A MANUAL OF ELEMENTARY PROJECTION DRAWING, INVOLrVING THREE DIMENSIONS. DESIGNED FOR USE IN HIGH SCHOOLS, ACADEMIES, ENGINEERING SCHOOLS, EITC.; AND FOR THE SELF-INSTRUCTION OF INVENTORS, ARTISANS, ETC. An Aire'Mivi!ionsDIV. I. ELEMIEN-TARY PROJECTIONS. DIV. II. DETAILS OF CONSTRUCTIONS IN MASONRY, WOOD, AND MIETATL. DIV. III. RUDIMENTARY EXERCISES IN SHADES AND SHADOWS. DIV. IV. ISOMETRICAL AD) OBLIQUE, OR PICTORIAL PROJECTIONS. DIV. V. ELEME-NTARY STRUCTURAL DRAWING. BY S. EDWARD WARREN, C. E. PRIOFESSOR OF DESCRIPTIVE GEOMETRY, ETC., IN THE MASS. INSTITUTE OF TECHNOLOGY, AND AUTHOR OF A SERIES OF WORKS ON DESCRIPTIVE GEOMETRY AND STEREOTOMY. Fou)rth IEdition, Revised and Enlarged. NEW YORK: JOHN WILEY & SON, PUBLISHERS9 15 ASTOR PLACE, 1872. Entered aceordlng to Act of Congress, in the year eighteen hundred and sixty-one, by S. EDWARD WARREN, In the Clerks cffice of the District Court of the United States for the Northern District ao New York. CONTENTS. NOTE TO THE FOURTH EDITION........................................ Vi I trTI1'AcE............................. Vii PR['I]LIMINARY NOTES............................................ X DIVISION FIRST. PROJECTIONS. CHAPTER I. —First P1rinciples. I.-The purely geometrical or rational thleory of projections...... 1 ~ II.-Of the relations of lines to their projections............ ~ III.-Physical theory of projections..................5........ ~ IV.-Conventional mode of representing the two planes of projection, and the two projections of any object upon one plane; viz,, the plane of the paper................................. 5 ~ V.-Of the conventional direction of the light; and of the position and use of heavy lines................................. 6 ~ VI.-Notation................................... 7 ~ VI.-Of the use of the method of projections..................... 8 CHAPTER II. —Pqojection of lines. Problems in right projectiona; and ii,n oblique projection, shozwing tvo sides of a solid righit anigie. (Thirtytwo problems.) ~ I.-Projections of straight lines...... 9 ~ II. —Right projections of solids............................. 12 ~ III.-Projections showingo two sides of a solid right angle......... 14 ~ IV.-Special Elementary Intersections and Developments........ 23 DI)IVISION SECOND. DETAILS OF CONSTRUCTIONS IN BIASONRY, WOOD, AND METAL. CH &PTER I.-C'osstructions in,Masonr y. i I. —General definitions and principles applicable both to bricek ailnd stone work.......... 32 IV CONTENTS. PAGE ~ II.-Brick work............................................ 32 ~ III.-Stone work............................................ 35 CHAPTER II. —Constructions in Wood. ~ I. —-General remarks. (Explanation of Scales.).................. 3 ~ II.-Pairs of timbers whose axes make angles of 0~ with each other. 40 ~ III.-Combinations of timbers whose axes make angles of' 90~ with each other...................................... 43 ~ IV.-Miscellaneous combinations. (Dowelling, &c.). 46 ~ V.-Pairs of timbers which are framed together obliquely to each other......................................... 41 ~ VI. —Combinations of timbers whose axes make angles of 180~ with each other............................................ 48 CHAPTER III.- Constructions in MVetal.................................... 53 Cage valve of a locomotive pump. Metallic packing for stuffing boxes, &c. 54 DIVISION THIRD. RUDIMENTS OF SHADES AND SHADOWS. CHAPTER I. -Shadows. ~ I.-Facts, Principles, aud Preliminary Problems.................. 59 ~ II.-Practical Prablems. (Twelve.)............................ 63 CHAPTER II.-Shading......................................0......... 0 (Hexagonal prism; cylinder; cone, and sphere.) DIVISION FOURTH. ISOMETRICAL DRAWING AND CABINET PROJECTIONS. CHAPTER I.-First Principles of Isometrical Drawing...................... 7 i' II.-Problems involving only isometric lines...................... 81 " III. —Problems involving non-isomnetrical lines.............. 84 " IV.-Problems involving the construction and equal division of circles in Isometrical Drawing................................ 8E (Approximate construction of the isometrical drawing of a circle.) (Nineteen problems.) V. -Oblique, pictorial, or cabinet projections.................... 95 CONTENTS. IV PAGl DIVISION FIFTH. ELEMENTARY S RUCTURAL DRAWING. CHAPTER I.-Stone structures....................................... 101 A brick segmental arch................................ 101 A semi-cylindrical culvert, with vertical cylindrical wing. walls............................................... 103 CaPTER II.- Wooden structures...................................... 107 A king post truss............................... 107 A queen post truss-bridge.......................... 108 An architectural model with shadows...................... 1 10 CIwTER IH I.-ron constructions.................................... 114 A railway track-frog, chair, &c......... 114 a hydraulic ram......... 116 NOTE TO THE FOURTH EDITION. Tias edition is improved chiefly by the extension of the chapter on oblique, or pictorial projections, otherwise called cabinet projections, with the addition of a new plate. Numerous minor corrections, and new or improved paragraphs have been scattered throngh the work, as suggested by further experience. Yet it is not expected that this edition, however improved, can supersede either careful and thinking study on the part of the willing student, or ample, repeated, cancd vtried -instruction on the part of the willing teacher. AIany minds require variously changed and amplified statements of the same thing before they are ready to exclaim heartily, "I see it now;" and the teacher must be ready to meet, with many forms of instruction, the conditions presented to him by different minds. A very moderate collection of such objects as are illustrated in Division II., and such as can be made by a carpenter or turner, or in machine, gas-fitting, or pattern shops, will very usefully supplement the text, and add to the pleasure and benefit derived by the student; and will be better than an increased size of the boolk, which is meant to be rather suggestive than exhaustive, or to be too closely followed, in respect to the examples chosen for practice. Finally, the author's chief desire in relation to this, as well as the rest of his " elementary works," is, to see them so generally used in higher preparatory instruction as to give due place to higher studies in the same department in the strictly Technical Schools. R. P. I., TROY, July, 1871. PREFACE. EXPERIENCE in teaching shows, that correct conceptions of the forms of objects having three dimensions, are obtained with considerable difficulty by the beginner, from drawings having but two dimensions, especially when those drawings are neither "natural"-that is "pictorial"-nor shaded, so as to suggest their form; but are artificial, or "conventional," and are merely "skeleton," or unshaded, line drawings. Hence moderate experience suggests, and continued experience confirms, the propriety of interposing, between the easily understood drawings of problems involving two dimensions, and the general course of problems of three dimensions, a rudimentary course upon the methods of representing objects having three dimensions. This much may be said in defence of Division I. of the present volume. Experience again proves, in respect to the drawing of any engineering structures that are worth drawing, that it is a great advantage to the draftsman to have —lst, some knowledge of the thing to be drawn, aside from his knowledge of the methods of drawing it; and 2d, practice in the leisurely study of the graphical construction of single members or elements of a piece of framing, or other structure. The truth of the second of the preceding remarks, is further apparent, from the fact that in entering at once upon the drawing of whole structures, three evils ensue, viz.-lst, Confusion of ideas, arising from the mass of new objects (the many different parts of a structure) thrown upon the mind at once; 2d, Loss of tine, owing to repetitirn of the same detail many times in v1ii PRE Ci. the same structure; and 3d, TCWaste of dra'w'ings, as well as of titne, through poor execution, which is due to insufficillt previous practice. Hence Division II. is filled up with a tolerably liberal and comprehensive collection of details of structures, each one of which affords a useful problem, illustrative of some one feature of conslructivert arznd of graphical procedure. Here it may be said, that the department of instruction, part of which is embodied in this work, has involved in it a subdepartment, viz., that of Descripttive Constructions, i.e. a popular descriptive study of the ways in which many common things are made arnd put together, or in which they operate. The classification of Chap. II., Div. II., is believed to be a happy one, as readily suggesting various forms of framings. Since my object is not so much to acquaint the student solely with the works of the authors or artizans of other nations, as to give them just those specimens of American methods of construction which are not usually found in books, I must refer them to foreign works for examples of foreign modes of construction-not that such examples should be disparaged, but because it is intended that this work shall add something to our stock of book knowledge of the subjects of which it treats, rather than merely repeat what has been before published. Division III. contains a short special study of Shadows, taken up in the form of the study of a few individual problems of a practical character, assuming that the light comes only in the simple conventional direction, commonly agreed upon by draftsmen, and saying nothing of the possibility of allowing the light to be taken in any other direction. The problems of this Division are intended to be sufficient, when fully understood, to enable the student to construct accurately, or by a sufficiently near approximation, all the shadows he will meet with in Geometrical Drawing, till he shall pursue a general course on shades and shadows. Division IV. contains a sufficient exposition of the few and PREFACE. ix simple principles of Isometrical _projection and drawing, to inmeet all the ordinary wants of the student or draftsman. Division V. includes examples of a few simple structures, to fulfil the threefold purpose of affording occasion for learning trhe names of parts of structures; for practice in. the combination of details into whole structures; and for profitable review practice in execution. Finally, since experience has dictated so much of this preface, it shall dictate also these closing remarks. First: Classes will generally be fbund to take a lively interest in the subjects of this volume-because of their freshness to most learners, as new subjects of interesting study-because of the variety and brevity of the topics, which causes frequent change of the subject of study (such change being thought quite desirable by most learners)-and because of the compactness, neatness, and beaut) of the volume which is formed by binding together all the plates of the course, when they are well executed. Second: As to the use of this volume, it is intended that there should be formal recitations of the problems in the 1st, 3d, and 4th divisions, with graphical constructions of a selection of the same or similar ones; and occasional interrogations mingled with the graphical constructions of the practical problems of the remaining divisions. Remembering that excellence in mere execution, thoughl highly desirable and to be encouraged, is not, at this stage of the student's progress, the sole end to be attainedl, a small nunber of graphic exercises will be sufficient. The student may, in place of a tedious course of finished drawings, be called on frequently to describe, by the aid of pencil or blackboard sketches, how he would construct drawings of certain objects-either thos:e given in the several Divisions of this volume, or other similar ones proposed by his teacher. In a course of school education, the full delivery to the student of the greatest number of distinct ideas, with the least amount of merely manual labor, should be the teacher's general aim. PRELIMINARY NOTES. As beginners not seldom find peculiar difficulties at the outset of the study of projections, the removal of which, however, makes subsequent progress easy, the following special explanations are here prefixed. I. Figs. 1, 2, 3, 5, and 15, of P1. I. are pictorial diagrams, usedfor illustration in place of actual models. Thus, in Fig. 3, for example, MIMt represents a horizontal square cornered plane surface, as a floor. MGV represents a vertical square cornered plane surface, as a wall, which is therefore perpendicular to MHt. P represents any point in the angular space included by these two planes. Pp represents a line from P, perpendicular to the plane MHt, and meeting it at p. Pp' represents a line from P, perpendicular to thlle plane MGV, and meeting it at p'. Then Pp and Pp' are called the projecting lines of P. The point p is called the horizontal projection of P, and p' the vertical projection of P. The projecting lines of any point or of any body, as in;Fig. 1, are perpendicular to the planes, as MHt and MGV, which are called planes of projection. II. In preparing a lesson from this work, the object of the student is, by no means, to commit to memory the figures, but to learn, from the,first principles, and subsequent explanations, to see in thesefigures the realities in space which they represent; so as to be able, on hearing the enunciation of any of the problems, to solve it from a clear understanding of the subject, and not "by rote" from mere menmory of the diagrams. The student will be greatly aided in so preparing his lessons, by working out the problems, in space, on actual planes at right angles to each other, as on the leaves of a folding slate, when one slate is placed horizontally and the other vertically. In the construction of his plates, he should also test his knowledge of the principles, by varying the form of the examples, though without essentially changing their character. A MANUAL OF ELEMENTARY PROJECTION DRAWING. DIVISION FIRST. PROJECTIONS. CHAPTER I. FIRST PRINCIPLES. ~ 1. The porely Geometrical or Rationval Tzeory of Projections. ]. ELEMIENTARY P1OJJECTION DRAWING is an introduction to Descriptive Geometry, and shows how to represent simple solids, singly and in combination, upon plane surfaces, yet so as to show their real dimensions. 2. Since a plane has but two dimensions, only plane figures, that is figlr'es having but two dimensions, can be placed in, or parcallel to, a plane; and hence no more than two dimensions can be shown in both their true size and relative position upon a plane. But tllelre is a vast number of geometrical magnitudes, as prisms, pyranlids, cylinders, cones, spheres, etc., which have three dimensions, and the study of which gives rise to problems involving three dimensions. It appears, then, that there are problems of three as well as of two dimensions, and also that more than one plane sunrface is necessary to enable us to represent, both completely and accurately, bodies having three dimensions. 3. What, then, is the number and the relativeposition of the planes which will enable us to represent all the dimensions of any geometrical solid, in their real size, on those planes? To assist in answering this question, reference may be made to P1. I. Fig. 1. Let ABCOFED be a regular square-cornered block, whose length is AB; breadth, AD; and thickness, AC; and let MN be any horizontal 2 FIRST PRINCIPLES. plane below it and parallel to its top surface ABED. If now friom the four points A, B, D, and E, perpendiculars be let fall upon the horizontal plane MIN, they will meet it in the points a, b, d and e. By joining these points, it is evident that a figure —abed —will be formed, which will be equal to the top surface of the block, and will be a correct representation of the length and breadth of that top surface-i-. e. of the length and breadth of the block. Similarly, let iMP be a vertical plane, behind the block and parallel to its firont face, ABCF. Fromn A, B, C, and F let fall perpendiculars to the vertical plane MP, and connect the points, a', b', c', and f, thus found. The figure, a'c'b'f', so found, will evidently be equal to the front face ABCF of the block, and will therefore correctly represent the length and thickness of the block. 4. From the foregoing article, the following fundamental princi ples are deduced. First: Two planes, at right angles to each other, are necessary to enable us to represent, fiilly, the three dimnensions of a solid. Second: In order that those dimensions shall be seen in their true size and relative position, they mu1st be parallel to that plane on which they are shown. Third: Each plane shows two of the dimensions of the solid, viz., the two which are parallel to it; and that dimension which is thus shown twice, is the one which is parallel to both of the planes. Thus AB, the length. and AD, the breadth, are shown on the plane MN; and AB, the length again, and AC, the thickness, are shown on the plane MP. 5. From Art. (3), the following definitions naturally arise. The point a, P1. 1. Fig. 1, is where A would arrive if thrown, that is, projected, vertically downwards along the line Aa. Likewise, a' is where A would be, if thrown or projected along Aa', firom A to a', perpendicularly to the plane IMP. Hence a is called the horizontal projection of A; and a' is called its vertical projectio?. Also, conversely, A is said to-be horizbntcally projected at a, and vertically projected at a'. The plane MN is thence called the horizontalplane of projection; and the plane MP, the verticalplane of projection. Aa and Aa' are called the two projecting lines of the point A, relative to the p]lanes of projection, MN and iMP. Hence we have this definition. If from any given point. a per. pendicular be let fall upon a plane, the point where that:pl/pendicular meets the plane will be the projection of the posit rmpon the plane, and the perpendicular will be the projecting line of' th-eipoint. Similar remarks apply to any numbor of points or to obj4cdts limited by such points; and to their prqjections upon any othel planes FrIST PRINCIPLES. 3 of projection. Thus abcd is the horizontal projection of the block ABCD, and a'b'c/' is its vertical projection, and, generally, the projecting lines of objects are perpendicular to the planes of projection employed. Finally, the intersection, as MR, of a horizontal, and a vertical plane of projectionis the ground line for that vertical plctze. 6. The preceding definitions enable us to express a fourth fundamental principle more briefly than could otherwise be done; viz., The height of the vertical projection of a point above the ground line, is equal to the height of the point itself, in space, a3bove the horizontal ptne; and the perpendicular distance of the horizontal projection of a point from the ground line, is equal to the perpeadnicaular distcance of the point itself, in fro2nt of tlie vertical piaene. Thus: P1. I., Fig. 1, aa"=Aa' and a'a"=Aa. 7. The preceding principles and definitions are the foundation of the subject of projections, but, by attending carefully to P1. I., Fig. 2., some usefull elementary applications of them may be discovered, which are frequently applied in practice. PI. I., Fig. 2, is a pictorial model of a pyramid, Vcdeg, and of its two projections. The fcee, Vcd, of the pyramid, is parallel to the vertical plane, and the triangle, Xcab, is equal and parallel to Vcd, and a little in front and at one side of it. By first conceiving, now, of the actual models, which are, perhaps, represented as clearly as they can be by mere diagrams,in Pl. I., Figs. 1 and 2; and then by attentive study of those figures, the next two articles may be easily understood. ~ II- Of the Relations of Lines to their Projections. B. JRelations of single lines to their projections. a. A vertical line, as AC, Pl. I., Fig. 1, has, for its horizontal projection, a point, a, and fbr its vertical projection, a line a'c'. perpendicular to the ground line, and equal and parallel to the line AC, in space. b. A horizontal line, as AD, which is perpendicular to the vertical plane, has, for its horizontal projection, a line, ad, perpendicular to the ground line, and equal and parallel to the line, AD, in space; and for its vertical projection a point, a'. c. A horizontal line, as AB, which is parallel to both planes of projection, has, for both of its projections, lines ab and a'b', which are parallel to the ground line, and equal and parallel to the line, AB, in space. d. A horizontal l'ine, as BD, which makes an acute angle Kwith the vertical plane, has, for its horizontal projection, a line, bd, which makes the same angle witll the g'roullnd line that the line, BD, 4 FIRST PRINCIPLES. mlakes with the vertical plane, and is equal and parallel to the line itself (BD); and has for its vertical projection a line b'a', which is parallel to the ground line, but shorter than BD, the line in space. e. An oblique lines as BC, P1. I., Fig. 1, or Vd, P1. I., Fig. 2, which is parallel to the vertical plane, has, for its vertical projection, a line b'c', or v'd', which is equal and parallel to itself, and for its horizontal projection, a line ba or vcd, parallel to the ground line, but shorter than the line in space. f. An obliqute line, as Vg, P1. I., Fig. 2, which is oblique to both planes of projection, has both of its projections, v'd' and vg, oblique to the ground line, and shorter than the line itself. g. An oblique line, as All, PI. I., Fig. 1, which is oblique to both planes of projection, but is in a plane ACDH, perpendicular to both of those planes, has both of its projections, a'c' and ad, perpendicular to the ground line, and shorter than the line itself. h. A line, lying in either plane of projection, coincides with its projection on that plane, and has its other projection in the ground line. See cd —c'd', the projections of cd, P1. I., Fig. 2. 9. Jlemark. A general principle, which it is important to be perfectly familiar with, is embodied in several of the preceding examples; viz. When any line is parallel to either plane of projece tion, its projection on that plane is equal and parallel to itself, and its projection on the other plane is palallel to the ground line. 10. The preceding remark serves to show how to find the true length of a line, when its projections ale given. When the line, as Vg, PI. I., Fig. 2, is oblique to both planes of projection, its length, Vg, is evidently equal to the hypothenuse of a right-angled triangle, of which the base is vg, the horizontal projection of the line, and the altitude is Vv, the height of the upper extremity, VY, above the horizontal plane. When the line, as All, P1. I., Fig. 1, does not touch either 1_lane of projection, it isevidently equal to the hypo. thenuse of a ri-ght-angled triangle, of which the base, CH, equals the horizontal projection,- ad, and the altitude, AC, equals the dsiference of the perpendiculars, Aa and Hid, to the horizontal plane. In the same way, it is also true that the line, as Yg, P1. I., Fig. 2, is the hypothenuse. of another right-angled triangle, whose base equals the vertical projection, v'd', and whose altitude equals the difference of the perpendiculars, Vv' nnd d'yg, from the extremities of the line to the vertical plane of projection. 11. Relations of pairs of lines to tzheir projections. These rela. tions, after the full notice now given of the various positions of single lines, may be briefly expressed as follows. FIRST PRINCIPLESo a. A pair of lines which are eqptal and parallel in space, aned also parallel to a plane of projection, as AB and CF, P1. I., Fig. 1, or Vc and Xa, P1. I., Fig. 2, have their projections on that plane-a'b' and c'f', P1. I., Fig. 1, or v'c' and x'a', P1. I., Fig. 2 —equal and parallel-to each other, and to thie lines in space. b.. A pair of lines which are equal ancld parallel in space, but not.pcrallel to aplane of projection, will have their projections on that pIlane equal and parallel to each other, but not to the lines in splace. c. Parallel lines gmake equal angles with either plane of projection; hence it is easy to see that lines not parallel to each otheras Vd and Vc, or Vg and Ve, P1. I., Fig. 2-but which make equal angles with the planes of projection, will have equal projections on both planes-i.e. v'd'-v'c' and vg=ve, also vd=vc. III. —Physical Theory of Projections. 12. The preceding articles comprise the substance of the purely geometrical or rational theory of projections, which, strictly, is suflicient; but it is natural to take account of the physical fact that the magnitudes in space and their representations, both address themselves to the eye, and to inquire from what distance and in wchat direction the magnitudes in P1. I., Figs. 1 and 2, must be viewed, in order that they shall appealr just as their projections represent them. Since the projecting lines, Q, regarded as rays, reflected from the block, Fig. 1, to the eye, are parallel, they could only meet the eye at an infinite distance in front of the vertical plane. Hence the vertical projection of an object represents it as it would appear to the eye, situated at an infinite distance from it, and looking in a direction perpendicular to the vertical plane of projection. Likewise, the projecting lines, S, show that the horizontal projection of an object represents it as seen from an infinite distance above it, and looking perpendicularly down upon the horizontal plane. Thus, the projecting lines represent the clirection qf vision, which. is perpendicular to the plane of projection considered. g IV. —Conventional ifocle of representing the two Planes of Projection, and the two Projections of any Object upon one plane -viz. the Plane of the Pcaper. 13. In practice, a single flat shreet of paper represents the two planes of projection, and in the following manner. The vertical plane, MV, P1. I., Fig. 3, is supIposed to revolve backwards, as 6 FIRST PRINPCIPLES. shown by the ares ru and Vt, till it coincides with the horizontal plane produced at MA t t G. Hence, drawing a line firom right to left across the papel, to represent the ground line, MIG, all that part of the paper above or beyond such a line will represent the vertical plane of projection, and the part below it the horizontal plane of projection. 14. Elementary Rational Geometry teaches that two lines, as PP', Pp, P1. I., Fig. 3, which intersect, fix the position of the plane containing them. It also teaches that if fiom a point in space, as P, two lines P —P, P-p, be drawn-one and the other perpendicular to one and the other of two planes, tMV and MH, which are at. right angles to each other —then the plane, PP' p'2p, of these lines, will be perpendicular to the two given planes, and will intersect those planes, in lines P'-p- and p —pv, which will be perpeadicular to the intersection, MG, of the planes, at the same point, p". This conclusion being reached, it can now be expressed more briefly by saying, that the plane of the projecting lines of any point, intersects the planes of projection, in lines which are pepTendicular to the ground line at the sacne 2oint. 15. Now, when the vertical plane revolves about the line MG, as an axis, the point p', being in the axis, must remain fixed, and any other point, as P', must revolve in a plane perpendicular to MG. Hence the line P' —p will revolve-with the plane MV —to the position p — p", still remaining perpendicular to the ground line. Hence we have this fundamental principle of graphic constructions, via. The two projections p, p', P1. I., Figs. 3 or 4, of a oint in space, are always in a line, pp', perpendicular to the ground line. ~ V.- Of the Conventional Direction of the Light; and of the Position and Use of Hteavy Lines. 16. Without going into this subject fully, as in Div. III., it is sufficient to say here that, as one faces the vertical plane of projection, the light is assumed to come from behind, and over the left shoulder, in such a direction that each of its projections, but not the light itself, makes an angle of 450 with the ground line, as shown in P1. I., Fig. 6. And note that the light is sqpposed to turn with the observer, as he turns to face any other vertical plane. 17. The practical effect of the preceding assumnp-tion in reference to the light, is, that upon a body of the formn and position shown in P1. I., Fig. 5, for example, the top, front, and left hand surfacesi. e. the three seen in the Fig.