8) at 45o to the horizontal and vertical dimeters. The circle is divided into 8 equal sectors 36 To divide a circle into 12 equal parts Draw the two diameters 1–7 and 4–10, perpendicular to each other. With 1 as a centre and radius = R (= radius of the circle), cut two arcs at 3 and 11 on the circle. Similarly, with 4, 7 and 10 as the centres and the same radius, cut arcs on the circle respectively at 2 and 6, 5 and 9, and 8 and 12. The points 1, 2, 3, etc., give 12 equal divisions of the circle. 37 To draw a normal and a tangent to an arc or circle at a point P on it With centre P and any convenient radius, mark off two arcs cutting the arc/circle at C and D. Obtain QR, the perpendicular bisector of arc CD. QR is the required normal. Draw the perpendicular ST to QR for the required tangent. 38 Tangent to a given arc AB (or a circle) from a point P outside it. Join the centre O with P and locate the midpoint M of OP. With M as a centre and radius = MO, mark an arc cutting the circle at Q. Join P with Q. PQ is the required tangent. Another tangent PQ’ can be drawn in a similar way. 39 Common external tangent to 2 circles Given circles are with radii R1 and R2 and centers O and P respectively Draw a circle with radius R1-R2 and center O Draw a circle with dia. OP cutting the circle with radius R1-R2 at T Draw a line OT extended cutting the circle with radius R1 at A Draw a line PB parallel to OA with B lying on the circumference of circle with radius R2 Line AB is the required tangent Circle with radius R1 Circle with radius R2 A B T O P Circle with radius R1-R2 NOTE: PT is a tangent from point P to the circle with radius R1-R2 See N. D. Bhatt pg. 88, 89 40 Common internal tangent to 2 circles Given circles are with radii R1 and R2 and centers O and P respectively Draw a circle with radius R1+R2 and center O Draw a circle with dia. OP cutting the circle with radius R1+R2 at T Draw a line OT cutting the circle with radius R1 at A Draw a line PB parallel to OA with B lying on the circumference of circle with radius R2 Line AB is the required tangent T Circle with radius R2 Circle with radius R1 A O P B Circle with radius R1+R2 41 See N. D. Bhatt pg. 89 Line parallel to another line Example 4.24 To draw a line parallel to a given line AB and at a given distance R from it. Solution Refer Fig. 4.30. 1. Draw a perpendicular bisector of the line AB, cutting it at M. 2. Set off MN = R. Draw PQ perpendicular to MN at N. PQ is parallel to AB. 42 Draw an arc (radius R) touching 2 given lines AB and AC are the given lines Draw a line PQ parallel to and at a distance R from AB Draw a line EF parallel to and at a distance equal to R from AC intersecting PQ at O With O as center and radius R draw the arc touching to 2 lines 43 Draw an arc (radius R2) touching a given line and another arc CASE I • AB is the given line • Draw a line parallel to AB at a distance R2 • With O ac center and radius R1R2, draw an arc EF cutting the line at P • With P as center and Radius R2, draw the required arc CASE II • AB is the given line • Draw a line parallel to AB at a distance R2 • With O as center and radius R1+R2, draw an arc EF cutting the line at P • With P as center and Radius R2, draw the required arc 44 Finding the center of an arc Draw 2 chord of the arc (CD and EF in this case) Draw perpendicular bisectors of CD and EF intersecting each other at O. O is the required center. 45 Curve (given radius) joining 2 other curves Draw arcs with radius R1 + R3 (center O) and R2 + R3 (center P) intersecting at Q. Draw arcs with radius R1 - R3 (center O) and R2 + R3 (center P) intersecting at Q. Draw arcs with radius R3 – R1 (center O) and R3 - R2 (center P) intersecting at Q. With center Q draw an arc with radius R3 joining the 2 curves. With center Q draw an arc with radius R3 joining the 2 curves. With center Q draw an arc with radius R3 joining the 2 curves. 46 Geometric Constructions Drawing a Square Given the inscribed circle, draw two diameters at right angles to each other. The intersections of these diameters with the circle are the vertexes of an inscribed square. Given the circumscribed circle, use the T-square and 45º triangle and draw the four sides tangent to the circle. 47 Geometric Constructions Drawing a Hexagon Given the inscribed circle, draw vertical and horizontal center lines and the diagonals (AB) and (CD) at 30º or 60º with the horizontal. With the 30º x 60º triangle and T-square, draw the six sides of the hexagon. Given the circumscribed circle, draw vertical and horizontal center lines . With the 30º x 60º triangle and T-square, draw the six sides of the hexagon tangent to the circle. 