"TE^ ^^ Digitized by tine Internet Archive in 2008 with funding from IVIicrosoft Corporation http://www.archive.org/details/descriptivegeomeOOyounrich BuGineenng Science Series DESCRIPTIVE GEOMETRY ENGINEERING SCIENCE SERIES EDITED BY DUGALD C. JACKSON, C.E. Professor of Electrical Engineering Massachusetts Institute of TEcnNOLOQY Fellow and Past President A.I.E.E. EARLE R. HEDRICK, Ph.D. Professok of Mathematics, Univeksitv of Missouri Member A.S.M.E. DESCRIPTIVE GEOMETRY BY GEORGE YOUNG, Jr. PROFESSOR OF ARCHITECTURE CORNELL UNIVERSITY AND HUBERT EUGENE BAXTER ASSISTANT PROFESSOR OF ARCHITECTURE CORNELL UNI^^ERSITY ^ • /»\ L. Tr '• . • i^*" o '* ' * » ' THE MACMILLAN COMPANY 1921 All rights reserved (^A^i>l PRTNTEP IN THE FNITI'D RTATEB OF AMERICA Engineering Library' Copyright, 1921, bt the macmillan company. Set up and electrotypeA Published October, 192 1. bA^ Kortoooti 5Prf33 J. S. Gushing Co. — Berwir-k & Smith Co. Norwood, Mass., U.S.A. PREFACE In adding to the already long and excellent list of texts on Descriptive Geometry, the ^^Titers have not attempted to present any new abstract material nor even to include all of the standard problems. On the other hand, they have sought to avoid a presentation which dwells solely on the practical side of the subject. The value of Descriptive Geometry in developing the imagi- nation, through the ability to visualize, has, for a long time, been fully recognized, not only in the schools, but also by men who have won recognition and maturity in the practice of their professions. The average technical student, however, while eager in the pursuit of any subject which seems to offer an immediate and definite return, is reluctant to apply himself to a subject whose value is not at once apparent. In preparing this text, the attempt has been made to hold the student's attention by means of introductory paragraphs and other explanatory matter intended to show the relation of the principles under discussion to structural work. At the same time the treatment of the various subjects is kept purely abstract in order to avoid, in so far as is possible, the tendency of the whole subject to degenerate into practical rules and formulas. Finally, there has been added a set of exercises designed to show the application of the abstract ideas to concrete, work-a-day problems. These applications have been grouped apart from the abstract problems in order to make it possible to use them with more or less freedom as the case may require, and also to emphasize the secondary and dependent character of such problems. V 451780 vi PREFACE Believing that the chief value of Descriptive Geometry lies in its imaginative quality, the ^\Tite^s have attempted so to present the subject matter as to encourage intuitive rather than rigidly formal methods. For this reason the usual terminology has occasionally been avoided. The use of first quadrant solu- tions has been adopted as being more natural, easier to visualize, and in no wa}^ incompatible with actual practice. Some parts of the book, notably the chapters on Curves, Topography, and Other Methods of Projection, have been intro- duced mainly as reference matter, though they might be in- cluded as a part of the regular course where time permits. More than the usual amount of attention has been given to the elementary principles of projection, in order to lay a very thorough foundation for later w^ork. Exercises intended for classroom work or review are distributed through the text, and a parallel set, intended for drafting-room w^ork, is given in the appendix. Each set of exercises covers the same ideas and in the same order, so that they may be used interchangeably. In preparing the course on which this text is based, the writers have freely used the standard works, particularly those of Pillet, Warren, AYillson, Smith, and Anthony and Ashley. The use of a limited number of models has been found advan- tageous. The method has been to require the student to draw a freehand sketch, in plan and elevation, from the model and to complete his working drawing from the sketch, without further reference to the model. CONTENTS Chapter Page I Purpose and Scope .1 II Projection of Points .10 III Projection of Lines 20 IV Projection of Points and Lines in the Second, Third AND Fourth Quadrants ...... 42 V Plane Figures and Solids ...... 53 VI Problems Dealing with Points, Lines, and Planes 66 VII Curved Lines 119 VIII Curved Surfaces . 153 IX Applications for Chapters I to V . . . . 204 X Applications for Chapter VI .... . 212 XI Applications for Chapters VII and VIII . . . 226 XII Shades and Shadows ....... 242 XIII Other Methods of Projection ..... 262 Appendix 275 GENERAL EXPLANATIONS AND INSTRUCTIONS Notation. The system of indication and notation shown in the accompanying cuts is used throughout the text. Students are expected to use this system for all exercises. It should be recognized, however, that this notation is merely a means to an end. Some system is needed in the earlier problems in order to make the answers definite and complete ; in the later prob- lems, a good system simplifies the actual processes of the solution. While system and systems are valuable aids, they should never be allowed to take the place of a thorough comprehension of the problem and a visualization of the facts. The system here shown is that followed generally throughout the text. It has been changed occasionally to suit the requirements of indi- vidual illustrations. Exercises. The problems in the body of the text are intended for classroom or home work. In general, they are arranged in groups, each group treating a single idea. They are intended to be laid out on 8''xl0^'' paper, the standard loose leaf note- book size, with margins, etc., as shown on page xii. The problems given in the appendix cover the same ground as those in the text, but are intended for drafting-room work. They are arranged to go either on 8"xl0^'' sheets, or on 15''X22'' sheets (standard *' Imperial " size, cut in two), as shown on page xii. It will be noticed that the 15''x22'' sheet will accommodate four of the smaller drawings, thus making it possible to use the larger size only, if desired. All exercises can be adapted to blackboard work by changing the unit of measure. ix X GENERAL EXPLANATIONS AND INSTRUCTIONS .Ij^, P Isometric of, point, line and plane. \ - V Orthographic projection fn^ of above construction. Description of Quantities. A point is indicated by a small (lower case) letter, and its distances from the tliree planes of projection (profile, horizontal, and vertical) are given in that GENERAL EXPLANATIONS AND INSTRUCTIONS xi NOTATION AND INDICATION THE CO-ORDINATE PLANES Horizontal plane - H Vertical ■• V Profile •• P Auxiliary horizontal plane H' vertical " _ . V' profile " P' Origin (Intersection of H.Vond P) A Horizontal axis - _ . . _ . HA Vertical •• VA Profile •• - PA The four quadrants . I.H.IU.IZ POINTS (Smoll letters) Point, "q" in space -- a " moved to new psitions . - _ . . _ . . a', a" Projections of "a" on H.V and P _______ a. c^a'* " " "n' " any plane, e.g. RorV _____ n.n^ " " moved points . _ _ _ . - _ . . _ b',d' LINES y Visible Invisible v Given or required (projections) _ _ _ a b Auxiliary ___________ k 9 Traces ( piercing points ) . - . . _ . «» «^^ Projecting lines ..-___-- --- = --- - Showing movement and direction . - d- -»- *d' PLANES (Capitol letters) ^ .^ . . , H ^^sM^^ A Iraces, given or required . _ . . . H — ■ ^^ A Auxiliary plane ^ ^^ xii GENERAL EXPLANATIONS AND INSTRL'CTIONS PLATE LAYOUTS lot" - lO* C^OO The smaller (8"Mok") sheet i5 for note -book or home work . The larger one(i5'V22') is for drafting-room use . The subdivisions of the latter are the same size as the small sheet, making the problems adaptable to either. order and in units of i". Thus a (14, 6, 3) indicates that the point a is 14 units (1|") from P, 6 units {f") from //, and 3 units (f") from V. When the point described is in a quadrant other than the first, a Roman numeral is used to indicate this fact ; thus b (III, 17, 4, 9) means that the point b is in the third quadrant, 17 units from P, 4 from H, and 9 from I'. GENERAL EXPLANATIONS AND INSTRUCTIONS xiii A line is described by any two of its points, for example e (4, 7, 3) / (IV, 16, 3, 7) ; or by one point and its direction. A normal-case plane is described thus: 7^(17, T''30°r, T^'45° /), which means that the traces of the plane T meet on HA, 17 units from P, the horizontal trace making an angle of 30° with HAy to the right, while the vertical trace makes an angle of 45° with HA to the left. Special cases are described in full. Presentation. The most important items under this head are accuracy and clearness. It is recommended that the solu- tions of problems, except in special cases, be not inked, but that clean pencil work be encouraged. In general, indicate all construction lines very lightly, and given or required lines more strongly. Distances, angles, etc., are not indicated on draw- ings unless they are specified. Be careful to letter each point properly, particularly in the earlier problems. :"/ DESCRIPTIVE GEOMETRY CHAPTER I PURPOSE AND SCOPE 1. Introduction. When several persons are to cooperate in making an object or in erecting a structure, the work of all must be controlled by a preconceived plan, so that all may work inteUigently toward a definite end. For very simple work, the plan can be carried in the mind and the workers can be directed by word of mouth. In most cases, however, a full description of the object or structure is committed to paper. In this way, it is possible to foresee more fully all complications which may arise ; a greater degree of clearness and accuracy is possible ; and a record may be kept for reference and for reproduction. Some features of the proposed structure can be described most easily in a written specification. Thus, the quality of materials and the details of workmanship may be described in words. But all matters pertaining to size, shape, disposition of parts, and general appearance may be shown much more readily by means of drawings. In fact many parts such as moldings, ornament, etc., which would defy a description in words, can be explained easily by a small and simple drawing. The making and reading of these descriptive or working drawings is based on a line language, as complete in its way as any word language. Descriptive Geometry is the science which underlies this line language, just as grammar is the science which underlies word language. 1 '2'.: /'''A-;:.: \'Pesg.riptive geometry [i, § i One may acquire a superficial but usable knowledge of a foreign language without understanding its grammar. One may also be able to make useful working drawings without under- standing the underlying geometric principles. But if any lan- guage, whether of w^ord or of line, is to be used with confidence and skill, its underlying science must be thoroughly mastered. ^Yorking drawings usually represent an object w^iich is to be made for the first time. Nothing exactly like it exists. The person making the drawing must therefore depend solely on his mental image of the object that is to be made. If he is de- signing a building, he must be able to see his phantom structure from many points of view. He must be able mentally to go through and around it, and to imagine its appearance in differ- ent lights and seasons. In order to produce anything worth w^hile the architect or engineer must have first of all imagination and the ability to visualize. But he must also have some means of recording his ideas, so that others may be able accurately to carry them out. The study of descriptive geometry develops the power of visu- alization to a marked degree. Its principles furnish the basis for recording an idea in the form of accurate and complete drawings. For these reasons the study of descriptive geometry is taken up early in schools of architecture and in schools of engineering. When any study has two aspects, one immediately applicable to real problems, the other more or less abstract (and hence in general more difficult), there is a strong temptation to neglect the abstract in favor of the so-called " practical." But an idea is worth more than its bald statement. Learning to think is more important than learning facts. The chief value of de- scriptive geometry lies in the cultivation of the imagination, and not in the acquisition of tricks of drawing, or of methods for the solution of special problems. Students are urged to keep this idea in mind while studying the problems which follow. No problem should be left behind I, §2] PURPOSE AND SCOPE until there is formed in the mind a clear and satisfactory picture of the conditions and of the solution. 2. Kinds of Drawing. The fundamental problem of all drawing lies in representing three-dimensional objects on a two- dimensional sheet of paper. Two methods are commonly employed in making such representations. One method is known as perspective drawing, the other as projection drawing. _-Oit Perspec+ive- FiG. 4 In perspective drawing, the idea is to produce a more or less exact representation of the object as it appears to the eye, including the well-known foreshortening effect of distance. This foreshortening (see Fig. 4) is due to the variation between the visual angles a and a'. Its effect is to reduce remote parts and to crowd the details there. It complicates the making and reading of the drawings. Consequently, perspective drawings are ordinarily used only for pictorial purposes. The other method, projection drawing, arbitrarily eliminates the natural foreshortening, so that all parts which are of the DESCRIPTIVE GEOMETRY [I, §2 Fig. 5 Perspective same size are so represented in the drawing. This destroys the pictorial effect, but it makes the drawing more useful for struc- tural purposes. Although there are several methods of pro- jection drawing, only the two most usual ones (ortho- graphic and isometric), will be considered here. Isometric projection (see Fig. 6) shows the object in all three dimensions on one drawing. It is useful to a limited extent in showing complicated details not otherwise easily explained.^ Orthographic projection (see Fig. 7) shows but two di- mensions (one face) of an object on a single drawing. At least two, usually three, and often more, such drawings are needed fully to describe an object. However, the greater clearness of such drawings and the rapidity and ease with which they can be made, render them far more useful than isometric or per- spective drawings. For this reason, orthographic pro- jection is used for the most part in this book. For those illus- trations in which a pictorial quality is desired, isometric draw- ^ For a more extended treatment of isometric projection and other similar methods, see Chapter XIII. Fig. 6 I, §3] PURPOSE AND SCOPE ings have been used, in general, though CavaHer projections or perspective drawings have been substituted in a few cases. 3. The Theory of Plan Drawing and of Building from Plans. ^ The size and shape of any three-dimensional object is determined by its bounding planes or surfaces, and these in turn by their Orthographic Fig. 7 Projeciion bounding lines. Any line is determined by any two of its points. Hence the size and shape of any object are fixed by the interrelations of its limiting points. For example, in Fig. 7 the roof surface A is defined b}^ the lines ab, be, cd, and da. These lines in turn are determined by the four points a, b, c, and d. These four points then limit or define the surface A. In starting a set of working drawings, the draftsman first ^ ^ The word "plans" is sometimes used to indicate particular projections of an object (see Fig. 7) and sometimes to indicate the whole set of draw- ings. Here it is used in fhe latter sense. 6 DESCRIPTIVE GEOMETRY [I, § 3 draws some lines to represent certain of the salient lines of the proposed structure (such as the axes in Fig. 7). Then, measuring from these lines as bases of reference, he locates the more important points and lines and finally the details of the pro- posed structure. Proceeding thus from point to point, draw- ings are finally evolved which fully describe the position of every point and line with reference to one another and with reference to the first determined axes. AVith these drawings as a guide, the structure they represent can be made an\n^'here, at any time. Since the representation of a point is so fundamental, it is worth while to study it carefully at the start. The purpose of such a study is to describe a method by which the location of a point, and from this the location of other points and lines, can be described so clearly that they can be materialized into a perfectly definite structure. The location of any point in space may be described most readily by means of its relations to three fixed planes of reference. For example, a mountain peak may be located by its latitude, its longitude, and its altitude. A building may be located by reference to two of the lot lines and to the curb level. The position of a suspended light in a room may be fixed definitely by stating its distance from the end, from the side walls, and from the floor. In the latter case, if the distance from the floor alone is given, the light may be an^^'here in a plane parallel to the floor and at the given height. If the distances from the floor and from one wall are given, the light may be anywhere on a line parallel to the floor and the wall But when its distances from all three planes are given, its location is definitely fixed. The walls and the floor form three fixed planes of reference. Such planes are called the datum planes for locating the light. These datum planes, or planes of reference, may be chosen at will. Ordinarily, however, each plane is taken perpendicular to the other two (as in the cases cited above), one being hori- zontal, and the other two vertical. This will be done in this I, §3] PURPOSE AND SCOPE book except where otherwise stated. The object to be drawn is conceived to be placed within the solid angle formed by these planes. Each salient point of the object is then projected H = the horizontal plane of projection F=the vertical plane of projection P = the profile plane of projection '' = horizontal projection of a av = vertical projection of a j^jQ_ g aP = profile projection of a ^ aa'' =horizontal projecting line aa'^^ vertical projecting hne aaP= profile projecting line on each plane bv means of a line passing through the point perpendicular to^he plane. (See Fig. 8.) The foot of such a perpendicular is called the projection of the point. 8 DESCRIPTIVE GEOMETRY [I, § 3 When a sufficient number of points have been projected, the projections on each plane are connected by Hnes. The projec- tions of the object thus obtained form descriptive drawings. Since the projecting lines to each one of the planes are all parallel, the relative positions of the various points are main- tained in the projections. (Compare and contrast Figs. 5 and 7.) Thus foreshortening and distortion are avoided. It follows that the process of drawing plans is essentially that of projecting points on arbitrarily selected planes, while the process of building consists in establishing the planes of reference as given by the plans and projecting the points, in a reverse sense, back into space. 4. Special Applications. All of the various methods of drawing in a more or less pictorial way, such as perspective drawing, isometric drawing, shades and shadows, etc., are special applications of descriptive geometry. The determination of the actual size and shape of the individ- ual parts of a complex structure (for example, the cutting of the stones for a vault) , so that each may fit perfectly into its proper place, is a problem that becomes increasingly important with the present-day tendency toward shop production. The principles of descriptive geometry are a sure guide for all such work. All the rules, formulas, and practice of pattern making depend ultimately on the same principles. In the following chapters the more important theorems of descriptive geometry will be presented in a purely abstract fashion, and then some of them will be applied to the solution of concrete problems. It is hoped that through the frequent cross references, the student will come to realize that no problem stands . alone, that the whole subject is coherent, and that it cannot be mastered piecemeal. 5. Historical Note. As in the case of most sciences, the main ideas of descriptive geometry were in daily use long be- fore its principles were formulated into a separate science. I, § 5] PURPOSE AND SCOPE 9 The early architects, engineers, and craftsmen worked from plans that were, from a modern point of view, imperfect and incomplete. The rules and methods used in making the plans and in carrying out the work were passed on from generation to generation by word of mouth, chiefly as guild secrets, which con- sisted largely of mere rules and formulas. It remained for the celebrated French mathematician, Gaspard Monge (1746-1818), to formulate into a distinct science the prin- ciples already in use. The way was then prepared for rapid, sure, and permanent progress. The development of modern practice, infinitely more efficient and more flexible than the old, was thus made possible. CHAPTER II PROJECTION OF POINTS 6. Introduction. In the previous chapter it was shown, by the use of Fig. 8, how the projections of a given body may be found. Usually it is easier to visualize the process of projec- tion when it is applied to a body than when it is applied to a single point. This chapter is devoted to a study of point projection. i7A= horizontal axis yA= vertical axis PA = profile axis ew Fig. 9 7. Fundamentals of Point Projection. Figure 9 shows the method of orthographic projection applied to a single point. It should be carefully studied in order to fix thoroughly in the mind a few fundamental relations. (See footnote, p. 7.) The figure shows the point a in space, projected on the planes //, V, and P. The points a^, a^, and a^ are, respectively, the horizontal, vertical, and profile projections of a. If now the point is removed, while the projections are retained, the point a 10 II, § 8] PROJECTION OF POINTS 11 itself can be exactly restored at any time to its original position, since that position is recorded by the projections. The projections of a point are thus a record or a representation of the point in space. Such a representation of a point is quite as distinct from the point itself as a sign board is distinct from the town toward which it points. The projections of a point bear exactly the same relation to a point as the plans of a building bear to the building itself. Referring again to Fig. 9, observe that the point a, together with its projections a'", a^, and aP, determine a right parallelepiped. It is obvious that the opposite sides are equal. Hence the distances of a^ from HA and PA represent the distances of a from V and P, a- from HA and VA represent the distances of a from H and P, aP from PA and VA represent the distances of a from H and T^. In other words the location of a point with respect to T" and P is shown on H ; its location with respect to H and V, on P ; and its location with respect to H and P, on V. Thus no single projection can locate a point definitely with respect to all three reference planes. (Compare the H projection of a and a' in Fig. 9.) But any two projections, taken together, do definitely locate a point in space, since the two taken together show its location with respect to all three of the planes of reference. 8. Relation to Plan and Elevation. ^Mien a solid body is to be drawn, as in Fig. 8, relations similar to those of § 7 hold true. The V projection shows relations of height and length. The P projection shows relations of height and width. The // projection shows relations of length and width. Since any two of the projections taken together show points in their three-dimensional relations, it is quite possible to show fully the size and shape of a simple body by projecting it upon only two planes. "When it becomes necessary to fix the position of the body in space, however, the third plane of reference must 12 DESCRIPTIVE GEOMETRY [11, § 8 be used, even though the projection on this third plane may not actually be drawn. Moreover, when a body becomes more com- plex and many points have coincident projections on one or another plane, as in Fig. 8, it becomes necessary to draw three or more projections in order to avoid confusion. Simplicity and clearness are highly desirable in this work. (See Fig. 7.) The horizontal projection of a body is commonly called the plan. It shows the flat or level relations. The vertical projec- tions are called elevations, since they show relations of height. Direc+ion of view Fig. 10 9. Representation on Paper. To outline a structure of any magnitude, a point at a time, on three mutually perpendicular planes of projection erected in space, would be far too tedious and quite impracticable. Methods must be found for dealing with points in wholesale quantities, a line at a time or a plane at a time, and for representing them on flat sheets of paper. We shall consider first the problem of flattening out the planes, so as to show the projections on a two-dimensional surface. In order to record all three projections on a single flat sheet, it is necessary to imagine the vertical and profile planes rotated about their axes, until they lie in the same plane with the II, § 10] PROJECTION OF POINTS 13 horizontal plane (//). The simplest form of this operation is shown in Fig. 10, which shows the V and P planes separated along VA and turned away from each other into the plane H. Let us now see what happens to the projections of a point during this operation. It will be noted that as V revolves on HA as an axis, the T^ projection of a, a'', describes an arc of a circle whose plane coincides with the plane of a, a^, a^\ Hence the line a^'xa^ becomes a straight line perpendicular to HA. By the same reasoning the line a^za'' becomes a straight line per- pendicular to A P. Figure 1). shows the flat drawing which results from the flattening process illustrated in Fig. 10. 10. Connections. It will be noted that the erected planes consist of three flat surfaces, while the flat drawing is divided V a". Ix H 1 A j'l z 1 < -D Fig. 11 into four parts by the axis lines. No projection will appear on one of these spaces (VAV), since it has no equivalent in the erected planes. Across this space the circular arc yy is drawn with its center at A. The significance of this arc is frequently misunderstood. It signifies that before the planes were flattened, the lines a^'y and a^y (Fig. 11) touched one another at y, i.e. that the two points y represent the single point y in Fig. 10. 14 DESCRIPTIVE GEOMETRY [II, § 10 A line drawn at an angle of 45° to VA would serve the same purpose as the arc, or the entire line a^^yya^ could be omitted. The only necessary condition is that a^z = a''x. 11. Alternative Openings for the Profile Plane. Let the student now cut out a card or sheet of stiff paper and draw D D H I I ' 1 ^ V A < Fig. 12 on it the projections of an unsymmetric object, as shown in Fig. 12. Now let the card be placed on a table, face up, and let the parts marked V and P be bent upward, creasing the card along the lines EA and AF. There will now be three planes, carrying the projections of the given object. In flat- tening out these planes, V is simply turned backward into U ; while P may be turned into U by tipping it either toward or away from the object which is being projected. Next let P be separated from E by cutting along AF, let it be fastened to V along AV , and let the solid angle be erected as before. The flattening process now may be started by turn- ing F OYi AV as an axis, into V . Both planes are then folded backward into //. In this operation, the P plane may be folded II, § 12] PROJECTION OF POINTS 15 toward or away from the object to be drawn. Thus there are four ways in which the P plane may be opened. When P is turned into V and then both planes rotated into H, it is clear that the two elevations are seen side by side. When P is turned at once into H, however, the plan and the end elevation are shown side by side, while the front and the end elevations are separated. It seems more logical, and is actually easier in drawing, to have the two elevations closely related. When P is turned towards the object, the end elevations will appear as if viewed through P ; whereas when P is turned aioay from the object, the elevation shows the object reversed, the left- hand side appearing at the right of the drawing. This is of es- pecial importance when the two ends are not alike, so that both must be drawn. Thus, of the four possible methods of rotating the P plane, it would seem best to use the one giving the most natural result, viz. : the elevations side by side and the end elevations appearing as if viewed through the plane. In other words, the best method is first to fold P toward the object into V and then to turn V (and P) backward into H. Although the preceding method will be used most frequently, it is often desirable to use some one of the others. Consequently the student should familiarize himself with all four methods, and with placing P either to the right or to the left of the object. The P plane may be used on either one or on both sides of the object, and nearer or farther from it, as circumstances may require. 12. Normal and Special Cases. We shall classify the subject matter so that the majority of cases fall into one class, while special cases that occur less frequently form other classes. The solution of problems is more rapid and more accurate with such a classification. Thus, in trigonometry, multiples of 90° are special cases of angles. In Fig. 9, the general case is that of any point lying within the solid angle formed by H, V, and P, since such points have 16 DESCRIPTIVE GEOMETRY [II, § 12 the same general properties and constitute the great majority of all possible points.^ We shall call them iionnal-case points. But if a point lies in one or more of the planes of projection, one or more of its projections will have special characteristics. Such a point will be called a special-case point.^ The ability to recognize special cases and to make use of their special properties is often the key to the best solution for a problem. The auxiliary points, lines, etc., that are used in solving problems are usually special cases, and their special properties make them particularly useful. 13. Recapitulation of Chapter II. The following principles with reference to the projection of a point may now be stated : 1. Three planes of reference, and at least two projections are neces- sary to record the location of any point. 2. To show the size and shape of a body, hut not its exact location in space, one of the planes of projection may he omitted. 3. The projections of a point on any two planes having a common axis lie in the same straight line perpendicular to that axis. 4. The vertical projection of a point is as far from HA as the actual point in space is from H. 5. The horizontal projection of a point is as far from HA as the actual point in space is from V. 6. The vertical and horizontal projections of a point are as far from AV and AP as the actual point in space is from P. 7. The profile projection is as far from AP as the point is from H, and as far from AV as the point is from V. 1 It may occur to the student that many points in space may lie wholly outside the angle between the planes of projection. Such cases are dis- cussed in Chapter IV. 2 Undoubtedly some difficulty will be experienced in the use of these terms when applied to lines and planes. No two persons will classify them in quite the same manner. It is felt, however, that the use of these terms permits a most desirable, though somewhat vague, grouping of the subject matter, which will be found to have justified itself in the long run. II, § 13] PROJECTION OF POINTS 17 8, // any two projections of a jjoint are known, the third can be established. (Result of 4, 5, 6, and 7, above.) 9. Two projections of a point may be assumed at random, within the restrictions laid down in 3 above. For there is some point in space corresponding to any two projections that may be selected. PRELIMINARY WRITTEN EXERCISES 1. Prove that two projections of a point lie on the same straight line perpendicular to the axis, 2. Prove that the projections of a point are at the same distances from the axis as the actual point in space is from the planes of reference. 3. Describe the special cases of points that may occur, and tell the distinguishing characteristics of their various projections. 4. Describe in writing the following points. (See the description of quantities, p. x.) -ic: d- d^ b^ Fig. 13 V ''h^ A 5. Describe in writing the following points. V ■,d^ U»' Id'' :<-h xqh FiQ. 14 18 DESCRIPTIVE GEOMETRY [II, § 13 EXERCISE SHEET I [Note, For the typical layout for exercise sheets, applying to all exercises in the text, except a few which are separately described, see p. xii of the general explanations.] Take P at the right, 20 from the right border line, turned away from the given points, into F. Draw all three projections of the following points. Point Distance from Point Distance from P H V P H V a 55 15 19 / 27 6 8 b 49 6 g 20 c 43 9 2 h 15 3 8 d 36 14 i 9 9 7 e 32 12 12 3 3 12 12 EXERCISE SHEET II Take P at the left, 4 from the border line, turned toward the given points, into H. Draw all three projections of the following points. Point Distance from Point Distance from P H V P H V a b c d e 16 23 30 36 42 11 8 10 4 5 15 10 20 10 / g h i J 48 53 60 65 72 11 2 11 10 2 20 10 II, § 13] PROJECTION OF POINTS 19 EXERCISE SHEET III Take P at the left, 4 from the border line, turned toward the given points, into V. Draw all three projections of the following points. a (12, 8, 10) d (38, 13, 12) g (34, 0, 0) j (72, 13, 12) h (19,23, 6) e (46, 8, 0) h (60,11, 8) k (50, 0,15) c (28, 17, 19) / (53, 19, 4) i (66, 5, 8) I { 0, 10, 18) EXERCISE SHEET IV Take P at the right, 20 from the border line, turned away from the given points, into H. Draw all three projections of the following points. 1. a (55, 17, 9). Move a 9 to the right. Call the new position a'. 2. 6 (41, 6, 14). Move h upward until it Ues in the same horizontal plane as a. 3. c is 34 from P, 10 from H, and lies on a plane passing through HA and bisecting the dihedral angle between H and V. 4. d lies in the same horizontal plane as c, but is 9 nearer P and 3 farther from V. Keeping d at the same distance from H, swing it about c as a center until it lies in the same line perpendicular to V as does c. 5. e is in a straight line parallel to HA with d but at 19 from P. Rotate it about its o\vn H projection till it lies in H, keeping it in a plane parallel to V. 6. / lies on HA, 3 from P. Move / upward 14. Call this position /'. Move the point forward {i.e. toward the observer) until it is 5 in front of V. Call this position /". Now move it again, this time 5 to the left, and mark this position /'". CHAPTER III PROJECTION OF LINES 14. Generation of a Straight Line. If any point, as a, Fig. 15, is moved to a new position, a' , the projections also move to new positions, as shown. During its motion, the point a has generated the Une aa' . Simi- larly the motions of the projec- tions have generated the lines a'a''" and a^a'^. Also, the pro- jecting lines {aa^ and aoF) of the point a have generated the pro- jecting planes {aa^a'^ a' and (wya'" a') of the line aa' . 15. Straight Line Determined by Two Points. One and only one straight Hne can be drawn through two points, i.e. any two points determine a straight line. Hence the projections of any two points determine the projections of a line. The two points that de- termine a line do not neces- sarily limit the line. They merely fix its position in space. The line is thought ^\^ of as extending indefinitely in both directions. 16. Assuming a Line. Any two lines drawn on a sheet of paper, one above and the other below HA will represent the 20 Fig. 15 H- A Fig. 16 Ill, § 17] PROJECTION OF LINES 21 projections of some line in space. This fact is readily seen by reference to Fig. 16. Let d'h^ and ay¥ be any two lines drawn as above. Consider now Fig. 17, where these pro- jections are shown in an isometric drawing. Through a^h"^ pass a plane perpendicular to V, and through a^h^ a plane per- pendicular to H. These planes will intersect. The line of intersection, ah, is the line whose projections are a^h"^ and aJ'h^. 17. Length and Slope. It is evident from Fig. 15 that when a line is parallel to H and V, each of its projections is equal in length to the line, and is parallel to HA. Figure 18 shows a line ah which is parallel to V but inclined to H. It is evident that the V projection is equal in length to the given line ah, and that the length of the H projection is less than the length of the line. Moreover, the angle 6 is equal to the angle 0. Now imagine the line ah to be rotated in the plane aa^h^h, using a as a center. During the rotation, the length of the V projection, a^h^, will remain constant and equal to ah. The length of the H projection will vary with the cosine of the varying angle. The limits of this variation are, when 0=0°, a^h^—ahj and when = 90°, a^h^=0. 22 DESCRIPTIVE GEOAIETRY [III, § 17 Figure 19 shows a line which is inclined to both planes of projection. It will be evident immediately that the projections are each shorter than the line itself, and that the angles a'' and 0!" between the projections and HA are greater than the angles o6=a line in space a^6^= horizontal projection a''6'' = vertical projection aP6P = profile projection Fig. 19 abb^a^, ahh^'a'', and abbPaP are the horizontal, verti- cal, and profile, projecting planes a and /3, which the line makes with the planes of projection. 18. Normal and Special Cases. The previous article will suggest the natural separation between the normal and the special cases of lines. All lines not parallel to one or more of the planes of projection can be considered as normal cases. Among the lines parallel to one or more of the planes of pro- jection a number of special cases can be distinguished. Ill, § 20] PROJECTION OF LINES 23 19. Method of Study. An opened book cover may be used to represent the planes of projection. Place the book on the table with its binding turned away from the observer. Open the top cover to a vertical position. Use a pencil to represent a line, and use the book and cover to represent H and V. 20. General Theorems. The following simple theorems are merely stated, the proofs being left as exercises for the student. 1. One projection does not determine a line. 2. Two projections definitely determine a line. 3. If two projections of a line are known, the third can he found. 4. Any two lines draum at random, on different planes of projection, represent the projections of some line in space. Compare the above theorems with numbers 1, 8, and 9 of § 13. 5. When a line is parallel to V, (a) its V projection is equal in length to the line itself, (b) the angle between the V projection and HA is equal to the angle between the line and H. 6. When a line is perpendicidar to V, (a) its V projection is a point, (b) the projection on H is the same length as the lijie and perpendicular to HA. State theorems similar to 5 and 6 for the planes H and P. [Note. A line which is perpendicular to any plane of projection is necessarily parallel to the other two planes, if, as is usual, the planes of projection are perpendicular to each other.] 7. When a line is not parallel to a plane of projection its true length and slope are not shown by its projections. 8. The projection of a line is never longer than the line itself. 9. The angle between the V projection of a line and HA is never less than the angle between the line and H. EXERCISE SHEET V Take P at the right, 20 from the border, and opened away from the given lines, into V. Draw the projections of the following lines. 1. Line ah : a (54, 5, 9) ; h (44, 8, 2). 2. Line cd : c (40, 12, 7) ; d is nearer to V and H than is c, and 31 from P. 24 DESCRIPTIVE GEOMETRY [III, § 20 3. Line cj : e, on V, 28 from P ; /, on H, 20 from P. 4. Line ^/; : perpendicular to, but not touching, V ; 19 above H ', and 17 from P. 5. Line i; : lies on // and runs from i (14, 0, 10) backward and to the right, to j, which is 8 from P . 6. Line kl : parallel to, and 4 from, P ] k m nearer V and farther from H than is I. 21. Visualizing a Line. The student now should have gained some facility in visualizing a line. Let him then try to describe ^ a line whose projections are given. Figure 20 shows the pro- jections of a line ah which H 1 ■ ! A rises and recedes from V, as it passes from left to right. If the position of a line is not readily grasped from its projections, each projection should be considered separately. By taking succes- sive points along the H projection, and comparing their relative distances from HA, the slope of the actual line with regard to V can be determined. The same process with the F projection gives the slope of the actual line with regard to H. After a little practice the projections of a line will suggest its position and its slope with regard to the planes of projection. EXERCISE VI Describe in writing the following Hnes. Fig. 20 Ill, § 23] PROJECTIOX OF LINES 25 22. Limiting Projections of a Plane Figure. Another exer- cise of value consists in studying the possible variations in the projections of plane figures. Try a square. Place it in various positions with respect to H, V, and P, and note the projections in the limiting cases and the law of variation between the limits. EXERCISE SHEET VII Take P at the right, 18 from the border line and turned away from the objects, into V. Profile projections are required for #3 only. Use notation carefully. (1) Draw the H and T' projections of a square, abed, 10 on a side, lying in a plane 13 above H, and with the back left-hand corner at a (55, 13, 4). The side ah is parallel to V. Rotate the figure toward H on ad as an axis, until it lies in a vertical plane. Draw its projec- tions in this position. Now with ah' as an axis, swing the square through 45° to the right and toward V, and draw its projections in this position. (2) The center of a circle, 12 in diameter, is at o (30, 10, 8). The plane of the circle is parallel to V. The horizontal diameter is eg, and the vertical diameter is fh. Rotate the figure on eg, the top traveling away from T', until it hes in a horizontal plane. Draw its projections in this position. (3) (Use the profile plane.) With the line j (15, 4, 5), /: (4, 4, 5) as a base, construct an equilateral triangle whose plane is perpendicular to both V and P. On jk as an axis, swing the triangle upward until it is parallel to T'. Draw the projections in this position and in two intermediate positions. Use a different kind of lino for each position. 23. True Length of a Line. In § 17 it was pointed out that the projections of a normal-case line do not show either the true length of the line in space or its true inclination with respect to the planes of projection. Ordinary plans and elevations make frequent use of just such lines to indicate structural parts. Thus, in Fig. 199 a, p. 219, the line ac indicates the hip line of a four- pitched roof. It is only after the true length and the slope of such members are known that the piece can be made. Hence special methods are devised for determining these quantities. Two general methods are used for determining the true lengths 26 DESCRIPTIVE GEOMETRY [III, § 23 Ill, § 24] PROJECTION OF LINES 27 and the slopes of normal-case lines. In each method the fundamental idea is to arbitrarily reduce the normal-case line to a special case (parallel to a plane of projection), and to find its projections under the altered conditions. 24. First Method for Finding True Length. The first method consists in swinging the line about until it is parallel to one of the planes of projection. But since on paper we deal w^ith the projections of the line and not with the line itself, it is necessary to have a definite method of procedure for this operation. The general method can be explained best by reference to a definite case. In Fig. 22 a, the normal-case line ab is to be swung around into a plane parallel to T , let us say. In order to define the motion of the line and to follow it in projection, let the lower end a be kept in its original position throughout the operation, and let the angle a, which the line makes with H, be maintained. If the line is revolved in this manner through 360°, each of its various positions will correspond to an element of a cone whose apex is at a and whose base is parallel to and at the distance bb^ from H. AYhile the line ab is thus generating a cone in space, the horizontal projection of b is generating on H sl circle whose center is at a^ and whose radius is equal to a^b^, the horizontal projection of ab. Indeed, this circle is the horizontal projection of the base of the cone. At the same time, the vertical projection of b is moving in a line parallel to HA, since b is always at the same distance from H. It will be seen that for every position of ab the length of the horizontal projection is the same, while that of the vertical pro- jection varies as the angle between ab and V varies. Now a^b^ never shows the true length of ab ; but there are two positions of a^'b"" which do show it, viz., those which correspond to ab at the particular instant when it is parallel to V. These positions cor- respond to the elements of sight of the generated cone. In Fig. 22 b, one of the two possible special-case positions of ab is marked ab'. The V projection of ab' {a^'b'^') is the true length of ab. 28 DESCRIPTIVE GEOMETRY [HI, § 24 Fig. 23 Turning now to Fig. 23, the operation is repeated in projec- tion. The // and V projections of ab are given and it is re- quired to find the true length of ab and the true angle which it makes with H. A short arc is swung from a^ as a center, using a radius equal to a'^b^. This arc represents a portion of the cir- cumference of the base of the cone in horizontal projection. Next a line {a^b'^) is drawn through a^ parallel to HA. This line repre- sents the horizontal projection of ab when it is parallel to V. It is necessary now to pass to the V projection. A line (b^x) is drawn through the V projection of b, parallel to HA. This line repre- sents the V projection of the base of the cone. The F projection of the point b^ will be found somewhere on this line. x\lso it will be directly above the H projection (6'^) already found (3, § 13). Hence, if a line is drawn through b'^. perpendicular to HA, its intersection with b^x, (6''), will determine the V projection of the point b when the line ab is parallel to V. Connecting this point with a'' we have the line which is the V projection of ab w^hen it is parallel to V, and which shows the true length of ab. The angle a is, of course, the real angle })etween ab and //. 25. The True Angle with V. The true angle that a line makes with V may be found, if required, by swinging the line into a plane parallel to //, instead of parallel to F, as was done in' § 24. Such a case is shown in Fig. 24. Here the point b is kept stationary while a is rotated in a Fig. 24 Ill, § 26] PROJECTIOX OF LINES 29 plane parallel to V. The generated cone in this case has its base parallel and close to V, while the apex (b) is farther from V. The line b^a'^ shows the true length and the angle is the true angle with T^. 