sltfi-G trrtfetJiM t & Series of Crxts on topics in IHnQtncrrino; ISbitrt by laarlc foagmmrti |^£*Drtck DESCRIPTIVE GEOMETRY THE MACMILLAN COMPANY NEW YORK • BOSTON - CHICAGO • DALLAS ATLANTA • SAN FRANCISCO MACMILLAN & CO., Limited LONDON • BOMBAY • CALCUTTA MELBOURNE THE MACMILLAN CO. OF CANADA, Ltd. TORONTO DESCRIPTIVE GEOMETRY BY ERVIN KENISON ASSOCIATE PROFESSOR OF DRAWING AND DESCRIPTIVE GEOMETRY MASSACHUSETTS INSTITUTE OF TECHNOLOGY AND HARRY CYRUS BRADLEY A.SSI8TAMT PROFESSOR OF DRAWING AND DESCRIPTIVE GEOMETRY MASSACHUSETTS INSTITUTE OF TECHNOLOGY Ncto ©ork THE MACMILLAN COMPANY 1918 All rights reserved Copyright, 1916 and 1917, By THE MACMILLAX COMPANY. Set up and elcctrotyped. Preliminary edition published October, 1916. Complete edition, October, 1917. ICormooU ^rcss J. S. Cushing Co. — Berwick & Smith Co. Norwood, Mass., U.S.A. PREFACE This book represents a teaching experience of more than twenty years on the part of both of the authors at the Massa- chusetts Institute of Technology. In presenting such a book for the consideration of other institutions and of the public at large, it is well to note the relation of this course in de- scriptive geometry to the general study scheme, in order that the point of view of the authors may be better understood. In any professional drawing of an engineering or architec- tural nature, especially if the element of design occurs, the draftsman or designer must see clearly the conditions in space. This has been well expressed by saying that he must be able to think in three dimensions. Such an ability is natural to but few. Fortunately, however, it is a power which can be acquired, more or less readily, by the great majority. This power can be gained, and it usually is by the so-called " prac- tical " draftsman, by the simple process of making and re- making working drawings. But the authors believe that the same power can be acquired much more rapidly, and when acquired can be more forcefully and efficiently applied in de- signing, through the study of descriptive geometry. The point of view in this text is therefore that of the drafts- man. Mathematical formula; and analytic computations have been almost entirely suppressed. It has been found that the students readily apply their knowledge of the theoretical mathematics to a finished drawing. For example, they make trigonometric computations from drawings with considerable facility. On the other hand, in applying even the simplesl principles of solid geometry during the construction of a drawing the student, is often anything bul facile. The method of attack throughout this book is intended to be that which v vi PREFACE shall most clearly present the actual conditions in space. Wherever experience has shown that a simple plan and ele- vation are not amply sufficient for this purpose, additional views or projections have been introduced freely, correspond- ing to the actual drafting practice of making as many side views or cross sections as may be needed. In the matter of arranging the views in a practical drawing, there has been, still is, and probably always will be, contro- versy. Architects, who must embellish their drawings with shades and shadows, as well as some engineers, prefer to place the plan below the elevation, according to the so-called first quadrant projection. Entire text-books, noticeably the older ones, have been prepared with all the objects and all the theory studied in the first quadrant. Other engineers, me- chanical especially, insist that the only proper placing of the views is to have the plan above the elevation, according to the method of the third quadrant. In consequence, some recent text-books have been issued with all the work in the third quadrant. The authors have no desire to take sides in this controversy, but believe in giving the student practice in both of these methods. Hence some of the problems are studied in the first quadrant, others in the third. It has been found that the student has no greater difficulty in working in one quadrant than in the other, while considerable additional power is gained, and a clearer insight into space in general is obtained, by not restricting the work to a particular corner of space. A requirement of practical drafting is that the construction shall be confined to the limits of the drawing board. There is always a limit beyond which points cannot be made avail- able. A similar requirement, namely, that all the points used in the construction must be within the limits of the figure, is rigidly insisted upon in the authors' classes. This frequently involves an auxiliary construction not embodied in the gen- eral theory of the problem. Numerous auxiliary constructions of this kind are given throughout the book. The additional grasp of the subject obtained by being obliged thus to force PREFACE vil his way through whatever difficulties may arise, has been found of great value to the student in his subsequent work. The amount of ground covered by this book is that which is considered sufficient to enable the student to begin the study of the technical drawings of any line of engineering or architecture. It is not intended to be a complete treatise on descriptive geometry. Detailed exposition of such branches as shades and shadows, perspective, stereographic projection, axonometry, the solution of spherical triangles, etc., will not be found. The student is prepared by the present text, how- ever, to take up any of these subjects. The notation adopted in this book is a modification of that first suggested and used by Professor William Watson of the Institute of Technology (around 1870). The modifications have been made from time to time by the authors in the course of their teaching experience. The authors have exam- ined many books ; and when opportunity offered, they have questioned students coming from other institutions in their search for a clear system of notation. While the system here given is not above criticism, the authors are firmly of the opinion that it is the best that has ever come to their attention. Although this book has been prepared with the needs of one particular institution in mind, it is hoped that it will not be found wanting in general interest. ERVIN KENISON, HARRY CYRUS BRADLEY. Massachusetts Ifstiti n of Technology, September, 1917. CONTENTS PAGE Introduction 1 Descriptive Geometry — Definition — Visualization — Practical Applications. I. Elementary Principles : the Coordinate Planes . 3 Orthographic Projection — Orthographic View — Projection on a Single Plane — Plane of Projection — Choice of the Planes of Projection — The Horizontal and Vertical Coordinate Planes — Names of the Views — Plan — Elevation — Position of the Observer — The Ground Line — The Four Quadrants — Relation be- tween an Object and Its Projections — Relation between Two Projections of an Object — Notation — Actual Projections. II. Projections of the Point and of Simple Solids . 10 Notation of the Point — Projections of Isolated Points in All Four Quadrants — Projectors — Visualizing the Quadrants — Uses of the Quadrants — Distances from H and V — Special Positions of the Point — Projections of Simple Solids — Visibility of Solid Objects — Illus- trative Projections of the Prism, Pyramid, Cylinder, Cone, Sphere, Torus. III. Represf;ntation of the Straight Line: Traces . 19 The Straight Line — Its Projections — Notation — Vis- ualization — Slope — Special Positions — The Profile Line — A Point Lying in a Line — Traces of a Line. Problem 1. To find the traces of a straight line . . 24 Problem 2. Given the two traces of a straight line, to find ils projections ........ 25 IV. Simple Shadows ........ 26 Shadows — Shadow of a Point — Opaque and Trans- parent Planes — Shadow of a Line — Of Surfaces — Of a Convex Solid. x CONTEXTS PAGE V. Representation of the Plane ..... 29 The Plane — Its Representation — Traces — Notation — Visualization — Slope — Special Positions — Edge View. VI. The Profile Plane of Projection .... 33 The Profile Plane — A Profile Projection — Relation of Profile Projection to H- and F-Projections — Profile Projections in All Four Quadrants — Profile Projection of the Straight Line — Problems on the Profile Line — Its True Length — Its Traces on H and V — Projections of a Point Lying in the Line — Profile Trace of a Line — Of a Plane — Left-Side Views — Conventional Placing of Views — Second Method of Obtaining Pro- file Projections. VII. Secondary Planes of Projection .... 46 Secondary Planes of Projection — The Profile Plane as a Secondary Plane — Secondary Ground Lines — Prin- ciples of Secondary Projection — Simplification of Prob- lems — End View of a Line — Edge View of a Plane — Secondary Projections of Solids — The Two 7-Pro- jections Compared — Oblique Secondary Planes. VIII. Revolution of a Point : True Length of a Straight Line : Applications ..... 52 Revolution of a Point about a Straight Line — Axis of Revolution Perpendicular to H or V — Axis Lying in a Given Plane — True Length of a Line — Angles which a Line Makes with H and V — Converse Prob- lems. Problem 3. To find the true length of a line (Two methods) 54, 56 Problem 4. To find the projections of a line of definite length, when its slope, the angle which it makes with one coordinate plane, and the direction of its projection on that plane, are known ....... 58 Problem 5. To find the projections of a line making given angles with H and V 60 CONTENTS Xi PAGE IX. Some Simple Intersections : Developments . . 62 Simple Intersections of Solids of Revolution — Visibility of the Intersection — A Plane Perpendicular to H or V Intersecting a Cone, Sphere, Torus, Hyperbolic Spindle — Developments — Working Method for Finding the True Length of a Line — Development of a Solid with Plane Faces — Projection of a Prism Whose Long Edges Make Given Angles with H and V. X. Lines in a Plane : Parallel Lines and Planes . 71 Intersecting and Parallel Lines — Parallel Lines in Space — Intersecting Lines in Space — Test for Intersecting Lines — A Line in a Plane — A Plane Containing a Given Line — Principal Lines of a Plane — To Pro- ject the Principal Lines — A Plane Containing a Line Parallel to H or V — Parallel Planes — A Line Parallel to a Plane — A Plane Parallel to a Line — The Plane Determined by Two Intersecting Lines, Two Parallel Lines, a Line and a Point not on the Line, Three Points not in the Same Straight Line — Planes Parallel to Lines or to Other Planes — Use of Auxiliary Lines in Finding the Traces of Planes. Problem 6. To find the plane which contains two given intersecting or parallel lines ...... 79 Corollary I. To find the plane which contains a given line and a given point ....... 82 Corollary II. To find the plane which contains three given points not in the same straight line ... 82 Problem 7. To find the plane which contains a given line and is parallel to a second given line .... 84 Problem 8. To find the plane which contains a given point and is parallel to each of two given lines . . 80 Problem 9. To find the plane which contains a given point and is parallel to a given plane ..... 86 XI. Perpendicular Lines and Planes .... 91 Perpendicular Lines — Perpendicular Lines Whose Pro- jections are Perpendicular — A Line Perpendicular to a Plane — Test for Perpendicularity of a Line and Plane — Perpendicular Planes — Lines of Maximum Inclina- xii CONTENTS PAGE tion to H and V — Planes Perpendicular to a Given Line or Plane. Problem 10. To find the plane which contains a given point and is perpendicular to a given line ... 95 Problem 11. To find the plane which contains a given line and is perpendicular to a given plane .... 96 XII. Intersection of Planes and of Lines and Planes : Applications ........ 98 Intersecting Planes — Line of Intersection of Two Planes — Intersection of a Line and a Plane — Shortest Dis- tance from a Point to a Plane — Projection of a Point or Line on a Plane. Problem 12. To find the line of intersection of two planes 98 Problem 13. To find the point in which a straight fine intersects a plane ........ 103 Problem 14. To find the shortest distance from a point to a plane .......... 105 Problem 15. To project a line on an oblique plane . . 106 XIII. Intersection of Planes and Solids Bounded by Plane Faces ........ 108 A Plane Determined by Two Lines — Intersection of a Line with a Plane Determined by Two Lines — Inter- section of Two Limited Plane Surfaces — Visibility — Intersection of a Plane and a Pyramid — Of Two Solids Bounded by Plane Faces — Of a Sphere and a Prism — Of a Sphere and a Right Cylinder — Of a Cylinder and a Torus. XIV. Problems Involving the Revolution of Planes . 125 A Line Lying in a Given Plane — A Point in a Plane — Revolution of a Plane about an Axis Perpendicular to H or V — Distance between Two Parallel Planes — Angles between an Oblique Plane and the Coordinate Planes — Planes making Given Angles with H and V. Problem 16. Given one projection of a line lying in a plane, to find the other projection ...... 125 Corollary I. Given one projection of a point lying in a plane, to find the other projection .... 128 CONTEXTS xiii PAGE Corollary II. To find the second projection of a line lying in a plane when the general solution fails, partially or wholly 129 Corollary III. To find a line of maximum inclination of a plane .......... 129 Problem 17. To find the perpendicular distance between two parallel planes ....... 131 Corollary. To find the perpendicular distance from a point to a plane ........ 132 Problem 18. To find the angles which an oblique plane makes with H and V ...... 132 Problem 19. Given one trace of a plane, and the angle which the plane makes with either H or V, to find the other trace of the plane ...... 134 Problem 20. Given the angles which a plane makes with H and V, to find the traces of the plane .... 135 XV. Other Problems Involving the Revolution of Planes 139 Revolution of a Plane about One of Its Traces — Re- volved Position of a Point in a Plane — Working Rule — Solution of Plane Problems by Revolution of the Plane — Angle between Two Intersecting Lines — Angle between a Line and a Plane — A Rectilinear Figure Lying in a Plane Problem 21. To find the position of a point lying in a plane, when the plane is revolved into // or V about the corre- sponding trace . . . . . . • b • 139 Corollary. To find the revolved position of a line lying in a plane, when the plane is revolved into // or V about the corresponding trace ...... 142 Problem 22. To find the angle between two intersecting lines 143 Corollary. To find (he projections of the bisector of the angle between two intersecting lines .... 145 Problem 23. To find the angle between ;i line and a plane 147 Problem 24. Given one complete projection, and three points iii the older projection of a plane polygon, to complete the projection ...... 142 xiv CONTENTS PAGE Corollary I. To find the true size and shape of a plane figure .......... 150 Corollary II. To find the projection of the line which bi- sects one of the interior angles of a plane polygon . . 151 XVI. Miscellaneous Problems of the Line and Plane 152 Distance from a Point to a Line — Distance between Two Lines — Angle between Two Planes — Application. Problem 25. To find the shortest (perpendicular) distance from a point to a line ....... 152 Problem 26. To find the shortest distance between two lines not in the same plane, and the projections of their common perpendicular ....... 155 Problem 27. To find the angle between two planes . . 158 Problem 28. Given the horizontal traces of two planes, and the angle each plane makes with H, to find the line of intersection of the two planes . . . . .162 Corollary. To find the angle between the given planes 163 XVII. Counter-revolution of Planes .... 164 Counter-revolution of a Point, a Line, a Plane Figure — A Line of Given Length Perpendicular to a Plane — Projections of a Prism or Pyramid Whose Axis is In- clined to Both H and V. Problem 29. Given the position of a point lying in a plane after the plane has been revolved into H or V about the corresponding trace, to find the projections of the point 164 Corollary. To counter-revolve a plane polj T gon . . 166 Problem 30. At a given point in a plane, to draw a line which shall be perpendicular to the plane and of given length 168 XVIII. Tangent Lines and Planes : General Prin- ciples .......... 174 Curves — Plane Curves — Space Curves — Projections of Curves — Tangent Lines — Tangents to Plane Curves — Tangent Planes to Curved Surfaces — Determina- tion by Means of Tangent Lines — Rectilinear Ele- ments — The Normal — Determination of Tangent Planes by Means of the Normal. CONTEXTS XV XIX. Tangent Planes to Cones and Cylinders . . 178 The Cone and Cylinder — Definitions — Representation — Projections of Cones and Cylinders — Projections of a Point in the Surface. — Tangent Planes to Cones and Cylinders at a Given Point in the Surface — Through a Given Point without the Surface — Parallel to a Given Line. Problem 31. To pass a plane tangent to a cone at a given point in the surface ....... 184 Problem 32. To pass a plane tangent to a cone through a given point without the surface. (Two results.) . . 189 Problem 33. To pass a plane tangent to a cone parallel to a given line. (Two results.) ...... 192 Problem 34. To pass a plane tangent to a cylinder at a given point in the surface ...... 194 Problem 35. To pass a plane tangent to a cylinder through a given point without the surface. (Two results.) . 196 Problem 36. To pass a plane tangent to a cylinder parallel to a given line. (Two results.) ..... 198 XX. Tangent Planes to Double Curved Surfaces of Revolution ......... 201 Double Curved Surfaces of Revolution — Represen- tation — Meridians — Principal Meridian Plane — Parallels — Projections of a Point in the Surface — — The Sphere — The Torus — Tangent Planes to Double Curved Surfaces of Revolution at a Given Point in the Surface — Tangent Planes Which Contain a Given Line. Problem 37. To pass a plane tangent to a sphere at a given point in the surface ....... 206 Problem 38. To pass a plane tangent to a double curved surface of revolution at a given point in the surface . 207 Problem 39. To pass a plane tangent to a sphere through a given line without the surface ..... 212 Problem 40. To pass a plane tangent to a double curved surface of revolution through a given line. (Special cases.) . . . . . . . . . .216 xvi CONTENTS XXI. The Intersection of Curved Surfaces by Planes 219 Classification of Curved Surfaces — Ruled Surfaces — Surfaces of Revolution — Other Curved Surfaces — In- tersection by a Plane of a Ruled Surface — Of a Sur- face of Revolution — Of Any Curved Surface — Method by Means of Secondary Projections — Visibility of the Intersection — A Rectilinear Tangent to the Curve of Intersection — Development of a Curved Surface. XXII. Intersection of Planes with Cones and Cylin- ders . 222 Intersection of a Cone or Cylinder with a Plane — A Line Tangent to the Section — Development of the Curved Surface and Tangent Line — Intersection of a Pyramid or Prism with a Plane. Problem 41. To find the intersection of a cone and plane 222 Problem 42. To find the intersection of the frustum of a cone and a plane ........ 232 Problem 43. To find the intersection of a cylinder and a plane .......... 236 Problem 44. To find the intersection of a pyramid or prism and a plane ......... 244 XXIII. Intersection of Planes with Surfaces of Revolution ......... 247 Intersection of a Surface of Revolution and a Plane — Points Determined by Meridian Planes — A Line Tan- gent to the Intersection — Number of Curves in the Case of a Torus Intersected by a Plane. Problem 45. To find the intersection of a double curved surface of revolution and a plane ..... 247 XXIV. The Intersection of Curved Surfaces by Curved Surfaces ....... 256 Intersection of Two Curved Surfaces — The Auxiliary Surfaces — Visibility of the Intersection — A Line Tangent to the Curve of Intersection. CONTENTS xvii XXV. The Intersection of Cones and Cylinders with Each Other 258 Intersections of Cones and Cylinders — Of Two Cones, Two Cylinders, a Cylinder and a Cone — Auxiliary Planes Which Cut Elements from Each Surface — Num- ber of Curves — General Cases — Special Cases — Limiting Planes Doubly Tangent — Parallel Elements — Infinite Branches — Intersection of Cylinders and Cones of Revolution. Problem 46. To find the intersection of two cylinders . 260 Problem 47. To find the intersection of a cylinder and a cone 270 Problem 48. To find the intersection of two cones . . 272 XXVI. The Intersection of Various Curved Surfaces 278 Intersection of Two Curved Surfaces of Revolution — Of a Sphere and a Cone — Of a Sphere and a Cylinder — Of Any Two Curved Surfaces. Problem 49. To find the intersection of any two surfaces of revolution whose axes intersect ..... 278 Corollary. To find the intersection of two surfaces of rev- olution whose axes are parallel ..... 280 Problem 50. To find the intersection of a sphere and a cone 282 Problem 51. To find the intersection of a sphere and a cylinder 284 Problem 52. To find the intersection of any two curved sur- faces. (General case.) 286 DESCRIPTIVE GEOMETRY INTRODUCTION 1. Descriptive Geometry. Descriptive Geometry is an ap- plied science which treats of the graphical representation of lines, planes, surfaces, and solids, and of the solution of prob- lems concerning size and relative proportions. Thus it lies at the foundation of all architectural and mechanical drafting. While the study of descriptive geometry does not require extended mathematical knowledge, and while its operations are not strictly mathematical, the best results cannot be obtained without some acquaintance with the principles of plane and solid geometry. Descriptive geometry differs from analytic geometry of three dimensions in that the solutions are based, not on algebraic equations, but on drawings. In a certain sense, this greatly increases the scope of descriptive geometry. By the graphical processes, objects of any form whatever, no matter how irregu- lar, or whether any equations for them exist or not, may be treated. On the other hand, descriptive geometry resembles analytic geometry in that the drawings, like the equations, are merely representations of the conditions in space. By the methods of descriptive geometry the solution of any problem involving three dimensions consists of three distinct processes, as follows : (1) Representation of the lines, planes, surfaces, or solids in space by correspond ing plane figures. (2) Solution of the problem by the use of the plane figures. (3) Determination of the relation in space which corresponds to this solution. B 1 2 DESCRIPTIVE GEOMETRY [§ 1 In order that these processes may lead to a successful result, it is evident that it must be possible to pass without ambiguity from the object in space to its representations, and also with- out ambiguity from the representations to the object in space again. 2. Visualization. The process of passing from the repre- sentations to the object in space is a purely mental one. It is, therefore, apt to be ignored by the student at the beginning of his course, to his detriment later on. The process is called " visualizing " or " reading " the drawing, and is absolutely essential in any practical application of descriptive geometry. (See § 3.) Because of its importance, both in theory and in practice, considerable emphasis will be laid on visualization in the present text. The student is advised to accustom him- self early to look upon his drawings as representations of conditions in space, and to work out his problem by consider- ing the space relations involved, rather than by the (plane) geometrical relations existing in the drawing itself. 3. Practical Application of Descriptive Geometry. In the de- sign of all engineering and architectural structures, as machines, bridges, buildings, etc., as well as in many of the less preten- tious mechanic arts, there comes a time when the forms and arrangements of the various parts must be considered. The problem then becomes one which is solved, either wholly or in part, by the graphical methods of descriptive geometry. Moreover, the drawings, once made, take the place of written description, and form the language by means of which the designer conveys his ideas to the builder or mechanic, who, after " reading " (visualizing) them, can build or construct the work. In addition to the usefulness of descriptive geometry in its practical applications, it is the conviction of the authors that its fundamental and educational value lies in its unique power to develop the mental concept, or visualization, the importance of which has already been emphasized. CHAPTER I ELEMENTARY PRINCIPLES — THE COORDINATE PLANES 4. Orthographic Projection or Orthographic View. The terms orthographic projection and orthographic view, or more simply projection and view, are used to denote a drawing which repre- sents, in accordance with a certain artificial and conventional system of vision hereafter described, some object, surface, line, point, or combination of these, that has a definite position in space. Such a drawing may be made on any plane surface, as, for example, a sheet of paper or a blackboard. 5. Projection on a Single Plane. In Fig. 1, let Q represent any horizontal plane in space, and A a square right prism with >' 2 3 > 1 1 1 7 A 1 X 7 i / 5 r~v:/ /a Fig. 1. its bases parallel to the plane. If the prism be looked at from above, at right angles to the plane, the arrows represent- ing the lines of sight, the square top, 1-2-3-4, will be the only part seen. If this square be now imagined to drop vertically until it lies in the plane Q in the position 5-6-7-8, the latter would be the orthographic projection of the prism on the plane Q. The plane Q is called a plane of projection. In this case the projection is a top view of the given object. 3 4 DESCRIPTIVE GEOMETRY [I, § 5 In looking at any object naturally, it is a familiar fact that the lines of sight all converge to the eye. In orthographic projection, however, this is not the case ; instead, the lines of sight are all assumed to be parallel, and every view or projec- tion is made on this basis. On account of this difference, the resulting views are more or less unlike those seen with the natural eye. With natural vision, the same object, placed at varying distances from the eye, appears smaller when farther away, while in orthographic projection the size of the view or projection is not affected by the distance of the object from the observer. Figures 2-10 show pictorially the projections of objects of simple form. Figures 2, 3, 4, 5, 6, and 9 are projections on a Fig. 2. horizontal plane, and are top views. Figures 7, 8, and 10 are projections on a vertical plane, and are front views. In each case the plane Q is a plane of projection. From these figures it is evident that one view of an object is not sufficient to determine its size and shape, hence two or more views or projections, requiring as many planes of pro- jection, must be used. 6. Choice of the Planes of Projection ; the Coordinate Planes. The simplest and most advantageous angle between the planes is a right angle. At any given point On the earth's surface, there are two natural mutually perpendicular directions, which must always be considered in the applied arts and sciences : the horizontal or level, as shown by the surface of still water, and the vertical or plumb, as shown by a freely falling body. I, §6] ELEMENTARY PRINCIPLES Fig. 7. Fig. 6. Fig. 9. Fig. 10. 6 DESCRIPTIVE GEOMETRY [I, § 6 The planes of projection are chosen, • accordingly, the first horizontal, the second perpendicular to the first, and therefore vertical. These two planes are known as the horizontal and vertical coordinate planes, respectively, and will be designated by the letters H and V. At any given place the horizontal plane is always fixed in direction, since all horizontal planes are parallel. But any plane perpendicular to the horizontal is vertical. Hence the direction of the vertical plane is to some extent arbitrary. 7. Names of the Views. A view or orthographic projection made on a horizontal plane of projection, H, is known as a hori- zontal projection, usually abbreviated to JJ-projection. This term is equivalent to the expression top vieiv which was used above, and which may still be employed. In practice, however, it is usually called the plan. Thus, the terms top view, plan, and horizontal projection are synonymous. Similarly, a view made on a vertical plane of projection, V, is called a vertical projection, or a F-projection, or a front view, or an elevation. 8. Position of the Observer with Reference to the Coordinate Planes. When making a vertical projection or front view, the observer is in a position squarely facing the F-plane, and either above or below H, as the size and position of the object may require. When making a top view or if-projection, the position is as if the observer were to stand on or above the -fiT-plane, facing the F-plane, then to bend forward and look vertically down upon the given object and the horizontal plane. Except in rare instances, not considered here, the F-plane is never viewed from the back or further side, nor the iT-plane from underneath. 9. The Ground Line and the Four Quadrants. Let the coordi- nate planes be chosen as in Figs. 11 and 12. The line of intersection, OL, of these planes is known as the ground line — a somewhat unfortunate name, since it is derived from but one of the properties of the line, and ignores another equally important property, as will be seen later. The ground line divides each of the coordinate planes into two parts. I, § 11] ELEMENTARY PRINCIPLES Although a pictorial representation can show only a limited portion of each plane, the coordinate planes are supposed indefinite in extent. Hence they divide the whole of space into four quadrants, numbered I, II, III, and IV, as follows : Quadrant I : above H and in front of (or on the near side of) V. Quadrant II : above H and behind (or on the far side of) V. Quadrant III : below H and behind V. Quadrant IV : below H and in front of V. 10. The Relation between an Object and Its Projection. In Fig. 1, the projection 5-6-7-8 of the top of the prism might be obtained by extending the vertical edges of the prism until they intersect the plane of projection, Q. The point 5 is the pro- jection of point 1, 6 of 2, and so on. In general, the (ortho- graphic) projection of a point on any plane may be defined as the foot of a perpendicular from the point to the plane. The projection of any object is composed of the projection of all its points, found in the same manner. 11. The Relation between Two Projections of an Object. Let a rectangular card, abed, be placed as shown in Fig. 11. The card is in the first quadrant, above II, in front of and parallel to V. The long edges are vertical, the short edges parallel to H. The projection of the card on F~is the equal rectangle a v b v c v d v , and on // the straight line a*6*c*d*, equal in length to the short edge. The perpendiculars dd" and dd h deter- mine a plane which is perpendicu- lar to both II and V, and therefore bo their line <>(' intersection, GL. The same is true for the two perpendiculars from every other corner of the card. Hence the tivo projections of any jwint, together with the point itself, must always be in the same plane Fig. 11. 8 DESCRIPTIVE GEOMETRY [I, § 11 perpendicular to GL. This is one of the fundamental relations of orthographic projection. 12. Notation. In this book, projections on the horizontal coordinate plane, H, will be denoted by the small letter, h, attached as an exponent to the letter or character which denotes the actual point, line, or object projected. Projections on the vertical coordinate plane, V, will be similarly denoted by the Fig. 11 (repeated). Fig. 12. use of a small letter, v, as an exponent. Thus, in Fig. 11, point a is one corner of the card in space ; the projection of a on H is lettered a h ; of a on V is called a". In Fig. 12, the object, A, a rectangular block placed in the third quadrant, is projected on H as A h , on V as A". Further statements in regard to the notation will be made from time to time, when necessary. 13. Monge's Method. The use of two mutually perpen- dicular coordinate planes, as in Figs. 11 and 12, enables us to represent objects of three dimensions by plane figures. It is sometimes called Monge's method, after the originator. Drawings, however, are not made ordinarily on two drawing surfaces at right angles to each other. A single drawing surface (paper, cloth, board, etc.) is taken to represent both H and V. 14. Projections. In the pictorial representations, Figs. 1-12, the planes of projections have been shown limited in extent, in order to give a clear idea of their position. Actually, the I, § H] ELEMENTARY PRINCIPLES 9 planes are unlimited in extent (§ 9), so that in making pro- jections no outlines for the coordinate planes are shown. The projections or views of the card of Fig. 11 are given in Fig. 13. When looking toward V, the horizontal plane will be a v i ib □: □* d h c h Fig. 13. 3 A" Fig. 14. seen edgewise, and is represented by GL ; the F~-projection of the card will be the rectangle a v b v c v d v , as in Fig. 11. Looking down on the i/-plane, the P"-plane will be seen edgewise, and is usually represented by the same line, GL, previously used as the edge view of H. The card will also be seen edgewise, and will project on H in the straight line a h d h b h c h . The projections, plan and elevation, of the rectangular block of Fig. 12 are given in Fig. 14 at A h and A v respectively. Figure 15 shows the projections of the object of Fig. 3. Fig- ure 16 shows the projections of the hollow ring of Fig. 2. ^3 Elev. Plan Plan ] Elev. Fig. 15 Fig. l(i Since no ground line is given in Figs. 15 and 10, the quadrants in which these objects are placed are not known. However, this does not affect the size or the shape of the projections (§ 5). CHAPTER II PROJECTIONS OF THE POINT AND OF SIMPLE SOLIDS 15. Notation of the Point. An isolated point in space will be denoted, in general, by a small letter. A point lying in a line, or in the boundary of a surface or solid, will be denoted either by a small letter or by a number, according to convenience. The projection of any point on H or V will be designated by the suitable use of the small letters h and v as exponents (§ 12). 16. Projecting Isolated Points. I Let a, Fig. 17, be a point situated 1 in Quadrant I. By inspection of ^° h >v »j,iv\ the figure, it is seen that this point projects on // in front of GL, and on V above GL. Let the points b, c, cl, be placed in Quadrants II, III, IV, respectively. Visualizing in a similar manner, and using the notation of § 15, we may make the following table : Quadrant I : a h lies in // in front of GL. a v lies in V above GL. Quadrant II : b h lies in H behind GL. b v lies in V above GL. Quadrant III : c h lies in H behind GL. c* lies in V below GL. Quadrant IV : d h lies in H in front of GL. d v lies in V below GL. By the method of the single ground line, Fig. 13, the drawing is divided into two parts by the line GL. The upper part represents both V above GL and H behind GL, the lower part represents both H in front of GL and V below GL. In this method, the lines joining a h and a", b h and b v , etc., must be per- pendicular to GL. The resulting projections are given in Fig. 18. 10 II, § 18] POINTS AND SIMPLE SOLIDS 11 17. Projectors. The line aa" or bb v , Fig. 11, which projects the point on the plane of projection, is called a projector. A line like a v a h , Fig. 13 or Fig. 18, which connects the two pro- jections of a point in the drawing, also is called a projector. These two uses of the word rarely cause confusion ; if necessary 1 II III IV i b * i I I - 1 1 • l h i -t-d Fig. 18. to distinguish, however, the terms space projector, in space, and ruled projector, in the drawing, may be employed. 18. Visualizing the Quadrants. The propositions of § 16 can be reversed, and the quadrant in space in which any point lies can be told at once from its projections. To do this merely by memorizing the relations of the two projections to the ground line is not visualizing in any sense, and is of little or no value. As the first step in visualizing, let us inquire which side of each of the coordinate planes is represented by the drawing surface. Regardless of the position of the point or object rep- resented, this is always the side of the plane which is nearer to an observer placed in the first quadrant. The same idea is expressed by saying that the //-plane is always viewed from above, and the F-plane from in front (§ 8). Since the point may be on either side of the coordinate plane while the drawing is always viewed from the same side, it follows that it maybe necessary to visualize the point in either of two ways : (1) nearer to the observer than its projection, that is, above or in front of the drawing ; (2) beyond its pro- jection, that is, below or behind the drawing and seen as if looking through a transparent surface. 12 DESCRIPTIVE GEOMETRY [II, § 19 19. Uses of the Quadrants. In Figs. 19-22 an isolated point is shown placed, successively, in the four quadrants. The re- marks and deductions apply equally well to any object lying wholly within the quadrant considered. First Quadrant (Fig. 19). The point is on the nearer side of the drawing in each projection. This quadrant is much Fig. 19. used in architectural work and occasionally in engineering work. On account of the comparative ease with which aids to visualization may be placed over, rather than under, the drawing, this quadrant is very generally used in the first presentation of a problem in descriptive geometry. Second Quadrant (Fig. 20). The point is on the nearer |b h Fig. 20. side of the 77-projection, but on the farther side of the F-pro- jection. The principal use of this quadrant is in the subject of perspective. Third Quadrant (Fig. 21). The point lies beyond the plane of the drawing in each projection, and must be imagined II, § 20] POINTS AND SIMPLE SOLIDS 13 as if seen through transparent planes. This quadrant is used more frequently in practical drafting than any of the others. . J Yd ~! ch W\ G 1 1 ! I di- k_lc L \ 1 This is analogous to the id* +d Fig. 21. Fourth Quadrant (Fig. 22). second quadrant, with H and V reversed. It is little used in practice. 20. Views of a Prism. The plan and elevation of a hex- agonal right prism placed in the third quadrant are given in Pig. 23. The prism is in a vertical position, its upper base in the if-plane. Visualize the solid in space. The distance y, the length of the prism, is the height of one end above the other. The distance x is the dis- tance between the parallel sides, or the thickness of the prism. This illustrates one of the funda- mental principles of orthographic projection : an elevation, or V-pro- jection, shows heigh ts, or distances up and down ; a plan^ or 1 1- projection, Fig. 23. shows distances from front to back. Fig. 22. 1 1 G I- t 1 y 1 A tf 1 L. X .-J 14 DESCRIPTIVE GEOMETRY [II, § 21 21. Distances of a Point from H and V. Let it be required to visualize the point m, Fig. 24. First, suppose the drawing paper to represent H; then m* is ignored, while GL becomes the //"-projection of the V coordinate plane. Hence m* shows that the point m in space is not only in ~J front of V, but also at the distance x from V. Now suppose the drawing paper to repre- sent V; then m* is ignored, GL becomes the F-projection of the incoordinate plane, and m" shows that the point m is at the j v distance y below H. FlG ^ Conversely, if the distances of the point in space from H and V are given, the pro- jections can be located at the given distances from GL. Which projection is determined by which distance? 22. Special Positions of the Point. In Fig. 25, the point e lies in H, and in front of V; the point / lies in H, and behind V; i r +e n Fig. 25. the point g lies in V and above H; the point j lies in both H and F, that is, in the ground line. Visualize each of these points. 23. Projections of Simple Solids. Solids bounded by plane faces, such as prisms, pyramids, wedges, frustums, etc., are projected by drawing the projections of the straight lines which form the edges of the solid. Only simple positions of these solids can be considered at present. II, § 25] POINTS AND SIMPLE SOLIDS 15 24. Visibility of Solid Objects. In viewing a solid, no matter from what point of view, only a portion of its surface is visible, while the rest is invisible. In any projection of a solid, there- fore, there are usually both full and dotted lines, which repre- sent respectively visible and invisible edges. The correct representation of these visible and invisible edges is an essen- tial part of the projection of any solid. The student's facility in determining visibility is, to a considerable extent, a measure of his understanding of the problem. In visualizing any object or objects, an /T-projection, or plan, is always viewed from above ; a P~-projection, or elevation, from hi front (see § 8). This is true, whatever be the relative posi- tions of these projections with respect to the ground line, or to each other. In a drafting office, the relative position of plan and eleva- tion is usually prescribed by the custom of the office, so that it is known by the position on the sheet which view is to be read as an //"-projection, and which as a P"-projection. In studying the theory, however, this is not the case ; objects may be placed in any position in any quadrant, and some indication as to which is the l/-p rejection and which the V* projection must always be given. For the present, we shall do this by the nota- tion, using the index letters A and • (§ 12). 