48 To construct a regular hexagon of given side length With any point O as centre and radius = AB, draw a circle. Starting from any point (say A) on the circle, mark off the five arcs of radius = AB consecutively cutting the circle at B, C, D, E and F. Join A, B, C, D, E and F for the required hexagon. Principle: The distance across opposite corners in a regular hexagon = 2 x side length AD = 2 x AB 49 CONSTRUCTION OF A POLYGON Draw side AB of specified length Draw a perpendicular BP at B such that BP = AB Draw a straight line joining A and P P 6 With B as center and radius AB draw arc AP 5 4 Draw a perpendicular bisector of AB to meet the line AP at 4 and arc AP at 6 Locate point 5 as the midpoint of 4-5 A square of side AB can be inscribed in the circle with center 4 and radius A4 A B 50 Polygons of different number of sides on same construction • Similarly a hexagon of side AB can be inscribed in the circle with center 6 and radius A6 8 • Mark points 7, 8, 9 on the perpendicular bisector such that 5-6 = 6-7 = 7-8 = 8-9 and so on 7 5 • A heptagon of side AB can be inscribed in the circle with center 7 and radius A7 • An octagon of side AB can be inscribed in the circle with center 8 and radius A8…and so on P 6 4 A B 51 Drawing a pitch circle and marking the holes Pitch circle: circle on which lies certain features e.g. the centers of smaller circles or holes. Fig. shows 8 holes drawn on a pitch circle in a square plate •Draw the pitch circle •Since there are 8 holes, the angle between the lines joining their centers to the center of the pitch circle will be 45o •Divide the pitch circle into 8 parts by drawing lines from its center at 45o to the adjacent one C •The points of intersection of these lines and the pitch circle are the centers of the required holes Pitch circle •Draw the holes with specified diameter 52 Geometric Constructions Bisecting an angle Given the angle (BAC) to be bisected. Strike the large arc (R) at any convenient radius. Strike equal arcs (r) with a radius slightly larger than half (BC) to intersect at (D). Draw line (AD) which bisects the angle. 53 Geometric Constructions Transferring an angle Given the angle (BAC) to be transferred to the new position at (A’B’). Use any convenient radius (R) and strike arcs from centers (A) and (A’). Strike equal arcs (r) and draw side (A’C’). 54 DRAWING LINES Find the Centre of an Arc 1. 2. 3. Select three points A, B and C on the arc and join AB and BC Bisect AB and BC. Fine the intersection point of the bisecting lines/bisectors. That is the centre of the arc. 55 DRAWING LINES Inscribe a Circle in a Triangle 1. 2. 3. Bisect angle ABC and angle BAC. Fine the intersection point of the bisecting lines/bisectors. That is the centre of the circle. The radius of the circle is the length of a perpendicular line on any of the sides of the triangle drawn from the centre of the circle. 56 DRAWING LINES Circumscribe a Circle on a Triangle 1. 2. 3. Bisect sides AC and BC. Fine the intersection point of the bisecting lines/bisectors. That is the centre of the circle. The radius of the circle is the length of a line joining any one of the vertices of the triangle to the centre of the circle. 57 DRAWING LINES Draw a Hexagon • To draw a regular hexagon given the distance across flats Draw a circle having a diameter equal to the distance across flats. • Draw tangents to this circle with a 60° set square to produce the hexagon. 58 DRAWING LINES Draw a Hexagon • To draw a regular hexagon given the distance across corners, draw a circle having a diameter equal to the distance across corners • Step off the radius round it to give six equally spaced points. • Join these points to form the hexagon. 59 Loci of Points & Application of Loci • The Ellipse, Parabola, Hyperbola • Curves: Roulettes or Cycloidal curves Cycloids, Epicycloids, Hypocycloids, Involutes, Spirals and Helices 60 III(a)- 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. Hyperbola Ellipse Parabola Section Plane Through Generators Section Plane Parallel to Axis. Section Plane Parallel to end generator. 61 III (a) - Conic Sections - Engineering Curves – Ellipse, Parabola and Hyperbola ELLIPSE PARABOLA HYPERBOLA 1.Concentric Circle Method 1.Rectangle Method 1.Rectangular Hyperbola (coordinates given) 2.Rectangle Method 2 Method of Tangents ( Triangle Method) 3.Oblong Method 4.Arcs of Circle Method 3.Basic Locus Method (Directrix – focus) 2 Rectangular Hyperbola (P-V diagram - Equation given) 3.Basic Locus Method (Directrix – focus) 5.Rhombus Metho 6.Basic Locus Method (Directrix – focus) Methods of Drawing Tangents & Normals To These Curves. 