-are illuminated, while the other three faces of the body are in the shade. FIRST PRINCIPLES. 7 18. The practical rule by which the direction of the light, and its effect,are indicated in the projections, is, that all those visible ecldes of the body in space, which divide the light from the dark surfaces, are made heavy in projection. 19, To illustrate: The edges BC and CD of the body in space, Pl. I., Fig. 5, divide light from dark surfaces, and are seen in looking towards the vertical plane, and hence are made heavy in vertical projection, as seen at b'c' and c'd'. BK and KF divide illuminated firom dark surfaces, and are seen in looking towards the horizontal plane, and are therefore made heavy in the horizontal projection, as shown at bkc and kf. 20. By inspection, it will be seen that the following simple rule in reference to the position of the heavy lines on the drawings, mlay be deduced, as an aid to the memory. In all ordinary four-sided prismatic bodies, placed with their edges respectively parallel and perpendicular to the planes of projection, or nzearly so, the right haznd lines, and those nearest the ground lize, of both projections, are made heavy. 21. Heavy lines are of considerable use, in the case of line drawings particularly, in indicating the forms of bodies, as will be seen in future examples. ~ VI. —Notation. 22. Under the head of Notation, two points are to be considered, the manner of indicating the various lines of the diagram, and the lettering. As will be seen by examining P1. I., Figs. 1, 2 see Ve, eg, &c.-and 5, the visible lines of the object represented are indicated by full lines; lines of construction and invisible lines of the object, so far as they are shown, are made in dotted lines. The intersections of auxiliary planes with the planes of projection, called traces, are represented by broken and dotted lines, as at PQP', Pi. I., Fig. 16. 23. Unaccented letters indicate the horizontal projections of points. The same, with one or more accents, denote their vertical projections. The simple rule of thus always lettering the saime 2point with the same letter, wherever it is shown, affords a key to every diagram, as will be shown as the course proceeds. The projections of a body are geometrically equivalent to the body itself, since they show its fbs'im, position and dimensions. Hence objects are considered as named by naming their projections. Thus, the point pp2' /means the point whose projections are p and p'; the line ab-a'b' means the one whose projections are ab and a'b'. 8 PFIRST PRINCIPLES. 24. In the practical applications of projections, " horizontal pl o. jections" ale usually called "plans," and "vertical projections,"' " elevcatiofns." lBefore entering upon the study of the subsequent constructions, the terms "perpendicular" and. "vertical" should be clearly distinguished. "Perpendicular" is a relative term, showing that any line or sulface, to which it isk applied. is at right angles to some other line or surface. "Vertical" is an absolute term, at any one place, and applies to any line or surface at right angles to a level, as a water surface. A vertical line, L, is perlpendicular to all hori. zontal lines which intersect it, but if the entire systeml of linles thus related were inclined, so that all should be oblique, L would still be perpendicular to all the rest, though no longer vertical. ~ VII. — Of he Use of the Mfethod of Projections. 25. Under this head it is to be noticed, that all drawings are made to serve one or the other of two purposes, i.e. they are made for use in aidinfg workmen in the construction of works; or inl rendering intelligible, by mueans of drawings, the realformn and size of some existing structure e; or else, they are made for ornament, or to embellish our houses andcl gratify our tastes, and to show the app2arent forms and relative sizes of olj ects. 26. Drawings of the former kind are often called, on account of the uses to which they are applied, " mechanical" or "working" drawings. Those of the latter kind are commonly called pictures; and he-re it is to be noticed that if " working," ldrawings alre to show the true, and not the celOapparent, proportions of all parts of an object, they must, all and always, conform. to this one rule, viz. All those lines ohich are equal and similarly situated on the object, mutst be eqcual and similarly situated on the drawing. But, as is now abundantly evident, drawings made according to the method of projections, do conform to this rule; hence their use, as above described. CHAPTER II. PROJECTIONS OF LINES: PROBLEMS IN RIGHT PROJECTION; AND IN OBLIQUE PROJECTION, SHOWING TWO SIDES OF A. SOLID RIGHT ANGLE. 27. The style of execution of the following problems is so simple, and so nearly alike for all of them, that it need not be described for each problem separately, but will be noticed from time to time. In the solution of problems, lines are considered as unlimited, and may be produced indefilnitely in either direction. I. —Projections of Straight Lines. 28. PROB. 1. To construct the projections of a vertical straight line, 1- inches long, whose lowest point is ~ an inch from the horizontal plane, and all of whose points are - of an inch from th]e vertical plane. Remarks. a. The remaining figures of P1. I. are drawn just half the size indicated by the dimensions given in the text. It may be well for the student to make them of full size. b. Let MG be understood to be the ground line for all of the above problems, without further mention of it. 1st. Draw, very lightly, an indefinite line perpendicular to the ground line, P1. I., Fig. 7. 2d. Upon it mark a point, a', two inches above the ground line, and another point, b', half an inch above the -ground line. 3d. Upon the same line, mark the point a,b, three-fourths of an inch below the ground line. Then a' b' will be the vertical, and ab the horizontal projection of the required line. (8 a) 29. PROB. 2. To construct the projections of a horizontal line, I1 inches long, 1 inches above the horizontal plane, perpendicular to the vertical plane, and with its furthermost point-from the eye -- of an inch from that plane. P1. I., Fig. 8, in connection with the full description of the preceding problem, will afford a sufficient explanation of this one. Remark. It often happens that a diagrlam is made more intel. 10 PROJECTIONS OF LINES. ligible by lettering it as at ab, P1. I., Fig. 7, and at c'd', PI. I. Fig. 8, for thus the notation shows unmistakably, that ab or c'd' are not the projections of points but of lines. 30. PROBLEMS 3 to 8, inclusive, need now only to be enunciated, with references to their constructions, in P1. I. Fig. 9 shows the projections of a line, 24 inches long, parallel to the ground line, 1 inches from the horizontal plane, and 1 inch from the vertical plane. Fig. 10 is the representation of a line, 2 inches long; parallel to the horizontal plane, and 1 inch above it; and making an angle of 300 with the vertical plane. Fig. 11 represents a line, 21 inches long, paralleI to the vertical plane, and 14 inches from it, and making an angle of 60~ with the horizontal plane. Fig. 12 gives the projections of a line, 1 inches long, lying in the horizontal plane, parallel to the ground line, and 14 inches from it. Fig. 13 shows the projections of a line, 11 inches long, lying in the vertical plane, parallel to the ground line, and 1 inch from it. Fig. 14 indicates a line, 24 inches long, lying in the vertical plane, and making an angle of 60~ with the horizontal plane. 31. Projections of the revolution of a point about an axis. When a point revolves about an axis, it describes a circle, or are,g whose plhne is perpendicular to the axis. Thus a point, revolving about an axis which is perpendicular to the vertical plane, describes an arc, parallel to that plane. The vertical projection of such an arc is an equal are. Its horizontal projection (6) is a straight line parallel to the ground line. Thus, P1. I., Fig. 15a, Ca'represents a perpendicular to the vertical plane, V. The point, A, by revolving a certain distance about this axis, describes the are AB; whose vertical projection is the equal are, a'b'; and whose horizontal projection is ab, a straight line parallel to the ground line. Likewise, briefly, in Figs. 15b and 15c, XY is a vertical axis. The point A, revolving about it, describes a horizontal arc, AB; whose horizontal projection, ab, is an equal arc; and whose vertical projection,a'b. is a straight line parallel to the ground line. 32. PROB. 9. To construct the projections of a line which is in a plane perpendicular to both planes of projection, the line being oblique to both planes of projection. Plate I., Fig. 15, represents a PROJECTION OF LINES. 11 model of this problem. AB represents the line in space; ab its horizontal projection; a'b' its projection on the vertical plane MP'; annd A'B' its projection on an auxiliary vertical plane PQP'; which is parallel to AB, and perpendicular to the ground line. Hence A'B'_-AB. Now in making these three planes of projection coincide with the paper, taken as the horizontal plane of projection, the plane PQP' is revolved about P'Q as an axis, till it coincides with the primitive vertical plane, MP', produced, as at P'QV",:and then the united vertical planes, MP'V", are revolved backward about MH' as an axis into the horizontal plane. In the first revolution, A' describes, according to the last article, the horizontal arc, A'a", (:31) about nm as a celttre, and whose projections are a ""a"', leaving its centre at Q, and ma". Also B' describes the horizontal arc, B'b", about n as a centre, and whose projections are b'"'b", whose centre is Q, and nb'/. Thus we see that two or more do e rent vertical projectionzs, as a' and a", -of the same point, are in the same parallel, a'a", to the ground linej that is, they are at the same height above that line. Hence a" is at the intersection of a'a", parallel to the ground line, MIH', with a"'c", perpendicular to NH'. 33. a. Notice further that a""'b"" is the horizontal projection of A'B', and that it coincides with the projection of ab upon PQP'. Likewise, that amrn is the vertical projection of A'B', and that it coincides with the projection of a'b' upon PQP'. b. Note that Bb', for example (6), is equal to bt, and that bt-b"n the distance of the auxiliary vertical projection, b", of B, from the trace, ol axis, P'Q, of the auxiliary plane. c. Note that a"b" shows the true length and direction of AB; that is, the angles made by a"b" with H'Q and P'Q, respectively, are equal to those made by AB with the planes of projection. 34. To construct P1. I., Fig. 15, in projection. See P1. I., Fig. 16, where, to make the comparison easier, like points have the same letters as in Fig. 15. Supposing the length and direction of the line given, we begin with a"b", which suppose to be 2" long, and to make an angle of 600 with the horizontal plane. Suppose the line in space to be 1I inches to the left of the auxiliary vertical plane P'QP then a' b', its vertical projection, will be perpendicular to the ground line, between the parallels a"c' and b"b' (32), and 1iI inches fiom P'Q. The horizontal projection, ab, will be in a'b' produced; b"n-b"'b"" are the two projections of the arc in which the point b" revolves back to its position, n —b"", in the plane P'QP, and b"b- nb' is the line in which nb"" is projected back 12 PROJECTION OF LINES. to its primitive position b/b. Therefore, b is at the intersection of bt"'b with a'b' produced. a is similarly found, giving ab as the horizontal projection of the given line. Art. 32 shows sufficiently how to find the length a"b" if ab-a'b' we'e given. EXAMPLE. Construct the figure when a',b is the highest point. 35. Execution. The foregoing problems are to be inked with very black ink; the projections of given lines, and the ground line, in heavy full lines; and the lines of construction in fine dotted lines as shown in the figures. Lettering is not necessary, except for purposes of reference, as in a text book, though it affords occasion for practice in making small letters. On the other hand, poorly executed lettering disfigures a diagram so much that it should be made only after some previous practice, and then carefully; making the letters small, fine, and reygular. ~ II.-Right Projections of Solids. Remark. The term "right projection " becomes significant only when it refers to bodies which are, to a considerable extent, bounded by straight lines at right angles to each other. Such bodies are said to be drawn in right projection when their most important lines, and faces, are parallel or perpendicular to one or the other of the planes of projection. 36. PROB. 10.-To constru'ct the projections of a vertical right prism, having a square base; standing upon the horizontal plane, and with one of its faces parallel to the vertical plane. PI. II., Fig. 17. Let the prism be 1 inch square, 1- inches high, and I of an inch from the vertical plane. I st. The square ABEF, ~ of an inch from the ground line, is the plan of the prlism, and strictly represents its upper base. 2d. A'B'C'D', 1~ inches high, is the elevation of the prism, and strictly represents its fiont face. In this, and in all similar problems, it is useful to distinguish the positions -of the points, lines, and faces, in words; as upper and lower; front and backh; right and left; just as is done in speaking of the bodies which (23) the projections represent. Thus, lst. AA' is the front, upper, left hand corner of the prism. 2d. EF-A'B' is the back top' edge; BF-D' is the lower right band edge; each corner of the plan is the horizontal projection of a vertical edge; etc. DI V..I..LI dl 2 ~~~~~~~~~~~v; —-—.........go'' y di~~~~~~~~~~~~, 14 ~~-~~~~~~~~_ IN _:Nfi.-. —'A.~! i~ I / ix,, \ ///' /', ~'v.~-.. o 1'- ---------—.-..... —-------..'.......1..-' i.I -'a /'' -~.-"-,~ —--—,l c! dfL j A P A'" ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~~~~~~~~~ t M-l I _ _ _ C-1 -------------------------— ~~~~~~~~~~ — __-, ___- _ C / - —. b -, —- -... t -- - - - - - - -- - - - - - - - - - --------- -. —-: —— /-. —--- a' - OL~~~kb ~ ----------— bi-'. —-',, —, —- J; -, Ie~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ B~~~~ b;~~~~~~~~~~~~~~~~ ~~~~~~~~~~~~~~~~..-, —........:....'.... a~~~~~~~~~~~~~~~~~~~~~~~ aY ________ 12 18'_________ ~~:~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~,.',. —~',~~a l / ---— T-!;~' ~.i~.... x' / i / ~:-..,',:;,,: i/,,' /,,,,,, aii~b _T'.I`B.H.... / r ~I''Mjr np E: ig ihs 55 Pr P?,~.'. PIR(JECTIONS OF SOLIDS. 13 3d. AE-A'C' is the left hand face; etc. 37. PROB. 11.-To construct the plan and two elevations of a prism having the 2xoportions of a brick, andpl1aced with its length parallel to the ground line. Plate II., Fig. 18. Ist. abcd is the plan, - of an inch broad, twice that distance in length, and 3 of an inch from the ground line, showing that the prism ill space is at the same distance from the vertical plane of projection. 2nd. a'b'e~f' is the elevation, - of an inch thick, and as long as the plan; and 7 of an inch above the ground line, showing that the prism in space is at this height above the hoirizontal plane. 3rd. If a plane, P'QP, be placed perpendicular to both of the irincipal planes of projection, and touching the right hand end of the prism, it is evident that the projection of the prism upon such a plane will be a rectangle, equal, in length, to the width, b(d, of the plan, and, in height, to the height, bIf' of the side elevation. This new projection will also, evidently, be at a distance from the primitive vertical plane, i.e. from P'Q, equal to dQ, and at a distance from the horizontal plane equal to Qf'. When, therefore, the auxiliary plane, P'QP, is revolved about P'Q into the primitive vertical plane of projection, the new projection will appear at 4th. doc"' is the horizontal, and b'c" the vertical projection of the arc in which the point db' revolves into the primitive vertical plane. bha"', b'a', are the two projections of the horizontal arc in which the corner bb' of the prism revolves. EXAMPLE.-Let the auxiliary plane PQP' be revolved about PQ into the horizontal plane. a"c" will then appear to the right of PQ and at a distance from it equal to Qb'. 38. PROB. 12.-To construct the two projections of a cylinder which stands upon the horizontalplane. P1. II., Fig. 19. The circle AaBb is evidently the plan of such a cylinder, and the rectangle A'B'C'D'its elevation. Observe, here, that while the elevation, alone, is the same as that of a prism of the same height, Fig. 17, the plan shows the body represented, to be a cylinder. Any point as a in the plan, is the horizontal projection of a vertical line lying on the convex surface, and called an element. AA'C', and B —B'D', which limit the visible part of the convex surface, are called the extreme elements. 39. As regards execution, the right hand line B'D' of a cylinder to PROJECTIONS OF SOLIDS. or cone may be made less heavy than the line B'D', Fig. 17; and in the plan, the semicircle, aBb, convex towards the ground line, and limited by a diameter ab, which makes an angle of 450 with the ground line, is made heavy, but gradually tapered, into a fine line in the vicinity of the points a and b. 40. PROB. 13. —TO construct the projections of a cylinder whose axis is placed parallel to the qrouzd line. P1. II., Fig. 20. Let the cylinder be 1- inches long, - of an inch in diameter, its axis - of an inch firom the horizontal plane, and - an inch from the vertic;l plane. The principal projections will, of course, be two equal rectangles, gehf and a'b'c'd', since all the diameters of the cylinder are equal. The centre lines, g'h' and ab, are made at the same distances from the ground line, that the axis of the cylinder is from the planes of projection (6). The end elevation, knowing its radius, which is equal to half of the diameter g/e, or a'c', of the cylinder, may be made by revolving the projection of its centre a,g', only, upon PQP', around P'Q as an axis. 41. Standing with a horizontal cylinder before one, with its axis lying from right to left, and parallel to the ground line, one of its eleinents is its highest one, that is the highest above the ground, or the horizontal plane; another is the lowest; another, the foremost, that is the one nearest to onIe, and another the hinzcdmost, or the one furthest from one. Transfering the same terms to the projections of the same elements, by (23) we have ab-a' b' —b" [the three prlojections of] the highest element; ab-c'cl' —d", the lowest element; gh-g'h'-h", the foremost element; and ef-g'h' —f", the hindmotost element. In inking, the end elevation, b"f"d", is made heavy at 9?.f"d"1)p, and tapered into a fine line in the vicinity of n and p; because, by (16) when the observer turns to face the plane PQP', looking at it in the direction hg (12), the light turns with him. 42. We have now three ways of distinguishing the projections of a horizontal cylinder from those of a square prism of equal dimensions. First, by medium instead of fully heavy lines on ef and c'dt. Second, by the lettering of the principal elements, as just explained. Third, and most clearly, by the circular end elevation. ~ III -Projections showi~ng two sides of a Solid Bight Angle. 43. A solid right angle is an angle such as that at any corner of PROJECTION, OF SOLIDS. 15 a cube, or a square prism, and is therefore bounded by three plane right angles. When two faces of such a body are seen at once, they will be seen obliquely, and neither will appear in its true size. Hence only one of the projections of the object will show two of its dimensions in their real size. Hence, we must always mtake, first, that projection, whichever it be, which shows two dinmesions in their real size. 44. PnoB. 14.-To construct the plan acnd two elevations of a vertical prism, with a square base; resting on the horizontal plane, and having its vertical faces inclined to the vertical plane of 2rojection. PI. II., Fiigs. 21-22. Ist. ABCG is the plan, which must be made first (43) and with its sides placed at any convenient angle with the ground line. 2d. A'B'D'E' is the vertical projection of that vertical face whose horizontal projection is AB. 3d. B'C'E'F' is the vertical projection of that face whose horizontal projection is BC. This completes the vertical projection of the visible parts of the prism, when we look at the prism in the direction of the lines CF', &c. 4th. Let gb be the horizontal trace of an auxiliary vertical plane of projection, which is perpendicular to both of the principal planes of projection. In looking perpendicularly towards this plane, i.e. in the directions Gg, &c., AG and AB are evidently the horizontal projections of those vertical faces that would then be visible; and the projecting lines, Gg, Aa, and Bb determine the widths ga and ab of those faces as seen in the new elevation. Now the auxiliary plane gb is not necessarily revolved about its vertical trace (not shown), but may just as well be taken up and transferred to any position where it will coincide with the primitive vertical plane; only its ground line gb must be made to coincide with the principal ground line, as at H"E". Hence' making H"D" and D"E" respectively equal to ga and ab, and by drawing HU"G", &c., the new elevation will be completed. 45. Tile two elevations-P1. II., Figs. 21, 22-appear exactly alike, but the faces seen in Fig. 22 are not the same as the equal ones of Fig. 21. The different projections of the same face may be distinguished by nmrks. Thus the surfaces marlked # are the two elevations of the same face of the prism; the one marked ~ is visible only on the first elevation, and the one marked x is visible only on the second elevation-Fig. 22. 16 PROJECTIONS OF SOLIDS. 46. P1. II., Fig. 23, represents a small quadrangular prism in two elevations, the axis being horizontal in space, so that the left hand elevation shows the base of the prism. In the practical applications of this construction, the centre, s, of the square projection is generally on a given line, not parallel to the sides of the square. Hence this construction affords occasion for an application of the pl)oblem: To draw a square of given size, with its centre on a given line, and its sides not parallel to that line. The following solution should be carefully remembered, it being of frequent application. Through the given centre, s, draw a line, L, in any direction, and another, L', also through s, at right angles to L. On each of these liles, lay off each way from s, half the length of a side of the square. Through the points thus formed, draw lines parallel to the lines L and L' and they will form the required square whose centre is s. 47. PROB. 15. — o construct the plan and several elevations of a vertical hexagonal prism, which rests upon the horizontal plane of projection. P1. II., Figs. 24, 25, 26. The distinction between bodies as seen perpendicularly, or obliquely, becomes obscure as we pass from the consideration of bodies whose surfaces are at right angles to each other. Figs. 24 and 25 show a hexagonal prism as much in right projection as such a body can be thus shown, btit, as in both cases a majority of its surfaces are, considered separately, seen obliquely, its construction is given here. In Fig. 24 the hexagonal prismn is, as shown by the plan, placed so that two of its vertical faces are parallel to the vertical plane of projection. Observe that where the hexagon is thus placed, three of its faces will be visible, one of them in its real size, viz., BC, B'C'F'G', and that the extreme width, E'H', of the elevation, equals the diameter, AD, of the circumscribing circle of the plan. This is therefore thle widest possible elevation of this prlism. Notice, also, that as BC equals half of AD, while AB and CD are equlal, and equally inclined to the vertical plane, the elevations, A'F' and G'D', of these latter faces, will be equal, and each half as wide as the middle face. This fact enables us to construct the elevation of a hexagonal prism situated as here described, without constructing the plan, provided we know the width and height of one face of the prism. This last construction should be remembered, it being of frequent and convenient application in the drawing of nuts, bolt-heads, &c., in machine drawing. ID IL. L PL.. 20. I~ ~~'...e A........' El~ ~ ~~i L 3 2'" -A'C I is' I IF' ", T lb- 1311. 0 2L @' 6"'i 43" i4; I]E~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~' C,' H~~~~~~~~~~~i 11"53XN'_ b~~~~~~~~~~~~4! ~ l~r g ~~F..r~ r~r!, I~~~~~~~I' 113 ______'a — sI ~~ 93', ~ —' —",S~~~~~i I x 1 I S 1) _~~~~~~~_____? j-.... FG 1liK I""I" I" a " a'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ - ~ - --.... — II A~ i'! /::............. ~! CC"\jj —- - 1' / B 97,~~~~~~ PROJECTIONS OF SOLIDS. 1 7 48. P1. II., Fig. 25 shows the elevation of the same prism on a plane which originally was placed at ib, and perpendicular to the horizontal plane; whence it appears, that if a certain elevation of a hexagonal prism shows three of its faces, and one of them in its fiill size, another elevation, at right angles to this one, will show Ibut two faces, neither of them in its full size; the extreme width. I"B", of the second elevation being equal to the diameter of the inscribed circle of the plan. This is therefore the narrowzest possible elevation of this prism. 49. P1. II., Fig. 26 showts the elevation of the same prism as it appears when projected upon a vertical plane standing on jb"', an(d then transferred to the principal vertical plane, at Fig. 26. In this elevation, none of the faces of the prism are seen in their true size. The auxiliary vertical plane, on jb"', could have been revolved about that trace, directly back into the horizontal plane, causing the corresponding elevation to appear in the lines Dr!, &c., produced to the left ofjb" as a ground line. Elevations on auxiliary vertical planes can always be made thus, but it seems more natural to see them side by side above the principal ground line, by transferring the auxiliary planes as heretofore described. 50. Fig. 27 represents two elevations of a hexagonal prism, placed so as to show the base in one elevation, and three of its faces, unequally, in tile other. The centre of the elevation which shows the base, may be made in a given line perpendicular to o'g', by placing the centre of the circumscribing circle used in constructing the hexagon, upon such a line. Having constructed this elevation, project its points, ab, &c., across to the other vertical plane, P', which is in. space perpendicular to the plane, P, at the line, o'g'. By representing the elevation on P' as touching o'g', we indicate that the prism, touches the plane, P, just as the elevation in Fig. 24, indicates that the prism there shown rests upon the horizontal plane. 51. PROB. 16. —To constrrct the plan and tawo elevations of a pile of blocks of equal widcths, but of different lengths, so placed as to form a symmetrical body of uniform width. P1. III., Fig. 28. Here for example afg is the plan of the lowest step; kbe is that of the middle step, and cclh that of the upper step (43). The auxiliary vertical plane of projection, perpendicular to the horizontal plane at k"f"'", is made to coincide with the principal vertical plane by direct revolution. The point a"'a"", the projec. iS PROJECTIONS OF SOLIDS. tion of aaG' on the auxiliary vertical plane, revolves in a horizontal aric, of which a"'a" is the horizontal, and a""a" the vertical projection (3.1), giving a"/, a point of the second elevation. Other points of this elevation are found in the same way. This figure differs from Figs. 18 and 20, of' Plate II., only in presenting more points to be constructed. If the student finds any difficulty with this example, let him refer to those just mentioned, and to first principles. ExAMIPLE.-Construct an elevation on a plane parallel to caf. 52. PROB. 17. —To construct the vertical projection of a vertical circle, seen obliqutely. P1. III., Fig. 30. Let BF be the givenz projection of the circle. It is required to find its vertical projection, A'B'D'F'. For this purpose, the circle must be first brought into a position parallel to a plane of projection, since we can then make both of its projections, and hence can then take both projections of any point upon it. Let the circle be made parallel to the vertical plane. To do this, it only need be revolved about any vertical axis. In the figure, the axis is the vertical tangent, F-f'F'. After this revolution, the projections of the circle are bF-b'c'F'h'. Now taking any point on this circle, as ac', it returns about the axis F-/'F' in the horizontal are aA —a'A' (31), giving A' by projecting A upon a'A'. Likewise bb' returns in the are bB-b'B to BB'; and cc', which is vertically under caa', returns in the arc cC-c'C' to CC'. Thus all the points A',B', C', etc., being found and joined, we have A'B'C'H', the required oblique elevation of the vertical circle FB. Ex.MPLEs.