25. Illustrative Examples. The stu- dent should visualize the solids projected in Figs. 26-32, aided by the following hints. Notice carefully the visibility in each case, as shown by the full and dotted lines (§ 24). Figure 26. First quadrant. A square right prism. The base is in H, and the lateral edges are vertical. The //-projec- tion shows the true size of the base, the distance of each lateral edge from V, and the angle between each lateral face and V. shows the altitude of the prism. Fig. 26. The F-projection 16 DESCRIPTIVE GEOMETRY [II, § 25 Figure 27. Third quadrant. A rectangular right prism. The base is parallel to H, and the lateral edges are vertical. The //-projection shows the true size of the base, and its position (angles, distances) with respect to F The F-projection shows the altitude of the prism and its distance from H. Figure 28. Fourth quadrant. A triangular right prism. The base is parallel to V, and the lateral edges are per- pendicular to V. The F-projection shows the true size and shape of the base, and its position with respect to H. Note that the base is not a regular (equilateral) triangle. The //-projection shows the length of the prism and its distance from V. Figure 29. Second quadrant. A square right pyramid. The base is parallel to H. The //-projection shows its size and position. The F-projection shows the distance of the base from H, and the altitude of the pyramid. Figure 30. Third quadrant. A pentagonal right pyramid. The base is parallel to H. The apex is below the base, so that the pyramid is inverted. The base is a regular pentagon. One edge of the base is perpendicular to F, so that one of the lateral faces projects on Fas a straight line. Figure 31. First quadrant. A frustum of a regular hexag- onal pyramid. The lower base of the frustum is in H, the upper base is parallel to H. In order to aid in the visualization of the frustum, the complete pyramid is shown in light dotted lines. Figure 32. Quadrant indeterminate. An irregular triangular pyramid. This solid may be constructed by assuming any four points in space, and then connecting each point with the other three, thus forming a solid bounded by four triangular faces. In visualizing this pyramid, any one of the four points may be taken as the vertex, and the other three points as the corners of the base. In Fig. 32, no ground line is shown; hence the distances of the object from H and F cannot be told. Nevertheless, the projections represent a pyramid of definite size and shape (compare § 14, Figs. 15 and 16). II, § 25] POINTS AND SIMPLE SOLIDS 17 Fig. 27. Fig. 28. Fig. 29. Fig. 30. Fig. 31. Fig. 32. 18 DESCRIPTIVE GEOMETRY [II, § 26 26. Curved Solids. In projecting solids bounded partly or wholly by curved surfaces, the projection cannot always be made wholly of edges of the solid, but may consist partly or wholly of the apparent outline or contour of the object. A few typical curved solids are represented in Figs. 33-38. Fig. 33. Fig. 34. Fig. 35. Fig. 36. No ground line is shown, but the objects are all supposed to be in the first quadrant : A — A circular right cylinder. B — A circular right cone. cm C — A sphere. D — A hollow hemi- sphere, or hemispherical bowl. E — A torus. This is the solid generated by revolv- ing a circle about an axis situated in its own plane, but outside the circle. F — A spool, consisting of a cylindrical body, coni- cal ends, and with a cy- lindrical hole lengthwise through it. CHAPTER III REPRESENTATION OF THE STRAIGHT LINE — TRACES 27. The Straight Line. A line is the path of a moving point, and is not necessarily straight. Yet in ordinary use, the term line, by itself, and without anything to imply the contrary, always means a straight line. 28. Projections of the Straight Line (Fig. 39). Let cd be any straight line, and Q any plane of projection. Then as a point Fig. [59. moves along the line from c to d, its projection will move along the plane from cfl to d". The following propositions are evident : 1. TJie projection of a straight line is a straight line. 2. Tlie projection of any point in the line lies in the projection of the line. Since any point in space is definitely determined when its projections on //and Tare known, it follows that, in general, any two straight lines assumed at random, one in // and one in V, are the projections of one and only one straight line in space. Certain exceptions will be noted in § 34. 29. Notation of the Straight Line. A straight line of definite length will be given by the two points at its extremities, as the line cd (c h d h , cd v ). A straight line (if indefinite length will be denoted by a single capital letter. This letter alone denotes the line in space, and the letter with a suitable index indicates a projection. Thus we shall say tin- line A, or the line Ii(B h , B"). 19 20 DESCRIPTIVE GEOMETRY [III, § 30 30. Visualizing the Straight Line. The following example illustrates a general method of visualization. A single projec- tion, as we have seen, is not sufficient to locate a point or line in space. By a constant comparison of one projection with the other, however, a sufficient number of facts are brought out to complete the mental picture of the conditions in space. Consider the line A, Fig. 40. In visualizing or " seeing " the line in space, either the //"-projection or the F-projection may be used first as a basis. As the natural position of the drawing surface is horizontal, let the //-projec- tion be chosen first. Then in viewing the line A from above, the line in space is either directly over or directly under A h , or else partly above and partly below. Visualizing in this direction alone, how- ever, can give no idea whether the line is parallel or oblique to the //-plane. By now looking toward V, and seeing the line in space projecting as A v , the fact is revealed that the left-hand end of the line is the higher, and that, reading from this end, the direction or slope of the line in space is downward to the right. Returning to the if-projec- tion, and reading along the line in the same direction, namely, from left to right, the line is seen to slope away from the observer, that is, backward. Finally, by comparison of the two projections, we find that the left-hand end of the line is in the first quadrant, the right-hand end in the third, the central portion in the second quadrant (§§ 16, 19). 31. The Slope of a Line. In determining the slope of a line, the directions up and down, right and left, both of which are shown by the ^projection, rarely trouble the student. The directions forward and backward, however, which are alwaj r s shown by the //"-projection, often cause confusion. This diffi- culty should disappear by considering the method of projecting a solid object ; for in projecting any actual object, the side nearest the observer is always considered to be the front of the object (see §§ 20, 24). A point which is farther away from HI, § 32] STRAIGHT LINES 21 Fig. 41. the observer is behind one which is nearer ; so that the direc- tion from front to back is away from the observer. This is shown in Fig. 41 by the arrow A, which points away from the observer, and therefore points backward. The arrow B in this figure points toward the ob- server, and is pointing forward. Points in the first and fourth quadrants are always in front of points in the second and third quadrants ; but points on the same side of the F"-plane must be read by the relative positions of their .^-projections, the lower side of an iZ-projection being always the front. The slope of the line A, Fig. 40, would be described as downward, backward, to the right. The slope may also be given as upward, forward, to the left, since it is im- material in which direction the line is read. 32. Additional Examples. Line B, Fig. 42. Eead in the manner already described, it will be found that this line is oblique to H and V, passes through the fourth, third, and second quadrants, and slopes upward, backward, to the right (or downward, forward, to the left). Line A, Fig. 43. The F-projection is parallel to the ground line, so that every point of the line is the same dis- tance from II. The line A is parallel to II, oblique to V, passes through the second and first quadrants only, Flr; - 43 - and slopes forward to the right (backward to the left). Note. Lines parallel to //are known as ^-parallels ; lines parallel to V are known as F-parallels. A line parallel to // may also be called a horizontal line : but the corresponding term, vertical line, is limited to a line perpendicular to H. h- and F-parallels are very important lines. Fig. 42. 22 DESCRIPTIVE GEOMETRY [HI, § 33 Fig. 45. 33. Special Positions of the Line. The student should satisfy himself of the truth of the following propositions by direct visualization. 1. If a line A intersects the ground line, A h and A v intersect on GL, and in the same point as A (Fig. 44). 2. Conversely, if A h and FlG - 4i - jiy intersect the ground line in the same point, the line A intersects the ground line. Through how many quadrants does such a line pass ? 3. If a line A is parallel to H, A v is parallel to the ground line; and conversely, if A v isimrallel to the ground line, A is parallel to H (Fig. 43). 4. If A is parallel to V, A h is parallel to the ground line; and conversely, if A h is parallel to the ground line, A is parallel to P"(Fig. 45). 5. If A is perpendicular to H, A v is perpendicular to the ground line, and A h is a point; conversely, if A h is a point, A is perpendicular to H. (See line A, Fig. 46.) 6. If A is perpendicular to V, A h is perpendicidar to the ground line, and A" is a point; conversely, if A" is a point, A is perpendicular to V. (See line B, Fig. 46.) 7. If A lies in H, A h coincides with A, while A v coincides icith the ground line (Fig. 47). The statements of the analogous and converse propositions are left to the student. 34. Exceptional Cases. While in general (§ 28) the two projections of a line may be assumed in any direction at random, certain exceptions are illustrated by the following examples : ■ ■B v + A n Fig. 46. Fig. Ill, § 34] STRAIGHT LINES 23 (a) (Fig. 48). The projection A h is perpendicular to the ground line, while the projection A v is not. Assume on A h any point c h , and find c v by projecting to A v . The projector c h c v coincides with A h , hence c v is always the same point on A", no matter where c h is chosen. Now let d° be any point on A v except C. The projector from d v is parallel to A h , and there ■ ■cT Fig. 48. • ■rh Fig. 49. o T cT l e h Fig. 50. is no point d h to correspond with d°. Hence there is no line in space to correspond with the given projections. An analogous case arises if A v is taken perpendicular to the ground line, while A h is not. (b) (Fig. 49). The projections, B h and B v , are both perpen- dicular to the ground line, but at different points. Let c h be any point on B h ; the projector from c h is parallel to B v , and there is no point on B" to correspond with c\ Similarly, if d v is any point on B v , the projector from d" is parallel to B h , and there is no point d h . Therefore there is in space no line B corresponding to these two projections. (c) (Fig. 50). The projections O 1 and O v are both perpendicu- lar to the ground line, and at the same point. Let a/" be any point on C h . The projector from a h coincides with O, and the particular point, a", on C", which corresponds with a h cannot be determined. The line C is therefore indeterminate. Bui if the line be projected by means of two of its points (§ 29), it at once becomes determined. Thus the line de, Fig. 50, is a definite line, lying in a plane perpendicular to the ground line. 24 DESCRIPTIVE GEOMETRY [III, § 35 35. The Profile Line. A plane which is perpendicular to the ground line is known as a profile plane, and any line lying in a profile plane is termed a profile line. We have just seen that such a line cannot be projected in the same way as the general straight line. Hence problems in which such lines occur will usually — but not always — call for particular solu- tions. The simplest method of dealing with a profile line is usually by means of an additional plane or projection. The solution of cases involving such lines will be deferred, in general, until Chap- ter VI, where this topic is considered. 36. A Point Lying in a Line (Fig. 51). It follows at once from the second proposition of § 28, that if a point lies in a line, the projections of the point lie in the projections FlG - 51 - of the line. This condition is suffi- cient if the line does not lie in a profile plane. (See § 35.) 37. Traces of a Line. Of all the points in a straight line, the two in which it pierces the planes of projection are con- sidered the most important. These points are called the traces of the line. The horizontal trace is the point in which the line pierces H, and will be designated as the point s ; the vertical trace, in which the line pierces V, will be called t. Problem 1. To find the traces of a straight line. Analysis. The solution of this problem depends on direct visualization. General Case. Line not lying in a profile plane. Construction (Figs. 52, 53). The horizontal trace. Looking towards V, H is seen edgewise as GL, the line A appears as A", hence the line A will pierce H at the point seen as s", where A v crosses GL. "While this projection conveys no idea of the distance of the point in // from V, it does single out, to Fig. 52. Ill, § 38] STRAIGHT LINES 25 the exclusion of every other point, the H piercing point of the line. The actual position of the point in H is found at s h by the projector s l 's h . The vertical trace. Looking down toward H, V is seen edgeAvise as GL, A is seen as A h , hence A is seen to pass through V at the point t h . The actual location, t", in V, is found on A v by means of the projector t h t v . Special Case. A profile line (no figure). The general solution evidently fails ; the solution will be given in Chapter VI. Problem 2. Given the two traces of a straight line, to find its projections. Analysis. Each trace is a point lying in a coordinate plane ; and for such a point, one projection is identical with the point, the other projection is in the ground line. Construction (Figs. 52 and 53). The traces s h and V are given ; s v and t h are found by projection on GL ; s* and t h determine A h , s v and V determine A v . If s h and t v lie in the same projector (Fig. 54), the required line is a profile line. Note that this is a definite line, since two points, s and t, are known. 38. Exceptional Cases. The solution to Problem 2 shows that, in general, a straight line is uniquely determined by its two traces. The only exception is the case in which s and t fall together as a single point on the ground line. The required line can then be any line which passes through this point; in order to determine the line, some other condition must be given. The following questions on traces are left to the student : 1. How must a line be placed so as to have but one trace ? 2. Hew must a line be placed so as to have no truer ? 3. How must a line be placed so that its horizontal (or ver- tical) trace is indeterminate ? -■t v X Fig. 54. CHAPTER IV SIMPLE SHADOWS 39. Shadows. The shadow of a point on a plane is the point in which a ray of light passing through the given point is in- tercepted by the plane. The conventional ray of light is a line sloping downward, backward, to the right, with both projections inclined at 45° with the ground line. Such a ray is shown by the line L, Fig. 55. 40. Shadow of a Point. The shadow on V of the point a, Fig. 55, is the point, a a in which the ray of light, L, passing through a, intersects V. The shadow on H is the point, a s , where L pierces H. It will be noted that the points a t and a s are the traces (§ 37) of the line L on V and H respectively. Hence the shadow of a point on the coordinate planes is found by drawing through the point a line representing the ray of light, and then finding the traces of this line. Fig. 55. 41. Opaque and Transparent Planes. In order that an actual shadow may be cast, it is necessary that the plane receiving the shadow be opaque. Thus, in Fig. 55, if V be con- sidered opaque, a t is the actual shadow of the point a ; but if V be considered transparent and H opaque, a s will be the actual shadow. In work with shadows, it is customary to place the object in the first quadrant, and to consider both //and 'Fas opaque. Since the two projections of the ray of light make equal angles with the ground line, it follows that the actual shadow 20 IV, § 43] SIMPLE SHADOWS 27 of a point must always fall on the nearer coordinate plane. In the case of a solid object, the actual shadow may fall wholly on H, wholly on V, or partly on both planes, according to the size and position of the object. 42. Shadow of a Line. The shadow of a line will consist of the shadows of the points which compose the line. If the line is straight, only the shadows of the two ends of the line need be cast. In Fig. 56, the actual shadows of both points a and b fall on V ; hence the shadow of the line ab is the line a t b n lying wholly in V. In Fig. 57, the actual shadow falls partly on // and partly on V. To find this shadow, Ave may find the com- plete shadow on each plane, regarding the other plane as trans- parent, and then take the actual portion of each shadow. But the two shadows, a s b s and a t b t , one in // and one in V, must in- tersect in a point, e, in GL, since this line is the intersection of the planes // and V. Hence we may find one complete shadow, as a b, note the point e in which ajb, crosses GL, and draw from e to the actual shadow a t . 43. Shadows of Surfaces. Shadows of surfaces and solids are found by extending the methods just given for the point and straight line. The only other case which will be taken up at this time is that of a convex solid bounded by plane faces. 28 DESCRIPTIVE GEOMETRY [IV, § 44 44. Shadow of a Convex Solid. A frustum of an irregular triangular pyramid (Eig. 58) has six corners and nine rectilinear edges. Not all of them cast shadows which lie on the boundary of the complete shadow. Find the shadows of the six corners 1, 2 ... 6 on one of the planes, as II, and draw the largest convex %—* Fig. 58 polygon possible, with three or more of these shadow points as vertices. In Fig. 58, the shadow is the polygon 1 4 2 4 5 s 6,4 4 1j> the point 3, being omitted because it falls inside the polygon. The part l s e/4 s on H in front of GL is an actual shadow ; the remaining part, e2, s bJo s f, shows the portion of the shadow intercepted by V. Find the actual shadows on Voi the points 2, 5, and G ; then e2 l 5fi l f is the part which falls on V, and completes the actual shadow. CHAPTER V REPRESENTATION OF THE PLANE 45. Representation of the Plane. It has been shown (§ 28) that a straight line of indefinite extent can be represented by- two orthographic projections on H and on V. This method, however, cannot be extended to the representa- tion of a plane of indefinite extent, since the projections of all the points in the plane would, in general, cover the whole plane of projection. It has been shown, however (§ 38), that a line of indefinite extent may also be determined by its points of intersection with the planes of projection. This method is capable of extension to the representation of a plane. 46. Traces of a Plane. A plane of indefinite extent will, in general, cut H and V in two straight lines, which meet in the point in which the given plane cuts the ground line. These lines of intersection of the plane with It and V are called, respectively, the horizontal and vertical traces of the given plane. Conversely, two arbitrary straight lines, one in H and one in V, which intersect the ground line at the same point, determine a plane of which the arbitrary lines are the horizontal and ver- tical traces. Thus a definite method of representation of the plane has been found. This method of representation fails only in the case in which the plane contains the ground line, when the two traces will coincide with each other and with the ground line. In this case the plane will not be determined until some additional condition is imposed. Compare this with the analogous case of a line having coin- cident traces, § 38, 29 30 DESCRIPTIVE GEOMETRY [V, § 47 47. Notation of the Plane. A plane in space will be denoted by a single capital letter, taken usually from the middle or end of the alphabet. The horizontal trace of a plane Q, being the line of intersec- tion of H and Q, will be called HQ. The vertical trace, or intersection of V and Q, will be called VQ. The form HQ is used in preference to the form QH, since the latter, in speech, might be confused with Q h , the //"-projection of a line Q, which is by no means the same thing as the //-trace of a plane Q. 48. Visualization of the Plane. The visualization of a plane from its traces is an entirely different process from reading the projections of a line or of a point. The traces of a plane oblique to H and V are not projections of the plane in the ordinary sense. Let HQ and VQ, Fig. 59, be the horizontal and vertical traces of a given plane Q. V / / / \/ V V /%> \ ^ / / N / / \/ / Fig. 59. Fig. 60. I Since the plane is indefinite in extent and cuts the ground line, it must pass into all four quadrants. Since the two traces are oblique to GL, it follows that the plane determined by them is oblique to both H and V. 49. The Slope of a Plane. The slope of an oblique plane is de- scribed in the same terms as the slope of a line, and may perhaps be best understood by reference to the planes of Figs. 59—63. V, § 49] REPRESENTATION OF THE PLANE 31 Plane Q in Fig. 59, shown pictorially in Fig. 60, would be said to slope downward, forward, and to the left (or upward, backward, to the right). In Fig. 61 the slope is downward, backward, to the left (ab- breviated to D.B.L.). Fig. 61. In Fig. 62, the slope is downward, forward, to the right (D.F.R.). r /// p. \ t» V V f s Fig. (J2. In Fig. 63, plane Q, perpendicular to H, would be said to slope simply backward to the right (B.R.). VR J. LJ" ^ A, VR & Fig. 63. It is usually easier to visualize the slope of a plane by con- sidering, at first, one quadrant only (the first or third). The plane can then be visualized as extending into the other quadrants. 32 DESCRIPTIVE GEOMETRY [V, §50 50. Special Positions of the Plane. The student should satisfy himself of the truth of the following propositions by direct visualization. 1. If a plane, Q, is parallel to H, there is no H-trace (i.e. HQ does not exist), while VQ is parallel to the ground line. 2. Conversely, if Q has no H-trace, VQ must be parallel to the ground line, and Q is parallel to H 3. Similarly, if Q is parallel to V, VQ does not exist, while HQ is parallel to the ground line; and conversely. 4. If Q is parallel to the ground line, HQ and VQ are also parallel to the ground line ; and conversely. 5. If Q is perpendicular to H, VQ is perpendicular to the ground line ; if Q is perpendicular to V, HQ is perpendicular to the ground line ; and conversely in each case. 6. If Q is perpendicular to both H and V, that is, perpen- dicular to the ground line, HQ and VQ fall together in the same perpendicular to GL. 51. Edge View of a Plane. The particular position of a plane in which it becomes perpendicular to one of the coordinate VR & Fig. 63 (repeated). planes is an important one to visualize. If Q is perpendicular to H, then the //-projection of every point in Q must fall in HQ, which thus becomes an actual projection of the plane. The trace HQ is, in fact, an edge view of the plane Q (Fig. 63). It should be noted that a plane parallel to one of the coordinate planes is necessarily perpendicular to the other, and that the single trace of such a plane is its edge view (plane R, Fig. 63). CHAPTER VI THE PROFILE PLANE OF PROJECTION 52. The Profile Plane. The profile plane, when used as a plane of projection, will be designated by the letter P. This plane has already been defined as a plane perpendicular to the ground line (§ 35)", that is, P is perpendicular to both H and V. Hence the three planes H, V, and P form a system of mutually perpendicular planes (Fig. 64), intersecting in a / A V / / ■/ / / / / HV / / i 1 * // / / / / A / U/~ ti-?-t x / 4-i 3 i i i / "7^ / t 7 H / 1 L_ Fig. 04. common point o, and intersecting each other by pairs in three mutually perpendicular lines, II V, IIP, and VP, all passing through o. The intersed ion, IIV, of //and Fis the ground line. 53. A Profile Projection. Let the model of some familiar object, such as a cabin, be placed on the horizontal plane in tlic first quadrant, l''i.^ r . <*>l. Let the front and hack be parallel to the F-plane, and the ends parallel to P. Let a view be taken D ;j3 34 DESCRIPTIVE GEOMETRY [VI, § 53 looking in the direction of the arrow, that is, from right to left. In such a view, the whole surface of P will be seen, while the Fig. m (repeated). F-plane will be seen edgewise as a vertical line, the //-plane edgewise as a horizontal line, and only the right-hand end of the cabin will appear. The actual end view is shown in Fig. 65. No boundary is shown in this figure for the P-plane, since it, like the H- and /\ P Fig. 65 F-planes, is of indefinite extent. The quadrants will appear as indicated by the numbers, and must always so appear when P is viewed from the right. In this end or profile view of VI, § 54] THE PROFILE PLANE 35 the cabin, both heights and widths, or, in other words, all dis- tances perpendicular to // or to V are shown in their true size. Hence in this view the actual distances from the object to H and to V are necessarily shown. 54. Relation of the Profile Projection to the H- and 7-Pro- jections. Figure 66 shows the construction of the end view, or ric Showing \J ht hand end A P < X * A' /rv"~: Front 1 1 Back 1 £ •< H 1 \ \ \ i 4- \ \ \ \ 3 A h \ \ \ < V Fig. 66. profile projection, of the cabin from given horizontal and vertical projections. The distance x, from the edge view of V to the end of the cabin, may be chosen at will, as this evidently has no effect on the shape or size of the end view of the cabin, or its distances from // and V. The edge views of // and V, together with the end view of the cabin, might also be placed at the right of the plan and elevation, instead of the left; and would preferably be so placed if the object were given in the third quadrant. This would locate the views in accordance with the usual custo t placing the view of the right-hand end of an object to the right of its front view. For a consideration of a view of the left-hand end, see §§ 62 and 63. 36 DESCRIPTIVE GEOMETRY [VI, § 55 55. A Second Example. Let it be required to draw the plan, elevation, and side elevation (profile projection) of a piano packing case placed in the third quadrant, with its long edges parallel to both H and V. The dimensions of the case and its distances from H and V are supposed to be v "x s \ 2 \ \ \ \ \ \ \ \ \ i Plan H ' 4 3 i i 7 . t : j View <— m — > of right hand end Elevation A" Fig. 67. known. The end view is drawn first at A" (Fig. 67), located as shown in the third quadrant, and at the given distances, k from H and m from V. The plan and elevation may then be readily constructed as shown. It is important to see clearly why all the long edges appear visible both in plan and elevation. 56. Profile Projections in All Four Quadrants. Figures 68-71 show a regular triangular prism placed successively in the four quadrants. The long edges of the prism are parallel to both H and V, and in each quadrant the upper front lateral face makes an angle of 15° with //. In each position the right- hand end is shown in the profile view. The student should visualize the prism for each of the four positions, noting espe- cially the visibility of the edges lettered B v and C\ lettered also in the profile view as the points B" and C p . VI, § 56] THE PROFILE PLANE 37 2d Quadrant, Fig. 68. Fig. 69. 3d Quadrant \ \ \ \ 4-th Quadrant "I r ~ r l ,15' Fia. 70. Fig. 71. 38 DESCRIPTIVE GEOMETRY [VI, § 57 Q. > b p b v a"7 V 1 l i 1 i 1 2 N i 1 • b h I H : 1 1 1 S i \ ,4 a. I 3 1 s 1 s a h Fig. 72. 57. Profile Projection of the Straight Line. The profile pro- jection of any straight line may be obtained from its hori- zontal and vertical projec- tions by finding the profile projections of any tAvo points in the line (Fig. 72). An important case of the profile projection of a line arises when the line is par- allel to P, that is, when the line is a profile line. By means of its profile projec- tion we can solve the follow- ing problems for such a line, which cover solutions previously omitted (§§ 36, 37). 58. Problems on the Profile Line. Given the horizontal and vertical projections of a profile line, to find (a) its true length ; (b) its traces on H and V ; (c) the projections of a point lying in the line. (a) The true lesgtii. Analysis. Since the line is parallel to P, its profile pro- jection will be equal to its true length. (b) The traces ox H and V. Analysis. These traces will appear in the profile projection as the points in which the line cuts the respective edge views of H and V. Construction for (a) and (b) (Fig. 73). Let ab(a h b h , a v b v ) be the given profile line. If the profile plane of projection is not given, it may be assumed anywhere, perpendicular to the ground line. Find the profile projections of the points a and b. Then a p b p is the profile projection of the line, and is equal to the true length of ab. Produce a p b" indefinitely. Then the line is seen to pierce i/at s p , and Fat t". Find the horizontal and vertical projections of the points s and t ; the point s (s h , s v ) is the H-tra.ce, and the point t (t h , t v ) is the P~-trace. VI, § 58] THE PROFILE PLANE 39 A second example is shown in Fig. 74. This is the more common construction, the profile plane of projection being taken directly through the given line. (c) The projections of a point on the line. Analysis. If the profile projection of the line be found, we Fig. 73. can project a point in the ^-projection to the P-projection, and from this to the F-projection ; or vice versa. Construction (Fig. 75). Let ab (a h b h , a v b") be the given line. Find a p b". Now if c h is given, project from c* to c p in a"b p ; and from c p to C in r . If c" is given, c* may be found by reversing the process. The line o.b may be pro- duced indefinitely, and the as- sumed point may be beyond the limits of a and b. Thus, if (I" is taken. d h is found as shown, and the point d lies on the line produced into the third quadrant. Or, d h may be given, and d" found. 40 DESCRIPTIVE GEOMETRY [VI, § 59 59. Profile Trace of a Line. The general line, oblique to H, V, and P, will pierce all these planes. Hence, besides the horizontal trace s, and the vertical trace t, it will have a profile trace, u, which must lie in the profile projection of the line. Given the horizontal and vertical projections of a general line, to find its profile projection and profile trace. Analysis. The profile projection of any straight line can be obtained from its horizontal and vertical projections by finding the profile projections of any two points in the line (§ 57). The profile trace can be found by noting the point in which the given line pierces the edge view of P. Construction (Fig. 76). A h and A v are the given projections. Find the points s and t, the H- and P"-traces of the line A (Prob. 1, § 37). Project s to s p in GL, and t to t p in VP; s» and V determine the required profile projection A p . Since Fig. 76. HP and VP are both edge views of P, we have u h and u v as the H- and F-projections of the point u ; from u h and u v rind u p , the actual profile trace. As a check, u p must lie in A". Since any two of its projections are sufficient to determine the line A, we have the following proposition : Corollary. Given A p and either A h or A"; to find the remaining projection and the three traces of the line. VI, § 60] THE PROFILE PLANE 41 Analysis. The desired result may be obtained by reversing the preceding construction. The construction is left to the student. 60. Profile Trace of a Plane. A plane not parallel to P will intersect it, and the line of intersection is known as the profile trace of the plane. Fio. 77. A general plane Q is shown in Fig. 77, intersecting H, V, and P, in the traces HQ, VQ, and PQ. It is evident that HQ and VQ intersect on HV (the intersection of H and V, or ground line), II Q and PQ intersect on IIP, and VQ and PQ Fig. 78. intersect on VP. Whence, if HQ and VQ are given, PQ may be found as shown in Fig. 78. 42 DESCRIPTIVE GEOMETRY [VI, § 60 The profile trace of a different plane, R, is found in Fig 79. Fig. 79. The profile trace of a plane parallel to the ground line is found in the same way, as shown in Fig. 80. This is an ini- V / vs *y CL > r H \ n I "V HS Fig. 80. portant case, for the plane S is perpendicular to P, and conse- quently the profile trace is an edge view of the plane S. Fig. 81. In Fig. 81 the given plane T passes through the ground line? VI, § 61] THE PROFILE PLANE 43 and is determined by the coincident traces HT, VT, and the point c lying in the plane. To tind the profile trace PT, find first the profile projection c p , of the point c. Then, as in the preceding case, PT is an edge view of T, and therefore passes through c p and o. It will be seen that the plane T is definitely determined by its three traces HT, VT, and PT. Instead of being determined by passing through the point c, the plane T might also have been determined by stating the quadrants through which it passes, and the angle between T and either H or V. Then the profile trace, PT, could have been located accordingly. From Figs. 78-81 it is evident that the three traces of a plane are so related that if any two are given, the third can be found. The cases in which the given traces are the profile trace and either the horizontal or vertical trace, to find the remaining trace, are left to the student. 61. Planes Parallel to H or V. A plane parallel to II or V has no trace on one of these planes, and the student sometimes has difficulty in constructing the profile trace. Such a plane may be regarded as a limiting case of the plane. S, Fig. 80 ; but the result may be obtained by direct visualization, as fol- lows. Since the plane is perpendicular to P, the profile trace will be its edge view. Then the profile projection shows the PQ vq Fio. 82. Fig. 83. edge views of both // and V, to the proper our of which the plane will show parallel. A plane parallel to Ilia shown in Fi;-, r . 82 ; one parallel to V in Fig. 83. 44 DESCRIPTIVE GEOMETRY [VI, § 62 62. Left-side Views. Hitherto, the profile projection has been obtained in just one way. Hence the student might suppose that the method of projecting on the profile plane is as invariable as that of projecting on H or V. In practice, how- ever, this is not the case. The method thus far adopted, when applied to an actual object, always gives a view of its right side ; whereas a view of the left side may be decidedly prefer- a+ -> ef+- + a° Fig. 84. able. For this purjiose, P is viewed from its left side, that is, from left to right, and then projected so as to bring this side into the drawing surface. This method is shown for a point in the first quadrant in Fig. 84. That this method is directly opposite to that previously used is shown also by the line ab, Fig. 85. Since the H- and F-p rejections of a and b show that z 1 / 3 4 t 1 / i I i b h - — " i / i i b v - 'b' / Fig. 85. these points are in the first and fourth quadrants respectively, the quadrants must appear in the profile projection as marked. In Fig. 86 the profile trace of the same plane Q is obtained, first, by looking from right to left, and then by looking from VI, § 64] THE PROFILE PLANE 45 left to right. Note that, since HQ is at 45° with the gi-ound line, PQ coincides with VQ in the tirst figure, and that the Fig. 8(>. second figure is much the clearer, especially if some further construction is to be made with the plane Q. Hereafter we shall view P from either side ; although, other things being equal, the direction from right to left will be pre- ferred. If the quadrants are numbered in every profile projec- tion, there should be no confusion, whichever view is used. 63. Conventional Placing of Left-side Views. When the pro- file projection of a solid object is made so as to show its left side, it is customary to place this projection at the right of the front elevation when the object is located in the first quadrant, and at the left of the front elevation when the object is located in the third quadrant. This is the reverse of the method of placing the view of the right side, as shown in Figs. 68 and 70. (See § 54.) Actual objects are rarely placed in the second or fourth quadrants, so that there is no fixed custom for locating such views. 64. Second Method of Obtaining Profile Projections. Profile projections may be constructed also by treating the profile plane as a secondary plane of projection, by the methods of the next chapter. Such a construction is less common, however, than that given above, since the resulting position of the profile projection is not so generally satisfactory. CHAPTER VII SECONDARY PLANES OF PROJECTION 65. Secondary Planes of Projection. In practical work, views of objects are often wanted in other directions than those which are obtained by the use of the horizontal and vertical coordinate planes. In fact, it often happens that an actual object cannot be adequately represented by a simple plan and elevation. In the theory, also, an additional projection in a suitably chosen direction may give readily a solution otherwise difficult to ob- tain. Such views, or projections, are obtained on secondary planes of projection, taken perpendicular to either H or V, and making any angle whatever with the other coordinate plane. Secondary planes of projection are by preference taken per- pendicular to H, for while any plane perpendicular to H is vertical, a plane perpendicular to V and oblique to H is neither vertical nor horizontal, and therefore does not conform to either of the natural directions (§ 6). 66. The Profile Plane as a Secondary Plane. A very impor- tant secondary plane of projection is the profile plane, which is perpendicular to both H and V. This plane has just been discussed in detail in the preceding chapter. Indeed, some writers consider this plane as a third coordinate plane of equal rank with the horizontal and vertical coordinate planes. In the present chapter, the profile plane will be considered as a special case of a plane perpendicular to H. 67. Secondary Ground Lines. The method of projecting on a secondary vertical plane is shown in Figs. 87 and 88. The secondary plane V x intersects H in a secondary ground line, G X L X , which may be at any angle with the original ground line GL. Since the projectors aa v and ac j !\ ' ' A i i i i 1 • ! b»i I. V U Fig. 97. pendicular to V, the circular path projects on 7as a circle, and on if as a straight line parallel to GL. 76. Revolution of a Point about an Axis Lying in a Given Plane (Fig. 98). Let us consider a point a, which is to be revolved Fig. 98. about an n x is X, lying in a plane Q. The circular path • it the revolving point will project on Q as a straight line, be, perpendicular to the axis X. It follows that if the point a be revolved into the plane Q, its revolved position, b or c, will be at a distance he or ce from the axis X, equal to the true distance ae, of the point from the axis. 54 DESCRIPTIVE GEOMETRY [VIII, § 77 77. The True Length of a Line. Let ab be a straight line of definite length. If placed parallel to any plane of projection, as Q, Fig. 99, the projection of a q b q will evidently be equal hi length to ab. If, however, the line is placed at any angle with Q, the length of the projection will be shorter than that of the line. (See the line cd, Fig. 39.) But by revolution about a Fig. 99. suitable axis, a line can always be brought into such a position that the true length will appear. 