!ll!l!lllli i i I ll '![ nitjijiiiiii liiiipi li 353 L85 Itutiiuiiuriiuii '-\ ^ — -^. Class Gopyiight^I". COPYRIGHT DEPOSrc AN OUTLINE COURSE IN Mechanical Drawing WITH VARIOUS PLATES, DIAGRAMS AND KINDRED PRINTED MATTER. BY WARREN S. LOCKE, HEAD OF MECHANICAL DEPARTMENT OF THE RHODE ISLAND SCHOOL OF DESIGN. THIRD EDITION. PROVIDENCE, R. I. Copyright 1903. s ^ I wo GoDi»s Hece«»ec j MAY 25 1908 I 5 '5 3 9 '1 i COPY B. PRESS OF E. L. FREEMAN «: SON, PROVIDENCE, R. I, INDEX. INTRODUCTION , 5 PREE-HAND DRAWING 7 PLANE GEOMETRY 12 Problems in Plane Geometry 16 SOLID GEOMETRY ^6 Square Root , 31 Cube Root 31 ORTHOGRAPHIC PROJECTIONS 33 Revolutions 36 True Length of a Line 38 Problems in Projections 38 CONIC SECTIONS. = 40 INTERSECTIONS and DEVELOPMENTS 44 ISOMETRIC PROJECTIONS 50 LINEAR PERSPECTIVE 55 WORKING DRAWINGS 61 DESIGN 68 CAMS 70 Face Cams 71 Square Cams 75 Edge Cams 77 Leaders 79 GEARING 81 Single Curve Gearing 86 Rack to Mesh with Single Curve Gear 89 Single Curve Gears having less than 30 teeth ... 92 IV. INDEX. Bevel Gear Blanks 94 Epicycloidal Gears g8 Proportions of Gear Wheels 100 Teeth of Gear Wheels loi Strength of a Tooth 102 Comparison of Teeth 103 Gear Designing , 105 Sample Plate of Involute Teeth 108 TABLES AND DIAGRAMS. Strength of Materials 110 Factor of Safety . 113 Horse Power of Shafting 114 Strength of Belting 115 Decimal Equivalents 118 Handle Table 118 Proportions of Bolts and Nuts iig Alphabets 120 Areas of Circles 121 Circumferences of Circles 123 INTRODUCTION. This book is not a treatise on Mechanical Drawing, but an outline course on the subject. A mastery of its contents will not make the student an accomplished draughtsman, but it is hoped that such mastery will help him a long way toward that desired result. The book is chiefly for that class of students who, from age or condition, are not able to take the time necessary to go through a complete course of study with the view of making themselves mechanical draughtsmen. The difficulty of gauging the needs of this body of students makes the success of any such effort problematical ; nevertheless, this trial is made w^ith the hope of ultimate success. In attempting to meet the want referred to, the style of w^riting at- tempted, is forcible rather than elegant, direct rather than graceful. While much is lost by adopting this style, it is hoped that more is gained. *i FREE-HAND DRAWING. The student, like the child, must *' creep before he can walk," that is, do some preparatory work before he can stride along in his chosen path. In this pre- liminary work, the foundation, care should be taken. Study and work thoroughly, rather than fast. Make clear to your mind each subject as it is taken up, and by what is understood that which follows will be made easier. Before writing became common, the saying, " The pen is mightier than the sword," was accepted by the world. In these days, when time is very valuable, draw^ings are generally used in all kinds of business. Drawing is the universal language, and the ability to make good drawings is of more value, to-day, than any mere physical power. More or less elaborate drawings are used to illustrate form, fancy, or thought, wherever man communicates v^ith his fellow^. The art of making pictures has grow^n from rude outlines made on bark of trees with charcoal to inagnificent oil paintings in splen- did colors. It is not the object of this book to teach this art, but it will be well for the student of Mechanical Drawing to learn to make outline drawings, first, with- out instruments, other than pencil and paper. The ability to make these free-hand drawings is useful and valuable, always cultivating the eye and hand for more complicated work. Representations of simple forms s MECHANICAL DRAWING. should be attempted at first, as even these may prove difficult to the beginner. We must learn to know what .0U7' eyes see a?id then to represent that correctly. I. Figure i represents three "views," top, side, and end of a simple block. The student should begin ^with some such sim- ple form, and learn to see and represent it correctly. To do this, place the block on the level of the eye, in such a position that only four lines, bounding one plane of the block, can be seen. These simple drawings having been made, the block should be placeel lower and an attempt made to draw a '^ perspective'' of it. Fig. 2 is an accurate perspective of the block, three of w^hose faces are sliown in Fig. i. While the repre- sentation is correct, quickly recognized and simple, there iire some details that will repay study. First, it is stand- ing on a level surface because certain lines are vertical. Second, the lines that form the bottom and top do not, in themselves, appear as level lines, while we know^ they must be. Third, the angles made at the meeting of the lines are not right angles — the corners do not appear as being square, though we kno\\" they are so. If the stu- dent sees and understands these three points, and also the fact that the drawing — Fig. 2 — is correct, he has taken a good step in the right direction. It will not be arrived at in an hour, and many times a day the attention .MECHANICAL DRAWING. 9 should be fixed on a form of definite outlines and an attempt made to see and represent it. This training should continue until the student can draw, with fair speed and accuracy, such simple inanimate objects as he handles in his business, It will often happen that the ability to do this will save time and money to employer or employed, or both. The draughtsman is lame with- out it. The fundamental principles of perspective being thus understood, the student may continue to practice free-hand drawing both to his pleasure and profit, but for the purpose of accomplishing the work laid out in this Course, we may leave free-hand drawing for the next step. This may well be Geometrical Drawing, for the science which treats of Position and Extension must help us to correct such inaccuracies as we have now noticed in work where the eye is the only judge of correctness. INSTRUMENTS. Before we take up Geometrical Drawing we should know something of instruments and their use. The skill of hand already acquired will be of great use in handling instruments, and the accuracy of eyesight, necessary in making measurements. Accii7'ate vieasiire- ments are absolutely necessary to good Geometrical and Mechanical Drawing. A complete outfit for Mechanical Drawing may be bought for $15.00, and half this sum will buy instruments sufficient for most plain work. The following is a satisfactory outfit : One drawling board, one T-square, two triangles, one scale, two rub- bers, pencils, and case of instruments. The following are good sizes : Board, twenty-four by thirty-two inches, T-square with thirty-inch blade, a six-inch rub- ber 45^ triangle, a ten-inch rubber 30-and-6o'^ triangle, lO MECHANICAL DRAWING. a twelve-inch steel scale, pencil and ink eraser, one each, four thumb tacks, four and six H. Faber's drawing pen- cils, one case instruments, containing one iive-and-one- half-inch compasses, with j^en, pencil and needle points, one five-inch spring dividers, two right line pens, three- inch spring dividers, three- inch spring bow pencil and three-inch spring bow pen. This may come well within $15.00, and be composed of goods that will w^ear for many years. In using the T-square it should be placed on the board w^ith its head held against the left hand side. All hori- zontal lines should be drawn with the pencil held against the upper side of the blade of the square as it is thus held. Perpendicular lines should be drawn with the combined help of T-square and triangle. With the 45° and the 30-and-6o° triangles the following angles, w^ith a horizontal, may be obtained: 15°, 30*^, 45^, 60^, 75 and 90*^. The student will profit by a facility in placing jiis triangles to make these angles. In the Geometry which follows, a method is given of dividing an angle in halves, so that the student should be able to make any desired angle. The most convenient scale is a steel scale, twelve inches long, graduated to sixteenths on one edge and thirty-seconds on the other. It is made by Brown & Sharpe, and sold everywhere for $1.25. The compasses are for making circles and arcs of cir- cles. In drawing a circle never take hold of a point of the dividers, but always hold the compasses by the head •or handle. The joints in the stocks are used to bend the points to nearly right angles with the paper when draw- ing large circles. In using the right-line pen always hold it in a plane at right angles to the paper. Always draw, never push the pen. This last is a good rule also in using the pencil, Avhich should be sharpened to a chisel edge rather than to a point. This sort of '' point'' MECHANICAL DRAWING. I I is also best for the pencil used in the compasses. It will wear longer and make a better line. Much care should be taken in all these operations, for these are details of the art of drawing, which is composed of details. Ten times more time is lost erasing incorrect lines than is lost in careful decision as to whether or not they will be right. In drawing, it is very clearly seen that " haste makes waste.'