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OBJECTIVES, 1. To develop in students graphic skill for, communication of, Concepts,, Ideas and, Design of engineering products, and expose them to existing national standards, related to technical drawings.
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Concepts and conventions, 1. Importance of graphics in engineering, applications., 2. Use of drafting instruments., 3. BIS conventions and specifications., , 4. Size, layout and folding of drawing sheets., 5. Lettering and dimensioning.
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What is Engineering Drawing?, 1. ENGINEERING DRAWING IS THE LANGUAGE OF, ENGINEERS., 2. By means of Engineering Drawing one can, express the shape, size, finish etc. of any object, accurately and clearly.
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Methods of Expression, There are three methods of writing the graphic, languages., 1. Freehand, 2. With hand-held instruments, , 3. By computer
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Methods of Shape Description, 1. Two Dimensional Drawings or Plane Geometrical, drawings., 2. Three Dimensional Drawings or Solid, Geometrical drawings.
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Plane Geometrical drawings, Plane Geometrical drawing is the drawing which, represents the objects having two dimensions., Ex : Representing square, triangle etc. on a, drawing sheet.
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Solid Geometrical drawings, Solid Geometrical drawing is the drawing which, represents the objects of three dimensions., Ex : Representing cone, sphere etc. on a drawing, sheet.
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Engineering Drawing Instruments, 1. Drawing board, 2. Drawing sheets, 3. Mini-drafter/Drafting machine, , 4. Instrument box
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Engineering Drawing Instruments, 5. Set-squares (45° triangle and 30° - 60° triangle), 6. Protractor, 7. Scales (celluloid/card-board - M1, M2. . . . Me), , 8. Drawing Pencils (HB, H and 2H Grades)
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Engineering Drawing Instruments, 9. Eraser, 10. Clips or Adhesive tape (cello-tape), 11. Sharpener and Emery paper, 12. French curves
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Drawing board
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Mini-drafter/Drafting machine
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Instrument box
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Set-squares
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Drawing Pencils
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Drawing Pencils, 1. HB - (Soft grade) Used for drawing thick outlines, like borderlines, lettering and arrow heads., 2. H-Used for finishing lines, outlines, visible lines, and hidden lines., 3. 2H - (Hard grade) Used for construction lines,, dimension lines, centre lines and section lines.
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French curves
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Standards, 1. I.S.O- International Organization for, Standardization, 2. I.S.I-Indian Standards Institution, , 3. B.I.S -Bureau Of Indian Standards
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Sheet Layout, BORDER LINE, , 20, , 5, , FILING MARGIN, , 5, , 185, , 65, , TITLE BLOCK, , 5
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Title Block, 185, 130, , 65, , 15, , 7, , 10, , 5, , 10, , 5, , 10, , 5, , 10, , AAMEC KOVILVENNI, RAVI SIVAPRAKASH, ROLL NO : 01, I B.E.-MECH, , SECTION : A, , ENGINEERING, , DATE : 02/08/2014, , 5, , 10, , SHEET NO : 01, , 5, , 10
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Folding of Drawing Sheets
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Types of Lines, Type of line, , Illustration, , Application, , Pencil Grade, , Continuous thick, , Visible outlines., , H, , Continuous thin, , Dimension lines, leader lines,, extension lines, Construction lines, & Hatching lines., , 2H, , Continuous thin wavy, (drawn free hand), , Irregular boundary lines, short, break lines., , 2H
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Types of Lines, Type of line, , Illustration, , Application, , Pencil Grade, , Continuous thin with zigzag, , Long break lines., , 2H, , Short dashes, , Invisible edges., , H, , Long chain thin, , Centre lines, locus lines., , H, , Long chain thick at, ends and thin, elsewhere, , Cutting plane lines., , 2H & H, , -----------------------------
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Types of lettering, i. Vertical Letters, a) CAPITAL (UPPER CASE) LETTERS., b) Small (lower case) letters., ii. Inclined Letters (inclined at 75 to the horizontal), a) CAPITAL (UPPER CASE) LETTERS., b) Small (lower case) letters.
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Lettering Standards, 1. Standard heights for lettering are 3.5, 5, 7 & 10, mm., 2. Ratio of height to width, for most of the letters is, approximately 5:3., , 3. However for M and W, the ratio is, approximately5:4
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Lettering Standards, Different sizes of letters are used for different, purposes:, 1. Main title - 7 (or) 10 mm, 2. Sub-titles - 5 (or) 7 mm, , 3. Dimensions, notes etc. - 3.5 (or) 5 mm.
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Lettering Standards, NOTE:, 1. Vertical letters are preferable., , 2. Guide lines -2H pencil, lettering -HB pencil
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Lettering Standards, NOTE:, 3. Spacing between two letters 1/5th of the height, of the letters., 4. Space between two words 3/5th of the height of, the letters.
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Exercise :, Write free-hand, In single stroke vertical (CAPITAL&, lower case) letters, the following:, 1. Alphabets and Numerals (Heights 5, 7 & 10 mm)., 2. “DRAWING IS THE LANGUAGE OF ENGINEERS”, (Heights 5 & 7 mm).
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Dimensioning, 1. An Engineering drawing should contain the, details regarding the sizes, besides giving the, shape of an object., 2. The expression of details in terms of numerical, values regarding distances between surfaces etc.,, on a drawing by the use of lines, symbols and, units is known as dimensioning.
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Anatomy of a dimension, HIDDEN LINE, 10, EXTESION LINE, , 25, , LEADER LINE, , DIMENSION LINE, , 45, , CENTER LINE, , DIMENSION
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General Principles, 1. All dimensions should be detailed on a, , drawing., , 2. No single dimension should be repeated except, where unavoidable., 3. Mark the dimensions outside the drawing as far as, possible.
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General Principles, 4. Avoid dimensioning to hidden lines wherever, possible., 5. The longer dimensions should be placed outside, all intermediate dimensions, so that dimension lines, will not cross extension lines.
