r 
 
 REESE LIBRARY 
 
 of Till-. 
 
 UNIVERSITY OF CALIFORN 
 
 
 IA 
 
 ;[-, 
 
 H 
 
MECHANICAL 
 
 AND 
 
 ENGINEERING DRAWING 
 
FIRST PRINCIPLES 
 
 OF 
 
 MECHANICAL 
 
 AND 
 
 ENGINEERING DRAWING 
 
 A COURSE OF STUDY ADAPTED TO THE SELF-INSTRUCTION OF 
 
 STUDENTS AND APPRENTICES TO MECHANICAL 
 
 ENGINEERING IN ALL ITS BRANCHES 
 
 AND FOR THE USE OF TEACHERS IN TECHNICAL AND MANUAL 
 INSTRUCTION SCHOOLS 
 
 BY 
 
 H. HOLT-BUTTERFILL, M.E. 
 
 3 
 
 FORMERLY A MEMBER OF THE INSTITUTION OF MECHANICAL ENGINEERS AND 
 INSTITUTION OF NAVAL ARCHITECTS 
 
 WITH UPWARDS OF 350 DIAGRAMS JN ILLUSTRATION OF THE 
 PRINCIPLES OF THE SUBJECT 
 
 LONDON: CHAPMAN AND HALL, LIMITED 
 
 1897 
 
HICHARD CLAY & SONS, LIMITED, 
 LONDON & BUNQAY. 
 
PREFACE 
 
 THE greater part of the subject matter of this book appeared in a 
 series of articles in the Mechanical World. The purpose in writing it 
 is so fully explained in the Introduction that a Preface is hardly 
 required. As the forms given to the various parts of a machine or 
 engine are on analysis invariably found to be combinations of certain 
 geometrical solids, a knowledge of how each of these should be drawn 
 when in any position should be first acquired by the student draughts- 
 man. To this end a series of problems is given in the following pages, 
 commencing with the construction of those simple geometrical figures 
 which form the surfaces of the solids which give shape to mechanical 
 details, and subsequently the method adopted in representing the solids 
 themselves, singly and in combination. 
 
 As no amount of copying "drawings" of mechanical details will 
 ever give the student a knowledge of the reasons why they are made 
 to take the special forms given to them, so in the earlier stages of the 
 study of mechanical drawing it is impossible for him to acquire the 
 power to draw the simplest solids in different positions correctly 
 without a knowledge of the principles of " Orthographic Projection," 
 which is the basis of the representation of all solid objects. In this 
 part of the subject an extended series of problems is given, the solution 
 of which should enable the student to draw any simple object without 
 further help. 
 
 In the method of studying the contents of this work, the student is 
 advised to take the different parts of the subject in the order in which 
 they are arranged, as he will thereby be led to acquire a mastery of it 
 in a way that will impress upon his mind the connection that each 
 part bears to that which follows. The order of study may not be that 
 usually followed, but it is such as an association of many years with 
 draughtsmen and students has proved to the author to be the best for 
 the acquisition of the preliminary knowledge necessary to the successful 
 practice of the draughtsman's art. 
 
 This work is not intended as a treatise on either Plane or Solid 
 Geometry, but as much of these subjects is given as will be required 
 by the student ta attain to an easy comprehension of the first principles 
 of mechanical drawing as herein exemplified. Their actual application 
 to the delineation of machine elements and engine details may possibly 
 form the subject of a further work. 
 
 H. HOLT-BUTTERFILL. 
 
 Greenwich, 1897. 
 
CONTENTS 
 
 Introduction. THE VALUE OF A KNOWLEDGE OF DRAWING TO THE STUDENT 
 
 PAGE 
 
 CHAPTER I 
 
 THE TOOLS AND MATERIALS REQUIRED BY THE STUDENT 
 Drawing-Board Tee-Square Adjustable Bladed Square Set-Squares 
 Pencils Drawing - Pins Paper Kubber Ink Drawing Instru- 
 ments 1 11 
 
 CHAPTER II 
 MECHANICAL AND FREEHAND DRAWING: THEIR DIFFERENCE 
 
 AND USES 
 
 The meaning of Freehand Drawing How objects are made visible 
 What a Perspective is How a Perspective Drawing is obtained 
 The use of a Perspective Drawing to the workman An Orthographic 
 Projection, and how obtained The meaning of Plan and Elevation 12 16 
 
 CHAPTER III 
 
 PRACTICAL GEOMETRY AND MECHANICAL DRAWING 
 The meaning of the term "Geometry" The difference between Plane and 
 Solid Geometry Definition of Geometrical terms used in the work 
 Plane Geometrical Figures 1722 
 
 CHAPTER IV 
 
 PLANE GEOMETRY PROBLEMS 
 
 To divide a straight line into two equal parts To erect a perpendicular 
 to a given straight line To let fall a perpendicular to a straight 
 line To bisect a given angle To draw a line parallel to a given line 
 To draw an angle equal to a given angle To draw a line making 
 an angle with a given line ... ... ... ... ... ... 23 27 
 
viii CONTENTS 
 
 PAGE 
 
 CHAPTER V 
 
 PLANE GEOMETRICAL FIGURES 
 
 To construct an equilateral triangle, an isosceles triangle, a scalene triangle 
 To construct a square, a rectangle, a rhombus, a rhomboid, a tra- 
 pezium, a regular pentagon, a hexagon, a regular octagon 
 
 CHAPTER VI 
 
 ORTHOGRAPHIC PROJECTION 
 
 The Planes of Projection The difference between a vertical and a per- 
 pendicular plane The relative position of the planes of projection 
 The projections of a point and a straight line The projections of a 
 line inclined to the planes of projection ... ... ... 35 41 
 
 CHAPTER VII 
 
 PROJECTION OF PLANE FIGURES 
 The Projection of the Triangle The square The pentagon and the 
 
 hexagon, etc. 42 46 
 
 CHAPTER VIII 
 THE PROJECTION OF SOLIDS 
 
 Definitions of the Plane Solids The cube, the prism, the pyramid, etc. 
 Models of the Solids necessary to the Student for a thorough know- 
 ledge of their projection Elevations of objects given to find their 
 plans Meaning of section, side elevation, sectional plans and eleva- 
 tions 47 58 
 
 CHAPTER IX 
 PROJECTION IN THE UPPER PLANE 
 
 The front elevation given, to find the side elevation The sectional eleva- 
 tions of the solid and hollow cube, prism and pyramid ... ... 59 70 
 
 CHAPTER X 
 
 PROJECTION FROM THE LOWER TO THE UPPER PLANE 
 The plans of objects given, to find their elevations and sectional elevations 71 76 
 
 CHAPTER XI 
 
 LINING-IN DRAWINGS IN INK 
 
 The kind of lines to be used The direction in which the light is supposed 
 to fall on the object represented To find the angle that the rays of 
 light make with the planes of projection Why different qualities of 
 lines are used in Mechanical Drawing The importance of correctness 
 in their application How ink for lining-in a drawing should be 
 made And how to fill the drawing-pen 77 84 
 
CONTENTS IX 
 
 PAGE 
 
 CHAPTER XII 
 
 THE PROJECTION OP CURVED LINES 
 
 The definition of a curved line The front elevation of a curved line being 
 given, to find its side elevation and plan How to find the projections 
 of a line of double curvature The projections of combined curved and 
 right lines The plan of a circular plate being given, to find its eleva- 
 tion To draw an ellipse ... ... ... v 85 97 
 
 CHAPTER XIII 
 
 THE PROJECTION OF SOLIDS WITH CURVED SURFACES 
 The definitions of the cylinder, the cone, and the sphere The plan of a 
 cylinder given, to find its elevation in various positions The plan 
 of a cone given, to find its elevation in different positions ... ... 98 102 
 
 CHAPTER XIV 
 THE PROJECTION OF THE SECTIONS OF A CYLINDER 
 
 The elevation of a cylinder given, to find its sectional elevation and 
 
 plan 103105 
 
 CHAPTER XV 
 
 THE PROJECTION OF THE CONIC SECTIONS 
 
 The definitions of the sections of a cone The plan and elevation of a cone 
 being given, to find its sectional projection -To find the true form of 
 any section of a cylinder or cone The sections of a sphere and their 
 projections Definitions of the subsidiary solids of revolution The 
 lining-in in ink of solids with curved surfaces How the light falls 
 upon them .... ... .. 106119 
 
 CHAPTER XVI 
 THE PROJECTION OF OBJECTS INCLINED TO THE PLANES OF 
 
 PROJECTION 
 
 To find the projection of a point, and line lying on an inclined plane The 
 projection of plane figures when inclined to the planes of projection 
 The projection of a solid when inclined to the planes of its projection 
 The projections of the solid and hollow cube, the pyramid, and 
 cone, when inclined to the planes of their projection The projec- 
 tions of a six-sided nut, when inclined to the planes of projection ... 120 141 
 
 CHAPTER XVII 
 
 THE PENETRATION AND INTERSECTION OF SOLIDS 
 The penetration of prisms by prisms at right angles to each other The 
 
 penetration of prisms having their axes inclined to each other . . . 142152 
 
X CONTENTS 
 
 CHAPTER XVIII 
 
 THE INTERSECTIONS OF PLANE SOLIDS (continued) 
 The penetration of a prism by a pyramid The penetration of pyramids 
 
 by pyramids ... 153163 
 
 CHAPTER XIX 
 
 THE INTERSECTIONS OF SOLIDS 'HAVING CURVED SURFACES 
 The intersections of equal-sized cylinders at right angles to each other 
 The intersection of unequal-sized cylinders The intersection of in- 
 clined cylinders The intersection of the cylinder and cone The 
 intersection of cones by cylinders and cones The intersection of the 
 sphere by prisms and pyramids The intersection of the sphere by 
 the cone and cylinder 164 186 
 
 CHAPTER XX 
 
 THE DEVELOPMENT OF THE SURFACES OF SOLIDS 
 What a development means What is a developable surface The develop- 
 ment of plane-surfaced solids The development of the surface of a 
 pyramidal-shaped solid, having a curved surface The development 
 of the oblique pyramid The development of the surface of a right 
 cylinder, and the frustum of a right cylinder The development of 
 the surface of an oblique cylinder, and of a right and oblique cone 
 The development of the surface of the sphere and hemisphere ... 187 211 
 
INTEODUCTION 
 
 THE FIRST PRINCIPLES OF MECHANICAL AND ENGINEERING 
 
 DRAWING 
 
 IT being incumbent on every one who aspires to become a really 
 efficient Engineer, that he should possess a thorough practical know- 
 ledge of the Mechanical Draughtsman's art, we would in the outset 
 of an attempt to explain the fundamental principles which govern its 
 operations, observe, that the inducement to undertake such a task is 
 the desire to place within the reach of every earnest engineering student 
 and apprentice, a means of enabling him to read and to make such 
 drawings as are placed before him in an engine factory to work from, 
 and to prepare him for the subsequent study of engine and machine 
 design. 
 
 It is assumed by the majority of engineering students and appren- 
 tices, that the drawing practised in the Drawing Office will be taught 
 them upon their first admission to it, but an experience of many years 
 in some of the principal offices in England, has made the writer alive 
 to the fact, that so far as the "principles " which underlie the practice of 
 the draughtsman's art are concerned, absolutely nothing is taught the 
 student, and that if he ever acquires a knowledge of them, it will be 
 by his own unaided study, independent of any drawing-office help. 
 With a view, then, to the acquisition by the student of this all- 
 important knowledge, in the best possible way, we have in the following 
 pages formulated a method of imparting it, which from practical 
 experience as a draughtsman, and teacher, we have found answers 
 every requirement. Whether that method is an improvement on any 
 now adopted, is left to those who earnestly follow its exposition to 
 determine. 
 
 Before proceeding with that exposition, we would, however, put 
 before the student, some facts bearing upon the study of drawing (and 
 Mechanical Drawing more particularly), which may help him to 
 appreciate the necessity that exists for his acquiring the ability to 
 draw, if he desires to rise in his profession. Without wishing in the 
 least to under-estimate the great worth of a really first-class skilled 
 workman, who may have little or no knowledge of drawing, it is still 
 
xii INTRODUCTION 
 
 a fact very generally admitted that just in proportion to the knowledge 
 of drawing possessed by one workman over his fellow, so is he superior 
 to him ; and it follows that those ignorant of that art must hold a 
 lower position as workmen, than those having a knowledge of it. 
 
 The utility of the power to draw may not present itself to the 
 mind of the workman on its first suggestion to him, but a little thought 
 about the matter will soon make it clear that it has a much closer 
 connection with his daily work than he had any idea of. Neither 
 spoken nor written language can at all times convey ideas that we wish 
 to impart to another, and recourse must be had to some other means, 
 more especially if those ideas relate to the form and position of material 
 substances. To assist us in making our meaning clear, we must make 
 use of what has been aptly called the "language of mechanics," or 
 Drawing a language which appeals at once to the eye for the truth 
 of its assertions, and which enables us, without further assistance, to 
 judge of the form, appearance, and dimensions of bodies. 
 
 To the intelligent mechanic, a real power of drawing is a priceless 
 advantage, as it enables him to either reproduce a true representation 
 of forms, that upon a casual inspection may have made an impression 
 on his mind ; or, on the other hand, to transfer to paper what he may 
 have conceived, but which has not as yet had any existence. Many 
 a valuable invention has been lost to posterity through the want of the 
 power to draw, on the part of the would-be inventor. 
 
 Again, a knowledge of the graphic art is now demanded of all who 
 are in any way connected with mechanical constructions of any kind, 
 and no one can now hope to obtain any position of trust that an 
 engineer fills, who has not acquired the power of correct drawing. It 
 was long a fallacy with many, that draughtsmen were born, not made ; 
 that although a youth, or a man, may be taught to write or copy 
 letters the law did not hold good as regarded drawing. This fallacy 
 has happily gone the way of many others, and it is now held that those 
 who will give to the study of Drawing the necessary concentration of 
 thought, coupled with persistent effort, will undoubtedly attain its 
 mastery and achieve success. 
 
FIRST PRINCIPLES 
 
 OP 
 
 MECHANICAL AND ENGINEERING DRAWING 
 
 CHAPTER I 
 
 THE TOOLS AND MATERIALS REQUIRED BY THE STUDENT 
 
 1. Drawing-Board. As all drawings of Mechanical and Engineering 
 subjects are made on flat surfaces, and as the most suitable material 
 on which such drawings may be made is paper, the first requisite of 
 the student is a drawing-board, on which to lay or stretch the paper. 
 The board should be made of well-seasoned pine, of a convenient size 
 say 23 in. by 17 in., which will take half-a-sheet of Imperial paper, 
 leaving J- in. margin all round J in. in thickness, and fitted at the 
 back, at right angles to its longest side, with a couple of hardwood 
 battens, about 2 in. wide and J in. thick ; the use of these battens 
 being to keep the board from casting or winding, and to allow of its 
 expansion or contraction through changes of temperature. This latter 
 purpose, however, is only effected by attaching the battens to the back 
 of the board in the following manner : At the middle of the length of 
 each batten which should be 1 in. less than the width of the board 
 a stout well-fitting wood screw is firmly inserted into it, and made to 
 penetrate the board for about J in., the head of the screw being made 
 flush with the surface of the batten. On either side of this central 
 screw two others, about 3J in. apart, are passed through oblong holes 
 in the battens, and screwed into the body of the board until their 
 heads are flush with the central one ; fitted in this way the board itself 
 can expand or contract lengthwise or crosswise, while its surface is 
 prevented from warping or bending. 
 
 The working surface of the board or its front side should be 
 perfectly smooth, but instead of being quite flat it should have a very 
 slight camber, or rounding, breadthways, this latter feature in its con- 
 struction being to prevent the possibility of a sheet of paper when 
 stretched upon its surface having any vacuity beneath it. The four 
 edges of the board need not form an exact rectangle, as much valuable 
 time is often wasted in the attempt to produce such a board ; but it 
 
 B 
 
2 FIRST PRINCIPLES OF 
 
 will answer every purpose of the draughtsman so long as the adjacent 
 edges at the lower left-hand corner of it are at right angles to each 
 other, or square. To produce really good work in the shape of a 
 mechanical drawing, one perfectly straight edge only is required on a 
 drawing-board, and that the left one, which is always known as the 
 working edge ; but for the convenience of being able to draw a long 
 line across the board at right angles to its lower edge, this edge is made 
 truly square with that on the left side of the board. 
 
 Fig. A 
 
 A further improvement in such a drawing-board as above described 
 is made by cutting lengthways a series of narrow grooves in the back 
 of it and inserting in its working edge a strip cf ebony, to help in 
 keeping it true, and to serve as a guide to the stock of the drawing 
 square. Such an improved board is shown in Fig. A. There are 
 other kinds of drawing-boards in use ; but as the one described has 
 stood the test of many years' service, and finds most favour in 
 drawing offices, a detailed description of them is not necessary here. 
 A reason for giving at such length a description of the kind of 
 drawing-board so universally in use in modern engineering drawing 
 offices is that it may be the means of inducing students and apprentices 
 capable of handling joiners' or patternmakers' tools to make such a 
 board for themselves, which, if made of good well-seasoned pine, free 
 from knots and shakes, will retain its specially good features for years. 
 Those, however, who may be unable to accomplish such a feat, may 
 purchase such boards at a reasonable price from manufacturers of 
 drawing materials, who make them a speciality. 
 
 2. Tee-Square. The next most important adjunct to the drawing- 
 board is the drawing- or tee-square. Some inexperienced youths, and 
 even those of larger growth, have a notion that anything will do for a 
 tee-square ; but, if correct work is to be done, the tee-square is as import- 
 ant to the draughtsman as the drawing-board. It need not, however, 
 be an expensive one, provided a knowledge of what constitutes a really 
 serviceable and efficient tool is possessed by its intended user. As its 
 
MECHANICAL AND ENGINEERING DRAWING 
 
 Vi 
 
 
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 Fig. 1 
 
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 Fig. 3 
 
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 i^7^\vv^ ^x?<5kv\<^ 
 
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 Fig. 
 
 Fig. 4 
 
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 Fig. 6 
 
4 FIRST PRINCIPLES OF 
 
 name implies, it is an instrument in the form of the letter T, the two 
 parts of it being known as stock and blade, the horizontal component of 
 the letter being the stock, and the vertical one the blade. To form the 
 square, the two parts are joined together in such a way as to make them 
 exactly at right angles to each other ; the stock, which is applied to the 
 working edge of the drawing-board, being about one-third the length of 
 the blade, and about three times its thickness. 
 
 The manner, however, in which the stock is united to the blade 
 determines its adaptability or otherwise to the use made of it by the 
 draughtsman. In some the stock is rectangular in section, and the 
 blade morticed into it, as in Fig. 1. In others the blade is dovetailed 
 and let into the stock for the whole of its thickness, as in Fig. 2 ; or 
 morticed, as in Fig. 1, but fitted with a tongue-piece the length of the 
 stock, as in Fig. 3. Neither of these plans is to be recommended ; 
 they involve unnecessary work and care in fitting during their manu- 
 facture, are more liable to get damaged in their usage, and are 
 practically imperfect as a tee-square in some of its essential require- 
 ments. To be perfect in construction, a tee-square should be as light 
 as is consistent with its necessary strength and stiffness of parts ; it 
 should be made of suitable material, easily manufactured, put together, 
 and repaired, and withal as truly correct as it is possible to be made. 
 Such a square is represented in Fig. 4 ; it has a taper blade, which is 
 generally about double the width where secured to the stock as it is at 
 the tip. Its tapering form serves two purposes, the primary one being 
 that'it adds strength and stiffness to the blade and prevents its buck- 
 ling a common fault with all parallel-bladed squares and the other, 
 its excess of width at the stock, prevents it from rocking, and gives 
 amjAe room for securing it to the stock. The blade is also easily and 
 correctly fitted to the stock, and has also one great advantage over all 
 the others in that the set-squares used with it are far more easily manip- 
 ulated than is possible with any of the three previously referred to. 
 
 3. In cases where many parallel lines have to be drawn, of lengths 
 beyond the capabilities of ordinary set-squares, and in directions other 
 than square with, or parallel to, the working edge of the drawing- 
 board, it is convenient to have for use an adjits'able-bladed tee-square, 
 or one whose blade can be set at any desired angle. The blade of such 
 a square should be tapered as in Fig. 4, but shaped at its wide end as 
 in Fig. 6, and have a stock wide enough to allow for the surface 
 required in the washers of the fittings necessary to make the blade 
 adjustable. These fittings, though requiring to be well made and 
 neatly finished, are not expensive or difficult to make, as they consist 
 merely of two washers, a square-necked bolt, and a fly-nut, articles 
 that any one capable of making a pair of calipers could supply himself 
 with. Fig. 5 shows a section of these fittings, which are generally 
 made in brass. The top and bottom washers A, B, are slightly dished 
 on their faces to ensure contact with blade and stock, and the spread 
 of the wings of the fly-nut is such as to give sufficient leverage for 
 a good grip. 
 
 Reference is here made to an adjustable-bladed square, as one may 
 possibly be required later on by the student ; but there is no present 
 
MECHANICAL AND ENGINEERING DRAWING 5 
 
 necessity for the provision of such a tool, as all lines that may be 
 required other than those drawn with the tee- and set-squares in 
 conjunction, are easily put in by a proper manipulation of the set- 
 squares, which will be explained in due course. 
 
 Set-Squares. Of the set-squares used conjointly with the tee- 
 square, those of 45 and 60 are all that are required by the student in 
 the earlier stages of study. A 6-in. 45 and an 8-in. 60 set-squares 
 are the most useful sizes. Framed ones, well made, of foreign manu- 
 facture, may now be obtained at a reasonable price, but the kind most 
 generally in use are made of vulcanite. Those, however, of this 
 material made with the middle part cut out to imitate framed wooden 
 ones should be avoided, as they are very liable to fracture at the 
 angles, and it is impossible to repair them. 
 
 The other requirements of the student of mechanical graphics, 
 apart from what are known as instruments, are some pencils, drawing- 
 pins, rubber, paper, and ink. A few words descriptive of the qualities 
 that should obtain in each of these articles, that satisfactory work may 
 be done, will be of advantage to him. 
 
 Pencils. The present great demand for pencils has, notwith- 
 standing the millions that are annually made and sold, added few to 
 the number that are specially suited to the wants of the mechanical 
 draughtsman. Many erroneously assume that any sort of pencil will 
 suit a learner. No greater mistake can be made. If he is to acquire 
 a draughtsman's habit of work, his first necessity will be a good, 
 serviceable, reliable pencil one that is neither too hard nor too soft, 
 and that will retain a good point for a considerable time. The pencils 
 now generally used in drawing offices are of Faber's make, which can 
 be had of different degrees of hardness from H to six H's, the cedar 
 covering of the lead being hexagonal in form, instead of round. But 
 such pencils are too expensive for students' use. A good, serviceable 
 pencil, made by Cohen, and known as the " Alexandra H pencil," has 
 been in use by the writer for some years, and costs about half the price 
 of Faber's. They are, however, of the ordinary round form, which is 
 inimical to the draughtsman, it tending to cause them to be constantly 
 rolling off his board and damaging their points. To obviate this, the 
 writer's practice is to cut a flat side on the pencil throughout its whole 
 length, taking care not to bend the pencil in doing this for fear of 
 breaking the lead. If neatly done a perfectly flat side is produced, 
 which serves as a guide to the way in which the pencil should be 
 pointed and held, and will prevent any tendency to rolling, even if the 
 drawing-board is much inclined. To do away, however, with the 
 necessity for constantly sharpening the pencil, and thereby reducing 
 its length at every such operation, pencils with movable leads have 
 been in use in drawing offices for some years. They are far to be 
 preferred, as the part of the pencil which is held by the fingers never 
 alters in length, and the lead can be used to the last quarter of an 
 inch. This kind of pencil is known as "Faber's artist's pencil," is 
 hexagonal in outside form, and thus partly prevented from rolling. 
 The acme of perfection in this class of pencil has, however, only lately 
 been introduced, the part for holding the lead being triangular in 
 
6 FIRST PRINCIPLES OF 
 
 section, which renders it easy to hold without turning in the fingers, 
 and rolling off the drawing-board is impossible. It is made by Hardt- 
 muth, of Vienna, but can be purchased of any photographic chemist 
 or artists' colourman. 
 
 4. Drawing-Pins. In the study of mechanical drawing in its 
 
 earlier stages, and even in the making of working drawings for shop 
 use, it is not necessary or essential that the paper on which the 
 drawing is made should be secured to the drawing-board in any other 
 way than by pinning it. This is effected by the use of drawing-pins. 
 There are, however, several kinds of drawing-pins to be had, and their 
 variety is often the cause of difficulty in choosing, to the uninitiated 
 user. A pin that would answer well the purpose of the free-hand 
 draughtsman in putting a sheet of paper on his drawing-board, might 
 be the very worst that a mechanical draughtsman could possibly use. 
 The former, not needing a tee-square in the practice of his art, if he 
 does not stretch his paper, pins it down to his board with any drawing- 
 pins that are at hand. These may possibly be pins with heads a 
 sixteenth of an inch thick, beautifully milled on their edges and 
 perfectly flat on their under and upper sides. Such pins would be 
 shunned by any mechanical draughtsman who wished to keep the 
 edges of his tee- and set-squares intact and free from notches. Pro- 
 jecting the whole thickness of their heads from the surface of the 
 paper, they would foul the edges of the tee- and set-squares and cause 
 damage. The only kind of drawing-pin a mechanical draughtsman 
 should use should have a head as thin as possible, without cutting at 
 its edge, slightly concave on the under side or that next to the paper, 
 and only so much convexity on its upper surface as will give it 
 sufficient central thickness to enable the pin to be properly secured to 
 it. There is neither sense nor reason in making the head of a drawing- 
 pin half-an-inch in diameter if its circumferential edge does not bear 
 on the paper when its pin is as far into the board as it will go. The 
 purpose of the pin is to keep its head from rising from the surface of 
 the paper, and it need only be long enough and strong enough to effect 
 this. It is better practice to use four small, good-holding drawing- 
 pins as shown in Fig. B, along the edge of a sheet of paper, than one 
 large, clumsy, badly-made pin at each end of it. Suitable drawing-pins 
 which answer every purpose required of them by the draughtsman are 
 now to be obtained for half-a-crown per gross. 
 
 5. Paper. As the student from the very commencement of learn- 
 ing to draw should study to acquire the good draughtsman's habits of 
 work, and as one of these is the making of clear, sound lines in his 
 drawing, whether in ink or pencil, it is advisable that he should 
 accustom himself to draw on fairly-good paper. It is not meant by 
 
MECHANICAL AND ENGINEERING DRAWING 7 
 
 this that such paper as Whatman's is recommended for use in his 
 preliminary work, but rather to guard him against purchasing soft, 
 spongy paper, which will not stand the application of indiarubber for 
 erasing, or of ink for lining in, without damaging its surface or causing 
 the ink to run. Drawing-papers are of two kinds viz., hand-made and 
 machine-made. The former is the best, but is expensive ; while the 
 latter is made in great variety, and, as a consequence, of varying 
 quality. Most students, in learning to draw, require a frequent use of 
 the rubber ; therefore a tolerably hard-faced paper is desirable. Since 
 the advent of so much drawing as now obtains, a new special make 
 of hard-faced, close-textured cartridge paper has been produced for 
 students' use. Lt costs abo.ut 2#. per imperial sheet, and is very 
 suitable for 'the mirposS. ""'For more," advanced work there is, to the 
 writers knowledge, nothing that will compare with Whatman's smooth 
 double-elephant paper, which takes the finest line either in ink or 
 pencil. 
 
 Rubber. For cleaning drawing-paper, a piece of soft, grey vulcanized 
 rubber should be used, as it will not injure the surface of the paper if 
 properly applied. Its only drawback is the appearance at times in it 
 of small specks of some hard substance like coke-dust, which find their 
 way into it during the process of manufacture ; these, however, are 
 easily removed when detected. For erasing any portion of a line in 
 pencil, a piece of prepared white vulcanized rubber is the best small 
 rectangular pieces of this material are now to be had of any artists' 
 colourman. What are called pencil-erasers, or rubber-sticks-, are now 
 in common use amongst draughtsmen for the same purpose. They are 
 made in the form of a large square pencil, with rubber inserted in the 
 body of it. To use it the wood is cut away, as is done in pointing an 
 ordinary pencil, exposing the rubber, and it is then applied to the 
 pencilled line with a to-and-fro motion of the hand, pressing lightly, 
 until the line disappears. 
 
 Fig. C 
 
 Ink. A further and all-important requisite to the student draughts- 
 man is ink with which to line in his drawings after they have been 
 carefully put in in pencil. We say this is an " important " requisite, 
 because so much depends on its quality. It is generally known as 
 India or China ink. The definition given of it in standard dictionaries 
 viz., "a substance made of lampblack and animal glue" is no doubt 
 answerable for the large amount of a material made and sold in 
 Britain under its name. Pure India or China ink is only made in 
 those countries, because the special wood from which it is produced is 
 found only in those regions; therefore in purchasing ink for use en 
 
8 
 
 FIRST PRINCIPLES OF 
 
 drawings, the only way to ensure its being the real article is to obtain 
 it from a bond fde importer. The best mathematical instrument- 
 makers are generally importers of it. It is sold in hexagonal sticks, 
 as shown in Fig. C, and is expensive, but small oval and round sticks 
 of it are to be had costing about a shilling each. 
 
 6. Before noticing the few "instruments" that are necessary when 
 commencing the study of mechanical drawing, we think it advisable to 
 show, in a combined sketch (Fig. D), the special tools viz., drawing- 
 board, tee- and set-squares recommended, that the student may note 
 the position they each should assume when in use. The tee-square 
 should only be used in the two positions indicated by its outline in full 
 and dotted lines. In the latter it will seldom be required. All lines at 
 right angles to its edge, when in the first position, should be drawn 
 with the 60 set-square applied, as shown. The 45 set-square is placed 
 as it would be applied when a line is required at that angle near the 
 left edge of the sheet of paper. Care should be taken, when drawing 
 by lamp or gaslight, that the light is in such a position as to cause little 
 or no shadow to be cast on the paper by the edges of the tee- or set- 
 squares. This is important, as such shadows often cause errors in lining 
 in, whether in ink or pencil. We may mention that the drawing-board 
 and tee-square recommended for use are known as "Stanley's 
 Improved," they having been introduced many years ago by Mr. W. F. 
 Stanley, of London. 
 
 Fly. D 
 
 Instruments. The Drawing Instruments required by the student 
 draughtsman are few in number, and should be acquired as the neces- 
 sity for their use arises. No greater mistake can be made than that of 
 purchasing a " box of instruments," as it generally contains some articles 
 that are never required, and is wanting in those that are necessary for 
 the special kind of drawing practised. All that the" student requires for 
 use for some time is a pair of 6-in. compasses with a pen-and-pencil leg, 
 
MECHANICAL AND ENGINEERING DRAWING 9 
 
 pen-and-pencil bows, a ruling- or drawing-pen, and a set of drawing- 
 scales. For future service, everything depends on a .proper choice in 
 their purchase, more particularly if their use is to be continuous ; 
 and as we assume throughout this work that the student has little, 
 if any, previous knowledge of the subject, it is especially necessary 
 that he should know what constitutes a good serviceable instrument, 
 as the possession of inferior ones will be a constant source of annoyance 
 to him. 
 
 Fig. 7 
 
 7. In giving the characteristics of a good instrument, it is of the 
 first importance to understand the use to which it is applied. With 
 draughtsmen, a pair of compasses and a pair of dividers serve two very 
 different purposes, and are therefore differently constructed, but their 
 names and uses are often misunderstood. " Compasses " are never used 
 for dividing, nor are "dividers" applicable to compass- work. Beginners 
 should therefore note that the former are specially intended for putting 
 in circular lines in pencil or ink, and that the proper and only use of 
 the latter is the division or measuring-off of lines and spaces. These 
 separate and distinct purposes give at once a clue to their proper form 
 and construction. They are both instruments with two movable legs, 
 joined together by a forked end, and secured by a pin and washer, as 
 shown in Figs. 7 and 8 at A, A. The compasses, however, being used 
 to draw circular lines, or lines described about a point everywhere equi- 
 distant from it, should have jointed legs, one with a knee-joint at B, and 
 the other with a socket, as at C, to enable it to be easily removed 
 and replaced by the ink- or pencil-points D, E, Fig. 10, when required. 
 The purpose of the knee-joints shown at B in the compasses, and b b 
 in the pen and pencil points, is to enable the lower parts attached to 
 them to be adjusted perpendicular to the surface of the paper, in 
 order to obtain a truly circular line, and to allow both nibs of the 
 inking-point to bear fairly upon it. 
 
 Dividers, which are not necessary to the student for some time 
 forward in his study, should have legs of equal length, but without 
 joints, as in Fig. 8, their lower parts being made of steel of triangular 
 section to within f in. of the ends, which should be gradually worked 
 off into nicely-rounded points, as shown. This latter feature is one that 
 
 Fig. 8 
 
 should obtain in the points of compasses, bows, etc. Triangular-pointed 
 instruments should never be used, as their points act the part of a 
 
10 
 
 FIRST PRINCIPLES OF 
 
 rimer, cutting their way through the paper into the drawing-board, 
 making unsightly holes, and causing them to describe anything but true 
 circles. 
 
 Fig. 9 
 
 Pen-and-yencil Bows are compasses intended for putting in smaller- 
 circles and circular arcs. Single- jointed ones, such as are shown in Fig. 
 9, will serve all the present wants of the student, if well made. The 
 socket in the pencil-bow should be tubular, and of a size to take leads, 
 and not lead-pencils. As these two instruments will be much oftener 
 used than any other, it is advisable that the student should supply him- 
 self with the best to be afforded, as they will amply repay any present 
 outlay. 
 
 What are known as " half sets," shown in Fig. 10, are now specially 
 made by drawing-instrument makers, for the use of students. They 
 comprise compasses, lengthening bar, pen and pencil point, and knife 
 key, and are a very serviceable outfit if well made. 
 
 Pencil Point D. 
 
 Fig. 10 
 
 In selecting the foregoing instruments, care should be taken that 
 they are all sector- jointed with double-leaves and well made ; there 
 should be no shake or slackness in any of them, and they should be 
 equally stiff in the joints at any point from being full open to closing. 
 The test for a pair of compasses is to open out their legs well apart and 
 then to fold each lower half-leg together if the points meet each other 
 truly, they are correct in the joints ; if they cross one another, the joints 
 are not properly made. 
 
 Drawing or Ruling-pens are of two kinds viz., those made with a 
 jointed nib, as in Fig. 11, and those without a joint, as in Fig. HA. The 
 former, though more expensive, is to be preferred, on account of the 
 facility in cleaning and sharpening ; but the latter is a very serviceable 
 
MECHANICAL AND ENGINEERING DRAWING 
 
 11 
 
 pen, if well made and finished. It will be observed in the sketch of the 
 first, that the under or fixed nib is much straighter and thicker than the 
 hinged one ; this is so made to resist the pressure of the hand upon it 
 when drawn along the edge of the tee- or set-squares. In all ordinary 
 pens the nibs are of equal thickness, and the hand-pressure tends to 
 close them and prevent the flow of the ink ; but by providing a stout 
 springless inner nib this tendency is overcome. The stem or handle of 
 this pen, it will be noticed, is squared, to indicate how it should be held 
 by the fingers when in use. 
 
 Fig. 11 
 
 X> 
 
 Fig. lla 
 
 The Drawing-scales recommended for present use by the student are 
 a set of three lately introduced by Messrs. Jackson Bros., of Leeds, 
 made of pliable varnished beech, and giving twelve scales of the 
 standard units of measurement generally used in engineering drawing. 
 They are decidedly to be preferred to any cardboard-scale, as the dividing 
 is well done and there is no tendency to double up or get dirty by use. 
 When the student acquires a more perfect knowledge of the use 
 of instruments and scales, he can add to his stock already in possession 
 whatever is necessary, always bearing in mind that a good tool in 
 the hands of one who knows how to use it will invariably do 
 better w^ork, and is to be preferred to one of inferior quality ; in 
 the meantime, those herein recommended are all-sufficient for present 
 requirements. 
 
CHAPTER II 
 
 MECHANICAL AND FREEHAND DRAWING : THEIR DIFFERENCE AND USES 
 
 8. BEFORE proceeding with an exposition of the principles on which 
 the practice of mechanical drawing is based, it is necessary that the 
 student who is assumed to have no previous knowledge of the subject 
 should thoroughly understand the radical difference, in character and 
 application, which exists between it and that kind of drawing known as 
 " freehand." 
 
 The generic term " drawing," strictly speaking, is the art of 
 representing objects on a surface generally flat by means of lines 
 showing their forms and general contour, independent of colour or 
 shading ; for the latter, without form, would be meaningless and 
 incapable of expressing anything. Freehand drawing is the practice of 
 the art of drawing by means of the hand, the eye alone controlling and 
 guiding the tool or instrument used for delineation. The hand guided 
 by the eye can, however, only picture or draw what is seen from one, 
 position at a time ; for were it otherwise, a distorted view of the object 
 would be the result, as its appearance to the eye from one point of view 
 would be different to that from any other. 
 
 All objects are made visible to the sense of seeing by the agency 
 of light, whether natural or artificial, for without light it would be 
 impossible to distinguish one object from another. To the artist or 
 draughtsman, light is a stream of matter given off by a luminous 
 body, travelling from its source in thin straight lines or rays to 
 the object illumined, from which it is reflected or transmitted in the 
 same way to his eye. What is seen, or is apparent to his sense of sight, 
 he depicts or draws on his paper. If he changes his position with 
 respect to the illumined object, he sees it differently, and obtains a 
 different view of it ; each such view, if correctly drawn, is known 
 as a " perspective," and would agree with that obtained in the following 
 manner. 
 
 In the diagram (Fig. 12) let HP represent a flat surface, such 
 as a piece of ground or a floor, exposed to sunlight, and VP a sheet 
 of glass set up on HP, in a vertical position. At any distance to 
 the left of YP, and parallel to it, is erected a piece of fencing OO, 
 1 laving its top and bottom edges parallel to HP, and its side edj?es 
 
 12 
 
MECHANICAL AND ENGINEERING DRAWING 
 
 13 
 
 1- 
 
 Fig. 12 
 
14 FIRST PRINCIPLES OF 
 
 perpendicular to it. At a given distance to the right of VP, and 
 perpendicular to HP, a staff S, surmounted by a small rectangular 
 plate of any opaque material, and pierced with a sight-hole is fixed ; 
 the height of this sight-hole from HP being supposed to equal 
 that of an observer's eye from the ground. The sheet of glass 
 VP being transparent, it is evident that the spectator, on looking 
 through the sight-hole, will see the whole of the piece of fencing, 
 and can judge of its appearance from the position occupied by his eye. 
 If he wish for a record of this appearance he can obtain it by drawing 
 on the glass what he sees through the sight-hole. The view he 
 would get would be a perspective of the original object OO, or 
 the fence. But its contour or outline on the glass, although similar, 
 would be much smaller than its original. How much smaller, 
 would entirely depend upon the distance between the eye at sight- 
 hole, the sheet of glass VP, and the fencing OO. It is evident that 
 the nearer VP is to OO, the eye remaining in the same position, the 
 larger would be its image or picture upon VP, and the converse of this 
 would obtain were the conditions reversed. 
 
 It will be seen from the diagram that the perspective view of the 
 original object is obtained by finding where the luminous or visual rays 
 represented by broken lines proceeding from its principal points, 
 are intercepted on VP in their passage to the eye, and then joining 
 such points by right lines as in the original. Now, as these visual rays, 
 or " projectors," are the means by which the view of the object is 
 projected or thrown on VP, such a view is called a " projection," and in 
 the special case we are considering a " perspective projection." In such 
 a delineation it is apparent that all rays proceeding from the visual 
 points in the object form a pyramid, the vertex of which is the point 
 where they meet in the eye ; and from this fact it will at once be 
 seen that a perspective drawing of an object can serve no other 
 practical purpose than that of showing its appearance when viewed 
 from a certain fixed position, for its boundary lines altering with 
 the altered position of the spectator, it is difficult to determine their 
 actual lengths, as they only bear a relative proportion to their originals. 
 As they cannot be measured with an ordinary rule or scale, it would 
 be impossible to construct a machine or erect a building from such 
 drawings. In perspective drawing, HP in the diagram is known as 
 the horizontal or ground plane, and VP the perspective or piciure plane, 
 which latter is always supposed to be transparent, although actually 
 represented by the artist's sheet of drawing-paper or canvas. 
 
 9. As, then, a perspective, or freehand drawing, does not fulfil 
 the requirements of the workman, in that he cannot determine at sight 
 the actual form, dimensions, or arrangement of the piece of work he is 
 called upon to execute, some other method of delineation becomes a 
 necessity. This want is supplied in what is generally known as a 
 " mechanical drawing," or a drawing obtained by the correct application 
 of the principles of a kind of projection called " orthographic," which 
 gives results differing widely from that already explained, in that it 
 affords a means of at once determining the actual form, size, and 
 disposition of every part of the object represented, and gives an adept 
 
MECHANICAL AND ENGINEERING DRAWING 15 
 
 in the application of its principles the power to commit to paper the 
 entire design of a machine or engine, that will enable the engineer 
 or machinist to determine at sight whether the stationary and working 
 parts of one or the other are disposed in such a way as to meet the 
 requirements for which they were designed. In fact, a mechanical 
 drawing is the only efficient way of describing by means of lines,- 
 properly disposed according to fixed rules, the actual construction and 
 arrangement of a piece of mechanism. 
 
 As " orthographic " projection is the basis of mechanical and 
 engineering drawing, its difference as compared with perspective 
 projection must be understood before the study and application - of 
 its principles are entered npon. An important consideration in 
 connection with either kind of projection is, that the bodies, or objects, 
 whose forms it is wished to depict on paper, are in all cases assumed 
 to be illumined by solar light, and have the power of reflecting or 
 throwing off the light that is cast upon them. As the source of 
 light or the sun is at a comparatively infinite distance from all 
 objects illumined by it, its rays will not sensibly diverge, or approach 
 each other, but may be regarded as exactly parallel among themselves. 
 Then if, instead of the rays from an illumined object being reflected 
 so as to converge in the eye as in perspective projection they be 
 conceived as travelling from the object in parallel lines, till intercepted 
 on a plane surface at right angles to themselves, and the points of 
 interception be joined by straight or curved lines, the representation 
 thus formed on that surface will be an " orthographic " projection 
 of the original object. In this case the visual or projecting rays, 
 being always parallel to each other and perpendicular to the surface 
 on which they are projected, form a prism ; and it follows, that, 
 however far that surface is from the object, its representation remains 
 the same, and the projected length of all its lines parallel to that 
 surface will be of the same length as in the orginal, and therefore their 
 exact dimensions can be at once ascertained. 
 
 It will be understood from this explanation that instead of the eye 
 being stationary and viewing the object from one point alone, as in 
 perspective, it is in orthographic projection supposed to move in such, 
 a way as to be directly opposite to each of the principal points of the 
 object, the projecting rays from it being always perpendicular to the 
 plane on which its image is projected. It is manifest, however, that in 
 this way only one projection of an object is obtained; but as any 
 solid body has more than one dimension, it becomes evident that more 
 than one view of it must be given before its other dimensions can be 
 ascertained. To this end it is usual to determine its projections on 
 two planes, which are always at right angles to each other, and from 
 these correct and definite ideas as to its shape and dimensions may 
 at once be obtained. 
 
 10. To illustrate the foregoing diagrammatically, let HP (Fig. 1 3) 
 be a horizontal plane, and YP another plane at right angles or 
 perpendicular to HP. At any distance from YP, and in front of 
 it, a rod R is set up perpendicular to HP, supporting on its upper end 
 a bar F of rectangular section and a given length. Yisual rays or 
 
16 
 
 MECHANICAL AND ENGINEERING DRAWING 
 
 projectors parallel among themselves and perpendicular to VP are 
 shown proceeding from the corners A, B, C, I), of the bar penetrating 
 VP in a, 5, c, d. As the edges of the original object, or the 
 rectangular bar F, are all straight, it follows that if a, b, c, d on VP 
 be joined by straight lines, an orthographic projection of the face 
 A, B, C, D, of the bar will have been obtained, which will, on 
 measurement, be found to be an exact counterpart of it. But this 
 projection only gives the length and depth of the bar; and as it is 
 necessary to know its other dimension, or width, a view showing that 
 dimension must be obtained. Now it is evident that a view of the 
 bar, looking at it from above and in the direction of the arrow, will 
 
 Fig. 13 
 
 supply the information required. If, then, visual rays, or projectors, 
 proceed as before from the four corners of the face of the bar seen 
 from above, to the plane HP below, they w r ill penetrate that plane at 
 the points a, b', e,f, and these points being joined as before as the 
 same conditions obtain there is produced on HP an orthographic 
 projection of the top face of the bar which determines its width. 
 With these two projections, or views of the original, it will be seen that 
 a workman could produce any number of such bars without the 
 assistance of a model or other guide. To distinguish the two pro- 
 jections of the same object, the one obtained on VP is known as an 
 " elevation " or vertical projection, and that obtained on HP is called 
 a " plan " or horizontal projection. 
 
CHAPTER III 
 
 PRACTICAL GEOMETRY AND MECHANICAL DRAWING 
 
 As it has been necessary, in explaining the difference between a 
 .mechanical and a freehand or perspective drawing, to make use of 
 terms which pre-suppose a knowledge of geometry by the student which 
 he may not possess, and as it is advisable to take nothing for granted 
 in the exposition of our subject, it will be necessary at this stage to 
 define the meaning of the geometrical terms that will be made use 
 of as we proceed, and to show /now, by a special combination of 
 lines, those geometrical figures are constructed which form the 
 surfaces of objects whose delineations are subsequently to be obtained 
 by orthographic projection. 
 
 The term "geometry," in its generally-accepted sense, means the 
 science or knowledge of magnitude reduced to system, and has to do 
 with the measurement of lines, surfaces, and solids. It has, like other 
 sciences, two sides or branches, one " theoretical," which demonstrates 
 or proves its principles, and the other " practical," or that which 
 applies those principles to construction. Theoretical geometry, or 
 Euclid, will seldom be referred to in the course of this work, as 
 most of the demonstrations used are self-evident; but practical 
 geometry a sub-divison of which is the basis of our subject must 
 be understood by the student to such an extent as will enable him 
 to work out the problems that will arise in the exposition of it. 
 
 The two parts into which practical geometry is divided are : Plane 
 geometry, which has reference only to the solution of questions relating 
 to points, lines, and figures, situated in one plane ; and solid geometry, 
 which shows by special representations on two or more defined planes, 
 the relations of the points, lines, and surfaces of bodies having length, 
 breadth, and thickness. 
 
 We would, in passing, guard the student, on his entering on the 
 study of geometrical drawing, against w r asting valuable time in 
 working out the problems many of which will be of no use to 
 him given in most text-books on the two subjects of plane and 
 solid geometry, as all that is absolutely necessary for him to know 
 in connection with either will be explained to him as occasion arises. 
 
 As we cannot form a conception of the magnitude of any material 
 
 17 c 
 
18 FIRST PRINCIPLES OF 
 
 object without reference to one or more of its dimensions, and as each 
 of these involves the idea of extension in some direction, the word 
 length, or its representative, "a line," would appear to be the first 
 term used in geometry requiring definition, but as a line can have 
 no existence till it is generated or drawn, our first term must be that 
 of the generator, or " point." We therefore define 
 
 A point, as having no magnitude, that it is used to denote 
 "position" only, and is represented geometrically by a dot or mark 
 made by any pointed instrument, such as a pen, pencil, etc. 
 
 A line, as the path made by a point moving over a surface. 
 It may be straight, crooked, or curved, according to the direction in 
 which the point travels or moves. 
 
 A straight line, as the shortest path that can be made by a point 
 moving from one position to another, or the nearest distance between 
 two points, as the line A between points 1 and 2. 
 
 A crooked line, as the path of a point that has changed its direction 
 after moving in a straight line for a given distance 1 to 2, as the 
 line B from 1 to 3. 
 
 A curved line, as the path of a point that continually changes its 
 direction, as the line C from 4 to 5. 
 
 If the path of a moving point changes in such a way as to enclose a 
 certain amount of surface, then the enclosed surface is called a " figure," 
 and the path its boundary line, as in Fig. 14, where the "point path" 
 from a, through b, c, d, defines the form of the figure. 
 
 If a point move continuously in such a way as to be always at a 
 given distance from some fixed point, then the surface enclosed by the 
 " point path " becomes the figure called a " circle," as the continuous 
 line ABC (Fig. 15), any point in which is equi-distant from D, which is 
 called its centre. 
 
 It is evident from the foregoing that two straight lines cannot enclose 
 a surface, or form a figure, but that one such line in combination with 
 a curved or a crooked line will effect this, as shown in Figs. 16 and 17, 
 where we have in one case a straight line and a crooked one, and in 
 the other a straight and a curved line, combined to form figures. 
 