26. Variations of Preceding Methods. In either of the above cases, either end of the line might have been chosen as the stationary end. If the true length only is wanted, the angles not being re- quired, either method may be used. Figure 25 shows the above variations on the general method applied to a single case. The first operation con- sists in swinging the line parallel to V, keeping the end a stationary. This is shown in Fig. 25 in solid lines. The second operation consists in swinging ab parallel to H, using b as the still point. The true lengths found by any two meth- ods should agree. The sum of the angles a and never exceeds 90°.^ 1 The proof is as follows. (See Fig. 26.) Let a'-'l^ be any line. The angles 1 and 2 are the angles which the given line makes with H and T^ respectively. To prove that 1 + 2 cannot exceed 90°. In the right triangle a'^a'^b", the side a^'a'' < a'-'h". The right triangles a^'a^b^ and a'^b^b^ have a common hypotenuse ; hence the one having the greater base will contain the smaller angle adjacent to the base. Thus 2 < 3. But. 1 + 3 = 90°, therefore 1 + 2 < 90°. When ab is parallel to P, a^fe" coincides with a'"a^. In that case the angles 2 and 3 are identical and 1 + 2 = 90°. But 1 + 2 cannot exceed 90°. Fig. 26 30 DESCRIPTIVE GEOMETRY [III, § 27 27. Second Method for Finding True Length. The second method, as well as the first, depends on having the line parallel to a plane of projection. It differs from the first method, however, in that the plane is brought to the line, rather than the line to the plane. The student is already familiar with the idea of projecting a point on three planes. (See Chapter II, especially 3, 8, § 13.) When the projection of a point 15 found on a new plane . the nofahon corres- ponds to that of the plane y/hen using V When using V. Fig. 27 The only extension of that idea which is required here is that the third projection may be made on a plane perpendicular to //, but not necessarily perpendicular to V. In fact, the new plane may be chosen parallel to any given line, thus making that line a special-case line with reference to the new plane. The operation described above is illustrated in Fig. 27. The line ab and its projections, a^b^ and a^'fe^, are given. It is required to find the true length and the angle of inclination to //. A new Ill, § 27] PROJECTION OF LINES 31 plane (F') is chosen perpendicular to H, and parallel to ah. The line is projected on this plane in the usual manner, and the plane folded into H. The projections thus found will show the required facts. Figure 28 shows the same operation in projection. Starting with the given H and T^ projections {a!^h^ and a'-'h^'), the auxiliary axis H' A' is drawn parallel to a^h^ to represent the axis line be- tween the new plane and H. The projections of a and h on this new plane will lie on lines passing tlirough a^ and h^ and per- pendicular to H'A' (3, § 13). Let two such lines {a^r and 6^5) be drawn, and let them be of indefinite length. The height of the points a and h (in space) from H will govern the distances of the new {V) projections from H'A'. (Com- pare, in Fig. 27, bb^, b'y, and b^'y'). If then we measure off x'a^' and y'b^' equal to xa" and yb^ respectively, we will have determined the V pro- jections of the points a and 6, and hence of the line ab. This projection shows the true length of the line (a'"'6'0 and the true angle of inclination to H, 0, as required. Since three planes of projection are used in the operation, three projections result. Two are elevations and one is a plan. The plan is common to the two elevations. The process of mak- ing the new projection is sometimes more easily visualized if the student turns the drawing about until H'A' is horizontal, thus placing the new plane in the position that V occupied for- merly. Note the arrows in Fig. 27. Figure 29 shows the same method applied to finding the true 32 DESCRIPTIVE GEOMETRY [III, § 27 length and the angle of inclination to V. In this case the new plane must be taken perpendicular to V and parallel to the given line, so as to show the true angle with V. Of course the true lengths as determined in Figs. 28 and 29 should agree. 28. Comparison of the First and Second Methods for Finding True Lengths. When the methods of §§24 and 27 are carefully compared, it will be noticed that the second method is simpler, both in concep- tion and in execution. However, both of them depend on the same idea, i.e. that of arbitrarily altering the given conditions by moving either the line or a plane of projec- tion so that the conditions become those of a special case. Sometimes one method is more desirable, and sometimes the other. The student should be able to use either method with facility. The principle of the second method, viz., that it is always possi- ble, given a plan and elevation, to construct a new elevation projected from a new point of view, is used in many ways in descriptive geometry and in its application to working drawings. Fig. 29 EXERCISE SHEET VIII In Exs. 1-5^ take P at the right, 16 from the border line and turned away from the objects, into V. Draw the profile projections in prob- lems 4 and 5 only. In Exs. 1-4, use the first method. 1. Given a (60, 6, 7) and b (51, 13, 11), find the true length of ab and the angle it makes with H. 2. Given c (44, 7, 2) and d (39, 11, 6), find the true length of cd and the true angle it makes with V. Ill, § 29] PROJECTION OF LINES 33 3. Given e (33, 10, 14) and/ (25, 4, 6), find the true length of ej and the true angle it makes with V, by revolving it about its center point g. 4. Given h (22, 12, 4) and i (14, 22, 11), find the true length of hi and the angles it makes with H and V, by swinging the line about i as a center until it is parallel to P. 5. Given y (11, 10, 6), k (11, 2, 6), I (3, 3, 12), m (5, 8, 10^), draw all three projections of the quadrilateral jkhn and find its true shape. EXERCISE SHEET IX Take P at the right, 16 from the border line and turned away from the objects, into H. Use P in problem 5 only. Use the second method. 1. Given a (60, 5, 7) and h (52, 11, 5), find the true length of ab and the angle it makes with H. 2. Given c (46, 7, 6) and d (38, 9, 11), find the true length of cd and the angle it makes with T'. 3. Given e (32, 3, 10), / (29, 10, 13), and g (24, 5, 18), find the true shape of the triangle efg by finding the true lengths of its sides. 4. Given h (19, 8, 3) and i (11, 11, 5), find the true length of hi and the angles it makes with V and H. (Two operations.) Compare results. 5. Given i (9, 2, 14) and k (3, 11, 5), find the true length oi jk and the angle it makes with P. 29. To Draw the Projections of a Line of Given Length, Which Makes Given Angles with H and V. This problem is the converse of that stated in §§ 24-26. In the former case, the projections are given and a true length and two true angles are required. In the present case, the length and angles are given and the projections are required. It is important to study this problem, since the method can be used in solving many other problems whose converse solutions are well known. The method proposed is as follows. (1) Carry through a solution of the converse problem (in this case that of § 24), without reference to the quantities involved, i.e. in the present case, assume a line at random, without thought as to its actual length or inclination. (2) Trace backwards the steps leading to this solution, in the reverse order. Thus, the method and order of procedure for the required case may be established. 34 DESCRIPTIVE GEOMETRY [III, § 29 For example, let it be required to find the projections of a line 1" long, which makes an angle of 45° with H, and an angle of 30° with V. Let the solution be based on the ** first method " for finding true length (§ 24). The operation consists in swing- ing the line into parallelism with V (let us say), then projecting it on V, and thereby determining its true length and slope. Applying this method in the reverse sense (Fig. 30 a), let us draw the projections of a line 1'' long, making an angle of 45° with H, and parallel to V, as ab'. If this line is now swung away ih H av ^ ib"^ ^o- Fig. 30 a Fig. 30 6 J'h -■^ from F, as in the first method, b'" will move toward z"-', while b'^ moves in a circular arc toward z^. Meanwhile the line itself is making constantly changing angles with V, some one of which is the required angle. Therefore the required V pro- jection of b will be somewhere on the line b'''z% and the required H projection will be somewhere on the arc b"'z''. Again, turn- ing to Fig. sob, draw the projections of a line (ab") 1" long, making an angle of 30° with V, and parallel to H. When this line is swung around away from H, the end marked b" will trace out the line b'^'x^ in horizontal projection, and the arc b^'^x" in vertical projection. The required projections of 6" Ill, § 29] PROJECTION OF LINES 35 will be somewhere on the line and the arc just mentioned. It is not possible to say, either in Fig. 30 a or in 30 b, just where, on the lines or arcs, the required projections will be found. But by combining the two drawings, as shown in Fig. 30 c, the arc b'h^ and the line b'^^x^ will intersect. Since the required H pro- jection of the swinging end of the line must be on both the line and the arc, it must lie at the point of intersec- tion, b^. Likewise the V projection is found to fall at b\ Note that if the construction is accurately drawn, the line 6^6^ will be perpendicular to HA. This fact may be used as a test of accu- racy. More than one solution is possible for the given data. Fig. 30 c EXERCISE SHEET X Take P at the right border line. No P projections are required. 1. The point a (72, 6, 4) is given. The Une ab is 10 units long and slopes upward and forward from a, making an angle of 30° with H and an angle of 45° with V. Find its projections. (Two solutions.) 2. The point c (48, 4, 8) is given. The line cd, 13 units long, slopes upward, forward, and away from P, making an angle of 45° with H and an angle of 15° with T^. Find its projections. (One solution.) 3. The point c (38, 16, 18) is given. The line ef is 8 units long and makes an angle of 15° with H and an angle of 60° with V. Find its projections. (Four solutions.) 4. The point g (27, 5, 6) is given. The line gh slopes upward and forward, making an angle of 60° with H and an angle of 30° with F. Draw the projections. 5. The point i (11, 5, 6) is given. The hne ij is 10 units long and makes an angle of 60° with H and 45° with V. Explain the result. 36 DESCRIPTIVE GEOMETRY [III, § 30 The given distance 30. To Measure a Given Distance from a Given Point Along a Given Line. In Fig. 31, let the line az be given of indefinite length, and let it be re- .^v ^ quired to mark off on that line a given distance, ac, from a. Select any other point on the line, as b. Find the true length of a6 (§ 24), as shown by a!"})'^. On this line measure off the re- quired distance from a^ (as al'c'^). Now rotate the line back to its first position. The projections of point c will move back along paths parallel to those traced by h^ and 6^' in the first rota- tion. Thus the projections c^ and c^ are determined. The point c in space is on ah, at the required distance from a. EXERCISE SHEET XI Take P at the right border line. No P projections are required. 1. Given a (75, 6, 4) and 6 (66, 12, 8), locate c on ah \" from a. 2. Given d (60, 10, 6) and e (50, 5, 13), locate / on '6'). From the known P projections of h, j, and /, locate the H and V projections of these points on a^b^ and a^'b^'. Now cb and gc will be parallel to H and T^ ; hence their H and V projections will show their true lengths, and will be parallel to HA. Draw these lines, mark off their end points, and con- nect the points so found. II. A slightly different construction must be used if the plane of the figure is in- clined toward all three of the planes of projection. For example, let it be re- quired to draw the projections of a square with one side on H, inclined at an angle of 45° to T', and with the plane of the figure inclined at an angle of 60° to //. (See Fig. 52.) Start with the projection on a V plane, perpendicular to the plane of the figure as indicated by the axis H'A'. The line a^-V, drawn at an angle of 60° to H'A', and in length equal to a side of the square, is such a projection. One side of the H projection (aV) is known in length and in direction ; hence that side can be drawn. The angles of the square will project in their true size, by § 36. The H projection can now be com- pleted by drawing a perpendicular through b'', thus determining b^ and d^. Knowing the plan, the elevation may be drawn by projecting the points upward and measuring the heights from the F' projection, as in § 27. .B"d^' 56 DESCRIPTIVE GEOMETRY [V, § 47 47. Plane Solids. Solids that are determined by plane surfaces are called plane solids. There are two general types, regular and irregular. The regular solids are those whose Tetrahedron Cube Octahedron Fig. 53 Dodecahedron Icosahedron faces are regular polygons, so joined together that the angles between the planes are equal, and that no one of the planes, if prolonged, will cut into the solid. Besides the regular solids ^,. ,.^ there are prisms, pyramids, and an infinite number of other irregular shapes. 48. The Regular Solids. There exist in all just five regular solids. (See Fig. 53.) They are named as follows : (1) The tetrahedron, bounded by four equilateral h triangles. (2) The hexahedron, or cube, bounded by six squares. (3) The octahedron, bounded by eight equi- lateral triangles. (4) The dodecahedron, bounded by twelve regular pentagons. (5) The icosahedron, bounded by twenty equilateral triangles. 49. Development. If the surface of a given solid can be flattened out on a plane without distortion, that surface is said H 3'd' /K zz A ^ / ' \ / • \ b" i \ii(o^ ^/" d' \j/ Fig. 54 V, § 49] PLANE FIGURES AND SOLIDS 57 to be developable. The making of collapsible paper boxes illus- trates the point. A sphere could not be made in the same way. This idea is important in the sheet metal trade. In Fig. 54, for example, the plan and elevation of a square pyramid are shown. To develop the surface, let each of the sloping faces be rotated outward, until they fall in the plane of the base. The figure a^e — d^f is the required development. Fig. 56 Fig. 57 If the pyramid is placed with one of its triangular faces on the plane of development, and then is rolled over so that each face in turn is brought into this plane, the development appears as in Fig. 55. Figures 56 and 57 show the developments of an octahedron and an icosahedron, respectively. 58 DESCRIPTIVE GEOMETRY [V, § 50 50. To Construct the Projections of a Tetrahedron. In whatever position a tetrahedron may be placed with reference to the planes of projection, some of its edges will be normal-case lines ; hence they will be foreshortened in projection. The problem of drawing the projections of such a figure is therefore not simple. * Let it be required to draw the projections of a regular tetra- hedron 1" on each side and resting on H. (See Fig. 58.) The plan can be drawn from considerations of symmetry. The elevation is started by locating a^'6^ and c^, which are known to lie in HA. The determination of the height of the point o presents some difficulty, since it is connected to the points so far determined by lines that will be foreshortened in projection. The principle of the following solution is that of § 24. Let ao be rotated about a, until it is parallel to V. Its plan is now at a^o'^. The V projection of o' will lie on o'^'z. Moreover, the V projection of the rotated line will show its true length. Therefore o' will lie on an arc swung from a"-', with a radius of 1" (as 1-2). The intersection of this arc with o'^'z will give V, § 51] PLAXE FIGURES AND SOLIDS 59 the point o''. Now if the line is swung back to its correct position, o'' will trace a horizontal line, and o'' will be on this line and directly above o^. Having determined o% connect it with a'', c'', and b'\ The projections of the tetrahedron are now determined. From the H and V projections the P projection can be determined if it is required. 51. To Construct a Solid from Its Development. For the purpose of illustration a dodecahedron will be used. The dT' Fig. 59 general method for this prol^lem is to work from the known converse solution. (See § 29.) In this case the development of one half of the solid (consisting of six pentagons) will be as shown by the dotted lines in Fig. 59. The outside figures will be rotated about the lines joining them to the central figure until corresponding points (as / and g, k and j, etc.) fall together. This operation will be seen to be exactly the reverse of the process for making a development. To follow this operation in detail, think of the development in Fig. 59 as on the H plane. Let the central figure remain 60 DESCRIPTIVE GEOMETRY [V, § 51 on H, as above, while the figures surrounding it are rotated up- ward, swinging on the side which touches the base as an axis. Each of the vertices will move in a circular arc which is pro- jected on plan in a straight line perpendicular to its axis of revolution, as shown in the figure by the dotted lines j^2^, k^2^, /1\ fl^, etc. For any two corresponding points, as / and g, these arcs are of the same radius and are struck from the same center ; hence ---dT' 'Fig. 59 (Repeated) corresponding parts have equal elevations. Therefore the points 2^, \^, etc., are real points of intersection for these arcs. Before the rotation was started, the lines hf and })g had the point h in common. If the points / and g have been brought together at 1, these lines coincide throughout. The V projection of the points 1, 2, etc., can be easily found by following the rotation as projected on a V plane parallel to the plane of the arc. In this way we can determine the heights of each of the required points, e, h, etc. When all the polygons are thus rotated into position, the cup-shaped figure V, § 52] PLANE FIGURES AND SOLIDS 61 shown by the soHd lines in Fig. 59 is obtained. This is the lower part of the dodecahedron. The upper part is found by rotating the faces downward, making a cap-shaped figure as shown by solid lines on Fig. 60. This figure shows the com- plete projection of the solid found by placing the cap on the cup. Fig. 60 52. Prisms. A right prism may be defined as a solid gen- erated by the motion of a plane polygon which moves without rotating, in a direction perpendicular to its own plane. If this generating polygon is a regular polygon, the prism is called regular, and the line through the center of the polygon perpen- dicular to its plane is called the axis of the prism. Then we may say that the generating polygon always remains perpendicular to the axis of the prism ; that any plane perpendicular to the axis cuts from the prism a section equal to the generating polygon, and that no other plane will cut out of the prism a section exactly equal to that one. In what follows, we shall be interested chiefly in regular prisms. In other cases, however, any line perpendicular to the generating polygon may be substituted for the axis in the preceding statements. Since a section made by a plane perpendicular to the axis is unique, it is used to describe the prism. Such a section is called a right section, or sometimes simply the section, of the prism. 62 DESCRIPTIVE GEOMETRY [V, § 53 53. To Draw the Projections of a Right Prism. If it is desired to draw the projections of a right prism, the right section as well as the length and the slope of the axis must be given. The '. b' b'^ Fig. 61 V, § 53] PLAXE FIGURES AXD SOLIDS 63 projections of the axis can be constructed from the principles of § 29, and then the projections of the ends can be found from § 46. Finally, if the edges are drawn to connect the ends, the projections will be complete. For example, let it be required to draw the projections of an hexagonal prism, f in diameter and H" long, with its axis inclined at angles of 30° and 45° to H and T", respectively. In Fig. 61, aW' and a^'h^' are the projections of such a line, determined by the method of § 29, as shown in the half-scale drawing {a). On the full-scale drawing, choose an auxiliary plane of projection V parallel to this line, as indicated by the axis line H' A' . Find the projection of the axis of the prism on this plane (§ 27), as shown by a^'^b^'. Now the bases of the prism will be planes perpendicular to the new plane of projection. Hence they will project on it in lines perpendicular to a^'h'' . Draw such lines of indefinite length {cd and cf) . It is now necessary to fix the definite points on this line which represent the corners of the hexagon. In order to avoid confusing the drawing, imagine that the generat- ing hexagon has been slipped out along the axis till its center is at x''\ Next let it be rotated till it is parallel to V\ Then its projection can be drawn as shown at 1-6. By projecting these points back to the lines cd and ef we can locate the projections of the required corners at l^'-6^'' and 'jvf_i2vf 'pj^p fj projections of these points lie on lines drawn through the I ' projections and perpendicular to H' A' . Since the prism has been so placed that the points 6 and 2, also 3 and 5, lie on lines parallel to H,^ the projection 2''6'' will be equal to 2-6 and symmetrical with respect to the projection of the axis, a^h^. Thus the plan is easily completed. The elevation on V can be constructed from the points thus found on plan together with the heights shown on the V elevation. ^ The more general case is given in Art. 97. 64 DESCRIPTIVE GEO:\IETRY [V, §54 54. Intersection of Right Prisms — Special Case.^ When two prisms are drawn as shown in Fig. 62 a, it is not at once apparent where they intersect. The problem of determining the Hne of intersection becomes very simple when it is re- membered that the line of intersection will be that line which joins the points where the edges of one solid pierce the faces of the other. Since one solid, in Fig. 62 a, is composed entirely of vertical and horizontal planes, the obvious method is to find the points («) b* (Jb) Fig. 62 in which each edge of the inclined prism intersects some face of the upright prism, using the principle of § 31. In this case we find the trace of a line on the plane bounding the prism instead of finding its trace on H or V. In Fig. 62 a, the trace of the edge cd is found by noticing that the intersection of c^d^ and ^^5^ is the point k^. This point is the H projection of the trace of cd on the plane 4-5-11-10. Drawing a line through k^ perpendicular to HA till it intersects ^ The more general case is given in § 104. V, § 54] PLANE FIGURES AND SOLIDS 65 c^'d", we find the T^ projection of A*, which is the trace of cd on the prism. By the same method, the traces of the other lines may be found. When these traces are properly joined, the intersection as shown in Fig. 62 b is determined. EXERCISE SHEET XVIII Take P at the left border. No profile projections are required. 1. Assume a card in the shape of a hollow rectangle 1" wide and 11" long. The inner rectangle is |" wide and IJ" long. This figure is placed with one corner at the point a (8, 0, 12). The short side ah runs forward and to the right. The Une ab lies in H and makes an angle of 30° with HA. The plane of the figure inclines toward V and makes 45° with H. Draw its H and V projections. 2. Draw the H and V projections of a regular tetrahedron 2" on a side, with its base parallel to H. The back edge of the base is the line joining the points i (52, 4, 5) and j (68, 4, 5). EXERCISE SHEET XIX Place the sheet with the long dimension vertical, and with the wide margin at the left. Locate HA at a distance 3" from upper border line. Draw the H and V projections of a solid bounded by 6 squares and 8 equilateral triangles. Each side of each of these squares and of each of the triangles is 1" long. The sohd rests on a triangle as a base, the center being placed at the center of the lower part of the sheet. EXERCISE SHEET XX Take P at the left. No profile projections are required. 1. The point a (20, 0, 12) is the center of the base of a right hexag- onal prism, 2|" in diameter and If" high. The point b (17, 0, 14) is the center of the base of a regular tetrahedron, 3" on a side. Find the line of intersection of the solids. 2. Assume an irregular quadrilateral prism with its faces vertical, and an irregular triangular prism whose edges are normal-case lines. Find the fine of intersection of the sohds. The exact layout is left to the student. [Note. Having completed the foregoing work, the student can profitably take up the problems of §§ 179-183 and 202-210.] CHAPTER VI PROBLEMS DEALING WITH POINTS, LINES, AND PLANES 55. Introduction. The preceding chapters cover the essential principles of orthographic projection. By the use of these principles only, a large part of all ordinary drafting operations may be performed. The great majority of cases deal with objects composed of plane surfaces, and the resulting intersec- tions, which are straight Hues and points. Cases arise constantly, however, in which it is necessary to establish relations of parallelism, perpendicularity, etc., between lines and planes, and to record the results in projection. The present chapter will be devoted to the further study of points and lines with reference to their relations to planes, and to the representation of plane surfaces and their intersections. 56. Representation of Planes. In making working drawings, the planes dealt with are usually definite in extent and are repre- sented by the projections of their bounding lines or points. Thus, in Fig. 7, the plane A is defined by the lines ab, be, cd, and da. It is frequently necessary, however, to extend these planes indefinitely. Hence we must start with the conception of planes of indefinite extent, and we must find a means of rep- resentation which can be used in connection with the methods of point and line projection which we have studied before. From solid geometry we know that a plane is determined : (1) by any three points, and (2) by any two intersecting or parallel lines. Figure 63 a shows a part of a plane, X, and three points, a, b, and c, lying in X. The projections of these points might be said to determine the plane, X. No other plane could be determined by these projections, but any other three 66 VI, § 56] POINTS, LINES, AND PLANES 67 points chosen at random, in X, not on the same straight line, would determine the plane equally well. Hence the projections of three points do not constitute an entirely satisfactory de- scription of the plane, since the choice of these points is not unique. In the same way, it can be shown that the projections of intersecting or parallel lines, lying in X, would not constitute a unique description. However, if the plane is considered as of in- definite extent, it will intersect the planes of projection in two lines (Fig. 63 6, XX\ XX'*), which do constitute a unique description of the plane. These lines are called the traces of the plane. (Compare this with the definition of the traces of a line, § 31.) Since the plane is indefinite in extent, it will extend into all four quadrants (Fig. 64). The traces below and behind HA are merely continuations of those above and in front of HA. Fig. 64 68 DESCRIPTIVE GEOMETRY [VI, § 56 WTien the planes of projection are opened or flattened out, these traces become merely two lines intersecting on HA, as V Fig. 65 Fig. 66 shown in Fig. 65. When a profile plane of projection is used there will be a third trace, which intersects the other traces, one on each of the profile axes. (See Figs. 66 and 67.) The Fig. 67 profile trace is little used, however. Hereafter, unless otherwise specially noted, the phrase traces of a plane will be taken to mean the traces on the H and V planes. VI, § 59] POINTS, LINES, AND PLANES 69 57. Assuming a Plane. It will be evident at once that any two lines intersecting on HA represent the traces of some plane ; hence to assume a plane it is sufficient to draw any two lines intersecting on HA. It may be well to note, however, as a precautionary measure to guide one's sense of visualization, that the angle X^ X X^, in Fig. 65, is not equal to the corresponding angle in Fig. 64, that is, to the one which exists in space. This point is brought out more fully in Fig. 68, which is drawn to show the opening of the planes with reference to these angles. 58. Normal and Special Cases. For our purposes special-case planes may be defined as those that are perpendicu- lar or parallel to a plane of projection. It will be a helpful exercise for the student to draw the traces of as many special-case planes as occur to him. In doing this start with the limiting cases, viz., (1) when the plane is perpendicular to HA, and (2) when the plane contains HA. Draw also the traces of the normal cases which are in nearly the same position as (1) and (2) above ; and of several intermediate cases. Also study each special case with reference to possible variations and limits. 59. The Use of Isometric Projections. In the study of plane problems, the power of visualization is heavily taxed. A free use of isometric sketches similar to Figs. 63, 64, and 67 will be remarkably helpful in this respect. In the end, it will be found that the making of such sketches has created the ability to visualize without such help. For this purpose quickly drawn freehand sketches are quite satisfactory if they are kept in fairly correct proportion. 70 DESCRIPTIVE GEO:\IETRY [VI, § 60 60. The Fundamental Relations of Point, Line, and Plane. A point can be represented by its projections only (§7). A line may be represented by its projections, or by its traces (§§ 14, 15, and 31). A plane can be represented satisfactorily only by its traces (§ 56). These facts should be clearly recognized, since many of the important principles and methods of the following problems depend upon them. Figure 69 shows a point a, on the plane X, and Figure 70 shows the same point in projection. For normal-case points. Fig. 69 Fig. 70 on normal-case planes, the projections of the points do not fall on the traces of the planes. Moreover, there is no simple re- lation between the projections of any point and the traces of the plane which contains it, beyond the fact that a^ (Fig. 70) must lie within the angle X"" XA, and a!" must he within the angle AXX^. Let the student work out this idea in its application to points in the second, third, and fourth quadrants. In certain special cases, one or both of the projections of a point may lie on the traces of the plane. Let the student work out a few such cases. VI, § 60] POINTS, LINES, AND PLANES 71 Figures 71 and 72 show the projections of a normal-case line, aby lying in a normal-case plane, Y. As long as the line is limited, and is described merely by means of the normal-case Fig. 71 points a and h, no particular relations appear to exist between the traces of Y and the projections of ab. However, the traces of the line, m and n, being special-case points, do have a definite relation to the traces of the plane. It will be seen that the H trace of the line must fall in the H trace of the plane ; and the V trace of the line must fall on the V trace of the plane. Why ? 72 DESCRIPTIVE GEOMETRY [VI, § 61 61. Special Cases. Special-case lines or special-case planes may show definite relations between the projections of the line Fig. 73 and the traces of the plane. One such case is of sufficient Fig. 74 importance to deserve particular notice. Figure 73 shows a line lying in the plane Z, and parallel to //. It can be proved readily VI, § 63] POINTS, LINES, AND PLANES 73 that the H projection of such a line is parallel to the H trace of the plane, and that the V projection of the line is parallel to HA, as shown in Fig. 74. Let the student prepare a proof of the above statement, and work out the other special cases in which there are definite relations between the projections of the line and the traces of the plane. 62. Use of the Profile Plane. In Fig. 75, let the lines ah and cd be given, and let the traces of the plane determined by VI x" X' a" b^ \ i)a''b'* nf \' c" d^ \cMf 1 1 A nr\ a^ i b^ / ' / Ic*' d^ • X. x^ p Fig. 75 them be required. The given lines have no H and V traces ; hence it is necessary to use a profile plane. Find the profile traces of the lines {m. and ^0- The profile trace of the required plane will pass through these points, and its intersections with the profile axes ( Xi and X^) will determine the required H and V traces, X^X^ and X'X". 63. Choice of Auxiliary Lines and Planes. In the problems that follow, it will be necessary to make frequent use of auxiliary 74 DESCRIPTIVE GEOMETRY [VI, § 63 lines and planes, i.e. lines or planes arbitrarily introduced into the problem in order to arrive at a solution.^ For the purposes of §§ 65 and 67, where an infinite number of solutions is possible, any one of an infinite number of auxiliary lines may be chosen. When the conditions of the problem are prescribed more and more closely, however, it becomes more and more necessary to use good judgment in choosing the auxiliary lines and planes so that the more rigid conditions of these problems may be fulfilled. When only one solution of a problem is possible, the choice of an auxiliary becomes quite restricted. Moreover, the judicious choice of auxiliary lines and planes often may simplify materially the necessary drawing. Such a case is treated in § 78. It follows that the student should make a careful study of the way in which his choice affects any given solution, and thus strengthen his ability to select instinctively the best possible auxiliary lines and planes for any given problem. 64. To Assume a Line Which Lies in a Given Plane. In Fig. 76, let the plane Z be given and let it be required to assume a line which lies in Z. i Fig. 76 ^ This idea of arbitrarily introducing a foreign element into a problem in order to arrive at a solution is frequently used in mechanics, biology, chemistry, medicine, and other sciences, as well as in all branches of pure mathematics. VI, § 66] POINTS, LINES, AND PLANES 75 Since the traces of any such line are known to He in the traces of the plane, any two points chosen at random, one on ZZ^^ the other on ZZ"^, will be the traces of a line in Z. Let the point m be chosen on ZZ^'. This point lies in V \ therefore its // projection lies in E.A, at m^. In the same way n^ and iV^ are the projections of some point in the H trace of Z, By con- necting tny-ny and n^m^^ we may determine the projections of the line mriy which lies in Z. • 65. To Assume a Point Which Lies in a Given Plane. This problem is complicated by the fact mentioned in § 60, that a point has projections only, whereas a plane has traces only. For this reason it is necessary to employ an auxiliary line. The line, ha\ing both projections and traces, can be used to establish relations between the point and the plane. It is sufficient, then, to assume any line in the plane (§ 64). Any point on such a line will lie in the given plane. For this purpose, such a line as is shown in Fig. 73 is frequently found most convenient. 66. To Pass a Plane through a Given Line. In Fig. 77, let ah be the given line. Determine its traces, c and d. Through FiQ. 77 these points draw any two lines that intersect on HA (XX^ and XX^). These lines will be the traces of some plane 76 DESCRIPTIVE GEOMETRY [VI, § 60 passing through ab. An infinite number of such planes can be drawn. 67. To Pass a Plane through a Given Point. Pass an auxil- iary line through the given point, then pass a plane through the line (§ 66). This plane will contain the given point. Here again the line parallel to a plane of projection is usually most convenient (Fig. 73). 68. To Find the Traces of the Plane Determined by Two Given Intersecting Lines. Let the lines ab and cd, in Fig. 78, Traces of a Plane Determixed by Two Intersecting Lines Fig. 78 be given and let it- be required to find the traces of the plane determined by them. It should be noticed that the lines must VI, § 68] POINTS, LINES, AND PLANES 77 intersect in order to determine a plane (§ 33). The given lines intersect at e. The traces of the given lines will lie in the required traces of the planes (§ 60) ; therefore if the traces of ab {n and q) and of cd {m and y) are found (§ 31), these traces will determine the traces of the required plane X. If the work has been done carefully, the lines thus determined will meet on HA. This furnishes a test of accuracy of draftsmanship. Isometric Projection Corresponding to Fig. 78 Fig. 79 Figure 79 shows the space figure that corresponds to Fig. 78, in isometric projection. Sometimes it may be inconvenient or impossible to find all four traces of the given lines. In this event, determine two traces on either H or V ; draw the trace of the plane through 78 DESCRIPTIVE GEOMETRY [VI, § 68 these points and continue it to HA. The remaining trace of the line together with the point just found on HA will determine the other trace of the plane. 69. Plane Determined by Three Points. Let any three points in space be given, and let it be required to find the traces of the plane determined by them. It is sufficient to draw two inter- secting lines through the three points and proceed as above. However, since the three points determine three lines, any two of which intersect, some room is left for the exercise of good judgment in selecting the pair that will give the easiest solution. EXERCISE SHEET XXI Take P at the left, 20 from the border line, and turned away from the objects, into V. 1. Draw the H, V, and P traces of the plane R ^ (22, 72" 45° I, 2. Given the points a (IV, 33, 20, 2), b (IV, 54, 20, 2), c (IV, 33, 4, 7), and d (IV, 54, 4, 7). Find the H, V, and P traces of plane S determined by the Hnes ab and cd. EXERCISE SHEET XXII ^ [Note. Unless otherwise indicated, P will hereafter be taken at the right border line and no profile projections will be required.] 1. Given the plane R (75, i^'' 45° r, R"" 90°), draw the projections of two Hnes lying in R ; one parallel to and 1 1 from H, the other parallel to and 6 from V. Letter these lines ab and cd. 2. Given the plane S (31, Sf" 30° I, S" 45° I), locate a point e on S and at a distance 7 from H and 3 from V. 3. Given the points/ (18, 9, 4) and g (9, 3, 12). Through /gr pass three planes, T, U, and W, the plane T being perpendicular to H. 1 For explanation of the notation used in describing planes, see Descrip- tion of Quantities, page xiii. 2 Sheets XXII to XXXVII consist of three problems each, so spaced that #1 from any sheet will combine readily with #2 and #3 from any other sheet or sheets. VI, § 71] POINTS, LINES, AND PLANES 79 EXERCISE SHEET XXni 1. Given the points a (III, 72, 4, 11) and 6 (57, 6, 5). Through ab pass the plane X intersecting HA at 70. 2. Given the points c (48, 6, 10) and d (37, 6, 2). Through cd pass the plane Y intersecting HA at 29. 3. Given the point e (15, 8, 6). Through e pass the plane Z meet- ing HA at 3. EXERCISE SHEET XXIV 1. Through the point a (65, 5, 4) pass two lines as follows. The H projection of one line makes an angle of 30° to the left and the V projection an angle of 60° to the right with HA. The H projection of the other makes an angle of 60° to the right and the V projection an angle of 45° to the left with HA. Draw the traces of the plane X determined by these hnes. 2. Given the line joining the points h (46, 11, 2) and c (34, 3, 14), and the point d (II, 41, 2, 5). Draw the traces of the plane Y deter- mined by the point d and the line ab. 3. Given the points e (23, 2, 12), / (20, 11, 4), and g (14, 2, 4). Draw the traces of the plane of the triangle efg. 70. Parallel Planes. If two planes are parallel, their cor- responding traces are parallel. Let the student prove this by using the theorem of solid geometry concerning the lines of intersection of two parallel planes with any third plane. 71. Line Parallel to a Plane. In general, when a line is parallel to a plane, the projections of the line are not parallel to the traces of the plane. Exceptions to this statement may be noted in the cases of planes whose traces are both perpendicular to HA. Another exception is the case in which both the line and the plane are parallel to HA. If a line is parallel to a plane and also to H or V, one pro- jection of the line w^ll be parallel to one trace of the plane. This case is similar to that shown in Fig. 73. 80 DESCRIPTIVE GEOMETRY [VI, § 72 72. To Pass a Plane through a Line and Parallel to Another Line. From solid geometry we know that a line and a plane are parallel if the plane contains any line which is parallel to the given line. In Fig. 80, let the lines ah and cd be given. Let it be required to pass a plane through ah, parallel to cd. Choose any point e, on ab, and through it draw the auxiliary line fg, parallel to cd. The two lines ah and fg determine the plane Z (§ 68). This plane is parallel to cd, since it contains a line parallel to cd. Fig. 80 73. To Pass a Plane through a Point and Parallel to a Given Line. The analysis and construction are so nearly identical with that of § 72 that no further explanation is needed. The student should recognize, however, that the problem is indeterminate, as are those of §§ 66 and 67. 74. To Pass a Plane through a Given Point and Parallel to a Given Plane. The traces of the required plane will be parallel to the traces of the given plane (§ 70). An auxiliary line, pass- ing through the given point, is needed to determine the required plane (§ 67). In this case the problem is simplified by choosing VI, § 741 POINTS, LINES, AND PLANES 81 as the auxiliary a line parallel to H and to the given plane. Refer to §§ 63, 61 and the last paragraph of § 71. In Fig. 81, let the plane Z and the point a be given, and let it be required to pass a plane through a, parallel to Z. Draw the H projection of the auxiliary line through a^ and parallel to ZZ^, as nhn^. The V projection of such a line (since the Hne is parallel to H) will be parallel to HA, and will pass through FiQ. 81 a^, as shown by n'-'m''. The V trace of ??m (??') will lie in the V trace of the required plane. Hence a line drawn through 71^ parallel to ZZ% will be one of the required traces, Y Y". Through 7 draw YY^ parallel to ZZ\ EXERCISE SHEET XXV 1. Given the plane M (75, M^ 60° r, M" 45° r) and the point a (61, 6, 8). Through a draw a line parallel to M. 2. Given the plane .V (45, A'^ 75° I, N^ 45° r) and the point b (39, 10, 12). Through h draw a line parallel to N. 3. Given the points c (III, 25, 11, 19), d (III, 16, 15, 11), e (20, 2, 14), and / (8, 9, 4). Through the line ef pass a plane parallel to cd. EXERCISE SHEET XXVI 1. Given the points g (III, 77, 6, 17), h (III, 70, 6, 8), j (68, 1, 11), and k (60, 12, 3). Through the line jk pass a plane Q parallel to gh. 82 DESCRIPTIVE GEOMETRY [VI, § 74 2. Given the points I (51, 3, 12), m (42, 8, 5), and n (36, 7, 7). Through n pass the plane R parallel to the Une Im. 3. Given the points q (IV, 16, 7, 14), r (IV, 24, 18, 9), and o (II, 12, 9, 3). Through o pass the plane S parallel to the line qr. EXERCISE SHEET XXVII 1. Given the plane X (80, X^" 60° r, Z" 45° r) and the point a (63, 7, 6). Through a pass the plane Y parallel to X. (Use a normal-case auxiliary line.) 2. Given the plane Z (45, Z^ 30° r, Z^ 60° and the point h (III, 38, 8, 7). Through b pass the plane W parallel to Z. (Use a special- case line.) 3. Given the plane U (3, m 30° Z, C/'^ 45° I) and the point c (IV, 16, 5, 7). Through c pass the plane T parallel to U. 75. The Line of Intersection of Two Planes. The line of intersection of two planes contains all points common to the two planes. If the traces of two planes intersect, the point of Fia. 82 intersection is a point common to the two planes ; hence it will be in the line of intersection of the planes. Two such points are necessary to fully determine the line of intersection of any two planes. VI, § 76] POINTS, LINES, AND PLANES 83 In Fig. 82 are shown two planes, A^ and Y. All points on X X" lie in X. All points on Y 7^ lie in Y. Therefore the point a, which lies in both the lines, is in both the planes, and hence it is on their line of intersection. The same is true for the point b. Figure 83 shows the same operation in projection. The V traces of X and Y are extended to their intersection at o^. H This point lies in V, and hence a^ lies on HA. In the same way b^ and b^ are found by using the H traces of the given planes. Hence the lines a^b^ and a"6'' are the projections of the line of intersection of the planes X and Y. [Note. The relation between such a line as ab, above, and its projections seems to be difficult for the average student to visualize. Quite a common mistake consists in connecting a* and b^ to show the line of intersection. A little study will show that such a line is mean- ingless on a projection drawing. It is true that the line in space does connect these points; but in the projection the V projection of one point and the H projection of another cannot determine a line.] 76. Special Cases. \Mien the traces of the given planes do not meet within the limits of the drawing, the method of § 75 for 84 DESCRIPTR^ GEOMETRY [VI, § 76 finding the line of intersection cannot be applied directly. However, a solution usually can be obtained within the limits of the drawing, by using a well chosen auxiliary plane. Case L In Figs. 84 and 85, the plane Z is taken as the auxiliary plane and is erected parallel to V. This plane con- stitutes, in effect, a new V plane on which the traces of X and FiQ. 84 y intersect at the point d. These traces, when projected on the original V plane, will intersect within the limits of the drawing, making the following solution possible. Referring to Fig. 85, the plane Z is passed parallel to F and cutting X and 7. The points h and c are seen to lie on the Hues of intersection between the given planes and Z. The V pro- jections of U and cd will be parallel to Y 7^ and XX- (§ 61). Thus \ > — Y X Fig. 87 Note that by choosing the auxiliary plane parallel to V, we have, in effect, merely shifted V forward, temporarily, in order to fix points otherwise beyond our reach. The principle involved can best be understood by making a sketch like Fig. 85. Fold this sketch as shown in Fig. 88. In this manner HA is brought down to ZZ^, eliminating from the drawing all the space between. Disregarding the lines YY^ and XX^, it will be seen that the V plane of projection has been brought forward to the position occupied by Z in Fig. 84, and that, by using the axis line thus established, the problem can be solved by the method of § 75. VI, § 77] POINTS, LINES, AND PLANES 87 Cases 2 and 3 are merely slight modifications of the above principle. This device of temporarily shifting the position of one or both of the planes of projection in order to confine a solution Fig. 88 IS to a limited space is of great convenience in many cases. It i sometimes spoken of as '' shifting the axis fine " or *' choosing a new plane of projection." EXERCISE SHEET XXVIU Find the lines of intersection of each of the following pairs of planes. 1. G (76, & 60° r, G'' 45° r) and H (57, m 30° l, H- 60° /). 2. J (50, J" 45° r, J^' 90°^ and K (30, /v" 60° /, K'' 30° /). 3. L (18, L^ 30° r, L- 75° /) and M (8, M^ 60° r, M- 45° r). EXERCISE SHEET XXIX Find the lines of intersection of each of the following pairs of planes. 1. N (74, A^'' 45° r, N^ 75° I) and 0, parallel to HA ; O^ is 16 above and 0" is 9 above HA . 2. P (50, P^ 821° r, P^' 75° r) and Q (28, (?^ 75° I, Q" 75° Z). 3. R, parallel to FA, R^ is 5 below and R' is 27 above HA ; and S, parallel to HA, S^" is 17 below and S'' is 8 above HA. EXERCISE SHEET XXX Find the lines of intersection of each of the following pairs of planes. 1. T (79, Th 45° r, T" 75° r) and U (61, T'' 30° /, V 75° r). 