- st. Let the circle be revolved about its vertical diameter HD, or any vertical axis between F and B. 2d. About any vertical axis; in the plane BF produced; or only parallel to it. 3d. Let the circle be made perpendicular to the vertical plane, and oblique to the horizontal plane. 53. PRoB. 18. — o construct the projections of a cylinder whose convex surface rests on the horizontal plane, and whose axis is inclilned to the verticcal plcce. PI. III., Fig. 31. As may be learned friom Fig. 19, Pl. II., the projection of a right cylinder upon any plane to which its axis is parallel, will be a rectangle. Therefore let CSTV, P1. III., Fig. 31, be the plan of the cylinder. Since it rests upon the horizontal plane, q'/t', in the ground line, is the vertical projection of its line of contact with that PROJECTIONS OF SOLIDS. 19 plane, and p'A' is the vertical projection of pA, the highest element of the cylinder, as it is at a height above the ground line, equal to the diameter, TV, of the cylinder. The vertical projection of either base may be found by the last problem. In the figure, the left hand base is thus found, and the construction, being fully given, needs no further explanation. 54. The vertical projection of the right hand base TV is found somewhat differently. It is revolved about its horizontal diameter, \TV —T'V', till parallel to the horizontal plane. It will then appear as a circle, and a line, as 9n"n, will show the true height of n above the diameter TV. So, also, o"o will show the true distance of o belw TV. Therefore the vertical projections of the points n and o, will be in the line n-n', perpendicular to the ground line, and at distances above and below T'V', the vertical projection of TV, equal, respectively, to nn" and oo". Having, in the same manner, found r' and t', the vertical projections of two points whose cominon horizontal projection t-r is assumed, as was n-o, the vertical projection of the base TV can be drawn by the help of the irregu1ar curved rutler. 55. In the execution of this figfure, SV is made slightly heavy, and TV fully heavy, and the portion, n'T't', of the elevation of the right hand base, and the small portion,D'Wu', of the left hand base, arle made heavy. Suffice it to say: First. That a part of the convex surface is in the light, while the right hand base is in the dark. Second. n'T't' divides the illuminated half of the convex surface, firom the base at the right, which is in the dark; and D'u' divides the illuminated left hand base from the visible portion of the darkened half of the convex surface (18-20). EXAMPLE. Let the axis of the cylinder be parallel to the vertical plane, only. 56. PROB. 19. To construct the two projections of a right cone, with a circular base in the horizontalplane; and to construct et]ther projection of a line, drawn from the vertex to the circum..ference of the base, havcing the other projection of the same line given. PI. III., Fig. 32. _Remark. When the axis of a cone is vertical, perpendicular to the vertical plane, or parallel to the ground line, the cone is shown in right projection as much as such a body can be, but as all the straight lines upon its sulfate are then inclined to one or both p!anes of projection, the above problem is inserted here among problems of oblique projections. 20 PROJECTIONS OF SOLIDS. Let VB be the radius of the circle, wlich, with the point V, is the horizontal projection of the cone. Since the base of the cone rests in the horizontal plane of projection, C'B' is its'vertical projection. Since the axis of the cone is vertical, V', the vertical projection of the vertex, must be in a perpendicular to the ground line, through V, and may be assumed, unless the height of the cone is given. V'C' and V'B', the extreme elements, as seen in elevation, are parallel to the vertical plane of projection, hence their horizontal projections are CV and BV, parallel to the ground line (8 e). Let it be required to find the horizontal projection of any element, whose vertical projection, V'D', is given. V is the horizontal projection of V', and D', being in the circumference ot the base, is horizontally projected at D, therefore VD is the hoi-. zontal projection of that element on the fiont of the cone, whose vertical projection is V'D'. V'D' is also the vertical projection of an element behind VD, on the back of the cone. Having given, VA, the horizontal projection of an element of the cone, let it be required to find its vertical projection. V' is the vertical projection of V, and A, being in the circumference of the base, is vertically projected at A'. Therefore V'A' is the required vertical projection of the proposed line. In inking the figure, no part of the plan is heavy lined, and in the elevation, only the element V'B' is slightly heavy. EXAMPLES. —st. Construct three projections of a cone placed as the cylinder is in Prob. 13. 2d. As the cylinder is in Prob. 18. 57. PROB. 20. To construct the projections of a right hexagonal prism; whose axis is oblique to the horizontal plane, and parallel to the verticalplane. P1. III., Figs. 33, 34. 1st. Commence by constructing the projections of the same prism as seen when standing vertically, as in Fig. 33. The plan only is strictly needed, but the elevation may as well be added here, for completeness' sake, and because some use can be made of it. 2nd. Draw J'G",- making any convenient angle with the ground line, and set off upon it spaces equal to G'J', J'H', and J'I', from Fig. 33. 3rd. Since the prism is a right one, at J", &c., draw perpendiculars to J"G", make each of them equal to J'C', Fig. 33, and draw F"C", which will be parallel to J'G", and will complete the second elevation. 4th. Let us suppose that the prism was moved from its first PROJECTIONS OF SOLIDS. 21 position, Fig. 33, parallel to the vertical plane, and towards the right, and then inclined, as described, with the corner, CJ', of the base, remaining in the horizontal plane. It is clear that all points of the new plan, as B"', would be in parallels, as BB"', to the ground line, through the primitive plans, as B, of the same points. It is equally true that the points of the new plan will be in perpen. diculars to the ground line through the new elevations B1', &c., of the sanme points (15), hence these points B"', &c., will be at the intersections of these two groups of lines. Thus, A"' is at the Intersection of AA"' with A"A"'; C"' is at the intersection of CC!' with C"C"'; K"' is at the intersection of IDK"' with 1"K"', &c. 5th. B"'C"', F"'E"', and G"''", being the projections of lines of the prism which are parallel in space, are themselves parallel. A similar remark applies to C"'D"', A"'F, and H"'G"', Observe, that as the upper or visible base is viewed obliquely, it is not seen in its true size, F"'C"' being less than FC, Fig. 33; so that this base A"'C"', E"', does not appear in the new plan as a regular hexagon. 58. PRon. 21. To conzstriect the projections of tlhe jprismn, given in, the previous problem, when its edges are inclined to both planes of projection. P1. III., Fig. 34a. If the prism, P1. IIl., Fig. 34, be moved to any new position, such that the inclination of its edges to the vertical plane, only,,shall be changed, the inclination of its edges to the horizontal plane of projection being unchanged, the new plan will be merely a copy of the second plan, placed in a new position. Let the particular position chosen be such that the axis of the prism shall be in a plane perpendicular to the ground line, i.e. to both planes of p)ojection; then the axis of symmetry, C"'G"', of the second plan, will take the position C""G"", and on each side of this line the plan, Fig.. 34 a, will be made, similar to the halves of the plan in Fig. 34. As the prism is turned horizontally about the corner J", and then transferred, producing the result that the inclination of its axis to the horizontal plane is unchanged, all points of the third elevation, as A""', C""', &c., will be in parallels to the ground line through A", C/, &c., and in perpendiculars to the ground. line, through A"", C"", &c. By examination of this solution, and by inspection of Figs. 34 and 34a, it appealrs that a change in the position of the axis, with reference to but one plane of projection at a time, can be 22 PROJECTIONS OF SOLIDS. represented directly from projections already given; also that a curve, beginning with the first plan, and traced through the six figures composing the three given pairs of projections in the order in which they mfust be made, would be an S curve, ending in the third elevation. 59. Erxecution. —The full explanation of the location of the heavy lines cannot here be given. The careful inquirer may be able to satisfy himself that the heavy lines of the figures, as shown, are the pr)ojections of those edges of the prism which divide its illuminated fiom its dark surfaces. 60. PROB. 22. To construct the projections of a regular hexagornal pyramid, whose axis is inclined to the horizontal plane only. P1. III., Figs. 35, 36. 1st. Commence, as with the prism in the last problem, by repre. senting the pyramid as having its axis vertical. 2nd. D raw a"d", equal to a'd', and divided in the same way. At n", the middle point of a"d", draw n"V" perpendicular to a"d", and make it equal to n'V', which gives V" the new elevation of the vertex. Join V" with a", b", c", and d", and the new elevation will be completed. 3rd. Supposing the same translation and rotation to occur to the primitive position of the pyramid, that was made in the case of the prism (57, 4th), the points of the new plan, Fig. 36, will be found in a manner similar to that shown in Fig. 34. V"' is at the intersection of VV"' with "V"V'; c"' is at the intersection cc"' with c"c"'; d"' is at the intersection ofdd"' with cl"d"', &c. 4th. The points, a"'b"'c"'... f"', of the base, are connected with V"', the new horizontal projection of the vertex, to complete the new plan. If the pyramid were less inclined, the perpendicular V"V"' would fall within the base, and the whole base would then be visible in the plan. As it is,.f "'a"' and a"'b"' are hidden, and tlherefore dotted. 5th. The heavy lines are correctly placed in the diagram; also the partially heavy lines, which are all between V"'d"' and the ground line, but the reasons for their location cannot here be given, beyond the general principle (18-20) already given. 61. PROB. 23. To construct the projections of the regular hexagonal pyramid, when its axis is oblique to both planes of projec. tion. P1. III., Fig. 36a. Suppose the pyramid here shown to be the one represented in . CL' c 7I.7.VH' Vt - -V A / /, /9,,~f 19~~~~~~~~~~~~~~~~~~~~~~~:~:............ V ~ ~ ~ K ~ p ~~~,,,~~~~~~I /~7,,,,A A_~,. _V,,, 72~ ~~~ ~~~~~~+!~,, 1:a, 1-01, ~'-9_ /[~,~ ~~ ~ ~~ ~ ~~~ ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~.~, f,):]:,' - I " D \ // --.,,, ~.' -'.....~~c A,?'......,,z. 73,P'?... ~~~~~~~~~~~~~~~~~~~~~~~~~~~~.__o III~~~~~~~~~~~~~-,J.I /A ( - ELEMENTARY INTERSECTIONS. 23 figures 35 and 36, and suppose that it has been turned about any vertical line as an axis. Then,,first, every point of it will move horizontally; second, every point will hence remain at the same height as before; third, therefore, the inclination of all the edges to the horizontal plane will be unchanged; and hence, fourth, the new plan, Fig. 36a, will be only a copy of the second plan, Fig. 36, placed so that its axis of symmetry, V""d"", shall make any assumed angle with the ground line. By (second) and (15) the points, as V"", of the third elevation, will be at the intersection of parallels to the ground line, through the corresponding points, as V", of the second elevation, with perpendiculars through the samne points, as V"/", seen in the third plan. Observe that the two points vertically projected. in c", being at the same height above the ground line, will appear in the third elevation at c""' and e""', in the same straight line, through c", and parallel to the ground line. (32). Remembering also that lines which are parallel in space must have parallel projections, on the same plane, c""'d""' will be parallel to f""'ca"'", &c. The heavy lines are indicated in the figure. ExAMrPLE.-Constiruct Fig. 36a from Fig. 36, without a new plan, by taking a new vertical plane with its ground line parallel to V""d"", and revolving it directly back as mentioned in (49). IV. —Special Elementary Intersections and _Developments. 62. The positions of other planes, than those of projection, are indicated by their intersections with the planes of projection. These intersections are called traces. A plane can cut a straight line in only one point; hence, if a plane cuts the ground line at a certain point, its traces, both being in the plane, must meet in that point. Ill PI. I., Fig. 5, bBb'k' is a plane per9pendicular to tAe ground line, MG, and, therefole, to both planes of projection, and we see that its two traces, bk' and b'k', are perpendicular to the ground line at k'. Likewise in P1. I., Fig. 15, Aaa't is a plane perpendicular to the ground line MQ, and its traces at and a't are perpendiculfar to MQ. That is: if a plane is perpendicular to the ground line, its traces will also be perpenclicular to that linle. This is seen in regtllar projection, in P1. I., Fig. 16, where PQ is the horizontal trace, and P'Q, the vertical trace, of such a plane. In P1. I., Fig. 5, FKfk is a plane, parallel to the vertical plane, and it has only a horizontal trace, fk, which is parallel to the ground line. The same is true for all such planes. Likewiso, 24 ELEzMENPXrrY INTERSECTIONS. ABa'b' is a orizontalplane. All such planles have only a vertical race, as a'b', parallel to the ground line. In Pl. I., Fig. 2, the plane Vv'd'd is perpedicular only to the certicag plaze, and, as the figure shows, the horizontal trace only, as dd', of such a plane, is perpaendicar to the ground line. Also the angle v'd'b', between the vertical trace, v d', and the ground line, is the angle made by the plane with the horizontal plane. In like manner, it can easily be seen that, if a plane be perpendicular ony to the horizontal plane, as in case of a partly open door, its vertical trace only (the edge of the door at the hinges) will be perpendliczuar to the grouad izew, and the angle between its holiniontal trace and the ground line, will be the angle made by the plane with the vertical plane of projection. Finally, if a plane is oblique to both planes of projectionz, both of its traces will be oblique to the ground line, and at the same point. Thus, PL. I., Fig. 6, may represent such a plane, having LF fbi its horizontal, and L'F for its vertical trace. All the principles just stated can be simply illustrated by taking a book, half open, for the planes of projection, and either o-f the triangles for the given movable plane; and when clearly understood, the following problems can also be easily comprehended. Pnon. 24. —2Tofd the curve of intesrsection of a cylinder with a plane. P1. IIL. A., Fig. 1. Let the cylinder, ADBG-,A'B", be vertical, and the cutting plane, PQP', be perpendicular only to the vertical plane. All points in such a plane must have their vertical projections (that is, must be vertically projected) in the vertical trace, QP', of the plane, but the required curve must also be embraced by the visible limits, A'A" and B'B'", of the cylinder. Hence, a'b' is the vertical projection of this curve. Again, as the cylinder is vertical, all )points oal its convex surface must be hlorizontally projected in ADBG. Hence, this circle is the horizontal projection of tle required curve. PRoB. 25. —To revolve the curvefound in the lastproblem, so as to show its true size. When a plane revolves about any line in it as an axis, every point of it, not in the axis, moves in a circular arc, whose radii are all perpendicular to the axis. The representation of the revolution is much simplified by taking the axis in, parallel to, or perpendicutar to, a plane of projection (31). ELEMENTARY INTERSECTIONS. 25 Let AB-a'b', the longer axis of the curve, and which is parallel to the vertical plane of projection, be taken as the axis of revolution. The curve may then be revolved till parallel to that plane, when its real size and form will appear. Then, at c', d', &c., the vertical projections of C and H, D and G, &c., draw perpendiculars, as c"h", to a'b', and make c'c"=c'h"-=nC. Proceed likewise at d', &e., since the lines, as nC, being parallel to the horizontal plane, are seen in their true size in horizontal projection; and join the points a'h"g", &c., which will give the required true form and size of the curve of intersection before found. EXAMPLE.-This curve is an oval, called an ellipse. Its true size could have been shown by revolving its original position about DG as an axis, till parallel to the horizontal plane. The student may add this construction to the plate. PROB. 26.-To develope the portion of the cylinder, P1. III. A., Fig. 1, below the cutting plane, PQP'. The convex surface of a cylinder is wholly composed of straight lines, called elements, parallel to its axis. The convex surface of a cone is composed of similar elements, all of which meet at its vortex. Hence, each of these surfaces can evidently be rolled upon a plane, till the element firlst placed in contact with the plane, returlns into it again. The figure, thus rolled over on the plane, is called the development of the given convex surface, and its area equals the area of that surface. Suppose the cylinder to be hollow as if made of tin, and to be cut open along the element B'b'. Then suppose the element A'c' to be placed on the paper, as at A'a', Fig. 2, and let each half' be rolled out upon the paper. The part ADB will appear to the left of A'a', and the part AGB, to the right. The base being a circle, perpendicular to the elements, will develope into a straight line B'B", Fig. 2, found by making A'c=AC, Fig. 1, cd=CD, Fig. 1, &e., and A'h:AH, Fig. 1, &c. B'B" may also, for convenience, be A'B', Fig. 1, produced.. Then the parallels to A'a', through c, d, &c., will be developments of elements standing on C, D, &c., Fig. 1, and by projecting over upon them, a' at a', c' at c' and Ah' d' at d' and g',... B' at b' and b", and joining the points, the figure B'B"b"a'b', will be the required development of the cylinder. 1Renark.-If, now, a flat sheet of metal be cut to the pattern just found, it will roll up into a cylinder, cut off obliquely as by the plane PQP'. By making the angle-P'QA' of any desired size, the corresponding flat pattern can be made as now explained. 26 ELEMENTARY INTERSECTIONS. PROB. 27.-.To,find the intersection of a vertical cone, with a plane, per2pendicular to the vertical plane of projection. P1. III. A, Fig. 3. Let V-ADBC be the plan, and A'B'V' the elevation of the cone, and PQ and P'Q the traces of the given cutting plane; whose horizontal trace, PQ, shows it (62) to be perpendicular to the vertical plane. For the reasons given in Problem 25, a'b' will be the vertical projection of the required curve. The convex surface of the cone not being vertical, the horizontal projection of the intersection will be a curve, which must be found by constructing its points as follows..First. The method by elenents. Any line, as'V'E', is the vertical projection of two elements whose horizontal projections are VE and VF (Prob. 19). Therefore e', where it crosses the vertical projection, atb', of the intersection, is the vertical projection of two points of the required curve. Their horizontal projections, e and f, are found by projecting e' down upon VE and VF. Other points can be found in the same manner, except d and y, since the projecting line d'd coincides with the elements VD and VG. The horizontal projections of a' and b' are a and b. Seconcl. The method by circular sections. Let M'N' be the vertical trace of a horizontal auxiliary plane through d'. This plane will cut from the cone the circle nztl —G2 cn g, on which X can be projected at d and g, the points desired. Other points of' the horizontal projection can be found in the same manner. eemzcarks.-a. The curve acdbg —a' is an ellipse, whose longer axis is the line ab —ab'b, whose true length is a'b'; and whose shorter axis is the line pq-p', whose true length is pq, and which bisects pq. Since an ellipse has thus two axes of symmetry, it should be sketched by finding somne arc of the irregular curved ruler, which will fit a quarter of it, as ap. Then use the same arc to draw each of the other three quarters. b. On the cylinder, l', the middle of a'b' is on the axis O-clcl'. That is, the centre of the ellipse cut from a cylinder, is on t1he axis of the cylincder. Not so, however, with the cone; p', the middle of a'W', is not on V'D', the vertical projection of the axis, but is on the side of it towards the lowest point, bb', of the curve of intersection. On account of the acuteness of the intersections at9p and q, these points can better be found as were d and g. EXAMPLES. —1st. To make the horizontal projection less circular than in the figure, let the cone be quite flat, as at AVB, Fig. 9, and with a' near the vertex, and b' quite near the base. ELEMENTARY INTERSECTIONS. 27 2d. Find the true size of the curve by either of the ways indicated in Prob. 25, also by revolving the plane PQP', contaiining it, either, about PQ as an axis, into the horizontal plane; or, about P'Q as an axis, into the vertical plane. In the former case, it is only to be remembered that e'Q, for example, shows the trul distance of ee' from PQ; anid, in the latter case, that ek, for exam. ple, shows the true distance of ee' fi'om the vertical trace P'Q (6). PROB. 28.-To develope the convex surface of a cone, P1. III. A., Fig. 4, together with the curve of intersection, found in the last problem. -First. If the element VB —V'B' be placed in contact with the paper at V'B', and if the cone be then rolled, upon the paper till this element returns into it again, as at V'B", the development, V'B'B", will be made. As all the elements are equal, and as the vertex is stationary, the development of the base will be the are B'B", with a radius equal to V'B', the cone's slant height, and of a length equal to the circumference ADBG. Tbils length is found, as in case of the cylinder, by taking equal arcs of the base, so small that their chords shall be sensibly equal to thenm, and laying off those chords from B', on the are B'B", till B3" is located. Thus, BE being one eighth of ADBG, its length is laid off as at B'e" eight times to find B". Second. To show the curve, acdby —a'b', on the development, consider that only the extreme elements, as V13-V'B', show their true length in projection. Hence, the points between a' and b' must be revolved around the axis of the cone, into these elements, in order to show their true distances from the vertex. This axis being vertical, the arcs of revolution will be horizontal, and will therefore be vertically projected in the horizontal lines c'u, d'n, &e., and Vu, Vn, &c., will be the true distances of c', d', &c., from the vertex. Hence, makeV'a'" —V'a'; Vu', and V'u"'=V'u;YV'n', an(I V'n" =Vzn, &c, and the curve b'ct"b" will be the development of the intersection of the plane PQP' with the cone. 1?Remalrk.-The remarks made upon the development of the cylinder equally apply here. PROB. 29. —To find thme intersection of a velrtical cylinder with two horizontal ones; their axes being in a plane parallel to the vertical plane oof projection. P1. II[. A, Fig. 5. ABE-A'B'A"B" is the vertical cylinder, and MNQR —O'P'O' P. the lower horizontal cylinder. >28 ~ EELEMENTARY INTERSECTI ONS. First. To find the highest and lowest, and foremost and hind. most points of the intersection. Since the horizontal cylinder is the smaller one, it will enter the vertical cylinder on one side, and leave it on the other, giving two curves; but as one cylinder is vertical, and the intersection, being common to both, is on it, the horizontal projections of both curves are known at once to be CAE and DBF. Now A is the horizontal projection of both the highest and lowest points of CAE. Their vertical projections are a" and a'. Also C and E are the horizontal projections of the fore. most and. hindmost points, and c', on M'N', midway between O and 0", is the vertical projection of both of them. (41.) In like manner b', b' and d' are found. Second. To find other intermediate points. Take the two points whose horizontal projection is G, for example. They are on the horizontal elements, one on the upper, and the other on the lower half of the horizontal cylinder, and whose horizontal projection is ST. But to find their vertical projections, we must revolve one of the bases, as MQ, till parallel to a plane of projection. Let this base revolve about its vertical diameter, O-O'O", till parallel to the vertical plane, when OM" —O'M.'"O" will be the vertical projection of its fiont half. In this revolution the points, S, revolve to S/, and will thence be vertically projected at U' and S'. In counter revolution, these points return in horizontal arcs to u' and s', and u'v' and s't' are the vertical projections of the elements ST. Hence, project G, and also H, at g' and g#, h' and h/, and we shall have four more points of intersection. Any number of points can be similarly found. ExAMPLES.-lst. The last four points could as easily have been found by revolving the base MQ about.the horizontal diameter IQ MI', till parallel to the horizontal plane. This construction is left for the student. 2dl. If the axes did not intersect each other, as at II', the points C and E would not be equidistant from OP, and would not have one point, c', for their vertical projection, and the vertical projection of the back half, as AE, of each curve would be a dotted line, separate firom the same projection of the friont half. The student may construct this case, also that where one of the elements, MN, or QR, does not intersect the vertical cylinder. 3d. The horizontal cylinder, Fig. 6, shows that when the two cylinders, placed as before, are of equal diameter, the vertical projections of their curves of intersection are straight lines. Hence, each of the curves themselves is contained in a plane, th'at is, it is ELEM:ENTARY INTERSECTIONS. 29 a "plane curve." This figure, if regarded separately, as a plan view, therefore may represent the plan of the intersection of two equal semi-circular arches, and the curves, KL and AY, of intersection, will be ellipses. The curves on the cylinders in Fig. 5 cannot be contained in planes. Such curves are said to be of double curvature. 4th. By developing the cylinders, in Figs. 4 and 5, as in Fig. 2, the patterns may be found which will give intersecting sheet metal pipes, when rolled up in cylindrical form. The student should construct these developments, also the case in which the vertical cylinder should be the smaller one. PROB. 30.-Tofind the intersection of a hbrizontal cylinder with a vertical cone. P1. III. A, Fig. 7. Let ABV be the vertical projection of a cone, and let the circle with radius oa, be an end view of the cylinder; its axis, o'o", intersecting A'Y', that of the cone. Let PQ be the vertical trace of a second vertical plane, perpendicular to the ground line, as in PI. I., Figs. 15 and 16, and let V'E'D be the vertical projection of the cone, and G'G'N'N" that of the cylinder, on this plane. In this construction, therefore, two vertical projections are employed, instead of a horizontal and vertical projection, for any two projections of an object are enough to show its form and position. This will more readily appear by turning the plate to bring VAB below PQ, when PQ will be the ground line, the right hand projection a plan, and the left hand one an elevation, like Fig. 5. Now to find the intersection. Speaking as if facing the vertical plane of projection, represented by the paper to the left of PQ, after revolving that plane about PQ into the paper, AV-A'V' is the foremost element, and a' is found by projecting a across upon A'V'. Next, DV is the right hand projection of two elements, whose left hand projections are EV and DV. We therefore project G at g' and e'. To find intermediate points. Assume any element FV, draw FF" perpendicular to AB, then make an are, AF", of the plan of the cone's base, and make A'F' = A'H' OFF". Then V'F' and M'IT' will be the left hand projections of the two elements project. edinFV. Then projectf atf'and h' on these elements, andg'a'h/'e' will be the visible part of the intersection. Its right hand projee. tion is afG, wheref and G are, each, the projectiou of two points on opposite sides of the cone. 30 ELEMENTARY INTERSECTIONS. EXAMIPLE.