A
 
 Course in Descriptive Geometry and Stereotomy, 
 
 BY S. EDWARD WARREN, C.E., 
 
 PROFESSOR OP DESCRIPTIVE GEOMETRY, ETC., IN THE MASS. INSTITUTE OF 
 
 TECHNOLOGY, BOSTON. FORMERLY IN THE RENSSELAER 
 
 POLYTECHNIC INSTITUTE, TROY. 
 
 The following works, published successively since 1860, have been well received by all the sci- 
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 paration of his works which entitle them to a favorable consideration. 
 
 I. ELEMENTARY WORKS. 
 
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 tion of Teachers, Artisans, Builders, etc. 
 
 1. ELEMENTARY FREE- 
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 4. DRAFTING INSTRU- 
 
 MENTS AND OPERATIONS. 
 
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 Object* of Two Dimensions (Pavements, 
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 mentary Esthetics of Geometrical Draw- 
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 5. ELEMENTARY PROJEC- 
 TION DRAWING. Third edition, 
 revised and enlarged. In five divisions. 
 I. Projections of Solids and Intersections.* 
 II. Wood, Stone, and Metal details. 
 III. Elementary Shadows and Shading. 
 IV. Isometrical and Cabinet Projections 
 (Mechanical Perspective). V. Element- 
 ary Structures. This and the last volume 
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 mended for the use of all higher Public 
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 . ELEMENTARY LINEAR 
 PERSPECTIVE OF FORMS 
 AND SHADOWS. With many 
 Practical examples. This volume In com- 
 plete in itself, and differs from many other 
 elementary works in clearly demonstrat- 
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 rules of perspective are based, without 
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 Scientific Course, preparatory to the fully Professional Study of Engineering, etc. 
 
 III.-HIGHER LINEAR 
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 I.-DEXCRIPTIVE 
 
 KTK Y. Adapted to Colleges nn>l pur- 
 poses of liberal education, as well as Tech- 
 nical Schools and Technical Education. 
 Purt I. Surfaces of Revolution. The 
 Point, Line, and Plane, Developable Sur- 
 faces, Cylinders and Cones and the Conic 
 Sections, Warped Surfaces, the Hyperbo- 
 lold, Doable-Curved Surfaces, the Sphere, 
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 i.r:.u* or MI\I>I:S AND 
 
 Ml I DOWN. liM'liKlinirawHcrimK'' 
 of prolili-m , .in. I a thorough discussion of 
 the principle* of shading. One vol. 8vo. 
 With numeroui plate*. Cloth. 860 
 
 V.-8TERKOTOMY. Problems 
 In Stone-cutting. With cuts and 10 fold- 
 ing plates. 8vo, cloth , 2 60
 
 GENERAL PROBLEMS 
 
 SHADES AND SHADOWS. 
 
 FORMED BOTH BY PARALLEL AND BY RADIAL RAYS ; AND 
 
 SHOWN BOTH IN COMMON AND IN ISOMETRICAL 
 
 PROJECTION: TOGETHER WITH THE 
 
 THEORY OF SHADING. 
 
 S. EDWAED WARREN, C.E., 
 
 PBOF. OP DESCRIPTIVE GEOMETRY, ETC., IN THE RENSSELAER POLYTECHNIC INSTITUTE; AND AUTHOP 
 
 OP "ELEMENTARY PLANE PROBLEMS;" "DRAFTING INSTRUMENTS, ETC.;" "ELEMENTARY 
 
 PROJECTION DRAWING;" "ELEMENTARY LINEAR PERSPECTIVE;" 
 
 AND " DESCRIPTIVE GEOMETRY." 
 
 NEW YORK: 
 JOHN WILEY & SON, 
 
 15 ASTOR PLACE. 
 1875.
 
 Entered according to Act of Congress, in the year 186T, by 
 JOHN WILEY & SON, 
 
 In the Clerk's Office of the District Court of the United States for the So them District 
 
 of New York. 
 
 JOHN F. TROW & SON, PRINTERS, 
 805-313 EAST i3TH ST., NBW YORK.
 
 CONTENTS. 
 
 GENERAL TABLE ix 
 
 PREFACE xi 
 
 BOOK I. Construction of Shades and Shadows 1 
 
 PART L IN COMMON PROJECTION , 1 
 
 SERIES L By parallel rays 1 
 
 General Principles 1 
 
 1. Definitions and classification of problems 1 
 
 2. Graphical representation of Shades and Shadows 5 
 
 DIVISION I. Shades and Shadows on Ruled Surfaces 7 
 
 CLASS!. " " PLANK " 7 
 
 L Shades 7 
 
 PROBLEM I. To determine, first whether the line of intersection 
 of two oblique planes is, or is not, a line of shade ; and, 
 second, to determine, by inspection, the compound line of 
 shade on a given plane sided body 8 
 
 PROBLEM II. To determine the edges of shade on a pyramid, 
 whose axis is oblique to both planes of projection. Two 
 methods 9 
 
 II. Shadows 11 
 
 PROBLEM IIL To find the shadow of a straight line on the hori- 
 zontal plane of projection 11 
 
 PROBLEM IV. To construct the shadow of a square abacus upon a 
 square pillar, and of both, on the horizontal plane of projec- 
 tion 12 
 
 PROBLEM Y. Having the projections of a ray of light, upon two 
 vertical planes at right angles to each other ; to find its projec- 
 tions on those pknes, after revolving it 90 about a vertical 
 axis 13 
 
 PROBLEM VI. To find the projections of the shadows on the parts 
 of a timber framing, shown in two elevations, on two vertical 
 planes at right angles to each other 14
 
 IV CONTENTS. 
 
 PAGB 
 
 PROBLEM VII. To construct the shadow of a circle on the vertical 
 
 plane of projection 16 
 
 PROBLEM Till. To construct the shadow of a straight lii e on a 
 
 plane whoso traces are parallel to the ground line 19 
 
 PBOBLESI IX. To find the shadow of a chimney, situated on the 
 
 end portion of a hipped roof, upon the end and side roofs. . 19 
 PROBLEM X. To find the shadow of a shelf and brackets upon a 
 
 vertical plane 22 
 
 CLASS H. SHADES AND SHADOWS ON SINGLE CCRVED SURFACES, IN 
 
 GENERAL 23 
 
 SECTION L On Developable Single Curved Surfaces 23 
 
 I. Shades 23 
 
 PROBLEM XI. To construct the elements of shade on any cylinder 
 
 or cone 24 
 
 First. On a horizontal right semi-cylinder 24 
 
 Second. On any oblique cylinder 24 
 
 Third. On a cylinder, whose axis is parallel to the 
 
 horizontal plane only 24 
 
 Fourth* On a cone 25 
 
 Fifth, On two intersecting cylinders 26 
 
 H. Shadows 27 
 
 PROBLEM XII. To construct the shadow of the projecting head of 
 
 a cylinder upon the cylinder 27 
 
 PROBLEM XIII. To construct the shadow of the edges of shade 
 
 of a cross upon a cylinder, both bodies being seen obliquely. 28 
 PROBLEM XIV. To construct the shadow cast by the upper base 
 
 of a hollow oblique cylinder upon its interior. 30 
 
 PROBLEM XV. To construct the shadow of the upper base of a 
 vertical right cone, upon the lower nappe of the same 
 
 cone 32 
 
 PROBLEM XVI. To construct the shadow cast by the vertical cylin- 
 der (Prob. XI., Fifth) upon the horizontal cylinder, and by 
 
 the horizontal cylinder upon the vertical cylinder 34 
 
 1. Of the element of shade of the horizontal 
 
 cylinder on the vertical cylinder 35 
 
 2. Of the upper base of the vertical cylinder 
 
 upon the horizontal cylinder 36 
 
 8. Of the element of shade of the vertical cylin- 
 der upon the horizontal cylinder 36 
 
 PROBLEM XVII. To find the axes of an elliptical shadow, two of 
 
 whose conjugate diameters are known 37 
 
 PROBLEM XVIII. Having a vertical semi-cylinder with its meridian 
 plane parallel to the vertical plane of projection, it is required
 
 CONTENTS. V 
 
 PACK 
 
 to find the shadow cast on its base and visible interior, by 
 its edge of shade, and by a vertical semicircle, described on 
 
 the diameter of its upper base 3C 
 
 1. The shadow of the vertical edge 39 
 
 2. " " front semicircle 39 
 
 3. " on the upper base circle 39 
 
 4. " by auxiliary spheres 39 
 
 SECTION IL Shades and Shadows on Warped Surfaces 43 
 
 L Shades 41 
 
 PHOBLEM XIX. To construct the curve of shade on a hyperbolic 
 
 paraboloid 42 
 
 PHOBLEM XX. To find the curve of shade on the common oblique 
 helicoid, in the practical case of the threads of a triangular- 
 threaded screw 45 
 
 1. Description of the screw 45 
 
 2. To construct any particular element. (Two 
 
 methods.) 46 
 
 Points of shade on assumed helices 46 
 
 First. By planes of given declivity. , 4*7 
 
 1. On an outer helix. 
 2._ " inner " 
 
 Second. By helical translation 48 
 
 Points of shade on assumed elements. Discussion 53 
 
 PROBLKM XXI. To construct points of the curve of shade of a 
 
 conoid. Discussion 55 
 
 II. Shadows 58 
 
 PROBLEM XXII. To find the several shadows cast by a triangular- 
 threaded screw upon itself 60 
 
 First. Of the nut 60 
 
 1st. Of any unknown point of an edge 60 
 
 2d. Of any assumed point of an edge 62 
 
 Second. Of the curves of shade 63 
 
 1st. On an assumed element. 63 
 
 2d. On an assumed helix 64 
 
 3d. On another branch of the curve of shade .... 64 
 
 Third. Of the outer helix 64 
 
 1st. On an assumed elemont 64 
 
 2d. On any helix 65 
 
 3d. On particular elements 65 
 
 Fourth. The shadow on the horizontal plane 66 
 
 1st. The determination of the lines casting this 
 
 shadow 66 
 
 2d. The Shadow of any point 66 
 
 PROBLEM XXIII. To determine, in general, the shadows of a ver- 
 tical right conoid 66
 
 VI CONTENTS. 
 
 PAGB 
 
 DIVISION n. Shades and Shadows on Double Curved 
 
 Surfaces 68 
 
 I. Shades 68 
 
 PROBLEM XXIV. To find the curve of shade of the sphere, using 
 
 only one projection of the sphere 09 
 
 PROBLEM XXV. To find the curve of shade on an ellipsoid of re- 
 volution 11 
 
 PROBLEM XXVI. To find the projections of the curve of shade on 
 
 an ellipsoid of three unequal axes 72 
 
 PROBLEM XXVII. To construct the curve of shade upon a torus. 75 
 
 1. Four points on the visible boundaries 75 
 
 2. The highest and lowest points 75 
 
 3. Intermediate points 76 
 
 PROBLEM XXV1IL To find the curve of shade on a picdouche. . . 76 
 
 1. Preliminary points of shade 77 
 
 2. Intermediate points, by tangent cones 78 
 
 3. by tangent spheres 79 
 
 II. Shadows 81 
 
 PROBLEM XXTX. To construct the shadow of the front circle of 
 
 a niche upon its spherical part 82 
 
 1 . First Solution. General Method 82 
 
 2. Second Solution. Special Method 83 
 
 3. Point where the shadow leaves the spherical 
 
 part , 84 
 
 4. Third Solution. Special Method. Discussion. 
 
 Various constructions of e : 84 
 
 PROBLEM XXX. To find the shadow of the upper circle of a pie- 
 douche upou its concave surface 88 
 
 1. The highest and lowest points of shadow. ... 88 
 
 2. Any intermediate points. Discussion 88 
 
 III General Problem in Review of Shades and Shadows. 
 
 determined by Parallel Rays 91 
 
 PROBLEM XXXI. To find the shades and shadows of the shaft 
 
 and base of a Roman Doric Column 92 
 
 PROBLEM XXXII. To find the shades and shadows of the capital 
 
 and shaft of a Roman Doric Column 95 
 
 SERIES II. Shades and Shadows determined by diverging rays 97 
 
 SECTION I. General Principles 97 
 
 " II. Problems involving diverging rays 99 
 
 PROBLEM XXXIIL To find the shadow of a semi-cylindrical 
 
 abacus, upon a vertical plane through its axis 99
 
 CONTENTS. Vll 
 
 PAGE. 
 
 PROBLEM XXXIV. To find the curve of shade on a sphere, the 
 
 light proceeding from an adjacent point 100 
 
 PBOBLEM XXXV. Having a niche, whose base is produced, form- 
 ing a full circle ; and a right cone, the centre of whose cir- 
 cular base coincides with the centre of the base of the niche, 
 it is required to find the shades and shadows of this system, 
 when illumined by an adjacent point 101 
 
 PAET IL SHADES AND SHADOWS IN ISOMETRICAL PROJECTION 103 
 
 Section L General principles 103 
 
 PROBLEM XXXVI. To find the angle made by the isometrical ray 
 
 of light with the isometrical plane of projection 104. 
 
 Section n. Shades and shadows on isometrical planes 105 
 
 PROBLEM XXXV II. Having given a cube, with thin plates pro- 
 jecting vertically and forward, in the plane of its left hand 
 back face, to find the shadows of the edges of these plates 
 
 upon the cube and its base 105 
 
 Section TTT Shades and shadows on non-isometrical planes 106 
 
 PROBLEM XXXVni. To find the shadow of a hexagonal cupola 
 on a coupled roof, one face of the cupola making equal angles 
 
 with two adjacent walls of the house 106 
 
 Section IV. Shades and Shadows on single curved surfaces 107 
 
 PROBLEM XXXIX. To find the elements of shade on an inverted 
 hollow right cone, the shadow on its interior, and the shadow 
 
 of one of its elements of shade on an oblique plane 107 
 
 Section V. Shades and Shadows on double curved surfaces 108 
 
 PROBLEM XL. To find the curve of shade on a sphere 108 
 
 PROBLEM XLI. To construct the curve of shade on a torus 109 
 
 BOOK IL The finished execution of shades and shadows 112 
 
 CHAPTER I. THEORY AND CONSTRUCTION OF BRILLIANT POINTS ; AND OP 
 
 GRADATIONS OP SHADE 112 
 
 SECTION I. Preliminary general principles 112 
 
 1. Geometrical conditions for the adequate graphical representa- 
 tion of form 112 
 
 2. Of the physical conditions for the visibility of bodies, and an 
 
 adequate representation of their forms 113 
 
 3. Of brilliant points and lines 117 
 
 SECTION II. The construction of brilliant points and lines 118 
 
 1. Brilliant elements of planes 118
 
 Vlll CONTENTS. 
 
 PAGE 
 
 PROBLEM XLII. To find the brilliant point of a plane which re- 
 ceives light from a near luminous point 119 
 
 2. Brilliant dements on developable surfaces 119 
 
 PROBLEM XLIIL To find the brilliant element on a cylinder, illu- 
 
 minated by parallel rays. (Two solutions). 119 
 
 PROBLEM XLIV. To find the brilliant point of a cylinder which is 
 
 illuminated by diverging rays 121 
 
 PROBLEM XLV. To find the brilliant point on a cone which is illu- 
 minated from a near luminous point 122 
 
 3. Brilliant points on warped surfaces 123 
 
 PROBLEM XL VI. To find the brilliant element of a warped surface 
 
 when illuminated by parallel rays r 123 
 
 PROBLEM XLVII. To construct the brilliant point on a screw, 
 
 when illuminated by parallel rays 124 
 
 4. Brilliant points on double curved surfaces 125 
 
 PROBLEM XLYIII. To find tho brilliant point on any double-curved 
 
 surface, when illuminated by parallel rays 125 
 
 PROBLEM XLIX. To find the brilliant point on a sphere, illuminat- 
 ed by parallel rays 126 
 
 PROBLEM L. To find the brilliant point on a piedouche 126 
 
 CHAPTER n. OF THE REPRESENTATION OP THE GRADATIONS OP LIGHT 
 
 AND SHADE.. 127
 
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 PREFACE. 
 
 WHEREFORE thus pursue a shadow, some may ask, seeing the title of 
 this volume, and the complexity of some of its figures. 
 
 We will endeavor to reply. 
 
 The study of Shades and Shadows is an application of the general 
 problems of Descriptive Geometry, in connexion with a few physical 
 principles ; and, of Descriptive Geometry, no one, who has occasion to 
 be conversant with forms, singly or in combination, can know too 
 much, either for practical purposes, or as a promoter, in its peculiar 
 way, of mental power. 
 
 In particular, having in a previous work on the General Problems of 
 Descriptive Geometry, disposed of the Projections, Intersections, Tan- 
 gencies, and Developments of geometrical forms in the abstract, the 
 way is prepared for the extended direct application of the second and 
 third of these operations, involving incidentally that of the first, to the 
 concrete geometrical subject of Shades and Shadows. For a Shade is 
 always bounded by a line of contact, and a Shadow by one of intersec- 
 tion ; and before either can be found, the body casting, and the surface 
 receiving the shadow, must be given by their projections. 
 
 The easy logical sequence of " General Descriptive " Problems, and 
 " Shades and Shadows " as a theoretical branch of applied " Descrip- 
 tive," is, therefore, one point of interest attaching to the study of the 
 latter subject. 
 
 Further, there is beauty in the idea, that for any distance and direc- 
 tion of the source of light, and for any form and position of the bodies 
 casting and receiving shadows, the mind can know, and the hand can 
 execute those shadows truly. 
 
 The utility, however, of delineated shades and shadows, in rendering 
 working drawings at once not only more beautiful, but more intelligible, 
 because more nearly conformed to reality, is the chief ground of inte-
 
 xil PREFACE. 
 
 rest in the study of them. The student, therefore, who has experienced 
 any degree of pleasure in the pursuit of the general problems of 
 " Descriptive," may, without probable disappointment, promise himself 
 added enjoyment in the study here introduced to him. 
 
 Once more : The many interesting theoretical and practical particu- 
 lars in which the science of optics is found related to the special study 
 of shades and shadows, lend interest to the latter, as affording a field 
 for observation and reflection to be enjoyed in common by the artist, 
 the physicist, and the geometrical draftsman. Indeed, it is hoped that 
 Book II. of this volume, especially Chapter II., on the execution of 
 shades and shadows, will be found worthy of the attention of artists. 
 
 A word now as to the plan of this work. 
 
 Great care has been bestowed on the statements of general principles, 
 which, in any effective study of the subject, must be mastered as fully 
 as the problems. 
 
 Again : with his mind properly guarded against perplexity, by due 
 comprehension of a simple, and manifestly rational, classification of 
 the problems of a proposed course, the student will gain little effective 
 assistance from the study of a protracted series of similar problems. 
 
 One good characteristic problem, under each of the heads pointed 
 out in the preceding general table, would therefore be, at least in 
 theory, sufficient. 
 
 The student having mastered these representative problems, would 
 then readily refer any new problem, met by him, to its appropriate 
 'group ; and would then solve it by an intelligent use of the general 
 principles applicable to all problems of that group. 
 
 But it is usually difficult for the beginner to dispossess himself 
 entirely of the idea that problems, which differ considerably in out- 
 ward form, do not differ likewise in essential principle, and method of 
 solution. Besides, there are often indirect and special methods oi 
 solution which really need separate exemplification ; hence, a liberal 
 number of special problems, some of them of quite a practical cast, 
 have been scattered through this work. 
 