 A surface is a magnitude that has extension in two directions only 
 viz., lengthwise and crosswise. Its dimensions with one or two 
 exceptions are given as length and breadth. 
 
 A plane surface, is one that a perfectly straight edge will touch 
 or coincide with if applied to it in any direction. A mathematical or 
 perfectly true plane does not exist it can only be imagined. 
 
 Parallel straight lines are the point paths of two lines on a plane 
 surface that are everywhere equi-distant from one another, as the lines 
 A and B. 
 
 Converging straight lines are the point paths of two lines on a plane 
 surface, which, if continued, meet and cross each other as the lines C, D. 
 When the paths or lines increase their distance from each other as they 
 leave the meeting point, they are said to diverge. 
 
 An angle is formed when two straight lines meet each other in a 
 point, as D meets F in d (Fig. 18). If the inclination of one line to 
 the other be such that the angles are equal on both sides of the meeting 
 
MECHANICAL AND ENGINEERING DRAWING 19 
 
 
20 FIRST PRINCIPLES OF 
 
 point, then the angle formed by the lines D and F is a right angle. If 
 they are not equal, as in the meeting of E and F in d, then the smaller 
 of the two, or angle a, will be an acute angle, and the larger, or angle b, 
 will be an obtuse angle. And as the line D makes equal angles on both 
 sides of it with the line F, the two lines D and F are perpendicular to 
 each other. 
 
 11. As angles cannot be measured without a knowledge of the parts 
 and divisions of the circle (Fig. 15), we must, before giving further 
 definitions of plane figures, explain these. The boundary line ABC of 
 this figure is called its circumference. Any straight line drawn through 
 D, its centre, and touching the circumference on both sides, is a diameter. 
 Half of such a line is called a radius. Any portion of the circum- 
 ference, such as from A to B, would be an arc, and a straight line 
 joining A and B the chord of that arc ; the space enclosed by the arc 
 AB and its chord is called a segment, and that by the arc AC and the 
 two radii AD, CD, a sector. One quarter of the whole figure or circle 
 is a quadrant, and one half of it a semi-circle. For the measurement of 
 angles, arcs, chords, etc., the circumference of every circle is supposed 
 to be divided into 360 equal parts called degrees, which are indicated 
 when speaking of them by attaching a small circle to the right of the 
 number stated as 30, 60, etc. (or 30 degrees, 60 degrees, etc.). A 
 quadrant, therefore, contains 90 degrees, and a semi-circle 180 degrees. 
 The size of any angle is determined by the number of degrees contained 
 in the arc subtending the angle, described about the angular point d, 
 as a centre (Fig. 18). The complement of the angle a is the number of 
 degrees it is wanting to make it a right angle, and its supplement is the 
 number of degrees contained in the angle b. 
 
 Circles are concentric when they have the same centre as in Fig. 19, 
 and eccentric when their centres are different, as in Fig. 20. 
 
 A tangent is a straight line which touches a circle or a curve in one 
 point, and when produced does not cut it, as in Fig. 21. 
 
 Tangent circles and tangent curves are those which touch each other 
 in one point, but do not cut as in Fig. 22. 
 
 The point of contact is that point where a tangent touches a circle 
 or a curve, or where two tangent curves touch, as a in Fig. 22. 
 
 lla. In continuing our definitions of plane figures, we take first 
 those constructed with the least number of straight lines that will 
 enclose a space, which is three. Such figures are called triangles or 
 three-angled, and are named according to the disposition of their sides 
 and quality of their angles. 
 
 An equilateral triangle has equal sides and equal angles, which are 
 all acute, as in Fig. 23. 
 
 An isosceles triangle has two sides equal and two of its angles always 
 acute, the third angle being acute or obtuse, dependent on the length of 
 its third side, as Figs. 24 and 25, the latter having one obtuse angle at a. 
 
 A right-angled triangle has one of its angles a right angle, the other 
 two being acute, as Fig. 26. 
 
 A scalene triangle has three unequal sides, as Fig. 27 ; an obtuse- 
 angled triangle has one obtuse angle, as Fig. 28 ; and an acute-angled 
 triangle has all its angles acute. 
 
MECHANICAL AND ENGINEERING DRAWING 
 
 Fig. 35 Fig. 36 Fig. 37 Fig. 38 
 
9-7 
 
 MECHANICAL AND ENGINEERING DRAWING 
 
 Of four-sided figures bounded by straight lines 
 
 A square has all its sides equal, and its angles right angles, as in 
 Fig. 29. A rectangle has opposite sides pairs, and parallel, and its 
 angles right angles, as in Fig. 30. A parallelogram, or rhomboid, has 
 two pairs of parallel sides, as Fig. 31. A rhombus has all its sides 
 equal, two of its angles being acute, as Fig. 32. A trapezoid has only 
 two sides parallel, as Fig. 33. A trapezium is an irregular figure of 
 four sides, none of which are parallel, as Fig. 34. 
 
 A regular polygon is a figure having all its sides and angles equal. 
 One of five sides is a pentagon, as Fig. 35. One of six, a hexagon, as 
 Fig. 36. One of eight sides, an octagon, as Fig. 37. An irregular 
 polygon is a figure whose sides and angles are unequal. 
 
 The centre of a polygon is a point that is equi-distant from its sides 
 and its angular points. A polygon is circumscribed when all its 
 angular points touch a circle described about it, as in Fig. 38. 
 
 A circle is inscribed in a polygon when its circumference touches all 
 the sides of the polygon, as in Fig. 39. 
 
 Fig. 39 
 
 Fig. 40 
 
 Fig. 41 
 
 A right or straight-lined figure is described about a circle when all 
 the sides of the figure touch the circumference of the circle, as in Fig. 
 40 ; and a straight-lined figure is inscribed in another such figure when 
 the angular points of the inscribed figure are upon the sides of the 
 figure in which it is inscribed, as in Fig. 41. 
 
 A straight line joining the opposite angular points of any four- 
 sided figure is a diagonal. A square or a rectangle has its diagonals of 
 equal length. In a rhombus, rhomboid, trapezium, and trapezoid the 
 diagonals are unequal. 
 
 As an apprenticeship to any mechanical trade cannot be served in 
 a factory or workshop without the apprentice, as he advances in know- 
 ledge and skill in it, being often called upon to line out his own work, 
 it is necessary that he should be able to draw any of the above-described 
 plane geometrical figures on the material he works in, with a straight- 
 edge, a scriber, and a pair of shop compasses, instead of the tee- and 
 set-squares, etc., of the draughtsman. With this fact in view, we shall 
 give in the solution of each of the following problems the simplest 
 possible method of construction, it being undesirable to burden the 
 memory of the student with the many ways of solving them to be 
 found in text-books. 
 
CHAPTER IV 
 
 PLANE GEOMETRY PROBLEMS 
 
 12. As the lines forming the boundaries and determining the forms 
 of the plane geometrical figures previously described, have a certain 
 relative position, it is necessary, before attempting to construct the 
 figures themselves, that we know how to draw geometrically, lines 
 having any defined relation to each other. As this knowledge is 
 generally imparted in the form of problems, with their solutions, we 
 shall adopt the same plan ; but in explaining the constructions do 
 riot confine ourselves to any orthodox method where a simpler one 
 may be used. The student will remember that in solving the sub- 
 sequent problems, only the tools mentioned in the last paragraph 
 of the previous chapter are to be used, as the assistance of either 
 a drawing-board or squares is inadmissible. As it is not always 
 possible to apply a rule or a scale to a line, when we wish to sub-divide 
 it into parts, our first problem is 
 
 Problem 1 (Fig. 42). To divide a given straight line into two equal 
 
 parts. 
 
 Now, if the given line is near the edge of the material on which it is 
 drawn, a different method of construction must be used to that which 
 would be possible if the line were some distance from that edge. In 
 the former case proceed as follows : With a distance greater than half 
 the length of the given line AB, as a radius, and with A and B as 
 centres, describe arcs cutting each other in C, and with a still larger 
 radius than before, and from the same centres describe arcs cutting 
 each other in D ; then the point E, where a straight-edge laid exactly 
 on C and D crosses the line AB, is the middle of the given line, and 
 divides it into two equal parts. In the latter case, with the distance 
 greater than half AB (which may be gauged by eye) as radius, and 
 from A and B as centres, describe arcs on both sides of the line AB, 
 cutting each other in C and F. Then the straight-edge applied to C and 
 F will give E in AB as its point of section, dividing it into two parts 
 of equal length. 
 
 23 
 
24 FIRST PRINCIPLES OF 
 
 Problem 2 (Fig. 43). At a given point C, in a straight line, to erect 
 
 a perpendicular. 
 
 Here the point may be near the middle of the line or near the end 
 of it. If the former, as at C, in the line AB, and if AB be near the 
 edge of the material, proceed as follows : Set off from C, on either side 
 of it, equal distances, as CD, CE, and from D and E as centres, with a 
 radius greater than half the distance between D and E, draw arcs 
 cutting each other in F, then a line drawn through F and C will be 
 perpendicular to AB. If the given point is near the end of the line 
 and the edge of the material, as A in BD (Fig. 44), then from any 
 point a, above BD, and with a radius equal to a A, describe an arc 
 CAT, passing through A, and cutting BD in T. Draw a line from 
 T through a, and produce it till it cuts the arc in C. A line from C 
 through A will be perpendicular to BD at A. 
 
 Problem 3 (Fig. 45). From a given point A, above a straight line 
 BC, to let fall a perpendicular to that line. 
 
 Here the point may be nearly over the middle, or over the end of 
 the given line. If in the first position, with any radius greater than 
 the distance from the point A to the line BC, describe an arc cutting 
 BC in D and E, and from points D and E as centres, with a radius 
 greater than half the distance between D and E, draw arcs cutting each 
 other in a and b ; then a line drawn through the given point A and 
 the intersections of the arcs in a and b will be the required perpen- 
 dicular. If the point is nearly over the end of the given line, as b in 
 Fig. 46 is over AB, from b, draw a line intersecting AB in C, and 
 bisect it in S ; with SC as radius and S as centre, describe an arc 
 cutting AB in D, join b and D, and the line will be perpendicular to 
 AB. The student will notice that the construction in the second cases 
 of Problems 2 and 3 is similar. This arises from the fact that the line 
 drawn to the given point has in each case to be at right angles to the 
 given line, and as the angle in a semi-circle is always a right angle, the 
 problem is to draw a semi-circle that shall contain the three angular 
 points of a right-angled triangle, one of which is the given point in the 
 problem. 
 
 Problem 4 (Fig. 47). To bisect (or divide into two equal parts) a given 
 
 angle. 
 
 When speaking of an angle, it is usual to name it by affixing either 
 a single letter at the angular point, or a letter to each of its lines and 
 the angular point, the one denoting the latter being always the second, 
 In the problem, let BAG be the given angle. With any convenient 
 radius set off from A equal distances on BA and CA in the points D 
 and E, and from these points, with a radius greater than half the 
 distance across from D to E, draw arcs intersecting in F ; a line 
 through F and A will bisect the angle BAG. This construction, it 
 
MECHANICAL AND ENGINEERING DF AWING 
 
 25 
 
 Fig. 
 
 O . 
 
 Fig. 4o 
 
 Fig: 44 
 
 Fig. 4$ 
 
 Fig. 4$ 
 
 Fig. 47 
 
 c 
 
 Fig. 48 
 
 Fig. 49 
 
 Fig. 50 
 
 Fig. 52 
 
 -JB 
 
 =2? 
 
 L' 
 
26 FIRST PRINCIPLES OF 
 
 will be seen, is tantamount to bisecting a line from D to E, and 
 drawing a line through its bisection and point A, the only requisite 
 condition being that the two points D and E in the lines forming the 
 angle must be equi-distant from the angular point A. 
 
 Problem 5 (Fig. 48). To draw a line parallel to a gicen line at a given 
 
 distance from it. 
 
 Here it is evident that if from any two points C and D in the given 
 line AB, arcs be drawn, of a radius equal to the given distance the two 
 lines are to be apart, and a line EF be drawn tangent to those arcs, then 
 the line EF will be parallel to the given line AB. This is the simplest 
 possible solution of the problem, involving the least work, but requires 
 care in drawing the parallel line exactly tangent to the arcs. Another 
 solution, requiring much more work in the construction, is the following : 
 At the points C and D, in line AB (Fig. 48), erect two perpendiculars 
 to AB, and set off on each of them from C and D the distance the 
 parallel lines are to be apart. Through the two points obtained draw a 
 line, and it will be parallel to the given line AB. 
 
 Problem 6 (Fig. 49). Through a point P, to draw a line parallel to a 
 
 given line AB. 
 
 With P as a centre and any convenient radius, describe an arc EC, 
 cutting the given line AB in C, and from C as a centre, with the same 
 radius, draw an arc through P, cutting AB in D. Set off the distance 
 PD on the arc EC, and through P and E draw a line ; it will be parallel 
 to the given line A B. 
 
 Problem 7 (Fig. 50). To draw an angle equal to a given angle A. 
 
 This means that two lines are to be drawn having the same inclina- 
 tion to each other that two given lines have. We must therefore first 
 find the inclination of the given lines. To do this we have only to 
 draw on the given angle an arc of any convenient radius, with A as 
 centre, such as BC. The length of its chord is the distance subtended 
 by the lines forming the angle at the radius AB or AC. If, then, from 
 point a, in the line DE, and with a radius equal to AB, we describe 
 an arc be, and from c set off a distance on be equal to the chord of the 
 arc BC, then a line drawn through b and a will make the same angle 
 with DE that AB does with AC in the given angle, which solves the 
 problem. 
 
 Problem 8 (Fig. 51), To draw a line making a given angle say 60 
 with a given line. 
 
 The solution of this problem involves the relation that the radius of 
 any circle has to the chord of an arc which subtends an angle of 60 in 
 the circle. To solve it, let AB be the given line, and C a point in it at 
 which it is desired to draw a line making an angle of 60 with AB. 
 
MECHANICAL AND ENGINEERING DRAWING 27 
 
 From C, as centre, and with a convenient radius, draw the arc DE, 
 cutting AB in E ; from E with the same radius cut DE in D, then a 
 line drawn through D and C will make an angle of 60 with the line 
 AB. If the circle were completed with the same radius, it would be 
 found, on stepping the radius round it, that it exactly divides it into 
 six equal parts, and as every circle for geometrical purposes (as before 
 explained) is divided into 360, one-sixth of the circle must contain 60, 
 or the angle which the two lines in the problem have to make with 
 each other. Knowing this specific relation subsisting between the 
 radius and the chord of an arc of 60 of a circle, we are enabled to lay 
 down any angle with the assistance of a " scale of chords," which will 
 be found on one of the set of drawing-scales previously recommended. 
 To show its use, let us take, for example 
 
 Problem 9 (Fig. 52). To draw a line, making an angle of, say, 70, 
 with a given line at a given point in it. 
 
 Let AB be the given line, and a the given point in it. From the 
 zero point, on the extreme left of the scale of chords, and with a radius 
 in the compasses equal to the distance from that point to the one marked 
 60 with the arrow over it on the scale, draw with a, on the line AB 
 as a centre, the arc be, cutting AB in c, and from c as a centre, with a 
 radius equal to 70 on the scale of chords, cut the arc be in . A line, 
 drawn through b and a will make, with the given line AB, an angle of 
 70; and so with any other angle, always remembering that from zero 
 to 60 on the scale of chords is the radius with which the first arc in the 
 construction is to be drawn. 
 
CHAPTER Y 
 
 PLANE GEOMETRICAL FIGURES 
 
 13. IT may be noted, before passing on to the construction of the 
 plane geometrical figures which form the surfaces of the plane solids 
 whose projections we shall next show how to obtain, that as the angles 
 most generally chosen for the surfaces of mechanical details are those 
 which contain some multiple of 5, it is not necessary to use even a 
 scale of chords in laying them down on paper or other material, as most 
 of them can be obtained by simple geometrical construction, which has 
 fewer chances of error than even measuring from a scale. A few of 
 such angles are 15, 30, 45, 60, 75, 120, 135, etc., and are thus 
 obtained : For 30, bisect 60 ; for 15, bisect 30 ; for 45, bisect 90 ; 
 for 60, use radius ; for 75, add 15 to 60 ; for 120, mark off radius 
 twice ; for 135, take 45 from a semi-circle. With these simple con- 
 structions committed to memory, and the use of a scale of chords for 
 any angle not easily obtained otherwise, the student will be able to lay 
 down any angle that may be required. We may now proceed with the 
 construction of plane figures, taking first 
 
 Problem 10 (Fig. 53). To construct an equilateral triangle on a given 
 
 base. 
 
 (Note : The base of any triangle is that side of it on which it stands ; 
 the vertex, the point immediately over the base ; and the altitude the 
 height of the vertex from the base.) With the given base AB as a 
 radius, and from A and B as centres, describe arcs cutting each other 
 in C, the vertex, join AC and BC, and the triangle is constructed. If 
 the altitude only be given as CD (Fig. 54) : Then, as the sum of the 
 angles of any triangle are together equal to two right angles, or 180, 
 and as the triangle required is equi-angular, the angle at its vertex will 
 be one-third of 180, or 60. To construct it, draw EF, GH through 
 C and D at right angles to CD, and from C, with any convenient 
 radius, describe a semi-circle cutting EF in a and c ; with the same 
 radius, and from a and c as centres, cut the semi-circle in d and e, 
 draw lines through Cd and Ce, and produce them to meet GH in g and 
 h, then gCh is an equilateral triangle having an altitude CD. 
 
 28 
 
MECHANICAL AND ENGINEERING DRAWING 
 
 29 
 
 c F 
 
30 FIKST PRINCIPLES OF 
 
 Problem 11 (Fig. 55). To construct an isosceles triangle, the base AB 
 and one of the equal sides CD being given. 
 
 With CD as a radius, and from A and B as centres, draw arcs inter- 
 secting in a, join aA and B, and the triangle is constructed. If the 
 base AB and the altitude ab are given (Fig. 56) : Bisect the base 
 AB in b, and at b erect a perpendicular and make it equal to ab, join 
 aA and &B, then AB is the required isosceles triangle. 
 
 Problem 12 (Fig. 57). To construct a scalene triangle, the sides being 
 
 given. 
 
 Take the longest side AB for the base, and with the shortest as a 
 radius, and from B as a centre, describe an arc ; then with the length 
 of the third side as radius, and from A as centre, cut the arc described 
 from B in b, join b and A, and b and B, then A6B is the required 
 triangle. 
 
 Problem 13 (Fig. 58). To construct a square on a given line AB as 
 
 a side. 
 
 Erect at A a perpendicular to AB, and from it cut off AC equal to 
 AB then from C and B as centres, and with AB as radius, draw 
 arcs intersecting at D, join C and D and B and D, and the square is 
 constructed. If the given line be a diagonal and not a side : Bisect 
 the diagonal AB (Fig. 59) in a, by a perpendicular b, a, c, and from a 
 set off ab, ac, equal to a A, or aB, join A6, 6B, Be, cA, and the square 
 is constructed on the given diagonal AB. 
 
 Problem 14 (Fig. 60). To construct a rectangle, the length of two 
 adjacent sides being given. 
 
 Let the line AB be one of those sides. At A erect a perpendicular 
 to AB, and cut off from it in C, a length equal to the other given side ; 
 from B as centre, and with a radius equal to AC, draw an arc, and 
 from C as centre, with a radius equal to AB, draw another intersecting 
 the first in D, join CD and DB, and the required rectangle is con- 
 structed. 
 
 Problem 15 (Fig. 61). To construct a rectangle, a diagonal AB and 
 one side BC being given. 
 
 As the diagonal of a rectangle divides it into two right-angled 
 triangles, if it is made a diameter, and on it a circle is described, the 
 circle will contain the two right-angled triangles which will form the 
 rectangle sought. Therefore, bisect the given diagonal AB in a, and 
 from a, with B as radius, describe the circle ABCD ; from B as a 
 centre, and with BC as radius, cut the circle in C, and from A, with 
 the same radius, cut it in D, join ACBD, and it is the required 
 rectangle. 
 
MECHANICAL AND ENGINEERING DRAWING 31 
 
 Problem 16 (Fig. 62). To construct a rhombus, one of a pair of 
 opposite angles and length of a side being given. 
 
 Let AB be the length of given side, and C the given angle ; at A 
 make the angle BAD equal to angle C, and the side AD equal to AB ; 
 from B and D as centres, with AB as radius, draw arcs intersecting in 
 E ; join EB and ED, and ADEB will be the required rhombus. If a 
 diagonal AB and length of a side AC be given (Fig. 63) : Then, if from 
 A and B as centres, with a radius equal to AC, arcs be struck cutting 
 each other in C and D, and lines be drawn joining A and B to C and 
 D, the figure ACBD will be the required rhombus. 
 
 Problem 17 (Fig. 64). To construct a rhomboid, the lengths of two 
 adjacent sides and one of a pair of its opposite angles being given. 
 
 Let AB be one (the longest) of the adjacent sides, and E one of 
 the opposite angles. At A make the angle CAB equal to the angle E, 
 and cut off AC equal to the shorter adjacent side. From C, with AB 
 as radius, describe an arc, and from B, with AC as radius, describe 
 another cutting the first in D, join ACDB, and it is the required 
 rhomboid. If a diagonal AB (Fig. 65) and the lengths of two adjacent 
 sides be given : Then, with the length of one of those sides as a radius, 
 and from A and B as centres, describe arcs on opposite sides of AB, 
 and from the same centres, with the length of the other adjacent side 
 as radius, describe arcs cutting those first drawn in C and D, join AC, 
 CB,BD,DA, and it will be the required rhomboid. 
 
 Problem 18 (Fig. 66). To construct a trapezium, the length of its 
 sides and one of its angles being given. 
 
 Let AB be the base of the figure or side on which it stands, and C 
 the given angle. At A in AB make DAB equal to the angle C, and 
 let AD equal the length of that side of the figure ; with the length of 
 the opposite side as radius, and from B as centre, describe an arc, and 
 from D as centre, with the length of the fourth side as radius, strike 
 an arc cutting the last in E, join ADEB, and the required trapezium is 
 constructed. 
 
 14. In the construction of the preceding plane figures, the lengths 
 of one or more of their sides, with their relation to each other, are 
 previously known or determined by the given problem. In the case of 
 a regular polygon, the data generally given are its kind, and the length 
 of a side, or a given circle within which it is to be inscribed. The 
 ordinary solution in such cases involves the remembering of certain 
 specific constructions which are liable to be forgotten when most 
 needed. All that is absolutely required to be known for the con- 
 struction of any regular polygon, is the relative position of any two of 
 its adjacent sides, and in certain cases the length of one of them. 
 
 The relative position, or, in other words, the angles made by any 
 two adjacent sides of a regular polygon, are easily determined. The 
 
32 FIRST PRINCIPLES OF 
 
 exterior angle, or that formed by one side with the other produced, is 
 always equal to 360 divided by the number of the sides of the polygon, 
 and the interior angle, or that formed by the meeting of the two 
 adjacent sides, is 180 minus the exterior angle. The angle at the 
 centre (or central angle) of a regular polygon is equal to the exterior 
 angle. With these simple facts committed to memory, the student or 
 apprentice can, with a scale of chords now generally found on all 
 pocket rules, lay down at once on his work any regular polygon 
 having either an odd or an even number of sides. To apply these facts 
 we will take 
 
 Problem 19 (Fig. 67). To construct a regular pentagon with a given 
 
 length of side. 
 
 Here 360 + 5 equals 72, the exterior angle; and 180 ^ 72 = 
 108, the interior. Let AB be the given side, produce it (say to the 
 left) at A, draw the line AC, making an angle of 72 with AB produced, 
 and of a length equal to AB ; bisect AB and AC by perpendiculars 
 intersecting in S, then S is the centre of the circumscribing circle. 
 Describe it, and from C, with AB as a distance, set off on it the points 
 D, E, join CD, DE, EB, and ACDEB is the required pentagon. 
 
 If the pentagon has to be inscribed in a given circle, then from its 
 centre which will be the centre of the pentagon draw any radius 
 as SA (Fig. 67) at S, draw a line making with SA an angle of 72, 
 and cutting the circle in B, join A and B, then AB is one side of the 
 required pentagon ; set off the distance AB from A or B round the 
 circle, and it will give points C, D, E ; join ACDEB, and the pentagon 
 is constructed in the given circle. 
 
 Problem 20 (Fig. 68). To construct a regular Jiexagon with a given 
 
 length of side. 
 
 Here 360 -=- 6 equals 60, and 180 - 60 = 120. Let AB be 
 the given side, produce it, and draw AC, making with AB produced 
 an angle of 60 ; make AC equal to AB, bisect them by perpendiculars 
 intersecting in S, which is the centre of the circumscribing circle ; 
 describe it, and set off the distance AB round it from C, in points 
 D, E, F, join CD, DE, EF, FB, and the required hexagon is constructed. 
 If a hexagon has to be inscribed in a given circle, the central angle 
 will be 60 ; this angle laid down with the centre of the circle as the 
 angular point will give A, B (Fig. 68), points in the circle, and the 
 line joining them will be a side of the hexagon ; step this length round 
 the circle in points C, D, E, F, join AC, CD, etc., and the required 
 hexagon is inscribed in the given circle. As the side of a hexagon is 
 the chord of an arc of 60, and is equal to the radius of the circumscribing 
 circle, that radius set off round the circle will divide it into six equal 
 parts, and if the points of division be joined by right lines they would 
 form the inscribed hexagon as before. 
 
MECHANICAL AND ENGINEERING DKAWlNG 
 
 A. 
 
 ' 
 
 ftp $9. 
 
34 MECHANICAL AND ENGINEERING DRAWING 
 
 Problem 21 (Fig. 69). To construct a reyular octayon, with a given 
 
 lenyth of side. 
 
 Here 360 -f 8 = 45; and 180 - 45 - 135. Let AB be the 
 given side. Produce it in both directions, and at A and B draw lines 
 AE, BF, of the same length as AB, and making with AB produced 
 angles of 45 ; bisect the angles formed at A and B, and their inter- 
 section at S will be the centre of the circumscribing circle. With SA 
 or SB as radius, describe this circle, step AB round it from E to F in 
 the points 1, 2, 3, 4; join E 1, 2, 3, 4 F, and the required octagon is 
 constructed. To inscribe an octagon in a given circle : Draw two radii 
 (Fig. 69) at an angle of 45 to each other, and they will cut the circle 
 in points A and B ; join AB, and it will be a side of the octagon. Its 
 length stept round the circle will give the same points as in the previous 
 construction ; join them, and an octagon will be inscribed in the given 
 circle. 
 
 15. The same principle of construction as used in the last three 
 problems is applicable to any regular polygon, whatever may be the 
 number of its sides ; but in practice it is preferable to subdivide the 
 sides of those we have given if the division will give the required 
 number of sides than to lay down an independent construction, the 
 chances of not obtaining the exact length of the side of the polygon 
 required increasing as the number of sides increase. On paper, and 
 with the assistance of tee- and set-squares, many of the figures already 
 given can, of course, be easily and quickly constructed ; but, as before 
 observed, the ability to draw them without such aids is absolutely 
 essential, when we consider the calls often made upon the workman 
 for the practical application of such knowledge. 
 
 As figures, or solids, having more than eight sides or plane surfaces 
 are seldom met with in mechanical construction, and as those we have 
 given include all that form the surfaces of the plane solids intended to 
 be used as objects for projection, we shall now proceed to show how 
 their projections are obtained. 
 
CHAPTER VI 
 
 ORTHOGRAPHIC PROJECTION 
 
 16. A CAREFUL study of the preceding chapters, and the solution 
 of the problems contained in the two last, will have prepared the 
 student for entering upon that more important part of our subject 
 viz., "Orthographic Projection," or that special kind of delineation 
 which, when applied to the representation of mechanical subjects, en- 
 ables the engineer or machinist to determine at sight the actual dimen- 
 sions and arrangement of any part of an engine or machine. As, how- 
 ever, a part of a piece of mechanism is but a compound of simple forms 
 made up of what are known as plane solids and solids of revolution 
 alone or combined it is at once manifest that to be able to draw any 
 part of a machine, the would-be draughtsman must first master the 
 delineation of its component parts, and as these resolve themselves into 
 solids, with either plane or curved surfaces, having straight or curved 
 lines for their boundaries, the question of their ultimate accurate 
 representation as a whole becomes one of the correct projection in the 
 first stage of the study, of the lines bounding the surfaces of solids ; 
 and as straight lines and flat surfaces are more easy of projection than 
 curved ones, we commence this part of the subject by an illustration of 
 its principles in the projection of points, straight lines, and the simple 
 figures which form the surfaces of those plane solids used in giving 
 shape to machine and engine details. 
 
 By a reference to the latter part of Chapter II., it will be noted 
 that to obtain the views of an object required for the purposes of 
 manufacture its, projections are determined on two planes, at right 
 angles to each other that is, their relative positions are as shown in 
 Fig. 13 ; that lettered VP being a plane assumed to be vertical, and 
 the other HP a horizontal plane perpendicular to VP. These two 
 planes are called the vertical and horizontal "planes of projection," 
 and will throughout the exposition of the subject of " projection " be 
 denoted by the letters VP and HP. 
 
 17. The student should be particular to note the precise difference 
 of meaning existing between a " vertical " line or plane and one that is 
 " perpendicular." One line or plane may be perpendicular to another 
 line or plane, and yet neither of them be vertical. A vertical line is a, 
 
 35 
 
36 FIRST PRINCIPLES OF 
 
 plumb line, or the position a weighted line assumes when freely 
 suspended. A horizontal line is one which is parallel to the horizon, 
 and, therefore, perpendicular to a vertical one. A "vertical plane," 
 then, is one with which a plumb line will coincide, and similarly a 
 "horizontal plane" is one parallel to the earth's surface taken as a 
 2)lane, and is at right angles to the vertical. 
 
 A.j)lane, strictly defined, is nothing more than a perfectly flat "sur- 
 face," without any reference to substance ; but as it cannot be dealt 
 with for explanatory purposes without being assumed to be material 
 and inflexible, it will, when spoken of, or used for that purpose in this 
 work, be considered as having such a thickness as would be repre- 
 sented by a line. Assuming this, the edge view of a plane will, under 
 any circumstance of position, be a perfectly straight line. If, then, two 
 planes intersect or meet each other at an angle, as the " planes of pro- 
 jection " we are about to deal with do, their meeting will be in a line, 
 which forms a boundary or dividing line between them, and is called 
 
 Fig. 70 
 
 the " intersecting line " of the planes. This line will throughout the 
 subject have IL for its distinguishing letters. 
 
 Knowing, then, the true relative position of the " planes of pro- 
 jection " on which we wish to obtain the representations of an object, 
 we will first proceed to find the projections of a " straight line " in 
 different positions with respect to those planes. Let its position at 
 first be perpendicular to the VP. 
 
 Here, as the thing to be projected is a " line " having ends or points, 
 before we can obtain its projections we must first know how to find 
 those of a "point." Let, then, A on the left in the diagram Fig. 70 be 
 a point in space, such as a small bead invisibly suspended, and let it 
 be required to find its vertical and horizontal projections that is, its 
 projections on the VP and HP. 
 
 To obtain these, we have to find the points in the VP and HP 
 where a visual ray or projector perpendicular to each of the planes, and 
 drawn through A, would penetrate them. This, it will be seen in the 
 diagram, is in a in the VP, and a 1 in the HP, and therefore they are 
 
MECHANICAL AND ENGINEERING DRAWING 37 
 
 the required projections, a being an elevation or vertical projection of 
 A, and a its plan or horizontal projection. If it were required to find 
 from its projections the position of the original point A with respect to 
 the YP and HP, then perpendiculars to those planes let fall from its 
 projections a and a would intersect in A, giving it as the position of 
 the original point. 
 
 Knowing how to obtain the projections of a point, we shall now be 
 able to find the projections of a straight line. 
 
 1st. Let the line AB (Fig. 70) be perpendicular to the VP. 
 
 Here AB being perpendicular to the YP, will be parallel to the 
 HP; therefore, from its position with respect to the YP, its projec- 
 tion on that plane will become a point a, as the eye being directly 
 opposite the end of it, the visual ray or projector proceeding from the 
 eye will travel along the line itself, coinciding with it, and penetrate 
 the YP in a, then a is the "elevation" of the line AB. To find its 
 " plan " or projection on the HP, let fall projectors perpendicular to 
 the HP from both ends of AB, and the points a b,\ where these pro- 
 jectors penetrate the HP, will be projections of the ends A and B of 
 the line AB, and if a b be joined, then ab will be the plan or hori- 
 zontal projection of the original line. 
 
 2nd. Let CD (Fig. 70) be the given line, and let it be perpendicular 
 to the HP, and its projections required. 
 
 In this case CD is parallel to the YP, and its projection on that 
 plane will be obtained by letting fall from C, D its ends, projectors to 
 the YP, and the points c d, where these fall on that plane, will be the 
 vertical projections of C and D in it ; then, if c and d be joined, c d 
 is the elevation of the line CD. As the given line is perpendicular 
 to the HP, its plan will be a point obtained by producing a visual ray 
 passing through and coinciding with CD itself, until it penetrates the 
 HP in d'. 
 
 3rd. Let EF be the given line (Fig. 70), and let it be parallel to both the 
 VP and the HP, and its projections required. 
 
 Here EF being parallel to both planes, by letting fall projectors 
 from E and F to the YP, we obtain points e and/*, and to the HP 
 points e and f, then ef and e'f being joined will be the required 
 projections. It will be noted here that the projections of the original 
 line EF are two lines of the same length as their original. This is 
 owing to the relative positions of the original line, and the planes on 
 which its projections were required. Had the line been in any other 
 position with respect to those planes, a different result would have been 
 obtained, as will be seen by the following problem. 
 
38 
 
 FIRST PRINCIPLES OF 
 
 4:th. Let Gil be the given line, and let it be parallel to the VP, but inclined 
 to the HP, and its projections required. 
 
 Here GH being parallel to the VP, its vertical projection or eleva- 
 tion is found by letting fall projectors from G and H to the VP, giving 
 points g and h, which, when joined, will be a line of the same 
 length as GH ; but its horizontal projection, obtained by letting fall 
 projectors from the same points G and H on to the HP, giving g'h', will 
 be found to be projected into g'h', a much shorter line than its original. 
 
 The diagram Fig. 70, the student must note, is drawn in what is 
 known as " quasi-perspective," and is adopted as a simple and ready 
 means of showing the two planes of projection in their relative positions, 
 and the positions of the lines given in the foregoing problems in relation 
 
 V 
 
 H 
 
 Fig. 71 
 
 to those planes. It is in no sense an orthographic projection, diagram, 
 although used to explain the application of the principles of that kind 
 of projection. 
 
 1 7. To convert the actual relative positions of the two planes of 
 projection, as shown in the diagram Fig. 70, into the positions they 
 occupy on the sheet of drawing-paper when laid on his board, the 
 student has to suppose the " upper " plane, or that we have named the 
 VP, turned backwards on the IL (intersecting line) as a hinge, until it 
 is on the same level with the " lower " one or the HP, the two planes 
 thus becoming one flat surface, as in Fig. 71, with the IL dividing them, 
 and the plans and elevations of the lines in the problems shown on them 
 as obtained by projection. 
 
 Assuming that the student has found no difficulty in understanding 
 the explanations already given of the way in which the projections of a 
 line when it is in either of the suggested positions, with respect to the 
 
MECHANICAL AND ENGINEEEING DRAWING 39 
 
 planes of projection, are obtained, there are yet two other positions 
 that a line may occupy with respect to those planes, whose projections 
 we must know how to find before we can proceed with the projection of 
 plane figures. One of those positions is that of a line inclined to both 
 the YP and the HP. We have shown, in Figs. 70 and 71, that if a 
 line be parallel to one plane and inclined to the other, its projection on 
 the plane to which it is parallel will be a line equal in length to the 
 original, and on the one to which it is inclined its projected length will 
 depend upon the angle the given line makes with its plane of projection, 
 This will be made still clearer by the demonstration of the problem 
 where 
 
 5tk. A line AB is inclined to both the VP and HP (Fig. 72), and its pro- 
 jections are required. 
 
 Let the given line at first be parallel to the VP, and perpendicular 
 to the HP. In this position its projection on the HP will be a point, 
 as a, and on the VP a line AB, at right angles to the IL. While 
 keeping AB parallel to the VP, conceive it to swing round to the right 
 On A as a joint, until it makes any desired angle with the IL ; or say 
 until B has moved into the position b, its elevation bA in this position 
 is a line inclined to the IL, of the same length as AB, but its plan, 
 obtained by letting fall from b a projector perpendicular to the HP, or 
 IL, in c, gives ac as its projection on the HP, or a line less than half 
 the length of its original. It is evident from this that the projected 
 length of a line is entirely dependent upon its angle with the plane of 
 its projection, for if the motion of the line AB in this case were con- 
 tinued until it coincided with the IL, its projected length adj and its 
 original length AB, would then become equal. 
 
 But so far the given line is only inclined, as at Ab, to one of the 
 planes of projection, the HP ; for although we have moved it from its 
 assumed first position that is, perpendicular to the HP to that of 
 making an angle bAD with it, it is still parallel to the VP. Let it also 
 be inclined to that plane, say 45. For distinctness, let C be a new 
 position of A on the IL ; at C draw Ct, at an angle of 45 with the IL 
 and equal to ac, or the projected length of Ab in the HP ; then Ct will 
 be a plan of Ab when at 45 to the VP, and at the angle, b Al) with the 
 HP. To obtain its elevation, draw from, t a projector perpendicular to 
 the IL, and from b another parallel to it, to cut the one from t inp, join 
 C and p, and the line Cp will be the elevation of the original line AB, 
 inclined to both planes of projection. Here it will be noticed that the 
 original line AB, in addition to its having been moved on A as a joint 
 from B to b, has also, while making the angle bAd with the HP, been 
 swung round on A through 45. 
 
 Now to make this matter of the projection of inclined lines still 
 more clear, as much depends on the student having a thorough grasp of 
 this first part of the subject. We will assume that the two projections, 
 Cp in the VP, and Ct in the HP, are given, and it is required to find the 
 real length of the line of which they are the projections. 
 
 Here the line Ct is the plan or horizontal projection of the line Cp, 
 
40 
 
 FIRST PRINCIPLES OF 
 
 the latter being a line having one of its ends C, in the HP, and the other 
 end p a given distance above that plane. C/> is also the projected length 
 of the hypothenuse (or longest side) of a right-angled triangle, having 
 Ct for its base, and a line equal to the vertical height of p from the HP 
 for its perpendicular. With these two sides given, we can find the third 
 side, or the actual line of which Cp is the projection. Therefore, at t in 
 the line Ct,, and perpendicular to it, draw a line indefinitely, and from it 
 cut off in h, a length tli equal to the height that p in the line Cp is above 
 the HP or IL, join C and h ; then Ch is the real length of the original 
 line, of which Gp and Ct are its projections. This is self-evident, for if 
 the right-angled triangle Gth, which may be assumed to be lying on the 
 HP, with its base line coinciding w^ith Ct, be raised to a vertical position, 
 moving on Ct as a hinge, its base and hypothenuse will then be coincident 
 with Ct, and its third side lit is a vertical line perpendicular to the HP 
 represented by the point t. 
 
 Fig. 72. 
 
 The other position a line may have, with respect to the planes of its 
 projection, is that of being parallel to the HP, but making an angle with 
 the YP. Putting this in the form of a problem, we will say 
 
 Qtk. Let a given line be parallel to the HP, but inclined to the VP, and its 
 projections required. 
 
 In this case, let the given line at first be perpendicular to the VP ; 
 its elevation when in that position in the VP will be a point as e, and 
 its plan a line EF at right angles to the IL. But as EF is perpendicular 
 to the VP it is parallel to the HP. While keeping it so, let it be con- 
 ceived to swing on its end F as a joint in its direction of the arrow, 
 until it makes any desired angle with the VP or IL, or until, say, E 
 
MECHANICAL AND ENGINEERING DRAWING 41 
 
 has moved into the position f ; its elevation in that position is found by 
 drawing a projector through f, perpendicular to the IL, and a line 
 through e parallel to it to cut the projector from/" in g, then the line eg 
 is the vertical projection of EF when making the angle LF/" with the 
 VP. Here it is again seen that the projected length of a line, although 
 parallel to one of its planes of projection, is determined on the other 
 plane by the amount of its inclination to that plane ; for had the given 
 line EF in this case been moved through any greater or less angle than 
 the one assumed in the diagram, its projected length in the VP would 
 have been greater or less than eg, directly in proportion to its altered 
 position with respect to the VP. If EF had been swung so far round 
 on F until it had coincided with Ff, then its projection in the VP 
 would be eg', or a line equal in length to the given line EF. 
 
 18. In the foregoing problems in this chapter, we have given all the 
 positions which it is possible for a line to occupy with reference to its 
 planes of projection, and we could at once proceed to the projection of 
 plane figures, were it not necessary at this stage that the student should 
 thoroughly understand the true significance of the line lettered IL in 
 previous and all future diagrams throughout this work. 
 
 This line, we have already shown, is the line of intersection of the 
 two "planes of projection"; but it is much more than this. The VP 
 and HP, being for all the future purposes of the student draughtsman 
 represented by the one flat surface of his sheet of paper, with the IL 
 either shown on it or assumed to be there, dividing its surface into two 
 planes, it becomes when shown in on a drawing or explanatory diagram 
 at one and the same time the representative, not only of the IL, but 
 of the VP and HP as well, for it is a plan of the VP and an elevation of 
 the HP, and as these it is a datum line from which heights above the 
 HP, or distances from the VP, may be measured or set off. 
 
 These facts, it will be seen, are verified by a reference to Fig. 72. 
 Here the IL is for all the figures, a plan of the VP, showing the line Ab 
 by its plan to be parallel to the VP, and in front of it, at a distance 
 equal to Aa. Similarly C, the lower end of the line Cp, is shown 
 touching the VP, while its upper end p, projected into t in the HP, 
 stands out from the VP a distance equal to tt', and above the HP a 
 height equal to t'p. Then again Fe is the height of the line EF above 
 the HP, the end F touching the VP in e, and the end E being a distance 
 equal to EF from the VP when in the first position, and a distance equal 
 to fJi when in the second. In the two last cases, it will be seen that the 
 IL is a plan of the VP and an elevation of the HP. With the foregoing 
 explanation of the projection of lines thoroughly digested, the student 
 should have no difficulty in finding the projection of plane figures, to 
 which we now proceed. 
 
CHAPTER VII 
 
 PROJECTION OF PLANE FIGURES 
 
 19. KNOWING how to find the projections of a line having any given 
 position with respect to the planes of its projection, w r e can now proceed 
 to the projection of those straight-sided plane figures which form the 
 surfaces of the solids used in giving shape to machine details. As the 
 same principles apply to the projection of all plane figures, whether their 
 sides are few or many, it is only necessary that their application should 
 here be shown in the case of one of .each class of figure chosen. Com- 
 mencing with that figure having the least number of sides the triangle 
 we shall give as additional subjects for projection the square, rectangle, 
 pentagon, and hexagon, or those which usually form the sides and ends 
 of the plane solids we have before referred to. Our first problem in this 
 subject is 
 
 Problem 22 (Fig. 73). The line a c is tlie plan of an equilateral 
 triangle, with its base resting on the IIP, and parallel to the VP ; it 
 is required to find its elevation or vertical projection. 
 
 Here, as the base of the triangle is parallel to the YP and its 
 plan is represented by a line, the triangle itself is parallel to the VP, 
 and therefore perpendicular to the HP. To find its elevation, through 
 a and c draw projectors perpendicular to the IL, cutting it in A and C ; 
 through A and C with the 60 set-square draw lines intersecting in B, 
 join A, C, and the figure ABC is the required vertical projection of the 
 triangle of which the line a c is the plan. The projection, it will be seen, 
 is an equilateral triangle, for all the sides of the triangle being by the 
 conditions of the problem parallel to the VP, they are projected on that 
 plane into lines of the same length. 
 
 Problem 23 (Fig. 74). To find the elevation of the triangle obtained 
 in the jrrevious jyroUem when it is inclined 4& to the VP, its plan 
 being the line a c, as before. 
 
 Here the IL may be considered as a plan of the VP. If we draw 
 a line (with the 45 set-square) of a length equal to a c, at an angle of 
 
 42 
 
MECHANICAL AND ENGINEERING THAWING 
 
44 FIRST PRINCIPLES OF 
 
 45 with the IL, that line will be a plan of the triangle ABC when at 
 that angle with the VP. Bisect ac in b, through b draw a projector 
 perpendicular to the IL, and from B, in Fig. 73, another parallel to it, 
 cutting the one drawn from b in b'. Then, from a and c draw projectors 
 to the IL, cutting it in a and c, join a'b', ac, c'b' ; the figure a'b'c is 
 the required elevation of the given triangle inclined 45 to the YP. In 
 this case it will be noticed that the elevation obtained is an isosceles 
 triangle, resulting from the altered position of its original with respect 
 to the YP, the plane of its projection. The line ac is bisected in b to 
 find the plan of the vertex B of the triangle ABC, Fig. 73 ; and the 
 vertical projector through this bisection b determines, by its intersection 
 with a parallel one through B, the elevation b' of the vertex in its new 
 position. 
 
 Problem 24 (Fig. 75). The line a c is the elevation of an equilateral 
 triangle having its base touching the VP and parallel to the HP ; to 
 find its plan or horizontal projection. 
 
 The position of the original figure, of which the line a c is the eleva- 
 tion, is in this case the converse of that in Fig. 73. Here a c being a 
 line in the YP parallel to the IL, the triangle whose projection it is 
 must be parallel to the HP. To obtain its plan let fall from a and e, 
 its ends, projectors perpendicular to the IL, cutting it in A and C, and 
 through AC with the 60 set-square draw lines intersecting in B, then 
 the figure ABC is the required plan of the triangle of which a c is the 
 elevation. 
 
 Problem 25 (Fig. 76). Let the triangle obtained in the last problem 
 be inclined to the HP at 45, its base resting on that plane at right 
 angles to the VP, and one angular point touching the VP ; to find 
 its projections in that position. 
 
 Assume the position of the triangle at first to be perpendicular to 
 both the YP and HP ; its elevation will then be a line perpendicular to 
 the IL, equal to the altitude of the triangle, as Cp ; and its plan, a line 
 AC, also perpendicular to the IL and equal to the base of the triangle. 
 If, then, the triangle be moved on AC as a hinge (to the right) through 
 45, its elevation at that angle is found by drawing with the 45 set- 
 square through C a line CB equal to Cp. To find the plan, bisect AC 
 in a, and through a draw a line parallel to the IL : let fall from B a 
 projector perpendicular to the IL to cut the line drawn through a in 6, 
 join Cb and A5, and the figure CAb is the projection of the triangle 
 ABC (Fig. 75) when inclined at 45 to the HP. 
 
 Problem 26 (Fig. 77). Given a straight line b c parallel to the IL as 
 the plan of a square resting with one of its sides on the IIP ; to find 
 its elevation. 
 
 The plan of the square being a line parallel to the IL, the square 
 itself will be parallel to the YP and perpendicular to the HP ; there- 
 
MECHANICAL AND ENGINEERING DRAWING 4o 
 
 fore, from b and c draw projectors perpendicular to the IL, cutting it 
 in A and D ; set off on one of them a distance AB equal to be in the 
 plan, and through B draw BC parallel to the IL ; join AD, and the 
 figure ABCD is the required elevation of the square whose plan is the 
 line b c. 
 
 Problem 27 (Fig. 78). To find the elevation of the square obtained in 
 the last problem when it is inclined at 30 to the VP, its plan being a 
 line b c, as before. 
 
 Draw in the HP (with the 60 set-square) a line b c the plan of 
 the square making an angle of 30 with the IL ; and from b and c 
 draw projectors cutting the IL in a' and d', make a'b' equal to be, and 
 through b' draw b'c parallel to ad', then the figure a'b'c'd' is the elevation 
 required. 
 
 Problem 28 (Fig. 79). A square with one of its sides touching the 
 VP is represented in elevation by a line b c parallel to the IL ; to find 
 its plan. 
 
 From b and c let fall projectors perpendicular to the IL ; make AB 
 equal to be, and '.through B draw BC parallel to the IL, join ABCD, 
 and it is the required plan of the square. 
 
 Problem 29 (Fig. 80). The square ABCD obtained in the last prob- 
 lem is inclined at 60 and 30 to the HP, with one of its sides 
 touching the VP and an adjacent side the HP ; to find its plan 
 and elevation in those positions. 
 
 At B in the IL, draw BC, BC' (both equal to AB) with the 60 set- 
 square, making angles of 60 and 30 with the IL ; then BC, BC' are 
 the required elevations of the square at those angles. From BcZd', draw 
 BA, dc, d'c, perpendicular to the IL, and each equal to BC ; and Ace 
 parallel to it ; then B Acd ; EAc'd' are the plans of the given square 
 when inclined at 60 and 30 to the HP. The projections of the rect- 
 angle are not given, as the method of obtaining them is in all respects 
 the same as that for the square, but allowing for the difference of length 
 of adjacent sides. 
 