-By developing the cone and the cylinder, patterns could be made for a conical pipe entering a cylindrical one. 63. Observing that, in every case, the auxiliary planes are made to cut the given curved surfaces in the simplest manner, that is, in straight lines or circles, we have the following principles. To cut right lines, at once, firom two cylinders, as in Fig. 5, aplane must be parallel to both their axes. To cut a cylinder and cone, at once, in the same manner, as in Fig. 7, each plane must contain the vertex of the cone, and be parallel to tkhe axis of tfhe cylin der. To cut elements at once fiom two cones, a plane must simply contain both vertices. EXAMPLES. — st. Thus, in Fig. 10, all planes cutting elements, both from cone VV', and cone AA', will contain the line VAB, hence their traces on the horizontal plane will merely pass through B. Thus the plane BD cuts fromthe cone, V, the elements V'cta-Va, and V'c'-Vc; and finom the cone, A, the elements A'D' —Ad, and A'd'-AD. The student can complete the solution, the remainder of which is very similar to the two preceding. 2cl. To find the intersection of a sphere and cone, Pi. III. A., Fig. 11, auxiliary planes may most convenientlybe placed in two ways. First, horizontally. Then each will cut a circle fiom the sphere, and one from the cone; whose horizontal projections will be circles, and whose intersections will be points of the intersection of the cone and sphere. Second, vertically. Then each plane must contain the axis of the cone, from which it will cut two elements. It will also cut the sphere in a circle, and by revolving this plane about the axis of the cone till parallel to the vertical plane, as in Prob. 17, the intersection of the circle with the revolved elements, see Prob. 27, may be noted, and then revolved back to their true position. The student can readily make the construction, after due familiarity with precedingproblems has made the apprehension of' the present article easy. PROB. 31.- TO find the intersection of a vertical hexagonal prism with a sphere, whose centre is in, the axis of the prism. P1. III. A, Fig. 8. Let O-ABC be part of the sphere, and DGHK the prism, showing one face in its real size, and therefore requiring no plan (47). Draw dg parallel to AC, and the are eAf with O as a centre, and through e andf: This arc is the real size of the intersection of the middle face of the prism with the surface of the sphere. All the faces, being equal, have circular tops, equal to ehf; but, being c"zI- ~Z ~l- - fZ~- --— p~ ___ ---- L1TA B") 9 AjF -~i.~'I /! i -;"',1'-x,a -~;' 0', ____;-,. —'_____/___-'" k;'t.:-' 0. J' i —!- r~'B d c B- h I ('' -/ / \ j''~~~~~~~~~~~....,,' \~~~ II' C"N1'I ", jI /,- ~~~ ~'%'',,,I,(j ~..-, /.__:'_ —.,,.....'-....................... tl:~~~~~~~~~~~~~~~~~~~~~!l I~~~NS~ I 7 - i............... ". -- --- l- 4 —.. /,'. -' -''_ __"__'".~'- I"/:''.,' / ___,,,.' -i _.., i_' I'!~ ~ ~~~,,....,,,' /,,, / < ~~~~~~~~~~~~~~~1 i:~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ r' ~/ / ir "',,,'V ~ ~ ~~ ~ ~ ~ ~~~~ ~ ~ ~ ~ ~ ~ ~~~~~~~~~~~~~~~-' \' "'. /,'~,'fj It AJ~~~~~~~~~~~~~~~~~~~~~~' u I/i K'Ii F _ -- ~~~~~~~~~~~~~~~~~~~; —' — K-" —~,~~ ar'~~ ~ ~ ~ -'''::, —'.-''..... — -.. ~ —- 1 i -, I iii ~~A I r - I e S ~ ~~~~~~~ // I!- i -:i i __ ____ ___ _ __ ____ __.__ _ I... A C~. O D i: _=-_______:____ _____= -. — ii_.___._._j jj /,,,~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~Te',. 1?iip[n.."i.~L't.:,.P:' i~:. ELEMENTARY INTERSECTIONS. 31 seen obliquely, they would be really elliptical in projection. It is ordinarily sufficient, however, to represent them by circular arcs, tangent to hn, the horizontal tangent at h, and containing the points d and e, andf/ and g, as shown. Remark. —The heavy lines here, show the part of the prism within the sphere, as a spherical topped bolt head. To make 1)d=EF, draw Od at 450 with AC, to locate dg. To make the spherical top flatter, for the same base DG, take a larger sphere, and a plane above its centre for the' base of the prism. PROB. 32.-To construct the inter.section of a vertical cone with a vertical hexagotal prism; both having the same axis. P1. III. A, Fig. 9. Let VAB be the cone, and CFGH, the prism, whose elevation can be made without a plan (48), since one face is -seen in its real size. The semicircle on cf is evidently equal to that of the circumscribing circle of' the base of the prism, and ct is the chord of two thirlds of it. Then half of ct, laid off on either side of 0, the middle of CF, as at On, will give np,the projection of the middle face EDd after turning the prism 90~ about its axis. This done, np will be the height, above the base, of the highest point at which this and all the faces will cut the cone. A vertical plane, not through the vertex of a cone, cuts it in the curve, or " conic section," called a hyperbola. The vertical edges of the prism cut the cone at the height Ff, hence, drawing the curves, as dse, sharply curved as at s, and nearly straight near d and e, we shall have a sufficiently exact construction of the required intersection.:Remark.-The heavy lines represent the part of the prism within the cone, finished as a hexagonal head to an iron " bolt," such as is often seen in machinery. The horizontal top, hy, of the head, may be drawn by bisecting pr at g. To make Cc=ED, as is usual in practice, simply draw Oc at an angle of 450 with AB, to locate ef. Bymaking VAB-_30~ perhaps the best proportions will be found. 64. In the subsequent applications of projections in practical problems, the ground line is very generally omitted; since a knowledge of the object represented makes it evident, on inspection, which are the plans, and which the elevations. DIVISION SECOND. DETAILS OF CONSTRUCTIONS IN MASONRY, WOOD, AND METAL CHAPTER I. CONSTRUCTIONS IN MASONRY. ~ I. —General Definitions and Principles applicable both to _Brick and Stone-oork. 65, A horizontal layer of brick, or stone, is called a coterse. The seam between two courses is called a coursing-yoint. The seam between two stones or bricks of the same course, is a vertical or heading-joilnt. The vertical joints in any course should abut against the solid stone or brick of the next courses above and below. This arrangement is called breaking joints. The particular arrangemenlt of the pieces in a wall is called its bond. As far as possible, stones and bricks should be laid with their broadest surfaces horizontal. Bricks or stones, whose length is in the direction of the length of a wall, are called stretchers. Those whose length is in the direction of the thickness of a wall, are called headers. ~ II.-Brick Work. 66. If it is remembered that bricks used in building have, usually, an invariable size, 8"x 4"X2" (the accents indicate inches), and that in all ordinary cases they are used whole, it will be seen that brick wvalls can only be of certain thicknesses, while, in the use of stone, the wall can be made of any thickness. Thus, to begin with the thinnest house wall which ever occurs, viz. one whose thickness equals the length of a brick, or 8 inches; the next size, disregarding for the present the thickness of mortar, would be the length of a brick added to the width of one, or equal to the width of three bricks, making 12 inches, a thickness employed in the partition walls and upper stories of first class houses, or the CONSTRUCTIONS IN MASONRY. 35 outside walls of small houses. Then, a wall whose thickness is equal to the length of two bricks or the width of four, making 16 inches, a thickness proper for the outside walls of the lower stories of first class houses; and lastly, a wall whose thickness equals the length of two bricks added to the width of one; or, equals the width of five bricks, or 20 inches, a thickness proper for the base. ment walls of first class houses, for the lower stories of few-storied, heavy manufactory buildings, &c. 67. In the common bond, generally used in this country, it may be observeda. That in heavy buildings a common rule appears to be, to have one row of headers in every six or eight rows of bricks or courses, i.e. five or seven rows of stretchers between each two successive rows of headers; and, b. That in the 12 and 20 inch walls there may conveniently be a row of headers in the back of the wall, intermediate between the rows of headers in the face of the wall, while in the 8 inch and 16 inch walls, the single row of headers in the former case, and the double row of headers in the latter, would take up the whole thickness of the wall, and there might be no intermediate rows of headers. c. The separate rows, making up the thickness of the wall in any one layer of stretchers, are made to break joints in a horizontal direction, by inserting in every second row a half brick at the end of the wall. 68. Calling the preceding arrangements common bonds, let us next consider the bonds used in the strongest engineering works which are executed in brick. These are the English bond and. the Flemish bond. The English Bond. —In this form of bond, every second course, as seen in the face of the wall, is composed wholly of headers, the intermediate courses being composed entirely of stretchers. Hence, in any practical case, we have given the thickness of the wall and the arrangement of the bricks in the front row of each course, and are required to fill out the thickness of the wall to the best advantage The Flemish Bond.-In this bond, each single course consists of alternate headers and stretchers. The centre of a header, in any course, is over the centre of a stretcher in the course next above or below. The face of the wall being thus designed, it remains, as before, to fill out its thickness suitably. 69. EXAMPLE 1. To represent an Eight Inch Wall in English Bond. Let each course of stretchers consist of two rows, side 3 34 CONSTRUCTIONS IN MASONRY. by side, the bricks in which, break joints with each other horni zontally. Then the joints in the courses of headers, will be distant half the width of a brick from the vertical joints in the adjacent courses of stretchers, as may be at once seen on constructing a diagram. 70. Ex. 2. To represent a Twelve Inch Wall in English Bond. See P1. IV., Fig. 37. In the elevation, four courses are shown. The upper plan represents the topmost course, and in the lower plan, the second course from the top is shown. The courses having stretchers in the face of the wall, could not be filled out by two additional rows of stretchers, as such an arrangement would cause an unbroken joint along the line, ab, throughout the whole height of the wall-since the courses having headers in the face, nzzust be filled out with a single row of stretchers, in order to make a twelve inch wall, as shown in the lower plan. In order to allow the headers of any course to break joints with the stretchers of the same course, the row of headers may be filled out by a brick, and a half brick-split lengthwise-as in the upper plan; or by two three-quarters of bricks, as seen in the lower plan. 71. Ex. 3. To represent a Sixteen Inch Wall in English Bond. The simplest plan, in which the joints would overlap pro. perly, seems to be, to have every second course composed entirely of headers, breaking joints horizontally, and to have the intermediate courses composed of a single row of stretchers in the front and back, with a row of headers in the middle, which would break joints with the headers of the first named courses. If the stretcher courses were composed of nothing but stretchers, there would evidently be an unbroken joint in the middle of the wall extending through its whole height. 72. Ex. 4. To represent an Eight Inch Wall in Flemish Bond. P1. IV., Fig. 38, shows an elevation of four courses, and the plans of two consecutive courses. The general arrangement of both courses is the same, only a brick, as AA', in one of them, is set six inches to one side of the corresponding brick, B, of the next course -measuring from centre to centre. 73. Ex. 5. To represent a Twelve Inch Wall in Flemish Bond. P1. IV., Fig. 39, is arranged in general like the preceding figures, with an elevation, and two plans. One course being arranged as indicated by the lower plan, the next course may be made up in two ways, as shown in the upper plan, where the grouping shown at the right, obviates the use of half bricks in every second course, ~:j'. l(;!.led'!' c U:!ly'S l,!'~u. IJdbu ~.!qr'~.1 ___~~~oa __ ___ __ I - i I 01 ~~~~~~~, __1___II__'_ _ _ _ _ _ _ _ _ _ - I 1I Z -~~~~-~ - XI TI l [ I _.; Y 1L _ _ _ _ _ _ _ q II I J I I _ 1 K! -- _l_ - - -. _______ 3~~~ t}0 i, L1 l ) i I AJ I u I.....Ir F-' —-. i I I I AI) j J - i I J -r —-- j''r-l ] ] — l] I tjl — r —l — | -— T z' ----— 0 —-lI ----:, tX ~~~ —- 1 4< - - I 0~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ ~~~~~~~~~~~~~.. ____ __._ _ I / --- X~,t IJI- tt ~.tu CONSTRUCTIONS IN MASONRY. There seems to be no other simple way of combining the bricks in this wall so as to avoid the use of half bricks, without leaving open spaces in some parts of the courses. 74. Ex. 6. To represent a Sixteen Inch Wall in Flemish Bond. P1. IV., Fig. 40. The figure explains itself sufficiently. Bricks may not only be split crosswise and lengthwise, but even thicknesswise, or so as to give a piece 8 x 4 x 1 inches in size. Although, as has been remarked, whole bricks of the usual dimlensions can only form walls of certain sizes, yet, by inserting fiagments, of proper sizes, any length of wall, as between windows and doors, or width of pilasters or panels, may be, and often is, constructed. By a similar artifice, and also by a skilfuil disposition of the mortar in the vertical joints, tapering structures, as tall chimneys, are formed. ~ III.-Stone Wobre. 75. The following examples will exhibit the leading varieties of arrangement of stones in walls. EXAMPLE 1. Regular Bond in Dressed Stone. P1. V. Fig. 41. Here the stones are laid in regular courses, and so that the middle of a stone in one course, abuts against a vertical joint in the course above and the course below. In the present example, those stones whose ends appear in the front face of the wall,. seen in elevation, take up the whole thickness of the wall as seen in plan. The right hand end of the wall is represented as broken down in all the figures of this plate. Broken stone is represented by a smooth broken line, and the under edge of the outhanging part of any stone, as at n, is made heavy. 76. Ex. 2. Irregular Rectangular Bond. P1. V., Fig. 42. In this example, each stone has a rectangular face in the front of the wall. These faces are, however, rectangles of various sizes and proportions, but arranged with their longest edges horizontal, and also so as to break joints. 77. That horizontal line of the plan which is nearest to the lower border of the plate, is evidently the plan of the top line of the ele. vation, hence all the extremities, as a', b', &c., of vertical joints, found on that line, must be horizontally projected as at a and b, in the horizontal projection of the same line. 78. Ex. 3. Rubble Walls. The remaining figures of P1. V., represent various forms of "rubble" wall. Fig. 43 represents a wall of broken boulders, or loose stones of all sizes, such as are found abundantly in New England. Since, of course, such stones 386 CONSTRUCTIONS IN MASONRY.t Nvwould not fit together exactly, the "chinkls" between them are filled with small fragments, as shown in the figure. Still smaller irregularities in the joints, which are not thus filled, are represented after tinting by heavy strokes in. inking.' Fig. 44 represents the plan and elevation of a rubble wall mnlade of slate; hence, in the plan, the stones appear broad, and in the elevation, long and thin, with chink stones of similar shape. Fig. 45 represents a rubble wall, built in regular courses, which gives a pleasing effect, particularly if the wall have cut stone corners, of equal thickness with the rubble courses. 79. _Ececution.-The graphical construction of objects being now more an object of study than the mere style of executioln employed, the graining of the stone work on P1. V. is omitted. Plate V. may be, 1st, pencilled; 2cl, inked in fine lines; 3d, tinted. The rubble walls, having coarser lines for the joints, may better be tinted, before lining the joints in ink. The plans need not be tinted. Also, in case of the rubble walls, sudden heavy strokes may be made occasionally in the joints, to indicate slight irregularities in their thickness, as has already been mentioned. The right hand and lower side of any stone, not joining another stone on those sides, is inked heavy, in elevation, and on the plans as usual. The left hand lines of Figs. 43 and 44 are tangent at various points to a vertical straight line, walls, such as are represented in those figures, being made vertical, at the finished end, by a plumb line, against which the stones rest. The shaded elevations on P1. V., may serve as guides to the depth of color to be used in tinting stone work. The tint actually to be used, should be very light, and should consist of gray, or a mixture of black and white, tinged with Prussian blue, to give a blue gray, and carmine also if a purplish gray is desired. Remarks. —c. A scale may be used, or not, in making this plate. The number of stones shown in the width of the plans, shows that the walls are quite thick. b. Rubble walls, not of slate, are, strictly, of two kinds; first, those formed of small boulders, used whole, or nearly so; and second, those built of broken rock. Each should show the broad. est surfaces in plan. DI')J II. PL.V: Div.~ ~ ~ ~ it -7 — 45 -J7; I t t................................. Ii m:.......... 2=.-...a.or&Ka....J.ft d..... P ceN y........................ X. ~ ~ ~ ~ ~ ~ ~ i~iiiiiiiiiiiiiiiiii::::r:::-_-=...........i~ —:~-~:II i::ll~~'illIlliiii~ili ii::::::::~:::~:c:~:i~:~: —~~~~~~:=_ — ~~~::::~St::=:e::::~~~~~;::::::::::::-1-=: —- 44 ~~~~~~~~~~.........:::,::::,~ ~ ~~~~~~~~~~~~~~~~~...... CHAPTER II. CONSTRUCTIONS IN WOOD. ~ I. —General Remarks. 80. Two or more beams may be framed together, so as to make any angle with each other, from 0~ to 180~; and so that the plane of two united pieces may be vertical, horizontal, or oblique. 81. To make the present graphical study of framings more fully rational, it may here be added, that pieces may be framed with reference to resisting forces which would act to separate them in the direction of any one of the three dimensions of each. Followillg out the classification in the preceding article, let us presently proceed to notice several examples, some mainly by general description of their material construction and action, and some by a complete description of their graphical construction and executionz also. 82. Two other points, however, may here be mentioned. First: A pair of pieces may be imiecdiately framed into each other, or they may be intermediactely framed by "bolts," "keys," &c., or both modes may be, and often are, combined. Second: Two combinations of timbers which are alike in general appearance, may be adapted, the one to resist extension, and the other, compression, and may have slight corresponding differences of construction. 83. Note.-For the benefit of those who may not have bad access to the subject, the following brief explanation of scales, &c., is here inserted. (See my 1" Drafting Instruments and Operations.") Drawings, showing the pieces as taken apart so as to show the mode of union of the pieces represented, are called "-Details." Sections, are the surfaces exposed by cutting a body by planes, and, strictly, are in the planes of section. Sectional elevations, or plans, show the parts both in, and beyond, the planes of section. Drawings are made in plan, side and end elevations, sections and details, or in as few of these as will show clearly all parts of the object represented. 84. In respect to the instrumental operations, these drawings are 38 CONSTRUCTIONS IN WOOD. supposed to be " made to scale," from measurements of models, or firom assumed measurements. It will, therefore, be necessary, before beginning the drawings, to explain the manner of sketching the object, and of taking and recording its measurernents. 85. In sket;ching the object, make the sketches in the same way in which they are to be drawn, i.e. in plan and elevation, and not in perspective, and make enough of them to contain all the measlirements, i.e. to show all parts of the object. In measuring, take measurements of all the parts which are to be shown; and not merely of individual parts alone, but such connecting measurements as will locate one part with reference to another. 86. The usual mode of recording the measurements, is, to indicate, by arrow heads, the extremities of the line of which the figures between the arrow heads show the length. 87. For brevity, an accent (') denotes feet, and two accents (") denote inches. The dimensions of small rectangular pieces are indicated as in P1. VI., Fig. 50, and those of small circular pieces, as in Fig. 51. 88. In the case of a model of an ordinary house fiaming, such as it is useful to have in the drawing room, and in which the sill is represented by a piece whose section is about 2 inches bv 3 inches, a scale of one inch to six inches is convenient. Let us then describe this scale, which may also be called a scale of two inches to the foot. The same scale may also be expressed as a scale of one foot to two inches, meaning that one foot on the object is represented by two inches on the drawing; also, as a scale of 6, thus, a foot being equal to twelve inches, 12 inches on the object is represented by two inches on the drawing; therefore, one inch on the drawing represents six inches on the object, or, each line of the drawing is d of the same line, as seen upon the object; each line, for we know from Geometry that sutifaces are to each other as the squares of their homologous dimensions, so that if the length of the lines of the drawing is one-sixth of the length of the same lines on the object, the area of the drawing would be one thirty-sixth of the area of' the object, but the scale always refers to the relative lengths of the lines only. 89. In constructing the scale above mentioned, upon the stretched drawing paper, 1st. Set off upon a fine straight pencil line, two inches, say three times, making four points of division. CONSTRUCTIONS IN WOOD. 39 2d. Number the left hand one of these points, 12, the next, 0, the next, 1, the next, 2, &c., for additional points. 3cd. Since each of these spaces represents a foot, if any one of them, as the left hand one, be divided into twelve equal parts, those parts will be representative inches. Let the left hand space, from (12) to (0) be thus divided, by fine vertical dashes, into twelve equal parts, making the three, six, and nine inch marks longer, so as to catch the eye, when using the scale. 4th. As some of the dimensions of the object to be drawn are measured to quarter inches, divide the first and sixth of the inches, already found, into quarters; dividing two of them, so that each may be a check upon the other, and so that there need be no con tinual use of one of them, so as to wear out the scale. 5th. When complete, the scale may be inked; the length of it in fine parallel lines about -' of an inch apart. 90. It is now to be remarked that these spaces are always to be called by the names of the dimensions they represent, and not according to their actual sizes, i. e. the space from 1 to 2 represents a foot upon the object, and is called a foot; so each twelfth of the foot from 12 to 0 is called an inch, since it represents an inch on the object; and so of the quarter inches. 91. Next, is to be noticed the directions in which the feet and inches are to be estimated. The feet are estimated from the zero point towards the right, and the inches from the same point towards the left. Thus, to take off 2' —5/ fiom the scale, place one leg of the dividers at 2, and extend the other to the fifth inch mark beyond 0, to the left; or, if the scale were constructed on the edge of a piece of card-board, the scale being laid upon the paper, and with its graduated edge against the indefinite straight line on which the given measurement is to be laid off, place the 2' or the 5' mark, at that point on the line, from which the measurement is to be laid off, according as the given distance is to be to the left or right of the given point, and then with a needle point mark the 5" point or the 2' point, respectively, which will, with the given point, include the required distance. 92. Other scales, constructed and divided as above described, only smaller, are found on the ivory scale, marked 30, &c., meaning 30 feet to the inch when the tenths at the left are taken as feet; and meaning three feet to the inch when the larger spaces -three of which make an inch-are called feet, and the twelfths of the left hand space, inches. Intermediate scales are marked 40 CONSTRUCTIONS IN WOOD. 35, etc. Thus, on the scale marked 45, four and a half of the larger spaces make one inch, and the scale is therefore one of four and a half feet to one inch, when these spaces represent feet; and of forty-five feet to one inch, when the tenths represent feet. In like manner the other scales may be explained. So, on the other side of the ivory, are found scales marked.-: &c., meaning scales of a inch to one foot, or ten feet, according as the whole left hand space, or its tenth, is assumed as representing one foot. Note that X of an inch to a foot is - of a foot to the inch, ~ of an inch to ten feet, is 16 feet to an inch, &c. 94. Of the immense superiority of drawing by these scales, over drawing without them, it is needless to say much: without them, we should have to go through a mental calculation to find the length of every line of the drawing. Thus, for the piece which is two and a half inches high, and drawn to a scale of two inches to a foot, we should say —2 inches —2~ of a foot=-5- of a foot. One foot on the object=two inches on the drawing, then -2j of a foot on the object= —! of 2 inches= —— =- of an inch, and 2-5 of a foot (= 21 inches)= — s of 2 inches= 2 of an inch. A similar tedious calculation would have to be gone through with for every dimension of the object, while, by the use of scales, like that already described, we take off the. same number of the feet and inches of the scale, that there are of real feet and inches in any given line of the object. ~ II.-Pairs of Timbers whose axes make angles of 00 with each other. The student should be required to vary all of the remaining constructions in this DIVISION, in one or more of the following ways. First, by a change of scale; Second, by choosing other examples from models or otherwise, but of similar character; or, Third, by a change in the number and arrangement of the projections employed in representing the following examples. 95. EXAMPLE 1. A Compound Beam bolted. P1. VI., Fig. 46. MJechanical Construction. The figure represents one beam as laid on top of another. Thus situated, the upper one may be slid upon the lower one in the direction of two of its three dimensions; or it may rotate about any one of its three dimensions as an axis. A single bolt, passing through both beams, as shown in the figure, will prevent all of these movements except rotation about the bolt as an.axis. Two CONT'SUCTIONS IN WOOD. 41 or more bolts will prevent this latter, and consequently, all movement of either of the beams upon the other. A bolt, it may be necessary to say, is a rod of iron whose length is a little greater than the.aggregate thickness of the pieces which it fastens together. It is provided at one end with a solid head, and at the other, with a few screw threads on which turns a "nut," for the purpose of gradually compressing together the pieces through which tho bolt passes. 