 With this modification of what was at first pointed out as a theoreti- 
 cally sufficient course, it is hoped that the present effort will be accept- 
 able to professors and artists, and to classes in scientific collegiate 
 courses, and in the Polytechnic Schools ; especially to those who have 
 already become familiar with the rudiments of Shades and Shadows, as
 
 PREFACE. Xlli 
 
 presented in the author's "Elementary Projection Drawing," or in 
 similar works. 
 
 As to methods of study, none is better, after reading the text till it 
 is fully understood, than repeated rehearsals of the problems, and prin- 
 ciples, by the student, with the book closed ; the figures having been 
 made upon a slate or portable blackboard, such as every student should 
 have, and in any form which will truly represent the essential opera- 
 tions of the problems, without regard to the particular forms of the 
 figures in the book. 
 
 The recitations may properly consist of mingled interrogation upon 
 the principles, and explanations of the problems ; the latter, both from 
 the figures in the book, and from blackboard constructions ; also from 
 the finished constructions on the student's plates, when time enough is 
 devoted in the class-room to their construction, to make due interroga- 
 tion upon them possible. It will, however, probably appear on trial 
 that no test equals the regular class recitation, in that thoroughness 
 which is due to the student in his training. 
 
 It is a peculiar pleasure to add, that this expensive volume now 
 appears, through the every way timely and pleasant kindness of Students 
 of the Institute, to the Author, and, as he truly hopes, no less to them- 
 selves, in making up a liberal subscription in aid of its hitherto delayed 
 publication.
 
 7 
 
 
 
 SHADES AND SHADOWS. 
 
 Book 1. 
 
 CONSTEUCTION OF SHADES AND SHADOWS. 
 
 PART I. 
 IN COMMON PROJECTION. 
 
 SERIES I. 
 
 SHADES AND SHADOWS FORMED BY PARALLEL RATS. 
 
 GENERAL PRINCIPLES. 
 1. Definitions, and Classification of Problems. 
 
 1. Light proceeds from its source in uninterrupted straight 
 lines, called rays, except when diverted by reflection or refrac- 
 tion, or arrested by opaque bodies ; as is sufficiently proved by 
 the impossibility of direct vision through a bent tube. 
 
 2. If the source of light be a point, the volume of rays pro- 
 ceeding from it, and intercepted by an opaque body, will be a 
 cone, having the luminous point for its vertex. If this point be 
 at an infinite distance from the opaque body, the cone becomes a 
 cylinder, and the rays of light will be sensibly parallel. The latter 
 case is the one considered in the present series of problems, and 
 is substantially the same as that of bodies illuminated by the 
 Sun, for, on account of his great distance from the Earth, the 
 solar rays to all points of any terrestrial object, are sensibly parallel. 
 
 3. Let there now be given, in the following order of successive 
 position, a 'Source of light, an opaque point, and an opaque sur- 
 face. The opaque point will intercept one of the rays from the 
 source of light The path of the ray, beyond the opaque point, 
 
 1
 
 2 GENERAL PROBLEMS. 
 
 will therefore be a line of darkness, and its intersection with the 
 opaque surface will be a dark point. 
 
 The line of darkness is the shadow in space of the given point ; 
 the dark point on the opaque surface is the superficial shadow, 
 or, simply, the shadow of the same point. 
 
 4. The shadow of a point upon any surface is, therefore, tho . 
 intersection of a ray (meaning the indefinite path of the ray) 
 drawn through the point, with the given surface. 
 
 5. Any plane containing a ray, is a plane of rays ; for any 
 number of rays can be drawn in the plane, and parallel to the 
 given ray, and these rays will, collectively, constitute a plane of 
 rays. Hence a plane of rays can be drawn through a straight 
 line having any position in space. For the plane containing 
 this line, and a ray through any point of it, will be the required 
 plane of rays. Furthermore, since two lines determine a plane, 
 but one plane of rays can be passed through a given line, unless 
 the line be parallel to the rays of light. 
 
 6. A line being made up of points, its shadow will consist of the 
 shadows of all its points. Eays of light being parallel, the rays 
 through all the points of a straight line, will form a. plane of rays 
 (5), unless the line coincides with a ray. The shadow of a straight 
 line is, therefore, in general, the line of intersection of a plane 
 of rays passed through it, with the surface receiving the shadow. 
 When the line coincides with a ray, its shadow is its own point 
 of intersection with the given surface. 
 
 When a line is parallel to any plane, its shadow on that plane 
 is parallel to the line itself. 
 
 The shadows of parallel lines, on the same plane, are parallel. 
 Also the shadows of the same line on parallel planes are parallel ; 
 only one of them, however, can be real. 
 
 7. The parallel rays through all points of a curve, will form a 
 cylinder of rays, unless the curve be plane, and coincides with a 
 plane of rays. Hence the shadow of a curved line is, in general, 
 the intersection of the cylinder of rays having the curve for its 
 directrix, with the given opaque surface. When the curve is 
 plane, and in a plane of rays, its shadow is the line in which 
 that plane intersects the opaque surface. 
 
 8. The shadow of a surface is a surface, unless it is a plane 
 surface and coincides with a plane of rays, when its shadow will 
 be a line, as in the case of a plane curve lying in a plane 
 of rays.
 
 SHADES AND SHADOWS. 3 
 
 9. The shadow of a solid is always a surface, which is known 
 when its boundary, called its line of shadow, is found. Let the 
 given body be circumscribed by parallel tangent rays. These 
 rays will form a circumscribing tangent cylinder of rays. The 
 curve of contact of this cylinder and the body, is evidently the 
 boundary between the side of the body towards the source of 
 light, and the part which is opposed to the light. The former is 
 the illuminated part of the body, the latter is the shade of the 
 body, and the line of contact is called the line of shade. 
 
 10. Therefore 1st. From all the portion of space within the tan- 
 gent cylinder of rays, and beyond the given body, light is exclud- 
 ed, and that portion is called the shadow in space, of the body. 
 2nd. The area, bounded by the intersection of the cylinder of 
 rays with the surface receiving the shadow, is the shadow of the 
 body on that surface. 3d. The line of shadow is thus the shadow 
 of the line of shade. 4th. The line of shade on a body must 
 therefore be known, before its line of shadow can be found. 
 
 11. Any one point of a curve of shade upon a body, is the 
 point of contact of one element of the tangent cylinder of rays ; 
 that is of one ray : and the intersection of tliat ray, with any 
 surface beyond the given body, is the shadow of that one point of 
 the curve of shade. Both shades and shadows are found, in 
 practice, according to this statement, i.e. by separately finding 
 single points and joining them. 
 
 12. The constitution, and mode of construction, of the line of 
 shade of a body, will depend on the nature of the surface of 
 that body. The line of shade on a plane sided body, is com- 
 posed of those edges, collectively, which divide its light, from its 
 shaded surfaces. On a developable single curved surface, it con- 
 sists of the elements of contact of tangent planes of rays. On 
 a warped surface, it consists of a succession of points, one on 
 each of a succession of elements, these points being the points 
 of tangency of planes of rays. On a double curved surface, it 
 consists strictly of the curve of contact of a circumscribing tan- 
 gent cylinder of rays. 
 
 The problems of shades may therefore be arranged in the order 
 of the four kinds of surfaces as just named. 
 
 13. Every problem of shadows, evidently reduces to finding 
 the shadow of a point on a given surface ; that is, to finding the 
 intersection of a straight line (ray) with the surface. The 
 method of finding this intersection depends on the nature of the
 
 4: GENERAL PROBLEMS. 
 
 given surface, hence the problems of shadows may be arranged in 
 the same way as those of shades may be. 
 
 But the construction of the shadows of straight lines being 
 generally a little simpler than that of the shadows of curves, 
 the surface receiving the shadow being the same in both cases, 
 the former may be made introductory to the latter, in each of the 
 four principal groups of problems of shadows. 
 
 14. Rehearsing the principal conclusions thus far reached, we 
 have the following : 
 
 The curve of shade of a body is the curve of contact of a tangent 
 cylinder of rays, with that body. The curve of shadow of the 
 same body, is the curve of intersection of the same cylinder with 
 the surface receiving the shadow. 
 
 Any one point of the curve of shade, is the point of contact of one 
 element of the cylinder, and this element is a ray of light. The 
 intersection of the same ray, or element, with the surface receiving 
 the shadow, is the shadow of this same point in the curve of shade. 
 
 15. Here, expanding the last article somewhat, we observe 
 that the line of shade is either straight or curved, hence the sur- 
 face of rays passed through it and whose intersection with any 
 other surface is the line of shadow will always be a plane when 
 the line of shade is straight, and single curved when the line of 
 shade is a curve. In the latter case the line of shade will be the 
 directrix of the single curved surface of rays. When the rays 
 are parallel, the single curved surface of rays will be cylindrical ; 
 when they are diverging, it will be conical, having the luminous 
 point for its vertex. Hence, for all combinations of straight and 
 curved lines of shade, with parallel rays, and rays diverging 
 from a point, the curve of shadow defined as found by the 
 problems of orthographic projection will be the intersection of a 
 plane, a cylinder, or a cone, with the surface receiving the shadow. 
 
 16. The case, in which the source of light is a luminous poin t, 
 either at a finite or infinite distance, is the only one which needs 
 to be considered. For, by way of a simple example of simul- 
 taneous illumination from several points, let a sphere be exposed 
 to the light from a finite luminous straight line. Each point of 
 such a line would be the vertex of a separate tangent cone of 
 rays, whose circle of contact with the sphere would be the circle 
 of shade due to that cone. Hence, only so much of the sphere 
 os received no light from either extremity of the luminous line, 
 would be totally in shade ; and only so much as received light
 
 SHADES AND SHADOWS. 5 
 
 from every point of the line would be totally illuminated. The 
 intermediate portion would be in partial shade, which would 
 become more and more complete as the total shade should be 
 approached. In order, therefore, to secure both simplicity and 
 precision in the diagrams of shades and shadows, the source of 
 light is assumed to be a point. 
 
 2. Graphical Representation of Shades and Shadows. 
 
 17. In passing to the graphical construction of shadows, it is 
 to be observed, that, in representing them in orthographic projec- 
 tion, the planes of projection, the projecting lines, and the use of 
 these, are similar to the same things, and their use, as found in 
 any other problems of orthographic projection. 
 
 18. The three things given in space (3) for the production of a 
 real shadow, are given in projection, in representing that shadow ; 
 only, as the source of light is supposed to be indefinitely remote, 
 it is only virtually represented, by the projections of the parallel 
 rays proceeding from it. 
 
 If the source of light were a near luminous point, its projec- 
 tions would be given, together with rays proceeding from it in 
 diverging lines. 
 
 19. The ordinary rules of orthographic projection, concerning 
 given and auxiliary, hidden and visible lines and planes, are 
 observed in problems of shades and shadows. Moreover, visible 
 shades or shadows are represented, either by a tint of indian ink 
 or by parallel lines. Hidden shades or shadows are represented 
 only by their boundaries, which are made in dotted lines ; plain, 
 or fringed, as in PI. L, Fig. 5, for greater distinctness. 
 
 20. In constructing shadows merely as a geometrical exercise, 
 the main object is to indicate clearly their position. Neatness 
 and distinctness are, therefore, all that is required in their exe- 
 cution ; hence, to save time, they are represented in flat or uni- 
 form tints. In making finished drawings, the shades and shadows 
 are done in graduated tints, according to their actual appearance. 
 
 21. In industrial applications of problems of shades and sha- 
 dows, the light is usually taken to correspond with the body 
 diagonal of a cube, two of whose faces coincide with the planes 
 of projection. That is, the light coming from the left, forward 
 and downward, makes an angle of 35 16' with the planes of
 
 6 GENERAL PROBLEMS. 
 
 projection, and its projections make angles of 45 with the 
 ground line. 
 
 In the following general problems, the light is taken indif- 
 ferently in any direction, since the general methods of solution 
 are not altered by so doing. 
 
 22. Methods of solution are distinguished as General and /Spe- 
 cial, also as Direct and Indirect. 
 
 General methods consist of such operations as apply equally 
 well, in theory, though not always with equal practical conve- 
 nience, to whole classes of problems. 
 
 ^Special methods are such as are founded on the distinguishing 
 peculiarities of individual problems, or groups of problems. They 
 are therefore as numerous as those peculiarities, and hence afford 
 the natural field for the exercise of ingenuity in their discovery 
 and application. 
 
 Direct methods are those which immediately and obviously 
 conform to the usual definition of a shade or a shadow ; as when 
 a point of shade is found as the point of contact of a given ray 
 with a given surface. 
 
 Indirect methods are those in which auxiliary magnitudes are 
 employed, to avoid the laborious constructions which direct 
 methods may sometimes occasion ; as when we find a point of 
 shadow on some irregular surface, by noting where the easily 
 found shadow upon some secant plane meets the intersection of 
 that plane with the surface.
 
 SHADES AND SHADOWS. 
 
 SERIES I. 
 
 SHADES AND SHADOWS FORMED BY PARALLEL RAYS 
 
 DIVISION I. 
 SHADES AND SHADOWS ON RULED SURFACES. 
 
 23. The line of shade on any ruled surface is obtained by find- 
 ing, either by inspection, or by construction by means of tangent 
 rays or planes of rays the edges, elements, or points which 
 constitute that line of shade. 
 
 24. A shadow cast by a body upon any ruled surface, is deter- 
 mined by finding an element of that surface, and a point of the line 
 of shade of the body, both of which shall be in the same plane 
 of rays. A ray through the point of shade will then intersect 
 the element of the ruled surface in a point of the required shadow. 
 
 CLASS I. 
 
 SHADES AND SHADOWS ON PLANE SURFACES. 
 I. Shades. 
 
 25. The shade of an opaque plane, or plane figure, is the whole 
 of one side of such plane or plane figure. The former being of 
 indefinite extent, can have no line of shade. That of the latter 
 is its perimeter. 
 
 26. The line of shade of any plane-sided body, as a pyramid, 
 consists of those edges, collectively, which divide its illuminated 
 from its unilluminated surfaces (12). This line of shade can 
 often be determined by inspection, having given the relative 
 position of the body and the rays of light. 
 
 The two following problems will show the method of finding 
 lines of shade on plane-sided bodies, when they cannot be deter- 
 mined by inspection.
 
 8 GENERAL PROBLEMS. 
 
 PROBLEM I. 
 
 To determine, first, ivhether the line of intersection of two oblique 
 planes is, or is not, a line of shade ; and, second, to determine, by 
 inspection, the compound line of shade on a given plane-sided body. 
 
 First: Principles. After finding the projections of the line of 
 intersection of the given planes, by Prob. V., Des. Geom., assume 
 any point on that line, and draw the projections of a ray of light 
 through it. If the ray, thus given, pierces either plane of pro- 
 jection between the traces of the given planes on that plane, it 
 shows that both planes are illuminated, both being pierced by 
 rays, and hence that their intersection is not a line of shade. On 
 the contrary, if the ray pierces either plane of projection outside 
 of the traces of the given planes on that plane, it shows that the 
 intersection of the given planes is a line of shade. 
 
 Construction. PI. I., Fig. 1. Let RS be the ground line, 
 PRP' one of the given planes, PSP' the other, and PfP'e' 
 their line of intersection, n, n' is an assumed point of this inter- 
 section, through which the ray, nan'a', parallel to the given 
 ray, rr', is drawn. This ray pierces the horizontal plane at 
 a, a', indicating, according to the principles of this solution, that 
 PfP'e' is a line of shade. 
 
 Remarks. a. Hence the projections of the visible portion of 
 the unillumiuated plane, PSP', are shaded, and PfP'e' is inked 
 as a heavy line. 
 
 b. The above construction is applicable to the case of any 
 pyramid, in a simple, or oblique position, when we wish to deter- 
 mine its line of shade. In such cases, unless the pyramid rests 
 upon a given plane, it will first be necessary to determine the 
 traces of its faces on some plane, whose intersections with the 
 rays through points in the edges of those faces will also be con- 
 structed. 
 
 27. As a second illustration of this problem, see PI. IV., Fig. 
 14, where PQP' and PRP' are the given oblique planes. PV 
 and Pb are the projections of their intersection, and m,m is a 
 point on that line, mnm'n' is a ray of light through m,m', 
 and it pierces the horizontal plane at n, outside of both planes. 
 Jlenc'-, when Ph P'a' is regarded as the edge of the salient die- 
 ilral angle, indicated by the portions, QP and RP, of the traces, 
 it is an edge of shade for the direction, mnm'n' of the light.
 
 SHADES AND SHADOWS. 9 
 
 If PQP' as now shown, and that part only of PBP', extending 
 in front of the edge P6 PV, are considered as the surfaces 
 exposed to the light, the left side of each plane is then the one 
 considered, and both of these surfaces are in the light. 
 
 The next problem affords a more practical example of th 
 general constructions just given. 
 
 28. Second. The lines of shade on the abacus and pillar. See 
 PI. I., Fig. 4, where the projections of a square pillar and its 
 abacus, resting on the horizontal plane, are given, together with 
 the direction of the light, Cm D'ra'. 
 
 Here it is evident that t t"t' ; C C'D'; the upper edges 
 CE and E A ; the vertical edges at A and q ; AB A/B', and 
 BC B'C', are the edges which constitute the compound line of 
 shade of the given body. They are therefore (9 11) the edges 
 which cast shadows on the pillar, or on other surfaces ; and they 
 are identical with the lines, which, in geometrical line-drawings, 
 are inked as heavy lines. 
 
 PROBLEM II. 
 
 To determine the edges of shade on a pyramid, whose axis is oblique 
 to both planes of projection. 
 
 Construction. PI. IV., Fig. 15. To make the problem quite 
 definite, let Gr"L" be the ground line of an auxiliary vertical 
 plane, parallel to the axis of the pyramid, which we will suppose 
 to be a triangular one, in order to avoid needless repetition of the 
 same operations of construction. Let the plane of the base be 
 perpendicular to the axis, and let a"c" be its trace on the auxi- 
 liary vertical plane. Let ABC be the true size and form of this 
 base, seen when its plane is revolved about a"c", as its trace, into 
 the auxiliary vertical plane, and let D be the projection of the 
 axis on the plane of the base. Then by counter-revolution, A,B 
 and C return in arcs, projected in Aa", etc., perpendicular to 
 a"c", to the points a", b" and c ' ; and D, the foot of the axis, to 
 d". Then make d"v", perpendicular to a"c", equal to the axis 
 of the pyramid, and join v" with a", b" and c" to complete the 
 auxiliary projection. 
 
 To make the horizontal projection, make aa", etc., equal to 
 Aa", etc., and perpendicular to the ground line Gr"L" ; and v
 
 10 GENERAL PROBLEMS. 
 
 abc will be the horizontal projection of the pyramid. The vertical 
 projection, v' a'b'c', is found, as in all similar cases, by pro- 
 jecting up v b, etc., in perpendiculars to GL, and by making 
 the heights of the points v', &', etc., above the ground line GL, 
 equal to their heights v"k, b"h, etc. above G"L". 
 
 Now, to test any edge of the pyramid, to see if it be an edge 
 of shade, or not. 
 