 20. The method of obtaining the projections of a pentagon and 
 hexagon in different positions with respect to the VP and HP is fully 
 shown in Figs. 81 88, and will not need explanation further than to 
 say that the elevations in Figs. 81 and 85 shown with reference letters 
 in capitals must be drawn first, according to the construction rules 
 given in Problems 19 and 20, before their plans can be found ; in all 
 other respects the procedure is the same as in the previous figures. 
 
 To make himself thoroughly conversant with the application of the 
 principles of projection to the delineation of plane figures, which it is 
 most important he should be before passing on to the projection of 
 solids, the student should draw all the foregoing plane figures at least 
 twice, making them three times the size here given. 
 
48 
 
 MECHANICAL AND ENGINEERING DRAWING 
 
 -TJ 
 
CHAPTER VIII 
 
 THE PROJECTION OF SOLIDS 
 
 21. ASSUMING that the student has followed the advice given in the 
 last paragraph of Chapter VII., and thoroughly mastered the elementary 
 principles of projection which we have expounded in it and Chapter 
 VI., we can now proceed to apply those principles to the delineation 
 of the simple geometrical solids of which engine and machine details 
 are invariably made up. These solids are of two kinds viz., plane, 
 and circular or curved. The first-named have all their surfaces plane 
 figures, the projections of which the student already knows how to 
 obtain, and the second includes all solids whose bounding surfaces are 
 all curved, or plane and curved combined. What are known as the 
 simple solids are the cube, the prism, the pyramid, the cylinder, the 
 cone, and the sphere, the first three being plane solids, and the others 
 circular or curved solids. 
 
 A cube (Fig. 89) is a solid having six equal sides or faces, all of 
 them squares. 
 
 A j^'ism (Fig. 90) has two ends or bases parallel to each other, each 
 being equal and similar figures ; its sides are rectangles. 
 
 A pyramid (Fig. 91) has one base, its sides being triangles, with 
 their vertices meeting in one point a, called the apex of the pyramid. 
 (As it is advisable for the student to confine himself for the present to 
 the study of the projection of plane solids, we defer any consideration 
 of the circular ones until after the projection of curved lines in any 
 position is understood by him.) 
 
 47 
 
 OF THK ' r ^\ 
 
 UNIVERSITY 
 
48 
 
 PRINCIPLES Otf 
 
 Regular prisms and pyramids have "regular" figures for their 
 bases. 
 
 The axis of a prism, or pyramid, is an imaginary line joining the 
 centres of the bases of the former, and the centre of the base and the 
 apex of the latter, as the dotte'd lines AA in Figs. 90 and 91. 
 
 A right prism, or pyramid, has its axis perpendicular to its base, as 
 Figs. 90 and 91. If its axis is inclined to its base, the prism or pyramid 
 is oblique, as in Figs. 92 and 93. 
 
 A truncated pyramid, or prism, is the part of the solid left when its 
 upper part is cut away by a plane, and is called a frustum. The cutting 
 plane may be either parallel or inclined to the base of a pyramid, but 
 only inclined to the bases of a prism. In Figs. 94 and 95, A and B 
 are frustums. 
 
 Fig. 92 
 
 Fig. 93 
 
 Fig. 94 Fig. 95 
 
 As all the sides of a prism are parallel to its axis, the edges of the 
 sides connecting its bases are perpendicular to the bases in a " right " 
 prism, and inclined to them in an " oblique " one. 
 
 In a right pyramid all its sides are isosceles triangles, and its axis is 
 perpendicular to its base. If the base is a regular figure and the axis 
 perpendicular to it, the sides of the pyramid will all be equal and 
 similar isosceles triangles, but if the axis be inclined to the base, then 
 the sides become unequal triangles and the pyramid an oblique one. 
 
 Both prisms and pyramids are named according to the figures of 
 their bases. If the base is a triangle, square, pentagon, hexagon, or 
 octagon, then the solid becomes a triangular, square, pentagonal, hexa- 
 gonal, or octagonal pyramid or prism, as the case may be. 
 
 There are other plane solids, such as the tetrahedron, octahedron, 
 etc., etc., but such forms are seldom adopted by the engineer or machinist 
 in his constructions. Their special features may, however, be studied 
 to advantage by the student draughtsman in his spare time. 
 
 22. The working out of the problems in Chapter VII. will have 
 shown how the projections of the figures chosen are obtained, and as 
 they form the surfaces of the solids used in giving shape to machine 
 details, our next step is to show how to obtain the projections of such 
 solids in any given position. Now, as a sketch is often a much more 
 satisfactory means of explaining a method of procedure than many 
 words, we give in Fig. 96 a perspective view of the planes of projection, 
 and the construction or working lines, showing how the plans and 
 elevations of a few simple objects (all bounded by plane surfaces) are 
 
MECHANICAL AND ENGINEERING DRAWING 49 
 
 obtained, which will, we think, materially assist the student in his study 
 of the application of the principles involved. The objects chosen for 
 the illustration of these principles are simple prismatic solids, or a com- 
 bination of such, and only require for the comprehension of the method 
 of their projection, such a knowledge of principles as the student from 
 what has gone before should have now acquired. 
 
 The diagram is so plain that it hardly requires explanation ; but as 
 it is important that the procedure in obtaining the projections should 
 be thoroughly understood, we will at the risk of repetition endeavour to 
 make it, if possible, still more intelligible. The constructions to the 
 left of the diagram are a repetition of the three first problems in Fig. 
 70, but show more fully, by means of the arrows, the direction of the 
 projectors or visual rays with respect to the YP and HP. After the 
 very full explanation given in Chapter VI. of the way of obtaining the 
 projections of a straight line in any position, nothing more need be said 
 in reference to it here than that the positions of the lines given in those 
 problems are the positions of the edges which are all straight lines 
 of the prismatic solids given in the diagram as the subjects for 
 projection. 
 
 The first solid, whose projections in the VP and HP of the 
 diagram are figured 1, 1, is that of a right prism (of any material 
 substance say wood) ; its bases or ends are rectangles, square with 
 its sides, and its position with respect to the planes of projection is 
 such, that its sides are perpendicular and its ends parallel to the VP, 
 its upper and under sides are parallel to the HP, while its other 
 two sides are perpendicular to it ; its ends being parallel to the 
 VP are also perpendicular to the HP. As the sides and ends of 
 the solid are plane rectangular surfaces, and in known position with 
 respect to the VP and HP, their projections on those planes are 
 obtained in the same way as those of any plane figures of the same 
 form, and in the same position. 
 
 23. At this stage in our subject it will have become apparent 
 to the would-be draughtsman that he must either be possessed of a 
 perfect knowledge of the forms of the solids he is attempting to 
 delineate, or have models of them to guide him in his delineations; 
 in other words, he must have either a true conception or a possession 
 of the object he wishes to draw, for it is evident from the diagram 
 we are proceeding to explain, that without a model of the object to 
 be delineated, or its conceived counterpart, neither plan nor elevation 
 of it could be obtained. This is one great reason why an earnest 
 student of mechanical drawing should at this stage in his study 
 possess himself of a convenient set of models of the solids enumerated 
 in a previous paragraph, for no greater mistake can be made in the 
 study of the "projection of solids" than that. of making a servile 
 copy of any diagram or drawing to be found in text-books on this 
 subject. With this slight digression which has been necessary in 
 the interest of students of projection we proceed with the explanation 
 of our diagram. 
 
 The form and dimensions of the solid having been predetermined, 
 and its position with respect to the VP and HP known, its 
 
 E 
 
50 FIRST PRINCIPLES OF 
 
 projections plan and elevation are obtained as shown. Projectors, 
 or visual rays (shown by the dotted lines in the digram), are let 
 fall perpendicularly, as shown by the arrows, from the principal points 
 in the object which are the ends of the lines or edges bounding 
 it upon the VP and HP. The points on these planes where 
 the projectors fall are each the plan, or elevation, as the case may 
 be, of the original line end, or edge end, and these points or 
 projections being joined by what their originals are connected with 
 viz., straight lines give the plan and elevation required. The 
 other objects in the diagram, whose projections are numbered 2, 2 ; 
 3, 3 ; 4, 4, are all prismatic in form, and represent a carpenter's 
 pencil, a hollow wooden tube, and the lower end of a square post. 
 Their projections are all obtained in the same way as those just 
 explained, numbered 1, 1, so there is 110 necessity for their 
 demonstration. The actual rendering of the true projections of the 
 four objects chosen for illustrating the subject of this chapter will 
 now be given with the sheet of drawing-paper as one plane surface. 
 
 To obtain the correct projection of the four objects shown in the 
 diagram Fig. 96, we must assume that we have their exact models 
 before us, and are able to draw in the VP on the sheet of paper 
 a full-size view of each of their ends such, in fact, as are shown at 
 A, B, C, D, Fig. 97. Each of these views is an end elevation, 
 or vertical projection, of the " original object " or model, and will be 
 found to agree with those shown in the diagram Fig. 96, on the 
 " Vertical Plane of Projection," and numbered 1, 2, 3, 4. The height 
 of these views above the IL on the sheet of paper is immaterial ; 
 but whatever it is, it should be understood that the objects represented 
 by A, B, C, D are the same height above the HP as these end views 
 of them are above the IL of the VP and HP. Having these end 
 views given, and knowing by measurement the length of the models, 
 we can find their "plans" or projections in the HP. 
 
 Now A, B, C, D are the front end views of objects in front of 
 the VP, and as the objects are a given length, their back ends must 
 either be assumed to touch the VP, or to be a given distance- from 
 it. Assume them to be as in the diagram Fig. 96 viz., a certain 
 distance from the VP ; in this case the ends will all be in a straight 
 line that distance from the IL, for, as before shown, the IL is a " plan " 
 of the VP. Taking, then, the IL as a datum line, set off from it the 
 distance it is intended the back ends of the objects shall be from 
 the VP, and through the point set off draw with the tee-square a faint 
 line parallel to the IL, as b, e; then as the objects are assumed to 
 be of the same length, and their front ends (or those nearest the 
 eye) are parallel to their back ends, set off from the line b, e, a distance 
 equal to the length of the objects, and at this distance draw another 
 faint line f, e, parallel to b, e ; then the plans of the front ends of 
 A, B, C, D will, in the HP, be in the line /, e, and of their back ends 
 in the line b, e. From the points 1, 2 of A, in the VP, let fall 
 projectors perpendicular to the IL into the HP, cutting the faint 
 lines b, e and/, e in points 1', 1 ; 2', 2 ; join these by straight lines as 
 shown, and the plan, or horizontal projection of the object, whose 
 
MECHANICAL AND ENGINEERING DRAWING 
 
 51 
 
 \ 
 
 Fig. 96 
 
52 FIRST PRINCIPLES OF 
 
 end elevation is A in the VP, is obtained ; for two of the sides am 
 the two ends of the object being vertically disposed to the HP, am 
 their edges only seen from above, their horizontal projections becom< 
 severally the lines 1', 1 ; 2', 2, for the sides, and 1', 2'; 1, 2, for thi 
 ends. The exact shape of the object's upper surface as viewed ii 
 the direction of the arrow, is truly shown in the plan by the dispositioi 
 of its bounding edges obtained by projection. 
 
 As the plans of the other three original objects represented ii 
 elevation by B, C, D, in Fig. 97, are obtained in the same way as th< 
 plan of A, all the necessary construction lines for obtaining then 
 being shown in, nothing further need here be said of the method o: 
 their projection, it being advisable that the student should think ii 
 out for himself that it may be the better remembered. In thif 
 exercise in projection, which we will call Sheet 1, the elevation oj 
 objects being given, and their plans required, the subject is callec 
 " Projection from the Upper to the Lower Plane " ; the examples giver 
 should be drawn to as large a scale as a half-imperial sheet of drawing 
 paper will admit of, leaving a fair margin all round ; the four 
 elevations A, B, C, D should be disposed in a row (suitably spaced] 
 the long way of the paper, and the lengths of the objects may be 
 assumed to be such as will occupy about two-thirds of the space 
 between the IL and the lower edge of the sheet of paper. 
 
 24. In our explanation of the method of obtaining the actual 
 projections of the simple objects shown in perspective in the diagram 
 (Fig. 96), we purposely ignored the fact that all the objects depicted 
 were not " solid " in the sense of their having perfect solidity, although 
 prismatic in form. Our reason for this was to prevent confusion in 
 the mind of the student on his first introduction to the study of this 
 part of our subject ; but as it will be necessary for him to know how 
 to obtain other views of objects than their mere outside plan and 
 elevation (as shown in Figs. 96 and 97), we will, before proceeding 
 further, explain what those views are, as they will in part be required 
 for the completion of the problems given in Sheet 2. 
 
 On referring to the object in the diagram (Fig. 96), whose 
 projections in the VP and HP are figured 2, 2, it is stated in the 
 description of it that it represents a carpenter's pencil, or an article 
 composed of two material substances wood and plumbago. In 
 manufacturing such an article it would be important to know how 
 much of each of the substances which go to make a pencil is to be 
 put into it. To decide this, we must have such a view of it as would 
 show how far the lead or plumbago extends into the wood covering 
 it. Such a view is called a " section," and is obtained by supposing 
 the pencil to be cut horizontally, in the case before us right through 
 the middle of its depth from end to end, the cutting plane dividing 
 the pencil into two halves. This cutting, when the upper part of the 
 pencil is removed, will show at once the extent of the lead, as the 
 "plane of section" will have passed through it. How to give a view 
 of the section of the pencil after being cut will be shown further 
 on. The principle involved in obtaining it is that of assuming the 
 plane of projection or the sheet of drawing-paper to be transparent 
 
MECHANICAL AND ENGINEERING DRAWING 
 
 D 
 
 a 
 
 L 
 
 H 
 
 i 
 I 
 1 e 
 
 Sheet 1 F'uj. 
 
 -I/ 
 
 D 
 
 c 
 
 . 
 
 Sheet 2 Fiy. 97 
 
4 FIRST PRINCIPLES OF 
 
 with the object beneath it, the visual rays from all its principal points 
 being projected on that plane in points which are afterwards joined 
 by right lines. Such a view when obtained is called a "sectional 
 plan," the plane of its projection being the original HP, the position 
 of the object only with respect to that plane having been reversed. 
 
 A similar plan of the object whose projections are figured 3, 3, 
 in Fig. 96, will have quite a different appearance to that given of 
 it in Fig. 97. It is, we are told, a hollow wooden tube, whose 
 thickness is shown in its end elevation on the VP. When looked at 
 from above, in the direction of the arrow, its appearance is the same as 
 if it were merely a rectangular prism of solid material throughout, 
 with its sides inclined to the HP : but if a cutting plane be caused to 
 pass through it horizontally from end to end, it would then be seen 
 (on the removal of the part cut off) that its interior is hollow, and 
 its "sectional plan" something very different from that before obtained. 
 What that view would be we shall see later on. 
 
 25. Now, in addition to the simple plan and front elevation of 
 an object being given, it is necessary, before its exact shape and 
 construction can be understood, to have one or more " side views " 
 or side elevations of it, for there is hardly anything in mechanical 
 constructions, at any rate which is of the same shape when viewed 
 from the side and end. To show how such views are obtained, let VP 
 and HP in the sketch (Fig. 98) represent the two planes of projection, 
 as before, in their normal position that is, at right angles to each 
 other and let it be assumed that the YP is in two parts, A and B, 
 hinged together at c d, and that the part A is capable of being swung 
 round on c d until it is at right angles to the part B. When so swung 
 we have virtually three "planes of projection," two of which are 
 vertical and one horizontal, and each at right angles to the other. 
 With planes in these positions it is evident that three different 
 projections of an object may be obtained on them. Let the object 
 be, say, a simple prismatic solid, as S, and let its position be such that 
 two of its sides are respectively parallel to the two vertical planes 
 A and B; then the view obtained on A, looking in the direction 
 of the arrow s, will be a side elevation of S; and that on B, in the 
 direction of /, a front elevation ; the plan P of the object in the 
 HP being obtained as shown by the projectors to that plane. If, then, 
 the part A of the VP, with its obtained side view of S, be swung 
 back on c d into its original normal position with respect to the part B, 
 we should have in the VP a front and side elevation of the original 
 object S, and in the HP the plan P or view obtained when 
 looking in the direction of the arrow t. On turning down the VP on 
 the IL as a hinge until it and the HP become, as before explained, 
 the one flat surface of the sheet of drawing-paper, the three true 
 projections viz., the front and side elevations and the plan of S 
 will appear as shown in Fig. 99. In that figure the assumed 
 motion of the part A of the VP into the position of being at right 
 angles to the part B, is shown by the dotted arcs and arrows, the 
 winged arrow indicating the assumed motion of the plane A, and the 
 barbed ones the transference of the projections. 
 
MECHANICAL AND ENGINEERING DRAWING 
 
 55 
 
 i<. 98 
 
 H 
 
 Fig. 09 
 
56 FIRST PRINCIPLES OF 
 
 From the foregoing explanation it will be seen, on reference to Fig. 
 98, that a sectional front or side elevation of an object may be ob- 
 tained in the same way as a simple elevation, for it is only necessary 
 to assume the object as cut through by a vertical section plane, such as 
 that shown in the figure, and the part cut off by it nearest the eye 
 removed ; the view then obtained when looking in the direction of the 
 arrow s w r ould be the elevation required. 
 
 26. Having explained at some length the specific meaning of a 
 sectional plan, and of a front, and side, and " sectional elevation " of 
 an object, we will now revert to the problem of showing how to obtain 
 by actual projection the sectional plans of objects, taking for our pur- 
 pose those given in the diagram Fig. 96. The first on the left is 
 assumed to be a beam of wood of rectangular section, shown with its 
 two widest sides parallel to the HP, and its narrow ones perpen- 
 dicular to it. As a horizontal section of a beam in such a position 
 would give in plan a similar projection to the one already shown, and 
 lettered in the diagram Fig. 96 " Plan of Original Object," we will, for 
 the better practice afforded, assume the beam to have its sides inclined 
 to the HP, as shown at A, Fig. 97, Sheet 2, and the cutting or section 
 plane x y to pass through it horizontally when in that position from 
 end to end. We will assume also that all the sides of the beam are 
 coloured green. The problem then is to obtain by projection a plan 
 of the beam when the upper part cut off by the section plane x y is 
 removed. 
 
 Now it is evident on looking at the beam from above that the 
 surface exposed by the cutting plane, showing the nature of its 
 material, will have two edges at b, c, parallel to each other and at right 
 angles to the YP, and in length equal to that of the beam ; and at a 
 an edge parallel to that at b, the surface between these latter edges 
 being that untouched by the cutting plane, and therefore coloured 
 green ; the lower edge of the beam at d, not being seen from above, 
 will have no counterpart in the plan, and the two ends of the beam 
 being square with its sides, and therefore parallel with the VP, will 
 be represented in plan by lines in that position. Therefore, having 
 drawn in in the YP of the sheet of paper, as at A, Sheet 2 the 
 end elevation of the beam with its sides at the intended inclination to 
 the HP, let fall from the points , b, c projectors perpendicular to the 
 IL into the HP, set off on the projector let fall from a a distance 
 from the IL equal to that the inner end of the beam is assumed to be 
 from the YP which is arbitrary and from this point a", on the 
 same projector, set off a distance a", ', equal to the length of the 
 beam ; through a", a draw lines parallel to the IL cutting the projec- 
 tors from , 5, c, in a", b", c", ', b', c f ; join these points by straight 
 lines as shown, and the required projection is obtained. As the sur- 
 face bounded by the parallel lines b", b', c", c, and b", c", b', c, is the 
 plan of the section of the beam exposed after cutting, it is indicated by 
 drawing lines with the 45 set-square, as shown. 
 
 The projection of the sectional plan of the " carpenter's pencil," 
 shown in elevation at B, Sheet 2, will present no difficulty to the 
 student, as the procedure is clearly shown ; there is, however, one 
 
MECHANICAL AND ENGINEERING DRAWING 57 
 
 point in reference to it to which his attention is to be directed. The 
 pencil being made of two kinds of material, this is indicated in the 
 projection by drawing the sectional lines referred to in the last prob- 
 lem in opposite directions across the materials, as shown. This point 
 will, however, be more fully enlarged upon later on. 
 
 From the projection of the pencil we pass on to that of the hollow 
 tube, whose elevation is given in Sheet 2 in the YP at C, the section 
 plane being shown by the line x y. .Now in the plan of this object, 
 given in Fig. 97, Sheet 1, it will be noted that as there are only 
 three side edges, and the two ends, seen from above, its "plan" is 
 obtained by the projection of these edges into the lower plane or HP, 
 and joining them up, as shown. But in the problem before us the 
 upper part of the tube is assumed to be cut away, leaving the section 
 of two of its sides exposed, together with part of its interior. To find 
 its plan under these conditions we proceed as follows : From a and b 
 in the elevation C, let fall projectors perpendicular to the IL into the 
 HP ; and as the tube is assumed to be of the same length as the beam 
 and pencil, and its back end the same distance from the YP as their 
 ends are, at this distance from the IL, and parallel to it, draw the line 
 a" b", cutting the projectors from a and b in those points ; and parallel 
 to this line and at a distance a" a equal to the length of the tube from 
 it draw the line a 1 b'. We have so far obtained the bounding lines of 
 the plan sought. For the plan of the parts of the tube cut by the 
 section plane x y, let fall projectors from points 1, 2, 3, 4 in the eleva- 
 tion, cutting the lines a" b", a b', in the plan in points 1', 2', 3', 4', 1", 
 2", 3", 4" ; join these points by lines as shown, and the surfaces 
 between each of these pairs of lines will be the plan of the parts of the 
 sides of the tube seen from above when the upper part cut off by the 
 plane x y is removed. By its removal, however, a part of the in- 
 terior of the tube is now seen when viewed from above, and this 
 must be indicated in the plan. The part seen is the angle formed by 
 the meeting of the two bottom inside surfaces of the tube immediately 
 over the point lettered c, and as these are plane surfaces, their inter- 
 section forms a line the plan of which is found by letting fall a pro- 
 jector from c cutting the lines a" b", a 9 b', in c" c, and joining them by 
 a straight line. With the section lining of the parts cut by the plane 
 x y t the " sectional plan " of the " original object," or hollow tube C, is 
 completed. 
 
 The projection of the sectional plan of the object represented in 
 end elevation by D, in Sheet 1, would give little useful practice to the 
 student if kept in the same position as there shown, all its surfaces 
 being either parallel or perpendicular to the HP, and therefore result- 
 ing in a very trifling change in the plan (got by a horizontal section of 
 it) from that previously obtained ; it is consequently shown with its 
 principal surfaces inclined to the HP, thereby giving a more difficult, 
 but more useful, problem in projection. With the original object in 
 this altered position the student will at once be struck with the 
 identity in appearance of its elevation with that of the hollow tube in 
 the last problem, and he will perhaps be momentarily puzzled to under- 
 stand how two apparently similar end views are the vertical projections 
 
58 MECHANICAL AND ENGINEERING DRAWING 
 
 of two such different objects. Here we have an instance showing 
 the absolute necessity of a pre-conceived knowledge of the form of 
 the object to be delineated, or the possession of a model of it. Having 
 either, little difficulty would be experienced in fully understanding the 
 similarity of the two views in tne diagram. The object represented 
 by D in elevation (Sheets 1, 2) is merely a combination of two prisms 
 of different dimensions cut from one solid piece of material, from which 
 a portion is cut off by a plane such as a saw-blade throughout its 
 whole length, the cut being a horizontal one ; and the problem is to 
 show the actual appearance of the remaining part of the post when 
 looked at from above, after the part cut off is removed. 
 
 To do this, let x y in the elevation D (Sheet 2) be the section 
 plane, as before ; let fall projectors from a and d into the HP, and 
 assuming the back end of the object to be the same distance from the 
 VP as those of the pencil and tube, draw parallel to the IL a line 
 cutting the projectors from a and d in a" d" ; from a", on the pro- 
 jector from a, set off a" d, the length of the thick end of the post 
 which is arbitrary and through d draw a faint line a' d f parallel to 
 a" a" ; then from b and c in the elevation D let fall projectors into the 
 HP, cutting the line drawn through a' d', in b' c, and from these last 
 points set off on their projectors in the points b" c" the length of the 
 small part of the post ; then a line drawn through b" c", parallel to 
 a" d", will give the bounding lines of the plan of the post. For the 
 sectional part of it, let fall projectors from points 1 and 4, and 2 and 3, 
 in the elevation D, the former to cut the line a" d" in points V 4', and 
 the latter the faint line d d' in points 2' 3'; through 1' 4' draw lines 
 parallel and equal to d' d ; and through 2' 3' lines parallel and equal 
 to b' b" ; the surface enclosed by the last drawn lines is that made by 
 the cutting plane x y, and is indicated as such by the section lining. 
 As the post is of solid material, the edges formed by the meeting of its 
 under sides in e and /will not be seen from above, and are, therefore, 
 not shown in the projection. 
 
CHAPTER IX 
 
 PROJECTION IN THE UPPER PLANE 
 
 27. HAVING carefully worked out the problems given in Sheets 1 
 and 2, and studied, with the assistance of Figs. 98 and 99, and the 
 descriptive matter in connection with them, the principles involved in 
 obtaining the sectional plans and elevations of objects, the student 
 should find no difficulty in solving the problems in the projection of 
 solids which are now to follow. 
 
 The first subject we take it being the simplest of all the plane 
 solids is the " cube " ; but although simple, it is necessary that the 
 student should fully comprehend the specific relations of its various 
 faces and edges to each other, before starting to find its projections. 
 This solid is defined as one having six equal sides or faces, all of them 
 
 ;D 
 
 Fig. 09a 
 
 squares. It follows from this that adjacent sides must be at right 
 angles to each other and opposite faces parallel ; the adjacent and 
 opposite edges of the solid having the same relative positions. Bear- 
 ing these facts in mind, the projections of the cube are easily ob- 
 tained. To put the original object on the paper as required in the 
 problems, it is necessary to note the difference between the diagonal of 
 a face of a cube, and a diagonal of the cube itself. The first is a line 
 joining the opposite corners of any one of its faces (as d, Fig. 99), and 
 the latter is an imaginary line joining the vertices of any two of its 
 opposite solid angles, or those formed by the junction of three adjacent 
 
 59 
 
GO FIRST PRINCIPLES OF 
 
 faces of the solid, as the dotted line D in Fig. 99. Our first problem 
 in the projection of this simple solid is 
 
 Problem 30. Given the front elevation of a cube, with the diagonal 
 of one of its faces parallel to the VP, and perpendicular to the HP, 
 to find its side elevation, or a view of it looking in the direction of 
 the arrow x. To solve it 
 
 Draw in on the left side of the sheet of paper, with the 45 set- 
 square, the square A, B, C, D (Fig. 100), with its diagonal AC at 
 right angles to the IL, as shown. This figure will be the front 
 elevation of the given cube in the position stated in the problem. To 
 find its side elevation, through points A, B, C, draw projectors of in- 
 definite length parallel to the IL, and as the back face of the cube is 
 parallel to its front one and the YP, draw a line at right angles to the 
 IL, cutting the projectors from A, B, C, in a, b, c ; this line will re- 
 present the back face of the cube. Then, as the edges of a cube are all 
 of the same length, set off from a, on the projector drawn through it, a 
 distance a a', equal to the length of a side of the cube, as AB, and 
 through a' draw the line a' b' c parallel to a b c ; the four bounding lines, 
 a c, c c, c a, a' a', and the line b V when joined up as shown give the 
 side elevation (Fig. 101) of the cube when looked at in the direction of 
 the arrow x. Next 
 
 Problem 31. Let the cube be cut by a plane through its diagonal AC, 
 and let it be required to give a side elevation of it when the part to 
 the left of the cutting plane is removed. 
 
 Here the part of the cube to the left of the diagonal AC being 
 removed, only four edges of the part left are seen. These are the top 
 edge from point A to the point beyond it nearest to the VP, and the 
 corresponding bottom edge from point C ; also the two edges of the 
 front and back faces of the cube, cut through by the section plane. 
 Therefore, in the projector drawn through A, Fig. 100, at any con- 
 venient point (say d), draw, as in Fig. 102, at right angles to the IL, 
 the line d e, and parallel to it, at a distance equal to the length of the 
 side of the cube, the line d' e ; join the points d d', e e, as shown, and 
 the required projection is obtained. Again 
 
 Problem 32. Let the original object the cube be cut by a section 
 plane, as SP, and a side elevation of it be required, after the part 
 cut off to the left of the cutting plane is removed. 
 
 In this case the section plane cuts through two adjacent sides of 
 the cube, leaving parts of those sides, as SA and PC, in view. To 
 obtain the projection required, at any convenient point (say./"), in the 
 projector drawn through A, Fig. 100, draw a line fg perpendicular to 
 the IL, and parallel to it at a distance ff, equal to the length of a 
 side of the cube, the line fg ; through the points AS, PC, and parallel 
 to the IL, draw the lines ff, ss, pp ', gg', and the required projection 
 is obtained. As in Fig. 102, the section obtained is that produced by 
 
MECHANICAL AND ENGINEERING DRAWING 
 
 61 
 
 X 
 
 L V 
 
62 FIRST PRINCIPLES OF 
 
 the cutting of the cube through the plane of one of its diagonals, it is 
 evident that no greater section could be got, and therefore its whole 
 surface must be section-lined, as shown. In Fig. 103, however, as the 
 cutting plane SP leaves a portion of the faces AB, BC of the cube 
 untouched, section-lining is only required on the part of the cube 
 actually cut. through by that plane, as shown. 
 
 28. As some further problems in connection with the projections of 
 the cube will follow, it will be well at this point to more fully explain 
 the significance of the lines obtained in the three preceding projections, 
 as upon their correct comprehension depends the ease or difficulty with 
 which subsequent ones will be found. 
 
 As in Fig. 100, the points A, B, C, D are the front ends of four 
 edges of the cube perpendicular to the YP, and parallel to the HP, 
 any edge produced by a section or cutting plane passing through the 
 cube in a direction perpendicular to the VP, and making any angle 
 with the HP, will be a line parallel to the HP, and at right angles to 
 the lines representing in projection the two faces of the cube which are 
 parallel to the YP. And as these faces are distant from each other 
 a space equal to the length of a side of the cube, the projection of 
 these faces will in all cases be lines parallel to each other at that 
 distance apart, their projected lengths being dependent upon the angle 
 the cutting or section plane makes with the HP. 
 
 This reasoning will be verified on applying it to the three projec- 
 tions given in Figs. 101, 102, and 103. In these figures the lines 
 ac, a'c ; de, d'e ; fg, fg', are the projections of the front and back 
 faces of the cube, the three parallel lines aa bb', cc in Fig. 101 being 
 the projections of the three edges of the cube at right angles to those 
 faces, and the equal parallel lines dd', ee the projections of the top 
 and bottom edges of the cube, of which A and C in Fig. 100 are the 
 front ends. In Fig. 103 the boundary lines of the projection are the 
 same as in Fig. 101, but the part of the cube exposed by the action of 
 the section plane SP, is that between the parallel lines ss, pp' 9 and 
 has to be section-lined, as shown. Had all the faces of the cube been 
 coloured, only those parts of the two seen when looking in the direc- 
 tion of the arrow x viz., from the edge at S to that at A, and from 
 P to that at C would show of that colour ; the surface exposed by 
 the section being, of course, that of the material of which the cube is 
 made. The faces of the cube parallel to the YP in Fig. 100 are, of 
 course, in Fig. 103 seen only as lines, as aty^, f'g', and as the cube is 
 of solid material the edge of it at D, directly opposite to that at B, will 
 not be seen in the side elevation. 
 
 As further problems in the projection of solids in the "upper 
 plane," we give those shown in Figs. 105, 106, and 107, where the 
 original object or cube, Fig. 104, is cut by section planes 1, 2, 3, 
 making different angles with the HP, but all of them perpendicular to 
 the YP. As there is really no material difference of procedure in 
 obtaining such projections of the solid from that already so fully 
 explained in the previous problems, it is not necessary here to go 
 through the process in detail, as the construction lines given in con- 
 nection with each figure are sufficient to enable the student to work 
 
MECHANICAL AND ENGINEERING DRAWING 
 
 63 
 
 
 
 &. 
 
 I 
 
 I 
 
 1 
 
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 V" 
 
 to 
 
 
 ^ i 
 
 1 
 
 I 
 
 1 
 
 i 
 
 1 
 
 I 
 
 1 
 
 I 
 
 1 
 
 1 
 
 1 
 
 j 
 
 rf- 
 
 '<*> 
 
 s 
 
 1 
 
 I I 
 
 vk 
 
 "K 
 to. 
 
 ^ ^ 
 
 I 
 
 1 1 
 
 fc 
 
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64 FIRST PRINCIPLES OF 
 
 out the problems without further explanation. He has only to beai 
 in mind, as before advised, the relationship of the faces and edges oi 
 the original object the cube to each other, and knowing this, nc 
 possible difficulty should be met with in obtaining the required projec 
 tions. A couple of sheets of paper should at the least be devoted tc 
 these projections, varying the direction of the plane of section in each 
 problem, so as to thoroughly master any apparent difficulty that mighl 
 arise in any similar problem in the future. 
 
 29. Advancing from the simple to the more difficult, we again take 
 the cube as the original object, but instead of it being solid through 
 out, as in the last problems, it is now hollow and of an equal thickness- 
 of material all over. To indicate this, in drawing in on the sheet oi 
 paper a front or other elevation of the object, we have to resort to the 
 use of " dotted " lines. In mechanical drawings especially, such lines- 
 are invariably used to indicate those parts of an object not directly ir 
 sight, and it will be found, as we proceed in the study of this special 
 kind of drawing, that although appropriated more especially to this 
 purpose, their use is indispensable to the draughtsman, for by theii 
 proper application an insight is given on the mere inspection of s 
 drawing into the internal structure of the object depicted. In tht 
 case before us, we indicate by their use the thickness and disposition oj 
 the material of which the object is made, for, being opaque, its out ware 
 appearance would be the same whether solid or hollow. Drawing in 
 then, as in Fig. 100, Sheet 3, the front elevation of a cube with one oj 
 its diagonals perpendicular to the IL, and indicating by dotted lines a.' 
 shown in Fig. 108, Sheet 4, the thickness of the material of which it if 
 made, we proceed to obtain by projection the various elevations required 
 Let the first be 
 
 Problem 33. The front elevation of a hollow cube being given, to fine 
 its side elevation, or a view of it when looked at in the direction oj 
 tJie arrow x. 
 
 For convenience, letter the four corners of the front face of th< 
 cube Fig. 108 as in Fig. 100, and obtain a side elevation of it as ir 
 Fig. 101, lettering it in the same way. The two views, so far, an 
 identical; but to show that the cube represented by Fig. 108 is hollow 
 set off from points b, b' the thickness b 5', b 5" of the cube equal to 
 in Fig. 108. Through 5', 5" draw dotted lines parallel to ac, ac' ; anc 
 from points 3, 4 (Fig. 108) draw projectors to cut the dotted line; 
 drawn through 5', 5" in points 3, 3' ; 4, 4' ; dot in the lines betweer 
 the last-named points, and the required elevation (Fig. 109) ii 
 obtained. Next 
 
 Problem 34. Let the cube Fig. 108 be cut by a plane passing throug) 
 its diagonal AC, and an elevation of it be required, ivhen the par 
 to t/ie left of the cutting plane is removed. 
 
 Proceed as before to obtain, as in Fig. 102, the bounding line: 
 dd', d'e'y e'e, e'd of the section ; set off at 5' 5" from de, d'e tin 
 
MECHANICAL AND ENGINEERING DRAWING 
 
 65 
 
66 FIRST PRINCIPLES OF 
 
 front and back faces of the cube the thickness of its sides, and 
 through 5' 5" draw lines parallel to those sides ; through points 3, 4 in 
 Fig. 108 draw projectors to cut the parallels drawn through 5' 5" in 
 points 3' 3", 4' 4" ; the surface included between the inner and bound- 
 ing lines of the figure is that exposed by the cutting plane AC in 
 passing through the cube, and is indicated by cross or section-lining. 
 As the removal of the part of the cube to the left of the cutting plane 
 AC exposes its interior, the angle formed within it at point 5, by the 
 meeting of the sides AD, DC, will be seen as a line between points 
 5" 5" in Fig. 110. With the drawing in of this line as shown, the 
 required sectional elevation is complete. Again 
 
 Problem 35. Let the cube Fig. 108 be cut "by a plane 8P, and let it 
 be required to give a side elevation of it, when the part cut off to the 
 left is removed. 
 
 Proceed as in Fig. 103, Sheet 3, to obtain by projection the top and 
 bottom edges, and the front and back faces of the cube, and letter 
 them as before. Through points S P in the section plane Fig. 108, 
 draw projectors parallel to the IL, cutting the line f g (Fig. Ill) in 
 points sp, and through these points parallel to ff' draw the lines s's", 
 pp. Then to show the thickness exposed by the cutting plane in 
 passing through the sides AB, BC of the cube, through points 1, 2 
 (Fig. 108), and parallel to the IL, draw in Fig. Ill the lines IT', 
 2 '2". Parallel to, and at a distance from the front and back faces 
 fg, f'g', equal to the thickness t of the cube, draw the lines 1', 2' ; 
 1", 2" : cross-line the surface between the inner and bounding lines of 
 the section as shown, and as in the case of the section obtained in 
 Fig. 110 the interior of the cube is exposed, a line will be seen at the 
 junction (point 5) of its two inner inclined surfaces ; with the drawing 
 of this line 5' 5" the required sectional elevation is complete. 
 
 As all the construction lines and the reference letters in the 
 further examples given in Figs. 112 to 115 are shown in, it is left 
 to the student to work them carefully out without further explan- 
 ation. No difficulty need be experienced with either of the sectional 
 projections, provided due thought is given, as before advised, to 
 the relative positions of the faces and edges in the original object. 
 Throughout the problems, the views required are in all cases those of 
 the part of the object left, when that to the left of the cutting plane 
 is removed. 
 
 30. Having satisfactorily worked out the problems in Sheet 4, the 
 student will now be able to proceed with the following, which will require 
 on his part a closer study of the construction of the original object 
 than before, for although it is still in the form of a cube, the per- 
 foration of its sides by openings, as shown in Fig. 116, Sheet 5, will 
 involve greater attention to the method of procedure in obtaining its 
 projections than has before been necessary. This, however, is to be 
 expected in drawing, as in every other art worthy of study or 
 acquisition. 
 
 On an inspection of Fig. 116, it will be seen that the cube is in the 
 
MECHANICAL AND ENGINEERING DRAWING 67 
 
 same position with respect to the VP and HP as in the two last sheets 
 of problems that is to say, the diagonals of its front and back faces 
 are respectively perpendicular and parallel to the HP, the faces them- 
 selves being parallel to the YP. Each side of the cube, however, 
 instead of being, as in the last problems, solid, has now a square hole 
 through it, the sides of the holes being parallel to the sides of the cube. 
 It is therefore possible to see right through the cube from any of its 
 six sides. The consideration of these simple facts in connection with 
 the original object will be found of service in the attempt to obtain 
 any required projection of it. Let the first be 
 
 Problem 36. Given the front elevation of a hollow cube, with square 
 openings in a central position in each of its sides, and a diagonal 
 of its front face perpendicular to the HP, to find its side elevation 
 when viewed in the direction of the arrow on tlie left. 
 
 First draw in the front elevation of the cube, showing the thick- 
 ness of the material by dotted lines, as in Fig. 108, Sheet 4. Letter the 
 corners of it A B C D as before. Divide each of the edges AB, BC, 
 of the cube into three equal parts, in the points 1, 2, 3, 4, and through 
 these points draw faint lines parallel to the edges BC, AB respectively 
 across it. The square abed formed by the intersection of these four 
 lines gives the opening on that face of the cube, and the one directly 
 behind it, and the position of those on the other four faces is shown by 
 the parts of the same four lines which cross the thickness of the sides 
 of the cube at the points 1 to 8 in their edges. On putting in the 
 square a b c d in full lines, and dotting in those last referred to, the 
 elevation of the cube shown in Fig. 116, Sheet 5, is that specified in the 
 problem. 
 
 To give the side elevation required, proceed to obtain first, as in 
 Fig. 109, an elevation of the cube without openings in its sides. Then, 
 to show the openings that will be seen when looking in the direction of 
 the arrow x, we have to remember that only two faces of the cube (of 
 which AB, BC, are the front edges) are in sight, and that therefore 
 only the two openings figured 1, 2; 3, 4 are seen. Now, as the sides 
 of these openings are one-third the length of the sides of the cube, and 
 as the four edges of these openings figured 1, 2, 3, 4 will be seen of 
 their actual lengths, to show them in the position they occupy on the 
 cube, set off in points y z from the lines ac, a'c' (Fig. 117) the front 
 and back faces of the cube the distance (a 1 or b 3) that the edges of 
 the square opening abed are from the edges of the cube. Through 
 these points draw short lines in the upper and lower half of the figure 
 parallel to ac a'c", and from points 1, 2, 3, 4 (Fig. 116), and parallel 
 to aa or cc (Fig. 117) draw lines cutting those drawn through y z in 
 1' 1", 2' 2", 3' 3" 4' 4". The outer edges of the two openings seen from 
 x will thus have been obtained. Two of their inner edges, one at e and 
 another at f, are also seen, the first at e being that of the top side of 
 the upper opening, and that at f the corresponding edge of the lower 
 opening. Projecting these points over on Fig. 117, and drawing in the 
 lines they represent, will give their elevation. To complete the view of 
 
68 FIRST PRINCIPLES OF 
 
 the cube required, we have yet to show the position of the two open- 
 ings in its front and back faces, represented in Fig. 1 1 6 by the square 
 abed. This is found by drawing in in dotted lines in Fig. 117, the 
 projection of the two corners a and c of that square ; the other corners 
 b and d being directly behind the edge of the cube at B, and coinciding 
 with it, will not be seen in the required elevation. 
 
 Figs. 118 and 119 are sectional side elevations of the same original 
 object, the first being that of the cube cut by a plane passing through 
 ts diagonal AC, and the second by a similar plane at SP parallel to 
 that diagonal. With the foregoing explanation, showing how the out- 
 side view of the cube is obtained, and the assistance of Fig. 110, 
 together with that given by having all the principal projectors shown 
 in for each figure, the attentive student should be able to obtain 
 without further aid or assistance the two elevations of the cube shown 
 in the figures referred to. 
 
 In Fig. 120 is given an elevation of the same hollow cube in the 
 same position as that shown in Fig. 116; but the sectional side eleva- 
 tions of it required by planes cutting through it at 1, 2, 3, and shown 
 in Figs. 121, 122, and 123, although found in exactly the same way as 
 Figs. 118 and 119, necessitate much closer attention on the part of the 
 student in obtaining them, the section planes being purposely drawn in 
 such directions as to make the resulting projections a test of his ability 
 in applying the principles which have previously been so fully ex- 
 plained to him. 
 
 In obtaining the three sections required, the only likely difficulty 
 to be met with may possibly occur at that part of each projection where 
 the cutting plane crosses an opening in one or other of the sides of the 
 cube. No. 1 section plane, it will be noticed, is shown to cut three 
 such openings viz., the one immediately in front and its fellow one at 
 the back of the cube, and the lower opening at f ; No. 2 plane cuts 
 through the upper left-hand opening at 6, the lower right-hand one at 
 d, and the front and back ones, dividing the cube into two equal parts ; 
 and No. 3 plane cuts through the upper right-hand opening at g, and 
 across the corners of the front and rear openings. 
 
 31. In showing the parts of the sides of the cube in section in the 
 three views, care must be taken to note especially where the section 
 plane enters and leaves the solid parts of the object, and crosses the 
 open parts, for these points determine what parts are in section and 
 seen, and what are not. In Fig. 121 the upper part of the projection 
 is shown as being solid right across, because the section plane cuts 
 through a solid part of the cube's side. The same is seen in the lower 
 part of Fig. 123 ; but in the opposite parts of each of these figures the 
 section plane has cut through an opening in that side of the cube, and 
 this is indicated by the gaps shown. The same reasoning will explain 
 the projection obtained at Fig. 122, only in this instance the section 
 plane has equally divided the four openings it passes through. With 
 the foregoing explanation, the three sectional elevations of the original 
 object should be obtained without difficulty, projectors being shown in 
 the front . elevation of it (Fig. 1 20), partly drawn in, from all the 
 important points cut through by the section planes. 
 
MECHANICAL AND ENGINEERING DRAWING 
 
 I 
 
 m 
 
70 MECHANICAL AND ENGINEERING DRAWING 
 
 As the side elevations of the prism and the pyramid are obtained in 
 the same way as those of the cube so fully shown in Sheets 3, 4, and 
 5, a few problems in their projection are given in Sheet 6, for the 
 student to solve without further aid than that afforded by the con- 
 struction lines shown in the diagrams. 
 
 Fig. 124 is the elevation of a square prism, assumed to be solid in 
 the first instance, with its base resting on the HP, and two of its sides 
 adjacent ones respectively parallel and perpendicular to the VP. 
 It is required to give side elevations of the prism looking in the 
 direction of the arrow x when cut by the section planes A and B. 
 Then, assuming the prism to be hollow, but with closed ends, as shown 
 by the dotted lines in Fig. 124, side elevations of it are required when 
 cut by the same section planes as before. 
 
 Fig. 125 is the elevation of a square pyramid, with its axis 
 vertical, and two adjacent edges of its base respectively parallel and 
 perpendicular to the VP. Side elevations as in the previous prob- 
 lems are required of the sections produced by the cutting planes C, D : 
 first, assuming the pyramid to be solid, and then hollow. In obtaining 
 the projections of the pyramid, the student must not forget that as 
 each of its sides are triangles, their width at any height from the base 
 varies with that height, and must be found accordingly. 
 
 After some further practice in " projection in the upper plane," by 
 devoting a sheet or two of paper to finding the projections of the cube, 
 prism, and pyramid, cut by planes drawn in other directions than those 
 given in the problems, the student will be enabled to enter upon the 
 next stage in advance in our subject. 
 
CHAPTER X 
 
 PROJECTION FROM THE LOWER TO THE UPPER PLANE 
 
 32. IT has already been shown in Figs. 70, 71, that if a point in the 
 HP be the plan or horizontal projection of a line, then the line is a 
 straight one, perpendicular to the HP and parallel to the VP, and 
 that its " elevation " is obtained by drawing through the given point 
 a straight line in the VP, perpendicular to the IL, of a length equal 
 to that of the line represented by the point. It is also shown in the 
 same figures that if a line be represented in plan or in the HP by 
 a line of the same length as its original, perpendicular to the IL or 
 VP, then the elevation of that line will be a point in the VP, where 
 the foot of the projector drawn through the original line touches the 
 VP. These two cases embrace the principles involved in finding by 
 projection the " elevation of an object when its plan is given," or pro- 
 jection from the lower to the upper plane. 
 
 Having previously worked out the problems of finding the 
 " elevations " of any of the four plane geometrical figures from their 
 "plans," we proceed now to show how the elevations of solid objects, 
 whose sides or faces are plane figures, are obtained when their plans 
 are given. As previously advised in the case of the cube, prism, etc., 
 no difficulty will be met with in obtaining these projections, if the 
 relations of the several sides of the original objects to each other are 
 previously understood. Our first problem in this part of the subject 
 is 
 
 Problem 37. Given the plan of a rectangular slab of solid ma- 
 terial, with its vertical sides inclined to the VP, to find its 
 elevation. 
 
 Let the rectangle ABCD, Fig. 126, Sheet 7, be the plan of the slab 
 in the position stated in the problem. Thus shown, all its sides and 
 ends are rectangular plane surfaces, the upper and under ones being 
 assumed to be parallel to the HP. The parts of it that will be seen, 
 looking in the direction of the arrow x, will be the end AB and 
 the side BC. The points A, B and C are the upper ends of the corner 
 edges, or lines, which have a length equal to the thickness of the slab ; 
 therefore to find its elevation, through points ABC, draw projectors 
 Aa, B6, Cc perpendicular to the IL into the upper plane or VP. If the 
 
 71 
 
72 FIRST PRINCIPLES OF 
 
 slab is resting on the HP, set off its thickness on the projector from A, 
 from the IL, as at a, and through a' draw a line parallel to IL 
 between a and c ; then Fig. X will be the elevation required. 
 
 But if the slab is not resting on the HP, but is assumed to be 
 some distance above it, then on the projector from A set off from the 
 IL the assumed distance that the top surface of the slab is from the 
 HP, say at a ; through a draw a line parallel to the IL cutting the 
 projectors from B C in be; then a b c will be the elevation of the 
 points A B C in the plan, and the line drawn through them that of the 
 top surface of the slab. Set off from a, on its projector towards the 
 IL, the thickness of the slab ; through this point draw a line parallel 
 to a be, or the IL, and Fig. Y will be the . required elevation. The 
 corner D of the slab will of course not be seen in the elevation ; but 
 to show that the form and position of the slab are understood by the 
 student, he should indicate its position by a dotted line, as shown. 
 