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Illustration of principles of, dimensioning, , 20, , 20, , 1. Place the dimensions outside the views., Note:, Dimensions of diameter, circle and radius may be, shown inside., , 50, 50, , Not Correct, , Correct
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Illustration of principles of, dimensioning, 2. Place the dimension value above the horizontal, line near the middle., , 50, , 50, 50, , Not Correct, , Correct
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Illustration of principles of, dimensioning, , Not Correct, , 50, , 50, , 50, , 3. Dimensioning a vertical line., , Correct, , 50
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Illustration of principles of, dimensioning, 4. When on overall dimension is shown, one of the, intermediate dimensions should not be given., , 50, , 20, 70, , Not Correct, , 50, 70, , Correct
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Illustration of principles of, dimensioning, 5. Overall dimensions should be placed outside, intermediate dimensions., , 90, 50, , 20, , 50, , 20, 90, , Not Correct, , Correct
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Illustration of principles of, dimensioning, 6. Arrange a chain of dimensions in a continuous, line., , 90, 50, 20, , Not Correct, , 50, , 20, , Correct, , 90
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Illustration of principles of, dimensioning, 7. Arrowheads should touch the projection lines., , 20, , Not Correct, , 20, , 20, , Correct
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Illustration of principles of, dimensioning, 8. Dimension lines should be placed at least 6 to 10, mm away from the outlines., , 20, , 20, , 6, , 50, , 50, , Not Correct, , Correct, , 6
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Illustration of principles of, dimensioning, 9. Dimensions are to be given to visible lines and, not to hidden lines., 50, , 50, , Not Correct, , Correct
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Illustration of principles of, dimensioning, 10.Centre line should not be used as a dimension, line., , 50, , 50, , Not Correct, , Correct
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Illustration of principles of, dimensioning, , 20, , 20, , 11.Do not repeat the same dimension in different, views., , FRONT VIEW, , L.S. VIEW, , Not Correct, , Correct
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Illustration of principles of, dimensioning, 12.Dimensioning from a centre line should be, avoided except when centre line passes through, the centre of a hole or a cylinder part., , 20, , 50, , Not Correct, , Correct
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Illustration of principles of, dimensioning, 13.Indicate the depth of the hole as notes written, horizontally., , 10, 25DEEP, , 10, 25DEEP, , Not Correct, , Correct
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Illustration of principles of, dimensioning, 14.Locate holes in the proper view., , 50, , Not Correct, , 50, , Correct
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Illustration of principles of, dimensioning, 15.Diameter and radius symbols should be placed, before the values., 50, , 50, , Not Correct, , Correct
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Illustration of principles of, dimensioning, 16.Dimensions are to be given from visible lines and, not from hidden lines., 50, , 50, , Not Correct, , Correct
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Methods of Dimensioning, 1. Unidirectional System (preferable), 2. Aligned System
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Unidirectional System, 1. In this method dimensions shall be horizontally, so that they can be read from the bottom of the, sheet., 2. Here the dimension lines may be interrupted, near the middle for the insertion of dimensions.
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Aligned System, 1. In aligned system, dimensions shall be placed, parallel to (i.e., aligned with) and above the, dimension lines, preferably in the middle and not, by interrupting the dimension lines., 2. Here the dimensions can be read from the, bottom or from the right side of the drawing.
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Arrow heads
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Geometrical Construction, 1. Geometrical construction of lines, arcs, circles,, polygons and drawing tangents and normal form, the basics of Engineering drawing.
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Points, 1. A point represents a location in space or on a, drawing, and has no width, height and depth., 2. A point is represented by the intersection of two, lines., , ., , .
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Lines, 1. A straight line is the shortest distance between, two points and is commonly referred as “Line”., 2. It as length and no width., , ., , .
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Angles, 1. An Angle is formed between two intersecting, lines., 2. A common symbol for angle is
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Triangles, 1. A Triangle is a plane figure bounded by three, lines, and the sum of the interior angle is always, 180°., 2. A right angle triangle has one 90°angle.
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Triangles
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Quadrilaterals, 1. A Quadrilateral is a plane figure bounded by four, lines., 2. In this quadrilaterals if the opposite sides are, parallel, the quadrilateral is called parallelogram.
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Quadrilaterals
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Polygons, 1. A Polygon is plane figure bounded by number of, straight lines., 2. If the polygon has equal angles and equal sides, and if it can be inscribed in or circumscribed, around a circle, it is called as Regular polygon.
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Circles and Arcs, 1. A Circle is a closed curve and all points of which, are at the same distance from a point called the, center., 2. Circumference refers to the distance around the, circle., 3. If number of circles of circles have a same center,, they are called as Concentric circles.
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Circles and Arcs
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Bisect a line, 1.Set a Compass width to a approximately, two thirds the line length., The actual width does not matter., 2.Using a straight edge,, draw a line between the points, where the arcs intersect., , P, , Q
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To Bisect a Circular Arc, D, , C, , B, , A, , E
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Divide a Line into number of equal, parts, , 1. Draw a straight line AB., 2. Draw a line AC at any convenient acute angle with, AB., 3. Set the divider to a convenient length and mark off, seven spaces on AC. Let the points obtained be, l’,2’,3’,4’,5’,6’,and7’., 4. Join 7’to the point B., 5. Draw lines through points 1’, 2’, 3’. 4’, 5’ and 6’, parallel to 7’B to meet AB at points 1, 2, 3, 4, 5and, 6 respectively. These points divide AB in equal, length.
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Divide a Line into number of equal, parts, A, , 1, 1’, , 2, 2’, , 3, 3’ 4’, , 4, , 5, , 5’, , *An acute angle is less than 90°, , 6’, , 6, , 7’, , B, , C
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To Construct an Isosceles Triangle, 1. Mark a point P that will become one vertex of the, triangle., 2. Mark a point R on arc. PR will be the base of the, triangle., 3. Draw the base PR of the triangle., 4. With Points P and R as centres and radius R, equal, to the length of the sides, draw intersecting arcs to, locate the vertex (top point) of the triangle., 5. PQR is an isosceles triangle with the desired, dimensions.
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To Construct an Isosceles Triangle, Q, , P, , R
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To Construct an Equilateral Triangle, 1. Mark a point P that will become one vertex of, the triangle., 2. Mark a point Q on either arc to be the next, vertex., 3. Without changing the width, move to Q and draw, an arc across the other, creating R, 4. Draw three lines linking P, Q and R
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To Construct an Equilateral Triangle, Q, , P, , R
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Construct a regular pentagon, First Method, 1. Draw a line AB equal to the given length of a side., 2. Draw a line AB equal to the given length of a side., Extend the side BA and mark P such that AP=AB, 3. Divide the semi-circle into 5 equal parts (for, pentagon) by trial and error method and name, the points as 1,2,3,4 and 5 starting form P.