* The first 7'eqinsite of a drawing, is ACCURACY, the second, neatness. After these, in a sheet of drawings, comes, third, arrangement. With careful practice one may learn to draw fast, but no one can make good, reliable drawings in a hurry. PLANE GEOMETRY, DEFINITIONS. Geometry is the Science of Position and Exiensio7i. A Point has merely position, witliout extension. Exteiision has three dimensions : Lengthy Breadth and Thickness, A Line has only one dimension, namely, length. All lines are either straight or curved. Note. — In this book the word "line" is understood to ni'^^an a straight line, unless otherwise specified. A Surface has t\vo dimensions : Length and breadth. A Solid has the three dimensions of extension : length, breadth and thickness. The Position of a Point is determined by its Direction and Distance from any known point. A Plane is a surface in which any two points being taken, the straight line joining those points lies wholr/ in that surface. • Axiom. A straight line is the shortest distance be- tween two points. THE ANGLE. An Angle is formed by t\vo lines meeting or crossing each other. MECHANICAL DRAWING. 1 3 The Vertex of the angle is the point where its sides meet. The magnitude of the angle depends solely upon the difference of direction of its sides at the vertex or amount they are spread apart. The magnitude of the angle does not depend upon the length of its sides. When one straight line meets or crosses another, so as to make the two adjacent angles equal, each of these angles is called a Right angle, and the lines are said to be perpendicular to each other. Thus the angles ABC and ABD (Fig. 3) being equal, are right angles. An Acute angle is one less than a right angle, as K (Fig- 3)- An Obtuse angle is one greater than a right angle, as ^(Fig.3)- Parallel Lines are straight lines which have the same Direction, as AB, CD (Fig. 3). Parallel lines cannot meet, however far they are pro- duced. I POLYGONS. I A plane figure is a plane terminated on all sides by lines. If the lines are straight, the space which they contain is called a rectilineal figure, or polygon (^P Fig. 3). The polygon of three sides is the simplest of these figures, and is called a triangle ; that of four sides is Z3\\cd 3. quadrilateral ; that of five sides, a / thus found would be at equal dis- tances from A and B. c. The problem is impossible, when the distance between the known points is greater than the sum of the given distances or less than their difllerence. MECHANICAL DRAWING. 1/ d. If the required point is to be at equal distances from the known points, its distance from either of them must be greater than half the distance between the known points. 2. Problem. To divide a given straight line AB into two equal parts ; that is, to bisect it. Solution. Find by § ^, a point C, above the line, at equal distances from the extremities A and B. Find also another point D^ below the line, at equal distances from A and B. Through C and D draw the line CD^ which bisects AB at the point E. 3. Problem. At a given point A^ in the line BC^ to erect a perpendicular to this line. Solution. Take the points B and C at equal distances from A J and find a point B> equally distant from B and C. Join AD and it is the perpendicular required. 4. Problem. From a given point A, above the straight line BC, to let fall a perpendicular upon this line. Solution. From y^ as a centre, with a radius suffi- ciently great, describe un arc cutting the line BC in two points B and C ; nnd a point D below BC^ equally dis- tant from B and C, and the line AD is the perpendicular required. 5. Proble?n. To draw a perpendicular to a line at one end. Solution. Let AB be a horizontal line. With A as centre and radius AB draw an arc. With same radius and centre B draw arc, cutting first one at D. With D as centre and same radius draw arc over A. Draw line through B and D, meeting the last arc at E. Line EA will be perpendicular to line AB, ♦2 1 8 MECHANICAL DRAWING. 6. Prohlevi. To make an arc equal to a given arc AB^ the centre which is at the given point C. Solution Draw the chord AB, From any point JD as a centre, with a radius equal to the given radius CA^ describe the indefinite arc FH. From 7^ as a centre, with a radius equal to the chord AB^ describe an arc cutting the arc FH in H^ and we have the arc FH = arc AB. 7. Proble7n. At a given point A^ in the line AB^ to make an angle equal to a given angle K. Solution. From the vertex K^ as a centre, with any radius, describe an arc IL meeting the sides of the angle ; and from the point ^ as a centre, by the preceding prob- lem, make an arc BC equal to IL. Draw AC^ and we have angle BA C = angle A'. 8. Problem. To bisect a given arc AB. Solution. Find a point D at equal distances from A and B. Through the point D and the centre C draw the line CD.^ which bisects the arc AB at F. 9. Problem. To bisect a given angle A. Solution. From A as a centre, with any radius, de- scribe an arc BC, and by the preceding problem, draw the line AF to bisect the arc BC, and it also bisects the angle A, 10. Proble/n. Through a given point A, above a given straight line BFCy to draw a straight line parallel to the line BFC. Solution. Join FA, and, by Problem 7, draw AF>, making the angle FAD = AFC, and AF> is parallel to BFC, MECHANICAL DRAWING. I 9 11. Proble7?i, Two sides of a triangle and their in- cluded angle being given, to construct the triangle. Solution. Make the angle A equal to the given angle, take AB and AC equal to the given sides, join BC^ and ABC is the triangle required. 12. Problem. The three sides of a triangle being given, to construct the triangle. Solution. Draw AB equal to one of the given sides, and, by § I, find the point C at the given distances AC and BC irovci the point C, join AC and BC^ and ABC is the triangle required. Note. — The problem is impossible, when one of the given sides is greater than the sum of the other two. 13. Problem. The adjacent sides of a parallelogram and their included angle being given, to construct the parallelogram. Solution. Make the angle A equal to the given angle, take AB and AC equal to the given sides, find the point j9, by § I, at a distance from B equal to AC^ and at a distance from C equal to AB. Join BD and DC, and ABCD is the parallelogram required. Note. — If the given angle is a right angle, the figure is a rectangle; and, if the adjacent sides are also equal, the figure is a square. 14. Problem. To find the centre of a given circle or of a given arc. Solution. Take at pleasure three points, A., B^ C, on the given circumference or arc ; join the chords AB and BC, and bisect them by the perpendiculars Z^^and PG; the point O in which these perpendiculars meet is the centre required. 20 MECHANICAL DRAWING. n 15. Problem. Find by the same construction as i Problem 14 a circle, the circumference of which passes through three given points not in the same straight line. 16. Fi^oblein. Through a gi^'en point to draw^ a tangent to a given circle. Solution. If the given point A is in the circumference, draw the radius CA^ and through A draw AZ> perpendic- ular to CA, and AZ> is the tangent required. 17. Prohleju. Through a given point to draw tan- gents to a given circle. Solution. If the given point A is without the circle, join it to the centre by the line AC j upon ^Cas a diam- eter describe a circumference cutting the given circum- ference in M and N ; join AM and AN^ and they are the tangents required. 18. Problem. To inscribe a circle in a given triangle ABC. Solution. Bisect the angles A and B by the lines AO and BO^ and their point of intersection O is the centre of the required circle, and a ^perpendicular let fall from O upon either side is its radius. Note. — The three lines AO, BO and CO, which bisect the three angles of a triangle, meet at the same point. 19. Problem. To divide a given straight line AB into any number of equal parts. Solution. Suppose the number of parts, for example. Is six. Draw the line AO.^ making an acute angle w^ith line AB ; take ^Cof any convenient length and apply it six times to AO. Join B and the last point of divis- ion, D., by the line BD. Through the last point of division but one draw a line (see Problem 10) parallel to BD. This line will cut AB.^ at E^ one-sixth of the distance from B toward A. Apply EB six times to AB. MECHANICAL DRAWING. 21 20. Problein, To find a mean proportional between two given lines. Solution. Draw^ the straight line ACB, making AC equal to one of the given lines, and BC equal to the other. Upon ACB as a diameter describe the semicircle ADB. At C erect the perpendicular CD, and CD is the required mean proportional. 21. Problem. To divide a given straight line ACB at the point C in extreme and mean ratio ^ that isj so that w^e may have the proportion : AB : AC ^ AC : CB. Solutio7i. At end B erect the perpendicular BD equal to half of ACB, Join AD, take DE from D on AD equal to BD, and AC equal to AP, and C is the required point of division. REGULAR POLYGONS. Definitio7is. A regular polygon is one which is at the same time equiangular and equilateral. Hence the equilateral triangle is the regular polygon of three sides, and the square the one of four. An equilateral polygon is one which has all its sides equal ; an equiangiclar polygon is one which has all its angles equal, 22. Problem. To inscribe a square in a given circle. Solution. Draw two diameters, AB and CD, perpen- dicular to each other ; join AD, DB, BC, CA ; and ADBC is the required square. 23. Problem. To inscribe a regular hexagon in a given circle. Solution. Take the side BC of the hexagon equal to the radius AC oi the circle, and, by applying it six 22 MECHANICAL DRAWING. times round the circumference, the required hexagon BCDEFG is obtained. 