2. X (49, X''60°r, XM5° r) and TT, parallel to //A. JV^ is 6 below and TT^^' 11 above HA. 3. Y (23, 7'' 60° r, F'' 90°) and Z (3, Z^ 45° ?, Z^' 75° 0- 88 DESCRIPTIVE GEOMETRY [VI, § 78 78. To Determine the Trace of a Given Line on a Given Plane. In this case, of course, the line does not lie in the plane. Hence the traces of the line on H and V bear no definite relations to the traces of the plane. Thus, the old difficulty of establishing relations between a line (shown only by its projections) and a plane (shown only by its traces) is encountered again (§ 60). In Fig. 89, it can be seen merely by visualization that the plane Z and the line ah are not parallel. Hence they must Intersect. But the point of intersection (the trace of ah on Z) cannot be determined so easily. Here again the solution is obtained by the use of an auxiliary plane;, which this time should be passed through the line. From Fig. 90, which is a Cavalier projection of the case shown in Fig. 90 Fig. 89, it will be seen that the line of intersection of the auxiliary plane Y, with the given plane Z (cd), contains the required trace, ?', of ah on Z. VI, § 78] POIXTS, LINES, AND PLANES 89 Fia. 91 In Fig. 91, the operation is carried out in projection. The plane Z and the line ah are given. The auxiliary plane Y is passed through the line ah (§ 66). The hne of intersection of Z and Y , cd, is next determined (§ 75) . This Hne is seen to intersect the given line ah at i, which is the required trace of the line ah on the plane Z. It will be seen that i^ and i^ are determined independently, so that if the construction has been accurately drawn, they should lie on the same line perpendicular to HA. By a judicious choice of the auxiliary plane, the construction often can be simplified materially. In Fig. 92, using the same line and plane as be- fore, the auxiliary plane has been passed through ah, and perpendicular to H. The H trace, IT'', will pass through a^h^, while the V trace, YY'\ will be perpen- dicular to HA. Thus the determination of the auxiliary plane is greatly simplified. Also one projection of the line of intersection between Z and Y, c^d^, will coincide with a^h^. The V projection gives the required point i. Fig. 92 90 DESCRIPTIVE GEOMETRY [VI, § 78 By comparing Figs. 91 and 92, the simplification brought about by choosing an auxiHary plane which is perpendicular to a plane of projection will be evident. Sometimes it is better to choose the auxiliary plane perpendicular to one, and some- times to the other of the projection planes. The proper choice will depend upon the relative positions of the line and the plane. Applications of this problem will be found in § 188. EXERCISE SHEET XXXI 1. Find the trace of the hne joining the points a (72, 2, 12) and 6 (62, 9, 5) on the plane Z (58, Z^ 60° /, Z'' 45° T). Use a normal-case auxihary plane. 2. Find the trace of the line joining the points c (46, 12, 7) and d (III, 31, 9, 3) on the plane Y (49, Y^ 45° r, 7" 60° r). Use a special- case auxiliary plane. 3. Find the trace of the line joining the points e (20, 10, 7) and / (8, 2, 7) on the plane X (18, X^ 60° I, X- 30° r). EXERCISE SHEET XXXH Take P at the right border hne, turned toward the objects, into V. Use P only when necessary. * 1. Find the intersection of the Hne joining the points g (72, 5, 6) and h (60, 11, 10) with the plane W (75, W^ 45° r, TF" 90°). 2. Find the intersection of the Hne joining the points i (46, 13, 10) and j (III, 32, 9, 2) with the plane U, which is parallel to HA, U^ being 6 below and U'' 10 above HA. 3. Find the intersection of the line joining the points k (20, 18, 9) and I (20, 2, 9) with the plane T, which passes from II to IV, through HA. and makes an angle of 60° with H. 79. Line Perpendicular to a Plane. If a line is perpendicular to a plane, the projections of the line are perpendicular to the traces of the plane. This will be seen by reference to Fig. 93. The line ah is perpendicular to the plane 7. The H projecting plane Z is shown passing vertically downward through it. This plane Z is perpendicular to // and also to 7, since it contains ab, which is perpendicular to 7. Since both // and 7 are perpendicular to Z, the line of inter- VI, § 81] POINTS, LINES, AND PLANES 91 section between H and Y, Y Y^, is perpendicular to Z, and hence to every line in Z passing through the point c. There- fore Y y* is perpendicular to a^b'^. Fia. 93 By the same reasoning it can be proved that YY"" is perpen- dicular to a'"6''. 80. Perpendicular Planes. Let the student visualize for himself the truth of the following statements : (1) "When two planes are perpendicular, their traces are not, in general, perpendicular. (2) The limiting conditions occur (a) when the line of inter- section of the planes is parallel to a plane of projection, — in this case the traces on that plane of projection are parallel ; and (6) when the line of intersection of the two planes is per- pendicular to a plane of projection, — in this case the traces on that plane of projection are perpendicular. (3) An exception to the preceding statement occurs when one of the planes is perpendicular to a plane of projection, as in Fig. 93. 81. To Pass a Plane Perpendicular to a Given Line. It is merely necessary to draw the traces of the required plane perpendicular to the projections of the given line. u 92 DESCRIPTIVE GEOMETRY [VI, § 82 82. To Pass a Plane through a Given Point and Perpendic- ular to a Given Line. Pass any plane perpendicular to the given line (§ 81). Then pass a plane through the given point, parallel to the plane just drawn (§ 74). After following the above analysis of the problem, and solving an example or two, the student will doubtless discover a way to simplify the construction. 83. To Pass a Plane Parallel to a Given Plane and at a Given Distance from It.^ The distance between two parallel planes is measured along a line perpendicular to both planes. Assume a point on the given plane (§ 65). Through the assumed point draw a line perpendicular to the given plane (§ 79). On this line, measure off the given distance from the starting point (§ 30). Through this last point pass a plane parallel to the given plane (§ 74). 84. To Determine the Distance from a Point to a Plane. The required distance is measured along a line through the point and perpendicular to the plane. First Method. One method, which employs no new operations, is as follows. (1) Pass a line through the point and perpen- dicular to the plane (§ 79). (2) Find where this line pierces the plane (§ 78). (3) Find the true length of the line between 1 Method of study. The problems of §§ 83-88 are quite typical of many relatively complex problems, in that they involve a number of steps in analysis and solution, each one of which, in itself, is quite simple. If the student allows himself to become confused, or if he does not thoroughly understand each of the underlying problems, he is lost. On the other hand, if the foundation work has been done thoroughly, and if attention is centered on one step at a time, both the analysis and the construction may be built up quite easily. In working up such a problem, it is best to go through a complete analysis, starting with the given conditions, erecting the necessary auxiliary lines and planes, and performing all the necessary operations visually. Place nothing on paper until the conditions and the method of solution are seen distinctly. In this way the mind is freed of the detail of each individual operation and is left free to grasp the important principles involved. This having been done, the detailed working out of the construction can follow, and attention can be centered on the detail of each operation without confusing the main issue. VI, § 84] POINTS, LIXES, AND PLANES 93 the point given and the piercing point thus found (§ 24). Second Method. An- other method is shown in Figs. 94 and 95. The point a and the plane Z being given, it is required to find the distance from a to Z. Pass the auxihary plane, Y, through a and perpendicular to ZZ^. This plane will contain the required perpendicular between a and Z, and will intersect Z Fia. 95 in the line he. Now let Y be rotated about YY^, into H, so as to show he and a in their actual relations. This rotation is shown in Fig. 95. The point h falls at h'^, which is located in Fig. 94 by making h^h'^ equal to h%'' and the angle c^h^h'^ equal to 90°. The point a rotates to a'^. The perpen- dicular a'^d'^ can now be drawn and from it d^ and d'' can be located by projecting back on eh. 94 DESCRIPTIVE GEOMETRY [VI, § 85 85. To Determine the Distance from a Point to a Line. The required distance is measured along a line passing through the given point and perpendicular to the given line. Pass a plane through the given point and perpendicular to the given line (§ 82). Find where the given line pierces the plane thus found (§ 78). Connect this point with the given point, and find the true length of the line thus determined (§ 24). 86. To Determine the Common Perpendicular between Two Lines. The common perpendicular between two normal-case lines is not easy to visualize nor to de- termine (§ 36). But it is easy to con- struct a line which is perpendicular to a plane (§ 79). Moreover, if any two lines are given, as ab and cd, in Fig. 96, and if a plane is passed through ab and parallel to cd, then any perpen- dicular from cd A to this plane will measure the distance between the two lines, and one of these perpendiculars will be the com- mon perpendicular between the two lines. These considerations furnish the basis for the following solution. Referring to Fig. 96, the common ^^" perpendicular between ab and cd being required : (1) Pass the plane Z through ab and parallel to cd (§ 72). The line ef, parallel to cd and intersecting ab at any point, as g, was used as an auxihary line. (2) Draw the hue ch,. perpendicular to Z (§ 79), and find where ch pierces Z (§ 78), namely, at j. In the same manner determine dk, the perpen- dicular through d and touching Z at k. (3) By joining j and k VI, § 87] POINTS, LINES, AND PLANES 95 the projection of cd on Z is obtained. This Hne contains the foot of every perpendicular dropped from cd on Z. (4) The point where jk crosses ab, namely, ???, will be the foot of the common perpendicular between cd and ab. (5) Draw mn parallel to cj, and intersecting cd at 7i. This line is the required common perpendicular. (6) If the distance between the lines is required, find the true length of the line mn (§ 24). 87. Special Case of Parallel Lines. If the given lines in § 86 are parallel, the solution may be simplified as follows. (1) Pass a plane perpendicular to the two lines. (2) Find where each line pierces this plane. (3) Join the points thus found, obtaining the common perpendicular. If the two lines lie in the same plane, and are not parallel, the construction is obviously impossible. EXERCISE SHEET XXXIII 1. On the plane M (73, ^/'^ 60° r, M" 45° r) locate a point a, 60 from P and 7 from H. Draw ab perpendicular to M and If" long. 2. Given the points c (50, 11, 6) and d (38, 5, 9), through d pass a plane A'' perpendicular to cd. 3. Given the points e (24, 20, 6), / (16, 11, 9), and g (III, 12, 8, 9), through g pass a plane O perpendicular to ef. EXERCISE SHEET XXXIV 1. Draw a plane R parallel to the plane Q (68, Q^^ 30° I, Q" 60° and I" from Q. 2. Given the points h (47, 12, 2), i (IV, 30, 2, 8), and j (36, 10, 12), find the shortest distance from j to the line hi. 3. Given the plane T (7, T'' 60° I, T" 30° I) and the point I (15, 10, 8), find the shortest distance from I to T. EXERCISE SHEET XXXV 1. Given a (63, 3, 10), b (47, 15, 3), c (48, 2, 2), and d (41, 5, 4), find the common perpendicular between ab and cd. 2. Given e (25, 22, 8), / (16, 7, 3), and g (13, 24, 7), through g draw a line parallel to ef and find the common perpendicular between this line and ef. 96 DESCRIPTIVE GEOMETRY [VI, § 88 88.^ The Angle between Two Planes. The angle between two planes is the angle cut from the planes by another plane passed perpendicular to their line of intersection. In Fig. 97, the plane Z is perpendicular to mm. According to the preceding definition, the angle hac is the angle between the planes E and 7. Any plane, as Z, cuts from tw^o intersecting planes an acute and an obtuse angle (6ac and hoK). It is usual to speak of the acute angle as the angle between the planes. It should be recognized that this is the greatest angle that can be cut from Fig. 97 these planes by a third plane perpendicular to E. For in- stance, the plane X cuts out a smaller angle. 89. The Angle between a Line and a Plane. The angle between a line and a plane is the angle between the line and its projection on that plane. In Fig. 97, the angle dcj is the angle between dr and U since X is perpendicular to //; and d is the angle between gj and }'. The method of determining such an angle in projection is given in § 98. 90. The Line of Greatest Declivity of a Plane. The line of greatest declivity of a plane A with respect to another plane B is that line in the plane A w^hich makes the greatest possible 1 Arts. 88-90 inc. arc introduced merely as a review of some of the fundamental definitions in solid geometry. VI, § 911 POINTS, LINES, AND PLANES 97 angle with the plane B. It is the line cut out of ^ by a plane perpendicular to both A and B, and hence perpendicular to their intersection. Thus in Fig. 97, ab is the line of greatest declivity of the plane Y with respect to H. Let it be required to draw the line of greatest declivity of the plane Y with respect to H in Fig. 98 a. Pass a cutting plane Z perpendicular to YY^. In space, these planes have the same relations as the planes Z and Y in Fig. 93, and their line of Fig. 98 a intersection will be the required line of greatest declivity. This line, ab, is determined by the intersections of the traces, as in § 75. The same plane Y, is shown in Fig. 98 6. Here the line of greatest declivity with respect to V is shown. EXERCISE SHEET XXXVI Draw the Hne of greatest decHvity of each of the following planes with respect to H and also with respect to V. Determine the true angle that each plane makes with H and also with V. The planes are : X (76, Z'^30°r, XM5°r), Y (36, F'' 30° ^, FM5° r), and Z parallel to HA. ZZ^ is 12 above and ZZ^ is 16 below EA. 91. To Rotate a Point about the Trace of the Plane in Which it Lies. — Rabattement. A problem which occurs frequently in the making of w^orking drawings consists in finding the true shape of a figure whose projections are known and whose plane is given. 98 DESCRIPTIVE GEOMETRY [VI, § 91 When it was required to find the true length and slope of a line (§ 24) the method consisted in reducing the line to a special case. Here also the most natural method will be to bring the given figure parallel to, or into coincidence with, a plane of projection. Place one of the acute angles of a triangle in the opened cover of a book, one of its edges touching the cover while the other rests on the book. Consider any three points Fig. 99 on this plane as representing a plane figure, which it is required to rotate into a position parallel to, or into coincidence with, H. The purpose of the operation is to determine the true shape of the imagined figure; hence the relative positions of the points must not be changed during the rotation. Under these conditions, it will readily be seen that the most natural axis of rotation is the edge of the triangle which rests on H, and which is the // trace of the plane containing the figure. VI, § 91] POINTS, LINES, AND PLANES 99 ax In order to study the laws governing such a rotation, let us refer to Fig. 99, which shows the point a, lying in the plane Z. This point a is being rotated about the trace ZZ^ as an axis, into H. During the operation, the radius of rotation, ay, generates a plane perpendicular to ZZ''. The lines ay and a'^y are equal, since they are radii of the same circle. Now this radius ay is the h^-potenuse of a right triangle aa^y. It is this triangle that gives the key to the whole solu- tion ; hence its relations to all parts of the problem should be thoroughly understood and com- pletely visualized. It should be noticed that the hypotenuse is a line of greatest declivity of the given plane with respect to H. Moreover, the base of this triangle is the distance from the H projec- tion of the point to the H trace of the plane, a^y, while the altitude of the triangle is the height of the point in space, aa'', which in turn is equal to the distance from the V projection of a to HA, a^x. Figure 100 shows this operation in projection, the point a being in the plane Z. It is required to rotate it about ZZ^ as an axis, into H. The line a^c is drawn perpendicular to ZZ^. Since this line is the H projection of the path of the point during rotation, the required point w^ill lie somewhere on it. In order to determine the distance of the required point from y, lay out the right triangle 1 at any convenient location on the drawing. Make it equal to the right triangle aa^y in Fig. 99, i.e. with its base equal to a^y and its altitude equal to a^x. The purpose of this triangle is merely to obtain the length of the hypotenuse, which is then measured off from y, along the line a^c. This gives the required point, a'^. 100 DESCRIPTIVE GE0:METRY n, § 91 After a few examples have been solved, the actual construction of such a triangle as 1 may be omitted altogether, the required distance being determined directly with the dividers. In Fig. 101, the point b, lying on the plane Z, is rotated about the trace ZZ', into V. The opera- tion is the same in principle as that shown in Fig. 99. The process described in this A article, that of rotating a point about the trace of a plane in which it lies, is called rabattement. 92. Special Cases. Many prob- lems involving rabattement may be shortened considerably by recognizing, in this connection, the special properties of points on the traces of the rotating plane. In Fig. 102 the points a and 6, and the plane Z are shown in Cavalier projection, as is also the rabattement of the line ah Fig. 101 Fig. 102 and that of the vertical trace ZZ'. The point a itself lies in the axis of rotation and therefore will not move during the rotation. It will be noticed also that the rabatted position of h lies on a line VI, § 93] POINTS, LINES, AND PLANES 101 {h^h"'), passed through h^ perpendicular to ZZ'^, and that the dis- tance Zh'^ is equal to Z6^'. It should be noted particularly that the line ah, its H pro- jection, a'^b^, the rabattement, a^h'^, and the H trace of the plane, ZZ^, all pass through the point a^, which is the H trace of the given line. In Fig. 103, the above operation is repeated in projection. It is required to rabatte the special-case point h, into H. Draw the line h^h'^ through h^, and perpendicular to ZZ^. Swing an arc from Z, with a radius equal to Zh''. Where the arc and line Fig. 103 intersect is the required point h'^. The point just determined may be connected with a^, giving the rabattement of the line ah ; or with Z, giving the rabattement of the V trace of Z. It should be noted that the rabattement of all points on Z, and within the first quadrant, will lie within the angle Z^'^ZZ^\ 93. To Find the True Shape of a Plane Figure by Rabatte- ment, the Projections and the Trace of the Plane Containing it Being Given. I. The General Method. The figure may be rotated into H, without distortion, by merely rabatting each of the ver- tices separately (§ 91), and then connecting the rabatted points. 102 DESCRIPTIVE GEOMETRY [VI, § 93 VI, § 94] POINTS, LIXES, AND PLANES 103 II. An Important Simplification. The preceding method may be simpUfied by the use of the principle of § 92. Let the triangle cde, lying in the plane Z, be given, as shown in Figs. 104 and 105. In Fig. 105 let the side ce be extended to its H and V traces {g and h). Next rabatte the V trace of the plane (ZZ^), using the point h (§ 92). The Hne gW^ is the rabattement of gh, and the points e'^ and c'^ can be estab- lished on it by perpendiculars through e^ and c^. Next, the rabattement of d is found by connecting c'^ with/'', and drawing the perpendicular through d^. This completes the triangle. A check on the accuracy of the construction may be had by extending the side of the triangle not used above {de), to the H trace of the plane, both in the projection and in the rabatte- ment. These extensions should meet on ZZ^. Again, the V projection of any one of the sides may be extended to ZZ% and the point of intersection may be rabatted to some point on Zh'^ ; the rabattement of this point should meet an extension of the rabatted side. 94. To Find the True Angle between Two Lines. Find the traces of the plane determined by the lines (§68). Rabatte the lines into F or F (§ 93). EXERCISE SHEET XXXVII 1. Given the point a at distances 59 from P and 12 from H, and a point h at distances 63 from P and 6 from T^; and given that both points lie in the plane X (77, X'' 30° r, X^ 45° r). Rabatte a into H and b into T^. 2. Given the point c at distances 40 from P and 4 from H, and lying in plane Y (43, Y^ Q0° I, F''30°r). Rabatte c into H. Also rabatte YY^ into H. 3. Given the plane Z (14, Z'' 60° r, Z^ 45° r), and four points e, f, g, and h, all of which Ue in Z : e is in III, 24 from P and 7 from H, rabatte e into H ; f is in I, 7 from P and 5 from V, rabatte/ into T^ ; g is in II, 14 from P and 7 from T^ rabatte g into H ; A is in IV, 16 from P and 8 from F, rabatte h into V. 104 DESCRIPTIVE GEOMETRY [VI, § 94 EXERCISE SHEET XXXVIII 1. The triangle abc lies in the plane S (72, S'' 45° r, S'' 30° r) ; o is 54 from P and 8 from i/ ; 6 is 62 from P and 2 from // ; c is 50 from P and 3 from H. Find the true shape of the triangle by rabatting into //. 2. The line joining the points d (29, 2, 13) and e (15, 9, 3) is inter- sected at/, IV' from d, by /^, where g is the given point (25, 13, 5). Find the true angles between these lines by rabattement. 95. Counter-Rabattement. The converse problem, counter- rabaitement, as its name implies, consists in finding the pro- jections of a point, given its rabattement and the traces of the plane in which it is to lie. I. First Method. Counter-Rabattement by Use of THE Line of Greatest Declivity. Given a"^ (Fig. 106), as the rabattement of some point in Z ; to find its projections after it has been turned back into Z. Pass a plane through a"^ and per- pendicular to ZZ^. It will cut from Z a line of greatest declivity which will contain the point a (in space). The H projection of this line of greatest declivity is mhi^. The V projection, not being needed, is not drawn. Between this line, its H projection, and V, lies a right tri- angle, iVn^m^, on whose hypotenuse, nm,, lies the point a in space ; and on whose base, n^jnJ', lies a^. A triangle of this kind, aal'y, is shown in Fig. 99. Let this triangle be rotated on ?i"# as an axis, into F, giving ii^n^m'^ as its true shape. The point a will be found on this line, at a distance from m'^ equal to a"''m^, the radius of rotation in rabattement, as shown at a'\ Fig. lOG VI, § 95] POINTS, LINES, AND PLANES 105 This point {a''') being established, we know the height of the point a in space ; hence a^ will lie on a horizontal through a'''. The distance from m'' to a^, which Avill lie on m^7i^, as pointed out above, will be equal to m'^a'^. It is now easy to locate a^ by measurement. In the construction used to draw Fig. 106, the circular arc a'^a^ was used to make this measure- ment. The point a" can now be found by erecting a perpen- dicular from a^. I. Second Method. Counter-Rabattement by Use of AN Auxiliary Line Parallel to H. Given h'^ (Fig. 107) as the rabattement of some point in Z; to find the projections of h. Find the rabattement ZZ'^ of ZZ' (§ 92). Through h'^ draw the ra- battement of a line in Z parallel to H. The rabattement will be paral- lel to ZZ^ and will pass through h'^ {b'^k'^). Since the point k (in space) is on ZZ"", it will have its rabatte- ment at k'^ and its H projection on HA. During counter-rotation, any point, as k'^, will generate on ^ a straight line perpendicular to the line ZZ^, and the H projection of the point will be somewhere on that line. Draw such a line through k'^. Its intersection with HA will determine k^, and a vertical line drawn from k^ to ZZ' will give k\ Through k^ and /:'' draw the projections of the auxihary line, k^^^ and Z:^p''. The required projections of the point h will lie on these lines. As we have showm in the preceding para- graph for A•^ h^ will lie on a line through h'^, perpendicular to ZZ^. A vertical Hue through this point to h'p'' gives the V projection of h. 106 DESCRIPTIVE GEO:\IETRY [VI. § 96 96. To Find the Projections of a Given Figure When it Lies in a Given Plane. This problem requires merely an extension and application of the principles of § 95, and constitutes the general case of the problem stated in § 46. In Fig. 108, let the plane Z be given and let it be required to find the projections of a regular pentagon lying in Z. First Fig. 108 rabatte the trace ZZ"" into //, using the point a (§ 92). If the pentagon is to lie wholly within the first quadrant, as is usual, its true shape may be drawn within the angle Z^'^Z Z^, which includes the rabattement of all points on Z which lie in the first quadrant. The exact placing of the figure within this space will de- pend upon the position which it is expected to take on the VI, § 97] POINTS, LINES, AND PLANES 107 plane Z, when it has been turned up into that plane. In this case, let it be so drawn that one side in space will be parallel to //. The rabatted position of that side will then be parallel to ZZ\ as shown by r^2'^, Fig. 108. It is now possible to counter-rabatte each of the vertices, b}' either of the methods of § 95. Usually, however, a simpler solution is made possible by varying the methods to suit the circumstances, and by dealing with lines instead of with points. Let the side l'^2'^ be extended to the rabatted trace ZZ"'^. The point of intersection, 6'^^, is the rabattement of the V trace of the line. The counter-rabattement of the line now follows the second method of § 95. Thus P^'' and 1*2'' can be estab- lished. Next let the side 2'^3''' be extended to c^ (its H trace). The H projection of 2-3 passes through (^ and 2^ which have been found. These two points establish the line 2-3. The point 3^" lies on this line and also on a perpendicular to ZZ^, drawn through 3'^. The V projection of 2-3 is found by drawing a line through c^ and 2^ and erecting a perpendicular from 3^". In a similar manner the other sides can be found. In the figure, the auxiliary lines drawn to determine the sides 4-5 and 5-1 are omitted for the sake of clearness. 97. To Construct the Projections of a Plane Solid, One Face of Which is in Contact with a Given Plane. This problem is the general case of the one of which a special case was stated in § 53. The analysis is quite simple, consisting principally in an application of the method of § 96. First, the base is counter- rabatted upon the given plane (§ 96). Next a side (in the case of right prisms), or the altitude (in the case of pyramids), is drawn, perpendicular to the plane (§ 79). The proper distance is then marked off on this line, thus fixing the apex (in the case of a pyramid), or a point on the top face (in the case of a prism) (§ 30). These points are then connected with the base pre- viously determined. 108 DESCRIPTIVE GE0:METRY m, § 97 In Fig. 109, let the plane A' be given, and let it be required to construct the projections of an equilateral triangular prism of given height, with one triangular face in contact with A". The plane is rabatted into H, using the point a, on its T trace, and giving the line A'A^'"'''. AVithin the area thus defined, the true Fig. 109 shape of the base {c'^h'^d'^) is arranged in the desired position, and then counter-rabatted upon A^ (§ 96). Next, one of the vertices, h, is selected, and the projections of a line of indefinite length perpendicular to X are drawn through the projections of the point, to represent one of the edges of the prism {hH:"^ and h^k^) (§ 79). On this line the given height is marked off {hm) (§ 30). The point m thus found is one of the vertices of the top VI, § 98] POINTS, LINES, AND PLANES 109 face of the prism. Lines through c and d, parallel to hm, and other lines through m and parallel to he and hd, will complete the required projections. EXERCISE SHEET XXXIX \. Given the plane Z (75, Z^2>0°r, Z''4o°r). Construct the pro- jections of a regular hexagon inscribed in a 2" circle and lying in Z with one corner touching H. 2. Given the plane Y (26, Y'' 30° I, F«60°r). Construct the pro- jections of a 5 pointed star inscribed in a 2" circle and lying in Y. Let the center of the circle be Ih" from YY"" and YY^. Start by ra- batting the H trace toward the right into V. EXERCISE SHEET XL Take HA 2\" below the top border hne. Given the plane Z (59, Z^ 45° r, Z^ 30° r), and using a scale of |" = 1", draw the projections of a hollow brick, 2" X 4" X 8", resting with one of its 4" X 8" faces in contact with Z. Let one of the 8'' edges of this face make an angle of 30° with ZZ^. Keep the entire soHd within the first quadrant. 98. Angle between a Line and a Plane. In order to determine the angle between a given line and a given plane (§ 89), first find the projection of the given line on the given plane. This may be done by selecting any point on the line, drawing a per- pendicular from this point to the plane (§ 79), finding where this perpendicular pierces the plane (§ 78), and connecting the point thus found with the trace of the given line on the given plane. The angle thus found will be the projeciion of the required angle. The true size of the angle may be found by the method of § 94. EXERCISE SHEET XLI 1. Given the plane Z (79, Z^ 30° r, Z^ 45° r), and the points a (75, 13, 16) and 6 (67, 8, 11). Find the angle between ah and Z. 2. Given the plane Y (19, Y^ 45° r, F" 60° D, and the points c (28, 8, 15) and d (16, 8, 8). Find the angle between cd and Y. no DESCRIPTIVE GEOMETRY [VI, § 99 99. To Draw the Traces of a Plane, Given One Trace and the Angle of Inclination to H or V. I. First Method. In Fig. 110, let the trace ZZ^ be given and let the required angle of inclination to H be 60°. It is required to determine the V trace, ZZ^ Imagine a cutting plane, 7, to have been passed perpen- dicular to ZZ^, at any convenient point. This plane will cut from Z a right triangle, the base of which is YaJ", the acute angle at a being the true angle between Z and H, or 60°. Im- agine this triangle to have been rotated into F, on Y Y'' as an axis, a^ moving to a'^. It will now be seen in its true size, the hypotenuse making an angle of 60° with HA. Drawing the hypotenuse, the other apex, h, is located at h . This is a point on the V trace of Z, and the required trace is now determined by the points Z and fe". [Note. The alternative solution given below is perhaps more difficult to visuahze than that just given. It is, however, an excellent introduction to § 100.] VI, § 100] POINTS, LINES, AND PLANES 111 II. Second Method. In Fig. Ill, let the trace ZZ^ be given and let the required angle of inclination to V be 45°. It is required to determine the V trace ZZ^. Imagine a cutting plane Y to have been passed perpendicular to the (unknown) V trace of Z, at any convenient point. The V trace cannot be drawn now, but the B. trace of such a plane will be perpendicular to II A. Draw YY^ to represent this auxiliary trace. If the right triangle cut from Z by this auxiliary plane Y is now imagined to have been revolved into H about Yh^ as an axis, the hypotenuse will pass through h^ and will Fig. Ill make an angle of 45° with EA. Now, if the triangle thus found is revolved backward on the same axis, the point a'^ will trace the arc shown on F. The required V trace, ZZ'', will be tangent to this arc. [Note. Compare the two solutions above and see how readily (6) may be evolved from (a) by working backward.] 100. To Draw the Traces of a Plane Making Given Angles (a and ^3) with H and V. In this problem an auxiliary sphere with its center on HA may be used to advantage. Imagine such a sphere and the required plane passed tangent to it. Now any cutting plane passed through the center of the sphere perpendicular to the E. trace of the given plane will cut a great circle from the sphere and a line of greatest declivity from the plane. The line and the circle will be tangent. ^ 112 DESCRIPTIVE GEOMETRY [VI, § 100 In Fig. 112, let adc be the plan and ahc be the elevation of that part of the auxiliary sphere which lies in the first quadrant. Any cutting plane passed as described above will be shown by a F trace perpendicular to HA passing through 0, as 0''0''. Imagine the quarter circle and its tangent, which are cut by this plane from the quarter sphere and the required plane, Fig. 112 respectively, to have been revolved Into V on 0^0' as an axis. The quarter circle will coincide with cb, and the tangent, being a line of greatest declivity of the required plane, will make the given angle a with HA, since It here appears at Its true incHnation. Draw such a line, me^. Its Intersection with 0''0\ e^ establishes one point on the required V trace. In a similar manner,/'^ is established as one point on the required H trace. VI, § 101] POINTS, LINES, AND PLANES 113 If now the triangle moe'-' is rotated back on oe^ as an axis, the point d will trace the arc m^m'^. The required H trace is tangent to this arc (§99, II), and can be so drawn, passing through/'^. The required V trace is tangent to w'^n'^ and passes through e\ By means of a proof similar to that in the footnote to § 26, it may be proved that the sum of the angles which a plane makes with H and V cannot be less than 90°, nor more than 180°. [Note. It is worth while to compare the general method used in above problem with that of § 29.] EXERCISE SHEET XLII 1. The plane Z cuts HA at the point (77, 0, 0). The trace ZZ" makes an angle of 45° r with HA, and Z makes an angle of 75° with H. Find ZZ^. 2. The plane Y cuts HA at (57, 0, 0), the trace YY'' makes an angle of 45° r with HA, and Y makes an angle of 45° with H. Find YY^. 3. The plane X cuts HA at the point (25, 0, 0), the trace XX^ makes 45° r with HA, and .Y makes an angle of 30° with H. Find XX^. EXERCISE SHEET XLIH Using auxiliary spheres 1-|" in diameter with centers at the points {1) (65, 0, 0), {2) (40, 0, 0), and (5) (14, 0, 0), respectively, draw the traces of the planes that make the following angles with H and V. (1) The plane R; 45° with H and 60° with F; (2) The plane S ; 45° with H and 30° with F; (3) The plane T ; 30° with H and 60° with F. EXERCISE SHEET XLIV 1. Through the point a (60, 10, 5), pass as many planes as possible that make angles of 45° with H and 60° with F, respectively. [Note. When the first plane is found, use an auxiliary cone with its apex at a.] 2. Given the points c (22, 6, 2) and d (14, 2, 5) ; through cd pass as many planes as possible, making angles of 30° with H. 101. To Determine the Angle between Two Planes. The problem of determining the angle between two plane surfaces (§ 88) occurs frequently in detailing structural work. Several practical applications are given in Chapter X. 114 DESCRIPTIVE GEOMETRY [VI, § 101 In Fig. 113, let the planes X and Y be given, and let it be required to find the angle between them. Any plane passed perpendicular to ah will cut from X and Y a triangle similar to cde, and the true shape of such a triangle will show the required angle between the given planes, i.e. dee. Before attempting the operation in projection, it will be well to fix in mind the geometric relations involved. Since the plane of the triangle dee (the auxiliary plane) is perpendicular to a6, oc is perpendicular to ah. Also de (the H trace of the auxiliary plane) is perpendicular to a^h (the projection of ah) (§ 79). Moreover, the true shape of the triangle dee can be established if the base, de, the point o, and the altitude oc, are known. We can now take up the opera- tion in projection, as shown in Fig. 114. Let the planes X and Y be given, and the H projec- tion of the line of intersection {a^h^) be found (§ 75). The V projection of this line will not be needed ; hence it is not drawn. The H trace of the auxiliary plane described above will be perpendicular to a^h^. Since it can be chosen anywhere along ah, any line perpendicular to a%^, as ZZ^, will represent the trace of such a plane. The points d and e are the vertices of the triangle which measures the required angle between the planes ; and the point o lies at the foot of a perpendicular through the apex. Compare Fig. 113. As soon as the length of this per- pendicular has been found, the triangle, and hence also the re- quired angle between the planes, will be established. Turning to Fig. 113, the required perpendicular, oc, is seen to lie in the triangle aa%, as well as in the triangle dee. Figure 115 shows this triangle and the line oc. If now aa^h is rotated Fig. 113 VI, § 101] POINTS, LINES, AND PLANT:S 115 on a^'b as an axis into //, the line oc will be seen in its true length and drawn perpendicular to a'b as shown by oc\ Returning to Fig. 114, draw a'^a'^ ^rpendicular to a^b^ and make it equal to a'a!". The triangle is the true shape the corresponding triangle in Figs. 113 and 115; ./^ and the point in the base foot of the cular oc'^, now be drawn. gives the true length of the altitude of the triangle dee. Fig. 114 With the known base de, the point o, and the altitude just found, the true shape of the triangle dee can be constructed {dc"^e). This gives the required angle fi between X and Y. This whole construction centers around the line oc, which is common to the triangles ad'b and dee. The first operation (drawing the line ZZ^) merely locates a definite auxiliary- plane, thus fixing the points d, c, and o. The second opera- tion (rabatting the triangle aa^b into H) Fig. 115 gives the true length of oc. The third operation places oc in its true relation to de, and mav be thought of as resulting from the rabattement of dee. 116 DESCRIPTIVE GEOMETRY [VI, § 101 EXERCISE SHEET XLV Find the angle between each of the following pairs of planes. 1. Z (75, Z^ 60° r, Z- 45° r) and Y ijyh, Y>^ 45° Z, Y- 45° /). 2. X (50, X^ 60° r, X- 45° r) and 17 (30, IP 60° Z, IF- 45° l). 3. f/ (25, C/''60°r, C7« 75° r) and F (5, FM5° Z, 7- 75° 0- (See §76.) EXERCISE SHEET XLVI Find the angle between each of the following pairs of planes. 1. (75, O'^ 90°, O- 30° r) and P (60, F^ 45° I, P- 90°). 2. Q (45, Q^ 90°, Q- 90°) and R (30, 7^'' 60° I, R^ 30° Z). 3. S (13, 5'' 45° r, S^ 45° and T (3, T^ 30° Z, T" 30° I), 102. To Find the Points in Which a Line Pierces a Plane Solid. The faces of the solid are planes determined by the edges. Find the traces of the plane faces (§68). Then find the point in which the given line pierces each of these planes (§ 78). If the given R" line pierces the solid, two of the points deter- mined above will lie within the limits of the projections of the solid. They are the required points. Wlien the given solid is so placed that the traces of the bound- ing lines fall outside the limits of the drawing, it is difficult to determine the traces of the bound- ing planes. Figure 116 shows such a case. The tetrahe- dron 1-4 and the line ah are given ; the traces of Fig. 116 ah on 1-4 are required. VI, § 104] POINTS, LINES, AND PLANES 117 Pass the H projecting plane, R, through ah. The traces of the lines 1-4, 2-4, 3-4 on R are at once apparent on plan, at c^, d!", e^. The V projections of c, d, and e are found by projecting up- ward from c^, d^y e^, to the lines P-4^ etc. The line of inter- section between the tetrahedron and R is therefore the broken line c'rf'e^. The points / and g, where ah crosses cde, are the required traces of ah on the tetrahedron. 103. To Find the Line of Intersection between a Plane and a Plane Solid. The points in which the edges of the solid pierce the plane will determine the line of intersection. Hence if the piercing point of each of the edges be found (§ 78) and if these points are connected by straight lines, the required line is determined. Another method would be to find the line of intersection between the planes bounding the solid and the given plane (§ 75). Sometimes one of the above methods is better, sometimes the other. Occasionally the two may be combined in the same problem. The selection will depend on the exact conditions in the given case. 104. To Find the Line of Intersection of Two Plane Solids. [Note. A special case of this problem is discussed in § 54. The general case here analyzed is that for solids in any position. In execu- tion the method here given often becomes quite cumbersome. In such cases the slicing method given in § 149 may be used.] If it is remembered that the edges and faces of plane solids are lines and planes, it will be seen that this problem is merely an application of principles already worked out in §§ 75, 78, 102, and 103. In a particular problem, it is only necessary to carry out these principles. The traces of the planes bounding the sohds may be found and the lines of intersection of these planes worked out, one at a time ; or the traces of the edges of one solid on the faces of the other may be found ; or the two methods may be combined. 118 DESCRIPTIVE GEOMETRY [VI, § 104 EXERCISE SHEET XLVII 1. The points a (75, 0, 6), 6 (63, 0, 6), and c (69, 0, 16i) define the base of a right triangular prism, 2" high. Find the points (m and n) in which the hne joining the points d (62, 1, 14) and e (77, 20, 7) pierces this prism. 2. The points/ (48, 0, 13), g (45, 0, 9), h (40, 0, 15), and i (44, 0, 20) define the base of a quadrilateral prism. The points / and j (45, 12, 4) determine one edge. Find the points (o and p) where the Une joining the points k (51, 14, 17) and I (37, 2, 3) pierces this prism. 3. The point q (14, 0, 7) is the center of the base of a regular tetra- hedron, 1*" on a side (§50). Find the line rstu, in which the plane Z (30, Z^ 30° r, Z" 30° r) cuts this solid. EXERCISE SHEET XLVIII The points a (30, 0, 2), 6 (30, 0, 14), and o (40, 0, 8) define the base of a regular tetrahedron, while the points d (33, 0, 14), e (43, 0, 11), and/ (41, 0, 17) define the base of a triangular prism, of which the line eg (36, 7, 0) is one edge. Fmd the line of intersection between the solids. Mark it h-o. [Note. At this point §§ 183-190 can be profitably studied.] CHAPTER VII CURVED LINES 105. Introduction. Though most of the hues in any structure are straight, the use of curves is so frequently necessary that the designer and draftsman must always be prepared to use them with confidence and skill. His problem is twofold : first, to select from among all the possible curves the one best suited for the purpose in hand ; and second, to make drawings from which his conception may be executed precisely by another person. Facility in handling curves can come only as the result of a knowledge of the principles governing the generation of curves in general, and from practice on some specific cases. In the following articles no attempt will be made to discuss in detail a large number of curves, nor to cover any one of them com- pletely. For reference, the more useful rules are given for the curves most frequently encountered in actual practice. A few of the examples are included merely to show the construction of curves that are typical of certain groups. Many other curves that are used occasionally in special classes of work are to be found in books dealing with those special fields. 106. Generation. A point moving in a constant direction generates a straight line ; but if the direction of motion of the point is constantly changing, a curved line is generated. A curve may be considered as made up of a series of adjacent points.^ Any two adjacent points will determine a minute straight line, which is sometimes called an element oi the curve. ^ See the footnote on page 120. 119 120 DESCRIPTIVE GEOMETRY [VII, § 106 The direction of this Hne will give the instantaneous direction of the generating point when it passes along the curve through this position. Thus, in Fig. 117, the curve a-h may be regarded as made of elements which, enlarged, are represented at a'-h'. The instantaneous direction of the curve at d is represented at d'e', and is indicated by the angles and <^'. Fig. 117 A curve is sometimes defined as being generated by a point which moves so that three consecutive positions of the generat- ing point do not lie, in general, on a straight line.^ 107. Tangent and Normal. If any element of a curve is extended in its own direction, as dj, Fig. 117, the straight line ^ The conception of a curve as presented in §§ 106-107, i.e. that it consists of co7isecutive points, no three of which, in general, lie on the same straight line, cannot be defended as rigidly exact, since between any two points there are an infinite number of other points. Moreover, even ad- mitting the conception as a possible one, there would exist at a point of inflexion (Fig. 118) three consecutive points all in the same straight line. Therefore the use of the word consecutive may need further explanation. Consider a curve and two points P and Q, on the curve. A tangent at P is said to contain two consecutive points of the curve, understanding thereby that it is the limiting position of a secant through P and Q when Q approaches P as a limit. The word consecutive is thus used in the sense that Q can come as close as we please to P, along the curve. The same general conception, open to the same general objections and explicable on the same general basis, occurs in §§ 108, II, 143, 158, 161, and elsewhere. It may be said in favor of the use of this conception that it has been found to aid the student in visualizing and classifying curved lines and surfaces, since the classifications are consistently based on the same conception and are made to depend on geometric relations that are easily visualized. VII, § 108] CURVED LINES 121 so produced is tangent^ to the curve. Such a Hne evidently contains two consecutive points of the curve and its direction is the same as the instantaneous direction of the curve at the point of tangency. Several cases of tangency are illustrated in Fig. 118. The word tangent is derived from the Latin word tangere FiQ. 118 (to touch), which expresses quite closely the fundamental idea involved. Any line perpendicular to a tangent, at the point of tangency, is said to be normal to the curve. 108. Classification. I. Regular and Freehand Curves. When the generat- ing point of any curve changes its direction in obedience to a definite law, the curve is called a regular curve. In the usual case, the controlling law fixes some relation that must be maintained between the moving point and one or more fixed points or lines. Such examples as the circle, ellipse, etc., are readily recalled. When a curve is drawn according to no definitely stated law% but only by the hand and eye, the curve is said to be a freehand 1 This conception of a tangent implies that a curve be regarded as made up of a series of adjacent points as described in § 106. Fundamentally the same idea appears again in § 143. 