 Our next object is that of a solid with inclined sides, and the 
 problem i; 
 
 Problem 38. Given the plan of a frustum of a square pyramid with 
 its base on the HP, and its base edges inclined to the VP ; to 
 find its elevation. 
 
 Let Fig. 127 be the plan of the frustum in the position specified, 
 A B C D being the four corners of the base, and let its height be equal 
 to the length of the side ab of its upper surface, shown in plan. Find 
 the elevation of the points A B C in the plan, by projectors to the IL, 
 cutting it in points A', B', C'. On the projector from A produced, set 
 up from the IL the height of the frustum in point a", and through it 
 draw indefinitely a line parallel to the IL ; then the points a', b', c in 
 this line, where the projectors from a, b, c in the plan cut it, will be the 
 elevations of the three corners of the upper face of the frustum, seen 
 when looking at it in the direction of the arrow x. Join a A', b' B, 
 c'C' by straight lines, as shown, and the required elevation is obtained. 
 Proceeding from the simple to the more difficult, our next problem 
 is 
 
 Problem 39. Given the plan of the frustum of a square pyramid 
 resting on the HP, and surmounted by a cube, its sides being inclined 
 to the VP ; required its elevation. 
 
 Let Fig. 128 be the plan of the combined solid, in the position 
 given in the problem, and let the height of the frustum be equal to the 
 length of a side of the cube. Then, having found the elevation of the 
 frustum, as in the last problem, find by projectors from points 1, 2, 3 
 in the plan the elevation of the three corners of the cube, seen when 
 looking in the direction of the arrow. On the projector from point 
 1 in the plan set off from the upper face of the frustum, or the line 
 a' If c, the height of the cube, and through the point 1' draw the line 
 1' 2' 3', and the required elevation is obtained. The elevation of any 
 of the regular plane solids from their plan is found in the same way. 
 For example 
 
MECHANICAL AND ENGINEERING DRAWING 
 
74 FIRST PRINCIPLES OF 
 
 Problem 40. Let Fig. 120, Slieet 8, be the plan of an hexagonal prism 
 with its axis vertical and its base on the HP, and let its height be 
 eqw.il to twice the diameter of the inscribed circle of its base ; required 
 its elevation. 
 
 In the position in which the object is standing with respect to the 
 VP, three of its sides and four of its vertical edges will be seen. Find 
 the elevation of these edges by projectors through the points A B C D ; 
 on the one through A set up the height of the prism equal to twice ', 
 and at this height and parallel to the IL, draw a line cutting the 
 projector from D in d, and the required elevation is obtained. Again 
 
 Problem 41. Let Fig. 130 be tJie plan of a square pyramid, its axis 
 vertical, its base resting on the HP, and its height equal to twice 
 t/te length of the diagonal of its base ; find its elevation. 
 
 As the solid is resting with its base on the HP, the elevation of its 
 three base corners that will be seen viz., A B C will be found in 
 a b c on the IL. Its apex is the point p in the plan where the two 
 diagonal lines which are the plans of its side edges intersect. Find 
 the elevation of the axis by a projector through p ; set off on this from 
 the IL upwards the height of the pyramid in the point p, and join p 
 by right lines with points a' b' c on the IL, and the required elevation 
 is obtained. 
 
 The " sectional " elevation of an object is obtained from its plan by 
 similar methods. 
 
 Problem 42. Let Fig. 131 be the plan of a solid cube resting 
 with one of its faces on the HP, and let it be cut by a plane *S7 J 
 perpendicular to the HP, and an elevation of it be required, when 
 the part X, cut off by the plane, is removed. 
 
 Projectors being drawn from the points A S P C, into the upper 
 piane, as shown, and the height of the cube set off from the IL in point 
 a' ; on the one drawn through A, a line through a, parallel to the IL, 
 cutting the projector from C in c and the cross-lining of the part cut 
 through by the section plane, completes the elevation required. 
 Again 
 
 Problem 43. Let Fig. 132 be the plan of a hollow square pyramid, 
 its height being twice tJie length of a side of its base, and a sectional 
 elevation of it on the line SP be required. 
 
 As the cutting plane passes through the axis of the pyramid, its 
 section will be a triangle. First find the elevation of this triangle, 
 making its altitude the given height stated ; then to show the thickness 
 of the material cut through by the plane, find the elevation of the 
 points 1, 2 in the plan on the IL, and through these points 1' 2', and 
 parallel to the sides of the triangle, draw lines meeting in the axis in 
 
MECHANICAL AND ENGINEERING DRAWING 
 
76 MECHANICAL AND ENGINEERING DRAWING 
 
 the point a, cross-line the part cut by the section plane as shown, and 
 the required sectional elevation is obtained. 
 
 From the foregoing problems it will be seen that to obtain the 
 elevation of an object from its plan alone, without other important 
 data, would be impossible, as the heights and conformation of its 
 different surfaces not seen in the plan must be known before its correct 
 elevation can be attempted. 
 
 It being necessary, at this stage of the student's progress in the 
 study of our subject, that he should know something about the correct 
 lining-in of his work in ink, we shall next explain the proper applica- 
 tion of the different kinds of lines used in mechanical drawings, that 
 he may be able to practise in spare moments on the sheets of drawings 
 in pencil the results of his study, which it is assumed he has preserved. 
 
CHAPTER XI 
 
 LINING-IN DRAWINGS IN INK 
 
 33. ALTHOUGH the student, up to the present stage in his study, has 
 not been called upon to draw anything to scale which necessitates a 
 greater amount of exactness in the use of his pencil and instruments 
 than he may yet have exercised he should still have acquired a sufficient 
 ability in their manipulation to enable him to put in fairly sound and 
 fine lines when necessary. But as the permanence and practical use of 
 a drawing especially one of any engineering subject are matters of 
 necessity and great importance, it must be committed to paper in a 
 better medium for its preservation than that offered by the use of 
 plumbago. The lead-pencil is only employed for the rapid committal 
 to the paper of the ideas embodied in the drawing, but for the preser- 
 vation of these, and for constructive purposes, the design must be 
 fixed in some coloured pigment or ink. As previously stated at 
 the commencement of this work, China or India ink is the special 
 pigment used by the mechanical draughtsman for this purpose, and it is 
 now intended to explain the proper application of the different kinds 
 of lines used in inking or lining-in an outline mechanical drawing. 
 
 We may state at the outset in this part of our subject that in what 
 are known as ordinary, workshop, or " shop drawings," only one kind 
 of line is used in inking them in, and that is a firm, sound, black line, 
 about -fa of an inch in thickness. In all good modern workshops, no 
 workman is allowed to decide by measurement with his rule the dimen- 
 sions of any part of a piece of mechanism shown on a drawing, as all 
 such parts are, or should be, dimensioned with figures on the drawing 
 itself before it leaves the drawing office. But for the purposes served by 
 a drawing in outline, or one not intended for shop use, different thick- 
 nesses of lines are necessary to enable it to be properly read and 
 understood. These lines are generally of three degrees of thickness, 
 and are defined as fine, medium, and shadow or thick lines. Their use, 
 however, without some well-defined rule of application, would be futile ; 
 as the very reverse effect of that intended would be produced by the 
 incorrect use of either of them. 
 
 34. Now, as the proper application of these lines is directly con- 
 cerned with the effect caused by light falling on an object, it is a 
 matter of importance in this special kind of drawing that a uniform 
 
 77 
 
78 FIRST PRINCIPLES OF 
 
 rule be adopted with respect to the direction in which the light is 
 supposed to fall upon the object represented. With the free-hand 
 draughtsman or artist, this direction is optional, as he can adapt it to 
 the way he thinks most conducive to effect in showing up any particu- 
 lar object in his picture ; but with the mechanical draughtsman, as his 
 drawings are not representations of objects as they appear to the eye, 
 but are projections obtained by parallel rays from all parts of them 
 falling upon certain planes having definite relations to each other, but 
 represented by his sheet of paper, he has to adopt some rule of illu- 
 minating the visible surfaces of his objects, in accordance with the system 
 he uses in projecting their outlines. 
 
 Although the illuminant is the sun, and its light is diffused equally 
 around, it is generally assumed that we see objects by light coming from 
 above and behind us ; but it is evident that if the light shone directly 
 from behind, the spectator would be in his own light, and part of the 
 object would be in shade. The light must then be assumed to come 
 either from the right side or the left. As a rule, the rays of light are 
 always assumed to come over the left shoulder of the draughtsman in 
 jporo&Z lines, and to strike the planes of projection or the VP and HP 
 of the drawing at an apparent angle of 45 with the IL, or inter- 
 secting line of those planes. The actual direction of the rays is 
 graphically shown in the diagram Fig. 133. 
 
 Let VP and HP represent the two planes of projection, and the 
 line IL the intersecting line dividing them. In these planes draw in 
 ABCD to represent the elevation, and a'b'c'd' the plan of a cube. 
 In the position shown, the front and back faces of the cube are parallel 
 to the VP, and all the others perpendicular to it. Through C in the 
 elevation, and c in plan, draw lines EC, ec' y making angles of 45 with 
 the IL ; then EC and ec will represent the plan and elevation of a 
 ray of light and the apparent direction in which it falls upon the VP 
 and HP. The actual direction, or path of the ray, is from the upper 
 anterior or front corner of the cube at A, to the lower posterior or back 
 corner of the cube behind C. In other words, the ray of light is 
 assumed to travel in a direction coinciding with the diagonal of the 
 cube, drawn between point A and the point beyond C. 
 
 To find the actual angle that this ray of light makes with the 
 planes of projection : At point A in the elevation of the cube erect at 
 A, a perpendicular to AC \ on this set off from A a length Ad equal to 
 a side of the cube ; join d and C ; then the angle ACd is that made 
 by the ray of light with the planes of projection VP and HP. For 
 if the right-angled triangle dAC be supposed to turn on its base AC as 
 a hinge until its plane coincides with AC, then the angular point d 
 will coincide with A, and the hypothenuse dC of the triangle dAC 
 will become, as before stated, the path of the ray of light. The angle 
 made by this ray with the VP and HP will be found to be, both by 
 measurement and calculation, one of 35 16". 
 
 To make the rule adopted by the mechanical draughtsman as to 
 the assumed direction of the light falling on a body still more clear, 
 let the student cut out in stiff paper a model, allowing a strip for gum- 
 ming, as in sketch A, of the right-angled triangle dAC, Fig. 133; 
 
UNIVERSITY 
 MECHANICAL AND ENGINEERING DRAWING V %&= 
 
 r: 
 
 A _ 
 
 H 
 
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 Jiiteva frost- <? 
 
 2*0156. 
 
 
 
 
 
 
 
 
 
 
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 J } fa, f <f 
 
 -ff 
 
 
 
 
 
 
 f r 
 
 
 
 
 
 
 
 
 
 <2 
 
 
 LJ, 
 
 
 '^f 
 
 
 C 
 
 
 
 
 J} 
 
 
 
 
 
 
 z, 
 
80 FIRST PRINCIPLES OF 
 
 fold over the strip on the model, on line AC, until it is at right angles 
 to the triangular part, gum the strip on its under side and apply the 
 model to the elevation of the cube, making the lines AC on each 
 coincide. Then prepare a similar model of the triangle and apply it 
 to the plan of the cube on the line a c. With the sheet of paper on 
 which the plan and elevation of the cube are drawn, fiat on the draw- 
 ing-board, the triangular models will both stand perpendicular to the 
 plane of the paper. If now the VP be bent over forwards on the IL 
 as a hinge until it is at right angles to the HP, the direction of the 
 ray of light with respect to both planes of projection will at once be 
 seen, for the hypothenuses of the triangles with which the ray of light 
 coincides are parallel to each other, and make equal angles with the 
 YP and HP. It will also be proven by the aid of the model, that a 
 line drawn in the YP and HP at an angle of 45 with the IL, cor- 
 rectly represents the elevation and plan of a ray of light when making 
 an angle of 35 16" with those planes. 
 
 35. We have now to explain why a distinction should be made in the 
 quality of the lines used in inking-in an outline mechanical drawing. 
 It will be readily understood on looking at any illumined object that 
 there are some boundaries in its surfaces on which the light will 
 fall direct, and others on which it will not so fall, but will cause them 
 to cast a shadow on some other surface beyond them. Of the former, 
 some are often so much in the light that the edges bounding the sur- 
 faces are almost imperceptible; all such must of necessity be repre- 
 sented by fine lines. Those on which the light does not fall direct, but 
 which cast a shadow on some more distant surface, are represented by 
 thick lines. A medium thickness of line is only used under exceptional 
 conditions of position of the bounding edges of surfaces ; but of their 
 use we shall speak later on. For the present we shall deal only with 
 fine lines, and thick or shadow lines, and their application to plane- 
 surfaced solids, or such as the student has already drawn-in in pencil. 
 Knowing all the forms of the solids which have been taken as subjects 
 for projection, their proper inking-in should present no difficulty. All 
 that has to be determined in connection with the pencilled-in lines is 
 what each represents that is to say, is it an edge or bounding line on 
 which the light falls direct, or does it cast a shadow 1 This decided, 
 the line can be drawn-in in ink. 
 
 Taking the two projections of a cube given in the diagram Fig. 
 133 as an explanatory example let it be determined how they should 
 be inked in. To begin with the elevation, and knowing the direction 
 in which the light falls upon the object, it is evident that on two of its 
 front edges AB and AD the light is falling direct ; their representa- 
 tive lines AB and AD must therefore be fine ones. It is equally 
 obvious that the edges BC and DC will cast a shadow on the YP 
 against which the cube is resting, therefore the lines BC and DC, repre- 
 senting these edges, must be thick, or shadow lines. Three other edges 
 of the cube are also affected by the light viz., those of which A, B 
 and D are the front ends ; but this cannot be indicated in the ele- 
 vation. The cube whose plan is shown in the diagram is for explan- 
 atory purposes set forward some distance from the IL, and must not 
 
MECHANICAL AND ENGINEERING DRAWING 8l 
 
 therefore be taken as that of the cube given in elevation. In this 
 case it is assumed to be standing with one of its faces on the HP. In 
 that position two of its top edges cud' and a'b are directly in the 
 light, and will therefore have to be fine, lines, 'while the corresponding 
 opposite edges b'c and c'd', as they will cast a shadow on the HP, 
 must be thick, or shadow lines. Of the "cast shadows" thrown on 
 the YP and HP by the cube in the positions shown, we shall speak 
 later on. 
 
 36. To show the importance of correctness in the use of fine and 
 thick lines in an outline drawing, we give in Fig. 135 a very simple 
 drawing which may be made, by a different application of such lines, to 
 represent totally different objects. The figure being drawn-in altogether 
 in fine lines may represent anything; say a square thin plate of any 
 material, with a small square lined-in on its surface. To give an 
 idea of substance or thickness in the plate, this would be effected 
 by thickening the lines from A to B and B to D, as in Fig. 136. The 
 inner square would represent a " recess " in the plate by thickening the 
 lines ac and cd, as in Fig. 136 ; but if the opposite sides, ab, bd, of this 
 small square were thick lines, then this square would represent a 
 "projection" on the surface of the plate or slab as shown in Fig. 137. 
 That Figs. 136 and 137 are plans or horizontal projections of objects is 
 evidenced by the position of their shadow, or thick lines, the shadows 
 cast by their projecting edges shown in indicating the direction in 
 which the light falls upon them. The original drawing, Fig. 135, may 
 be either a plan or elevation, there being nothing in its lining to show 
 which is intended. The student should be careful to note that even in 
 this simple drawing of only eight lines, the reversing of the positions 
 of the thick lines bounding two of the sides of the inner square 
 produces not only a different appearance in the object, but gives quite a 
 different reading as to its construction. 
 
 37. From the foregoing explanation of the use and application 
 of a fine and thick line in defining the meaning of an outline drawing, 
 it will not surprise the student to find that an alteration in the position 
 of an object with respect to the light falling upon it may, and possibly 
 will, affect its lining-in. To show the effect of such an alteration 
 in position, let E and P in the diagram Fig. 134 be the elevation 
 and plan of a cube, with four of its faces vertical, and making 
 angles of 45 with the VP. In this position the light falls wholly and 
 directly upon the vertical face Ae Ce, leaving its parallel and opposite 
 one wholly in shade, while the face eB e'D and its opposite one are 
 in the plane of the light. In this case the lining-in of both plan and 
 elevation of the cube becomes quite different to that of the same objects 
 when differently placed with respect to the light falling upon them. 
 
 As in the previous figure, in elevation, the under side of the cube is 
 assumed to be some distance above the HP, but in plan it is supposed 
 to have that side resting upon it. The correct lining-in of the two 
 views in accordance with their altered position will then be as 
 follows : In elevation, the top front edges Ae, eB, forming one 
 line, being directly in the light, must be a fine line. The vertical 
 edge AC, being "in the plane of the light," must also be a fine line. 
 
 G 
 
82 FIRST PRINCIPLES OF 
 
 The edge ee is in the plane of the light, but it being so near to 
 becoming one that would either cast a shadow or be fully in the light, a 
 medium thickness of line would be most appropriate. As the vertical 
 edge BD of the cube would cast a definite shadow on any surface 
 behind it, it must of course be a thick line. The same applies to the two 
 lower front edges of the cube Ce e'D, forming one line, which must also 
 be a thick one. Practice among mechanical draughtsmen varying as to 
 the appropriate use of a medium thickness of line in a drawing, it may 
 be remarked in the case here presented that if "colour" were used 
 to assist the eye in determining the form of the object represented 
 in elevation by E in the diagram, the tint would be dark near the edge 
 ev, and gradually lighten off as it approached the edge BD, although 
 the latter would cast a shadow on any surface behind it. For this 
 reason a medium thickness of line seems best suited to the position. As 
 to the lining-in of the plan P of the cube, it will at once be seen from its 
 position that only one edge, or that between points f and , will cast 
 a shadow, and it must therefore be made a thick line. The other three 
 two in the plane of the light and one having light falling direct upon 
 it must be fine lines. 
 
 38. To prevent the misreading by the student of some engravings 
 of mechanical subjects in outline still to be met with, it is as well here 
 to observe that in England some draughtsmen and engravers line-in 
 both plans and elevations of objects as if they were situated in the same 
 plane, and cast their shadows in the same directions ; but that this 
 is contrary to the principles upon which their projections were obtained 
 may at once be seen from a study of the shadows cast by the cube 
 in plan and elevation in Fig. 133. As the planes of projection are 
 at right angles to each other, it would be impossible for the shadows 
 to fall in the same direction in both planes, unless the light fell upon the 
 objects represented in two totally different directions at one and the 
 same time, which would be absurd. In all representations of objects, 
 therefore, throughout this work, where any difference in the lining is 
 shown, we shall recognize the existence of the two planes of projection, 
 in their pn*oper relative positions, and apply the rules for such lining 
 in accordance with the principles before explained. 
 
 As the study of the projection of solids having curved surfaces will 
 be the next division of our subject that will occupy attention, a few 
 words only are here necessary with reference to the lining-in of such 
 objects in ink. As a general rule, only one kind of line the fine line 
 is applicable to the outline representation of the curved surface of the 
 solids of this class the cylinder, the cone, and the sphere with which 
 we shall deal, consequent on the fact that the particular part of the 
 curved surface of each which would cast a shadow on any adjacent 
 surface, falls within the bounding lines of its representation, and cannot 
 be expressed by a line, as in the case of bodies having plane surfaces. 
 We shall, however, before completing our explanation of the projection 
 of such solids, give all necessary information as to inking them in. 
 
 39. Before leaving this part of our subject a few words are 
 necessary for the guidance of the student in reference to the mixing 
 and usage of the ink with which he will line-in his drawings. Although 
 
MECHANICAL AND ENGINEERING DRAWING 83 
 
 liquid ink for this purpose is now to be purchased, it is advisable for 
 the student to know how to make suitable ink for himself. He will not, 
 of course, without some considerable practice, succeed in making really 
 serviceable ink, but such as he will be able to produce will sufficiently 
 serve his present purpose. 
 
 To prepare the ink for use, let the student provide himself with a 
 medium-sized colour saucer sold in nests of half-a-dozen by any artists' 
 colourman and having procured a stick of good India-ink and a 
 camel' s-hair water-brush, proceed as follows : Charge the brush with 
 clean water, and by its means put as much of it in the saucer as would 
 half -fill an egg-spoon. With the stick of ink held vertically, and with 
 one end in the saucer, rub it in a circular direction, applying a gentle 
 pressure at first, until the water has attained a dark brown colour and 
 begun to thicken ; continue rubbing, gradually adding water in drops 
 from the water-brush, until a sufficient quantity of liquid ink is made. 
 To test its blackness and consistency, blow steadily into the saucer : if 
 the ink is a good black all through, the bottom of the saucer will not be 
 seen; but if too thin, it will come in sight, and the rubbing must be 
 continued. For the ink to be in working order it should be neither too 
 thin nor too thick. To bring it to a proper consistency, about a quarter 
 of an hour's rubbing will be required. If it is found at any time to 
 have thickened by evaporation of the water in it, add more in drops 
 by the use of the water-brush, keeping it well stirred while doing so. 
 
 Some writers on this subject advise the student to always use fresh- 
 rubbed ink, taking care to wash out the saucer before making it. This 
 is by no means the practice in many of the largest drawing offices 
 in the kingdom. On the contrary, good ink is never wasted, but is 
 improved by being worked up afresh. The writer's practice for years 
 has been to keep by him in his instrument-box a home-made india- 
 rubber pestle, similar in shape to a diminutive Indian club. With the 
 use of this and a few drops of clean water added as required in the ink- 
 saucer, he is able to produce an ink of equal density and colour 
 throughout, and which does not contain a particle of anything held in 
 suspension in it that would be liable to clog the drawing-pen. 
 
 Special unglazed saucers are now to be purchased for the rapid 
 making of India-ink, but these the writer finds very extravagant sub- 
 stitutes for patience and the expenditure of a few minutes of time. A 
 more suitable saucer would be one made of glass of the same shape as 
 now used, but with its inner surface finely ground. Through want of 
 patience on the part of the draughtsman, and the application of too 
 much pressure while rubbing, the ink is liable to break away from the 
 stick in small lumps, which are never thoroughly reduced to the liquid 
 state, but cause constant annoyance by clogging the pen. With a 
 properly-ground glass saucer this tendency of the ink-stick to fray away 
 in particles would be avoided. One great mistake made by many 
 draughtsmen is the practice of dipping the rubbing end of the ink-stick 
 in water before starting to make their ink. By so doing it becomes 
 soddened, and when dry will be found to be cracked all over at that 
 end, and ready to break away in pieces on the least pressure being 
 applied. The rubbing end of the ink-stick should always be wiped dry 
 
84 MECHANICAL AND ENGINEERING DRAWING 
 
 directly it is done with, if it is wished to keep it in good order for use ; 
 and the ink saucer should only be uncovered while filling the pen. This 
 latter operation is best effected by capillary attraction, and not by the 
 use of a brush. When more ink is required, first pass a piece of soft 
 wash-leather between the nibs of the pen, to remove any deposit, and 
 before dipping in the ink draw it through the lips, which will give 
 sufficient moisture to cause the ink to flow freely between its nibs, and 
 so fill it. The use of a brush is objectionable, on the ground of its 
 being liable to pass into the pen any sediment that may have collected 
 in the ink-saucer. 
 
 With reference to the drawing-pen and how to use it the compass 
 or bow pens will not at present be required the student will remember 
 the explanation of its construction given at the commencement of this 
 work, and from that will understand that in using it, it must be 
 held nearly upright, keeping its strong nib in contact without pressure 
 with the edge of the tee- or set-square ; that all horizontal lines must 
 be drawn from left to right, and vertical ones upwards, or from bottom 
 to top of the board. The drawing-pen when purchased should be set 
 for use, and its satisfactory manipulation will soon be acquired if due 
 care is taken not to damage its points. 
 
CHAPTER XII 
 
 THE PROJECTION OF CURVED LINES 
 
 40. UP to the present stage in our subject we have purposely 
 refrained from dealing with figures having curved bounding lines, or 
 solids with curved surfaces, knowing how important it is for the student 
 to thoroughly understand the projection of straight-lined plane figures 
 and plane-surfaced solids, before attempting anything in which a curved 
 line or surface occurs. Assuming that he has mastered all that has now 
 been explained, he will be the better able to apply himself to that part 
 of the subject to which we next proceed viz., the projection of solids 
 having " curved " surfaces. As this, however, involves a knowledge of 
 the projection of curved lines, this must first be acquired. 
 
 From our definition Chapter III. of a curved line viz., that it is 
 the path of a point which is continually changing its direction it will 
 at once be manifest that its projection cannot be obtained in the same 
 way as that of a straight line, in that all that is here required is to find 
 the projections of its ends and join them by a right line. As the 
 generating point of a curved line is moving in a different direction at 
 all parts of its path, it follows that the only way in which its projection 
 can be obtained is by getting the projection of that point in several of 
 its positions, and joining them by a line passing through them, which 
 will be the projection sought To obtain this, however, it is first 
 necessary to have a perfect knowledge of the original curved line 
 and its varying direction, as it may be a line lying wholly in a plane, 
 or, on the other hand, one of such a character that no part of it would 
 coincide with or lie in a plane. For a first example in curved line 
 projection we take as our problem 
 
 Problem 44 (Fig. 138). Given the front elevation AB of a curved 
 line in a plane parallel to the VP, it is required to find its side 
 elevation and plan, or views of it when looked at in the direction of 
 the arrows. 
 
 Here the line being a simple curve, it is evident, on looking in the 
 direction of the arrow on the left, that its projection will be a straight 
 line ab, No. 1, as all the points in the curve are in one and the same 
 
 85 
 
86 FIRST PRINCIPLES OF 
 
 plane. For the same reason, the plan of this curve will be the straight 
 line a'b'j No. 2, obtained by letting fall projectors from its two ends 
 AB into the HP, and joining them by a right line. 
 
 Now, let the plan of this curved line or ab', No. 2 be assumed 
 to make an angle of 40 with the YP, and its front and side elevation 
 be required. To obtain these, we must know the position of some 
 points in the curve with respect to a fixed datum line, from which 
 measurements may be taken. Let the points in the curve of which 
 ab' is the plan be A, 1, 2, 3, B, in the elevation, and let the line CB, 
 drawn through B at right angles to the IL, be the datum line, which 
 is assumed to be in the same plane as AB. Through A, 1, 2, 3, B, 
 draw lines parallel to the IL, cutting CB in points C, 1', 2', 3', B. Now, 
 the line a"b' in the HP is the plan of the curved line AB in elevation, 
 moved through an angle of 45 with the YP, the point a" being nearest 
 the eye, and b' farthest from it. To find the plans of the intermediate 
 points 1, 2, 3, in the curve AB, set off from b' in the plan the distances 
 CA, 11', 22', 3 3', in elevation in the points a", x, y, z. Through these 
 draw projectors -into the YP, cutting those from A, 1, 2, 3, in c, d,f, g. 
 Then a line drawn through these points to B will be the required 
 elevation of the curve. 
 
 To find its side elevation, take a new datum line C'B', No. 3, parallel 
 to CB, and produce the projectors drawn through A, 1, 2, 3, B, in the 
 original curve line AB, No. 1, to cut it in points C', 4, 5, 6, B' ; at 
 each of these set off in points c, d', f, g', the distances measured from 
 the datum line that their corresponding points a", x, y, z, in the plan 
 No. 2 are from a' b', measured on the projectors drawn through them ; 
 then a line drawn through the points so obtained, as shown in No. 3, 
 will be the side elevation of this curved line in its new position. 
 
 If the given line of which the projections are required is one of 
 double curvature, but still in one plane, its projections are obtained in 
 the same way as those of a simple plane curve. As an example we 
 will take a very common double-curved line known as the ogee moulding ; 
 and let the problem be 
 
 Problem 45 (Fig. 139). Given the front elevation of a line of double 
 curvature AB, No. 1, to find its plan and side elevation, the plane of 
 the curve "being parallel to the VP. 
 
 In this case, as every point in the curve is at an equal distance 
 from the YP, its plan will be a straight line a'b', No. 2, parallel to the 
 IL, and its projected length will equal the distance between projectors 
 let fall from its ends A and B into the HP. Its side elevation will 
 also be a straight line ab, at right angles to the IL, for as every point 
 in it is equi-distant from the YP, it will coincide with or lie in a plane 
 which we are told in the problem is parallel to the YP, and therefore 
 perpendicular to the HP. 
 
 To find the elevation of the line AB when the plane in which it 
 lies makes any given angle with the YP, proceed as follows : 
 
 Choose any convenient points other than its two ends in the 
 curve as 1, 2, 3, 4, No. 1. Through these let fall projectors into the 
 
MECHANICAL AND ENGINEERING DRAWING 
 
 I 
 
 87 
 
88 
 
 FIRST PRINCIPLES OF 
 
 HP to cut ab', No. 2, the plan of AB in the points 1', 2', 3', 4'. At 
 b', in this plan, draw a line making with ab' an angle equal to that 
 say 45 the plane containing the curve line is to make with the YP 
 Transfer by arcs struck from b' the points 1', 2', 3', 4' in ab' to 
 a"b', and through these draw projectors into the YP, cutting those 
 drawn through A, 1, 2, 3, 4, No. 1, to the datum line CB in c, d, e,f, 
 g ; a curve drawn through these points, as shown, will be the elevation 
 required. In this way the elevation of any curved line lying in a plane 
 making any angle with the YP, may be found, for the line, or the plane 
 in which it is supposed to lie, may be assumed to rotate round the 
 datum line as an axis, carrying with it all its points, the position of 
 which with respect to the YP may always be easily determined, and 
 from them the projections of the lines in which they lie may be found. 
 41. To find the projections of a compound curved line, or one no 
 part of which would coincide with or lie in a plane, the process is some- 
 what more difficult, as its appearance in two directions at right angles 
 to each other must be known, or given, before a third view of it can be 
 obtained. As an example 
 
 Fig, 140 
 
MECHANICAL AND ENGINEERING DRAWING 89 
 
 Let AB, No. 1 (Fig. 140), be the elevation of such a curved line, 
 and ab, No. 2, another elevation of the same line, at right angles to 
 the first. It is evident that such a line could not coincide with a 
 plane, yet if the relation of any points in it with the planes of its 
 projection be known, or can be determined, then the plan or horizontal 
 projection of the line can be found. Let, then, the line ab, No. 2, be 
 the front elevation of the line, the VP being behind it, and AB, No. 1, 
 its side elevation, looking in the direction of the arrow x. Then the 
 end A or a of the line is nearest the YP, or farthest from the eye ; 
 and the end B or b the converse. To prove this, through A, No. 1, 
 and b, No. 2, draw the lines Ap, cb, perpendicular to the IL ; then Ap 
 will represent an end elevation or edge view of the YP looking in the 
 direction of the arrow x and cb that of a plane at right angles to the 
 YP. With the assistance of these as datum planes we can determine 
 the distance of any point in the line AB from each of them, and thus 
 obtain its horizontal projection. A line drawn through a series of 
 points thus found will be the plan of the curved line sought. 
 
 To find it, choose a suitable number of points in the given line AB, 
 No. 1, say at 1, 2, 3, 4 ; project these over on to ab, No. 2, and produce 
 the projectors to meet the datum plane cb, as shown ; the exact position 
 in plan of any of these points can now be fixed. From a, 1', 2', 3', 4', 
 b, No. 2, let fall projectors into the lower plane, or HP, and on these, 
 in No. 3, set off from the IL the distances that the corresponding 
 points in AB, No. 1, are from Ap ; through the points thus found draw 
 the line a' b', and it will be the plan of the given compound curved line 
 sought. By letting fall similar projectors from the points in AB, No. 
 1, and setting off on them the distances that their corresponding points 
 in ab, No. 2, are from the datum plane cb, a similar plan of the com- 
 pound curved line will be obtained, as shown at a f B', No. 4, to that 
 found in No. 3, but at right angles to it, as AB, No. 1, is at right 
 angles to ab, No. 2. 
 
 From the foregoing explanation the student should be enabled to 
 apply the principles involved to the projection of any single curved 
 line, whether simple or compound ; but to show the actual appearances 
 the same curved line will assume in projection when occurring in 
 different positions, we give one problem in this connection before pass- 
 ing on to the projection of curve-bounded figures. 
 
 By way of making the problem more interesting, and of practical 
 application, we will take an ordinary curve such as is given to the 
 foot of a square or six-sided column combined with a straight line, 
 and let the problem be 
 
 Problem 46. Given the elevation of a combined curved and right 
 line parallel to tJie VP, to find its plan and elevation when it makes 
 ang given angle with the VP or HP. 
 
 Let ABC, No. 1 (Fig. 140a), be the elevation of the combined line. 
 Being parallel to the YP, every part of it is equi-distant from it. Its 
 plan, in this position, will be the straight line ab', No. 2, parallel to 
 the IL, its length being determined by projectors from A and B in the 
 
90 FIRST PRINCIPLES OF 
 
 elevation. Produce BC, No. 1 the straight part of the line verti- 
 cally to c', and let c'C be assumed as an axis round which the curved 
 line can be turned through any given angle ; and also as a datum line 
 from which measurements may be taken. In AB the curved part of 
 the line take any convenient points 1, 2, 3, 4; and through them and 
 A draw projectors parallel to the IL, to meet the axis, or datum line, 
 in c', 1', 2', 3', 4'. Now, suppose ABC to rotate on the axis line c'BC, 
 from its normal position parallel to the VP through any given 
 angle ; then its plan at that angle will still be a straight line, but it 
 will be at an angle with the IL or with its plan when in its first 
 position equal to that it has been turned through. Let this angle 
 be one of 60. Then to find its elevation in this new position, proceed 
 as directed in solving Problem 45 by first projecting over on to db in 
 plan the points in the curve, transferring them by arcs to a"b', and 
 from thence by projectors into the upper plane, cutting those drawn 
 through A, 1, 2, 3, 4, B, in the points shown ; then a line through a' 
 and these points to B will be the elevation of the curve in its new 
 position. 
 
 Next, let this newly-found elevation of the combined line be inclined 
 to the HP in such a way that the axial line and with it the part BC 
 of the combined line makes an angle of 45 with the HP or IL ; and 
 let its plan when in that position be required. 
 
 To put on the paper the new elevation, No. 3, in the required posi- 
 tion, draw in the axial line at the required angle, and to it transfer the 
 points in the same line, No. 1 ; through these draw ordinates as shown, 
 and on them set off in points a, 1, 2, 3, 4, the corresponding lengths 
 of the same lines in No. 1, and draw in the curve. To obtain its plan 
 in this position, let fall projectors from the several points in it in the 
 elevation, and through the corresponding points in the plan, No. 2, 
 draw projectors parallel to the IL, cutting those let fall from No. 3 ; 
 a line drawn through the points of intersection of these projectors will 
 give a 1 BC, No. 4, the plan of the line required. 
 
 In the foregoing, Nos. 1 to 4, the axial line about which the com- 
 bined line has been supposed to rotate has in each case been assumed 
 to be parallel to the VP. Let it now be supposed, while still inclined 
 to the HP at the same angle as before, to be also inclined to the VP 
 at an angle say, of 45, and let an elevation of the combined line in 
 this position be required. 
 
 To obtain this, we must first show in on the paper, the plan of the 
 combined line in its new position with respect to the VP or IL, 
 before we can proceed with its elevation. This means that the last- 
 obtained plan of the combined line must be moved from its present 
 position viz., that of having the assumed axial line c' C parallel to the 
 VP to one in which that line will make an angle of 45 with the VP 
 or IL ; and is tantamount to swinging the line on its end C, as a pivot, 
 through an arc of 45 in such a way that each point in it while being 
 swung is kept at the same height in its plane of rotation above the 
 HP. This new position of the last-found plan of the line is therefore 
 correctly shown, by drawing a line the assumed axial one at an angle 
 of 45 with the IL, and transferring to it, as was done in the case of 
 
MECHANICAL AND ENGINEERING DRAWING 
 
 91 
 
 1 
 
92 FIRST PRINCIPLES OF 
 
 No. 3, the various points, ordinates, etc., of No. 4, as given in No. 5. 
 Its elevation is then obtained by producing the projectors drawn 
 through the points a', 1, 2, 3, 4, BC, No. 3, and intersecting them by 
 those drawn through the corresponding points in the plan No. 5. A 
 line drawn through the points where these projectors intersect will give 
 the line a'BC, No. 6, which is the required elevation. 
 
 42. To familiarize himself with the principles involved in this very 
 interesting part of our subject viz., the projection of curved lines in 
 any position the student should accustom himself to make wire 
 templets, of copper or lead, bent to the exact shape of the lines 
 given for projection. Such models, when posed before him in the 
 positions stated in the problems, would materially assist him in 
 obtaining the correct projections required, as he would be able to 
 follow the movements of any points in them in the different positions 
 they assume. Practice with such models is all the more advisable in 
 view of the more difficult problems that will have to be mastered in the 
 projection of solids having curved surfaces, when inclined, or otherwise, 
 to the planes of their projection. 
 
 After the foregoing full explanation of the application of the 
 principles of projection to curved lines, the projection of curved- 
 bounded figures, to which we now pass, will present little or no 
 difficulties. 
 
 As the simplest possible curve-bounded figure is the circle, its 
 projections claim first attention. We have defined such a figure as the 
 path described by a moving point that is always at the same distance 
 from a fixed point, around which it moves. When such a figure is cut 
 out of any material substance, such as a piece of metal plate, or card, it 
 is called a disc. Looked at in various directions, it assumes different 
 forms. The exact form taken is obtained by projection. Our first 
 problem in connection with it is 
 
 Problem 47. Given the plan of a circular plate, to find its front 
 
 and side elevation. 
 
 Let the straight line AB, Fig. 141, No. 1, be the given plan. Now 
 as the edge view of a disc, looked at from above, is a straight line, we 
 see that if AB is such a view, the plate must be perpendicular to the 
 plane of its projection, or the HP, and being parallel to the IL every 
 part of its surface must be parallel to the YP ; therefore its elevation 
 will be a circle and have its bounding-curved line everywhere equi- 
 distant from its centre. To find this bisect AB in C, and through C 
 draw a projector into the upper plane at right angles to the IL. Take 
 any convenient point in this projector at a greater distance from the 
 IL than half AB say a and through a' draw a line parallel to the 
 IL. With a' as a centre and AC, No. 1, as a radius, describe the circle 
 acbd, No. 2, which will be the front elevation of the circular plate. If 
 correctly drawn, vertical projectors through A and B in the plan should 
 cut the diametral line drawn through the centre a' of the circle in the 
 points a and b in that line. Then as the surface of the plate is parallel 
 
MECHANICAL AND ENGINEERING DRAWING 
 
 93 
 
 to the YP, the side elevation of it will be the straight line cd', No. 3, 
 parallel to cd, No. 2, obtained by projectors from c and d, as shown. 
 
 Next, let the line AB, Fig. 142, No. 1, be the given plan of the 
 circular plate, and its elevation be required. 
 
 Here the plate being still represented in plan by a line, is perpen- 
 dicular as in the previous case to the HP, but being inclined to the 
 YP its elevation becomes an ellipse, or a curve-bounded figure in which 
 points in its bounding line are not all the same distance from its centre. 
 To find the elevation, bisect AB as before in C, and through C draw Cc', 
 at right angles to the IL. Choose any point in Cc, as a', and through it 
 draw a line parallel to the IL. Set off in c and d from a on the pro- 
 jector through C, a distance equal to AC or CB, and through A and B 
 draw projectors into the YP, cutting the line through a! in a, b ; then 
 acbd, No. 2, will be four points in the ellipse, the line between c, d being 
 its major axis, and that between a, b its minor axis. Divide AB in the 
 plan into any number of equal parts, say six, in the points 1, 2, C, 3, 4 ; 
 through these at right angles to AB draw ordinates to the semi-circle. 
 Project points 1, 2, 3, 4, in AB on to the minor axis ab in the elevation, 
 and through them draw lines parallel to cd. Set off on these on either 
 
94 FIRST PRINCIPLES OF 
 
 side of ab, the length of the corresponding ordinates to the semi-circle in 
 No. 1, and through the points thus obtained draw the curve of the ellipse 
 as shown in No. 2, which will be the required elevation of the circular 
 plate in its new position. 
 
 43. For the better comprehension of this last problem by the student, 
 before passing to the next a few words in further explanation of its 
 solution is desirable. It has been shown in previous problems in the 
 projection of curved lines that if the position of points in the curve can 
 be determined the problem is virtually solved. In the case of the circle 
 in Fig. 141, it would have been quite possible to have first found any 
 number of points in the bounding line and then to have joined them 
 by a line passed through them, and thus have produced the circle, but 
 the use of the compass obviated the necessity. In the problem before 
 us, the necessity exists for finding exactly several points in the required 
 elevation of the object in the readiest possible way. The original object 
 being a circle, a regular figure as distinguished from an irregular one 
 whose surface is readily divisible into parts, it is apparent that by 
 turning down one half of that surface on the line AB in No. 1 as a 
 diameter, and drawing lines upon it at right angles to AB, through any 
 points in the semi-circle, we can at once find the actual length of those 
 lines and transfer them to their vertical projections in No. 2. The 
 solution of this problem may also be reasoned out in another way. The 
 original line AB, No. 1, is in fact as was shown in the previous 
 problem a curved line, although presented to the eye as a straight 
 one, every point in it, on either side of that at C, being farther from the 
 eye the farther it is to the right or left of C. As the actual length of a 
 line drawn across the circle at C, through its centre to the opposite 
 edge, is known to be equal to AB, or twice Cc, so can the distance 
 across the circle through any other point in AB be ascertained in the 
 same way. For, as the ordinate Cc is measured at right angles to AB, 
 so any other, as 2, 2', is equal to half the distance from the point 2' 
 across the circle to the corresponding opposite point in its edge. 
 With this explanation the following problem should present no 
 difficulty in its solution. 
 
 Problem 48 (Fig. 143). Given tlie plan of a circular plate as a 
 circle ; to find its projections in t/ie VP and HP when making a 
 given angle with either plane. 
 
 Let the circle ABCD, No. 1, be the given plan, the diametral line 
 AB being parallel to the IL, and the similar line CD perpendicular to 
 it. Bisect each quarter of the circle in the points 1, 2, 3, 4. As 
 shown, the surfaces of the plate are evidently parallel to the HP. Let 
 it now be inclined to that plane at an angle of 60, keeping the 
 diametral line AB parallel to the HP, and the point C uppermost, or 
 nearest to the eye. To obtain its elevation when in that position, an 
 edge view of it looking in the direction of the arrow x must first be 
 given, which will be a line equal in length to CD, inclined to the IL at 
 the required angle. Draw in this line CD as in No. 2, and bisect it in 
 the point B, which will be the end of the diametral line AB, nearest 
 
MECHANICAL AND ENGINEERING DRAWING 
 
 95 
 
 the eye in side elevation, A being the point beyond it across the 
 circle. Through B, No. 2, draw a projector parallel to the IL, and in- 
 tersect it in points a, b (No. 3) by projectors from A and B in No. 1. 
 Transfer points 2 and 4 in No. 1 to 2 and 4 in No. 2, and through 
 them draw projectors parallel to that through B ; then from 3, 4, in 
 No. 1, intersect by projectors which will pass through 1, 2 those 
 drawn from 2, 4, No. 2, in I', 2' 3', 4', No. 3. Through the eight 
 points thus found, draw the closed curved-bounding line shown, and 
 the figure or ellipse will be the front elevation of the circle No. 1 in 
 plan when inclined at an angle of 60 to the HP, and 30 to the VP. 
 
 To find the plan of the plate in this position, let fall projectors 
 from the points in No. 2 into the HP, and their intersection by those 
 drawn through C, 2, B, 4, D, No. 1, will give points through which 
 the ellipse No. 4 which is the plan of the circular plate is to be 
 drawn. It is here seen that the form assumed by a circular plate 
 or a circle entirely depends upon the direction in which itr is 
 viewed, alternating, as it may be made to do, from that of being a true 
 circle or any form of ellipse to that of a perfectly straight line. 
 The student will note what may appear to be a discrepancy in the 
 lettering of the last-obtained projection of the original object, No. 1, 
 but he will find that it is correct when he remembers that the view of 
 it given in No. 2 is that of No. 1 looked at in the direction of the 
 arrow -jc t and therefore turned through an angle of 90. If a disc, 
 
96 
 
 FIRST PRINCIPLES OF 
 
 similar in size and similarly figured, be cut out in card and laid upon 
 No. 1 with a pin through the centres, it will be found, on turning the 
 card disc through an angle of 90, that all the points on it will have 
 shifted through that angle, bringing A and B where C and D now are, 
 thus showing the projection No. 4 to be correct. 
 
 44. Before leaving this part of our subject, as elliptical figures 
 will be very frequently required, we here give a simple method of 
 drawing such a figure, which is sufficiently near the true form for all 
 the practical purposes of the draughtsman, and which obviates the 
 
 necessity of finding all the points in the curve. Premising that the 
 student understands that an ellipse differs from the circle in having its 
 diameters of unequal length, and that the sum of the distance of any 
 point in its curved-bounding line, from two fixed points called its 
 foci in its longest diameter is always the same, then draw an ellipse 
 by the method above referred to : 
 
 Let the line AB, No. 1 (Fig. 144), be its major or longest diameter 
 or axis, and CD, drawn at the mid-length of and at right angles to 
 AB, its minor or shortest diameter or axis. Anywhere on the sheet 
 of drawing paper draw two lines, as ac, be, No. 2 making an angle 
 with each other say 30- intersecting at c. On a c set off with the 
 compasses from c, half the length of AB, No. 1, and describe the arc 
 de, and on be with half CD as radius, and from the same centre, 
 describe the arc fg, join df by a right line, and parallel to it draw 
 lines through e arid y, cutting ac in point 1, and be in point 2. Then 
 on the major axis AB, No. 1, set off from A and B the distance cl, 
 
MECHANICAL AND ENGINEERING DRAWING 97 
 
 No. 2, in the points II', and with it as a radius from II' as centres, 
 draw arcs of about 30 through A and B. On the minor axis CD set 
 off from C and D, the distance c2, No. 2, in the points mm', and 
 with it as a radius and m m' as centres draw through C and D arcs 
 similar to those drawn through A and B. For intermediate points 
 take a straight-edged slip of paper SP and mark on its edge with a fine- 
 pointed pencil in points 1 and 3, half the longer axis aA, or aB, No. 
 1, and in point 2 measured from point 1 half the shorter axis aC, 
 or aD. If, then, this slip of paper be laid on the figure No. 1 and 
 moved over it in such a way that point 2 in its edge is always on the 
 major axis, while point 1 is on the minor, point 3 will describe, if 
 carried through a whole revolution, a perfect ellipse. As, however, 
 only a few points are required between the ends of the arcs already 
 drawn, these can be marked off from the edge of the slip of paper by 
 moving it in the desired consecutive positions. 
 
CHAPTER XIII 
 
 THE PROJECTION OF SOLIDS WITH CURVED SURFACES 
 
 45.. THE particular solids with which we have here to deal are 
 those known as " solids of revolution," which are assumed to be 
 " generated," or produced, by the revolution of certain geometrical 
 figures round one of their bounding lines as an axis ; the other lines 
 in their circling round this axis enclosing a space assumed to be 
 material of a certain form, dependent upon that of the generat- 
 ing figure. A "solid of revolution" therefore implies an object 
 having curved-bounding surfaces. Of such solids there are three 
 primary ones of which machine details are made up viz., the cylinder, 
 the cone, and the sphere the subsidiary ones being generated by 
 figures which are portions of plane sections of the first two primary 
 ones. 
 
 A Cylinder is a solid generated by the revolution of a rectangle 
 about one of its sides as an axis. If ABCD, Fig. 145, be the rect- 
 angle, then if it be caused to revolve round AB as an axis, making one 
 complete circuit, the sides AD, DC, CB will generate the solid called a 
 " cylinder," having a height, or length, equal to DC, a diameter equal 
 to twice AD, and its ends E, F, circular flat surfaces, at right angles 
 to its axis AB. 
 
 A Cone. If one half of the rectangle divided diagonally Fig. 
 146, be removed, leaving the right-angled triangle ABC remaining ; 
 then, if this triangle be revolved on AB as an axis, as before, the solid 
 generated will be a "cone," as in Fig. 146, its base DC being a circular 
 fiat surface at right angles to its axis AB. 
 
 98 
 
MECHANICAL AND ENGINEERING DRAWING 99 
 
 A Sphere. Again, if the two lines AC, BC of Fig. 146 be removed 
 and a semi-circle be described on AB as a diameter, as in figure 147, 
 then on revolving this figure AB as an axis, as before, the solid gene- 
 rated will be a " sphere," as in Fig. 147. 
 