62 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. 63 ELLIPSE Problem 1 :Draw ellipse by concentric circle method. Take major axis 100 mm and minor axis 70 mm long. BY CONCENTRIC CIRCLE METHOD 3 2 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. 4 C 1 2 3 5 4 1 5 A B 10 10 6 9 8 D 9 7 6 7 8 64 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 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 side part.along with lower half of the rectangle.Join all points in smooth curve. It is required ellipse. ELLIPSE BY RECTANGLE METHOD Problem 2 Draw ellipse by Rectangle method. Take major axis 100 mm and minor axis 70 mm long. 4 D 4 3 3 2 2 1 1 A B C 65 ELLIPSE BY OBLONG METHOD Problem 3: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. 4 4 3 3 2 2 1 A 1 B 66 PROBLEM 4. MAJOR AXIS AB & MINOR AXIS CD ARE 100 AMD 70MM LONG RESPECTIVELY .DRAW ELLIPSE BY ARCS OF CIRLES METHOD. STEPS: 1.Draw both axes as usual.Name the ends & intersecting point 2.Taking AO distance I.e.half major axis, from C, mark F1 & F2 On AB . ( focus 1 and 2.) 3.On line F1- O taking any distance, mark points 1,2,3, & 4 4.Taking F1 center, with distance A-1 draw an arc above AB and taking F2 center, with B-1 distance cut this arc. A Name the point p1 5.Repeat this step with same centers but 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/ ELLIPSE BY ARCS OF CIRCLE 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 points (F1 & F2) remains constant and equals to the length of major axis AB.(Note A .1+ B .1=A . 2 + B. 2 = AB) p4 p3 C p2 p1 F1 1 2 3 4 O B F2 D 67 ELLIPSE PROBLEM 5. DRAW RHOMBUS OF 100 MM & 70 MM LONG DIAGONALS AND INSCRIBE AN ELLIPSE IN IT. BY RHOMBUS METHOD 2 A STEPS: 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 smaller diagonals. 4. Mark points 1,2,3,4 as four centers. 5. Taking 1 as center and 1-A 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. B 4 3 C D 1 68 ELLIPSE PROBLEM 6 :- 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. { ECCENTRICITY = 2/3 } STEPS: 1 .Draw a vertical line AB and point F 50 mm from it. 2 .Divide 50 mm distance in 5 parts. 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. 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. DIRECTRIX-FOCUS METHOD A ELLIPSE 45mm (vertex) V F ( focus) B 69 PARABOLA RECTANGLE METHOD PROBLEM 7: A BALL THROWN IN AIR ATTAINS 100 M HIEGHT AND COVERS HORIZONTAL DISTANCE 150 M ON GROUND. Draw the path of the ball (projectile)- STEPS: 1.Draw rectangle of above size and divide it in two equal vertical parts 2.Consider left part for construction. Divide height and length in equal number of parts and name those 1,2,3,4,5& 6 3.Join vertical 1,2,3,4,5 & 6 to the top center of rectangle 4.Similarly draw upward vertical lines from horizontal1,2,3,4,5 And wherever these lines intersect previously drawn inclined lines in sequence Mark those points and further join in smooth possible curve. 5.Repeat the construction on right side rectangle also.Join all in sequence. This locus is Parabola. . 6 6 5 5 4 4 3 3 2 2 1 1 1 2 3 4 5 6 5 4 3 2 1 70 Problem no.8: Draw an isosceles triangle of 100 mm long base and 110 mm long altitude. Inscribe a parabola in it by method of tangents. Solution Steps: 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 PARABOLA METHOD OF TANGENTS C B 71 PARABOLA PROBLEM 9: 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. DIRECTRIX-FOCUS METHOD PARABOLA SOLUTION STEPS: 1.Locate center of line, perpendicular to 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 those draw lines parallel to AB. 3.Mark 5 mm distance to its left of P and name it 1. 4.Take O-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. (FP1=O1) 5.Similarly repeat this process by taking again 5mm to right and left and locate P3P4. 6.Join all these points in smooth curve. A P1 O (VERTEX) V 1 2 3 4 F ( focus) P2 B It will be the locus of P equidistance from line AB and fixed point F. 72 Problem No.10: Point P is 40 mm and 30 mm from horizontal and vertical axes respectively. Draw Hyperbola through it. Solution Steps: 1) Extend horizontal line from P to right side. 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 part [P-B] also at 1,2,3,4 points. 5) From horizontal 1,2,3,4 draw vertical lines downwards and 6) From vertical 1,2,3,4 points [from P-B] draw 40 mm horizontal lines. 7) Line from 1 horizontal and line from 1 vertical will meet at P1.Similarly mark P2, P3, P4 points. 