24. Probleni. To describe a regular decagon in any circle. Solution. Divide the radius of a (three-inch) circle in extreme and mean ratio. (Problem 21.) The longer part is equal to one side of the regular decagon required. Apply it ten times to the circum- ference, and join the points by straight lines, making the decao^on. Make a pentagon by joining the alternate vertices of the decagon. 25. Problem. To circumscribe a circle about a given regular polygon ABCD^ &c. Solution. Find, by Problem 14, the circumference of a circle w^hich passes through three vertices, A^ B, C ; and this circle is circumscribed about the given polygon. 26. Problem. To inscribe a circle in a given regular polygon ABCD.^ &c. Solution. Bisect tw^o sides of the polygon by perpen- diculars, the point of intersection is the centre of the required circle. The sides of the polygon become tangents to the circle. 27. Problem. To inscribe a pentagon in a circle. Solution. Draw" a diameter AB and a radius CD per- pendicular to it. Bisect BC at E. With centre at E and radius ED draw arc DF^ — 7^ being on AC, With D as a centre, and DE as a radius, draw arc EGj meet- ing the circumference at G. Draw line DG. It is one side of the required pentagon. 28. Problem. To construct a regular polygon of any number of sides in a circle (Approximate inethod) . MECHANICAL DRAWING. 23 Solution. Draw a diameter AB and divide it into as many equal parts as there are sides in the required poly- gon — say eight. With A and B as centres, and radius AB draw arcs intersecting in C. Draw line from C through the second point of division oi AB to meet the circumference at D. AD is one side of the required polygon. AREAS. Definitions. Equivalent figures are those w^hich have the same surface. The area of a figure is the measure of its surface. The ////// of surface is the square whose side is a linear unit ; such as a square inch or a square foot. The area of a square is the square of one of its sides. A parallelogram is equivalent to a rectangle of the same base and altitude. The area of a parallelogram is the product of its base by its altitude. Parallelograms of the same base are to each other as their altitudes ; and those of the same altitude are to each other as their bases. All triangles of the same base and altitude are equiva- lent. The area of a triangle is half the product of its base by its altitude. Every triangle is half of a parallelogram of the same base and altitude. The circumference of a circle is equal to 3. 141 6 multi- plied by its diameter^ or r. D. The area of a circle is equal to 3. 141 6 multiplied by the square of its radius., or iz R^. The area of a trapezoid is half the product of its alti- tude by the sum of its parallel sides. 24 MECHANICAL DRAWING. 29. Problem. To construct a square equal to three times a given square. Solution. Extend one side of the square until side and extension together, equal a diagonal of the square. From extremity of extension, to farthest corner of square, draw a line. This line is equal to a side of the required square. 30. Problem. To make a square equivalent to the sum of two given squares. Solution. Construct a right angle C ; take CA equal to a side of one of the given squares ; take CB equal to a side of the other ; join AB, and AB is a side of the square sought. A square may be found equivalent to a given triangle ^ by taking for its side a mean proportional betw^een the base and half the altitude of the triang^le. A square may be found equivalent to a given circle^ by taking for its side a mean proportional between the radius and half the circumference of the circle. PRACTICAL PROBLEMS IN PLANE GEOMETRY. Note. — Divide the sheet into four equal parts and put a problem in each. Make an angle of 75-' and bisect it. Draw six -inch parallel Wne^^ four inches apart. Divide a seven-inch line into ;////;/^. The <:7jc/> of the cone is the line drawn from the vertex to the centre of the basis. Fig. L. A right cofie is one, the axis of which is perpendicular to the base. The right cone may be considered as generated by the revolution of a right triangle about one leg as an axis ; the other leg generates the base, and the hypothenuse generates the convex surface. The area of the convex sicrface of a right cone is half the product of the circumference of the base, by the side. We may reduce this rule to a formula, if we let the surface be represented by S^ diameter of the base by D and the side by H\ as follows : ^'==^71 DII\ As before, it must be remembered that the quantities w^ritten together are multiplied together. The volume or solidity of a cone is one-third of the product of its base by its altitude or height. Let H represent the height. The area of the base is 3. 141 6 x R^ ; hence the volume of a cone is K = y^ r. R^ H. The volume of any pyramid is one-third of the product of its base by its height. Pyramids or cones of equivalent bases and equal alti- tudes are equivalent. Any pyramid or cone is a third part of a prism or cjdinder of the same base and altitude. THE SPHERE. DEFINITIONS. A Sphere is a solid terminated by a curved surface, all the points of which are equally distant from a point within, called the centre. *3 30 MECHANICAL DRAWING. The sphere may be conceived to be gen- erated by the revolution of a semicircle about its diameter. The radius of a sphere is a straight line Pig. E. dravsm from the centre to a point in the sur- face ; the diameter or axis is a line passing through the centre, and terminated each w^ay by the surface. All the radii of a sphere are equal ; and all its diam- eters are also equal, and double the radius. Every section of a sphere made by a plane is a circle. The section made by a plane which passes through the centre of the sphere is called a great circle, Any other section is called a s??iall circle. The radius of a great circle is the same as that of the sphere, and therefore all the great circles of a sphere are equal to each other. The centre of a small circle and that of the sphere are in the same straight line perpendicular to the plane of the small circle. The area of the surface of a sphere is the product of its diameter by the circumference of a great circle. The surface of a sphere is equivalent to four great circles. The surfaces of spheres are to each other as the squares of their radii. From the foregoing w^e may deduce the following formulas : Surface of a sphere ■=l S ■=i 4 r. R^ :=i r. D'^^ for since the area of one circle is r. R^^ the surface of a sphere is ^ X T. R = 4- R\ The solidity or volume of a sphere is one-third of the product of its surface by its radius. The surface = S -=1 4 r. R multiplied by i/^ i? is F =r {411 R) X i^ ^ = I r i?', or Volume of sphere — V — Ye ^ D\ MECHANICAL DRAWING. 31 For the convenience of the student, rules for extract- ing the Square and Cube Roots are here inserted. SQUARE AND CUBE ROOT. RULE. To extract the square root of any number : 1. Beginning with the units figure, point off the ex- pression into periods of two figures each. 2. Find the greatest square in the number expressed by the left hand period, and write its square root as the first figure of the root. 3. Subtract this square from the part of the number used, and to the remaindei unite the next two terms of the given number for a new dividend. 4. Double the part of the root already found for a trial divisor ; and by it divide the new dividend — less the last figure — and write the quotient as the next figure of the root. Also, write it at the right of the trial divisor, the combined figures making the true divisor. 5. Multiply the true divisor by the last figure of the root and subtract the product from the new dividend. 6. If there are any more terms of the root to be found, unite with the remainder the next two terms of the given number, and take for a trial divisor, double the root already found, and proceed as before. RULE. To extract the cube root of a number ; I. Beginning with the units figure, point off the ex- pression into periods of three figures each. 32 MECHANICAL DRAWING. 2. Find the greatest cube in the number expressed by the left hand period, and write its cube root as the first fio^ure of the root. 3. Subtract the cube from the part of the number used, and with the remainder unite the next three figures of the given number for a new dividend. 4. Take three times the square of the part of the root already found, with two ciphers annexed, for a trial divisor, and by this divide the new dividend, and write the quotient as the next term of the root. 5. To the trial divisor add three times the first term of the root^ with a cipher annexed, multiplied by the last term, also the square of the last term of the root. 6. Multiply this sum by the last term of the root^ and subtract the product from the new dividend. 7. If there are more terms of the root to be found, unite with the remainder the next three fiofures of the given number, take for a trial divisor three times the square of the part of the root now found, and proceed as before. ORTHOGRAPHIC PROJECTIONS. All Mechanical Drawing Is founded on Mathematics — principally on Plane, Solid and Descriptive Geometry. We have now some knowledge of Plane and Solid Geometry. Descriptive Geometry is that branch of Mathematics w^hlch has for its object the explanation of the methods of representing, by drawings, all geometrical magni- tudes ; also, the solution of problems relating to these magnitudes in space. Drawings are so made as to present to the eye, situ- ated at a particular point, the same appearance as the magnitude or object itself, were it placed in the proper position. The representations thus made are the pro- jections of the object. The planes upon which these projections are usually made are \h^ planes of projection. The point at which the eye is situated is the point of sight. Definition. When the point of sight is in a perpen- dicular, drawn to the plane of projection, through any point of the drawing, and at an Infinite distance from this plane, the projections are Orthographic. This result Is reached, physically, if we suppose the eye to be as large as the object and placed in the perpen- dicular referred to, and at any convenient distance. Definition. When the point of sight is within a finite distance of the drawing, the projections are Scenographic^ commonly called the Perspective. 