122 DESCRIPTIVE GEOMETRY [VII, § 108 curve. It is impossible even to cite examples from this class as the lack of any fixed or definite characteristics precludes an effective nomenclature. In general, freehand curves are more varied in character and more subtle in form than are regular curves. For this reason they are much used for ornamental work, especially where models or full size patterns can be used to explain them. For large work, however, such as arches, vaulting, etc., the use of regular curves is to be preferred because of the greater ease and precision possible both in the drawing and in the execution. II. Plane and Space Curves. Any curve, whether regular or freehand, which lies wholly within a single plane, is called a plane curve. This class includes all curves generated by a point moving in a plane or by a plane cutting a curved surface. A space curve may be thought of as generated by a point which constantly passes from one plane to another, i.e. no four consecutive positions of the generating point lie, in general, in the same plane. ^ This class includes curves generated by a point moving on a curved sm'face ^ or by the intersection of curved surfaces." 109. Projections. Plane curves are projected in the same manner as other lines (§§ 14 and 15). The projecting lines of the points on a curve are elements of a projecting cylinder (Fig. 119).^ The intersection of this cylinder with a plane of projection gives the projection of the curve. This projection may vary between the true shape of the curve (a'-'b^'c'') and a straight line (a^b^c^). The projections of any curve lying in a normal-case plane, as def, will be curved, but not equal to the curve in space. Space curves (Fig. 120) have plane curves for their projections. No matter how a space curve may be turned, the projecting ^ See footnote, page 120. 2 In certain special cases plane curves may be generated in this manner. 3 The word cylinder is here used in the general sense as explained in § 145. VII, § 109] CURVED LINES 123 cylinder can never degenerate into a plane (as with abc, Fig. 119). Hence the projections of a space curve are always curved. For the same reason, these projections can never show the true shape of the curve. The true form of such a curve can be seen only by visualizing it from its projec- tions, or by building a three-dimensional model. FiQ. 120 124 DESCRIPTIVE GEOMETRY [VII, § 109 Let the student prove that if any two curves are drawn on H and V, at equal distances from P, these curves are the pro- jections of some space curve. Tangents. The projections of tangent Hues will be tangent to the corresponding projections of the curve. But the con- verse of this proposition is not necessarily true, as lines which appear to be tangent in projection may lie in different planes. Normals. The projections of a normal are not, in general, perpendicular to those of the tangent (§ 36). 110. Methods for Drav/ing Curves. Curves may be drawn in many different ways. The following methods deserve mention. I. By Continuous Motion of a Guided Point. The generation of a circle by compasses is a familiar example. In general, this method calls for some arrangement of pivoted or movable arms, links, wires, or some similar device (see Figs. 126, 131, 132, and 142). Considerable ingenuity can be used to devise ways and means, which are adapted to the work in hand, and which are at the same time based on correct mathematical principles. The generation of curves by con- tinuous motion is of frequent occurrence both in the drafting- room and in actual construction. II. By Establishing Points on the Curve. Several points that lie on the required curve are located, and are con- nected by a freehand curve. In skillful hands this method is commonly the quickest, and it is quite as satisfactory as any other. This is the exact counterpart of the method used by a plasterer in " spotting " or ''screeding " an arch. In drafting- room practice, particularly for inked work, the freehand curve is sometimes abandoned in favor of a curve drawn along one of the usual forms of irregular curves, such as a French Curve or a Ship Curve. 111. By Approximation. It is sometimes possible to ap- proximate a complex curve, such as an ellipse or a spiral, VII, § 111] CURVED LINES 125 by means of circular arcs (see Figs. 136, 137, 138). A curve, especially an ellipse, drawn in this manner is usually unpleasant in appearance, as it appears flattened at the points where the arcs are tangent.^ Therefore such a curve is but little used for executed work. In the drafting-room, however, this method may be used advantageously to indicate a curve which is otherwise fully described. Moreover, by this method the use of the compass- pen for inking is made possible. 111. To Develop a Curve. The general idea of development as explained in § 49 is that of flattening or straightening out a complex surface into a plane. I. Plane Curve. Following the general idea stated above, a curve can be developed by rolling it out on a straight line. In the case of a circle, this development can be obtained by direct computation, using the tt relation. In the case of curves where no such relation is known, the develop- ment may be obtained by the use of instru- ments. In Fig. 121, the curve a-h is to be developed on the straight line a-h\ Let the dividers be set to any convenient small distance and stepped off on the curve from a to g. Next step off the same distance on the straight line a-g\ Now by adding the. small remainder gh at g'h\ the total length of the curve is measured along the straight line. This process is also called rectifying the curve. Fig. 121 ^ By using a sufficient number of known points, the method of §§ 120, 127, and 130, IV, can be made to give a very close approximation of the true curve ; but the construction is tedious. 126 DESCRIPTIVE GEO:\IETRY [VII, § 111 II. Space Curve. Any space curve may be developed into a plane curve by making it the directrix of a cone or cylinder/ the surface of which is then developed, along with the given curve, as explained in § 157 and as illustrated in a I ^ L_d_e ^ ^ x s \ , Z 1 Q^ b*' c^ d*- e*" z a b c .d e (3) Fig. 122 Fig. 122. The plane curve thus determined may then be rectified to a straight line, as shown in Fig. 122, 3, if desired. 112. The Conic Sections. Curves of the group known as the conic sections occur frequently in optics, acoustics, mechanics, and other sciences. If a cone is cut by a plane, the resulting line of intersection will be one of the curves of this group (see § 145) ; ^ The words cone and cylinder are used in the sense explained in § 145. VII, § 112] CURVED LINES 127 hence the name conic sections, or sometimes simply conies. The group consists of the elHpse, the parabola, and the hyperbola (including the circle and straight line as special cases). If the conies are considered as being traced by a point moving in a plane, methods for their generation can be stated that do not involve the use of an actual plane cutting an actual cone. The principle of gencratioji applying to all conies is that the generating point shall move so that its distance from a fixed point shall bear a constant ratio to its distance from a fixed Eccentricity of the curve Fig. 123 hne. The fixed point is called the focus, the fixed line the directrix, and the ratio is called the eccentricity. (See Fig. 123.) Variation in the shape of the different conies comes from a variation in the eccentricity. When the eccentricity is unity, the curve is known as the parabola. When the eccentricity is less than unity the curve is an ellipse. When the eccentricity is greater than unity the curve is an hyperbola. Variation in size comes from increasing or decreasing the distance between focus and directrix. 128 DESCRIPTIVE GEOMETRY [VII, § 113 113. General Construction for Any Conic. In Fig. 124, let the focus /, and directrix Y, be given, and let it be required to construct a conic having an eccentricity equal to n/m (ex- pressed at scale by the lines n and m). Set the proportional dividers to the ratio n/m. With this setting determine the point a which divides fo in the given ratio. Draw any line hs parallel to OY. Set the long leg of the divider to the distance oh. Fig. 124 With the short leg at this setting strike an arc from / as a center, cutting hs at c. The point c is one point on the required curve. Another line, dh, yields another point, e. There are of course corresponding points below the axis. As many points as are needed may be found in this manner, so that the curve may be drawn as accurately as may be required. Since n/m is Qi etc., are on the required parabola. This construction is based on the same idea as that of § 124, II (1), for the ellipse, and § 130, III, for the hyperbola. 1 Practical difficulties that arise in fastening the thread and in getting a thread that will not stretch make this method more of theoretical than of practical value. This criticism also applies to the methods of §§ 124, I and 130, V. 2 See footnote, page 129. 132 DESCRIPTIVE GEOMETRY [VII, § 117 117. To find the Focus of a given Parabola. In Fig. 128, choose any point m on the curve. Draw vm perpendicular to the axis. On this Hne, lay off np equal to twice vn. Draw pv, cutting the curve at r. From r drop a perpendicular to the axis. The foot of this perpendicular, /, is the required focus. 118. To draw a Tangent to a given Parabola. I. At a Given Point on the Curve. First method. In Fig. 128, let a be the given point of tangency. Draw ab perpendicular to the 'axis. Lay off vc equal to vb. The required tangent is the line ca. Second method, using the focus. Let d be the re- quired point of tangency. The focus, /, can be de- termined by § 117. Lay off vz = vf, and draw the directrix, xy. Connect d with the focus, /. Draw dg parallel to the axis and bisect the angle gdf. This bisector de is the required tangent. Compare with §§ 125, I and 130 a. . 11. Through a Point Outside the Curve. Let s, Fig. 128, be the given point. With 5 as a center and sf as a Fig. 128 VII, § 120] CURVED LINES 133 radius, draw a circle tfw intersecting the directrix at t and w. Perpendiculars to the directrix drawn from these points will intersect the parabola at q and u which are the required points of tangency. 119. To draw a Normal to a given Parabola. First method. One obvious method consists in first drawing a tangent, then drawing a perpendicular to the tangent at the point of tangency. Second method. Another method, using the focus, follows. (See Fig. 128.) Let h be the point at which the normal is to be drawn. Draw hj per- pendicular to the axis. Lay off jk equal to twice vf. The required normal passes through k and h. 120. To approximate a Parabola by Arcs of Circles. Establish any number of points that are known to lie on the parabola (Fig. 129, points Vy 1, 2, 3, etc.), by the method of §§ 115, II, or 116. Draw the normals at these points (§ 119). The axis is a normal at the vertex. Prolong these normals to their intersections at a, h, and c. With a radius equal to av strike the circular arc v-\, from the center a. Move the center to h, and using the radius h-\, strike the arc 1-2. Continue for each center.^ Note. c f = focus X2,"L)3 etc., II 0X15. 2b II xf . 3C II yf ietc. Fig. 129 1 See footnotes to §§ 110 and 113. Also note that the principle here used may be applied to the construction of an ellipse (§ 127, III) or of an hyperbola. 134 DESCRIPTR^ GEOMETRY [VII, § 121 121. The Ellipse. An ellipse is generated by a point which moves so that its distance from a fixed point bears to its dis- tance from a fixed line a constant ratio less than unity. (See also § 112.) For typical form and nomenclature for the ellipse see Fig. 130. The symmetry of the curve gives two foci and two directrices. For any point on the curve, as d, df/dp, and V and v' = vertices. hj and ge = conjugate diomeiers . hj being parallel to tanqents at g and e. / nf = cv. kf : kl ; : df : dp: : df : dp' ; : vf : vo < 1 . = eccentricity.. df + clf= vf+vf'= vv". df and df'= focal distances of d. Fig. 130 df/dp' are equal to the eccentricity of the curve. Also cf/eo is the same ratio. The ellipse may also be defined as a curve generated by a point which moves so that the sum of its distances from two fixed points remains the same ; that is to say, in the figure, df-^df = hf+hf = vf+vf = the major axis. The shape of the ellipse will vary with the eccentricity and with the distance between focus and directrix. As the eccen- tricity approaches zero, the curve approaches a circle or a point. VII, I 124] CURVED LINES 135 As the eccentricity approaches 1, the curve approaches its axis, or a parabola. 122. To construct an Ellipse, given the Focus, Directrix, and Eccentricity. This is a special case of § 113, which see. 123. To find the Foci of an EUipse. From § 121.it will be seen that in Fig. 130, nf-\-nf' = vv' . Therefore if an arc is struck from n as a center, with a radius equal to half the major axis, the points / and f, where this arc cuts the axis, will be the required foci. 124. To construct an Ellipse, given Major and Minor Axes. I. By Continuous Motion. (1) First meihod. In Fig. 131, let ah and vv' be the axes. By § 123, locate the foci, / and /'. Fig. 131 Put pins at the foci and loop around them a continuous string, the length of which equals 2fv. A pencil, p, sliding in the loop will describe the required ellipse. For large size rough work, such as gardening, this method is satisfactory, but for small or accurate constructions, the difficulty of keeping the string taut but without stretching is too great. Compare with § 115, 1. (2) Second method. A method which gives better results than the former, is the construction by means of a trammel ^ 1 A trammel is a mechanical dev'ice for constructing curves by continuous motion. The term is sometimes applied to beam compasses, though not usually applied to compasses of smaller size. 136 DESCRIPTIVE GEO^IETRY [VII, § 124 as shown in Fig. 132. Thin strips are tacked down along the axes ab and vv\ A lath pmn is bored as shown, p77i being equal Fig. 132 to ao and p?i equal to vo, pegs inserted at w? and n, and a pencil is placed at p. The pegs being inserted in the grooves between the strips, the motion of the lath will cause the pencil to de- scribe an ellipse. e^ - ^^^-1-^ ^^ — ^/—^T ,*' 11. By Estab- LisHixG Points. (1) Firsi method. In Fig. 133, let ab and cd be the given axes. Construct a rectangle efgh on these diameters. Divide ao and ae into the same num- ber of equal parts as shown. Draw d 1, d 2, etc., and cl', c2',etc. The intersections x, y, etc., lie on the required curve. This method resembles that of §§ 116 and 130, III. It applies to either the ellipse or the circle. It may be used without change when conjugate diameters (see pr =pz. psll rz. V '» Z'k= t\= ab: X ^- \k/ Fig. 133 VII, § 1251 CURVED LINES 137 Fig. 134 Fig. 130) are given instead of the main axes, except that the rectangle efgh will be replaced by a parallclo^^ram as in Fig. 134. (2) Second method. In I (2) above, a piece of paper, marked at points corre- sponding to m, 71, and p, may be substituted for the lath. If the paper is placed so that m and n fall on the axis lines, the point p is a point on the eUipse. (3) Third method. In Fig. 135, let db and eg be the major and minor axes. Draw the inscribed and circumscribed circles, and the radii oj, ok, etc. Draw verticals j/, kk\ etc., and hori- zontals J"ff k"k' y etc. The intersec- tions /, k', etc., are on the required ellipse. 125. To draw a Tangent to an El- FiG. 135 lipse. I. At a Point on the Curve. (1) First method. Sup- pose the ellipse drawn as in Fig. 135. Let k' be the given point of tangency. Draw mn and kl tangent to the inner and outer circles at k" and k. The line ml is the required tangent. 138 DESCRIPTIVE GEOMETRY [VII, § 125 (2) Second virfhod. In Fig. 133 another construction is shown. Let it be required to draw a tangent to the elHpse at p. Connect p with the foci, z and z'. Extend z'p to r making pr equal to pz. Bisect the angle rpz. The bisector pm is tangent to the elhpse. The center point of zr determines the bisector. Compare with §§118 and 130 a. II. Through a Point Outside the Curve. In Fig. 133, let q be the given point. With g as a center and qz as a radius, draw the circle zjk. "With z' as a center and a radius equal to ah, draw an arc intersecting the circle zjk at j and k. Draw z'j and z'k, intersecting the ellipse at u and v, which are the required points of tangency. 126. To draw a Normal to an Ellipse. Using the construction described in § 125, I, (2), and shown in Fig. 133, a line ps through p parallel to rz will be normal to the ellipse. 127. To approximate an Ellipse by Arcs of a Circle. IMany rules for approximating to an ellipse have been proposed. (See § 110, III.) Of course any such rule will give a curve which varies more or less from a true ellipse. Some of the following methods give a closer approximation than others. Some are useful only for curves that have a certain ratio of major to minor axis. Some are much more tedious than others. In general the greater the number of centers used, the closer the approximation, and the greater the labor involved. I. Four Centers (Fig. 136). Given the axes cm' and bb'. Lay off ac^oh, and oc' =oc. Bisect cc' at d, lay off cc^cd, and of = oe. The points /' and e' are symmetric to / and e. The four points serve as centers from which arcs can be struck, which will be tangent to one another, and which will approximate an ellipse. This method is not useful for narrow^ curves, but it may be executed rapidly. When one half of this curve is used for an arch, the arch is called a three-centered arch. II. Eight Centers (Fig. 137). Given the axis ab and the semi-axis oc. Draw the rectangle CLefb. Connect a with c, and VII, § 127] CURVED LINES 139 draw ep perpendicular to ac. Lay off or = oc, and on ar as a diameter, construct the semicircle ajlr. Draw the radius jg, perpendicular to ab. Lay off ok = h j, and with pk as a radius, and with pas a center, swing the arc mkm'. Extend oc to I. Using a and h as centers, and a radius equal to ol, strike arcs cutting mkvi' at m and m'. The required centers are at n, m, p, m', and n\ This method is applicable to ellipses of nearly all propor- tions. An arch following this curve is called a five-centered arch. -41 \\f/ Fig. 137 i 140 DESCRIPTIVE GEOMETRY [VII, § 127 III. Numerous Centers (Fig. 138). The major and minor axes are given. The first step is to locate the foci, / and /' (§ 123). Next locate a number of points, a, b, c, etc., which are known to lie on a true ellipse (§ 124, II). At the points thus located, draw normals to the ellipse, ah, hi, etc. (§ 126). The points 1, 2, 3, etc., where these normals intersect, may be used as centers for striking the curve. The construction for any normal, as cj, is made by making eg equal to cf, and drawing g~\'-- cj parallel to gf. If g and / are known, the line cj can be drawn parallel to gf without actually drawing the line gf, thus simplify- ing the drawing and shortening the work. 128. To draw the Projections of a Circle inclined in any Manner to H and V. The key to this problem lies in the fact that, in whatever position a circle may be placed with respect to H (or V), some one of its diameters will be parallel to H (or V), and hence will be projected on H (or V) in its true length. This being the greatest possible projection of any diameter, it must form the major axis of the ellipse which is VII, § 128] CURVED LINES 141 the required projection of the given circle. By the same token, that diameter of the circle which is a line of greatest declivity of the plane in which the circle lies will have the least possible projection of any diameter ; and hence will be the minor axis of the projection. Fig. 139 In Fig. 139, let it be required to draw the projections of a circle lying in the plane X, with its center at o. Rabatte the V trace, X A^^', and the center o, into H (§§ 91 and 92). Draw a circle of the required size with its center at o'^. Draw the diameters which are parallel and perpendicular to X X^. When 142 DESCRIPTIVE GEO:\IETRY [VII, § 129 these lines are rotated back into X, they will give the major and minor axes of the H projection, since one will be parallel to //, while the other is a line of greatest declivity of the plane. The lines a^b^ and c^d^ are the counter-rabattements as seen on H (§ 95). The ellipse may now be drawn by any of the methods of § 124. The F projection of these axes being found (§ 95), they are conjugate diameters of the required ellipse (§ 124, II, (1)). Let the student determine the diameters of the rabatted circle which, by counter-rabattement, will give the major and minor axes of the V projection. 129. The Hyperbola. An hyperbola is generated by a point which moves so that its distance from a fixed point bears to its distance from a fixed line a constant ratio, greater than unity. (See also § 112.) As in the case of the ellipse, there are two foci, two directrices, and two axes, but the foci lie outside the directrices, and the curve is composed of two symmetrical parts, branching off to infinity. An important property of the hyperbola is that the difference between the focal distances of any point on the curve is always the same, and is equal to the distance between the vertices. This property is closely parallel to a similar property of the ellipse stated in § 121. The hyperbola is met with so infrequently in practice that only a few constructions will be given. 130. To Construct an Hyperbola. I. The General Construction. The method described in § 113 applies to the hyperbola as well as to all other conic sections. II. A Construction Based on the Difference of Focal Distances. This method is shown in Fig. 140. Given the foci F and F' and the vertices v and v\ From F and F' as centers, with radii greater than Fv, swing arcs aa and a'a\ Add to the radius just used an amount equal to vv\ and swing arcs bb and b'b' from the centers F' and F. The intersections c and cf are points on the hyperbola. VII, § 130 a] CURVED LINES 143 III. Another method, similar to that of §§ 116 and 124, II, 1 is shown by the dotted Hnes in Fig. 140. IV. The hyperbola may be approximated by circular arcs, following the method used in §§ 120 and 127, III. 3 2 I Fig. 140 V. A method similar to that of §§ 115, 1, and 124, 1, may be devised for drawing an hyperbola by continuous motion. 130 a. Tangent and Normal to the Hyperbola. The tangent to an hyperbola bisects the angle between the focal radii of any point. Thus, in Fig. 140, the angle F'pm being made equal to the angle mpF, the line vip is tangent to the hyperbola at p. To draw a normal, produce Fp to r making pr = pF'. Draw r F', and bisect it at n. The line pn is the required normal. (Com- pare with §§ 118, II; 125, I, (2) and 126.) 144 DESCRIPTIVE GEOMETRY [VII, § 131 131. The Sinusoid. The curve shown in Fig. 141, is called a sinusoid or a sine curve. In this curve the ordinates (vertical distances from the horizontal axis) are proportional to the sines of the abscissae (horizontal distances from the origin). This curve is introduced to call attention to a curve of contra- flexure, i.e. one in which the direction of curvature changes periodically. It will be noticed that as the curve approaches the axis, its curvature becomes quite flat while the sharpest point is farthest from the axis. At the axis, the upper and lower curves have a common tangent. WTiere brackets, moldings, etc., are designed with outlines composed of reverse curves, even when freehand curves are used, the principle of flattening the parts as they approach their common tangent, is adhered to. In the figure, the distances 0-1, 1-2, 2-3, etc., are equal, and each is equal to 1/24 of the circumference of the circle from which the ordinates are derived. Flatter or steeper sine curves could be drawn by increasing or decreasing the ratio between 0-1, 1-2, etc., and 0-1', l'-2', etc. All such sine curves belong to the general family of harmonic curves. 132. Link Motion Curves. — The Lemnlscate. Figure 142 shows three links, AB, CD, and BDE, pivoted at A, B, C, and D. The pivots A and C are stationary, while B and D are free to move. Obviously B can move only in a circular path about ^ as a center. Also D is confined to a circular path about C. Moreover, B and D must always be equally distant from one another, the constant distance being fixed by BD. Thus the points B and D have a simple circular motion. But VII, § 132] CURVED LINES 145 the motion of any other point on BDE is a compound motion. To trace the path of E, let the circular paths of B and D be drawn, giving the circles a-e-g-v and g'h'q' . Divide a-v into any number of parts. In the figure sixteen equal parts were used, with a few intermediate points for accuracy. Assume that D is moved toward /, passing successively through the EO = DB-CO AB = AC CD'A&:: 2:3 Fig. 142 other points. Meanwhile B is traveling along q'h'g' ; at the distance BD from the points /, g, h, etc., successively. The positions of B which correspond to/, g, h, etc., may be found by striking arcs from these points with a radius of BD. The intersections of these arcs with g'b'q' give the points /, g', etc., which mark successive positions of B. Through corresponding points (/ and f, g and g', etc.) draw Hues equal to BE. The extremities of these lines give the points /i, gi, etc., which are on the required curve. It will be noted that the path of B is confined to somewhat less than half of the circle g'b'q'. 146 DESCRIPTIVE GEOiVIETRY [VII, § 132 The particular curve shown in Fig. 142 is a lemniscate because of the proportion between the Unks indicated in the figure. Interest- ing variations may be found by varying these proportions or by tracing the path of some other point on the free Hnk. The prin- ciple of curvature at points of contraflexure, noted in connection with the sinusoid, will be seen to hold good for this curve as well. Similar problems arise in connection with hinging of certain types of doors, casements, etc. 133. The Cycloid. The cycloid is a curve traced by any point on the circumference of a circle which is made to roll on a straight line, as in the case of a wheel rolling on a rail. 7 a* JTI 12 Fig. 143 In Fig. 143 the circle 1-12 rolls on the base vi-n. Let it be required to trace the path of the point a. After one complete revolution of the circle, the point a will again be in contact with the base line, at a distance from its original position equal to 27rr. This locates the point a^^. The center o moves forward on a level Hne oo^^. As the points 2, 3, 4, etc., come in contact with the base, at 2^ 3', etc., the center moves to o^, o^ etc., directly above. When 2 is at 2', the lowering of 2 will have resulted in an equal raising of a. At the same instant the distance of a from o is equal to the radius of the circle. Thus, by swinging arcs of radius equal to that of the generating circle, from the centers o\ cr, o^, etc., cutting the lines of level, 2-12, 3-11, etc., the points, a}, o?, a^, etc., on the cycloid are estab- lished. VII, § 134] CURVED LINES 147 The cycloid Is a special case of the family of curves called trochoids. This family includes all curves generated by a point on a circle which rolls in contact with another circle. In the case of the cycloid the base circle is of infinite radius. If the radius of the base circle is finite, and the rolling circle is outside of the base circle, the curve generated is an epicycloid. If the rolling circle is within the base, the curve is an hypo- cycloid. Various other trochoids result from tracing the paths of points within or outside the circumference of the rolling circle. 134. The Involute of a Circle. This curve is generated by unwrapping a thread from a circular drum. In Fig. 144 let Fig. 144 the circle 1-12 be a plan of the cylinder from which the thread is to be unwound, starting at 1. ^^^len one complete turn has been made, the thread will be tangent to the circle at 1 and the end of the string will be at a distance from 1 equal to the cir- cumference of the circle, as shown by the line 1-m. At other points on the circumference, the thread will always be tangent to the circle, the distance from the point of tangency being equal to a proportionate part of the circumference. 148 DESCRIPTIVE GEOMETRY [VII, § 134 Of course this curve may consist of as many turns as may be desired, the general shape being that of a spiral. The circle from which the curve is evolved is called the evolute. Any plane figure may be used as an evolute to form a spiral curve, which is called its involute. 135. The Ionic Volute. This curve, as commonly con- structed, is the involute of a diminishing square. The con- FiG. 145 (After Vignola ) struction is completely shown in Fig. 145. It is composed of mutually tangent circular arcs. The centers for these arcs are located in the order shown on the large scale insert in the figure, the small circles indicating the centers for the outer curve, while the short cross lines indicate the corresponding centers for the inner curve. 136. Other Spirals. The possible variety of spirals is infinite. In general, there is a center from which the generating VII, § 137] CURVED LINES 149 point moves outward in increasing convolutions. The law governing the generation is based on the angle through which the generator moves and the distance it attains from the center. In Fig. 146, the distances 0-1, 0-2, etc., are directly propor- tional to the angles 9i, 02, etc. The resulting curve is the spiral of Archimedes. (See also § 137.) Other spirals may be generated by making the distance from the center proportional to the square, cube, or other power of the angle. Again the distance may be inversely propor- tional to the angle. 137. The Helix. The helix is a space curve (§§ 108 and 109) and, like all other curves of this class, it cannot be drawn on a flat sheet of paper. It may, however, be repre- sented by its projec- tions. A helix is gener- ated by a point mov- ing on the surface of a right circular cylinder, in such a manner that a uniform angular velocity is combined with a uniform velocity paraHel to the axis. In Fig. 147, ahcd is the elevation and 1, 5, 8, 12 is the plan of a right circular cylinder. The generating point starts from 1, moving to the left around the cyhnder and rising at such a rate that one complete revolution is made in the height. The construction is obvious. The Hghter line shows a helix that makes two revolutions in Fig. 146 150 DESCRIPTIVE GEOMETRY [VII, § 137 the same height and moving in the opposite direction. On the front of the cyHnder, this curve ascends toward the right, and is said to be a right-hand hehx, with a pitch of 1/2 p. It will be seen that the instantaneous direction of a helix, with respect to //, is constant ; i.e. the tangents to a helix make uniform angles with the base plane. ^Vhen this constant angle is 45°, the F projection of the curve will be the sinusoid of § 131. Examples of helixes are the thread on a screw or bolt, the hand rail on a spiral stair, etc. Another type of helix can be generated by a point moving on the surface of a cone, follow- ing the same laws as above. Such a curve is called a conical helix. Its hori- 11 . ■ h" zontal projection is a spiral of Archimedes. (See § 136.) EXERCISE SHEET XLIX [Note. Each of the following problems is intended to be solved on a sheet 8" X 10|", with margins as shown on p. xii.] The door shown in Fig. 148 is hinged at h and h'. At the bottom of the door is a roller, r, at each side. Fig. 148 -IiLj: VII, § 137] CURVED LINES 151 When the dooi" is opened, the point h' moves outward and upward as indicated by the arrow h", while the roller moves directly upward, in contact with the wall, as shown by the arrow r' . Determine the exact limits of the space not interfered with by the opening of the door. Scale V' = I'-O". EXERCISE SHEET L In Fig. 149 is shown a casement adjuster. The swivel A is fast to the sill, while the pivot B is fast to the sash. The rod C is pivoted at B and slides in A. The sash is hinged at D. Lay out the plan as shown, placing it on left hand half of a standard sheet divided ver- tically. Place hinge 3" below top border. Find : 1 (a) The path traced by the end of the rod when the sash is opened. 1 (6) The posi- tion of the sash when its swing is checked by the rod. 2. On right half of sheet, redraw the plan and let the adjuster be applied to the sash and sill at E and F, instead of as shown in Fig. B. Repeat the above solution. Is the placing described in 1 better or worse than that in 2? Scale 3" = I'-O". Elevation Fig. 149 Llevatiorv Fig. 150 EXERCISE SHEET LI Repeat Problem L using an in-swinging sash adjuster like that shown in Fig. 150. EXERCISE SHEET LII (a) Draw both of the projections of a helix generated on a cylinder 2" in diameter and 3" high, and draw a straight line tangent to the curve. (6) Draw both projections of the curve generated on a sphere 2" in diameter, by a point moving with uniform angular and vertical velocities, (c) Draw both projections of the curve generated in the same manner on a cone, 2" in diameter and 3" high. 152 DESCRIPTIVE GEOMETRY [VII, § 137 EXERCISE SHEET LUI (a) Trace the epicycloid formed by a point on the circumference of a circle of 7/8" radius which rolls on another circle of If" radius. (6) Trace the hypocycloid, using the same base circle, but letting the rolling circle be of 5/8" radius. EXERCISE SHEET LIV Taking the same circles as in Sheet LIII, trace the path of a point midway between the center and circumference of the moving circle. EXERCISE SHEET LV The derrick in Fig. 151 can turn on the vertical mast as an axis. The distance a and the angle 6 may be made to vary. Starting with ^ = 90'^ and a = 24' — 0", let the derrick be revolved through 225°. Meanwhile let w be raised till a = 2'— 0" and ^ = 30°. Let all motions be uniform. Trace the path of w. Scale i-" = I'-O". Fig. 151 •^777777Tr77i77777777777777TT7771-. Fig. 152 EXERCISE SHEET LVI A merry-go-round, or carrousel, like Fig. 152, is made to dip twice in a revolution, the extreme angle of inclination of the ring being 30°. Trace the motion of any point on the circumference, assuming vertical and angular velocities to be uniform. Scale i" = l'-0". EXERCISE SHEET LVII Place the sheet vertically with UA in the center. Given a (33, 30, 24), and 6 (24, 14, 8). Draw the projections of a circle, 3" in diameter, whose center is at 6, and whose plane is perpendicular to ah. ~" EXERCISE SHEET LVIII In the center of the sheet, placed horizontally, draw an ellipse whose major axis is 2" and whose minor axis is \" . Place the major axis vertically. Draw the involute (§ 134). Use § 124 (3) and § 125. CHAPTER VIII CURVED SURFACES 138. Introduction. The use of curved surfaces in the form of vaulting, domes, and moldings is one of the commonplaces of architectural practice. The sphere, the cylinder, and the cone, are the most usual for the excellent reason that they are the simplest. It is not so evident, however, why some of the other forms, notably certain warped surfaces, should be so rarely used, unless it be that designers and draftsmen, through a lack of knowledge of the possible forms and of the correct methods for generating them, are timid in attempting the unusual. A good knowledge of the fundamental ideas regarding generation of surfaces will certainly give the designer a broader means of expression, and an abstract study of surfaces will introduce him to some rarely encountered and inherently beautiful forms that will add much to his technical vocabulary and equipment. In using curved surfaces, some discrimination is necessary in selecting materials for the execution. Materials which are plastic while being worked, such as terra cotta, or plaster, may be used for nearly any describable surface. Bricks, be- cause of their rectangular, flat faces are less adaptable to small curved surfaces ; but for large areas the numerous joints allow of a close approximation to nearly any surface. The use of flat rectangular tiles to approximate a large variety of curved surfaces is a notable feature of the Guasfavino system of vaulting. Nearly anything can be done in stone ; but the cost of working is high. 153 154 DESCRIPTIVE GEOMETRY [VIII, § 139 139. Generation. Just as a line is generated by a moving point, a surface is generated by a moving line.^ The character of any surface will be determined by the following items. (1) The character of the moving line which produces it. This line is called the generatrix of the surface. It may be straight, or it may be curved in any manner. (2) The particular kind of motion imparted to the line. This motion may be a motion of revolution, or a motion of transposition, as defined below. Revolution implies an axis, i.e. a straight line about which the generatrix revolves. Each point of the generatrix de- scribes a circle, the plane of which is perpendicular to the axis and the center of which lies in the axis. The essence of such a motion lies in the fact that after one complete revolution the generatrix comes back to its first position. Further motion merely retraces the surface already generated. A surface described in this manner is called a surface of revolution. Such a surface may be readily executed by turning or spinning operations, as in a lathe or potters' wheel. Transposition is any motion other than revolution. Fre- quently it is a motion that does not recur, the moving line creating a surface that constantly increases as long as the motion is continued. In this type of motion the generatrix moves so as to constantly touch certain guiding lines, called directrices; or so as to maintain a fixed relation to a guiding surface, called a director; or both. Any one position of the generatrix is called an element of the surface. ^ We have seen how the motion of a generatrix (point or line) in a direc tion not contained within itself creates a quantity (line or piano) having one dimension more than its genefatrix. It is simple to see how a two- dimensional generatrix gives a three-dimensional solid. The student who may be curious to know what would happen if a solid could be moved in a direction not contained in itself is referred to "The Fourth Dimension Simply Explained," by Henry P. Manning; and to an interesting applica- tion of this idea in "Projective Ornament" by Claude Bragdon. VIII, § 140] CURVED SURFACES 155 Surfaces of transposition are executed by '' running " ; i.e. by slipping a cutting edge along guides and over the surface to be worked, as in running a plaster cornice ; or by forcing the material past a revolving knife, as in a molding machine ; or by dies or rolls, past which the material to be shaped is drawn. As an example of the use of the preceding terms, a sphere may be described as a surface of revolution whose generatrix is a circle that rotates about an axis which passes through its center and lies in its own plane. Again a right conoid (see Fig. 175), is a surface of transposition generated by a straight line which constantly touches both a given semicircle and a given straight line as directrices, while remaining parallel to a given plane director, H. Any surface of revolution may also be described as a surface of transposition. Thus a right circular cylinder may be gen- erated by a straight line revolving about an axis to which it is parallel ; or by a circle which slides along a straight directrix. In such a case as the sphere, the generatrix in a motion of transposition must be thought of as variable in size. 140. Classification. Since the character of any curved surface depends upon the character of the generatrix and the kind of motion imparted to it (§ 139), the grouping of such surfaces for the purpose of study must naturally be based on these two considerations. We shall classify surfaces fbst according to the generatrix. Any surface that can be generated by the motion of a straight line is called a ruled surface. All others are called double- curved surfaces. Another classification is by means of the motimi of the gen- eratrix. Any surface formed by rotating the generatrix as described in § 139, is called a surface of revolution. All others are surfaces of transposition. These two classifications are independent of one another. Thus there may be ruled surfaces 156 DESCRIPTIVE GEOIMETRY [VIII, § 140 of revolution or of transposition ; and there may be double curved surfaces either of revolution or of transposition. The characteristic difference between ruled and double- curved surfaces lies in the fact that while a ruled surface is composed wholly of straight-line elements, and therefore a straight line can be drawn through any point of the surface, a straight line cannot be drawn through an arhitrarily chosen point on a double-curved surface. In fnct, the more usual double-curved surfaces, for example those used as illustrations here, can contain no straight line whatever. 141. Tangents and Normals. If any curved surface, ss the sphere in Fig. 153, is cut by a plane P which passes through Fig. 153 a given point a, on the surface ; the tangent, tt\ to the curve of intersection is also tangent to the surface at the given point. Two such cutting planes passing through the same point on the surface would determine two intersecting tangent lines. These lines in turn would determine the plane which is tangent to the given surface at the given point. In the case of a ruled surface, a tangent plane can be de- termined as above or by a single tangent line and the element of the surface which passes through the given point. (See § 161 .) A line is normal to a curved surface when it is perpendicular to a tangent plane at the point of tangency. A plane is normal to a surface if it contains a line which is normal to the surface. 142. Representation. Curved surfaces lack the sharply limiting points and lines that are characteristic of plane solids. VIII, § 142] CURVED SURFACES 157 Therefore it is less easy to visualize their projections. However the fundamental idea is about the same. Projecting lines are passed tangent to the surface, forming in the aggregate, an envelope, which is tangent to the surface as shown in Fig. 154. The line of tangency between a surface and its envelope is called the line of sight. In general the envelope will take the form of a cylinder (see § 145 and Fig. 157) which may be com- bined with planes. Some surfaces, particularly the warped surfaces, can be shown best by drawing the projections of a number of elements (Fig. 177). The line of sight on any curved surface is the boundary between the parts of the surface that are visible and those that 'The envelope fVojecting lines Fig. 154 Projeciion of the sphere. are invisible in projection. It is well to note that some parts of a surface that are visible on the H projection may be invisible on the V projection, and vice versa. Also there are other parts that may be visible on both projections, and still others that may be invisible on both. Care should be taken, in drawing projections, to keep the visible and the invisible portions clearly separated in the mind. 158 DESCRIPTIVE GEOMETRY [VIII, § 143 143. Ruled Surfaces. Ruled surfaces vary widely in char- acter. Since all ruled surfaces have the same generatrix (§ 140), this variation must be the result of the kind of motion imparted to the generatrix. Whether in any given case the motion of a generatrix be one of revolution or one of transposition, its effect on the character of the surface generated can be best Upper b osejN Upper nappe Clement. Open cone. X/ / i\| Lower ^ nappe. • m 0=Anqle of inclination .^' Lower bo6e. Right circular cone. Right circular cylinder, Open , inclined cylinder. Fig. 155 studied by noting the relations existing between adjacent ele- ments. By adjacent elements is meant two elements between which there is no appreciable distance.^ All of the possible relations that can exist between adjacent elements can be sum- marized as follows : (I) Adjacent Elements in the Same Plane. (1) When more than hvo adjacent elements are in the same plane, the surface is plane or plane-sided. 1 See footnote, p. 120. VIII, § 145] CURVED SURFACES 159 (2) When only two adjacent elements lie in the same plane, the surface is curved. The two adjacent elements may be : (a) Parallel. The surface is a cylinder (Fig. 155). (6) All intersecting at the same point. The surface is a cone (Fig. 155). (c) Intersecting two by two (Fig. 171). Since the elements are adjacent, the points of intersection will be adjacent ; and since no three elements are in the same plane, no four of the points of intersection will be in the same plane. Hence these points of intersection will form a space curve, to which the ele- ments are tangents. The surface thus generated is called a convolute. (II) Adjacent Elements not in the Same Plane. The surface is some form of warped surface. In group (I) above, any two adjacent elements determine a plane. Hence the surfaces are developable for the same reasons as for prisms and pyramids (§ 49). Warped surfaces, however, are not developable since even their smallest imaginable elements are not planes (Fig. 175). 144. Developable Surfaces. Developable surfaces, i.e. all surfaces included under heading (I) of § 143, play an important role in building operations. Wherever sheet metal is used, wherever the designs used are set in mosaic or painted on canvas ; in fact, wherever a material, originally flat, is to be applied to a curved surface, it is important that the surface in question be developable. 145. The Cone and the Cylinder. Generation. A cone is generated by a straight line which always passes through a fixed point (the apex) while at the sa;me time it moves in contact with a curved directrix. The directrix may be any kind of a curve, either a plane or a space curve and it may be either open or closed. If the generatrix is unlimited, the cone will be composed of two equal nappes meeting at the apex and it will have no definite base. 160 DESCRIPTIVE GEOMETRY [VIII, § 145 Usually, however, only one nappe is drawn, and the elements terminate in a plane base. The right circular cone is a special case (Fig. 155). A cylinder is merely a special-case cone, the apex being at an infinite distance from the base. Thus all the elements are parallel, and there are no separate nappes. The directrix may be of any form. The right circular cylinder is a special case (Fig. 155). Tlie right section, sometimes called simply the section, of a cylinder is that section which is cut from the surface by a e' = e Fig. 156 a e* 4), h (28, 12, 19), pierces the cone. VIII, § 152] CURVED SURFACES 169 152. To Find the Line of Intersection between Cones and Cylinders. It is impossible to state any one best method. The choice of a trpe of cutting plane is all important. In b" d^ Fig. 164 general, the best plane to choose is that which cuts the simplest lines from the given surfaces. In Fig. 164, if the bases were circles, planes parallel to H would do very well ; but with elliptical bases, probably it would be easier to use planes parallel to the axes of both cyhnders (§ 72). In the figure a plane R is used. The V trace of R is not drawn, since the H trace is sufficient to establish the ele- ments ah and c^ on one cylinder, and cf and gh on the other. These elements give four points, 1-4, on the line of intersection. 