8) Repeat the procedure by marking four points 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. HYPERBOLA THROUGH A POINT OF KNOWN CO-ORDINATES 2 1 2 1 P 1 2 3 1 2 3 O 30 mm 73 HYPERBOLA P-V DIAGRAM Problem no.11: A sample of gas is expanded in a cylinder from 10 unit pressure to 1 unit pressure. Expansion follows law PV= Constant. If initial volume being 1 unit, draw the curve of expansion. Also Name the curve. 10 9 8 7 Form a table giving few more values of P & V P 10 5 4 2.5 2 1 V = C 1 = 2 = 2.5 = 4 = 5 = 10 = 10 10 10 10 10 10 Now draw a Graph of Pressure against Volume. It is a PV Diagram and it is Hyperbola. Take pressure on vertical axis and Volume on horizontal axis. PRESSURE ( Kg/cm2) 6 5 4 3 2 1 0 1 2 3 4 5 6 7 8 10 9 VOLUME:( M3 ) 74 HYPERBOLA DIRECTRIX FOCUS METHOD PROBLEM 12:- 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. { ECCENTRICITY = 2/3 } STEPS: 1 .Draw a vertical line AB and point F 50 mm from it. 2 .Divide 50 mm distance in 5 parts. 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. 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. A 30mm (vertex) V F ( focus) B 75 ELLIPSE TANGENT & NORMAL Problem 13: TO DRAW TANGENT & NORMAL TO THE CURVE FROM A GIVEN POINT ( Q ) 1. 2. 3. JOIN POINT Q TO F1 & F2 BISECT ANGLE F1Q F2 THE ANGLE BISECTOR IS NORMAL A PERPENDICULAR LINE DRAWN TO IT IS TANGENT TO THE CURVE. p4 p3 C p2 p1 A F1 1 2 3 4 O B F2 Q D 76 Ellipse Construction 1. Draw the axes AB and CD and draw circles (called auxiliary circles) on them as diameters. 2. Divide the circles into a number of equal parts, by radial lines through O. Each of the radial lines intersect the major and minor auxiliary circle. 3. Through the points where radial lines cut the major auxiliary circles drop vertical perpendiculars, and through the points where the radial lines cut the minor auxiliary circle draw horizontals to cut the verticals. These 77 intersections are points on the ellipse. ELLIPSE TANGENT & NORMAL Problem 14: 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 3. EXTEND THE LINE TO MEET DIRECTRIX AT T 4. JOIN THIS POINT TO Q AND EXTEND. THIS IS TANGENT TO ELLIPSE FROM Q 5.T O THIS TANGENT DRAW PERPENDICULAR LINE FROM Q. IT IS NORMAL TO CURVE. ELLIPSE T (vertex) V F ( focus) 900 N N Q B 78 T PARABOLA TANGENT & NORMAL Problem 15: TO DRAW TANGENT & NORMAL TO THE CURVE FROM A GIVEN POINT ( Q ) 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 TANGENT TO THE CURVE FROM Q 5. TO THIS TANGENT DRAW ERPENDICULAR LINE FROM Q. IT IS NORMAL TO T PARABOLA A VERTEX V 900 F ( focus) N Q N B T 79 HYPERBOLA TANGENT & NORMAL Problem 16 TO DRAW TANGENT & NORMAL TO THE CURVE FROM A GIVEN POINT ( Q ) A T 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 TANGENT TO CURVE FROM Q 5.TO THIS TANGENT DRAW PERPENDICULAR LINE FROM Q. IT IS NORMAL TO CURVE. (vertex) F ( focus) V 900 N N Q B T 80 III (b) - ENGINEERING CURVES : LOCI Point undergoing two types of displacements) INVOLUTE 1. Involute of a circle a)String Length = D b)String Length > D c)String Length < D 2. Pole having Composite shape. 3. Rod Rolling over a Semicircular Pole. CYCLOID 1. General Cycloid 2. Trochoid ( superior) 3. Trochoid ( Inferior) 4. Epi-Cycloid SPIRAL HELIX 1. Spiral of One Convolution. 1. On Cylinder 2. On a Cone 2. Spiral of Two Convolutions. 5. Hypo-Cycloid AND Methods of Drawing Tangents & Normals To These Curves. 81 DEFINITIONS CYCLOID: SUPERIORTROCHOID IF THE POINT IN THE DEFINATION OF CYCLOID IS OUTSIDE THE CIRCLE IT IS A LOCUS OF A POINT ON THE PERIPHERY OF A CIRCLE WHICH ROLLS ON A STRAIGHT LINE PATH. INFERIOR TROCHOID INVOLUTE: EPI-CYCLOID IT IS A LOCUS OF A FREE END OF A STRING WHEN IT IS WOUND ROUND A CIRCLE OR POLYGON SPIRAL: IT IS A CURVE GENERATED BY A POINT WHICH REVOLVES AROUND A FIXED POINT AND AT THE SAME MOVES TOWARDS IT. IF IT IS INSIDE THE CIRCLE IF THE CIRCLE IS ROLLING ON ANOTHER CIRCLE FROM OUTSIDE HYPO-CYCLOID. IF THE CIRCLE IS ROLLING FROM INSIDE THE OTHER CIRCLE, 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) 82 CYCLOID • The cycloid is the locus of a point on the rim of a circle rolling along a straight line. 83 CYCLOID Problem 22: Draw locus of a point on the periphery of a circle which rolls on straight line path. Take circle diameter as 50 mm. Draw normal and tangent on the curve at a point 40 mm above the directing line. 6 7 p6 p5 5 p8 4 p4 9 C1 C p3 C2 3 C3 C4 C5 p2 10 2 11 Solution Steps: 7) p1 12 P C6 C7 C8 C9 p9 C10 C11 40mm 8 1) 2) 3) 4) 5) 6) p7 p10 p11 p12 1 1’ 2’ C12 3’ 4’ 5’ 6’ 7’ D 8’ 9’ 10’ 11’ 12’ Q From center C draw a circle of 50mm dia. and from point P draw a horizontal line PQ equal to D length. Divide the circle in 12 equal parts and in anticlockwise direction, after P name 1, 2, 3 up to 12. Also divide the straight line PQ into 12 number of equal parts and after P name them 1’,2’,3’__ etc. From all these points on circle draw horizontal lines. (parallel to locus of C) With a fixed distance C-P in compass, C1 as center, mark a point on horizontal line from 1. Name it P 1. Repeat this procedure from C2, C3, C4 up to C12 as centers. Mark points P2, P3, P4, P5 up to P12 on the horizontal lines drawn from 1,2, 3, 4, 5, 6, 7 respectively. Join all these points by curve. It is Cycloid. 