78. First Method for Finding True Length. The projection of a line is equal to its true length when the line is parallel to the plane of projection (§ 77). The true length of any straight line may therefore be found by revolving the line until it is parallel to either H or V. Problem 3. To find the true length of a straight line. First Method. By revolving about an axis perpendicular to H or V. Analysis. Through either end of the line assume an axis perpendicular to H and (of necessity) parallel to V. Revolve the given line about this axis until the line is parallel to V. Then the T r -projection of the line in its revolved position will show the true length of the line. Or, assume the axis perpendicular to V and hence parallel to H, and revolve the given line until parallel to H. Construction (Fig. 100). Let ab (a k b h , a'b") be the given line. Assume an axis of revolution perpendicular to H to pass through the point a. Revolve the line about this axis until the /^-projection takes the position a"c h , parallel to OL. VIII, § 79] TRUE LENGTH OF A LINE 55 In this revolution, the angle which the line makes with the axis does not change. Consequently, the angle which the line makes with H, and therefore the length of the if-projection, remain constant. The point b revolves in the circular arc J(J h , J v ), lying in a plane parallel to 7/ (§ 75), and is found at the point c (c h , c"). The line in the position ac (a h c h , a^c") is parallel to V, and cfc" shows the true length. Fig. 100. Fig. 101. In a second example, Fig. 101, the axis of revolution is taken through the point b and perpendicular to V. The true length appears after the line has been revolved parallel to //. 79. The Angles Which a Line Makes with the Coordinate Planes. A line in the position ac (<< h c h , a v c v ), Fig. 100, which is parallel to V, shows in the ^projection the angle, a, which the line itself makes with //. The line ac is the revolved position of the line ab. In the preceding construction we saw that during the revo- lution the angle which the line made with H did not change. Hence a is the angle which the given line ab makes with II. That is, in addition to finding the true length of the line ab in Fig. 100, we have incidentally found the angle which this line makes with //. Similarly, in Fig. Kll. we found, ;is an incidental part of the construction, the angle (3 which the line ab makes with 1'. 56 DESCRIPTIVE GEOMETRY [VIII, § 80 80. Second Method for Finding True Length. The true length of a straight line may also be found by revolving the line about one of its own projections as an axis, until the line lies in a coordinate plane. Problem 3 (bis). To find the true length of a straight line. Second Method. By revolution about one of the projec- tions of the line. Fig. 102. Analysis (Fig. 102). Let ab be the given line, projected on any plane Q at a q b q . Revolve the plane abb q a? about a q b q as axis into Q. Then a falls at a r , where a r a" equals aa", and is perpendicular to a q b" (§ 76) ; b falls at b r , with b r b q equal to bb q and perpendicular to a q b q ; hence a r b r equals ab, and shows the true length of the line. Construction. The projection a h b h , Fig. 103, corresponds to a q b q , Fig. 102, and is taken as the axis of revolution. Since a* is the projection of the point a in space, the distance, aa h , of point a from the axis is equal to the distance of the point a from H, which shows in the ^projection as the distance from a v to GL. Hence to find the revolved position a r , make a h a r per- pendicular to a h b h , and the distance a h a r equal to a v e. Simi- larly, b h b r is perpendicular to a h b h , and is equal to b v f. Then a r b r shows the true length of the line ab. In Fig. 104 the F-projection is taken as the axis, and the line is revolved into V. The distances a v a r and b v b r are the distances of the points a and b, respectively, from V, and are equal to the distances from a h and b h to the ground line. !N T ote that not only the distances, but the directions, of the points a and b from the F-plane must be considered. The //-projection VIII, § 82] TRUE LENGTH OF A LINE 57 shows that a and b, in space, are on opposite sides of V, hence their revolved positions, a r and b r , must fall on opposite sides of the axis, a v b", which lies in V. Fig. 103. Fig. 104. 81. Traces of the Line. Let the line ab, Fig. 102, be pro- duced to intersect its projection a q b" in the point s ; then in revolving about a q b q as axis, the point s will remain fixed, and a r b T produced must pass through s. But since a"b" lies in Q, the point s must be the point in which ab pierces Q, or the Q-trace of ab. In the projection, Fig. 103, let a r b r be produced to meet a h b h produced at s ; then 8 must be the //-trace of the line ab, and should be the same point that is found by project- ing from the intersection of a v b v produced and GL, as in Problem 1 (§ 37). 82. Angles with the Coordinate Planes. The angle a, Fig. 102, between ab and a''//' is the angle which the line ab makes with the plane Q, and this angle is not changed by the revo- lution into Q. Hence, in Fig. 103, the angle a, between a h b h and a,.b r , is the true angle which the line ab makes with //. In Fig. 104, since the line ab was revolved into V, we have found the F-trace, t, where a r b r intersects a v b v . The angle, (3, between these lines is the angle which the line ab mukes with V. 58 DESCRIPTIVE GEOMETRY [VIII, § 83 83. Converse Problems. In each of the preceding methods for finding the true length of a line it is necessary to find the angle between the line and one of the coordinate planes. Conversely, it is easy to see that a line may be located, when certain angles are known, by reversing one of the preceding constructions. Problem 4. To find the projections of a line of definite length, when its slope, the angle which it makes with one coordinate plane, and the direction of its projection on that plane, are known. In order to give the line a definite position in space, it will be assumed, in addition to the above data, that one end of the line is located at a given fixed point a. It is evident geometrically that the position of the line in space is now completely determined. Analysis. If the angle which the line makes with H is given, reverse any construction of Problem 3 which gives the angle with H. Proceed similarly if the angle with V is the given angle. Fig. 100 (repeated). Fig. 101 (repeated). Construction. First Method (Fig. 100). [Reverse of First Method, Prob. 3, § 78.] Let the angle with H and the direction of the if-projection be given. Let the point a be the fixed end of the line. VIII, § 83] TRUE LENGTH OF A LINE 59 Place the line in the position ac (a v c v , a h c h ), parallel to V, making a v c v equal to the true length of the line, at the given angle with H, and sloping upward or downward from a as given. Take an axis perpendicular to H through the fixed end, a, and revolve the line until the //-projection takes the given direction a h b h . Find the corresponding projection a v b". Second Method (Fig. 103). [Reverse of Second Method, Prob. 3, § 80.] As before, let the angle, a, with II and the -A- / \ a L X Fig. 103 (repeated). Fig. 104 (repeated). direction of the H"-projection be given, and let the point a be the fixed end of the line. Through a' 1 draw a h s, indefinite in length, and in the given direction. Draw a h a r perpendicular to a h s, and make the distance a h a r equal to a v e. Draw a r s in- definite in length and at the angle a with a h s. In this case, the line is supposed to slope downward from the point a. Make, a r b r equal to the given true length of the line. Draw b r b h per- pendicular to ats, locating b h . Project from &* perpendicular to GL, and make the distance fb v equal to b h b r , thus locating &". The student should see if he can solve the problem when the angle, with P"and direction of the ^projection are given, by reversing the constructions of Figs. 101 and 104. 60 DESCRIPTIVE GEOMETRY [VIII, § 83 Problem 5. To find the projections of a line making given angles with H and V. In order to bring a definite solution it will be assumed that the true length, slope, and location of one end of the line are known. The general solution to the problem as stated above consists of four series of parallel lines, corresponding to the four possible slopes of a line (§ 31). Analysis. Consider the problem solved, as in Fig. 105, and let ab (a h b h , a v b v ) be the resulting line, with point a the fixed Fig. 105. end. Take an axis through a perpendicular to H, and revolve the line to the position ac (a h c h , a"c v ), parallel to V, where aV shows the true length of the line and its angle, a, with H (Prob. 3, First Method, § 78). Also take an axis through a perpendicular to V, and revolve the line to the position ad (a v cl v , a h d h ) parallel to H, where a h d h shows the true length of the line and the angle, /?, with V. The problem may now be solved by reversing this construction. Construction. Place the line in the position ac (a v c v , a h c h ) par- allel to Y, where a v c v is equal to the true length of the line and makes with GL the angle, a, which the line makes with H. VIII, § 83] TRUE LENGTH OF A LINE 61 Revolve this line about an axis through a perpendicular to H, by drawing the projections, J h and J v , of the arc in which point c revolves. Place the line also in the position ad (a h d h , a v d") parallel to H, where a h d h (equals a v c v ) equals the true length of the line and makes with GL the angle, /?, which the line makes with V. Revolve this line about an axis through a perpendicular to V, by drawing the projections, K" and K h , of the arc in which d revolves. The point b (b h , b v ), which determines the projections of the required line ab, is found at the intersection of the two paths of revolution J and A". Note that this point is independently determined in each projection; hence as a check on the work the projector b v b h should be perpendicular to GL. CHAPTER IX SOME SIMPLE INTERSECTIONS — DEVELOPMENTS 84. Simple Intersections of Solids of Revolution. In generat- ing the surface of a solid of revolution, each point of the gener- ating line describes a circle which lies in a plane perpendicular to the axis, and whose center lies in the axis (§ 74). If the axis of the surface be placed perpendicular to H (or V), these circles will project on H (V) in their true shape and size, and on V (H) as horizontal straight lines which are the edge views of the planes containing the circles (§ 75). The intersection of a surface of revolution by a plane may be found by drawing on the surface a sufficient number of circles, obtained by passing auxiliary planes perpendicular to the axis of the surface, and then finding the points in which these circles are intersected by the given planes. 85. Visibility of the Intersection. In this and subsequent parts of this book, whenever a solid is intersected by a plane, no part of the solid will be considered as cut away or re- moved unless so stated. In determining the visibility of the line of intersection, the cutting plane will be considered trans- parent and the solid opaque. The line of intersection will be visualized as a line drawn on the surface of the solid. Hence, the line of intersection, or any portion of it, will be visible in any given projection when the line lies on a portion of the sur- face which is visible in that projection ; and conversely. The visibility of the line of intersection, then, depends directly upon the visibility of the given solid (§ 24). 86. Examples. The only position of the cutting plane which will be considered at this time is an edge view. Then 62 IX, § 86] INTERSECTIONS — DEVELOPMENTS 63 the plane is perpendicular to either H or V. In the following examples, the axis of each surface is perpendicular to H, so that the auxiliary circles (§ 84) all lie in planes which are parallel to H. Note, in each example the visibility of the curve of intersection (§ 85). Example 1 (Fig. 106). A cone of revolution cut byaplane perpendicular to V. Since the whole convex surface of the cone is visible in plan, the entire curve is visible in plan. Fig. li ii J Fig. 107. The true size of the section is found by revolving the section until parallel to V about an axis parallel to VQ. The true width at any point, as 2'8', equals the true distance 2-8 which appears in the fl"-projection. Example - (Fig. 107). A roue of revolution cut by a plane parallel to V. The curve is visible in elevation, since it ia wholly on the front half of the cone. The true si/e of the .section appears in the ^projection. 64 DESCRIPTIVE GEOMETRY [IX, § 86 Example 3. (Eig. 108). A sphere cut by a plane perpen- dicular to V. Here the intersection, as seen in elevation, lies partly on the upper half of the sphere, and partly on the lower. Hence, in the plan one part of the curve is visible, and the other part invisible. The two points in which the curve changes from visible to invisible are necessarily on the outline of the sphere. Why ? The true size of the section is a circle whose diameter appears in the F-projection. Fig. 108. Fig. 109. Example 4 (Fig. 109). A torus (§ 26) cut by a plane per- pendicular to V. An auxiliary plane, such as M, cuts the gen- erating circle in two points. Hence, in this plane two circles lying in the surface of the torus can be drawn, one of diameter ad, and the other of diameter be. Similarly, two circles can be drawn in every plane perpendicular to the axis, except in IX, § 86] INTERSECTIONS — DEVELOPMENTS 65 those which pass through the highest and lowest points of the generating circles. The visibility of the curve in plan is determined in the same way as for the sphere, Fig. 108. The true size of the section is shown parallel to V, as in Example 1. Example 5 (Fig. 110). A hyperbolic spindle, cut by a plane perpendicular to V. In this case the surface that is cut is wholly hidden in plan, hence the curve is invisible. The true size of the section is shown parallel to V. Fig. 110. Fig. ill. Example 6 (Fig. 111). A torus, cut by two planes parallel to V. The X section, wholly on the hack of the surface, is invisible. The FTsection, entirely on the visible part of the torus, is wholly visible. The true, size of each section appears at once. 66 DESCRIPTIVE GEOMETRY [IX, § 86 Example 7 (Fig. 112). A hyperbolic spindle, cut by tv:o planes parallel to V. Each plane cuts a section composed of Fig. 112. two parts. The visibility of the curves should be evident from an inspection of the position of the cutting planes. 87. Developments. The development of a solid bounded by plane faces consists of a series of polygons of the true sizes and shapes of the various faces of the solid, so arranged, edge to edge, that the development may be folded up to reproduce the surface of the solid. If a face of a solid is a triangle, the true size of the face may be constructed from the true lengths of its three sides. If the face is a polygon of more than three sides, the true size and shape cannot be determined from the true lengths of the sides alone. Any plane polygon, however, can be divided into triangles by drawing certain diagonals, and the polygon can be constructed from these triangles. Hence the development of any solid bounded by plane faces can be obtained wholly by finding the true lengths of straight lines. IX, § 88] INTERSECTIONS — DEVELOPMENTS 67 88. Working Method for Finding the True Length of a Line. In the solution of many problems the true length of a line is found in order to be used as a radius for setting compasses or dividers. The angle made with H or V is not needed. Hence it is sufficient to construct two points whose distance apart is equal to the true length of the line. Let ab, Fig. 113, be any line, and let the true length of this line be found by revolving it parallel to V, keeping the upper Fig. 113 end of the line fixed (Prob. 3, First Method, § 78), thus giving a v c as the true length. Now the point c can be found by draw- ing through b v a horizontal line, intersecting the projector a v a h at e, and then laying off the distance ec directly with the dividers ecpial to u k b h . Hence the following working rule : Rule. Tlie true length of a straight line equals the hypothe- nuse of a right triangle whose base equals the length of the H-pro- jection of the line, and whose altitude in equal to the difference in elevation between the two ends of the line. If the lower end of the line is in the HT-plane, 6* will lie in OL, and the method becomes even simpler. Thus, the true length of the line ab, Fig. 114, may be found by laving off on OL the distance ec equal to a h b h ; then a v c is the desired true length. 68 DESCRIPTIVE GEOMETRY [IX, § 89 89. To Find the Development of a Solid with Plane Faces. Let the given solid be the frustum of an irregular four-sided pyramid, Fig. 115. Each face is a quadrilateral, which can be divided into two triangles by means of one diagonal. Never- theless it is easier to begin by finding the development of the triangular faces of the complete pyramid. Assuming that the true lengths of the various lines are found as needed, we proceed with them as follows (B, Fig. 115). On any convenient line, Fig. 115. lay off the distance 0-1 ; with and 1 as centers, and radii 0-2 and 1-2 respectively, strike arcs intersecting in point 2. This gives the development of the triangular face 0-1-2. With and 2 as centers, radii 0-3 and 2-3 respectively, locate point 3, thus obtaining the development of the face 0-2-3. In the same way obtain the faces 0-3^ and 0-4-1. The base, 1-2-3-4, of the pyramid is divided into triangles by the diagonal 1-3, and is plotted to join one of the edges previously located, as 2-3. The result thus far obtained is the development of the complete pyramid. To convert it into the development of the frustum, proceed as follows : IX, § 90] INTERSECTIONS — DEVELOPMENTS 69 On 0-1, 0-2, etc., of the development, measure the true lengths of the lines 1-5, 2-6, 3-7, 4-8, and draw 5-6-7-8-5. Draw one diagonal, as 5-7, of the upper base, plot this base by means of the two triangles thus formed, and join it prop- erly to one of the previous lines of the development. The development of the frustum is now complete. 90. To Construct the Projections of a Prism Whose Long Edges Make Given Angles with fl"and V (Fig. 116). Let it be required to draw the projections of a prism whose long edges shall make 30° with H and 45° with V. Place the prism first with its long edges parallel to H and making the given angle, 45°, with P". Assuming that the neces- sary sizes, slopes, and distances are given, we can construct Fig. ll«. these projections by using a secondary ground line, as in § 72 (B, Fig. 116). Find, by Problem 5, § 83, the projections of a line ab making the given angles with II and V (A, Fig. 116). Now the long edges of the prism as placed in B, Fig. 116, make the given angle with V. If the prism be revolved about any axis perpendicular to V, the edges will continue to make the same angle with V, while the projection of the prism on V 70 DESCRIPTIVE GEOMETRY [IX, § 90 will remain of the same shape and size. Let the prism be re- volved until the F-projections of the long edges are parallel to a v b v , as shown at C, Fig. 116. In making this revolution, the actual axis need not be used. It suffices to copy the F-projection obtained at B, making the long edges of the prism parallel to the direction a v b v obtained at A. But in this revolution, the ^-projection of the path of each moving point will be a horizontal straight line, regardless Fig. 116 (repeated). of the position of the axis of revolution. Hence the //"-projec- tion at C may be obtained by projecting vertically from the F-projection, and horizontally from the //-projection at B. As a check, the long edges of the projection thus obtained should be parallel to the direction a h b h found at A. The projections at C are the required projections. 91. Other Methods. The projections at B, Fig. 116, were made by knowing the angle which the long edges of the prism made with V, while the projections at C were made by know- ing the inclination of the F-projection of the edges. These two quantities might have been given directly, and projections B and C constructed by Problem 4, § 83. CHAPTER X LINES IN A PLANE — PARALLEL LINES AND PLANES 92. Intersecting and Parallel Lines. Two lines intersect when they pass through a common point. Two lines are parallel when they are everywhere equally distant. If two straight lines lie in the same plane, they either intersect or are parallel. This is not necessarily the case, however, with any two lines in space. 93. Parallel Lines in Space. If two lines in space are parallel, their projections on any plane are parallel. Let the two parallel lines A and B, Fig. 117, be projected on the plane Q by planes perpendicular to Q. Since the two Ftg. 117. projecting planes will be parallel, their intersections, A q and B q , with the plane Q will be parallel. In general, if lines A and l> are parallel, it is definitely shown by the simultaneous condition that A h is parallel to Il h , and A" is parallel to />'". Profile lines are an exception ; such lines may be tested by means of their profile projections. Thus, the profile lines ad and cd, Fig. IIS, are parallel, since their profile projections are parallel. 71 72 DESCRIPTIVE GEOMETRY [X, § 94 94. Intersecting Lines in Space. If two lines in space intersect, their corresponding projections intersect in the same projector. Let A and B, Fig. 119, be two intersecting lines. Since these lines intersect, there must be a point, c, common to each of them. Since c is a point of the line A, the Zf-p rejection, c h , Fig. 119. must lie on A h (§ 36). Moreover, since c is in the line B, c* must lie on B h . Hence c* is at the intersection of A h and B\ Similarly, the F-projection of c, c", must lie at the inter- section of A v and B". But since c h and c" are two projections of the same point, they must lie in the same projector. If two lines have a common projection (Fig. 120), they lie in the same plane, namely, the plane whose edge view coincides Fig. 120. with the common projection. Such lines are either intersect- ing, as in the figure, or parallel (§ 92). X, § 95] LINES IN A PLANE 73 95. Test for Intersecting Lines. Conversely, if the correspond- ing projections of two lines intersect in the same projector, the lines in space intersect. The only exception occurs when one of Fig. 121. the lines is a profile line ; thus, the line E, Fig. 121, does not intersect the profile line ab, since the point c, in which E pierces the profile plane containing ab, does not lie on the line ab. If the point of intersection is outside the limits of the draw- ing (Fig. 122), the test fails. In this case take any two points, d and e, on A, and any two,/ and g, on B. Then if A and B intersect, df and eg (or else dg and ef) will intersect. 74 DESCRIPTIVE GEOMETRY [X, § 96 96. A Line in a Plane. If a line lies in a plane, the traces of the line lie in the corresponding traces of the plane. Let a line A lie in a plane Q (Figs. 123 and 124). Then if A intersects H, it can evidently do so only in some point of Fig. 123. Fig. 124. the line in which Q intersects H ; and the intersections with H are the respective traces of the line and plane. Similarly for any other plane of projection. 97. To Locate a Line in a Plane. A line will lie in a given plane if the line passes through two points which lie in the plane. Thus, in Fig. 123, let us assume s h on HQ, and t v on VQ. These two points are the traces of some line lying in the plane Q (§ 96). The projections A h and A" may then be found by Prob- lem 2, § 37. Other methods of determiniDg a line so that it shall lie in a given plane will be given later. (See §§ 100, 133.) 98. A Plane Containing a Given Line. The proposition of § 96 enables us to pass a plane through a given line. Thus, in Fig. 123, let A(A h , A") be a given line, whose traces are the points s and t. A plane Q may be passed through A by drawing HQ in any direction through s*, and then drawing VQ through t v and the point in which HQ intersects the ground line. Conversely, VQ may be drawn first, in any direc- tion through f, and then HQ may be drawn through s h and the intersection of VQ with the ground line. An indefinite number of planes may be passed through a X, § 99] LINES IX A PLANE 75 given line. In particular (Fig. 125), we may draw a plane X, perpendicular to H, a plane Y perpendicular to V, and a plane S parallel to the ground line. (See also § 102.) Fig. 125. 99. Principal Lines of a Plane. For every plane there are two systems of lines which are of importance, and are called the principal lines of the plane. One of these systems is parallel to II, the other parallel to F. The horizontal principal lines are //-parallels, cut from the given plane by a series of planes parallel to //. The lines arc therefore parallel to the //-trace of the plane ; their //-projections are parallel to this trace, while their ^projections are parallel to the ground line. Visualize this state- ment, aided, if necessary, by Fig. 126. The vertical principal lines are cut from the plane by a series of planes parallel to V. They are F-paralhls, parallel to the F-trace of the plane. Their F-projections are parallel to this trace, their ff-projections are parallel to the ground line. These tun groups of lines are, in general, different. They reduce to the same set when, and onl\ when, the given plane is parallel to the ground line. /-^A v \ V / 7 Fig. 126. 76 DESCRIPTIVE GEOMETRY [X, § 100 100. To Project the Principal Lines of a General Plane. To draw the projections of a horizontal principal line of a general plane, Q, Fig. 127, assume the F-trace of this line as any point, Fig. 127. t", in VQ, and find t h on GL. Then, by § 99, A h passes through t h and is parallel to HQ, while A v passes through f and is parallel to GL. Similarly, a vertical principal line, B, may be drawn by first assuming its ZZ-trace, s, on HQ. 101. To Project the Principal Lines of a Plane Parallel to the Ground Line. To draw a principal line of a plane when the plane is parallel to the ground line, as Q, Fig. 128, find the Q. > VQ \fl u v A v / 1 / 1 [ \ \ U h A h HQ D. X Fig. 12S. profile trace, PQ, of the given plane. On PQ assume any point, u p , as the profile trace of the required principal line. Both projections of this line are parallel to the ground line. X, § 103] PARALLEL LINES AND PLANES 77 102. A Plane Containing a Line Parallel to H or V. Any plane, as Q or R, Fig. 129, which is passed through an H- parallel will have its //"-trace parallel to the //"-projection of the line. (Converse of § 100.) Similarly, any plane passed through a F-parallel will have its F-trace parallel to the F-projection of the line. Any plane, as Q, Fig. 128, which contains a line parallel to both H arid V will have both its H- and F-traces parallel to the ground line. (Converse of § 101.) 103. Parallel Planes. If two planes are parallel, their corre- sponding traces are parallel. Let Q and R be the given planes. Let Q intersect H in HQ, and R intersect // in HR. The figure is left for the student to draw. Now when two parallel planes, Q and R, are intersected by a third plane, H, the lines of intersection must be parallel. Hence HQ is parallel to HR. Similarly for any other plane of projection. VQ. 7 > HR N \ \ . \ \\ a. X A / VR HQ VS \ -HS Fia. 130. Conversely, if the //- and F-traces of two planes are parallel, the planes are, in general, parallel. The only exception occurs with planes parallel to the ground line. Such planes may be tested by means of their profile traces. Thus, in Fig. 130, the plane R is parallel to the plane Q, while the plane S is not. 78 DESCRIPTIVE GEOMETRY [X, § 104 104. A Line Parallel to a Plane. A straight line is parallel to a plane if it is parallel to some line lying in the plane. Thus the line A, Fig. 131, is parallel to the plane Q, since it is parallel to the line B which lies in Q. If the line B were not Fig. 131. drawn, however, it would be difficult to recognize the parallel- ism. Thus, in Fig. 132, both the lines A and B are parallel to the plane B, but this fact can hardly be seen by inspection. Fig. 132. Fig. 133. The fact that a line is parallel to a plane can be recognized at once only when the line is parallel to one of the systems of principal lines of the plane. In Fig. 133, it is evident that both lines C and D are parallel to the plane S. X, § 106] PARALLEL LINES AND PLANES 79 105. A Plane Parallel to a Line. Conversely, a plane is par- allel to a straight line if the plane contains a line which is parallel to the given line. Thus, in Fig. 131, since the lines A and B are parallel, the plane Q, or any other plane passed through B, is parallel to the line A. The only exception is the plane which contains both A and B. 106. A Plane Determined by Lines or Points. One plane, and one only, may be found which contains (a) two intersecting lines ; (6) two parallel lines ; (c) a line and a point not on the line ; (d) three points not in the same straight line. Problem 6. To find the plane which contains two given intersecting or parallel lines. / y\\ \^4 ^^P^ 1 1 1 V^ Fig. 134. Analysis. The trace of the required plane on any coordinate plane must contain the trace of each of the given lines (§ 96). The plane is found when its II and V traces are found. Construction (Fig. 134). Let A and B, intersecting at point c, be the given lines. Find the traces of A and B (Prob. 1, § 37). The required trace HQ is determined by the two iZ-traces s x and s 2 ; the trace VQ by the two P"-traces t x and L. Check. The H- and F-traces of a plane must intersect on the ground line (§ 46). Hence II Q and VQ, when produced, must intersect on GL. 80 DESCRIPTIVE GEOMETRY [X, § 106 A plane which contains two given parallel lines is shown in Eig. 135. The construction is entirely similar to that just given. Fig. 135. In Fig. 136 the IZ-trace of the given line B is too far removed to be used in the construction. The plane Q, containing A and Fig. 136. Fig. 137. B, is located by first drawing VQ through the F-traces of the given lines ; then HQ is drawn through the JT-trace of A and X, § 106] PARALLEL LINES AND PLAXES 81 the point in which VQ intersects the ground line. This con- struction gives no check on the work, but may be used in case of necessity. Other constructions which may be employed wheu traces of the given lines are not accessible will be given in § 108. We shall consider at present only those cases in which one or both of the given lines are parallel to H or to V. Special Case I. Let us suppose that one of the given lines is parallel to H or V. Let A, Fig. 137, be parallel to V. Then VQ is parallel toi" (§ 102), and passes through the F-trace of the line B. The trace HQ is determined, as in the general case, by the if-traces of the given lines. In Eig. 137, the point in which HQ and VQ intersect on GL is not available, but it is not necessary. Special Case II. If one of the given lines, as A, Fig. 138, is parallel to both H and V, that is, parallel to the ground line, hq Ha Fia. 138. A h a. r uf "N B h "* \ \ A" l ' / 1 i / ur ' J/. B" a. > i 7U1 VQ A Fig. 139. both HQ and VQ are parallel to the ground line (§ 102), one point on each trace being located by the line B. Special Case III. Suppose both given lines parallel to the ground line. Let A and B, Fig. 139, be the given lines; then HQ and VQ are both parallel to GL. To find a poinl on each, we may find first the profile trace of the plane Q on any assumed profile plane; this trace, PQ, will pass through the profile traces up and v./, of A and B respectively. 82 DESCRIPTIVE GEOMETRY [X, § 106 Corollary I. To find the plane which contains a given line and a given point. Analysis. Through the given point, draw an auxiliary line parallel to the given line, thus reducing the problem to one of two parallel lines. Or, assume any point on the given line, and draw an auxiliary line through this point and the given point, thus reducing the problem to two intersecting lines. The first method is the one usually adopted. Construction (Fig. 140). A is the given line, c the given point. The auxiliary line B is drawn through c parallel to A Fig. 140. (§ 93), and the required plane Q is passed through the lines A and B. In Fig. 141 the line A is parallel to V\ hence the direction of VQ is known to be parallel to A v (§ 102). But as neither A nor B has a F-trace, a point on VQ is determined by noting where HQ intersects GL. Corollary II. To find the plane which contains three given points not in the same straight line. Analysis. Connect any two points by a straight line, and through the third point draw a second line parallel to the first, X, § 107] PARALLEL LINES AND PLANES 83 thus reducing the problem to two parallel lines. Or, draw lines connecting any point with each of the other two, thus reducing the problem to one of two intersecting lines. ^k\ \ s >^ Cjf / 1 Xi / / 4 / 7 V ^> b v ; Fig. 141. Fig. 142. No figure for the general case is thought necessary. It is interesting to note that this device may be used for passing a plane through a profile line and a point. Thus (Fig. 142), let ab be the given line and c the given point. Draw the lines ac (aW, a v c v ), and be (b h c h , b v c v ), and pass the required plane Q through these lines. This avoids the necessity of finding the traces of the profile line ab, although its traces, if found, should lie on HQ and VQ respectively (§ 96). 107. A Plane Parallel to Lines or to Another Plane. A plane which is parallel to given lines or to another plane is deter- mined by the fact that a plane containing one of two parallel lines is parallel to the other line (§ 105). One plane, and one only, may be found which (a) contains a given line and is parallel to a second given line that is not parallel to the first line ; (b) contains a given point and is parallel to each of two given lines that are not parallel to each other ; (c) contains a given point and is parallel to a given plane. 84 DESCRIPTIVE GEOMETRY [X, § 107 Problem 7. To find the plane which contains a given line and is parallel to a second given line. Analysis. Through any point on the first given line draw an auxiliary line parallel to the second given line. The required plane is determined by the first line and the auxiliary line. Fig. 143. Construction (Fig. 143). Let it be required to pass the plane through the line A. Assume any point con J; through c draw the line D parallel to B. Pass the required plane Q through the lines A and D (Prob. 6, § 106). Fig. 144. X, § 107] PARALLEL LINES AND PLANES 85 Special Case I. Suppose the second line is parallel to II or V; then no auxiliary line is needed. Thus, in Fig. 144, let us find the plane which contains the line A and is parallel to B. The .ff-trace of any plane which is parallel to B must be parallel to B h (§ 104). Hence, find the traces of A. Draw HQ through s x parallel to B h ; draw VQ through t x and the point in which HQ intersects GL. Special Case II. Suppose that either the first or the second given line is a profile line. The general solution will apply to this case ; but if the problem be solved in this manner, a profile projection will be necessary. A simple construction, by which the use of a profile projection may be avoided, is as follows : Let ab and C, Fig. 145, be the given lines. Through one point of the profile line, as a, draw an auxiliary line D parallel to the line C; through the other point, b, of ab, draw a second Fig. 145. auxiliary line E parallel to C. Pass the plane Q through the two parallel lines D and E. Then if the required plane is to contain ab and be parallel to C, plane Q is the required plane. If, however, the required plane is to contain C and be parallel to ab, the plane It, passed through C parallel to plane Q (§ 103), is the required plane. 86 DESCRIPTIVE GEOMETRY [X, § 107 Problem 8. To find the plane which contains a given point and is parallel to each of two given lines. Analysis. Through the given point draw two auxiliary lines, one parallel to one given line, the other parallel to the other given line. The required plane is determined by the two aux- iliary lines. Fig. 14(5. Construction (Fig. 146). Let c be the given point, A and B the given lines. Through c draw the auxiliary lines, D parallel to A, and E parallel to B. Pass the required plane Q through the lines D and E (Prob. 6, § 106). The special cases of this problem are too similar to those of Problem 7 to require a detailed discussion. Problem 9. To find the plane which contains a given point and is parallel to a given plane. Analysis. Through the given point, pass a line parallel to the given plane. Through this line, pass the required plane parallel to the given plane. Construction (Fig. 147). Let Q be the given plane and a the given point. A line may be drawn through a and parallel to Q by drawing it parallel to either the horizontal or vertical principal lines of Q (§ 104). The line M, drawn through a, is parallel to the vertical principal lines of Q ; through M pass the required plane T, parallel to Q (§ 103). X, § 107] PARALLEL LIXES AND PLANES 87 In Fig. 148, the required plane T is found by means of the auxiliary line N, which is parallel to the horizontal principal lines of the given plane Q. Fig. 147. Fig. 148. Special Case" (Fig. 149). Let the given plane Q be parallel to the ground line. Then the auxiliary line M, drawn through a and parallel to the principal lines of Q, is parallel to GL (§ 101) and has no traces on H and V. Pass the auxiliary Fki. 149. profile plane P through the given point. Find the profile trace, PQ, of Q, and the profile projection, a p , of a. The profile trace, PT, of the required plane will pass through a" and be parallel to PQ (§ 103). From PT may be found IIT and VT. 88 DESCRIPTIVE GEOMETRY [X, § 108 108. Use of Auxiliary Lines in Finding the Traces of Planes. A straight line is determined : 1. When two of its points are known ; 2. When one of its points and its direction (such as parallel or perpendicular to another line) are known. Eor accuracy of construction, the second method of deter- mining a line is often better than the first. In any event, how- ever, a line is not determined until at least one point is known. The traces of the required planes in the preceding problems, being straight lines, have necessarily been located by one of these methods. It occasionally happens, however, that neces- sary points for locating these traces fall far outside any rea- sonable limits for the size of the figure, and recourse must be had to auxiliary lines which will locate points within reach. Example 1 (Eig. 150). Let it be required to find the plane which contains the lines A and B, so situated that only one Fig. 150. point, s 2 , on HQ, and one point, t 2 , on VQ, can be found. Assume any point, e, in the given line A ; through this point draw the line D, parallel to the given line B. Pass the plane Q through the parallel lines B and D. This plane must necessarily contain the line A, since A intersects both B and D, and is the required plane. In general, assume any point on one of the given lines. X, § 108] PARALLEL LINES AND PLANES 89 Through this point draw an auxiliary line parallel to the second given line ; the auxiliary line will lie in the plane of the given lines. Judgment must be used in selecting the line to which the auxiliary line should be made parallel, in order that this latter line may be of service in obtaining additional points. Example 2. Another method of obtaining a line lying in the plane of two given lines is shown in Fig. 151. Let A and B be the given lines. Assume any point, e, in A, and any point, /, in B. Then the line D, passing through e and /, will lie in the plane of the lines A and B, and the traces of D will lie in the traces of the required plane Q. As before, judg- ment must be used in selecting the points e and /. Fig. 151. Fig. 152. Example 3. When the given data are sufficient to determine one trace of the plane, a very useful auxiliary line is one of the principal lines (§ 99) of the plane itself. Let A and B, Fig. 152, be sufficient to locate VQ but not HQ. Assume a point in one of the given lines, as point c in line A. Through c draw a vertical principal line of Q, by making M v parallel to VQ, and M h parallel to GL. Then the line M lies in Q, and its H-tra.ce, s 3 , is a point in HQ. If HQ were the known trace, a horizontal principal line of the plane could be similarly drawn to locate a point in VQ' 90 DESCRIPTIVE GEOMETRY [X, § 108 Example 4. Another method of procedure, when one trace of the plane is known, is shown in Fig. 153. Let VQ be the known trace. Assume any point, t v , in VQ ; this point lies in V, and t h is in GL. Through the point t, draw the auxiliary line D parallel to one of the given lines, as A. Then the line D lies in the plane Q, and its //-trace, s*, is a point in HQ. Fig. 153. Fig. 154. In general, the auxiliary line D may be drawn parallel to either of the given lines A or B ; but in this case D should obviously be drawn parallel to A in order to obtain a result within the limits of the figure. Example 5. Let the lines A and B, Fig. 154, be sufficient to locate HQ but not VQ. A third method of obtaining an auxiliary line is to assume any point, as s, on HQ, and any point, as e, on either of the given lines A or B. Then the line I), joining s and e, lies in the plane Q, and its F-trace, t v , is a point in VQ. In some situations, in order to get a sufficient number of points, it may be necessary to employ more than one auxiliary line. This would have been the case in Figs. 152, 153, and 154, if in any one of these figures the intersection of HQ and VQ with the ground line had not been available. CHAPTER XI PERPENDICULAR LINES AND PLANES 109. Perpendicular Lines. If two straight lines are perpen- dicular to each other, the fact is not, in general, apparent from the projections of the lines. For example, consider the square prism represented in Fig. 94. The edges of this solid form Fig. M (repeated). three series of mutually perpendicular lines; but all these edges are oblique to V, and in the ^projection no right angles appear. In the II- and secondary ^projections, however, the pro- jected edges are at right angles. This shows that, under cer- tain conditions, it is possible to recognize perpendicular lines from their projections. 