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Construct a regular pentagon, First Method, 5. Join A2. Now A2 = AB = AE., 6. Join A3 and A4 and extend them., 7. With AB as radius and B as centre, draw an arc to, cut the extension of A4 at C., 8. With E as centre and same radius draw an arc to, intersect the extension of A3 at D., 9. Join BC, CD and DE. ABCDE is the required, pentagon.
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Construct a regular pentagon, D, E, 2, , 3, , 1, P, , C, 4, , A, , B, 5
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Pentagon (vertical)
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Construct a regular pentagon, Second Method, 1. Draw AB equal to one side of the pentagon., 2. Draw AB equal to one side of the pentagon. Extend, the side AB and mark P such that, AB = BP., 3. With B as centre and BA as radius draw a semicircle., 4. Divide the semi-circle into 5 equal parts (for, pentagon) by trial and error method and name the, points as 1,2,3,4 and 5 starting from P.
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Construct a regular pentagon, Second Method, 5. Join B2. Now AB = B2 = BC, 6. Draw the perpendicular bisectors of AB and BC to, intersect at O., 7. With O as centre and OA = OB = OC as radius draw, a circle., 8. With A and C as Centre's and AB as radius, draw, arcs to cut the circle at E and D respectively., 9. Join CD, DE and EA. ABCDE is the required, pentagon.
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Construct a regular pentagon, Second Method
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Problem, Problem 1 :, Construct a regular pentagon of 40mm side with, side (i) horizontal and (ii) vertical.
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Construct a regular hexagon, 1. Let AB be the given side., 2. With A and B centres and AB as radius, draw two, arcs to intersect at O., 3. With O as centre and AB as radius describe a, circle., 4. With the same radius and A and B as centres,, draw arcs to cut the circumference of the circle at, F and C respectively.
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Construct a regular hexagon, 5. With the same radius and C and F as centres,, draw arcs to cut the circle at D and E respectively., Join BC, CD, DE, EF and FA. ABCDEF is the, required hexagon.
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Hexagon (horizontal), E, , D, , O, , F, , C, , A, , B
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Hexagon (vertical) C, D, , O, , B, , E, , A, F
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Problem, 1. Construct a regular hexagon of side 35mm when, one side is, (i) horizontal and, (ii) vertical.
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CONIC SECTIONS
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CONIC SECTIONS, 1. The sections obtained by the intersection of a, right circular cone by cutting plane in different, positions relative to the axis of the cone are, called Conics or Conic Sections.
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Circular Cone, 1. A right circular cone is a cone having its axis, perpendicular to its base., 2. The Top point of the cone is called APEX., 3. The imaginary line joining the apex and the, centre of the base is called AXIS., 4. The Lines joining the apex to the circumference, of the base circle is called GENERATORS.
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Right Circular Cone, APEX, , GENERATOR, , AXIS, , BASE
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Definitions, 1. The conic sections can be defined in TWO WAYS :, a) By Cutting a right circular cone with a sectional plane., b) Mathematically, i.e., with respect to the loci of a, point moving in a plane.
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Cutting Planes
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Cutting Planes
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Definition of Conic sections by Cutting, a right circular cone with a sectional, plane, 1. Circle, When the cutting plane AA is perpendicular to the, axis and cuts all the generators, the section obtained is a, circle.
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Definition of Conic sections by Cutting, a right circular cone with a sectional, plane, 2. Ellipse, When the cutting plane BB is inclined to the axis of, the cone and cuts all the generators on one side of the, apex, the section obtained is an ellipse.
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Definition of Conic sections by Cutting, a right circular cone with a sectional, plane, 3. Parabola, When the cutting plane CC is inclined to the axis of, the cone and parallel to one of the generators , the, section obtained is a parabola.
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Definition of Conic sections by Cutting, a right circular cone with a sectional, plane, 4. Hyperbola, When the cutting plane DD makes a smaller angle, with the axis than that of the angle made by the, generator of the cone, the section obtained is a, hyperbola.
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Definition of Conic sections by Cutting, a right circular cone with a sectional, plane, 5. Rectangular Hyperbola or Equilateral Hyperbola, When the cutting plane EE is parallel to the axis of the, cone, the section obtained is a RECTANGULAR or, EQUILATERAL HYPERBOLA.
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Conic sections Defined, Mathematically-Conic Terminology, 1. Conic, It is defined as the locus of a point moving in a, plane in such a way that the ratio of its distance, from a fixed point to a fixed straight line is always, a constant. This ratio is called eccentricity.
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Conic sections Defined, Mathematically, 2. Ellipse, Ellipse is the locus of a point moving in a plane in, such a way that the ratio of its distance from a, point (F) to the fixed straight line (DD) is a, constant and is always less than 1.
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Conic sections Defined, Mathematically, 3. Parabola, Parabola is the locus of a point moving in a, plane in such a way that the ratio of its distance, from a point (F) to the fixed straight line (DD) is, a constant and is always equal to 1.
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Conic sections Defined, Mathematically, 4. Hyperbola, Hyperbola is the locus of a point moving in a plane in, such a way that the ratio of its distance from a point (F), to the fixed straight line (DD) is a constant and is, greater than 1., , 5. Focus, The fixed point is called the focus (F)., , 6. Directrix, The fixed line is called the directrix (DD).
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Conic sections Defined, Mathematically, 7. Eccentricity (e), It is the ratio =, distance of the moving point from the focus, Distance of the moving point from the directrix
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Conic sections Defined, Mathematically, 8. Axis (CA), The line passing through the focus and perpendicular to, the directrix is called axis., , 9. Vertex (V), It is a point at which the conic cuts its axis.
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Conic sections Defined, Mathematically
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Ellipse, Methods of Construction, 1. Eccentricity method, 2. Pin and String method, 3. Trammel method, 4. Intersecting arc or Arc of Circles or Foci method
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Ellipse, Methods of Construction, 5. Concentric Circles method, 6. Rectangle or Oblong method, 7. Parallelogram method, 8. Circle method (using conjugate diameters), 9. Four centers (approximate) method
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Problem 1, a)Construct an ellipse when the distance between, the focus and the directrix is 50mm and the, eccentricity is 2/3., (b) Draw the tangent and normal at any point P on, the curve using directrix.