 As the subsidiary solids before referred to are generated by parts of 
 certain sections of the cylinder and cone, their definition is deferred 
 until those sections have been found by projection. Taking the 
 "cylinder" as the first curved-surfaced solid as our object, the prob- 
 lem is 
 
 Problem 49 (Fig. 148). Given the plan of a cylinder with its axis 
 perpendicular to the UP, to find its elevation when its length is twice 
 its diameter. 
 
 Let the circle AB, No. 1, be the given plan; then its centre a will 
 be the plan of the axis of the cylinder. Find by projection the eleva- 
 tion of this axis a a. Assume the cylinder to be standing with one 
 end on the HP ; then, as its ends are in the same relative position as 
 the sides of the rectangle which generated them viz., parallel to each 
 other and one of them is on the HP, set off on the axis, from the 
 IL, the length of the cylinder in the point a, and through it draw a 
 line parallel to the IL. Now, in looking at the cylinder in the direc- 
 tion of the arrow in the plan No. 1, the visual rays will impinge upon 
 its surface from A to B ; at A and B the rays will be tangential, and 
 being at the same time perpendicular to the plane of projection, or the 
 YP, they will strike both sides of the cylinder in lines drawn through 
 A and B on its surface, perpendicular to the HP. Therefore through 
 A and B, No. 1, draw the lines AC, BD, No. 2, and the required 
 elevation is obtained. 
 
 Now, let the cylinder be inclined to the IIP, at an angle of 4$ its axis 
 being still parallel to the VP and its plan when in that position be 
 required. 
 
 First draw in the elevation of the cylinder in the given position, as 
 in No. 3. Its ends AB and CD are now inclined to the IL or HP, 
 and will in plan become ellipses as explained in Problem 40 
 because they are circular, but inclined to the plane the HP -on 
 which their projections are required. Now, in viewing the cylinder 
 in this position from above, or in the direction of the arrow, only one 
 of its ends, AB, will be seen, the other, CD, being by its inclination in- 
 visible. To find its plan when so inclined, first draw in the plan of its 
 axis a a ; on it obtain by projection the plan of the end AB, No. 3, 
 as shown by the ellipse Al, B2 ; through points 1 and 2, in this ellipse, 
 draw lines parallel to the IL, which will be plans of the sides of the 
 cylinder seen from above, or along the line a'a t No. 3 ; then find 
 by projection the plan of that part of the bottom edge of the cylinder 
 that will be seen, and the required plan will be obtained. 
 
 Again, let the cylinder, as in No. 5, have its axis inclined to the VP, at an 
 angle of 39, but parallel to the HP, and its elevation be required. 
 
 To obtain this, proceed as in the last case, by getting first the 
 
100 
 
 FIRST PRINCIPLES OF 
 
MECHANICAL AND ENGINEEEING DRAWING 101 
 
 elevation of the axis, and then finding by projection the two ends of 
 the cylinder upon it, and drawing in its sides as shown in No. 6. 
 For the projections of the " cone " the first problem is 
 
 Problem 50 (Fig. 149). Given the plan of a cone, having its axis 
 perpendicular to the HP, and an altitude equal to twice the diame- 
 ter of its base, to find its elevation. 
 
 Let the circle No. 1, Fig. 149, be the given plan of the cone ; its 
 centre a as in the cylinder will be the plan of its axis. Find by pro- 
 jection the elevation of this axis, and on it set off from the IL the 
 altitude of the cone in point a. Through a, the centre of the circle in 
 No. 1, draw AB parallel to the IL. Find by projection the elevation 
 of points A and B in the base of the cone on the IL, and join these 
 by right lines with point a on the axis a' a. The triangle A a B will 
 be the required elevation. 
 
 Next, let the base of the cone be inclined to the HP at an angle of 45, 
 keeping its axis parallel to the VP, and its plan be required. 
 
 Here, an elevation of the cone at the given angle must first be 
 drawn in, as in No. 3 ; then find the plan of its axis, which, being 
 parallel to the YP, will be a line parallel to the IL, as in No. 4. 
 Obtain by projection the plan of the base of the cone, which is an 
 ellipse, its major axis being CD arid its minor AB, No. 4 ; then find 
 the plan of the apex a, No. 3, of the cone in a, No. 4, and join it by 
 right lines to points C and D ; the required plan of the cone in the 
 position stated will thus have been obtained. 
 
 Again, let the axis of the cone be parallel to the HP, but inclined to the 
 VP, at an angle of 4^ V an d its elevation be required. 
 
 To obtain this, draw-in the plan of the cone in the position 
 stated, then find the elevation of its axis, and on this, by projection 
 from the plan, draw-in the base, which will be an ellipse ; join the two 
 ends of its major axis with the projected apex as shown in No. 6, and 
 it will be the required elevation. 
 
 As a "sphere," when looked at from any direction, is a circle, 
 when projected, we shall defer any problems in connection with it 
 until we come to consider the projections of its sections which will 
 follow those of the cylinder and cone, to which we next give attention. 
 
102 
 
 MECHANICAL AND ENGINEERING DRAWING 
 
CHAPTER XIY 
 
 THE PROJECTION OF THE SECTIONS OF A CYLINDER 
 
 46. FROM the definition previously given of a "cylinder," it 
 follows that any and every point in its curved surface is at the same 
 distance from its axis ; every plane section of it taken parallel to 
 the axis is a rectangle, while every section at right angles to its 
 axis or parallel to its ends is a circle. These are self-evident truths, 
 and require no graphic illustration to prove them. It is then with ita 
 sections not taken in either of the directions mentioned that we shall 
 now deal. Our first problem in connection with them is 
 
 Problem 51 (Fig. 150). Given the elevation of a cylinder, cut by 
 a plane making an angle ivith its axis; to find the sectional 
 elevation. 
 
 Let the cutting plane KP, No. 1, make an angle of 45 with the 
 HP, and be perpendicular to the VP, and the view required be that 
 in the direction of the arrow. To solve this problem, we must know 
 the distance between any point in the line of section on the front 
 side of the cylinder, and its corresponding point on the side of the 
 cylinder nearest the VP. To determine this, first find the plan of 
 the cylinder No. 2, and project over in No. 3 the elevation of the 
 lower part of the cylinder not affected by the cutting plane. Then 
 to find the contour of the section, choose any points 2, b, 3, in the 
 line KP; from these let fall projectors into the HP, cutting the 
 circle No. 2 in the points 2'2" ; 3'3". Similarly through points 1, 2, b, 3, 
 4, in KP, No. 1, draw projectors parallel to the IL through the axial 
 line a' No. 3, and on them set off from No. 2 the lengths of the lines 
 2'2" ; 3'3" ; then a curve drawn through the points thus obtained will 
 give the required projection. As the cylinder is standing vertically 
 on the HP, the plan of No. 3, although having its upper surface in- 
 clined to its base, will still be a circle ; but as that surface is in section, 
 it has to be cross-lined as shown in No. 4. 
 
 Next, let the direction of the cutting plane be KP, No. 5, and an elevation 
 and plan of the section be required, when the part to the left of tlie 
 cutting plane is removed. 
 
 103 
 
104 
 
 FIRST PRINCIPLES OF 
 
MECHANICAL AND ENGINEERING DRAWING 105 
 
 The portion of the cylinder left in this case is called an ungula, and 
 the required elevation and plan of the section are obtained in a similar 
 way to No. 3 and No. 4. First, to find the elevation : choose, as 
 before, a convenient number of points in the line KP ; from them 
 let fall projectors, cutting the plan of the cylinder No. 6, as was 
 done in No. 2. Similarly, from the same points draw parallel to 
 the IL projectors through the axial line No. 7, and on these set off, 
 in their proper order, the lengths of the corresponding lines found 
 in No. 6 ; a line drawn through the points thus found will give the 
 elevation or contour of the section made by the cutting plane KP, 
 in its altered position, which is a portion of an elongated ellipse. 
 The plan of this section of the cylinder shown in No. 8, differs 
 little from that of No. 3 ; the section plane KP, passing out of the 
 cylinder within its base at K, cuts off a portion of the base, as shown 
 by the line cc. 
 
 As the projections of all the possible sections of a cylinder are 
 obtained in the same way as the foregoing, further examples need 
 not be given, and we can now proceed to consider those of the cone. 
 
CHAPTER XY 
 
 THE PROJECTION OF THE CONIC SECTIONS 
 
 47. THE same preliminary remarks may be made in reference to 
 the simple sections of a " cone " as were made in the case of the 
 cylinder. From the definition given of- a cone, and its assumed mode 
 of generation, it will be at once seen that any section of it taken 
 through its apex to its base will be a triangle, while a section parallel 
 to its base or at right angles to its axis will be a circle. Admitting 
 these as self-evident truths, we have then to define those other 
 sections of the cone not taken in the directions mentioned, and to 
 show how their projections are obtained. 
 
 The three most important sections of a cone other than the two 
 mentioned are the " ellipse," the " parabola," and the " hyperbola." 
 
 If a cone be cut by a plane inclined to its axis, or its base, the 
 section is an ellipse, or a regular closed curve having diameters 
 differing in length. 
 
 If the section of a cone be made by a plane parallel to its axis, 
 but not coinciding with it, then the section is a hyperbola. 
 
 If a cone be cut by a plane parallel to its slant side, then the 
 section is a parabola. From its definition it will be readily understood 
 that the elevation of a cone when standing with its axis vertical is a 
 triangle, and its plan a circle. Before attempting the projection of any 
 of its sections, it must first be known how to find the plan, or 
 elevation either being given of a point on its surface, as this will 
 materially assist in finding its sections. 
 
 In Fig. 151 : If No. 1 be the plan of a cone, No. 2 its elevation, 
 and x in No. 1 a given point on its surface, it will be seen from its 
 assumed mode of generation that this point must lie somewhere in the 
 slant side of its triangular generator a'Ab, No. 2. Now every point 
 in this line it may be assumed to be made up of points in its 
 revolution about the axis a a of the cone describes a circle, and the 
 circle in which x must lie will have a radius equal to the distance 
 between it and a ; therefore in No. 1, with a as a centre and a x 
 as radius, describe a circle cutting the diametral line AB in x. Again, 
 through the given points x and a in No. 1, draw the line axe, cutting 
 the base AB of the cone in c. Find in No. 2 the elevation c of c 
 
 106 
 
MECHANICAL AND ENGINEERING DRAWING 
 
 107 
 
 No, 1, and from it draw the line c'a No. 2 ; then this line will be the 
 elevation of ca, No. 1, on the surface of the cone, and it is in it that 
 the given point x lies. To ascertain where, get the elevation of the 
 circle drawn through x in No. 1 by a projector from x', cutting the 
 slant side a A of No. 2 in point x', then the point #, where a line 
 through x parallel to AB cuts a'c No. 2, will be the required elevation 
 
 of the original point x given in No. 1, for the line last drawn may be 
 assumed to be the base of a cone of a lesser height than the given one, 
 x being a point in its base in the same position relatively that c or c is 
 in the base of the larger or given cone. By a converse process to that 
 here so fully explained, the plan of a point on the surface of a cone 
 may be found from its given elevation. We now proceed to the 
 projection of the conic sections, and as a first problem we take 
 
 Problem 52 (Fig. 152). Given the plan and elevation of a cone, 
 to find its sectional projections when cut by a plane, at an angle 
 of oO ivith its axis. 
 
 Let No. 1 and No. 2 be the given plan and elevation of the cone, 
 and LS, No. 2, the line of section. Draw in the axial line No. 3, and 
 
108 
 
 FIRST PRINCIPLES OF 
 
MECHANICAL AND ENGINEERING DRAWING 109 
 
 produce it into the HP. Choose any convenient points, as, 1, ^, 
 5, in LS, No. 1, and through them, parallel to the IL, draw lines to 
 cut the sides Aa, B&', of the cone. These, being parallel to its base, 
 will each be the edge-view of a circle or base of a small cone. To find 
 the distance through the cone at the points 2, 3, 4, in LS, No. 2 for 
 this is what is wanted to be known let fall from them projectors 
 into No. 1, parallel to the axial line aa ; then from a, No. 1, as 
 centre, and with radii equal to half the length of the lines drawn 
 through 2, 3, 4, No. 2, draw arcs, cutting the projectors let fall from 
 these points in 2'2", 3'3", 4'4", No. 1, the length between which 
 is the distance through the cone at their corresponding points in 
 No. 2. 
 
 Then, for the contour of the section, project over on to the axial 
 line No. 3, the points 1 and 5 in LS, No. 2, and from the other points 
 2, 3, 4, draw projectors parallel to the IL through the axis No. 3. On 
 these, set off the lengths 2'2", 3'3", 4'4", taken from No. 1, and 
 through the points thus found draw the curve of the ellipse. With 
 the radius a A, or B, No. 1, set off from the axial line on the IL, 
 No. 3, the base of the cone in CD ; join these with points 3'3", and 
 the required elevation is obtained. The plan of the section obtained 
 in No. 3, will be the ellipse, found by drawing a line through the 
 intersections of the arcs and double ordinates in No. 1. To show this 
 in its correct position as a projection of No. 3, transfer the points and 
 lines in it to No. 4, as shown, and it will then be seen that projectors 
 let fall from the points in the section No. 3 will fall upon their 
 transferred plans in No. 4. A line drawn through these points will be 
 the required plan of the cone when ,cut by the section plane LS, No. 2. 
 The student must remember that the view No. 3 being at right angles 
 to that of No. 2, the plan No. 4 is necessarily in the same position 
 with respect to No. 1. 
 
 Next, let the cone be cut by a plane parallel to its axis, and its 
 sectional projections be required. 
 
 48. In this case the projected section will be a true " hyperbola, ' ; 
 as the view obtained of it will be directly at right angles to the 
 cutting plane. Let No. 5 and No. 6, Fig. 152, be the given elevation 
 and plan of the cone, and LS, No. 5, the line of section. The simplest 
 way to obtain the required elevation will be to cut up the cone from 
 the point where the cutting plane enters it into horizontal sections, and 
 then find where the plane of section enters and leaves their horizontal 
 surfaces. To do this, draw lines at suitable distances apart, as 1 1', 
 2 2', 3 3', 4 4', parallel to the IL, as shown in No. 5. These will all be 
 edge-views of circular planes perpendicular to the VP. The circle 
 ACBD, No. 6, is the plan of the cone ; to find where the section 
 plane passes through it, produce the line LS, No. 5, and it will cut the 
 circle in a line p p parallel to CD. The part of the cone to the left of 
 the section plane may now be supposed to be removed, showing when 
 looked at in the direction of the arrow on the left the exact contour of 
 the section. To find this, all that is required to be known is the 
 
110 FIRST PRINCIPLES OF 
 
 width across the face of the section at the heights of the circular 
 planes 1, 2, 3, 4, No. 5. Draw in, as in No. 7, the side elevation of 
 the cone. Project over from No. 5 on to its axial line the point 1, 
 where the cutting plane first enters the side of the cone which will be 
 the highest point in the curve- through points 2, 3, 4, in LS, No. 5 ; 
 draw lines in No. 7, parallel to the IL or CD. In No. 6, with a as 
 a centre, and the half-lengths of the lines drawn through 2, 3, 4, 
 No. 5, as radii, draw arcs cutting the line pp', No. 6, in the points 
 2 2', 3 3', 4 4'; the distances between these points severally will be 
 the distances across the face of the section at the heights of their 
 corresponding points in No. 5. Set these distances off on the lines 
 drawn in No. 7 ; then a curve drawn through the points thus found 
 will be the required sectional elevation of the cone, cut by a plane in 
 the position given. This curve is a true hyperbolic, because the plane 
 of the section of the cone producing it is parallel to the cone's axis and 
 to that of the plane of its projection. The plan of this section, from 
 its position with respect to the HP viz., perpendicular to it becomes 
 a straight line, which, combined with the part of the cone not affected 
 by the cutting plane, will appear as shown in No. 8. 
 
 Again, let the cone be cut by a plane parallel to its slant side, and the 
 sectional projections, elevation, and plan be required. 
 
 Let No. 1 and No. 2 (Fig. 153) be the elevation and plan of the 
 cone, and LS the cutting plane, drawn parallel to its slant side Aa'. 
 Here the true form of the section will be a " parabola," but the plane 
 of the section being inclined to that of its projection, or the VP, the 
 view obtained of it, looking in the direction of the arrow, will be its 
 apparent and not its true form; but this point will be explained 
 further on. 
 
 To find the sectional elevation, first choose, as before, suitable 
 points, such as 1, 2, 3, 4, in the line LS, No. 1 ; through them and 
 point S draw lines parallel to AB the base of the cone to touch both 
 of its slant sides Aa', Ba'. These lines, as in the last problem, will be 
 edge- views of circular planes. From the points S, 1, '2, 3, 4, L, let fall 
 projectors on to the plan No. 2, parallel to the axial line a a. With 
 the half-lengths of the lines first drawn through S, 1, 2, 3, 4, as radii, 
 and a, No. 2, as centre, cut the vertical projectors let fall from these 
 points in IT', 2'2", 3'3", 4'4". Through the corresponding points in the 
 elevation No. 1 draw horizontal projectors through the axial line a' a, 
 No. 3, and on them set off the distances taken from No. 2 between 
 the points IT, 2'2", 3'3", 4'4"; then a line drawn through the points 
 thus found will give the elevation required. 
 
 The plan of this section, as shown at No. 4, is obtained in the same 
 way as the plan of the hyperbola No. 6 (Fig. 152) viz., by transferring 
 to No. 4 the points obtained by the intersection of the arcs and double 
 ordinates in No. 2, and through them drawing the parabolic curve, as 
 shown. 
 
 By the same method of procedure as shown in the three foregoing 
 problems, the projection of any possible plane section of the cone can be 
 obtained. Nor is the method confined to the sections made by a plane : 
 
MECHANICAL AND ENGINEERING DRAWING 111 
 
 it is equally applicable to those producing a curved sectional surface. 
 As an example in this direction, take the following as a problem- 
 
 Problem 53 (Fig. 154). Given the front elevation of a cone, and the 
 curve of the line of section, to find the sectional elevation of the same. 
 
 Let No. 1 (Fig. 154) be the given elevation of the cone, and LS the 
 curved line of section. Take, as in the previous problems, any con- 
 venient number of points in LS, No. 1, and through them draw lines 
 parallel to the base of the cone. Draw in the plan of the cone No. 2, 
 and let fall into it, parallel to the axis, projectors from the points taken 
 in LS, No. 1. With a, No. 2, as centre, draw arcs as before, cutting 
 the vertical projectors from the points in the curve in No. 1 in corre- 
 sponding points in No. 2. Then, to find the elevation of the curved 
 section, draw in the outline of the cone as in No. 3, and the projectors 
 from the points in LS, No. 1 ; upon these set off the length of the 
 double ordinates previously found in No. 2, and through the points so 
 obtained draw in the closed curve shown in No. 3. The plan of the 
 section found in No. 3 is obtained in precisely the same way as in the 
 previous problem- viz., by transferring the points found in No. 2 to 
 No. 4, and drawing through them the closed curve as shown therein. 
 
 49. In a previous paragraph reference is made to the question of 
 the true and apparent form of the section of a solid. As it is absolutely 
 necessary that the student should distinctly appreciate the difference 
 between these views of a solid, we give in Fig. 155 a graphic illustration, 
 showing wherein the difference exists. The solid chosen for its 
 explanation is the cylinder, it being the simplest for the purpose. 
 
 A cylinder ABCD (Fig. 155} is placed with the axis perpendicular to, 
 and one of its ends resting on, the HP ; the true form of a section of 
 it is required when cut by a plane SP, inclined to its axis at an angle 
 of 45\ 
 
 Now, a true, section of the solid on the line given should be such 
 that supposing the cylinder to be of material substance and cut 
 through on that line the section of it obtained by projection, if cut 
 out in paper and laid on the cut surface of the solid, should exactly 
 fit it. 
 
 In the projected section of a similar solid in Fig. 150, it is evident 
 that if the ellipse obtained were cut out in paper and applied to the 
 solid on the cut surface, it could not possibly fit it, as the majoi axis of 
 the projected ellipse would be found to be of a less length than that of 
 the line of section ; to find, therefore, the true section, a view of it must 
 be got directly at right angles to the plane of section, or in the direction 
 of the arrow y in Fig. 155. 
 
 This view may be obtained in two ways, either by direct projection 
 from the plane of section with the cylinder standing as it is in the 
 figure, or by moving the cylinder into such a position as to bring the 
 plane of section SP at right angles to both the VP and HP, and then 
 getting a side elevation of it. The latter of the two ways, as it gives 
 
 UNIVERSITY )) 
 
112 
 
 FIRST PRINCIPLES OF 
 
MECHANICAL AND ENGINEERING DRAWING 
 
 113 
 
 the student a fuller insight into the methods of procedure in such a 
 case, is the one adopted. 
 
 Now, as the cylinder is shown as vertical and the section plane 
 inclined at 45 to its axis, to bring this plane perpendicular to the HP 
 it must be moved through an angle of 45. For the purpose of the 
 projection it is assumed that the upper part of the cylinder above, the 
 section plane is removed. To bring the lower part into the required 
 position it must be turned through a vertical arc of 45. To do this, at 
 the point D in its base draw indefinitely to the right and left lines 
 making angles of 45 with the IL. On the one to the right, set off Dp 
 
 equal to DP ; and on that to the left, DC equal to DC the diameter of 
 the base of the cylinder ; at c draw cs equal to CS, the long side of the 
 cylinder, at right angles to cD ; join sp, then Dcsp will be the lower 
 end of the cylinder when the plane of section SP is vertical, or at right 
 angles to the YP and HP. The true form of the section can now be 
 found. First draw in No. 2 the plan of the cylinder in its vertical 
 position. Divide its diametral line AB into four equal parts in points 
 1, 2, 3 ; through these draw double ordinates and projectors to cut CD, 
 No. 1, in 1, 2, 3. Transfer by arcs points 1, 2, 3 in CD, No. 1, to cD, 
 and through these last draw lines parallel to cs to cut sp in points 
 
114 FIRST PRINCIPLES OF 
 
 1', 2', 3'. For the form of the section sp draw in, in No. 3, the axial 
 line a) a ; project over to it from sp, No. 1, the points s' and p, and 
 from 1', 2', 3' in sp draw projectors through a a, No. 3. On these set 
 off the length of the corresponding double ordinates drawn through 
 1, 2, 3, No. 2, and through the points thus found draw the ellipse 
 S2p2, No. 3, which is the required true form of the section at the line 
 SP, No. 1. 
 
 50. As opportunities will occur as we proceed with this part of our 
 subject of obtaining the true form of section of any solid in another 
 way than that just explained, we pass on to the sections of the third of 
 the primary solids with which we are dealing viz., the "sphere." 
 
 Now, as the sphere is generated by the revolution of a semi-circle 
 about its diametral line or chord as an axis, it follows that any plane 
 section of the generated solid must be a circle, whether it pass through 
 the axis or not, for any and every point in the curved-bounding line of 
 the semi-circle not only describes in its circuit round the axis a true 
 circle, but is at the same time at an equal distance from the centre of 
 the solid which is generated. It follows from this that all the sections 
 of a sphere, by planes passing through its centre, are equal circles, 
 having the same diameter and centre as that of the sphere. Such 
 circles are known as " great circles " of the sphere, and their projections, 
 if taken at right angles to the plane of section, are circles equal to their 
 originals. If, however, a sectional projection of a sphere is required at 
 an angle to the plane of section, then the projection becomes an ellipse, 
 and is obtained in the same way as that of any circular plane or disc 
 when at an angle to the plane of its projection. As all such sections of 
 a sphere would in projection be ellipses, only one problem is given in 
 this connection, which will show the difference between the sections of 
 the same sphere when cut by a plane at varying angles to its plane of 
 projection. The problem is 
 
 Problem 54. Given the plan of a sphere and the line of section, to find 
 its sectional elevation. 
 
 Let AB, No. 1 (Fig. 156), be the plan of the sphere, and SP the 
 line of section, the axis a or pole of the sphere being perpendicular to 
 the HP. Here SP being parallel to the axis a' and to the VP, the 
 section of the sphere by it will be a circle. To find its elevation, first 
 show in that of the sphere AB, No. 2. Project over point 1, in SP, 
 No. 1, to 1 in AB, No. 2 ; then in No. 2, with s as centre and s\ as a 
 radius, describe the circle 1, 2, and it will be the sectional elevation 
 required. For SP, No. 1, being parallel to the VP, the circular section 
 made by it will be concentric with the great circle AB, No. 2, and 
 therefore their centres will be coincident ; and as SP cuts the great 
 circle AB, No. 1, in points 1 and 2, the elevation of the section made 
 by SP will be a circle, having a diameter equal to the distance between 
 those points, as shown by its projection in No. 2. 
 
 Next, let the sphere be turned on its axis in the direction of the arrow in 
 No. 7, until the section plane SP is at an angle as in No. 3 with 
 the VP, and its elevation be required. 
 
MECHANICAL AND ENGINEERING DRAWING 
 
 115 
 
116 FIRST PRINCIPLES OF 
 
 Here, SP being inclined to the YP, the elevation of the section will 
 be an ellipse. To find it, first get the elevation of the sphere, as in No. 
 4, having a a for its axial line. Bisect the distance between 1, 2 in SP, 
 No. 3, in the point 6, and through it draw a projector to cut AB, No. 4, 
 in b' '; also project over points 1, 2 in SP, No. 3, to points 1, 2 in AB, 
 No. 4. Now the true section of the sphere made by SP is a circle of a 
 diameter equal to the distance between points 1 and 2 in that line, and 
 it is the elevation of this circular section inclined to the VP, as shown, 
 that is here required. It is left to the student without further explan- 
 ation to find the resulting elliptic section, as was done in Problem 48, 
 Fig. 143. 
 
 Again, let the problem be 
 
 Problem 55. Given the elevation of a sphere cut ~by a plane inclined 
 to the IfP, to find its sectioned projections. 
 
 Let No. 5, Fig. 156, be the elevation of the sphere, and SP the line 
 of section. To find the side elevation, first draw the outline of the 
 sphere, as in No. 7. Then as the true section of the sphere by SP is a 
 circular plane, find by Problem 48 its vertical projection when inclined 
 as in No. 5, which will be an ellipse as shown in No. 7. For the plan 
 of No. 7 in its correct position, proceed as was done in the case of the 
 cone No. 1, Problem 52, by finding the plan of the points in the contour 
 of the section, as in No. 6, and transfer them to No. 8 ; then a line 
 drawn through the points so found will give, as in No. 8, the required 
 sectional plan of No. 7. 
 
 In the same way as here shown can the projection of any possible 
 section of the sphere be obtained, no matter how the line of section be 
 drawn, or whether it is straight or curved, as the same rule holds good 
 in this case as in that of the cylinder and the cone viz., Find the 
 actual distance across the face of the cut surface of the section, from 
 any points on the surface of the solid in the line of section nearest 
 the eye, to those directly behind them on the opposite side farthest from 
 the eye, and the problem of its projection is solved, for the points 
 denoting the distances thus found have only to be put in on the paper 
 in their proper places, and a line drawn through them will give the 
 sectional projection sought. 
 
 51. Having shown how the sectional projections of the three primary 
 solids the cylinder, the cone, and the sphere are found, we now pass 
 on to the consideration of those subsidiary solids of revolution which 
 are generated by parts of the sections of the primary ones. 
 
 The particular solids of this class which assist in giving shape to 
 machinery details are the "spheroid," the " conoid," the " cylindroid," 
 the " spindle," and the " oval " ; and they are thus defined : 
 
 A spheroid is a solid generated by the revolution of a semi-ellipse 
 about one of its diameters. If about it* longer one, it is a prolate 
 spheroid ; if about its shorter one, an oblate spheroid. 
 
 A conoid is a solid generated by the revolution of half of a conic 
 section about its axis, and becomes an "ellipsoid," "paraboloid," or 
 
MECHANICAL AND ENGINEERING DRAWING 117 
 
 " hyperboloid," according as its generating figure is half an ellipse, 
 parabola, or hyperbola. 
 
 A cylindroid is a cylindrical solid having elliptical ends. 
 
 A spindle is a solid generated by the revolution of an arc or portion 
 of a curve cut off by a chord or double ordinate, around which it 
 revolves as an axis. It is circular, elliptic, parabolic, or hyperbolic, 
 according as the generating arc, or curve, is a portion of a circle, ellipse, 
 parabola, or hyperbola. 
 
 An oval is an egg-shaped solid, and may be conceived to be com- 
 pounded of half a sphere and a like portion of an elliptic conoid. 
 
 As all the above-defined solids are assumed to be generated by the 
 revolution of a particular figure about an axis, it follows that a cross- 
 section at right angles to its axis of any one of them will be a circle, 
 while a section throng Ji its axis will be a similar figure to that of which 
 its generator formed a half-part. An oblique section through the axis 
 of all of them would be a figure of elliptic form, while one which did 
 not cut the axis would partake of the form of its axial section. Under 
 these circumstances, as the finding of any required section of one or all 
 of them merely involves the application of methods of procedure already 
 explained, it is unnecessary here to further elaborate them. All that 
 is required by the student for the mastery of this part of our subject, 
 is to thoroughly appreciate the exact form and mode of generation of 
 the different solids enumerated. This attained, there will be no difficulty 
 in the solution of any problem that may arise in connection with their 
 projection. 
 
 Before concluding this chapter, and with it the subject of the 
 " Projection of Solids with Curved Surfaces," it will be necessary to 
 show, how the principles explained in Chapter XL, on the " Lining-in of 
 Drawings in Ink," are to be applied to representations of this class of 
 solids. 
 
 It has been shown in Chapter XL, that the projections of rays of 
 light, in the direction they are assumed (in Mechanical drawing) to fall 
 upon the object illumined, are parallel right lines in plan and elevation, 
 making an equal angle of 45 with the intersecting line of the planes 
 of projection. To determine then how to line-in the representation 
 of a curved-surfaced solid, we have to find how the light falls upon it. 
 
 Now to do this, let the circle cedb in No. 1, Fig. 156a, be the plan 
 of a cylinder, and the rectangle ABCD No. 2 its elevation. Through 
 , the axis in No. 1, draw lines at an angle of 45 with the IL, cutting 
 the circle in points b and c; and through b draw a line parallel to ca. 
 Also through b and c, in No. 1, draw lines in No. 2, perpendicular to 
 the IL, and produce them to meet AB, the top-end of the cylinder in 
 c and b. Now the lines drawn through b, and ca in No. 1, are the 
 plans of two planes of rays of light ; the former touching the cylinder 
 throughout its length in the line bb' ; and the latter striking it in the 
 line cc (No. 2). The line bb' is known as the " line of shade," as it 
 divides the cylindrical surface in the light from that part which is in 
 the shade, and is moreover the darkest part of the visible surface. The 
 brightest part of the cylindrical surface will be the "line of light" cc', 
 as it is here that the light shines directly upon it. 
 
118 FIRST PRINCIPLES OF 
 
 It follows from the foregoing considerations, that in lining-in the 
 representation of a cylindrical surface in ink, fine lines only should be 
 used, as either a medium or shadow line would be inappropriate, the 
 lightest, and darkest, parts of the surface of such a solid always falling 
 within the outlines of its representation, as shown at bb', and cc', in the 
 elevation No. 2, Fig. 156. The same rule also applies to either conical 
 or spherical surfaces, as shown in Nos. 4 and 6, in the figure where it 
 is seen that the " lines of shade " obtained in the same way as in the 
 case of the cylinder are well within the outlines of the solids, and 
 must therefore be put in, in fine lines. 
 
 It would, however, be correct to shade-line the end of a cylinder, 
 or the base of a cone if inverted as a part of their edges would 
 throw a shadow. To determine how much would do so : in No. 1 , 
 Fig. 156&, produce the line drawn through b, a to e ; and that through 
 c, a to d. Now as all the rays of light falling on the cylinder, and its 
 upper end AB, are parallel to the one drawn through c, a, d, it follows, 
 that the semi-circular edge ecb in No. 1 will be directly in the light, 
 and will cast no shadow, while the opposite similar edge edb will do so, 
 either on the HP or YP, according to its position with respect to 
 those planes and must therefore be shade-lined. But as the semi- 
 circular edge edb of the cylinder end would cast different intensities 
 of shadow, the shade-line from e through d to b must be of varying 
 thickness (as shown in the diagram), its thickest part being at d. 
 
 If the cylinder be hollow as represented in No. 1 then that part 
 of its inside top-edge, or the semi-circle 1, 2, 3, will throw a shadow on 
 the opposite inside surface, and must be shade-lined as shown, the 
 thickest part of the line being directly opposite the similar part of the 
 outer semi-circle edb. In the sphere and other entirely curved-surfaced 
 solids, no shade-lining is admissible in inking them in, fine lines being 
 the only proper ones to be used for the purpose. 
 
MECHANICAL AND ENGINEERING DRAWING 
 
 119 
 
CHAPTER XVI 
 
 THE PROJECTION OF OBJECTS INCLINED TO THE PLANES OF PROJECTION 
 
 52. FOR the more immediate purpose of representing anything in 
 process of manufacture in the workshop, it will be readily understood, 
 from what has already been explained in these chapters, that the only 
 drawings absolutely necessary to the workman are those in which the 
 plans, elevations, and sections of the objects he is called upon to make 
 are correctly shown. But as many of the elements which go to make 
 up a machine, or engine, are often inclined to the vertical and horizontal 
 positions, it is essential that the draughtsman should know how to 
 draw the plan or elevation of an object, however intricate, when inclined 
 to the YP or HP, at any angle other than a right angle. Before 
 showing how the principles of projection are applied in this part of our 
 subject, we would guard the student against accepting as true, the 
 notion entertained by some, that it is a different kind of drawing to 
 that which has preceded it, and that a knowledge of it is more difficult 
 to acquire. The projections of an object inclined to the planes of its 
 projection, certainly differ in appearance to those of it when not so 
 placed, but the method of finding such projections involves no new 
 departure in principle, as the " planes of projection " the VP and HP 
 are still in the same relative position as before viz., at right angles 
 to each other although the " object " is inclined to both. The " pro- 
 jections," therefore, of an object/ so placed, are nothing more than its 
 appearances when viewed at right angles to those planes. 
 
 To show this graphically, let A, No. 1 (Fig. 157), be the plan of a 
 wooden block, with its under side resting on a flat plate abed, lying on 
 the HP, and its sides inclined to the VP, or IL, as shown. A front 
 elevation or view in the direction of the arrow e of the block in this 
 position would be B, No. 3. Now, assume the plate with the block 
 upon it to be raised through an angle, moving on its edge afe, as a 
 hinge ; the elevation of the block in this position would then appear 
 like C, No. 3 ; for its upper face would become more and more visible 
 as the angle made between the plate on which it rests and the HP 
 increased, until the plate had attained a vertical position, when it 
 would again appear as shown in 'plan at A, No. 1 ; but as an elevation. 
 We therefore see that its " appearance " when inclined to the planes 
 
 120 
 
MECHANICAL AND ENGINEERING DRAWING 121 
 
 of its projection, although radically different to either its plan A, No. 1, 
 or its elevation B, No. 3, is still nothing more than an ordinary pro- 
 jection of the original object when in its changed position. This is 
 further shown in No. 2 (Fig. 157), where a side view is given of the 
 plate with the block upon it, such as would be seen when looked at in 
 the direction of the arrow/, No. 1. Then by the aid of a few projectors 
 from No. 2 and No. 1, it is seen that C, No. 3, and D, No. 4, are just 
 ordinary projections, differing little from those which have preceded 
 them, and certainly no more difficult to obtain than some which the 
 student has already projected or drawn. As the projections of the 
 object given in this figure form a problem for solution later on, we 
 defer showing how they are exactly obtained until we have, as in 
 previous instances, explained how the projections of simple " figures " 
 in similar positions are found. 
 
 53. Now, to facilitate the process of finding the projections of an 
 object when inclined to the VP and HP, we have to introduce into 
 the construction an imaginary inclined plane interposed between the 
 VP and HP, on which the object whose projections are required is 
 assumed to rest. This plane is represented in Fig. 157 by the line OP, 
 No. 2, and is the only new feature in our subject of study, it being not 
 a fixed plane like the VP or HP, but one that may assume any desired 
 angle with those planes. Such a view of an object as is shown in 
 No. 2 (Fig. 157) could of course be given without the assistance of the 
 line OP; but as it determines the inclination of the object to the 
 vertical or horizontal, it is of material help in finding its projections, 
 as will be seen as we proceed. 
 
 Throughout the problems that will be given in this part of our 
 subject the original object whose projections are required is to b3 
 always understood as resting on a horizontal plane, and the view first 
 given of it, is its pl:in or horizontal projection, such a A, No. 1 (Fig. 
 157), which shows all that would be seen of such an object when looked 
 at from above. 
 
 Now, to find the projection of a point situate in a plane which is 
 inclined to the VP and HP, we must know the angle made by the 
 plane in which the point lies with either the VP or HP. Our first 
 problem is 
 
 Problem 56. Given the plan of a point, it is required to find its 
 elevation when the plane in which it lies is inclined to tJie HP at an 
 angle of 30. 
 
 Let a (Fig. 158) be the plan of the given point, IL being the inter- 
 secting line. At O draw a line OP, making with the IL an angle of 
 30. Through the point a draw a projector into the VP. Set off on 
 OP from O, a distance O^', equal to that which point a is from the 
 IL. Through a draw a line parallel to the IL, cutting the projector 
 from a in plan in point a" ; then a" is the required elevation. In the 
 same way, the elevation of a line lying on a plane inclined to the VP 
 and HP is found, for we have only to find the vertical projection of its 
 ends, and join them by a right line, and the desired elevation is 
 obtained. 
 
122 FIRST PRINCIPLES OF 
 
 Problem 57. Given the plan of a line, to find its elevation when the 
 plane in which it lies is equally inclined to the VP and HP. 
 
 Let AB (Fig. 159) be the plan, IL being the intersecting line as 
 before. As the plane on which the line AB is to lie equally inclines to 
 the vertical and horizontal, its angle with either will be one of 45. 
 Therefore, at any point in the IL, say O, draw a line OP at that angle 
 with it. Now OP, we have before said, is a side view of the inclined 
 plane on which the object in this case a line is to lie ; the view we 
 shall obtain of the object in elevation will be one looking in the direc- 
 tion of the arrow x, and the position of the YP with respect to OP can 
 be either in front of or behind it. If in front, it would be represented 
 by a line drawn through O perpendicular to the IL ; if behind, then a 
 similar line through P. Generally, OP is assumed to be interposed 
 between the VP and HP, and therefore a side view of the three planes 
 would be correctly represented by the right-angled triangle PxO in the 
 diagram; the vertical projection or elevation of the inclined object 
 being made on Px or the VP, and the horizontal one, or plan, on Ox, 
 or the HP. 
 
 We can now proceed with the solution of the problem before us. 
 Having the inclination of the plane on which the line AB is lying, to 
 find a front elevation of it we require to know the exact position of its 
 two ends with respect to the foot of the plane on which it is lying. 
 Now, the intersection of two planes is a straight line ; therefore, as the 
 inclined plane OP in the figure intersects the HP in O, the plan of that 
 intersection will be a straight line parallel to the VP or the IL ; and 
 assuming the VP to be as far from the foot of the inclined plane OP as 
 x is from O, set off from the IL a distance equal to Ox, and through 
 the point so found draw the line DL parallel to the IL, and it will be 
 the plan of the intersection of the HP with the inclined plane OP. It 
 is from this line that all measurements of points in the plans of original 
 objects have to be taken, when giving their elevations on the inclined 
 plane OP. Then to find the required elevation of the line AB, set off 
 from O on the line OP in b and a, the distances A and B6, that its 
 ends A and B are from DL, and through them parallel to the IL draw 
 projectors to cut those drawn from A and B in the plan in the points 
 A'B' ; join these by a right line and it will be the elevation required. 
 
 Then, as lines are the boundaries of " figures," the projection of a 
 figure in a similar position with respect to the VP and HP is obtained 
 in the same way. One problem in this connection will be sufficient to 
 show the application of the principle. Let it be this 
 
 Problem 58. Given tJie plan of a triangle, and the inclination of 
 the plane OP ; to find tJie elevation of the trianyk. 
 
 Let ABC (Fig. 160) be the plan of the triangle, and the inclination 
 of the plane OP be 30 with the HP. Draw in the line OP at the 
 given angle with the IL and the line DL as in the last problem. Find 
 the position of the angular points ABC on the plane OP, and through 
 
MECHANICAL AND ENGINEERING DRAWING 123 
 
124 FIRST PRINCIPLES OF 
 
 them draw projectors parallel to the IL, cutting those drawn through 
 the corresponding points in the plan. The intersections of these pro- 
 jectors A'B'C' will be the elevations of the angular points of the 
 triangle ; join these by right lines and the triangle ABC' will be the 
 elevation required. 
 
 If the plan of the original figure in its new position be required as 
 well as its elevation, this is found in a similar way. Let the problem 
 be 
 
 Problem 59. Given the plan of a rectangular plate and the inclina- 
 tion of the plane, OP ; to find its projections. 
 
 Let A, No. 1 (Fig. 161), be the given plan of the plate, and the 
 inclination of OP be 30 with the HP. Draw OP, No. 2, at the given 
 angle with the IL, and the line DL as before. Number the corners of 
 the plate 1, 2, 3, 4, as shown in No. 1. Find the position of these 
 points on OP, No. 2, and from them obtain, by projection, the elevation 
 of the plate shown at B, No. 3. To find the plan of the plate in its 
 inclined position, let fall projectors from points 1, 2, 3, 4, in the line 
 OP, No. 2, perpendicular to the IL ; then the corners of the plate in 
 plan will lie in these projectors, and the problem is to decide where. 
 The line DL, below the original plan of the plate at A, No. 1, is, we 
 have shown, the plan of the intersection of the inclined plane OP with 
 the HP; then a line dl, drawn through O, No. 2, at right angles to 
 the IL, as shown in No. 4, will also be a plan of the same intersection. 
 As the distances of the points 1, 2, 3, 4, in A, No. 1, from the line DL, 
 were set off on OP, No. 2, the position of this plane with the plate 
 upon it will be at right angles to that of A, No. 1 ; therefore, its plan 
 No. 4 or the view of it looking in the direction of the arrow x, No. 2 
 will be in the same position with reference to A, No. 1. Then if the 
 points where the projectors drawn through 1, 2, 3, 4, in A, No. 1, cut 
 the IL, be transferred to the line dl, No. 4, and lines be drawn through 
 them at right angles to that line, their intersections with the projectors 
 let fall from 1, 2, 3, 4, No. 2, will give 1, 2, 3, 4, No. 4 ; join these by 
 right lines, and the required plan of the plate in its inclined position is 
 obtained. 
 
 54. As the projection of any plane figure, inclined to both the 
 planes of projection, is found in the same way as here shown, we can 
 without further example proceed with the application of the principles 
 involved, to the projection of solid objects similarly placed. Now, a 
 solid object, such as the block of wood taken for explanatory purposes 
 in Fig. 157, has not only length and breadth, but thickness also, and 
 each of these dimensions has its representative lines in the projections 
 that are obtained of it, when it is inclined to both the VP and HP. 
 The actual thickness of an object or its measurement from surface to 
 surface is the distance through it at right angles to the surface from 
 which it is measured. If the thickness be represented by a line, then 
 the line will be perpendicular to that surface, and its projections will 
 depend upon its position with respect to the planes of projection. 
 
 In Fig. 162, if the plan of a line be represented by the point a, 
 then, when the plane to which the line is assumed to be perpendicular 
 
MECHANICAL AND ENGINEERING DRAWING 
 
 125 
 
 Fiy. 160 
 
 Fiy. 161 
 
126 
 
 FIRST PRINCIPLES OF 
 
 is inclined, say, 30 to the horizontal, its side elevation on that plane 
 OP will be a'b. To find its front elevation, we draw a projector through 
 a perpendicular to the IL, and from a and b the two ends of the line 
 projectors parallel to the IL, to cut the one through a in a'b'; then a 
 line joining these points is the front elevation of the original line a'b 
 when inclined to both planes of projection. To apply this reasoning to 
 the projection of such an object as the ; rectangular block of wood in 
 Fig. 157, proceed as follows: First draw in its plan as at A, No. 1, 
 and number the four corners as shown. " Each of these will be the end 
 of a line perpendicular to the HP, of a length equal to the thickness of 
 the block. On OP, No. 2, drawn at an angle of 3Q with the IL, give 
 an elevation of the block, looking at it in the direction of the arrow/, 
 remembering the position of its corners and top surface with respect to 
 the plane on which it is assumed to be standing. Having drawn in 
 this view, proceed to find by projection that of C, No. 3. This is best 
 and most correctly done by numbering both top and bottom ends of 
 the lines representing the angular corners of the block in No. 2. The 
 
 Fig. 
 
 intersection of horizontal projectors drawn through these points, and 
 their corresponding vertical ones in No. 1, will give the position of all 
 the corners of the block that will be seen looking in the direction of 
 the arrow x in No. 2, and the joining of these by straight lines, as 
 shown in C, No. 3, will give the required elevation. 
 
 To find the " plan " of the block when inclined as shown, let fall 
 vertical projectors from all the corners of it seen from above in No. 2, 
 when looking in the direction of the arrow g ; transfer the points 
 where the vertical projectors from 1, 2, 3, 4, No. 1, cut the IL, to the 
 projector drawn through the foot of the inclined plane at 0, as shown 
 in Fig. 161 previously, and through these draw projectors to cut those 
 let fall from 1, 2, 3, 4, No. 2 ; then the intersections of these projectors 
 will be the plans of the corners of the block seen from above it, which, 
 on being joined by straight lines, will give D, No. 4, the plan of the 
 block, when inclined in the position shown in side elevation at No. 2, 
 and front elevation at No. 3. 
 
 55. As a test of the correctness of the two last-found projections, 
 all the lines in the original object A that are parallel to each other 
 
MECHANICAL AND ENGINEERING DRAWING 127 
 
 should be parallel in their projections, for if they are not, there is some 
 inaccuracy in projecting over corresponding points in the different 
 views, as no alteration of position of the original object, with respect to 
 the planes of its projection, can in any way affect its form, or the 
 relative position of its surfaces, from whatever direction it may be 
 viewed. These deductions apply, of course, to all " original " objects, 
 but they require to be more particularly remembered here, as in no 
 other part of the subject is there the same likelihood of making a false 
 projection of an object, as when it is inclined to both the planes of its 
 projection. 
 
 As no better examples of plane-surfaced solids, as objects for pro- 
 jection in this part of our subject, can be chosen than those whose 
 forms the student is already familiar with viz., the cube, prism, pyra- 
 mid, etc. they will be taken in the same order as before, and the first 
 problem is 
 
 Problem 60 (Fig. 163). Given the plan of a solid cube, with one of 
 its faces upon the HP ; to find its plan and elevations, together with 
 a sectioned plan and elevation, when the plane on which it rests is 
 inclined to the horizontal at an angle of 80. 
 
 Let the square 1, 2, 3, 4, No. 1, be the given plan of the cube, IL 
 being the intersecting line of the VP and HP, and DL a datum line 
 parallel to the IL, at any convenient distance below the plan of the 
 cube. At 0, say in the IL, draw a line OP, making with the IL an 
 angle of 30. Set off on OP, from 0, in points 1', 4', 3', the distances 
 that points 1, 4, 3 in No. 1 are from the line DL, measured at right 
 angles to it. Through 1', 4', 3', No. 2, draw lines perpendicular to OP, 
 and on that through 1' set off from OP a height I'l, equal to a length 
 of the side of the cube. Through 1, draw a line parallel to OP, cutting 
 the perpendiculars to it at points 4 and 3 ; and the view No. 2 will be a 
 side elevation of the cube looked at in the direction of the arrow x 
 in the plan when inclined to the HP at 30. 
 
 To find a front elevation of the cube at the same inclination, we 
 have to consider what parts of it will be seen in such a view. This 
 will evidently include the two inclined front faces and a top face, or 
 what would be seen looking in the direction of the arrow to the left of 
 No. 2. To find the projection of these faces, project over points 1, 
 1' ; 4, 4' in No. 2, on the projectors drawn through points 1, 2, 4, No. 
 1, cutting them in points 2, 2' ; 1, 1' ; 4, 4', No. 3 ; join these by right 
 lines as shown, and the projection of the two front faces of the cube 
 will have been found. To find the top face, project over in like man- 
 ner point 3 in No. 2, to cut the vertical projector through the corre 
 spending point in No. 1, and it will give point 3 in No. 3 ; join this by 
 right lines with points 2 and 4, and the required front elevation of the 
 cube in its inclined position is obtained. 
 