84 STEPS: DRAW CYCLOID AS USUAL. MARK POINT Q ON IT AS DIRECTED. CYCLOID Method of Drawing Tangent & Normal WITH CP DISTANCE, FROM Q. CUT THE 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 P C5 C6 C7 C8 N D 85 CYCLOID Problem 22: Draw locus of a point on the periphery of a circle which rolls on straight line path. Take circle diameter as 50 mm. Draw normal and tangent on the curve at a point 40 mm above the directing line. 6 7 p6 p5 5 p8 4 p4 9 C1 C p3 C2 3 C3 C4 C5 C6 p2 10 2 11 Solution Steps: 7) p1 12 P C7 C8 C9 C10 p9 C11 40mm 8 1) 2) 3) 4) 5) 6) p7 C12 p10 p11 p12 1 1’ 2’ 3’ 4’ 5’ 6’ D 7’ 8’ 9’ 10’ 11’ 12’ Q From center C draw a circle of 50mm dia. and from point P draw a horizontal line PQ equal to D length. Divide the circle in 12 equal parts and in anticlockwise direction, after P name 1, 2, 3 up to 12. Also divide the straight line PQ into 12 number of equal parts and after P name them 1’,2’,3’__ etc. From all these points on circle draw horizontal lines. (parallel to locus of C) With a fixed distance C-P in compass, C1 as center, mark a point on horizontal line from 1. Name it P 1. Repeat this procedure from C2, C3, C4 up to C12 as centers. Mark points P2, P3, P4, P5 up to P12 on the horizontal lines drawn from 1,2, 3, 4, 5, 6, 7 respectively. Join all these points by curve. It is Cycloid. 86 STEPS: DRAW CYCLOID AS USUAL. MARK POINT Q ON IT AS DIRECTED. CYCLOID Method of Drawing Tangent & Normal WITH CP DISTANCE, FROM Q. CUT THE 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 P C5 C6 C7 C8 N D 87 HYPOCYCLOID The curve produced by fixed point P on the circumference of a small circle of radius a rolling around the inside of a large circle of radius b. 88 EPICYCLOID The path traced out by a point P on the edge of a circle of radius a rolling on the outside of a 89 circle of radius b. What is an involute ? • Attach a string to a point on a curve. • Make the string a tangent to the curve at the point of attachment. • Then wind the string up, keeping it always taut. The locus of points traced out by the end of the string is called the involute of the original curve. • An involute can be described as the path traced by the end point of a string as it is unwound from a line, a polygon, or circumference of a circle 90 Involute of a line (AB): A B C 91 Problem: Draw involute of an equilateral triangle of 35 mm sides. B C A 35 3X35 92 Problem: Draw involute of a square of 25 mm sides C B D A 25 100 93 INVOLUTE OF A CIRCLE Problem no 23: Draw Involute of a circle of 40 mm diameter. Also draw normal and tangent to it at a point 100 mm from the centre of the 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 given circle. B will meet A after winding. 2) Divide D (AP) distance into 12 number of equal parts. 3) Divide circle also into 12 number of equal parts. P5 4) Name after A, 1, 2, 3, 4, etc. up to 12 on D line AP as well as on circle (in anticlockwise direction). 5) To radius C-1’, C-2’, C-3’ up to C12’ draw tangents (from 1’,2’,3’,4’, etc to circle). 6) Take distance 1 to P in compass P6 from point 1’ and mark it on tangent on circle (means one division less than distance AP). 7) Name this point P1 8) Take 2-P distance in compass and mark it on the tangent from point 2’. Name it point PP2.7 9) Similarly take 3 to P, 4 to P, 5 to P up to 11 to P distance in compass and mark on respective tangents and locate P3, P4, P5 up to P12 (i.e. P8 A) points and join them in smooth curve it is an INVOLUTE of a given circle. P3 P4 P2 P1 6’ 7’ 8’ 5’ 4’ 9’ c 3’ 10’ 2’ 11’ 12’ A P11 P9 P10 1’ 1 P 2 3 4 5 6 7 8 9 10 11 D 94 12 STEPS: DRAW INVOLUTE AS USUAL. Involute Method of Drawing Tangent & Normal INVOLUTE OF A CIRCLE MARK POINT Q ON IT AS DIRECTED. JOIN Q TO THE CENTER OF CIRCLE C. CONSIDERING CQ DIAMETER, DRAW A SEMICIRCLE AS SHOWN. Q MARK POINT OF INTERSECTION OF THIS SEMICIRCLE AND POLE CIRCLE AND JOIN IT TO 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 2 C 6 7 1 8 P8 1 2 3 4 5 6 P 8 7 D 95 Archimedean Spiral • Spiral of Archimedes is a spiral , a curve generated by a point that moves at a uniform speed along a straight line while the straight line rotates with uniform angular velocity about a fixed point. 