 First Method. Let the edge vb v'b' be taken. Find the 
 traces, on one plane of projection, of the planes of the faces of 
 which the given edge is the intersection. Then draw a ray 
 through any point of that edge, and if it meets the plane of the 
 traces outside of those traces, when the edge belongs to a salient 
 angle turned towards the light, the given edge is a line of shade. 
 Accordingly, vb v'b' pierces the horizontal plane at m, and 
 cb c'b' pierces it at n, giving mn, as the horizontal trace of the 
 plane of the face vbc v'b'c'. Again, va v'a' pierces the hori- 
 zontal plane at a, giving ma as the horizontal trace of the plane 
 of the face vab v'a'b'. Now a ray, bp b'p', through any point 
 bb' of the intersection, vm v'm', of these planes, pierces the 
 horizontal plane atjp, outside of both traces, ma and inn. But 
 the parts of these planes which are the faces of the pyramid, esti- 
 mated from the common edge, vb v'b', have ma and mO for 
 their horizontal traces. Hence the direction of the light evidently 
 strikes only the interior face of each of these planes, that is, the 
 face towards the axis of the pyramid ; so that the exterior sur- 
 faces vbc and vba are both in the dark, and vb v'b' is not a line 
 of shade. 
 
 If now we find 0, the intersection of the edge, vc v'c' with 
 the horizontal plane, aO is the horizontal trace of the plane of 
 the face vca v'c'a', and vcO is a salient edge, similar to Pb 
 PV in Fig. 14, so that the ray, cr cV, by piercing the hori- 
 zontal plane outside the traces On and Oa, shows that vc v'c' 
 is an edge of shade, which, indeed, was known as soon as it was 
 found that vb v'b' was not such an edge. 
 
 Second Method. Find the shadows of all the edges of the 
 pyramid on either plane of projection. Then, if, for example, 
 the shadow of vc v'c' upon the horizontal plane is further from 
 the pyramid, in the direction of the light, than that of vb v'b', 
 it would show that vc v'c', only, of those two edges, really 
 cast a shadow. Hence it would be shown that vc v'c' was an 
 edge of shade (26). That is, we have this principle: TJiosc
 
 SHADES AND SHADOWS. 11 
 
 edges, whose shadows are the boundaries of the real shadow of a body, 
 are the edges of shade of that body. 
 
 Bemark. The next problem, if not the preliminary general 
 principles already given, will enable the student to construct this 
 second method. 
 
 H. Shadows. 
 
 29. Here observe First : That whenever a plane is perpen- 
 dicular to either plane of projection, its entire surface, and all 
 contained in it, is projected in the trace of the plane on that 
 plane of projection. /Second: Having a point, p, which casts a 
 shadow on a plane, the projections of that shadow must be in 
 the projections of the ray (4) through p. 
 
 30. From the two foregoing principles we have the following 
 simple general method (22) of finding points of shadow on the 
 planes of projection, or on any planes which are perpendicular to 
 them. Find, by inspection, the intersection of the linear projection 
 of the plane receiving the shadow, ivith the corresponding projection 
 of the ray through any point casting a shadow upon it. This will 
 be one projection of the point of shadow sought. Its other projection 
 will be the intersection of the projecting line of the point found, with 
 the other projection of the same ray. 
 
 It is only necessary here to remember, that the linear projection 
 of both planes of projection, is the ground line. 
 
 PROBLEM III. 
 
 To find the shadow of a straight line, on the horizontal plane of 
 
 projection. 
 
 Principles. The shadow of a straight line on a plane, is the 
 intersection of that plane with a plane of rays passed 'through 
 the given line (5-6). This shadow will therefore be a straight 
 line. But two points determine a straight line, and two parallel 
 lines determine a plane. Hence if, through any two points of 
 the given line, we pass rays, their intersection with the given 
 plane will determine the shadow of the given line. Also, the 
 point in which the given line meets the given plane is a point of 
 the required shadow, for the line and its shadow are in the same 
 plane (of rays) and therefore intersect.
 
 12 GENERAL PROBLEMS. 
 
 Construction. PI. L, Fig. 2. Let ab a'b' be the given line, 
 and r r the direction of the light. By (30) the ray ac a'c', 
 drawn through any assumed point, as a, a' of the given line, 
 pierces the horizontal plane in the point whose vertical pro- 
 jection is c', and whose horizontal projection is therefore c. 
 Likewise the ray bd b'd' pierces the horizontal plane at d',d 
 (naming that projection first which is found first). Hence cd is 
 the shadow of ab a'b' on the horizontal plane. Also e,e', the 
 point in which ab a'b' pierces the horizontal plane, is a point 
 of its shadow on that plane, and hence will be found in dc pro- 
 duced. 
 
 Remark. In taking a strict view of the principles of this solu- 
 tion, and in similar cases, it should be observed that the rays, as 
 ah a'b', are to be regarded as cut from the plane of rays (6) 
 passed through the line, by secant planes of rays, which may 
 have any position, but are most readily conceived of as perpen- 
 dicular to one of the planes of projection. 
 
 31. When a line is perpendicular to a plane, its shadow, on that 
 plane, is in the direction of the projection of the light on that plane j 
 for the plane containing the line and a ray through any point of 
 it, is a plane of rays (5), perpendicular to the given plane, and is 
 therefore a projecting plane of all rays intersecting the line. 
 Hence its trace on the given plane, is both the shadow of the line 
 (6), and the projection of a ray, upon that plane. 
 
 This principle is of very frequent application to the shadows 
 of vertical lines upon the horizontal plane, and of co- vertical lines 
 (that is, of lines perpendicular to the vertical plane) upon the 
 vertical plane. 
 
 PROBLEM IV. 
 
 To construct the shadow of a square abacus upon a square pillar, 
 and of both, on the horizontal plane of projection. 
 
 [There being no new principles in this problem, we pass at 
 once to its graphical solution as a new illustration of the method 
 of (30) referring to (28) for the line of shade.] 
 
 Construction. PI. I., Fig. 4. Let the body be seen obliquely 
 as shown on the vertical plane, and let CmD'm' be the direc- 
 tion of the light. The point d,d', found by drawing the ray qd, 
 and projecting d at d', casts the point of shadow q,q' on the ver*
 
 SHADES AND SHADOWS. 13 
 
 tical edge at q ; q' being found at the intersection of this edge 
 with the vertical projection, d'q', of the raj through d,d'. From. 
 q' the shadow q'c' is parallel to the line ded'e', which casts it, 
 since that line is parallel to the vertical face, qc, of the pillar (6). 
 The ray through B,B' pierces the front of the pillar at </, whose 
 vertical projection is in the vertical projection, By, of the ray, at 
 g r . Hence c'g' is the shadow of the portion, eB e'B', of the 
 edge, AB A'B', of the abacus, upon the front of the pillar. 
 g'f, parallel to B'C', is the shadow of the portion B/-B/ of the 
 edge BC B'C', on the parallel front of the pillar. The ray 
 through //', passing beyond the pillar, pierces the horizontal 
 plane (30) at hf, A, and hi is the shadow of the portion, tt'tf', of 
 the edge of the pillar, which casts a shadow on the horizontal 
 plane. The portion JVf'W of BC- B'C' casts the shadow hk, 
 beginning at h and limited at k',k by the ray CkC'k. The 
 edge C C'D' casts the shadow km, limited by the intersection of 
 the rays Ck C'k' ancUC/n D'm' with the horizontal plane. 
 Similarly, CE has rar?, equal and parallel to itself, for its shadow ; 
 E A has, likewise, no for its shadow ; the vertical edge at A deter- 
 mines the shadow op, found as km was ; pr is the shadow of 
 Ad AW, and rq is the shadow of the vertical edge of the pillar 
 at q. 
 
 PROBLEM Y. 
 
 Having the projections of a ray of light, upon two vertical planes at 
 right angles to each other ; to find its projections on those planes, 
 after revolving it 90 about a vertical axis. 
 
 Principles. When we view any object in reference to a 
 given vertical plane, the direction of the light is regarded as 
 fixed ; and it is therefore fixed in projection, while we are find- 
 ing the shadow of the object on that plane. When, however, 
 we turn and face the object in a different direction, the light 
 may be supposed either to turn with us, so as to come in the 
 same relative direction with respect to the new projection that it 
 did with respect to the old; or it may be supposed to remain 
 absolutely fixed, as it is practically in Nature for any short 
 interval. A little reflection on the latter supposition, will show 
 that, in certain aspects of an object, only its shade could be seen ; 
 hence it is sometimes, if not generally, desirable to adopt the first 
 view, viz. that the light turns with us, so as to come in the same
 
 14: GENERAL PROBLEMS. 
 
 relative direction with respect to each new vertical plane of pro* 
 jection. This first view is adopted in the following 
 
 Construction. See PI. II., Fig. 6. Here let the given projec- 
 tions of the light, on the two vertical planes, be R'L' and R"L", 
 and let qpx' indicate the real position of the vertical plane of the 
 left hand projection. By an obvious construction, we find RL, 
 the horizontal projection of the given ray, R^L" RT/. Then, 
 as the observer turns 90 to face the plane qp, the light turns 
 with him, an equal amount, about any vertical axis, as at L, and 
 so appears in horizontal projection, as at rL, perpendicular to RL. 
 The vertical projections of rL are, obviously, r'l/ and r"~L". 
 
 We see now, by inspection, that, as the vertical planes, and 
 also the two positions of the light, as seen at RL and rL, are at 
 right angles to each other, pt=sm, and pq=mn. That is, x'y' 
 being the same for all the vertical projections of the ray, the new 
 vertical projections, r'L' and r"L", make the same angle with the 
 ground line as R'L' and R"L" do. So that, in practice, it is 
 only necessary to draw rT/ parallel to R"L", and r"L" equally 
 inclined with R'L', but in an opposite direction in order to 
 obtain the projections of the light as it proceeds in viewing the 
 side or left hand elevation. 
 
 When the projections, r? rT, of the light make the conven- 
 tional angle of 45 with the ground line, as in PI. II., Fig. 7, 
 the vertical projection, r""l"", of it on the right hand plane, as 
 it comes when the observer has turned, from viewing that plane, 
 to view the left hand one, simply inclines at 45 to the ground 
 line in the opposite sense, from r'l'. 
 
 PROBLEM VI. 
 
 To find the projections of the shadows on the parts of a timber fram- 
 ing, shown in two elevations, on two vertical planes at right angles 
 to each other. 
 
 Principles. The chief new principle in this problem is, that a 
 point of shadow may exist, geometrically, on a material surface 
 produced ; though, physically, only on the real portion of that 
 surface. 
 
 Construction. PI. II., Fig. 8. MJ-J'J" is a horizontal 
 timber, resting on the transverse timber AJ A'B'C', and 
 on the lateral braces khg'y and kJtnu. These braces are
 
 SHADES AND SHADOWS. 15 
 
 framed into the post LH A'H', in front of whicja is the front 
 brace GP GT'. The left hand elevation is supposed to be 
 made on a vertical plane at right angles to the plane of the paper. 
 Taking the light in this practical problem in the conventional di- 
 rection (21) GL G'l/ are the projections of a ray, as it proceeds 
 in viewing the object as seen on the right hand plane of projection. 
 
 This being established, we are prepared to construct the sha- 
 dows in Fig. 8. The point A, A' casts a shadow on the front 
 brace at a, whose other projection, a', is the intersection of the 
 projecting line, a a', with the other projection, AV, of the ray 
 Aa. A A'B', being parallel to the face of this brace, its shadow, 
 a' a", is parallel to A A'B'. Next, c,c' and L,L/, the shadows 
 of the points a,a" and G,G' in the front right hand edge of the 
 brace, are found precisely as a,a' was, and they determine LY, the 
 position of the imaginary shadow (in this figure) of this edge on 
 the plane of the front of the post produced. The shadow of the 
 back left hand edge,- which pierces the post at P,P', begins, 
 therefore, at P,P' and is parallel to LY, as seen at P'K. The 
 front of the lateral brace, g'y, being parallel to the front of the 
 post, the shadow of Ga G'a" upon it, will be parallel to LV. 
 One point of this shadow is b,b", found by drawing the ray ab 
 a"b" ; hence b"f, parallel to LY, is the required shadow. The 
 point A, B' casts the shadow 5,6', found like the preceding, on 
 the plane of the front of the lateral brace. Then 6Y, parallel to 
 B'C', and limited by the ray through C,C', is the shadow of AC 
 B'C' on the same plane. The portion of shadow on the brace, 
 bounded by CY, is cast by the edge, CJ C', of the transverse 
 timber, and b'b" is the shadow of A A'B' on the plane of the 
 front of this brace. Finally, for the right hand elevation, the edge 
 J 3' 5" casts the shadow n y on the lateral braces, found by 
 projecting over k", where a ray through any point of this edge 
 meets the plane of the front of these braces. 
 
 Proceeding to the side elevation, two planes of rays, containing 
 the edges whose projections are J and M, will include between 
 them the shadow in space (10) of the longitudinal timber JM 
 J'J". Their traces on the side of the transverse timber, will 
 therefore bound the shadow on that timber. A portion of one 
 of these traces is seen at Jk". The edge AB B' casts a shadow 
 on the side of the front brace at T', found by drawing the ray 
 g/rp/ (pro^ y^ According to the construction of all the other 
 points of shadow in this problem, T, its other projection, is at
 
 16 GENERAL PROBLEMS. 
 
 the intersection of the projecting line, T'T, with the other pro- 
 lection, AT, of the ray through A, B', as it comes according to 
 previous explanation (Prob. V.) when viewing the framing as 
 seen in the left hand projection. AB being parallel to the side 
 of the front brace, its shadow, tv, through T, is parallel to itself. 
 Lastly, the brace hk" g'y casts a shadow on the side of the post. 
 A little consideration of a model will show, that, when the brace 
 makes a less angle with the horizontal plane than the light does, 
 its left hand front edge and right hand back edge, as seen on the 
 left hand elevation, will cast shadows. When the former angle 
 is the greater of the two, the right hand front edge and left hand 
 back edge will cast shadows. When the two above angles are 
 equal, the front and back planes of the brace will be planes of 
 rays, and their traces, kk'" and gh, produced, will bound the 
 shadow of the brace. 
 
 In the present problem, the brace makes a less angle with the 
 horizontal plane than the light does, hence the edges, k'y kk" 
 and g'fhp, cast shadows on the post. The shadow of the former 
 edge begins at k, and of the latter at A, these being the points 
 where these edges pierce the post. Assume any point F, F 7 and 
 draw the ray F'H' FH, which, according to the previous con- 
 structions, pierces the plane of the side of the post at H',H. H& 
 is therefore the shadow of kk" ; and sh parallel to it is the shadow 
 of hp. 
 
 PROBLEM VII. 
 To construct t/ie shadow of a circle, on tfie vertical plane of projection. 
 
 Principles. The only essential difference between this and the 
 previous problems is, that, in case of a curve, a plane of rays can 
 generally contain but one or two points of it, while it may con- 
 tain the whole of a straight line. But we have seen that the 
 shadow of a straight line is practically determined by the sha- 
 dows of two of its points. Now the shadow of a point is found 
 in the same way, whether that point be on a straight line or on 
 a curve ; hence shadows of straight lines, and shadows of curved 
 lines, are not here distinguished as belonging to distinct classes 
 of problems. 
 
 According, therefore, to the method of (30), again, we have 
 the following
 
 SHADES AND SHADOWS. 17 
 
 Construction. PI. III., Fig. 10. Let the plane of the circle 
 be perpendicular to the ground line. Then the equal lines, AB 
 and C'D', perpendicular to the ground line, will be the projec- 
 tions of the circle. X,et Aa AV be the projections of a ray of 
 light. This ray meets the vertical plane at a', the shadow, there- 
 fore, of A, A', the foremost point of the horizontal diameter of 
 the circle. The ray B6 A 'b' determines the shadow of the 
 point B,A', at b' ; the rays Cc C'c' and Cc D'd f determine the 
 points of shadow, c' and cT, of the extremities of the vertical 
 diameter C D'C'. Assuming either projection, as E, of any 
 other two points on the circle, their other projections must be 
 constructed by some one of the methods of (Prob. XL., Des. 
 Geom.), E is the horizontal projection, as indicated by the con- 
 struction, of the two points E 7 and F'. Each of these, again, is 
 the vertical projection of two points, of which the two foremost, 
 one in each pair, are horizontally projected at G found by 
 making CG^CE. .These four points, G,E'; G,F' ; E,E' ; and 
 E,F' cast shadows at e', g', n f , and/', respectively. By joining 
 the points thus found, and tinting the inclosed figure, we shall 
 have the required shadow of the circle on the vertical plane. 
 
 Remarks. a. The lines a a! ; b b' ; C' c', and D' c?', are 
 tangents to the shadow, which is an ellipse. Also, a' b', and 
 c' d', are conjugate diameters, from which the axes can be 
 found, if desired, as commonly shown among problems on the 
 ellipse. 
 
 b. Observing that in PI. I., Fig. 4, km is the shadow of 
 a vertical edge at C on the horizontal plane of projec- 
 tion, and that in Plate III., Fig. 10, a'b' is the shadow of the 
 diameter AB A' on the vertical plane, we conclude from 
 inspection, "vhat was proved in (31), that whenever a line is 
 perpendicular to either plane of projection, its shadoio on that 
 plane is in the direction of the projection of a ray of light on that 
 plane. 
 
 33. PI. III., Fig. 11, illustrates the forms of the shadows of a 
 circle, whose plane is oblique to the direction of the light, upon 
 several planes variously inclined to the plane of the circle. 
 
 AB represents a vertical projection of a horizontal circle, and 
 ABGC the vertical projection of an oblique cylinder of rays 
 having the circle AB for its base, and parallel to the vertical 
 plane of projection. The traces drawn through C, are the 
 traces of a number of planes, perpendicular to the vertical plane 
 
 2
 
 18 GENERAL PROBLEMS. 
 
 of projection, and whose intersections with the cylinder of rays 
 are the shadows of the circle AB. 
 
 DC is a circular shadow, equal to AB, its plane, DC, being 
 parallel to AB. CF is an elliptical shadow less than the circle 
 AB, and whose transverse axis, projected at the point/ is equal 
 to the diameter of AB. CF being a plane of right section, con- 
 tains the least shadow of AB. Oil is an elliptical shadow 
 greater than AB, CH, its transverse axis being greater than AB, 
 and h, its conjugate axis, being equal to AB. Between CF, the 
 least, and CH, the greater shadow, there must be one equal to 
 the circle AB. This is found by making GCF=FCD when 
 CG=CD, and as all the diameters h, g,f, etc., are equal, CG is 
 another circular shadow, called the subcontrary section of the 
 cylinder, to distinguish it from the parallel section CD. Finally, 
 all shadows between the circular ones CD and CG, are ellipses, 
 less than, and all exterior to these are ellipses greater than the 
 circle AB. 
 
 34:. The foregoing problems being sufficient to illustrate the 
 method of (30) we have the following general method for the 
 more general case in which the plane receiving the shadow is 
 not perpendicular to either plane of projection. Pass planes of 
 rays, each of which will cut a point, p, from the line L, casting the 
 shadow, and a line, k, from the plane, Q, receiving the shadow. 
 The point in which a ray through the point, p, meets the trace, k, 
 will be a point of the shadoio ofljonQ,. 
 
 Remarks. a. When L is a straight line, the plane of rays 
 may always contain it, instead of merely cutting points from it, 
 and then the trace of the plane of rays on the plane Q, will be 
 the shadow of L. 
 
 b. The trace just mentioned, must, however, be constructed 
 by the above method, as seen in the line cd, Prob. III. Hence, 
 practically, planes of rays are not immediately passed through 
 straight lines ; i. e. without construction, unless those lines are 
 perpendicular to a plane of projection ; in which case the planes 
 of rays containing them will be so also, and their traces will be 
 at once known. (29.)
 
 SHADES AND SHADOWS. 19 
 
 PKOBLEM VIII. 
 
 To construct the shadow of a straight line on a plane whose traces 
 are parallel to the ground line. 
 