 To find the plan of No. 2, or a view in the direction of the arrow 
 shown above it, first find the plan of the top face of the cube, by letting 
 fall vertical projectors from points 1, 4, 3, in it, into the lower plane, 
 or HP; intersect these by horizontal projectors from points 1, 2, 3, 
 
128 FIRST PRINCIPLES OF 
 
 No. 1, in points 1, 2, 3, 4, No. 4; join these by right lines, and the 
 plan of the top face of the cube is found. For the plan of its two 
 inclined faces, seen from above, produce the horizontal projectors drawn 
 through the three corners 2, 3, 4 of the top face just found and by 
 vertical projectors from points 4' and 3', in No. 2, cut the produced 
 horizontal ones in points 2', 3', 4', No. 4 ; join points 2, 2' ; 3, 3' ; 4, 4' ; 
 and 2', 3' ; 3', 4', by right lines as shown, and No. 4 is the required plan 
 of the cube in its given inclined position. 
 
 56. Before proceeding to show how the sectional projections of the 
 cube are found, it is advisable here to save repetition later on to 
 explain the reasoning which has directed the process of solving the 
 first part of this problem. Taking the " original " object the cube 
 as the first subject of thought, we have to note its special form, and the 
 shape and relative position of its six faces and their bounding edges, as 
 upon the correct realization of these features depends in a great mea- 
 sure the truth or falsity of its projections. For whatever relative posi- 
 tion its sides have to each other in the original, they cannot, as long as 
 they remain uncut, have any other than the same relationship in their 
 projections. A second consideration is the position of the object, and 
 its faces, with respect to the plane on which it is assumed to be stand- 
 ing ; and a further one is that of the inclination this plane eventually 
 assumes with respect to the planes of projection. 
 
 Now, the cube is standing with one of its faces on the HP, thereby 
 making its four corner edges, figured 1, 2, 3, 4, No. 1, in the diagram, 
 at right angles to that plane, and the top face of the cube parallel to 
 it. It follows from this that however much the plane on which the 
 cube rests is inclined to the horizontal, the position of the faces and 
 corner edges of the cube with respect to that plane will remain exactly 
 as before its inclination. Therefore, in setting the cube on the inclined 
 plane OP in the diagram, all its vertical edges will be perpendicular to 
 OP, and its horizontal ones parallel to it. Then as to its projections 
 No. 3 and No. 4, although the cube itself in its new position No. 2 
 is inclined to the VP and HP, this has not in any way altered the 
 relative position of the bounding edges of its faces in the projections, 
 as the lines 1, 1' ; 2, 2'; 3, 3' ; 4, 4', are still parallel to each other, as 
 well as those representing the top and bottom edges of the cube. 
 
 To assist the student to a more complete realization of the position 
 of the " planes of projection " with respect to the original object, in 
 Fig. 163 let a line be drawn perpendicular to the IL through point 3' 
 of the cube in No. 2. This line would then represent an end elevation 
 of the YP, with the corner 3' of the cube touching it. If, then, the 
 inclined plane OP, with the cube upon it, together with this vertical 
 plane, be imagined to swing round on the HP through 90, the view of 
 the cube shown at No. 3 would be identical with its appearance on the 
 inclined plane when viewed at right angles to the YP. 
 
 57. With the foregoing explanation of the reasoning applied in 
 finding the projection of the cube in its uncut state when inclined to 
 the VP and HP at a given angle, there should now be no difficulty in 
 obtaining its sectional projections required by the problem, as these 
 are nothing more than similar views to those already found, with the 
 
MECHANICAL AND ENGINEERING DRAWING 
 
 129 
 
130 FIRST PRINCIPLES OF 
 
 exception that a portion of the solid is assumed to be removed, exposing 
 the cut surface. 
 
 The fact of the cube being in the same position as before, reduces 
 the problem to the finding of the projections of its exposed cut surface, 
 when the part to the left of the cutting plane is removed. For the 
 sectional elevation, draw in first in faint lines to the right of No. 2 a 
 similar view of the cube to that given in No. 3. The cutting plane ac, 
 it will be seen, passes through Jive of the faces of the cube, entering it 
 at c on its top face, passing through its two back and front faces, and 
 leaving it at a in its foremost edge. 
 
 Now, the position of the cube in No. 3, with respect to the VP, is 
 that of having its two front and back faces making equal angles with 
 it, and it follows from this that a plane passing through the cube, as 
 ac is assumed to do, will cut opposite faces of it in parallel lines. There- 
 fore to find the projection of the section in No. 5, made by ac in No. 2, 
 through c draw a horizontal projector to cut the back top edges of the 
 cube in c c, No. 5, and similar projectors through b and a, the former to 
 cut the side edges 2, 2' ; 4, 4' in b b', and the latter the front edge in a. 
 Then join by right lines in No. 5 the points ab, be, cc f , cb', b'a thus 
 found, and the space enclosed by them will be the sectional surface of 
 the cube cut by the plane ac. Put in, in full lines, the remaining 
 uncut part of the cube, and the complete view, as at No. 5, thus 
 obtained will be the required sectional elevation of the cube. 
 
 Then as this last-found view is one looking at the cube in No. 2, in 
 the direction of the arrow on its left, when the portion on that side 
 of the cutting plane ac is removed, the plan now required is a view 
 of this remaining part when looked at from above, and as the position 
 of No. 2 is at right angles to that of No. 5, the plan of this latter,- 
 which is what has now to be found, will be in the same position with 
 respect to No. 4. 
 
 Now, the simplest way to obtain this view is to make use of the 
 assumed VP, previously drawn in No. 2, touching the corner 3' of the 
 cube. As No. 5 is a view of the cube directly at right angles to that 
 at No. 2, this assumed VP will be behind it in No. 5, and a plan of 
 it in that position will be a line in the HP parallel to the IL. Draw 
 such a line as x y, No. 6 ; then all the points and lines seen in No. 2 
 from above can be set off from this line. The back lower corner of the 
 cube is touching it, and we know its position is behind 1', No. 5 ; 
 therefore let fall first a vertical projector from this point to x y, No. 6, 
 in point 3', and similar projectors from points 2' and 4'. Set off from 
 x y on the projector through 3', and similar projectors from points 2' 
 and 4'. Set off from x y on the projector through 3', the distance 
 3' 3 equal to that of point 3, No. 2, from the assumed VP, and on 
 those let fall from 2' 4', No. 5, in points 2' b, 4' b, the distances that 
 their corresponding points in No. 2 are from the same assumed VP. 
 Join points 2' and 4' with 3' by right lines, and parallel to them 
 draw two others indefinitely through point 3. The plan of the line 
 c'c, No. 5, is next required. This may be found in two ways : Either 
 by setting off from xy on the projector through 3', the distance that 
 c in No. 2 is from the assumed VP, and drawing a line through the 
 
MECHANICAL AND ENGINEERING DRAWING 131 
 
 points so found parallel to xy to cut those drawn through 3 in c'c, 
 No. 6 ; or by letting fall projectors from the ends c c of the corre- 
 sponding line in No. 5 to cut the same lines from 3 as before. The 
 plans of the two short inclined edges of the section between c'b' and 
 cb in No. 5 are found by joining the corresponding points in No. 6 by 
 right lines. On cross-lining the parts of the cube exposed by the 
 cutting plane in both views the required sectional plan and elevation 
 will be completed. 
 
 58. Now, to test the accuracy of these projections, the same reason- 
 ing holds good with reference to the sections thus found as with the 
 original uncut solid. For it will be found, if the sections are correctly 
 projected, on applying the set-squares or a parallel-ruler to any one of 
 the bounding lines of the sections, the edge of that section on the 
 opposite face of a cube is exactly parallel to it. This is as it should 
 be ; for if otherwise, the section obtained would be incorrect, for the 
 opposite faces of the cube being always parallel to each other, any plane 
 section of it will have parallel edges on opposite or parallel faces. 
 
 Before passing to the next problem, which is a variation of the last, 
 and somewhat more difficult of solution in that it involves a greater 
 amount of thought, it is necessary to point to the case of finding the 
 plan of the uncut cube, No. 4, Fig. 163. The student will remember 
 the reference made in a previous problem to the change of position which 
 takes' place in an original object when a plan of it is obtained from its 
 side elevation, as No. 4 is from No. 2 in Fig. 163. In this case, 
 the matter of finding the plan is simplified by the position the cube 
 occupies with respect to the YP ; its vertical faces making, as they do, 
 equal angles with it, enable the corners figured 3, 4, 1 in No. 1 to be 
 used as the representatives of 2, 3, 4, in finding the plans of its edges, 
 shown in No. 4, by the lines 2, 2 ; 3, 3' ; 4, 4'. Had the original 
 position of the cube, however, in No. 1 been any other with reference 
 to the VP than that actually occupied by it, then the plan of the cube 
 in No. 1 could not have been usetKta project from. The reason of this 
 will be seen in the next problem, to which we now proceed. 
 
 Problem 61 (Fig. 164). Given the plan of a solid cube, with its 
 vertical faces making unequal angles with the VP ; to Jind its 
 projections, and sectional projections when the plane on which it 
 rests is inclined to the HP at 4*5, and -the section or cutting plane 
 alters in direction. 
 
 Assuming that the student has worked out the previous problem, 
 and repeated it, with the section plane in varied directions, he will be 
 the better able to arrive at a solution of the one now presented to him. 
 As the first part of it is nothing more than a repetition of the process 
 adopted in the last case, applied to the altered position of the original 
 object, no further assistance is given than that indicated by the con- 
 struction lines shown, as the views almost explain themselves. No. 1 
 is the plan of the original object or cube in the given position ; No. 2 
 its side elevation when inclined at the given angle ; No. 3 its front 
 elevation ; and No. 4 its plan at the same inclination. Some of the 
 
132 
 
 FIRST PRINCIPLES OF 
 
MECHANICAL AND ENGINEERING DRAWING 133 
 
 reference figures are purposely omitted that the student may feel less 
 dependent upon such helps as he proceeds. 
 
 In views Nos. 5 and 6 (Fig. 164), the only point likely to cause 
 difficulty in a first attempt at getting the sectional elevation and plan 
 of the object will be at the change of the plane of section on the line 
 c'c. This, however, should be thought out by the student, as it only 
 involves the finding of the exact position of the two ends of this line 
 on opposite faces of the cube. If it is borne in mind that it is the inter- 
 secting line of two plane surfaces, and that one of its ends is repre- 
 sented in elevation c in No. 2 by a point, it is known at once that 
 it must be a line passing through c, parallel to the HP, and therefore 
 to the IL. Then, as it is parallel to the HP, its ends must be of 
 equal height from the bottom face of the cube. This height is evidently 
 the perpendicular distance between c and the plane OP ; or the depth 
 through the cube along the line c'c, square with its bottom face. 
 With these data, the position of its ends should be easily determined 
 without further guidance. Having found the elevation of this line, 
 its plan offers no difficulty, as it may be got in one of two ways 
 already explained. 
 
 As a test of the knowledge of our subject the draughtsman in 
 embryo should by this time have acquired, we give for solution, without 
 assistance other than that afforded by the few projectors shown in the 
 views of the object, the following problem 
 
 Problem 62 (Fig. 165). Given the plan of a hollow cube, with square 
 holes, centrally situated in all its sides ; to find its projections, when 
 the plane on which it rests inclines 30 to the horizontal, and its 
 sectional projections at the same inclination, when cut by a plane 
 along a given line. 
 
 The view No. 1 (Fig. 165) is the given plan of the cube, and the 
 line sp in No. 2 the line of section. The different views should be 
 drawn in on the sheet of paper in the order of their numbers, as they 
 are consecutive projections, each one of which, when correctly found, 
 enables that which follows to be more easily comprehended and drawn. 
 This problem, with the one that follows it, may be looked upon by the 
 young student as the pons asinorum in this particular part of the sub- 
 ject, which, when ably surmounted, may be regarded by him as one 
 landmark passed on the road to his success as a practical draughtsman. 
 
 59. The problem referred to in the last paragraph is that of finding 
 the projections of a skeleton cube, whose plan is given, resting on a 
 plane inclined at an angle to both planes of projection. 
 
 In Fig. 166 is given the several views that would be obtained of 
 such an object when in the assumed position, but without construction 
 lines or any explanation of the process of finding its projections ; the 
 problem being intended as a test of the student's ability in applying 
 the methods of procedure in its solution, which have previously been so 
 fully dealt with. The order in which the several views should be drawn 
 is the same as that shown in the three previous problems, Figs. 163, 
 164, and 165; and as the bounding edges of the different members of 
 
134 
 
 FIRST PRINCIPLES OF 
 
MECHANICAL AND ENGINEERING DRAWING 135 
 
136 FIRST PRINCIPLES OF 
 
 the original object are all straight ones, and their surfaces plane, the 
 finding of its projections should offer no difficulty to the learner who 
 has attentively followed the development of the subject to its present 
 stage. 
 
 The projections of the prism and pyramid, or other plane-surfaced 
 solid, when similarly inclined, should also be found without trouble, 
 their surfaces being in most cases plane figures having straight bound- 
 ing lines or edges. As, however, it is often necessary to obtain a view 
 of an inclined object, when turned through any given angle, to show 
 the method of procedure in such a case, the following problem is given, 
 as it combines the projection of both the prism and pyramid when 
 inclined to both planes of projection. 
 
 Problem 63 (Fig. 167). Given the plan No. 1, and elevation No. 2, 
 of a hexagonal pyramid, with a similar shaped prismatic base, to 
 find its projections ; first, when its axis is inclined to the HP, and 
 parallel to the VP ; and again ivhen it is inclined to both the VP 
 and HP. 
 
 If the inclination of the plane on which the given object is to rest 
 be, say, 30 to the horizontal, this at once determines the angle made 
 by its axis with the HP as one of 60, the one angle being the com- 
 plement of the other. Then the elevation No. 2 being given, to find 
 the projections first required, show as in No. 3 a view of the object on 
 a plane inclined to the HP at 30. This is nothing more than a 
 transfer of the view given in No. 2, directly to this plane, as the same 
 faces of the object seen in No. 2 will be seen in No. 3, the former view 
 having been obtained by direct projection from its plan No. 1, before, 
 and not after, its inclination, as in previous problems. The plan of No. 
 3, given in No. 4, is found, as before, by projection from No. 3 and 
 No. 1. 
 
 Then to find the projections of the object when inclined to both the 
 VP and HP, as required in the second part of the problem, assuming 
 the angle its axis is to make with the YP is known, we proceed as 
 follows : Draw in, in the HP, a line as ab, No. 5, making the same 
 angle with the IL that the axis of the solid is to make with the YP 
 say one of 45. To this line which is a plan of the axis of the solid 
 transfer the view obtained in No. 4, and from it, and No. 3, find by 
 projection that given in No. 6, then will the views Nos. 3, 4, 5 and 6, 
 Fig. 167, be the required projections of the original object, represented 
 in plan and elevation by Nos. 1 and 2 in the figure. In the same way 
 as here explained may the plan or elevation of any plane solid in any 
 desired inclined position be obtained. 
 
 60. Passing to the projection of solids bounded by curved surfaces 
 inclined to both the YP and HP, a similar combination of the cone and 
 cylinder is taken as the object for projection, the problem being 
 
 Problem 64 (Fig. 168). Given tJie plan No. 1 and elevation No. 2 
 of a cone having a cylindrical base, its jirojections are required, first 
 when it rests on a plane inclined to the IIP at 45, with its axis 
 
MECHANICAL AND ENGINEERING DRAWING 
 
 137 
 
138 
 
 FIRST PRINCIPLES OF 
 
MECHANICAL AND ENGINEERING DRAWING 139 
 
 parallel to the VP, and after it has been swung from that position 
 through a horizontal angle of 30. 
 
 The method of procedure in this case being the same as in the 
 previous problem, little difficulty should be found in obtaining the pro- 
 jections shown in Nos. 3, 4, 5 and 6 in the figure. As both portions 
 of the object the conical and the cylindrical have a curved surface, 
 greater care will be required in drawing in the ellipses, into which the 
 top and bottom edges of the cylindrical portion are projected, but with 
 this exception there is nothing to prevent correct projections of the 
 object being easily found if the instructions previously given in refer- 
 ence to curved-surfaced solids are carefully followed. 
 
 To assist the student in determining the direction of the axes of the 
 ellipses with respect to the IL, in the view required in No. 6 let him 
 assume the base of the given object to be cut out of a square prism, as 
 shown in dotted lines in No. 6, then the projection of the diametral 
 lines , b ; c, d, of No. 1 on the upper face of this prism represented in 
 No. 6 by the faint lines a'b' ; c'd', will be the axes of the ellipse 
 sought, and a', b f , c', d' will be four points through which the curve of 
 the ellipse into which the base of the cone is projected has to be 
 drawn. The lower edge of the cylindrical portion of the solid in its 
 new position is, of course, projected into a similar ellipse as the upper 
 edge, or base of the cone, but only half of it, as shown, will be seen. 
 
 As the projections of a sphere, in similar positions to the other solids 
 taken as examples, offer little useful practice beyond that of the correct 
 drawing of ellipses, we give as the concluding problem in this part of 
 our subject, one in which the application of its principles is practically 
 shown in the delineation of an ordinary "nut," in such positions as 
 frequently occur in actual machine details. The problem is 
 
 Problem 65 (Fig. 169). Given the plan and elevation of a six-sided 
 chamfered nut, to find its projections when the plane surface, on which 
 it lies is inclined to the planes of its jyrojection. 
 
 Before proceeding to find the required projections of the nut, it may 
 be noted in reference to such an article, that its flat sides at right angles 
 to its top and bottom faces are called " panes " ; that its height, or 
 thickness, is generally equal to the diameter of the bolt it is intended to 
 fit; its width across the "panes" or "flats" bears a certain fixed pro- 
 portion to this diameter, and the "chamfer," or cutting down of its 
 upper corners or angles, is usually at an angle of 45 with its top face. 
 
 To draw in the plan of a nut such as that shown at No. 1, Fig. 169, 
 it is necessary to know first what its width across the flats is to be. 
 This width for all nuts up to 6 in. diameter of bolt may be found in any 
 engineer's pocket-book. With the given width for the size of nut 
 proposed to be drawn as a diameter, describe a circle in the HP at a 
 convenient distance from the IL, and from the same centre a second 
 circle equal in diameter to that of the bolt for which the nut is intended. 
 Through the common centre of these circles draw lines indefinitely, 
 parallel and perpendicular to the IL. Then with the T-square and set- 
 
140 
 
 FIRST PRINCIPLES OF 
 
MECHANICAL AND ENGINEERING DRAWING 141 
 
 square of 60, draw lines tangential to the outer circle in the positions 
 shown, and the resultant hexagon, with its inscribed circles, will be the 
 plan of the intended nut when resting on the HP. 
 
 To find its elevation in this position, project over into the YP the 
 four corner edges of the nut seen when looked at in the direction of the 
 arrow in plan. These will, of course, be lines perpendicular to the IL. 
 Now, as the thickness, or height, of an ordinary nut generally equals the 
 diameter of its bolt, set off this diameter as a height from the IL 
 and through it draw a line parallel to the IL. This line will represent 
 the top face of the nut, and were it a plane-faced one and not chamfered, 
 its elevation would be completed by making this line to cut the four 
 perpendiculars previously drawn in. The chamfering is represented in 
 elevation by the three circular arcs and the two short lines at an angle 
 of 45, tangent to the outer arcs, the reason for which will be fully 
 explained when treating of screws, nuts, etc., later on. 
 
 The plan and elevation of the nut being thus given, its projections 
 as required in the problem are now to be found. Assuming the angle 
 of the plane surface on which the nut is supposed to rest to be one of 
 30 to the horizontal, draw in a line making with the IL such an angle. 
 Transfer to this line the view of the nut given in No. 2, and obtain by 
 projection its plan in this inclined position, as shown in No. 4. Then 
 to find its elevation when swung through a horizontal angle say of 45 
 draw in its axial line in the HP making that angle with the IL, and 
 transfer to it the plan of the nut found in No. 4, giving the view of it 
 shown in No. 5. From this plan, and the elevation No. 3, find by 
 projection the view given in No. 6, and the requirements of the given 
 problem will be fulfilled. 
 
 An infinite variety of problems might be given in this very interest- 
 ing part of our subject, but as they would only involve in their solution 
 the correct application of principles which have been fully explained, we 
 pass on to the elucidation of projection as applied to the " penetration " 
 and " intersection " of solids, a knowledge of which is of the first import- 
 ance to the would-be draughtsman. 
 
CHAPTER XVII 
 
 THE PENETRATION AND INTERSECTION OF SOLIDS 
 
 61. UP to the present stage in our subject, all the objects chosen as 
 examples to illustrate the application of the principles of projection 
 have consisted of simple solids, with either plane or curved surfaces or 
 both -and have each been treated independently. They have not, of 
 course, included all the elementary forms which, in combination, give 
 shape to machine and engine details ; but they have been such as will 
 enable the student to delineate correctly any subsidiary solid generally 
 derived from one of the primary ones which may enter into the con- 
 struction of a complete mechanical structure. 
 
 As in the details of such a structure, some of the solids, which have 
 so far been dealt with singly, are joined to, or made to penetrate, one 
 another, their surfaces by such junction or penetration produce a line or 
 lines which require accurate delineation in any drawing having preten- 
 sions to truthfulness. Such lines are, however, not only required for the 
 correct representation of the objects in combination, but they are neces- 
 sary to be known before the objects represented could be constructed in 
 any material. It is then to the solution of problems by the aid of 
 projection presented by such combinations, that attention is now to be 
 directed. 
 
 In determining the lines of intersection of two solids, the simplest 
 possible method of procedure so long as it gives correct results should 
 be aimed at ; and although plane-surfaced solids are probably not so often 
 met with in combination in mechanical details as they are in building 
 constructions, yet it is necessary that their intersections should be 
 thoroughly understood by the draughtsman. The introductory pro- 
 blems in this part of the subject will therefore consist of those in 
 which the intersections of plane-surfaced solids only are required to be 
 found. 
 
 62. One of the first facts realized by the student in commencing the 
 study of Projection, was that the intersection of two planes at any angle 
 produced a straight line. It follows from this that the intersections of 
 plane solids must also be straight lines, as the surfaces which inter- 
 sect are planes. This consideration, then, is the key to the solution of 
 any problem in which the intersections of plane-surfaced solids are 
 
 142 
 
MECHANICAL AND ENGINEERING DRAWING 
 
144 FIRST PRINCIPLES OF 
 
 concerned ; for if through two such intersecting solids a section plane 
 is passed, the points where that plane cuts the surfaces of both solids 
 will be at once shown. If, then, the two solids be imagined to be cut 
 8iiiin.Uaneov.sly by a number of planes, and the points of intersection of 
 their surfaces by those planes be noted, a line drawn through those 
 points will be the " line of penetration '' of one solid by the other. As 
 the plane solids usually met with in machine details are prisms and 
 pyramids, with their frustums, the problems in this connection will deal 
 first with prisms penetrated by prisms. The first is 
 
 Problem 66 (Fig. 170). Given the plan Xo. l y of two square pivsms 
 A and B, of equal length, one penetrating the other at right angles ; 
 to find their elevation and lines of penetration. 
 
 As shown in the plan, A is the penetrated, and B the penetrating 
 prism : the former having two of its sides parallel to the TP, while all 
 four sides of the latter are perpendicular to it. The lines of intersection 
 of the two, therefore, will only be seen on the front side of A, or that 
 nearest to the eye in elevation. As that side is parallel to the TP, and 
 the prism B penetrates it at right angles, the lines of penetration will 
 coincide with those forming the front end of that prism. Therefore, 
 find by projection the elevation of the prisms, in the positions shown in 
 plan, and the lines ah, be, cd, da, in No. 2 will although representing 
 the front end of prism B- be the lines of penetration sought. In this 
 problem, as the four sides of the penetrating prism B are planes of 
 intersection with the two sides or the one nearest to, and that .farthest 
 from the TP of the penetrated one A, the use of any section planes in 
 finding the lines of penetration is unnecessary. 
 
 , let the two prisms be moved on tlie ajcis of A as shoim in No. 3 
 until that of B makes an angle of 30 with the JJL, and their elevation 
 and lines of penetration be required. 
 
 Here, although the axes of the prisms are in the same relative posi- 
 tion as before, that of B, being inclined to the TP, brings the vertical 
 sides of both at an angle with that plane, and will, in elevation, show 
 as in No. 4 two sides of A and only one of B, and consequently but 
 one actual line of penetration, or that formed by the intersection of the 
 plane of the side a, 6, c, d of the prism B, with the side e, f t g, h of the 
 prism A. To show this line, draw a projector through point x in the 
 plan No. 3 into the TP, and it will cut the lines a, 6, c, d in No. 4 in 
 jc', y, which is the line sought. The other lines of penetration of A by 
 B cannot be seen, as they are covered by those representing the front 
 edges of B, and the back ones do not come into view at all. 
 
 To find the lines of penetration of two intersecting square prisms 
 when the axis of one of them is inclined to both planes of projection, 
 the procedure is as shown in Nos. 5, 6, 7, and 8, Fig. 1 70. First draw 
 in the elevation as at No. 5 of the prisms with the axis of the pene- 
 trating one B, parallel to the TP, and at the angle it is to make with 
 the HP, and find their plan in this position. Swing this plan, No. 6, 
 round on the axis of the prism A until that of B makes the required 
 
MECHANICAL AND ENGINEERING DRAWING 
 
 14*) 
 
146 FIRST PRINCIPLES OF 
 
 angle say of 45 with the VP or IL. Then with this plan, No. 7, 
 and the elevation, No. 5, find by projection the elevation of the prisms 
 in their new position, as shown in No. 8. Thus posed, but two lines of 
 penetration will be seen viz., where the prism B passes through the 
 side a, b, c, d of A, and are found by projectors drawn through e and 
 / in No. 5, and g and h in No. 7. The lines of penetration of A by B 
 on the side opposite to a, b, c, d shown in dotted lines are found by 
 similar projectors from points e',f in No. 5, and g' ' , h' in No. 7. 
 
 The next problem is a case in which a diagonal of the intersected 
 prism is in the same plane as that of a diagonal of the prism intersect- 
 ing it. 
 
 Problem 67 (Fig. 171). Given the plan of two square pi'isms of 
 equal length, penetrating each other at right angles, their axes and a 
 diagonal of each being in one and the same plane, to Jind their lines 
 of penetration. 
 
 First, let the axes of both prisms be parallel to the VP, then No. 1 
 (Fig. 171) will be their plan. From this, first find the elevation of the 
 prism A, making it equal in height to the length given in the plan 
 of B. At the middle of this height draw in the axial line of B, and on 
 it project over from the plan No. 1, the two ends of that prism. Now 
 the lines of penetration of A by B may be found in two ways, either by 
 projecting over to the elevation the points of intersection of the visible 
 edges of B with A, seen in plan, or by the use of section planes. By the 
 first method it will be seen at a glance, that in elevation only two sides 
 with their three edges of the prism B will be seen as penetrating A. 
 These are the top, bottom, and front edges, the two former of which 
 intersect the right and left vertical edges of A in points 1, 2, 3, 4. To 
 find where the front edge of B enters and leaves A, project over from 
 the plan of this edge the line ab No. 1, to its elevation ab No. 3, the 
 points 5, 6; join points 1, 5 ; 5, 3 ; 2, 6 ; 6, 4 by straight lines, and they 
 will be the lines of penetration sought. 
 
 63. To find the same lines by the use of intersecting planes, it will 
 be seen at once that if a plane be caused to pass through the axes of 
 both prisms parallel to the VP, it would intersect the prisms in the 
 points 1, 2, 3, 4, No. 2 ; and if a second plane, parallel to the first and 
 tangent to the edge ab of the prism B, be passed through A, it would 
 intersect the edge ab, No. 1, in the points 5 and 6, as the section of A 
 made by this plane would be the parallelogram cdef shown in faint lines 
 in No. 2, the points of intersection of the edge ab by it being 5 and 6, 
 which, joined to 1 and 3 and 2 and 4, give the same lines of penetration 
 as by the first method. It will be seen, on studying this figure, that 
 the resultant lines of penetration are nothing more nor less than the 
 intersections of pairs of planes at an angle, and it is in the judicious 
 application of such section planes that the whole art of finding the lines 
 of penetration of solids consists. It will, however, be evident that in 
 the problem just solved, and that which follows it, their use in practice 
 would be dispensed with, as the lines sought can be found at once by 
 simple projection from the plans of the objects ; but, as it is advisable 
 
MECHANICAL AND ENGINEERING DRAWING 147 
 
 that at this stage the student should understand the application and 
 use of such planes, they have been introduced thus early. 
 
 For our next problem, the same solids are taken as in the last, but 
 instead of both axes being parallel to the VP, that of the prism B is 
 inclined at an angle to it. This position is shown in plan in No. 3, 
 Fig. 171 a transfer of No. 1 at an angle to the VP or IL and it is 
 required to find the elevation of the prisms, and their lines of penetra- 
 tion when so posed. 
 
 To do this, first find by projection from No. 3 and No. 2, the eleva- 
 tion of the prisms as if they were entire. From the plan No. 3, draw 
 projectors from the points 1, 2, 3, 4, in it, where the edges of the prism 
 B that will be seen in elevation intersect those of A ; then straight 
 lines joining these points will be the lines of penetration sought, as 
 shown in No. 4. 
 
 64. In cases where the axes of the intersecting solids are not in the 
 same plane, it may happen that the lines of penetration cannot be found 
 by direct projection alone, as in the previous problems, but will necessi- 
 tate the use of section planes to discover them. As an instance of 
 this, take the following problem 
 
 Problem 68 (Fig. 171). Given the plan of two square prisms of equal 
 length, intersecting each other at right angles, but having their 
 axes in different planes ; to find their lines of penetration. 
 
 Let No. 5, Fig. 171, be the given plan of the prisms, the axis of A 
 being vertical and that of B horizontal, but inclined to the VP. Ob- 
 tain by projection the elevation of the prisms as if they were entire. 
 The two lines ab, be of penetration, of A by the prism B, that will be 
 seen in elevation, assuming it to enter A from the left, can be found by 
 direct projection from the plan, as in the previous problem. Those 
 through the front side of A to the right may be found in either of two 
 ways, by the use of a section plane. 
 
 First assume a p'ane passing vertically through the prism B, and 
 coinciding with the right-hand front side of prism A, a plan of which 
 would be the faint line xx, No. 5. The section of B by this plane 
 would be a triangle, its apex being the point d in the f rontmost edge 
 of the prism B, and its base, a line across the right end of that prism, 
 perpendicular to the HP. Find, by projection, the elevation of this 
 triangular section, as shown in No. 6, and the points 5, 6 in its sides 
 will be those where the two front sides of the penetrating prism B in- 
 tersect the vertical line 2 4, or the right edge of the prism A in passing 
 through it. Right lines joining points 5 and 6 with d the apex of the 
 triangle thus found, will be the lines of penetration required. 
 
 The same lines may be found by assuming a plane to pass vertically 
 through both prisms parallel to the axis of B at point 2 in plan ; or 
 through the line 2 4 in elevation. Find, by projection, the elevation of 
 the rectangular section of B made by this plane, and it will give the 
 points 5 and 6 in No. 6, as those where the two front sides of the 
 prism B intersect the vertical edge 2 4 of prism A. These points, joined 
 
148 
 
 FIRST PRINCIPLES OF 
 
MECHANICAL AND ENGINEERING DRAWING 149 
 
 by right lines to the vertical projection of point d in the plan, give the 
 same lines of penetration as before obtained. 
 
 65. Had the prisms in the foregoing problems been equal pairs, the 
 resultant lines of penetration would of course have been different. In 
 such case, when seen in elevation as in No. 6, Fig. 172, there would be 
 no lines of penetration at all, as the two visibles faces are in one and 
 the same plane, and therefore show no juncture. If, however, their 
 sides were inclined to the planes of projection their axes being still at 
 right angles and in one plane parallel to the VP then the lines of 
 penetration would be as shown in No. 2, Fig. 172, taking the form of 
 a cross, due to their surfaces intersecting each other equally. This 
 may be proved to exactness by the use of section planes. Let No. 1, 
 Fig. 172, be the plan of two equal-sized prisms A and B, intersecting 
 each other at right angles, and with their axes in one and the same 
 plane, and let it be required to find their lines of penetration. 
 
 From the position, shown in plan, in which the prisms are with 
 respect to each other, it is evident that a section plane passing vertically 
 through both their axes, would give equal rectangular sections of each, 
 and four points of intersection with the prism B ; a and b being those 
 of entering, and c and d those of leaving it. Then as the prisms are 
 equal, it follows that any number of parallel planes passing simultane- 
 ously through both of them would give equal sections and points of 
 intersection; lines, therefore, drawn through these points, taken in 
 order, will give the lines of penetration of one solid by the other. 
 
 To find these lines in the given problem, obtain by projection from 
 the plan No. 1, an elevation, as in No. 2, of the prisms as if they were 
 entire. Parallel to the axis of B in No. 1, draw the lines 1 1', 2 2', 3 3', 
 which will be the edge view, or plan, of the section planes to be used. 
 Find the elevation of each of the sections of the prism A, produced by 
 these planes, by vertical projectors from the points 4 4', 5 5', 6 6', in 
 No. 1 ; then as the prisms are equal, and the section planes are assumed 
 to pass through both, the horizontal one B will give similar sections to 
 that of A. To find their bounding lines, produce the axial line x y of 
 the prism B in No. 2 to the left, and on it draw in a half-end view of 
 that prism looked at in the direction of the arrow on the left. On this 
 set off from the line ef the distances that the lines 1 1', 2 2', 3 3', in 
 plan No. 1 are from the axis of B, and through the points thus found 
 draw vertical lines to meet the two sides of the triangles in the points 
 1 1', 2 2', 3 3'. Then parallel to the IL, and through these last-found 
 points, draw lines from end to end of the prism B, as shown in No. 2. 
 The points where these lines cut the corresponding vertical ones drawn 
 through points 4 4', 5 5', 6 6', in No. 1, will be points in the lines of 
 penetration of the two prisms. On drawing lines through these points, 
 they will be found to be straight ones, crossing each other at the point 
 2, where the two frontmost edges of the prisms intersect. Although 
 these lines might have been at once obtained by joining the points a d, 
 b c, in No. 2, by straight lines, it is advisable that the student should 
 be able to prove that they become straight, rather than take the matter 
 for granted. 
 
 If the axis of one of such equal prisms be inclined to the VP, 
 
150 
 
 FIRST PRINCIPLES OF 
 
MECHANICAL AND ENGINEERING DRAWING 151 
 
 the other remaining parallel to it as before, then the lines of penetration 
 will be as shown in No. 4 ; obtained by first finding the elevation of the 
 prisms as if they were entire, and then projecting over from No. 3 
 which is No. 1 with the axis of the prism B at the desired inclination 
 to the VP to the corresponding edges in No. 4, the points where the 
 edges of B that will be seen in elevation intersect those of the prism A. 
 The right lines joining these points as shown in No. 4 are then the 
 visible lines of penetration of the two prisms. 
 
 For the next problem a case is taken where, although the axes of 
 both prisms are still in the same plane, yet neither of them is vertical 
 or horizontal. 
 
 Problem 69 (Fig. 173). Two equal square prisms have their axes in 
 one and the same plane, and intersect each other at the middle of their 
 length; give the elevation and plan of the j^'isms, and their lines of 
 intersection when their axes are parallel to the VP, but inclined to 
 the HP at 52\\ 
 
 As there would be no visible lines of penetration of the two prisms 
 given in the problem, if their sides were parallel and perpendicular to 
 the plane in which their axes lie, it is assumed that that plane divides 
 each square prism into two equal triangular ones, and as their axes are 
 inclined to the HP, it is evident that their elevations must first be 
 drawn. Now the axes being in a plane which is parallel to the Y P, to 
 draw the elevation of the prisms in the position stated in the problem, 
 proceed as follows : 
 
 Draw in, in the VP, two lines at an angle of 52J with the IL, 
 intersecting each other. This may be done by using a " scale of chords " 
 or a " protractor," or with the set-squares of 45 and 60. If neither 
 of the former is to hand, take the set-square of 45, and at a convenient 
 point say a in the IL, draw a faint line at 45 with it ; through the 
 same point draw a similar line with the 60 set-square ; bisect the angle 
 formed by these lines, and the line of bisection will be at an angle 
 of 52J with the IL, as it is the mean betwen 45 and 60 ; for 
 
 = 52J. Through a convenient point as x in this line, let 
 
 fall a perpendicular to the IL, cutting it in point x ; make x b in 
 that line equal to x a, and from b through x draw a line indefinitely. 
 Then as the lines passing through x each make an angle of 52 J with 
 the IL, they will represent the axes of the required prisms, and from 
 their position with respect to the YP, their frontmost edges also. Set 
 off from x, in points 1, 2, 3, 4, half the length of the intended prisms, 
 and at those points draw lines at right angles to the axial ones ; 
 those drawn through 2 and 4, cut the IL at 2', 4' ; in these make 2 2", 
 4 4" respectively, equal to 2 2', 4 4'. Through 2' 2" and 4' 4" draw 
 lines parallel to the axial lines 2 x 3, and 4 a? 1, and the resultant rect- 
 angles will be the elevations of the prisms as if they were entire, resting 
 on the HP, on the two opposite corners, 2', 4', of their lower ends. 
 
 Having drawn in the elevation, the plan of the prisms may now 
 be found. Their axes being parallel to the YP, and in one plane, that 
 
152 MECHANICAL AND ENGINEERING DRAWING 
 
 plane will be correctly represented by a line as a b in No. 2 parallel 
 to the IL. It is evident such a plane will pass through the two 
 diagonals of the ends of the prisms seen from above in the direction 
 of the arrows which are parallel to the VP. Having found the plan 
 as shown in No. 2, the lines of penetration of the prisms are now required. 
 These in elevation will be the intersections at equal angles of the front 
 sides of the prisms. The points of intersection of the edges of the prisms 
 are those between which the lines of penetration will pass, and as the 
 prisms are equally inclined to each other, those lines will pass through 
 point x in No. 1 at right angles to each other, as shown. In the plan 
 No. 2, only two such lines will be seen viz., xy, and the one directly 
 behind it in elevation but which, on account of their position both 
 being in one plane merge into one line xyx', shown in No. 2. Now, 
 as such junctions of prisms as are shown in No. 1 (Fig. 173) often 
 occur with their axes in a plane at an angle with the VP, we give such 
 an instance in the following problem 
 
 Problem 70 (Fig. 173). driven the plan of the equal square prisms 
 with tlieir axes inclined to each other at the same angle as in the 
 last problem ; to find their elevation and lines of penetration, when the 
 plane of their axes is inclined to the VP at an angle of 30. 
 
 The position the prisms will now assume will be tantamount to 
 swinging the two together as shown in No. 1 (Fig. 173), on the corner 
 figured 2', of the left-hand one, through a horizontal angle of 30. If 
 then the plan of the prisms obtained in No. 2 be drawn in with the 
 line a b, making that angle with the IL, as shown in No. 3, then the 
 elevation of the prisms in this position may be obtained by direct 
 projection from No. 3 and No. 1, care being taken that the projectors 
 used in doing so are drawn from corresponding points in each. If due 
 consideration be given to the relative position of the edges of the prisms 
 to each other, no difficulty should be experienced in obtaining a correct 
 projection of them, as shown in No. 4. The lines of penetration in 
 elevation do not, of course, in this case show as being at right angles to 
 each other, although they are virtually in that position. 
 
 Nos. 5 and 6 in Fig. 173 show the application of the same method 
 of procedure to the case of the penetration of triangular prisms as 
 to those of square section ; which require no further explanation than 
 that afforded by the projectors shown in the diagram. One side of the 
 penetrating prism is shown as coinciding with and passing through the 
 axis of the penetrated one. In each case the cross-section of the prism 
 is an equilateral triangle, the size of the penetrating one being shown by 
 the end views given at A and B. As a multiplicity of similar examples 
 to the foregoing would only show the application of the same principles 
 in determining the lines of penetration of prisms by prisms, a few 
 problems will now be taken in which the intersections of pyramids 
 are involved. 
 
CHAPTER XVIII 
 
 THE INTERSECTIONS OF PLANE SOLIDS (continued). 
 
 66. BEFORE passing directly to the finding of the lines of penetration 
 of one pyramid by another, the solution of such a problem will come 
 much easier to the student if some preliminary practice is had with the 
 case of a pyramid penetrating a prism. As such a combination is often 
 used in practice, in giving form to simple ventilators, cowls, etc., we 
 give as problems in this connection those in which the two solids are 
 combined for such purposes. 
 
 The first combination, shown in plan and elevation in No. 1 and No. 
 2 (Fig. 174), is the simplest possible, it being that of a square prism, 
 with its axis vertical and sides parallel and perpendicular to the VP, 
 penetrated by a square pyramid, the axis of which is parallel to both 
 the VP and HP, and in the same piane as that of the prism. 
 
 From the relative position of the two solids, as shown in the figure, 
 it is evident that neither in plan nor elevation will any lines of penetra- 
 tion be visible in this case, as the two faces a and b of the prism, which 
 are penetrated by all four faces of the pyramid, are in plan and 
 elevation through being perpendicular to both planes of projection 
 each represented by one straight line only, with which the actual lines 
 of penetration coincide, and therefore cannot be seen. Had the faces of 
 the prism, however, been in the least inclined to the VP still remain- 
 ing perpendicular to the HP, then one line of penetration only would 
 have come into view, dependent upon the inclination of the axis of the 
 pyramid to the VP. 
 
 For all purposes where combinations of solids are used to give form 
 to mechanical details, it is usual in practice to so combine them that 
 their intersections shall be symmetrical. That is to say, that when the 
 form of the article to be made requires that one solid should penetrate 
 another, the penetration is made in such a way that an axial plane 
 passed through the solids shall divide each of them into equal parts. 
 This, although it necessitates the axes of both solids being in one and 
 the same plane, does not necessarily prevent either the axes or sides 
 of the solids from being inclined to the VP or HP. As instances we 
 may take the following cases put in the form of problems 
 
 153 
 
154 
 
 FIRST PRINCIPLES OF 
 
MECHANICAL AND ENGINEERING DRAWING 155 
 
 Problem 71 (Fig. 174). Given the plan of a square prism, with its 
 axis vertical and its sides equally inclined to the VP, penetrated by 
 a 'square pyramid having its axis parallel, and its opposite sides 
 inclined to one of the planes of projection and perpendicular to 
 the other ; to find the elevation and lines of penetration of two 
 solids. 
 
 Let No. 3 (Fig. 174) be the plan of the two intersecting solids. 
 Then from the definition of a prism, and of a pyramid, it will at once be 
 manifest that an axial section or one along the line AC</ of these 
 solids will in elevation give a rectangle for the prism, and a triangle for 
 the pyramid. Obtain these sections by projection from the plan ; and 
 the points 1, 2, 3, 4 in that of the pyramid will be the limiting ones in 
 the penetration. Complete the elevation of the prism by drawing in its 
 front vertical edge. To determine the intersections of the front side of 
 the pyramid fgf, No. 4, with the two inclined vertical sides of the 
 prism seen in elevation, find by projection the elevation of the lines 
 formed by the meeting of those sides. Now the points a and b, in the 
 line fg in No. 3, are evidently the plans of the lines required. There- 
 fore, through them draw projectors into the VP, and they will cross the 
 front face of the pyramid between the points 5, 6, and 7, 8 in its top 
 and bottom edges, and will give the lines of penetration sought. These 
 being put in, in full, and joined up to / and /', and a' respectively, will 
 complete the solution of the problem. 
 
 As shown in a previous problem, the lines of penetration just found 
 can also be determined by assuming a section plane to pass vertically 
 through the prism on the line of the front face of the pyramid giving 
 the rectangular section a bed, No. 4 shown in faint lines, intersect- 
 ing the triangular section first found in the points 5, 6, 7, 8 ; and 
 thereby giving the lines of penetration sought. 
 
 The second case of the penetration of a prism by a pyramid is that 
 in which the prism is in the same position with respect to the VP 
 as before ; but the pyramid penetrates it in such a way that an axial 
 plane passing through both, divides each solid into two equal ones 
 having triangular bases. 
 
 Problem 72 (Fig. 174). A square prism, in the same position as in 
 the last problem, is penetrated by a square pyramid having its sides 
 equally inclined to both planes of projection ; required their plan and 
 elevation and lines of penetration. 
 
 With one exception, the plan of the solids in the positions given in 
 the problem will be similar to that given in No. 3 (Fig. 174). The 
 exception is caused by the altered position of the pyramid. The triangle 
 efg in No. 3 represented & flat surface perpendicular to the VP one 
 of the faces of the pyramid whereas in No. 5, the pyramid having 
 been turned on its axis through 90, has brought two of its sides 
 into view in both plan and elevation equally inclined to the planes 
 of their projection, and therefore requiring a third line to represent 
 
156 
 
 FIRST PRINCIPLES OF 
 
MECHANICAL AND ENGINEERING DRAWING 157 
 
 their junction. This line or edge in No. 5 will be a a ; that part of it 
 where it passes through the prism being out of sight. From this plan 
 No. 5, find by projection the elevation No. 6 of the solids as if they 
 were entire. For the lines of penetration that will be visible, draw 
 projectors from the points 1, 2, in the line b a in No. 5, to cut b' a" in 
 No. 6 in the points 1', 2' ; then lines drawn from these last-found points 
 to 3, 3' ; 4, 4' ; or where the upper and lower edges of the pyramid 
 enter and leave the prism, will be the lines of penetration required. 
 67. Although, as stated in a previous paragraph, it is usual in 
 practice to make the intersections of solids in the design of mechanical 
 details symmetrical, cases may sometimes arise in which this symmetry 
 cannot be adhered to. Through the exigencies of position of the solid 
 penetrated, it may be impossible to bring the axes of both into one and 
 the same plane, although at right angles to each other. To meet such 
 a case the following problem will show the method of procedure in 
 determining the lines of penetration. 
 
 Problem 73 (Fig. 175). Given the plan of a square prism, pene- 
 trated by a square pyramid, their axes being in different planes ; re- 
 quired the elevation and lines of penetration of the solids. 
 
 From No. 1 (Fig. 175), which is the given plan of the solids, it is 
 seen that the axis d, a' of the pyramid passes through the prism in a 
 direction parallel to two of its opposite sides, but some distance in 
 front of its axis. To find the lines of penetration when in such a posi- 
 tion, obtain by projection the elevation of the solids as if they were 
 entire, both of them being of the same dimensions as in the previous 
 problems. As a vertical plane passed through the axis of the pyramid 
 would divide it into two equal triangular ones, and give an isosceles 
 triangle for the section made by the cutting plane, it is evident that 
 any vertical plane passing through the apex of the pyramid would give 
 a similar though not equal triangular section. 
 
 Now, in looking in the direction of the arrow in No. 1, it is seen 
 that the frontmost edge a' a of the pyramid enters the left visible face 
 of the prism at point 1, and leaves it on the right face at point 2. 
 These points projected over to the corresponding edge of the pyramid 
 in No. 2, give 1', 2". To find where the vertical edge bb' of the prism 
 seen in elevation in No. 2 enters and leaves the two front faces of 
 the pyramid, draw (in No. 1) a line through a', its apex, and the 
 corner b of the prism, and produce it to cut the base line a c of the 
 pyramid in the point 3. We have then to find by projection the eleva- 
 tion of the section thus made, to enable us to obtain the points re- 
 quired. Taking this line, 3 b a, as the edge view, or plan, of a verti- 
 cal plane passed through the pyramid, it will give a triangle for the 
 section. To determine it we must first get an end view (in elevation) 
 of the base of the front half of the pyramid. Now, its whole base is a 
 square, with diagonals equal in length to the line c d a in No. 1 one 
 of them being vertical and the other horizontal. To draw the half- 
 base required, find the elevation of the vertical diagonal of the base of 
 the pyramid, represented in No. 2, by the faint line drawn between the 
 
158 FIRST PRINCIPLES OF 
 
 points d and d' t they being the top and bottom ends of that line ; 
 produce the axis a a of the pyramid, and on it draw the half-square 
 d x d' as shown. 
 
 Now the assumed section plane, represented by the line 3 b a in 
 No. 1, cuts through the base of the pyramid at the distance d 3 from 
 its vertical diagonal. To find the line on the half -base so made, 
 through d in No. 1 draw a line indefinitely, parallel to the IL ; with d 
 as centre and radius d a, draw an arc, cutting the line through d pro- 
 duced, in the point x, and from the same centre, with a radius equal to 
 d3, describe an arc cutting xd in 3'. As the line xd, No. 1, is a plan 
 of the half -square d x d', No. 2, swung round on its diagonal d, d' as a 
 hinge, until it is parallel to the IL, on this half-square or front half- 
 base of the pyramid can now be found the exact length of the base 
 of the isosceles triangle produced by the cutting plane, and from it the 
 section itself, and the lines of intersection of the sides of the prism and 
 pyramid. 
 