34 MECHANICAL DRAWING. The Student should gain a sound knowledge of Ortho- graphic projection before attempting Perspective, hence our attention will be directed to the former for the present. In Orthographic project- ions, three planes of pro- jections (sometimes two suffice) are used, at right angles to each other, one horizontal and the other two vertical, called respect- ively the horizontal and ver- tical planes of projection, and denoted by the letters H and V, Let a rectangular cross (Fig. 5) be imagined self- suspended near an upper corner of a room, or between three sheets of paper, placed in a similar position, name- ly, at right angles to each other ; the three principal dimensions, length, breadth and thickness, of the cross being each perpendicular to one of the sheets of paper — which serve as the three planes of projection. As indi- cated by the dotted lines, let perpendiculars be drawn from the principal points of the cross to each plane of projection. Let the two vertical sheets be now laid down on 2 table, keeping the top of the cross in line on both. Now (Fig. 4) w^e have the three projections of the cross or one plane in the manner in which it is proper to repre- sent them as Orthographic projections. It is easily seen that neither of these projectiofis is a correct representation of the cross as we see it, and, also, that collectively the three projectiojis truly represent the cross in length, breadth and thickness. Here, then, is the value of this method of representing objects. MECHANICAL DRAWING. 35 All that these projections need to make them working drawings^ are the dimensions in figures. The projection on H is called the Flan^ and the two on V and P^ are called Elevations, H A V P The representation (in Fig. 5) is in Perspective. Fig- ures I and 2 represent the whole principle in the same manner. GROUND LINE. It is to be observed that the line w'hich (in Fig. 5) is made by the intersection of H and V^ is preserved in Fig. 4. It is called the Ground Line. The representa- tion of the object above the Ground Line is called an Elevation, and the one below is called the Plan, of the object. Returning now to Fig. 5, it will be seen that the Plan is drawn upon the plane you look up upon, and the elevations upon planes you look upon horizontally. After a little experience, the Ground Line becomes as 36 MECHANICAL DRAWING. imaginary as the Equator, but like the latter serves its purpose. 'At^mWQ)"0 Figure 6 represents in Plan and Elevation a triangular prism ; Fig. 7, a rectangular prism ; Fig. 8, a square pyramid ; Fig. 9, a hexagonal pyramid ; Fig. 10, a right cylinder, and Fig. 11, a cone. These names, objects and representations should be kept in mind, for they w^ill be referred to many times. This subject of Orthographic projections is the most important of all subjects to the mechanical draughts- man. He uses it a thousand times to one of any other method of representation, and should be proportionally well acquainted w^Ith it. In all that follows in this book this acquaintance will be cultivated, until, it is hoped, any line in space may be comprehended and drawn. The subjects which immediately follow, are especially chosen to develop facility in making Orthographic pro- jections, as well as to gain accurate knowledge of various solids and their combinations. 1 1 1 z 3- <\ < 1/ A 3 > 6 A L REVOLUTIONS So far we have confined ourselves to projections of objects placed at right angles to the planes of projection, MECHANICAL DRAWING. 37 but It will be easily understood that in making drawings of machines or houses, we shall find many lines which are not so related to natural planes of projection already described. For an example, let us take the rectangular prism (Fig. 7) just used. To assist us in the right way, we figure the corners of the front side (Fig. 12). In the plan the figures double, but make no confusion so long as we have the elevation to look at. Tip the prism, now, so that the base line 3, 4 will make an angle of 30^ with the ground line, keeping the plane of the face 1,2, 3, 4, parallel to the ground line. To make a second plan of the prism in this second position : As the different representations or views of an object are supposed always to be in positions perpen- dicular to each other, the corner i, for example, will be found in a perpendicular to ground line. As the prism was not inclined to the vertical plane, the desired corner will be found in a line through i of the plan, parallel to the ground line. The perpendicular from i in the second elevation, and the parallel from i in the first plan meet, making i of the second plan. In the same way seven other corners are found and. the new plan finished. We have now a plan and elevation of the prism as it is inclined to the GL. Move this second plan to the right and Incline it to the GL at an angle of 45^. Now from the third plan draw a perpendicular, and from the second elevation draw a parallel, from corner i. The point in which these two lines meet is corner 1 in the third elevation. One by one, seven other points may be found, completing the elevation of the prism as It ap- pears inclined to both planes of projection. This work brings us to a point where we may attempt the problem : " To find the true length of a line." 3S MECHANICAL DRAWING. TRUE LENGTH OF A LINE. RULE. Revolve one of the projections of the line until it is parallel to the Ground Line. Construct the other pro- jection of the line to agree with the second position. This construction will be the true length of the line. EXPLANATION. Let ///, //, represent the horizontal, and ;;/' ;/' the vertical, projections of a line. To proceed according to the rule : Revolve ;;/, ;/, about ;;/ until ;/ is at ;/" and ;//, //" is parallel to G. L. In this revolution it is supposed that the angle of the line with the hori- p^^. p zontal plane is not changed. If this be so, ;/' has not changed its height above the horizontal plane, and its new position is to be found at the intersection of a parallel from ;/' and a perpendicular from ;/", or at N. ;;/' not having moved, the true length of the line must be ;;/' N. For practice, take the two projections of the line i, 4, in the third plan and elevation of the prism just drawn in " revolutions." Revolve the elevation of the line and construct the plan. The accuracy of the work may be tested by the first elevation of the line. PROBLEMS. 1. Find the true length of a line whose elevation ap- pears as a line 4' long, inclined 45° to the Ground Line, and whose plan has an angle of 30° to the G. L. 2. Draw two views of a rectangular prism 3 ' X 2' x i and give the true length of its diagonal. MECHANICAL DRAWING. 39 3. Draw two projections of a hexagonal pyramid, having a base inscribed in a circle whose diameter is three inches, and whose altitude is five inches. Give the true length of the centre line of one side. 4. Draw a hexagonal pyramid exactly like that of Problem 3, and give the true length of the line which joins the middle point of the middle line of a side, to the centre of the base. CONIC SECTIONS. The Conic Sections are so called because they are sec- tions of a cone. We have had a definition of a plane. Imagine two such surfaces passed through a solid, at a distance from each other of less than the thousandth part of an inch. The slice of the solid between the planes is termed a Section. It is also called a la77iina^ or a slice section. Also, we often use the term Section when but one plane is passed. The Conic Sections are taken from a right cone and are, the Triangle, Circle, Ellipse, Parabola and Hyper- bola. The Triangle is a section cut from a cone by two planes passed through the apex cutting the base. A Circle is a section of a right cone cut at right angles to the axis. The Ellipse is a curved section cut at any angle to the axis, large enough to cut both sides of the cone. The Hyperbola is a curved section cut from the cone parallel to the axis and perpendicular to the base. The Parabola is a curved section cut from a rio^ht cone parallel to one of the sides, as it appears in elevation. Of these Sections, the triangle and circle are in con- stant use, in drawing ; the ellipse in frequent use ; the parabola occasionally, and the hyperbola but rarely. MECHANICAL DRAWING. 