 Construction. PI. L, Fig. 3. as a's' is the given line, and 
 MK and M'K' are the traces of the given plane. OTO' is a 
 vertical auxiliary plane which contains the given line, as a's', 
 and intersects the given plane in the line nT n'O'. Hence, 
 from Des. Geom. (Prob. VI.), the given line intersects the given 
 plane at s',s, which (Prob. III.) is one point of the required 
 shadow. LJI/ is another vertical auxiliary plane, containing 
 the ray ab a'b', drawn through the point a,a' of the given line 
 as a's'. This plane intersects the given plane in the line LJ 
 fe'L', hence the ray through a,a f intersects the given plane at 
 b',b, which by (34) is therefore another point of the required 
 shadow. Hence bs b's' is the required shadow. 
 
 Remarks. a. The auxiliary planes might have been placed 
 perpendicular to the vertical plane of projection. In that case, 
 the horizontal projections b and s of the points of the shadow 
 would have been found first. 
 
 b. Let this problem be solved thus, and also when the given 
 plane has any oblique position. 
 
 PKOBLEM IX. 
 
 To find the shadow of a chimney, situated on the end portion of a 
 hipped roof, upon the end and side roofs. 
 
 Principles. It only need be noticed here, first, that as most 
 of the edges of the chimney are perpendicular to one, or the 
 other, of the planes of projection, the auxiliary planes of rays 
 may all contain edges of the chimney, and be also perpendicular 
 to one of the planes of projection; and second, that when a 
 plane, P, is perpendicular to either plane of projection, all 
 lines in that plane, P, will be projected on the plane of projec- 
 tion, to which it is perpendicular, in its trace on that plane 
 (29). 
 
 Construction. PI. I., Fig. 5. URR"T K'T' is the horizon- 
 tal plane of the eave-trough, intersected by the roof in the lines
 
 20 GENERAL PROBLEMS. 
 
 TE, EG, and GU. GIU GTT' is the front roof, EIG GT 
 the end roof, perpendicular to the vertical plane of projection, 
 and EIT GTT' is the back roof; xw is the chimney-flue, 
 dve a horizontal section of its outside surface, and ABD A'C'D' 
 is its abacus. 
 
 The shadow of the abacus on the front of the chimney, ia 
 found as in Prob. IY. 
 
 To find the shadows on the roof: First, the shadow on the 
 front roof. C'Y' is the vertical trace of a plane of rays through 
 EC C', and perpendicular to the vertical plane of projection. 
 According to principle second, O'Y' is the vertical projection of 
 its trace on the front roof. Y' Y"Y is its trace on the 
 horizontal plane of the eave-trough ; hence, projecting 0' at O 
 we have OY and OY", for the horizontal projections of its 
 traces on the front and back roofs. Drawing the ray BA 
 whose vertical projection is C'Y' and projecting h at h', the 
 point A, h' is the shadow of B,C', and hO /I'O', the shadow of 
 BC C' on the front roof. Drawing a similar plane of rays, 
 BT/, its traces on the roof are F'L' FL", parallel to O'Y'- 
 OY", a line from L parallel to YO, and the short line, paral- 
 lel to GE, across the end roof at F. The ray B^ B'L' meets 
 the trace FL" at g,g', giving hg li'g' for the shadow of B B'C'. 
 From g,g f the shadow of AB A'B', on the parallel front roof, 
 is the parallel line, gfg'f. Drawing the ray /K /'K', we 
 find the portion BK B'K' whose shadow is gfg'f. 
 
 Second. The shadow on the end roof, ec is the trace, on this 
 roof, of a vertical plane of rays through the vertical edge, e 
 e"e', of the chimney. This plane cuts the edge, AB A'B', of 
 the abacus in the point a,a', through which, drawing the ray 
 a'c' oc, and projecting c' at c, we find ec, the shadow of the 
 portion e e"e' of the edge of the chimney. Above e' this edge 
 is in the shadow of the abacus on the chimney. Joining c,c' 
 and f,/'; cfc'f is the shadow of aK a'K'. Likewise, du is the 
 shadow of the vertical edge at d. By projecting u at u', and by 
 drawing the vertical projection of a ray through u', we could 
 find the portion of the edge from d,d'j upwards, which casts the 
 shadow du d'u'. 
 
 Third. To find the shadow on the back roof and eaves plane. 
 A'o' is the vertical trace of a plane of rays parallel to the plane 
 C'Y'Y". Its trace on the back roof is NV oN. It intersects 
 the edge, d d'v', of the chimney at d,v', through which, draw-
 
 SHADES AND SHADOWS. 2l 
 
 ing a ray, v'o' dp, we find' pjp' the limit of the shadow of d 
 d'v' on the back roof. This point p,p' is also by construction 
 the shadow of r,A' where the ray p'v' pd meets the edge 
 AD A' of the abacus. Drawing the ray or", rr" is determined 
 as the portion of this edge, which casts the shadow po p'o'. 
 The remaining portion, Dr", casts the parallel shadow on o', a 
 part of the trace of the plane of rays AV on the plane of the 
 eave-trough, and limited by the ray Dn AV. Drawing the 
 ray Y6, we find the limit of that portion of the edge BC C' 
 which casts a shadow on the back roof. &C casts the equal and 
 parallel shadow, Yy, on the parallel plane of the eave-trough. 
 Likewise ym is equal and parallel to CD C'D', and is limited 
 by the rays Cy and Dm. Then mn is the shadow of D A'D', 
 which completes all the required shadows. 
 
 Remarks. a. The boundary of the shadow on the back roof, 
 as seen in vertical projection, is fringed, to give it greater dis- 
 tinctness. 
 
 b. To locate either projection of a point on the side roofs, one 
 projection of such a point being given. Let H be the horizontal 
 projection of a vertical rod on the front roof. Draw Hg parallel 
 to the trace, GU, of the front roof; q is vertically projected at 
 q 1 , and j'H', parallel to GT', is the vertical projection of gH ; 
 hence H'H" is the vertical projection of the rod at H. This 
 construction would be used in locating the intersection, with 
 either front or back roof, of a chimney situated on either of those 
 roofs. (See Des. Geom. Prob. X.) 
 
 c. To find the shadow of the rod, H H'H", pass a vertical 
 plane of rays through it ; HJ, parallel to ~Bg, is the horizontal 
 projection of the trace of this plane on the front roof. Project 
 J at J' and H'J' will be the vertical projection of the same 
 trace. The ray H"A'" Kh" meets this trace at h"',h", giving 
 HA"- TL'h'" for the projections of the shadow of the rod. This 
 construction could have been employed in finding all points of 
 the shadow of the chimney, instead of making the planes of 
 rays perpendicular to the vertical plane. Observe that in the 
 construction given for the chimney, we find the horizontal pro- 
 jections of points, as h, first ; while in the construction of h"h'" 
 we find h" f first. 
 
 35. In dismissing the subject of shadows on planes, it may 
 be noticed, that it is sometimes advantageous to consider the 
 ray of light, itself, as making an angle of 45 with one of the.
 
 22 GENERAL PEOBLEMS. 
 
 planes of projection. For, suppose such a ray to be passed 
 through the uppermost point of a vertical line, and to make an 
 angle of 45 with the horizontal plane. Evidently, the distance 
 from the foot of the line to the intersection of the ray with the 
 horizontal plane which would be the shadow of the line 
 would then be equal to the height of the line. Hence, without 
 having any vertical projection of the line, its height may be 
 known from its shadow. Or, which amounts to the same thing, 
 the height of a point above the horizontal plane will be known 
 by the distance of its shadow from its horizontal projection. 
 
 The same essential principle is true in more general terms. 
 For, by knowing the direction of the light, we can find the rela- 
 tive lengths of vertical lines and their shadows, and this relation 
 being known, we can find the height of all points of an object, 
 having given only its plan and shadow. The same principles 
 are applicable to constructions made on the vertical plane of 
 projection. To illustrate, let us take the following example : 
 
 PEOBLEM X. 
 To find the shadow of a shelf and brackets upon a vertical plane. 
 
 Construction. PL II., Fig. 9. Let AB, taken as the ground 
 line, be the horizontal trace of the given vertical surface ; let 
 CD C'D' be the lower front edge of the shelf, and let cl c'V 
 be one of the brackets. 
 
 Let E'F be the vertical projection of a ray, drawn in any 
 direction, relative to the ground line. It is sufficient to under- 
 stand that this ray makes an angle of 45 with the vertical 
 plane. Then the shadow, c'd', of the edge en c' will be parallel 
 to E'F and equal to en. Next, d'e' can be drawn, parallel to 
 ca cV, and limited by the ray a'e', and the shadow of the 
 unshaded part of a a'V will be the parallel line e'k'. 
 
 Likewise, taking any point //, in the edge CD C'D' of the 
 shelf, make the ray //*' equal to fg : then h'k' } parallel to C'D' f 
 will be the shadow of CD C'D'. " 
 
 Remark. Though not necessary, the horizontal projection of 
 the ray may be found as follows: Make GA=E'F, then will 
 GF E'F be the two projections of a ray which makes an angle 
 of 45 with the vertical plane of projection.
 
 SHADES AND SHADOWS. 23 
 
 CLASS II. 
 
 SHADES AND SHADOWS ON SINGLE CURVED SURFACES, IN 
 GENERAL. 
 
 SECTION I. 
 On Developable Single Curved Surfaces. 
 
 1. Shades. 
 
 36. The line of shade on a developable surface, consists of 
 those reclilineUr elements along which planes of rays are tangent 
 to the surface, since such a line would evidently separate the 
 illuminated from the dark portion of the surface. Since the 
 tangent planes are planes of rays, they may be regarded as 
 parallel to a given ray, in space. Hence the general method for 
 the construction of the elements of shade on a developable sur- 
 face, resolves itself into the operations of (Des. Geom. Probs. 
 53, &c.) " To construct a plane, parallel to a given line, and tan- 
 gent to a developable single curved surface" Since two such planes 
 can generally be found in any case, there will usually be two 
 opposite elements of shade. 
 
 37. When the indefinite surface is limited by one or more 
 plane intersections, taken as bases, the complete line of shade 
 will consist of the elements of shade on the convex surface, toge- 
 ther with those portions of the circumferences of the bases which 
 divide light from dark portions of the entire body. 
 
 38. If a plane of rays be tangent to a developable surface 
 along one of its elements, then any secant plane will cut from the 
 surface a curve, and from the plane of rays, a line, tangent to the 
 curve at one point of the element of shade. Moreover, if the secant 
 plane be perpendicular to the tangent plane of rays, the line cut 
 by it from the latter will evidently be the projection of the light 
 upon the secant plane, since the tangent plane, when thus per- 
 pendicular to the secant plane, becomes a projecting plane of 
 rays upon the latter plane.
 
 24 GENERAL PROBLEMS. 
 
 PROBLEM XL 
 To consiruct ike elements of shade on any cylinder or cone. 
 
 Constructions. First. -To construct the element of shade on a 
 horizontal right semi- cylinder. PI. IV., Fig. 16. Let Yb be the 
 trace of one of two vertical planes of projection upon the plane 
 of the paper taken as the second vertical plane, and at right 
 angles to the former one. Let the former plane be perpendicular 
 to the common axis of the cylinder, and cylindrical abacus, shown 
 in the figure, and let the latter plane contain the axis of the 
 cylinder. The projections will then be as in the figure. Let 
 De DY be the projections of a luminous ray. De is also the 
 trace, on the right hand plane, of a plane of rays perpendicular 
 to that plane ; e is therefore the right hand elevation of its 
 element of contact with the cylinder, that is, of the element of 
 shade of the cylinder. E'e" is therefore the left, or linear elevation 
 of the same element. 
 
 Likewise, the parallel tangent plane of rays at G, determines 
 the element of shade, G G^G', on the abacus. 
 
 /Second. To construct the elements of shade on a cylinder whose 
 axis is oblique to both planes of projection. PI. III., Fig. 13. ABF 
 A'G' is the lower base of the cylinder. Ibe FJ' is its upper 
 base, Cb and M</ are its extreme elements, as seen in horizontal 
 projection, and A'F and G'J', the extreme elements as seen in 
 vertical projection. The axis of the cylinder pierces the hori- 
 zontal plane at K, the centre of the lower base. The ray L 
 a'L', through a point a,a' of the axis, pierces the same plane at 
 L, hence KL is the horizontal trace of a plane containing the axis 
 and a ray. From (5) such a plane is a plane of rays through 
 the axis. The tangent planes of rays will be parallel to the one 
 just found, hence their horizontal traces will be parallel to KL. 
 BB" and FF" are these traces, and their points of tangency, B 
 and F, with the lower base, are where their elements of contact 
 the elements of shade pierce the horizontal plane. Hence Bw 
 BV, and F/* F/', are the required elements of shade; one 
 projection of each being visible. 
 
 Tliird. To construct t/ie elements of shade on a cylinder whose 
 axis is horizontal, but oblique to the vertical plane of projection. PI. 
 IV., Fig. 17. In order to construct the vertical projection of a 
 cylinder, thus seen obliquely, it is necessary to have, first, its
 
 SHADES AND SHADOWS. 25 
 
 projection on a plane perpendicular to its axis. Then let NZH 
 be the horizontal projection of the cylinder, and take O"C", 
 perpendicular to its axis, for the ground line of a new vertical 
 plane, parallel to the base NZ. This base will then have the 
 circle H"O"6" for its auxiliary vertical projection, and that of the V j 
 cylinder. Now, any point as Z', in the principal elevation, is inj 
 a projecting line, ZZ', and at a height, Z'O', from the ground 
 line, equal to the height of the same point, as seen at Z", above 
 the ground line O"C". 
 
 Having thus found the projections of the cylinder, we may- 
 assume any two projections of a luminous ray, and find the third. 
 Let FL F'L' be the assumed projections of a ray. L,L' is pro- 
 jected on the auxiliary plane at L"; I/, and L" being at equal 
 heights above their respective ground lines. Likewise G" and 
 F', both projected from F, are at equal heights above their 
 respective ground lines. GK'L" is then the auxiliary projection 
 of the ray. " Hence also T"H" and M"J", parallel to G"L", are 
 the traces, on the auxiliary vertical plane, of tangent planes of 
 rays perpendicular to that plane. Hence H" HI and J" KJ 
 are two projections of the two elements of shade of the cylinder. 
 KJ is again vertically projected at K'J' at a height, equal to 
 J'V, above the ground line C'O', or by projecting J into the 
 principal vertical projection of the base at J', and drawing J'K' 
 parallel to the ground line. The former construction is obviously 
 more accurate in practice. 
 
 Otherwise : omitting all the operations of the last paragraph, 
 the ray, NO N'O', may be projected into the plane, NZ, of the 
 base of the cylinder, by projecting 0,0' at o",o'". Then, con- 
 sidering that N,N' is its own projection on the plane NZ, we 
 have No" Wo'" to represent the projections of the ray on this 
 plane. Hence draw a line tangent to the base, and parallel to 
 NV", and by the principle of (38) J' will be its point of tan- 
 gency with the base N'J'Z', and therefore a point of the element 
 of shade J'K'. 
 
 Fourth. To construct the elements of shade on a cone. PI. III., 
 Fig. 12, and PI. V., Fig. 18. 
 
 Principles. A plane is tangent to a cone along an element, 
 and all the elements contain the vertex of the cone, hence a plane 
 of rays tangent to a cone will contain the ray through its ver- 
 tex. Therefore the trace of this plane on any given plane, will 
 contain the intersection of that ray with this given plane, and,
 
 26 GENERAL PROBLEMS. 
 
 by (38) the point of contact of this trace with the section of the 
 cone, made by the given plane, will be a point of the element of 
 shade. 
 
 Constructions. In PI. III., Fig. 12, draw a ray, VE Y'E', 
 from the vertex, V, V, of the cone. It pierces the horizontal 
 plane, which is here the plane of the cone's base, at E. Then 
 ET is the horizontal trace of one of the two tangent planes which 
 can be drawn, and TV T'V is the element of shade determined 
 by it. In PI. V., Fig. 18, as the ray, YN Y'E' pierces the 
 horizontal plane within the cone's base, on that plane, no lines 
 can be drawn from N, tangent to the base. This indicates that 
 no planes of rays can be drawn tangent to the cone. The lower 
 nappe, therefore, would be wholly illuminated, but. for the pre- 
 sence of the upper nappe, which casts a shadow upon it. 
 
 Fifth. To construct the elements of shade on two intersecting 
 cylinders, wJiose aoces are at right angles to each other, in a plane 
 parallel to the vertical plane of projection, one of these axes being ver- 
 tical. PI. V., Fig. 19. The elements of shade on the vertical 
 cylinder are found at once, by drawing tw T o vertical planes of 
 rays, tangent to that cylinder. A/ is the horizontal trace of 
 such a plane, coinciding with the horizontal projection of a ray, 
 and giving the element of shade A A'A". 
 
 The elements of shade on the horizontal cylinder, are found 
 by revolving the plane of either of its bases either into, or paral- 
 lel to, one of the planes of projection. Let the plane uDv ' be 
 revolved about its vertical trace Di/, into the vertical plane of 
 projection. The centre n,ri of the base contained in this plane, 
 will then revolve in the arc nn'" n'n", to n". The circle, with 
 n" as a centre, and radius n"T"=n'E', will then be the revolved 
 position of the base uy eV, and the projection of the entire 
 horizontal cylinder upon the plane uT)v'. Now, T"7c" is the 
 trace on the plane of the revolved base, of a plane of rays per- 
 pendicular to that plane and tangent to the horizontal cylinder, 
 T"&" being drawn in any assumed direction, if the principal pro- 
 jections of a ray are not given. T", and the point V" diametri- 
 cally opposite, are then the auxiliary projections of the elements 
 of shade of the horizontal cylinder. By a counter revolution, in 
 which these points, considered as the extremities of these ele- 
 ments of shade, revolve in the arcs T'Y , t"e and Y'V v"p, 
 we find ell e'W for the lower front element of shade, and 
 pV v'V for the upper back element of shade.
 
 SHADES AND SHADOWS. 27 
 
 Remark. If the principal projections of the light were given, 
 T"&", instead of being assumed, would be constructed as the 
 auxiliary projection of the light upon the plane uDv', by an 
 obvious construction. 
 
 n. Shadows. 
 
 39. As in the case of planes, so in that of single curved sur- 
 faces, when all their elements are parallel, the projection of the 
 surface will be a line, when it is perpendicular to a plane of pro- 
 jection. The projection of the perpendicular plane will, how- 
 ever, be a straight line, while that of the single curved surface 
 will be a curve ; viz. its curve of right section. As cylinders 
 alone, among single curved surfaces, have parallel elements, they 
 only can have all their elements perpendicular to the same plane, 
 and hence they only can be projected in a line. 
 
 40. For all cases of shadows upon cylinders whose axes are 
 perpendicular to a plane of projection, we have, from the pre- 
 ceding article, the following special method. Determine, by inspec- 
 tion, the intersection, M, of a ray through any point, P, of the line 
 casting the shadow, with the linear projection of the cylinder. This 
 will be one projection of the shadow of P. Its other projection will 
 be Hie intersection of the line perpendicular to the ground line, Hirough 
 M, with the otiier projection of the ray through f. 
 
 The two following problems illustrate this method. 
 
 PROBLEM XII. 
 
 To construct the shadow of the projecting head of a cylinder upon 
 
 the cylinder. 
 