91 92 DESCRIPTIVE GEOMETRY [XI, § 110 110. Perpendicular Lines Whose Projections Are Perpendicular. If two lines in space are mutually perpendicular, and are pro- jected on any plane of projection parallel to one of the lines, the two projections will he perpendicular to each other. There is one and only one exception. The plane of projec- tion must not be perpendicular to the second line ; for in this case the line would project as a point, and a point cannot be said to be perpendicular to a line. Let the line A, Fig. 155, be parallel to H. Let Q be a plane perpendicular to A and consequently perpendicular to H. In this position of the line and plane, it is evident that A h and HQ are perpendicular to each other. Now let B be any line drawn in the plane Q. Since A is perpendicular to Q, the lines A and B must be mutually perpendicular. Also, since Fig. 155. Fig. 15li. Q is perpendicular to H, B h must coincide with II Q ; that is, the projections A h and B h are perpendicular to each other. It may be noted that the line B need not necessarily intersect the line A, since any line lying in Q is considered to be perpen- dicular to A. What is true of the if-plane of projection must be true of any plane of projection, hence the proposition is established. 111. A Line Perpendicular to a Plane. If a line is perpen- dicular to a plane, any projection of the line is perpendicular to the corresponding trace of the plane. Let the line A, Fig. 156, be perpendicular to the plane Q. Then A will be perpendicular to any line lying in Q, in par- ticular to the horizontal principal line B. But B is parallel to XI, § 113] PERPENDICULAR LINES AND PLANES 93 H; hence A h is perpendicular to B h , by § 110. Moreover B h is parallel to HQ. Hence A h is perpendicular to HQ. Similarly for any other plane of projection. 112. Test for Perpendicularity of a Line and a Plane. Con- versely, if the horizontal and vertical projections of a line are perpendicular to the corresponding traces of a plane, the line and plane are mutually perpendicular. The only exception occurs when the plane is parallel to the ground line, for the projections of any profile line on H and V are perpendicular to the H- and F-traces of such a plane. Such cases may be tested by the use of the profile projection. Thus, in Fig. 157, the line ab is found to be perpendicular to the plane Q. 113. Perpendicular Planes. If two planes are mutually perpen- dicular, the fact is not, in general, evident from the traces of the planes. Thus, in Fig. 158, the line A is perpendicular to the plane Q. Hence any plane, as R or S, passed through A, is perpendicular to Q ; but this relation cannot be seen directly from the traces of the planes. Fig. 158. Fig. 15it. But let the line A, Fig. 159, be perpendicular to the plane Q, and let us pass through A the plane T perpendicular to //. 94 DESCRIPTIVE GEOMETRY [XI, § 113 Since HT coincides with A h , and A h is perpendicular to HQ, it follows that HT is perpendicular to HQ. Hence we have the proposition : If two planes are mutually perpendicular, and one of them is perpendicular to a coordinate plane, the traces of the two given planes on that coordinate plane are perpendicular. 114. Lines of Maximum Inclination to H and V. Let Q be any plane oblique to H. Those of its lines which are perpen- dicular to HQ have the peculiarity that they make with H a greater (acute) angle than any other lines in the plane. For this reason they are called the lines X. of maximum inclination to H of the \ plane Q. Let M, Fig. 160, be one V \ob °f these lines. Since M is perpen- V/\ dicular to HQ, it follows that M h \ \^ must also be perpendicular to HQ. N. ^ / A plane is determined when one >^ J / of its lines M (M h , M") of maximum y^ inclination to H is given. For sup- ^y pose that s and t, Fig. 160, are the „ ,_. traces of this line. Then the Fig. 160. //-trace, HQ, of the required plane passes through s and is perpendicular to M h , while the F-trace, VQ, is determined by t and the point in which HQ cuts the ground line. A similar analysis applies to lines of maximum inclination to F 115. Planes Perpendicular to a Given Line or Plane. One plane, and one only, may be found which contains a given point and is perpendicular to a given straight line. The point may be on the line, or at any distance from it. One plane, and one only, may be found which contains a given straight line and is perpendicular to a given plane. An exception occurs when the line is itself perpendicular to the given plane, in which case an infinite number of planes may be found. XI, § 115] PERPENDICULAR LINES AND PLANES 95 Problem 10. To find the plane which contains a given point and is perpendicular to a given line. Analysis. Through the given point draw an auxiliary line which shall be perpendicular to (but does not necessarily inter- sect) the given line. Through this line pass the required plane perpendicular to the given line. Construction (Fig. 161). Let c be the given point and A the given line. Through c draw the auxiliary line M, making M h perpendicular to A h and M v parallel to OL. Then the line M is parallel to H, and consequently perpendicular to A (§ 110). Find the F-trace, t, of M. Through t draw VQ perpendicular b- \na CL h a. N % - 1 S ^ > Fig. 161. Fig. 102. to A v , and from the point in which VQ intersects GL draw HQ perpendicular to A h . Then Q is perpendicular to A (§ 112) and is the required plane. Note that the auxiliary line M is a horizontal principal line of the required plane Q. The construction could also be effected by using an auxiliary line parallel to V; such a line would be a vertical principal line of Q. Special Case. The general method fails when the given line is a profile line. Let ab, Fig. 102, be the given line, and c the given point. The required plane Q will be parallel to the ground line, hence perpendicular to P. Find the profile projection of the line ab and the point c. Through c" draw the edge view and profile trace, PQ, of Q, perpendicular to a p b p . From PQ find J/Q and VQ. 96 DESCRIPTIVE GEOMETRY [XI, § 115 Problem 11. To find the plane which contains a given line and is perpendicular to a given plane. Analysis. Through any point of the given line draw an auxiliary line perpendicular to the given plane. Any plane containing this line will be perpendicular to the given plane (§ 113). Hence the required plane is the plane determined by the given and auxiliary lines. Construction (Fig. 163). Let A be the given line and Q the given plane. Assume any point c on the line A, and through Fig. 163. Fig. Ifi4. c draw the auxiliary line B perpendicular to Q (§ 112). Find the plane T, containing the lines A and B (Prob. 6, § 106) ; this is the required plane. Special Case I. If the given line is a profile line, the gen- eral method applies ; but the use of a profile projection may be avoided as follows. Let ab, Fig. 164, be the given line, and Q the given plane. From each end of the given line draw the respective auxiliary lines D and E, each perpendicular to Q. Pass the required plane T through the parallel lines D and E (compare Fig. 145). Special Case II (Fig. 165). Let the given plane, Q, be parallel to the ground line, and let A be the given line. As in XI, § 115] PERPENDICULAR LINES AND PLANES 97 the general case, assume any point c in the line A, and draw the line B perpendicular to Q. Then the required plane, T, is the plane determined by the lines A and B. To find the traces of T we need the traces of these two lines. The traces of A are readily obtained (Prob. 1, § 37). To find the traces 0- H* X Id V / ^C /\: vol ¥ x c 0. > r V Fig. 105. of By we note that B is a profile line and that we know only one point, c, in this line. Nevertheless, B is a definite line in space, since its direction, perpendicular to the plane Q, is known. Hence, find the profile projection, c p , of the point c, and the profile trace, PQ, of the plane Q. Then the profile projection, B p , of line B is drawn through c p perpendicular to PQ. From B" we can find the //- and F-traces of line B. These traces, together with the traces of line A, determine the traces of the required plane T. CHAPTER XII INTERSECTION OF PLANES AND OF LINES AND PLANES — APPLICATIONS 116. Intersecting Planes. Let two planes Q(HQ, VQ) and R (HR, VR) intersect in a line A. The figure will be left for the student to draw. One point of the intersection will be determined if we find where a line in one plane intersects a line in the other. A second point may be similarly found by the intersection of a second pair of lines. Evidently any such pair of lines cannot be chosen at random, for in that case they probably would not intersect. The //-traces of the two planes, however, as they are both in H, will in general intersect. Likewise the two F"-traces will in general intersect. But the intersection of the i7-traces will be the //"-trace of the required line .1 (§96), and the intersection of the p^traces will be the F-trace of A. The projections of the line A may then be determined (Prob. 2, § 37). 117. A Point in the Intersection of Two Planes. The general method of obtaining a point in the line of intersection of two planes is as follows : Let two planes Q and R, as before, inter- sect in a line A, and let X .be any plane not parallel to A. Then X will intersect Q in a line xq, and will intersect R in a line xr. Since these lines are in the same plane, X, and can- not be parallel (why ?), they intersect in a point, xqr, which lies in all three planes, A", Q, and R. Hence xqr is a point in the line of intersection, A, of Q and R. 118. The Line of Intersection of Two Planes. The line of intersection of two planes becomes known: (1) when two points of the line are known ; (2) when one point and the direction of the line are known. (Compare § 108.) Problem 12. To find the line of intersection of two planes. Analysis. The traces of the required line of intersection lie at the intersection of the corresponding traces of the given 98 XII, § 118] INTERSECTION OF PLANES 99 planes (§ 116). The line may now be determined from its traces (Prob. 2, § 37). Construction, General Case (Fig. 166). Let Q and R be the given planes. The construction should be evident from the analysis. Fig. 1G0. Fig. 167. A second example, with planes differently situated, is given in Fig. 167. The general case may fail because of (a) parallel lines ; (6) intersections inaccessible ; (c) points s and t coincident. Special Case I. Suppose that one pair of traces is parallel (Fig. 168). Let Q and R be the given planes, with HQ and 11 R parallel. The intersection of VQ and VR gives the F-trace, t, of the required line of intersection A. Consider the planes Q, R, and //. Since the intersections of H with Q and R are parallel, H must be parallel to the line of intersection, A, of these planes. Hence A h passes through t h and parallel to HQ and II R, while A" passes through t" and is parallel to GL. The line of intersection is thus seen to be a horizontal principal line (§ 99) of each of the given planes. If the F-traces of the given planes were parallel, while the //-traces intersected, the resulting line of intersection would be parallel to V, a vertical principal line of each of the gives planes. Fig. Ifi8. 100 DESCRIPTIVE GEOMETRY [XII, § 118 Q. I Ha VR \ A h u h \ \ If both the H- and F"-traces of the given planes are respec- tively parallel, the planes are parallel (§ 103), except in the case in which each plane is parallel to the ground line. Special Case II. Suppose that both planes are parallel to the ground line (Fig. 169). Let Q and R be the given planes. One point of the line of intersec- tion may be obtained by finding its profile trace on any profile plane of projection. Let P(HP, VP) be any assumed profile plane. Find the profile traces PQ and PR of the given planes (§ 60). Their intersection gives u p , the profile trace of the re- quired line of intersection A. From u" obtain u h and u v , one point in the line A. A second point is not necessary, since this line must be parallel to both H and V (Case I), that is, parallel to the ground line. Special Case III. One plane is parallel to H or V, the other is oblique (Fig. 170). Let Q and R be the given planes, \N HR > \ v ' A v y\ 9 / \ VQ. a. > Fig. 109. Fig. 170 the plane R being parallel to H. The plane R therefore inter- sects Q in a horizontal principal line of Q (§ 99). If the plane R be taken parallel to V, the intersection be- comes a vertical principal line of Q. XII, § 118] INTERSECTION OF PLANES 101 This is an important case, since, on account of the simplicity of the intersection, planes parallel to H or V are often intro- duced as auxiliaries when the general case fails. Special Case IV. The traces are not parallel, but one or both pairs fail to intersect within the limits of the drawing. Let Q and R, Fig. 171, be the given planes. The intersection of the P~-traces of these planes gives the point t, as in the general case. Since HQ and HR do not intersect within reach, pass the auxiliary plane, X, parallel to H. Planes X and Q intersect in the line M (Case III) ; planes X and R intersect in the line X; the lines M and N intersect in the point c, which is a point in the intersection of the planes Q and R (§ 117). The points t and c determine the required line of intersection A. In Fig. 172, neither pair of traces intersect within the limits of the figure. Two auxiliary planes, X and Y, are used, each of which locates a point on the line of intersection A. The auxiliary planes may be parallel to either H or V; in the figure they are taken parallel to V. Fig. 172. Fig. 17:5. Special Case V. Both planes intersect the ground line at the same point (Tig. 173). Let Q and R be the given planes. The general r-ase fails because the points sand t are coincident. A point, c, in the required line of intersection, A, may be determined by means of the auxiliary plane X parallel to V, as in Fig. 172. An auxiliary plane parallel to H may be used if prefer icd. 102 DESCRIPTIVE GEOMETRY [XII, § 118 Special Case VI. One of the planes contains the ground line (Fig. 174). Let Q and R be the given planes. R passes through the ground line, and is determined by the quadrants through which it passes and its angle with H. As in Special Case V, the points s and t are coinci- dent on the ground line. An additional point in the required line of intersec- tion may be found by the use of an auxiliary profile plane of projection, as- sumed anywhere except through the coincident points s and t. Find the profile traces, PQ and PR, of the given planes (§ 60). These traces intersect in c p , which is the profile projection of a point in the required line of intersection. 119. The Intersection of a Line and a Plane. Let a line A (Fig. 175) intersect a plane Q. The point of intersection will R is in quad- rants I %- 3, and makes 60° with H Fig. 174. Fig. 175. be determined if we find where A intersects a line in plane Q. This line cannot, however, be any line chosen at random in Q, for such a line will probably not intersect A. Let a plane, X, be passed through the line A. Then X will intersect Q in a line, T. This line Y is a line in the plane Q, which is inter- XII, § 119] INTERSECTION OF PLANES 103 sected by A at the point c. Hence c is the required point in which A intersects Q. The solution thus depends directly upon the preceding problem. Problem 13. To find the point in which a straight line intersects a plane. Analysis. Through the given line, pass any auxiliary plane. Find the line of intersection of the auxiliary plane with the given plane, and note the point in which this line intersects the given line. This is the required point of intersection of the given line and plane. General Method. Construction (Fig. 176). Let A be the given line, and Q the given plane. Through A pass any auxiliary Fig. 176. plane, as X (§ 98). Find the line of intersection, B, of X and Q (Prob. 12, § 118). The line A intersects the line B in the point c, the required point of intersection of A and Q. Check. The projection c h is determined by the intersection of A h and B h ; C is independently determined by the intersec- tion of A v and B". But c* and c" are two projections of the same point, and hence must lie in the same projector. Since the auxiliary plane X may be any plane passed through the line A, ease of construction often depends upon a judicious choice of this plane. Ordinarily, the simplest con- struction results when the auxiliary plane is taken perpen- dicular to H or V (see the planes X and Y, Fig. 125). 104 DESCRIPTIVE GEOMETRY [XII, § 119 Usual Method. Construction (Fig. 177). Let A be the given line and Q the given plane. The auxiliary plane X is taken through A and perpendicular to H. Planes X and Q intersect in the line B. The projections A v and B" intersect in c v , one projection of the required point. The projections A h and B h are coincident, so that their intersection is indeterminate. Conse- quently the projection c h must be located by projecting from c v . The auxiliary plane X might otherwise have been taken per- pendicular to V, in which case c h would be determined by the direct intersection of two projections, while c" would need to Fig. 177. Fig. 178. be located by projection from c\ The usual method does not give the check on the construction which appears in the general method. In applications, however, the gain in simplicity usually more than compensates for this. Special Case. The given plane is perpendicular to H or V. Let the given plane Q, Fig. 178, be perpendicular to V. ISTo construction is necessary, since VQ is an edge view of the plane, and in the ^projection the point in which the line A pierces the plane appears directly. The //-projection of this point is obtained by projecting from the F-projection. 120. The Shortest Distance from a Point to a Plane. The shortest distance from a given point to a given plane may be obtained by dropping a perpendicular from the point to the plane, and then measuring the length of this perpendicular. XII, § 120] INTERSECTION OF PLANES 105 Problem 14. To find the shortest distance from a point to a plane. Analysis. From the given point drop a perpendicular to the given plane. Find the foot of the perpendicular, that is, the point in which the line pierces the given plane. Obtain the true length of the perpendicular. Note. Observe tbat this solution is a direct application of the previous Problem. Construction (Fig. 179). Let a be the given point, and Q the given plane. From a draw the indefinite line, O, perpendicular to Q (§ 111). Find the point, b, in which C intersects Q Fig. 179. Fig. 180. (Prob. 13, § 119) ; in the figure, this is done by using the aux- iliary plane X, perpendicular to II. Then a h b h and a v b v are the projections of the required shortest distance, the true length of which may be found by Problem 3 (§ 78 or § 80). Special Case i Fig. 180). The given plane Q is parallel to the ground line. The required perpendicular from a is evi- dently a profile line, and may be drawn directly in the profile projection as soon as the profile views of the given point and and plane are obtained. In (ids case it is not necessary to have the H- and ^projections of the perpendicular in order to know its actual length ; these projections are, however, usually considered a part of the problem, and are obtained bj project- ing from the profile view. 106 DESCRIPTIVE GEOMETRY [XII, § 121 121. The Projection of a Point or Line on a Plane. The projection of a point on a plane is the foot of the perpendicular dropped from the point to the plane. This definition is not confined to the coordinate planes of projection, but applies to any plane in space. However, when a point is projected on to some oblique plane represented by its traces on H and V, the projection must in turn be represented by its projections on H and V. Thus, in Figs. 179 and 180, the point b (b h , b T ) is the projection of the point a (a h , a v ) on the plane Q (HQ, VQ). The projections of a straight line may be obtained by pro- jecting any two of its points, and then connecting these pro- jections. Problem 15. To project a line on an oblique plane. Analysis. Erom any two points on the given line, drop per- pendiculars to the given plane. Eind the points in which Fig. 181. these lines intersect, respectively, the given plane, and connect these points. Construction (Fig. 181). Let it be required to project the line ab on the plane Q. From a draw the line C (C h ,C v ) per- pendicular to the plane Q (§ 111), and find the point, e, where C intersects Q (Prob. 13, § 119). From b draw the line D perpendicular to Q, and find point /, where D intersects Q. Then the line ef (e H f h , e v f v ) is the projection of ab on Q. A second example is given in Fig. 182, the lettering and explanation being the same as for the preceding figure. The XII, § 121] INTERSECTION OF PLANES 107 fact that the projection ef crosses the given line ab shows that this line intersects the plane Q at the point n, where ab and ef intersect. Note that the projections n h and n v are independ- Fig. 1S2. ently determined, and should check by lying in the same projector. Special Case. In Fig. 183 the given line A is known to be parallel to the given plane Q (§ 104). Hence, to project A on Fm. 183. Q it is sufficient to project one point of A, as for example, point c, which projects at point c. Then the required line B is drawn through e parallel to the line A. CHAPTER XIII INTERSECTIONS OF PLANES AND SOLIDS BOUNDED BY PLANE FACES 122. A Plane Determined by Two Lines. A plane is com- pletely determined when any two of its lines, not necessarily its traces, are known (§ 106). Hence, if a plane be given by means of any two intersecting or parallel lines in space, it is not always necessary, nor even desirable, to find its traces in the solution of a problem in which such a plane occurs. 123. The Intersection of a Line with a Plane Determined by Two Lines. Let it be recpiired to find the point in which the Fig. 184. line C, Fig. 184, intersects the plane of the intersecting lines A and B. Instead of finding the traces of the plane containing A and B (Prob. 6, § 106), and then finding the point in which intersects this plane (Prob. 13, § 119), let us proceed at once, as in the usual method of Problem 13, to pass through the line 108 XIII, § 123] INTERSECTION OF PLANES 109 C an auxiliary plane perpendicular to H .(or to V). The line HX, coincident with C h , is the TZ-trace and edge view of such a plane. The plane X intersects the line A in point j (see Fig. 178) and the line B in point k. The line D, connecting j and k, must therefore be the line of intersection of the plane X with the plane of the lines A and B. The projection D v intersects C" at e v , which, for the same reasoning as that given in Problem 13, must be one projection of the point in which C intersects the plane of A and B. Finally, e h is found by pro- jecting from e v . Note that in this solution the position of the ground line is not essential. It may therefore be omitted. Figure 185 shows how to apply this method to find the point in which the profile line cd pierces the plane of the in- tersecting lines A and B. Pass through cd a profile plane. The plane of A and B intersects the profile plane in the line ab. By means of a profile projection it is readily found that the lines ab and cd intersect in the point e, the required piercing point. In this figure the ground line is not omitted, since it is necessary in order to find the profile projection. Nevertheless the positions of the projections e* and e v are independent of the position of the ground line. 110 DESCRIPTIVE GEOMETRY [XIII, § 124 124. The Intersection of Two Limited Plane Surfaces. Let it be required to find the intersection of the triangle abc, Fig. 186, with the plane, indefinite in length, but limited in width by the parallel lines J and K. The intersection can be found, without rinding the traces of either plane, by applying the preceding method, as follows. Using the auxiliary plane X which contains J and is perpendicular to H, we find that the line J intersects the plane of the lines ab and be in the point d. Using the auxiliary plane Y, containing K, this line inter- sects the plane of the lines ac and be in the point e. There- fore the plane JK intersects the plane abc in tlie line de. 125. Visibility. If both the plane surfaces of Fig. 186 are considered to be opaque, each of them must hide a portion of the Fig. 186. Fig. 187. other. The visibility of either projection must be determined by means of information obtained from the other projection. Thus, to determine the visibility of the Zf-projection, take any point in which intersect the projections of two lines not in the same plane. For example, consider the point where K h inter- sects b h c h . This is actually the projection of two points, n in XIII, § 126] INTERSECTION OF PLANES 111 the line be, and o in K. Project to the ^projection, obtaining n v and o". From these projections we see that the point o is higher than the point n ; that is, iT passes above be. Therefore, ha the iZ-projection, K h , which contains o h , is the visible line at the point under consideration. The other points in the ^-projection where the projections of lines not in the same plane cross each other can be tested in the same way. This process finally results in the visibility shown in the figure. We may reason also as follows : the line K has been found to be above the triangle at the point o. Now K intersects the triangle at the point e. Therefore beyond e the line K passes out of sight under the triangle, so that at the point where K h intersects a h c h , the latter must be the visible line ; and so on around. The visibility of the P"-projection is similarly determined. Begin at any point where the two projections of lines not in the same plane cross each other, as for instance where J v inter- sects a v c v . Project to the JET-view to find which line is in front of the other. In this case, point r in ac is in front of point q in J, therefore a v c v is visible, and J v is hidden by the triangle. And so on until the complete visibility is found. 126. The Intersection of Two Planes Limited in One Direction. Figure 187 represents the intersection of two planes, each limited in width but indefinite in length. This case presents one new feature over Fig. 18(>. Using auxiliary planes per- pendicular to H, the line J intersects the plane AB in the point d. But the line K does not intersect the plane AB within the limits of its extent. If, however, the plane AB extended indefinitely beyond B, K would intersect this plane in the point e, outside of B, and cle (d"e v , d h e h ) would be the line of intersection of the plane .IK with such a plane. Therefore, the portion df of the line de, which lies within the limits of the plane AB, must be the intersection required. The visibility of the two projections is obtained as explained above. 112 DESCRIPTIVE GEOMETRY [XIII, § 127 127. The Intersection of a Plane and a Pyramid. Before be- ginning to find the intersection, visualize the solid, to deter- mine its visible faces. Then when the first line of intersection is found, if it be on a visible face, make it a full line. If it is on an invisible face, make it a dotted line ; and so on for the complete intersection. The evident advantage of this is that the student will be able to visualize his work more readily and more completely as he goes along. The construction of the Fig. 188. intersection requires merely the application and extension of the preceding method. In Fig. 188 let it be required to find the intersection of the limited plane JK with the faces of the triangular pyramid XIII, § 127] INTERSECTION OF SOLIDS 113 oabc. In the plan, all the faces are visible except the base abc ; in the elevation oab and obc are visible, while oac is not visible. In beginning to find the intersection, take the lines J and K separately. Select any face of the pyramid, as oab. Find the point 4 where J intersects the face oab, by using the plane X perpendicular to V. Similarly find 9, the point where K intersects the same face. The line 4-9 is then the inter- section of plane JK with face oab, and is made a full line in both views, as oab is a visible face in both plan and elevation. Taking another face of the pyramid, as oac, and proceeding as before, K is found to intersect at point 10, while J intersects at the point 11, in the plane of oac, but outside the face itself. Here it is well to note that while the actual face of the pyramid is of limited extent, the plane of the face is unlimited. Joining 10 with 11, the part from 10 to 12 is the actual intersection of JK with the face oac, is a full line in plan, and dotted in ele- vation. The line J has been found to enter the pyramid at point 4, but the point where it comes out has not yet been found. By inspection of the plan, J" will necessarily come out on face obc. The construction gives 5 as the point where J pierces this face. The intersection of KJ with the face obc will be the line 5-12, for as 12 is on the edge oc of the pyramid, it must be a point common to the intersections on the adjoining faces oca and ocb. The line 5-12 is visible in both plan and elevation. It should be observed that since point 12 is the intersection of plane JK with the edge of the pyramid, it might have been obtained directly by finding where oc intersected the plane JK. This would have been a convenient construction in case the point 11 luid fallen outside the limits of the drawing. Completive the Visibility. With the intersection itself prop- erly lined in with lull and dotted line, the visible portion of each edge of the pyramid or plane can usually be determined l>v inspection. In case of doubt, however, the relative position of two edges which apparently intersect in one view may be determined l>\ projecting the point of apparent intersection to the other view (§ 125). 114 DESCRIPTIVE GEOMETRY [XIII, § 128 128. The Intersection of Two Solids Bounded by Plane Faces. By further extending the preceding methods, the intersections with each other of the surfaces of any two solids bounded by plane faces — a broken line or lines usually known as intersec- tion of the solids — can be found. The detail of the construc- tion varies in each particular case, but involves nothing which has not been already explained. In order not to get lost in the construction, however, it is necessary to proceed in a system- atic and orderly manner ; therefore, for the benefit of stu- dents who may wish to follow through such a construction before attempting one for themselves, we shall give the entire detail of one typical construction. The two solids chosen, Fig. 189, are a triangular right prism whose long edges are the lines J, K, and L, and a wedge, with a rectangular base adbe, and an edge cf parallel to the plane of the base. Before starting to draw, we should notice from the //-projection that the ends of the prism JKL are not con- cerned in the intersection at all. Therefore it will be well to begin by finding where the lines J, K, and L, considered sepa- rate^, intersect the other solid. We may notice also, if we wish, that the face def of the wedge will not be intersected by the prism. Again, it is evident that the line ad will not inter- sect the prism. This information, however, is not of particular importance at the start. Before beginning to draw, it is well also to determine the visible faces of each solid considered alone. Then, when any line of the intersection is found, it can be visible only when it is the intersection of two visible faces, one of each solid. The visi- bility of the intersection being thus determined, and the visible faces of each solid taken alone being known, the visible lines in the combined figure usually can be determined by inspection. In the case in hand, the F-projection shows that ad is the highest edge of the wedge, hence ad is visible in plan. There- fore tf/and de are also visible. It follows that the faces adfc, dihh. and dfe are visible in plan. The plan shows cf as the extreme back edge, hence in elevation this edge is invisible. This means that adeb is the only visible face in this view. For XIII, § 128] INTERSECTION OF SOLIDS 115 Fia. 189. — The Intersection of Two Solids Bounded by Plane Faces. 116 DESCRIPTIVE GEOMETRY [XIII, § 128 the other solid, «/, the highest edge, is visible in plan, so the faces JL and JK are seen in plan. In elevation, «/A"and KL are the visible faces. To find the actual intersection. Choose any face of the wedge, as adeb, and find where all three edges K, J, and L in turn intersect this face. The plane X, perpendicular to H, passed through K, intersects adeb in the line 2-3, which crosses K (seen in elevation) in point 6 ; K intersects adeb then in point 6. In the same way J is found to intersect adeb in point 12 ; join- ing 6 and 12 we have the intersection of face KJ with adeb. Since both surfaces are visible in both plan and elevation, the line 6-12 is a full line in each view. The line L is found to intersect adeb produced in point 18. Join 12-18 and 6-18, retaining only the portions 12-19 and 6-23, and the intersec- tion of the prism KJL with face adeb is complete. The line 12-19 will be invisible in elevation and visible in plan, while 6-23 will be visible in elevation and invisible in plan. The intersection of the two solids must evidently continue from the points 23 and 19 on to the adjoining face bcfe, hence choose next this face and find where J, K, and L intersect. Plane X through K intersects this face in the line 4-3, almost if not quite parallel to K. This shows that K, and therefore also J and L are nearly parallel to the face bcfe, so their points of intersection are inaccessible. Therefore choose next another face, as abc. Line K is found to intersect at point 5, and J at point 16 in the face produced. Joining these points, the part 5-17 is the actual intersection of JK with the face, and is invisible in both views. Line L does not actually intersect the face, but will intersect the face produced. The plane Z through L intersects the edges ac and be extended in the points 28 and 27, and this line when produced will cut L in 29. Manifestly, however, it would be a very inaccurate construction to use points so close together as 27 and 28 when the line must be produced much beyond either point, and such a construction should not be used without some check on its accuracy. In this case, such a check is readily available. The planes X, Y, and Z are parallel, and their intersections with the plane abc XIII, § 128] INTERSECTION OF SOLIDS 117 Fig. 189. — The Intersection of Two Solids Bounded by Plane Faces. 118 DESCRIPTIVE GEOMETRY [XIII, § 128 must be three parallel lines. Hence, in extending 27-28 to L at 29, care should be taken to make the line parallel to the intersections 1—4 and 7-10. Lines K, J, and L intersect the face abc or the face produced in points 5, 16, and 29 respec- tively. Joining these points, the parts 5-17 (already noted), 21-22, and 5-26 form the complete intersection of KJL with this face of the wedge. From the points 17 and 21 the inter- section must continue on to the adjoining face acfd. A little reflection will show that J must actually intersect this fare. The construction determines the point as 11. As point 17 is in the plane JK and also acfd, 11-17 joined will be the inter- section of JK with face acfd, and likewise 11-21 is the intersec- tion of JL with the same face of the wedge. There remains to be found the intersection on the face bcfe. No further points need to be found, however, for as 26 and 23 are both in this face and also in the plane KL the line connecting them will be the intersection of KL with the face. Similarly 22-19 is the intersection of JL with the face, which completes the intersec- tion of the two solids. Having shown the visibility of the intersection in both views as soon as found, notice that the portions of the edges of the solids which are finally visible may be readily determined by inspection. (The student should take pains to satisfy himself as to the correctness of the visibility of each edge as shown.) As a check, start with any point of the intersection, as 5 ; by tracing 5-17-11-21-22-19-12-6-23-26-5, we return to the starting point. In this case there is but one continuous inter- section. With different positions of the solids, however, there may be two intersections. This happens, for example, when one solid completely penetrates the other. 129. The Intersection of a Sphere with Another Solid. Sup- pose that we have two intersecting solids, the first a sphere, the second any solid bounded by plane faces. Pass an auxil- iary plane parallel to H, so as to cut both solids. This plane will cut from the sphere a circle, and from the other solid a polygon of three or more sides. Both the circle and polygon will project in true shape and size in the J7-view, where any XIII, § 128] INTERSECTION OF SOLIDS 119 Fig. 189. — The Intersection of Two Solids Bounded by Plane Faces. 120 DESCRIPTIVE GEOMETRY [XIII, § 129 points of intersection may be noted. These points will evi- dently lie on the line or lines of intersection of the two solids. An auxiliary plane parallel to V may be similarly used, the sections cut from the two solids appearing in true shape and size in the F-projection. In any particular problem, therefore, the solution would be effected by planes parallel to H, or to V, according to convenience. Indeed, it is often desirable to use auxiliary planes in each of these directions. The above method of solution can be applied to special cases of the intersection of a sphere with some of the curved solids. For example, let the curved solid be a circular right cylinder with its axis perpendicular to H. Planes parallel to H will cut circles, while planes parallel to V will cut rectangles from this cylinder. Hence the analysis previously given can be applied to this case. In general, if two intersecting solids are such, and so placed that planes parallel to H or to V will intersect both solids in simple, readily determined figures, such as straight lines or circles, the intersection of the solids may be found by the use of such planes as auxiliaries. A few simple examples are given in the following Articles. 130. The Intersection of a Sphere with a Prism (Fig. 190). Given the sphere and the triangular right prism abc-def, both placed in the first quadrant. It is evident that the intersection is formed by the three vertical sides of the prism, therefore the plan of the intersection coincides with the plan of the prism. The intersection of each vertical side of the prism is here found as explained in § 84. (See Figs. 108 and 109.) Let us consider, for example, the face cbfd. The points 1 and 5 are located at once in the elevation on jB", the other pro- jection of circle B. The plane R cuts from the sphere the circle R. How is its diameter found? The vertical side of the. prism intersects this circle in the points 2 and 8. The other points, 3, 4, 6, 7, are found in a similar manner, and the ellipse drawn. It may be noted that the ellipse in elevation has its major axis vertical and equal in length to l*-5*, while its minor axis is equal to 1 V -5 V . XIII, § 130] INTERSECTION OF SOLIDS 121 The intersections of the other two sides of the prism are found in a similar manner, as in Fig. 190. The F"-projection of the first intersection is a com- plete ellipse, but the other two are partial ones, with the points 10 and 11 in common. Visibility. In any case of the intersection of two sur- faces, not only the visible part of the intersection should be shown, but also the visible edges and the outlines of the surfaces, as in § 129. Visible Inteksections. The face first considered, cbdf, is the back of the prism, hence its intersection with the sphere is wholly invisible in eleva- tion. The other two faces are each visible in elevation. In each case as much of the inter- section as lies on the front or visible half of the sphere will be visible in elevation, that is, 10 to 13 and 11 to 14 on the right-hand curve, 10 to 16 and 11 to 17 on the left-hand curve. Visible Edges axd Out- lines. The edge ae of the prism is the front edge, and intersects the sphere at the points 10 and 11 on the front half. Hence ae is visible from a v to 10", and from 11" to e". Since the edge cd is tangent to the sphere at point 5 on the back half, cd disappears on V where it crosses the outline of the sphere. Hence the outline of the sphere is visible from the left toward the right as far as points 16" and 17". Likewise, the edge bf passes behind the sphere, and the outline of the sphere is visible from 13* around to 14". d c Fig. 190. 