41 Fig. G. To learn, practically, what these curves are and how to get them from a cone, proceed as follows : Make a plan and elevation of a cone, having a base four and a half inches in diameter and six and one- half inches high. In the elevation draw the elevations of the Sections, as represented. The plans of but two of the Sections are given, for more would make confusion. It will be easily understood that J^ G is the base of the hyperbola, and I? E the base of the triangle. These are in their true lengths. The altitudes of all the Sections are given in the ele- vation of the cone ; that of the hyper- bola, for example, is H L. If, now, we erect a perpendicular, equal to H Z, to the middle of a base line equal to F G^ the points corresponding to F^ L and G will be three important points in the hyperbola. If, now, we want more points, upon which to construct the curve, w^e proceed as follows : Pass a plane through the cone parallel to the base, cutting the hyperbola. It will cut a circle from the cone, as N^ O. This circle is on the convex surface of the cone. As the hyperbola is also upon its surface they must intersect at A. Drawing, now, the plan of circle N O from centre X, we find that the width of the circle, at A^ is B C. But the two curves intersect at A. Hence B C is also the width of the hyperbola at a distance, H A^ from the base. Set off FT A on the axis of the hyperbola, from the base, and draw^ a line parallel to the base. On this line set offj^ S, each side, from the axis. These points will be points in the hyperbola. More *4 43 MECHANICAL DRAWING. planes must be passed, more circles drawn, and more points obtained to be accurate, especially near the top of the curve. The parabola and ellipse are obtained in the same v^^ay. In the parabola care must be taken to set off R P on the axis and the perpendicular from j^, that is, Y Z as the width, of the parabola at height R P. X E D \^ the plan of the triangle. It is to be drawn '' full size," but no directions are given, as the student is expected to work out for himself the major part of these Sections. A knowledge gained by wo7'k is retained and used, when frequently careful instruction is forgotten. As the ellipse occurs frequently in drawing and requires expensive instruments for its delineation, the following approximate method, by arcs of circles, is given : Draw major axis A B, and half minor axis CD. Complete the rect- angle A B X a. Draw A D. Draw a b, per- pendicular to^ D, Ex- tend C n to e b. With radius CD draw arc D f. With diameter A f draw semicircle A e /, Draw radius g h. Lav off h i on C b \.o j. Pig. H. Through / draw an arc with b as centre. From A as centre and radius C e draw an arc, cutting arc through j at k. Through k draw b /, and through ;;/ draw k o ; in is a centre for A o^ k centre for o /, and b^ for ID. A o I D \^ one-quarter of the ellipse. The parabola may be described as follows : Suppose the parabola to have a base C D and an alti- tude A B. Extend A B to E, making B E equal to B A, n. y^^^^i — ,^\ X y C / }\/^"^^ $ I \ 1 y^ 7 1 /i> MECHANICAL DRAWING. 43 Draw E C and ED. Divide C E and E D into any number of equal parts, numbering one from the top and the other from the bottom. Join i , i , — 2, 2, — 3, 3, &c. These lines will all be tangents to the parabola. With a French curve the para- bola may be drawn. In drawing an irregular curve with a French curve in this way, be sure that the instrument touches three of the points through which the curve is to be drawn. Intersections and Developments. As the memory will easily recall, there are many lines •seen on a manufactured article, or on a drawing of it, that are not lines of any individual part of the article, hut lines that occur where two or more forms intersect or join each other. Such lines are called intersection lines. H Intersections is the name given to that part of geometrical •drawino: tliat treats of the in- tersection lines and their cor- rect delineation. It will be seen at ^nce that a thorough knowledge of geo- metrical solids will be neces- sary to the student wdio desires to take a full course in inter- isections. On the other hand, Ave all have a fair understanding of many geometrical forms that meet our eyes in com- bination every day, and with these we will make a beginning. Let A B, C D^ represent two projections of a hexagonal prism, and ^,3, 6, H, i, 4, two projections of a hex- agonal pyramid, passing through, or ifitersecting the prism. To learn what the intersection line will be, proceed as follows : D ^h N ^^^^^^ f — ^ i.-. Fig. J. MECHANICAL DRAWING. 45 First, number the angles of the base of the pyramid so that the position of a line in both projections may be easily noted. The line \-E intersects the prism at Z, whose elevation is at K^ on line i H. The line 6—E intersects the prism at J/, whose elevation is at N j hence the in- tersection of the face Z^,-i,-6, of the pyramid, with the prism, is the line K N. ^-E is directly under d-E^ so its elevation w^ill be, at O. By similar reasoning, the intersecting point on /\—H is at jP, and K-N-O-P is the intersection line. It will easily be seen that the similar line on the other side of the prism is constructed in the same way. It will be good practice for the student to copy the plan of the combination, with the line \E at an angle of 15^ with the Ground Line, and then construct the elevation. We will suppose a triangular prism passed through a right cylinder (Fig. 13.) First make a plan and tw^o elevations of the cylinder. In the middle of the elevation which is projected from the plan, draw the end elevation of the prism. Now draw the plan, assuming that the prism projects from the cylinder at either side, and that centre line of prism and cylinder coincide in D, Ey C. 46 MECHANICAL DRAWING. In the side elevation it is evident that there will be an intersection line ; that none appears in the plan and end elevation is evident because in the plan it coincides with the outline of the cvlinder, and in the end elevation it coincides with the outline ot the prism. We can lay out the top and bottom lines and ends, on the side elevation, from the other views. The point where the lower line of the prism pierces the cylinder is found as follows : In the plan draw the line Z, iV, perpendicular to the diameter Z, O^ through the point J/, where the prism pierces the cylinder. Lay off the distance Z, O from X to K, the latter being the desired j^oint. The top point of the intersection line is on the cir- cumference. To find other points in the intersection line, pass the planes i and 3 and proceed in the same Avay as in the case of the bottom line. It is now required to develop the surface of the cylin- der. In Mechanical Drawing this means to draw an equivalent plane figure. This may be illustrated by fitting the surface of the cylinder with a covering of paper. When this paper is unrolled and spread on a table, we have a surface equivalent to that of the cylin- der. It is no^v required to outline, in this development of the cylinder, the hole that the prism makes. Let A^ H (Fig. 13) represent the development of the cylinder. Let a perpendicular at D^ represent the axis of the prism. Lay off^ the arc Z, J/, developed as a straight line on each side of the axis, making the line R^ P . The third corner is found on the axis at the alti- tude of the prism from R^ P. Intermediate points are laid off from the axis on traces of the planes i and 2 in the same manner. MECHANICAL DRAWING. 47 We have had the intersection of plane surfaces with plane surfaces, plane surfaces with curved surfaces, and now we have curved with curved surfaces. r r k "^ 1 Fig. 13 A illustrates the intersection of two cylinders. The method of finding the intersection line is as fol- lows : In the plan pass the plane i ' through the cylin- ders parallel to both axes. It will cut a rectangle from each. \ or number of teeth in a gear divided by the diametrical pitch equals diameter of the pitch circle. The diameter of the pitch circle of a wheel hav- ing 60 teeth, 12 P, would be, consequently, five inches. 86 MECHANICAL DRAWING. (5) ~^ = ^y or, add 2 to the number of teeth in a wheel and divide the sum by the diametrical pitch, and the quotient will be the luhole diamete}^ of the gear or the diameter of the addendum circle. The diameter of gear blank for a gear of sixty teeth, 12 F^ would be, conse- quently, 5 yv inches. (6) D = -P^ or number of teeth divided by diameter of pitch circle, in inches, gives the diametrical pitch. (7) ^^ = P-, or add 2 to the number of teeth, divide by whole diameter and quotient will be diametrical pitch. PD' = ^V, or pitch circle diameter multiplied by dia- metrical pitch equals number of teeth in the gear. (8) Formula (i) may be transposed, -^^ =P. SINGLE CURVE GEARS. Single curve teeth are so called because their working surfaces have but one carve, which forms both face and flank of tooth sides. This curve is, approximately, an involute. In a gear of 30 teeth or more, this cui*ve may be the single arc of a circle, whose radius is one-fourth the radius of the pitch circle. A fillet is added at tlie bottom of the tooth, to make it stronger, equal in radius to one-sixth the widest part of tooth space. A cutter made to leave this fillet has the advantage of Wearing longer than it would if brought up to a corner. In gears«having less than thirty teeth this fillet is made the same as just given, and the sides of teeth formed with more than one arc, as will be shown farther on. Having calculated the data of a gear of 30 teeth, |" circular pitch we proceed as follows : 1. Draw pitch circle and point it off into parts equal to one-half circular pitch. 2. From one of these points, as at B, (see plate MECHANICAL DRAWING. 