96 Geometric Constructions Solids Solids bounded by plane surfaces are called polyhedra. The surfaces are called faces. If the faces are equal regular polygons the solids are called regular polyhedra. 97 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 And radius of directing circle i.e. curved path, 75 mm.Also draw normal and tangent on the curve at 110mm from the centre of directing circle. 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 360º. c9 c8 3) Construct angle with radius c7 OC and draw an arc by taking O as c6 center OC as radius and form sector of angle . c5 9 4) Divide this sector into 12 8 7 number of equal angular parts. And Rolling circle or 6 generating circle from C onward name them C1, C2, c4 C3 up to C12. 5 5) Divide smaller circle (Generating c3 circle) also in 12 number of equal 4 parts. And next to P in anticlockwDirecting circle ise direction name those 1, 2, 3, up 3 c2 to 12. 6) With O as center, O-1 as radius 2 3’ draw an arc in the sector. Take O-2, 4’ 2’ O-3, O-4, O-5 up to O-12 distances c1 1 5’ with center O, draw all concentric 1’ arcs in sector. Take fixed distance CP in compass, C1 center, cut arc of 1 6’ 12’ at P1. C P Repeat procedure and locate P2, P3, O P4, P5 unto P12 (as in cycloid) and 11’ OP=Radius of directing circle=75mm 7’ join them by smooth curve. This is PC=Radius of generating circle=25mm EPI – CYCLOID. 8’ 10’ θ=r/R X360º= 25/75 X360º=120º 9’ c1 c1 0 c1 1 2 10 11 12 θ 98 Problem 17: A circle of 50 mm diameter rolls on another circle of 175 mm diameter and outside it. Draw the curve traced by a point P on its circumference for one complete revolution.Also draw normal and tangent on the curve at 125 mm from the centre of directing circle. Draw a horizontal line OP of 87.5 mm and draw an arc with O as centre and PO as radius Draw a horizontal line CP of 25 mm and draw a circle with C as centre and CP as radius. θ=(OP/PC) X 360º = (25/87.5) X 360º = 102.8º ≈103º Divide the rolling circle in 8 equal parts Also divide the angle in 8 equal parts using angle bisectors Directing circle Rolling circle or generating circle C P θ=103º O 99 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 rolling circle 50 mm and radius of directing circle (curved path) 75 mm. Also draw normal and tangent on the curve at a point 40mm from the centre of directing circle Directing circle Solution Steps: 1) Smaller circle is rolling here, inside the larger circle. It has to rotate anticlockwise to move ahead. 2) Same steps should be taken as in case of EPI – CYCLOID. Only change is in numbering direction of 12 number of equal parts on the smaller circle. 3) From next to P in clockwise direction, name 1,2,3,4,5,6,7,8,9,10,11,12 4) Further all steps are that of epi – cycloid. This is called HYPO – CYCLOID. 10 11 12 6 5 4 c6 3 c7 c8 c9 c10 c11 c12 c5 c4 2 3’ c3 2’ 4’ c2 1 1’ 5’ c1 θ 12’ 6’ C P 11’ Rolling circle or generating circle 9 8 7 7’ O 8’ 10’ 9’ OP=Radius of directing circle=75mm PC=Radius of generating circle=25mm θ=r/R X360º= 25/75 X360º=120º 100 Problem 28: A point P moves towards another point O, 75 mm from it and reaches it while moving around it once. Its movement towards O being uniform with its movement around it. Draw the curve traced out by point P. SPIRAL Important approach for construction Find total angular and total linear displacement and divide both in to same number of equal parts. Total linear movement 75 mm. Total angular movement 360º 2’ With OP radius & O as center draw a circle and divide it in EIGHT parts. Name those 1’,2’,3’,4’, etc. up to 8’ Similarly divided line PO also in EIGHT parts and name those 1,2,3, starting from P. Take O-1 distance from OP line and draw an arc up to O1’ radius vector. Name the point P1 Similarly mark points P2, P3, P4 up to P8 And join those in a smooth curve. It is a SPIRAL of one convolution. P2 3’ P1 1’ P3 P4 4’ O P5 7 6 5 4 3 2 1 P7 P P6 7’ 5’ 6’ 101 Draw an Archemedian spiral of one convolution, greatest and least radii being 115mm and 15 mm respectively. Draw a normal and tangent to the spiral at a point 65 mm from the pole. Important approach for construction! Find total angular and total linear displacement and divide both in to same number of equal parts. Angular displacement =360º, Linear displacement = 100mm 3’ Solution Steps 1. With PO & QO radii draw two circles and divide them in twelve equal parts. Name those 1’,2’,3’,4’, etc. up to 12’ 2 .Similarly divided line PQ also in twelve parts and name those 1,2,3,-- as shown. 