 PI. IY., Fig. 16. The projections of this body have been fully 
 described in the first case of the preceding problem. The right- 
 hand edge, G"A', of the head, above the element of shade, G 
 G'G", casts the shadow required. Drawing the projection, Cc?, 
 of a ray, on the plane containing the linear projection, bdE, of 
 the cylinder, d is at once found, as one projection of the shadow 
 of C,C' upon the cylinder. Its other projection, d', is the inter- 
 section of the perpendicular, d d f , to the trace (ground line) F6, 
 with the other projection, C'ef, of the ray through 0,0'. The
 
 28 GENERAL PROBLEMS. 
 
 other points of shadow are found in exactly the same way, as is 
 evident by inspection. The tangent ray, De D'e', determines 
 that point of shadow, e,e', in which the shadow is lost in the 
 element of shade, e e'e". 
 
 PROBLEM XIII. 
 
 To construct the shadows of the edges of shade of a cross upon a 
 cylinder, both bodies being seen obliquely. 
 
 Principles. PI. IY., Fig. 17. The three projections of the 
 cross and cylinder, of the ray of light, and of the elements of 
 shade of the cylinder, having been already constructed, as 
 explained in Prob. XI. (Third), we may proceed at once with 
 the construction of the shadow, after having, first, determined, 
 by inspection, the edges of shade of the cross, and, second, the 
 limits of that part of the body of the cross, whose edges of shade 
 cast shadows on the cylinder. 
 
 The edges of shade must, evidently, here be determined with 
 reference to a single direction of the luminous ray in space; that 
 is, we must find the projections of those edges, on the three 
 planes of projection, which in space really cast shadows, when, 
 the light comes in the single direction projected in FL FT/ 
 Gr"L". These edge* are, in the present case, B"D" BD B'D' ; 
 the one diagonally opposite, through YY"; then c" c"'P c'P', 
 and the one diagonally opposite, G" FG F'G' ; also X"G" 
 XG X'G' ; its diagonally opposite edge, c"w" c'"w ; next 
 XV XP XT' ; and its diagonally opposite edge, Fw G"*, 
 and the horizontal edge at D",D,D'. 
 
 The limits of that portion of the cross, which casts a shadow 
 on the cylinder, are found by drawing the two tangent planes of 
 rays, whose traces are IVT'J" and T"H", and which determine 
 the elements of shade on the cylinder (Prob. XI. Third). These 
 planes contain those points of the cross, which cast shadows on 
 the elements of shade, where the shadow disappears from the 
 cylinder. 
 
 Construction. The point M",M,M', cut from the cross by the 
 plane M"J", casts a shadow on the element of shade, at a point 
 whose auxiliary projection is J", whose horizontal projection 
 may be found by projecting 3" into the horizontal projection, 
 Ma, of the ray through M",M, at a ; and whose vertical pro-
 
 SHADES AND SHADOWS. 29 
 
 jection may then be found by projecting a into the vertical pro- 
 jection, MV, of the same ray, at a'. Otherwise : we may con- 
 struct the vertical projection, J'K', of the element of shade, on 
 which the shadow of M",M' is known to fall, as explained in 
 (Prob. XI. Third}. Then the intersection, a', of this element 
 with the ray, vertically projected in MV, is the vertical projec- 
 tion of the required point of shadow, which may be then pro- 
 jected into the horizontal projection, Ma, of the same ray, 
 at a. 
 
 Thus every point of shadow may be found, first in horizontal 
 projection, and then projected up ; or first in vertical projection, 
 and then projected down. Also, a may be projected up into 
 J'K', instead of into M'a' ; or, a' being first found, as above, it 
 may be projected down into JK, to find a. Indeed this is. prac- 
 tically, the most exact construction. 
 
 As all the other points of the shadow are found precisely as 
 just described, we shall only point out those, which, in the 
 present figure, it is desirable to find. This will assist in solving 
 the problem, when the relative position of the bodies, the light, 
 and the planes of projection, are slightly changed. 
 
 The shadow on the foremost element, b" vN v'X', is cast 
 by the point, R", R,R', in the ray &"R", and falls at &",&,&'. 
 
 In the figure, the cross is made tangent to the cylinder along 
 the edge c" c'"P, which is therefore its own shadow. The 
 highest point, S", of the edge B"X", which casts a shadow, is found 
 by drawing the ray c"S". Its shadow is c",c,c'. The frag- 
 ment, S"X", casts the shadow, cT, on the front of the short arm 
 of the cross. 
 
 The shadow on the highest element, is found by drawing the 
 ray p"Q,' f , which determines the point Q",Q,Q', whose shadow 
 is p"&e'. The edge XV XP casts the shadow Pd P'd' 
 the edge XGr, the shadow from d to the point just beyond e ; the 
 edge GF, the shadow from the latter point to o ; the edge through 
 F and w, the shadow op, and the edge through c" and w", the 
 shadow pc'", partly invisible. The portion G"T" UT, of UD, 
 casts the shadow gf, and the portion, V"^ of the lower left hand 
 edge, the shadow Teh. 
 
 JfemarJcs. a. The portion of shadow cast on the horizontal 
 plane in front of the cylinder is shown. It is found as in Prob. 
 III. The shadow of the cylinder, and of the part of the cross 
 which is beyond the plane H"T", can be found on the horizontal
 
 30 GENERAL PROBLEMS. 
 
 plane in the same manner, and will add to the beauty of the 
 figure, especially when shaded in graduated tints. 
 
 b. If the light were to make a greater angle with the horizon- 
 tal plane than the edges parallel to X"c", as seen in the auxiliary 
 projection, the edges, as X" XT, would be edges of shade. 
 
 c. Those edges are inked as lines of shade, in the auxiliary 
 elevation, which are so, in reference to the single direction of the 
 luminous ray, shown in this figure. 
 
 41. The general method for rinding shadows on oblique cylin- 
 ders, and other single curved surfaces, is this : Pass planes of 
 rays so that each shall cut the line casting the shadoiv in a point P, 
 and the single curved surface in a straight element, E or in some 
 other simple section. The point, S, where tJie ray through P meets 
 the element, E, is tJie shadow of P upon E, and is therefore one point 
 of Oie shadow of the given line upon the given surface. 
 
 Remarks. a. For the cylinder; a plane of rays must be 
 parallel to the axis, in order to intersect the cylinder in straight 
 elements. 
 
 b. For the cone ; a plane must contain the ray through the 
 vertex, in order to intersect it in straight elements. 
 
 c. The two following problems illustrate this general method. 
 
 PROBLEM XIY. 
 
 To construct the shadow cast by the upper base of a hollow oblique 
 cylinder upon its interior. 
 
 Principles. The method of (41) applied to this problem gives 
 the following principles of solution. Any secant plane, parallel 
 to the plane containing the axis and a ray, will cut two recti- 
 linear elements from the cylinder, one of which will be towards 
 the source of light. The upper extremity of this element casts a 
 point of shadow on the opposite element in the same plane. The 
 elements contained in the same tangent planes of rays, coincide 
 in one, which is the element of shade. Hence the upper extre- 
 mity of this element is also a point of shadow on the interior of 
 the cylinder. 
 
 Constructions. PI. HI., Fig. 13. According to the foregoing 
 principles, f,f, and u,u' ; the upper extremities of the elements
 
 SHADES AND SHADOWS. 31 
 
 of shade (Prob. XL Second) are the points where the required 
 shadow on the interior begins. 
 
 To find any other point of this shadow. The plane of rays 
 whose horizontal trace, MET, is parallel to the horizontal trace, 
 NEL, of the plane of rays through the axis, contains the elements 
 M<7 and He. The point, g,g', of the former element, casts its 
 shadow on the latter element, at h,h. This point may be found 
 by drawing the horizontal projection, gh, of a ray, noting the 
 point h, and then projecting it, either into the vertical projection, 
 g'h', of the ray, or into the vertical projection, D V, of the ele- 
 ment containing it. Otherwise : the same point may be found 
 by first drawing the vertical projection, g'h', of the ray through 
 g,g\ noting its intersection, /?', with the vertical projection, DV, 
 of the element containing it, and then projecting it either into 
 gh, the horizontal projection of the same ray, or into He, the hori- 
 zontal projection of the element containing h,h r . 
 
 To find the shadow cast by any particular point, assumed in 
 advance, on the upper base, as n,n'. Draw the element, n N, 
 containing this point, and through N, its intersection with the 
 horizontal plane, draw the trace, NE, of a plane of rays, and 
 then proceed as before, to find 0,0', the shadow of n,n f . 
 
 Conversely, to find the shadow cast upon any element, assumed 
 in advance, as Cb C'J'. Through the intersection, C, of this 
 element, with the horizontal plane, draw the trace CE, parallel 
 to NE, of the plane of rays which cuts from the cylinder the ele- 
 ment Er, towards the source of light ; whose upper extremity, r, 
 casts the shadow, 5,5', on the assumed element bG J'C'. 
 
 The lowest point, 0,0', of the shade, is evidently on the ele- 
 ment, EC? E'cT, contained in the plane of rays through the axis, 
 because the two elements contained in this plane are the farthest 
 apart, being diametrically opposite. 
 
 Remarks. a. It does not strictly belong to this problem to 
 construct the shadow of the cylinder on the horizontal plane. It 
 is however shown in the figure, and is bounded by BB", the 
 shadow of the element of shade, Eu B V ; the semi-circle, B"F", 
 whose centre is L, and which is the shadow of udf, and FF", the 
 shadow of the element of shade, FfFf. 
 
 b. The vertical projection of the shadow on the interior of the 
 cylinder is invisible. 
 
 c. It happens in the present figure, that the horizontal pro- 
 jection, /ma, of the same shadow is a straight line. This shows
 
 32 GENERAL PROBLEMS. 
 
 that tliis shadow is a plane curve, whose plane happens, in this 
 instance, to be vertical. The shadow being a plane curve, and 
 also the intersection of two cylinders having a common base, 
 /JI JT, viz. the given cylinder, and the cylinder of rays 
 through its upper base we infer, what is actually true, that the 
 intersections of such cylinders, having, as they do, two common 
 tangent planes, are plane curves. 
 
 d. The last remark serves as a simple illustration of the use of 
 the exact constructions of Descriptive Geometry as a means of 
 research in discovering theorems, particularly of form and posi- 
 tion, relating to geometrical magnitudes. 
 
 42. In the following problem, besides an illustration of the 
 general method of (41) there is an illustration of the special method 
 called the method of one auxiliary shadow, which is as follows: 
 
 Where the shadow of a line on any surface meets the intersec- 
 tion of that surface with a second surface, is a point of the sha- 
 dow of the given line upon that second surface. 
 
 Thus : where the shadow of a staff, upon the ground, meets 
 the intersection of the ground with a house, is where the shadow 
 of the staff upon the house begins. 
 
 PROBLEM XV. 
 
 To construct the shadow of the upper base of a vertical right cone, 
 upon the lower nappe of tiie same cone. 
 
 Principles. According to (41) the upper base is the line cast- 
 ing the shadow, the lower nappe is the single curved surface 
 receiving the shadow, and any secant plane containing the ray 
 through the vertex, and cutting the base, is a secant plane of 
 rays containing two elements of the cone. The intersection of 
 each of these elements with the upper base, is, in general, a point 
 casting a shadow on the other element. 
 
 Construction. PI. V., Fig. 18. AGL and A'H' V' A"H" 
 are the projections of the cone, and VN V'E', of the ray 
 through the vertex, which pierces the horizontal plane at N,E', 
 and the plane of the upper base at D'",D". As all the planes of 
 rays are to contain this ray, their traces on the plane of the lower 
 base will all pass through N, and those on the upper base, 
 through D'", as seen in horizontal projection.
 
 SHADES AND SHADOWS. 33 
 
 To find the points of shadow, which fall on the circumference of 
 the lower base of the cone. These points are found by the special 
 method of (42). The ray VJ V"J', through the centre of the 
 upper base, pierces the horizontal plane (the plane of the lower 
 base) at J'. Then a circle with J as a centre, and radius JL 
 equal to V"H", that of the upper base, will be the shadow of the 
 upper base on the plane of the lower base. LG is an arc of this 
 shadow, and L,L' and G,G', its intersections with the lower base, 
 are the required points of shadow on the circumference of that 
 base. Other points might be similarly found. 
 
 Ifemark. It will be observed that, by this method, we neces- 
 sarily find, neither the elements nor the rays, containing these 
 points of shadow ; nor, consequently, the points casting them. 
 To find the elements, connect L,L' and G,G' with the vertex 
 V,V. To find the rays, draw the projections of rays through 
 these points and note their intersections with the corresponding 
 projections of the upper base. The latter intersections will be 
 those points of the upper base, whose shadows are L,L' and G,G'. 
 
 To find other points of the required shadow. These are found by 
 the general method of (41). Let A 1ST I be the horizontal trace of 
 any secant plane of rays. It cuts from the cone the two ele- 
 ments IYB, and AH A'V'H". The point B,B" of the former, 
 in the upper base, casts a shadow on the latter at n f ,n, found by 
 drawing the vertical projection B' V of a ray, and projecting its 
 intersection, n f , with A'V, at n, on AY, which last point may 
 also be found as below. 
 
 Again : let AF be the horizontal projection of the trace of a 
 secant plane of rays, upon the plane of the upper base. This 
 plane cuts the upper base at A,A", and the cone in the element 
 FVM H'V receiving the shadow of A, A", at a,a'. This point 
 was found by drawing the horizontal projection, Aa, of a ray, 
 and projecting a into M'V at a'. The point, a, might, instead, 
 have been projected into the vertical projection, A'V, of the 
 same ray. It might also have been found as n,ri was. (See a,a' 
 in PI. IV., Fig. 17.) 
 
 "When the secant plane of rays, as at EK, is perpendicular to 
 the vertical plane of projection, the vertical projections of the 
 elements, and rays, contained in it, are confounded together in 
 its vertical trace ED". Hence we have not, in this case, the 
 choice of methods of procedure, given above, but must find, 
 first, the horizontal projection of the point of shadow contained 
 
 3
 
 84 GENERAL PROBLEMS. 
 
 in this plane. Thus : the plane ENK cuts from the cone the 
 elements EV and KVD ; the latter of which determines the 
 point D,D" on the upper base, which casts a shadow on the 
 former at b,b' ; found only by drawing the ray D&, and projecting 
 b into the vertical projection, E'V, of the element EV, at b'. 
 
 The highest point of the shadow, is in the vertical meridian 
 plane of rays, CVJ. The point 0,0", cut from the upper base 
 by this plane, casts a shadow on the element CY C'V, in the 
 same plane, at e,e'. 
 
 Through the points now found, the projections of the curve of 
 shadow can be sketched. In the horizontal projection, the upper 
 nappe conceals the shadow. In the vertical projection, the whole 
 of the front of the upper nappe, and all of the lower nappe, above 
 the line of shadow rib'Gr, are shaded, being visible portions of 
 the darkened surface of the cone. 
 
 Remark. If the light had made a less angle with the hori- 
 zontal plane, than that made by the elements of the cone, there 
 would have been elements of shade on the cone and a shadow 
 cast by the upper circle on the interior surface. It is recom- 
 mended that the problem should be solved under these con- 
 ditions. 
 
 43. In the following problem occurs an illustration of the 
 special method of (40), of the special method of (42), and of a new 
 special method, which may be called the method by one indefinite 
 rectilinear projection of the shadow, known by inspection. Thus the 
 shadow of a vertical straight line, upon any surface, will be 
 straight, as seen in horizontal projection. The method itself, as 
 applied in finding any other projection of the shadow, is this. 
 Having tfie intersection of the given projection of the shadow with any 
 element of the surface on which it falls, project that point upon OIA 
 oilier projections, one or more of the same element. 
 
 PROBLEM XVI. 
 
 To construct Hie shadow cast by the vertical cylinder (Prob. XL, 
 Fifth] upon the horizontal cylinder, and by the horizontal cylinder 
 upon tfte vertical cylinder. 
 
 Principles. The lines of each cylinder, which cast the sha- 
 dows, are their elements of shade, and the portions of the cir- 
 cumferences of their bases, which are curves of shade, as found
 
 SHADES AND SHADOWS. 35 
 
 iu (Prob. XL, Fifth). The shadow of the horizontal cylinder 
 upon the vertical one, will be found by the special method of 
 (40). The shadow of the upper base of the vertical cylinder on 
 the horizontal cylinder will be found by the special method of 
 (42), and the shadow of the element of shade of the vertical 
 cylinder, upon the horizontal cylinder, will be found by the spe- 
 cial method of (43). 
 
 Constructions.l . PI. V. Fig. 19. To find the shadow of the 
 lower front element of shade, eH e'H, of the horizontal cylinder, 
 upon the vertical cylinder. This shadow begins at t,t', where this 
 element of shade pierces the latter cylinder. Through any point, 
 X,X', of the element of shade, pass a ray Xx XV. This ray 
 meets the vertical cylinder in a point, whose horizontal projec- 
 tion is x, and whose vertical projection is x ', the intersection of 
 the perpendicular, x x', to the ground line, with the vertical pro- 
 jection, XV, of the ray. In the same manner, s,s', the shadow 
 of the extremity of the element of shade, is found ; also, r,r', the 
 last visible point of shadow on the vertical cylinder, which is 
 cast by the point, q,q', of the left hand base, found by drawing 
 the ray, rq r'q', backwards from r, the horizontal projection of 
 the extreme left hand element of the vertical cylinder. Through 
 the points now found, t'x's', the line of shadow, cast by ete't', 
 may be sketched ; also, #V, the shadow of the portion, eqe'q', 
 of the base, just mentioned. 
 
 2. To find the shadow of the upper base of the vertical cylinder, 
 upon the horizontal cylinder. Assume any horizontal plane, as 
 the one whose vertical trace is W, and containing elements of 
 the horizontal cylinder, in which it is supposed that points of the 
 required shadow will fall. This plane cuts two elements from 
 the latter cylinder, the hindmost of which is projected at W 
 V". By making a counter-revolution, W v"p, of V", to 
 v',p, we find ^Y, the horizontal projection of this element. O'D' 
 OD is the ray through the centre of the upper base, and it 
 pierces the plane W at D',D. A circle described with D as a 
 centre, and radius equal to OT, is the shadow of the upper base 
 on the auxiliary plane VV ; then d, the intersection of this 
 shadow with the element, pV, cut from the cylinder by that 
 plane, is the horizontal projection of a point of the shadow of the 
 upper base of the vertical cylinder upon the horizontal cylinder. 
 By projecting d, at d\ into the vertical projection, W, of the 
 element in which it lies, we have both projections of this point
 
 36 GENERAL PROBLEMS. 
 
 of shadow. The point Z>,&' is found in the same manner on the 
 element nib B'Z/ contained in the auxiliary horizontal plane 
 whose vertical trace is B' b'. In like manner the point c,c', on 
 the highest element, wF E'F', is found ; the shadow of the cen- 
 tre of the base OT, on the horizontal plane containing the high- 
 est element, being at C',C. 
 
 3. To find the shadow of the front element of shade, A A 'A", 
 of the vertical cylinder, upon the horizontal cylinder. By (43) we 
 have, at once, the straight line AK, coinciding with the horizontal 
 trace of a vertical plane of rays through A A' A", for the hori- 
 zontal projection of this shadow. It begins at A,A', where the 
 element of shade, A A' A", pierces the upper half of the hori- 
 zontal cylinder. Any other point, as g', of the vertical projec- 
 tion, is thus found. Assume any element, as ^'J', whose projec- 
 tion on the end elevation is g". By a counter-revolution, the 
 element through g" is found in horizontal projection at Jg. 
 Project g, its intersection with the shadow, AK, into the vertical 
 projection, J'g', of the assumed element, and g' will be determined 
 as required. Likewise a,a', on the highest element, and //', on 
 the back element of shade, produced, are found. The curve 
 h'a'f, which, being an oblique plane section of a cylinder, is an 
 ellipse, can next be sketched through the points now found. To 
 find how far this shadow is real, draw the vertical projection, 
 A'V, of the ray through the highest point A, A", of the element 
 of shade A A' A", note its intersection, o', with the indefinite 
 shadow h'a'f, and project o' at o. Then Ao h'g'o' is the definite 
 shadow of A h'A." upon the horizontal cylinder. At 0,0' 
 begins the shadow, ocbd o'c'Vd', of the upper base, which is lost 
 in the shade of the cylinder at d,d'. 
 