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Solution, 1. Draw a vertical line DD’ to represent the directrix. At any, point A on it draw a line perpendicular to the directrix to, represent the axis., , 2. The distance between the focus and the directrix is 50, mm. So mark F1, the focus such AF1 = 50 mm., 3. Eccentricity = 2/3 i.e., 2 + 3 = 5. Divide AF1 into 5 equal, parts using geometrical construction and locate the, vertex V1 on the third division from A. Now V1F1/V1A =, 2/3.
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Solution, 4. Draw a perpendicular line at V1. Now draw 45° inclined, line at F1 to cut the perpendicular line drawn at V1., Mark the cutting point as S. Or V1 as centre and V1F1 as, radius cut the perpendicular line at S., , 5. Join A and S and extend the line to Y., From F1 draw a 45° line to intersect the line AY at T., From T erect vertical to intersect AA' at V2, the another, vertex. V1V2 = Major axis.
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Solution, 6. Along the major axis, mark points 1, 2... 10 at, approximately equal intervals. Through these points, erect verticals to intersect the line AY (produced if, necessary) at 1', 2',...10'., , 7. With 11' as radius and F1 as centre draw two arcs on, either side of the axis to intersect the vertical line drawn, through 1 at P1 and Q1., 8. With 22' as radius and F1 as centre draw two arcs on, either side of the axis to intersect the vertical line drawn, through 2 at P2 and Q2.
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Solution, Draw the tangent and normal at any point P on the curve, using directrix., 12. Mark a point P and join PF1., 13. At F1 draw a line perpendicular to PF1 to cut DD’ at Q., Join QP and extend it. QP is the tangent at P., 14. Through P, draw a line NM perpendicular to QP. NM is, the normal to the ellipse at P.
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Engineering Applications, The shape of an ellipse is used for making, 1. Concrete arches, 2. Stone bridges, 3. Glands, 4. Stuffing boxes, 5. Reflectors used in automobiles etc.
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Exercise, Problem 2:, a) Construct an ellipse when the distance between, the focus and the directrix is 60mm and the, eccentricity is 3/4., (b) Draw the tangent and normal at any point P on, the curve using directrix.
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Exercise, Problem 3:, a) Construct an ellipse given the distance of the, focus from the directrix as 60 mm and eccentricity, as 2/3., (b) Draw the tangent and a normal to the curve at a, point on it 20 mm above the major axis.
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Exercise, Problem 4:, a) Construct an ellipse when the distance of the, focus from the directrix is equal to 5 cm and the, eccentricity is 3/4., (b) Draw the tangent and normal at any point P on, the curve using directrix.
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Exercise, Problem 5:, a) Draw the locus of a point P moving so that the, ratio of its distance from a fixed point F to its, distance from a fixed straight line DD’ is (i) 3/4, (ii) 1 and (iii) 4/3., (b) Point F is at a distance of 35 mm form DD’. Draw, a tangent and a normal to each curve at any point, on it.
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Construct a Parabola, Problem 6 :, Construct a parabola when the distance between, focus and the directrix is 50mm. Draw tangent and, normal at any point P on the curve.
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Construct a Parabola, 1. 1. Draw a vertical line DD’ to represent the directrix. At, any point A on it draw a line perpendicular to the, directrix to represent the axis., 2. The distance between the focus and the directrix is 50, mm. So mark F the focus such AF = 50 mm., 3. For parabola the eccentricity is always equal to 1. So, mark the mid-point of AF as V (vertex) Now VF/VA = 1., 4. Draw a perpendicular line at V. Now draw 45° inclined, line at F to cut the perpendicular line drawn at V. Mark, the cutting point as S. Or V as centre and VF as radius, cut the perpendicular line at S.
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Construct a Parabola, 5. Join A and S and extend the line to Y., 6. Along the axis AA’ mark points 1, 2...5 at approximately, equal intervals. Through these points erect verticals to, intersect the line AY (produced if necessary) at 1', 2',...5'., 7. With 11' as radius and F as centre draw two arcs on, either side of the axis to intersect the vertical line drawn, through 1 at P1 and Q1., 8. With 22' as radius and F as centre draw two arcs on, either side of the axis to intersect the vertical line drawn, through 2 at P2 and Q2.
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Construct a Parabola, 9. Repeat the above the obtain P3 and Q3…P5 and Q5, corresponding to, 2,3, ... 5 respectively and draw a, smooth parabola passing through P5,... P1, V,Q1,Q5., Draw the tangent and normal at any point P on the curve, using directrix., 10. Mark the given point P and join PF., 11. At F draw a line perpendicular to PF to cut DD’ at Q. Join, QP and extend it. QP is the tangent at P., 12. Through P, draw a line NM perpendicular to QP. NM is, the normal to the parabola at P.
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Exercise, Problem 7:, Draw a parabola given the distance of the focus, from the directrix as 60 mm. Draw tangent and, normal at any point P on the curve.
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Exercise, Problem 8:, Draw the parabola whose focus is at a distance of, 40 mm from the directrix. Draw a tangent and a, normal at any point on it.
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Exercise, Problem 9 :, Draw a parabola when the distance of focus from, the directrix is equal to 65 mm. Draw a tangent, and a normal at any point on it.
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Exercise, Problem 10:, A fixed point is 55 mm from a fixed straight line., Draw the locus of a point moving in such a way, that its distance from the fixed straight line is, equal to its distance from the fixed point. Name, the curve. Draw a tangent and a normal at any, point on it.
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Engineering Applications, Parabola is used for, 1. Suspension bridges, 2. Arches, 3. Sound and Light reflectors for parallel beams, such as search lights, machine tool structures etc.
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Construct a Hyperbola, Problem 11 :, Construct a hyperbola when the distance between, the focus and the directrix is 40 mm and the, eccentricity is 4/3. Draw a tangent and normal at, any point on the hyperbola.
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Construct a Hyperbola, 1. Draw a vertical line DD’ to represent the directrix. At any, point A on it draw a line perpendicular to the directrix to, represent the axis., 2. The distance between the focus and the directrix is 40, mm. So mark F the focus such that AF = 40 mm., 3. Eccentricity = 4/3 i.e., 4 + 3 = 7. Divide AF into 7 equal, parts using geometrical construction and locate the, vertex V on the third division from A. Now VF/VA = 4/3.