122 DESCRIPTIVE GEOMETRY [XIII, § 131 131. The Intersection of a Sphere with a Right Cylinder. Example 1 (Fig. 191). Given a sphere and a circular cylin- der. The plan of the intersection is evidently the arc of the circle 1*5*7*, and a study of this plan view will show that the intersection consists of one continuous curve. The points 1 and 7, shown in plan on the outline of the sphere, are projected directly to 1" and 7*. The points 4, 10, 6, 8, must lie on the Fig. 191. Fig. 192. great circle of the sphere in elevation, and are projected at once to 4", 10", 6" and 8". Other points are found by means of aux- iliary planes parallel to V, as in the preceding figure. The plane T, passed through c h , gives the points 3" and 11", in which XIII, § 131] INTERSECTION OF SOLIDS 123 the intersection is tangent to the outline of the cylinder. The visibility of the curve and of the outlines of the surfaces is determined as previously explained. Example 2 (Fig. 192). Given a sphere and an elliptical cylin- der. The intersection and the visibility are found as in Ex- ample 1. Notice that here two parts of the curve are visible. Example 3 (Fig. 193). Given a sphere and a circular cylin- der. In this case an inspection of the plan serves to show that the intersection will consist of two curves, each entirely distinct from the other. Each curve in elevation will be symmetrical with respect to the horizontal center line of the sphere (why'.'). The solution is left to the student. Fig. 193. Vir,. mi. Example 4 (Fig. 194). Given a sphere and a circular cylin- der. In this example it may not be so easy to vizualize the intersection by considering the plan view. Nevertheless a little Btudy will show thai there must be two closed figures having a common point a. Asa matter of fad flic int.! lection when found in elevation will he shaped like an irregular figure 124 DESCRIPTIVE GEOMETRY [XIII, § 131 8, the curve crossing itself (not tangent), at a. It is called a one-curve intersection. The solution is left to the student. 132. The Intersection of a Cylinder and a Torus. Another example is shown in Fig. 195. An auxiliary plane, as Q, par- allel to H, cuts the cylinder in the elements A and B, and the torus in the circles C and D, giving six points, 1, 2, ... 6, of the intersection. The chosen secant planes, parallel to H, should Fig. 195. include one through the contour element E of the cylinder ; one through the highest element, G ; one through the center of the torus ; and one tangent at the bottom of the torus. The points 7 and 8 of the intersection evidently lie in the plane X, which passes through the center of the torus parallel to V. CHAPTER XIV PROBLEMS INVOLVING THE REVOLUTION OF PLANES 133. A Line Lying in a Given Plane. One method of locating a line so that it will lie in a given plane is to assume the traces of the line in the corresponding traces of the plane. This has already been explained in § 97. (See Fig. 123.) Fig. 123 (repeated). Another method is to assume one of the projections of the line. As soon as either the H- or the F-projection of a line is assumed, the line becomes definite from the condition that it shall lie in the given plane. Problem 16. Given one projection of a line lying in a plane, to find the other projection. Analysis. The other projection of the line will be deter- mined if two points on it are found. Whenever it is practi- cable, the simplest points to consider are the two traces, s and t, which lie, respectively, in the horizontal and vertical traces of the given planes (§ 90). One trace of the line will be at the intersection of the given projection with the corresponding trace of the plane, the other projection of this point being in GL. The other trace of the line will be in the other trace of the plane, and is determined by a projector drawn from the intersection of the given projection with GL. 125 126 DESCRIPTIVE GEOMETRY [XIV, § 133 Construction (Fig. 196). Let the plane Q and the projection A h be given. The intersection of A h and HQ is the if-trace of the line A (§ 96). The C-projection, s v , of this point lies in GL. The intersection of A h and GL is the .ff-projection of the Fig. 196. F-trace of the line (Prob. 1, § 37). The actual trace, f, lies in VQ. The projection A" is now determined, since two points, s v and f, are known. Similarly, the projection A h may be found if A v is given. Fig. 19 A second example is given in Fig. 197. The lettering and explanation are the same as for Fig. 196. Special Case I. Suppose that the given projection is par- allel (a) to the corresponding trace of the plane ; or (b) to the ground line. In either event the line should be recognized as one of the principal lines of the plane (§ 99), parallel to H or V as the case may be. Thus, in Fig. 198, if A h is given parallel to HQ, A v is parallel to GL ; or if A v is given parallel to GL, then A h is parallel to HQ. The line A in this case is a horizontal XIV, § 133] A LINE IX A PLANE 127 principal line of Q, having its F-trace at the point t (t h , t"). Similarly, if A v is given parallel to VQ, A h is parallel to GL, and vice versa, the line being a vertical principal line of Q. Fig. 198. Special Case II. Suppose that both the plane and the given projection of the line are parallel to the ground line. Whichever projection of the line is given, the other projection will also be parallel to the ground line. Let Q, Fig. 199, be the given plane, and suppose A h to be given. One point will be 0. A h h \ u \ i A v VQ. /"' a. > F G. 1!)9. sufficient to locate A\ This point may suitably be the profile trace (§ 59) of the line, on any assumed profile plane, as P. Find first the profile trace, PQ, of the given plane (§ 60). Then from w* we can project to u" on PQ, thence to a v on VP. The projection A" passes through //". tion will locate A h if A* is given. The reverse (if this construe- 128 DESCRIPTIVE GEOMETRY [XIV, § 133 Special Case III. Suppose that the given plane is perpen- dicular to H or V. Let the given plane Q, Fig. 200, be perpen- dicular to H. Then if A v is given, in any position except perpendicular to GL, A h coincides with HQ, since this trace is an edge view of the plane Q (§§ 51 and 98). In the exceptional position, shown by B° perpendicular to GL, the .//-projection, B h , reduces to a point on HQ. Fig. 200. Fig. 201. Conversely, if A h is given coincident with HQ, A v may be taken at random in any position except perpendicular to GL, and the case is indeterminate. But if the given plane Q, Fig. 201, be perpendicular to H, and if A h be assumed not coincident with HQ, the solution becomes impossible, for there are no points in the plane Q to correspond with the assumed projections c h , d h , or in fact with any point of A h except the intersection, e h , of A h and HQ. The results are similar if the given plane is perpendicular to V. Corollary I. Given one projection of a point lying in a plane, to find the other projection. Analysis. Through the given point pass any line lying in the plane. Find the other projection of the line. This pro- jection must contain the required projection of the given point. Construction (Fig. 202). Let Q be the given plane, and asr sume a h to be given ; it is required to find a". Through a* draw the //-projection, M h , of some line in the plane Q. In general, the simplest construction results when the assumed line is a principal line of the plane ; hence M h is here drawn parallel to HQ. Find M v (Special Case I). Then u" lies on XIV, § 133] A LINE IN A PLANE 129 M v and is obtained by projecting from a h . The reverse of this construction will locate a* when a" is given. Fig. 202. Fig. 203. Corollary II. To find the second projection of a line lying in a plane when the general solution fails, partially or wholly. Let Q, Fig. 203, be the given plane, and let a h b h be given so that it cannot be produced to intersect either HQ or GL within the limits of the figure. Then a" and b v , or in general any two points on the line, may be located by Corollary I. Corollary III. To find a line of maximum inclination of a plane. Let Q, Fig. 204, be the given plane. A line of maximum inclination to 77 is perpendicular to HQ (§ 111). Hence assume A h perpendicular to HQ, and hnd A v by the general method. Fig. 201. Fig. 205. In Fig. 205, the line 7> is in the plane R, and is a line of maximum inclination to V. The projection 11" is taken per- pendicular to VR, and then B h is found. K 130 DESCRIPTIVE GEOMETRY [XIV, § 134 134. Revolution of a Plane About an Axis Perpendicular to H or V. As an illustration of the general case of this problem, let it be required to revolve the plane Q, Fig. 206, through an angle, «, about the line A, perpendicular to H, as an axis. The axis pierces Q at the point c (c A , c"). This point is found most readily by observing that since A h is a point, c h coincides with A h . Then c h is one projection of a point lying in plane Q, hence c° is the corresponding projection (Prob. 16, Cor. 1, § 133). In any revolution of Q about A, the point c will re- main fixed. Now a plane is determined when a point and a line are known ; hence, having the fixed point c, it will be necessary to revolve but one line of Q. This line may con- veniently be the //-trace, IIQ, of Q, which is now revolved about c h as a center to the required position HR. The F-trace, VR, is then determined by the fact that the plane R must con- tain the point c. Consequently R contains the line N, drawn through c and parallel to HR. (See Example 3, § 108.) Let now the plane Q, Fig. 206, be further revolved, until the //-trace takes position HS, perpendicular to the ground line. Fig. 206. Fig. 207. In this new position, the plane will be perpendicular to V. Hence the F-trace, VS, will be an edge view of the plane, and must therefore pass directly through c". A plane may be revolved until perpendicular to H by using XIV, § 135] THE REVOLUTION OF PLANES 131 an axis perpendicular to V. This is done in Fig. 207, where the plane Q is revolved about the line A, perpendicular to V, until the P"-trace of the plane, in its new position VR, is perpendicular to the ground line. Then HR is an edge view of the plane, and passes through c\ In this figure, the axis A, besides being perpendicular to V, is taken lying in H } which simplifies the hnding of the point c, in which A intersects Q. As in the revolution of a straight line (§§ 78, 79), a plane, when revolved about an axis perpendicular to H, maintains always its original angle with H. If the axis is perpendicular to V, the angle between the revolving plane and V does not change. 135. The Distance between Two Parallel Planes. As a simple application of the revolution of planes about axes perpendicular to H or V, we will consider the following Problem. Problem 17. To find the perpendicular distance between two parallel planes. Analysis. Assume an axis perpendicular to H (or V), and revolve both planes about this axis until they are perpendicular to V(oxH). The perpendicu- lar distance between the planes will then appear. Construction (Fig. 208). Let Q and R be the given planes. Assume the axis A perpendicu- lar to H, and for convenience lying in V. Revolve both planes about the axis A until they are perpendicular to V (§ 134). Then the new F-traces, VQi and VR U are the edge views of the planes, and the perpendicular distance, p, be- FlG ' 208 ' tween these traces is equal to the distance between the planes. 132 DESCRIPTIVE GEOMETRY [XIV, § 135 Corollary. To find the perpendicular distance from a point to a plane. Analysis. Take an axis perpendicular to H (or V) through the given point. About this axis, revolve the plane until it is perpendicular to V (or H). Then, since the given point on the axis does not move, or change its relation to the plane, the perpendicular distance appears directly in the V (or H) projection. The construction is left to the student. The problem is the same as Problem 14 (§ 120), a different method of solution hav- ing now been found. 136. The Angles between an Oblique Plane and the Coordinate Planes. The angle which a given plane Q, oblique to both H and V, makes with either coordinate plane, for example H, may be found : (a) by drawing in the plane a line of maximum inclination to H (Prob. 16, Cor. 3, § 133) and then finding the angle which this line makes with H (§ 79 or § 82); (6) by finding the edge view of the plane on a secondary plane of projection taken perpendicular to HQ (§ 70) ; (c) by revolving the plane about an axis perpendicular to H until the plane is perpendicular to V (§ 134). The various constructions resulting from these different methods resemble each other to a considerable extent, and are essentially the same, so that it is needless to take them all up in detail. The method chosen in this text is thought to be the one which best shows the connection between this problem and the two that follow. Problem 18. To find the angles which an oblique plane makes with H and V. Analysis. To find the angle between the given plane and H, assume an axis of revolution lying in P"and perpendicular to H. Revolve the plane until perpendicular to V, when the angle between the new F-trace and the ground line will be the angle the plane makes with H (§§ 134, 136). Similarly for the angle between the given plane and V. XIV, § 136] THE REVOLUTION OF PLANES 133 Construction (Fig. 209). Let Q be the given plane. Assume the axis A, lying in V and perpendicular to H. Revolve Q about A to the position R (HR, VR), perpendicular to V (§ 134). Then the angle between VR and GL is the angle a, which Q makes with H. Similarly, by revolving about the axis B, lying in H and perpendicular to V, we obtain the angle (3, which Q makes with V. It will be seen in Fig. 209 that the traces HR and VS, perpendicular to GL, are not essential to the required angles « and /3, and that these traces may therefore be omitted. This is done in Fig. 210, in which are found the angles a and fi, which the given plane Q makes with H and V, respectively. Fig. 209. Fig. 210. Q. I Ha \ \ \ * / a / Y va Q. > Fig. 211. Special Case. Let the given plane Q, Fig. 211, be parallel to the ground line. Find the profile trace of ^>on any assumed profile plane. This trace, PQ, is an edge view of the plane (S 60), and gives the required angles, a with //and (3 with V, directly. 134 DESCRIPTIVE GEOMETRY [XIV, § 137 137. Planes making Given Angles with H and V. The pre- ceding problem gives rise to two converse problems, in which it is required to find planes which make given angles with the coordinate planes. Problem 19. Given one trace of a plane, and the angle which the plane makes with either H or V, to find the other trace of the plane. There are two cases, which are stated below. Analysis. The required result may be obtained by revers- ing the construction of Problem 18. Case I. Given one trace and the angle which the plane makes ivith the same coordinate plane. Construction (Fig. 212). Let HQ and the angle «, which Q makes with H, be given. Assume the axis A, lying in P~and perpendicular to II. Revolve the given trace HQ about this axis to the position IIR, perpendicular to GL. Complete the plane in this position by drawing VR at the given angle, a, Fig. 212. Fig. 213. with H. Then R intersects the axis A at the point c (C, c*). Let R now be supposed to be revolved back about the axis A until IIR coincides with HQ ; the point c remains fixed. Since c" is in V, VQ must pass through c", and is determined by this point and the point in which HQ intersects GL. The given trace HQ may also be revolved to the position HR U giving a second point ut when the sum of these angles is greater than 90°, and each angle is less than 90°, four possible non-parallel results may be found. Thus, having found the plane Q, Fig. 215, making a with // and (3 with V, make on' = on. Then it is evident by symme- try that the plane T, drawn with VT through c and n', and HT through (I and //', makes the same angles. Also, make oc' = oc, and draw VX through n and c'. Then VX makes the same angle with d L as does VQ, hence taking HX coincident with HQ, plane A' makes with // and V the same angles as does Q (compare Fig. 212). A fourth result, plane Y (VY, HY) is obtained by symmetry from plane X. 138 DESCRIPTIVE GEOMETRY [XIV, § 137 There are other methods of obtaining the three additional results after the first plane Q is obtained. Thus, a distance od' = ocl might be laid off below the ground line, and //-traces drawn from d' to n and n'. P>ut after four non-parallel results are obtained, all further results will be parallel to some plane already found. This is because there are but four possible slopes of an oblique plane : (1) downward, forward, to the right ; (2) downward, backward, to the right ; (3) downward, forward, to the left ; (4) downward, backward, to the left. (See § 49.) The slope of a plane is the same as the slope of its lines of maximum inclination to H. These lines, if desired, may be found by Problem 16, Cor. 3, § 133. Special Case. The sum of a and ft is 90°. The plane is then parallel to the ground line, and its profile trace is its edge view. Hence (Eig. 216), assume any profile plane of projec- tion. Draw two non-par- allel profile traces making the given angles with H and V, and find the corre- sponding H- and F"-traces. Only two results are possi- ble, corresponding to the two possible slopes : down- ward and forward, and downward and backward (or upward and backward, upward and forward). Other special cases, which may be solved by inspection, result when (1) either a or ft is 0°, in which case the other angle must be 90° (one result) ; (2) either « or ft is 90°, the other angle being greater than 0° and less than 90° (two results) ; (3) a = ft = 90° (one result). These cases are left to the student. Fig. 216. CHAPTER XV OTHER PROBLEMS INVOLVING THE REVOLUTION OF PLANES 138. The Revolution of a Plane about One of Its Traces. A very useful revolution of a plane is that in which the plane is revolved about one of its own traces as an axis. For example, the plane Q may be revolved about HQ as an axis until Q coincides with H. The figure is left for the student to draw. Any points, lines, or figures contained in Q Avill be carried by this revolution into the coordinate plane., where they will appear in their true sizes and relations. Problem 21. To find the position of a point lying in a plane, when the plane is revolved into H or V about the corresponding trace. Note. Since only one projection of a point lying in a plane can, in general, be assumed, it may be necessary to find the second projection by tbe use of Problem 10 (§133), before attempting to solve the present X>roblem. Analysis. Suppose point a lying in plane Q, to be revolved about IIQ until it lies in //. The figure is left to the student. The process is the same as described in § 70, and shown pic- torially in Fig. 08. The statement of § 76 may be modified to meet the present situation, as follows. If a point a Lying in plane Q, be revolved about IIQ as an axis, the circular path of the revolving point will project on //as a straight line per- pendicular to IIQ. If the point a be revolved into //, its revolved position will be at a distance from H(J equal to the true distance of the point from HQ. An analogous statement applies to the revolution of a point about VQ into V. 139 140 DESCRIPTIVE GEOMETRY [XV, § 138 Construction (Fig. 217). Let a, lying in the plane Q, be the given point, and let it be required to revolve a about HQ into H Erom a h draw the line a h e h perpendicular to HQ, and pro- duce the line indefinitely. This line is the //-projection of the circular path of the revolving point, the center of revolution being the point e (e h ) lying on HQ. The line a h e h is one pro- Fig. 217. Fig. 218. jection of a line lying in Q (a line of maximum inclination to H, § 11-4). Find a v e v (Prob. 10, § loo), and the true length, a v e u of the line ae (Prob. 3, § 78). The distance cVe^ is the actual distance of the point a from the center of revolution e h . Lay off e h a r = eCe^, thus giving the required revolved position n r . Note. The revolved position a T represents the revolution of Q about HQ through an ancjle greater than 90°. The distance a v ei may be laid off from e h in the opposite direction if preferred (that is, on the same side of HQ as the projection a*), representing the revolution of Q through an angle less than 90°. It is customary to place the revolved position where it will be as far from the other parts of the figure as possible. In Fig. 218, the point a, lying in Q, is revolved about VQ into V. The construction is entirely analogous to that for revolving about HQ into H, and is left without further explana- tion to the student. Working Rule. Returning to Fig. 217, the revolved posi- tion, a T , is obtained by laying off from e h , on the perpendicular XV, § 138] THE REVOLUTION OF PLANES 141 a k e h , a certain distance, namely, the true length of the line ae. The situation of this line is such that it is not necessary to draw the complete construction, or even the F-projection, a v e v , in order to find the true length. For, it will be seen that the true length, a v e u is the hypothenuse of a right triangle, of which the base is e u f= a h e h , and the altitude a v f. We may therefore state the following working rule : To revolve point a, lying in the plane Q, about HQ into H, proceed as follows. From a h draw a line perpendicular to HQ. Then the revolved position, a T , lies on this perpendicular, at a distance from HQ equal to the hypothenuse of a right triangle, of which the two sides are the distances a h to HQ, and o* to GL. This hypoth- enuse can be found directly with the dividers, as follows : Take the distance a h e h ; set it off from / to point e x on the ground line ; span the distance from e x to a v . Similarly, to revolve a point lying in a plane about the V- trace of the plane into V (Fig. 218), find the hypothenuse of a right triangle, of which the two sides are the distance from the F-projection of the point to the F"-trace of the plane, and from the if-projectlon of the point to the ground line. This hy- pothenuse is the true distance of the revolved point from the F"-trace of the plane. Special Case. The trace of the plane about which the revolution is made is perpendicular to the ground line. Let HQ, Fig. 219, be perpendicular to GL, VQ being at any angle. The plane Q / is therefore perpendicular to V, so that the revolution of the given point occurs in a plane which is parallel to V. The path of the revolution must then project on V in its true shape, that is, as a circular arc, and the revolved posi- tion, a r , can be obtained at once by projecting from the intersection of ground line. / / Fig. 219. this arc and the 142 DESCRIPTIVE GEOMETRY [XV, § r38 The ease with which the revolved position is obtained in this particular situation of the plane suggests another method for solving the general case, since an auxiliary plane of pro- jection can always be assumed perpendicular to either trace of Fig. '220. any given plane (see § 70). Such a construction is shown in Fig. 220, and is left without further explanation to the student. Corollary. To find the revolved position of a line lying in a plane, when the plane is revolved into H or V about the corre- sponding trace. Since a straight line is determined when two of its points are known, this problem can always be solved by revolving two points of the line by the method already given. However, a simpler method can usually be found, as in the following examples. (1) Let the line A, Fig. 221, which is to be revolved about VQ into V, intersect VQ in the point t. Revolve any point of A, as c(c h , c v ) to c r . If now we attempt to revolve the point t, this point, being on the axis of revolution, will not move. Hence the revolved position of the line, A r , passes through c r and t. (2) Let the line A, Fig. 222, which is to be revolved about VQ into V, be parallel to VQ. Revolve any point c, of A, to c r . Every other point of A is the same distance from the axis XV, § 140] PROBLEMS OF REVOLUTION 143 VQ, and will revolve just as far. Hence, the revolved position A r passes through c r and is parallel to VQ. Fig. 221. Fig. 222. 139. Solution of Plane Problems by Revolution of the Plane. By extending the preceding problem, any number of points or lines lying in a plane can be revolved into one of the coordinate planes, and the revolved position will show the true relation existing between them. The problem thus becomes a very useful auxiliary in the solution of a certain class of problems. In revolving about either trace of a plane, as for example about HQ into H, it should be noticed that, while the F-pro- jections of the points and lines concerned must be considered, no use is made of the U-trace of the plane, VQ. Similarly, when revolving about VQ into V, no use is made of HQ. Hence, when a plane is introduced as an auxiliary, for the ex- press purpose of revolving about one of its traces, it is usually unnecessary to find more than one trace of the plane. 140. The Angle between Two Intersecting Lines. The angle between two intersecting straight lines is a plane angle. It is measured in the plane that is determined by the two lines. Problem 22. To find the angle between two intersecting lines. Analysis. Find the plane containing the given lines. About one trace of this plane, revolve the lines into the corresponding coordinate plane, when the true angle between them will appear. 144 DESCRIPTIVE GEOMETRY [XV, § 140 Construction (Fig. 223). Let A and B be the given lines. Find either trace of the plane containing these lines, for example the IT-trace, HX (Prob. 6, § 106). Revolve both lines Fig. 223. Fig. 224. about HX into H. To accomplish this most readily, revolve their point of intersection, o, to o r (Prob. 21, Working Rule, § 138). Then A T passes through o r and the trace s 1} while B r passes through o r and the trace s 2 (Corollary, § 138). The angle, 8, between A r and B T is the true angle between the lines. A similar construction is shown in Fig. 224, in which the P-trace, VX, of the plane containing A and B is used. The angle, 8, between A r and B r , is the true angle between the lines. Special Case. One line is parallel to H, and the other is parallel to V. The angle may be found by the general method, but the following is simpler. Analysis. Revolve either line about the other as axis, until both are parallel to the same coordinate plane. Construction (Fig. 225). Let A and B be the given lines. Choose one of them, as A, for the axis. Since A is parallel to H, B must now be revolved about A until B, also, is parallel to H. Assume any point c(c h , c v ) in the line B, at a reasonable distance from the intersection, o, of the lines. Through c* draw an indefinite line perpendicular to A h . Take the distance o v c v in the compasses, and with o h as center strike an arc across XV, § 140] PROBLEMS OF REVOLUTION 145 Fig. 225. this perpendicular, giving c r . Then o h c r is B revolved, B r , and the angle between B r and A h is the true angle 8, between the given lines. Proof of the Construction. Since A is parallel to H, the revolution of point c about A will project on H as a straight line perpendicular to A h (§ 76). Hence c revolved, c r , must lie some- where on the perpendicular to A h through c\ Also, when the portion oc of the line B is revolved until parallel to H, it will appear in true length. Since the point o in the axis A does not move, o h c r should equal the true length of oc. But o h c r was made equal to o v c v , which is the true length of oc, since the line B is parallel to V. Hence the construction is correct. Corollary. To find the projections of the bisector of the angle between two intersecting lines. Let it be required to find the bisector of the acute angle. In Fig. 223, draw first the actual bisector, 0„ in the revolved position, bisecting the angle between A r and B r . Xow C T necessarily passes through o r , and in this case intersects the trace IIX in the points/. Imagine the plane Xto be counter- revolved (revolved back) about IIX to its original position, taking with it the lines A, B, and C. The lines A and B, and their intersection, point o, will resume their former positions. The point 8 3 k , being on the axis of revolution HX, will not move. Hence the //-projection, C* of the line C must pass through o* and .% h . To find the ^projection of C, note thato* must be projected to o", while N,\ being an ff-trace, lies in H, and musl project to Sj" in GL. II' ace <' \ determined bj " and -V- The second example, Fig. 224, cannot be solved, however, in this way. As soon as the bisector, C r , i drawn in the revolved position, it is seen that the intersection of 0, with the trace I'.V 146 DESCRIPTIVE GEOMETRY [XV, § 140 is beyond the limits of the figure. One point of C" is o", and one point of C h is o h . It remains to find a second point in each projection. Assume a point, e r , in C r . Through e r draw the revolved position, L r , of a line lying in the plane of the given lines, by making L T parallel to the revolved position of either of the given lines A or B ; in this case L T is drawn parallel to A T . Since the line L is in the plane X, the intersection, t 3 v , of L r and VX is the F-trace of the line L, and t 3 h is in GL. Also, since the line L is parallel to the line A, through t z v draw L v parallel to A 1 ', and through t 3 h draw L h parallel to A h . Now e is a point in the line L. Hence revolve back from e r per- FlG - 22i Orated). pendicular to VX, and find e v on L" ; jjroject from e" and obtain e h on L h . Since e is a point also in the line C, we have C v determined by o v and e", and C h determined by o h and e\ 141. The Angle between a Line and a Plane. The angle which a line makes with a plane is equal to the angle between the line and its projection on the plane. Hence we only need to find the angle between two intersecting straight lines. Problem 23. To find the angle between a line and a plane. First Analysis. Project the given line on the given plane. Find the true angle between the given line and its projection ; this is the angle required. Second Analysis. From any point of the given line, drop a perpendicular to the given plane. Find the true angle between the given line and the perpendicular ; this is the complement of the required angle. In general, the second analysis gives the simpler construc- tion, and is the one usually adopted. Construction (Fig. 226). Let A be the given line and Q the given plane. Assume any point e (c\ c") on A. From c draw XV, § 141] PROBLEMS OF REVOLUTION 147 the line B (B h , B v ) perpendicular to the plane Q (§ 111). Find the true angle, S, between the lines A and B (Prob. 22, § 140). Then y, which is the complement of S, is equal to the true angle between line A and plane Q. Fig. 226. Fig. 227. In Fig. 227, the given plane Q is parallel to the ground line. The general case still applies, but there is an extra detail in the construction. As before, assume any point c in the line A. From c drop the perpendicular B to the plane Q. Find the angle, S, between the lines A and B. Then y, the complement of 8, is the angle required. As a necessary step in finding the angle S, we need one trace, as VX, of the plane containing the lines A and B. To determine VX we need the F-trace of each of these lines. The F"-trace of A is readily found (Prob. 1, §37). But B is a profile line, of which hut one point, c, is known. Nevertheless, the line B is a definite line, since it is perpendicular to Q. To find the F-trace of B, proceed as fol- lows : Find the profile projection, c ,p , of c; also the profile trace, PQ, of Q. From <•" draw the profile projection of B, B p perpendicular to PQ. Then the F-trace of B is the point t 2 , in which B" intersects VP. (See Prob. 11, Special Case 2, § 115.) 148 DESCRIPTIVE GEOMETRY [XV, § 142 142. A Rectilinear Figure Lying in a Plane. A plane poly- gon of more than three sides cannot be assumed by assuming at random the projections of a number of points in space, and then connecting these points by straight lines. For, as soon as three points not in the same straight line have been assumed, a plane is determined ; thereafter some consideration is neces- sary to assure us that the remaining points lie in the plane determined by these first three points. Problem 24. Given one complete projection, and three points in the other projection of a plane polygon, to complete the projection. First Analysis. Lines which intersect, or are parallel, are necessarily in a plane. Hence, by producing the sides of the given complete projection until they intersect, or by drawing in this projection lines which are parallel to some of its sides or which intersect the sides or each other, auxiliary lines are ob- tained by means of which the second projection can be built up. In particular, if the figure is a quadrilateral, the intersection of its diagonals is a point in the plane of the figure. Fig. 228. Fig. 229. Second Analysis. Both projections of three points are known. Find the traces of the plane containing these points. For the remaining points, one projection of each point, and the traces of the plane containing it, are known ; hence the second projection can be found. XV. I 142] PROBLEMS OF REVOLUTION 149 Construction by First Analysis (Fig. 228). Let the ^-projec- tion. a k b k c k d k , and a r b T c T of the U-projeetion be given. Since this figure is a quadrilateral, we may draw the diagonals a h c k and b h d k , intersecting in the point e\ In the F-projection draw a r c r . Locate e T on this line by projecting from e*. Draw b r e r , produce it, and locate on it d T by projecting from d k . Otherwise (Fig. 229), through d h draw d k f k parallel to b k c k . and intersecting a h b h at /*. Project from /* to / r in a r b T . Through f T draw f T d T parallel to b T c T . Then the line dt\ since it passes through the point /in ah, and is -parallel to be. lies in the plane of the quadrilateral. Locate d z by projecting from d h , A second example is given in Fig. 230. Let the complete F-projection, a T b T c T d T e T . and the 17-projection of three points, a*, &*, and c\ be given. In this poh - and d T e T can both be produced to meet ^^ > -. > ,s> Fig. 244. Fig. 245. the included angle, 0, between these lines is the angle between the given planes. Note. The student may discover that T r Z, and also c, are superfluous in this construction. Special Case II. One of the given planes is parallel to II or V. This case falls under the solution just given ; a still simpler solutionis shown in Fig. 245. The given plane R is parallel to II. Therefore the angle between planes Qand B is equal to the angle, 6, which Q makes with //, the latter angle being found by Problem 18, § 136. 146. Application of the Preceding Problem. As instances in which the angle between two planes is wanted in actual con- struction, we may cite the case of two intersecting pitch (slop- ing) roof's, or of two intersecting masonry walls with battered (sloping) faces. The way in which the problem usually appears is as a corollary to the following problem. M 162 DESCRIPTIVE GEOMETRY [XVI, § 146 Problem 28. Given the horizontal traces of two planes, and the angle each plane makes with H, to find the line of intersection of the two planes. Note. For the problem as stated, four results are possible, in general. In addition to the above data, it will be assumed that the slopes of the planes are known in order to limit the problem to one result. This would be the case in the practical applications of the problem. Analysis. One point in the line of intersection is located at the intersection of the given traces of the planes. To locate a second point, draw in each plane a line parallel to H, so that these lines will intersect. The necessary and sufficient condi- tion for their intersection is that these two lines shall be at the same distance from H. Construction (Fig. 246). Let HQ and HR be the given H- traces. Let plane Q make the angle a l with H, and R the angle a 2 with H. In any convenient position, draw a ground line, GyLi, perpendicular to HQ, and draw the F-trace and edge view, V X Q, of Q, making the given angle, a„ with G X L X . Draw a second ground line, G t L 2 , perpendicular to HR, and draw the F-trace and edge view, V 2 R, of R, making the given angle, «2> with Gr 2 L 2 . Assume any arbitrary distance, d. Draw the XVI, § 146] PROBLEMS ON THE LINE AND PLANE 163 F"-trace and edge view, V x W, of a plane W, parallel to H and at the distance d above H, also the F"-trace and edge view, V 2 W, of this same plane, that is, draw V\ IF and V 2 W parallel respec- tively to GiL x and G 2 L 2 , an d at the assumed distance d. The planes Q and W intersect in a line M, one projection of which is the point, Mf, where FilFand ViQ intersect. Project from M{" and obtain M h , parallel to HQ. The planes R and W in- tersect in a line N, a projection of which is the point N 2 V , where V 2 IF intersects V 2 R. Project from N 2 V and obtain JV h , parallel to HR. The projections M k and N h intersect in b h , which is an actual intersection in space, since M and N are both in the plane W. This point is common to both Q and R. Hence the //-projection, A h , of the required intersection of Q and R, is drawn through b h and the intersection, s, of HQ and HR. To locate the line A further, assume a third ground line, 6r 3 L 3 , coincident with A h . A ^projection, b 3 v , of point b, is located by making the distance from b z v to G 3 L 3 equal to the distance d, since this is the distance from H of any point in the lines M and N. The projection, A 3 V , of A passes through b./ and the point s. The line A is now definite, since we have its //"-projection, A h , and the angle « 3 (between A 3 V and G 3 L 3 ) which this line makes with H. Corollary. To find the angle between the given planes. In finding the angle between two planes by Problem 27, the construction already made in Fig. 24G is a part of the construc- tion of that problem ; namely, the //-projection, A h , of the line of intersection of the given plane, and the projection, Af, of this line, using a secondary ground line coincident with A h . To complete the construction, draw, through any convenient point o on A h , the trace HZ perpendicular to A\ Find the perpen- dicular distance, oc 3 ", from o to A 3 V . Locate c r on A h by making the distance c r o equal to oc 3 ". From c, draw to the intersect ions, s 2 and 8 8 , of HZ with HQ and HR respectively. The angle, 0, between these lines is the required angle between the planes Q and R. CHAPTER XVII COUNTER-REVOLUTION OF PLANES 147. Counter-revolution. A point, line, or any number of lines lying in a plane may be revolved about one of the traces of the plane into H or V. The basis for all such revolutions is the revolution of a single point, given in Problem 21, § 138. It has further been shown (see Prob. 22, Corollary, § 140 ; Prob. 24, Corollary 2, § 142 ; Prob. 25, § 143) that the revolved position of a line lying in a plane can be given or assumed, and the plane counter-revolved, that is, revolved back about its trace so that the projections of the line can be found. In all these solutions, however, the counter-revolution was made to depend upon a previous direct revolution of lines already existing in the plane. We shall now take up the prob- lem of a direct counter-revolution, starting only with the revolved position of a point and the traces of the plane ; in other words, the direct reverse of Problem 21, § 138. Problem 29. Given the position of a point lying in a plane after the plane has been revolved into H or V about the corresponding trace, to find the projections of the point. Analysis. Suppose the point a, lying in the plane Q, to have been revolved about HQ into H. The figure is left to the stu- dent. The path of the revolving point is the arc of a circle, which will project on Fas a straight line perpendicular to HQ, and on any plane perpendicular to HQ in its true shape and size. Referring to the various solutions of Problem 21, § 138, it is seen that the solution given in Fig. 220 shows these two projections of the path of the revolving point. The con- struction of Fig. 220 may therefore be reversed to give the solution of the present problem. Construction (Fig. 247). The plane Q is given by its traces HQ and VQ, and a r is given as the position of a point which 164 XVII, § 147] COUNTER-REVOLUTION OF PLANES 165 has been revolved about HQ into H. Assume a secondary ground line G X L X perpendicular to HQ. Assume a point, f, in VQ ; project to t h in GL. Find the secondary projection, t x v , of t, and through this point draw the trace and edge view V X Q (§ 70). Project a r to G X L X . With o as center, revolve this Fig. 247. point to a x v in V X Q. Then a x v is the secondary projection of the point a. Project from a x v perpendicular to G X L X , and from a r perpendicular to HQ. Both of these lines must pass through a h , which is thus determined. To find a", we now have given a* as one projection of a point lying in the plane Q. Through a h draw M h parallel to HQ; this is one projection of a horizontal principal line of Q. The other projection, M", is parallel to GL; o" lies in M' (Prob. 16, § 133). Check. The distance of a v from GL equals the distance of a x v from G x Ij^ (§ 67). A Second Result. The point a, thus found, lies above II. The given revolved position may also he the revolved position of a point b, lying below II. To obtain this result, utter pro- jecting from the given revolved position to G X L U revolve about o to b x v , which lies in V { Q produced below G X L X . Then pro- ceed as for the point a. As before, there is a cheek on the construction ; the distance from b" to GL equals the distance from b x " to G X L X . 