3. Take O-1 distance from OP line and draw an arc up to O1’ radius vector. Name the point P1 4. Similarly mark points P2, P3, P4 up to P12 And join those in a smooth curve. It is a SPIRAL of one convolution. 2’ 4’ P3 P2 P4 5’ P1 P5 N c Q P6 6’ O 12 1110 9 8 7 6 5 4 3 2 1 P11 P7 P10 P8 P9 7’ C=(Rmax-Rmin)/No. of convolutions in radians = (115-15)/3.14 X 2 =15.92 1’ 11’ 8’ 9’ 10’ 102 P 12’ Draw an Archemedian spiral of one and half convolution, greatest and least radii being 115mm and 15 mm respectively. Draw a normal and tangent to the spiral at a point 70 mm from the pole. Important approach for construction Find total angular and total linear displacement and divide both in to same number of equal parts. Total Angular displacement 540º. Total Linear displacement 100 mm 3’ 15’ 1 Draw a 115 mm long line OP. 16’ 4’ 2 Mark Q at 15 mm from O 3 with O as centre draw two circles with OP and OQ radius 4 Divide the circle in 12 equal divisions and 17’ 5’ mark the divisions as 1’,2’ and so on up to 18’ P 5 5 Divide the line PQ in 18 equal divisions as 1,2,3 and so on upto 18 6.Take O-1 distance from OP line and draw an arc up to O1’ radius vector. P6 18’ 6’ Name the point P1 7.Similarly mark points P2, P3, P4 up to P18. 8. And join those in a smooth curve. It is a SPIRAL of one and half convolution. 2’ 14’ P3 P2 P4 P1 P15 1’13’ P14 P16 P13 P17 P18 P12 Q 18 O 16 14 12 10 8 6 4 P 12’ 2 P11 P7 P10 P8 7’ P9 C=(Rmax-Rmin)/No. of convolutions in radians = (115-15)/3.14 X3 =10.61 8’ 9’ 11’ 10’ 103 Spiral. Method of Drawing Tangent & Normal SPIRAL (ONE CONVOLUSION.) 2 P2 3 P1 Q Difference in length of any radius vectors 1 Constant of the Curve = Angle between the corresponding radius vector in radian. P3 = P4 4 O P5 7 6 5 4 3 2 1 P7 P6 7 5 6 P OP – OP2 /2 = OP – OP2 1.57 = 3.185 m.m. STEPS: *DRAW SPIRAL AS USUAL. DRAW A SMALL CIRCLE OF RADIUS EQUAL TO THE CONSTANT OF CURVE CALCULATED ABOVE. * LOCATE POINT Q AS DISCRIBED IN PROBLEM AND THROUGH IT DRAW A TANGENTTO THIS SMALLER CIRCLE.THIS IS A NORMAL TO THE SPIRAL. *DRAW A LINE AT RIGHT ANGLE *TO THIS LINE FROM Q. IT WILL BE TANGENT TO CYCLOID. 104 SPIRAL of two convolutions Problem 28 Point P is 80 mm from point O. It starts moving towards O and reaches it in two revolutions around.it Draw locus of point P (To draw a Spiral of TWO convolutions). IMPORTANT APPROACH FOR CONSTRUCTION! FIND TOTAL ANGULAR AND TOTAL LINEAR DISPLACEMENT AND DIVIDE BOTH IN TO SAME NUMBER OF EQUAL PARTS. 2,10 SOLUTION STEPS: Total angular displacement here is two revolutions And Total Linear displacement here is distance PO. Just divide both in same parts i.e. Circle in EIGHT parts. ( means total angular displacement in SIXTEEN parts) Divide PO also in SIXTEEN parts. Rest steps are similar to the previous problem. P2 P1 3,11 1,9 P3 P10 P9 P11 4,12 16 13 10 P4 P12 8 7 6 5 4 3 2 1 P P8 8,16 P15 P13 P14 P7 P5 5,13 P6 7,15 105 6,14 Problem No.7: A Link OA, 80 mm long oscillates around O, 600 to right side and returns to it’s initial vertical Position with uniform velocity.Mean while point P initially on O starts sliding downwards and reaches end A with uniform velocity. Draw locus of point P Solution Steps: Point P- Reaches End A (Downwards) 1) Divide OA in EIGHT equal parts and from O to A after O name 1, 2, 3, 4 up to 8. (i.e. up to point A). 2) Divide 600 angle into four parts (150 each) and mark each point by A1, A2, A3, A4 and for return A5, A6, A7 andA8. (Initial A point). 3) Take center O, distance in compass O-1 draw an arc upto OA1. Name this point as P1. 1) Similarly O center O-2 distance mark P2 on line O-A2. 2) This way locate P3, P4, P5, P6, P7 and P8 and join them. ( It will be thw desired locus of P ) OSCILLATING LINK O p 1 p1 p2 p3 p4 2 3 p5 A4 4 5 p6 A3 6 7 A8 p8 p7 A5 A2 A6 A1 A7 A8 106 OSCILLATING LINK Problem No 8: A Link OA, 80 mm long oscillates around O, 600 to right side, 1200 to left and returns to it’s initial vertical Position with uniform velocity.Mean while point P initially on O starts sliding downwards, reaches end A and returns to O again with uniform velocity. Draw locus of point P Op 16 15 1 14 Solution Steps: ( P reaches A i.e. moving downwards. & returns to O again i.e.moves upwards ) 1.Here distance traveled by point P is PA.plus A 12 AP.Hence divide it into eight equal parts.( so total linear displacement gets divided in 16 parts) Name those as shown. 