87 Single Curve Gear) draw radius to pitch circle, and upon this radius describe a semicircle ; the diameter of this semicircle being equal to radius of pitch circle. Draw addendum, working deptli and whole depth cir- cles. 3. From the point B^ where semicircle, pitch circle, and outer end of radius to pitch circle meet, lay off a inis. 6 -.4^9|!yDc/n, GEAR, 30 TEETH, J"CIRCULAR PITCH, . P'= = f"or .75" N= =30 P = 4.1888' t = .375" 8 = .2387' D" - .4775' Sf/ = .2762"' D"+/ - .5150" D' = 7.1610' D = 7.7384' SINGLE CURVE GEAR. 88 MECHANICAL DRAWING. distance on semicircle equal to one-fourth of the radius of pitch circle, shown in the figure at B A, This is laid off as a chord. 4. Through this new point A, upon the semicircle, draw a circle concentric to pitch circle. This last is called the base circle^ and is the one for centres of tooth arcs. In a certain system of single curve gears, the diameter of this circle is .968 of the diameter of pitch circle. 5. With dividers set to one-quarter of the radius of pitch circle, draw arcs forming sides of teeth, placing one point of the dividers in the base circle and with the other point describing an arc through a point in the pitch circle that was made in laying off the parts equal to one-half the circular pitch. Thus with A as centre, an arc is drawn through B. 6. With dividers set to one-sixth of the widest part of tooth space, draw the fillets for strengthening teeth at the roots. These fillet arcs should just join the whole depth .circle to the sides of teeth already described. Single curve or involute gears are the only gears that will run at varying distances of axes, and transmit un- varying velocity. This peculiarity makes involute gears especially valuable for driving rolls or any rotating pieces, the distance between whose axes is likely to be changed. The assertion that gears crowd harder on bearings when of involute than w^hen of other forms of teeth, has not been proved in actual practice. It is an excellent practice to cut out the drawings of a pair of gears, that have been made on Bristol-board, and test their accuracy in running together. MECHANICAL DRAWING. 89 RACK TO MESH WITH SINGLE CURVE GEARS HAVING 30 TEETH AND OVER. This gear is made precisely the same as the one last described. It makes no difference in which direction the construction radius is drawn, so far as obtaining form of teeth and making gear is concerned. Here the radius is drawn perpendicularly to pitch line of rack and through one of the tooth sides, B. A semicircle is drawn on each side of the radius of the pitch circle. The points A and A ' are each one-fourth the radius of the pitch circle distant from point B^ and correspond to the point A in the last figure. In this construction draw the lines BA and BA '. These lines form angles of 75i° (degrees) with radius BO. Lines BA and BA' are called lines of pressure. The sides of the rack teeth are mide perpendicular to these lines. A Rack is a straight piece having teeth to mesh with a gear. A rack maybe considered as a gear of infinitely long radius. The circumference of a circle approaches a straight line as the radius increases, and when the radius is infinitely long any finite part of the circum- ference is a straight line. The pitch line of a rack, then, is a straight line just touching the pitch circle of a gear meshing with the rack. The thickness of teeth, addendum, and depth of teeth below pitch line are cal- culated in the same manner as for a wheel. A rack to mesh with a single curve gear of 30 teeth or more is drawn as follows : 1 . Draw straight pitch line of rack ; also draw ad- dendum line, working depth line and whole depth line, each parallel to the pitch line (see figure). 2. Point off the pitch line into parts equal to one- half the circular pitch, or = t. 90 MECHANICAL DRAWING, WHOLE DEPTH LiNE RACK 3^ CIRCULAR FITCH. RACK TO MESH WITH SINGLE CURVE GEAR HAVING 30 TEETH AND OVER. MECHANICAL DRAWING. 9 1 3. ThroLig.h these points draw lines at an angle of 75i^ with pitch lines, alternate lines inclined in opposite directions* The left-hand side of each rack tooth is perpendicular to the line^^. The right-hand side of each rack tooth is perpendicular to the line jB A' , 4. Add fillets at bottoms of teeth equal to one-sixth of the width of spaces between the rack teeth at the ad- dendum line. ANGLES FOR RACK TEETH. To get the proper angle for rack teeth, draw a semi- circle on a line A B. With centre A, and radius equal to one-quarter of A B draw arc cutting semi-circle at C. A straight line through A C will form an angle of 75|^ with the line A B. To get the angle for sides of a tool for planing out rack teeth proceed as follows : On line O B describe a circle. From B lay off on the circumference chords B A and BA\ each equal to one-fourth of O B. Angle A OA' is 29^ — the proper angle for the point of the tool. Make end of rack tool .31 of circular pitch, and then round the corners of the tool to leave fillets at the bottom of rack teeth. Thus, if the circular pitch of a rack is i4^", and we multiply it by .31, the product, .465", will be the width of tool at end, for rack of this pitch, before the corners are taken off. GEARS AND RACKS TO MESH WITH GEARS HAVING LESS THAN 30 TEETH. The construction of the rack is similar to that just described. (See Figure Gear 3 P.^ i3 teeth in mesh with rack) . The cur\^e on face of tooth, or that part outside of pitch circle, may be constructed as for a gear having 30 ^2 MECHANICAL DRAWING. teeth or more, but i\\Q fla?7ks, or curve of tooth inside of base circle, are drawn as follows: In gears of 12 and 13 teeth the flanks are parallel for a distance, inside the base circle, of one-third of a diameter pitch {Yz s) gears of 14, 15 and 16 teeth, one-fifth s; 17 to 20 teeth, one- sixth s. In gears of more than 20 teeth the parallel con- struction is omitted. T/ie involute tooth of this gear of 12 teeth is drawn as follows : Draw three or four tangents to the base circle, iV jj\ kk Il\ letting the points of tangency on the base circle i\ j\ k\ /', be one-fourth of the circular pitch apart ; the first point, /', being distant from / one-fourth of radius of pitch circle. With dividers set to one-fourth the radius of pitch circle, placing one point on /', draw the arc a' tj; with one point onJ\ radius^', drawy'/^y with one point on k' draw kL Should the addendum circle be outside of / the tooth side may be completed with the last radius //'. The arcs a ' /, ij\ j'k^ kl^ together form a very close ap- proximation to a true involute from the base circle ij'k'l\ The exact involute for gear teeth is the curve made by the end of a band when unwound from a cylinder of the same diameter as the base circle* With diameter equal to the distance between the ends of two adjacent involutes, where they meet the base cir- cle, draw a circle about centre of gear. Lines from these points tangent to the ciicle form part of the flanks of teeth. From the whole depth circle, draw fillets with radius equal to y^ widest tooth space. These will butt into the parallel lines about y^ s from the base circle. This method is conventional, depending upon the judgment of the designer, to eflect the object to have spaces as wide as practicable just inside base circle and then strengthen flank with as large a fillet as will clear SINGLE CURVE GEAR, 2 IN MESH WITH RAC N =12 ■p'= 1.57' t = .7854' S = .500" = 1.000' g+/= .5785 O'+/ = 1.078* D *= 7r ^is. 10. 94 MECHANICAL DRAWING. addenda of any gear. If flanks in any gear will clear addenda of a rack, they ^vill clear addenda of any other gear, except internal gears. An internal gear is one having teeth on the inner side of a rim or ring. The foregoing operation of drawing tooth sides, al- though tedious in description, is very easy of practical application. The faces of teeth of rack are rounded ofl" by an arc or radius of 1 1^ pitch, \vith centre in working depth line. BEVEL GEAR BLANKS. The pitch of Bevel Gears is always figured at the largest pitch diameter. Most Bevel Gears connect shafts that are at right angles to each other. The following directions apply to any angle, but the sketch is made with axes at right angles. Having decided upon the pitch, numbers of teeth and angle of shafts : (The sketch is made for gears i . i and 2.2 inches diameter.) Draw axes AOB^ COD^ Fig. i8. At a distance from AOB, equal to one-half the diameter of the gear, draw a line parallel to AOB. At a distance from C01\ equal to one-half the diameter of the pinion, draw^ a line par- allel to COD. From the point ;/, where these two parallels meet, draw perpendiculars to AOB and COD. On these perpendiculars lay ofl^ the pitch diameters, tj of the gear, and 7nn of the pinion, the point / being com- mon. From j\ n and ;// draw lines to O These lines give size and shape of pitch cones, and are called Cone Pitch Lines. Through points ;;/, / and j\ draw lines ///x, iy and jz perpendicular to Cone Pitch Lines. On these lines, from cone pitch line, lay ofl' distances MECHANICAL DRAWING. 95 for addenda, working depth and whole depth of teeth. From the points so obtained, draw lines to the centre O. These lines give the height of teeth above Cone Pitch Lines, and the whole and w^orking depths of teeth. The teeth become smaller as they approach O and be- come nothing at that point. It is quite as well never to have the length or face of teeth, ;;/;;/' longer than one- third the distance 0/n, nor more than two and a half times the circular pitch. Having decided upon the length of face, draw limit- ing lines in\x\ i'J' and /'z\ We have now the outline of section of g-ears throu^-h their axes. A straio^ht line drawn throuo^h the laro^est diameter of the teeth, perpendicular to axis of the gear, is called the largest diameter. In practice, these diam- eters are obtained bv ineasurinc^ the drawinsf. To obtain data for teeth, w^e need onlv make draw^ins" of section of one-half of each gear. We first draw centre lines AO, BO and lines ^// and cd^ then gear blank lines as in the case just described. (See Fig. 17.) To obtain shape of teeth in bevel gears, we do not lay them of! on pitch circles in same way as in spur gears. :B -h v?l ^ PINION 1C TEETH. ( \ GEAR, 24 TEETH. 