166 DESCRIPTIVE GEOMETRY [XVII, § 147 Note. The student does not always see readily why the F-projections a v and b v should be located by means of auxiliary lines in the plane Q, since the distances of these points from GL appear at once in the second- ary projection. Indeed, the projections a v and b v can be located by trans- ferring from the secondary projection the distances of a\ v and bi v from G\L\. But if this method be employed, not only the distances, but the directions of these points from G\L\ must be considered, and there is a chance for error. The method of locating a" and b v by means of the auxiliary lines M and N is free from this ambiguity. In Fig. 248, the given revolved position, a r , is that of a point which has been revolved about VQ into V A secondary Fig. 248. ground line, G'L', is taken perpendicular to VQ, and the con- struction is entirely analogous to that of Fig. 247. Two results are possible, namely, points a (a v , a h ) and b (b v , &*). Corollary. To counter-revolve a plane polygon. Analysis. Find the projections of each corner of the polygon according to the preceding method. Construction (Fig. 249). The given revolved position, 1,2,3 r 4 r , is that of a square which has been revolved about HQ into H. As in the case of a single point, this may be the revolved posi- tion of two figures lying in Q. To limit the construction to one result, therefore, it is given that point 1 is the highest corner of the square. The projections of each corner of the square are obtained by the method of Fig. 247, using a second- ary ground line perpendicular to HQ. Xote especially point 3. XVII, § 147] COUNTER-REVOLUTION OF PLANES 167 Since the edges of the square are lines lying in the plane Q, the traces of these lines must lie in the traces of the plane (§ 96). This fact furnishes a series of checks on the work, as, for example, that 2 r 3 r and 2 A 3 h intersect HQ in the same point, namely, the iT-traee of this line, and that this trace projects to GL in the point where 2 V 2>" intersects QL. Similarly, l r I r and 1*4*, if produced, must intersec! HQ in the same point. In fact, the counter-revolution may be effected largely by the use of the traces of the various lines, l>ut ;it leasl one point must be counter-revolved as in the general problem. The counter-revolution of a plane figure is the reverse of Problem 24, Corollary 1, § 142, and should be compared with it. 168 DESCRIPTIVE GEOMETRY [XVII, § 148 148. An Auxiliary Problem. As an application of the pre- ceding corollary, we shall construct the projections of a right prism or a pyramid resting on an inclined plane, using the polygon which is located in the plane as the base of the solid ; or, what will result in the same construction, we shall construct the projections of a right prism or pyramid, the axis of which is inclined to both H and V. As, however, the axis of the pyramid, or the long edges of the prism, will be a line or lines perpendicular to the plane of the base, it will be necessary first to solve the following problem. Problem 30. At a given point in a plane, to draw a line which shall be perpendicular to the plane and of given length. First Analysis. A line perpendicular to a plane and indefi- nite in length may be drawn by making the projections of the line perpendicular to the traces of the plane (§ 112). Through the given point draw a line perpendicular to the plane. Assume a second point on this line. Find the true length of the line between the assumed and given points. On this true length, measure from the given point the length of line desired. Ob- tain the projections of this line by reversing the construction for finding the true length. Construction by First Analysis (Fig. 250). Let a, lying in Q, be the given point. Draw the line ac (a h c h , a 1 ^"), indefinite in length and perpendicular to Q. Assume a point c (c*, C) on XVII, § 148] COUNTER-REVOLUTION OF PLANES 169 this line, and find the true length, a v c r v , of etc (Prob. 3, § 78). Measure off a v b r v equal to the true length of the required line. From b r v find the projections b v and b k by reversing the con- struction for finding the true length. Then a h b h and a v b" are the projections of a line of the given length. Note. Any method of finding the true length of ac may be reversed, all leading to the same point b. Second Analysis. Any plane perpendicular to the given plane will be parallel to the required line. Hence assume a secondary plane of projection perpendicular to either trace of the given plane. Find the secondary projection of the given plane and point on it. Since the required line is parallel to this plane of projection, the secondary projection of the line may be drawn at once, perpendicular to the secondary trace of the plane and of the given length. From the secondary projection of the line the projections on if and F~may be found. Construction by Second Analysis (Fig. 251). Let a, lying in Q, be the given point. Assume Gri-Li perpendicular to HQ. Vui. -T.l. Project a to a x v , and through this point draw the Fi-trace and edge view, V\Q. Draw ,". Project from l> h to 6", making the distance from b v to GL equal to the distance from b { v to G\L X . Draw a v b". As a check, this should be perpen- dicular to VQ. Then a*6* and a v b v are the required projections. 170 DESCRIPTIVE GEOMETRY [XVII, § 149 149. A Prism or Pyramid Whose Axis Is Inclined to Both H and V. The method of obtaining the projections may best be shown by considering concrete examples. Example 1. Draw the projections of a right pyramid. Tlie axis is 1|" long, makes 45° with H and 30° with V, and slopes (from the apex of the pyramid) downward, backward, to the right. The base is a regular pentagon, inscribed in a circle of 1" radius. Hie lowest edge of the base is parallel to H. The center of the base is f" above H and 1" in front of V. Scale HaJf Size Fig. 252. Construction (Fig. 252). The first step in the construction is to find the plane which contains the base of the pyramid. The data for this plane are not given directly in the statement of the problem. But, since the plane of the base of the pyramid is perpendicular to its axis, we derive at once, from XVII, § 149] COUNTER-REVOLUTION OF PLANES 171 the data given for the axis, that this plane makes the comple- mentary angles, 45° with H and 60° with V, and slopes down- ward, forward, to the left. Construct the plane X, using these angles and slope (Prob. 20, § 137). We might next find a point, a, in this plane, distant |" above H and 1" in front of V. To do this, draw the line L, every point of which is so located ; that is, draw If par- allel to GL and |" above it, and L h parallel to and 1" in front of GL. Find the point, a, where L pierces X (Prob. 13, § 119). In the figure, however, in order to render more clear the subsequent steps of the construction, point c (c*, c") is located at the given distances from H and V, and through this point the plane Q is passed parallel to X (Prob. 9, § 107). Revolve c about HQ into H, obtaining c, (Prob. 21, § 138). "With c r as center, draw a circle of 1" radius. Inscribe in this circle the regular pentagon l r 2 r 3 r 4 r o r , so located that the side nearest HQ is parallel to it. This pentagon is the revolved position of the base of the pyramid. Assume a secondary ground line, GiL u perpendicular to HQ. Project c to c x v , and draw V X Q through cf. Using the edge view ViQ, counter-revolve the pentagon into the plane Q (Prob. 29, Corollary, § 147). To obtain the apex of the pyramid, we must draw from c a line 1 J" long and perpendicular to Q. Draw from c? the line 'Y'o,", perpendicular to ViQ; make the length of this line 1;". Draw from c* perpendicular to HQ] project from of t" o* on this line. Project from o* to o v , making the distances of o v and of from their respective ground lines equal. Note as a check whether o v c v is perpendicular to VQ. (Compare Fig. 251.) Complete the projections of the pyramid, making visible edges full, and invisible ones dotted. In the figure, the projection of the entire pyramid on the secondary plane of projection is completed, as well as the pro- jections on // and V. 172 DESCRIPTIVE GEOMETRY [XVII, § 149 Example 2. Draw the projections of a right prism, resting with its lower base in a given plane Q. The prism is If" long. Tlie base is a square of 1" sides. TJie lowest corner of the base is \" above H, f " in front of V. Two edges of the base make equal angles with H and V, and slope dowmvard, backward, to the right. Note. A plane, like W, Fig. 253, which has coincident traces parallel to GL, will slope downward and backward, and will evidently make equal angles, namely, 45°, with H and V. Any line lying in such a plane will make equal angles with H and V. Construction. Let Q, Fig. 254, be the given plane on which the base of the prism is to rest. Draw the line L (L v , L h ), \" Fig. 253. above H and J" in front of V. Find the point in which the line L intersects the plane Q (Prob. 13, § 119). This locates the point 1, the lowest corner of the base. Pass through the line L a plane sloping downward and backward, and making 45° with //and V; since the line L is J" above 77, the trace, HW, is drawn \" behind L h . Find the line of intersection of the planes Q and W. This is si, and is determined by the point s, where HW intersects 1IQ, and the point 1 already found. The line si lies in Q, makes equal angles with H and V, and slopes downward, backward, to the right. Hence si determines one side of the base of the prism. Revolve si about IIQ into H to the position sH r (Prob. 21, Corollary, § 138). With s h l r (produced) as one side, draw the square l r 2 r 3 r 4 r , making each side 1" long ; this square is the revolved position of the base of the prism. Note that, since point 1 is to be the lowest corner, no position of the square, other than that shown, is possible. XVII, § 149] COUNTER-REVOLUTION OF PLANES 173 Assume CrjL, perpendicular to HQ; locate V X Q by means of some point, as e, and counter-revolve the square (Prob. 29, Corollary, § 147). As a check, note that the point 2 must fall on the line si, produced, already determined. The long edges of the prism are lines If" long and perpen- dicular to Q. Draw these lines, l x v 5^, 2 1 r 6 1 ", etc., in the secondary projection ; from this projection find the projections on H and V (Prob. 30, Second Analysis, § 148). Complete the projections of the prism as shown. An alternative method for finding the long edges of the prism is also given in Fig. 254. From point 1 draw the line (l h f h , L'/'j perpendicular to Qand of unknown length. Find the true length 1"./V Make 1*5! the required length, L§"; from 5 t find the projections 5' and 5* (Prob. 30, First Analysis, > 1 18). The remaining long edges, 2-6, 3-7, and 4-8, may now be drawn equal and parallel to 1-5. CHAPTER XVIII TANGENT LINES AND PLANES — GENERAL PRINCIPLES 150. Curves. A curve may be defined as a line, no portion of which is straight. A plane curve is one which lies wholly in a plane. If no part of the curve is plane, the curve is a space curve, or curve of three dimensions. Familiar examples of plane curves are circles and ellipses. An example of a space curve is the helix, as shown by a screw thread or a spiral spring. Space curves result usually, though not necessarily, when two curved sur- faces of any kind intersect each other. 151. Projections of Curves. The projection of a plane curve on any plane of projection is, in general, a curve. If, however, the plane of the curve is perpendicular to the plane of projec- tion, the projection of the curve is a straight line. The pro- jection of a space curve on any plane is always a curve. Conversely, if one projection of a curve is a straight line, the curve is a plane curve. But if both projections of a curve are curves, the actual curve may or may not be a plane curve, so that no general rule can be given. 152. Tangent Lines. At any point in a curve, a straight line tangent to the curve can be drawn. The pro- jection on any plane of the curve and its tangent are tangent to each other. (See Fig. 255.) An excep- tion occurs if the plane of projection is perpendicular to the rectilinear tangent, since the latter then pro- jects as a point. 174 Fig. 255. XVIII, § 154] TAXGEXT LIXES AXD PLANES 175 153. Tangents to Plane Curves. A straight line tangent to a plane curve lies in the plane of the curve. Hence, if it is required to draw a straight line tangent to a plane curve, from a point not on the curve, there is no solution unless the given point lies in the plane of the curve. In other words, a straight line tangent to a plane curve cannot be drawn from a point in space which lies outside the plane of the curve. Fig. 256. Fig. 257. For example, Fig. 256 represents the projections of a plane curve C (§ 151), and of two points a and b. A line tangent to the curve C cannot be drawn from the point a, since it is obvious that a is not in the plane of the curve. Why ? On the other hand, the point b lies in the plane of the curve, and from it two tangents (not shown in the figure) may be drawn to the curve C. 154. Tangent Planes to Curved Surfaces. Let S, Fig. 257, represent any curved surface. Let a be any point on the sur- face, and let T be the tangent plane to the surface at the point a. Let Cbe any curve drawn on the surface through the point a. If drawn at random, C will probably be a space curve, al- though it may be taken so as to be a plane curve. Let ./ be the line tangent to the curve C at the point a. Then ./ will be tangent to the surface 8, and will lie in the tangenl plane T. Let D be a second curve lying on & and passing through 176 DESCRIPTIVE GEOMETRY [XVIII, § 154 a. Then a line K, tangent to D at a, will also lie in the plane T. It follows that the tangent plane to a surface at a point on it may be defined as the plane which contains all the straight lines tangent to the given surface at the given point of the surface. 155. Determination of Tangent Planes by Means of Tangent Lines. Two intersecting straight lines determine a plane. Suppose the given surface to be such that two curves of known properties can be drawn on the surface through the required point of tangency. The tangent plane is then determined by finding the plane which contains the two rectilinear tangents to these curves. In general, the curves used will be plane curves. If the surface is such that a straight line, as E, Fig. 257, can be drawn on the surface through the given point a, the line E will lie in the tangent plane T. Such a straight line which lies wholly on the surface is called a rectilinear element of the surface. In this case, but one tangent line in addition to the element E is necessary to determine the plane T. Curved surfaces exist in which two rectilinear elements of the surface can be drawn through a given point in the surface. For such surfaces the tangent plane at any point is the plane containing the two rectilinear elements passing through the point. 156. The Normal. The normal to a surface at any point on it is the straight line perpendicular to the tangent plane at that point. For certain curved surfaces, for example, the sphere shown in Fig. 258, the normal can be determined readily from known properties of the surface. Hence the tangent plane T at the point a can be found by drawing first the normal N, and then passing through the point a the plane T perpendicular to the line N. XVIII, § 157] TANGENT LINES AND PLANES 177 157. Determination of Tangent Planes by Means of the Normal. It is to be noted that the tangent plane T is absolutely deter- mined in space by means of the normal N, since but one plane can be passed through a given point a perpendicular to a given line N. But in a projection drawing, where the plane T is not determined until its traces HT and VT are found, addi- tional lines are necessary. This is shown in Fig. 259. The point a represents the given point in some given surface, and the line N rep- resents the normal at that point. The tangent plane T is passed through the point a perpendicular to N. But al- though the projections N h and N v give the directions of HT and VT respec- tively (§ 112), the normal alone fur- nishes no point on either of these traces. Hence at least one line, as J, lying in the tangent plane, is required in the projection drawing to supply a necessary point, as s. (See Prob. 10, § 115.) In determining the traces of a tangent plane by the use of the normal, therefore, the working method consists in finding, in addition to the normal, one line which is tangent to the given surface. The required tangent plane is then passed through the tangent line perpendicular to the normal. Fig. 259. CHAPTER XIX TANGENT PLANES TO CONES AND CYLINDERS 158. The Cone and Cylinder. Definitions. The terms cone and cylinder are variously used, both in mathematical and in popu- lar language, to denote either surfaces, or solids bounded in part by these surfaces. As a surface, a cone may be defined as generated by a mov- ing straight line, indefinite in extent, which always passes through a fixed point in space, and is so guided, by a curved line or otherwise, that a plane is not formed. A cylinder dif- fers from a cone in that the generating straight line, instead of passing through a fixed point, remains always parallel to its initial position. In both of these surfaces, any position of the generating straight line is known as a rectilinear element, or simply as an element, of the surface. The solids understood by the terms cone and cylinder are familiar objects, and do not need to be defined here. It should be remembered, however, that the axis of a cone or cylinder is not necessarily at right angles to the base. 159. Representation of the Cone and the Cylinder. We shall consider at present only the solid forms of these objects, and shall confine ourselves to those in which the base is either a circle or an ellipse. The projections (C h , C") of a general cone, in which the base is an ellipse, and the axis oblicpie to the plane of the base, are given in Fig. 2C0. We shall not attempt to solve problems, however, when the cone is projected as in Fig. 260. Since the base of the cone is a plane curve, it can be projected as a straight line (§ 151). In Fig. 260, let B (HB, VB) be the plane of the base, found by passing a plane through any three points of the curve (Prob. 6, Cor. II, § 106 ; construction not shown). Assume a second- 178 XIX, § 160] TANGENT PLANES 179 ary ground line perpendicular to HB ; then the plane B will project edgewise as ViB (see Fig. 92, § 70). Consequently, in Fig. 200. the secondary projection, CV, of the cone, the base will pro- ject as a straight line. Since this transformation always can be effected, we shall discuss further only those cases of cones and cj'linders in which one projection, at least, of the base is a straight line. This results when the plane of the base is perpendicular to, or coincident with, one of the coordinate planes. 160. Projections of Cones and Cylinders. We shall place these objects in the first or third quadrants only (see § 19). Projections of cones are given in Figs. 201-265. These fig- ures represent the following objects : Fig. 2G1. Cone in the third quadrant; base in a plane per- pendicular to V; base visible. Note that if we attempt to read these as the projections of a cone in the first quadrant, the base should be invisible. As this is not the case, the coin' cannot be in the first quadrant. In this and similar cases, the student will often be expected to determine the quadrant from the given visibility of the projection. 180 DESCRIPTIVE GEOMETRY [XIX, § 160 Fig. 262. Cone in the third quadrant; base in a plane per- pendicular to H ; base invisible. Fig. 261. Fig. 262. Fig. 263. Cone in the first quadrant; base in H; base invisible. Fig. 264. Cone in the third quadrant; base in V; base visible. Fig. 263. Fig. 264. Fig. 265. Cone of revolution; base a circle in the P-plane. While these are the projections resulting from projecting XIX, § 160] TANGENT PLANES 181 this object, so placed, they are, by themselves, and without further information, ambiguous. They should be supple- mented by the profile projection, and the quadrant in which the object lies must be indicated in some manner, as for example, by lettering the vertex. In projecting a cylinder, only one base will, in general, be shown, the cylinder being left indefinite in ex- tent in one direction. Cylinders are projected in Figs. 266-270, which rep- resent the following objects : Fig. 266. Cylinder in the third quad- rant; base in a plane perpendicular to V; base visible. As in the correspond- ing case of the cone (Fig. 261), an Fl «- 265. attempt to read these views as the projections of a cylinder in the first quadrant fails. Fig. 266. Fig. 267. Fig. 267. Cylinder in the third quadrant ; base in a plane perpendicular to II; base invisible. 182 DESCRIPTIVE GEOMETRY [XIX, § 160 Eig. 268. Cylinder in the first quadrant ; base in V; base invisible. Fig. 269. Cylinder in the third quadrant ; base in H ; base visible. The elements of this cylinder are all parallel to V. Fig. 268. Fig. 269. Fig. 270. Fig. 270. Cylinder of revolution ; base a circle lying in P. Like the cone of revolution (Fig. 265), these projections are ambiguous, and similar re- marks apply. 161. Projection of a Point in the Surface of a Cone or Cylin- der. Only one projection of a point lying in a conical or cylindrical surface can be as- sumed. The other projection of the point may be found by means of an element of the surface, as follows. Let a cone be given as in Fig. 271, and let a v be an as- sumed projection. The ele- ment E, passing through a, Fro- 271. must also pass through the vertex o. Hence draw E" through a v and o". Produce E v to meet the F-projection of the base XIX, § 161] TAXGEXT PLAXES 18.; at c". Project from c" to c* in the if-projection of the base. Then E h is drawn by connect- ing c* and o*. Find a* in E h by projecting from a". A second result is possible, namely, point b lying in the element F. In general, an assumed pro- jection will represent two points. But let k v be assumed in the outside or contour element L. Then k v is the projection of but one point in the surface. A similar construction for the cylinder is shown in Fig. 272, the assumed projections being the double point a r b v , and the single point k v . In Fig. 273, the assumed projections are a h , b h . The base of the cone lies in P, which necessitates the use of the P-pro- jection of the base. (Compare Fig. 265.) A similar construc- tion applies to a cylinder whose base lies in P. Fig. 272. 184 DESCRIPTIVE GEOMETRY [XIX, § 162 162. Tangent Planes to Cones and Cylinders. At every point in the surface of a cone or cylinder, a plane tangent to the sur- face can be drawn. Planes tangent to cones and cylinders can also be passed to fulfill certain other conditions. Thus, the tangent planes may be required to contain a given point outside the surface, or to be parallel to a given straight line. In determining these tangent planes, we shall make use of the following propositions. (a) Through every point in the surface of a cone or cylinder, a rectilinear element can be drawn, and this element will lie in the tangent plane at that point (jj 155). It is a property of both the cone and cylinder that a plane tangent at any point in a given element is tangent at every point in the element. Hence, if the element of tan- geney is known, a second line in the tangent plane may be drawn tangent to the cone or cylinder at any point in this element. A convenient point is generally the point in which the element of tangency intersects the base. (c) If the base of a cone or cylinder lies in H. any line tan- gent to the base lies in H (§ 153), and thus becomes the H- trace of a plane tangent to the surface. Similarly if the base lies in Vot P. (d) Every plane tangent to a cone contains the vertex. Problem 31. To pass a plane tangent to a cone at a given point in the surface. Analysis. The plane is determined by the element which passes through the given point, and a line tangent to the base at the point where this element intersects the base (§§ 155, 162, &). Construction. Case I. 77/<= base of the cone lien in H or V. Example 1 (Fig. 274). The base of this cone lies in H. Let a (o h . a'), lying in the element E (E h . E*) be the given point in the surface. Since the required tangent plane contains the element E, find the traces s x and t-, of E. Then the Z7-traee. HQ. of the tangent plane passes through s lf and is determined XIX, § 162] TANGENT PLANES 185 by the fact that HQ must be tangent to the base of the cone (§ 162, c). The trace VQ is now determined, since it must pass through t x and the point in which HQ intersects GL. Let it also be required to find the tangent plane at the point b, lying in the element F. The Zf-trace, HT, of this plane, Fig. 274. passes through the //-trace, s 2 , of F, and is tangent to the base of the cone. The F-trace, VT, passes through the V-trace of F, but this point is inaccessible. Recourse must be had to some auxiliary line lying in the plane T, and so situated that its F-trace is accessible. (See § 108.) As HT is known, a hori- zontal principal line of the plane may be drawn through any point of the line F (§ 108, Ex. 3). A convenient point is the given point b. Through 6* draw U parallel to HT; through b v draw L v parallel to GL. Then the line L lies in the plane T, and the F-trace, t 2 , of L is a point in VT. 'Example 2 (Fig. 275). The base of this cone lies in V. Tangent planes are passed at the points a and b. For the 186 DESCRIPTIVE GEOMETRY [XIX, § 162 plane Q, tangent at a, the trace VQ is drawn tangent to the base of the cone through the F-trace of the element E. Since the i7-trace of the element E is not available, a point on HQ is found by means of the auxiliary line L, a vertical principal Fig. 275. line of Q, drawn through the given point a. The 77-trace of L lies in HQ. The plane T is tangent at the point b. The trace VT is drawn tangent to the base of the cone, through the "F-trace of the element F passing through b. In this case VT does not intersect GL within reach, neither is the iJ-trace of F obtain- able. Hence two points on HT must be obtained by the use of auxiliary lines. Two convenient lines are the vertical principal lines, M and N, of T, drawn, one through the given point b, the other through the vertex, o, of the cone, since the vertex is in every tangent plane (§ 162, d). The .ff-traces, s 4 and s 5 , of these two lines determine HT. XIX, § 162] TANGENT PLANES 187 Case II. Hie base of the cone lies in P (Fig. 276). The base of this cone is a circle, and the cone is the symmetric form known as the cone of revolution. Let a (a h , a v ) be the given point. Point a lies in the element E, which intersects the plane of the base in the point c, the profile trace of the line. Through the actual trace c p draw the profile trace, PQ, of the tangent plane, tangent to the profile projection of the base. From PQ are found the points, t 2 on VQ, and s 2 on HQ (§ 60). .Since E lies in Q, find also, if possible, the II- and F^traces oi E. In this case the P~-trace, t u may be found, and this is sufficient; for VQ passes through t 2 and t u and 1IQ through s 2 and the point in which VQ intersects OL. The plane Tis tangent at the point !>, lyin^ in the element F. This element intersects ' parallel to A. Line 11 intersects the plane X in the point c. From c draw the tangents, J* and K, to the base of the cone. Plane (J is passed through B and the tangent ./, while plane T is passed through />' and the tan- gent 1\. No auxiliary lines arc necessary. The planes Q and Tare the required tangent planes. o 194 DESCRIPTIVE GEOMETRY [XIX, § 162 Note. It will be seen in Figs. 280 and 281 that if the line B, parallel to A, intersects the plane of the bases of the cone within the circumfer- ence of the base, no tangent lines will be possible. Consequently, there will be no solution. This will happen when the line A is parallel, or nearly so, to the axis of the cone. A special case, in which one, and only one, solution exists, will occur when the line B, parallel to A, is found to coincide with an element of the given cone. Problem 34. To pass a plane tangent to a cylinder at a given point in the surface. Analysis. As in the corresponding problem with the cone (Prob. 31), the tangent plane is determined by the element which passes through the given point, and a line tangent to the base at the point where this element intersects the base. Construction. Case I. The base of the cylinder lies in H or V (Tig. 282). The construction is entirely similar to that for the corresponding case of the cone (Prob. 31, Case I), and it should not require an extended explanation. The plane Q, tangent at the point a, contains the element E. No auxiliary line is needed. XIX, § 162] TANGENT PLANES 195 The plane T is tangent at the point b, lying in the element F. A necessary point, t 3 , in VT, is here determined by the auxiliary line L. Case II. The base of the cylinder lies in P (Fig. 283). The given cylinder is a cylinder of revolution, with a circular base lying in P. Obtain, if not already given, the profile projection of the base. For the plane Q, tangent at the point a, obtain s F h HQ. a. i u h " 1 \ 1 »' 1 \ >^r ' \ \ VQ. \!a* E v t uvyu p >. VT / • \ - - ( -+ J F v b v J ^\j y HT -. (L - _ > \<> Fig. 28 ; the profile trace, n p , of the element E containing a. Through u v draw the profile trace, PQ, tangent to the base. This trace furnishes the point s on HQ and the point / on VQ. In this case no other points are necessary, since the plane (J is evidently parallel to GL. The plane T (PT, HT, VT), tan- gent at the point b in the element F, is obtained in a similar manner. Case III. Tlie base of the cylinder does not lie in H, V, or P (Fig. 284). The construction is entirely similar to thai Eor the corresponding case of the cone (Prob. 31, Case III), and should be compared with it. A detailed explanation will Dot be given. 196 DESCRIPTIVE GEOMETRY [XIX, § 162 The required planes are Q, tangent at the point a, and T, tangent at the point b. Fig. 284. Problem 35. To pass a plane tangent to a cylinder through a given point without the surface. (Two results.) Analysis. Through the given point pass a line parallel to the elements of the cylinder. Since the recpiired tangent plane contains some element of the surface, this line must lie in the required tangent plane. Find the point in which this line pierces the plane of the base of the cylinder. From this point draw a line tangent to the base. Pass the required tan- gent plane through the tangent line and the line first drawn. In general, two tangent lines, and hence two tangent planes, are possible. Construction. Case I. The base of the cylinder lies in H or V (Fig. 285). The base of the given cylinder lies in H. The given point is a. Through a draw the line B (B h , B") parallel XIX, § 162] TANGENT PLANES HT 197 to the elements of the cylinder. From this on the construc- tion is the same as for the corresponding case of the cone (Prob. 32, Case I). The required planes are Q and T. Case II. Tlie base of the cylinder does not lie in II or V (Fig. 286). The base of the given cylinder lies in the plane Z, perpendicular to II. Through the given point a, draw the Fig. 286. 198 DESCRIPTIVE GEOMETRY [XIX, § 162 line B parallel to the elements of the cylinder. Then proceed as in the corresponding case of the cone (Prob. 32, Case II). The resulting planes are Q and T. Problem 36. To pass a plane tangent to a cylinder parallel to a given line. (Two results.) Analysis. No point of the required tangent plane is known in advance, as in the corresponding problem with the cone (Prob. 33) ; hence the result must be accomplished by indirect methods. Through the given line, pass a plane parallel to the elements of the cylinder. This is possible, since all the elements are parallel. This plane will then be parallel to Fig. 287. the required tangent planes, and may be moved, parallel to itself, until tangent to the cylinder. The manner of accom- plishing this is best shown in the construction. Construction. Case I. Tlie base of the cylinder lies in H or V (Fig. 287). The base of the given cylinder lies in V. The given line is A. Through A pass the plane X parallel to the elements of the cylinder. To do this, assume any point, c, in A ; through c draw the line B, parallel to the cylinder ; pass the plane X through the lines A and B. (See Prob. 7, § 107.) Since the base of the cylinder lies in V, VQ is now XIX, § 162] TANGENT PLANES 199 drawn tangent to the base parallel to VX; HQ is then drawn parallel to HX. The traces VT and HT of the second required tangent plane are obtained in a similar manner. Case II. The base of the cylinder lies in P (Fig. 288). The base of the given cylinder is a circle whose center is o ; find the P-projection of this circle. As in Case I, through the given line A pass the plane X, parallel to the elements of the cylinder. Since the base of the cylinder lies in P, find the profile trace, PX, of this plane. Draw PQ and PT parallel to PX, tangent to the base of the cylinder ; these are the profile Fig. 288. traces of the required tangent planes. From PQ and PT are obtained the //-and F-traces of the plain's, HQ and ///'being parallel to HX, while VQ and FT 7 are parallel to VX. Cask III. The hose of the cylinder tines not lit in //, V, or P (Fig. 289). The base, of the given cylinder lies in the plane Zjj perpendicular to V. The given line is A. Through this line is passed, as in the previous cases, the plane X. parallel to the elements of the cylinder. The next step is to draw a tine, J", which shall be tangent to the base of the cylinder and at 200 DESCRIPTIVE GEOMETRY [XIX, § 162 the same time parallel to the plane X. This line must lie in the plane, Z, of the base of. the cylinder (§ 153) ; it must also be parallel to some line in the plane X (§ 104) ; hence J will be parallel to the line of intersection of the planes X and Z. Find this line of intersection, L (Prob. 12, § 118). Draw the line J tangent to the base of the cylinder parallel to the line Fig. 289. L. Through J pass the required tangent plane, Q, parallel toX A second result, plane T, is found by means of the tangent line K, also parallel to L. Here VT is not located either by the F-trace of K or the point where HT intersects GL. The construction used is to assume some point, as e, on K. Through e draw the line M, parallel to the elements of the cylinder. Then the line M lies in the plane T, and the F-trace t is a point on VT. CHAPTER XX TANGENT PLANES TO DOUBLE CURVED SURFACES OF REVOLUTION 163. Double Curved Surfaces of Revolution. A double curved surface of revolution is a surface formed by the revolution of any curve about any straight line as an axis, provided the resulting surface is such that no straight lines can be drawn on it. Familiar solids whose surfaces are double curved surfaces of revolution are the sphere, the various ellipsoids, and the torus (§ 26). 164. Representation of Double Curved Surfaces of Revolution. In solving problems involving double curved surfaces of revo- lution, we shall place the axis of the sur- face perpendicular to one of the coordinate planes. If not so given, the transforma- tion can be effected by new planes of pro- jection. In Fig. 290 is shown a general case, a vase whose outer surface is a double curved surface of revolution. The axis is placed perpendicular to //. 165. Meridians. The F-projection, Fig. 290, shows the true shape of the section made by a plane containing the axis of the surface. Such a section is called a meridian section, or simply a meridian. Any plane which contains the axis is called a meridian plane. All meridians are alike, and the surface is usually considered as formed by the revolution of its meridian section. 201 Fhi. 290. 202 DESCRIPTIVE GEOMETRY PCX, § 165 Fia. 290 (repeated). The meridian which forms the outline of a projection of the surface, in Fig. 290 the F~-projection, is called the principal meridian. The plane which contains this meridian is called the principal meridian plane. The principal meridian plane is al- ways parallel to one of the coordinate planes ; in Fig. 290 it is parallel to V. 166. Parallels. Let a double curved sur- face be formed by the revolution of its meridian section about the axis. Each point of the generating meridian describes a circle lying in a plane perpendicular to the axis (§ 74). These circles are called parallels of the surface. The projection of a double curved sur- face of revolution on a plane perpendicular to its axis will consist of one or more cir- cles, which are the projections of particular parallels of the surface (§§ 75, 84). For example, see the .ff-projection of the vase, Fig. 290. 167. Projections of a Point in a Double Curved Surface of Revo- lution. Through every point of a double curved surface of revolution a circle which is a parallel of the surface (§ 166) can be drawn. A point will lie in the surface if its projections lie in the corresponding projections of a parallel of the surface. With the axis of the surface perpendicular to either H or V, one projection of the parallel will be a circle, the other projec- tion a straight line (§ 75). For example, consider the ellipsoid shown in Fig. 291, where the axis of the ellipsoid is perpendicular to H. Let a h be a given projection of a point in the surface. Through a h draw the iaT-projection, a circle, R h , of the parallel passing through a. Find the F-projection of the circle B, namely, the straight line M v . Project from a h to a v in R v ; then point a(a h , a") lies in the surface. The given ^-projection may also represent a point b, lying on the symmetric parallel S (S h , S v ). XX, § 168] SURFACES OF REVOLUTION 203 Let C be given. This point lies on the parallel T, whose F-projection, T", is a straight line through c". The //-projec- tion of this parallel is the circle T*. In T h is found c h by pro- jection from c v . There is also a second result ; namely, the point d (d", d h ) If a surface is wholly convex, as in this example, any point assumed in either projection represents two points in the sur- FlG. 291. face, unless the assumed projection lies in the contour (outline) of the surface. Thus, if e" is assumed in the ^projection, 3f", of the principal meridian (§ 105), the single //-projection, e h , lies in the straight line M h > which is the //-projection of the principal meridian. If /* be assumed in the H"-projection, /"', of the greatest parallel (§ 1 <><>), /" lies in the straight line P . the ^projection of this parallel. 168. The Sphere. The sphere possesses the unique property that every plane section is a circle. All planes which contain 204 DESCRIPTIVE GEOMETRY [XX, § 168 the center of the sphere intersect the surface in circles of the same size, known as great circles. Any diameter of the sphere may be taken as its axis. In the solution of problems involv- ing the sphere, advantage is usually taken of some or all of these properties. The solution thus becomes a particular solu- tion, and the sphere, in consequence, is not a good surface to use to illustrate the general case of a double curved surface of revolution. In the problems of tangencies which follow, we shall treat the sphere as a particular case, apart from the general problem. Fig. 292. Fig. 293. On account of the perfect symmetry of the sphere, and the possibility of taking any diameter as an axis, students often have more difficulty in visualizing the relation of points on the surface than with a less symmetric surface. This difficulty usually disappears by visualizing first a hemisphere, as shown in Fig. 292. This figure represents the front half of a sphere, bounded by the great circle 0(0", C h ), and the portion of the surface represented by each projection appears more clearly than when the entire sphere is given, as in Fig. 293. 169. The Torus. The surface of a torus (Fig. 294) will usually be taken to illustrate the general case of a surface of revolution. This surface is divided by the circles C and D XX, § 170] SURFACES OF REVOLUTION 205 into two portions : the so-called outer portion, doubly convex, where a tangent plane contains but a single point of the sur- face, and an inner portion, concavo-convex, where a tangent plane at any point also intersects the surface. Certain properties of this surface have already been given. Thus in § 86, it has been shown that any plane perpendicular to the axis between C and D cuts the surface in two circles A Fig. 294. and B, which are parallels of the surface. Hence, if the sur- face is given as in Fig. 294, points assumed in the F-projection may represent, according to their position, one, two, three, or four points lying in the surface. A point assumed in the H- projection, however, as in the ellipsoid, Fig. 291, can never represent more than two points in the surface. 