2.Link OA goes 600 to right, comes back to A A13 11 original (Vertical) position, goes 600 to left and returns to original vertical position. Hence total angular displacement is 2400. Divide this also in 16 parts. (150 each.) Name as per previous problem.(A, A1 A2 etc) 3.Mark different positions of P as per the procedure adopted in previous case. and complete the problem. p1 p2 p3 p4 2 13 p 3 12 A4 5 4 11 p6 5 10 A10 A14 6 9 7 A9 A15 A3 8 A p8 A8 A16 p7 A1 A7 A5 A2 A6 107 ROTATING LINK Problem 9: Rod AB, 100 mm long, revolves in clockwise direction for one revolution. Meanwhile point P, initially on A starts moving towards B and reaches B. Draw locus of point P. 1) AB Rod revolves around center O for one revolution and point P slides along AB rod and reaches end B in one revolution. 2) Divide circle in 8 number of equal parts and name in arrow direction after A-A1, A2, A3, up to A8. 3) Distance traveled by point P is AB mm. Divide this also into 8 number of equal parts. 4) Initially P is on end A. When A moves to A1, point P goes one linear division (part) away from A1. Mark it from A1 and name the point P1. 5) When A moves to A2, P will be two parts away from A2 (Name it P2 ). Mark it as above from A2. 6) From A3 mark P3 three parts away from P3. 7) Similarly locate P4, P5, P6, P7 and P8 which will be eight parts away from A8. [Means P has reached B]. 8) Join all P points by smooth curve. It will be locus of P A2 A1 A3 p1 p2 p6 p5 A P 1 2 3 p7 p3 p4 A7 4 5 p8 B A4 6 7 A5 A6 108 Problem 10 : Rod AB, 100 mm long, revolves in clockwise direction for one revolution. Meanwhile point P, initially on A starts moving towards B, reaches B And returns to A in one revolution of rod. Draw locus of point P. ROTATING LINK A2 Solution Steps 1) AB Rod revolves around center O for one revolution and point P slides along rod AB reaches end B and returns to A. 2) Divide circle in 8 number of equal parts and name in arrow direction after A-A1, A2, A3, up to A8. 3) Distance traveled by point P is AB plus AB mm. Divide AB in 4 parts so those will be 8 equal parts on return. 4) Initially P is on end A. When A moves to A1, point P goes one linear division (part) away from A1. Mark it from A1 and name the point P1. 5) When A moves to A2, P will be two parts away from A2 (Name it P2 ). Mark it as above from A2. 6) From A3 mark P3 three parts away from P3. 7) Similarly locate P4, P5, P6, P7 and P8 which will be eight parts away from A8. [Means P has reached B]. 8) Join all P points by smooth curve. It will be locus of P The Locus will follow the loop path two times in one revolution. A1 A3 p5 p1 p4 A P p8 p2 1+7 2+6 p 6 +3 5 4 +B p7 p3 A7 A5 A6 109 A4 Problem 28: A link OA, 100 mm long rotates about O in anti-clockwise direction. A point P on the link, 15 mm away from O, moves and reaches the end A, while the link has rotated through 2/5 of a revolution. Assuming that the movements of the link to be uniform trace the path of point P. θ= 2/5 X 360º = 144º Total angular movement = 144º Total linear movement = 85 mm To divide both of them in equal no. of parts ( say 8) 5’ 6’ 4’ 7’ 3’ P6 P7 8’ P5 2’ P4 P8 P3 1’ P2 144º P1 O 15 P 1 2 3 100 4 5 6 7 8 110 A Logarithmic Spiral: If a point moves around a pole in such a way that The value of vectorial angle are in arithmatic progression and The corresponding values of radius vectors are in geometric progression, then the curve traced by the point is known as logarithmic spiral. A3 A2 P3 A1 P2 θ θ P1 θ O A P Let OA be a straight line and P be a point on it at radius vector OP from O. Now let the line moves at uniform angular speed to a new position OA1 ,at vectorial angle θ from OA and the point moves to a new position P1 , at radius vector OP1 from O. The line now gradually moves to the new position OA2, OA3 at vectorial angle θ and the point to P2 and P3 , at radius vectors OP2 and OP3 respectively. In Logarithmic spiral OP3/OP2 =OP2/OP1=OP1/OP 111 Problem37: In a logarithmic spiral, the shortest radius is 40mm. The length of adjacent radius vectors enclosing 30º are in the ratio of 9:8 Construct one revolution of the spiral. Draw tangent to the spiral at a point 70 mm from it. First step is to draw logarithmic scale. Draw two straight lines OA & OB at angle of 30º. B P12 Mark a point P on OA at 40 mm from O. Calculate OP1 such that OP1/OP = 9/8. => OP1 = 45 mm Mark a OP1 on OB at 45 mm from O. Join P with P1. P4 P3 Draw an arc of radius OP1 from OB to OA. P2 O P5 Draw a line parallel to PP1 from P1 on OA to intersect OB at P2. P Repeat the steps to get the points P3,P4 and so1 on up to P12. P1 30º 40 P P2P3 P4 P P5 6 P7 P8 P9 P10 P11 P1P2 P3 P4 P5 P6 P7 P8 P9 P10 P11 P12 P6 P P12 P7 P11 P8 P9 P10 112 A