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Construct a Hyperbola, 4. Draw a perpendicular line at V. Now draw 45° inclined, line at F to cut the perpendicular line drawn at V. Mark, the cutting point as S. Or V as centre and VF as radius, cut the perpendicular line at S., 5. Join A and S and extend the line to Y., 6. Along the axis AA’ mark points 1, 2... 5 at approximately, equal intervals. Through these points erect verticals to, intersect the line AY (produced if necessary) at 1', 2',...5‘.
Page 144 :
Construct a Hyperbola, 7. With 11' as radius and F as centre draw two arcs on, either side of the axis to intersect the vertical line drawn, through 1 at P1 and Q1., 8. With 22' as radius and F as centre draw two arcs on, either side of the axis to intersect the vertical line drawn, through 2 at P2 and Q2., 9. Repeat the above the obtain P3 and Q3…P5 and Q5, corresponding to 3.. 5 respectively and draw a smooth, hyperbola passing through P5, P4.... P1 V, Q1,… Q5.
Page 145 :
Construct a Hyperbola, Draw the tangent and normal at any point P on the curve, using directrix., 10. Mark a point P and join PF1., 11. At F1 draw a line perpendicular to PF1 to cut DD’ at Q., Join QP and extend it. QP is the tangent at P., 12. Through P, draw a line NM perpendicular to QP. NM is, the normal to the hyperbola at P.
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Engineering Applications, Hyperbola is used in, 1. Design of Channels etc., 2. The expansion curve (p-v diagram) of a gas or, steam is represented by a Rectangular Hyperbola.
Page 148 :
Exercise, Problem 12:, Draw a hyperbola when the distance between its, focus and directrix is 50 mm and eccentricity is, 3/2. Also draw the tangent and normal at a point, 25 mm from the directrix.
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Exercise, Problem 13:, Draw a hyperbola when the distance between its, focus and directrix is 50 mm and eccentricity is, 5/3. Also draw the tangent and normal at any, point on the hyperbola.
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Exercise, Problem 14:, Draw a hyperbola given the distance of the focus, from the directrix as 55 mm an eccentricity as 1.5.
Page 153 :
Cycloid, 1. A cycloid is a curve generated by a point on the, circumferences of a circle as the circle rolls along, a straight line., 2. The rolling circle is called the generating circle, and the line along which is rolls is called the, directing line or base line.
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Cycloid, NOTE :, 1. When a circle makes one revolution on the base, line it would have moved through a distance =, circumference of the rolling circle., 2. This circumference should be obtained by, geometrical construction.
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Problem 1, A coin of 40 mm diameter rolls over a horizontal, table without slipping., A point on the circumference of the coin is in contact, with the table surface in the beginning and after, one complete revolution., Draw the cycloidal path traced by the point. Draw a, tangent and normal at any point on the curve.
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Solution, 1. Draw the rolling circle of diameter 40mm., 2. Draw the base line PQ equal to the circumference, of the rolling circle at P., 3. Divide the rolling circle into 12 equal parts as, 1,2,3 etc., 4. Draw horizontal lines through 1,2,3 etc., 5. Divide the base line PQ into the same number of, equal parts (12) at 1’, 2’, 3’…etc.
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Solution, 6. Draw lines perpendicular to PQ at 1’, 2’, 3’ etc to, intersect the horizontal line drawn through C, (called the locus of centre) at C1, C2 ….etc., 7. With C1, C2 etc as centres and radius equal to, radius of rolling circle (20mm) draw arcs to cut, the horizontal lines through 1, 2, …etc.at P1,, P2….etc., 8. Draw a smooth curve (cycloid) through P, P1,, P2…etc.
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Solution, To draw normal and tangent at a given point D, 9. With D as centre and radius equal to radius of the, rolling circle, cut the line of locus of centre at C’., 10.From C’ draw a perpendicular line to PQ to get, the point E on the base line. Connect DE, the, normal., 11.At D, draw a line perpendicular to DE and get the, required tangent TT.
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Problem 1
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Applications, 1. Cycloid is used in the design of gear tooth system., 2. It has application in conveyor for mould boxes in, foundry shops and, 3. some other applications in mechanical, engineering.
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Exercise, Problem 2 :, Draw a cycloid formed by a rolling circle 50 mm in, diameter. Use 12 divisions. Draw a tangent and a, normal at a point on the curve 30mm above the, directing line.
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Exercise, Problem 3 :, A circle of 40 mm diameter rolls on a straight line, without slipping., In the initial position the diameter PQ of the circle is, parallel to the line on which it rolls., Draw the locus of the points P and Q for one, complete revolution of the circle.
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Exercise, Problem 4 :, A circle of 40 mm diameter rolls on a horizontal line., Draw the curve traced out by a point R on the, circumference for one half revolution of the, circle., For the remaining half revolution the circle rolls on, the vertical line., The point R is vertically above the centre of the, circle in the starting position.
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Exercise, Problem 5 :, A circle of 40 mm diameter rolls on a Straight line, without slipping., Draw the curve traced out by a point P on the, circumference for 1.5 revolution of the circle., Name the curve.Draw the tangent and normal at a, point on it 35mm from the line.
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Epicycloid, 1. Epicycloid is a curve traced by a point on the, circumference of a circle which rolls without, slipping on the outside of another circle.
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Problem 6, Draw an epicycloid of rolling circle 40 mm (2r),, which rolls outside another circle (base circle) of, 150 mm diameter (2R) for one revolution. Draw a, tangent and normal at any point on the curve.
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Solution, 1. In one revolution of the generating circle, the, generating point P will move to a point Q, so that, the arc PQ is equal to the circumference of the, generating circle. is the angle subtended by the, arc PQ at the centre O.
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Solution, 2. Taking any point O as centre and radius (R) 75 mm, draw, an arc PQ which subtends an angle = 96° at O., 3. Let P be the generating point. On OP produced, mark PC, = r = 20 mm = radius of the rolling circle. Taking centre, C and radius r (20 mm) draw the rolling circle., 4. Divide the rolling circle into 12 equal parts and name, them as 1, 2, .3 etc in the CCW direction, since the, rolling circle is assumed to roll clockwise., 5. O as centre, draw concentric arcs passing through 1, 2, 3,, . . . etc.