170 DESCRIPTIVE GEOMETRY [VIII, § 152 In the case shown in Fig. 165, planes passed through both apexes will cut elements from both cones. An auxiliary line passed through the apexes, as vin is used to determine the planes (§ 66). In the figure, one plane, W, was used as an example. The V trace is omitted, as in Fig. 164. Fig. 165 In Fig. 166, the position of the objects is such that an auxiliary projection on a V plane, parallel to the base of the cylinder, is found most convenient. Once this projection is found, planes passed through the apex and parallel to elements of the cylinder, as X, will cut straight lines from each surface. VIII, § 152] CURVED SURFACES 171 Fig. 166 172 DESCRIPTIVE GEOMETRY [VIII, § 152 When both surfaces are surfaces of revolution and their axes intersect, as in Fig. 167, concentric spheres with their centers at the intersection of the axes can be used. This case is more fully treated in § 177, II. Fig. 167 It will be seen from the above examples that much ingenuity can be exercised in selecting an auxiliary plane or other surface that suits the given conditions. A wise selection is possible only when the surfaces to be dealt with are thoroughly under- stood and a number of cases have been worked out carefully. VIII, § 152] CURVED SURFACES 173 [Note. Rapidity and accuracy in intersection problems can be attained only if a careful system of notation is used. It is well also to think out carefully not only the kind of plane to be used, but also the most advantageous spacing, whether regular or irregular. Often some points can be determined by inspection; where possible, this should be done. Visualize the required line as completely as possible. If one surface is symmetrical, start from the center and work both ways. After drawing an auxiUary plane, determine the required points both on plan and elevation before another plane is drawn. In each of the preceding cases, the Hne of intersection is tangent to the elements of sight. This is true for any problem in intersections where the entire surface is cut.] EXERCISE SHEET LXII 1. The line a (70, 0, 13), h (54, 16, 13) is the axis of a U" right circular cylinder. The line c (56, 0, 14), d (73, 13, 14) is the axis of a 1" right circular cylinder. Find the line of intersection of the surfaces. 2. The Hne e (30, 0, 8), / (6, 20, 13) is the axis of an elhptical cone, the base of which is a 2" circle in H. The hne g (12, 0, 5), h (34, 18, 18) is the axis of an elliptical cylinder, the base of which is a Ij" circle in H. Find the line of intersection of the surfaces. EXERCISE SHEET LXIII 1. The point a (62. 0, 12) is the center of a 2\" circle in H. This circle is the center of the base of a cone whose apex is at h (72, 23, 3). A right circular cone, 1|" in diameter and If" high, has the center of its base at c (59, 0, 13). Find the line of intersection of the surfaces. 2. The line d (13, 0, 21), e (28, 22, 6) is the axis of a cyhnder whose base is a cu-cle If" m diameter in H. The line/ (30, 0, 19), g (16, 22, 5) is the axis of a cyhnder whose base is a circle 1\" in diameter in H. Find the line of intersection of the surfaces. EXERCISE SHEET LXIV Place sheet vertically, HA in the center. The point a (28, 0, 20) is the center of the base of a right cu-cular cone, 4" in diameter and 4^" high. The line h (52, 16, 20), c (4, 16, 20) is the axis of a right circular cylinder 3" in diameter. Find the line of intersection of the surfaces. 174 DESCRIPTIVE GEOMETRY [VIII, §153 153. To Pass a Plane Tangent to a Cone through a Point on the Surface. (See § 141.) I. Base of Cone on H or on V. In Fig. 168, let the point a be determined on the surface of the cone as in § 148. Let it be required to pass a plane through a tangent to the cone. Any tangent plane which contains the point a will also contain the element he, which passes through a. The H trace of the tangent plane is tangent to the cone base at h^, as ZZ^. The V trace passes through the V trace, d^\ of the element of contact. II. Base of Cone not z>^;" ON H NOR ON F. Find the element of contact as before. Pass an auxiliary H (or V) plane through the given point and find its line of intersection with the cone. Draw a line in this plane which is tangent to the line of inter- section, and which passes through the given point. This tangent and the ele- ment of contact will deter- mine the required plane. Sometimes it is easier to extend the cone to H (or to V) and to proceed as in I above. The idea is the same in either case. III. Cylinders. The same construction will suffice for cylinders if it is kept in mind that a cylinder may be considered as a cone with its apex at an infinite distance. 154. To Pass a Plane Tangent to a Cone (or a Cylinder) through a Point Outside. Every tangent plane must pass through the apex of the cone. Hence a line passed through the apex and the given point will be in the required plane. The traces of the required plane pass through the traces of this line. Fig. 168 VIII, § 155] CURVED SURFACES 175 // the cone rests on H, the H trace of the required plane will be tangent to the base of the cone. Two tangent planes may be passed through any point outside the surface. // the hose of the cone does not rest on H, an auxiliary plane parallel to //, and cutting the cone can le used, as in § 153, II. // the given surface is a cylinder {i.e. a cone with its apex at an infinite distance), the only change in the preceding method consists in drawing the guide line through the given point / parallel to an element. 155. To Pass a Plane Tangent to a Cylinder (or a Cone) and Parallel to a Given Line. Such a plane must contain an element of the cylinder; hence it will be paral- lel to all of the elements. In Fig. 169, let ab be the given line. Through any point on ab, as c, draw a line parallel to an element of the cylinder, as cd. The lines ab and cd determine a plane, X. Any plane parallel to X, the H trace of which is tangent to the base of the cylinder as Y or Z, will contain an element and will be parallel to ab. If the given surface is a cone, a guiding line passed through the apex and parallel to the given line may be used. The traces of this line give one point on each of the traces of the required plane. The H trace must also be tangent to the base of the cone. If the inclination of the given line toward H is less than that of the cone, there are two possible tangent planes. Fig. 169 176 DESCRIPTR^ GEOMETRY [VIII, § 155 If the inclination of the given Hne is equal to that of the cone, there is but one tangent plane. If the inclination of the line is greater than that of the cone, the construction is impossible. 156. To Pass a Line Tangent to a Cone or a Cylinder. Fol- lowing the general methods heretofore employed, two different methods are apparent. (a) Pass a plane cutting the given surface in a curve. Draw a line i7i that p^.ane tangent to the curve (§ 141). Q)) Pass a plane tangent to the surface, and draw in that plane the required tangent line. EXERCISE SHEET LXV 1. Given a 45° right circular cone, If" in diameter, resting on H and with the center of its base at the point a (67, 0, 10). Determine two points, h and c, on the surface of the cone, 64 from P and 3 from H. Draw the traces of the planes X and F, which are tangent to the cone at the points h and c. 2. The point d (40, 10, 11) is the center of a \\" circle parallel to V. This circle is the base of an oblique cone whose apex is at the point g (33, 4, 3). Determine the points e and /, which He on the cone, 14 from H and 11 from V. Through e and/, draw the planes V and W, tangent to the cone. 3. The line h (4, 0, 5),i (11, 13, 16) is the axis of an oblique cylinder whose base is a 1" circle in H. Through the point k (13, 5, 10), draw the planes T and U, tangent to the cylinder. EXERCISE SHEET LXVI 1. Given a 60° right circular cone, \\" in diameter, resting on H and with the center of its base at a (70, 0, 11). Given also the lines h (59, 0, 7), c (65, 13, 11), and d (74, 23, 18), e (75^, 15, 23). Draw as many planes as possible which are tangent to the cone and parallel to either 6c or de. 2. The line/ (45, 0, 11), g (39, 12, 22) is the axis of an oblique cyhnder whose base is a l\" circle in H. Through the point h (33, 8, 4) draw two planes tangent to the cylinder. 3. The line/ (16, 0, 17), k (8, 16, 8) is the axis of an obHque cyhnder whose base is a Ij" circle in H. Through the point I (14, 17, 1) draw YIII, § 157] CURVED SURFACES 177 three lines, Im, In, and lo, which are tangent to che cyHnder at points ^", 1", and 1|", respectively, above H. (See § 156 b.) 157. To Develop a Cone. I. Right Circular Cone. If a cone is placed on a plane with one element of the cone in contact with the plane, and the cone is then rolled out on the plane, the original element mean- while remaining fixed, the resulting figure is the development of the cone (§ 49). Since all the elements of the cone are of equal length, and the apex does not move, the development will be a sector of a circle. The radius of this sector will be the slant height of the cone. The actual length of the arc bounding the sector is equal to the circumference of the cone base. In making the drawing the length of this arc may be stepped off with the dividers, as in rectifying a cu^ve (§ 111), or the angle of the sector may be computed by proportion. Let the student prove that the angle of the sector given by developing a right circular cone is equal to 360° cos a, where a is the angle of inclination of an element of the cone to the base. II. Oblique Cone. Between any two elements, drawn near together (as oa, oh, Fig. 170), the surface of the cone ap- proximates a triangle. By finding the true lengths of its sides, any such triangle can be drawn as part of a develop- ment, as at oab. This method is a more or less close 178 DESCRIPTIVE GEOMETRY [VIII, § 157 approximation, depending on the care and the number of elements used. III. Lines on the Surface. Any line on the surface of a cone can be shown on the development by finding the true lengths of the intersected elements and transferring these to the development. If it is required to trace the shortest path between two points on a surface, locate the points on the development, draw a straight line between them cutting various elements, locate these points of intersection on the surface, and draw a curve through them. IV. Cylinders. The principles and methods are the same as for the cone. V. To Develop a Space Curve. (See also § 111, II.) Con- sider the given curve as the generatrix of a cone or cylinder. Draw the projections of the surface. This surface may then be developed and the given curve traced on the developments, as in III, above. If required, this curve may then be rectified into a straight line as in § 111. EXERCISE SHEET LXVH [Note. In this sheet, and in several hereafter, no special location is given for the objects to be drawTi. In such cases the student should carefully plan the layout from the start, to make sure that all the re- quired work can be done without overlapping, and that the finished sheet will be attractively arranged. 1 1. Draw the projections of a 60° right circular cone, whose base is 3" in diameter. 2. Develop the cone. 3. Midway on the center ele- ment of the development, draw a |" square and trace the involute (§134) of this square on the development. 4. Transfer the involute just drawn from the development to the projections of the cone. 158. The Developable Helicoid. I. Type Characteristics. As a single representative of the class of curves defined in § 143, 1 (2, c) the developable helicoid will be chosen for study. VIII, § 158] CURVED SURFACES 179 Such a surface is generated by a straight Hne generatrix that moves so that consecutive positions intersect at adjacent points, no four of which taken in succession He in the same plane.^ In Fig. 171, let ae be a series of such points. Then it is evident that bg and ch are Hues in the same plane, as are also ch and di. But if the points a and d are not in the same plane, it is evident that the lines bg and di are not in the same plane. Nevertheless the surface afje is developable, since it is com- posed of plane elements, as in the case of the pyramid. If the points a and b, b and c, etc., are made to approach one another, keeping their relative positions, the broken line ae approaches a space curve to which the lines bg, ch, etc., are e d Fig. 171 tangent ; and the series of plane surfaces approaches a develop- able ruled surface. This evidently means that such a surface may be generated from any space curve. Such a surface as described above, being a developable ruled surface, is most closely related to the cone and cylinder in its construction, but it bears a visually closer resemblance to some of the warped surfaces (§ 161) in that it has no simple visual outlines. In projection the shape may be revealed by Hmiting the surface between fixed planes which the surface intersects, and by drawing regularly spaced elements on plan and elevation. 1 See footnote, p. 120. 180 DESCRIPTIVE GEOMETRY {VIU, § 158 11. Special Characteristics. The developable helicoid is generated by the tangents to a helix (§ 137). If the tangent is considered as extending in both directions from the curve, the surface which is generated is one of two nappes. If the tangents extend in but one direction, the surface is of a single nappe. It is the surface that results from unwrapping a string wound about a cylinder and following a helix (Fig. 172). Some of the more interesting characteristics of this surface can be seen in Fig. 172 (2). Here the surface is limited by a pair of horizontal planes, the spacing of which is equal to the (2) b h One noppe. Two nappes. Fio. 172 pitch of the helical directrix. The tangent at a is continued to the limiting planes, giving the line be. From the manner in which a helix is generated, it is known to be a curve of constant inclination. Hence the inclination of the tangent, i.e. the angle abd, is the same for any point on the helix. Therefore the lines be, efy gh, etc., are equal, and each of them is equal to the length of the helix. Moreover the H projection of any one of these lines (fe) will be tangent to the // projection of the cylinder, and the length of the projection (gf) will be equal to the circum- ference of the circle (glcx). The curve gcf (or cbh), as well as any section cut from the surface by any horizontal plane, is an involute of the circle (§ 134) which the plane cuts from the cylinder. These facts may be visualized without formal proof. VIII. § 159] CURVED SURFACES 181 159. To Draw the Projections of a Developable Helicoid of Two Nappes. Let the directing cylinder and one turn of the helical directrix ahcde be given, as shown in Fig. 173. Draw the involute (§ 134) of the lower base, I''— 4\ and that of the upper base, w^ — z^. Next draw tangents equally spaced around Fig. 173 the base circle, and extending to the involutes, as X^h^y^, 2''cV, etc. These lines are the H projections of elements of the required surface (corresponding to fg in Fig. 172 (2)). The V projections of 1, 2, 3, 4 are in HA, and those of ic, x, y, z are in the plane of the upper base, WA\ The elements of both nappes can now be drawn on the V projection, as a^'2% 182 DESCRIPTIVE GEOIVIETRY [VIII, § 159 If accurately determined, the V projections of the elements should all be tangent to the given V projection of the helical directrix. Or if the V projection of the helix has not been drawn, these tangents will determine it. 160. To Develop the Developable Helicoid of One Nappe. Consider the surface shown in Fig. 174 a as an approximation (6) Fig. 174 to a developable helicoid. The lines ab, be, etc., are equal, and the angles abc, bed, etc., are equal. But the lines aj, jk, etc., are not equal, nor are the angles ajk, jkl, etc. If the surface is developed, the line a-i becomes a broken line, the component parts of which are equal in length and in change of direction. The line a-q becomes an irregular broken Hne whose elements VIII, § 161] CURVED SURFACES 183 increase in length. The angles between them increase in magnitude as the line progresses. If, by decreasing the dis- tances ab, be, etc., the surface approaches a developable helicoid, the development (Fig. 174 c/) will be bounded by a circular arc ai-ii and the curve ai-gi. It can be shown by trigonometry that the radius, a*, of the inner curve (Fig. 174 c/) is equal to R/cos-O, where R is the radius of the cylinder and 6 the angle of inclination of the helix. A graphical construction for determining this ratio is shown in Fig. 174 e. It is also evident that the length of this arc ajiii must be equal to the length of the helix a^fi^ in Fig. 174 6. The required length may be found by developing the original helix as shown in Fig. 174 c, and measuring the distance thus found along the circumference ajiii Fig. 174 c/. From the manner in which the surface is generated, it is evident that the length of any element f7i^ (Fig. 174 b) is equal to the length of the curve fd^a^. Hence, in the development, the length of any tangent to the circular arc/i7?i must be equal to the length of the arc fidiOi. From this consideration, the outer curve is seen to be the involute (§ 134) of the inner curve. If accurately constructed, the length of the curve aiUiQi will be found to equal that of a^n}q^ in Fig. 174 b. EXERCISE SHEET LXVIII 1. The point a (67, 0, 20) is the center of the base of a right circular cylinder 1" in diameter and 2" high. Draw the projections of the helix and the developable helicoid, making one turn about the cyUnder. With the point b (14, 0, 10) as the center of the inner curve, draw the development of the surface just drawn. 161. Warped Surfaces. I. Generation and Character- istics. As noted in § 143, warped surfaces are generated by the motion of a straight line generatrix, the one neces- sary condition being that adjacent positions of the generatrix shall not lie in the same plane. 184 DESCRIPTIVE GEOAIETRY [VIII, § 161 A glance at Fig. 175, which represents a typical warped sur- face, will show that adjacent elements (as ab and cd) are not parallel ; neither can they intersect, since they are both parallel to the same plane. ^ That portion of a warped surface which lies between any two elements can not be approximated by a single plane figure, as is the case with cones and cylinders (§ 157, II). The details, 1 and 2, in the figure, show how such a portion of a warped surface can be approximated by two triangles that are not in the same plane. Thus the smallest imaginable part of a warped sur- face is not a plane, and it follows that the whole surface is non- developable. An approximate development of a warped surface may be made by triangulation, as indicated in the figure and as explained in § 157, II. 11, Generation by Revolu- tion (§139). A ruled surface of revolution is formed by re- volving a straight line about an axis. The generatrix must bear one of three possible relations to the axis. It may be (a) paral- lel to, (b) intersecting, or (c) not in the same plane as the axis. In cases (a) and (6), the resulting surfaces are the cylinder and cone. Case (c) gives an hyperboloid of revolution, which may be defined as a surface generated by a straight line, revolving about another line not in its own plane. Since these three are the only possible cases of revolution of a straight line, and since two of them give developable surfaces, it is evident that the hyperboloid of revolution is the only possible warped surface of revolution. III. Generation by Transposition (§ 139). The motion of the straight line generatrix is controlled by lines or planes. 1 See footnote, p. 120. VIII, § 162] CURVED SURFACES 185 The more common arrangements of the directrices are as follows : (1) three lines, either straight or curved, or straight and curved Hues in combination ; or (2) two lines either straight or curved, or straight and curved lines in combination, and a plane director to which the generatrix maintains a fixed relation. The number of ways in which these factors can be combined is very great. Hence the possible number of warped surfaces of transposition is practically unlimited. A few typical cases have been selected for study. Once these are mastered, no difficulty will be found in handling other cases. IV. Time as a Factor ix Generation. Sometimes it is convenient to think of the generation of a warped surface as being controlled by a time factor, or velocity factor, which replaces one of the geometric factors (line or plane) mentioned above. Thus any two lines, together with a definite rate of motion of the generatrix along each, constitute a satisfactory method of defining a surface which is often more convenient. For example, in Fig. 177, the surface is generated by the line ac sliding along ah and cd, in equal spaces of time and at uniform velocities. (See also § 201.) V. Tangency. a line tangent to a warped surface is de- termined as explained in § 141. A plane is said to be tangent to a warped surface when it contains an element of the surface and a tangent line. By reference to Figs. 176 and 177 6, it will be seen that any such plane will intersect the surface to which it is said to be tangent. But in such a case as Fig. 175, a plane so determined will be tangent in the usual sense. 162. Representation. In purely theoretical discussions, warped surfaces, as well as planes, cones, etc., are supposed to be of indefinite extent. They may, however, be limited by the directrices or in other ways. A definitely limited surface sometimes can be shown in projection by projecting its outline, as in the circular stairs shown in Fig. 178. In most cases, however, a better conception is given if some of the elements 186 DESCRIPTIVE GEOMETRY [YIII, § 162 of the surface are drawn. When this is done, as in Fig. 176, the drawing shows the modehng of the surface to a certain extent. Another possible method of representing a warped surface is given in § 219. 163. The Hyperboloid of Revolution. I. General Proper- ties. This surface, which is the only possible warped surface of revolution (§ 161), is generated by revolving a straight line about an axis not in its own plane. As in the case of every surface of revolution, any section perpendicular to the axis will be a circle. That point of the generatrix which is nearest the axis will generate the smallest possible circular section, called the circle of the gorge. From the gorge, the surface swells out as the generatrix recedes from the axis. The general form of the surface is shown in Fig. 176. VIII, § 163] CURVED SURFACES 187 Sections made by passing planes through the axis are hyper- bolas. Hence the same sm'face might be generated by an hyperbola revolving about its axis. (Compare with the parab- oloid, § 167.) II. To Draw the Projections of an Hyperboloid of Revolution. In Fig. 176, let us suppose given, the axis and one position of the generatrix, as ab. A perpendicular from the axis to the generatrix, as gh, establishes the circle of the gorge on plan, while the distances, ga, gh, establish the radii of the bases. The projecting planes of all the elements will be tangent to the projecting cylinder of the gorge circle. Hence the H projection of every element will be tangent to the gorge, as a^b'', c^d^, etc. The V projections are found by projecting a^f c\ e'\ etc., to the upper limiting plane, and 6\ d^J\ etc., to the lower plane as shown, and then connecting the proper points. III. Double Ruling. It will be noticed that the element ab and the line jk have the same // projection. These lines lie in the same vertical plane, are inclined at the same angle toward H, and intersect at o. Corresponding points on these lines are symmetrically situated with respect to the axis. Hence the same surface will result from the rotation of either line. A surface that may be thus generated by the motion of either of two generatrices is said to be double-ruled. (See also § 164.) IV. To Locate a Point on the Surface. Let the H or V projection of the required point be assumed at will, within the limits of the surface. If the H projection is assumed, pass an element through the assumed projection tangent to the gorge. Next determine the V projection of the element, and then project the point from the plan up to this line. If the V projection of the point is assumed, pass an H plane through the assumed projection. This plane will cut a circle from the surface. This circle will appear as a line on V, Find the projection of this circle on H, and project down to it from the assumed V projection of the point. 188 DESCRIPTIVE GEOMETRY \yni, § 164 164. The Hyperbolic Paraboloid. This surface is generated by a line that slides along two straight directrices not in the same plane, and remains parallel to a plane director. It is represented in Fig. 177, as limited by its directrices, the lines ab and cd. The plane H is the plane director. Successive elements of the surface lie in successive planes parallel to H, and these planes divide the directrices pro- portionally, as at 1, 2, 3 and at c,f, g, etc. Hence lines joining 1-f, 2-/, Z-g, etc., are elements of the surface. It will be evident that if any two lines are divided into the same number of equal parts, and if the points thus determined are connected in order, the lines thus determined will be parallel to some plane. Hence these lines will be elements of an hyperbolic paraboloid. In Fig. 177 6, the lines ac and hd of Fig. 177 a are redrawn. In Fig. 177 a, these lines were elements of the surface. In Fig. 177 h, the same lines are divided proportionally at 1, 3, 5 VIII, § 165] CURVED SURFACES 189 and at 2, 4, 6. The lines 1-2, 3-4, etc., are elements of an hyperbolic paraboloid, as pointed out in the previous paragraph. By drawing the lines I'e', 2'j' , etc., corresponding to \ e, 2/, etc., in Fig. 177 a, it can be shown that the points x, y, etc., lie on both sets of lines. (Let the student complete the proof.) Thus either set of lines gives elements belonging to the same surface. Hence the surface is a double-ruled surface (§ 163). Circular stairs. (Helicoid ) Arch in a cylindrical well. ( Conoid ) PlAH Hood for a small fireplace. (Conoid.) SKew arch. (Corne de Vache) Fig. 17^ The recessed Marseilles Gate, (Irregular.) 165. Other Warped Surfaces. The two surfaces described above are interesting chiefly because they illustrate so well the characteristics of warped surfaces in general. In practice they are rarely encountered. Some of the forms that are niore usual in construction work, and which bear some resemblance to cones, cylinders, etc., are shown in Fig. 178. The cotzozc? forms have one straight and one curved directrix and a plane director. 190 DESCRIPTIVE GEOMETRY \yiU, § 165 The helicoid which is shown in the figure is the one known as the right helicoid: it has a heUx and its axis as its directrices and a plane H as its director, to which all the elements are parallel. Obhque heUcoids have the same directrices, but instead of a plane, they have a conical director. The Cornede Vache (Fig. 178) has for its directrices two equal circles in parallel planes, and a straight line perpendicular to the planes of the circles and passing through the center point of a line joining the centers of the circles. The generatrix moves in contact with these three lines. In the Marseilles gate (Fig. 178), the surface over the door, between the elements ac and bd, is generated by a straight line moving on the curves ab and cd, and along the straight line xy. The surface between ac and e has for directrices the curves e and ce and the line xy. The two surfaces have a common element in ac. EXERCISE SHEET LXIX 1. Given the lines a (66, 0, 12), b (66, 22, 12) and c (56, 22, 15), d (76, 0, 15). Draw the hyperboloid of revolution generated by rotat- ing cd about ab. Locate the points e and /, which are on the surface just drawn and are 2 from H and 14 from V. 2. Using the lines g (49, 21, 20), h (30, 4, 2) and i (28, 21, 25), j (50, 4, 3) as directrices, and fl" as a plane director, draw an hj^jerbohc paraboloid. 3. Using the lines g' (24, 21, 20), i' (3, 21, 25), and f (25, 4, 3), h' (5, 4, 2) as directrices, retrace the surface of exercise 2. Find the point m, in which the Une k (21, 24. 14), I (4, 12, 23) pierces the surface. EXERCISE SHEET LXX 1. Draw the projections of the warped surface which has three linear directrices as follows: (a) a straight Hne parallel to HA and passing through the point a (49, 0, 10) ; (b) a semi-ellipse lying in a plane perpendicular to HA. and ha\'ing its center at a, the minor axis being If" long and lying in H, while the semi-major axis is If" long; (c) a semi-circle 1|" in diameter, in a plane perpendicular to HA and with its center at b (60, 0, 10). 2. Imagine a right circular cyhnder 2^" long with its axis perpen- dicular to V. Let its diameter be 1|", and the center of the rear base VIII, § 166] CURVED SURFACES 191 be at c (29, 15, 4). Now let the front base be moved downward and to the right till its center is at d (12, 6, 24). During this motion of the front base let it be rotated in its own plane, in a clockwise direction, through 120°; the rear base meanwhile remaining fixed. During the above movements, let each element remain straight and always con- necting the same points on the circular bases, but varying in length. Draw the projections of the resulting warped surface. EXERCISE SHEET LXXI Place the sheet vertically, with HA in the center. Draw the pro- jections of an hyperboloid of revolution whose axis is the line a (29, 0, 20), b (29, 26, 20), and whose generatrix is the line c (46, 26, 25), d (12, 0, 25). Draw also the projections of a conoid whose directrices are the line e (46, 0, 8), / (46, 17, 8), and a semicircle 4j" in diameter with its center at g (14, 0, 24). The plane of the semicircle is per- pendicular to e^ and H is used as a plane director. Find the line of intersection of the given surfaces. 166. Double-Curved Surfaces. The chief characteristics of double-curved surfaces are given in §§ 138-142. They are as follows: (1) The generatrix is a curve. ^ (2) Such a surface is not covered by straight lines. (See p. 156.) (3) These surfaces may be generated either by transposition or by revolution. By far the most important of the double-curved surfaces are those which are generated by revolution. The simplest case is that of the sphere, formed by rotating a circle on its diameter. In the same class are balusters, vases, and most of the products of the lathe, the potter's wheel, and the metal spinner, excepting only such as take the form of a cone, cylinder, or hyperboloid of revolution (§ 161). When a curved generatrix follows a curved directrix with a motion of transposition, a double-curved surface is formed. This case covers a wide variety of surfaces, not all of which have distinct names. An example is shown in Fig. 186. 1 The case of surfaces generated by a line which changes form are not considered here. Thus a cord which changes position while in vibration would have many positions in which it would be straight, while in most positions it would be curved. A surface so generated would contain many straight lines, though the surface as a whole would not be ruled. 192 DESCRIPTHT: geometry [VIII, § 167 167. Typical Double-Curved Surfaces of Revolution. A circle, when rotated on a diameter, generates a sphere. An ellipse, rotated on its diameter, gives a spheroid. AVhen the major axis of the ellipse is used as the axis of rotation, the surface generated is called a prolate spheroid. When the minor axis is used, the resulting surface is an oblate spheroid. Paraboloids and hyperboloids are generated in a similar manner by the revolution of parabolas or hyperbolas. One of the possible hyperboloids is the warped surface described in § 163. A circle which revolves about an axis wholly outside itself generates an annular torus, the typical ring form. Other forms of the torus are given by revolving other closed curves. The bases of classic columns, and some of the capitals, contain various forms of the torus, some of which have special names. 168. Sections. Every point on the generatrix of a surface of revolution describes a circle, the plane of which is perpen- dicular to the axis of revolution. Hence any plane passed perpendicular to the axis will cut from such a surface a circle, the center of which is on the axis. This fact gives a basis for the solution of many of the problems which arise in connection with double-curved surfaces. Any plane passed through the axis of a double-curved surface of revolution cuts the surface in a curve identical with the generatrix. Such a plane is called a meridian plane ; the section is called a meridian section. 169. Representation. Normally, double-curved surfaces are drawn with the axis parallel to one of the planes of projection. When in this position, one projection consists of a meridian section, while the other is a circle representing the largest diameter, together with other circles showing the size of the surface at sections where the curvature changes (Fig. 179). The projections of a sphere consist of two equal circles. The plan shows the upper half of the sphere as visible, while the elevation shows the forward half as visible. Onl^' the upper front quarter of the surface is shown as visible on both projec- VIII, § 170] CURVED SURFACES 193 tions, and the lower rear quarter is not visible on either. This is typical of most double-curved surfaces of revolution. After sufficient practice, the beginner will be able to distinguish clearly between visible and invisible points on a surface. 170. To Locate a Point on the Double-Curved Surface of Revolution. In attempting to locate a point on a plane surface (§ 65), and again on a ruled sur- face (§ 148), it was found neces- sary to first determine a line on the surface. The same necessity will be found in the present problem. In this case the line most often used is a circle, since that is the simplest line that can be drawn on a double- curved surface. In Fig. 179 are shown the projections of a prolate sphe- roid. Let it be required to locate a point on the surface. One projection of the required point may be assumed at will, within the given limits of the surface. Let x'" be assumed, as shown. The point x lies in the horizontal plane P. If it is to lie on the spheroid, it must be on the line of intersection be- tween the spheroid and P. This line of intersection is seen as a circle on plan, the diameter of which is ah. The H projection of X lies on this circle and below .t^, i.e. either at x^ (on the front half) or at x'^ (on the rear). Again the assumed projection may be chosen on plan, as at y^. The point y lies on the surface circle y^m, which can be located on elevation either in the plane M or A^ The V projection of y will be found either at y" or y'"". Fig. 179 194 DESCRIPTIVE GEOMETRY [VIII, § 171 " 171. To Draw a Line Tangent to a Double-Curved Surface of Revolution at a Given Point on the Surface. If a plane cuts from the given surface any curve, any line in the cutting plane and tangent to the curve is tangent to the surface (Fig. 153). In Fig. 180, let it be required to pass a line tangent to the paraboloid at the point x. First let a meridian plane M be Fig. ISO passed through x. This plane will cut a parabola from the surface. The projection of this parabola could be determined by the method of § 149, and the tangent could be drawn at once. It will be found easier, however, to rotate the entire con- struction (toward the right, in the figure) on the axis ah, until the cutting plane, M, is parallel to V . The curve of inter- VIII, § 173] CURVED SURFACES 195 section between the paraboloid and the plane M will then appear as the meridian section 6^.c'^£^^ and the point x will move to x'. The line c'x\ drawn tangent to the meridian curve, is in the same plane with the curve. Hence it is tangent to the parabo- loid. If now the cutting plane is rotated back to its original posi- tion, the point c' will move to c and x' to x, the tangent line being ex. The lines ex and e'x' are elements of a right circular cone whose apex is at o, and whose base passes through x. The Hne just determined does not constitute the only possible solution to the problem. Many lines may be drawn tangent to such a surface at any given point. Any cutting plane whatever through the given point would yield a curve of inter- section to which a tangent line could be drawn. Perhaps the easiest solution for this problem results from the use of a hori- zontal cutting plane, since such a plane gives a circular line of intersection. In general, the student should be familiar with the solution using either a right section ( H cutting plane) or a meridian section, as above. 172. To Pass a Plane Tangent to a Double-Curved Surface of Revolution, at a Given Point on the Surface. Draw any two tangent lines through the given point, as explained in § 171. These lines will determine the required tangent plane. In the special case of the sphere, we may draw the radius terminating at the given point and we may pass a plane per- pendicular to this radius and through its extremity (§ 82). Such a plane will be tangent to the sphere. 173. To Draw a Right Circular Cone with a Given Apex and Tangent to a Given Sphere. The axis of the cone will pass through the center of the sphere, and the line of contact will be a circle. The simplest possible view of the surfaces will be that one which is projected on a plane parallel to the axis of the cone. 196 DESCRIPTIVE GEOMETRY [VIII, § 173 r / In Fig. 181, let the sphere whose center is c, and the apex of the cone, o, be given. Find the projections of o and c on a V plane taken parallel to oc. These are shown at o"' and c''. The elements of sight of the cone as seen on this new plane will be tangent to the element of sight of the sphere, and the circle of contact between the surfaces will appear as the straight line a'^'b^' It is now necessary to establish the circle of contact on the H and T" projections of the sphere, which " will appear as ellipses. On the H projection, the minor axis will be projected in a line parallel to V, at b^a''. The major axis is perpendicu- lar to ¥a^, at its center, and its length is equal to a^'b^' as given on the V projection. The H projection of the circle of con- tact can be com- pleted by drawing an ellipse through a^d^'b^e^. The V projection can be estab- lished point by point from the H and V projections, or by determining the V projections of the axes ab and de, which will be conj\igcite diameters of the required ellipse (§ 121). Fig. 181 VIII, § 174] CURVED SURFACES 197 The elements of sight of the cone can now be drawn through o and tangent to the circle of contact. If carefully drawn, the elements of sight, the circle of contact, and the projection of the sphere will pass through a common point. 174. To Draw the Projections of a Double-Curved Surface of Revolution in any Position. I. The Envelope Method. When a surface of revolution is placed so that its axis is parallel to a plane of projection, one line of sight is a meridian section, while the other is a right section (§ 169). If the axis is in- clined to the planes of projection, the lines of sight shift to new positions, which are not easily determinable. The method used below to determine the lines of sight is called the envelope method. (See § 142 and Fig. 154.) This method is used in many of the more complex problems in shades and shadows. Figure 182 shows a spheroid with its axis inclined toward H and V. A right section, as ab, is projected on H as the ellipse a^b^. The lines ea and fb are passed tangent to the spheroid, perpendicular to H, and touching the right section at a and b. The points a and b are on the line of sight of the spheroid and a'' and b^ are points on the H projections of the spheroid. Now let the spheroid be considered as a surface of transposition (§ 139), generated by a variable right section, ab. Successive positions of the right section give ellipses as their H projections. Successive positions of the projectors ca and fb generate a cylinder (or an envelope) which incloses the spheroid. The Fig. 182 198 DESCRIPTR E GEOMETRY [VIII, § 174 intersection of the envelope with H is a curve, tangent to the successive projections of the right section, which determines the projection of the surface. Fig. 183 VIII, § 175] CURVED SURFACES 199 II. The General Method. In Fig. 183, let it be required to draw the projections of the surface shown at (a) when its axis is incHned to H and V as indicated by the hne he. The general method which will be followed ^ consists in draw- ing the projection of the given surface on a V plane, parallel to the given position of the axis, as shown at ih) in the figure ; and then using this projection as a basis to construct the re- quired H and V projections. In the figure, the line V2'' is the V projection of a circular right section of the surface. The // and V projections of this circle may be determined by the method of § 128, as shown at 1''4'' and 1^4^ For any other section, as 5-6, and for as many others as may be desired, the projections may be found in the same manner The common tangents to these ellipses give the projections of the required surface. Where any portion of the surface is a true cone or cylinder (as between .1 and J5, C and D, in the figure), it is only necessary to determine the limiting sections, the tangents between them being straight lines. The line of sight of one portion of the surface may fall within the projection of another part, as at Z. This merely means that the surface here resembles a cone, the projection of the apex falling within that of the base. [Note. In attempting the solution of such a problem as the above, the student should find both the projections of a given right section before attempting to determine anything with regard to the next one. A common tendency is to determine aU the H projections first. This usually leads to endless confusion and mistakes.] 175. To Find the Line of Intersection between a Plane and any Surface of Revolution. The general considerations apply- ing to the intersection of surfaces (§ 149) suffice for this case. Since all right sections of the surface are circles, the auxiliary planes used will be those that will cut right sections. 1 Tliis method is identical in principle with that of § 53 (for prisms) and § 146, II (for cones). 200 DESCRIPTIVE GEOMETRY [VIII, § 175 Figure 184 shows the intersection between a plane A', and an annular torus. It will be noticed that there are points of tangency between the lines of sight of the torus and the curve of intersection. Any such curve of intersection is usually tangent to the sight lines in two points or more. The special case of the preceding problem which in- volves a cutting plane perpendicu- lar to H or V is X^ frequently used in later problems. 176. To Find the Points in which a Line Tierces a Surface of Revolu- tion. This prob- lem admits of the same solution that was used in § 151, I. The auxiliary plane will usually be chosen through the line and per- pendicular to H or V. The intersection of this plane with the given surface can be found as in § 175. The points where this curve meets the given line are the required piercing points. 177. To Find the Line of Intersection between any Two Sur- faces of Revolution. (See also § 149.) The methods given below will apply to either double-curved or ruled surfaces of revolution. Three cases may be distinguished, depending on the relations that exist between the axes of the given surfaces. VIII, § 177] o URVED SURFACES 201 I. Parallel Axes. In this case planes perpendicular to the axes will cut circles from each of the surfaces. II. IxTERSECTixG AxES. It is often difficult to find a plane which cuts l)oth surfaces in curves which are sufficient!}' simple to give an easy solu- tion. A simple solu- tion results, however, if spheres are uecd instead of planes for cutting the surfaces. If any surface of revolution is cut by a sphere, the center of which lies on the axis of the surface, the line of inter- section is a circle. Therefore, if two sur- faces of revolution have axes that in- tersect, the point of intersection may be used as the center of a series of cutting spheres. Each of these spheres will cut a circle from each of the surfaces and the intersections of these circles give points on the required line of intersection. In Fig. 185, let the vase form and cone of revolution be given, and let the axes intersect at o. \Yith o as a center, generate a sphere, as shown by abed. This sphere will cut from the cone a circle which is shovv'n on elevation at a'-'C. From the vase Fig. 185 202 DESCRIPTIVE GEOMETRY [VIII, § 177 it will cut a circle shown on elevation at b^'d''. These circles intersect at two points x and x\ which coincide on the elevation. On plan, the circle cut from the vase is shown at x^'^b^x^d^. The points on the required line of intersection, x and x\ are now fully determined. Other spheres, concentric vrith tlie one just drawn, will yield other points on the required curve. III. Axes not Ixtersectixg AND NOT Parallel. In this case, it is not possible to state any solution that will be best for all cases. Any cutting surface which gives a simple intersection with one surface will probably give a less simple one on the other surface, rendering the solu- tion tedious, but a solution is always possible according to the general principles explained in § l-t9. 178. To Draw the Projections of the Serpentine. The serpen- tine is a surface generated by a sphere which moves so that its center follows a given helical directrix. Figure 186 shows the projec- tions of the serpentine when the axis of the helix is perpendicular to H. Various positions of the spherical generatrix are drawn. These are connected by a common tangent as explained in § 174. Fig. 186 VIII, § 178] CURVED SURFACES 203 EXERCISE SHEET LXXII 1. The point a (66, 12, 12) is the center of a sphere 2j" in diameter. Locate the point h, which lies on the surface of the sphere, 2|" above H and 2" in front of V. Through b draw a diameter of the sphere. Call its opposite end c. Through h draw a line tangent to the sphere. Through c pass a plane tangent to the sphere. 2. The point / (40, 12, 10) is the center of a prolate spheroid, whose major axis is vertical and 3" long. Find the line of intersection be- tween the given surface and the plane Z (51, Z^ 60° r, Z" 60° r). 3. A circular arc passes through the points g (13, 22, 9), h (19, 11, 9), and y (13, 0, 9). Let this arc be rotated on gj as an axis, forming a double-curved surface of revolution. Find where the Hne k (5, 20, 4), I (22, 4, 13) intersects the surface. EXERCISE SHEET LXXIII Place the sheet vertically. Take HA 4" below the top border line. Draw the projections of the vase form shown in Fig. 208, using as an axis the line a (44, 24, 32), b (26, 6, 2). Let the length of the vase be about 3". EXERCISE SHEET LXXIV 1. Given the annual torus whose inside diameter is 1|" and whose outside diameter is Si". Let the center be at a (64, 12, 14), and let the plane of the top be horizontal. Given also a right circular cyHnder, 2|" in diameter, resting on H, with the center of its base at b (60, 0, 16). Find the line of intersection. 2. Given an annular torus described in (1) above. Let the center be at c (23, 12, 14). Given also a right circular cone, 3" in diameter and 3^" high. Let the center point of the axis of the cone be at c, and let the axis be parallel to V and inclined at an angle of 30° to H. Find the line of intersection. CHAPTER IX APPLICATIONS BASED ON CHAPTERS I TO V i 179. Introduction. The conditions of a concrete problem are often more easily seen, and its solution is more readily per- formed, than are those of an abstract problem of a similar nature. For this reason, one is always tempted, when commenc- ing problems similar to those in this and in the following chapters, to discard the fundamental ideas on which the solutions are based, to proceed with more or less vague and disjointed methods, and to be satisfied with intuitive solutions. These soon become set rules or formulas. When this happens, the time spent on such problems is worse than lost. If, however, the student seeks to trace back each step in the solution of a concrete problem to its underlying geometric principle, and to build his work on perfectly general laws, much benefit may be derived from solving problems of this nature. 