170. Tangent Planes to Double Curved Surfaces of Revolution, (a) A plane tangenl to a double curved surface of revolution is tangent, in general, at but a single point. Exceptions occur, however; for example, in the torus, Fig. 294, the planes which contain the circles C and D are tangent at every point in these circles. 206 DESCRIPTIVE GEOMETRY [XX, § 170 (6) Through every point in the surface two curves of known properties can be drawn, namely, the meridian (§ 165) and the parallel (§ 166) which pass through the given point. (c) Planes which are tangent to points lying in the same parallel of the surface intersect the axis at the same point. (d) The normal (§ 156) at any point of the surface lies in the meridian plane passing through the point. (e) Every normal to the surface intersects the axis. (/) Normals to points lying in the same parallel of the sur- face intersect the axis at the same point. Fig. 295. Problem 37. To pass a plane tangent to a sphere at a given point in the surface. Analysis. Through the given point, draw two circles lying in the surface. Draw tangent lines, one to each circle, at the given point. Find the plane determined by the two tangent lines. (See § 155.) Construction (Fig. 295). Let a (a h , a v ) be the given point. Through a we can draw two circles which have simple projec- XX, § 170] SURFACES OF REVOLUTION 207 tions, namely the circle C (C A , C v ) parallel to Fand the circle I) (D h , D v ) parallel to H. Through a draw the Hue J (J h , J") tangent to the circle C, and line K (A" \ K v ) tangent to the circle D (§ 152). The required tangent plane, T, is the plane determined by the lines J and A"(§§ 102, 106). A second result is the plane Q which is tangent at the point b (b h , b v ) and obtained in a similar manner. Problem 38. To pass a plane tangent to a double curved surface of revolution at a given point in the surface. First Analysis. The meridian and the parallel which pass through the given point are two known curves of the surface (§ 170, b). At the given point, draw tangent lines, one to each of these curves. Find the plane determined by the two tangent lines (§ 155). Second Analysis. Draw the normal to the surface at the given point. Pass the required tangent plane through the point perpendicular to the normal (§ 156). Third Analysis. Revolve the surface about its own axis until the given point is in the principal meridian plane. In this position the tangent plane will show edgewise as a line tan- gent to the outline of the surface at the given point. Counter- revolve the surface together with the tangent plane to the original position. The actual construction, if made by the first analysis, must often be supplemented by the use of auxiliary lines to obtain points within the limits of the drawing. It has also been shown, in § 156, that while the normal alone determines the plane in space, auxiliary lines must be used to locate the traces in the drawing. P>ut by the third analysis, it is possible in nearly every case to determine one trace of the tangent plane neatly and rapidly. We shall therefore begin the solution by the third analysis; then, to find the second trace of the plane, introduce as auxiliaries one or nunc of the lines indicated by the first and second analyses. Construction (Fig. 296). The surface chosen to illustrate this, the general case, is that of a torus. Let a (a*, a ") be the 208 DESCRIPTIVE GEOMETRY [XX, § 170 given point in the surface. Revolve the torus about its axis until a is in the principal meridian plane Y. The path of a will be the parallel C, and the resulting position a r (a r h , a r v ). At a r v we can draw by inspection the edge view VT X of the V ' D v \ \K |/fc L v y \^ s ^ \> xk ,/t. X Fig. 296. tangent plane at this point. The if-trace of this plane is HT^. Counter-revolving the torus to the first position, HT X takes the final position HT (reverse of § 134). The tangent plane, as shown at VT 1} intersects the axis at the point / (/", f h ), which becomes a fixed point for all planes tangent at any point in the parallel C (§ 170, c). Hence, connecting / and a, we XX, § 170] SURFACES OF REVOLUTION 209 have the line K (K v , K h ), evidently tangent to the meridian E and lying in the tangent plane at a. The vertical trace, VT, of the required tangent plane is now determined by the V- trace, ^ of K, together with the point in which HT intersects GL. In this case the line «/, tangent to the parallel C at a, might be used instead of the line K. Let b (b h , b"), Fig. 296, be a second point. Let b be revolved into the principal meridian at b r (b r h , b r "). At b r v we may draw by inspection the edge view VQ X of the plane tangent at this point. This edge view does not intersect the axis within reach, as was the case with point a; but the plane Q t inter- sects H in the ff-trace HQ X . AVe can therefore obtain HQ by revolving HQ X about the axis of the torus, as was done for point a. The line L, tangent to D at b, is one line in the required tangent plane Q. This gives one point, t i} in VQ. But since HQ does not intersect GL within the figure, VQ is still undetermined. Let us try the normal. Draw first the normal, N T , at the point b r v . This normal intersects the axis at the point g v . Then the normal at the point b passes through #(§ 170,/). Connecting^ and b v gives the F-projection, N". Since the plane Q is perpendicular to the normal N, VQ is now drawn through ? 4 perpendicular to N" (§ 11-5). Where is the //-projection of the normal A".' Note that 'this projection would not be needed in any event, since it would merely give the direction of HQ. A second example, involving some additional features of construction, is given in Fig. 297. The given point is so chosen in the F-projectiou thai it represents lour points in the surface (§ 169). Two tangent planes are shown, one at an outside point, ami one at an inside point. Tin- plane V is tan- gent at point a. This point lies in the parallel /«,' I /''. / >. ■> that it revolves into t be principal meridian plane at n (a r k , a I. The trace HQ is obtained by revolving HQ X about the axis of the torus, as in Fig. 296. The meridian tangent A', and the line. J tangent to the parallel h\ give tin' P-traces /, and /.., thus locating VQ. The normal, N, to the plane Q, although not needed, is shown in the F-projection ; if here furnishes a 210 DESCRIPTIVE GEOMETRY [XX, § 170 check on the construction, since N v should be perpendicular to VQ(% 112). The plane T, Fig. 297, is tangent at the point c. This point lies in the parallel F (F h , F") ; hence it revolves into the prin- cipal meridian at c r (c r h , c r v ). At c r v draw the edge view, VT U of the tangent plane, also the normal, U T , to this plane. Then HT X revolves to HT. Since VT X intersects the axis at 3", the line M (M v , 3I h ) is a line in the required tangent plane at c. But the line M gives no accessible point on VT; neither does the line L, drawn tangent to the circle F at point c. Draw the P"-projection, U v , of the normal to the plane T through 4" and C (§ 170,/). Note that now but one point on VT is nec- essary. Why ? Since the line M lies in the plane T, let the point 5 (5 A , 5") be some assumed point on M. Through the point 5 draw the line X (X k , X") parallel to the line L (§ 108, Ex. 1). Find the F-trace, t- , of X. Draw the recpiired trace, VT, through t perpendicular to U v . 171. Tangent Planes which Contain a Given Line. A plane tangent to a double curved surface of revolution may, under certain conditions, be determined by the fact that the plane must contain a given straight line. For example, with sur- faces like the sphere and the ellipsoids, it is evident that the necessary condition is that the line shall not intersect the sur- face ; then, in general, two tangent planes are jwssible. But, whatever the form of the surface, if a solution exists in space, the traces of the plane can be found. We are not prepared to take up at this time the general case of a plane through any line tangent to any double curved surface of revolution. The solution requires the use as an auxiliary of a surface whose properties have not yet been dis- cussed. Particular solutions can be found, however, for the following cases. (1) The given surface a sphere, the line general, in any position not intersecting the sphere. (2) Any double curved surface of revolution, with the line in particular positions with respect to the axis of the surface. XX, § 171] SURFACES OF REVOLUTION 211 £ I c n Fiu. '.'97. 212 DESCRIPTIVE GEOMETRY [XX, §171 Problem 39. To pass a plane tangent to a sphere through a given line without the surface. Analysis. This analysis is shown pictorially in Fig. 298. Let A be the given line, and o the center of the given sphere. Let Q and T represent the required tangent planes. Pass an auxiliary plane, X, perpendicular to A, through the center of the sphere. This plane intersects A in the point c. Since X contains the center of the sphere, the in- tersection with the sphere is a great cir- cle. This circle evi- dently contains the points of tangency of the planes Q and T. From c draw the line E tangent to the great circle. This can be done, since c and the circle are in the same plane, X (§153). Then E will lie wholly in the tan- gent plane Q, which is now determined, since we know two lines, A and E, lying in it. The plane T is similarly determined as the plane which contains the given line A and the tangent line F. An alternative method for determining the plane T is also shown in Fig. 298. From o draw the line N perpendicular to F. This line lies in the plane X, and passes through the point of tangency of F and the sphere. The line N is thus the normal to the tangent plane T. Hence T may be determined as the plane which contains the given line A and is perpendicular to N. We may observe in passing that it is not possible, in general, to pass a plane through one line perpendicular to Fig. 298. XX, §171] SURFACES OF REVOLUTION 213 another line, but the relative positions of the lines A and JV are such in this case that it may be done. Construction (Fig. 299). Let A be the given line, and o the center of the given sphere. Find the plane, X, which con- tains o and is perpendicular to A (Prob. 10, § 115). Find the point, c, in which A intersects X (Prob. 13, § 119). Since the Fm. 299. plane X is oblique to both // and V, the great circle in which X intersects the sphere will project as an ellipse in each new. To avoid drawing these ellipses, revolve the plane X about one of its traces into // or V (§ L38). Lei X be revolved about HX into H. Tin; point c revolves bo c, (Prob. 21, Working Rule, § 138); and the point o revolves to <>,. The great circle lying in X will now appear in true shape and size, 214 DESCRIPTIVE GEOMETRY [XX, § 171 and may be drawn at once, with o r as center. From c r draw the two tangents, E r and F T , to this circle. These are two lines lying in the plane X (see the Analysis), and their pro- jections may be obtained by counter-revolving the plane X (§ 147). The tangent E r intersects HX at s 3 *, whence s 3 v is on Fig. 299 (repeated). GL, and the projections E h , E v are determined by the fact that the line E passes through the point c (reverse of Fig. 221, § 138). Pass the required tangent plane Q through the lines E and A (Prob. 6, § 106). The tangent line F does not intersect HX within reach. In- stead of attempting to find the projections of F, let us deter- mine the tangent plane by means of the normal. Through o T draw N r perpendicular to F T . The line N lies in the plane X XX, § 171] SURFACES OF REVOLUTION 215 (see the Analysis) ; hence the intersection of N r and HX is the //-trace of N. Using this trace and the point o, determine the projections, JV* and N v , of N. The required tangent plane T contains A and is perpendicular to N. Let s x and t x be the traces of A. Through s x draw HT perpendicular to JV* ; through t x draw VT perpendicular to N" (§ 112). Fig. 300. Special Case. Tlie given line is parallel to H or V. Let the given line A, Fig. 300, be parallel to 77. Assume a secondary ground line, 0\L\, perpendicular to A h . The line A will then project as a point, A? (§ 70). Project the center of the sphere to of, and draw the sphere. The edge views, V X Q and V\T, of the required tangent planes may now lie drawn directly through A{. The plane Q (HQ, VQ) requires no further explanation. The trace HT is readily obtained, but for Fran auxiliary line is necessary. The line used here is /;, parallel to the line A. The projection /»',''. a point, is assumed in V\T. From />',' ;l "■ obtained B h and B v . The F-trace, t 2 , of B locates a point in I"/'. 216 DESCRIPTIVE GEOMETRY [XX, § 171 Problem 40. To pass a plane tangent to a double curved surface of revolution through a given line. (Special Cases.) Case I. The given line intersects the axis of the surface. Analysis. Using the point in which the given line intersects Fig. 301. the axis as a vertex, circumscribe the given surface by a cone of revolution. Pass the required tangent planes through the given line tangent to the auxiliary cone. Construction (Fig. 301). The given surface is an oblate spheroid, with its axis perpendicular to H. The given line A intersects the axis at the point c (c*, c ! ). Prom c* draw tan- gents to the P"-projection of the given surface; these tangents represent the auxiliary cone in the ^projection. The base of this cone on H is the circle B h . The if-traces, HQ and HT, of the required tangent planes pass through the JY-trace, s u of A, and are tangent to the circle B h . (Compare Prob. 32, § 162.) XX, § 171] SURFACES OF REVOLUTION 217 The F-traces, VQ and VT, pass through the U-traee, t u of A. A second point on VT is here obtained by finding the F-trace, t 2 , of the element, E, hi which the plane T is tangent to the auxiliary cone. Case II. TJie given line is parallel to the axis of the surface. Analysis. The axis of the given surface being placed, as usual, perpendicular to one of the coordinate planes, for ex- ample H, the .//-projection of the given line will be a point. The ^-traces of the tangent planes will therefore be their edge views, and the planes may be drawn by inspection. We pro- ceed similarly if the axis of the surface is perpendicular to V or P. No figure to illustrate the construction is deemed necessary Case III. The given line is perpendicular to, but does not fntersect, the axis of the given,, surface. First Analysis. Let the axis of the given surface be per- pendicular to H. Then the given line is parallel to H. Re- volve the line about the axis of the surface until the line is perpendicular to V, and projects as a point thereon. The edge views of tangent planes may now be drawn by inspection. Revolve these planes about the axis of the surface until they contain the original position of the given line. We proceed similarly if the axis of the surface is perpendicular to For P. Second Analysis. The given line may be projected as a point by choosing a secondary plane of projection perpendicular to the line. This plane necessarily will be parallel to the axis of the given surface. Hence the surface may be projected readily, and the edge views of the tangent planes may be drawn by inspection. The traces of the planes can then be obtained from the edge views. Since the met hod of the second analysis aecessitates the '-1111 struction of an additional projection oJ the given surface, if is usually easier to employ the firsl analysis. The following con- struction is made by the first analysis. Construction (Fig. 302). The given surface is thai of a aolid torus, only the outer (doubly-convex) portion of the curved 218 DESCRIPTIVE GEOMETRY [XX, § 171 surface being retained. The axis of the surface is perpen- dicular to V. The given line A (A v , A h ) is parallel to V. Re- volve A about the axis of the torus to the position A x (Af, A x h ), where the ^-projection, A x h , is a point. Through A x Fig. 302. draw the tangent planes & (HQ 1} VQi) and T x (HT U VTJ. Re- volve Qi and T x about the axis of the torus to the positions Q (VQ, HQ) and T ( VT, HT), so that each plane contains the line A (§ 134). Then Q and T are the required tangent planes. CHAPTER XXI THE INTERSECTION OF CURVED SURFACES BY PLANES 172. Classification of Curved Surfaces. Curved surfaces may be divided into three classes : (a) Ruled surfaces, on which a straight line, or rectilinear ele- ment, can be drawn through any point of the surface ; (b) Surfaces of revolution, on which a circle, lying in a plane perpendicular to the axis, can be drawn through every point of the surface ; (c) All other curved surfaces, on which, in general, neither straight lines nor circles can be drawn. Certain surfaces, for example the cone and cylinder of revolution, belong to both classes (a) and (b). 173. The Intersection of a Ruled Surface and a Plane. Select a sufficient number of the rectilinear elements of the surface. Find the points in which these straight lines intersect the given plane. These points lie in the required intersection. In the process of finding where a straight line intersects a plane, it is usual to puss through the line an auxiliary plane perpendicular to // or V. (See Prob. 13, § 119.) Hence the elements of the ruled surface are often chosen by passing through the surface a series of planes perpendicular to // or V, these same planes being then used as auxiliaries for finding the points in which the resulting elements intersect the secant plane. 174. The Intersection of a Surface of Revolution and a Plane. An auxiliary plane taken perpendicular to the axis of the given surface will cut one or more circles from the surface, and, in general, a straight line from the given pli Points of intersection, if any, of the straighl line with the circle or circles, will lie in the required intersection of the plane and the surface. 219 220 DESCRIPTIVE GEOMETRY [XXI, § 174 In order that the circular sections of the given surface may be projected readily, the axis of the surface should be perpen- dicular to one of the coordinate planes. (Compare §§ 84, 86, 164.) 175. The Intersection of Any Curved Surface and a Plane. In general, the intersection of a surface and a plane can be found by drawing on the given surface any series of curves which are known to lie in the surface, and then finding the points in which these curved lines intersect the given plane. If the curves which can be drawn in the surface are plane curves, the planes of these curves may be used as auxiliary planes, and the process of finding the intersection is similar to that for a surface of revolution. 176. Method by Means of Secondary Projections. Any plane may be seen edgewise by projecting on a suitable secondary plane of projection (§ 70). A general method of finding the intersection of any surface or solid with a plane is, therefore, to make a secondary projection of the surface and the secant plane so that the plane appears edgewise. This reduces one projection of the required intersection to a straight line. Since this method involves the construction of a new pro- jection of the given surface, as well as of the secant plane, it will be found of doubtful advantage in the case of surfaces consisting of rectilinear elements. The method is, however, a good one with surfaces of revolution, whose projections, from any point of view perpendicular to the axis, are alike. 177. Visibility of the Curve of Intersection. In finding the intersection of a plane with a curved surface or solid, the plane will be considered to be transparent, and in general no portion of the surface or solid will be removed. The various portions of the resulting line of intersection, therefore, will be visible or invisible, according as the}^ lie in a visible or an invisible portion of the surface. (Compare § 85.) 178. A Rectilinear Tangent to the Curve of Intersection. At any point in the intersection of any curved surface and a XXI, § 179] CURVED SURFACES AND PLANES 221 plane, a rectilinear tangent to the intersection may be drawn without knowing or considering the properties of the curve of intersection. For the tangent line must lie in each of two planes : first, in the given secant plane, since every tangent to a plane curve lies in the plane of the curve (§ 153) ; and second, hi the plane tangent to the given surface at the point in question (§ 154). Hence, if the plane tangent to the surface at the given point can be found, the line of intersection of this plane with the given secant plane will be tangent to the curve of intersection at the given point. 179. Development. A curved surface is developable when it can be unrolled into a plane, without extension, compression, or distortion of any kind. Only surfaces which consist of rectilinear elements are developable ; and of these surfaces, only those forms in which every two consecutive elements lie in the same plane, that is, either intersect or are parallel. Cones and cylinders are developable, and are the only forms of developable curved surfaces commonly used in practical work. CHAPTER XXII INTERSECTION OF PLANES WITH CONES AND CYLINDERS 180. The Intersection of a Cone or Cylinder with a Plane. The cone and the cylinder are ruled surfaces. Hence the method of finding points in the intersection is that described in § 173. In the three problems which follow, in addition to finding the curve of intersection we shall draw a tangent line to some point of the intersection (§ 178), and develop the curved sur- face, showing the section and the tangent line. Problem 41 . To find the intersection of a cone and plane. Case I. The base of the cone lies in H or V. A. The Intersection. Analysis. See § 173. Construction (Fig. 303). The base of the given cone lies in H. The given secant plane is Q. Eight elements have been chosen on the cone, distinguished by the numbers 1 to 8 around the base. Find the points in which these elements intersect the plane Q by passing through them auxiliary planes perpen- dicular to H or V (Prob. 13, Usual Method, § 119). The planes here chosen are the planes X, Y . . . X, perpendicular to H. The intersection of planes X and Q is the line A. The intersection of A v and the F-projection of the element 0-3 locates 13", one point of the required intersection. The intersection of planes Z and Q is the line C, the projection C being parallel to VQ. Since plane Z contains two elements of the cone, the construction locates two points, 11 on element 0-1 and 15 on element 0-5. But the planes Y, M, and X are so situated that in attempting to find their lines of intersection with Q, only one point in each line can be found within the limits of the figure, namely, the points b", d", and e" on the ground line. 222 XXII, § 180] PLANE AND CONE 223 B. The Common Point. Since the planes X, T, Z, M, and N are perpendicular to H, it follows that the lines of inter- im [Q, 303 section A, B, C, etc., will have their H"-projections coincident with HX, II Y, IIZ, etc., respectively. But the lines of inter- 224 DESCRIPTIVE GEOMETRY [XXII, § 180 section A, B, C, etc., are all lines in one plane, namely, the plane Q. Now if a number of lines known to be in the same plane appear to intersect in a common point in one view, it can only be because the lines actually intersect in space. Hence in the other view the lines must also pass through a common point. In the figure, the iZ-projection of this point is n h , while n v lies on A". The other F'-projections, B v , D", and E v , may now be drawn through n v and the points b v , d v , and e v on GL, already determined. The intersections of B v , D v , and E v with the elements lying in the corresponding auxiliary planes locate the remaining points in the curve of intersection. C. Visibility of the Curve of Intersection. (See § 177.) In the jff-projection, the point 11, on the highest element, is evidently visible. This determines the visible side of the curve, which will become invisible at the points 13 and 17, which lie on the boundaries of the visible surface of the cone. Similarly, in the F"-projection, the point 17, on the extreme front element, is visible. The visible side of the curve, beginning at 11, passes through 17 and ends at 15. D. A Line Tangent at a Given Point in the Curve of Intersection. Analysis. See § 178. Construction. Let 12, lying on the element 0-2, be the given point. Find the plane S(HS, VS), which is tangent to the cone at the point 12 (Prob. 31, § 162). Find the intersec- tion of S with the given plane Q (Prob. 12, § 118). This line of intersection T (T h , T v ) is the required tangent line. As a check, it should pass through the point 12, and show tangent to the curve of intersection in each projection (§ 152). E. Development of the Curved Surface of the Cone. Analysis. The entire curved surface, between the vertex and the base, should be developed first. Any other line drawn on the surface can then be put in. The conical surface is divided by its elements into pieces bounded by two straight lines and a curve. These pieces are approximately plane triangles, the approximation being closer the nearer the elements are to- gether. The curved surface may then be developed as if the XXII, § 180] PLANE AND CONE 225 cone were a many-sided pyramid, the triangular faces being built up from the true lengths of their sides. (See § 89.) Pro. 303 (repeated). Construction. The given conical Burface is divided info eight triangular pieces by the chosen elements <> L, '_'... 226 DESCRIPTIVE GEOMETRY [XXII, § 180 0-8. The true lengths of the elements may be found by the rule of § 88. The true lengths of 1-2, 2-3, etc. ap- pear directly in the //-projection of the base, the usual approximation being that the chord equals the arc. As- suming the true lengths found as needed, begin the develop- ment by laying off, on any convenient line, O'-l' equal to the true length of 0-1. With 0' and 1' as centers, radii respectively the true lengths of 0-2 and 1-2, strike arcs inter- secting at 2'. Working from 0' and 2', obtain 3' in a similar manner, and so on. Draw a smooth curve through 1', 2' . . . 8', 1". To locate the section 11, 12 . . . 18, 11 in the development, note that 11' may be located by measuring from 0' the true length of 0-11, or from 1' the true length of 1-11. By com- paring Figs. 113 and 114, § 88, it will be seen that in this case the true length of 1-11 is the more easily found. To con- tinue, locate 12' from 2' by using the true length of 2-12, 13' from 3' by the true length of 3-13, and so on. Draw a smooth curve through the points 11', 12', 13' . . . 18', 11". F. The Tangent Line in the Development. Analysis. The tangent line at any point in the section makes a definite angle with the element passing through that point. This angle must appear in its true size in the development. Construction. The line T intersects the element 0-2 at the point 12, and makes with it the angle 2-12-s. Let us con- nect the point 2 on the element 0-2 with the i/-trace, s, of the tangent line T. The line is already drawn, the //-projection, 2 h -s h , being a portion of the //-trace, HS, of the tangent plane at the point 2, while 2"-s v lies in OL. We thus have found a plane triangle, 2-12-s, one of whose angles is the required angle 2-12-s. Points 2 and 12 are already in the develop- ment at 2' and 12', respectively. Find the true lengths of 12-s and 2-s. Using 12' and 2' as centers, strike arcs inter- secting at s'. Then 2'-12'-s' is the true size of the angle 2-12-s. Consequently s'-l2' is the required development, T ', of the tangent line T. As a check, it should be tangent to the curve ll'-12'-13'. Incidentally, since the line s'-2' is the XXII, § 180] PLANE AND CONE 227 development of a line tangent to the base, s'-2' is tangent to the curve l'-2'-3'. Fio. 303 (repeated). It is geometrically possible to layoff the triangle -' 12' «' on the other side of 2'-12'. But by visualization of the ]><>si- 228 DESCRIPTIVE GEOMETRY [XXII, § 180 Fig. 304. XXII, §180] PLANE AND CONE 229 tion of the point s relative to the cone, as shown especially in the .^-projection (s on the side of element 0-2 aivay from 0-1) it will be seen that the position of s' as given in the develop- ment is the only one which represents the point. The remaining cases of this problem differ in details of con- struction, but do not differ in essential principles, from the first case. They will therefore not be taken up so fully, but all important variations will be noted. Case II. The base of the cone lies in P. Construction (Fig. 304). The given cone is one of revolution lying in the third quadrant. The given secant plane is Q. Eight elements have been chosen, equally spaced on the P-pro- jection of the base. The points in which the chosen elements intersect the plane Q are found by means of auxiliary planes perpendicular to H. The plane X, passed through the element 0-3, intersects Q in in the line C, the F-projection C v being parallel to VQ. The F'-projection, n", of the common point is found where 0° inter- sects the projector o h o v . The remaining lines of intersection are drawn through n v and the points a v , b v , d v , e v , on GL. As soon as these lines are drawn, it is seen that some of the points of the intersection lie beyond the limits of the cone. To Find the Points where the Curve of Intersection Leaves the Base. Find the profile trace, PQ, of Q (§ 60). .Since this is a line in the plane of the base, the intersections of PQ with the profile view of the base, namely 19 p and 20 p , are the profile projections of two points in the curve of intersection. The line T is tangent to the intersection at point 14. This line is the intersection of the given plane Q with the plane S, the latter being passed tangent to the cone at the point 14. The development, 0'-l'-2'-3' . . . 8'-l"-0', of the entire curved surface is the sector of a circle, since all the elements are of the same true length. The lengths 1/-2', 2'-3', etc., are equal to 1 P -2 P , 2 P -3 P , etc. Such a point as 13' may be located equally well by making 3-13' equal to the true length of 3-13, or by making 0'-13' equal to the true length of 0-13. 230 DESCRIPTIVE GEOMETRY [XXII, § 180 The point 19' is found by making 2'-19' equal to 2 P -19 P . The point 20' is found by making 6'-20' equal to 6 P -20 P . To develop the tangent line T, connect some point on T with some point on the element 0—4, thus forming a triangle. The points chosen are the "F-trace, t, of T, and the extrem- ity, 4, of the element. The triangle 4-14-£ is then plotted in the development at 4'-14'-£', by using the true length of its sides. Then £'-14' is the development of T, and is tangent to 13'-14'-15'. But since t-A was not tangent to the base, the development t'-A' will not be tangent to 3'-4'-5'. Case III. The base of the cone does not lie in H, V, or P. Construction (Eig. 305). The method of finding points in the intersection does not differ from that previously explained. We may note, however, that the auxiliary planes are taken here perpendicular to V. Consequently the F"-projection n" of the common point coincides with o v , while n h is found on the projector o v o h . The line T, tangent at the point 16, is the intersection of Q with the plane R, tangent at point 16. The development requires the true lengths of the elements, and the true lengths of the segments 1-2, 2-3, etc., of the base. Since neither end of any element lies in a coordinate plane, the true lengths of the elements must be found by the method of Fig. 113, § 88. The true size of the base does not appear in either projec- tion, and must be found before the true lengths of its segments become known. The true size is found here by revolving about the diameter l"-5", which is parallel to V. Then the true size is shown by the points 2 r , 3 r , etc. Therefore, in the development, l'-2' equals l»-2 r , 2'-3' equals 2 -3 r , etc. The developed tangent is obtained from the triangle 6-16-s, the points 6 and s being chosen as convenient points on the element and tangent line, respectively. The development 7" is tangent to 15'-16'-17', but the development s'-6' is not tan- gent to 5'-6'-7'. XXII, §180] PLANE AND CONE 231 Fig. 305. 232 DESCRIPTIVE GEOMETRY [XXII, § 180 Problem 42. To find the intersection of the frustum of a cone and a plane. Analysis. See §§ 173, 177. The general principles involved, are the same as in the preceding problem. Construction (Fig. 306). The given frustum is placed in the first quadrant, with its lower base on H, its upper base parallel to H. The given secant plane is Q. A. To Locate the Elements. Contour elements like 1-11 and 4-14 are readily obtained by noting certain tangent points. The arcs l*-4* and 11*-14* may then be divided in any equal ratios by the points 2 h , 3*,- 12*, 13*, which will determine the elements 2-12, 3-13. Or, we may note that the plane J, per- pendicular to H, which contains the element 5-15, must also contain the element 3-13. The frustum in Fig. 306 has been specially chosen. It is not in general possible to project eight elements with so few lines in each view. B. The Auxiliary Line. Let us attempt to find the points in which the eight chosen elements intersect the plane Q, by passing through them planes perpendicular to H. Take for example the plane X, containing the element 8-18. The intersection of HX and HQ locates one point, a, in the line of intersection of these planes. A second point cannot be located by the use of VX. A similar situation is found for the planes Y, Z, etc. Even if some of the lines of intersection could be found by means of the F"-traces of the planes, it would not assist in finding the other lines of intersection, since the com- mon point of the preceding problem is not available. In this case, had we begun by taking the auxiliary planes perpendicu- lar to V, we would have been confronted by a similar situation. The method given below is a general one, and may be used when finding the intersection of any ruled surface with a plane. Let L (L h , L v ) be any arbitrarily chosen line lying in the plane Q. For convenience, this line is usually taken as one of the principal lines (§ 99) of the plane. Other conditions gov- XXII, §180] PLANE AND CONICAL FRUSTUM 233 x v y z v j- K w c Fig. 30G. 234 DESCRIPTIVE GEOMETRY [XXII, § 180 erning the choice of the line L should be apparent after observing the use which is made of it. The line L intersects the plane X in the point a (see Fig. 178, § 119). But L lies in the plane Q. Hence the point x lies in both the planes X and Q, consequently is a point in their line of intersection, A. The F-projection, A v , must therefore pass through x v in L v , and is determined by x v and the point a v , previously noted in the ground line. Similarly, the line L intersects the plane Y in the point y, which must lie in the line of intersection of T"and Q; and so on. The F-projections of the lines of intersection of the auxiliary planes with Q being thus determined, the points on the curve of intersection are noted as in the preced- ing problem. C. A Line Tangent to the Curve of Intersection. Analysis. See § 178. Construction. The point chosen is 27, lying on the element 7-17. The plane S is passed tangent to the frustum at this point (Prob. 31, § 162). The required tangent line, T, is the line of intersection of S with the given plane Q. D. Development of the Curved Surface. Analysis. The portion of the curved surface between any two adjacent ele- ments is approximately a plane quadrilateral, and is assumed to be such. Any quadrilateral is divisible by either of its diagonals into two plane triangles, the true sizes and shapes of which may be plotted from the true lengths of the sides. The above method of development is a general one, and is known as development by triangulation. For the most accurate results, each quadrilateral should be divided into triangles by using the shorter diagonal. Construction. In any convenient position, lay off V— 11' equal to the true length of 1-11. Triangulate 12' from 1' and 11', using the true lengths of 1-12 and 11-12. The true length of 1-12 may be found with the compasses without actually draw- ing either projection of the line, while the true length of 11-12 is 11 A -12\ Triangulate 2' from 1' and 12', just located, using the true lengths of 1-2 and 2-12. Similarly, triangulate 3' from 2' and 12', then locate 13' from 3' and 12', and so on until XXII, §180] PLANE AND CONICAL FRUSTUM 235 x"y" z v j v K v L v Fig. 306 (repeated). 236 DESCRIPTIVE GEOMETRY [XXII, § 180 the development is complete. Draw the developed elements 2'-12', 3'-13', etc., as fast as they are obtained, and note as a check whether they appear to converge to a point as the ele- ments of a correctly developed cone should do. E. The Development of the Curve of Intersection. The points 21', 22', 23', etc., of the curve of intersection are here found by making 1/-21' equal to the true length of 1-21, 2'-22' equal to the true length of 2-22, and so on. They might also be found by making ll'-21' the true length of 11-21, 12'-22' that of 12-22, etc., but in this case the former set of true lengths are the easier to obtain. Why ? F. The Development of the Tangent Line. Xote the triangle 27-Z-17, one of whose sides lies along the tangent line T, and a second side along the element 7-17 containing the given point 27. Triangulate t' from 17' and 27', using the true lengths of 11-t and 27-t. Then f'-27' is the required development, T', of the tangent line T. Problem 43. To find the intersection of a cylinder and a plane. Analysis. See §§ 173, 177, 178. The same general principles apply as in the preceding two problems. Case I. Tlte base of the cylinder lies in H or V. Construction (Fig. 307). The base of the given cylinder lies in H. The given cutting plane is Q. Eight elements of the cylinder have been chosen. The points in which these elements intersect the plane Q are here found by passing through them auxiliary planes perpendicular to V. The construction is practically a repetition of that of the preceding problems. We should note, however, that since the elements of the surface are parallel, the auxiliary planes through these elements are parallel. Hence the lines of inter- section, A h , B h , etc., are parallel. This gives a method of drawing some of these lines when but one point of the line is available The resulting intersee- tiou is the curve 11-12-13 . . . 18-11, The visibility of the section is obtained as explained in § 177. XXII, §180] PLANE AND CONICAL FRUSTUM 237 x'y" z v j- k v C Fiu. 30t> (repeated). 238 DESCRIPTIVE GEOMETRY [XXII, § 180 The line T ( T k , T v ) is tangent to the section at the point 16. This line is the intersection of the plane S, passed tangent to the cylinder at the point 16 (Prob. 34, § 162) with the given plane Q (§ 178). Development of the Curved Surface Between the Base and Section. First Analysis. The portion of the curved surface bounded by two adjacent elements and the included portions of the base and section is approximately a plane quadrilateral (a trapezoid). The method of development by triangulation used to develop the frustum of the cone, Prob. 42, may therefore be employed. Second Analysis. Obtain a right section of the cylinder, that is, a section made by a plane at right angles to the elements. Find the true size of the right section. The development of the right section is a straight line, whose length equals the cir- cumference of the right section. The developed elements will be lines at right angles to the developed right section. Any desired points on the elements can be located by using the true distances from the right section. In practical applications of the development of cylinders, the right section is usually known, being commonly a circle. Con- sequently development by means of the right section will be the method adopted here. Construction. Any plane at right angles to the elements will cut a right section, and if no other condition prevails the plane may be chosen at random. Such a plane is the plane R. How are HR and VR drawn ? (§ 112.) The intersection of R and the cylinder must now be found. The construction is similar to, and entirely independent of, that used for finding the intersection of plane Q. The result- ing section is 21-22-23 . . . 28-21. Since this is a figure lying in plane R, its true size may be found by revolving about either trace of R, say about HR into H (Prob. 24, Cor. 1, § 142 ; Prob. 21, Working Rule, § 138). The resulting true size is shown at 21 r -22 r - . . . 28 r -21 r . To begin the development, draw a straight line, 2P-21", in XXII, §180] PLANE AND CYLINDER 239 Fio. 307. 240 DESCRIPTIVE GEOMETRY [XXII, § 180 any convenient position. Make 21'-22' equal to 21 r -22 r ; 22'- 23' equal to 22 -23 r ; and so on up to 28'-21" equal to 28 r - 21 r . This straight line is the developed right section. Draw perpendiculars, representing elements, at 21', 22', etc. Make 21'-1' equal to the true length of 21-1 ; 22'-2' equal to the true length of 22-2 ; etc. The curve l'-2'-3' . . . 