 Remarks. a. When, as in this case, the front element of shade 
 of the vertical cylinder lies to the left of the point, I/, a minute 
 shadow will be cast by it on the lower half of the horizontal 
 cylinder near I/. When, however, the same element of shade 
 lies to the right of I/, a small portion of T" I/IF will cast a 
 small shadow on the vertical cylinder. 
 
 b. By constructing the base G'F in end elevation, which would 
 be a straight line equal to GT, and in GT produced, and by 
 constructing, in the same elevation, the element of shade 
 A A' A", both of the shadows on the horizontal cylinder could 
 have been found by the simple special method of (40), applied 
 very nearly as in Prob. XIII.
 
 SHADES AND SHADOWS. 37 
 
 PfiOBLEM XVII. 
 
 To find the axes of an elliptical shadow, two of whose conjugate 
 diameters are known. 
 
 Principles. PI. VI., Fig. 66. It has already been shown 
 (Elemen. Proj. Drawing, Div. IV., Chap. V.) and (Descrip. Geom.) 
 that, if the projecting lines of an object are oblique to a plane of 
 projection, surfaces parallel to that plane will be shown, still, in 
 both their true form and size ; and that lines perpendicular to 
 that plane will be shown as parallel lines, whose direction will 
 depend on that of the oblique projecting lines. 
 
 In case the projecting lines make an angle of 45 with the 
 plane of projection, as in " Cavalier Perspective " or Cabinet 
 Projection, the projection of a perpendicular to the plane of pro- 
 jection will be equal to the line itself; since the line, its projec- 
 tion, and its projecting line, taken together, will form an isos- 
 celes right-angled triangle. But if the projecting lines make any 
 other angle than 45 with the plane of projection, the projection 
 of the line will be longer or shorter than the line itself. The 
 latter case is shown in Fig. 66, which is a general oblique pro- 
 jection of a cube; FG, etc., being less than EF, the edge parallel 
 to the plane of projection. 
 
 To view this figure, now, with reference to the present problem, 
 the ellipse mQnp is the oblique projection of the circle inscribed 
 in that face of the cube which is perpendicular to its front face, 
 EFLR, in the line LR. Hence the figure also truly represents, 
 in a pictorial manner, the projection of the circle MPNQ, con- 
 sidered as vertical, upon the horizontal plane LRHK ; and, more- 
 over, if Oo, which thus represents an oblique projecting line, is 
 regarded as a ray of light, then mQnp will represent the shadow 
 of MPNQ on the plane LRHK. 
 
 Construction. Let inn and pQ, Fig. 67, be given conjugate 
 diameters of an elliptical shadow, represented by oblique projec- 
 tion, as diameters of the shadow of a vertical circle, upon the 
 horizontal plane which is represented by the part of the paper 
 below A'D' ; a parallel to mn, through Q, as in Fig. 66. Then 
 we have ora, equal and parallel to the radius of the original circle 
 casting the shadow, and Q, as the shadow of the foot of the ver- 
 tical diameter of the same circle. Then erect QP, perpendicular
 
 38 GENERAL PROBLEMS. 
 
 to AT)', and equal to mn, and OQ=|PQ is the radius of the 
 original circle. 
 
 Next, lines that are parallel in reality, as ON and a tangent at 
 P, are parallel in projection ; and conjugate diameters are, each, 
 parallel to the tangents at the extremities of the other. Hence 
 any diameters at right angles to each other, in MPNQ, will cast 
 shadows which will be conjugate diameters of the given ellipse. 
 We therefore seek a pair of such diameters, whose shadows shall 
 be conjugate diameters at right angles to each other, for these 
 will be the axes required. 
 
 Now, since the shadows of the indefinite radii of MPNQ begin 
 in A'D', these radii, and their rectangular shadows, must be in- 
 scribed in two semicircles, having a common diameter in A'D'. 
 Hence 00 is a chord of the circle A'OD'o, thus composed, and 
 gG, perpendicular to it at its middle point, meets A'D' in the 
 centre, G, of this circle. Then O A' and OD' are the rectangular 
 radii, whose shadows, A'o and D'o, are at right angles to each 
 other, and are each parallel to the tangents at the extremities of 
 the other. Hence, limiting them by the rays a A and c?D, and 
 making oB=oA, and oC=0D, we have AB and CD for the 
 required axes of the elliptical shadow mQnp. 
 
 Remark. The construction just given is often found among 
 plane problems on the conic sections, as it can readily be explained 
 by the principles of plane geometry. But, as required by the 
 spirit of the present subject, it is here explained by the principles 
 of projections and of shadows. 
 
 44. The problem of the Niche is a favorite one in Shades and 
 Shadows, owing to the considerable number of points of peculiar 
 interest connected with it. The niche is a familiar concavity in 
 the walls of halls, staircases, etc., and consists usually of a verti- 
 cal half cylinder of revolution, united by its upper base with a 
 concave spherical quadrant. The problem of the niche, as a 
 problem of shadows, consists in finding the shadow of the edge 
 of shade of the cylindrical part upon that part, and upon the 
 lower base; and the shadow of the front semicircle of the 
 spherical part, upon that part, and upon the cylindrical part. 
 
 45. Agreeably to the classification of (12, 13) it becomes de- 
 sirable to divide this problem, and to retain here the construction 
 of the shadows on the cylindrical part, only leaving the shadow on 
 the spherical part to be placed with shadows on other double-curv-
 
 SHADES AND SHADOWS. 3 ( J 
 
 ed surfaces. One of the topics of special interest, in this problem, 
 is the direct construction of that point of shadow which falls on 
 the curve of the upper base ; and there are several such construc- 
 tions, but which are applicable only when the spherical part is 
 included in the problem. Hence it becomes quite desirable tc 
 find also a direct construction of the point in question which 
 shall be applicable in the problem as given below, where the 
 spherical part is not recognised. Such constructions will be found 
 in the following problem. 
 
 PROBLEM XVIII. 
 
 Having a vertical semi-cylinder with its meridian plane parallel to 
 the vertical plane of projection, it is required to find the shadow 
 cast on its base and visible interior, by its edge of shade, and by a 
 vertical semicircle, described on the diameter of its upper base. 
 
 Principles. All the points of shadow, save that on the circum- 
 ference of the upper base (45), are found by the special method 
 of (40). 
 
 Constructions. PI. YL, Fig. 20. AcZB A'A^B'B" is the 
 vertical half cylinder ; AB A^F'B" is the vertical semicircle, 
 whose shadow is to be found, and Aa A"a" is a ray of light. 
 
 1. To find the shadows of the vertical line, A A 'A". Aa 
 is the horizontal trace of a plane of rays through the vertical line, 
 A A A/'. This plane will, therefore, cut the cylinder in an 
 element whose horizontal projection is a. Hence, by drawing 
 a' a", the vertical projection of this line, and the rays, a'b' and 
 A"a", we find the whole rectilinear shadow, a' a", on the cylin- 
 drical surface, also the portion A"6' of the line A' A", which casts 
 the shadow a' a". The shadow of A'b', on the base of the 
 cylinder, is Aa. 
 
 2. To find points of shadow cast by the semicircle AB 
 A"F'B". Cc G'c' is a ray, through any assumed point, CO', of 
 this semicircle. It pierces the cylinder at the point whose hori- 
 zontal projection is c, and vertical projection c' (40) ; cc' is there- 
 fore the required shadow of CO'. Other points of this shadow 
 are found in the same manner. 
 
 3. To find the point of shadow on the upper base of t/ie cylinder, we 
 will first use the following special method of auxiliary oblique projec- 
 tions. The ray OS C'S' pierces the plane of the upper base of the
 
 40 GENERAL PROBLEMS. 
 
 cylinder at S'S. If then the semicircle, AB A"FB", be re- 
 volved forward about AB A"B" as an axis, till it coincides with 
 the plane of the upper base, CO' will appear at C", and C"S 
 will be an oblique projection of the raj CS C'S', when the 
 projecting lines come directly forward and downward at an angle 
 of 45 with the horizontal plane. Observing the limitation of 
 C"S by CS, it may be described as being inscribed in a circle, 
 U SKC", whose centre, U, is found at the intersection of AB 
 with uU, the bisecting perpendicular of C"S. Now a parallel ray, 
 similarly inscribed in the circle E B AG", will evidently be the 
 similar representation of a ray which contains a point of the semi- 
 circle AG"B casting the shadow, and of the semicircle AcB 
 receiving the shadow. The point on Ac?B will be the point of 
 shadow sought. This ray, parallel to C"S, will be a homologous 
 side of a triangle similar to SCO", and similarly situated. From 
 these things, it follows that the two rays will be to each other as 
 the radii, UK and EA, of the circumscribing circles SKC" and 
 BAG". Constructing this proportion at qmo, we find mq for the 
 length of the parallel ray, cZG". Its extremity, d. is located by 
 drawing Ed parallel to US. Projecting d at d', gives dd' as the 
 required point of shadow on the circumference of the upper base 
 of the cylinder. 
 
 Remark. To give completeness to the figure, draw dG" parallel 
 to SO' 7 . It will be equal to qm. Then make the counter-revolu- 
 tion of AG"B to its primitive position, when G" will be found in 
 its primitive position at GG'. Then drawing the projections, Gd 
 and G'd', of the ray which determines dd', we shall have the point 
 GG', whose shadow is dd', and, in plan, the complete triangle 
 G"Gd, similar to C"CS. 
 
 4. By another special method, that of auxiliary spheres, the 
 ray CS C'S' pierces the plane of the upper base of the cylinder 
 at S'S, as before. A plane, perpendicular to this ray, at its middle 
 point rr', will intersect AB A"B" in the centre of a sphere of 
 which the ray is a chord. Knowing (Des. Geom. 53i) that the 
 vertical trace of this plane will be perpendicular to C'S', the line 
 r'T' rT, where r'T' is perpendicular to C'S', and rT parallel to 
 the ground line is a line of the plane, hence T'T is a point of 
 its trace on the plane of the upper base. This trace must be per- 
 pendicular to CS, hence it is the line TU. UU' is then the 
 centre of a sphere, whose radius is US, and in which the ray 
 CS C'S' is inscribed.
 
 SHADES AND SHADOWS. 41 
 
 To find the ray similarly inscribed in a sphere whose centre ia 
 E,E' and radius EA. 
 
 In the proportion UK : EA::CC" : GG" (=gG f ) we find, by a 
 construction like qmo, the fourth term, and locate it as at #G'. 
 Then draw the ray Gd GW, which intersects AdB A"B" a 
 dd' the point sought. 
 
 Remarks. a. By finding the shadow on the cylindrical sur- 
 face produced, as at//"', the intersection of the sketched line of 
 shadow, a"c'f, with A"B", gives d' approximately and indirectly. 
 Such constructions are less satisfactory than determinate inter- 
 sections, found by direct constructions, such as the two preced- 
 ing. 
 
 b. The straight shadow, a' a", and the curved one, d'c'a", are 
 tangent to each other at a" ; for a a! a," is the element of con- 
 tact of a vertical tangent plane at a, and the surface of the given 
 vertical cylinder, and AaA."a" is the element of tan gency of the 
 vertical plane of rays, Aa, with the semi-cylinder of rays having 
 the semicircle, AB A^F'B", for its base. Hence a a' a" is the 
 intersection of the tangent planes to the two cylinders, and is 
 therefore also the tangent line to their curve of intersection ; viz. 
 to the shadow a"c'd'f. (D. G. 170.) 
 
 SECTION H. 
 
 SHADES AND SHADOWS ON WAKPED SUKFACES. 
 
 I. Shades. 
 
 46. The line of shade on a warped surface will, in general, be 
 a curve of some kind. Its points must, therefore, be separately 
 found. But any point of a curve of shade may be regarded, 
 either as the point of contact of a tangent plane of rays, or as that 
 of a single tangent ray. But, again, since we naturally associate 
 tangencies of surfaces to surfaces, and of lines to lines, a tangent 
 line to a given surface at any point, is generally constructed a? a 
 tangent to some curve, lying on that surface and passing through 
 the point. 
 
 47. Here it is to be remembered that, as the consecutive ele- 
 ments of a warped surface are not in the same plane, a plane, K, 
 passed through any one of them, E, will, in general, intersect all
 
 42 GENERAL PROBLEMS. 
 
 the others on each side of it, forming a curve, C, whose inter 
 section with the element, E, is, as explained in Descr. Geom. 
 (212, 342), the point of contact of the plane K, which is thus a 
 tangent plane as well as a secant plane. 
 
 48. Hence we have two direct general methods (22) for deter- 
 mining any point of the curve of shade on a warped surface. 
 One of these methods will be given here, the other presently. 
 
 First Method. Construct the point of tangency of any plane of 
 rays, tangent to the warped surface, by passing a plane of rays 
 through any element, and noting the point of intersection, P, of thai 
 element, with the curve of intersection of the tangent plane with the 
 warped surface. The point of intersection, P, will he the point of 
 tangency of the plane of rays, and hence a point of the required curve 
 of shade. 
 
 49. When the warped surface is of the second order, either a 
 warped hyperboloid, or a hyperbolic paraboloid, the plane of rays 
 passed through any element, intersects the surface in a second 
 element, of the other generation (Des. Geom.), whose intersection 
 with the given element is the point of tangency of the given 
 plane. 
 
 In illustration of the method of (48) take the following simple 
 case. 
 
 PROBLEM XIX. 
 To construct tfie curve of shade on a hyperbolic paraboloid. 
 
 Principles. By taking the planes of projection as plane direc- 
 tors, and one of the directrices vertical, the construction will be 
 simphe. The horizontal traces of the planes of rays containing 
 the horizontal elements will be parallel to those elements, respec- 
 tively; and will cut the horizontal trace of the surface in points 
 of the elements which are parallel to the vertical plane. 
 
 Construction. PI. VI., Fig. 21. For the sake of defmiteness, 
 call the horizontal plane the plane director of the first generation, 
 and the vertical plane, the plane director of the second generation. 
 Let the vertical line A A" A' and the line A"G A'G, in the 
 vertical plane, be the directrices of the first generation. Then 
 divide these directrices proportionally, according to the proper- 
 ties of the surface, and, for convenience, let these parts be equal
 
 SHADES AND SHADOWS. 43 
 
 on each line. Then AA" A'; AC c"C'; AD cTI)', etc., 
 will be elements of the first generation. In order that the curve 
 of shade shall be on the visible face of the surface, or real, in 
 case of an opaque solid, formed and placed as in the figure, the 
 light should come as shown at A7 in horizontal projection. 
 Without constructing the intersection of the ray through AA' 
 with the horizontal plane, by means of a given vertical projec- 
 tion, let 7 be assumed as this intersection. The line 7ft, parallel 
 to A A", will then be the horizontal trace of the plane of rays 
 through AA". Then na, drawn from n, the intersection of the 
 trace, 7w, with the trace, AG, of the paraboloid, and parallel to the 
 ground line, represents the element of the second generation con- 
 tained in the plane A. In. Its intersection, a, with the element 
 A A", is the point of contact of the plane of rays, and hence a 
 point in the required curve of shade. 
 
 Since the vertical directrix at A was equally divided by the 
 horizontal elements, the rays through the points of division, 
 being parallel lines, meet the horizontal plane in points, on A7, 
 which would divide that line into equal parts, and which are also 
 points in traces of planes of rays through those elements respec- 
 tively. Hence 6p, parallel to AB, and le, parallel to AD, are, 
 for example, horizontal traces of planes of rays containing those 
 elements. These planes also contain, respectively, the elements, 
 bp and ed, of the second generation, which intersect the former 
 elements at b and d, two more points of the curve of 
 shade. 
 
 Other points of shade are similarly found. The vertical pro- 
 jection of b is at b' ] of d, at d, etc. AN is the horizontal pro. 
 jection of k'N', the first element below the horizontal plane. 
 IK is the horizontal projection, parallel to AG, of the trace of 
 the plane of rays through AG on the horizontal plane &'N'. 
 Then by drawing k#, as in the previous constructions, we find g, 
 the point of shade on AG. 
 
 Remarks. a. From the properties of this surface, the curve of 
 shade, that is, geometrically speaking, the curve of contact of a 
 cylinder with a hyperbolic paraboloid, is a parabola, and hence 
 is also a parabolic curve in both projections. 
 
 b. The problem, as here given, shows the shadow, as on an 
 awning or porch roof in an angle of a building, and of the form 
 of a simple hyperbolic paraboloid. 
 
 50. Second Method (48). Pass any secant plane of rays tiirough
 
 44 GENERAL PROBLEMS. 
 
 the warped surface and construct its curve of intersection with the 
 surface. Then draw a ray tangent to this curve of intersection. 
 The point of contact, thus found, will be the point of contact of the ray 
 with the surface, and hence a point of its required curve of shade. 
 
 Either of the above direct and general methods (48, 50) -can be 
 easily applied to any of the general warped surfaces, with or with- 
 out plane directors, and given by their elements, in constructions 
 which are too simple to need formal illustration. 
 
 51. When the warped surface is a hyperboloid, particularly 
 when it is a hyperboloid of revolution, the indirect special method 
 of auxiliary tangent surfaces is applicable. This method is indirect, 
 as involving an intermediate surface whose line of shade is more 
 readily found than that of the given surface (22), and is special, as 
 being applicable to certain particular surfaces (22). It may be 
 stated as follows : 
 
 Make any circular section, C, of the given ivarped surface of revo- 
 lution, the circle of contact of an auxiliary tangent cone, having the 
 same axis as the given surface, and whose elements of shade, E and E', 
 are easily found. 
 
 The point, T, at which the element of contact, E, of the auxiliary 
 cone with & plane of rays, intersects the circle, of contact, C, of the 
 cone and the given surface of revolution, is the common point of 
 contact of the plane, the cone, and the given surface. That is, it 
 is the point of contact of the plane of rays and the given surface, 
 and hence is a point of the curve of shade on the latter. 
 
 Hence, the essential principle of the above method, stated 
 abstractly, is this : Two surfaces, M and N, being tangent in a curve 
 of contact, K, the intersection, T, of the line of shade of N with the 
 curve of contact, K, is a point of shade, common both to M and T, 
 and hence a point of the curve of shade of M. 
 
 jRemark. The method just explained might be applied in con- 
 structing the curve of shade of a warped hyperboloid ; but as it 
 will be exhibited in connection with an analogous double curved 
 surface (Prob. XXVII.) it is here omitted.
 
 SHADES AND SHADOWS. 45 
 
 PROBLEM XX. 
 
 To find the curve of shade on ike common oblique helicoid, in the 
 practical case of the threads of a triangular-threaded screw. 
 
 52. Principles. Under this head are here arranged a few pre 
 liminary matters, which will prepare the way for attention to the 
 immediate object of the problem. 
 