180. Simple Plan and Elevation. (§§ 1 to 25.) In starting to lay out the plan and elevations of a simple building, such as that shown in Fig. 7, it is common practice to develop the plan quite separately from the elevations, often on different drawing boards. In order to keep the plan and elevations in agreement some common basis of reference is needed. For this purpose vertical planes are used that cut the building on its principal axes. These, seen on plan, give the main axis lines. (See Figs. 197, 211.) From their relations to these planes the positions of all points can be shown on plan. 1 Exercise sheets to ucconipany this chapter arc given on page 295. 204 IX, § 180] APPLICATIONS 205 Having correctly located the main points, say the four out- side corners, other points can be located on the plan from their known relations to these points or to the axes. For the purpose of locating heights (drawing elevations), a horizontal plane is used. This is generally taken at the level of the first floor, or at some other level arbitrarily chosen. Such a plane is called a datum plane. It is usually established by some permanent natural feature of the land, such as the level of the lake in Fig. 247, or by some artificially established mark, such as the top of the curbing in a city street. The datum plane is shown on the elevations as a horizontal line, called the datum line. This datum line is the first line to be drawn on an elevation. All heights are measured from it, either upward or downward. Next, in order to relate the plan and the elevations, an axis line is drawn. This line should represent one of the axis planes used in drawing the plan. From this line all horizontal lengths are measured, either toward the right or left. The correct development of the drawings is thus assured. It is usually wise to start with the plan. Develop it as far as is easily possible. Then pass to the elevation. Possibly this cannot be completed at once. Perhaps now the plan can be completed. Passing thus from one drawing to the other, the whole can finally be finished. The order in which the work is to be done should be carefully planned in advance. As any plan develops, the beginner should keep in mind which lines are normal-case lines and which ones are special cases, which ones show true lengths and inclinations, and which ones are foreshortened in projection. The fact that the planes of projection, as formally erected in problems of descriptive geometry, are apparently discarded, should not lead to a different conception of the fundamental facts involved. The floor or datum level is in fact what we 206 DESCRIPTIVE GEO:\IETRY [IX, § 180 previously called the H plane. The axis planes are in fact V or P planes. \Yhen working in this manner (purely as a matter of con- venience), the planes of projection are not so opened as to lie adjacent to the same axis hue, but they are placed at will, perhaps on different drawing boards. However, the fundamental relations between the reference planes are maintained. The V planes passed through the plan axis and the datum plane for levels correspond exactly to V, P, and H, as used in the abstract problem. 181. Sections. Plans and elevations are sufficient for the description of simple solids. When the shape of the sohd to be -SfCIIOII PUKE SECTION Fig. 187 shown is such as to involve reentrant angles or when the object is hollow and it is necessary to show the thickness of the shell or the interior arrangement, a different kind of drawing, called a section, is required. The necessity for making a section lies in the fact that in ordinary projections certain parts of the object are concealed by other parts. The method used is to remove the obstructing parts. The object, a section of which is to be drawn, is considered as cut by a plane, along the desired line (Fig. 187). One part of the object is then removed and the remaining part is shown as projected on the cutting plane. IX, § 181] APPLICATIONS 207 It is usual to indicate all solid parts which are cut by the section plane in some distinctive manner. Cross hatching, soKd black, and other indications are commonly used. Dif- ferent indications are used on the same drawing in order to show different materials. Section planes are commonly vertical planes so chosen as to coincide with one of the principal axes of the object, although much latitude is allowed in this respect. For special reasons section planes may be chosen in almost any position in order to fully explain the object in hand. x\ll parts of the object, both inside and outside, which lie beyond the section plane are projected on the section plane in the usual manner. When an object has been drawn in plan and in elevation, and a section is required, it is necessary first to determine where the section plane may be located in order to be most useful in describing the hidden parts. This having been decided, the trace of the plane is shown on plan. All horizontal distances for constructing the section can be found by projecting points back to this line. The vertical relations are established from the elevation. The section is thus built up point by point from plan and elevation in the same way that a profile projection is determined from known horizontal and vertical projections. Floor plans of a building constitute a special case of sections. Here the section plane is chosen in a horizontal position, cutting the walls at about the mid-height of doors and windows. The parts above being removed, a single floor is shown as pro- jected on plan, the walls being cross hatched as in a section. In making sections through a complicated structure it is sometimes necessary to use cutting planes that are not con- tinuous, one part being out of line with another, in order to explain a particular feature of the construction. When this is done, care must be taken to avoid overlapping of different parts on the section drawing. 208 DESCRIPTIVE GEOMETRY [IX, § 182 182. A Corner Rafter (§§ 27, 28). Figure 188 shows the corner of a show rafter cornice. It will be noted that the hip rafter is of different size and shape from the others but that the shapes are definitely related. Commonly the jack rafter is designed first and from it the shape of the hip rafter is de- termined. Any horizontal line (as A A' A) parallel to the wall, and touching each jack rafter at a given point, determines the corresponding point on the corner rafter. Fig. 188 In Fig. 189, let the plan (c), together with the section drawn above it, be given. Let the section (6) show the shape desired for the jack rafters. The plan shows the position of both the jack and corner rafters. Let it be required to determine the proper shape for the corner rafter. The method to be followed consists in finding the projection of the rafter on a vertical plane erected parallel to it (§ 27), as shown by H'A'. Lengths are projected from the plan, while heights can be measured from any datum plane (as BB) on the section. The first lines determined should be the top and bottom. Then X, § 182] APPLICATIONS 209 the ornamental part may be determined by locating a sufficient number of points on the curve to give it the same character as the curve from which it is derived. Fig. 189 The true shape of the corner rafter having been found as above, a valuable exercise will be found in determining the front ele- vation of the corner, as shown at (a). 210 DESCRIPTIVE GEOMETRY [IX, § 183 183. A Pipe Shaft (§§ 27, 31 to 3G, 4G, 53). Figure 190 shows a portion of the floor and walls of a building, and two points a and h, in isometric projection. The line ah is to be the center of a pipe shaft, 3' 0" square. The problem is to locate the holes required in the floor and wall, and to show the shape of each. In Figure 191, the above conditions are shown in plan and elevation. The method of procedure will be to find first the Fig. 190 lines which form the edges of the shaft, then to find where each of these lines pierces each face of the floor and walls, and finally to connect these piercing points so as to show the required holes. In this problem it will be wise to establish no formal axis line, but to use the front and rear faces of the wall as V planes, and the ceiling and floor as // planes. Let the line ab, Fig. 191, represent the axis of the shaft. This shaft is to have, as its right section, a square 3' 0" on each side. The projections of such a section, having its center at any point, o, on ah, may be constructed as described in § 46. IX, § 183] APPLICATIONS 211 Having established the four corners, c, d, e,foi such a section, the edges of the shaft, eg — fh, can be drawn parallel to ab. W ^b" Fig. 191 The traces of these lines (§31) on the surfaces of the wall and floor establish the required holes. [Note. At tliis point the student can also make an application of the principles of Chapters I to V to the elementary problems in shades and shadows, given in §§ 202 to 210.] CHAPTER X APPLICATIONS BASED ON CHAPTER VI » 184. Roofing the Plan. In many buildings the roof treat- ment is the most important single feature of the exterior design. The form depends on the size and shape of the plan. Even when the plan is fixed, considerable latitude of selection remains. The elements involved are usually plane surfaces, set at a given pitch, meeting so as to shed water. The simplest case is a rectangular plan with two gables shown in Fig. 192 a. When the width of the plan, a, the height of the wall, b, and the inclination of the roof, 6, are given, the height of the ridge, c, becomes fixed. Any three of these dimensions will fix the fourth. Every line drawn on the roof parallel to the eaves is parallel to the ground. Any other line on the roof makes an angle with the ground w^iich may vary between 0° and 0°. Figure 192 6 shows a typical method of covering a square plan. Here the eaves are carried all around the building. The four planes meet in salient angles at the My lines, and the ridge shortens to a iieak. The hip lines make an angle w^ith the ground which is less than the inclination of the roof. In Fig. 192 c, the same plan is roofed with four gables. The two ridges are the same height, and meet at a point of crossing, c, while the roof planes intersect in reentrant angles or valleys. The slope of the valleys is the same as that of the hips in Fig. 192 b. If the point of crossing, c, in Fig. 192 c, be raised slowly, the ridges, ca, cb, begin to slope downward to the walls ; the in- clination of the valleys becomes greater ; and the angle between the planes meeting on the valleys is changed. The roof assumes 1 Exercise sheets to accompany this chapter are given on pages 295-299. 212 X, § 184] APPLICATIONS OF CHAPTER VI 213 the general form of Fig. 192 d, a form not unusual for towers. If c is raised still more, at a certain height the ridges ca, ch. Ridge. (c) Pea If -^b (6) Hipped roof. c Fig. 192 and the valley cd fall in the same plane, giving a simpler roof, as shown in Fig. 192 e. Determine what will happen if the point c is carried still higher. An irregular plan like that shown in Fig. 192 / lends itself well to a picturesque treatment of the fayades. 214 DESCRIPTIVE GEOMETRY [X, § 185 Fig. 185. Intersection of Roof Planes. Figure 193 shows two planes, P and P^ each inclined to H at the same angle, 6. These planes are related in the same manner as two roof planes meeting at a hip. Let us choose a point o, on the line of intersection of these planes, and let 0^ be its projection. Next let us pass planes through 00^ perpendicu- lar to ah and 6c, giving the triangles oo^a and oo^c. These triangles are equal, hence o^a is equal to oV. Then the triangles o^ab and o^cb are equal and the angles 4> and 4>' are equal. Since the angle abc is cmy angle, it may be stated that when roofs of equal slope intersect, the hip or valley, as seen on plan, bisects the angle between the eaves. Usually, all roofs on a given build- ing have the same slope. AVhen different slopes are used, the plan of the hip or valley does not bisect the angle between the walls. 186. Roofing Plans. In drawing a set of floor plans and elevations, it is quite easy to indicate an ar range ment of roof planes w^hich, in ^act, cannot exist, or that would fail to shed water, or that would leave some portion of the plan uncovered. A carefully drawn roof plan will reveal any such fallacies. Figure 194 shows a case in point. The planes a, b, e, and d do not meet, either in a ridge or in a peak, but leave an open spot, e. In such cases a (relatively) flat section of roof is sometimes Fig. 194 X, § 187] APPLICATIONS OF CHAPTER VI 215 5"(ribe) Roof surface Fig. 196 used. The visualization of a roof plan is made easier by a shading Hke that in Fig. 195. When a vertical surface, such as the side of a chimney, intersects a roof so as to leave a level pocket behind, an artificial slope, called a saddle or cricket, is used. (Fig. 195.) 187. Rise and Run. The slope of a roof, while sometimes la'(run) — ^ stated by giving the angle in de- grees, is more often expressed in terms of rise and run, as shown in Fig. 196. The roof shown in the figure would be said to have a rise of 5'' in a run of 12". When the relation of rise and run are known, many relations of height can be deter- mined from the plan without recourse to the elevation. Relations of distance can be deter- mined on elevation with- out using the plan, or a plan and elevation can be developed without draw- ing a roof plan. Thus, in Fig. 197, the inclina- tion of the roof being 45°, the eave sp will intersect the roof B, at the point p, on the plan, which is just as far back of the eave inn, on the plan, as s is above mn on the elevation. 9 J -n" Fig. 197 216 DESCRIPTIVE GEOMETRY [X, § 188 188. To Determine the Point at which a Guy Wire Meets a Roof. In Fig. 198 are shown the plan and elevation of a Fig. 198 building with a smokestack. Three guy wires are to be fastened to the stack at the height aa. These wires shall make angles of 30° with the ground, and shall lie in the vertical planes X, § 188] APPLICATIONS OF CHAPTER VI 217 indicated on plan by OA, OB, and OC. It is required to determine where, if at all, these wires will meet the roof. It is necessary first to determine the projections of the wires. One wire lies in the plane OB. Its H projection, rh^, coincides with OB, where r is chosen as the point where the wire meets the stack, and s is amj point on the wire. Revolve the wire till it is parallel to V. Its plan is now r^s'\ Its elevation passes through r'' and makes an angle of 30° (the given angle) with the horizontal axis. Thus s' is located. If now the wire is swung back into its proper position, 5'^' moves to 5^', and s'^ moves to s^, giving r''s^' as the true elevation of the wire. This operation is an application of § 24. To find where the wire pierces the roof, we apply § 78. We require the trace of the wire on the roof. The auxiliary plane is chosen through the wire and perpendicular to H. It cuts the building in the broken line t-z. The point F, where this Hne crosses the wire, is the point in which the wire meets the roof. The line tuv is determined by starting with the plan. The line cb pierces the auxiliary plane 05 at a point directly above u^. By projecting upward from u^, the V projection of w may be found. The intersection of the auxiliary plane with the stack is a vertical line, hence t lies du-ectly below r and on the intersection between the stack and the roof. The point v is not so easily located on the elevation, since it lies in a line parallel to P, but it may be found without using a profile projection. Draw the line vh'\ parallel to the cave line. This line is the H projection of a line in the roof and parallel to H. The point v' is located on elevation by projecting up from the plan. Lastly, r^ is found on a horizontal line through v''\ The remaining points on t-z are found in a similar manner. The other two wires, OA and OC, are not shown in the figure. [Note. A few words with regard to the construction of this plan and elevation will assist in the apphcation of this problem to the one given on sheet LXXX on p. 299. All roofs are of equal slope, hence 218 DESCRIPTIVE GEOMETRY [X, § 188 all the angles, d, are 45°, and the line V'cH^ is a straight line (§ 185). The plan is easily completed, excepting for the dormers. Leaving this point, and passing to the elevation, the height of the ridge ecj can be determined from the known span and slope. In this case, the slope being 45°, the height g is equal to the distance h. Let the front walls of the dormers be in the same planes with the walls below. Then i = j ', and if k is determined at will, m must be made equal to it. Thus the plan and elevation of one dormer can be finished. Next lay out the plan of the dormer which is seen partly from the side, mak- ing it like the one just completed. The elevation of this one can be determined by projecting corresponding points up from the plan, and across from the elevation of the other dormer. One point which fre- quently gives trouble is n. This point lies on the corner, at the same height as p, which has already been determined.] 189. Framing for a Hipped Roof. In framing a timber roof, it is desirable to cut the timbers so that they will fit in their re- spective places, before starting to raise and assemble them on the roof. In order to do this, the various angles for the cuts (see Fig. 200 a) must be worked out from the plans. The present problem takes the case of a simple hipped roof and stand- ing gutter, and works out the various angles, etc., which must be known in order to get out the various parts. The roof, as shown in Fig. 199, consists of three planes, meeting in the lines ab, ac, and ad. It will be assumed that the timbering conforms exactly to these planes, on top, and to similar parallel planes below. This will mean that the hip and ridge timbers will have the general form shown in Fig. 200 b, though in an actual construction they probably would be square edged. In this latter case, the ridge and hips are slightly blunted in the timbering, but the boarding is brought to a real line of inter- section. Plax, Elevation, and Section. The layout of the plan is quite simple. The detail of the framing, as shown within the circle .1 in Fig. 199, should be carefully noted. The front left-hand corner of the roof is shown covered with boarding, and fitted with a standing gutter. The section shows the depth X, § 1S9] APPLICATIOXS OF CHAPTER VI 219 z'\n^y Fig. 199 220 DESCRIPTIVE GEOMETRY [X, § 189 of the timbers and the manner of fastening them to the walls, by means of a plate. In drawing the elevation, it should be borne in mind that the line a'd'-' is the V projection of the plane hade, and that since the line fg does not lie in this plane, but in acd, its F projection, /''<7', will not coincide with aUh'. Similarly, the lower edges of the hip, directly below ad and Jg, will show as two lines, of which one is dotted. In determining the depth of the hip rafter, it should be remembered that its bottom and top edges lie in the same planes with those of the common and jack rafters. Figure 7 4. Down cut. 5. heel cut. ?. Side cut. Fig. 200 200 c shows this fact. It will be noted that the line yz is common to both rafters, and that it is vertical. Now since the hip is set at a flatter pitch than is the jack, and since a vertical line on either is of the same length, it follows that the hip rafter must be cut from a wider plank than the jack. However, in the elevation, this extra width is not apparent, and the hip appears to be of the same width as the common rafter shown in the section. It is worth while for several purposes, in making the small scale plan and elevation indicated in Fig. 199, to study the parts within the circles A and B (Fig. 199) at a larger scale, as shown in Fi^. 201, since the detaih of these portions cannot be drawn accurately on a small scale. X, § 189] APPLICATIONS OF CHAPTER VI 221 Let us try to find : This angle is sliown This is also shown in Cuts and Angles. (See Fig. 200 a.) 1. The down cut for a common rafter. in its true size on the section. 2. The heel cut for a common rafter. true size on the section. 3. The side cut for a jack rafter. This angle is shown on plan as hNc^m^. This angle does not show^ in its true size. Let it be rotated on hm as an axis, until it is parallel to H. During the rotation, k'' moves to //'" and k^ moves to k'^. The true size of the angle is then shown at h^k'hn^. 4 and 5. The down cut and the heel cut of the hijp rafter. Draw an elevation of the hip on a V plane chosen parallel to the rafter (§§ 27, 53). The angles marked 4 and 5 are the true size of the required angles. 6. The top bevel of the hip rafter. Assume an H plane of reference, as shown by HA. Extend the roof planes to intersect this plane. The traces will be YT', ZZ\ and ZZ\ as shown. The line of intersection of the planes is an. Now proceed to find the angle between the planes, as in § 101. 7. The side cut of the hip rafter. This is the angle shown at fil in Fig. 201. The true size may be found by revolving the plane of the angle, as shown, till it is parallel to H. 222 DESCRIPTIVE GEOMETRY IX, § 189 H A b' a^ A \ -Tf- — X \ / \ / \/ /\ ■■-, Y" Y _iJj jO, \ 7 b^ ' t- \ \ ~ ~1^ oV - ' ' Y' \ \x' y^r- \2. \ d^ A \ Fig. 20C 8 and 9. The down cut and ihc side cut for gutter boards. Draw an enlarged plan and ele- vation as in Fig. 202. Rotate the sides osvp and ostq, succes- sively, on OS as an axis, until the required angles are parallel to //. This gives the angles marked 8 and 9 in the figure as the true sizes. 10. The angle between the gutter boards, at the miter. As shown in Fig. 202, the face of one of the boards, osvp, is a plane per- pendicular to I" and inclined to // ; while the face of the other l)oard, opxj, is a plane parallel to HA. The line of intersection of these planes is op. In Fig. 203, the planes .Y and Y, and their line of intersec- tion ab, reproduce the condi- tions of Fig. 202. The angle between X and Y being found, according to the principles of § 101, the equal angle which exists between the gut- ter boards is de- termined. In the figure, this angle is numbered 10. X, § 190] APPLICATIONS OF CHAPTER VI 223 190. A Sheet Metal Chute. The metal chute shown in Fig. 204 (1), is fastened to the I beam by means of I olts that pass cz. Fig. 204 through a cHp made from a bent plate. Given the size and ■SECTIOK Fig. 205 position of the chute and the beam, it is required to determine the developed pattern of the clip, and the angle of the bend in it. 224 DESCRIPTIVE GEOMETRY [X, § 190 In Fig. 206, that part of the chute on which the chp is to be fitted is redrawn at large scale, in plan and elevation. Let it be assumed that the clip is to be so designed that its edges lie in the same vertical planes with the edges of the beam b' flanges. The lines eg and fh are established on this basis, the lengths being taken at any reasonable amount, and gh being made parallel to ab. The projections of the bottom face of the clip are now established. The lower face of the clip is to coincide with the side of the X, § 190] APPLICATIONS OF CHAPTER VI 225 chute and with the top of the I beam ; hence the angle of the bend will be determined, if the angle between the planes of the top of the beam and the side of the chute is found. Let the top of the beam flange be taken as the H plane, establishing HA as shown. The H traces of ah and cd are at j and h re- spectively. These points determine the line jm which is the H trace of the side of the chute. The line hi, a part of jm, is the line along which the clip must be bent. Now let ZZ^ be drawn to represent the H trace of a cutting plane, passed perpendicular to jm. Such a plane will cut the side of the chute in the line no, and the angle ony ^_______^ will be the supplement of the required angle for V®"^ \ bending the clip. The true size of this angle is / ^^n; ] found by rotating the triangle nop on the side np hi /^ until o falls in H, at o^. (o'^o^ is perpendicular to / '? - o / ZZ^ and is equal to p^'o^\) yig. lu7 ' To find the development of the clip, let the lines kg and Ih be rabatted about jm, into H, giving e^fH^h'^g'^lc^ as the development of the under side of the clip. By rabatting s, in the same manner, the line y^s'^' can be established to show the line of contact between the edge, cd, of the chute, and the clip. In Fig. 207, the development of the clip is redrawn and the bolt holes are located. The exact location of the holes and the necessary allowances for bending are matters of technical detail that need not be discussed here. It should be noted, ^ however, that if the edges of the clip are to be perfectly vertical planes, the sides kg and Ih must be slightly beveled before bending. CHAPTER XI APPLICATIONS BASED ON CHAPTERS VII AND Villi 191. To Determine the Appearance of the Grain on a Turned Wooden Vase. Let a-h, Fig. 208, be the block from which the vase is to be turned. Let the annual rings be assumed to take the form of cylinders, as shown on plan by the concentric circles, 1-8. Let it be required to trace the emergence of these rings, on the surface of the vase, as shown on the elevation. Any horizontal cutting plane, as A, will cut the vase in a circle as shown at m^'-x-f'- on plan. On the front side of the vase, this surface circle cuts the cylinders forming the grain, at points j, k, J, and m. These points are determined on plan and projected to the eleva- tion. Other cutting planes would give other points. By joining, on the elevation, all such points which result from the same grain cylinder, the appearance of the grain can be established readily. Where a part of the vase consists of a ^•ertlcal cylinder, the grain lines will be vertical straight lines. The points where 1 Exercise sheets to accompany this chapter are given on pages 300 304. 226 Fig. 208 XI, § 192] APPLICATIONS OF CHAPTERS VII-VIII 227 any given line crosses the meridian section, as n and o, can be determined exactly by noting on plan the intersection of the meridian plane, M, and the grain cylinder under consideration, No. 5. The point of tangency between the surface circle A and the grain cylinder 6, as seen on plan at m^, indicates that on the elevation the grain line just touches the plane of the surface circle. 192. A Spiral Chute. Figure 209 shows a spiral chute made of sheet iron for lowering packages. The vertical side forms Fig. 209 a portion of a cylinder. The bottom is a right helicoid, gen- erated by an horizontal line connecting two helixes of different diameters. The slope of the inner helix is much steeper than that of the outer one. Light packages descend along the steep inner edge, while heavier ones, under the effect of centrifugal force, work themselves out toward the lower gradient of the outer edge. The drawing of the plan and elevation involves the construc- tion of several helixes (§ 137). A developed pattern may be constructed for the vertical side following the principles of § 157. The bottom, being a warped surface, may only be ap- proximated by development (§ 161). 228 DESCRIPTIVE GEOMETRY [XI, § 193 193. Intersecting Vaults. I. Given the Cross-Section of two Intersecting Barrel Vaults, to Determine the Line of Intersection. A barrel vault is a hollow cylinder, the cross section of which may be of any desired shape. The most usual form of section is semi- circular, though elliptical, three centered, and pointed arch forms are not uncommon. Figure 210 shows two intersecting semicircular barrel vaults, drawn without thickness. These vaults intersect in the space curves ahc and a'b'c'. The problem of determining such a curve, on plan and elevation, is merely that of finding the line of intersection of two cylinders, as described in §§ 149 and 152. Figure 211 shows the plan and two elevations of intersecting vaults similar to those shown in Fig. 210. Here the thickness is indicated. The line of intersection of the outer surface of the shell is determined first. Cutting planes parallel to H are used. One such plane, X, is shown in the figure. This plane cuts the large vault along two elements which are projected on elevation in the points d'' and /' . These elements are shown on plan at d^e'' and f^g^. The same cutting plane, X, also cuts the smaller, or trans- verse vault, along two elements. These elements are projected in the points h^ and k^ on the end elevation, and in the lines h'' and / and JcH^ on plan. These two pairs of elements are in Ae same plane, A^, and are not parallel. Hence they will intersect. The points of inter- section, m, n, o, and p, arc on the surfaces of both vaults, and XI, § 193] APPLICATIONS OF CHAPTERS VII-VIII 229 hence are points on the hne of intersection of the vaults. Other cutting planes, parallel to X, would give other points, helping to determine the curve amhnc. The line of intersection of the inner surface of the vaults is shown as a dotted Hne. This line is determined in the same manner as above. The cutting plane, X, locates the four Fig. 211 elements which are shown by the dotted lines. These elements in turn locate the four points qrst. The two curves thus determined are, in general, some form of space curve. When the intersecting vaults are semicir- cular, and their centers are at the same height, the plan of the hne of intersection is an hyperbola. If the vaults are of the same diameter, and their centers are at the same height, the line of intersection becomes a pair of ellipses, and the plan becomes a pair of straight lines. 230 DESCRIPTIVE GEOMETRY [XI, § 193 II. Given one Vault and its Line of Intersection with Another Vault, to Determine the Cross-section of the Second Vault. In Fig. 212, the intersecting vaults are shown as without thickness. Let the plan and the front elevation be given in full, and let it be required to establish the cross section as shown on the side elevation. The cutting plane, Z, determines the element mn on the larger vault and hf and dk on the smaller vault. These lines, pro- PUN Fig. 212 jected to the side elevation, locate the points h^ and d^ which are two of the points required to establish the curve a^h^c^dPe'^. 194. Mitering of Mouldings. A moulding is a combination of cylindrical and plane surfaces. Its right section or profile is therefore a combination of straight and curved lines. Plaster and cement mouldings are formed by a knife which is slipped along on straight guides and in contact with the plastic material. In running a wooden moulding, the knife revolves while the material is shoved past it, on a straight guide. XI, § 194] APPLICATIONS OF CHAPTERS VH-VIII 231 If a moulding is run on a curved surface, it becomes a combina- tion of cylindrical and double-curved surfaces. Since such cases are relatively rare, the following discussion is limited to mouldings having straight elements. The principal prob- lems on intersec- tions of mould- ings depend on the ideas stated in §§ 138-152. If a piece of moulding is cut by planes making equal angles to right and left with its elements, as shown in Fig. 213, the sections a and b are equal. Therefore the end pieces, A and B, placed together as shown at (2), will match exactly, forming a miter. In Fig. 213 (2), the elements of the moulding are turned Fig. 213 through 90°. This is the usual case, but any desired angle may be secured at the miter by varying the angle between the cutting planes and elements, as shown in Fig. 213 (3). If the cutting plane is inclined, not only toward the elements, but also toward the plane of the base, as shown in Fig. 214 (1), the mouldings will miter as shown in 214 (2). Besides mitering, mouldings may also be brought to an intersection by coping, as shown in Fig. 214 (3). The appear- ance as seen from the finished side is the same in either case. 232 DESCRIPTIVE GEOMETRY [XI. § 194 Fig. 215 Mouldings that have different profiles may be made to meet in various ways under varying conditions. The cases described in §§ 195-197 are selected as being typical. 195. To Find the Line of Intersection of Two Mouldings of Different Profiles. In Fig. 215, let the plan and the profiles 1 and 2 be given, and let it be required to find the plan and the elevation of the line of intersection. A series of horizontal cutting planes, A, By C, is used to estab- lish a corresponding series of elements on plan and on elevation. Thus the plane D determines the element m on moulding 2, and element m' on moulding 1. On plan, these two elements are seen to intersect at m"^. By projecting up from m"^ to the plane D, on the elevation, m"^ is determined. One point on the line of intersection is now estab- lished. The principle involved is identical with that of § 193. 196. Raking Mouldings. Fig- ure 216 shows a small section of a typical classic pediment. Two types of mitering are illus- trated. Mouldings a and h have the same profile, and miter on a 45° plane at the corner, in the usual manner. The joining XI, § 196] APPLICATIONS OF CHAPTERS VII-VIII 233 of c and d, however, involves a different principle. The level moulding, c, is made not only to turn the corner but also to change its inclination to H at the same time. The moulding c is called the level moulding, while d is called the rake moulding. The intersection of a level moulding with a rake moulding is usually managed in one of two ways. These ways are illustrated in Fig. 217. At (1), two blocks, m and n, are shown mitering in the usual manner. At (2), the same blocks have been tipped, rotating on cd, so that n is in the position of a raking moulding. The blocks have the same section as before. They still meet perfectly on the mitering plane, but the mitering plane is no longer at an angle of 45° to T^, and 90° to H. More- over, the horizontal and vertical faces of m are thrown into slop- ing positions. The effect is unpleasant, since the level moulding has a greater projection than the rake moulding, and the miter- ing plane does not bisect the angle of the wall. A more difficult but more effective method is shown at (3). Here the block m remains in the same position as in (1), and the mitering plane is passed in the usual manner. The cross section of the block ?i' is determined so that it will meet accurately with m, while following the required rake. In this case, vertical sur- faces on the level moulding intersect perfectly with vertical sur- faces on the rake, though the widths are different. Horizontal surfaces, however, do not intersect so well, as will be shown later. ^ 1 Classic examples involving rake moulding are usually worked out on the principle of Fig. 217 (3). It is interesting to note, in these examples (see Fig. 216), how carefully the intersection of all horizontal surfaces of the level moulding with corresponding surfaces on the rake moulding has been avoided. 234 DESCRIPTIVE GEO:\IETRY [XI, § 196 The method of determining the profile required for a rake moulding is quite simple when the angle of the wall, ejg, Fig. 216, is 90°. When the angle of the wall is not 90°, the solution is more complex. Both cases are given in the following article. 197. To Determine the Profile of a Rake Moulding. I. When the Angle of the Wall is 90°. In Fig. 218, let the profile of the level moulding be given as ahfg, and let the pitch of the rake be given as (9. Let the angle of the wall, fal*m^, and the trace oi the mitering plane, AA^, be drawn. Next draw the plan of the element aj and as many others as are deemed necessary. These elements intersect in the mitering plane at the points a, b, etc. The plan of the rake mould- ing can now be drawn from a^, h^, etc., and parallel to a^m^. The elevation of the element am is drawn next, at the required pitch, the other elements being parallel to it. Thus the plan and elc\n- tion of the rake moulding are determined. It remains to deter- mine the right section. Pass the plane T perpendicular to the elements of the rake moulding. This plane cuts the moulding in the required section, the V projection of which is Jq. Rabatte the points l-q into H, as shown in the figure. The points thus found determine the required curve. The right angle at/, on the level moulding, corresponds to an XI, § 198] APPLICATIONS OF CHAPTERS VII-VIII 235 obtuse angle on the rake moulding. In large surfaces this would probably be unpleasant. (See footnote, page 233.) II. When the Angle of the Wall is not 90°. The prin- ciple involved in this case is the same as for case I. The method differs only in detail. The plan and elevation of the rake moulding are established in the same manner as in the former case. Next a plane is passed perpendicular to the elements of the rake moulding (§ 81). This plane will cut the required right section from the moulding. Next find where each element of the moulding pierces the cutting plane (§78). By connecting the points thus found, curves are determined w^hich are the projections of the required right section. The true shape of the right section can then be found by rabattement (§ 91). Fig. 219 198. Domes and Pendentives. Usually a dome is some form of double-curved surface, spheres and spheroids being the more usual forms. The simplest case is that of a hemisphere sup- ported on a cylindrical case, as in the Pantheon at Rome. It is often necessary, however, to support a dome on a base which is a square, octagonal, or other form of prism. When this happens, some special means must be employed to make the transition between the dome and its supporting walls or piers. In Fig. 219, one of the simplest cases is shown. Here the diameter of the dome is made equal to the diagonal of the square base, and the planes of the base are allowed to pass upward, cutting the dome in the circular arcs, ab, he, etc. If 236 DESCRIPTIVE GEOMETRY [XI, § 198 it is desired to reduce the supporting walls to four piers, it can be done by substituting, in place of the walls, four arches, sprung between the piers, as shown in Fig. 220. So far as the dome is concerned, the shape is the same whether it is carried on walls or on piers and arches. An example of this type, slightly modified, is to be found in the Baptistery at Ravenna. Another method of supporting a dome over a square plan is shown in Fig. 221, Here the diameters of the dome and of •HALf PIAN LOOKING UP- Fig. 220 the square base are equal. This construction leaves the spaces ahc, etc., which are not covered by the dome, and the dome is not supported by the base along the lines ac, etc. The necessary coverings and supports are furnished by filling in these spaces in one of several ways. These filling and supporting parts are called the pendentives of the dome. In Fig. 221 the pendentives are parts of a sphere whose di- ameter is equal to the diagonal of the plan. (See dbe, Fig. 219.) The well-known dome of Santa Sophia in Constantinople is supported on pendentives of this type. Sometimes the support of a dome is in the form of an octagon, XI, § 199] APPLICATIONS OF CHAPTERS VII-VHI 237 carried to a square plan by means of plane, cylindrical or conical pendentives, or merely by corbeling. The jointing planes of stonework are kept normal to the exposed faces, in so far as possible, in order to avoid thin edges on individual stones. It naturally follows that the horizontal joints in a spherical dome are formed along the sur- faces of cones whose apexes are at the center of the sphere, while the vertical joints follow meridian sec- tions, as shown in Fig. 221. 199. The Decorative Treatment of Curved Surfaces. Decoration may be classed as geometrical when it is composed of regular lines, repeated in a set manner, as in diapers, frets, over all patterns, etc., or as free, when it represents figures or other natural forms, without repeats. Geometrical decoration usually follows some element or other char- acteristic line of the surface to be decorated ; hence it can be accu- rately shown at small scale in the usual manner. The more naturalistic the forms used, however, the more difficult is any adequate representation at small scale. The cartoons for free ornament must then be made at full size, by skilled artists. Hence there is a tendency to have such work done in studios, on flexible backings of paper or canvas, these being later set in place complete. Of course, this manner of execution is only possible when the surfaces to be treated are developable. (See also § 144.) In designing the decoration for a developable surface, the Looking down— ^Looking up- Fig. 22 i 238 DESCRIPTIVE GEOMETRY [XI, § 199 development is drawn first, the pattern being worked out on the flat, so that they fit it perfectly and repeat or work out all around as may be desired. For double-curved or warped surfaces, the decoration, if geometrical, may be designed by using a plan and elevation or section, though it is somewhat difficult. But for naturalistic Plan - looking up. Development of one half larger voult. Fig. 222 representations, it is necessary to w^ork on a three dimensional model, either at scale or in full size, unless a satisfactory ap- proximate development (§§ 157 and 161) can be made. 200. To Show a Geometric Pattern on a Barrel Vault. Let two intersecting barrel vaults be given, as shown in Fig. 222, and let their line of intersection, ahc, be found by § 193. Next let the development of both be drawn, as explained in § 157. It will be noticed that the outlines of the penetration, a'h'c'. XI, § 200] APPLICATIONS OF CHAPTERS VH-VHI 239 and a"h"c" , are not of the same form in development. Ob- viously the two lines must be of equal length. In selecting a unit for the repeated pattern, it should be borne in mind that the size chosen must be common divisor of the circumferences of the two vaults. In the figure, the diagonal of the square forming the pattern is contained six times in the smaller and eight times in the larger vault. Now let the desired pattern be drawn on the developments, and let guide lines, as x'y' and y'ci , be drawn to assist in transferring the pattern to the plan. These guide lines, being elements of the surface, can be located on the plan, as at xy and 'pq, and their intersections can then be used as guide points for drawing the pattern. In the illustration, the main lines of the pattern, being diagonal on the development, appear as helixes on the plan. It will be noticed that the pattern chosen does not work out well along the penetration line. This is due to the fact that the curves, a'h'c', and a"h"c" , Fig. 222, do not cut the patterns at corresponding points. ]\Ioreover, it is improbable that any set pattern could be devised that would connect properly along the given line of intersection. This difficulty can be minimized by introducing a plain band along the penetration, as at ah in the illustration. The particular pattern used in the illustration is the same as that in the Baths of Caracalla and in the Pennsylvania Terminal in New York. It would be possible to determine the cross section of the smaller vault in such a manner that a pattern could be worked out on it which would be similar to that on the larger vault, and the two patterns would intersect perfectly on the pene- tration. Such an attempt, applied to the case shown in Fig. 222, gives for the outline of the smaller vault a pointed curve, awkward in line and quite impossible to use. The shape of the vaults, being the more important, will ordinarily be de- termined without reference to the ornamentation. 240 DESCRIPTIVE GEOMETRY [XI, § 201 201. Warped Surfaces. In abstract problems, warped sur- faces can be described best by means of their directing lines or surfaces, in one of the combinations given in § 161. But in practice it is often convenient to substitute a time element in place of one of the space elements required to define the surface, by specifying the rate of motion of the generatrix. Thus Fig. 223 shows the plan and elevation of a recessed doorway with a segmental head. The curves ad and be can be used as directrices for a warped surface covering the recess. The lines ab and cd evi- dently must be elements of such a surface. The other elements may be determined by setting the condition that the rate of motion of the generatrix, be- tween the positions ab and cd, shall be uniform. This results in elements spaced at equal in- tervals along each curve, i.e. ae = cf=fg, and bh = hi=ij, etc. If such a surface is cut by any plane, as Z, which cuts all the elements, the line of intersection, ac, may be used as a directrix. This new directrix, taken with the other two, will define the surface quite as well as the rate of motion described above. However, the method using the rate of motion is usually more convenient. The problem which arises most frequently in connection with warped surfaces is to fix the character of the intervening surface when two directrices are given as limits. One such case has already been shown in Fig. 223. Another is shown in Fig. 224. Here a semicircle and a semi-ellipse in parallel planes constitute two directrices for the surface. The case is such as might arise in designing a splayed archway, when the height Fig. 223 XI, § 201] APPLICATIONS OF CHAPTERS VII-VIII 241 of the opening must be the same inside and out, both at the crown and at the springing hne. Four methods of generation are shown and others are pos- sible, of course. In order to assist in a clear conception of the difference between the surfaces shown, an auxiliary section has been drawn, inside the wall. Case III would be impossible if Direcirices •. A- ,,, ' • ■ - J'. ,; abc ,cief /Vr' r«\y_ abc , def ghi , kim _"/ v-^^^'.;, ghi , kirr, and op." /'!','.' i " '•~.vV' qnd -ox and H. /u' > ."A and uniform -''//',' 1 \i.,. , ' • I ,^^"" , , i\«^ rareoTmo^- ; :/''/;; :j ; ';>. Auxiliar:^ verticil ; >''//; ', '• VV. ion on each. , /,,'/. ' I , I ,'• . '' cuTTinq plane, ^r^ i ;'. / - ' f ' > !>• i ..^ X','; ' , \ - Y ' — 1-tr /'/ ' I ! .t _ *^^^ — ':h^ the crowns of the two curves were at different levels, but the other cases are possible for any combination of arches. It is not at all unusual to run one warped surface into another, the two having one common element but quite different direc- trices. One such case is shown in the covering over the recessed gateway in Fig. 178. When this is done care should be taken to insure that the surfaces join smoothly. This can be done by passing an auxihary plane through the part in question to see if it gives a smooth curve of intersection. CHAPTER XII SHADES AND SHADOWS » 202. Introduction. Except to the trained eye, the elevations of a building convey no adequate picture of the proposed structure, since the modeling, i.e. the projections and recessions from the main plane of the fa9ade, is nowhere shown on a single drawing. An actual building, viewed through a fog or in a very diffused light, shows much the same quality of flatness as an elevation drawing. But in sunlight the modehng at once becomes clear because of the contrast between unequally lighted surfaces. In preparing drawings for exhibition, it is usual to imitate, as clearly as may be, this effect of sunlight and shadow, in order to assist the layman to a correct understanding of the work in hand. Hence the ability to correctly indicate the shades and shadows on a drawling is an important part of the draftsman's equipment. This chapter is not intended to give a complete knowledge of the casting of shadows. It is intended merely to lay a sound foundation for a more extended course by emphasizing the fact that shadow-casting is merely a special application of De- scriptive Geometry. For this reason short cut methods have been avoided and in many cases the text is abridged by refer- ences to the basic problems that have been explained in Chapters I to VIII. 203. Fundamental Ideas and Definitions. The factors involved in any pro])lem in shades and shadows are: (1) a source of light; (2) a body intercepting the light; and (3) a 1 Exercise Sheets to accompany this chapter are given on pp. 304-309. 