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Solution, 6. O as centre and OC as radius drew an arc to represent, the locus of centre., 7. Divide the arc PQ into same number of equal parts (12), and name them as 1’, 2’, . . . etc., 8. Join 01’, 02’ . . . etc. and produce them to cut the locus, of centre at C1,C2. . etc., 9. Taking C1 as centre and radius equal to r, draw an arc, cutting the arc through 1 at P1. ‘Similarly obtain the, other points and draw a smooth curve through them.
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Solution, To draw a tangent and normal at a given point M:, 10. M as centre, end radius r = CP cut the locus of centre at, the point N., 11. Join NO which intersects the base circle arc PQ at S., 12. Join MS, the normal and draw the tangent perpendicular, to it.
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Solution
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Hypocycloid, Hypocycloid is a curve traced by a point on the, circumference of a circle which rolls without, slipping on the inside of another circle.
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Problem 7, Draw a hypocycloid of a circle of 40 mm diameter, which rolls inside another circle of 200 mm, diameter for one revolution. Draw a tangent and, normal at any point on it.
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Involutes, 1. An involutes is a curve traced by a point on a, string as it unwinds from around a circle or a, polygon.
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Problem 8, Draw the involute of a square of side 20mm.
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Solution 8, 1. Draw the square ABCD of side 20mm., 2. With A as centre and AB as radius, draw an arc to, cut DA produced at P1., 3. D as centre and DP1 as radius draw an arc to cut, CD produced at P2.
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Solution 8, 4. C as centre and CP2 as radius draw an arc to cut, BC produced at P3., 5. Similarly, B as centre and BP3 as radius draw an, arc to cut AB produced at P4.
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Solution 8, NOTE :, BP4 is equal to the perimeter of the square., The curved obtained is the required involute of the, square.
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Solution 8, To draw a normal and tangent at a given point M., 1. The given point M lies in the arc P3 P4., 2. The centre of the arc P3 P4 is point B., 3. Join B and M and extend it which is the required, normal., 4. At M draw perpendicular to the normal to obtain, tangent TT.
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Problem 9, A coir is unwound from a drum of 30 mm, diameter. Draw the locus of the free end of the, coir for unwinding through an angle of 360°., Draw also a normal and tangent at any point on, the curve.
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Solution 9, 1. Draw the given circle of 30 mm diameter (D)., 2. Divide the circle into 12 equal parts as 1, 2, 3 . . ., 12. Let P be the starting point i.e. one end of the, thread., 3. Draw a line PQ tangential to the circle at P and, equal to D., 4. Divide PQ into 12 equal parts as 1’, 2’, . . .12’., 5. Draw tangents at points 1, 2, 3 . . . etc. and mark, P1 , P2 ,…P12 such that 1P1 = Pl’; 2P2 = P2’; 3P3 =, P3’ etc.
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Solution 9, Tangent and normal to the involute of the circle at a, given point M:, 1. Draw a line joining M and the centre of the circle, O., 2. Mark the mid-point C on OM., 3. With C as centre and MC as radius describe e, semi-circle to cut the given circle at B., 4. Join MB, which is the required normal., 5. At M, draw a line perpendicular to MB, to get the, required tangent TT.
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Problem 10, Draw the involute of a circle of diameter 40 mm., Draw also a normal and tangent at any point on, the curve.
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Problem 11, Draw one turn of the involute of a circle 50 mm in, diameter. Draw a tangent and normal to the curve, at a point 80 mm from the centre of the circle.
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ORTHOGRAPHIC PROJECTION
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ORTHOGRAPHIC PROJECTION, PROJECTION, 1. The figure or view formed by joining, in correct, sequence, the points at which these lines meet, the plane is called the projection of the object. (It, is obvious that the outlines of the shadow are the, projections of an object).
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ORTHOGRAPHIC PROJECTION, PROJECTION
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ORTHOGRAPHIC PROJECTION, PROJECTORS, 1. The lines or rays drawn from the object to the, plane are called projectors.
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ORTHOGRAPHIC PROJECTION, PLANE OF PROJECTION, 1. The transparent plane on which the projections, are drawn is known as plane of projection.
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ORTHOGRAPHIC PROJECTION, TYPES OF PROJECTION, 1. Pictorial Projections, a) Perspective Projection, b) Isometric Projection, c) Oblique Projection, , 2. Orthographic Projection
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ORTHOGRAPHIC PROJECTION, 1. PICTORIAL PROJECTIONS, The projections in which the description of the, object is completely understood in one view is know, as Pictorial Projection.
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P, I, C, T, O, R, I, A, L, , P, R, O, J, E, C, T, I, O, N, S
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ORTHOGRAPHIC PROJECTION, 2. ORTHOGRAPHIC PROJECTIONS, ‘ORTHO’ means ‘right-angle’ and ORTHOGRAPHIC means right-angled drawing., When the projectors are perpendicular to the plane, on which the projection is obtained it is known as, Orthographic Projection.
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ORTHOGRAPHIC PROJECTION, 2. ORTHOGRAPHIC PROJECTIONS
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ORTHOGRAPHIC PROJECTION, 2. ORTHOGRAPHIC PROJECTIONS
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ORTHOGRAPHIC PROJECTION, Vertical Plane, Extend the rays or projectors further to meet a, vertical (Transparent) plane (V.P) located behind the, object.
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ORTHOGRAPHIC PROJECTION, Horizontal Plane, As front view alone is insufficient for the, complete description of the object, let us assume, another plane, called Horizontal plane (H.P.) hinged, perpendicular to V.P.
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ORTHOGRAPHIC PROJECTION, XY Line, The line of intersection of V.P. and H.P. is called, the Reference Line and denoted as XY.
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ORTHOGRAPHIC PROJECTION, TERMINOLOGY, 1. V.P. and H.P. are called as principle planes of, projection or reference planes., 2. They are always transparent and at right-angles, to each other., 3. The projection on V.P. is Front view, 4. The projection on H.P. is Top view
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ORTHOGRAPHIC PROJECTION, FOUR QUADRANTS, When the planes of projections are extended, beyond their line of intersection, they form Four, Quadrants or Dihedral Angles., POSITION OF THE OBSERVER, The observer will always be in the right side of, the four quadrants.
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TOP VIEW, , ALWAYS ROTATE H.P., CLOCKWISE TO OPEN-OUT, I QUADRANT, , II QUADRANT, Above H.P., Behind V.P., I QUADRANT, Above H.P., In front of V.P., , III QUADRANT, Below H.P., Behind V.P., IV QUADRANT, Below H.P., In front of V.P.