96. GCraphical Construction. Assuming for simplieity's sake in this and in most of these examples, that the timbers are a foot square, and having the givenr scale; the diagrams will generally explain themselves sufficiently. The scales are expressed fractionally, adjacent to the numbers of the diagrams. The nut only is shown in the plan of this figure. It is an error to suppose that the nuts and other small parts can be carelessly drawn, as by hand, without injury to the drawing, since- these parts easily catch the eye, and if distorted, or roughly drawn, appear very badly. The method is, therefore, here fully given for drawing a nut accurately. Take any point in the centre line, ab, of the bolt, produced, and through it draw any two lines, cd and en, at right-angles to each other. From the centre, lay off each way on each line, half the length of each side of the nut, say 3 of an inch. Then, through the points so fobund, draw lines parallel to the centre lines cd and en, and they will form a square plan of a nut iif on each side. In making this construction, the distances should be set off very carefully, and the sides of the nut rruled, in very fine lines, and exactly through the points located. From the plan, the elevation is found as in P1. II., Fig. 21. 97. Ex. 2. A Compound Beam, notched and bolted. P1. VI., Fic. 47. Mecheanical Construction. The beams represented in this figure, are indented together by being alternately notched; the portions cut out of either beam being a foot apart, a foot in length, and two inches deep. When merely laid, one upon another, they will offer resistance only to being separated longitudinally, and to horizontal rotation. The addition of a bolt renders the "compound beam," thus formed, capable of resisting forces tending to separate it in all ways. Thin pieces are represented, in this figure, between the bolt-head and nut, and the wood. These are circular, having a rounded 42 CONSTRUCTIONS IN WOOD. edge, and a circular aperture in the middle through which the bolt passes. They are called " washers," and their use is, to distribute the pressure of the nut or bolt-head over a larger surface, so as not to indent the wood, and so as to prevent a gouging of the wood in tightening the nut, which gouging would facilitate the decay of the wooCd, and consequently, the loosening of the nut. 98. Graphical Construction.-The beams being understood to be originally one foot square, the compound beam will be 22 inches deep; hence draw the upper and lower edges 22 inches apart, and from each of them, set off; on a vertical line, 10 inches. Through the points, a and b, so found, draw veryfai-nt horizontal lines, and on either of them, lay off any number of spaces; each, one foot in length. Through the points, as c, thus located, draw transverse lines between the faint lines, and then, to prevent mistakes in inking, make slightly heavier the notched line which forms the real joint between the timbers. The use of the scale of J-2 continues till a new one is mentioned. The fbllowing empirical rules will answer for determining the sizes of nuts and washers on assumed sketches like those of P1. VI., so as to secure a good appearance to the diagram. The side of the nut may be double the diameter of the bolt, and the greater diameter of the washer may be equal to the diagonal of the nut, pins twice the thickness of the washer itself. Execautionr.-This is manifest in this case, and in most -of the following examples, from an inspection of the figures. 99. Ex. 3. A Compound Beam, keyed. P1. VI., Fig. 48. J~iechanical Constraction. The defect in the last construction is, that the bearing surfaces opposed to separation in the direction of the length of the beam, present only the ends of the grain to each other. These surfaces. are therefore liable to be readily abraded or made spongy by the tendency to an interlacing action of the fibres. Hence it is better to adopt the construction given in P1. VI., Fig. 48, where the " keys," as K, are supposed to be of hard wood, whose grain runs in the direction of the width of the beam. In this case, the bolts are passed through the keys, to pre. vent them fronm slipping out, though less boring would be required if they were placed midway between the keys. In this example, the strength of the beam is greatly increased with but a very small increase of material, as is proved in mechanics and confirmed by experiment. 100. Graphical Constrzuctioqn.-This example differs from the last Bo slightly as to render a particular explanation unnecessary. The CONSTRUCTIONS IN WOOD.:43 keys are 12 inches in height, and 6 inches in width, and are 18 inches apart from centre to centre. They are most accurately located by their vertical centre lines, as AA'. If located thus, and fiom the horizontal centre line BB', they can be completely drawn before drawing ee' and nn'. The latter lines, being then pencilled, only between the keys, mistakes in inking will be avoided. _Execution.-The keys present the end of their grain to view, hence are inked in diagonal shade lines, which, in order to render the illuminated edges of the keys more distinct, might terminate, uniformly, at a short distance fiom, the upper and left hand edges. By shading only that portion of the right hand edge of each key, which is between the timbers, it is shown that the keys do not project beyond the front faces of the timbers. 101. Ex. 4. A Compound Beam, scarfed. P1. VI., Fig. 49.. ifechanical Construction. This specimen shows the use of a series of shallow notches in giving one beam a firm hold, so to speak, upon another; as one cdeep notch, having a bearing surface equal to that of the four shown in the figure, would so far cut away the lower beam as to render it nearly useless. 102. Graphical Constructionz. —The notches, one foot long, and twoo inches deep, are laid down in a manner similar to that described under Ex. 2. 103. Execution.-The keys, since they present the end of the grain to view, are shaded as in the last figure. Heavy lines on their right hand and lower edges would indicate that they projected beyond the beam. Remark.-When the surfaces of two or more timbers lie in the same plane, as in many of these examples, they are said to be "gflush" with each other. ~ III.- Combinations oqf Timbers, whose axes make angles of 90~ with each other. 104. The usual way of fastening timbers thus situated, is by means of a projecting piece on one of them, called a "tenon," which is inserted into a corresponding cavity in the other, called a "mortise." The tenon may have three, two, or one of its sides flush with the sides of the timber to which it belongs; while the mortise may extend entirely, or only in part, through the timber in which it is made, and may be enclosed by that timber on three or on all sides. [See the examples which follow, in which some of these cases are replresented, and from which the rest can be underi stoold.] 44 CONSTRUCTIONS IN WOOD. When the mortise is surrounded on three or on two sides, par. ticularly in the latter case, the firamed pieces are said to be "halved" together, more especially in case they are of equal thickness, and have half the thickness of each cut away, as at P1. VI., Fig. 52. 105. EXAMPLE 1. Two examples of a Floor Joist and Sill. (From a Model.) P1. VI., Fig. 53. _Mechanical Construction. A-A' is one sill, B-B' another. CC' is a floor timber framed into both of them. At the left hand end, it is merely " dropped in," with a tenon; at the right hand end, it is fiamed in, with a tenon and " tusk," e. At the right end, therefore, it cannot be lifted out, but must be drawn out of the mortise. The tusk, e, gives as great a thickness to be broken off, at the insertion into the sill, and as much horizontal bearing surface, as if it extended to the full depth of the tenon, t, above it, while less of the sill is cut away. Thus, labor and the strength of the sill, are saved. 106. Graphical Construction.-lst. Draw ab. 2c. On ab construct the elevation of the sills, each 2- inches by 3 inches. 3dc. Make the two fragments of fidor timber with their upper surfaces flush with the tops of the sills, and 2 inches deep. 4th. The mortise in A', is A of an inch in length, by 1 inch in vertical depth. 5th. Divide cd into four equal parts, of which the tenon and tusk occupy the second. and third. The tenon, t, is 3 of an inch long, and the tusk, e, I of an inch long. Let the scale of 6 be used. 107. -Execution.-The sills, appearing as sectionsin elevation, are shaded. In all figures like this, dotted lines of construction should be freely used to assist in " reading the drawing,)' i.e. in comprehending, fiom the drawing, the construction of the thing represented. 108. Ex. 2. Example of a "M' ortise and Tenon," and of "Halving." (From a MIodel.) P1. VI., Fig. 54. lWechanical Construction. In this case, the tenon, AA', extends entirely through the piece, CC', into which it is framed. B and C are halved together, by a mortise in each, whose depth equals half the thickness of B, as shown at B" and C", and by the dotted line, ab. Grcaphiccal Constru-ction.-iMake, lst, the elevation, A'; 2d, the plan; 3d, the details. B" is an elevation of B as seen when looking in the direction, BA. C" is an elevation of the left hand portion of CC', showing the mortise into which B is halved. The dimensions may be assumed, or found by a scale, as noticed below. 109. Fsxecution.-The invisible parts of the framing, as the halv. CONSTRIUCTIONS IN WTOOD. 45 ing, as seen at ab in elevation, are shown in dotted lines. The brace and the dotted lines of construction serve to show what separate figures are comprehended under the general number (54) of the diagram. The scale is I. From this the dimensions of the pieces can be found on a scale. 110. Ex. 3. A Mortise and Tenon as seen in two sills and a post. Use of broken planes of section. (From a Model.) P1. VI., Fig. 55. 3fechanical Construction.-The sills, being liable to be drawn apart, are pinned at a. The post, BB', is kept in its mortise, bb", by its own weight; m is the mortise in which a vertical wall joist rests. It is shown again in section near mn'. 111. Graphical Construction.-The plan, two elevations, and a broken section, show all parts fully. The assemblage is supposed to be cut, as shown in the plan by the broken line AA'A"A"', and is shown, thus cut, in the shaded figure, A'A'A"'rn'. The scale, which is the same as in Fig. 53, indicates the measurements. At B", is the side elevation of the model as seen in looking in the direction A'A. In Fig. 55 a, A's obviously equals A"s, as seen in the plan. 112. Execution.-In the shaded elevation, Fig. 55a, the cross-sec. tion, A'A"', is lined as usual. The longitudinal sections are shaded by longitudinal shade lines. The plan of the broken upper end of the post, B, is filled with arrow heads, as a specimen of a way, sometimes convenient, of showing an end view of a broken end. Sometimes, though it renders the execution more tedious, narrow blank spaces are left on shaded ends, opposite to the heavy lines, so as to indicate more plainly the situation of the illuminated edges (100). The shading to the left of A', Fig. 55a, should be placed so as to distinguish its surface firom that to the right of A'. 113. Ex. 4. A Mortise and Tenon, as seen in timbers so framed that the axis of one shall, when produced, be a diagonal diameter of the other. P1. VI., Fig. 56. Mfechanical Construction.-In this case the end of the inserted timber is not square, and in the receiving timber there is, besides the mortise, a tetraedron cut out of the body of that timber. 114. Graphical Construction.-D is the plan, D' the side elevation, and D" the end elevation of the piece bearing the tenon. F' and F are an elevation and plan of the piece containing the mor. tise. Observe that the middle line of D, and of D', is an axis of symmetry, and that the oblique right hand edges of D and D' are parallel to the corresponding sides of the incision in F'. 46 CONSTRUCTIONS IN WOOD. ~ IV. —Miscellaneous Combinations. 115. EXArIPLE 1. Dowelling. (From a Model.) PI. VI., Fig. 57. M~echanical Construction.-Dowelling is a mode of fastening by pins, projecting usually from an edge of one piece into corresponding cavities in another piece, as seen in the fastening of the parts of the head of a water tight cask. The mode of fastening, how. ever, rather than the relative position of the pieces, gives the name to this mode of union. The example shown in PI. VI., Fig. 57, represents the braces of a roof fiaming as dowelled together with oak pins. 116. Graphical Construction.-This figure is, as its dimensions indicate, drawn from a model. The scale is one-third of an inch to an inch. 1st. Draw acb, with its edges making any angle with the imagi. nary ground line-not drawn. 2d. At the middle of this piece, draw the pin or dowel, pp, - of an inch in diameter, and projecting 3 of an inch on each side of the piece, acb. This pin hides another, supposed to be behind it. 3d. The pieces, d and d", are each 2~ inches by 1 inch, and are shown as if just drawn, off from the dowels, but in their true direction, i.e. at right angles to acb. 4th. The inner end of d is shown at d', showing the two holes, 1. inches apart, into which the dowels fit. -Execution.-The end view is lined as usual, leaving the dowel holes blank. 117. Ex. 2. A dovetailed Mortise and Tenon. P1. VI., Fig. 58. Mechanical Construction. —This figure shows a species of joining called dovetailing. Here the mortise increases in width as it becomes deeper, so that pieces which are dovetailed together, either at right angles or endwise, cannot be pulled directly apart. The corners of -drawers, for instance, are usually dovetailed; and sometimes even stone structures, as lighthguses, which are exposed to furious storms, have their parts dovetailed together. 118.' Graphical Construction.-The sketches of this framing are arranged as two elevations. A bears: the dovetail, B shows the length and breadth of the mortise, and B" its depth. A and B belong'to the same elevation. LExecution.-In this case a method is given, of representing a hidden cut surface, viz. by dotted shade lines, as seen in the hidden faces of the mortise in B". 119. Leaving now the examples of pieces framed together at right angles let us consider: k~~ 1~ -~~~- -- ~j~ I~~~ji~~ Hi ~ p$Iq *-i K> d /.4.~ ~J~~~~~~~~~~~~~~~~~~~~~~~~,.... t —- - A A iai:~~~~~~~~~~~~~~~~~~~~~ V~~~~~~~~~~~~~~~~V'%/ i ~~li! N. ~.-~ __ U,'~Iz;:1. 1T —-— 1 -'!'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~'!_,~' 1 ~~~~~~~~~~~~~ ~ ~ ~~...........__.=.=========== _._..._..... K..','-' —'iv~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~....... ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~......................... ~~~~~~~~~~~~~~~~~~~~::,,:......? iii i'i:..... II)I h?~ h-.?i —-- i~ ~ ~ ~~-. —AI 7C~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~I i................~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~........../~,Wv~H ~~~~~~~~~~~~~~~~~~~~~~................,w Iill:,,r-''. 9 ]Lv Ni O.IcLT Biii'~'ie i:'ciiiiiii i'iii' -iiii!~i~i~11~i:~iiiiii v,!'; —--—' —-- 8 —----------— ~~j, -........i -- — t —----------— ~ —---— ~~~~~~~~~~~~~~~~~~~~~ —-.; i!. _ _',~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~________________ K ~~~~~~~~~~~/'OS'/N, LW~~~dIl IJII~nlU —'q~~;I i 1 —1.!,-' ~~s~~":: ~ ~ 5 — i —,rr~~~~~~' <", I,~~~~~~~~~~~~~~7 rei...... ~~~~~~~~~~~ ~~~~~~'!-l~'r!-_ _~_' l —l ~{kI CONSTRUCTIONS IN WOOD. 4-7 ~ V. —Pairs of Timbers which are frcamed together obliquely to each other. EXAMPLE 1. A Chord and Principal. (From a Model.) Pl. VII., Fig. 59. Mlechanical Construction.-The oblique piece (' principal") is, as the two elevations together show, of equal width with the horizontal piece (" chord,"' or "tie beam "), and is framed into it so as to prevent sliding sidewise or lengthwise. Neither can it be lifted out, on account of the bolt which is made to pass perpendicularly to the joint, ac, and is " chipped up" at pp, so as to give a flat bearing, parallel to ac, for the nut and bolt-head. 120. Graphical Construction.-lst. Draw pdce; 2d. Lay off de -13 inches; 3d. Make e'ea=-30~; 4th. At any point, e', draw a perpendicular to ee', and lay off upon it 9 inches-the perpendicular width of e'ea; 5th. Make ec= 4 inches and perpendicular to e'e; bisect it and complete the outlines of the tenons, and the shoulder aec'; 6th. To draw the nut accurately, proceed as in P1. VI., Fig. 46-47, placing the centre of the auxiliary projection of the nut in the axis of the bolt produced, &c. (46) (96). b represents the bolt hole, the bolt being shown only on one elevation. 121. Ex. 2. A Brace, as seen in the angle between a "post " and " girth." (From a Model.) P1. VII., Fig. 60. iJlechanical Construction. —PP' is the post, GG' is the girth, and B'B" is the brace, having a truncated tenon at each end, which rests in a mortise.'When the brace is quite small, it has a shoulder onl one side only of the tenon, as if B'B" were sawed lengthwise on a line, oo'. 122. Graphical Construction.-To show a tenon of the brace clearly, the girth and brace together are represented as being drawn out of the post. 1st. Draw the post. 2d. Half an inch below the top of the post, draw the girth 2- inches deep. 3d. From a, lay off ab and a each 4 inches, and draw the brace I inch wide 4th. Make ccl equal to the adjacent mortise; viz. I- inches; make de=1 inch, and erect the perpendicular at e till it meets be, &c. The dotted plrojecting lines show the construction of B" and of the plan. At e" is the vertical end of the tenon e. On each side of e", ale the vertical surfaces, shown also at ed. Let B' also be projected on a plane parallel to be. 1;23. Ex. 3. A Brace, with shoulders mortised into the post. P1. VII., Fig. 61. This is the strongest-way of framing a brace. For the rest, the figure explains itself. Observe, hoWever, that while in Fig. 59, the head of the tenon and shoulder is perpendi ~48 CCONSTRUCTIONS IN WOOL. cular to the oblique piece, here, where that piece is frained alto a vertical post, its head nt is perpendicular to the axis of the post. In Fig. 61, moreover, the auxiliary plane on which the brace alone is projected, is parallel to the length of the brace, as is shown by the situation of B, the auxiliary projection of the brace, and by the direction of the projecting lines, as rr'. P' is the elevation of P, as seen in the direction an', and with the brace removed. 124. Ex. 4. A "Shoar." P1. VIL, Fig. 62. Juechlanical Construction.-A " shoar" is a large timber used to prop up earth or buildings, by being framed obliquely into a horizontal beam and a stout vertical post. It is usually of temporary use, during the construction of permanent works; and as respects its action, it resists compression in the direction of its length. To give a large bearing surface without cutting too far into the vertical tim-'lr, it often has two shoulders. The surface at ab is made vertical, for then the fibres of the post are unbroken except at cb, while if the upper shoulder were shaped as at adb, the fibres of the triangular portion icao would be short, and less able to resist a longitudinal force. 125. The Graphical construction is evident fiom the figure, which is in two elevations, the left hand one showing the post only. Execution.-The vertical surface at ab may, in the left hand elevation, be left blank, or shaded with vertical lines as in P1. VI., Fir. 55a., VI.-Co(mbinations of Timbers whose axes make angeks of 1800 with each other. 126. Timbers thus fiamed are, in general, said to be spliced. Six forms of splicing are shown in the following figures. EXAMPLE 1. A Halved Splicing, pinned. P1. VII., Fig. 63. The mechcnical construction is evident fiom the figure. When boards are lapped on their edges in this way, as in figure 69, they are said to be " rabbetted." 127. Graphical Construction —After drawing the lines, 12 inches apart, which represent the edges of the timber, drop a perpendicular of 6 inches in length from any point as a. From its lower extremity, draw a horizontal line 12 inches in length, an'd friom c, drop the perpendicular cb, which completes the elevation. In the plan, the joint at a will be seen as a fhull line a'a", and that at b, being hidden, is represented as a dotted line, at b7 CONSTRUCTIONS IN WOOD. 49 Draw a diagonal, a'b", divide it into four equal parts, and take the first and third points of division for the centres of two pins, having each a radius of three-fourths of an inch. Fxecution.-The position of the heavy lines on these figures is too obvious to need remark. 128. Ex. 2. Tonguing and Grooving; and Mortise and Tenon Splicing. P1. VII., Fig. 64. Boards united at their edges in this way, as shown in P1. VII., Fig. 70, are said to be tongued and grooved. Drawing, as before, the plan and elevation of a beam, a foot square, divide its depth an, at any point a, into five equal parts. Take the second and fourth of these parts as the width of the tenons, which are each a foot long. The joint at a is visible in the plan, the one at b is not. Let a'c be a diagonal line of the square a'd. Divide a'd into three equal parts, and take the points of division as centres of inch bolts, with heads and nuts 2 inches square, and washers of14 inches radius. To place the nut in any position on its axis, draw any two lines at right angles to each other, through each of the bolt centres, and on each, lay off 1 inch from those centres, and describe the nut. Project up those angles of the nut which are seen; viz. the foremost ones, make it I.inch thick in elevation, and its washer I an inch thick. In this, and all similar cases, the head of the bolt, s, would not have its longer edges necessarily parallel to those of the nut. To give the bolt head any position on its axis, describe it in an auxiliary plan just below it. 129. Ex. 3. A Scarfed Splicing, strapped. P1. VII., Fig. 65. JMechanical Construction.-While timbers, fiamed together as in the two preceding examples, can be directly slid apart when their connecting bolts are removed; the timbers, firamed as in the present example, cannot be thus separated longitudinally, on account of the dovetailed form of the splice. Strapping makes a firm connexion, but consumes a great deal of the uniting material. 130. Graphical Construction.-After drawing the outlines of the side elevation, make the perpendiculars at a and a', each 8 inches long, and make them 18 inches apart. One inch from a, make the strap, ss', 21 inches wide, and projecting half an inch-i. e. its thickness-over the edges of the timber. The ear through which the bolt passes to bind, the strap round the timber, projects two inches above the strap. Take the centre of the ear as the centre of the bolt, and on this centre describe the bolt head 11 inches square. 4 50 CC-NSTLRUCTIONS IN WOOD. In the plan, the ears of the strap are at any indefinite distance apart, depending on the tightness of the nut, s. A fragment of a similar strap at the other end of the scarf, is shown, with its visible ends on the bottom, near a', and back, a.t r of the beam. Execuztion.-The scarf is dotted where it disappears behind the strap; and so are.the hidden joint at a', and the fiagment of the second strap, as shown in plan. These examples may advantageously be drawn by the student on a scale of J. Care must then be exercised in making the large broken ends neatly, in large splinters, edged with fine ones. 131. Ex. 4. A Scarfed Splice, bolted. P1. VII., Fig. 66. lfechanical Construction.-AA' is one timber. BB' is the other. Each is cut off as at aea", forming a pointed end which prevents lateral displacement. cfand cdg are the ends of transverse keys, wvhich afford good bearing surfaces. See the same on P1. X.; Fig. 105, which shows the arrangement plainly. 132. Gralphical Constrvction.-Let each timber he 1 foot squa'e and let the scale be 1 foot to I inch. Draw the outlines accordingly, and, assuming a', make a'b'=3 feet; drop the perpendicular b'c', and diraw a line of construction a'c'. From a' lay off 4 inches on acc' and divide the rest of acc' into three equal parts, to get the size of the eqzeal spaces c'k, fd and gvl, and at c' and k draw perpendiculars to a'ce, above it and 2 inches long. Divide a'k into two equal parts at d, and fiom. k and d lay off 2 inches on at'c towards a' and complete the keys, as shown, kf above a'c', and dg below it; also the joints v'g, clce, etc., of the splice, where a'v' is perpendicular to a'c' and 2 inches long. Now, in plan, project down a' at a and a" and draw ae and a"e at 60~ with a"c, and do the same with. c', as shown. Project up e to c' and draw e's' parallel to a'v' till it meets c'g produced. Then project down v' at v and v", and s' at s, and draw sv and sv", which, of course, will not be parallel to ae and a"e, since se-s'e' is shorter than avi-a'v'. Thus av se a"v" —a'v' s e' is the obliquely pointed end of the piece AA'. BB' is similarly pointed, as, shown. Add the bolts, nuts and washers, nn' and rnn', in any convenient position, which will complete the construction. Execution.-The keys are shaded. The hidden cut surfaces of the notches are shaded in dotted shade lines, and the hidden joints are dotted. 133. Ex. 5. A Compound Beam, with one of the component beams "fished." P1. VII., Fig. 67. 3fechaiical Constru CONSTRUCTIONS IN WOOD. 5 1 tion.-The mode of union called fishing, consists in uniting two pieces, end to end, by laying a notched piece over the joint and bolting it through the longer pieces. The figure shows this mode as applied to a compound beam, i. e. to a beam "built'" of several pieces bolted and keyed toge. ther. The order of construction is as follows-taking for a scale h of an inch to a foot. 134. Graphical Construction.-lst. The outside lines of the plan are two feet apart, the outside pieces are each 44 inches wide, and the interior ones 51 inches; leaving four inches for the sum of the three equal spaces between the four beams. 2d. Let there be a joint at a'. Lay off 3 feet, each side of a' for the length of the "fish."' 3d. The straight side of this piece is let into the whole piece, b, two thirds of an inch, and into a', 1 inch, making its thickness- 3 inches. 4th. At a', it is two inches thick, i. e. at a', the timber, a', is of its full thickness. The fish, c, is 2 inches thick for the space of one foot at each side of a'. The notches at d and e are each 1 inch deep, dd' and ee' each are one foot, and the notches at d' and e' are each 1 inch deep. The remaining portions of the fish are 1 foot long, and 3 inches wide. 5th. Opposite to these extreme portions, are keys, I foot by 3 inches, in the spaces between the other timbers, and setting an equal depth into each timber. 6th. In elevation, only the timber a' is seen-i foot deep. Four bolts pass through the keys. b"b =-5 feet, and b' is three inches from the top, and from ick', the left hand end of the fish. s'n"=5 feet, and n' is 3 inches above the bottom of the timber, and 9 inches from Hke. The circular bolt head is one inch in diamneter, and its washer 31 inches diameter and 4 an inch thick. The thickness of the bolt head, as seen in plan, is 4 inch. The nuts, nn, are 1 inches square, and 4 an inch thick, and the bolts are half an inch in diameter. 135. The several nuts would naturally be found in various positions on their axes. To construct them thus with accuracy, as seen in the plan, one auxiliary elevation, as N, is sufficient. N, and its centre, niay be projected upon as many planes-xy-as there are different positions to be represented in the plan, each plane being supposed to be situated, in reference to N, as some nut in the plan is, to its elevation. Then transfer the points on xy, &c., to the outside of the several nut-Nwaslhers, placing tihe projection of the centre lines of the bolts in the 1)lan, a lines of reference. 5 2 CONSTRUCTIONS IN WOOD. J]fxecution.-The figure explains itself in this respect. 136. Ex. 6. A vertical Splice. P1. VIII., Fig. 68. Mecha. nical Construction.-This splice is formed of two prongs at opposite corners of each piece, embraced by corresponding notches in the other piece. Thus in the piece B', the visible prong, as seen in elevation, is a truncated triangular pyramid whose horizontal base is abc-a'c', and whose oblique base is enc-e'n"c". Besides the four prongs, two on each timber, there is a fiat surface abfq —c'a'q', well adapted to receive a vertical pressure, since it is equal upon, and common to, both timbers. 137. Graphical Construction.-To aid in. understanding this combination, an oblique projection is given on a diagonal plane, parallel to PQ. 1st. Make the plan, acfq, with the angles of the interior square in the middle of the sides of the outer one. 2d. }Make the distances, as ce=2 inches and draw en, &c. 3d. Make c'n'=c'n" 15 inches, and draw short horizontal lines, n"e', &c., on which project e, &c., after which the rest is readily completed. 138. Execution.-Observe, in the plans, to change the direction of the shade lines at every change in the position of the surface of the wood. CHAPTER III. CONSTRUCTIONS IN METAL. 139. EXAMPLE 1. An end view of a Railroad Rail. P1. VII., Fig. 71. Graphical Construction.-lst. Draw a vertical centre line AA', and make AA'=34 inches. 