242 XII, § 203] SHADES AND SHADOWS 243 receiving surface on which the shadow falls. These factors are illustrated in Fig. 225. From the source of light L, issue rays that illuminate the sphere and the plane P, beyond it. Those rays of light which fall on the sphere are intercepted by it and do not fall on P. All such rays are included within a cone tangent to the sphere, whose apex is at L. The frustum Fig. 225 of this cone which lies between the sphere and P, is a dark space called the umbra. The line of tangency between the cone and the sphere is called the shade line, or separatrix, and all that part of the sphere which lies within the umbra is said to be in shade. The inter- section of the receiving surface with the umbra defines the shadow. It is well to note carefully the following distinction : a surface in shade is a part of the intersecting body which is turned away from the source of light, while a surface in 244 DESCRIPTIVE GEOMETRY [XII, § 203 shadow is a part of the receiving surface from which the Hght is excluded. (See Fig. 225.) In some cases (see Fig. 230), a complex body may be partly in shade and partly in shadow. In this book, we cannot deal extensively with the problem of intensity of illumination. Suffice it to say that the intensity of illumination will be greatest on those surfaces which are normal to the rays of light, and zero on those surfaces which are tangent to the light rays. All conceivable intermediate cases exist. Shaded surfaces are usually illuminated to some extent by reflected light. 204. Source of Light. In such a case as that shown in Fig. 225, it is evident that if the source of light is moved, the shadow Fig. 226 will change its form or its position or both. If the source is moved to an infinite distance, the tangent cone becomes a tangent cylinder, and the rays of light are all parallel. This is what happens, in effect, when the sun is taken as the source of light. In order to simplify the problem of casting shadows, as well as that of interpreting shadows, it is usual to assume the sun, in a fixed position, as the source of light. The conventional position selected is above, in front of and to the left of the object which is being drawn, the precise position being such that any ray of light follows the diagonal of a cube, whose faces are parallel to the planes of projection, as shown in Fig. 226'. It XII, § 205] SHADES AND SHADOWS 245 will be noted that each of the projections ^ of such a ray makes an angle of 45° with HA. Thereby two things are accom- plished; first, in drawing any ray of light either on plan or elevation, the 45° triangle may be used; and second, the shadow cast on any object by a projecting part will show, by its width, the exact amount of the projection. Thus in Fig. 230, the width of the shadow, 1, is equal to that of the projection, 2. In special cases, for special reasons, the position of the sun may be varied at will from the above, or any nearer source of light may be chosen. In the following problems the source of light is varied in order to avoid the tendency toward the use of rules and formulas. 205. Fundamental Analysis. The rays of light tangent to any object form an envelope (§ 142) for that object. The line of intersection between this envelope and any surface receiving light, is the outline of the shadow. In the case of sun shadows, the envelope is a cylinder, a prism, or some combination of cylinders and prisms. For near-by sources of light the envelope is a cone, a pyramid, or some com- bination of cones and pyramids. It follows that the determination of the shade and shadow for any object is a problem in Descriptive Geometry. Xo special rules or formulas are needed in any case. However, in many special cases, the solutions become complex and tedious. For such cases special short cut methods may be established. These methods, however, are properly the subject for a separate course in shades and shadows. For the problems which follow, it is sufficient to determine the umbra by lines, planes, cylinders, cones, etc., that pass through the source of light tangent to the body. The intersection between this umbra and the receiving surface is found by the usual methods for finding intersections of surfaces. 1 Let the student determine, both analyticallj' and graphically, the true angle that such a ray makes with H or T'. 246 DESCRIPTIVE GEOMETRY [XII, § 206 ] / 206. Shadow of a Point. I. When the Source of Light is Close to the Body. Let the // and V planes be the receiving surfaces. In Fig. 227, let p be any point in space and let / be a source of light. The shadow of p is found by passing a ray of light through / and ;;. This ray is represented by the line Ip, whose trace (§31) falls on V, at the point s. The portion of the ray of light between p and s represents the umbra, and s is the shadow of p, on V. If the source of light is moved to I', the shadow of p falls on H, at s'. II. The Sun Shadow on H or V. If the sun is taken as the source of light, the only change from the preceding method lies in the manner of drawing the light ray. In this case, each of the projections of the ray is drawn at 45° to HA, as shown by the dot and dash line, the shadow falling on H, at s". 207. Shadow of a Line. The shadow of a line is determined by the shadows of any two of its points. If the shadow falls across HA, the problem lies partly in a quadrant other than the first. In Fig. 228, let I be a Source of light, and let it be required to find the shadow of ab. The shadow of a falls on H, at a\ The shadow of b falls on V, at b\ The line connecting these two points will be broken at HA. The direction of the shadow line may be established by casting the shadow of b as it would fall on // if it were not intercepted by V. In the figure this point falls at b^'. The line a'//' is the shadow which ab would cast if V were a transparent plane. Of this line, the portion a^o is the visible shadow on H, cast by a part of the given XII, § 209] SHADES AND SHADOWS 247 line ah. The remainder of the Hne ab casts its shadow on V, in the Hne b'o. 208. Shadow of a Plane Solid. The shadow of a plane solid is determined by the shadows of its vertices or of its edges. In Fig. 229, the cube a-h is illu- minated from the point /. The method of casting the shadow is found in §§ 206 and 207. It will be noted that the shadows of a and g fall within the area put in shadow by the cube. Thus the shadow of any line terminating at a or g, as ad,fg, etc., also falls within the shadow and does not affect the outline of the shadow. Any such line, on the cube in space, lies between surfaces, both of which are in light or in shade. Lines on the cube which lie between one surface which is in light and another which is in shade, as cd, bf, etc., together constitute the shade line of the solid and the shadows of these lines together limit the shadow. 209. Shadow FalUng on Different Planes. If the surface of any solid contains projections or recessions from the main face, as shown in Figs. 230 and 231, then the parts farther back will receive, in part or in whole, the shadows cast by the projections. Let it be required to cast the sun shadow of the body shown in Fig. 230. The shadow of the line bg will fall on the surface, cdef, while the shadow of de falls on V. The shadows of cb and bg are determined in the usual manner 248 DESCRIPTIVE GE0:METRY [XII, § 209 except that the traces of the light rays are found on the plane cdcf, rather than on H or on V. It will be noted that the width of the shadow 1, is equal to the depth of the projection 2. Also, and for the same reason, 3=4 = 5. The preceding example, and the one which follows, should be studied carefully to fix in mind the characteristic shadow of: (1) a line parallel to HA ; (2) a Hne perpendicular to H ; and (3) a line perpendicular to F. Again notice that the umbra formed by a line is a plane, and that the traces of that plane constitute the shadow of the line. In Fig. 231, a more complex case is shown. The shadow of cd falls on a surface perpen- dicular to V, hence it is invisible on the ele- vation. The shadow of de is, in general, a continuation of a'c% but it is a broken line, owing to the irregular surface on which it falls. The shadow of the point h touches the shade line mn at h% and continues, in the shadow^ of that line, to h^'. Similar cases occur at /', f and k% and in general, wherever a shade line is crossed by a shadow. Such a point is known as a point of loss. The line f'g is a shade line, the umbra of which coincides wath the sur- face gopq ; hence this surface is dark. The rest of the shadow is easily found in the usual way. 210. Use of the Profile Projection. It is obvious that if the shadow of any object is to be deter- mined, all three dimensions of the object must be known. In Fig. 231 XII, § 210] SHADES AND SHAD(3WS 249 general, two drawings are required to give the necessary informa- tion. For the most part, shadows are used in connection with elevations or sections. The plan is used merely to assist in the work, and the shadows on it are not often drawn in. Sometimes it is convenient to use three or more drawings, or the shadows on all drawings may be required. Again, the plan may be discarded in favor of a second elevation or section. The particular method selected for a given problem is wholly a matter of convenience and of the desired result. The only necessary condition, com- mon to all cases, is that the drawings used must together fully v//////////////////////y//////////////. Vy Z Fig. 232 describe all three dimensions of the object. Shadows cannot be cast on an elevation without the help of a plan, another elevation, a section, or other equivalent data. In Fig. 232 are shown two elevations of a dormer window on a steep roof, illuminated from the point L. The shadow will fall on the roof, hence the traces of the light rays on the plane of the roof are required to determine it. Points 1, 5, and 6 lie in the roof plane, hence they will coincide with their own shadows. The shadows of the other points are found in the order in which they are numbered. Points 5, 7, and 8 are 250 DESCRIPTIVE GEOMETRY [XII, § 210 auxiliary points used to determine the directions of the shadows of the lines on which they occur, though they are not on the outline of the required shadow. 211. The Shadow of a Circle. A circle illuminated from a near-by point of light casts an umbra in the form of a frustum of a cone, the apex of which is the source of light. The shadow Fig. 233 of the circle on any plane will be the conic section cut by that plane from the umbra. In Fig. 233, the circle ahd is illuminated from the point p. The plane of the circle being parallel to V, the shadow on V will be a circle, since the circle itself and its shadow are parallel sections of the same cone. The center of the shadow is deter- mined by casting the shadow of o, which is found at o\ The radius of the shadow is determined by casting the shadow of any point on the circumference, as a. With the radius o'a" the shadow on V can be drawn. The shadow on // is determined by casting the shadows of XII, § 212] SHADES AND SHADOWS 251 several points on the circumference, as bed, and drawing a curve through the points thus found. The precise nature of this curve depends on the relative positions of the circle and the source of light. The shadows on V and H will intersect in HA, at X and y. 212. The Shadow of a Cone. I. When the Source of Light is Near the Cone. Let the cone ao and the source of light p be as shown in Fig. 234. The rays of light passing from p tangent to the cone, form two planes which pass through p and which are tangent to the cone. The intersec- tion of these planes with H and V will give a part of the required shadow. The balance of the shadow is cast by the base circle. To pass the planes through p tangent to the cone, an auxiliary line pa and an auxil- iary cutting plane H' A' were used, as described in § 154. In this manner the traces of the planes X and Y which limit the shadow were found. The shadow of the base circle was found as described in § 211. The shadow of the base circle should be tangent to the traces of the planes. The lines of tangency betWeen the cone and the plants X and Y (for example, ac) are the shade lines on the cone. The point c may be established by projecting back from c^ to c^. Fig. 234 252 DESCRIPTIM^ GE0:METRY [XII, § 213 II. The Sux Shadow of a Coxe. In Fig. 235, the sun shadow of the cone base is the circle 6*c*c?V, and the shadow of. the apex on the same plane is a^ Lines from a^\ tangent to the circle give the H traces of the shadow planes. The V traces of the shadow planes pass through the shadow of the apex. Let the student determine how the sun shadow of the right circular cone, resting on H, varies as the angle of inclination of the elements of the cone varies between 0° and 90°. 213. The Shadow of a Sphere. The shadow of a sphere cast by a near-by source of light can be found by passing a cone tangent to the sphere, as described in § 173, and then finding the intersection of this cone with the planes receiving the shadow. (See Fig. 225.) If the sun shadow is required, substitute for the cone in the para- graph above a cylinder whose elements are parallel to a light ray. 214. To Cast the Shadow of a Straight Line on a Curved Surface. I. First Method. The umbra of a straight line is a plane. This plane will cut the given surface in a curved line, which is the required shadow. If the source of light is a given point, pass a plane through the point and the given line, as in § 69, and find the inter- section of this plane with the given surface, as in § 149. It will frequently happen that the traces of the light plane do not fall conveniently for this solution. In that case the following method may be used. Fig. 235 XII, § 214] SHADES AND SHADOWS 253 11. Second Method. The method given here is analogous to the shcing method of § 149, and furnishes the basis for the solution of the following two problems. Let the hemisphere and the line ab, Fig. 236, be given, and let it be required to find the sun shadow of ab on the hemisphere. First cast the shadow of ab on H, as a'b'. Since the hemi- sphere rests on H, the points c and d, being common to the shadow and the surface, will be the terminal point: of the required curve. Now pass a cuttin;: plane parallel to H, as shown by H'A', and cast the shadow of ah on this plane, as a^' b'' . This plane cuts tht circle men from the hemisphere. The in- tersections, e and /, of this circle with the shadow a^'b^', give two more points on the re- quired curve. Other points are established by means of other cutting planes. The shade line ghk is established by using the projection on a r' plane parallel to a light ray as indicated at H" A" . Using this view, we draw the light ray hs tangent to the sphere, constructing the angle d as shown in Fig. 226. This line is the top element of a cylinder tangent to the sphere (§ 213). The circle of tangency between the sphere and the light cylinder is ghh, and its shadow in H is gsh. The yoint of loss (§ 209), q, casts its shadow at the intersection of a'b' and g^s^k^. 254 DESCRIPTIVE GEO:^IETRY [XII, § 215 215. To Cast the Shadow of a Curved Line on a Curved Surface. Let the curve be made the directrix of a cyhnder, or of a cone in the case of a near-by source of hght. This cyhnder or cone defines the umbra and its intersection with the given surface defines the shadow. In Fig. 237, let the hemisphere o, and the circle c, be given and let it be required to find the sun shadow of c on o. Light rays tangent to c at the usual in- clination define the umbra, whose intersection with o is determined as in §§ 149, 177. Vertical cutting planes are used in this instance. In the figure, one such plane is drawn, the resulting points on the shadow being a and b. The analysis used in the second method of § 214, namely that of casting the shadow of the given line on successive planes, may be used in this problem without change in the figure. Figure 238 shows a niche composed of a half of a cylinder and a quarter of a sphere and illuminated from the point L. The shadow within the niche is cast by the line abed. The shadow begins at the point of tangency, c, between the curve bed and a ray of light through L. Succeeding points on the shadow are found by casting the shadow of bed on a series of vertical planes, as explained in § 214. One such plane is drawn. Here the line cut from the niche is fghj and the shadow of abed on V is khin. The point e is the shadow of the point p. The line bg casts its shadow in the straight line b'q\ Fig. 237 XII, § 216] SHADES AND SHADOWS 255 r->in 216. The Shade and Shadow on a Double-Curved Surface of Revolution. The shade Hne, or separatrix, on a double-curved surface of revolution is the line separating the lighted from the unlighted portion of the surface, as shown in Fig. 225, and as defined in §§ 203 and 205. The cone of rays shown in Fig. 225 is easily determined in the special case of the sphere, L' as described in § 213. When the surface in question is more complex, the cone of rays is not so easily deter- mined and other methods become necessary. Three methods are given below, each of which has special merits, but none of which is satisfactory for every case. All the methods de- pend upon locating a large number of points of tangency between straight lines and h — « curves. Therefore the de- gree of accuracy for a given surface depends upon choosing a method which w^ill yield curves whose tangents can be readily located. Curves which turn sharply at the point of tangency are preferable. In most such problems, it probably will be necessary to use a combination of two or more of the methods described below. For sun shadows, special methods can be devised which are much less laborious than any given here, since the rays of light may be assumed to be parallel. These methods are to be found in any standard text on Shades and Shadows. However, the methods given here may be adapted to the case of sun shadows if desired. Fig. 238 256 DESCRIPTIVE GEOMETRY XII, § 216 i.r Fig. 239 XII, § 216] SHADES AND SHADOWS 257 To facilitate this study, the student should review §§ 212- 213. The problem of determining the shade and shadow of a cone is given in § 212, with three important possible cases. The shadow of a cylinder and that of a sphere are treated in § 213. It is also important to realize that the shade line on any double-curved surface of revolution will be a continuous curved line, whose shadow will determine the shadow of the surface. As an aid in visualization, it may be noted that any double-curved surface of revolution may be considered as approximated by a series of cone frustums and that the shade lines on these frustums will indicate, in a general way, how the shade line will appear on the surface of revolution. I. The Envelope Method. This method is based on the slicing method of § 149, and on the envelope method of § 174. It consists in first casting the shadow^, and working back from that to the required shade line. In Fig. 239, let the baluster be illuminated from the light /. Let a series of horizontal cutting planes, Q, R, S, etc., be used to determine a series of circular sections. Cast the shadows of these sections (§ 211) on the H plane, obtaining the circles whose centers are q%r%s%t% etc. Draw the envelope of these shadows. This envelope outlines the shadow of the baluster, and hence the shadow of its shade line. In order to determine the shade line on the baluster, the points of tangency between the envelope and the shadows of the circular sections, as a, b, c, etc., may be used. Pass a light ray through each such point and through /^. It will cut the corre- sponding circle on the plan of the baluster in a point which is on the line of shade, as a^',b^,c^. The points m and 7i, where the envelope line passes within the shadow, are points of loss (§ 209), on the baluster. This method will serve quite well for simple surfaces of small dimensions. For surfaces of slight curvature it is perhaps the best of the three methods given in this article. 258 DESCRIPTIVE GEOMETRY [XII, § 216 II. The Tangent Cone Method. This method depends upon the fact that if a cone is passed tangent to any double- curved surface, as in Fig. 240, the shade line on the cone, at the circle of tangency, will give a point s on the separatrix of the surface. In Fig. 241, the shade line on the torus is determined in this manner. The source of light is at /. Let the horizontal plane ^yl be passed through the torus, cutting a circle from the sur- face. Let this circle be the base of a cone which is tangent to the torus and whose apex is at o. Pass two planes, Q and R, through I and tangent to this cone (§ 154). (In the figure RR'' / J Fig. 240 and QQ are the traces of these planes on the plane A.) These planes give the shade lines oq and or, on the cone (§ 212). The points q and r are points on the shade line of the torus. For the lower part of the torus, the cones point downward, but the solution is the same in principle. Considerable in- genuity is possible in handling this method when the apex of a cone falls off the paper or on a level with /. If the shadow of the surface is required, it may be cast as in I above, or point by point from the shade line, or by a combina- tion of the two methods. This method gives a fair degree of accuracy, but it is not well adapted to cases like Fig. 242, where the surface casts a shadow on itself. XII, § 216] SHADES AND SHADOWS 259 Fig. 241 260 DESCRIPTIVE GEOMETRY [XII, § 216 FiQ. 242 XII, § 216] SHADES AND SHADOWS 261 III. The Secant Plane Method. In this method use is made of secant planes passing through the source of light and cutting the surface in a curved line, as in Fig. 242. A ray of light in such a plane and tangent to the curve of intersection, r, s, t, will determine one or more points, as s or t, on the separa- trix of the surface. In Fig. 242, the plane Z, passed through /, cuts the vase in the curve a, h-f (§ 175). The ray of light, Ic, drawn tangent to this curve determines the point c on the required separatrix, and also the shadow of c, at c, on the pedestal. If the shadow of the vase is required, it may be cast point by point, or by method I above. This method is a good one when the given surface is reentrant, and when the curvature is not too slight. CHAPTER XIII OTHER METHODS OF PROJECTION 217. Introduction. Orthographic projection makes use of three mutually perpendicular planes of projection and of pro- jecting lines perpendicular to these planes (§7). Because of the three-dimensional nature of the space in which we live, this is the simplest and most natural possible method of projection. The relation of perpendicularity between lines and planes is a unique relation and a most useful one Again, most of the bodies represented in projection are largely composed of three mutually perpendicular sets of planes. In § 8 it was shown that if it is desired merely to represent a body (as distinguished from locating it), one of the planes of projection may be dispensed with. Again in § 27 and frequently elsewhere, it was shown that for special purposes one of the planes of projection may be shifted at will. In Chapter XII, all the work on shades and shadows is done in orthographic projection, in the usual manner. However, the shadows thus determined are, in themselves, projecticns of the given bodies, arrived at by another method. In the case of sun shadows, the parallel light rays are, in effect, projecting lines having a special relation to the planes of projection. WTien a near-by source of light is used, the projectors are divergent. Any surface whatever may receive the projection. In § 221 it will be shown that by placing an object in a par- ticular position with regard to the planes of projection, a special and very useful kind of representation is obtained. In perspective projection the projectors diverge from a view- point, chosen at will, and the projection is made on a surface between this point and the body which is to be represented. 262 XIII, § 218] OTHER METHODS OF PROJECTION 263 The surface on which the projection is drawn is usually plane, though sometimes it is cylindrical, or (very rarely) spherical. Various other methods of projection have been devised for special purposes, such as for making maps of various kinds, and for military drawing, machine drawing, and other special forms of drawing. In each case the projecting surfaces and lines are so chosen as best to explain the particular features of the object to be represented. It becomes evident that the position or number of projecting planes, the relation between the projecting planes and project- ing lines, or the position of the object to be projected, may be varied at will for special purposes. In the present chapter a number of these special variations that are most generally used will be discussed. 218. One Plane Descriptive Geometry. Any point in space becomes fixed when its projection on one plane and its distance Orthographic. One plane. Fig. 243 from that plane are known. This gives rise to a system of projection in which one projection, usually the plan, is used exclusively. In this system the elevations of the various points are indicated by index numbers on the horizontal projection. In Fig. 243, the line ab is shown in orthographic projection and also in one plane projection. The figures 2 and 6, on the one plane projection, are the indexes of height. The H trace of ab may be found by extending a'^b^ beyond a^ a distance equal to J of its length. 264 DESCRIPTIVE GEO:\IETRY [XIII, § 218 Fig. 244 A plane may be shown by its H trace and one other point in the phine, or by a Hne of greatest dechvity of the plane. Thus in Fig. 244, the Hne oo^ and the point a^ describe the plane 0. It can also be described by drawing a line of greatest declivity (§ 90), a%^. Let it be required to find the point where the line d^c^*^, Fig. 244, pierces the plane 0. Extend d^c^^ to its trace, e^, as above described. Draw XX^, the trace of the plane X, of which c^c^^ is the line of greatest declivity. Now the line fg^, drawn parallel to A^Y'', lies in X at the height 8. Similarly, aY lies in at the height 8. Hence g^ lies in both X and 0, as does also h^. The line hY is then the line of intersection of A" and 0, and the point j is the trace of d^c^^ on 0. Other problems might be worked out, but this one will sufhce to indicate the general nature of the subject. The principles indicated above are frecjuently used in drawing complicated roof plans. By this means it is often possible to draw a complete roof plan before any accurate elevations have been made.^ 219. Lines of Level. From the considera- tions of § 218, it follows that a plane may be fully described by a series of parallel, equally spaced lines, carrying proper indexes, as shown for the plane A' in Fig. 245. 1 One textbook on Descriptive Geometry uses the method of one phine projection, with modifications, very freely. XIII, § 219] OTHER METHODS OF PROJECTION 265 These lines are called the lines of level of the surface. The actual slope of such a plane, with respect to //, may be shown by passing a cutting plane, Z, perpendicular to the lines of level. This plane intersects A^ in the line a%'^. Now let Z be rotated into //, on ZZ^ as an axis. The point a^ is in the axis of rotation, and does not move, but b"^ will rotate to a point, 6'^, distant 7 height units from 5^. The line a^6'" is now the rabattement of a line of greatest declivity of the plane X. Hence is the true angle between .Y and H. A section through Fig. 246 A' cut by any other vertical plane, as Y, can be found in the same way. The line c^c?'^ shows such a section, rotated into H. The angle cf)' is the angle between the line c^(P and H. Lines of level may be used also to describe planes intersecting in various ways, as shown by the roof plan in Fig. 194. It is also possible, and sometimes convenient, to describe certain curved surfaces by means of lines of level. Figure 246 shows a surface composed of a hemisphere and mitered cylinders described by lines of level, together with a section cut from the surface by a vertical plane Z. Without any index of heights, the surface shown may be either convex or concave toward the observer. A set of height indexes w^ould make this definite. 266 DESCRIPTIVE GEOMETRY [XIII, § 220 220. Topography. ^Yhen it becomes necessary to describe accurately a very irregular surface, such as a piece of land, lines of level are found very useful. A map on which lines of level are shown is called a topographical map, or a contour map, and the lines of level are usually spoken of as contour lines. Such maps may be drawn, of course, at any scale, and the intervals of height between contour lines may be chosen to suit the particular case, the vertical scale commonly being different Fig. 247 from the horizontal scale. Shorter intervals between the contour lines give greater accuracy, but they involve more labor, of course. The natural surface of the earth follows no geometrical law. Hence its lines of level are free-hand curves (§ 108). But where land has been graded into formal terraces, roads, etc., the surfaces become geometric and the contour lines may be straight lines or regular curves. Original plans for grading are indicated in this manner. XIII, § 2201 OTHER METHODS OF PROJECTION 267 Figure 247 shows a contour map which includes a lake, a stream, and two hills. Contours are indicated at five-foot intervals, starting from the lake level as the zero datum. The contour of the lake bottom is shown by dotted lines. Hill 60 is seen to be much steeper on the northwest face than elsewhere and drops toward the lake in a concave spur, while hill 47 descends on the east into the ravine through whirh the stream flows. Fig. 248 In Fig. 248 the same piece of land is shown as in Fig. 247, but it has now been graded so as to give a flat space at level 25. The front part of this space is terraced up while at the back it is cut into the hills. In making contour maps, the engineer starts from some fixed and known point as a bench-mark, and determines by stadia ^ the bearing, distance, and levels with reference to the ^ By means of the stadia it is possible to determine the distance and direction between two points, as well as their relative levels, at one setting of the instrument. 268 DESCRIPTIVE GEOMETRY [XIII, § 220 known point of as many points as can readily be seen from this position. The instrument is then moved to a new known position, and other points, hidden before but now visible, are located. From the data thus secured, the points are mapped and their hei^fht indexes are recorded. Contour lines are then Fig. 249 drawn through corresponding heights, interpolation being used freely. As an example of the type o" problem that arises in contour work, let the contours in Fig. 249 be given. Let it be required to show the cut and fill necessary to carry a level road along the face of the hill as shown, assuming the angle of repose for the embankments to be 30°. The road follows the 25 ft. con- tour closely. Henor, if the surface of the road is to be level, it will cut into the hill on one side and will need embankment on XIII, § 221] OTHER METHODS OF PROJECTION 269 the other. The faces of the cut and fill will be planes where the road is straight, and conical where the road curves. The angle between the planes or the cones and the surface of the road will be 30°. It is required to find the line in which these surfaces will meet the natural surface of the ground as shown by the original contours. Let the surface of the road at elevation 25 be taken as H. Pass the vertical plane Z perpendicular to the axis of the road. This plane will be cut by the planes of the cut and fill, in lines which make angles of 30° with H. Let these lines be rotated about ZZ^, into H. They will appear at ahcd. Locate e/fg'h'f, on a-d, at 10' height intervals, above and below the road. These points, rotated back into Z, give the points e-j, which lie in the planes of cut and fill and at the same level as the given contours. Through these points draw Hues of level, hh'h'\ gg'g"t etc., in the surfaces of cut and fill. The points in which these lines intersect contour lines of a corresponding height, as 1-12, define the limits of the cut and fill. It should be noticed that above the road the natural contours are cut away, while below the road they are covered under. Problems involving topography are simple in principle. The beginner's errors usually include streams which flow uphill or along the crest of a ridge, lakes which are more or less out of level, or terraces which intersect in impossible lines. These errors can be avoided only by close attention and clear visual- ization. 221. Isometric Projection. In isometric projection (§§ 2, 217) the projecting fines are drawn perpendicular to the plane of projection, but the body to be projected is placed in a special position, in order to produce a drawing which has some of the pictorial qualities of a perspective, but which can be more easily made, and which has some value as a working drawing. 270 DESCRIPTIVE GEOMETRY [XIII, § 221 If a cube is so placed in orthographic projection that each of the edges which meet at a given corner make equal angles with the V plane (Fig. 250) it is obvious that three of its six faces become visible in projection. Moreover, the projections of the angles, aoc, aoe and coc, are equal, and each is equal to 120°. Again, all of the edges are equal in projection, but the length of the projections is less than the true length of the edges of the cube. On the other hand, the face diagonals, ac, ae, and ec, are in a plane parallel to V, and hence they are projected in their true length. Finally, the diagonals ob, od, of, are lines of greatest declivity of their respective planes, and hence they are foreshortened more than the edges. ^ It is now evident that any object com- posed of three sets of mutually perpendicu- lar planes may be projected by drawing one st't of edges vertically, while the others make angles of 30° with the horizontal, and laying these edges off at a reduced scale. Now let the scale at which the drawing is made be arbitrarily increased (about 22%) so that all main edge lines and all lines parallel to them are drawn in their true length. The resulting drawing is an isometric projection. On it all lines parallel to the main edges (called isometric lines) are shown in their true length, and either vertically or at an angle of 30° to //. All other lines are distorted in length ; some are reduced in size, while others are magnified. Fortunately, the lines that are of natural length occur most frequenth'. 1 Let the true length of any side of the above cube be S. Then the true length of a diagonal is S\/2. Now remembering that the angle aoc = 120°, oab = 60°, etc., prove that the projected length of any side of the cube is equal to 0.81G47.S. The ratio, .816+, is called the isometric scale, and it represents the reduction in size of the edges of the cube. It may be shown also that for lines not parallel to the edges, the scale will vary from 0.5773 (for bo, of, od) to 1.000 (for ac, ce. ej). XIII, § 222] OTHER ^METHODS OF PROJECTION 271 222. Typical Case. Figure 251 shows the elevation and section of a flat arch in a battered wall. Let it be required to show the keystone in isometric projection. For the sake of clearness, the scale of the detail is enlarged. First draw around the given stone a parallelepiped which incloses it and coincides with its extreme faces,^ as shown by a-h. Draw this block in isometric projection, as a'-h', assuming a point of view above, in front, and to the right. The points j'k'l'm' are found Fig. 251 by measuring along the isometric lines of the block distances determined from the section and elevation. Any point w^ithin the block, as z, may be located by measuring along isometric lines, as h'x, xy, yz' , from some known starting point. The dis- tance between any two points can be found in a similar manner. Figure 251 (2) shows an isometric projection of the same stone in a different position. Here the bottom, rear, and left-hand faces appear as visible. 1 The French call this inclosing block the solide capable; a term which if translated literally gives a very good notion of its nature and use. 272 DESCRIPTIVE GE0:METRY fXIII, § 223 223. Curves in Isometric Projection. The general method for drawing curves in isometric projection consists in making an isometric drawing of a polygon w^hich may be inscribed in or circumscribed about the given curve, and using this polygon as a guide in tracing the curve, as illustrated in Fig. 252 a. A circle in isometric projection appears as an ellipse. It may be most easily drawn by use of a circumscribed square, as in Figure 252 b. The diameters, ab and cd, of the square are conjugate diameters (§§121, 124, II) of the ellipse, which may be constructed as shown in Fig. 134. Another construction, Fig. 252 a which approximates the required ellipse by arcs of a circle, (§ 110, III) may be made as follows. In Fig. 252 b draw ea, ec, fdy and fo. From c as a center swing the arc ac and from h as a center the arc cb. The other half of the curve has its centers at/ and g. 224. Curved surfaces in Isometric Projection. The iso- metric projection of a cone can be established by drawing the base, the apex, and two lines from the apex tangent to the base. The isometric projection of a cylinder is given by two bases with their common tangents. Since a sphere, in any position whatever, projects as a circle, its isometric projection will be a circle, exaggerated in size in the ratio 122%. The . isometric projection of any double- curved surface may be made by the Envelope Method (§ 174). XIII, § 225] OTHER METHODS OF PROJECTION 273 Warped surfaces are best shown in isometric projection by drawing isometric projections of the directrices and of a number of elements. 225. Oblique or Clinographic Projection. In oblique pro- jection, the object is placed with its main faces parallel to a plane of projection and is projected on that plane by means of lines inclined in any convenient manner so as to give the desired result. It will be recognized that a sun shadow is a special kind of oblique projection of the object casting the shadow. The plane of projection is usu- ally, though not necessarily, verti- cal. In the following description it will be so assumed and will be referred to as V. It can be seen readily that the oblique projection of any face of an object that is parallel to V will be the true size and shape of that face. Also the projection of any line perpendicular to V will vary as to length with the inclination of the projecting lines toward V, and will vary as to slope with the in- clination of the projecting lines toward H. Moreover, in projecting a line which is perpen- dicular to V, the slope of the projection and its length may be made to have an^^ desired values whatever by properly selecting the inclination of the projecting lines. In practice it is usually convenient to draw an oblique pro- jection without considering particularly the exact position of the projecting lines. A front elevation of the object is drawn as shown by the dotted lines in Fig. 253. Next any con- venient slope is chosen for the lines perpendicular to the front face. Thirty degrees is ordinarily satisfactory, and is the slope Fig. 253 274 DESCRIPTIVE GEOMETRY [XIII, § 225 used in the figure. In laying off distances on these lines, any proportion to the true dimensions may be used, depending on how fully it is desired to show the side faces. In the figure, a one to one ratio was used. The inclination and length ratio to be used for lines perpen- dicular to V, having been chosen, and the main face of the solide capable (§ 222) having been drawn, the projection is con- structed by measurement along lines parallel to the edges of the solid, as explained in § 222. In such a projection, the side faces can be exaggerated or suppressed by varying the slope and length ratio as noted above. But it should be noticed that the length of diagonal lines is distorted (as in isometric projection, § 221), and that if the exaggeration of the side faces is carried too far, it is apt to produce an unsatisfactory drawing. Figure 253 is drawn with a slope of 30° and a length ratio of unity. This particular case is called a Cavalier Projection and is produced by projecting lines making an angle of 45° with V. It will not ordinarily be wise to use either a greater slope or ratio than the one here indicated. APPENDIX The exercises on the following pages may be divided into two parts. Numbers I a to LXXIV b are parallel with the similarly numbered exercises in the body of the text of Chapters I to VIII. The rest of the problems are designed to accompany the text of Chapters IX to XIV. These exercises employ the same principles as are explained in the text, with sufficient variation of, the conditions to make the solutions worth while. Sheets XXII a to XXXVII a consist of three problems each, so spaced that number 1 from any sheet may be combined with number 2 and number 3 from any other sheet or sheets. The typical layout for exercise sheets, applying to all cases (except where otherwise specially noted), is given on p. xii of the general explanations. For Exercise sheets I a to XXI a the location of the profile plane is given with the statement of each set of exercises. For all other sheets, no profile projection is required, and P is taken at the right border line, unless otherwise specially mentioned. EXERCISE SHEET la (§13) [Note. For the typical layout for exercise sheets, applying to all exercises (excepting a few which are separately described), see p. xii of the general explanations.] Take P at the right, 6 from the border line, turned toward the given points, into H. Draw all three projections of the following points. a (68, 11, 16) / (35, 7,13) b (61, 0, 11) 9 (27, 0, 5) c (55, 3, 11) h (19, 8, 8) d (48, 17, 5) i (12, 12, 9) e (43, 0, 0) 3 ( 4, 7, 0) 275 276 DESCRIPTIVE GEOMETRY EXERCISE SHEET II a (§13) Take P at the left,. 20 from the border Une, turned away from the given points, into V. Draw all three projections of the following points. a ( 4, 10, 8) / (33, 10, 13) b ( 9, 4, 16) g (39, 0, 6) c (17, 0, 0) h (44, 18, 20) d (23, 13, 13) i (50, 8, 2) e (29, 18, 0) j (56, 13, 13) EXERCISE SHEET Ilia (§13) Take P at the left, 20 from the border line, turned away from the given points, into H. Draw all three projections of the following points. a ( 3, 5, 4) / (31, 8, 15) b ( 9, 12, 2) 9 (37, 15, 0) c (15, 10, 9) h (45, 10, 15) d (22, 19, 20) ^ (50, 16, 12) e (27, 0, 0) j (56, 0, 18) EXERCISE SHEET IV a (§13) Take P at the left, 2 from the border line, turned toward the given points, into V. Draw all three projections of the following points. a (23, 8, 20). Move the point 9 toward V and 5 toward H. Call this position a'. b is 29 from P, 4 farther from H, and 15 nearer V than is a. ]Move b downward 5, forward 10, and 6 to the right. Call these successive locations b', b", b'". c is 43 from P, 11 from V, and on a plane passing through HA and making an angle of 60° with H. d lies in the same line perpendicular to P as c, but 10 farther from P. With c as a center, rotate d until it lies directly under c, keeping d always the same distance from V. Mark the new position d'. e lies on V, 15 from // and 60 from P. / lies in the same horizontal plane as c, 1|" (actual) from e and 7 in front of V. EXERCISE SHEET Va (§20) Take P at the right, 2 from the border, and opened toward the given lines, into V. Draw the projections of the following lines. APPENDIX 277 Line ab : a (75, 16, 12) ; 6 (06, 6, 3). Line cd : cd is perpendicular to // ; d is at (62, 0, 9). Line ef : e (57, 0, 0) ; ef runs upward and to the right, / being 48 from P. Line gh : parallel to HA, 17 from V and 9 from H ; ^ is 44 from P, and h is 34 from P. Line ij : is in H, between 39 and from P ; running forward and to the right from i. Line kl : parallel to P; runs 'ownward and forward from k (27, 17, 2). Line mn : intersects F at w (23, 20, 0), and makes an angle of 45° with F. n and m are in the same horizontal plane. EXERCISE Via (§21) L escribe in writing the following lines : r EXERCISE SHEET Vila (§22) Take P at the right, 18 from the border line, and turned away from the objects, into F. Profile projections are required for #3 only. Use notation carefully. 1. The center of a square, abed, 10 units on a side, is at e (50, 10, 10). The plane of the figure is parallel to H and its sides make angles of 45° with F. The corner a is farthest from P, c is nearest P, and b is nearest F. Rotate the square through 45°, on bd as an axis, a going down and c up. Draw its projections in this position. 2. A circle, 14 in diameter, has its center at o (31, 10, 10). It lies in a plane parallel to F. Call the horizontal diameter fg. Rotate the figure on its vertical axis, through 45°, and draw its projections. [Note. Assume several points on the circumference as auxiliaries.] 3. Use the profile plane. A regular hexagon jklnmp has its center at s (12, 10, 10). The diameter jm is parallel to HA and the plane 278 DESCRIPTIVE GEO:\IETRY of the figure is parallel to V. Draw its projections in this position, then rotate it on jm as an axis lentil it lies in a plane making 45° with H. Draw its projections. EXERCISE SHEET Villa (§26) Take P at the left, 16 from the border line, and turned away from the objects, into V. Draw the profile projections in Exs. 2 and 3 only. Use the first method. 1. Given a (3, 16, 4) and 6 (10, 9, 9). Find the true length of ah and the true angle it makes with H. Swing about b. 2. Given c (12, 4, 5) and d (19, 13, 10). Find the true length of cd and the true angle it makes with V, by swinging about c until the line is parallel to P. 3. Given e (23, 11, 6) and/ (23, 20, 11). Draw all three projections of ef and find the true length of ef and the angles it makes with V and P, by swinging the line about / as a center. 4. Given g (36, 4, 8) and h (45, 10, 13). Keeping h stationary, find the true length of gh and the angle it makes with H. Check the true length by swinging the line parallel to H. Indicate the angle with V. 5. Given i (50, 5, 7), j (56, 11, 14), and k (60, 4, 5). Draw the projections of the triangle ijk and find its true shape. [Note. Find the true lengths of the sides individually and construct the triangle.] EXERCISE SHEET IX a (§27) Take P at the left, 16 from the border line, and turned away from the objects, into H. Use P in problem 4 only. Solve by the second method. 1. Given a (13, 12, 8) and b (21, 6, 11). Find the true length of ab and the angle it makes with H. 2. Given c (25, 5, 15) and d (33, 8, 7). Find the true length of cd and the angle it makes with V. 3. Given e (36, 15, 8), / (39, 12, 13), g (43, 8, 2), and h (45, 6, 6). Find the true shape of the quadrilateral. 4. Given i (4, 4, 3) and i (10, 14, 12). Find the true length of ij and the angle it makes with P. 5. Given k (50, 5, 8) and I (59, 9, 13). Find the true length of kl and the angle it makes with H. Use an auxiliary plane through the H projection of kl. APPENDIX 279 EXERCISE SHEET Xa (§29) Take P at the right border line. No P projections are required. 1. The point a (76, 5, 4) is given. The hne ab is 10 units long and slopes upward, forward, and toward P from a, making angles of 30° and 45° with H and V respectively. Draw its projections. 2. The point c (55, 4, 7) is given. The line cd is 8 units long, and slopes upward and forward from c, making an angle of 45° with H and an angle of 30" with V. Draw its projections. (^Two solutions.) 3. The point e (40, 15, 17) is given. The hne ef is 8 units long and makes angles of 45° and 15° with H and V, respectively. Draw its projections. (Four solutions.) 4. The point g (27, 4, 5) is given. The hne gh is 12 units long and makes angles of 45° with H and V, sloping upward and forvv^ard. Draw its projections. 