 For the triangular section, draw a projector into the VP through 
 point 3', in the line xd, No. 1, and it will cut the lines xd and xd' in 
 No. 2, in points 3, 3' ; the distance between which is the length of the 
 base of the triangular section. For its sides, cut by projectors drawn 
 through points 3, 3', parallel to the IL, the lines d a, a d', in points 4, 4', 
 and through these and the apex a draw faint lines as shown in No. 2 ; 
 then the points 5, 5', where these faint lines or edges of the tri- 
 angular section cut the vertical edge b b' of the prism, are two of the 
 points sought. Join 1' in the line a a' to these points by straight lines, 
 and they will be the lines of penetration on the left front face of the 
 prism. For those on its riyht face, find by projection the elevation of 
 the points 2 and 6 in the edges of the pyramid where they leave 
 that face, and join these as before by the straight lines 6 2', 2' 6', and 
 they will be the lines of penetration seen on that face. For the corre- 
 sponding lines on the back faces of the prism not seen, but shown 
 dotted the same method of determining them is employed as for those 
 that are directly in view. 
 
 68. As an example of the case of the penetration of a prism by a 
 pyramid, when the axes of the two solids are in one plane, but not at 
 right angles to each other, we give in Nos. 3, 4, 5, 6 (Fig. 175) the re- 
 sultant projections and lines of penetration of the solids when having 
 these relative positions. Nos. 3 and 4 show the plan and elevation of 
 the solids when the plane of their axes is parallel to the YP, and 
 therefore perpendicular to the HP ; and Nos. 5 and 6, those of the 
 same solids when that plane is inclined to the VP, but still in the 
 vertical position. As the required projections and lines of penetration 
 should now be found by the student without further assistance than 
 that given by the projectors in the diagram, we pass on to the de- 
 termination of the lines of penetration of one pyramid by another. 
 
 The first case is that of the penetration of a square pyramid by a 
 similar one of equal size, the axes of both being in one plane, parallel to 
 the VP, and at right angles to each other. 
 
 First, let the penetrated solid be in such a position that the 
 diagonals of its l/ase are respectively parallel and perpendicular to 
 
MECHANICAL AND ENGINEERING DRAWING 159 
 
 the VP, and those of the penetrating one at equal angles with both 
 the planes of projection. Then to find their lines of penetration, draw 
 in first the plan and elevation of two pyramids having equal bases 
 and altitudes with their axes at right angles. The combined triangle 
 and square shown in No. 1, and the three triangles shown in No. 2 
 (Fig. 176) will be the plan and elevation of the solids as if they were 
 entire in the given positions. 
 
 Now an axial section of both solids by a vertical plane along the 
 line a b, No. 1, will give a triangle for each, and points 1, 2, 3, 4, No. 
 2, as the extreme ones in the penetration. Any other vertical section 
 of the solids, taken through the apex b of the penetrating one, will 
 also give a triangle for the section of each. If then a plane, coinciding 
 with the front face fb of the penetrating pyramid, be passed through 
 the penetrated one, it will give in No. 2 the triangle c d e partly 
 shown in faint lines as the section ; the parts of its two sides 
 between d'c and d'e, shown in full being the lines of penetration 
 seen in elevation. To find those seen in plan, let fall from points 1, 3, 
 d', d" in No. 2, projectors to cut the corresponding front and back 
 edges of the pyramids in 1, d', d", 3, No. 1 ; join these points by 
 straight lines as shown, and they will be the lines of penetration of 
 the top face of the horizontal pyramid with the slant sides of the ver- 
 tical one. To find the similar lines on the under face of the same 
 pyramid, project over from No. 2, to the plan No. 1, the points 2, c, e, 
 
 4 ; join these by dotted lines, as shown, and they will be the ones 
 required. 
 
 Assuming the axial plane a b of the two pyramids to be at an 
 angle, say of 30 with the VP, but still perpendicular to the HP, and 
 an elevation of the solids in that position to be required ; then the 
 projection shown in No. 4 (Fig. 176) will be what would be obtained, 
 as the vertical pyramid has been moved round on its axis until the 
 axial plane of both solids makes the required angle with the VP. 
 The elevation and lines of penetration have then been found by direct 
 projection from the plan No. 3 and elevation No. 2. 
 
 The case in which two equal square pyramids penetrate each other 
 in such a way that both have one of their greatest plane sections in one 
 and the same plane, their axes being at right angles, is shown in Nos. 
 
 5 and 6 (Fig. 176). 
 
 69. Now, the greatest plane section of a square pyramid is evidently 
 one obtained by cutting it through from apex to base in such a way as 
 to divide its base diagonally into two equal triangles. Then, if two 
 such pyramids as A and B (Fig. 176), equal in size, and at right angles 
 to each other, be cut simultaneously by a plane passing through their 
 axes and cutting their bases diagonally, their elevation, assuming them 
 to be entire, and both axes parallel to the VP, will be as shown in No. 
 5, but without the lines of penetration 1, 2, 3 ; 4, 5, 6. To find these 
 lines, first draw in the plan (No. 6) of the pyramids as if entire. 
 Assume a vertical plane, tangent to the front edge x b of pyramid B, to 
 cut through A to its base. It would intersect that base in its two front 
 edges at points f^g, and its front slant edge in point i. Obtain by 
 projection the elevation of the triangular section/, i, g thus produced 
 
160 
 
 FIRST PRINCIPLES OF 
 
MECHANICAL AND ENGINEERING DRAWING 161' 
 
 shown in faint lines in No. 5 and it will be found to cut the front 
 edge x b of the prism B in points 2, 5 ; join these by straight lines to 
 points 1, 3 and 4, 6 previously found and they will be the required 
 lines of penetration of the two solids seen in elevation. 
 
 To find the plans of these lines, let fall projectors from points 2, 5 
 and 1, 4, in No. 5, to cut the corresponding edges of the prism B in 
 No. 6, and they will give points 1, 2, 4, 5 ; join these as before by 
 straight lines as shown, and they will be the plans of the two lines of 
 penetration 1, 2, 4, 5 previously found in No. 5. 
 
 On looking upon the two pyramids, in the direction of the arrow 
 shown in No. 5, two pairs of lines of penetration will be seen. These 
 are the two front ones 1, 2; 4, 5 ; and the two, 1, 2'; 4, 5', imme- 
 diately behind them, shown in No. 6. To show the position of the four 
 lines of penetration made by the two lower inclined sides of pyramid B, 
 let fall projectors from points 3, 6 in No. 5, to cut the corresponding 
 edge of the same pyramid in 3, 6 in No. 6 ; join these points by dotted 
 lines to 2, 2' ; 5, 5' respectively, and they will be the plans of the lower, 
 or return lines of penetration of the two solids. 
 
 Should the two prisms have such a relative position that their axes 
 are inclined to each other at some angle other than a right angle, in a 
 plane common to both, the procedure for finding their lines of penetra- 
 tion woul,d still be the same. As a test of its application, let the two 
 pyramids in the last problem penetrate each other in such a way that 
 the axis of the penetrating one is inclined at an angle of 30 to the 
 horizontal, while that of the penetrated remains vertical, and let it be 
 required to find their projection and lines of penetration when the axial 
 plane of the solids is parallel to the VP, and also when that plane is 
 inclined to the VP at a given angle. 
 
 As the solids in the first-named position have their axial plane 
 parallel to the YP, and the axis of one of them is inclined to the HP, 
 their elevation, as if they were entire, must first be drawn. Having 
 done this as shown in No. 1 (Fig. 177), from it find by projection their 
 plan also as if entire as given in No. 2. The points of intersection 
 of the upper and lower edges of the pyramid B with A are at once seen 
 in No. 1, to be 1, 2, 3, 4. Proceed as in the previous problem to find 
 the section of A, by a vertical plane passing through it tangent to the 
 front edge of B, and the points in that edge cut by it ; join these up 
 to the first found points by straight lines as before, and they will be 
 the lines of penetration in elevation. On finding the plans of these 
 lines as before explained and shown in No. 2, the first part of the 
 problem is solved. For its latter part, draw in as shown in No. 3 in 
 the figure, a repeat of No. 2, but having its axial line a b at the required 
 angle say 30 with the VP or IL ; and from it and No. 1 find by 
 projection the elevation of the pyramids and lines of penetration in 
 their new position. This view, if obtained in accordance with the pro- 
 cedure already explained in a previous problem, will give No. 4 (Fig. 
 177) as the result. 
 
 70. As a concluding problem in the penetration of plane-surfaced 
 solids, we give in No. 5 and No. 6 (Fig. 177) a case in connection with 
 pyramids which sometimes occurs in practice viz., that wherein a 
 
 M 
 
102 
 
 FIRST PRINCIPLES OF 
 
MECHANICAL AND ENGINEERING DRAWING 163 
 
 square pyramid is penetrated by a similar solid, the relative position of 
 the two being such that an axial plane passed through both gives a 
 similar section in each, but divides them into dissimilar solids. The 
 problem is 
 
 Problem 74 (Fig. 177). A square pyramid A } with its axis vertical 
 and base edges parallel and at right angles to the VP, is penetrated 
 by a similar pyramid B, having its axis horizontal and parallel to 
 the VP ; required the projections plan and elevation of the lines 
 of penetration of the two solids, when the diagonals of the base of the 
 penetrating one are respectively parallel and perpendicular to the VP. 
 
 First draw in the plan and elevation of the pyramids as if they 
 were entire in the position stated in the problem, and shown in Nos. 
 5 and 6 in the figure. The extreme points of intersection of the top 
 and bottom edges of the pyramid B, with the right and left faces of A, 
 are at once seen to be 1, 2, 3, 4, No. 5. Then to determine where the 
 front and back edges of B enter and leave A, assume a horizontal plane 
 to pass through both solids on the line of the axis a a of B, and the 
 parts cut off by it to be removed. The sections exposed will be a 
 triangle b a'b', for B, and a square in faint lines for A, shown in 
 No. 6. 
 
 Now the two sides, b a and b'a, of this triangle are the plans of the 
 front and back edges of the pyramid B, and as the square is a section 
 of the pyramid A made by the same cutting plane, it is at once seen 
 that the two horizontal edges of B enter the pyramid A at points 5, 5' 
 in Us front and back faces, and leave it at points 6, 6' in its right face. 
 Project over point 5 in b'a, No. 6, to 5 in a a, No. 5. Then to find the 
 direction of the lines of penetration which will meet in point 5 in 
 No. 5 get a side elevation of one-half of the pyramid A, using the 
 base line of B produced to p as an axis. On this line find by projec- 
 tion a half-section of the pyramid B, made by a plane passing through 
 points 1 and 3 on the left face of the pyramid A, which will be a 
 triangle, having the line 1'3' for its base, and its two sides cutting the 
 slant side p'o of the half-pyramid previously found in two points. 
 These projected over to the left edge of the pyramid A, and joined to 
 point 5 in a a' as shown, are the lines of penetration sought. As the 
 finding of the plans of these lines, as shown in No. 6, require only the 
 careful application of principles already explained, they are left to the 
 student to draw in without further assistance. 
 
 As the lines of penetration of prisms and pyramids by similar 
 solids, having any number of sides, are found by a similar procedure to 
 that so fully explained in this and the preceding chapter, the subject of 
 the penetration of solids having curved surfaces a most important one 
 t6 the engineering draughtsman will next be considered. 
 
CHAPTER XIX 
 
 THE INTERSECTIONS OF SOLIDS HAVING CURVED SURFACES 
 
 71. As the cost of any engine, or machine detail, depends in a great 
 measure upon the number of processes involved in its production, it 
 will readily occur to the student of engineering that the simplest possi- 
 ble form, and one requiring the least amount of machining, to fit it for 
 the purpose to which it is to be applied, will be that which is produced 
 by a revolving motion of the material to be acted upon by the cutting 
 tool which gives it shape. The product of this motion, a " solid of 
 revolution," has also a larger cross-sectional area within the same 
 bounding line than any plane-surfaced solid ; and for this reason, it 
 will be found on studying the construction of engines, machines, and 
 all ordinary steam generators, that the cylinder, the cone, and the 
 sphere all of them solids having curved surfaces enter much more 
 largely into their design than do any other of the solids. A knowledge 
 of the way in which their surfaces intersect when in combination is 
 therefore highly essential, not only to the draughtsman who is called 
 upon to design such details, but to the skilled workman who will pro- 
 duce them. 
 
 This knowledge is the more necessary, as upon the correct applica- 
 tion of the principles of projection to the determining of the actual 
 intersections of such solids, depends the truth or otherwise of the 
 results arrived at when finding the " developments " of their surfaces a 
 subject which immediately follows that now being considered, and one 
 which includes within its scope the determination of the exact shape 
 the material should take when the article to be produced is one made 
 entirely of metal plates. 
 
 The preliminary problems in connection with the intersections of 
 curved-surfaced solids are those in which cylinders are penetrated by 
 cylinders. The first is 
 
 Problem 75 (Fig. 178). Given the plan of two equal-sized right 
 cylinders, with axes at right angles in one plane parallel to the FP, 
 and penetrating each other ; to find their lines of intersection. 
 
 Now, from the way in which a right cylinder is generated, it is 
 known that an axial section of the solid is a rectangle, and any other 
 
 164 
 
MECHANICAL AND ENGINEERING DRAWING 
 
 1G5 
 
166 
 
 FIRST PRINCIPLES OF 
 
MECHANICAL AND ENGINEERING DRAWING 167 
 
 section parallel to an axial one is a similar figure. Therefore, if No. 1 
 (Fig. 178) be the given plan of the two solids, to find the lines of inter- 
 section of their surfaces, first get an elevation of them as if entire. As 
 only the front halves of the two cylinders will be seen in elevation, it 
 is with them that we have to deal. If, then, a series of vertical section 
 planes parallel to each other and the axial plane ap of the solids, be 
 assumed to pass simultaneously through their front halves, the points 
 of intersection of the sections of the cylinders by these planes will be 
 points in their lines of penetration. 
 
 To find these lines, draw in No. 1 (Fig. 178) parallel to the line ap, 
 as many lines say three at convenient distances apart, as it is in- 
 tended to use section planes. Find by projection from the points 11', 
 2 2', 3 3' where these lines cut the plan of the vertical cylinder the 
 sectional elevations produced by them, as shown in dotted linejs in No. 
 2. For the corresponding sectional elevations of the horizontal cylinder 
 by the same planes, on its axial line x y, No. 2 produced to the left 
 with point x as centre and x a as radius, describe the semi-circle a x'a. 
 This will be one-half of the end of the horizontal cylinder turned on its 
 vertical diameter a a as a hinge. In this semi-circle set off from x in 
 the line xx the distances that the lines 1 1', 2 2', 3 3' in No. 1 are from 
 the line ap, and through the points thus found draw lines parallel to 
 a a to cut the semi-circle in points 11', 22', 33'. Faint lines drawn 
 through these points from end to end of the horizontal cylinder, parallel 
 to x y, or the IL, will give the corresponding sections of it made by the 
 same planes as used in the vertical cylinder. Then the points where 
 the edges of the corresponding sections of both cylinders cut each other, 
 as shown in No. 2, are points in their lines of intersection, which, when 
 joined, will be found to be straight ones crossing each other at right 
 angles. 
 
 The student will note that in this case the cylinders being equal in 
 diameter and at right angles to each other, their intersections form true 
 mitres dividing them into four equal parts, each having semi-elliptic 
 sectional surfaces perpendicular to each other. 
 
 In the case of two equal cylinders intersecting at right angles, and 
 having the plane of their axes at an angle to the VP, but still in the 
 vertical position, their lines of intersection would be found in the same 
 way as shown in No. 2 ; the only difference in the result being that 
 such lines will be curved instead of straight, showing, as will be seen in 
 No. 4 (Fig. 178), that they become, through the altered position of the 
 solids with respect to the plane of projection, portions of ellipses. As 
 the finding of the elevation and lines of penetration of the two cylin- 
 ders, in their new position, offers to the student some practice in the 
 projection of ellipses, which will very frequently occur in this part of 
 the subject, no further assistance is given in this problem than that 
 afforded by the few projectors shown in the diagram. 
 
 72. The next case is one where the intersecting cylinders are not of 
 equal diameter, but still have an axial plane common to both. The 
 problem is 
 
168 
 
 FIRST PRINCIPLES OF 
 
MECHANICAL AND ENGINEERING DRAWING 1G9 
 
 Problem 76 (Fig. 179). Given the plan of two cylinders of unequal 
 diameter intersecting each other, with their axes at right angles and 
 in one plane parallel to the VP ; to find their elevation and lines of 
 penetration. 
 
 The plan of the two cylinders in the position stated in the problem 
 will be that shown in No. 1 (Fig. 179), the penetrating one being 
 horizontal. From this plan, find by projection the elevation of the 
 cylinders as if entire, making them both of the same length, with the 
 horizontal intersecting the vertical one at the middle of its height. 
 Then to find the visible lines of intersection of the solids in elevation, 
 draw in as in the last problem the plans of the intended vertical 
 section planes to be used ; and obtain by projection the elevation of 
 the sections of the vertical cylinder made by these planes. Also draw 
 in, in No. 2, on the axis of the horizontal cylinder produced, an end 
 view of the front half of that cylinder. On this, show in as in the 
 previous problem the lines made by the vertical section planes in 
 passing through that cylinder, and from the points in the semi-circle, 
 or end view, cut by these lines, obtain as before the front elevation of 
 the sections of the horizontal cylinder; then the points where these 
 sections cut the corresponding ones of the vertical cylinder, will be 
 points in the lines of penetration of one solid by the other. On joining 
 these by lines drawn through them, they will be found to be (as shown 
 in No. 2, Fig. 179) symmetrical curves; the highest and lowest points 
 in them being determined by the intersections of the bounding lines of 
 the two solids, at 1, 2, 3, 4, while the middle ones, a'b', are the vertical 
 projections of a b in No. 1, or where the frontmost part of the horizontal 
 cylinder is seen to enter and leave the vertical one. 
 
 Should the axial plane of the two cylinders be at any given angle 
 with the VP as in No. 3 instead of being parallel thereto, the lines 
 of their intersection would still be found in the same way as before 
 shown, but they will not be symmetrical curves as in the problem 
 just solved, but such as are seen in full lines in No. 4 (Fig. 179), 
 the dotted ones being the return lines of intersection of the penetrating 
 or horizontal cylinder, with the surface of the vertical one, which is 
 out of sight. As the correct finding of the intersecting lines shown in 
 full, and dotted in No. 4 of the two cylinders is but a matter of 
 careful projection by the student, no further assistance than that given 
 in the diagram will be required by him. 
 
 73. The next case is that of the intersections of unequal cylinders, 
 at an angle other than a right angle, and having their axes in different 
 planes at a given distance apart. The problem is 
 
 Problem 77 (Fig. 180). Two cylinders of unequal diameter, with 
 their axes in parallel planes, intersect each other at an angle of 30 ; 
 it is required to find the projections and lines of penetration of the 
 solids, when the axis of one of them is vertical, and the other paral- 
 lel to the VP. 
 
170 
 
 FIRST PRINCIPLES OF 
 
MECHANICAL AND ENGINEERING DRAWING 171 
 
 As one of the cylinders is inclined, an elevation of both as if they 
 were entire must first be drawn, as shown in No. 1. From this obtain 
 by direct projection the plan of the solids as given in No. 2. Then to 
 find the visible lines of intersection, assume parallel planes represented 
 by the lines 1, 2, 3, in No. 2, to pass vertically through both solids. An 
 elevation of their sections made by these planes will show where they 
 intersect, and will give the points through which the required lines will 
 pass. For the return lines of intersection, or those out of sight shown 
 dotted in No. 1 the same method of procedure is repeated as in 
 finding those which are seen. As a vertical plane would be tangent to 
 the extreme back surface of both cylinders, the point x in the elevation 
 No. 1 would be where the lines of penetration of the two solids would 
 cross. A plane passing through the vertical cylinder tangent to the 
 front surface of the inclined one gives the two points a b in plan and 
 elevation, as those where that surface enters and leaves the vertical 
 cylinder. Should the plane of the axis of the inclined cylinder, repre- 
 sented by the line a i in No. 2, be otherwise than parallel with the YP 
 say, as in No. 3 the lines of intersection of the two solids would be 
 found in the same way as that explained in Fig. 179, the difference in 
 their appearance as seen in No. 4 being the result of the changed 
 position of the solids with respect to the plane of their projection, the 
 vertical cylinder having been turned on its axis through a certain angle, 
 carrying the inclined cylinder with it. With the assistance of the few 
 projectors shown, and bearing in mind that each point in the lines of 
 intersection found in No. 1 has, by the turning of the vertical cylinder, 
 passed through the same angle horizontally, the student should be able 
 to find without trouble the actual lines due to the altered position of 
 the solids which are shown in No. 4 (Fig. 180). 
 
 As the intersections of cylinders in any position only require the 
 correct use of section planes as exhibited in this chapter to determine 
 them, no further examples are given in this connection, but we pass on 
 to the solution of one or two problems in which the intersections of the 
 cone and cylinder are involved. 
 
 Problem 78 (Fig. 181). A cylinder in a vertical position is pene- 
 trated at the middle of its height by a cone, having its axis horizontal 
 and parallel to the VP ; to find the lines of intersection of the two 
 solids when their axes are in one plane. 
 
 Let No. 1 (Fig. 181) be the given plan, and No. 2 the elevation of 
 the two solids ; then from their relative position it is evident that an 
 axial plane passed through both of them will give the rectangle abed, 
 and the triangle efg, as their respective sections. The extreme points 
 in the penetration of the cylinder by the cone will be 1, 2, 3, 4. Now, 
 although a series of vertical planes passed through both solids, parallel 
 to this axial one, would give rectangular sections of the cylinder, those 
 of the cone by the same planes, instead of being triangular, would be 
 hyperbolic, necessitating a great amount of careful projection in finding 
 them. To obviate this unnecessary trouble the same results may be 
 arrived at by a much more simple procedure. Instead of assuming a 
 
172 
 
 FIRST PRINCIPLES OF 
 
MECHANICAL AND ENGINEERING DRAWING 173 
 
 number of parallel planes to pass simultaneously through both solids, all 
 the points of the intersection required may be found by making the 
 cutting planes divergent, 
 
 74. In a previous chapter it was shown that all the sections of a cone 
 made by a plane passed through its apex to its base, gave triangles for 
 those sections. If, then, in the problem under solution, cutting planes 
 are assumed to pass through the apex of the cone in varying directions, 
 cutting it and the cylinder simultaneously, it is evident that all the 
 sections of the two solids will be similar to those obtained by the axial 
 section viz., rectangles and triangles, each of which will give four 
 points in the lines of intersection sought. 
 
 To find these points in No. 2 (Fig. 181), through /, the apex of the 
 cone in No. 1, draw the lines/1,/2, /3, and let them, with e/and hf, 
 be the plans of vertical section planes passing through both solids in the 
 directions shown. Then find by projection the rectangular sections of 
 the cylinder produced by the cutting planes, as shown in faint dotted 
 lines in No. 2. For the corresponding sections of the cone by the same 
 planes, on its base line eg in No. 2 as a diameter, and h as centre, 
 describe the semi-circle e x g, which will be the front half of the base of 
 the cone looked at from the left. On the line x h, measuring from h, 
 transfer the distances el, e2, e 3 in No. 1, and through the points thus 
 found draw lines parallel to eg, to cut the semi-circle in 1 1', 2 2', 3 3' ; 
 project these points over to e g, and through them draw faint lines to 
 the apex / The triangles formed by these lines will then be the 
 sections of the front half of the cone, made by the same planes that cut 
 through the cylinder, and the points of intersection of these respectively 
 will be the points in the lines of penetration sought. 
 
 Should the axis of the cone be inclined to the Y P, as in No. 3, that 
 of the cylinder still remaining vertical, and the lines of intersection be 
 required, the process of finding them is exactly the same as before. The 
 lines will, however, assume a different form of curve, due to the altered 
 position of the cone with respect to the plane of its projection, the Y P. 
 This alteration, it will be noted, brings the base of the cone into view, 
 but causes the surface of the cylinder penetrated by its small end to 
 pass out of sight. As the axis of the cone in the previous figure was 
 parallel to the YP, the lines of intersection of its back half with 
 the surface of the cylinder being directly behind those in front, could 
 not be shown ; but the cone being now inclined to the YP, it is possible 
 to indicate exactly the form of the return curve of penetration. This is 
 shown by the dotted line I, obtained by projecting over the points 4, 5,' 
 6, in i h, No. 3, and finding as before the corresponding sections of the 
 back halves of the cone and cylinder, which give the points required. 
 
 75. Had the penetrating solid been but a portion of a cone say a 
 frustum then the lines of intersection of the two may be got in two 
 ways. If the apex of the cone is accessible, the method already explained 
 would be the simplest and best ; but if the taper of the cone be only 
 slight, and its apex out of reach or at an inconvenient distance, then the 
 procedure would be as shown in Fig. 182. 
 
 First, let it be assumed that the axis of the penetrating solid is 
 parallel to the YP, and at right angles to that of the cylinder. Having 
 
 UNIVERSITY 
 
174 
 
 FIRST PRINCIPLES OF 
 
MECHANICAL AND ENGINEERING DRAWING 175 
 
 drawn in the plan and elevation of the solids as if entire, to find 
 their lines of intersection divergent planes are assumed to cut through 
 both, as in the last case ; but to ensure that the planes used will pass 
 through the apex of the cone of which the frustum forms part and 
 thus give true sections, the procedure in finding them is as follows : 
 
 In No. 2 (Fig. 182), on a 6, c c?(the ends of the frustum), as diameters, 
 describe semi-circles ; divide each into the same number of equal parts 
 say, six and project the points of division in them over to their 
 respective diameters (a b, c <f), and join them by faint dotted lines, as 
 shown. Then the surfaces included within the bounding lines 1 1', 1' 4', 
 4' 4, 4 1, and 2 2', 2' 3', 3' 3, 3 2, will be the true sections of the front half 
 of the frustum, made by the two section planes passing vertically 
 through it, at distances from its axis at either end equal to that which 
 the vertical lines drawn through points 1, 4, and 2, 3 in each semi-circle 
 are from the lines a b and c d respectively. 
 
 To find the sections of the cylinder by the same cutting planes, set 
 off at the corresponding ends of the frustum in the plan No. 1 the 
 distances that these planes are from the axial one pp' of the two solids, 
 and through the points 1 1', 2 2', thus found, draw faint lines cutting 
 the plan of the cylinder at e e, ff. Obtain by projection the elevation 
 of the sections of the cylinder at these lines, and the points where they 
 cross the corresponding sections of the frustum in elevation will be 
 points in the lines of intersection sought. To find the two points in the 
 cylinder's front surface, where the frustum penetrates and leaves it, 
 draw projectors through g g in the plan No. 1, and they will give g g' 
 in the elevation ; through these and the points of intersection of the 
 sections already found draw the curved lines e g h, e g' h, and they 
 will be the lines of penetration of the cylinder by the frustum of the 
 cone. 
 
 If the axis of the penetrating solid is inclined to the VP or HP, or 
 to both, that of the cylinder still remaining as before, as shown in No. 3 
 (Fig. 182), the lines of intersection of the two are found as in No. 4 
 (Fig. 181) viz., by direct projection from the plan No. 3 of the 
 solids in their new position, and from the elevation No. 2 ; the result 
 being the view given in No. 4. The return lines, or those on the surface 
 of the cylinder nearest the YP shown dotted, are found as indicated by 
 the projectors. 
 
 Taking the converse position of the two solids which frequently 
 occurs in practice or that where a cone is penetrated by a cylinder, 
 for a further problem in this connection, it would be solved on the same 
 principles, though in a different way. 
 
 Problem 79 (Fig. 183). A cone with its axis in a vertical position 
 is penetrated by a cylinder ; required the lines of intersection of the 
 solids in plan and elevation, when the axis of the cylinder is hori- 
 zontal, and parallel to the V P, and in the same plane as that of the 
 cone. 
 
 Draw in first the plan and elevation of the solids as if entire. Then 
 as the axes of both are in one and the same plane, an axial section 
 
176 FIRST PRINCIPLES OF 
 
 of them will at once give 1, 2, 3, 4 in No. 2 as the extreme points in 
 the penetration. To find others through which the lines sought will 
 pass ; at one end of the cylinder say the left give a side elevation of 
 the front halves of the two solids as shown. Assume the right-angled 
 triangle a a &, and the semi-circle on c d, to be half-sections of both 
 solids swung round on the axis of the cone until parallel with the "VP. 
 From these it will be seen that a part of the front and back surfaces 
 of the penetrating cylinder lie wholly within the cone. To find how 
 much, and thus determine the lines of intersection sought, proceed as 
 follows : 
 
 Through the vertex of the right-angled triangle a a 6, draw straight 
 lines to its base ; the first one tangential to the semi-circle on c d, and the 
 others cutting it at suitable points, as shown. These lines are the edge 
 views of the section planes passing through both solids, which will give 
 the points required, as the sections produced by them will be triangles 
 for the cone and rectangles for the cylinder. The rectangles are at once 
 found by drawing lines through the points in the semi-circle (cut by the 
 assumed section planes) parallel to the axis of the cylinder. 
 
 For the corresponding triangular sections of the cone, describe on its 
 base B C as a diameter the semi-circle B X C. On its axis, A A' 
 produced, set off from A' the several distances that the points 5, 6, 7, 8 
 in the base of the triangle a a b are from the point a in it. Through 
 the points 5', 6', 7', 8, thus found, draw lines parallel to B A' C, and from 
 where they cut the semi-circle find by projection the elevation of the 
 triangular sections of the cone, as shown in No. 2 ; then the points 
 where the corresponding sections of the two solids intersect are points 
 in their lines of penetration. The two points, x y in those lines or 
 where the plane drawn through a 8 in the side elevation is tangent 
 to the surface of the cylinder are found by drawing a line through as, 
 where the plane a 8 touches, and intersecting it with the triangular 
 section A x" y of the cone made by the same plane. 
 
 The plans of the lines of intersection 1x2 and 3 y 4 in No. 2 
 are most easily determined by finding horizontal sections of the cone 
 which will all be circles at several points in those lines in elevation, 
 and letting fall projectors from them on to the circular sections so 
 found ; then the points where the projectors cut these sections will be 
 those through which the required lines of penetration are to be drawn. 
 As the solids are directly at right angles to each other, their lines of 
 intersection are consequently symmetrical on either side of the cone, 
 both in plan and elevation. 
 
 76. As the procedure in finding the lines of intersection of a cone 
 and cylinder having their axes in the same plane, but not at right 
 angles, would be the same as in the last example, as any plane passing 
 through the apex of the cone to its base, and simultaneously cutting 
 through a cylinder parallel to its axis, however inclined, would still give 
 a triangle as the section of the former, and a rectangle as that of the 
 latter, and the points of intersection of the sectional surfaces so 
 produced as points in their lines of penetration, we pass on to the next 
 combination of solids employed in giving form to engine and boiler 
 details viz., that of cones with cones, a knowledge of their intersections 
 
MECHANICAL AND ENGINEERING DRAWING 177 
 
 being the more important, as combined conical surfaces enter into 
 almost all designs where metal plates are used as the material of con- 
 struction. As a problem in this connection take 
 
 Problem 80 (Fig. 184). A cone with its axis in a vertical position 
 is penetrated by another cone having its axis horizontal and parallel 
 to the plane of projection ; required the lines of penetration in 
 plan and elevation of the two solids in the position given. 
 
 Draw in first the plan and elevation of the two cones as if entire. 
 Then, as a series of planes passing through the apex of one of the cones 
 to its base would give triangles for the sections made of that particular 
 cone only, but hyperbolas for those of the cone in combination, it is 
 evident that the finding of the lines of intersection of the two solids 
 by the use of section planes in such a way would be a long process and 
 very liable to error. What has to be aimed at, is to so use a section 
 plane that it will, in the problem to be here solved, give similar 
 sections of both cones. To effect this, a triangular plane is assumed 
 to pass through the apices of both simultaneously, and to be swung 
 on one of its sides its hypothenuse as a hinge through varying 
 angles. 
 
 To apply this principle to the problem to be solved (in the elevation 
 of No. 1, Fig. 184), produce the base line Ac of the horizontal cone 
 indefinitely in both directions ; then draw a line through the apices of 
 both cones to cut the produced base line in the point x and the IL 
 in point x. Then will the produced base of the horizontal cone from 
 v to the IL at B be the perpendicular, the line B# the base, and that 
 drawn through the apices of the cones from x to x the hypothenuse, of 
 the right-angled triangular section plane for finding the lines of 
 intersection of the two solids. Such a plane in its normal position 
 or parallel to the VP and perpendicular to the HP at once gives 1, 
 2, 3, 4 in No. 1 as the extreme points in the lines of penetration. 
 To find others through which these lines will pass, the positions of the 
 triangular plane x'lox in its swinging on the line x'x as a hinge have 
 next to be shown. 
 
 To do this, describe on the base Ac of the horizontal cone as a 
 diameter, a semi-circle, then through the vertex x of the right-angled 
 triangle x~Bx, draw straight lines, the outermost one x'3 tangential 
 to the semi-circle, and the others, x'2, x'l, at about equal distances 
 apart. These lines are the end elevations of the perpendicular edge 
 of the triangular plane x'Ttx, when swung on its hypothenuse x'x as a 
 hinge. Its outer position, represented by the line x 3, is drawn 
 tangent to the front surface of the horizontal cone, to find the points 
 5 and 6 in the lines of penetration. To determine their exact position 
 and that of other points in these lines, describe on the base CE of the 
 vertical cone as a diameter, the semi-circle CDE ; transfer the points 
 1, 2, 3 on the IL by arcs, as shown, to 1', 2', 3', on the produced base 
 line of the horizontal cone, and through these last-found points draw 
 lines to x, intersecting the semi-circle last drawn in the points^ g, A, 
 and i,j, k. Find by projection from these points and those made by 
 
 N 
 
178 FIRST PRINCIPLES OF 
 
 the same section plane, in the semi-circle drawn on the base h c of the 
 horizontal cone, the corresponding vertical sections of the two cones, 
 and the points where they intersect will be points through which their 
 lines of penetration will pass, as shown in No. 1 (Fig. 184). To find 
 the plans of these lines, the simplest method, and one involving the 
 least confusion of lines in the diagram, is that shown in No. 2, where 
 the triangular sections of the vertical cone made by the planes x'3, 
 se'2, x'l, are first found, the points required being then projected over 
 from No. 1 to their corresponding sections in the plan, and lines of 
 penetration drawn through them, as shown in No. 2. 
 
 Should the axis of the penetrating cone be at an angle with the 
 YP, and an elevation of the solids with their lines of penetration be 
 required, this would be obtained by transferring the plan of them 
 already found in No. 2, to such a position in the HP as would bring 
 the axis of the horizontal cone to make the required angle with the 
 VP or IL ; the required elevation is then found by direct projection 
 from the plan of the solids No. 2 (in their new position) and the 
 elevation given in No. 1. 
 
 In finding such a view of the two solids, the student is to be 
 reminded that great care is necessary in projecting over the various 
 points in both plan and elevation, so as to ensure the correct 
 representation of the solids themselves, as well as their lines of 
 intersection. 
 
 77. As the same procedure would be followed in finding the lines 
 of penetration of two cones, when the axis of the penetrating one is 
 inclined to both the VP and HP the only difference being, that 
 the triangular section plane used in finding the points in the lines 
 of intersection of the solids in the first part of the solution, would not 
 be right-, but acute-angled the student is left to work out such 
 a problem himself, and we now pass to the finding of the lines of 
 penetration or intersection of the sphere, with such of the solids as are 
 generally used in combination with it. 
 
 Premising that it has not been forgotten by the student that any 
 plane section of the sphere when viewed at right angles to the plane of 
 section is a circle, and when seen at an angle to that plane, an ellipse, 
 it is assumed that the intersections of the sphere by any of the 
 plane-surfaced solids will present little difficulty, and therefore will not 
 necessitate more than two or three such problems being given for 
 solution. The intersections of the sphere by solids having curved 
 surfaces, being much more difficult, will require a close attention to 
 the procedure followed in determining them. The first problem, 
 therefore, in this connection is 
 
 Problem 81 (Fig. 185). A sphere is penetrated vertically and 
 Centrally by a square prism whose sides make equal angles with the 
 VP ; it is required to find the lines of penetration of the solids, 
 in plan and elevation. 
 
 Here, it is evident from the way in which the sphere is penetrated, 
 that its axis, and that of the prism, coincide. Therefore, to find the 
 
MECHANICAL AND ENGINEERING DRAWING 179 
 
 lines required, first draw in the plan of the solids, with the sides of 
 the prism at an angle of 45 with the IL or VP, as in No. 1. Then 
 find by projection the elevation of them as if entire. As the axes 
 of the sphere and prism are coincident, their lines of intersection in 
 plan are in the planes of the sides of the prism, and therefore coincide 
 with the four lines in No. 1 representing its plan. Then, in the 
 elevation No. 2, as the sides of the prism are equally inclined to the 
 vertical plane of projection or the VP the lines of penetration of 
 the sphere by the prism will all be portions of ellipses, or parts of 
 vertical plane sections of the sphere taken through it at the sides 
 or faces of the prism seen at an angle of 45. To draw in these 
 lines, find from the plan No. 1 the major and minor axes of the 
 ellipses into which the circular sections of the sphere are projected 
 in elevation, and by means of the paper trammel previously explained 
 find the points through which the lines of intersection are drawn, as 
 shown in No. 2 in the figure. 
 
 As the lines of intersection of a prism having any number of 
 sides with a sphere, are but a series of circular or elliptic arcs, or 
 both, obtained in the same way as those in the case of the square 
 prism, further examples of their penetrations are unnecessary ; and as 
 the bounding edges of the faces of a prism are parallel, the next 
 problem is a variant from this, or one in which the sides and edges 
 of the penetrating solid incline equally to its axis. The problem is 
 
 Problem 82 (Fig. 186). A sphere is penetrated by the frustum 
 of a square pyramid, having its base edges parallel and perpen- 
 dicular to the VP ; required the lines of penetration of the solids, in 
 plan and elevation, when their axes coincide and are in a vertical 
 tion. 
 
 First draw in an elevation of the frustum of a square pyramid, as 
 if entire, in the position given in the problem. At about the middle 
 of its height, and on its axial line, describe a circle to represent the 
 sphere to be penetrated ; then find by projection a plan of the solids, 
 as in No. 2 in the figure. Now, as only part of the front face of the 
 frustum is visible in the position given, the lines of penetration on 
 that face only will be seen in elevation. This face being inclined to 
 the vertical, it is evident that a section of the sphere taken at it, 
 although actually circular, will become elliptic in projection ; but in 
 this case very slightly so, due to the small inclination of the sides of 
 the frustum. 
 
 To draw in the portions of the ellipse, which are in fact the lines 
 of penetration required in elevation, its two axes must first be found. 
 One of them, so far as its position is concerned, will coincide with the 
 axis of the frustum ; and the other, which will be at right angles to it, 
 must be drawn through the point which is the actual centre of the 
 circular section of the sphere, made by a plane coinciding with any 
 one of the sides of the frustum. For convenience, take that side 
 which is on the right in No. 1, and bisect that part of it which is 
 contained between the points a, b. Through the point of bisection 
 
180 
 
 FIRST PRINCIPLES OF 
 
MECHANICAL AND ENGINEERING DRAWING 181 
 
 c draw a line perpendicular to the axis of the frustum, and on it set on; 
 from d where the axes intersect on either side, a distance equal 
 to c, a or c, b, which is half the major axis of the elliptic section. 
 For the minor axis, project over to that of the frustum the points a, b 
 in its right side or edge ; then with these lengths, and the paper 
 trammel, mark off a few points at either end, and through them draw 
 the elliptic arcs to the right and left edges of the front face of the 
 frustum, and they will be the lines of penetration of the solids seen 
 in elevation. 
 
 For the plans of these lines on the upper surface of the sphere, let 
 fall projectors from points 1, 2, in No. 1, to cut the corresponding slant 
 edges of the frustum (in No. 2) in points 1 1', 2 2' ; and for points 
 b &', 5 5', in the same diagram, set off from 01, on the two lines drawn 
 through it at right angles to each other, the distance that b or &', in 
 No. 1, is from the axial line or point 5 ; then curved lines drawn 
 through the points thus found, as shown in No. 2, will be the lines 
 of penetration on the upper surface of the sphere by the frustum. 
 The corresponding lines on its under surface, if required, are found in 
 the same way. If the sides of the frustum are equally inclined to the 
 VP, as in the case of the square prism in Fig. 185, and an elevation 
 of the solids in this position be required, then, as in previous problems, 
 transfer the plan No. 2 to the required position shown in No. 3, and 
 from it and the elevation No. 1 find by direct projection the view 
 given in No. 4. The dotted lines and projectors show how the major 
 axes of the elliptical portions of the intersections are obtained, the 
 line c c, in which the minor ones lie, being a projector drawn through 
 din No. 1. 
 
 78. As a sphere penetrated centrally by a cylinder or cone would 
 in each case give circles as the lines of intersection, becoming in 
 projection either straight lines, circles, or ellipses, according to their 
 positions with respect to the planes of projection, it is considered 
 unnecessary to give any problems for solution with the solids so 
 combined, as the truth of the statement is self-evident without 
 illustration. It must not, however, be forgotten by the student that 
 such combinations of the sphere with the cylinder or cone, or both, 
 more frequently occur in practice than any other; as, for instance, 
 in many kinds of cocks and valves, tee-pieces, handrails, stanchions, 
 boiler mountings, etc. As cases, however, arise where these three solids 
 are in such combination that their axes are not coincident, or even in 
 one plane, the remaining problems in this part of the subject will have 
 reference to instances in which these infrequent junctions occur. 
 
 Problem 83 (Fig. 187). A sphere is penetrated by a cylinder whose 
 axis is vertical and parallel to that of the sphere ; required the 
 lines of intersection of the solids, when their axes are a given 
 distance apart, in a vertical plane inclined at an cmgle to the VP. 
 
 Let No. 1 (Fig. 187) be a plan of the solids, the larger circle being 
 that of the sphere, and the line vp, drawn through its axis , and that 
 of the cylinder c, a plan of a vertical plane at an assumed angle of 
 
182 
 
 FIRST PRINCIPLES OF 
 
MECHANICAL AND ENGINEERING DRAWING 183 
 
 60 with the VP. Then as vertical sections of the sphere parallel to 
 the VP will be simply circles, and similar sections of the cylinder, 
 rectangles, it is evident that the visible lines of penetration of the two 
 solids may at once be found by assuming vertical planes to pass 
 simultaneously through them both, and drawing lines through the 
 points of intersection of the sections thus produced, giving the lines 
 required. Therefore, having the plan of the solids given in position, as 
 in No. 1 in the figure, from this obtain by projection an elevation of them 
 as if entire. Then draw in, parallel to the IL and through the front 
 halves of the sphere and cylinder, as many lines as it is intended to 
 use section planes ; find by projection elevations of the circular and 
 rectangular sections produced by them, and through the points of 
 intersection of these, draw in the lines of penetration as shown in 
 No. 2, Fig. 187. 
 
 The next problem is a combination of a sphere with a cons. 
 It is- 
 Problem 84 (Fig. 188). A sphere is penetrated by a cone having 
 its axis vertical ; required the lines of intersection of the solids 
 when their axes are not coincident, but parallel, and lie in a plane 
 which is inclined to the VP. 
 
 It may not have occurred to the student, in dealing with the 
 previous problems in which the sphere is involved, that it is quite 
 possible for its axis to lie in a vertical plane and yet be inclined at an 
 angle to either or both of the planes of projection. In the problem 
 for solution the axis of the cone being vertical, and that of the sphere 
 parallel to it, both will be vertical, and be represented by points in 
 plan. Therefore in No. 1 (Fig. 188) let the points c and s be the plans 
 of the axes of the two solids, c being that of the cone and s that of 
 the sphere, both being in a plane of which the line a b is the plan 
 at an angle of 60 with the VP. Then as the sphere is the solid 
 penetrated, with s as centre, and half the sphere's diameter as radius, 
 describe the circle to represent its plan, and with c as a centre and half 
 the diameter of the base of the intended cone as radius, describe a 
 second circle, dotting in that part of it covered by the sphere, as shown 
 in No. 1 in the figure. From this plan find by projection the elevation 
 of the two solids as if entire, making the height of the centre of the 
 sphere from the IL about half the height of the cone. Then to 
 find the curved lines of penetration the principal points in them 
 should first be ascertained. 
 
 For points 1, 2, 3, 4, in No. 2, get a sectional elevation of the 
 sphere on the line fg in No. 1. The section obtained, which is a 
 circle, having a diameter equal to fg, will be found to cut the sides of 
 the cone cd and ce, in 1, 2, and 3, 4. For points 5 and 6 the lowest 
 and highest in the curves of intersection in the two hemispheres 
 conceive the cone to be turned on its axis in the direction of 
 the arrow, carrying the sphere with it, until the line a b in No. 1 
 becomes parallel to the IL. In so doing, the axis s will have moved 
 to s", and the sphere itself to the dotted positions shown in No. 1 and 
 
184 
 
 FIRST PRINCIPLES OF 
 
MECHANICAL AND ENGINEERING DRAWING 185 
 
 No. 2, the latter giving points xy, which will determine the actual 
 positions of the points sought on the conical surface, with the solids, as 
 originally placed. To fix these, find in No. 2 the elevation cb' of the 
 line from c to b in No. 1, and to it project over parallel to the IL 
 the points (x y) last found, cutting it in 5 and 6, which are those re- 
 quired, the projecting over of x and y being tantamount to turning the 
 cone back on its axis to its original position. For other points in the 
 lines of penetration, take a few horizontal sections which will all be 
 circles of both solids, between those already found, and through their 
 points of intersection, in plan and elevation, draw lines, and they will 
 be the lines of intersection of the surfaces of the cone and sphere, as 
 required in the problem, and shown in No. 1 and No. 2 (Fig. 188). 
 
 79. The next problem will show that although sections of the two 
 solids might be taken in such a way that they would produce circles 
 for both cone and sphere, and give apparently an easy solution of it, it 
 is necessary for the draughtsman to exercise judgment in deciding 
 upon his method of procedure. The problem is 
 
 Problem 85 (Fig. 189). A sphere is penetrated by a cone horizon- 
 tally ; required the lines of penetration in plan and elevation^ 
 when the axes of both solids are in a plane parallel to the VP, and 
 at right angles to each other, but that of the cone not passing 
 through the centre of the sphere. 
 
 Draw in first, as if entire, the plan of a sphere penetrated by a 
 cone, having its axis parallel to the IL, and passing through that of 
 the sphere, as shown in No. 1 (Fig. 189). From this plan get an ele- 
 vation of the solids, showing the axis of the cone some distance above 
 the centre of the sphere. Now, if sections of the solids be taken by 
 planes parallel to the axis of the sphere which is assumed to be 
 vertical and at right angles to the VP, the sections would in reality 
 be circles for both solids, but only straight lines in projection in 
 both plan and elevation, which would be useless in determining their 
 intersections. 
 
 Again, sections taken through both solids parallel to the axis of 
 the cone, although giving circles for the sphere, would still give hyper- 
 bolas for the cone, which are not desirable. The simplest method in 
 solving the problem will therefore be to assume vertical section planes 
 to be taken through the apex of the cone in varying directions and 
 passing through both solids simultaneously. These would give tri- 
 angles for the cone and circles for the sphere; the latter, however, 
 being at an angle to the YP, would in elevation be projected into 
 ellipses. Therefore in No. 1 (Fig. 189) draw in, through a' the vertex 
 of the cone, to its base, as many lines say three a'\ a 2 a 3, as it 
 is intended to use section planes. Next find in No. 2 the elevation 
 of the triangular sections produced by these planes, as in previous 
 problems. Then, so far as the sections of the sphere are concerned, all 
 that is required is to find the points in the ellipses into which the 
 circular sections of the sphere are projected which intersect the 
 
186 MECHANICAL AND ENGINEERING DRAWING 
 
 triangular sections of the cone, for it is through them that the lines of 
 penetration of the sphere by the cone will pass. 
 
 To determine these points, there is no necessity to draw in the 
 elliptic projections, as they may be found very readily by the use of 
 the paper trammel before mentioned. Having numbered the section 
 planes, in plan and elevation, as in No. 1 and No. 2 in the figure, and 
 noted the points 4, 5, 6, 7, where they cut the great circle of the 
 sphere in No. 1, draw in, in No. 2 through line cs, which passes 
 through the centre of the sphere, the lines in which the major axes of 
 the ellipses will lie. One of these only viz., that from x in No. 1 is 
 drawn in, in the diagram, so as to prevent confusion. Then with half 
 the major and projected half of the minor axes of each ellipse there 
 being one for each section taken on the trammel, manipulated as 
 explained, all the points in the lines of intersection, except the 
 extreme ones, are found as shown in elevation No. 2. For the plans 
 of these lines that will be seen from above shown in No. 1- let fall 
 projectors from the points in the lines 1, 2, 3, 4, in No. 2, to cut the 
 corresponding ones in No. 1, and through the intersections draw the 
 curved lines as shown, which will be the ones required. 
 