2d. Make A'b=A'b'=2 inches. 3d. Mfake Ac=Ac'=l 1 inch. 4th. Describe two quadrants, of which c'd is one, with a radius of half an inch. 5th. With p, half an inch from AA', as a centre, and pd as a radius, describe an are, dr, till it meets a vertical line through e. 6th. DI)raw the tangent rs. 7th. Draw a vertical line, as pt, i an inch from AA', on each side of AA'. 8th. Bisect the angle rst' and note s', where the bisecting line meets the radius, pr, produced. 9th. With s' as a centre, draw the arc rtK 10th. At b and b' erect perpendiculars, each one fourth of an inch high. 11th. Draw quadrants, as g't, tangent to these perpendiculars and of one fourth of an inch radius. 12th. Draw the horizontal line tv. 13th. Make nt'=nv and describe the arc t'v.:.. 14th. Repeat these operations on the other side of the centre line, AA'. Emecution.-Let the construction be fully shown on one side of the centre line. 140. Ex. 2. An end elevation of a Compound Rail. P1. VII., Fig. 72. Mechlanical Construction. —The compound rail, is a rail formed in two parts, which are placed side by side so as to break joints, and then riveted together. As one half of the rail is whole, at the points where a joint occurs on the other half, the noise and jar, observable in riding on tracks built in the ordinary manner, are both obviated; also " chairs," the metal supports which receive the ends of the ordinary rails, may be dispensed with, in case of the use of the compound rail. In laying a compound rail on a curve, the holes, through which the bolts pass, may be drawn past one another by the bending of the rails. To allow for this, these holes are " slotted," as it is termed, i. e. made longer in the direction of the length of the rail. 141. Graphical Construction.-lst. BIake ty=4 inches. 2d. Bisect ty at u, and erect a perpendicular, ua, of 3- inches. 3d. 54 CONSTRUCTIONS IN METAL. Make, successively, -ur —t an inch; firom r to cb=2 iiellhes; fironim to nh= — of an inch; and to ge-21 inches. 4th. For the several widths of the interior parts, make bc — of an inch, and g and e each 8 of an inch, from ua; nh —ge, and ro=3 of an inch. 5th. To locate the outlines of the rail, make ms, the flat top, called the tread of the rail, =2 inches, half an inch below this, make the width, td, 3 inches; and make the part through which the rivet passes, 143 inches thick, and rounded into the lower flange which is 4 of anl inch thick. The rivet has its axis 15 inches from ty. Its original head, q, is conical, with bases of —say 1 inch, and a of an inch, diameter; and is half an inch thick. The other head, p, is made at pleasure, being roughly hammered down while the rivet is hot, during the process of track-laying. A thin washer is shown under this head. 142. Ex. 3. A "-Cage Valve," from a Locomotive Pump. P1. VII., Fig. 73-74.- Jechanical Gonstruiction;-This valve is made in three pieces, viz. the valve proper, Fig. 74; the cage containing it, Abb'; and the flange bb'c; whose cylindrical apertureshown in dotted lines-being smaller than the valve, confines it. The valve is a cup, solid at the bottom, and makes a water tight joint with the upper surface of the flange, inside of the cage. The whole is inclosed in a chamber communicating with the pump barrel, and with the tender, or the boiler, according as we suppose it to be the inlet or outlet valve of the pump. This chamber makes a water tight joint with the circumference of the flange bb.' Suppose the valve to be the latter of the two just named. The " plunger" of the pump being forced in, the water shuts the inlet valve, and raises the outlet valve, and escapes between the bars of the cage into the chamber, and from that, by a pipe, into the boiler. 143. Graphical Construction.-Scale full size. Make the plan first, where the six bars are equal and equidistant,with radial sides. Project them into the elevation; as in Prob. 14, DIv. I.; taking care to note whether any of the bars, as E, on the back part of the cage can be seen above the valve, CD, and between the front bars, as F and G. The diameters of the circles seen in the cage are, in order, from the centre, 1A, 2}, 215, and 4 inches. The thickness of the valve, Fig. 74, is A-I of an inch, its outside height 13 inches, and the outside diameter 2-x3 inches. The diameter of the aperture in the flange is 13 inches, its length ji of an inch, and the height of the whole cage is 3 -1 inches. 144. Execution.-Observe carefully the position of the heavy lines. The section, Fig. 74, being of metal, is finely shaded. CONSTRUCTIONS IN METAL. 55 145. Ex. 4. An oblique elevation of a Bolt Head. P1. VII., Fig. 75. Let PQ be the intersection of two vertical planes, at right angles with each other; and let RS be the intersection, with the vertical plane of the paper to the left of PQ, of a plane which, in space, is parallel to the square top of the bolt head. On such a plane, a plan view of the bolt head may be made, showing two of its dimensions in their real size; and on the plane above RS, the thickness of the bolt head, and diameter of the bolt, are shown in their real size. Below RS, construct the plan of the bolt head, with its sides making any angle with the ground line RS. Project its corners in perpendcliculars to RS, giving the left hand elevation, whose thickness is assumed. 146. The fact that the projecting lines of a point, form, in the drawings, a perpendicular to the ground line, is but a special case of a more general truth, which may be thus stated.-When an object in space is projected upon any two planes which are at right angles to each other, the projecting lines of any point of that object form a line, in the drawing, perpendicular to the intersection of the two planes. 147. To apply the foregoing principle to the present problem; it appears that each point, as a", of the right hand elevation, will be in a line, a'a", perpendicular to PQ, the intersection of the two vertical planes of projection. Remembering that PQ is the intersection of a vertical planeperpendicular to the plane of the paper —with the vertical plane of the paper, and observing that the figure represents this plane as being revolved around PQ towards the left, and into the plane of the paper, and observing the arrow, which indicates the direction -in which the bolt head is viewed, it appears-that the revolved vertical plane, has been transferred from a position at the left of the plan acne, to the position, PQ, and that the centre line tu", must appear as far from PQ as it is in front of the plane of the paperi. e. e"u" =-ecu,showing also, that as e-e' is in the plane of the paper, its projection at e" must be in PQ, the intersection of the two vertical planes. Similarly, the other corners of the nut, as c', n", &c., are laid off either from the centre line tzu", or from PQ. Thus v"c"-vc, or v"'c" = sc. The diameter of the bolt is equal in both elevations. 148. Other supposed positions of the auxiliary plane PQ may be assumed by the student, and the corresponding construction worked out. Thus, the primitive position of PQ may be at the right of the 56 CONSTRUCTIONS IN METAL. bolt head, and that may be viewed in the opposite direction from that indicated by the arrow. 149. Ex. 5. A," Step " for the support of an oblique tim. ber. P1. VII., Fig. 76. Mechanical Construction.-It will be frequently observed, in the framings of bridges, that there are certain timbers whose edges have an oblique direction in a vertical plane, while at their ends they abut against horizontal timbers, not directly, for that would cause them to be cut off obliquely, but through the medium of a prismatic block of wood or iron, so shaped that one of its faces, as ab-n'b', Fig. 76, rests on the horizontal timber, while another, as ac —e"d"c", is perpendicular to the oblique timber. To secure lightness with strength, the step is hollow underneath, and strengthened by ribs, rr. The holes, hfh", allow the passage of iron rods, used in binding together the parts of the bridge. These holes are.here prolonged, as at h, forming tubes, which extend partly or wholly through the horizontal timber on which the step rests, in order to hold the step steadily in its place. 150. -Remark. When the oblique timber, as T, Fig. 76 (a), sets into, rather than upon, its iron support S, so that the dotted lines, ab and cb, represent the ends of the timber, the support, S, is called a shoe. 151. Graphical Cionstruction.-In the plate, abc is the elevation, and according to the usual arrangement would be placed above the plan, e"n"c", of the top of the step. a'b'e is the plan of the under side of the step, showing the ribs, &c. A line through nnm is a centre line for this plan and for the elevation. A line through the middle point, r, of muz, is another centre line for the plan of the bottom of the step. Having chosen a scale, the position of the centre lines, and the arrangement of the figures, the details of the construction may be left to the student. 152. Ex. 6. A metallic steam tight "Packing," for the " stuffing boxes "of piston rods. Pi. VII., Fig. 77. Mechanical Construction.-Attached to that end, b, of Fig.. 77 (a), of a steam cylinder, for instance, at which the piston rod, p, enters it, is a cylindrical projection or " neck," n, having at its outer end a flange, ff, through which two or more bolts pass. At its inner end, at p, this neck fits the piston rod quite close for a short space. The internal diameter of the remaining portion of the neck is suffi. cient to receive a ring, rr, which fits the piston rod, and has on its outer edge.a flange, t, by which it is fastened to the flange, ff, on the neck of the cylinder by screw bolts. The remaining hollow CONSTRUCTIONS IN METAL. 57 space, s, between the ring or " gland," tr, and the inner end of the neck, is usually filled with some elastic substance, as picked hemp, which, as held in place by the gland, tr, makes a steam tight joint; which, altogether, is called the "stuffing box." ~153. The objection to this kind of packing is, that it requires so firequent renewals, that much time is consumed, for instance in rail. road repair shops, in the preparation and adjustment of the packing. To obviate this loss of time, and perhaps because it seems more neat and trim to have all parts of an engine metallic, this metallic packing, Fig. 77, was invented. ABC —A'B' is a cast iron ring, cylindrical on the outside, and having inlaid, in its circumference, bands, tt', of soft metal, so that it may be squeezed perfectly tight into the neck of the cylinder. The inner surface of this ring is conical, and contains the packing of block tin. This packing, as a whole, is also a ring, whose exterior is conical, and fits the inner side of the iron ring, and whose interior, fkc, is cylindrical, fitting the piston rod closely. 154. For adjustment, this tin packing is cut horizontally into three rings, and each partial ring is then cut vertically, as shown in the figure, into two equal segments. abcd —a'b'c'd',is one segment of the upper ring; efgh-e'ff"g'h', is one segment of the middle ring, and rjlcl-rjIk'n'l, is one segment of the lower ring. The segments of each ring, it therefore appears, break joints with each of the other rings. Three of the segments, one in each partial ring, are loose, while the other three are dowelled by small iron pins, parallel to the axis of the whole packing. 155. Operation.-Suppose the interior, fkc,, of the packing to be of less diameter than the piston rod, which it is to surround. By drawing, it partly out of its conical iron case, the segments forming each ring can be slightly separated, making spaces at ab, &c., which will increase the internal diameter, so as to receive the piston rod. When in this position, let the gland, tr, be brought to bear on the packing, and it will be firmly held in place; then, as the packing gradually wears away, the gland, by being pressed further into the neck, will press the packing further into its conical seat, which will close up the segments round the piston rod. 156. Graphical Construction.-Let the scale be from one half to the whole original size of the packing for a locomotive valve chest. C is the centre for the various circles of the plan, and DD', projected up from C, is a centre line for portions of the elevation. Cf — of an inch; Ce=- 1 inches; CA =lIt inches. A'n-2 inches, and A's'=- of an inch. The iron case being constructed 58 CONSTRUCTIONS IN METAL. from these measurements, the rings must be located so that f'p', fbr instance, shall be =-, of an inch; and then let the thickness, f'f", of each segment be l" of an inch. These dimensions and the con. sequent arrangements of the rings will give spaces between the segments, as at ab; of 8 of an inch; though in fact, as this space is variable, there is no necessity for a precise measurement for it. In the plan, there are shown one segment, and a fragment, abef, of another, in the upper ring; one segment and two fragments of the middle ring, and both segments of the lower ring, with the whole of the iron case. The elevation shows one of the halves of each ring, viz., acdo-a'c'd', the upper one; ef gh-e'f' h', the middle one; and rjkl —rj'k'l', the lower one. The circles of the plan are found by projecting the points as p. upon the diameter AC. Then by comparing Hc', II', and nnl', for example, we see how the vertical projections, as k'n', of the ends of the half rings are found. 157.. Execution.-The section lines in the elevation indicate clearly the situation of the three segments, abed, efgh, and ijkl, there shown. The dark bands on the case at t and t', indicate the inlaid bands of soft metal already described. The student can usefully multiply these examples by constructing plans, elevations and sections, fiom measucrement, of such objects as steam, water, and gas cocks, valves, or gates, railway joints, and chairs, and other like simple metallic details; actual examples of which can be easily obtained as models almost anywhere. ? rl 5 r j f X -11l -'7~L - — W- |,,z,< ---------- -- - 1 0 1I^ |_ _ / / -,.~.....I. I __'1':~j' j! ~J...'-' f' -.... < I''?"... — ==' < w19 _. t \,/ 0 * +? 1 —-X -- /0- -- -i 0> -0>.;.,, 0,,,q 0 t _ c, < / I, ~. =.z-.S,:'\'"i' "',..-'??;:? 1:','-~'~ I,., C_._~ /:$_ x'W%-.-q / 7'0V-'7 ~ K'>:f/ _"L _ o~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~r, ---- I~~~~~~~~~~~~~~~~~~~~~~~~~~~~I~l:- -i'j/'-.... -J~'"' I. / I; a~~-I -'' P~~~ ~~~~~~~~~~~ j l'X..... -........ a~~~~~~~~~~~~~.d ~,.. I' _ -to -~~~~~, ~ ~ ~ ~~~~ ~C ~, -— /, ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ "-7" "~ -i-~.. _j —~~~~~~~~~~~~~~~~~3.... -— 11I~ —-I 7-' ~2 \~~~~~~~~~~~~~~~ f''?~ ~~~~~~~~!-?. 1:,,:.I,, "';,.' /.....,.~.' —:':r -,'.. -' ~~~~~~~~~~~~~~~~~~ ~ ~~~~o'"'e.... ]; F, =/ % ~..o/ tl, —-,7! /~l, ""'"i \J j'i isO.~~~~~~~~~~~~~~~~~~~ii,~,~,;// %i(f1/,ii';! /~: - -; _; ~i 1, DIVISION THItRD. RTUDIMENTS OF SHADES AND SHADOWS. CHAPTER I. SHADOWS. ~ I.-Facts, Principles, and Preliminary Problkenms. 158. RAYS of light are here assumed to proceed in parallel strcight lines. Sensibly they do thus proceed from a source as distant as the sun, as is shown by the fact that the shadows of parallel straight lines upon plane sulfaces are straight and parallel; or theat we cannot see through a bent tube. One can learn by simple inspection, from the shadows on buildings, many facts about shadows, which follow directly from the preceding principle, and which point out the solution of those simple problems, in which the shadows are cast by straight lines upon planres. 159. It will be thus observed that the shadow of a vertical edge ab, P1. VIII., Fig. 78, of the body of the house will be a vertical line, a'b', on the front wall of the wing behind it; that the shadow of a horizontal line; as be-the arm for a swinging sign-which is parallel to the wing wall, will be a horizontal line, b/e', parallel to be; that a horizontal line, be, which is perpendicular to the wing wall, will have an oblique shadow, cb', on that wall, commencing at c, where the line pierces the wing wall, and ending at b', where a ray of light through b pierces the wing wall; and finally that the shadow of a point, b, is at b' where the ray bb', through that point, pierces the surface receiving the shadow. 160. Passing now to P1. VIII., Fig. 79, which represents a chifnney upon a flat roof, we observe that the shadows of be and cdlines parallel to the roof —are b'c' and c'd', lines equal to, and parallel to, the lines be and cd; and that the shadows of ab and ed are ab' and ed'-similar to the shadow cb' in Fig. 78 —i. e. commencing 60 SHADOWS. ait ci and e. where the lines casting them meet the roof, and ending at b' and d', where rays through b andl,d meet the roof. 161. The facts just noted may be stated as elementary general principles by means of which many simple problems can be solved. 1st. The shadow of a point on any surface, is where a ray of light drawn through that point meets that surface. 2d. The shadow of a straight line, upon a plane parallel to it, is a parallel straight line. 3d. In like manner, the shadow of a circle upon a plane parallel to it is an equal circle; whose centre, only, therefore need be found. 4th. The shadow of a line upon a plane to which it is perpendicular,will coincide with the projection of a ray of light upon that plane. Thus ha, Fig..78, being perpendicular to the horizontal plane, its shadow, aa', upon that plane, has the same direction as the horizontal. projection, aa', of the ray of light, bb'. Also, be, being perpendicular to the Vertical plane cb'a', its shadow eb' upon that plane is in the direction of cb', the projection of the ray bb' upon that plane; since be is a projecting line of b. 5th. The shadow of a line upon a surface may be said to begin where that line meets that surface, either or both being produced, if necessary. See a, Fig. 79. 6th. When the shadow, aa', of a line, as ba, Fig. 78, upon one surface, as the ground, meets the intersection of that and a second surface as at a', the point as a' is a point of the shadow, as a'b', of the same line upon that second surface. 7th. To the foregoing, we may add, that the shadow of a straight line, upon a curved surface, or of a circle, otherwise than as in (3d) will generally be a curve, whose. points, separately found by (1st) must then be joined. 162. In applying these principles to the construction of shadows, three things must evidently be given, viz.1st. The body casting the shadow. 2d. The surface receiving the shadow. 3d. The direction of the light. Further, all these must be given, not in reality, but by their projections. 163. The light is, for convenience and uniformity, usually assumed to be in such a direction that its projections make angles of 45~ with the ground line (16). We will here turn aside for a moment to note the direction of the light itself, corresponding to the given projections of it. See P1. VIII., Fig. 80. SHADOWs. 61 Let a cube be placed so that one of its faces, L'L6b, shall coincide with the vertical plane, and another face, L"caLL, with the horizontal plane. The diagonals, LL, and L"L1, of these faces, will be the projections of a ray of light, and the diagonal, LL,, of the cube will be the ray itself; for the point of which L' and L" are the projections must be in each of the projecting perpendiculars L'L'Vand L"L; hence at L, their intersection. Now the right angled tri-:ngles, LL"L, and LL'Ll, are equal, but not isosceles; hence the angles LL1L" and LLL', which the ray itself,LL,,makes with the planes of projection are equal, but less than 45~. Again: in the triangle LL1G, the angle LL,G is that made by the ray LL1 with the ground line, and is more than 45~. 164. To find a shadow is really to find its bounding edges, and these are the shadows of certain edges of the given body. These edges can often be found by inspection; and they are the same as the "heavy lines," or those edges which divide the illuminated part of the body from its shaded part. Thus the line of separation between the light and shade of a body must first befound, before finding its.shadow. A very common application of this, arises in finding the shadows of rectangular timbers. Unless the light coincides with some of the faces of' the timber, its diagonally opposite edges will always be the ones which will cast shadows. 165. Having now stated the elementary facts and principles concerning shadows themselves, we proceed to show how to find their projections, by which they are represented. Returning to Art. 161, its first principle (ist) shows that we have first to learn how to find where a given line, as a ray of liglht, pierces a given surface. It will be sufficient here to show how to construct the points in which a line pierces the planes of projection. We have therefore the following Prelimin.ary Problems. PROBLEM l st. To find where a line drawn through a given point, pierces the horizontal plane. The construction is shown pictorially in Fig. 1, and in actual projection in Fig. 2. Let A, Fig. 1, be the given point, whose projections are represented by a and a'. If a line passes through a point, its projections will pass through the projections of that point, therefore if AB be the line, ab and a'b' will be its projections. Now, first, the required point, C, being in the horizontal plane, 62 SEBrows. its vertical projection, c', will be on the ground line; second, the same point being a point of the given line, its vertical projection will be on the vertical projection, a'b'c', of the line, hence at the intersection, c', of that projection with the ground line; third, the horizontal projection, c, of the point,, which is also the point itself, C, will be where the perpendicular to the ground line from d' meets -abc, the horizontal projection of the line. This explanation may be condensed into the following rule. 1st. Note where the vertical projection of the given line, produced if necessary, meets the ground line. 2cd. Project this point into the horizontal projection of the line; which will give the required point. Fig. 2, shows the application of this rule in actual projection. ab -a'b' is the given line, and by making the construction, we find c, where it pierces the horizontal plane. a C G. 1.. 2. FIG. 1. FIG. 2 EXAMPLE.- If bd (both figures)'had been less than b'd, the line would'have pierced the horizontal plane behind the ground line, that is in the horizontal plane produced backwards. Let this con. struction be made, both pictorally, and in projection. PROBLEM 2d.-To fnd where a line, drawn through a given point, pierces the vertical plane. See Figs. 3 and 4. The explanation is entirely similar to the preceding. The required point, c', being somewhere in the vertical plane, its horizontal projection, c, will be on the ground line. It being also on the given line, its horizontal projection is on the horizontal projection, abc, of the line. Hence, at once, the rule. SHADOWS. 63 1st. Note where the horizontal projection of the given line meets the ground line. a,/ d'Xi \ b FIG. 3. FIG. 4. Ocl. Project this point into the vertical projection of the line, which will give c'-C the velrtical projection of the point, which is also the point itself. EXAMPLE.-Here likewise, if b'd be less than bd, the line will pierce the vertical plane below the ground line, that is in the vertical plane extended down.ward. Let the construction be made for this case. 166. The ground line is, really,-the horizontal trace of the vertical plane; also the vertical trace of the horizontal plane. Hence the traces of any other holizontal or vertical planes may be called the ground lines of those plcnes; and therefore the preceding constructions may be applied to finding the shadows on such planes, as will shortly be seen in the solution of problems. 167. In respect to the remaining principles of (161) it only remains to note that, as two points determine a straight line, it is sufficient to find the shadow of two points of a straight line, when its shadow falls on a plane;. But if the direction of the shadow is known as in (161-2d, 4th) or if we know where the line meets a plane sulrface receiving the shadow, (161 —5th) it will be sufficient to construct one poiiet of the shadow. ~ II. -Practical Problems. 168. PRPOB. 1. To find the shadow of a vertical beam, upon a vertical weall. P1. VIII., Figt. 81. Let AA' be the beam, let the vertial wal. P1 ~III~ Fig 81.Let A~!)ethe bam, lt th 64 SEIADOWS. vertical plane of projection' be taken as the vertical wall, and let the light be indicated by the lines, as ab —a"b'. The edges which cast the visible shadow are a —a'a"; ac-a"; and ce-a"e'. The shadow of ca-a'a" is a vertical line from the point b', which point is where the ray from ac-a" pierces the vertical plane. ab-a"b' pierces the vertical plane in a point whose horizontal projection is b. b' must be in a perpendicular to the ground line firom b (Art. 15), ~and also in the vertical projection, a"b', of the ray, hence at b'. b'b is therefore the shadow of a —a"a'. The shadow of ac-a" is the line b'd', limited by d', the shadow of the point c-a" (161). The shadow of ce-a"e' begins at d', and is parallel to ce —a"e', but is partly hidden. 169. 1xecution. —The boundary of a shadow being determined, its surlface is, in practice, indicated by shading, either with a tint of indian ink, or by parallel shade lines. The latter method, affording useful pen practice, may be profitably adopted. 170. PROB. 2. To find the shadow of an oblique timber, which is parallel to the vertical plane,!uponr a similar timber resting against the back of it. P1. VIII., Fig. 82. Let AA' be the timber which casts a shadow on BB', which slants in an opposite direction. The edge ac-a'c' of AA', casts a shadow, parallel to itself, on the front face of BB'; hence but one point of this shadow need be constructed. Two, however, are found, one being a check upon the other. Any point, aa', t aken at pleasure in the edge ac-a'c' casts a point of shadow on the front plane of BB', whose horizontal projection is b, and whose vertical projection (see Prob. 1) must be in the ray aWb', and in a perpendicular to the ground line at b; hence it is at b'. The shadow of ac-a-c' being' parallel to that line, b'd' is the line of shadow. cl', the shadow of c-c', was found in a similar manner to that just descLribed. It makes no difference that b' is not on the actual timber, BB'; for the face of that timber is but a limitedjphysicaltplane, forming a portion of the indefinite immaterial plane, in which bb' is found; hence the point b' is as good for finding the direction of the indefinite line of shadow, b'd', as is d', on the timber BB', for finding the real portion only of the line of shadow, viz. the part which lies across BB'. Here, bd is taken as a ground line (166). Observe that the back upper edge of AA', which is a heavy line, as seen in the plan, is so represented only till the intersection, near b-c', of the two timbers. SHtADOTS. 65 171. PROB. 3. To find the sha(tow of a fragnment qf a horizon. tat timber, poni the h/orizontal top of an ~abutmeit on which the timber rests. P1. VIII., Fig. 83. Let AA' be the timber, and B IMP the abutment. Thle vertical edge, a' —a"a', casts a sha(low, ab, in the direction of the horizontal projection of a ray (161, 4th), and limited by the shadow of the point a —a'. The shadow of a-a' is at bM', where the ray ab —a'b pierces the top of' the abutmelnt; b1 being evidently the vertical projection of this point, and b being both in a perpendicular, b'b, to the ground line and in ab, the horizontal projection of the ray. The shadow of ad —a' is bbY, parallei to ad-a', and limited by the ray a'b'-dcb". The shadow of de-a'e' is b"c, parallel to cle —a'e', and limited by the edge of the abutment (161, 2d). Here, a"b' is used as a ground line (166). 172. PRoB. 4. To flnd the shacow of an oblique timber, upon a horizontal timber into which it is framed. P1. VIII., Fig. 84. The upper back edge, ca-c'a', and the lower front edge through ee', of the oblique piece, are those which cast shadows. By considering the point, c, in the shadow of be, Fig. 78, it appears that the shadow of acc-.a'c', Fig. 84, begins at cc', where that edge pierces the upper surface of the timber, BB', which receives the shadow. Any other point, as aac', casts a shadow, bb', onl the plane of the upper surface of BB', whose vertical projection is evidently b', the intersection of the vertical projection, a'b', of the ray ab —a'b' and the vertical projection, e'c', of the upper surface of BB', and whose horizontal projection, b, must be in a projecting line, b'b, and in the horizontal projection, ib, of the ray. Likewise the line through ee', and parallel to cb, is the shadow of the lowrer front edge of the oblique timber upon the top of BB'. This shadow is real, only so far as it is actually on the top surface of BB', and is visible and therefore shaded, only where not hidden by the oblique piece. Where thus hidden, its boundary is dotted, as shown at eca. The point, bb', is in thQ plane of the top surface of BB', produced. BRemaark. —Thus it appears that when a line is oblique to a plane containing its shadow, the direction of the shadow is unknown till bound. Let this and the following figures be made much larger. ExAMPEs. —lst. Find the shadow when the oblique timber is more nearly vertical than horizontal. 