5. The point i (14, 4, 5) is given. Draw the projections of the hne ij 12 units long and making an angle of 45° with H and 60° with V. Explain the result. EXERCISE SHEET XI a (§30) Take P at the right border line. No P projections are required. 1. Given a (75, 12, 13) and b (66, 6, 9), locate c on ab 7 units from a. 2. Given d (60, 5, 7) and e (51, 14, 13), locate / on de 1" from e. Solve in two different ways, and check the results. 3. From g (32, 8, 5) the hne gh slopes away from H, V, and P, making an angle of 45° with H and 30° with V. The point i lies on 9h 36 from P. Find the length of hi. 4. Bisect the hne joining the pomts j (26, 13, 8) and k (18, 9, 13), and mark the middle point I. Notice the relation between segments of projections. 5. Given m (13, 7, 10) and n (4, 13, 19), trisect 77in at o and p. State the ratios between segments of the projections and those of the true length. EXERCISE SHEET XII a (§ 31) Take P at the right, 16 from the border line, and turned away from the objects, into V. Use P only when necessary. 1. Given a (59, 12, 2) and b (51, 3, 9), find the H and T' traces of ah, 2. Given c (46, 11, 6) and d (38, 2, 6), find the traces of cd. 3. Given e (33, 17, 9) and/ (24, 17, 9), find the traces of ef. 4. Given g (20, 12, 12) and h (20, 4, 4), find the traces of gh. 5. Given i (12, 5, 6) and j (4, 16, 10), find the traces of ij. 280 DESCRIPTIVE GEOMETRY EXERCISE SHEET XIII a (§33) Take P at the left, 16 from the border Une, and turned away from the objects, into V. Use P only when necessary. 1. From a, the line ab slopes away from P and V but towards H. From c the line cd slopes away from P and H but towards V. These lines intersect at e, which is 7 from H and 6 from V. Draw the projections between 3 and 13 from P. (Use the P projections as a test of accuracy.) 2. Given/ (16, 2, 7), g (26, 8, 12), h (17, 9, 9), and i (25, 3, 5). Do fg and hi intersect? 3. Given j (30, 5, 7) and h (39, 10, 13), draw 7nn intersecting jk at I, which is 7 units from k. From m, mn slopes away from P and toward V and H. 4. Given p (47, 5, 7), q (58, 11, 14), and r (50, 12, 13), draw ro intersecting pq at its middle point o. EXERCISE SHEET XIV a (§34) Take P at the left border line and turned toward the objects, into V. 1. Draw all three projections of the line joining a (18, 14, 9) and h (27, 7, 5) and those of another line cd which is parallel to ah but nearer H and farther from V than is ah. 2. Given e (31, 10, 4) and/ (41, 9, 10), draw the projections of gh which passes through i (34, 13, 12) and is parallel to e/. Find its V trace. 3. Through I, the middle point of the line joining / (46, 6, 8) and k (58, 14, 13), draw nm parallel to ef, and 1" long. 4. Through o, 70 from P, draw op and oq parallel to ah and jk re- spectively. Find the true size of the angle poq. (See Note, p. 40.) EXERCISE SHEET XV a (§§38-42) Take P at the left, 16 from the border line, and turned toward the objects, into V} 1. Draw all three projections of the following points. a (16, 16, 9) h (II, 20, 10, 7) c (III, 24, 10, 13) d (IV, 28, 20, 3) e (III, 32, 20, 10) / (36, 4, 12) g (II, 40, 22, 2) 2. Given the point h (IV, 45, 6, 15), move h parallel to HA until it is 5 farther from P. From here move it straight up 15. 3. Given the point i (IV, 60, 12, 3), move i straight towards point a of Ex. 1, till it is 5 nearer to P. * In describing the movement of P, the description will be understood to apply to the movement of that part of P which lies in the first quadrant, as heretofore. APPENDIX 281 EXERCISE SHEET XVI a (§43) Take P at the right, 16 from the border hne and turned away from the objects, into H. Draw all three projections of the following hnes. 1. Given a (II, 61, 4, 6) and b (IV, 53, 8, 3), find the traces of ab, and designate the quadrants through which it passes, 2. In the space 40-50 from P draw the projections of a line cd running from a point c in I, through II, to a point d in III, and desig- nate the parts in the different quadrants. Do not use the profile. 3. From e (IV, 36, 8, 12) draw ef, through I into II, making ef parallel to P. Indicate the parts that he in I, II, and IV. 4. Given g (II, 30, 5, 14) and h (II. 18, 9, 3), from j (28, 4, 10) draw ji to i, the middle point of gh. Find the true distance between j and the V trace of ji. EXERCISE SHEET XVII a (§44) Take P at the left, 16 from the border line, and turned toward the objects, into V. 1. Given a (II, 4, 9, 7) and b (IV, 12, 4, 10), draw all three projec- tions, and find the H, V, and P traces of ab. 2. Through d (III, 16, 13, 8) draw cd 12 units long, making an angle of 45° with H and an angle of 30° with V. From d, cd slopes away from P and V, toward H. 3. Given e (IV, 31, 4, 15), / (IV, 40, 7, 10). Find the true length of ef and the true angle which it makes with H, by the second method. 4. Through g (II, 58, 3, 13) draw gh parallel to cd and gi parallel to ab. Both h and i are in III. Find the true angle between gh and gi. EXERCISE SHEET XVIII a (§§46-^8) Take P at the left border line. No P projections are required. 1. Locate the point a (18, 0, 3). Draw ab in H, 1" long, forward and to the right from a and making an angle of 45° with HA. This line is one side of a regular hexagon whose plane makes an angle of 45° with H and slopes away from V. Draw its H and V projections. 2. Draw the projections of an octahedron If" on a side. One face of the soHd is parallel to H and 5 above it. The horizontal projection of one axis of the solid makes an angle of 30° with HA. Draw the H and V projections and also the projection on a vertical plane parallel to the axis described above. 282 DESCRIPTIVE GEOMETRY EXERCISE SHEET XIX a (§51) Place the sheet vertically and with the wide margin at the left. Locate HA 3" from the upper border line. Draw the H and V projections of a solid bounded by 4 regular hexagons and 4 equilateral triangles, each side of each face being 1" long. Place the solid with a hexagon as its base, at the center of the lower part of the sheet. EXERCISE SHEET XX a (§54) Take P at the left, 20 from the border line, turned away from the objects, into V. 1. The points a (13, 0, 3), b (22, 0, 12), c (19, 0, 15), and d (7, 0, 9) determine the base of a right quadrilateral prism, 2|" high ; axis vertical. The points e (4, 16, 4), / (4, 5, 10), and g (4, 10, 16) deter- mine the base of a right triangular prism 3" long; axis horizontal. Find the line of intersection of the prisms. 2. At the right of the above problem, draw the development of each of the prisms at one half of the former scale. Show the line of intersection on the development. EXERCISE SHEET XXI a (§§55-62) Take P at the left, 20 from the border line, and turned away from the objects, into H. 1 . Draw the H, V, and P traces of the plane R ' (20, R^ 30° I, R'^ 45° . 2. Given the points a (II, 30, 3, 17), b (II, 52, 3, 17), c (II, 30, 11, 7), and d (II, 52, 11, 7). Find the H, V, and P traces of the plane S con- taining the lines ab and cd. EXERCISE SHEET XXII a (§64) [Note. Unless otherwise indicated, hereafter P will be taken at the right border line and no profile projections will be required.] 1. Given the plane R ' (75, R^ 90°, R'' 30° r) ; on R draw a line ab parallel to and 6 from V, and draw a line cd parallel to and 6 from H. 2. Given the plane S (48, S^ 45° r, .S" 60° r) ; on S locate a point e at 35 from P and 8 from V. 3. Given the points / (17, 3, 13) and g (7, 7, 3) ; through fg pass three planes T, U, and W, the plane T being perpendicular to H. 1 For explanation of the n(jtation used in describing planes, see descrip- tion of quantities, page xiii. APPENDIX 283 EXERCISE SHEET XXIII a (§ C6) 1. Given the points a (III, 74, 8, 10) and b (58, 3, 13) ; through ab pass the plane X, intersecting HA at 60. 2. Given the points c (44, 2, 8) and d (34, 11, 8) ; through cd pass the plane Y, intersecting HA at 50. 3. Given the point e (16, 5, 9) ; through e pass the plane Z, inter- secting HA at 21. EXERCISE SHEET XXIV a (§68) 1. Through the point a (68, 3, 7) pass two lines. The H projection of one line makes an angle of 60° to the left and the V projection an angle of 30° to the right with HA. The H and V projections of the other line make angles of 45° to the right and 60° to the left, respec- tively. Draw the traces of the plane X determined by these lines. 2. Given the line joining the points b (45, 15, 3) and c (33, 15, 15) and the point d (40, 6, 8). Draw the traces of the plane Y containing the point d and the line be. 3. Given the triangle whose vertices are/ (18, Q,2), g (10, 12. 5), and h (6, 4, 13) ; find the traces of the plane Z which contains the triangle. EXERCISE SHEET XXV a (§71) 1. Given the plane M (58, M'' 30° Z, M" 4:5° I) and the point a (III, 69, 13, 5) ; through a draw a line parallel to M. 2. Given the plane N (35, .V^ 45° /, N^ 75° r) and the point b (II, 46, 4, 10) ; through b draw a line parallel to A''. 3. Given the points c (III, 25, 5, 25), d (III, 17, 20, 6), e (15, 5, 3), and / (5, 11, 13) ; draw the plane O through the line ef parallel to the line cd. EXERCISE SHEET XXVI a (§72) 1. Given the points g (78, 18, 15), h (70, 18, 6), j (72, 13, 2), and k (64, 3, 8) ; through the line jk pass the plane Q parallel to the line gh. 2. Given the points I (51, 14, 4), m (42, 3, 8), and n (38, 9, 6); through n pass the plane R parallel to the line Im. 3. Given the points ?? (Ill, 24, 9, 7), q (III, 16, 3, 7), and o (0, 6, 7) ; through o pass the plane S parallel to the line pq. 284 DESCRIPTrV^E GEOMETRY EXERCISE SHEET XXVII a (§74) 1. Given the plane T {57, T^^ 45° /, T' 30° I) and the point a (71, 5, 7) ; through a pass the plane U parallel to T. (Use a normal case auxiliary line.) 2. Given the plane W (50, W^ ^o° r, TT^'-' 60° r) and the point 6 (III, 43, 5, 9) ; through h pass the plane X parallel to W. (Use a special case auxiliary line.) 3. Given the plane Y (4, Y^ 75° r, F" 45° and the point c (IV, 13, 11, 5) ; through c pass the plane Z parallel to Y. EXERCISE SHEET XXVIII a (§75) Find the line of intersection of each of the following pairs of planes. 1. G (74, G'' 30° r, G" 75° r) and H (56, m 60° I, H^' 45° /). 2. I (49, I'' 90°, I^ 45° r) and K (30, K"^ 45° I, K^ 60° /)• 3. L (22, L'' 75° r, L" 60° r) and M (10, JP 30° r, M^' 45° Z). EXERCISE SHEET XXIX a (§§ 76, 77) Find the line of intersection of each of the following pairs of planes. 1. N (59, iV* 15° I, iV" 60° r) and 0, parallel to HA ; O'' is 9 below and 0"" is 12 above HA. 2. P (49, P'^ 60° r, P" 82^° r) and Q (28, Q^ 90°, Q^' 75° Z). 3. i?, parallel to HA ; J?'' is 14 below and R" is 17 above HA ; and S, parallel to HA ; S^ is 27 above and *S^' is 4 above HA. EXERCISE SHEET XXX a (§§75-77) Find the line of intersection of each of the following pairs of planes. 1. T (74, T"^ 60° r, T^' 45° r) and L^ (64, U^ 60° r, C/^' 75° r). 2. X (32, X''60°Z, .Y"30°0 and W, parallel to i/A. W is 11 below and TF" is 6 above //A. 3. Y (23, y" 45° r, 7" 75° r) and Z (5, Z^ 60° Z, Z" 90°). EXERCISE SHEET XXXI a (§78) 1. Find the trace of the line joining the points a (74, 7, 10) and h (III, 57, 9, 3) on the plane Z (77, Z'' 45° r, Z^' 60° r). Use a normal-case auxiliary plane. 2. Find the trace of the line joining the points c (47, 7, 10) and d (III, 30, 9, 3) on the plane Y (50, Y^ 45° r, 7" 60° r). Use a special- case auxiliary plane. Compare result with that of Ex. 1. 3. Find the trace of the line joining the points c (18, 8, 2) and / (7, 8, 6) on the plane X (10, X'^ 30° Z, X-^^ 60° r). APPENDIX 285 EXERCISE SHEET XXXII a (§78) Take P at the right border line, turned toward the objects, into V. Use P only where necessary. 1. Find the trace of the line joining the points g (75, 9, 12) and h (62, 15, 19) on the plane W (59, W" 90°, T7^ 60° I). 2. Find the trace of the line joining the points i (III, 48, 15, 14) and J (IV, 35, 4, 8) on the plane U, which is parallel to HA, JJ^ being 12 below, and f/^' 9 above HA . 3. Find the trace of the line joining the points k (20, 8, 5) and I (IV, 10, 4, 1) on the plane T, which passes from I to III throuTh HA, and makes an angle of 30° with H. EXERCISE SHEET XXXIII a (§§79-82) 1. On the plane M (60, M^ 45° I, M^ 60° I), locate a point a 72 from P and 6 from V. Draw the line ab 1^" long and perpendicular to M. 2. Given the points c (45, 6, 3) and d (32, 13, 10) ; through c pass a plane A^ perpendicular to cd. 3. Given the points e (II, 25, 5, 13), / (II, 16, 11, 9), and g (IV, 10, 7, 15) ; through g pass a plane perpendicular to ef. EXERCISE SHEET XXXIV a (§85) 1. Draw R parallel to (? (61, Q'' 45° r, Q^ 60° r), and I" from Q. 2. Given the points h (III, 52, 2, 3), i (30, 12, 13), andj (43, 11, 9) ; find the shortest distance from j to the line hi. 3. Given the plane T (23, T^ 30° r, T^ 45° r) and the point m (13, 6, 10) ; find the shortest distance between m and T. EXERCISE SHEET XXXV a (§87) 1. Given the points a (65, 15, 2), 6 (IV, 54, 5, 12), c (55, 1, 17), and d (40, 7, 2) ; find the projections and true length of the shortest line between the lines ab and cd. 2. Given the points a (21, 18, 16), b (8, 5, 2), and c (III, 11, 2, 13). Through c draw the line cd, parallel to ab, and find the common per- pendicular between ab and cd. EXERCISE SHEET XXXVI a (§90) Draw the line of greatest declivity of each of the following planes with respect to H, and also with respect to V. Determine the true angle that each plane makes with H, and also with V. The planes are X (60, X^ 45° I, X^ 60° I) ; V (50, V^ 45° I, V- 30° r) and Z, parallel to HA. ZZ^ is 8 below and ZZ^ 20 above HA. 286 DESCRIPTIVE GEOMETRY EXERCISE SHEET XXXVII a (§91) 1. Given the point a at distances 71 from P and 10 from H, and the point 6 at distances 75 from P and 7 from V ; and given that both a and b lie in X (60, X'' 45° I, X" 60° I). Rabatte a into H and b into V. 2. Given the point c at distances 39 from P and 10 from H and lying in the plane Y (44, Y^ 45° r, F^' 75° I). Rabatte c into H. Also rabatte FF" into H. 3. The four points e, f, g, and A all lie in Z (14. Z'' 30° Z, Z"^ 45° ; e is in I 23 from P and 6 from H, rabatte e into V ; /is in III, 3 from P and 6 from H, rabatte/ into H ] (7 is in II, 15 from P and 4 from F, rabatte g into F; /^ is in IV, 12 from P and 9 from V, rabatte h into H. EXERCISE SHEET XXXVIII a (§§93,94) 1. The four points a, b, c, and d he on the plane S (46, 5^^ 30° I, S" 45° I) ; a is 70 from P and 12 from H ; 6 is 62 from P and 11 from H; c is 55 from P and 5 from H ; c? is 60 from P and 3 from H. Find true shape of quadrilateral determined by these points by rabatting into H. 2. Given the points e (30, 3, 7), / (17, 12, 8), and g (21, 3, 4). Find the true angle efg by rabattement into V. EXERCISE SHEET XXXIX a (§96) 1. Given the plane TF (72, TF^' 30° r, TF'' 45° r) ; draw the projec- tions of a six-pointed star, inscribed in a 2" circle and lying in TF. 2. Given the plane X (19, V' 60° r, V^' 75° ; draw the projec- tions of an equilateral triangle 1|" on a side lying in X with one side parallel to F. EXERCISE SHEET XL a (§97) Given the plane Z (59, Z^ 45° r, Z" 30° r), and using a scale of f " = 1", draw the projections of a hollow brick, 2" X 4" X 8", resting with one of its 2" X 4" faces in contact with Z. Let one of the 4" edges of this face make an angle of 15° with ZZ''. Keep the entire soUd within the first quadrant. EXERCISE SHEET XLI a (§98) 1. Given the plane X (75, X'' 30° r, X" 45° r^) and the points e (71, 13, 7) and / (57, 5, 7) ; find the angle between ef and X. 2. Given the plane \V (21, IF'' 45° r, IF" 45° I) and the points g {31, 0, 15) and h (18, 13, 8) ; find the angle between gh and W. APPENDIX 287 EXERCISE SHEET XLII a (§99) 1. The plane Z cuts HA at the point (77, 0, 0). The trace ZZ^ makes an angle of 60° r with HA and Z makes an angle of 75° with Y . Find ZZ^. 2. The plane Y cuts HA at the point (50, 0, 0). The trace FF^ makes an angle of 30° r with HA and Y makes an angle of 30° with H. Find YY^. 3. The plane X cuts HA at the point (3, 0, 0). The trace XX^ makes an angle of 60° I with HA and X makes an angle of 45° with HA. Find XXk EXERCISE SHEET XLIII a (§100) Using auxiliary spheres \\" in diameter with their centers at the points (i) (65, 0, 0), {2) (40, 0, 0), and (3) (14, 0, 0), respectively, draw the traces of the planes that make the following angles with H and 7. 1. The plane R\ 45° with Y and 60° with H. 2. The plane S ; 30° with 7 and 60° with H. 3. The plane T ; 30° with H and 30° with 7. EXERCISE SHEET XLIV a (§100) [Note. When the first plane is found, use an auxiliary cone with its apex at a.] 1. Through the point a (60, 5, 10), pass as many planes as possible that make angles of 45° with 7 and 60° with H, respectively. 2. Given the points c (23, 2, 13) and d (16, 8, 3) ; through cd pass as many planes as possible, making angles of 45° with 7. Is there an impossible case for this problem ? EXERCISE SHEET XLV a (§101) Find the angle between each of the following pairs of planes. 1. Z (75, Z^ 60° r, Z- 30° r) and Y (55, F'^ 30° Z, F- 60° l). 2. X (50, X'^ 60° r, Z- 30° r) and W (35, W' 60° r, W- 45° I). 3. V (25, f/*60°r, C7M5° r) and 7 (5, 7^90°, 7''45°Z). (See Art. 76.) EXERCISE SHEET XL VI a (§101) Find the angle between each of the following pairs of planes. 1. (75, 0^ 15° r, O" 90°) and P (55, P^ 90°, P" 30° I). 2. Q (46, Q'^ 60° I, Q' 60° Z) and R (34, P'"* 45° r, P" 45° r). 3. S (S'' is parallel to and 12 above HA ; iS^ is 12 below HA) and T (15, T^ 45° r, T" 60° r). 288 DESCRIPTIVE GEOMETRY EXERCISE SHEET XLVII a (§§102,103) 1. The point a (68, 3, 11) is the center of the base of a cube whose edge is I4" long. The base is parallel to H and one of its body diagonals is parallel to V. Find the points where the line joining the points h (77, 1, 8) and c (58, 13, 17) pierces this cube. Mark them d and e. 2. The points / (47, 0, 2), g (47, 0, 14), and h (37, 0, 8) define the base of a regular tetrahedron (§50). Find the points k and I in which the line joining the points i (50, 0, 15) and j (36, 10, 2) pierces the tetrahedron. 3. The points m (15, 0, 12), n (8, 0, 12), o (13, 0, 18), and p (6, 0, 18) define the base of a quadrilateral prism. The points m and q (21, 11, 2) determine one edge of the prism. Find the line of inter- section of the plane Y (29, Y^ 45° r, 7" 45° r) and the given prism. EXERCISE SHEET XL VIII a (§104) Place the sheet vertically with wide margin at the left. Given the tetrahedron a (46, 6, 4), h (8, 2, 21), c (14, 25, 28), d (25, 27, 39), and the pyramid e (31, 29, 19), / (46, 0, 22), g (41, 0, 4), h (16, 0, 16), i (21, 0, 32). Find the Hne of intersection of the soHds. __ [Note. Exercise sheets XLIX a to LVIII a are given as applications of the principles of Chapter VII, but without special reference to specific articles. Each exercise is intended to be solved on a standard sheet, 8" X 10^". The exact layout for each sheet is to be determined by the student.] a « b ?♦-- -fl ' 1 . . 1^2." ii 1 ^\-^ ,\ Fig. 255 Cv EXERCISE SHEET XLIX a The garage door shown in Fig. 255 is hung from two trolleys, pivoted at a and h. The trolley a moves on the track ab, while h moves on 6c, as shown by the arrows. Determine the clear- ances needed for the operation of the door. Scale, l" = l'-0". Fixed pin EXERCISE SHEET XLIX b Figure 256 shows the hinged front of a writing desk, supported when open by a slotted metal rod at the side. De- 3.*-- \ 'Jy/}^////////M Fig. 256 APPENDIX 289 termine the path of travel traced by the end of the rod, as the desk closes. Also show the entire area over which any part of the rod moves. Scale, 3" = 1' — 0". EXERCISE SHEET La Figure 257 shows the path followed by the front wheel of a motor- cycle equipped with a side car. Let A = 4' — 6"; B = 2' — 6"; C = 3' - 6". Let both wheels of the motorcycle follow the same path. Trace [ the wheel track left by the side car. a Scale, h" = 1' - 0". ' ,^'- :<^i^-L EXERCISE SHEET LI a ^S Figure 257 shows the path followed by the front wheel of a motorcj'cle " Fiq, 257 equipped with a side car. Let A = 4' - 0"; 5 = 3'- 0"; C = 4' - 6". Let it be assumed that the front wheel moves at a uniform velocity and that the rear wheel moves as slowly as is consistent with the motion of the front wheel. Trace the path of both wheels of the •motorc3'cle and of the wheel of the side car. Scale, h" = V - 0". EXERCISE SHEET LII a Draw the front elevation of an elUptical arch with a span of 24' — 0" and a rise of 8' — 0". Scale, j" = V — 0". Use the approximate construction of § 127, IIL Let the arch consist of 19 stones, of equal width on the intrados, and of equal depth. Joint lines between stones are made normal to the curve of the arch. Let the wall be 2' — 0" thick. Draw also a plan of the arch looking up, and a development of the soffit. On the elevation, for purposes of comparison, trace the outline of the extrados as given by the methods of § 127, 1 and § 127, II, as well as that of § 127, III. EXERCISE SHEET LII b Place the sheet vertically. HA, 3" from the top border. Dr.iw the curve traced by a point on the circumference of the base of a right circular cone, which is so placed that an element is in contact with H. Let the cone roll on H, keeping the apex in a fixed position. Let the diameter of the cone be 2j" and the altitude be 2f ". 290 DESCRIPTIVE GEOMETRY EXERCISE SHEET LIII a • Draw a set of pivoted links similar to Fig. 142, making CD = 2" ; AB = U"; CA = 4", and DB = something between 2\" and ^". Trace the path of a point on DB, 1" from D; also of a point on DB, extended 1" beyond B. EXERCISE SHEET LIV a Let a circle, 2" in diameter, roll on a straight line. Trace the path of a point which lies on a radius of the circle, 1" outside the circum- ference, and which moves with the circle. A! EXERCISE SHEET LV a The window sash shown in Fig. 258 swings on the hinges A A. Determine how the back of the sash must be cut to allow it to swing open. Scale, one half full size. EXERCISE SHEET LVI a Place the sheet vertically. In the center of the sheet draw a circle, 3" in diameter, and draw Fig. 258 its vertical axis AB. Trace each of the following curves. (1) The curve generated by a point so moving that its distance from AB i^ constantly equal to its distance from the circle. (2) When its distance from AB is constantly equal to twice its distance from the circle. (3) When its distance from a border line is constantly equal to its distance from the circle. (4) When its distance from a corner of the sheet is constantly equal to its distance from the circle. EXERCISE SHEET LVII a The points a (38, 10, 2), 6 (12, 24, 9), c (8, 2, 23) determine the plane of a parabola which has its vertex at a and which also passes through b. Draw the projections of the curve. APPENDIX 291 EXERCISE SHEET LVIII a In the center of the sheet, and with their axes placed vertically, draw two intersecting circular arcs each having a 1^" radius, and with their centers 2j" apart. Draw the involute of the curve thus estab- lished. EXERCISE SHEET LIX a (§§145-148) (1) The line a (54, 15, 15), h (61, 6, 5) is the right-hand terminating element of an open cylinder, similar to the one in Fig. 155. Draw the projections of the cylinder. Indicate at least six elements. Let all invisible lines be dotted. (2) Given the plane Z (38, Z^ 45° I, Z''45°r). Draw the projec- tions of a right circular cone, Ij" in diameter and 1^" high, which has its base in contact with Z. Let the cone point downward. (3) The point e (12, 0, 12) is the center of the lower base of a double right circular cone, the base of which is 2" in diameter and whose elements make angles of 45° with H. Locate the following points on the surface of the cone ; /, 16 from P and 4 from H; g, S from P and 12 from H; h, 11 from P, 9 from H, and 17 from V. EXERCISE SHEET LX a (§§ 149, 150) 1. The point a (68, 0, 13) is the center of a free-hand spiral curve similar to that in Fig. 146. Draw about two turns within a space about two inches square. Let this curve be a right section of a cylinder. Find the line of intersection between this cylinder and the plane Z (69, Z''60°Z, ZM5°r). 2. The point b (33, 0, 14) is the center of a 2" circle in H. The point c (10, 24, 4) is the apex of an elliptical cone having the above circle for its base. Determine the right section of the cone. EXERCISE SHEET LXI a (§151) 1. The point a (70, 16, 16) is the center of a circle 2" in diameter and which has its base in a plane perpendicular to HA. This circle is the base of a cone whose apex is at b (46, 2, 3). The line m (69, 4, 25), n (52, 11, 2) pierces the cone. Find the piercing points. Note: In this problem a profile view of the cone base is useful but no space has been provided for it on the sheet. Let the base circle be conceived as being rotated on its vertical axis, and let it be so drawn, using a'" as a center. If indicated by a light line, this projection can be used as needed without confusing the drawing. 292 DESCRIPTn^ GEOMETRY 2. The points o (28, 0, IG), p (14, 22, lOj are the termini of the axis of a cylinder whose base is a 2\" circle in H. The line q (29j 3, 10), r (9, 19, 17) pierces the cj-linder. Determine the piercing points. EXERCISE SHEET LXII a (§152) 1. The point a (61, 0, 14,) is the center of the base of a 60° right circular cone, 3" in diameter. A right circular cylinder, \\" in diameter and 2\" high, rests on H with the center of its base at h (56, 0, 18). Find the line of intersection. 2. The point c (24, 0, 14) is the center of the base of a cone like the one in (1) above. The point d (36, 0, 22 j is the center of a l\" circle in H, which is the base of a cylinder whose axis passes from d to e (12, 16, 9). Find the line of intersection. EXERCISE SHEET LXU h (§ 152) [Note, All four cones on this sheet are 60°, right circular cones with bases 2\" in diameter.] 1. Find the line of intersection between a cone with its apex at a (20, 12, 0) and another with the center of its base at h (17, 15, 0). The axis of each cone is horizontal, and perpenchcular to T'. 2. Find the line of intersection between a cone having the center of its base at c (69, 0, 12) and its axis vertical, with another cone which is tangent to H along the hne d (76, 0, 12), e (56, 0, 12). EXERCISE SHEET LXUI a (§152) 1. The point a (66, 0, 11 j is the center of the base of a 2\", 60°, right circular cone. The line c (.58, 10, 16), d (73, 10, 7) is the axis of a 2n" right circular cone. Draw the Une of intersection. It will be found convenient to use a V plane passed parallel to the base of the latter cone. 2. The point e (26, 0. 10) is the center of the base of a right circular cone, 2f" high and 2^" in diameter. A right circular cylinder, If" in diameter, has its axis in the line/ (21, 7, 18), g (34, 7, 5). Find the line of intersection. EXERCISE SHEET LXIV a (§152) The line a (15. 8. 16), 6 (32, 26. 16) is the axis of a 3" right circular cone. The point c (28, 0, 16) is the center of the base of a right cir- cular cone, 3" in diameter and 4:\" high. Find the hne of intersection by the concentric spheres method. Place the sheet vertically. APPENDIX 293 EXERCISE SHEET LXV a (§§ 153, 154) 1. The line a (69, 4, 10), 6 (62, 17, 23) is the axis of a cyHnder whose base is a 1" circle parallel to H. Determine a point c, on the cylinder, 8 from H and 16 from V. Through c draw a plane Z, tangent to the cylinder. 2. The point d (34, 0, 10) is the apex of a 60° right circular cone whose axis is perpendicular to H. Through e (39, 4, 5) draw the planes X and Y, tangent to the cone. (An auxiliary H plane passed through the base of the cone will be found useful.) 3. The line/ (14, 7, 0), q (4, 4, 16) is the axis of a cone whose base is a IV' circle in V. Through h (2, 18, 7) draw the planes V and TF, tangent to the cone. (In determining the traces, do not overlap prob- lem 2.) EXERCISE SHEET LXVI a (§§ 155, 156) Draw a cjdinder and two cones like those for sheet LXV a. 1. Pass a plane Z, tangent to the cylinder and parallel to the line a (61, 10, 4), 6 (53, 4, 5). 2. Through the point c (44, 7, 20) ; draw three Hnes cd, ce, and c/, each tangent to the cone. 3. Through g, which lies on the cone, 8 from H and 6 from V , draw a line hj, tangent to the cone. EXERCISE SHEET LXVH a (§157) In the upper left-hand corner of the sheet, draw a right circular cyhnder, 1" in diameter and 2" long, in a vertical position. Let this cyHnder be intersected by an equal cylinder in a horizontal position; and let this latter be intersected by still another which is inclined at 45° to H. Let the length of all three cylinders be equal and let all the axes intersect. Draw the development of each. On the develop- ments, and also on the elevation, draw a helix, inclined at 30° to the axes of the cylinders, and which is continuous on all three parts. EXERCISE SHEET LXVIH a (§§158-160) Draw the projections of a developable helicoid of two nappes, based on a cylinder \\" in diameter and 3" high. Place the sheet vertically with HA in the center. Locate the center of the base of the cylinder at a (28, 0, 28). Draw carefully to bring out the separate nappes, and to show the visible and invisible portions. 294 DESCRIPTIVE GEOMETRY EXERCISE SHEET LXIX a (§§163,164) Draw the projections of a If" cube placed with one body diagonal, parallel to HA. Rotate the cube on this diagonal and draw the pro- jections of the surface that results. Before concluding the exercise, the student should account for the surface generated bj'- each of the 12 edges. EXERCISE SHEET LXIX b (§§163,164) Repeat sheet LXIX a, using a right parallelepiped, 1" X 2" X 3". EXERCISE SHEET LXX a (§165) Place the sheet vertically and divide into four equal parts. The larger sheet will give better results. Draw the plan and elevation of four different warped surfaces, including at least one whose generation is dependent on a time factor. Show the directrices of each surface. Let at least one of the surfaces be cut by a plane perpendicular to H or V and trace the line of intersection. EXERCISE SHEET LXXI a (§§ 164, 165) Place the sheet vertically with HA in the center. Draw the line a (28, 0, 20), 6 (28, 36, 20). Let ah be the axis of a right heHcoid (§ 165), making two turns in the height. Let the lines c (16, 0, 38), d (50, 36, 38), and e (9, 36, 2), / (46, 0, 2) and the plane H be the directrices of an hyperboHc paraboloid. Find the line of intersection of the surfaces. EXERCISE SHEET LXXII a (§§166-172) Place the sheet vertically. Take HA in the center. Draw the projections of a torus generated by rotating a 1|" sphere about a vertical axis passing through m (36, 10, 20). Let the center of the sphere be l\" from the axis. Locate the points o and o', on the torus, 24 from P and 13 from V. Find the traces of the planes Z and Z' which are tangent to the torus at o and o'. Draw two lines tangent to the torus and one which pierces the torus at o and at another point, q, to be determined. EXERCISE SHEET LXXIII a (§173) Place the sheet vertically, HA 3|" from the top. Let a (34, 11, 11) be the center of a 2|" sphere and b (46, 25, 24) be the apex of a cone tangent to the sphere. Draw the projections. Extreme accuracy of construction is necessary if the points of tangency are to check properly on cone and sphere. APPENDIX 295 EXERCISE SHEET LXXIV a (§§175-177) Place the sheet vertically, HA in the center. Let a (25, 0, 18) be the center of the base of a 45° right circular cone 4" m diameter. Let b (34, 8, 20) be the center of an annular torus formed by rotating a 1|" sphere about a vertical axis through b, the radius of rotation being 1|" to the center of the sphere. Find the line of mtersection. EXERCISE SHEET LXXIV b (§§175-177) Make up a sheet similar to Fig. 185, using a vase form like that shown m Fig. 183, intersected by a right circular cylinder. Draw the figures as large as possible. EXERCISE SHEET LXXV (§181) Construct two sections through the building shown in Fig. 7, at about three times the scale of the figure. Let one section be cut by a plane perpendicular to the longitudinal plan axis and the other by a plane perpendicular to H and at 45° to the same axis. [Note. Figures 194, 197, 198 may be used for a further study of sections in a manner similar to the above.] EXERCISE SHEET LXXVI (§182) Construct a plan and elevation of a corner rafter similar to Figs. 188 and 189, but at three times the scale. Design a different outline for the jack rafter and increase or decrease the pitch of the roof. Deter- mine the proper shape for the hip rafter. This problem may be varied by requiring a reentrant corner to be drawn. EXERCISE SHEET LXXVII (§183) Construct a plan and elevation similar to Fig. 191, but at three times the scale. Find the openings required for a shaft, the right section of which is an equilateral triangle, If" on a side. EXERCISE SHEET LXXVIII (§§184-187) Use a large sheet, placed vertically and divided into four parts as shown on p. xii. Omit HA lines. Construct an elevation and a roof plan similar to Fig. 197 for each of the cases given in the table below. Use jV" sscale. The dimensions in the table are in feet and are to be used to replace the letters in Fig. 197. 296 DESCRIPTIVE GEOMETRY APPENDIX 207 Letters on Figure 197 Case ahcdefgh 1 40 20 20 40 25 25 10 10 2 40 20 20 40 12 30 8 12 3 40 20 20 40 25 10 12 8 4 Using dimensions as in 2, above, let the low wing be revolved on until the angle is 60°. [XoTE. Donners similar to those in Figs. 192 / and 198 may be added if desired. Let all roofs have a slope of 45° and the extreme overhang in anj' direction be 2' — 0".] EXERCISE SHEET LXXIX (§§184-187) Use a large sheet placed horizontally. 1. Figures 259, 260 show a floor plan and two elevations of a build- ing with a somewhat complicated roof. Draw a roof plan an place of the floor plan in the figure), four elevations, and two sections at ■g^" scale. Sufficient mformation is given to make it possible to con- struct the required drawings. Particular attention should be given to points A and B, in working out the roof. 2. Cast sun shadows on the Dlan and front elevation. 6i' 'z. N W-hD— E t-» All eaves project i-s' All roofs e"1hic.k.. .^ Pitch as shown below. ROOf PUM. 'I -* Fig. 260 298 DESCRIPTIVE GEO.METRY All wolls and roofs I'-oMhick. All roofs slope A5' and pro- ject 2' beyond walls. fT. is' above ground for all main roob; 3?' for tower. P'iG. 261 APPENDIX 299 EXERCISE SHEET LXXX (§188) Use a large sheet, placed vertically. Figure 261 shows a building with a tall chimney. Draw the plan and elevation at I" scale. Four guy wires are to be attached to the stack. They shall be in planes perpendicular to H and at 45° with F, and passing through the center of the chimney. The wires are to make -angles of 45° with H. Find where the wires will be attached to the building and the angles between the wires and roof. EXERCISE SHEET LXXXI (§189) Use a large sheet placed vertically. Draw the plan, ele- vation, and section of a hipped roof frame similar to Fig. 199, except that the end plane of the roof shall be inclined at 45° to H while the side planes remain at 30°. Deter- mine all 10 angles as described in § 189. Let the total width be 11' - 0"; rafters 3" X 8". Use f " scale. EXERCISE SHEET LXXXII (§ 189) Use large sheet placed vertically. Draw plan, eleva- tion, and section and determine all 10 angles as described in § 189 for the hipped roof frame shown in Fig. 262. Use |" scale. EXERCISE SHEET LXXXIII (§190) Draw the plan and section of a sheet metal chute similar to Fig. 204, #2, at f" scale. Draw the plan, elevation, and development of the clip at 3" scale and determine the angle of the bend, after the manner of Figs. 206 and 207. Use large sheet. Rafters is" c-c All timbers 3" thick. Roof planer moke 30° with H. Fig. 262 300 DESCRIPTIVE GEOMETRY EXERCISE SHEET LXXXIV (§§191, 192) Use large sheet placed horizontally and divided vertically through the center. 1. (On left half of the sheet.) Draw the plan and elevation of a vase form similar to Figs. 183, 185, or 242, using as large a scale as convenient. Assume the vase to be turned from a stick of wood with the grain centered as in Fig. 208. Use not less than 10 annual rings. Find the grain pattern on the elevation. 2. (On right half of the sheet.) Draw the plan and elevation of a spiral chute like Fig. 209, at ^" scale. Let the story height be 10' — 0" in the clear; floor thickness 1' — 0"; diameter of center support 0' — 8" ; diameter of outside of chute 7' — 0" ; vertical height of edge 0' — 9". Develop outer and inner helixes at l" scale. Develop well lining at ^" scale. Draw plan, elevation, and approximate de- velopment of Y^j of one turn of the helicoidal bottom at H" scale. Figure all details to show accurate sizes. EXERCISE SHEET LXXXV (§193) Use large sheet. Figure 263 shows the layout. 1. Find the line of intersection of both the inner and outer surfaces of the vaults. Fig. 263 APPENDIX 301 2. (a) The line of intersection of two vaults is given ; also the section of the larger vault ; find the right section of the smaller vault. (b) Assume an arbitrary hne of intersection on plan and find the corresponding section of the small vault. 3. A spherical dome intersected by two vaults is given. Find the line of intersection both on plan and elevation. 4. A spherical dome and two intersecting vaults. Find the line of intersection on plan. EXERCISE SHEET LXXXVI (§§194-197) Use large sheet or four small ones. Layout is given on Fig. 264. 1. Find the line of intersection of the membered mouldings, both on plan and elevation. Fig. 264 2. A vertical wall and a ceiling sloping at 45° to H meet in a corner. A picture moulding passes from the vertical wall to the sloping ceiling and miters in the corner on a plane which bisects the angle between the wall and ceiling. Determine the section for the rake moulding and show a full plan of the mouldings including the line of intersection. 3. Mouldings on a pediment, mitering at 45°. Determine the section for the raking moulding. 302 DESCRIPTIVE GEOMETRY 4. A double raking moulding, mitering on the bisecting plane. Determine the section of the raking moulding. EXERCISE SHEET LXXXVII Develop patterns for the execution in sheet metal of the membered mouldings of Exercise LXXXVI, #1. EXERCISE SHEET LXXXVIII Investigate the variation in the right section of the rake moulding in Exercise LXXXVI, #3, as the angle of inclination of the raking mould- ing varies between 0° and 90°. EXERCISE SHEET LXXXIX (§198) Use either a large or a small sheet. Vary scale to suit. Draw the plan, elevation, and two sections of a spherical dome similar to Fig. 221, but at a larger scale. Let the dome be supported on an hexagonal drum and spherical pendentives. EXERCISE SHEET XC f§ 199) Use two small sheets. L Draw a half plan and elevation of a spherical dome, about 6" in diameter. On the plan draw a pattern like that in Fig. 222, making E L E V AT I H Fig. 205 S ICT I N Z-Z APPENDIX 303 the diagonals of the large squares 1". Project the pattern on the elevation. 2. After studying the above problem with reference to the dis- tortion of tlie pattern on the lower part of the dome, develop a different pattern, based on equally spaced meridian sections and horizontal sections at variable spacings. EXERCISE SHEET XCI 1. Figure 265 shows a circular tower. The central portion is covered by a spherical dome and the circular passage is covered by a vault in the form of an annular torus. The door openings are connected by, and form a part of, a conical vault. Draw one half of the plan, an elevation, and a section at j\" scale, showing all penetration hues. 2. Change the conical vault in the above problem to the conoidal form shown in Fig. 178, and repeat. EXERCISE SHEET XCII Figure 266 shows the plan of a recessed doorway. Let the open- mgs A and B each be spanned by a semicircular arch and the space Fig. 266 between be covered by a warped surface. Draw a plan and elevation at h" scale, indicating stone jointing on the basis of 9 voussoirs for each arch. Let the maximum dimension of any stone be 2' - 9". Show three directrices for the warped surface. 304 DESCRIPTIVE GEOMETRY i' 6 EXERCISE SHEET XCIII Figure 267 shows the plan of a semi- circular headed doorw^ay. 1. Determine how far the door may be opened before it interferes w4th the ^. Fig. 267 arch over the door. Scale 1" = 1' - 0". 2. Determine a curve for the arch soffit that vdll allow the door to be opened through 90°. 3. Determine an arrangement that will permit of opening the door through 180°. EXERCISE SHEET XCIV (§200) Draw the projections of two intersecting barrel vaults like those shown in Fig. 222, making the larger 48' in diameter and the smaller 40'. Use i" scale. Draw on the vaults a pattern similar to that in the illustration. EXERCISE SHEET XCV (§§202-207) 1. A light is situated at the point I (77, 21, 19). Cast the shadow of the line a (74, 13, 8), h (63, 5, 12). 2. Cast the shadow of the line c (45, 11, 7), d (45, 11, 0), as cast from the point I in problem 1 above. 3. Cast the shadow of the Hne e (28, 22, 3), / (28, 10, 3), as cast by a light at l' (2, 26, 26). 4. Cast the shadow of the line g (19, 12, 22), h (11, 12, 13), as cast from the point l' in problem 3 above. EXERCISE SHEET XCVI (§§202-207) Cast the sun shadows of the following Hnes. 1. The line a (73, 15, 6), b (73, 0, 6). 2. The hne c (61, 20, 4), d (50, 14, 4). 3. The line e (43, 14, 11), / (43, 14, 0). 4. The line g (33, 20, 8), h (26, 13, 1). 5. The line k (20, 2, 20), I (6, 12, 3), and another line which passes through j (17, 15, 7) and bisects kl. EXERCISE SHEET XCVU (§208) 1. The point p (76, 23, 24) is a source of light casting the shadow of a regular hexagon with its center at a (66, 9, 11), and lying in a plane parallel to H. APPENDIX 305 2. Cast the sun shadow of a regular octahedron, 1|" on a side, and with ihe center of the top face at 6 (28, 17, 14). EXERCISE SHEET XCVIII (§209) 1. The point a (53, 27, 27) is a source of Hght and the soHd shown in Fig. 268 is located with the center of its base at b (58, 0, 7). Cast the shadow of the solid on H and V and also on the soUd itself. ■ o ^T^ 1' J, 2i 8 i.L :4- I Fig. 268 5-6 3*-6" 6" f 9r- FiG. 269 2. The steps shown in Fig. 269 are located with the point c at (27, 0, 14). Cast the shadows as from the source of light given above. Scale, f" = r. EXERCISE SHEET XCIX (§209) Repeat Sheet XCVIII, casting the sun shadows of the given objects. EXERCISE SHEET C (§210) Draw two elevations of a dormer window similar to Fig. 232, but at a larger scale. Change the source of light. Cast the shadows. EXERCISE SHEET CI (§210) Repeat Sheet C, casting the sun shadows. 306 DESCRIPTIVE GEOMETRY EXERCISE SHEET CU (§211) A circle 2" in diameter has its center at a (39, 10, 12), and its plane parallel to H. A source of light is at h (72, 28, 28). Cast the shadow of the circle. EXERCISE SHEET CIII (§211) 1. The point a (70, 15, 7) is the center of a circle 2" in diameter whose plane is parallel to V. Cast the sun shadow. 2. The point h (46, 8, 24) is the center of a circle 2" in diameter whose plane is parallel to V. Cast the sun shadow. 3. The point c (26, 8, 8) is the center of a circle 2" in diameter whose plane is parallel to P. Cast the sun shadow. EXERCISE SHEET CIV (§212,1) 1. The point a (31, 12, 28) is a source of light. The point h (40, 17, 15) is the apex of a cone whose base is a circle, parallel to H, and 2" in diameter. The center of the base is at e (50, 4, 10). Find the shade and shadow of the cone. 2. The point c (31, 28, 28) is a source of hght. The point d (15, 16, 8) is the apex of a right circular cone whose base is a circle in H and 1^" in diameter. Find the shade and shadow of the cone. EXERCISE SHEET CV (§212,11) The points a (66, 5, 14), h (35, 5, 14), and c (14, 5, 14) are the base centers for three right circular cones, whose bases are each 2" in diameter and parallel to H. The first is 2" high, the second is f " high, and the third is |" high. Cast the sun shadows of all three cones. EXERCISE SHEET CVI (§213) The point a (68, 19, 21) is a source of light. The point c (53, 8, 10) is the center of a sphere \\" in diameter. Cast the shadow of the sphere. EXERCISE SHEET CVII (§213) Place the sheet vertically, HA in the center. The point c (40, 15, 16) is the center of a sphere 2" in diameter. Using the sun as a source of hght, determine the shade line and cast the shadow of the sphere on H and V. EXERCISE SHEET CVIII (§214) Place the sheet vertically, HA 3^" from the top. The point a (54, 24, 25) is a source of light; h (38, 10, 8) is the center of a sphere 1^" APPENDIX 307 in diameter; c (50, 11, 16) d (36, 20, 12) is a line. Determine the shade Une and shadow of the sphere, also the shadow of cd on the sphere and on the planes of projection. EXERCISE SHEET CIX (§214) Cast the shadows on the band stand shown in Fig. 270, when the source of light is at x. Scale, Y' = 1' — 0". EXERCISE SHEET CX (§215) Cast the shadows on the band stand shown in Fig. 270, when the source of light is at y. Scale, I" = V — 0". Fig. 270 308 DESCRIPTIVE GEOMETRY EXERCISE SHEET CXI (§216) Determine the shade hne and shadow, also the shadow on //, for the double-curved surface in Fig. 271. Use a point source of light on the left border line, 11^" above H and 9|" in front of V. Fig. 271 EXERCISE SHEET CXII (§216) With the sun as the source of light, determine the shade line and shadows, also the shadows on H and V, for the double-curved surface in Fig. 272. Fig. 272 APPENDIX 309 EXERCISE SHEET CXIII (§216) With the sun as the source of Hght, determine the shade line and shadows, also the shadows on H and F, for the baluster in Fig. 273. Place the center of the base on H, V — 6" from V and 1' — 6" from the left border line. Scale, 3" = 1' - 0". EXERCISE SHEET CXIV (§216) Cast the sun shadows on a warped doorway recess, similar to Fig. 223. In drawing the doorway, let the front arch be semicircular, 7' — 0" in diameter; and let the arch over the door be segmental, 3' — 0" radius and 4' — 0" span. Let the depth of the recess be 2' — 6". Select any convenient line for the third directrix of the warped surface. Scale, I" = 1' — 0". Small sheet. Fig. 273 EXERCISE SHEET CXV (§216) Cast the sun shadows on Fig. 274. Scale, \" = 1' — 0". EXERCISE SHEET CXVI (§220) Fig. 274 Figure 275 gives the eleva- tions of various points on a certain piece of land and the course of a stream crossing it. Draw the contour map at a scale of 1" = 20' - 0", indi- cating the contours at 5' in- tervals. Let it be required to construct a road, 20' wide, on a level grade between points A and B. Assume the natural slope of the soil to be 30°. Show the cuts and embank- ments and the lines of cut and mi. 310 DESCRIPTIVE GEO:^IETIlY EXERCISE SHEET CXVn (§220) Using Fig. 275 as above, draw the contours and show a road 20' wide between points C and D. Let it be built on a 12 per cent grade. Show the lines of cut and fill and draw the contours for the finished embankments. EXERCISE SHEET CXVIII (§222) Draw the isometric projection of the Celtic Cross in Fig. 276. Scale, f" = 1' - 0". Let the thickness of the shaft be 1' — 0". EXERCISE SHEET CXIX (§§223, 224) Divide the sheet into three parts and draw the isometric projections of the following curved sur- faces : (1) A right circular cone, 2¥' in diameter and 4" high ; (2) a right hehcoid, 2" in diameter and 5" high; (3) a sphere, 2f" in diameter. EXERCISE SHEET CXX (§224) Draw the isometric projection of one of the following vase forms : Figs. 239, 241, 242, 208, 271, 272, or 273. The envelope method will be found useful. EXERCISE SHEET CXXI (§224) Draw the isometric projection of Fig. 221. UNIVERSITY OF CALIFORNIA LIBRARY This book is DUE on the last date stamped below. Fine schedule: 25 cents on first day overdue 50 cents on fourth day overdue One dollar on seventh day overdue. £NQ DEC ^^ W JUN 4 £ 1S48 7 NEERING LiBFtA«¥ iio\ DEC y- 1948 tEti i 5 "^AY 2 2 IS50^ JUL 31 1 LD 21-100m-12,'46(A20l|si6)4120 Tb <:0 7oo 454780 ^r.gtneering Library UNIVERSITY OF CALIFORNIA LIBRARY i;l II,