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ORTHOGRAPHIC PROJECTION, FIRST ANGLE PROJECTION, when the object is situated in first quadrant,, that is in front of V.P. and above H.P. the projection, obtained on these planes is called First Angle, Projection.
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ORTHOGRAPHIC PROJECTION, FIRST ANGLE PROJECTION, 1. The object lies in between the observer and the, plane of projection., 2. The front view is drawn above the XY line and, the. top view below XY. (Here, above XY line, represents V.P. and below XY line represents H.P.).
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ORTHOGRAPHIC PROJECTION, FIRST ANGLE PROJECTION, 3. In the front view, H.P. coincides with XY line and, in top view V.P. coincides with XY line., 4. Front View shows the length (L) and height (H) of, an object. Top View shows the length (L) and, breadth (B) or width (W) or thickness (T) of it.
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ORTHOGRAPHIC PROJECTION, THIRD ANGLE PROJECTION, In this the object is situated in Third Quadrant., The planes of projection lie between the object and, the observer., The front view comes below the XY line and the top, view above it., The top view above the XY line.
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V.P., , TOP VIEW, , X, , H.P., , FRONT VIEW, , Y
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ORTHOGRAPHIC PROJECTION
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ORTHOGRAPHIC PROJECTION, AUXILIARY VERTICAL PLANE (A.V.P.), 1. Auxiliary vertical plane is perpendicular to both, V.P. and H.P., 2. Front view is drawn by projecting the object on, the V.P.
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ORTHOGRAPHIC PROJECTION, AUXILIARY VERTICAL PLANE (A.V.P.), 3. Top view is drawn by projecting the object on the, H.P., 4. The projection on the A.V.P. as seen from left of, the object and drawn on the right of the front, view, is called Left side view.
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AUXILIARY VERTICAL PLANE (A.V.P.)
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ORTHOGRAPHIC PROJECTION, How to draw the Side View?, 1. Rotate the A.V.P. In the direction of the arrow, shown, so as to make it to coincide with the V.P., 2. Looking the object from the left, the left side, view is obtained and drawn on the right side of, the front view.
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AUXILIARY VERTICAL PLANE (A.V.P.)
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ORTHOGRAPHIC PROJECTION, RULE: In First Angle Projection, 1. A.V.P. is positioned on the right side of the V.P. to, obtain the left side view., 2. A.V.P. is positioned on the left side of the V.P. to, obtain the right side view.
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ORTHOGRAPHIC PROJECTION, RULE: In Third Angle Projection, 1. (In Third Angle Projection, A.V.P. is positioned on, the right side of the V.P. to obtain the right side, view and vice-versa.)
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FREEHAND SKETCHING
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Freehand Sketching, 1. A freehand sketch is a drawing made without the, use of drawing instruments., 2. It is not drawn to scale, but should be in good, proportion as accurately as possible by eye, judgment.
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Freehand Sketching, 3. A freehand sketch should, contain all the, necessary details such as dimensions and actual, shape., 4. HB pencil preferable.
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Sketching a circle
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Sketching an ellipse
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Problem 1
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Problem 1
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Problem 2
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Problem 2
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Problem 3
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Problem 3
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Problem 4
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Problem 4
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Problem 5
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Problem 5
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Problem 6
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Problem 6
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Problem 7
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Problem 7
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Problem 8
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Problem 8
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Problem 9
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Problem 9
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Problem 10
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Problem 10
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Problem 11
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Problem 11
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Problem 12
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Problem 13
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Problem 14
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Problem 15
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Problem 16
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Problem 17
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Problem 18
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Problem 19
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Problem 20
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Construction of Scales
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Scales, 1. It is not always possible or convenient to draw, drawings of an object to its actual size., 2. Drawings of very big objects like buildings,, machines etc. cannot be prepared in full size., , 3. Drawings of very small objects like precision, instruments, namely watches, electronic devices, etc.
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Full size scale, 1. If we show the actual length of an object on a, drawing, then the scale used is called full size, scale.
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Reducing scale, 1. If we reduce the actual length of an object so as, to accommodate that object on drawing, then, the scale used is called reducing scale., 2. Example :, a) large machine parts, b) Buildings, c) Bridges, d) Survey maps, e) Architectural drawings etc.
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Increasing or Enlarging scale, 1. Drawings of small machine parts, mechanical, instruments, watches, etc. are made larger than, their real size. These are said to be drawn in an, increasing or enlarging scale., NOTE :, The scale of a drawing is always indicated on, the drawing sheet at a suitable place either, below the drawing or near the title thus “scale 1 :, 2”.
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Representative Fraction (R.F), 1. The ratio of the drawing size of an object to its, actual size is called the Representative Fraction,, usually referred to as R.F., , R.F =, , Drawing size of an object, Its actual size, , (in same units)
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Reducing scale R.F, 1. For reducing scale, the drawings will have R.F, values of less than unity. For example 1 cm on, drawing represents 1 m length., 1 cm, R.F =, 1 X 100 cm, , 1, =, 100, , <1, , (in same units)
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Increasing or Enlarging scale R.F, 1. For drawings using increasing or enlarging the R.F, values will be greater than unity. For example,, when 1 mm length of an object is shown by a, length of 1 cm., 1 X 10 mm, R.F =, 1 mm, , =, , 10, 1, , = 10 > 1, (in same units)
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Metric Measurements, 1., 2., 3., 4., , 10 decimeters (dm), 10 meters (m), 10 decameters (dam), 10 hectometer (hm), , = 1 meter (m), = 1 decameter (dam), = 1 hectometer (hm), = 1 kilometer (km)
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Types of Scales, 1. Simple or Plain scales, 2. Diagonal scales, 3. Vernier scales
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Simple or Plain scales, 1. A plain scale is simply a line which is divided into, a suitable number of equal parts, the first of, which is further sub-divided into small parts., 2. It is used to represent either two units or a unit, and its fraction such as km an hm, m and dm, etc.
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Simple or Plain scales, NOTE :, 1. Before constructing a scale, it is necessary to, know: (a) Its R.F.,, (b) Maximum length to be measured and, (c) Divisions it has to show., 2. If the length of scale and distance to be marked, are not given in the problem, then assume the, scale length = 15 cm.
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Problem 1, Construct a plain scale to show meters when 1, centimeter represents 4 meters and long enough, to measure upto 50 meters. Find the R.F. and, mark on it a distance of 36 meters.