2d. Let the oblique timber ascend to the right. 173. PROB. 5. To find the shadow of the side wall of a fiight of steps upon the faces of the steps. P1. VIII., Fig. 85. The steps can be easily constructed in good proportion, without measure 66 SHADOWS. ments, by making the height of each step two-thirds of its width, taking four steps, and making the side walls rectangular parallelopipedons. The edges, aa" —a' and a —ra', of the left hand side wall are those which cast shadows on the steps. Tile former line casts horizontal shadows, as bb"-b', parallel to itself, on the tops of the steps (161, 2d), and shadows, as be"-b'c', on the fronts of the steps, in the direction of the vertical projection, a'd', of a ray of light (161, 4th) —firom the upper step down to the shadow of the point aa'. Likewise, the edge a-ra' casts vertical shadows, as g —y'h', on thefronts of the steps (159), and shadows on their tops, parallel to ad, the horizontal projection of a ray (161, 4th) firom the lower step, up to ddl', the shadow of aa'; which is therefore, where the shadows of aa —a', and a-ra', meet. a'd' is the vertical projection of all rays through points of aa"-a'; hence project down b', etc., to find the parallel shadows, b"b, etc. Likewise project up g, etc., to find the shadows, g'h', etc., ad being the horizontal projection of all rays through points of a —ra'. EXAMPLES.-lst. Vary the proportions of the steps and the direction of the light; and in each case, find the shadows; as above. 2cd. By preliminary problems 1st and 2d, find directly which surface of which step is pierced by the ray through the corner aa'. In the figure, it is the top of the third step, at dd'; found by projecting d',,on trial, into the horizontal projection of the ray ad-a'd'. 174. PROB. 6. To find the shadow of a short cylinder, or washer, t01on the vertical faoce of a board. P1. VIII., Fig. 86. Since the circular face of the washer is parallel to the vertical face of the board, BB', its shadow will be an equal circle (161, 3d), of which Nwe have only to find the centre, 00'. This point-will be the shadow of the point CC' of the washer, and is where the ray CO-C'O' pierces the board BB'. The elements, rv-r' and tuc-t', which separate the light and shade of the cylindrical surface, have the tangents r'r" and t't" for their shadows. These tangents, with the semicircle t"r", make the complete outline of the required shadow. 175. PROB. 7. To find the shadow of a nut, upon a vertical suwrface, the nuet hcaving any1 position. P1. VIII., Fig. 86. Let a'ce' —ace be the projections of the nut, and BB' the projections of the surface receiving the shadow. The edges, a'c' —ac and ce-c'e', of the nnlt cast shadlows parlallel to themselves, since they are parallel to the surface which receives the shadow. a'r&'-an SHADOWS. 6X are the two projections of the ray which determines the loint of shadow, nn'; c'o'-co are the projections of the ray used in find. ing oo', and e'r' —er is the ray which gives the point of shadow, rr'. The edges at aa' and ee', which are perpelldicular to BB', cast shadows a'n' and e'r', in the direction of the projection of a ray of light on BB'. (See cb', the shadow of cb, P1. VIII., Fig. 78.) 176. PRon. 8. To find the shadow of a vertical cylinder, on a verticalplane. PI. VIII., Fig. 87. The lines of the cylinder, CC', which cast visible shadows, are the element a-a"a', to which the rays of light are tangent, and a part of the upper base. The shadow of a —ala', is gg', found by the method given in Prob. 1. At g', the curved shadow of the upper base begins. This is found by means of the shadows of several points, bd', cc', dd', &c. Each of these points of shadow is found as g was, and then they are connected by hand, or by the aid of the curved ruler. It is well to construct one invisible point, as ut', of the shadow, to assist in locating more accurately the visible portion of the curved shadow line. 177. PROB. 9. To find the shadow of a horizontal beam, upon the slantingface of an oblique abutment. PI. VIII., Fig. 88. The simple facts illustrated by P1. VIII., Figs. 78-79, have no reference to the case of surfaces of shadow, other than vertical or horizontal. But fhey illustrate the fact that the point where a shadow, as aa', Pl. VIII., Fig. 78, on one surface,ineets another surface, is a point (a') of the shadow a'b' upon that surface. Thus this problem may be solved in an elementary manner by proceeding indirectly, i. e., by finding the shadows on the horizontal top of the abutment and on its horizontal base. The points where these shadows meet the front edges of this top and this base, will be points of the shadow on the slanting face, gne. By Prob. 3 is found ge, the shadow of the upper back edge, ax a'x', of the timber, AA', upon the top of the abutment. c, the point where it meets the front edge, ec, is a point of the shadow of AA' on the inclined face. By a similar construction with any ray, as bp —bp', is found qp, the shadow of c'b' —db upon the base of the abutment; and q, where it intersects nq, is a point of the shadow of db-c'b' on the face, qne. The point, dd', where the edge, db-c'b', meets dcc, is another point of the shadow of that edge; hence dq-d!q' is the shadow of the fiont lower edge, db-c'b', on the inclined face of the abutment. The line through cc', parallel to dq-d'q', is the shadow of the upper back edge, ax-a'x', and com Oletes the solution. 6 8 SHADOWS. 178. PROB. 10. To find the shcaow of a pair of horizontal timn bers, which are inclined to the verticalplane, upon that plane. P1. VIII., Fig. 89. Let the given bodies be situated as shown in the diagram. In the elevation we see the thickness of one timber only, because the two timbers are supposed to be of equal thickness and halved together. As neither of the pieces is either parallel or perpendicular to the vertical. plane, we do not know, in advance, the direction of their shadows. It will therefore be necessary to find the shadows of two points of one edge, and one, of the diagonally opposite edge, of each timber. The edges which cast shadows are ac —a'c' and ht —eh', of one timber, and ed-e'k and mv —a'm" of the other. All this being understood, it will be enough to point out the shadows of the required points, without describing their construction. (See bb', Fig. 81.) bb' is the shadow of aa', and dld is the shadow of cc'; hence the shadow line b' —d is determined. So uu' is the shadow of hh'; hence the shadow line u'w' may be drawn parallel to b'd'. Similarly, for the other timber, ff' is the shadow of ee', and oo' of mm'. One other point is necessary, which the student can construct. The process-might be shortened somewhat by finding the shadows of the points of intersection, p and r, which would have been the points p" and r" of the intersection of the shadows, and thus, points common to both shadows. 179. PROB. 11. To find the shadow of a pair of horizontal tim. bers, which intersect as in the last problem, upon the inclined face of an oblique abutment. P1. VIII., Fig. 90. The opening remarlks of Prob. 9 apply equally well to this present one; and the methods of solution being the same in both, only the peculiar features of the present; problem will be noticed. Since the timber, BB', lies in the same direction with the light, the intersection of its two vertical faces with the top of the abutment are its shadows on that top, and oo' and vv', where those shadows.meet cn-c'n', the front edge of the abutment, are points of the shadow on the longer oblique face of the abutment. The points r' and t', on the lower edge of the same slanting face, are similarly found, and o'v'rt' is the required shadow of BB'. The shadow line, en, of the piece AA' on the top of the abutment is found as was ge in Fig. 88. n —n' is a point of the shadow of AA' on the inclined face of the abutment. It is necessary to find two points of the shadow of its lower friont edge, bc —b'c'. cc' is one of these points, being the point where the edge be-b'c' meets the edge cn-c'n' of the abutment. In attempting to find a point J) I-~5 III. P LVIII[~~. // I' e' ___~ —~ —~-! — B... ~~~~~~~~~~~~~/ 88. a t?-I.,, _,Z d;"~~~~~~~~~~~~~~, -, ur, CI C'i~~~~~~~~~~~~~~~~~~~s c~~~~~~~~~~~~~~~~~~~~~~~~~~~~-~ ~:~/?! ~ 2ie ~b'"., a~~~~~'~~~~ I, ac _ - - _ _ _'. -~ ~ bB 2- -- -" —-'.-f I.....U! 0. -~c I b 21 ID`':X~ ~ ~ ~~~';">m: -: 1i a 8 >l ve< a; C 8/ - - _ _ _ _ _ _ _ _ _ _ _ _ ei' ~ ~ ~ ~ ~.:d' l! >.-'k', "''" "" I - _ _ _ _ ______ _ _ _ _ _ -- - j Ths M~ajorS}'napp Eni.&hlt!. t.. Sb Prk PlacNT.. SHADOWS. 69 of shadow on the lower edge, t'r', of the abutment by finding an indefinite shadow line, like pq, Fig. 88, on its lower base, we meet with the inconvenience of finding this shadow to be very fair out of the central portion of the figure, so that a point, as q, in Fig. 88, will not be below the ground line. To meet this case, take an intermediate horizontal plane, as s'd', and find its intersection, s'd' —sq, with the abutment, by projecting down the intersection of s'd' with any corner edge, upon the plan of the same edge, and by drawing through the point so found parallels, as sq, to the lower edges of the abutment. Now, bb' is a point of the lower front edge of AA'. The ray bd-b'd', drawn through it, pierces the plane s'd' at d', whose horizontal projection is in rd at d; and ds is the indefinite shadow of the lower front edge of the timber on the same plane; then ss' is a point in the required shadow on the inclined face of the abutment; and by drawing c's' and n'p, parallel to it, we have the required shadow in vertical projection. The horizontal projection of the shadow is omitted, to avoid confusing the diagram. It is contained between a line joining c and s, and a line through n, parallel to cs. 180. PROB. 12. To find the shadow of the floor of' a bridge pepon a vertical cylindrical abutment. P1. VIII., Fig. 91. The line ag-a'g' is the edge of the floor which casts the shadow. bdg —-b'e'g'n is the concave vertical abutment receiving the shadow. gg', where the edge ag-g'a' of the floor meets this curved wall, is one point of the shadow. f is the horizontal projection of the point where the ray, ef —ef', meets the abutment; its vertical projection is in e'f', the vertical projection of the ray, and in a perpendicular to the ground line, through f, hence -atf'. Similarly we find the points of shadow, dd' and bb', and joining them with f' and g', have the boundary of the required shadow. Observe, that to find the shadow on any particular vertical line, as b —"b', we draw the ray in the direction b —a; then project a at a', &c. iRenark.-The student may profitably exercise himself in changing the positions of the given parts, while retaining the methods of solution now given. For example, let the parts of the last problem be placed side by side, as two elevations, giving the shadow of a vertical wall on a horizontal concave cylindrical surface; or, let the timbers, Fig. 89, be in vertical planes, and let their shadows then be found on a horizontal surface. CHAPTER HI SHADING. 181. Yotes.-l. It is not proposed here to go largely into the theory of shading in general, but to make a special study of each one of the four practically useful exercises here given, and to notice only such principles as are involved in their elaboration. 2. The general problem for the author has been, on the one hand, to give nothing "' by rote," or arbitrarily, but rather on rational grounds; and on the other hand, to discover methods so elementary, and solutions so simple in form, as to render the problems of this DIVISION of the course perfectly intelligible to the student at this early stage of his general course of graphical study. 182. EXAMPLE 1. To shade the elevation of a vertical, right hexagonal prism, and its shadow on the horizontadl plane. P1. IX., Fig. 92. Let the prism be placed as represented, at some dis. tance from the vertical plane, and with none of its vertical faces parallel to the vertical plane. The face, A, of the prism is in tile light; in fact, the light strikes it nearly perpendicularly, as may be seen by reference to the plan; hence it should receive a very light tint of indian ink. The left hand portion of the face, A, is made slightly darker than the right hand part, it being more distant; for the reflected rays, which reach us from the left hand portion, have to traverse a greater extent of air than those firom the neigh. borhood of the line tt', and hence are more absorbed or retarded; since the atmosphere is not perfectly transparent. That is, these rays make a weaker impression on the eye, causing the left hand portion of A, from which they come, to appear darker than at tt', 183. 1Remarks.-a. It should be remembered that the whole of face A is very light, and the difference in tint between its opposite sides very slight. b. As corollaries from the preceding, it appears:first, that a surface parallel to the vertical plane would receive a uniform tint throughout; and, second, that of a series of such surfaces, all of SHADING 71 which are in the light, the one nearest the eye would be lightest, and the one furthest fiom the eye, darkest. c. It is only for great differences in distance that the above effects are manifest in nature; but drawing by projections being artificial, both in respect to the shapes which it gives in the drawings, and in the absence of surrounding objects which it allows, we are obliged to exaggerate natural appearances in some respects, in order to convey a clearer idea of the forms of bodies. d. The mere manual process of shading small surfaces is here briefly described. With a sharp-pointed camel's-hair brush, wet with a very light tint of indian ink, make a narrow strip against the left hand line of A, and soften off its edge with another brush slightly wet with clear water. When all this is dry, commence at the same line, and make a similar but wider strip, and so proceed till the whole of face A is completed, when any little irregularities in the gradation of shade can be evened up with a delicate sable brush, cdamp. only with very light ink. 184. Passing now to face B, it is, as a whole, a little darker than A, because, as may be seen by reference to the plan, while a beam of rays of the thickness ntp strikes face A, a beam having only a thickness pv, strikes fhce B; i. e., we assume,,first, that the actual brighAt. ness of a flat surface is proportioned to the number of rays of light which it receives; and, seconzcd, that its calparent brightness is, other things being the same, proportioned to its actual brightness. Also, the p.art at ac-c' being a little more remote than the line t-t', the part at a —a' is made a very little darker. 185. The face C is very dark, as it receives no light, except a small amount by reflection fiom surrounding objects. This side,. C, is dcarkest at the edge a-a' which is nearest to the eye. This agrees with experience; for while the shady side of a house near to us appears in strong contrast with the illuminated side, the shady side of a remote building appears scarcely darker than the illuminated side. This fact is explainecd as follows: The air, and particles foating in it, between the evse and the dark surfhace, C, are in the light, and reflect some liglht in a direction from the dark surflce C to the eye; and as the air is invisible, these reflections capgear to come from that dark surface. Now the more distant that surface is, the greater,will be the body of illuminated air between it and the eye, and therefore the greater will be the amount of light, apparently proceeding from the surface, and its conseqxuent apparent brightness. That is, the more distant a surface in the dark is, the 42 SHADING. lighter it will appear. It may be objected that this would make o:lt the remoter parts of illuminated surfaces as the lighter parts. But not so; for the air is a nearly perfect transparent medium, and hence reflects but little, compared with what it transmits to the opaque body; but being not quite transparent, it absorbs the reflected rays from the distant body, in proportion to the distance of that body, making therefore the remoter portions darker; while the very weak reflections from the shady side are reinforced, or replaced,by more of the comparatively stronger atmospheric reflections, in case of the remote, than in case of the near part of' that shady side. Thus is made out a consistent theory. In relation, now, to the shadow, it will be lightest where furthest from the prism, since the atmospheric reflections evidently have to traverse a less depth of darkened air in the vicinity of de-d' than near the lower base of the prism at abc. 186. Ex. 20. To shade the elevation of a ver'tical cylinder. PI. IX., Fig. 93. Let the cylinder stand on the horizontal plane. The figures on the elevation suggest the comparative depth of color between the lines adjacent to the figures. The reasons for so distributing the tints will now be given. The darkest part. of the figure may properly be assumed to be that to which the rays of light are tangent; viz., the vertical line at &t', firom which the tint becomes lighter in both directions. The lightest line is that which reflects most light to the eye. Now it is a principle of optics that the incident and the reflected ray amake equal angles with a perpendicular to a surface. But, nC is the incidellt ray to the centre, and Ce the reflected ray fiom C to the eye (12). Hence d, which bisects ne, shows where the incident ray, ed, and reflected ray, dq, make equal angles with the perpendicular (normal) dC, to the surface. Hence d-d' is the lightest element. 187. Remark.-The question may here arise, "If all the light that is reflected towards the eye is reflected from dc-as it appears to be-how can any other point of the body be seen?" To answer this question requires a notice —first, of the difference between polished and dull surfaces; and, second, between the case of light coming wholly in one direction, or principal4y in one direction. If the cylinder CC', considered as perfectly polished,'were deprived of all reflected light fiom the air and surrounding objects, the line at d —d' would reflect to the eye all the light that the body would remit towards the eye, and would appear as a line of brilliant light, while other parts, remitting no light whatever, would be totally SHADING. 73 invisible. Let us now suppose a reflecting medium, though an imperfect one, as the atmosphere, to be thrown around the body. By reflection, every part of the body would receive some light from all directions, and so would remit some light to the eye, making the body visible, though faintly so. But no body has a polish that is absolutely perfect; rather, the great majority of those met with in engineering art have entirely dull surfaces. Now a dull surface, greatly magnified, may be supposed to have a structure like that shown in P1. IX., Fig. 97, in which many of the asperities may be supposed to have one little facet each, so situated as to remit to the eye a ray received by the body directly fiom the principal source of light. 188. Having thus proved that a cylinder placed before our eyes can be seen, we may proceed with an explanation of the distribution of tints. b is midway between d and t. At b, the ink may be diluted, and at e-e', much more diluted, as the gradation fiom a faint tint at e to absolute whiteness at d should be without any abrupt transition anywhere. The beam of incident rays which falls on the segment dn, is broader than that which strikes the equal segment de; hence the segment nd is, on the elevation, marked 5, as being the lightest band which is tinted at all. The segment unr, being a little more obliquely illuminated, is less bright, and in elevation is marked 3, and may be made darkest at the left hand limit. Finally, the segment rov receives about as much light as rn, but reflects it within the very narrow limits, s, hence appears brighter. This condensed beam, s, of reflected rays would make rv the lightest band on the cylinder, but for two reasons; first, on account of the exaggerated effect allowed to increase of distance from the eye; and, second, because some of the asperities, Fig. 97, would obscure some of the reflected rays from asperities still more remote; hence rv is, in elevation, marked 4, and should be darkest at its right hand limit. 189. The process of shading is the same as in the last exercise. Each stripe of the preliminary process may extend past the preceding one, a distance equal to that indicated by the short dashes at the top of Fig. 94. When the whole is finished, there should be a uniform gradation of shade from the darkest to the light.est line, free from all sudden transitions and minor irregularities. 190. Ex. 3~. To shade a right cone standing upon the horizontal plane, together with its shadow. PI. IX., Fig. 95. The shadow of the cone on the horizontal plane, will evidently be bounded 7- 4 SHADING. by the shadows of those lines of the convex surflace, at wlich- tile ligl)t is tangrent. The vertex is commnon to both these lines, anl casts a shadow, v"'. The shadows being cast by straight lines of the conic surface, are straight, and their extremities must be in tlhe base, being cast by lines of the cone, which meet the horizontal plane in the cone's base; hence the tangents v"'t and v"'t" are tht boundaries of the cone's shadow on the horizontal plane, and the lines joining t' and t" with the vertex are the lines to which th( rays of light are tangent; i. e., they are the darkest lines of thle shading; hence tv, the visible one in elevation, is to be vertically projected at t'v'. The lightest line passes from vv' to the middle point, y, between n and p in the base. At q and at p a change in the darkness of the tint is made, as indicated by the figures seen in the elevation. In the case of the cone, it will be observed that the various bands of color are triangular rather than rectangular, as in the cylinder; so that great care must be taken to avoid filling up the whole of the upper part of the elevation with a dark shade. 191. Ex. 4~. To shacte the elevactio of a sphere. PI. IX., Fig. 96. It is evident that, if there be a s3ystem of parallel rays, tangenlt to a sphere, their points of contact will form a great circle, perpendicular to these tangents; and which will divide the light from the shade of the sphere. That is, it will be its circle of shade. Each point of this circle is thus the point of contact of one tangent ray of light. If now, parallel planes of rays, that is, planes parallel to the light, be passed through the sphere, each.- of them will cut a circle, great or small, from the sphere, and there will be two rays tangent to it on opposite sides, whose points of contact will be points of the circle of shade. In the construction, these parallel planes of rays will be taken perpendicular to the vertical plane of projection. Next, let us recollect that always, when a line is parallel to a plane, its projection on that plane will be seen in its true direction. Now BD' being the direction of the light, as seen in elevation, let BD' be the trace, on the vertical plane of projection-taken through the centre of the sphere-of a new plane perpendicular to the vertical plane, and therefore parallel to the rays of light. The projection of a ray of light on this plane, BD', will be parallel'to the ray itself, and therefore the angle made by this projection with the trace BD' will be equal to the angle made by the ray with the vertical plane. But, referring to P1. VIII., Fig. 80, we see that in the triangle SIADING. 7s LL'L, containing the angle LLL' made by the raly LL with the vertical plane, the side L'L1 is the hypothenuse of the triangle L'b L,, each of whose other sides is equal to LL'. Hence in P1. IX., Fig. 96, take any distance, Be, make AB perpendicular to BD', and,n it lay off BD= Be, then make BD' =Dc, join D and D', and DD' will be the projection of a ray upon the plane BD', and BD'D will be the true size of the angle made by the light with the vertical plane; it being understood that the plane BD', though in space perpendicular to the yertical plane, is, in the figure, represented As revolved over towards the right till it coincides with the vertical plane of projection, and with the paper. 192. We are now ready to find points in the curve of shade. oo' is the vertical projection of a small circle parallel to the plane BD', and also of its tangent rays. The circle og'o', on oo' as a diameter, repre8ents the same circle revolved about oo' as an axis and into the vertical plane of projection. Drawing a tangent to og'o', parallel to DD', we find g', a visible point of the curve of shade, which, when the circle revolves back to the position oo', appealrs at g, since, as the axis oo' is in the vertical plane, an are, g'g, described about that axis, must be vertically projected as a straight line. (See Art. 31.) In a precisely similar manner are found h, k, m, and f. At A and B, rays are also evidently tangent to the sphere. Through A, f, &c., to B the visible portion of the curve of shade may now be sketched. 193. The most highly illuminated point is 90~ distant fiom the great circle of shade; hence, on ABQ, the revolved position of a great circle which is perpendicular to the circle of shade, lay off k'Q =qB —the chord of 90~, and revolve this perpendicular circle back to the posiIiou qq', when Q will be found at r'. But the brilliant point, as it appears to the eye, is not the one which receives most light, but the one that reflects most, and this point is midway, in space, between r' and r, i. e. at P, found by bisecting QB, and drawing RP; for at R the incident ray whose revolved position is Qr, parallel to ID', and the reflected ray whose revolved position is rB, make equal angles with R'R, the perpendicular (normal) to the surface of the sphere. 194. In regard now to the second general division of the problem -the distribution of tints; a small oval space around P should be left blank. The first stripe of dark tint reaches from A to B, along the curve of shade, and the successive stripes of the same tint extend to BqA on one side of Bk/A, and to ocw, on the otLer. Then take SHADING. a li:ghter tint on the lower half of the next zone, and a still lighter one on its upper half (2) and (3). In shading the next zone, use an intermediate tint (3-4), and in the zone next to P a very light tint on the lower side (4), and the lightest of all on the upper side (5). After laying on these preliminary tints, even up all sudden transitions and minor irregularities as in other cases. Errors in Shadcing, Relative darkness of the light and shade, etc. The most frequent faults to guar'd against are —lst. A brush too wet, or too long applied to one part of the figure; giving a ragged or spotty appearance. 2d. Outlines inked in black, whereas the form and outlines of actual objects are indicated, not by black edges, but by contrast of shade only, with edges lighter than other parts. 3d. liuch too little contrast between the shading of the parts in the light and those in the dark. The former should generally, as on a cylinder, be very mnuch lighter, or not one quarter as dark as the parts in the dark. This is confirmed by examining photographs of objects illuminated in the manner supposed in this chapter. There arejive methods of shading, as follows: 1st. Softened wet shading. This is done with a large brush, quite wet, and is applicable to large figures. 2d. Softened dry shading. This is done with a brush, nearly dry, and is applicable to figures of the size of those on P1. IX., or not much larger. 3d. Shading by superposed Jfat tints. This method is very neat and effective for large figures, not to be closely examined. It consists of the preliminary stripes of the 2d method evenly and smoothly done, without softening their edges (183 d). 4th. Stippling, or dot shading. This is done with a fine pen as in making sand in topographical drawing. 