5 P. \ P =5. N =18 and 24 P' = .628" r = .209' t = .314" 8' = .133' S- = .200" .. 0" = .266" D"= .400" §'+/ = .16b' S + /= .231" D"-h/ = .298" D" + f = .431' DPig. XV BEVEL GEARS, FORM AND SIZE OF TEETH MECHANICAL DRAWING. M A line running from a point on cone pitch line to centre line of a bevel gear, perpendicular to this cone pitch line, is the radius for circle upon which to draw outlines of teeth at this point. Hence Ac is the geometrical pitch circle radius, for large end of teeth, and AW the geometrical pitch radius for small end of teeth of wheel. To avoid confusion, the distance A^c' is transferred to Ac' . For the pinion we have the geometrical pitch circle radius Be for large end of teeth, and the radius Be for small end of teeth. Transfer distance B'e' to line ^.'■"".* About A^ draw arc cnm^ and upon it lay off spaces equal to the thickness of tooth at pitch line, and draw outlines of teeth as previously described. We have now the shape of teeth at large end, repeat this operation with radius Be about B, and we have form of teeth, at large end of pinion. Upon arc of radius A 'e ' we get shape of teeth of small end of gear, and upon arc of radius B'c' we get shape of teeth at small end of pinion. The sizes of tooth parts at small end may be taken directly from the diagram, or they may be calculated as follows : Dividing the distance Oc\ which, for example, may be 2 inches, by Oe^ which inay be three inches, we get Yl or .666 for a ratio. Multiplying outside sizes by .666, we get the corresponding inside sizes. Thickness of teeth at outside being .314 inch, ^ of it gives us .209 inch as thickness of teeth inside. * Tredgold's method from Rankine, App. Mech. p. 448. 9 98 MECHANICAL DRAWING. FORMULA AND DRAWING FOR LAYING OUT BEVEL GEAR BLANKS. In finishing bevel gear blanks it is necessary to have the face and back angles. The following formulae will be found useful. They may be determined easily by aid of cut on this page. If X = the number of teeth in pinion, and P = the circular pitch: X P Then D = — = .3183 XP. Tangent of angle S = No. of teeth in pinion -5- No. of teeth in gear = D -^ D . . , B = n + {.62,6 PcosS.) X = .siS^P. £ = .368 p. ^+90°= IV. 0=^S+£. Tan. E^.zi^iP -^H^K-T-H. Tan. R=z,2,6ZF -^H= L-^H. CO CO 1 1 1 1 1 1 1 1 1 1 1 1 1 1 Q ^ a-.7d ixj ^^ >>. ^^->.Jo'r--^ " l\\\\\^^\^. V \ N. r^ ^^ "^^ ---., r 3jy^ |\\\\\\\\\|\N \N. rv.^ ^1 "^x ^=<. '^ ^~- ^^^,.\^K\\\rK\-^. .^ X --. --.. '^-im* \u\\'\k-\^ ■■ \ \ X S ^> "^^ >^ "^k^^ W\ -- 1^^ *^x. \V\\\\\\\\\ V \ \ ^. "\ < N. -.. MV\V\\^A \ \ k H \. \ ^x. K.!' MunM^H \ ^ K 1 \ ^^ >J ^j ^'' 1 \\\\\\\\\\\ \ \ \ \l\ ^ K ^1 1 X \\\\A\\ \\ \ \ \ \ \ 1\ \ \ ^.j ' \U\V\\\\^ V \ \ ^ 'X \, H ^ji; \v \\\^ >\ \ ' ^ '^ \ \ "^ > ^^^ U \\\\V\N N \ \ \ k \ X X X ■^ -^ JUVaA-i 'A \- V \ \ ^ - ^^ ^^ ^ \i\\\\^ \ \ ■ \ \ \ \ \ \ ^Ai \\V\\\\ \ N \ ^ \J ^ ^^ ^ ^' ^^ \\\\^\\\^ \ ^ ^^ ^^ \ "^ \ T' ^ \\\\i\^ \ ^- \ \ ^ \ \ \ K ^-j Al\\\\ \ \ \ N^ ^, \ \ \. 1 \ aAWW \ \ V \ V ^ \ ^ ^^' \\\r Mv \ \ ^' ^ \ \ ^ ^ ^- V W y \ \ ^ \ ^ \ '"^ \ ^ X \ \j\\\ W \ \ \ ^ M \ ^ ^^ ^ . ±1; ^1s\\^ v^ \ \\^ \ \ ^:iriti: ^ A\ ^\ \ \ \ > > V \ \ \i ^ ^r r ^\\ V^ \^ \ V ^ ^ \ M M ^ \u w \ w N \ \ \ \ ^. ^v ^ \>M\ M \ V '> \ \ \ \ \ \U'^r'L^^ \ \ \ 1 ,\ jN V " \ 1 ^ m Si ^nx^^r (fi-^ (' }/: - «' ^^^I M^ L r\ /if 4 / i>| 6B U4 ^dr 0^ Oz ^V^^'\'\ \ \ 1 I M^j ^ ^ " \ U \ \ M \ K ^^ '. \ \ \ ilttl3 t^-^ 3t^ X ^ ^^ \t \ Jiin3^i^i^± L ^^ ^L-_ s - ^ ^ MECHANICAL DRAWING. 1^5 follovv^ing the horizontal line rc23resenting power and the perpendicular line for revolutions per minute, it is found the nearest diagonal line is one representing 5J^" in diameter. Or, supposing there is a shaft ^J^" in diam- eter and its speed is 170 revolutions per minute, and it is required to find the horse power which it will safely transmit. Running along on the horizontal line of 1 70 revolutions per minute until the diagonal is reached, representing 71^" in diameter, at this intersection is found a perpendicular representing the horse power, which is 1,155. STRENGTH OF LEATHER BELTING. The width of the belts should always be a little less than the face of the pulley ; both are to be determined by the power to be transmitted and the velocity of move- ment. For light work a single thickness only is neces- sary, but for belts from prime movers and in other places where great power is to be transmitted, double belts are used. For single belts embracing half the pulley, with a A^elocity of 600 ft. per minute, one horse power can be transmitted for each inch in width of belt, with a max- imum stress on the belt of 50 pounds and pressure on the journals of about 85 pounds per inch of width of belt. J. T. Henthorn, M. E., has given the following for- mula for the strength of double belts, per inch in width, in which Z> is the diameter of the pulley in feet, 7? the revolutions per minute, and JT. P . the horse power : 450 This formula gives .7 H. P . for a belt one inch wide running on a one foot (diam.) pulley at 100 revolutions per minute, double that for the same belt on a 2 ft. pul- ley at 100 revolutions, triple on a 3 ft. pulley, etc. ii6 MECHANICAL DRAWING. HORSE POWER PER INCH OF WIDTH. 1234567 S 9 10 II 12 13 14 15 Ph 12: c c > [I|;^^Diameter of Pulleys shown by Diagonals. The diagram made from this formula is used as fol- low^s : To find the horse power that can be transmitted by a tw^elve inch belt on a nine feet pulley running at 190 MECHANICAL DRAWING. I 17 revolutions per minute : Following the line represent- ing 190 revolutions and the line representing the 9 feet pulley, vs^e find that they meet on the perpendicular num- bered 12. This means that for this pulley at this speed each Inch width of belt w^IU transmit 12 horse power; hence a 12-inch belt w^Ill transmit 12 x 12 == 144 horse powder. To find the width of belt necessary to transmit 50 H, 7^. on a 4 ft. pulley running at 210 revolutions : Inspection of the dlagrain shows that a belt i inch wide on this pulley at this speed w^ill transmit 6 H. P.^ whence (50 -j- 6) a belt 8^ Inches wide will be neces- sary to transmit 50 H. F . on this pulley at this speed. DIAGRAM FOR CONE PULLEYS. Circles M and N represent the largest and smallest sizes required on the pair of cones. C is the distance between centers when in working position. Draw tan- gent M N, With center A^ located as per drawing, draw circle tangent to line M N. Assume satisfactory sizes on one cone and draw tangents to circle A^ to de- termine diameters on other cone. One illustration is given. For angles greater than 18° see Kent, p. 875. ii8 MECHANICAL DRAWING. < > a 1-3 < o IT) in \o XT) vn m u~) in in in in in in in in in in in in in in in in ' r-. c^ r^ O *H CO rl- vn o CO qn O M CO T^ ino oo C\ O M CO Th c^ c^ IT) O M in QO M ^ r^ O Tt r-« O coo O C) in C^ C^ inco CN CO cn CO Tt Th Th in in in O O o r^ r^ t^ i^ cc GO GO C^ o o 1 1 i 1 1 II i II 1 II 1 1 II 11 1 1 1 1 1 <^ t-H CO U-) r^ O M CO in r-- On i-i c^ in r-^ O »-i CO in r-- C> i-i CO M C^ C< C^ (N M CO CO CO CO CO tJ- -* -^ rj- Tt- in in in in in O O in in in u~ in in in in in ir> U-) in in vn in in in in in in in 04 r^ (N r-- O M M 00 1 II II II 1 II ^ II 1 1 II II II II II M M CO '^ LnO r- M CO in r^ C> i-i CO in M CO in l^ hH 1— 1 1-1 < xn P U Q O incj iH O Ooo l>.Omcooi m O Ot^O tj-cooi »^'^'^"^'^COCOCOCOCOCOCO04 CO04 04 C\D r^GO OO ^-1 c^ coTt-ino r^co OO m oi cOTj-ino r^co oo xTk \n \n \ri \r^\Q \0 \Q \0 '^ <^ "^ "-O '^ <^ ii^t^t^r^r^t^r^r^t^t^oo O Tt r^ O in 04 CO CO 04 04 M M CO O O O CO v-i O O Oi^in04cO inM Ot^ino4 OO cocO m O^ (^ O"^ Oco CO CO i^ r^ r^ r^o ^ ^ \r^ \c^ OOOOOOOOOOOOOOOO CO OO'-'C^JCO-rhinOt^co OO'-iC^cO'^inOr^cO OO i-iOlco-^ 04 04 COCOcOcocOcOCOcOcocO'^-^Tj-ri-'d-'^'^'^-^J-'^ininininin ^ Q 1^ ?i> t^ O C4 04 M M M 04 04 O o o 04 04 04 O r^ rj- o^ o o-^ GO ino4 Ocoino4co t|-i-i voo CO GO CO r^ r^ r^o o o r^ in CO M CO O co in in in in Th -^ -rj- M M M M 04 cOTj-inO t^co OO I-" CHXi^>PeHc/^p^a(iHO^§H:ii^ H-,wH WOfofiqQop5 ^ ^ i S ^8-0 flL > o u J J i.r u o ■J m r [ ] IV m ~H ■ 1 1 Ijfl m H t 1 ■ jl 1 .UH i^-i I 1 ■■■ ■ T ■ z 1 I X I ■ ■^ ■ |J ] ■ 1 l' 1 ■ ■ ■ ■ 1 \ ■ ■ 1 «i« " ■ " ■-■-■ 1 ■■■ ■ I 1 w I bJ-l 1 il3 i ! pi 1 1 r r 1 I 1? ljl ^ ■ 1 . ■ Ti3 MECHANICAL DRAAVING. 121 AREAS OF CIRCLES. Area = 3. 1416 X Radius^. Diam. Area. ! Diam. Area. 1 Diam. Area. Diam. Area. .04908 y% 20.629 13 132.73 2 989.8 5 TIT .0767 y 21.647 Vz 143.14 36 IOI7.9 v% .11045 Y% 22.690 14 153.94 2 1046.3 1 .15033 K 23.758 2 165.13 I 37 1075.2 'A .19635 Y^ 24.850 15 176.71 '\ 2 1 104. 5 9 TIT .24850 Y 25.967 2 188.69 1 ^ II341 'A .30679 7/ /8 27.108 16 201.06 2 1 164. 1 1 1 Til" .37122 6 28.274 2 213.82 39 II94.6 Va, .44179 Y?> 29.465 17 226.98 ' 1 2 1225.4 » 3 T15" .51849 y 30.68 2 240.53 1 40 1256.6 74 .60I32 Y% 31.92 18 254 47 i 2 1288.2 1 5 T^ .69029 Y^ 33.183 2 268.80 : 41 1320 2 I .7854 Y^ 34-471 19 283 53 i 2 1352.6 ^M .99402 Y 35.785 2 298.65 42 1385.4 ^H 1.2272 7' /8 37.122 20 314.16 I 1 2 I418.6 i^ 1.4849 7 38.484 2 330.06 : 1 "^3 1452.2 i^ I. 7671 !» 39.871 21 346.36 1 2 i486. 