 Should the axis of the cone be inclined to the VP, still remaining 
 parallel to the HP, and the lines of intersection of the two solids be 
 required, when in this position, proceed as in previous cases to transfer 
 the plan No. 1 to such a position in No. 3 as will bring the axis of the 
 cone to make the required angle with the IL, and then from it and the 
 elevation No. 2 obtain by direct projection the view shown in No. 4. 
 
 From the problems which have been given in this part of the sub- 
 ject, the student will be able to appreciate the endless variety of posi- 
 tions the solids in combination may be made to assume, but as it is not 
 our object to multiply examples, but to give only such combinations as 
 are likely to occur in practice, we pass on now to the last section in 
 the application of the principles of projection viz., the " Development 
 of the Surfaces of Solids," a subject of the highest importance, not 
 only to the draughtsman, but to all who have to shape or fashion any- 
 thing constructed of sheet metal. 
 
CHAPTER XX 
 
 THE DEVELOPMENT OF THE SURFACES OF SOLIDS 
 
 80. To determine the exact form of the surface of any solid, 
 whether it be plane or curved, it is necessary to obtain what is known 
 as its " development " ; or, in other words, the particular shape its sur- 
 face will assume when laid out flat, supposing it possible that it can be 
 so treated. 
 
 For such a surface to be " developable " it must be one, on every 
 part of which a sheet of any flexible but non-elastic material can be 
 made to lie when bent, without leaving any hollow spaces. Should 
 this not be possible, then the surface is non-developable. 
 
 From this definition, it will at once be seen, that all ju>?awe-surfaced 
 solids are developable, while of those having curved surfaces, only the 
 cylinder, and the cone, with their frustums, fall within the same 
 category ; the sphere, with the spheroids, ellipsoids, and many other 
 solids of revolution, having surfaces which will not coincide with a 
 plane when laid out flat, but would tear or crease, being non- 
 developable. 
 
 The figure of the developed surface of every solid, which when bent 
 will cover it at any and every point, is called the "envelope " of that 
 surface. 
 
 In the cylinder and cone, as well as in every other solid of revolu- 
 tion, any line drawn on their surfaces, in the same plane as their axes, 
 is called a "meridian." If such a line is straight, the surface is de- 
 velopable, but if curved it is non-developable. 
 
 A surface which is generated by the motion of a straight line is 
 called a " ruled " surface, and may be either developable or non- 
 developable ; if the latter, it is a " twisted " surface, in that it cannot be 
 laid out on a plane without being torn. A ruled surface may, how- 
 ever, be curved, and developable, and yet form no part of a cylinder 
 or a cone, as will be shown later on. 
 
 The finding of the developments of plane-surfaced solids involving 
 no difficulty, few problems are necessary in connection with them, as it 
 will be seldom that any will occur in the practice of the student, with 
 which he will not be able successfully to grapple. The first is the 
 simplest possible, and hardly requires demonstration, but as it shows 
 
 187 
 
188 
 
 FIRST PRINCIPLES OF 
 
 the method of procedure in finding the development of the surface of 
 any plane solid, it is here given. 
 
 Problem 86. (Fig. 190). Given the plan of a cube; to find the 
 development of its surface. 
 
 Let a b c d in the diagram be the plan of the cube. Then as its six 
 sides are all of them equal squares, with every two adjacent ones al- 
 right angles, all that is required in finding its development, is to con- 
 ceive each of its vertical faces turned down on its lower edge as a 
 hinge, until it lays flat on the HP, one of the faces in its turning 
 carrying the top face abed with it. 
 
 Fig. 190 
 
 To show this graphically, in Fig. 190, produce the four sides of 
 the square abed indefinitely ; then from its four corners, set off on 
 the produced lines at a', b', c', d', a length equal to any one side of the 
 cube, and through the points thus found, draw lines parallel to its 
 sides, as shown. From b' in a b produced, set off a length b'x, equal 
 to that of the edge a 6, and at x draw a line parallel to b'y, and the 
 development of the cube will be complete. The surface enclosed 
 within the bounding lines of the diagram will then be the " envelope " 
 of the given solid, for it is its whole surface laid out flat, and will, if 
 cut out of a sheet of paper, and folded over on the lines a c, c d, db, 
 b a, and b y, until each adjoining surface is at right angles, exactly 
 cover all the cube, without leaving any vacant space between it and 
 them. 
 
 81. As the development of the whole of the surface of a square 
 prism would merely be a repetition of that given of the cube, in the 
 
' Y^ OF THE 
 
 UNIVERSIT 
 
 MECHANICAL AND ENGINEERING DRAWING 
 
 next problem only a part of a prism is taken, as the solid whose de- 
 velopment is required. 
 
 Problem 87 (Fig. 191). -Given the elevation of a truncated square 
 prism, to find the development of its surface. 
 
 Fig. 191 
 
 Let No. 1 (Fig. 191) be the side elevation of the prism, standing on 
 its base in a vertical position. As its base is square, its vertical sides 
 will all be of the same width ; therefore on the line c d produced inde- 
 finitely, draw in first an elevation of the prism, looking at it in the direc- 
 tion of the arrow. From c and c in It, set off in points 2, 2' a distance 
 equal to cd in No. 1, and through them draw vertical lines, and cut 
 them in points 3 and 3', by a projector from a. Join b' 3 and b 3'. 
 For the base, and top surface of the prism, produce the lines c'b', c b, 
 in both directions indefinitely. Then with c as centre, and c'2 as 
 radius, cut b'c produced in point 1 ; and from 6', with b' 3 as radius, 
 cut the same line in point 4, and complete the rectangle 4& (above) 
 and the square c'l (below) as shown. From point 2 in cd produced 
 set off a length 2'y, equal to c2', and through point y draw yx, 
 parallel to 2' 3' ; join 3' and x, and the development is completed. If 
 the resultant figure (the envelope of the prism) be cut out in paper, 
 and treated in the same way as that of the cube, it will be found to fit 
 the solid exactly. 
 
 82. As the surface of an oblique prism is often found to be more 
 difficult of development by the student draughtsman than that of a 
 right one, the next problem will show how it may be correctly found. 
 
190 
 
 FIRST PRINCIPLES OF 
 
 Fiy. 192 
 
LIECHANICAL AND ENGINEERING DRAWING 191 
 
 Problem 88 (Fig. 192). Given the side elevation of an oblique prism ; 
 to find the development of its surface. 
 
 . Let abed, No. 1, Fig. 192, be the given elevation of the prism, with 
 its end c d resting on a horizontal plane. In this position, it is evident 
 that. an ordinary front elevation of it, as of that of the frustum in No. 
 2, Fig. 191, would be of no service in this case, as the bounding edges 
 of its sides, being inclined to the horizontal, would not give actual but 
 only apparent lengths in projection. To find the actual sizes of all the 
 sides and ends of the solid, and their relative position to each other, on 
 a flat surface, a view directly at right angles to one of the sides of the 
 prism is necessary. 
 
 Now this view may be found in two ways. The prism may be 
 turned on its horizontal edge at c, as on a hinge, until its inclined edge 
 a c becomes vertical as shown in dotted lines and its development 
 found when so posed. Or, it may be found at once from the prism in 
 its inclined position, with less chances of error, by a projection of its 
 bounding surfaces, taken when looked at in the direction of the arrow 
 x ; such a view being tantamount to assuming the IL of the plane of 
 projection to be drawn through c, at right angles to the edge c a of the 
 prism, and its surface laid out flat on the VP. To find the develop- 
 ment as shown in No. 2, Fig. 192, proceed as follows : At right angles 
 to a c in No. 1 in the figure and through c, draw a line indefinitely. 
 At any convenient point, as c in that line, draw through c' a line 
 parallel to c a in No. 1. Then as all the side edges of the prism are 
 parallel to each other, set off on the line drawn through c in No. 1 from 
 c', the distances measured at right angles that those edges are apart ; 
 arid through the points in c c' produced thus found, draw lines indefi- 
 nitely, parallel to that first drawn from c', or to c a No. 1. 
 
 Now it is evident that the four side (or inclined) edges of the 
 prism will, when the surface is unfolded, lie in the lines last drawn, for 
 if the prism No. 1 as a solid, is laid with its edge ca coinciding with 
 the line ca in No. 2, and rolled over to the right its edges would 
 fall upon the lines drawn parallel to ca. 
 
 To complete the development, project over from No. 1 at right 
 angles to ac the points ac ; bd; to cut the corresponding edges in 
 No. 2; in a'c', ac, a"c"; and b'd", bd. Join a'b', ba, a a"; and c'd', dc, 
 c c", by straight lines as shown. For the two ends of the prism which 
 is square in cross-section produce the line b d in No. 1 in both direc- 
 tions, and from b and c?, set off in e and /, a length equal to a b or c d. 
 Project over e,/to e'f in No. 2, and through them draw lines at right 
 angles to b'd", to cut b c?, which completes the development required. 
 The envelope of the prism thus produced will, when folded over on the 
 edges represented in dotted lines, be found to cover without vacuities 
 all the surfaces of the given prism. 
 
 83. As the development of the surface of any prism whether right 
 or oblique having any number of sides, may be found as above shown, 
 a pyramid is taken as the next object for its surface development. The 
 problem is 
 
192 
 
 FIRST PRINCIPLES OF 
 
MECHANICAL AND ENGINEERING DRAWING 193 
 
 Problem 89 (Fig. 193). Given the elevation of a square pyramid; to 
 find the development of its surface. 
 
 Let ABC, No. 1, in Fig. 193, be the elevation of the pyramid. As 
 all its sides are alike, and incline equally to its apex, it is first necessary 
 to know the actual length of one of its side edges. To find this, at B 
 in No. 1, draw a line perpendicular to the IL. With B as centre, and 
 BA the length of a side of the pyramid as radius, describe an arc 
 cutting the perpendicular line in a. From B, set off in &, a length 
 equal to half BC, and join a, b ; then aB will be the length of a side of 
 the pyramid, and a b that of one of its edges. 
 
 For the development, at a convenient distance from No. 1, draw a 
 line perpendicular to the IL, and project over to it in a the point a in 
 No. 1. Then with a as centre, and a b in No. 1 as radius, describe an 
 arc indefinitely, which will cut the IL in points b'c. From &', set off 
 on the arc in &", a length equal to 6V, and from c' the same length 
 twice in c" and b"; join b'b'"; c'c" ; c"b f "; and draw lines from a' to these 
 points. For the base, construct a square on b'c as shown, and the 
 development is complete ; the figure enclosed within the boundary lines 
 of No. 2 being the envelope of the given pyramid. 
 
 A development of the same surface of the form shown in the small 
 diagram No. 3, may also be found by turning down the four sides of 
 the pyramid on their respective base edges. If, however, such an 
 object if large were made in metal plate, this development would 
 involve a great waste of material, and necessitate more seams than are 
 required. 
 
 84. As the frustum of a hollow pyramid is often combined with 
 parts of other solids in metal plate constructions, a problem in finding 
 the development of such a surface is next given. 
 
 Problem 90 (Fig. 194). Given the elevation of the frustum of a 
 hollow square pyramid; to find the development of its surface. 
 
 Let ABCD, No. 1, Fig. 194, be the elevation of the frustum. Pro- 
 duce its edges AB and CD, till they meet in w, and find the actual 
 length of one side, and an edge of the pyramid of which it is a part, as 
 in the last problem. Then draw in, as in No. 2, the development of 
 the base edges of the frustum. For its top edges, through A and C in 
 No. 1, draw faint lines across the front face of the pyramid, parallel to 
 its base BD, and with arcs struck from B as centre, transfer the points 
 A and x in B& to the vertical line Ba', and project them over to C'A' 
 in the line a'b. Then from a in No. 2, set off on the corresponding 
 lines of the pyramid, the distances that C', A', are from a in the line 
 a'b in No. 1 ; join the points thus found by straight lines as shown in 
 No. 2, and they will be the development of the edge required. By 
 connecting A'B'; A"B" at the ends of the diagram by lines as 
 shown, the development of the whole surface of the hollow frustum will 
 be completed. 
 
 85. As solids of a pyramidal form, with curved, in place of plane 
 
 o 
 
 
194 
 
 FIRST PRINCIPLES OF 
 
 sides, are often used in giving shape to mechanical details, the solution 
 of the following problem will show how the development of the surfaces 
 of such solids is found. 
 
 Fig. 195 
 
 Problem 91 (Fig. 195). Given the plan and elevation of a square 
 pyramidal-shaped solid ; to find the development of its surface. 
 
 Let ABCD, No. 1, Fig. 195, be the plan, and AB, No. 2, the 
 elevation of the solid, its axis aa' being perpendicular to its base. 
 Divide the side B, No. 2, into any number say five of equal parts, 
 and through 1, 2, 3, 4, the points of division, draw lines parallel to 
 AB and BT). Through a- the axis in No. 1, draw a line indefinite!}^, 
 cutting BD in x ; from x set off in this line the equal parts that <xB, 
 
MECHANICAL AND ENGINEERING DRAWING 195 
 
 No. 2, is divided into, and through the points of division draw faint 
 lines parallel to BD. Then the intersection of these, by lines drawn 
 parallel to AB through the points in the diagonals BC and AD, cut 
 by those let fall from 1, 2, 3, 4, in B, No. 2, will give b, c, d, e ; and 
 1), c, d', e ; through which, from B and D to a, the curved lines shown 
 are drawn. The surface enclosed by them and the line BD will be the 
 development of one side of the given solid. As its axis is perpendicular 
 to its base, and all its faces are alike, the development found of one 
 face, -if repeated on the other base edges (AB, AC, CD) of the solid, 
 will, with the base itself, give the complete development required. The 
 difference between the - apparent and real surface of a side of the solid 
 is shown in No. 2 in the diagram, where the full lines give the apparent 
 surface when bent to its shape, and the dotted ones the same surface 
 laid out flat. 
 
 The solid here dealt with, has been chosen to show that although it 
 has compound curved surfaces convex and concave combined with a 
 flat one its base yet it can all be made, if necessary, though at a 
 great waste of material, out of a single flat plate, its sides being after- 
 wards bent to the shape required. 
 
 86. As the whole, or a part of the surface of another plane solid, 
 the " oblique pyramid " (which often contributes in giving form to plate 
 and other metal structures), is rather more difficult of development than 
 that of a " right pyramid " on account of its frequent great inclination 
 from the vertical the solution of the next problem will show how its 
 correct development may be found with the least number of construc- 
 tion lines. 
 
 Problem 92 (Fig. 196). Given the elevation of an oblique square 
 pyramid, to find the development of its side surfaces, when its axis 
 is inclined 45 from the vertical; also the development of the surface 
 of a frustum of the same pyramid, of a given vertical height. 
 
 Let ABC, No. 1 (Fig. 196), be the elevation of the pyramid. As all 
 its sides with their edges are inclined to both the VP and HP, it is 
 evident that the actual shape of the whole or any part of its surface, 
 cannot be found by direct projection, or from the elevation alone, 
 without a plan of the solid. Now although an oblique pyramid differs 
 from a right one, in having its axis inclined to its base it must be 
 remembered that any section of it parallel to its base is still as in the 
 case of a right pyramid of the same/orm as its base. Bearing this in 
 mind, there will be no difficulty in solving the given problem, more 
 particularly the latter part of it. 
 
 The elevation of the pyramid being given, find by projection a plan 
 of it, as shown in full lines in No. 2, Fig. 196, keeping the axial line 
 Av a convenient distance from its base BC, in No. 1. Then to find the 
 development of the surface required: in No. 3, Fig. 196, draw a line 
 (indefinitely) at right angles to BC the base, and a projector from its 
 apex parallel to that base, to cut it in A'. On the line through A' set 
 off from A', the length of the longest side AB of the pyramid in a, 
 and through a draw a line at right angles to A'a. From a', set off to 
 
196 
 
 FIRST PRINCIPLES OF 
 
 ---^ 
 
MECHANICAL AND ENGINEERING DRAWING 197 
 
 right and left in B and B', half the length of one edge of the base 
 of the pyramid, then the line B'B will be the development of the edge 
 B'B in No. 2. Join B and B' with A' by straight lines, then will 
 the triangle A'B'B be the development of the longest face of the 
 pyramid. 
 
 For the adjacent faces BA'C, and B'A'C' : with BB' as radius and 
 B,B' as centres, describe arcs to right and left, and from A as centre 
 with the actual length of the edge AC in No. 1 as radius, cut the arcs last 
 drawn in C and C'. Join AC' and A'C by right lines, and the triangles 
 A'C'B' and ABC will be the development of the front and back faces 
 of the pyramid. For the shortest face or that to the right in No. 1 
 of the pyramid in No. 3, with C as centre, and CB as radius, describe 
 an arc, and from A' as centre, with AC as radius, cut that arc in C' ; 
 join AC" with a right line, and the development of the four faces of the 
 pyramid when laid out flat is complete. 
 
 The actual length of A'C in No. 3 is the hyj)othenuse of a right- 
 angled triangle, of which the line AC in No. 2 is the base, and the 
 vertical height of the pyramid or Ax in No. 1 the perpendicular. 
 This hypothenuse is shown by a dotted line drawn from A to y in 
 No. 1, the length xy being equal to AC in No. 2. 
 
 For the development of the frustum B&cC of the pyramid, as its top 
 edges are parallel to the base BC, it is only necessary to set off the 
 length Ab in No. 1 from A' in No. 3, in the point a, and through it 
 draw a line parallel to B'B, cutting A'B', and AB in b' and b; then 
 parallel to BC ; B'C' ; and CC' ; draw b c, b'c, and c'c", and the develop- 
 ment is complete. The dotted arcs in No. 3 show the direction the 
 developed side and end surfaces would move in to form the covering of 
 the solids. 
 
 87. The foregoing examples of the development of the surfaces of 
 plane solids being sufficient to show the principle on which they are 
 obtained, we pass on to the consideration of those which are bounded 
 by developable curved surfaces, such as the cylinder and cone, with 
 their frustums. 
 
 Problem 93 (Fig. 197). Given the plan and elevation of a right 
 cylinder; tojind the development of its curved surface, and that of 
 a given point, and line on that surface. 
 
 Let the circle No. 1, Fig. 197, be the plan of the cylinder, the rect- 
 angle bcde, No. 2, its elevation, and the line a a its axis. Also let x and 
 y be two given points in its front surface. Now, a cylinder has been 
 denned to be a solid generated by the revolution of a rectangle about 
 one of its sides as an axis ; or its surface may be conceived to be 
 generated by the motion of a straight line around another which is 
 fixed; the former being always parallel to the latter, and at a given 
 distance from it, thereby causing every point in it to lie in the 
 cylindrical surface throughout its motion. Th:s surface being generated 
 in one complete revolution, it is evident that it may be developed or 
 laid out flat, by causing, as it were, the solid to give up its surface while 
 rolling on a plane through one revolution. 
 
198 
 
 FIRST PRINCIPLES OF 
 
 Jt: 
 
 \ 
 
 J 
 
 Fig. 197 
 
 \6 
 
 \*_ 
 
 71 
 
 ^ 
 
 o 4- v o 
 
 A, 
 
MECHANICAL AND ENGINEERING DRAWING 199 
 
 To show this graphically : in No. 2, Fig. 197, produce indefinitely in 
 one direction say to the right parallel to each other, the lines b d and 
 c e, then divide half the circle in No. 1 into any number of equal parts 
 say eight and number them as shown. Project over each point thus 
 found to ce No. 2, and through them draw meridians parallel to be. On 
 ce in No. 2 produced, set off from e the eight equal parts into which the 
 semi-circle 8, 4, is divided, number them as shown, and repeat the 
 divisions to point f. Through points 1 to 8, and f, draw lines parallel 
 to the side de of the cylinder ; then will the rectangle defy be the 
 development of the surface of the cylinder b c d e ; for its length ef or 
 dg, equals the circumference, and its width de or gf, the height of the 
 cylinder ; and if it be conceived to roll on a plane with that part of its 
 surface represented by the line de in No. 2 touching it when starting, it 
 is evident that it will roll over the surface included in the rectangle 
 defg, and at the completion of one revolution, the line de (if drawn on 
 the cylinder's surface) would be found to fall exactly on the line gf, as 
 the meridians 1 to 16 assumed to be drawn on that surface would 
 each fall on its corresponding one drawn on the development. 
 
 By laying out the surface of any solid of revolution (if developable) 
 in a similar way to that of the cylinder, it will be apparent that the 
 development of any point or line on its surface in any position, or direc- 
 tion, may be soon found by the help of a few meridians. 
 
 For instance, let xy be two points on the front surface of the 
 cylinder, and it is wished to find their position on its envelope. First 
 let fall projectors from x and y in No. 2, on to the plan of the front 
 face of the cylinder in No. 1 ; they will be found to cut it in xy ; ,the 
 first between points 5 and 6 ; and y between 2 and 3 in the semi- 
 circle 1 8, 4, 0. Then on ef (the developed edge of the base of the 
 cylinder) set off from points 3 and 5 in it, the distances that xy are 
 from the corresponding points in the semi-circle 8, 4, ; and through 
 these draw faint lines parallel to de. Then the points x",y" in No. 3, 
 where these lines are cut by projectors drawn through x and y in No. 2 
 parallel to ef, are the positions of those points on the envelope of the 
 cylinder. 
 
 Again, if a line be given in position on the surface of a cylinder, 
 drawn, say between the points x and y, and its development is required, 
 then, as a line is the path of a moving point, take any convenient points 
 in the given line xy, and draw meridians through them. Proceed, as in the 
 case of the points x, y, to find the position of these meridians in No. 3 
 the development of the cylinder and then through the points chosen in 
 the given line x y in No. 2, draw projectors to cut the new meridians 
 found in No. 3. A line drawn through the points of these intersections 
 will be the development required. 
 
 88. Assuming the converse of the last problem, or one where the 
 development of a line is given, and its position and delineation on the 
 solid itself is required, then to show these, we have in effect to re-cover 
 the solid, with the envelope that contains the given line. Let the 
 problem be 
 
200 FIRST PRINCIPLES OF 
 
 Problem 94 (Fig. 197). Given a straight line eg, drawn on the 
 envelope defy of the cylinder ; required the projection of that line 
 on the solid itself. 
 
 To find this, divide the length of the given envelope into any 
 number say 16 of equal parts, and assume the circle of No. 1, and tin* 
 rectangle defg of No. 2, Fig. 197, to be the plan and elevation of the 
 c} 7 linder whose envelope is given. Divide the circumferential line of 
 No. 1 into the same number of equal parts, that the length of the 
 envelope is divided into, and through the points of division draw 
 meridians as in No. 2. On to each of these meridians in consecutive 
 order, project over the points where the given line eg cuts its corre- 
 sponding meridian drawn in No. 3. Eight of these from 1 to 8 will 
 be on the front side of the cylinder in No. 2, the remaining eight being 
 on its rear side. 
 
 Now, if through the points in the meridians 1 to 8 in No. 2 pro- 
 jected over from No. 3, a line be drawn, it will be that part of the 
 projected line required, which will be on the front side of the cylinder. 
 For the part on its rear side, project over the remaining eight points 
 9 to 16 in the line eg in No. 3, on to the corresponding meridians in 
 No. 2, and through them draw a dotted line it being out of sight and 
 the spiral line thus produced will be the projection of the straight line 
 e g in No. 3, when the surface on which it is drawn is wrapped round 
 the cylindrical solid represented in elevation by No. 2, Fig. 197. 
 
 In the same way as above shown, may be found by projection 
 the delineation on the solid itself of any line drawn on its envelope, the 
 method of doing it being merely the converse of that adopted in finding 
 the development of a given line drawn on a solid. 
 
 89. As frustums of a cylinder constantly occur in pipe connections, 
 when it is desirable to alter the direction, or lead of a pipe, the solu- 
 tion of the next problem will show how a plate has to be shaped to 
 form such a connection. 
 
 Problem 95 (Fig. 198). Given the elevation of the frustum of a 
 right cylinder, the plane of section being at an angle with the axis of 
 the solid ; to find the development of the surface of the frustum. 
 
 Let abed, No. 1, Fig. 198, be the elevation of the frustum. Here, 
 as the plane of section ad of the cylinder is at right angles to the VP, 
 the height of any point in the edge of the section nearest the VP will 
 be the same as that of a point directly in front of it, in the opposite 
 edge ; it is therefore only necessary to give a plan of the front half of 
 the frustum, to enable the whole of its development to be found. 
 Consequently, on the base b c of the frustum as a diameter, describe a 
 semi-circle and divide it into any number of equal parts say eight 
 and through the points of division draw in the meridians on the 
 frustum. Next produce its base line b c indefinitely, and set off on it 
 from c, the eight equal parts into which the semi-circle on be is divided. 
 Through the points of division, draw lines parallel to the sides of the 
 
MECHANICAL AND ENGINEERING DRAWING 201 
 
 frustum, and number them as shown. Then from each of the points in 
 the line ad, where the meridians drawn on the front side of the frustum 
 intersect it, draw projectors parallel to b c produced, to cut its corre- 
 sponding meridian in No. 3. A line drawn through the points of 
 intersection of the projectors and meridians, will then give the develop- 
 ment of the front half of the frustum. 
 
 For the back half, produce the projectors drawn through the points 
 1 to 7 in ad, to cut the corresponding meridians on the back face of the 
 frustum ; then a line drawn through their intersections as shown in No. 
 2, will give the development of the top edge of the frustum. For the 
 side and bottom edges, draw in, in full, the lines d c, d'c, c c, and the 
 required development is complete. 
 
 90. As parts of the surface of an oblique cylinder sometimes enter 
 into the design of boiler flues, uptakes, air shafts, etc., it is necessary 
 that the difference between its development and that of a right cylinder 
 should be understood by the student. 
 
 The particular difference between the two solids, the right and the 
 oblique cylinder, is, that in the former its axis is perpendicular to its 
 bases or ends ; whereas in the latter it is inclined to them ; the ends in 
 both cases being circular and parallel to each other. There is, however, 
 another important difference, which affects their developments ; viz., that 
 a section of either, taken parallel to its ends, is a circle; while a 
 section at right angles to the axis of either is a circle for the right 
 cylinder, and an ellipse for the oblique one. This difference in cross- 
 section, it may interest the student to know, is the reason why all cir- 
 cular vessels or pipes intended to withstand an internal pressure when 
 in use, are right cylinders, while those employed as mere conduits for 
 the passage of air, smoke, or light gases, not under pressure, may be 
 made in whole or in part of oblique cylinders. The solution of the fol- 
 lowing problem will show how the development of the surface of an 
 oblique cylinder is found. 
 
 Problem 96 (Fig. 199). To find the development of the surface of 
 anoblique cylinder of a given diameter and length, and hctvi ny its 
 axis inclined at a given angle. 
 
 Let the inclination of the axis of the cylinder be 60 ; then to find 
 the development of its surface, we must first draw its elevation. 
 Assuming it to be resting on one of its ends on a horizontal plane, 
 take the IL as that plane, and at any convenient point in it, as a in 
 No. 1, Fig. 199, draw a line making with the IL an angle of 60. With 
 a as centre, and half of the intended diameter of the base of the 
 cylinder as radius, describe a semi-circle cutting the IL in b and c ; 
 and through those points draw lines parallel to a a'. On the axial 
 line, set off from a the intended length of the cylinder, and through 
 the point thus given, draw the line d c parallel to b c ; then d bee will be 
 the elevation of the cylinder. 
 
 Next, divide the semi-circle into any number say eight of equal 
 parts, and through each point of division draw a line perpendicular 
 to be to cut b c in the points 1, 2, 3, etc.; and through these draw 
 
202 
 
 FIRST PRINCIPLES OF 
 
 Fi(J. 199 
 
 Fig. 200 
 
MECHANICAL AND ENGINEERING DRAWING 203 
 
 faint lines parallel to the axis a a. At any point, s in bd, draw the 
 line s I at right angles to a a, and produce it indefinitely. From the 
 points in s I where the meridians 1, 2, 3, etc., cross it, set off on each 
 of these respectively, the lengths of the ordinates in the semi-circle 
 measured from b c, and through the points thus found draw the semi- 
 elliptic line as shown. 
 
 Then, for the development No. 2, on the produced line through s I 
 in No. 1, set off from I to #', the distances that the points 1, 2, 3, etc., 
 in the semi-ejlipse s4, No. 1, are from each other, and through the 
 points thus found, draw lines parallel to the side e I c of the cylinder. 
 These lines will be the meridians shown on the front surface of the 
 cylinder when it is laid out flat ; for those on its near face, set off the 
 same spaces from s to I' as for the front face, and draw in the meri- 
 dians as shown. For the development of the top and bottom edges, 
 draw projectors through points 1, 2, 3, etc., in them, in No. 1, to cut 
 the corresponding lines in No. 2 ; a continuous line drawn through the 
 points of intersection at both ends of the meridians 1, 2, 3, etc., will 
 be the edges required. If the extreme meridians, e'l'c', be put in 
 full, as shown, the figure bounded by the lines ede, cb c', and e I c, 
 e'l'c', will be the complete development of the curved surface of the 
 oblique cylinder dbce, No. 1. 
 
 If the student now compares this development with that of a por- 
 tion of the surface of the right cylinder, lying between the two 
 parallel but oblique section planes ad and ef the latter shown in 
 dotted lines in No. 1, Fig. 198, he will note that, though apparently 
 similar, they are actually different in form and in outline, caused by 
 one solid being perfectly circular in cross-section, while the other is 
 elliptic. 
 
 91. From the development of the surface of the cylinder, and its 
 frustums, we proceed to that of the cone. From the definition of this 
 solid " as one generated by the revolution of a right-angled triangle 
 about its perpendicular side as an axis," it will be understood at once, 
 that, although all meridians drawn on its surface are, like those on a 
 cylinder, actually of the same length, yet in the representation of 
 them in elevation they are all, with the exception of the extreme 
 bounding ones of the figure foreshortened, and of varying length, de- 
 pendent upon their position. As errors are very likely to occur in 
 any attempt to develop the surface of a cone, if this fact is not borne 
 in mind, it is here referred to, to prevent a false development from 
 being made by the student. As a first problem in connection with 
 the cone and its development, we take the following 
 
 Problem 97 (Fig. 200). Given the elevation of a right cone ; to find 
 the development of its curved surface. 
 
 Now as the curved surface of a right cone may be conceived to be 
 generated by the revolution of a straight line round an imaginary verti- 
 cal axis, one end of that line moving in a circle described about the 
 axis as a centre, while the other is in the axis itself, it follows that if 
 a solid so generated be laid on a plane, and be caused to roll through 
 
204 FIRST PRINCIPLES OF 
 
 one revolution, all the points in its base edge will together form the 
 arc of a circle halving the apex of the cone as its centre, and the length 
 of its slant side as radius the length of the arc so struck or produced 
 being equal to the circumference of the base of the rolling cone. If, 
 then as in the case of the cylinder the cone be conceived to give off 
 its surface during one complete revolution, while rolling on a plane, 
 the surface of that plane actually rolled over by it, will be its de- 
 veloped surface. To show this graphically will be to solve the problem 
 above given. 
 
 Let ABC, Fig. 200, be the elevation of the cone, A being its apex, 
 the line BC its base, and that through Aa its axis. Then to find the 
 development of its curved surface, with a as centre, and half BC as 
 radius, describe a semi-circle, which will be a plan of the front half of 
 the cone. Divide this semi-circle into any number of equal parts say 
 eight and number them 1, 2, 3, etc. With A the apex of the cone 
 as centre, and AC its slant side as radius, describe from C indefini- 
 tely the arc of a circle, and on it from C, set off in C', tivice the num- 
 ber of equal parts that the semi-circle on BC is divided into. Join C' 
 with A, and the sector included by the two radii AC, AC', and the 
 arc CBC', will be the development required. 
 
 From the foregoing it will at once be seen, how the surface of a 
 frustum of a right cone may be readily found, for it is only neces- 
 sary to set off vertically from the base line BC of the cone the intended 
 height or depth of the frustum, through the point thus given to draw 
 a line parallel to BC, as b c in Fig. 200, and with A as a centre, and 
 Ac as radius, to describe an arc from c, cutting AC' in c', then the 
 figure included between the concentric arcs CC', c c, and the lines C'c, 
 Cc, will be the developed surface of the frustum. 
 
 If the line of section of a right cone be inclined to its base, as s I in 
 Fig. 200 is to BC, then the development of the frustum's curved 
 surface is found as follows : - - Through the points of division 
 1, 2, 3, etc., of the semi-circle on BC, draw lines perpendicular to BC 
 to cut it in 1', 2', 3', etc. ; join these with A, the apex of the cone, by 
 right lines, and they will be meridians on its surface at these points. 
 Then, from the points set off on the arc drawn from C to C', draw 
 lines to A, and they will be the development of the meridians both on 
 the front and back surface of the cone. Number these from C to B', 
 and B' to C' as shown. On to the slant side, AC of the cone, project 
 over parallel to BC, the several points where the line s I is cut by 
 meridians drawn from points 1', 2', 3', etc., in BC to A ; then from A 
 as centre, and the distances of the points just found in the slant side 
 AC from it as radii, cut the corresponding meridians drawn on the 
 development of the cone as shown, commencing on the one drawn 
 through AB', with A.S as radius, which will give the lowest point s' 
 in the curve. The surface enclosed by a continuous line drawn 
 through the points thus found, with the arc CB'C', and the lines 
 C/', C/, will then be the development of the surface of the frustum 
 required. 
 
 92. As the oblique cone and its frustums play as important a part 
 in giving shape to flues, uptakes, funnel bases, etc., as the oblique 
 
MECHANICAL AND ENGINEERING DRAWING 
 
 205 
 
 cylinder and pyramid, the solution of the following problem will show 
 how the development of their surfaces is found. 
 
 Problem 98 (Fig. 201). An oblique cone of a given diameter of 
 base and vertical height, has its axis inclined to its base at 45 ; 
 required its elevation and the development of its curved surface, 
 together with that of a frustum of the same solid of a given height. 
 
 Fig. 201 
 
 Assuming the cone to be resting with its base on the HP, to draw 
 its elevation, take any convenient point, a in the IL as a centre, and 
 with half the intended diameter of its base as radius, describe a semi- 
 circle cutting the IL in B and C. At a draw a line, making with BC 
 an angle of 45, and another perpendicular to it. On the perpen- 
 dicular line set off from a in a the intended vertical height of the cone, 
 and through a' parallel to BC draw a projector to cut the inclined line 
 from a in A, which will be the apex of the intended cone. Join 
 B and C with A by right lines, and the triangle ABC will be the 
 elevation. 
 
 To find the development of its curved surface, divide the semi- 
 circle drawn on BC into any number say eight of equal parts, and 
 from A let fall a vertical projector to the IL or base BC of the cone 
 produced to cut it in x. Now it is evident that the longest and 
 shortest meridians that can be drawn on the surface of the cone, will 
 be its bounding lines AB and AC ; then to determine the lengths of 
 any intermediate ones for that is what is required to be known 
 proceed as follows : 
 
 From x in BC produced, draw right lines to the several points of 
 division in the semi-circle, and take each of these lines as a radius, 
 
206 FIRST PRINCIPLES OF 
 
 and x as centre, and describe arcs to cut BC in points 1, 2, 4, etc. 
 Join each of these by right lines with A, the apex, then will they be 
 taken in order the actual lengths of meridians drawn on the front 
 surface of the cone, from A its apex to the points 1, 2, 3, etc., in its 
 base edge. The axis of the cone being in a plane parallel to the YP, 
 or plane of the paper, the meridians on its back surface will be of the 
 same length as those on the front one, being directly behind them. 
 
 93. To make the finding of the actual lengths of meridians on an 
 oblique cone's surface as clear as possible to the student, let him con- 
 ceive a right-angled triangular plane (having a constant altitude, or 
 perpendicular, as A x in the figure, but a varying base, of lengths equal 
 to the distances between x and the points 1, 2, 3, etc., in the semi- 
 circle on BC) to swing on its perpendicular edge A x as a hinge, 
 through the arcs drawn between the points 1, 2, 3, etc., in BC, to the 
 corresponding one in the semi-circle drawn on BC, it will then be 
 seen that the hypothenuse of such a trianglar plane would coincide 
 with, and be of exactly the same length, as a line drawn on the cone's 
 surface from its apex to the point in its base corresponding with that 
 of the particular plane taken. For instance, take the plane having a 
 base x3' in No. 1, then the line A3' will- be the actual length of 
 a meridian drawn on the surface of the cone from point 3 in its base 
 edge to its apex, and so wi.th any other of the planes. 
 
 Having thus found the length of any meridian drawn on the cone's 
 surface, its development offers no difficulty. To find it, draw a line in- 
 definitely in No. 2, Fig. 201, parallel to A x in No. 1, and project over 
 to it the point A ; on this line set off from A the distance AB, equal 
 to the length of AB, No. 1. On either side of B, No. 2, draw an arc of 
 a radius equal to the length between B and point 7 in the semi-circle 
 on BC in No. 1, and from A in No. 2, with the distance between A 
 and point 7 in BC, No. 1, as radius, draw arcs cutting those struck 
 from B in points 7, 7'; and join these points with A by faint lines as 
 shown. From 7 and 7' as centres, strike arcs of the same radius as 
 from B ; and from A with A6' in No. 1 as radius, cut the arcs last 
 drawn in points 6, 6'; join these points with A by right lines, and repeat 
 the process till AC in No. 1, as radius, is reached ; then join C and C' 
 with A, and the figure AC'BCA will be the complete development of 
 the whole surface of the oblique cone, of which ABC, No. 1, is the 
 elevation. 
 
 For the development of the surface of the frustum 6BCc, No. 1, of 
 the same cone, we have in effect to cut off from that just found so 
 much of the surface as will cover the smaller oblique cone in No. 1, 
 having the line of section be for base, and A for vertex. With the 
 meridians already drawn in, all that is required to be done is to 
 measure off on each of them respectively, its length from the vertex A 
 to the point in b c in No. 1, where that meridian crosses it, and 
 transfer it to its corresponding meridian in No. 2 ; then a continuous 
 line, as c'bc, drawn through the points thus found, will be the de- 
 velopment of the top edge of the frustum, giving the remaining lower 
 part of the figure in No. 2 as the required development of the surface 
 of the frustum 6BCc, given in elevation in No. 1, Fig. 201. 
 
MECHANICAL AND ENGINEERING DRAWING 207 
 
 94. Having shown how to find the development of the surfaces of 
 the cylinder and cone right or oblique with that of their frustums, 
 we have now to explain how the covering of a " sphere " is obtained. 
 Its surface, as a whole, has been previously stated to be non- 
 developable, but it does not necessarily follow that such a solid cannot 
 be covered. The impossibility of doing this with one sheet of material 
 is overcome in practice by dividing its surface into what are called 
 " gores," or figures which may be defined as made up of two spherical 
 triangles joined at their bases ; such a triangle differing from a plane 
 one in having all its sides curved they being arcs of great circles of 
 the sphere instead of being straight. 
 
 In previous problems in connection with the sphere, it was shown 
 that any section of it made by a plane passing though its axis, gave 
 a great circle or one equal in diameter to the sphere for that section ; 
 and from the fact, that such a solid is generated by the revolution 
 of a semi-circle about its diameter as an axis, it follows that any plane 
 section of it, at right angles to a great circle section, will be a circle 
 having a diameter proportioned to the distance that the cutting plane 
 producing it, is from the centre of the sphere. Then as the halves of 
 a sphere or hemispheres are equal solids, the covering of only one 
 half need be found. 
 
 The parts into which a sphere is supposed to be divided for the 
 purpose of finding its covering, are such as would be produced by a 
 series of vertical planes passed through its axis, dividing its equatorial 
 circumference into equal parts, and giving what are called " lunes " as 
 the resultant solids ; and it is the development of the surfaces of these 
 lunes which has to be found. The first problem in this connection is 
 
 Problem 99 (Fig. 202). Given the plan and elevation of a hemisphere ; 
 to jind its approximate covering, the axis, or pole, being assumed 
 to be vertical. 
 
 Let Nos. 1 and 2, Fig. 202, be the given plan and elevation of the 
 solid, resting with its base AB, on the HP, and its pole Pp, perpen- 
 dicular to it. First divide its surface in No. 1 into, say six equal parts, 
 by meridians AB ; 11'; 22'; passing through the pole P, and find 
 their elevations as in No. 2. Now the meridian lines passing through 
 P are the plans of meridian planes cutting the hemisphere into 
 six half-lunes, and as these are equal solids, having similar and equal 
 surfaces, it is only necessary to find the covering of one of them. 
 
 To do this, divide the quadrantal arc AP in No. 2, into, say 
 four equal parts, in points 3, 4, 5, etc., and through them draw 
 lines parallel to AB, to meet the opposite arc PB. Assume these lines 
 to be parallels drawn on the surface of the hemisphere, and find their 
 plans which are circles in No. 1 . Number the points where they cut 
 the meridians 2 2', 1 1', to correspond with their elevations. From the 
 point x, in the base of the hemisphere in No. 1, on the axial line fp 
 produced, set off a length x p, equal to the actual length of the arc AP 
 in No. 2 ; divide this into the same equal parts that AP is divided 
 into, and through the points of division, draw lines at right angles 
 
208 
 
 FIRST PRINCIPLES OF 
 
 Fig. 203 
 
MECHANICAL AND ENGINEERING DRAWING 209 
 
 to Pp. On either side of Pp, on these lines, set off in points 3 3', 4 4', 
 5 5', the actual lengths of the arcs drawn between the meridians 2 2', 
 1 1', in No. 1 ; and through them and p, draw curved lines. Make the 
 length of these measured along them from p to 1', 2 equal to p x, 
 and through 2, x, 1', draw the curved line shown. Then the spherical 
 triangle p2 1', will be the development of the surface of one-sixth of 
 the hemisphere. 
 
 The greater the number of gores, or parts the surface of the 
 hemisphere is divided into the nearer will they when properly bent 
 and joined together approach the true spheric form. 
 
 95. In practice, where a hemispherical surface is required, as, for 
 instance, in egg-ended boilers, tops of floating buoys, dome-shaped 
 coverings, etc., it would not be possible to make it of plates cut to 
 the shape of a perfect gore, as the plates could not be riveted together 
 at their upper or pointed ends. In such case, it is usual to cut the 
 gores short of the required length, and fit a dished circular crown 
 plate to which the upper ends of all the gores are riveted. In the 
 hemisphere given in the problem this crown plate would about equal 
 in diameter the smallest circle shown in the plan No. 1, necessitating 
 the gores being cut at their top ends, to the curve of an arc shown 
 dotted struck from p, of a radius 'to suit the crown plate. 
 
 In many curved surfaced structures of hemispherical form to save 
 expense and much labour in construction the covering adopted is that 
 of a spherical polyhedron, having a great number of faces ; each face 
 being tangent to a hemispherical surface, the touching part being a 
 straight line, coinciding with, and falling on a meridian, which divides 
 the face into two equal parts throughout its length. In the case of 
 the covering of a truly hemispherical surface, each gore would not only 
 have to be bent lengthwise, to the arc of a circle of the radius of the 
 sphere, but it would also require to be bent crosswise, or dished, which 
 would necessitate special care, so as to ensure the same curvature in all. 
 By adopting the polyhedron surface, the covering material only needs 
 bending in one direction lengthwise as each gore is a part of a 
 cylindric surface and therefore developable. To develop such a 
 surface proceed as follows 
 
 Let the circle ABCD, No. 4 (Fig. 202), be the plan of a hemisphere 
 to be covered by a surface of the form of a spheric polyhedron of 
 twelve faces or sides, and let the lines AB and CD which are at right 
 angles to each other be meridians on the hemisphere. Divide the 
 quadrant AC into six equal parts, and at the points of division draw 
 faint lines to the centre, or pole p. Also divide the arc CB into 
 the same number of equal parts, numbering the points of division 1, 2, 
 3, etc., as shown in No. 4. From p, draw radials through points 
 5, 6 ; and at C, a tangent to the circle, to cut them ; then will the 
 triangle p, 5, 6 in No. 4, be the plan of one of the gores of the 
 given surface. 
 
 To find its development, parallel to the diameter AB No. 4, and 
 through the points 1, 2, 3, etc., in the arc CB, draw lines to cut the 
 two radials p 5, p 6 ; produce the diameter CD indefinitely upwards 
 and on it from C, set off to p\ the equal parts into which the arc CB 
 
 P 
 
210 FIRST PRINCIPLES OF 
 
 is divided, and through the points of division draw lines parallel to the 
 tangent drawn through C. Then from the points of division in the 
 radials p 5, p6, cut by the lines drawn from points 1, 2, 3, etc., in the 
 arc CB No. 4, draw projectors parallel to CD produced, to cut the lines 
 drawn through it in the points 1 1', 2 2', 3 3', etc. Through these last- 
 found points and p, draw curved lines as shown, and the triangular 
 figure p'Q 5 will be the development of the gore required. 
 
 96. If a hemispherical or dome-shaped surface is of large area, 
 a different method of covering it would be resorted to. Instead of 
 dividing the spherical surfaces into gores, that of " zones " or circular 
 belts cut into convenient-sized .sections would be the form given to the 
 material. The method of finding the proper shape of such sections is 
 shown in Fig. 203, and is as follows 
 
 With p, in the given straight line AC in the figure as centre, and 
 with pA as radius, describe the circle ABCD. Let the upper half of 
 it, No. 1, be the elevation, and the lower half, No. 2, the plan, of the 
 front part of a given hemisphere, whose surface is required to be 
 covered by material in the form of zones or belts. Divide the arc AB, 
 of the semi-circle ABC, into any number of equal parts, say four ; and 
 at the points of the division 1, 2, 3, draw lines parallel to AC. 
 Produce the diametral line BD upwards indefinitely, and through 
 the points A, 1, in the arc AB, draw a line to cut this produced line in 
 x. Through points 1 and 2, and 2 and 3, draw similar lines to cut 
 BD produced in y and z. Then with seA, and sel, as radii, draw arcs 
 indefinitely, and repeat the process with y and z as centres, and yl, y2 ; 
 z2, z3, as radii. Now the surface enclosed between each of the pairs 
 of concentric arcs, struck from ,r, y, 2, as centres, are portions of the 
 surfaces of frustums of cones, having those centres as their apices, and 
 the lines AC, 11', 22', 3 3', as their bases ; the heights of the frustums, 
 and the slope of their sides, being determined by the number of parts 
 into which the arc AB of the semi-circle ABC is divided. 
 
 The, lengths of each of the covering pieces in the zones or belts are 
 determined by the number into which the belt is to be divided. Let 
 this number be four ; then to show their position and length on the 
 spheric surface, find by projection on the -plan of the front half of the 
 hemisphere in No. 2, the plans of the lines 1 1', 2 2', 3 3' in No. 1, which 
 will be semi-circles as shown. Now the radial line pD divides each of 
 the semi-circular belts into two parts, and as four such parts of each 
 belt will cover the hemisphere with the exception of the crown plate 
 the length of one of them in each, taken in order, will be the length 
 that each of the strips E, F, G, in No. 2, must be. Therefore on each 
 of the outer arcs drawn from a;, y, , set off the actual lengths in 
 points, c?, f, </, of their corresponding arcs in the quadrant Ajt?, D, 
 No. 2 ; and through d,f, </, draw lines radiating to x, y, , to cut the 
 inner arcs, or edges of each belt as shown. 
 
 In arranging the plates for covering such a surface, they would for 
 strengthening and other important reasons be made in practice to 
 " break joint," as shown by the short radiating thick lines in the half- 
 plan of the hemisphere No. 2 in the figure. 
 
 By one or the other of the methods shown and explained in 
 
MECHANICAL AND ENGINEERING DRAWING 211 
 
 Fi^s. 202-3 the coverings of the sphere oblate or plolate the ellipsoid, 
 paraboloid, and hyperboloid with their frustums may be found. 
 Combinations of parts of these solids, with those of the cylinder, 
 cone, and sphere, constantly occur in practice, and all that is required 
 of the student for the mastery of each case as it arises, is to make 
 himself thoroughly acquainted with the actual forms of the solids 
 which enter into combination, and then apply the principles which 
 have been so fully explained in this and previous chapters, and on 
 which their correct delineation depends. . 
 
 97. With the problems on the covering of the sphere, etc., the 
 subject of the " Development of the Surfaces of Solids " is concluded, 
 and with it the exposition of the principles of that special kind of 
 " projection " on which the art of Mechanical and Engineering Drawing 
 is based. The problems in each division of the subject might have 
 been considerably increased in number, but as such an extension would 
 have involved the expenditure of more of the student's time than the 
 subject warrants, as many have been given as will be found necessary 
 for all his future requirements. 
 
 A careful study of the foregoing first principles of the Mechanical 
 Draughtsman's art, and the conscientious working out of all the 
 problems furnished for their complete elucidation, will lead the student 
 to an easy comprehension of the method of their practical application 
 to the delineation of all kinds of machine elements, and engine 
 details, which may form the subject of a further work by the author 
 of the present one here concluded. 
 
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