5th. _Line shading. This is done by means of lines of graded size and distance apart; as in wood and lithographic mechanical engravings. The details of these methods are further explained in my "Drafting Instruments and Operations." DI)IVISI O N FO UR nTH. ISOMIETRICAL DRAWING) AND CABINET PROJECTIONS. CHAPTER I. FIRST PRINCIPLES OF ISOMETRICAL DRAWING. 195. FRoM what has been seen in the course on Projections (DIVISION FIRST), it appears that when two lines are equal, and both parallel to the planes of projection, as in P1. I., Figs. 1, 2, their projections are equal. This has also frequently been illustrated in the problems of DIVISION SECOND. 196. It has also been seen, that while a line viewed perpendicularly, i. e., a line parallel to the planes of projection, is seen in its true size,-one which is viewed obliquely, appears shorter than it really is. This has been repeatedly shown by comparison of the two projections of the edges of a nut, in the problems of DIVIsION SECOND. 197. From the foregoing, it is not difficult to see that if any two lines be equal and parallel to each other in space, their projections will also be equal and parallel to each other. This is easily proved, however; for as the lines in space are equal and parallel, the distance between the projecting lines of one of them is equal to the distance between the projecting lines of the other. Hence the projections of equal and parallel lines are equal and parallel-to each other —but not to the lines in space, when the latter are oblique to the planes of projection. 198. But it is not the parallelism, but the equal inclination that makes the projections equal; hence equal lines, equally inclined to the planes of projection, will have equal projections. Conversely, If equal projections are the projections of equal lines, those lines must be equally inclined to the planes of projection. 78 FIRST PRINCIPLES OF ISOMETRICAL DRAWING. 199. To apply the foregoing principles to the developrment of thie elements of Isometrical Drawing, let us take a cube, and begin by supposing that we see it in right vertical projection, as in P1. IX., Fig. 98. In this case, but one side of the cube will be visible, and that will appear as a square. Next, suppose the cube to be turned horizontally, until two of its vertical faces should become visible. Evidently it can be turned just so far as to show those two faces equally, as in P1. IX., Fig. 99, where a-ca' and b —b' are equal. If now, without turning the cube any more to the right or left, it should be turned forward about any axis parallel to the ground line, we should see the upper face in addition to the two faces seen in the last elevation, P1. IX., Fig. 99. It is also plain that the cube may thus be turned up just far enough to make these three visible faces appear equal, as in P1. IX., Fig. 100. Then the edge, c-c', Fig. 99, will appear equal to Ce" and Cf", Fig. 100. 200. Let us now observe carefully the. facts which result when the corner dd' is turned up just far enough to make the three visible faces of the cube appear equal. 1st. When the faces of the cube appear equal, the sides of those faces, i. e., the edges of the cube must appear equal; hence a-a', b-b', c-c', Fig. 99, for example, will be equal in the new projection, Fig. 100, but less than the edges themselves. 2nd. Being equal, both in projection and in space, these edges must be equally inclined to the plane of projection (198). 3rcld. These edges being equal, equally inclined to the plane of projection, and also equally inclined to each other —being at right angles to each other in space —their included angles will be equal in projection. (See PI. III., Fig. 35, the plan.) Hence, as a first step in constructing the new projection, lay off from any assumed point, C, P1. IX., Fig. 100, three lines making angles of 1200 with each other; observing that if one, Cn, be vertical, the other two, Cf" and Ce", will make angles of 30~ with a horizontal line through C, and can therefore be easily drawn with the 300 triangle. 4th. The diagonal e —e'f' of the upper base, P1. IX., Fig. 99, remaining parallel to the vertical plane, will still be seen in its true size; hence from C, PI. IX., Fig. 100, lay off half of the diagonal ef —ef' from Fig. 99, each way on the line EF, and at E and F erect perpendiculars, whose intersection with Ce" and Cef" will give the required diagonal in its true position, and consequently Ce" and Cfv, two sides of the required projection, and projections of ce —d'e' and f —d'f', Fig. 99. MIake now Cn=Ce"o FIRST PRINCIPLES OF ISOMETRICAL DRAWING. i9 5th. All the other edges of the cube are equal and parallel to Ce"' Cf:", and Cn. They are therefore readily drawn, and the required projection completed. This projection is called the Isometrical Prqjection of the cube. 6th. We see now, by inspection, that the perimeter, Df "onpe", of this projection is a regular hexagon, of whose circumscribing circle, C is the centre, and Ce", Cf", and Cn, are equidistant radii. It also appearsl' that Fig. 100 could have been constructed by taking e'f"=-ef from Fig. 99, as the side of an inscribed equilateral triangle, by then inscribing a hexagon in the circumscribing circle of that triangle, and by then drawing Ce", Cf", and Cn. Again, the three face diagonals, e'f", f'"n, and ne", being homologous lines of equal and similarly placed faces, are all parallel to the plane of projection, and therefore equal in projection. Finally, the body diagonal of the cube, through C, is projected at C, Fig. 102, the projection of the extreme front and back corners of the cube. But when a straight line is projected in a point, it is perpendicular to the plalne of projection. Hence in isom-etrical projection, the plane of' projection is penredicular to the body diagonal through the foremost corner of a cube. 201. Summing up in the form of definitions: 1st. Isometrical projection is that in which the three plane right angles, which form a solid right angle, appear equal in projection. 2nd. The three lines meeting at C are called isometric axes. 3rd. The point C is called the isometric centre. 4th. Planes containing any two of the isometric axes, or parallel to these axes. are called isometric planes. 5th. Lines parallel to the isometric axes are called isometric lines. Others are called non-isometric lines. 202. It is to be noticed, that while it is the three diagonals only of the faces of the cube, Fig. 100, that are seen in their real size, or on the same scale as that used for the ordinary plan and elevation, yet the edges are the lines of which the measurements are usually given, and which it is therefore desirable to represent on the same scale in all its projections. If, therefore, a figure be constructed, as in P1. IX., Fig. 101, similar to Fig. 100, but having its edges, Da, &e., equal to the edges of the plan in Fig. 99, such a figure will be called an Isometrical Drawinfg of the cube shown in Fig. 99, while it would be the strict isometrical projection of a larger and imaginary cube. This imaginary cube is to Fig. 101, as Fig. 99 is to its projection in Fig. 100. 80 FIRST PRINCIPLES OF ISOMETRICAL DRAWING. 203. Again, it is evident that it is the du'reczton of the projection of the edges of a solid riioht angle whic;h determines whether the projection is isometrical or not; hence, not merely cubes, but any rectangular objects, can readily be put into isornetrical projec, tion. 204. Recollecting that isometric lines of the object in space are at right angles to each other, and are seen in the isometrical cdrawing on the scale of the plan and elevation, and recollectiug that in most pieces of carpentry, in boxes, furniture, &c., the principal lines are at right angles to each other, we see at once the great convenience of isometrical drawing, which shows the true dimensions of objects in three directions on the same figure, and gives a clearer idea of the construction of objects than plans and elevations do. The disagreeably distorted appearance, however, of large or complicated structures, as buildings or machines, when shown in isometrical projection, causes this method to be confined, in its application, to small objects, details, &c. 205. It is often convenient to know in advance the extreme height of the isometrical drawing of a given object, in order that it may be better arranged upon the paper. See P1. X., Fig. 103, for example. Da being 10 inches-by scale-the distance of ca below a horizontal line throughiD, is the sine of 30~ to a radius of 10. Hence, as sin. 30~=-, that distance is 5 inches. In like Imn.a ner aC being 4 inches, C is 2 inches below a; and Cc being 3~ inches, c is 101 inches below D. 205a. As a final prelirninary, it is to be noted, thalt, as no planes of projection are commonly employed in is:)metrical drawing, the direction of the light is defined with respect to a cube as a standard body of reference. Thus, Fig. 102, the ray entering at a is sup. posed to leave the cube at I, and all other rays are parallel to tblip. CHAPTER II. PROBLEMS INVOLVING ONLY ISOMETRIC LINES. PROB. 1. To construct the isometrical drawing of a given cube; its heavy lines, and the shadows upon it, of two of its edges, produeed. 206. 1st Method. P1. IX., Fig. 98 represents the given cube. To make its isometrical drawing, Fig. 101, take any point, C, as a centre and describe a circle with a radius CD=ab inl Fig. 98. Cc being made vertical, lay off the radius three times each way from c, complete the hexagon thus determined and draw Ca and Cb with the 30~ triangle. 2d MfIethod. Draw the three axes, and lay off an inch on each of them. The remaining lines, being equal and parallel to these, are drawn with the 30~ triangle, placed in different positions, which will readily suggest themselves, on attempting the construction. 207. By 205 (a) three faces of the cube, P1. IX., Fig. 102, will be illuminated by light passing in the direction LL', from a to p. Two of these faces, aonC, and aCqD are visible. The under and right hand faces, onp.C, npqC, and pqDC, are in the dark, hence by the rule (18) the edges on, nC, Cq and qD are to be inked heavy. The shadows, of ab,=oa produced; and of ad, =Da produced. ab is perpendicular to aCqD, hence (161, 4th) its shadow will be in the direction of the projection of the light upon that face. But aq is evidently the projection of the ray, LL' upon aCqD; and as a is where ab meets aCqD (161, 5th), ab' is the shadow of ab on aCqD. In like manner, ad' is the shadow of ad upon aonC. Otherwise. A plane containing oa and ap, will contain rays through points of oab, and will also contain pq and Mwill therefore cut aCqD in the line aq. Hence rays fiom all points of ab will pierce aCqD in aq. Hence ab' is the shadow of ab. Again; Dap is the plane of rays, containing Dd and pn, and cutting the face actonC in the line an. Hence d' is the shadow of d, and ad', that of ad. 208. PROB. 2. To represent a prismatic block, as cut from a 82 PRIOBLEMIS INVOLVING ONLY ISOMEfRIC LINES. corner of the cube. PI. IX., Fig. 101. This problem involves the construction of points in the isometric axes. Let the cube now be supposed to be one whose edge is five inches long, and let it be drawn to a scale of one fifth, or of one inch to five inches. Suppose a piece, each of whose sides is a rectangle, to be cut from the corner nearest the eye. Let Ca'-2 inches, C'=3- inches, and Cc'= 1 inch. Lay off these distances from C, upon the axes, and through the points a', b', and c', draw isometric lines, as a'cd and b'd, which, by their intersections, will complete the portion cut from the cube. EXAMPLES. —It. Let a square panel be sunk in the centre of each visible face. 2de. Let a square opening in the centre of each face extend quite throicugh the cube. 3dl. Let there be a square tablet upon each face. 209. PROB. 3. To construct the isometrical cdrawing of a carpenter's oil stone box. P1. X., Fig. 103. This problem involves the finding of points which are ill the planes of the isometric axes. Liet the box containing the stone be 10 inches long, 4 inches wide, and 3 inches high, and let it be drawn on a scale of t. Assume C then, and make Ca=4 inches, Cb=10 inches, and Cc_ 3- inches, and by other lines aD, Db, &c., equal and'parallel to these, complete the outline of the box. Represent the joint between the cover and the box as being 1 inch below the top, aCb. Do this by making Cp —1 inch, and through p drawing the isometric lines which represent the joint. Suppose, now, a piece of ivory 5 inches long and 1 inch wide to be inlaid in the longer side of the box. Bisect the lower edge at d, make de= 1 inch, and ee'- =1 inch. Through e and e', draw the ftop and bottom lines of the ivory, making them 2,- inches long on each side of de', as at e't, e't'. Draw the vertical lines at t and t', which will complete the ivory. 210. To show another way of representing a similar inlaid piece, let us suppose one to be in the top of the cover, 5 inches long and 1I inches wide. Draw the diagonals ab and CD, and through their intersection, o, draw isometric lines; lay off oo'=21 inches, and 0oo"= of an inch, and lay off equal distances in the opposite directions on these centre lines. Through o', o", &c., draw isometric lines to complete the representation of the inlaid piece. 211. PROB. 4. To represent the same box (Prob. 3), with the cover removeld. P1. X., Fig. 104. td _)"",.,,, -r..............',,- ~ r —'~,~ —~.... ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~,,_....-_., to~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ 4~~~~~~. i,L~~~~~~~~~~~~~~~.I~~~~~~~, /7 —-- I - -cp — ~ —.~~~~~ K\ ~._-:-'...j_:.. ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~G / ~ ~ ~; —I _ -o - -Ii ~~~~-~~~~~;("t~~~~~~~~ /~~~~~~~i 0 ~ ~~~~~~~ IC_~ L — ~ ----- ------- -- -_____. _ -- ---- PROBLEMS INVOLVING ONLY ISOMETRIC LINES. 83 This problem involves the finding of the positions of points not in the isometric planes. Supposing the edges of the box to be indicated by the same letters as are seen in Fig. 103, and supposing the body of the box to be drawn, 10 inches long, 4 inches wide, and 2~ inches high; then, to find the nearest upper corner of the oil stone, lay off on Cb 12 inches, and through the point f, thus -found, draw a line, ff', paral. lel to Ca. On Jf', lay off 1 inch at each end, and from the points Ah', thus found, erect perpendiculars, as Ahn, each 4 of an inch lonag. Make the further end, z, of the oil stone 1' inches foomn the fiurther end of the box, and then complete the oil stone as showvn. To find the panel in the side of the box, lay off 2 inches firom each end of the box, on its lower edge; at the points thus found, erect perpendiculars, of half an inch in length, to the lower corners, as p, of the panel; make the panels 1.- inches wide, as at a21', and ~ an inch deep, as at pr, which last line is parallel to Ca; and, with the isonietric lines through r, p, p', and t, completes the panel. 212. TIe manner of shacdizng this figture will now be explained. The top of the box and stone is lightest. Their ends are a trifle darker, since they receive less of the light which is diffused through the atmosphere. The shadow of the oil stone on the top of the box is much darker than the surfaces just mentioned. The shadow of the fobiemost vertical edge of the stone is found in precisely the saine way as Nwas the shadow of the wire upon the top of the cube, P1. IX., Fig. 102. The sides of the oil stone and of the box, which are in the dark, are a little darker than the shadow, and all the surfaces of the panel atre of equal darkness and a trifle darker than the other dark surflaces. In order to distinguish the separate faces of the panel, when they are of the same darkness, leave their edges very light. The little light which those edges receive is mostly perpendicular to them, regarding them as rounded and polished by use. These light lines are left by tinting each surface of the panel, separately, with a small brush, leaving the blank edges, which may, if necessary, be afterwards made perfectly straight by inking them with a light tint. The upper and left hand edges of the panel, and all the lines corresponding to those which are heavy ill the previous figures, may be ruled with a dark tint. In the absence of an engraved copy, the figures will indicate tolerably the relative darkness of the different surfaces; 1 being the lightest, and the numbers not being consecutive, so that they may assist in denoting relative differences of tint. When this figure is thus shaded, its edges should not be inked with ruled lines in black ink, but should be inked with pale ink CHAPTER III. PROBLEMS INVOLVING NON-ISOMETRICAL LINES. 213. SINCE non-isometrical lines do not appear in their true size, each point in any of them, when it is in an isometric plane, must be located by too isometric lines, which, on the object itself, are at right angles to each other. Points, not in any known isometrical plane must be located by three such co-ordinates, as they are called, from a known point. 214. PROB. 5. To construct tile isometrical drawing of the scarfed splice, shown at P1. VII., Fig. 66. Let the scale be -I-, or three fourths of an inch to a foot. In this case, P1. X., Fig. 105, it will be necessary to reconstruct a portion of the elevation to the new scale (see P1. X., Fig. 106), where Ap-1=l2 feet,pA"-=6 inches, and the proportions and arrangement of parts are like P1. VII., Fig. 66. In Fig. 105, draw AD, 3 feet; make DB and AA' each one foot, and draw through A' and B isometric lines parallel to AD. Join A-B, and from A lay off distances to 1, 2, 3, 4, equal to the corresponding distances on Fig. 106. Also lay off, on BK, the same distances from B, and at the points thus found on the edges of the timber, draw vertical isometric lines equal in length to those which locate the corners of the key in Fig. 106. Notice that opposite sides of the keys are parallel, and that AV, and its parallel at B, are both parallel to those sides of the keys which are in space perpelldicular to AB. To represent the obtuse end of the upper tim ber of the splice, bisect AA', and make va-=Aa, Fig. 106, and draw Aa and A'a. Locate m by va produced, as at ar, Fig. 106, and a short perpendicular=rm, Fig. 106, and draw nzV and mnV'; VV' Wbeing parallel to AA', and arm being parallel to AV. To represent i;he washer, nut, and bolt, draw a centre line, vv', and at t, the middle point of Vn, draw the isometric lines tuz and ue, which will give e, the centre of the bolt hole or of the bottom of the washer. A point -coinciding in the drawing with the upper firont corner of the nutis the centre of the top of the washer, which may be made - of aln inch thick. P1'OBLEMS INV'OLVIN G NON-ISOME3LI0MTCAL LINES. 85 Tllhrough the above point draw isornetric lilies, rr' andii2', and lay off on them, friom the same point, the radius of the washer, say 2~ inches, giving four points, as o, through which, if an isometric square be drawn, the top circle of the washer can be sketched in it, being tangent to the sides of this square at the points, as o, and elliptical (oval) in shape. The bottom circle of the washel is seen throughout a semi-circumference, i. e. till limited by vertical tangents to the upper curve. On the same centre lines, lay off from their intersection, the half side of the nut, 1 () inches, and from the three corners which will be visible when the nut is drawn, lay off on vertical lines its thickness, 14 inches, giving the upper corners, of which c is one. So much being done, the nut is easily finished, and the little fiagment of bolt proejecting through it can be sketched in. EXAMPLE.- Construction of the nut when set obliquely. The nut is here constructed in the simplest position, i. e. with its sides in the direction of isometric lines. If it had been determined to construct it in any oblique position, it would have been necessary to have constructed a portion of the plan of the timber with a plan of the nut-then to have circumscribed the plan of the nut by a square, parallel to the sides of the timber-then to have located the corners of the nut in the sides of the isometric drawing of the circumscribed square. Let the student draw the other nut on a large scale and in some such irregular position. See Fig. 107, where the upper figure is the isometrical drawing of a square, as the top of a nut; this nut having its sides oblique to the edges of the timber, which are supposed to be parallel to ca. 215. PROB. 6. To make an isometrical drawing of an oblique timber framed into a horizontal one. P1. X., Fig. 108. Let the original be a model in which the horizontal piece is one inch square, and let the scale be -. Make ag and ah, each one inch, complete the isometric end of the horizontal piece, and draw ad, hn and gk. Let ab=-2 inches, bc=2~ inches, and draw bin. Let the slant of the oblique timber be such that if cd=2 inches, de shall be 3 of an inch. Then by (213) cle, parallel to ag, will give ce, which does not show its true size. Draw edges parallel to ce through b and qn. In like mannerf can be found, by distances taken fiom a side elevation, like P1. VII., Fig. 59. 216. PROB. 7. To make an isometrical crawing of a pyramid standing upon a recessed pedestal. P1. X., Fig. 109. (From a 86 P1ROB3LENIS INVOLVING N ON-ISOII METRICAL LINES. Model.) Let the scale be 1. Assuming C, construct the isometric square, CABD, of which each side is 4- inches. From each of its corners lay off on each adjacent side 14 inches, giving points, as e andf; firom all of these points lay off, on isometric lines, distances of' 3 of an inch, giving points, as g, k, and h. From all these points now found in the upper surface, through which vertical lines can be seen, draw such lines, and make each of them one inch in length, and join their lower extremities by lines parallel to the edges of the top surface. Isometric lines through g, m, &c., give, by their intersections, the corners of the base of the pyramid, that being in accordance with the construction of this model, and the intersection of AD and CD is o, the centre of that base. The height of the pyramid —l 5 inches -is laid off at ov. Join v with the corners of the base, and the construction will be complete. 217. To find the shadows on this model.-According to prin. ciples enunciated in DIVISION III., the shadow of f]E begins at A, and will be limited by the line hs, s being the shadow off, and the intersection of the rayfs with As. st is the shadow offF. According to (207), w is the shadow of v, and joining w with p and q, the opposite corners of the base, gives the boundary of the shadow on the pedestal. kac is the boundary of the shadow of gic on the face ku. The heavy lines are as seen in the figure. If this drawing is to be shaded, the numerals will indicate the darkness of the tint for the several surfaces —1 being the lightest. 218. PROB. 8. To construct the isometrical drawing of a wall in batter, with counterforts, and the shadows on the wall. P1. X., Fig. 110. A wall in batter is a wall whose face is a little inclined to a vertical plane through its lower edge, or through any horizon. tal line in its face. Counterforts, or buttresses, are projecting parts attached to the wall in order to strengthen it. Let the scale be onze fourth, i.e., let each one of the larger spaces on the scale marked 40, on the ivory scale, be taken as an inch. Assume C, and make CD = 2 inches and CI = 10 inches in the right and left hand isometrical directions. Make DF=6 inches, EF-1i. inches, FG-i=l inches, and GH=1- inches, and join H and E, all of these being isometrical lines. Next draw EC, then as the top of the counterfort is one inch, vertically, below the top of the wall, while EC is not a vertical line, make Fd —=one inch, draw de paral. lel to FE and ef, parallel to ElH. Make ef and Ca each one inch, at f make the isometric lillefg=] of an inch, and at a make ab=2 PROBLEMS INVOLVING NON-ISOMTRICAL LINES. 8 7 inches, and draw bg. Make bc= 1. inches, draw ch parallel to bg arld complete the top of the counterfort. The other counterfort is similar in shape and similarly situated, i. e., its furthermost lower corner, k, is one inch from I on the line IC, while HI is equal and parallel to CE. 219. To fizd the shadows of the counterfo rts on the face of the wall.-We have seen —P. VIII., Fig. 78-thllat iwhen a line is perpendicular to a vertical plane, its shadow on that plane is in the direction of the projection of the light upon the same plane, and from P1. IX., Fig. 102, that the projection, an, of a ray on the left hand vertical face of a cube makes on the isometrical drawing an angle of 600 with the horizontal line nn'. Now to find the shadow of ann. The fiont face of the wall-P1. X., Fig. 110-not being vertical, drop a perpendicular, fo, from an upper back corner of one of the counterforts, upon the edge ba, produced, of its base, and through the point o, thus found, draw a line parallel to CI. nzfoyp is then a vertical plane, pierced by ian, an edge of the further counterfort which casts a shadow; and n1p is the direction of the shadow of mnn on this vertical plane. The shadow of mzn on the plane of the lower base of the wall is of course paallel to, an, and p is one point of this shadow, hence pq is the direction of this shadow. Now ni, where mn pierces the actual face of the wall, is one point of its shadow on that face, and q, where its shadow on the horizontal plane pierces the same face, is another point, hence nq is the general direction of the shadow of mnn on the front of the wall, and the actual extent of this shadow is nr, r being where the ray mr pierces the front of the wall. From r, the real shadow is cast by the edge, mu, of the counterfort. r, the shadow of mi, is one point of this shadow, and s, where umn produced, meets'yn produced, is another (in the shadow produced), since s is, by this construction, the point where urn, the line casting the shadow, pierces the surface receiving the shadow. Hence draw srt, and urt is the complete boundaryof the shadow sought. The shadow of the hither counterfort is similar, so far as it falls on the face of the wall. Other methods of constructing this shadow may be devised by the student. Let t be found by means of an auxiliary shadow of uml on the plane of the base of the wall. Reimark.-In case an object has but few isometrical lines, it is most convenient to inscribe it in a right prism, so that as many of its edges, as possible, shall lie in the faces of the prism. CHAPTER IV. PROBLEMS INVOLVING THE CONSTRUCTION AND EQUAL DIVISION O0 CIRCLES IN ISOMETRICAL DRAWING. 220. PROB. 9. To make an exact construction of. the isometrzcat drawing of a circle. P1. X., Figs. 111-112. This construction is only a special application of the general problem requiring the con. struction of points in the isometric planes. Let P1. X., Fig. 111, be a square in which a circle is circumscribed. The rhombus-Fig. 112-is the isometrical drawing of the same square, CA being equal to C'A'. The diameters g'h' and e'f are those which are shown in their real size at gh and ej; giving gf, A, e, and f as four points of the isometrical drawing of the circle. In Fig. 1111, draw b'a', from the intersection, b', of the circle with A'D' and parallel to C'A'. As a line equal to b'a', and a distance equal to A'a' can be found at each corner of Fig. 111, lay off each way from each corner of Fig. 112, a distance, as Aa, equal to A'a', and draw a line ab parallel to CA and note the point b, where it meets AD. Similarly the points n, o, and r may be found. Having now eight points of the ellipse which will be the isometrical draw. ing of the circle, and knowing as further guides, that the curve is tangent to the circumscribing rhombus at g, ht, e andf, and perpen. dicular to its axes at b, n, o and r, this ellipse can be sketched in by hand, or by an irregular curve. 221. If, on account of the size of the figure, more points are desirable they can readily be found.. Thus; on any side of Fig. 111, take a distance as C'c' and c'd' perpendicular to it, and meeting the circle at d'. In Fig. 112, make Cc=C'c', aud make cd equal to c'd' and parallel to CD, then will d be the isometrical position of the point d'. 222. PROB. 10. To make an ap2proximate construction of the iso metrical drawing of a circle. P1. X., Fig. 113. By trial we shall find that an arec, gf, having C for a centre and Cf for its radius, will very nearly pass through n; likewise that an are c, with B for a centre, will very nearly pass througb r. These arcs will be tangents to the :Nail;,'"~~~~~~~~~~~~~~~~,~ ~ ~~~~~~~~~~~I H-~~~~~~~~~.....~_~..........:-..,.: ~r._.__!I,., /'.], ii _________________ ~~ ~ ~ ~__ i —. —---:' ================ =I === _____ ___ -~ ~~ ~ 1'i~','",', ~~~~I- I t _ Ii~~~~...-:':: I" — /