 1. Description of the screw. PI. VII., Fig. 22. Let the circle, 
 whose radius is AG", be the horizontal projection of a vertical 
 cylinder, called the cylinder, core, or newel, of the screw. Let 
 iy"T"T'" be any isosceles triangle, whose base, w'"T'", coincides 
 with an element of the cylinder, and which lies in a meridian 
 plane of the cylinder. Let this triangle have a compound 
 motion ; of rotation about the axis, A A' A", of the cylinder, 
 and of translation parallel to that axis ; and let each of these 
 motions be uniform. With such motions, each point of the 
 triangle, as w'", I'", or T'", will describe a helix (Des. Geom. 308) ; 
 the side Z"'T'", not shown, but corresponding to w'"l" f (dotted), 
 will generate an upwardly converging zone of an oblique heli- 
 coid, and the side w'"l'" (dotted) will generate a downwardly 
 converging zone of a similar helicoid. The entire triangle will 
 generate the volume called the thread of the screw. The heli- 
 coidal zone generated by l"T"' is called the upper surface of this 
 thread, and that generated by w'"l'" is its lower surface. The 
 curve, /"G-'aV, etc., generated by w'", is called an inner helix; 
 while the curve, l"'b"l", generated by I'", is called an outer helix. 
 
 All the inner helices are horizontally projected in the circle 
 G"am, and all the outer ones in the circle sqx. The vertical pro- 
 jection of a helix is found by dividing its horizontal projection 
 equally, and projecting the points of division upon the successive 
 equidistant horizontal lines, which mark its uniform ascending 
 progress. 
 
 The contours of the helicoidal surfaces are curves, apparently 
 tangent, as seen in vertical projection, to the helices. That por- 
 tion of the contour which bounds a single zone is, however, so 
 slightly curved that it is sufficiently accurate to represent it by a 
 straight line tangent to an outer and an inner helix, as at l"w'" 
 (the full line) and Z'"T'". By thus drawing all the visible con- 
 tours, the projections of the screw will be completed. The dotted
 
 46 GENERAL PROBLEMS. 
 
 line, l'"w'", whose horizontal projection is AS", is an asymptote 
 to the contour of the helicoid, being an element of the cone direc- 
 tor (Des. Geom. 333-4th) whose axis is A A' A", so that any 
 meridian plane cuts elements from it and from the helicoid, 
 which are parallel. 
 
 Tlie core of the screw is shown at V""E'" for a short distance 
 above the horizontal plane &"C", which cuts off the screw. The 
 intersection of the screw with this plane, is a spiral of Archi- 
 medes (D. G. 3396), and is found by dividing S"S'" into any 
 number, as eight, of equal parts, and the semicircle, S^'G^N"", 
 into the same number of equal parts ; then circles, with A as a 
 centre, drawn through the points of S^S 7 ", will intersect the 
 radii through the corresponding points of S"G-"N"" in points of 
 the spiral required. 
 
 2. To construct any particular element. First Method. Any 
 line, as Ae, is the horizontal projection of an element (D. G. 339), 
 using a, a moment, to mark a point on Ae. Project a at a', and 
 e at e', then aede' will be the projections of the portion of the 
 indefinite element Ae, which lies on the zone bounded by the 
 helices through the points S",S"" and S">'". 
 
 Second Method. Produce the element through S'^S"" and 
 S'",w'" (not shown in vertical projection) till it meets the axis, 
 A A' A", at a point which we will call 2. Then the element 
 through any point, as e,e', will meet the axis as far above 2, as e' 
 is vertically above S"". This follows from the uniformity of the 
 two motions of the element (1). 
 
 Remark. As the construction of a single point of the curve 
 of shade of the screw is somewhat lengthy, by any method, the 
 principles of each of the constructions now to.be explained, will 
 be given in immediate connection with the construction it- 
 self. 
 
 53. Methods ly assumed helices. These special methods 
 consist in finding the points of contact of planes of rays, which 
 are tangent at points on assumed helices. 
 
 The following will suffice for illustration: 
 
 First. The method by tangent planes of given declivity. 
 
 Principles. First. The axis of the screw being vertical, the 
 elements, and the tangents, at all points of the same helix, make, 
 each, a constant angle with the horizontal plane. Second. Hence 
 all the tangent planes to the screw, at points of the same
 
 SHADES AND SHADOWS. 47 
 
 helix, make a constant angle with the horizontal plane; they 
 being determined by the lines just mentioned. Third. Hence, 
 again, the angle between the element, and the line of greatest 
 declivity, in these tangent planes, is constant. This line of de- 
 clivity is perpendicular to the horizontal trace of the plane. 
 Fourth. A cone, having the axis of a helicoid for its axis, and 
 whose declivity is the same as that of all the tangent planes along 
 a given helix, may be taken as a cone director of the helicoid, 
 since its elements make a constant angle with those of the heli- 
 coid. Any plane, tangent to this cone will be parallel to one of 
 these tangent planes to the helicoid. Therefore, a plane of rays 
 being made tangent to this cone, a parallel plane of rays may be 
 drawn, tangent to the helicoid, and its point of contact will there- 
 fore be a required point of shade. 
 
 From these principles we have the following operations : Draw 
 any plane, P, tangent to the screw at a point on a given helix, 
 and note its declivity, L, taken in a meridian plane, and which will 
 give that of the cone director. Then, draw a plane of rays, 
 tangent to this cone. Its declivity, L', will be parallel to that 
 of the required tangent plane of rays, and in the same meridian 
 plane. Then the element, whose angle with I/ equals the angle 
 between L and the element of contact of P, will intersect the 
 helix at the point of contact of the plane of rays, that is, at a 
 point of shade. 
 
 Construction. PI. VII., Fig. 22. 1. Let the assumed helix he the 
 outer helix, S"eq S""e'. At any point, 0,0', of this helix, draw the 
 tangent line, oO, and the element as Ao. The tangent pierces 
 the horizontal plane of projection at O, where 00 is equal to the 
 arc osS'' (Des. Geotn. 317), and the element pierces the same plane 
 at F,F'; hence FO is the horizontal trace of the tangent plane to 
 the upper surface of the thread at 0,0'. Now AG, perpendicular 
 to FO, is the line of greatest declivity of this tangent plane, and 
 GAF represents the angle between this line and the element of 
 contact, AF, of the tangent plane. 
 
 Let the cone, whose axis is A A'A," and whose declivity is 
 equal to that of AG, be the cone director for the helicoidal upper 
 surface of a thread. AG meets the axis at the point A,F"; 
 then let AI A'T, having the same declivity as AG, be the gene- 
 ratrix of the cone director. The circle with A as a centre, and 
 radius AI, will be the base of this cone. Through its vertex, 
 A A", draw the ray AB A"B', which pierces the horizontal 
 plane of projection at B',B, hence BD is the horizontal trace of a
 
 48 GENERAL PROBLEMS. 
 
 plane of rays, tangent to the cone director on AD, which is also the 
 greatest declivity of this plane, and is the horizontal projection of 
 the line of greatest declivity of the parallel plane of rays, tangent 
 to the screw. Then lay off, on the outer helix, and from AD, an 
 arc 15, equal to u'"o, and the point b will be found as the point of 
 contact of a plane of rays, tangent to the upper surface of the 
 thread, at a point on the outer helix. Hence 5, and its vertical 
 projection, 5', are the projections of the required point of the curve 
 of shade on the upper surface of the thread and on this helix. 
 
 BC is the horizontal trace of another tangent plane of rays tc 
 the cone director of the upper surface of the thread, and CA ia 
 its element of contact, and line of greatest declivity. Therefore, 
 make C""q equal to w'"o, and q.q' will be another point of the 
 curve of shade of the upper surface of the thread, and on the 
 outer helix. 
 
 2. To find a point of the curve of shade, on the upper surface of a 
 thread and on the inner helix, S'"G"a w f "(j'a'. To simplify the 
 construction, take an auxiliary horizontal plane, p'J', as far 
 below rz/ra, on the element Ao, before used, as the horizontal 
 plane of projection is below 0,0'. Then, drawing AO, the line 
 riP will be the horizontal projection of a tangent at n,n', which 
 pierces the plane p'3' at P. For, by the properties of the 
 helix and its tangent, riP = nG"S'" and 00 = osS", while 
 G"S'":<wS"::An : Ao 
 ::AP: AO 
 
 hence wP: 00 ::AP: AO, 
 
 a proportion which is evidently constructed by limiting the 
 parallel tangents, at n and o, by the straight line AO. 
 
 The element, An', pierces the plane p'J' at p',p ; hence pP is 
 the trace, on the plane p'3, of the tangent plane to the thread at 
 njn! ; and A2, perpendicular to pP, is the line of greatest declivity 
 of this plane. Now, as before, BD is the horizontal trace of the 
 plane of rays, tangent to the cone director, and AD is the hori- 
 zontal projection of its line of declivity, and of that of the re- 
 quired parallel plane, tangent to the screw. Then make 4a equal 
 to Sn (53) and a,a' will be the required point of the curve of 
 shade of the upper surface of a thread and on the inner helix. 
 rf* is a similar point of the other curve of shade, on the back 
 part of the same zone. 
 
 Remarks. a. Points on other helices, such as intermediate 
 ones, can be similarly found.
 
 SHADES AND SHADOWS. 49 
 
 I. By going through with the preceding constructions, as 
 applied to the lower surface of the thread, it will be found that, 
 in passing from the element of contact to the line of declivity of 
 a tangent plane, the motion will be, in a (rotary) direction, around 
 A, opposite to the direction ou f ". Hence, DA, the line of decli- 
 vity of the tangent plane of rays to the cone director, being pro- 
 duced to.T, Ax is the horizontal projection of the declivity of this 
 plane, and of the parallel plane of rays tangent to the lower sur- 
 face of a thread. This being so, we proceed, according to the 
 former statement, to make xl equal to 1Z, and ym equal to 4# ; 
 then lin is a curve of shade on the lower surface of a thread. 
 Likewise, hj is the horizontal projection of a second curve of 
 shade on the same surface. 
 
 c. Each of the curves of shade, now found in horizontal pro- 
 jection, has as many vertical projections, and corresponding 
 curves in space, as there are helicoidal zones in the assumed por- 
 tion of the screw. Thus, ab is vertically projected at a'b' ', and 
 a"b" ; qr, at q'r' and in part near g' and s'; 1m, at I'm' and 
 I'm" ; and hj, at li'j' and in part at &"/', and m"' '. 
 
 d. If we conceive the entire helicoid, of which the upper sur- 
 face, as S""w'"a'e', of any thread is a zone, to be represented, 
 we may proceed as before to find other points of its curve of 
 shade. The point A,D', where the curve of shade meets the 
 axis, is the intersection of the axis with the element through 
 which the meridian plane of rays, RAK, is passed, since, as the 
 axis and this element are the two lines cut from the surface by a 
 plane of rays containing the supposed element, their intersection 
 is the point of tangency of the plane of rays (D. G. 342), and is 
 therefore a point of the curve of shade. We thus find baAjh 
 b'a'D' (the rest of the vertical projection not shown) for a com- 
 plete branch of the curve of shade on the helicoidal surface 
 bounded by a helix of a radius equal to AS". Likewise ImArq 
 I'm' is the branch of the curve of shade, on the similar heli- 
 coid of which the lower surface of a thread is a zone. 
 
 e. The point, A,D', is also the point of tangency of the plane 
 of rays RAK, from the particular properties of the helicoid, as 
 well as from those of warped surfaces in general, as explained in 
 the last remark. For the axis A A' A" is the limiting helix, 
 generated by that point of an element which is in the axis ; but, 
 being straight, it is its own tangent, throughout its whole extent; 
 hence, A,D', like other points of contact, e,e', etc., of tangent 
 
 4
 
 50 GENERAL PROBLEMS. 
 
 planes, is the intersection of a tangent to the helix, through A,D', 
 with the element through the same point. 
 
 54. Second. The method by helical translation of a tangent 
 plane. 
 
 Points of shade, on the lower surface of a thread, may be found 
 by the method of [53-2 (Rem. &)]. For the sake, however, of 
 illustrating a different construction, they will be found by the 
 special method just named. 
 
 Principles. Construct a simple tangent plane, P, to the 
 helicoid at any point of a given helix, find its trace, T, on any 
 horizontal plane, K; and note its intersection, T, with the axis of 
 the helicoid. A ray through I will pierce the plane K in the 
 trace, T, if P is a plane of rays. If this ray does not meet the 
 trace T, revolve it about the axis of the helicoid, when it will 
 generate a right cone, from which will be cut two elements, E and 
 E', by the plane P. By revolving either of these to coincide with 
 the ray, and by revolving the point of contact of P through an 
 equal angle, keeping it also on the helix, we shall have the point 
 of con tact of P, when translated, helically, so as to be a tangent 
 plane of rays. This point of contact will therefore be a point of 
 the required curve of shade on the given helix. 
 
 Construction. PI. VII., Fig. 22. Let the tangent plane be 
 constructed at the point S",'", and let its trace be found on the 
 horizontal plane S'K'. Thus S"A Z"'U, the element, through 
 S",l f ", of the helicoidal zone, forming the lower surface of the 
 thread considered, pierces the plane S'K' at S',S. The plane 
 S'K' being here taken at a distance above S'T" corresponding to 
 a quarter revolution of S'V" around the axis A A'A" we have 
 S"T"", equal to a quadrant of the circle S"#7, as the tangent at 
 S'T", piercing the plane S'K' at T"". Hence T""ST is the 
 trace, on the plane S'K', of the tangent plane at S",Z'". Now, 
 as this tangent plane contains the element "'U, it cuts the axis 
 at U, and we wish next to find whether a ray through U pierces 
 S'K' in the trace TT"". For lack of room on the plate, draw 
 AK UK', the position of a ray through A,TJ, after a revolution 
 of 180 from its primitive position, AR. (The angle K'UA' is 
 equal to the angle B'A"A'.) The revolved ray pierces the plane 
 S'K' at K',K, one point of the base, KQT, of the cone generated 
 by the revolved ray AK U'K. As this base does not contain 
 R, the ray in its primitive position, AR, does not pierce the
 
 SHADES AND SHADOWS. 51 
 
 plane S'K', in the trace, TT"", of the tangent plane. Hence this 
 tangent plane is not a plane of rays. But AT and AQ are 
 elements cut from the cone, just described, by the tangent plane, 
 and may be considered as two positions of the ray AR, after 
 being revolved about the axis, A A'A", till it becomes a line 
 of the tangent plane. Therefore, the ray being at T A, when the 
 point of contact of the tangent plane is at S'T", make S"h=7 5, 
 and project h at h', then will hji' be the point of contact of the 
 tangent plane after being revolved, with an ascending helical 
 motion about A A'A", till by the return of TA to being a ray 
 AR, it becomes a tangent plane of rays. Likewise, AQ being 
 the revolved position of the ray, RA, when Si", I'" is the point of 
 contact of the tangent plane, make S"l=S 5, and project I at I', 
 and 1,1' will be another point of contact of the tangent plane, 
 after moving, with a descending helical motion, till it becomes, 
 again, a tangent plane of rays. Hence, finally, h,ti and 1,1' are 
 points of the two curves of shade on the lower surface of a thread 
 of the screw. 
 
 In a similar manner j,f and m,m' could have been found. 
 hj h'j', being in front of the meridian plane SI, is visible in 
 vertical projection. Im I'm' is wholly invisible. 
 
 55. A little consideration of the properties of the helicoid,- 
 explained in Descriptive Geometry (Des. Geom. 339), will enable 
 the student to understand the following additional properties, 
 which are here applied in another construction of the required 
 curve of shade, by the special method, called the method by 
 assumed elements. 
 
 Principles. The development of a given portion of any helix 
 its axis being conceived to be vertical is the hypothenuse of 
 a right-angled triangle, whose altitude is the vertical height 
 between the extremities of the given portion, and whose base is 
 the development of the circular arc forming the horizontal pro- 
 jection of the same portion. 
 
 56. Portions of different helices on the same helicoid, but 
 included between the same meridian planes, may be called homo- 
 logous. Points in which they intersect the same element, may be 
 called homologous, or corresponding points. 
 
 Since all points of the generatrix of a helicoid ascend at the 
 same rate, the homologous portions of helices just mentioned, 
 will be of equal height, between their extremities ; that is, they
 
 52 GENERAL PROBLEMS. 
 
 will be included between pairs of equidistant horizontal planes. 
 Tangents to these helices at corresponding points, and limited by 
 the horizontal planes which include these helices, may be called 
 homologous tangents. Their horizontal projections will be par- 
 allel, and equal to the concentric circular arcs which are the 
 horizontal projections of the respective helical arcs to which they 
 are tangent. But these arcs, and hence their tangents, are pro- 
 portional to their radii ; hence the horizontal projections of these 
 tangents will terminate in a straight line through the horizontal 
 projection of the axis (53-2), which, with the tangents, and 
 their respective radii, will form similar triangles. 
 
 56. Once more, equidistant horizontal planes will intercept 
 equal portions of any element ; hence, if two such planes contain, 
 respectively, the point of contact and the trace of a tangent 
 plane through any element ; and if another pair of such planes 
 contain the point of contact and the trace of another tangent 
 plane through the same element, the distance between the trace 
 and the point of contact of one tangent plane will be equal to the 
 distance between the point of contact and the trace of the other 
 tangent plane, these distances being measured on the element 
 contained in both of these planes. Eecollecting, too, that the 
 helix coincides with its tangent when developed, we have the 
 following. 
 
 Construction. PI. VII., Fig. 22. Assume any element, Ae, 
 between the previously found points, a and 5, on the helices. 
 Making the tangent, eL, equal to the arc esS" ; the point L 
 is the intersection of this tangent with the horizontal plane of 
 projection (D. G. 317). Then H'H being the intersection of the 
 element Ae EV (52-2) with the horizontal plane, HL, is the 
 horizontal trace.of the tangent plane to the screw, at ee' (D.G. 347.) 
 
 Next, by drawing a ray (not . shown in vertical projection) 
 through A,E', it will pierce the horizontal plane of projection at 
 6 in the line AB, and 6HM will be the horizontal trace of a 
 plane of rays through the assumed element. 
 
 Now, if the trace, HM, of this plane of rays, were found on a 
 horizontal plane as far below its required point of contact, as the 
 trace, II L, is now below ee', this new trace would be parallel to 
 HM, the tangent at its point of contact would be parallel to eM, 
 and their distance apart, measured on Ae, would be equal to ell. 
 Hence the portion of the new tangent, cut off by the new trace, 
 would be equal to eM. But all tangents to helices at points on
 
 SHADES AND SHADOWS. 63 
 
 Ae, and included between pairs of equidistant horizontal planes, 
 will terminate in AL, as seen in horizontal projection (53-2) ; 
 hence, transfer M to N, on AL, by a line MN parallel to HA, 
 draw Nc, parallel and equal to dVl, and c,c' will be the point of 
 contact of the plane of rays, HM, with the screw, and on the 
 assumed element Ae. Hence c,c' is a point of the required curve 
 of shade. 
 
 Remark. By the above method, as well as by that of (53-2), 
 we can find the indefinite curve of shade on a complete helicoid. 
 
 Discussion. 
 
 FIRST. The meridian plane, RAK, which is tangent to the heli- 
 coid at A,D', makes an angle of 90 with the horizontal plane. 
 At a point, analogous to any point as e. but on the helix which 
 is at an infinite distance from the axis, the tangent eL would be 
 infinite, hence HL would then be parallel to eL ; that is, AH 
 would be both the line of declivity and the element containing 
 the point of contact of the tangent plane. Hence when the 
 element of contact of a plane, tangent to a helicoid, is also the 
 line of declivity of that plane, the^om^ of contact is on a helix 
 at an infinite distance from the axis. 
 