1 i>^ 2.0739 H 41.282 2 363.05 44 1520.5 i^/ 2.4053 Vs 42.718 22 380.13 2 1555.3 17^ 2.7612 y, 44.179 2 397.61 45 15904 2 3.I416 % 45.663 23 415.48 2 1625.9 % 3.5466 U 47-173 2 433-74 46 1661.9 K 3.9761 1/ /8 48.707 24 452.39 2 1698.2 yk 4-430I 8 50.265 2 471-43 47 1734-9 V2 4.9087 yi 51.849 I 25 490.87 2 1772. H 5.4JI9 H 53-456 1 2 510.70 : 48 1809.5 Ya. 5.9396 H 55.088 26 530.93 ' 2 1847.4 }i 6.4918 yi. 56.745 2 551.55 49 1885.7 3 7.0686 Vs 58.426 27 572.56 j 2 1924.4 % 7.6699 H 60.132 2 593.96 50 1963 5 \/ 8.2958 /'8 61.862 28 615.75 2 2002.9 Y% 8.9462 9 63.617 2 637.94 51 2042.8 1/ 9.6211 % 65-397 29 660 52 2 20S3. J^ 10.320 M 67.201 2 683.49 52 2123.7 ^/ 11.045 Vz 69.029 30 706.86 1 2 2164.7 7/ /8 11.793 'A 70.882 2 730.62 1 53 2206.2 4 12.566 j % 72.76 \ 31 754.77 2 2248. >^ 13.364 % 74.66 2 779.31 54 2290.2 y^. 14.186 7/ /8 76.59 32 804.25 2 2332.8 Y^ 15.033 10 78.54 1 2 829.58 55 2375.8 y^ 15.904 Vz 86.590 33 855.30 2 2419.2 Ys 16.800 ir 95033 2 881.41 56 2463. Y 17.72 K 103.87 1 34 907 92 2 2507.2 %. 18.665 12 113. 1 2 934.82 57 2551.7 5 19.635 Yz 122.72 , 35 1 962.11 2 2596-7 11 122 MECHANICAL DRAWING. AREAS OF CIRCLES.— Continued. Diam. Area. Diam. Area. 1 I 1 Diam. i Area. Diam. Area, 58 2642 . I 69 3739.3 2 4963.9 90 6361.7 2 2687.8 2 3793.7 80 5026.5 I ! 2 6432.6 59 2734. 70 3848.4 2 50S9.6 ; 91 6503.9 2 2780.5 2 3903.6 81 5153. 1 2 6575.5 60 2827.4 71 3959.2 2 5216.6 92 6647.6 2 2874.7 2 4015. I 82 5281. j 2 6720.1 61 2922.5 72 4071.5 2 5345.6 93 6792.9 2 2970.6 2 4128.2 83 5410.6 i 2 6866.1 62 3019. I 73 4185.4 2 5476. 94 6939.8 2 3067.9 2 4242.9 84 5541.78 2 7013.8 63 3117.2 74 4300.8 2 5607.9 95 7088.2 2 3166.9 2 4359.1 85 5674.5 2 7163. 64 3217. 75 4417.8 2 5741.4 ; 96 7238.2 2 3267.4 2 4477. 86 5808.8 2 7313.8 65 3318.3 76 4536.4 2 5876.5 97 7389.8 2 3369.5 2 4596.3 87 5944.7 1 2 7466.2 66 3421.2 77 4656.6 2 6013.2 i 98 7543. 2 3473.2 2 4717.3 88 6082. I 2 7620.1 67 3525.6 78 4778.3 2 6151.4 99 7697.7 2 3578.5 2 4839.8 89 6221. I 2 7775.6 68 3631.7 79 4901 . 7 /-» 6291.2 100 7854. 2 3685.3 MECHANICAL DRAWING. 123 CIRCUMFERENCES OF CIRCLES. C = 3. 1416 X Diameter. Diam. Circum. Diam. Circum. 1 Diam. Circum. Diam. Circum. y^ .7854 % 36.128 M' 71.471 38 119.38 Yi 1.5708 Y, 36.914 23 72.257 X. 120.95 ¥, 2.3562 12 37.699 y^ 73.042 39 122.52 I 3.1416 M 38.484 Vz 73.827 X 124.09 ^M 3.927 Vz 39.270 Ya 74.613 40 125.66 iVz 4.7124 Ya 40.055 24 75.398 y2 127.23 ^H 5.4978 13 40.841 X 76.184 41 128.80 2 6. 2832 M 41 .626 y2 . 76.969 X 130.37 2\i 7.0686 Vz 42.412 u 77.754 42 131.95 Vz 7.854 Ya 43.197 25 78.54 ^ 133.52 H 8.6394 14 43.982 X 79.325 43 135.9 3 9.4248 Ya 44.768 y2 80. Ill X 136.66 U 10.210 Yz 45.553 u 80.896 44 138.23 Vz 10.995 Y 46.338 26 81.681 'i 139.8 H II. 781 15 47.124 X 82.467 45 141.37 4 12.566 Ya 47.909 Y2 83.252 X 142.94 '^4 /4 13.352 Vz 48.695 u 84.038 '46 144.51 K 14.137 Y 49.480 27 84.823 M 146.08 H 14.922 16 50.265 X 85.608 47 147.65 5 15.708 M 51.051 Y2 86.394 y2 149.22 K 16.493 Yz 51.836 X 87.179 48 150.8 v^ 17.279 Y 52.622 28 87.965 X 152.37 H 18.064 17 53.407 X 88.75 49 153.94 6 18.849 Ya 54.192 K 89.535 X 155.51 H 19.635 ¥2 54.978 X go. 321 50 157.08 V2 20.420 Y 55.763 29 91 . 106 X 158.65 }i 21 .206 18 56.549 X 91 .892 51 160.22 7 21 .991 ^ 57.334 Y2 92.677 X 161.79 y^ 22.776 Vi 58.119 X 93.462 52 163.36 Yz 23.562 Y 58.905 30 94.248 X 164.93 H 24.347 19 59.69 Y2 95.819 53 166.5 8 25.133 M 60.476 31 97.389 Y2 168.07 K 25.918 Yi 61.261 X 98.96 54 169.64 Vz 26.703 U 62.046 32 100 53 X 171.22 /4 27.489 20 62.832 y2 102.10 55 172.79 9 28.274 U 63.617 33 103.67 X 174.36 M 29.060 Yz 64.403 ^ 105.24 56 175.93 ^2 29.845 Y 65.188 34 106.81 ^ 177.5 H 30.630 21 65.973 M 108.38 57 179.07 10 31.416 ¥. 66.759 ' 35 109.95 X 180.64 M 32.201 M 67 544 ^ X III. 53 58 182.21 Yz 32.987 Y 68.330 i 36 113.10 Vz ■183.78 Yat 33.772 22 69.115 y2 114.67 59 185.35 II 34.557 Y 69.9 2>1 116.24 )^ 186.92 Ji 35.343 i Yz 70.686 i X 117. 81 60 188.49 124 MECHANICAL DRAWING. CIRCUMFERENCES OF CIRCLES. Continued. DIam. 1 Circum. Diara Circum. ! Diam . 1 Circum. i Diain. Circum. ^ i 190.07 ' Yz 221.48 2 252.90 2 284.31 6i , 191.64 71 223.05 81 254.47 91 285.88 Vz 193.21 /2 224.62 1 2 256.04 2 287.45 62 194.78 1 72 226.19 82 257.61 92 289.03 M 196.35 ; 2 227.76 ! 2 259.18 2 290.6 63 197.92 1 73 229.34 ; 83 260.75 93 292.17 % 199.49 i 2 230.91 ' 2 262.32 2 293.74 64 201 06 74 232.48 1 , 84 263.89 94 295.31 Yz 202.63 2 234.05 2 265.46 ! 2 296.88 65 204.20 , 75 235.62 1 1 85 2D7.O3 95 298.45 2 205.77 2 237.19 1 2 268.61 2 300 . 02 66 207.34 ! 76 238.76 1 86 1 270.18 96 301.59 2 208.91 i 2 240.33 2 271.75 1 2 303.16 67 210.49 77 241.9 87 273.32 97 304.73 2 212.06 2 243.47 2 274.89 2 306.3 68 213.63 !. 78 245.04 88 276.46 98 307.88 K 215.2 2 246.61 2 278.03 2 309.45 69 216.77 79 248.18 89 279.60 99 311 .02 K 218.34 1 2 219.76 2 281.17 2 312. .59 70 219.91 80 251.33 90 282.74 100 314-16 MECHANICAL DRAWING. 12 POWER REQUIRED FOR MACHINE TOOLS. (From a paper read by F. B. Duncan before the Engineers' Society of Western Pennsylvania.) ENGINE LATHES. i6 in. ; motor power required, approximate, 2 H. P. at maximum. 18 in. X 6 ft. ; motor power required, 2.1 H. P. 36 in. X 10 ft. ; motor power required, 10 N. P. PLANERS. 10 X 10 X 20 ft. ; 3 tools, I X 1^ in. cut; cutting speed, 18 ft. ; planing 40-ton iron casting. H. P. required for cut, 26. 5 ; for return, 23.6; for reverse, 42.9. Ratio return, 3 to i. Motor, 30 H. P., belted to countershaft. 8 X 8 X 20 ft. ; 3 tools, f x i in. cut; cutting speed, 18 ft. ; planing 32-ton iron casting; H. P. for cut, 16; for return, 14.8; for reverse, 28.2. Ratio return, 3 to i. Motor, 25 H. P., belted to countershaft. 66 X 60 in. X 12 ft. ; 2 tools, | x -^^ in. cut; cutting speed, 21 ft, ; planing 4-ton openhearth casting. H. P. required for cut, 10; for return, 14; for reverse, 16. Ratio return, 3^ to I. Motor mounted on planer housing with 42-inch i, 500-pound flywheel, running at 400 revolutions, mounted on motor shaft ; flywheel used as driving pulley for return of platen. 10 ft. boring and turning mill; cutting tools, 2; cut, Jx ■^^ in. ; cutting speed, 20 ft. ; machining 3.5-ton casting ; H. P. required for cut, 8.6. Motor used, 12 H. P. Blotter; cut, f x -^-^ in.; speed of tool, 20 ft.; machining openhearth steel castings; power required, 6.98 H. P. Flat turret lathe; 1^ H. P. motor required. Gisholt tool grinder; speed, 1,600 to 1,800 rev. ; power re- quired, 7 for short periods, 4 on average. Motor used, 5 H. P. HORSE POWER FORMULAS. In an article by C. H. Benjamin, in March, 1899, MacJiinery, are the following formulas for computing the horse power re- quired to operate tools, where />F= weight metal removed per hour. Experiments with several lathes give: H. /^- = .035 W for cast iron. H. P. = .067 IV for machinery steel. Experiments with a Gray planer give: H. P. = .032 IV for cast iron. Experiments with a Hendey shaper give: ** H. P. = .030 IV^ for cast iron. For milling machines we have: H. P. = .14. IV for cast iron. JP. P. = .10 JV for bronze. H. P. = .30 TV for tool steel, In each case, the power required to run the tool, light, should be added 126 MECHANICAL DRAWING. TAP DRILLS FOR V THREADS. For taps .090 to i| inches. 1 bo w -o _e -a :=i rt _ 'u rt n:l Ui u u a V a. u "ti r^, 1 3' 64 » 1 ' 5 ' T6 .312' 18 15" 64 1 ' ¥ » 5 ♦ 16 f .375' 16 9 " 32 5 ' T6 f 7 ' T6 •437' 14 21" 64 23' 6¥ A' 1 ' .500' 12 3 " 8 13' 32 ' 1 ' 9 ' T6 .562' 12 7 " T6 1 5' J2 ♦ 9 ' 16 5 ' 8 .625' 1 ^^ 1 »» 2 1 7' 32 *' 1 1' T6 .687' ' II 1 7" 32 9 ' 16 ' 1 1' 1 6 3 ' .750' ' 10 38" 6¥ 5 ' 8 ' 3 ' ¥ 1 ' 8 .875' i 9 1 1" 1 6 45' 64 1' 8 I ' 1. 000' ' i 8 51" 6¥ 1 3' T6 l' ir 1. 125' > 1 ^ 29" 3 2 1 5' 1 6 l6¥ li' 1.250' ' i 7 tJL" -•■3 2 ItV ' xl 7' !«¥ If 1-375' ' i 6 Ti" ••■8 I A' ' T 2 5 ' !«¥ If 1.500' ' i 6 1 t15" lA' ' x33' l6¥ SIZES OF DRILLS FOR PIPE TAPS. 11 32 7 T^ 27 6¥ 45 6¥ 29 32 9 6¥ 2\ t23 ^32 _3_ 1 6 5 2VV 3i MECHANICAL DRAWING. I27 Conventional Forms for Cross-Seciionrng lead or Babbift. Cast Iron. Wrought Iron, Malleable Iron, Cast Steel Wrought Steel Nickle-- Steel. v^x70 M -^ 60 (0 o ^ N 5 ? 0) oo K3 ^ THREADS /»£•/? /NCH 0) • C/1 N '^ (a ^> s ^^ • Si i 5S 5 • M Cn is 8 • 09 01 • 00 is • to 1 (o>: :o: 01 2 u • 01 0) • 0\ o • 0) (A O 04 io -4 -"4 • ^ ^ • 3 1 i • oo » 1© • • (J • 00 oo 01 N 0) Ol 0) • o • 00 OO • 01 5 • Cn o 01 f4 • • •4) • 00 4^ ^ ' •X^ -K ^ 01 CO 01 00 01 oo M S -4 Cn N3 S -^ 1_ ^^V^ ^ • oi • i 00 00 s ^ Oi • o • 00 01 • o o 2 1 01 4k -4 -0 lA 00^ 9 OO is BOTTVM ar THRSAO N • 00 51 § * in • • Cn • • 5* is In 4 51 6i cn Si iJl J) 1 1 i i CO i • § & t 00 cn • g • • CO 01 oi 1 SI "i ^ -0 Cn ^jni ULU J OJ N ^) N „^ ,^ ^. ? S ^ ^ t :8 ^ 9\ o t (A O $ ^ (n o o 4^ BOLT \ 0) S 0) 0) «0 ^ -0 »3 4i 5 oi 00 OO 85 01 o 01 o 4k 0^ -4 ^ M t>J .^ - BOTTOM 0) :^ Oj ?0 o c^ CO 00 01 i g N o 5 hJ b CO o ? OF o ^ 01 «30 ^ .4 0) o 01 8 S h) Cn 9* oS $ 9\ (0 THRFAO. K) N) AT N 04 4^ 4^ N ^ o Oi 00 oo 0> 4> ^ O o o^ JJ ^ o% 4^ K3 fOOOO LBS i C 3 o ± -0 oo ± ^ ^ (J o 09 oo N -a 0) 0) -4 OO V o> CO P£RS^/M AT N 00 «4 0\ 00 -0 2? (A ?3 > 09 0^ 00 Oi (A Cn o cn ^ oo Cn (A iZSOO LBS, tu Oj (A o 01 -4 (A 0^ 00 00 f to 01 ^ 0> 4^ 00 00 (A PERSQJN, ^ AT i 00 1 00 (Pi (jy ?5 -0 -4 01 01 00 in 00 (A 00 01 01 01 00 -4 Oi 4^ nSOO LBS. PERSQ.m ^ 4T7600 O OO .^ C^ ^ (X> ^ Ol 4" (A N — ^ .^ .« LBS g ^ t J^ ^ ^ u ^3 o ^n S <0 5 W 01 OO 0^ 4^ --1 hJ oo Cn -4 lA PER 1 5 ^ h) U) o -► c3 M 4^ oj o 4^ ^ N ->( oo Cn oo SOJN. ^ 4^ o H > f5 CA •-4 (^ 4^ 0) w AT/0000 LBS. ± 2 o 0) 0> oo 4^ (0 4^ OO Cn o > § 4^ 00 CO cn Oi o o 5? 4^ CO PER Co l\J (A CO ~~ (0 r>j A ^ w 00 CO cn 0) } «„ «. LBS o o 4^ i 00 O) 00 ±5 (A Cn o 0) o o o s cr> ?2 0] -0 01 01 00 1 ro (0 PER S/^/M S3 I MAY 25 I9US / ■■■iiiiiiyiii LIBRARY OF CONGRESS 019 945 479 1