8'-l" thus obtained is the development of the base 1-2-3 . . . 8-1 of the cylinder. To obtain the development ll'-12'-13' . . . 11" of the sec- tion made by the plane Q, we may make the distance 21'-11' equal the true length of 21-11, 22'-12' the true length of 22-12, etc. Otherwise, we may make l'-ll' equal to the true length of 1-11, 2 / -12 / the true length of 2-12, etc. The latter method is preferable, as there is less chance for error. For example, compare the elements 3'-13' and 4'-14'. If 13' is measured from 3' and 14' from 4', the direction of measurement is the same ; but if 13' is measured from 23' and 14' from 24', the direction must be reversed. A construction for obtaining rapidly all the true lengths of the elements required in making the development is shown in the figure. Let the true length of one element, for example 1-11, be obtained by revolving the line parallel to V, as shown at l^-llj (Prob. 13, First Method, § 78). Then, since all the elements are parallel, the true length of any element can be found by projecting to the line l'-llx, produced if necessary. Thus, the true length of 8-28 equals l"-^!, and the true length of 28-18 equals 28J-18J. The developed tangent line, T", is determined from the tri- angle 6-16-s. The point s' is located by using its true distances, s'-16' and s'-6', from 16' and 6' respectively. Case II. The base of the cylinder lies in P. Construction (Fig. 308). The given cylinder is a cylinder of revolution lying in the third quadrant. It is intersected by the plane Q. Eight elements have been chosen, indicated by the numbers 1 to 8 on the P-projection. Points 12, 13, 14, 15 of the intersec- XXII, §180] PLANE AND CYLINDER 241 Pro. 307 (repeated). 242 DESCRIPTIVE GEOMETRY [XXII, § 180 tion, which lie within the limits of the cylinder, are found, as in the preceding examples, by finding where the chosen ele- ments intersect the plane Q. The plane Q intersects the left-hand base of the cylinder in the points 19 and 20. To find these points directly, as in the corresponding case of the cone, Fig. 304, note first that the pro- file trace of Q on the plane of this base would be determined from the points x and ?/, on HQ and VQ respectively (§ 60). The base which is already projected in profile is the right- hand base. Therefore project x and y to x x and y u and obtain the projected trace, P,Q. The points 19^ and 20 p , in which P X Q intersects the circle, are the profile projections of the re- quired points, which must then be projected back to the left- hand base. The line T is tangent to the section at the point 14. This line is the intersection of the given plane Q with the plane #, which is passed tangent to the cylinder at point 14. The development of the cylinder is obtained readily, since each base of the cylinder is a right section. The line l'-2'- 3' • • • 8'-l" is the development of the right-hand base, the distances l'-2', 2'-3', etc. being equal to 1*2", 2 P 3 D , etc. Re- placing the development of this base in line with its pro- jections, as shown, the points 12', 13', 14', 15' may be obtained by projection. The points 19' and 20' are obtained from the positions of 19* and 20 p . The development, Z", of the tangent line is obtained by locating in the development the triangle 4-14-i. Case III. The base of the cylinder does not lie in II, V or P. Construction (Fig. 309). The given cylinder lies in the third quadrant, and is intersected by the plane Q in the curve 11- 12-13 • • • 18-11. The construction involves nothing that has not been already explained. The line T, tangent at the point 12 of the curve of intersec- tion, is the line of intersection of the given plane Q with the plane S, passed tangent to the cylinder at point 12. The development is obtained by the use of a right section. This is the section 21-22-23 • • • 28-21, cut by the plane R, per- XXII, §180] PLANE AND CYLINDER 243 Fig. 308. 244 DESCRIPTIVE GEOMETRY [XXII, § 180 pendicular to the elements, but otherwise arbitrarily chosen. The true size of the right section, obtained by revolving about VR into V, appears at 21 r -22 r • • • 28 r -21 r , and its rectification or development at 21'-22' . • • 28'-21". The true lengths of the elements are obtained by projecting them all on to the true length of 5-15, obtained at 5 V -15 X by using the method of Prob. 3, § 80. The development of the tangent line is obtained by plotting the triangle 2-12-c, the point c being any arbitrary point on the line T. In determining the triangle which shall be used to develop the tangent line, one side must necessarily be a portion of the tangent line itself, and a second side must be a portion of the element. The third side of the triangle may be a line connecting any convenient points on each of the other two lines. Heretofore, the point chosen on the tangent line has always been one of its traces. This is not necessary, however, and in this case a better shaped triangle and con- sequently more accurate result is obtained by taking a dif- ferent point. 181. The Intersection of a Pyramid or Prism with a Plane. Although the pyramid and prism are bounded by plane in- stead of curved surfaces, the intersection of either of these solids with a general oblique plane can often be found most advantageously by the method of the preceding Article. Con- sequently the following problem may logically be introduced at this time. Problem 44. To find the intersection of a pyramid or prism and a plane. Analysis. Let a pyramid be given. Consider the lateral edges as the elements of a cone. Find the points in which these elements intersect the given plane, using the method of Prob. 41 or 42, § 180. Connect the points thus found by straight lines. Let a prism be given. Consider the parallel lateral edges as the elements of a cylinder, and use the method of Prob. 43, § 180. Connect the points found by straight lines. XXII, §181] PLANE AND CYLINDER 245 Fig. 309. 246 DESCRIPTIVE GEOMETRY [XXII, § 181 Construction (Fig. 310). The given solid is a pyramid. The points in which the lateral edges 0-1, 0-2, 0-3, 0^4 inter- sect the given plane Q are found by auxiliary planes X, Y, Z, Fig. 310. W perpendicular to H. The plane X (HX, VX) locates the common point n", after which the F'-traces of the remaining planes are not needed. The required intersection is the quadrilateral 11-12-13-14. CHAPTER XXIII INTERSECTION OF PLANES WITH SURFACES OF REVOLUTION 182. The Intersection of a Surface of Revolution and a Plane. The general method of obtaining points in the intersection is by means of auxiliary planes perpendicular to the axis of the surface, as has already been explained in § 174. Besides the points thus found, other points, lying on particular planes that contain the axis of the surface, are usually essential. The method outlined above applies to any surface of revolu- tion, not exclusively to double curved surfaces. We shall, however, apply it in the present chapter only to such surfaces. With certain double curved surfaces of revolution, noticeably the torus, the use of a secondary projection (§ 176) is often advantageous. This method does not affect the general prin- ciples, but makes these principles easier to apply. Problem 45. To find the intersection of a double curved surface of revolution and a plane. Analysis. See §§ 174, 176, and above. First Construction (without the use of a secondary projec- tion) (Fig. 311). The given surface is that of a circular spindle, formed by the revolution of the arc of a circle about its chord. The axis of the surface is placed perpendicular to //. An auxiliary plane perpendicular to the axis A, such as X(VX; there is no HX) intersects the given surface in the circle C, a parallel of the surface (§ 166), and the given plane Q in the line Ii (Prob. 12, Special Case III, § 118). The in- tersection of B h and C h determines the //-projections of two points in the required intersection. Other points are similarly obtained by other auxiliary planes perpendicular to the axis A. In the figure are shown the planes A',, symmetrica] with A", which gives the points 3 and I, and the plane Z of the greatest parallel E, winch locates the points 5 and 6. 247 248 DESCRIPTIVE GEOMETRY [XXIII, § 182 A. Visibility of the Curve of Intersection. In the i/-projection is seen all that part of the surface which lies above the plane Z of the greatest parallel. The points which divide the visible from the invisible portion of the curve are 5* and 6 A , already determined as in this plane. In the F-projection appears the half of the surface in front of the principal meridian plane T (HT, there is no VY; § 166). The points of division between the visible and invisible parts of the F-projection of the intersection lie in the plane Y, and should be obtained by the direct use of this plane as an auxiliary. B. Points Determined by Meridian Planes. The plane Y intersects the given surface in the meridian section which forms the outline of the P"-projection. The plane Y intersects the given plane Q in the line J, whose F-projection, J v , inter- sects the meridian in the points 7" and 8". These are points in the curve of intersection, and, as previously Doted, are the points of division between the visible and invisible portion of the curve in this projection. An inspection of the ZT-projection shows that the intersec- tion is symmetric with respect to the meridian plane M (HM, VM), which is perpendicular to Q. For this reason, the plane M is called the meridian plane of symmetry. The points of the curve which lie in M are important, and should be found by using M as an auxiliary. In this case they are the highest and lowest points of the curve. In any case they are points of maxima or minima, at which the tangent to the curve lies in a plane perpendicular to the axis of the surface. The plane M intersects the given surface in a meridian (§ 165), the counterpart of that appearing as the F-projection of the surface. This meridian projects on H as a portion of HM\ it is not projected on V, as the need of this projection can be avoided. The plane M intersects the given plane Q in the line K (K h , K v ) (Prob. 12, § 118). Since the plane M contains the axis, A, of the surface, the lines A and A" inter- sect in a point, o, whose F-projection is at o v . Note that J", previously determined as lying in the principal meridian plane XXIII, § 182] PLANE AND CIRCULAR SPINDLE 249 Y, must also pass through o", the point o being in fact the intersection of the axis A with the given plane Q. Revolve the plane M about the axis A until M coincides with the principal meridian plane Y. The line K, lying in M, will take the position K r (K r h , A"/). The revolution is here -^^V6> 0> hVi iL-r V >^^« A ^r/7 Sr J h HX/ i J^l ' 't J* 11 1 1 i i h \ j J \r ' sV ^ 1 1 1 1 1 , 11 S/\ 1, / 1 Ar/| 1 w J / 7 1 / ; y/ & ii 'i 1/ s_ / V / 8 VX i|/l/ 1 17 /ii/i i of! :'\ A i li \ F v VZ [p V| /// E v >6"1 111 '' ' / // / / 1 / vx, \ 1 / / i / v / a\^ '*>" j Fir,. 311. effected by using the //-trace, s, of A' and the fixed point, o, on the axis A. The meridian lying in M will coincide with the principal meridian, and therefore projecl Oil V as the contour of the surface. The intersection of A'/ with this con- tour at 9/ and 10/ determines the revolved positions of two points in the curve of intersection. The actual projections, 9", 10", 9 A , and 10 A , result from counter-revolving the line A". 250 DESCRIPTIVE GEOMETRY [XXIII, § 182 Second Construction (using a secondary projection) (Fig. 312). The given torus is to be intersected by the plane Q. To obtain an edge view of Q, assume a secondary ground line, Fig. 312. GiLi, perpendicular to HQ in any convenient location ; then proceed as in § 70. The secondary projection of the given torus is identical in form with the original P"-projection. Considering now the //-projection and the secondary V- projection as the given projections of the torus, and V\Q as XXIII, § 182] PLANE AND TORUS 251 the edge view of the secant plane, we have the same cor> ditions as in §§ 84, 85, and 86 ; in particular, a case similar to that shown in Fig. 109. Having obtained the points 1 to 10 in the //"-projection by the construction there given, the same points are projected to the original ^-projection (§§ 68, 72). A. Points Determined by Meridian Planes. Points on the meridian plane of symmetry M are needed to complete the //-projection. These and the points on the principal meridian plane Fare needed to complete the ^-projection. The meridian of the torus which lies in the plane of sym- metry M appears in true size in the secondary F-projection, it being the two circles which form a part of this projection. Since ViQ is an edge view, the points required appear directly at llj" and 12^, where \\Q intersects one of these circles. The F-projections, 11" and 12°, are found by transferring heights from the secondary projection. The points in the principal meridian plane Y are found as explained in the preceding case. The plane Y intersects Q in the line J, the V-projection J" being parallel to VQ. The meridian projects as the two circles which are a part of the Fprojection. The possible intersections, 13" and 14", of J v with these circles determine the required points. At 13" and 14" the projection of the curve of intersection is tangent to the meridian circle. B. Visibility ok the Curve ok [intersection. Tn the //-projection can be seen the upper half of the surface, lying above the plane of the greatest and least parallels. The points visible are 1, 2, 3, 4, 5, 6, 7, 8, 12. In the F-projection, only the front half of the outer portion of the surface can be seen. This visible portion is projected in the plan as the part in front of II Y and outside of circle D. A point like 6, for ex- ample, although in front of the principal meridian plane T, is nevertheless invisible in ^-projection, being hidden In a portion of the, surface still further in front. Beginning at the left, the visible portion of the F-projection begins at point 13, lying in the principal meridian plane P, extends through 9, 11, 10, 8, 1, and ends at point 2 in the highest parallel, J). 252 DESCRIPTIVE GEOMETRY [XXIII, § 182 C. A Line Tangent to the Intersection. A tangent line is drawn at the point 10. This line is the intersection of the given plane Q with the plane S, tangent to the surface at point 10 (§ 178). Find the tangent plane S by any conven- ient method (Prob. 38, § 170). Here only the H-tra,ce, HS, is found. The intersection of HS and II Q locates one point, .s, of the tangent line T. A second point is the given point 10 of the intersection, which is here used instead of finding the F-trace, VS, of the tangent plane. Fig. 313. 183. Points in Meridian Planes. Attention has already been called, in the preceding constructions, to certain properties of XXIII, § 184] PLANE AND TORUS 253 the points which lie in the principal meridian plane and the meridian plane of symmetry. These points are of sufficient importance to warrant a further discussion. In an actual con- struction, instead of finding these points last, as was done in explaining the preceding problem, it is better to find them at the outset. This is shown in Fig. 313, in which are found Fig. 314. only the points lying in the meridian plane of symmetry 3f, and the principal meridian plane Y, the construction being the same as that used in Fig. 312. The construction is repeated in Fig. 314 for the same planes M said Y. In this case we find that there are no points in the meridian plane of symmetry M. 184. The Number of Curves. The intersection of a plane and a torus always consists of either one or two closed curves. The number can be foretold as soon as the edge view of the secant plane is obtained. Thus, in Fig. 312, the edge view 254 DESCRIPTIVE GEOMETRY [XXIII, § 184 V X Q of the secant plane intersects but one of the circles lying in the meridian plane of symmetry M. By visualizing the position of the cutting plane with respect to the torus, it can be seen that the intersection will consist of but one curve, with the point 11 as its lowest point. In Fig. 313, the edge view ViQ intersects both circles lying in the meridian plane of symmetry M, giving four points, 1, Fig. 314 (repeated). 2, 3, and 4. Here, again by visualization, the intersection is seen to consist of two distinct curves, one wholly inside the other. The point 1 is the highest, and the point 4 the lowest point of the intersection ; no auxiliary planes need be taken beyond these limits. It may be noted in passing that these four points lie on the line of intersection, K, of the planes M and Q, and that their F-projections are here determined as lying on K v . (Compare Fig. 311.) XXIII, § 184] PLANE AXD TORUS 255 In Fig. 314, the edge view ViQ does not intersect either circle lying in the meridian plane of symmetry M, and, as previously noted, there are no points in this plane. Yet the plane Q obviously intersects the torus. Visualization here reveals the fact that the intersection in this case consists of two separate curves, symmetrically placed with respect to the plane M. It occasionally happens that the edge view of the given secant plane is found to be tangent to one or both of the circles lying in the meridian plane of symmetry. In this event, the plane is actually tangent to the torus, yet at the same time intersects the surface in a curve. A point of tangency of the secant plane gives a double point in the curve of intersection ; that is, a point in which Fig. 315. Fig. 316. Fig. 317. the curve crosses itself at an angle. According to whether the plane does or does not intersect the second circle lying in the meridian plane of symmetry, the resulting projection resembles Fig. 315 or 316, respectively. Each of these forms is considered a single curve. If the given secant plane is tangent to the torus at two points, there are two double points in the intersection (Fig. 317). In this case it can be proved analytically that the actual intersection consists of two equal circles. Summarizing, it may be stated that it' the edge view of the secant plane is found in the same relation to both meridian circles of the torus, that is, secant to both, or tangenl to both, or passing between the two without touching either, there will always be two curves. Any other position of the secant plane means one curve. CHAPTER XXIV THE INTERSECTION OF CURVED SURFACES BY CURVED SURFACES 185. The Intersection of Two Curved Surfaces. The inter- section of two curved surfaces is, in general, a space curve, or curve of three dimensions (§ 150). Points in this curve may be found as follows. Let Si and S 2 be the intersecting surfaces. Let X be any auxiliary surface, intersecting Si in a line C u and S 2 i u a line C 2 . Then the point or points of inter- section of Ci and C 2 lie in the line of intersection of Si and So- In order that the application of this method shall be a success, it is evident that the intersections of X with S x and S 2 must be simpler or more easily found than the intersection of the given surfaces S x and S 2 . 186. The Auxiliary Surfaces. Whenever possible, the auxil- iary surfaces employed are planes, so chosen as to intersect both the given surfaces in either straight lines or circles. This cannot always be accomplished, even when straight lines or circles can be drawn on each of the given surfaces. Spheres are employed occasionally as auxiliary surfaces. Their use is generally limited to cases in which they can be so placed as to intersect each of the given surfaces in circles. When no auxiliary surfaces can be found which will cut both given surfaces simultaneously in simple intersections, it is usual to employ planes. These are placed as advantageously as possible with respect to one of the given surfaces, but their intersections with the other given surface must needs be found by means of secondary auxiliary surfaces. 187. Visibility of the Curve of Intersection. When two sur- faces or solids intersect, both will be considered opaque, and no portion of either removed unless so stated. (Compare 256 XXIV, § 188] INTERSECTION OF CURVED SURFACES 257 §§ 85, 125 et seq.) In any projection, then, a point in the curve of intersection can be visible only when it lies on a visible portion of each of the given surfaces. Or, to put it another way, a point is invisible if it is on an invisible portion of either of the given surfaces. In any projection, let the visible portion of the curve of intersection be considered as traced by a moving point. Un- less the entire intersection is visible, which is possible but rarely happens, the moving point will soon reach a place where it can go no further on the curve, an invisible portion of the curve having been reached. This change must evi- dently take place on the contour or outline of the surface. Hence from this point on, the contour of the surface must be visible, and may be considered as the continuation of the path of the moving point. This is a matter often overlooked by students in determining the combined visibility of a pair of intersecting solids. 188. A Line Tangent to the Curve of Intersection. In general, a line tangent at any point in the intersection of two curved surfaces may be determined by finding the line of intersection of two planes. For the chosen point lies at the same time in each of the given surfaces, to each of which the line tangent to the curve is tangent. Then a plane passed tangent to either surface at the given point must contain the line in question, since a tangent plane contains all the lines tangent to the sur- face at that point (§ 154). Hence we may state the following rule : 1. Pass a plane tangent to each of the given surfaces at the given j»>ii//. 2. Find the line of intersection of these t'">> i>}. The second intersection does not exist, since the single element through point .s- does not intersect the cylinder. Hence the intersection consists of a single closed curve. Now let the cone and cylinder !>e placed as in Fig. .">-7, which differs from Fig. 320 in that the plane M, tangent to the cone along the parallel clement, lias now become <>ne of the auxiliary planes. The single intersection of the preceding 276 DESCRIPTIVE GEOMETRY [XXV, § 195 case now contains two infinite points, numbers 3 and 7. The result is that the curve of intersection, as traced on the sur- face of the cylinder, consists of two infinite branches, to which the elements 3 and 7 are asymptotic. These two branches of the intersection lie on opposite nappes of the cone. The preceding two examples do not exhaust all the possi- bilities of this case. They are, however, the most general ones, and, taken together, give a typical illustration of the formation of infinite branches. C. Two Cones. In the case of two cones, to discover the number of parallel elements, draw a third cone, whose vertex coincides with the vertex of one of the given cones, and whose elements are parallel respectively to the elements of the other. Then, corresponding to the number of possible points which two ellipses may have in common, we may find one, two, three, or four pairs of parallel elements, or every element of one cone may be parallel to an element of the other cone. The latter case is that already illustrated in Fig. 325, the intersection being a hyperbola. The possibilities of the other cases are quite numerous. As an extreme result, each curve of a normal two-curve intersection may be broken into two infinite parts, so that the complete intersection consists of four infinite branches, two branches on each nappe of each cone. Since the method of dealing with infinite points has already been shown for the case of the cylinder and cone, a further discussion will not be made. XXV, § 196] CYLINDERS AND CONES 277 196. The Intersection of Cylinders and Cones of Revolution. If we limit ourselves to cylinders and cones of revolution, usu- ally the case in practice, simpler solutions than those given in § 192 can often be found. It may be of advantage to take auxiliary planes so as to intersect one or both of the given surfaces in circles. For example, consider the intersection of a cylinder and cone, the axes of which are at right angles (not necessarily intersecting). Auxiliary planes can be passed per- pendicular to the axis of the cone, cutting the cone in circles, and the cylinder in straight lines (elements). If the axes of the surfaces intersect at some oblique angle, auxiliary spheres may be used. (See Prob. 49, Ex. 2, § 197.) A method available when one, at least, of the given surfaces is a cylinder, is to make a projection on a plane at right angles to the axis of the cylinder. The entire curved surface of the cylinder will then project as a circle, a part or all of which necessarily becomes one projection of the required intersec- tion of the given surfaces. CHAPTER XXVI THE INTERSECTION OF VARIOUS CURVED SURFACES 197. The Intersection of Two Curved Surfaces of Revolution. The intersection of two surfaces of revolution can be readily found when the axes of the surfaces are in the same plane, that is, either intersect or are parallel. Problem 49. To find the intersection of two surfaces of revolu- tion whose axes intersect. Analysis. With the point in which the axes of the surfaces intersect as center, describe a series of auxiliary spheres (§ 186). These spheres will intersect both given surfaces in circles. The intersections of the corresponding circles locate points in the required intersection. In order to get simple working projections of the auxiliary circles, one coordinate plane should be parallel to the two axes of the surfaces, the other coordinate plane perpendicular to one of the axes. (See the Construction.) Example 1. Construction (Fig. 328). The given surfaces are those of a cone and an ellipsoid of revolution, placed in the third quadrant. The plane Y (HY) which contains the axes of both surfaces is parallel to V. This plane is evidently a plane of symmetry of the required intersection, and contains two points, noted directly in the F"-projection at 1" and 2". The axes of the given surfaces intersect in the point o (o h , o v ). With o v as center, and any assumed radius, draw the outline, S", of an auxiliary sphere. This sphere intersects the cone in two circles, A and B, and the ellipsoid in the circle C, all projected in V as straight lines. The intersections of A" and B v with C v determine the F-projections of four points, 3, 4, 5, and 6, of the required intersection. To obtain the i7-projec- 278 XXVI, § 197] CONE AXD ELLIPSOID 279 tions of these points, note that the circle C projects as a circle, C h , in H. Find by projection 3*, 4*, o k , and 6 A in C. Other points are similarly obtained by the use of other Fig. 328. auxiliary spheres. Tin- only other sphere shown in the figure is T (T<), tangent to the cone. This sphere is tangenl to the cone in the circle D, ami intersects the ellipsoid in the circle E. The intersections of I) and E are the points 7 and 8, the lowest points of the curve of intersection. 280 DESCRIPTIVE GEOMETRY [XXVI, §197 Example 2. Construction (Fig. 329). The given surfaces are those of a frustum of a cone and a cylinder of revolution, placed in the third quadrant with their axes parallel to V. The common meridian plane Y of the two surfaces contains the two points 1 and 2. Other points are found by the use of auxiliary spheres whose centers are at the point of intersection o, of the axes. One sphere, S (S v ), is shown in the figure. This cuts the circle A from the cylinder, and the circle Cfrom the cone. The intersections of these circles are points 3 and 4 on the intersection. Other points are similarly obtained. Corollary. To Jind the intersection of two surfaces of revo- lution whose axes are parallel. Analysis. Pass auxiliary planes perpendicular to the axes of the surfaces. These planes will cut circles from each sur- face. The intersections of these circles determine points in the required intersection. In order to get simple working projections of the circles, the axes of the surface should be perpendicular to one of the coordinate planes. No figure for this case is deemed necessary. 198. The Intersection of a Sphere with Another Surface. When one of two intersecting surfaces is a sphere, the auxiliary sur- faces commonly employed are planes, since every plane section of a sphere is a circle. Simple projections of these circles, however, result only when the auxiliary planes are parallel to Hoy V. Cases of this kind have already been discussed (§§ 129, 131). But planes parallel to H ov F'may not always cut ad- vantageous sections from the second surface. In such cases, the planes are chosen with respect to the second surface, and various devices are adopted to avoid pro- jecting the sections of the sphere as ellipses. Two general methods are (a) revolution of the auxiliary planes ; (b) the use of a secondary projection. These methods will be shown in the following two problems. XXVI, § 198] CONICAL FRUSTUM AND CYLINDER 281 Fiu. 329. 282 DESCRIPTIVE GEOMETRY [XXVI, § 198 Problem 50. To find the intersection of a sphere and a cone. Analysis. Planes passed through the vertex of the cone will cut elements from the cone (§ 190), and circles from the sphere. Let these planes be taken perpendicular to one of the coordinate planes, say to H. Revolve each plane until parallel to V, carry- ing with it the elements cut from the cone, and the circle cut from the sphere. The circle will now appear in true shape and size. Xote the points of intersection of the revolved circle and ele- ments. Obtain their projections by counter-revolution. A convenient axis about which to revolve the planes is a line perpendicular to // through the vertex of the cone, since all the planes will contain this line. Construction (Fig. 330). The given surfaces are those of a cone whose vertex is o, and a sphere whose center is e. Let auxiliary planes be passed perpendicular to H through the vertex of the cone. Of these planes, the plane W has been selected as typical, and all the others omitted from the figure for the sake of clearness. The plane W appears in edge view at HW; the trace VW is not drawn, since it is understood that the plane is perpendicular to H. The plane intersects the cone in the elements 0-1 and 0-2. It intersects the sphere in a circle, of which 3 A -4 A is a diameter. The center of this circle projects at o h , the middle point of 3 A -4\ Let the plane W be revolved about an axis passing through o until it coin- cides with the plane Y, parallel to V. The elements 0-1 and 0-2 will now appear in ^projection as 0"-l r and 0"-2 r . The center, 5, of the circle is found at 5 r . Xote that, although we have not the ^projection, 5", of point 5, it is not necessary, since point 5 and the center, e, of the sphere are at the same distance above H. With 5 r as center, radius equal to 5 h -'S h or 5 h —i h , draw the circle C T , which represents the circle cut from the sphere by the plane W. In this case both elements inter- sect the circle. This gives four intersections, 6 r , 7 r , 8 r , 9 r , from which the projections of four points in the required inter- section are obtained by counter-revolution. Enough other points to determine the intersection are similarly obtained. XXVI, §198] SPHERE* AND CONE 283 *"ju. 330. 284 DESCRIPTIVE GEOMETRY [XXVI, § 198 Problem 51. To find the intersection of a sphere and a cylinder. Analysis. To cut elements from the cylinder, auxiliary planes must be taken parallel to its axis (§ 190). Let these planes be taken perpendicular to H. Assume a secondary plane of projection parallel to the axis of the cylinder. This plane will be parallel to all the auxiliary planes, so that on it all the circles cut from the sphere by the auxiliary planes will appear in true shape and size. Hence project the circle and elements lying in each auxiliary plane to the secondary plane of projection. Note the points of intersection. Project these points back to H and V. Construction (Fig. 331). The given surfaces are those of a sphere whose center is o, and a cylinder whose axis is A. Assume a secondary ground line G r ] L l parallel to the axis of the cylinder. The center of the sphere projects to o x v . To obtain the direction of the elements of the cylinder in the secondary projection, we may project the axis, as shown at A{°. Pass auxiliary planes perpendicular to II parallel to the axis A. The //-trace and edge view of one such plane is shown at II W; all the others are omitted in the figure for the sake of clearness. Plane W intersects the cylinder in two elements, E and F, and the sphere in a circle C. Find the secondary pro- jection of the elements and circle. The elements project as E{ and Ff, parallel to A L V . The circle projects as Ci", the center coinciding with of, while the diameter is obtained from the //-projection. We thus obtain four intersections, l x v , 2^, 3^, 4^". These are the projections of four points in the re- quired intersection from which the H- and F-projections may be found. A sufficient number of points to determine the inter- section may be found in a similar manner. In this example the intersection consists of two separate curves. 199. The Intersection of any Two Curved Surfaces. As has been mentioned in § 180, it is not always possible to find a series of auxiliary surfaces which will cut simultaneous simple sections from each of two given surfaces, even when it may be easily possible to cut simple sections from each surface when taken XXVI, § 199] SPHERE AND CYLINDER 285 separately. Such a case, for example, is that of two surfaces of revolution whose axes are not in the same plane, and which, from the nature of the surfaces, does not fall under any of the Fig. 331. cases already discussed. The intersection of two such surfaces, therefore, may well be taken as illustrative of the general method of finding the intersection of any two curved surfaces. 286 DESCRIPTIVE GEOMETRY [XXVI, § 199 Problem 52. To find the intersection of any two curved surfaces. (General Case.) Analysis. It is assumed that neither planes, spheres, nor other auxiliary surfaces can be found which will intersect simultaneously both of the given surfaces in straight lines or circles ; or at least that a sufficient number of points to deter- mine the required intersection cannot be found in this way. The solution is then effected by means of auxiliary planes (§ 186). Pass each plane so as to intersect, if possible, one of the given surfaces in straight lines or circles. Find the inter- section of this plane with the second surface, using secondary auxiliary planes for the purpose. Xote the points of inter- section of the two sections ; they are points on the required intersection of the given surfaces. In extreme cases, it may be necessary to find each section which the auxiliary plane cuts from the given surfaces by secondary auxiliary planes. Construction (Fig. 332). The given surfaces are those of a cone of revolution whose vertex is a, and a torus whose center is o. An auxiliary plane M (VM) can be passed through the center of the torus parallel to H, which will cut two circles from the torus, and one circle from the cone. "We note in the .ff-projection the intersection of E h , the circle lying in the cone, with the circles of the torus, and obtain points 1 and 2 of the required intersection. A plane N (HN) can be passed through the vertex, a, of the cone parallel to V, which will intersect the cone in two straight lines, the contour elements of the V- projection, and the torus in two circles. The P~-projection gives us six intersections, thus determining points 3, 4, 5, 6, 7, and 8. The points already found not being sufficient, and no other simultaneous straight line or circle intersections being possible, recourse must be had to the general method. This is illustrated by the plane R (HR). This plane is taken parallel to V, and intersects the torus in two circles, projected in Fat C" and D v . Plane R intersects the cone in a hyperbolic arc projected at F v . The curve F is found by using the secondary aux- iliary planes W, X, Y, and Z, parallel to H. (See Fig. 107, XXVI, § 199] CONE AND TORUS 287 § 86.) The intersections of F" with C v and D v locate four points, 9, 10, 11, and 12, of the intersection of the surfaces. I-'Ki. 332. Other points to determine this intersection completely may be similarly obtained. ' I 'HE following pages contain advertisements of a few of the Macmillan books on kindred subjects Electric and Magnetic Measurements BY CHARLES M. SMITH Associate Professor of Physics in Purdue University i2mo, 368 pp., §->..fo This work is the development of a course of lectures and laboratory notes which students have used in mimeo- graph form for several years. The equivalent of one year of general physics and some knowledge of the calculus are presupposed, hence the arrangement and treatment are not always those which would be followed with a class of beginners. Most of the laboratory exercises are de- scribed in such a way that particular types of apparatus are not demanded unless these are well known and gener- ally available. The methods described are for the most part standard ones, with such variations as many years of trial and sug- gestions from various instructors associated with the course have shown to be of value. THE MACMILLAN COMPANY Publishers 64-66 Fifth Avenue New York ANALYTIC GEOMETRY BY ALEXANDER ZIWET Professor of Mathematics in the University of Michigan And LOUIS ALLEN HOPKINS Instructor in Mathematics in the University of Michigan Edited by Earle Raymond Hedrick Flexible cloth. III., i2mo, viii + 369pp., $ 160 Combines with analytic geometry a number of topics, tradi- tionally treated in college algebra, that depend upon or are closely associated with geometric representation. 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THE MACMILLAN COMPANY Publishers 64-66 Fifth Avenue New York TRIGONOMETRY BY ALFRED MONROE KENYON Professor of Mathematics, Purdue University And LOUIS INGOLD Assistant Professor of Mathematics, the University of Missouri Edited by Earle Raymond Hedrick Trigonometry, flexible cloth, pocket size, long 12m o {xi-\- 132 pp.) with Complete Tables (xviii + 12 f pp. ) , $1.50 Trigonometry {xi-\- ij2 pp.) with Brief Tables (xviii -\- 12 pp.), $1.20 Macmillan Logarithmic and Trigonometric Tables, flexible cloth, pocket size, long i2>no (xviii + 12 j pp.) , $60 FROM THE PREFACE The book contains a minimum of purely theoretical matter. Its entire organization is intended to give a clear view of the meaning and the imme- diate usefulness of Trigonometry. The proofs, however, are in a form that will not require essential revision in the courses that follow. . . . The number of exercises is very large, and the traditional monotony is broken by illustrations from a variety of topics. 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THE MACMTLLAN COMPANY Publishers 64-66 Fifth Avenue New York ELEMENTARY MATHEMATICAL ANALYSIS BY JOHN WESLEY YOUNG Professor of Mathematics in Dartmouth College And FRANK MILLET MORGAN Assistant Professor of Mathematics in Dartmoi i Edited by Earle Raymond Hedrick, Professor of Mathematics in the University of Missouri Cloth, i2nio, J42 pp. A textbook for the freshman year in colleges, universities, and technical schools, giving a unified treatment of the essentials of trigonometry, college algebra, and analytic geometry, and intro ducing the student to the fundamental conceptions of calculus. The various subjects are unified by the great centralizing theme of functionality so that each subject, without losing its fundamental character, is shown clearly in its relationship to the others, and to mathematics as a whole. More emphasis is placed on insight and understanding of fundamental conceptions and modes of thoughl : less emphasis on algebraic technique and facility of manipulation. 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The same general tendency has led to the treatment of topics with a view toward bringing out their essential usefulness. Rigorous forms of demonstra- tion are not insisted upon, especially where the precisely rigorous proofs would be beyond the present grasp of the student. Rather the stress is laid upon the student's certain comprehension of that which is done, and his con- viction that the results obtained are both reasonable and useful. At the same time, an effort has been made to avoid those grosser errors and actual misstatements of fact which have often offended the teacher in texts otherwise attractive and teachable. Purely destructive criticism and abandonment of coherent arrangement are just as dangerous as ultra-conservatism. This book attempts to preserve the essential features of the Calculus, to give the student a thorough training in mathematical reasoning, to create in him a sure mathematical imagination, and to meet fairly the reasonable demand for enlivening and enriching the subject through applications at the expense of purely formal work that con- tains no essential principle. THE MACMILLAN COMPANY Publishers " 64-66 Fifth Avenue New Tork "01 THE LIBRARY UNIVERSITY OF CALIFORNIA Santa Barbara THIS BOOK IS DUE ON THE LAST DATE STAMPED BELOW. NOV 2^1886 ftrrn z m Series 9482 uc SOI ITHERN REGIONAL LIBRARY FACILITY AA 000 942 838 4