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Problem 1, size, 1cm, 1, ( in same units ) , , 1. R.F. Drawing, Actual size, 4 100cm, 400, 2. Length of scale = R.F. x maximum length to be, measured., Maximum length to be measured = 50 m (given), 1, 1, length of scale = 400 50 m 400 50 m 100 cm, 3. Draw a horizontal line of length 12.5 cm (L), 4. Draw a rectangle of size 12.5cm x 0.5cm on the, horizontal line drawn above., NOTE: Width of the scale is usually taken as 5 mm
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Problem 1, 5. Total length to be measured is 50m. Therefore, divide the rectangle into 5 equal divisions, each, division representing 10m., NOTE: 1. For dividing the length L into n number of, equal parts, use geometrical construction., 2. Use 2H pencil for the construction lines., 6. Mark 0 (zero) at the end of the first main division.
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Problem 1, 7. From 0, number 10,20,30 and 40 at the end of, subsequent main divisions towards right., 8. Then sub-divide the first main division into 10, sub-divisions to represent meters., 9. Number the sub-divisions. i.e. meters to the left, of 0., 10.Write the names of main units and sub-units, below the scale. Also mention the R.F., 11.Indicate on the scale a distance of 36 meters (3, main divisions to the right side of 0 + 6 subdivisions to the left of 0.
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36 m, , 5 mm, 10, , 5, , 0, , 10, , 20, METRES, , L = 12.5 cm, R.F =, , 1, 400, , 30, , 40
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DIAGONAL SCALE, 1. Plain scales are used to read lengths in two units, or to read to first decimal accuracy., 2. Diagonal scales are used either to measure very, minute distances such as 0.1 mm etc., or to, measure in three units such as dm, cm and mm.
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DIAGONAL SCALE, Divide AD into ten equal divisions of, any convenient length (5 cm), C, 88 ', , , , A8, , DC, , AD, , 88 ', , 8, , DC, , , , 10, , ; ButA 8 , , i .e . 88 ' , , 8, , 8, , AD, , D, 10, 9, , 10, , 8, , DC 0 . 8 DC 0 . 8 AB, , 7, , 10, , 6, 5, , 4, , 11’ equal to 0.1 AB, 22’ equal to 0.2 AB, ., ., ., 99’ equal to 0.9 AB, , 3, 2, 1, B, , A, , 5 cm
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Problem 2, The distance between two stations by road is 200, km and it is represented on a certain map by a 5, cm long line. Find the R.F. and construct a, diagonal scale showing a single kilometer and, long enough to measure up to 600 km. Show a, distance of 467 km on this scale.
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Problem 2, 5 cm, , 1. Determine R.F , , 200 km, 2. Calculate length of scale, L s R.F maximum, , length, , , , 5 cm, , , , 200 10 cm, 5, , 1, 4 10, , 1, 4 10, , 6, , 60 10 cm 15 cm, 5, , 6, , 3. Draw a rectangle ABCD of length 15 cm and, width between 40 to 50 mm., 4. Divide AB into 6 equal parts so that each part, may represent 100 km., 5. Divide A0 into 10 equal divisions, each, representing 10 km. Erect diagonal lines through, them.
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Problem 2, 6. Divide AD into 10 equal divisions and draw, horizontal lines through each of them meeting at, BC., 7. Write the main unit, second unit, third unit and, R.F., 8. Mark a distance of 467 km on the scale.
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467k m, , D, , C, , KILOMETRE, , 10, 8, , 6, 4, , 2, , B, , A, , 10, , 5, , 0, KILOMETRE, , 100, , R.F , , 200, 300, R.F = 1:4000000, , 5 cm, , , , 200 km, L s R.F maximum, , 5 cm, 200 10 cm, , length, , 5, , , , 1, 4 10, , 400, , , , 1, 4 10, , 6, , 60 10 cm 15 cm, 5, , 6, , 500
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Problem 3, Construct a diagonal scale of R.F. = 1:3200000 to, show kilometers and long enough to measure, upto 400 km. show distances of 257 km and 333, km on your scale.
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VERNIER SCALE, 1. Vernier Scale is a short scale used when a, diagonal scale is inconvenient to use due to lack, of space., 2. It consists of two parts, i.e., Main Scale (which is, a Plane Scale fully divided into minor divisions), and a Vernier Scale., 3. Vernier scale slides on the side of the main scale, and both of them are used to measure small, divisions up to 3 divisions like diagonal scales.
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VERNIER SCALE, 1. Least Count: It is the smallest distance that is, measured accurately by the vernies'Scale., 2. It is the difference between a main scale division, (m.s.d.) and a vernier scale division (v.s.d.).
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Problem 4, Construct a Vernier scale to read meters,, decimeters and centimeters and long enough to, measure up to 4 m. R.F. of the scale is 1/20. Mark, on your scale a distance of 2.28 m.
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Problem 4, 1. Least Count = Smallest distance to be measured =, 1 cm (given) = 0.01 m, 2. L = R.F. x Maximum distance to be measured =, (1/20) x 4 m = 20 cm, 3. Main Scale: Draw a line of 20 cm length., Complete the rectangle of 20 cm x 0.5 cm., Divide it into 4 equal parts each representing 1, meter., Sub-divide each part into 10 main scale divisions., Hence 1 m.s.d. = 1m/10 = 0.1 m = 1 dm.
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Problem 4, 4. Backward Vernier: Take 11 divisions on main, scale. Divide it into 10 equal parts on vernier, scale. So, 1 v.s.d. = 11 m.s.d./10= 11 x 1 dm/ 10= 0.11 m = 1.1, dm = 11 cm., Mark 0, 55, 110 towards left from 0 on the vernier, scale. The units of main divisions is METERS, subdivisions is DECIMETERS and vernier divisions is, CENTIMETERS
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Problem 4, 5. AB = (v.s.d x 8) + (m.s.d x 14), = (0.11m x 8) + (0.1m x 14) = (0.88+1.4)m
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2.28 m, 0.88 m, , 1.4 m, , CENTIMETERS, , 110, , A, , 55, , 0, , B, , 10, , 5, , 0, , 1, , DECIMETERS, AB = (v.s.d x 8)+(m.s.d x 14), = (0.11m x 8)+(0.1m x 14)=(0.88+1.4)m, , 2, METERS, , 3