 SECOND. To find tlie helix, II, the tangent planes for all points of 
 which make a given angle, /3, with the horizontal plane ; /3 being 
 greater than oc. Construct a cone, having the axis A A'A" for 
 its axis, and its vertex at the intersection of an element, E, of the 
 helicoid, with the axis, and let all its elements make an angle 
 equal to /3 with the horizontal plane of projection. Then con- 
 struct a plane, through the element E, and tangent to this cone. 
 This plane will make an angle p with the horizontal plane, and 
 its point of contact, on E, found as in (57 Const.), will be a point, 
 p, of the required helix. Having the point p, the helix can 
 readily be constructed. 
 
 THIRD. Now let /3 be the angle made by the rays of light with 
 the horizontal plane of projection, and oc, as before, the angle 
 made by the elements of the helicoid with the same plane. The 
 angle /3 may be greater than, equal to, or less than oc. In PI. 
 VII., Fig. 22, /3 is less than oc, and the curves of shade are open 
 curves with infinite branches. If ft were equal to oc, a tan- 
 gent plane of rays would be tangent to the helicoid at the inter-
 
 54 GENERAL PROBLEMS. 
 
 section of the element, AZ", with the infinitely remote helices, 
 one on each nappe of the helicoid. At the lower of these points, 
 b and q would unite; and at the upper one, I and h would unite. 
 When /3 is greater than oc, find the helix, H, described above, 
 and apply to it the method of (53-2), but giving to the elements 
 of the auxiliary cone an angle of declivity equal to /3. It will 
 then be found, that but one plane of rays can be drawn, tangent 
 on the helix H, and that the points q and b will unite on that 
 helix. Also I and h will similarly unite on a similar helix of 
 the upper nappe, and the curves of shade will be closed curves, 
 wholly within the helix H. 
 
 FOURTH. If we substitute for the triangular generatrix of the 
 thread of the screw, a square, having one of its sides in an ele- 
 ment of the cylindrical core of the screw (52), a square-threaded 
 screw will be formed, the upper and lower surfaces of whose 
 threads the axis being vertical will be right helicoids, having 
 the horizontal plane for their plane director. For all practical 
 cases the angle is not 0, hence for the case of the curve of 
 shade on the square threaded screw, /3 is greater than oc, and 
 the curve of shade would therefore be closed. Usually, also, , 
 and the rate of ascension of the threads, have such values that 
 the curve of shade would be wholly within the core ; hence, 
 practically, there is no problem of the curve of shade on the heli- 
 coidal surfaces of the square threaded screw. 
 
 The elements of shade on its cylindrical surfaces are readily 
 found. 
 
 58. Either of the direct general methods (48, 50) may be applied 
 in the construction of curves of shade on warped surfaces, but 
 owing to the ease with which a plane is drawn tangent to a 
 warped surface of the second order, the following indirect general 
 met/wd will be more convenient. 
 
 Observing that a hyperbolic paraboloid may be more easily 
 represented in any position than a warped hyperboloid, make 
 any element of the given warped surface the element of contact of an 
 auxiliary tangent hyperbolic paraboloid- The point of contact of a 
 tangent plane through this element, can easily be found, and it will be 
 the point of contact on both surfaces. (Des. Geom. 343.) 
 
 In illustration of this method the following problem is given.
 
 SHADES AND SHADOWS 55 
 
 PROBLEM XXL 
 
 To construct points of the curve of shade of a conoid. 
 
 Principles. Let the curved directrix of the conoid lie in the 
 horizontal plane, let its rectilinear directrix be perpendiculai to 
 the vertical plane of projection, and let this plane be taken as the 
 plane director of the conoid, and of the elements of the first 
 generation of the auxiliary paraboloid. Let the directrices of the 
 first generation of the paraboloid be the line of striction of the 
 conoid, and a tangent to its base at the foot of the assumed ele- 
 ment of the conoid, through which a plane of rays shall be drawn. 
 The horizontal plane of projectiQii will then be th.e plane director 
 of the elements of the second generation of the paraboloid. 
 
 Construction. PI. VIII., Fig. 24. MEK is the curved, and 
 A A" A' is the rectilinear directrix of the conoid. Let TB T'E' 
 be the element of the conoid, on which it is proposed to find a 
 point of the curve of shade. Through any point, as A", A', of this 
 element, draw the ray A"R A'R'. This ray pierces the hori- 
 zontal plane of projection at R, and the assumed element pierces 
 it at E,E', hence ERG is the horizontal trace of the plane of rays 
 through TE T'E'. GA', its vertical trace, is parallel to E'T', since 
 that element is parallel to the vertical plane of projection. 
 
 Now, making AF tangent to MEK at E, it follows that 
 AA" A' and AF are the directrices of the elements of the first 
 generation of the auxiliary paraboloid, which is tangent to the 
 conoid along the element ET E'T'. 
 
 F/i FA'A' is another element of the paraboloid, and it is cut by 
 the plane of rays, through ET E'T', at A 'A, which is, therefore, 
 a point in an element of the second generation. But the hori- 
 zontal plane of projection is the plane director of such elements, 
 AF being one of them ; hence A'T' is the vertical projection of 
 the element, and T' the vertical projection of its intersection with 
 ET E'T'. Projecting T' at T, we have T,T' as the intersection 
 of elements, one of each generation, of the paraboloid. These ele- 
 ments, being contained in the same plane, viz. the plane of rays, 
 T,T' is the point of tangency of this plane of rays with the para- 
 boloid. But, as the paraboloid and conoid are tangent to each 
 other throughout the element ET E'T', the point T,T' is the
 
 56 GENERAL PROBLEMS. 
 
 point of tangency of the plane of rays with the conoid, also ; and 
 hence is a point of the required curve of shade of the conoid. 
 
 Remarks. a. If the direction of the vertical trace, G'h', of the 
 plane of rays, were unknown, it would be found by drawing, 
 through any point of ET E'T', a horizontal line, as A"g A'g', 
 parallel to EG, whose intersection, g,g', with the vertical plane, 
 woulcf be a point of G'h'. 
 
 b. The line hT, being a horizontal line of the plane of rays, is 
 parallel to GE, the horizontal trace of that plane. Also, a vertical 
 line at A, being a directrix of the second generation of the parabo- 
 loid, AT necessarily meets FE at A. Hence to find T,T' we have, 
 practically, the following very simple construction, first, draw 
 EH. /Second, through A draw Ah parallel to ER. Third, note 
 T, the intersection of Ah arid ET, and project it at T' on ET'. 
 
 Any other points of the cuyve of shade may be similarly 
 found. 
 
 c. It follows, from the constructions now given, that either 
 projection of the curve of shade of a conoid may be found inde- 
 pendently of the other. 
 
 Discussion. 
 
 FIRST. Preliminary. This discussion may be conducted with 
 reference to the right conoid with a circular base, PL VIII. , Fig. 
 25, having the vertical plane of projection for its plane director, 
 and the line, AA" A', for its line of striction. 
 
 Let cR be the ray through the point c,A' of the element ec. 
 Then eR is the trace of the plane of rays through ec. By 
 (Rern. b) above, note a, where the tangent at e meets A"A'. 
 Then at, parallel to eR, meets ec at <, the point of shade in the 
 branch Ath, and determined by the tangent plane of rays eR. 
 
 The vertical projection of t will be on the vertical projection 
 of ec, and in the upper nappe of the conoid. By similar con- 
 structions, four branches of the curve of shade may be found, 
 whose horizontal projections are Ath; AFra; A"h, and Af'g. 
 
 SECOND. In proceeding with the discussion, the following cases 
 will appear. 
 
 First. The horizontal trace of a plane of rays through AA" 
 A', may intersect the base of the conoid. 
 
 1st. In any manner, as at BE", PI. VIII., Fig. 25. 
 2d. In the diameter AA" N'.
 
 SHADES AND SHADOWS. 57 
 
 Second. This trace may be tangent to the base of the conoid. 
 Third. The same trace may be wholly exterior to the same 
 base. 
 
 1st. At a finite distance. 
 2d. At an infinite distance. 
 
 TIIIRD. For the 1st form of the first case, it appears: 1. That 
 the points of shade, consecutive with A, A' and A", A', being 
 points of tangency of parallel rays, on elements consecutive 
 with the vertical elements at A and A"] A, A 7 and A", A' 
 are cusp points of the first species, as seen in horizontal projec- 
 tion. 
 
 2. That the element which pierces the horizontal plane at K,K' 
 is, in space, an asymptote to the branches Ah, on the upper, and 
 A"<7, on the lower nappe of the conoid. For, the tangent at K, 
 analogous to ea, will be parallel to A X/ A, and hence the points a' 
 and a", analogous to a, and determined by this tangent, will be 
 at infinite distances, on A A" in each direction from J. But 
 lines analogous to at, and through such points, a', a", and parallel 
 to the trace of the plane of rays through JK A'K', would meet 
 HK H/'K' only at infinitely distant points of shade, analogous 
 to t. Likewise, the element which pierces the horizontal plane 
 at H,H' is an asymptote to Am and A"Tc. 
 
 3. The points A,T,T" and A" are all vertically projected at 
 A', hence the branches Ah and A"g, each of which is partly on 
 both nappes of the conoid, form loops, as seen in vertical pro- 
 jection ; to which N'T' and E'K' are tangent, at the multiple 
 point T'. 
 
 4. All those parts of these curves of shade are real, or would 
 exist on the exterior of a conoidal opaque solid, like the figure, 
 from whose points rays would pierce the horizontal plane of 
 projection, outside of the base of the conoid. 
 
 For the 2d form of the first case, AA" and A'N' would 
 be the projections of a ray of light ; and the loops in the vertical 
 projection would disappear. 
 
 FOURTH. In the second case, the construction of (Fig. 24 
 Eem. b) will give a curve of one branch, in which T and T" will 
 unite at J, and loops will reappear in vertical projection, having 
 A'N' and H"K' for their tangents at the point J,T'. 
 
 In this case, the element JK A'K' will also be an element 
 of shade, being an element of tangency of a plane of rays, through- 
 out its whole extent. On this element, therefore, the conoid ia
 
 58 GENERAL PROBLEMS. 
 
 said to be a developable single curved surface. Along the ver- 
 tical elements at A and A" planes may also be tangent. 
 
 Remark. FIFTH. In the 1st form of the third case, the loops of 
 the vertical projection will again disappear, and two branches, 
 as seen in horizontal projection, will proceed from A to the right, 
 and two from A" to the left. 
 
 In the 2d form of this case, the ray of light will be horizontal. 
 The vertical projection will be of the same form as in the 1st 
 form of this case, but in horizontal projection, the two branches 
 from A, and the two from A", will coincide. 
 
 n. Shadows. 
 
 59. The fundamental condition for determining a plane, is, 
 that it shall pass through three points taken at pleasure. Equi- 
 valent derived conditions are, that it shall contain a point and a 
 straight line; or shall contain one straight line, and shall be paral- 
 lel to another. All the elements of a cone have a common point, 
 the vertex, hence, as already seen (41 5), it is easy to pass a sys- 
 tem of planes; each containing a given line, as a ray, and the 
 vertex; and these will be planes of rays, cutting the cone in rec- 
 tilinear elements. Likewise, all the elements of a cylinder have 
 a common direction : hence, as before seen (41 a), a system of 
 planes of rays can be readily passed through the various elements 
 of a cylinder. But let a plane be passed through a given ray, 
 and a given point of a warped surface ; only one element would 
 generally pass through that point, and that, generally, would not 
 happen to be in the supposed plane. Again, let a plane be passed 
 through a given ray and parallel to some given line; only one 
 element would generally be parallel to that line, and that element 
 would not commonly happen to be in the plane employed. 
 Hence we have the Theorem, that, as the elements of a warped sur- 
 face have neither a common point, nor a common direction, no simply 
 arranged system of planes of rays can be made to cut it, each in a 
 right-line element. 
 
 60. Hence (41) in its application to warped surfaces gives the 
 following general method. 
 
 Pass any secant plane of rays through any point lasting a 
 shadow, and, by constructing its intersections with a number of 
 elements, find its curve of intersection with the warped surface,
 
 SHADES AND SHADOWS. 59 
 
 Then, the intersection of a ray through the given point, with the 
 curve found, will be a required point of shadow. 
 
 61. So much construction for each point of shadow, would be 
 extremely tedious, hence we have the following inverse general 
 method. Pass a plane of rays through any assumed element of 
 the warped surface, and note the point, P, cut by it, from the 
 line casting the shadow ; then, the ray through this point will 
 pierce the assumed element in a point of shadow. 
 
 It is evident, however, that, unless the plane of rays happens 
 to be also a projecting plane, the construction of P will generally 
 be as tedious as that of the curve of intersection, in (60). Hence 
 various special methods, soon to be explained, are mostly era 
 ployed in finding shadows on warped surfaces. 
 
 62. Taking up the topics mentioned in shades on warped sur- 
 faces, and in the same order, we have, first, the shadow of the upper 
 base of a warped hyperboloid upon that surface, the construction of 
 which will be explained in a subsequent problem of double- 
 curved surfaces. 
 
 To find the highest and lowest points. Find the intersection of a 
 ray, drawn through the intersection of the meridian plane of rays 
 with the circumference of the upper base, with the meridian curve 
 contained in that plane. 
 
 This construction involves an easy application of the general 
 method of (60) whenever the warped hyperboloid is one of revolu- 
 tion, with its axis perpendicular to either plane of projection, for 
 then its meridian curve is readily found. 
 
 63. To find intermediate points. Employ the special method of 
 auxiliary shadows (42). Any horizontal plane, P, will intersect 
 the given surface, its axis being vertical, in an ellipse, or circle, 
 C, if the surface be one of revolution. The shadow of the upper 
 base on the plane P will be a curve equal to that base (33), and 
 the intersections of this shadow with C, will be points of shadow 
 falling on the given surface. 
 
 64. In PL VI., Fig. 21, the shadow of the curve of shade of the 
 hyperbolic paraboloid upon the horizontal plane, is the parabolic 
 curve L#, formed by the intersection of the traces of the succes- 
 sive tangent planes 7n, etc. L7, equal and parallel to aA, is the 
 shadow of aA, and 7 A is the shadow of A A" A'.
 
 60 GENERAL PROBLEMS. 
 
 PROBLEM XXII. 
 
 To find Hie several shadows cast by a triangular- threaded screw 
 
 upon itself. 
 
 65. These shadows are: 
 FIRST. The shadow of the nut. 
 
 1st. Of any unknown point on an edge of shade of the nut. 
 2d. Of some assumed point on such an edge. 
 SECOND. The shadow of the curves of shade. 
 1st. On an assumed element. 
 2d On an assumed helix. 
 3d. On another branch of the curve of shade. 
 THIRD. The shadow of the outer helix. 
 1st. On any assumed element. 
 2d. On any helix. 
 3d. On any particular element. 
 
 FOURTH. There may also be associated with the preceding, the 
 shadow of the entire screw and nut, on the horizontal plane of 
 projection. 
 
 1st. The determination, by inspection, of all the lines casting 
 
 this shadow. 
 2d. The construction of the shadow of any assumed point, 
 
 which will cast a shadow on the horizontal plane. 
 Each of the above topics, including both its principles and 
 construction, so far as these need be separately mentioned, will 
 now be taken up in the above order. 
 
 66. To find the shadow of the nut upon the screw. 
 1st. The shadow of any unknown point of the nut. 
 Principles. This shadow is found on the principle, that any line 
 
 L in a plane of rays, R, through an edge, E, of the nut, will meet 
 the intersection of B with a given surface, in a definite point of 
 shadow of some point in the edge, E ; though the line be not, 
 itself, a ray. 
 
 This, which is an inverse method (61) may be called the method 
 by assumed lines in planes of rays. It is usually applicable only 
 when the line E is straight, since a plane of rays can usually be 
 passed through a straight line only (6). 
 
 Construction. PL VIII., Fig. 2 J. The edges of the nut, which 
 cast shadows, are the lower ones, projected at JK and KL, and 
 portions of JI and LN", all of which are vertically projected in
 
 SHADES AND SHADOWS. 61 
 
 the line J'L'. Beginning with IJ, this edge, produced, pierces 
 the meridian plane of rays OAB at 0,0'. Let OA O'Q be the 
 direction of the light, then Q, the point where the ray OA O'Q' 
 meets the axis A A' A", is the intersection of this axis with a 
 plane of rays through IJ. Hence, any meridian plane of the 
 screw will cut a straight element from its surface, and a line from 
 the plane of rays. This line will pass through Q, and will inter- 
 sect the element in the shadow of some point of IJ. Thus, the 
 meridian plane, Am, cuts from the upper thread the line kj k'f, 
 and from the plane of rays, the line raA m'Q, which intersects 
 the former line at I', I, a point of the required shadow. 
 
 In like manner, the edge JK meets the meridian plane of rays, 
 OAB, at a,a'. Then through a,a' draw a ray, and Q" will be 
 determined as the intersection of a plane of rays, through JK, 
 with the axis, A A' A". The meridian plane, AB", now cuts, 
 from the edge JK, the point, B'^B'", and from the plane of rays 
 the line B"A B'"Q". This line meets the element ED 
 E'T)"', cut from the upper thread, at ri",n", the shadow of some 
 point of JL. 
 
 Remarks, a. The points, as e",e" f , where the shadow leaves 
 the screw, at its intersection with the helices, are all found by 
 the indirect special method, called the method of intersecting sha- 
 dows ; and which consists of the following easy operations. Con- 
 struct the shadow of the line, L, casting the shadow in this case the 
 edge of the nut upon any plane below. Find also, on the same 
 plane, the shadow of M, the line containing the required point of 
 shadow in this case the outer helix. Then the intersection of 
 the shadow of L with the shadow of M will be a point, through 
 which a ray will meet M in a point, m, and L, in a point, I ; and 
 ra will be the shadow of I. The point t,t' is thus found. 
 
 b. This construction is not fully shown at R,A", and e"e'". 
 By drawing the rays through these points, we find the points 
 Y,Y' and n,ri of which they are the shadows. Similarly, the 
 points which cast any ascertained point of shadow, may be found. 
 
 c. From Y,Y' the shadow of the nut falls on the next thread 
 below, beginning at q,q', the intersection of the ray through Y,Y' 
 with the shadow of the outer helix. 
 
 d. The point e,e f is the shadow of a,a', and is the intersection 
 of the ray ae a'e' with the element be b"c". This one point is 
 found by (41) as a meridian plane of rays cuts a straight element 
 from a helicoid.
 
 62 GENERAL PROBLEMS. 
 
 2d. To find the shadow of an assumed point of the n\it. 
 
 Let the point be an angle of the nut, as JJ'. Constructions 
 for the shadow of such a point prove, on examination, to be 
 quite numerous. The following are some of them. 
 
 First. By the most simple direct method. Pass a vertical plane 
 of rays through J,J'. Its intersection with the surface of the 
 thread, will be a curve, easily formed by joining the points in 
 which the plane intersects several assumed elements of the screw, 
 taken, for convenience, in the supposed vicinity of the desired 
 point of shadow. A ray through J,J' will meet this curve (not 
 shown) in the required point of shadow W, "W. 
 
 Second. By an obvious indirect method. Find the shadows of 
 points, which may be known to be on IJ and KJ produced. The 
 indefinite shadows of these edges, found as in (66 Is/), will inter- 
 sect at a point, W,~W', which will be the shadow of J, J x . 
 
 Third. By the following indirect method, interesting in theory, 
 but tedious in application. Make J,J' the vertex of a cone, 
 generated by the revolution of the ray through JJ', around a 
 vertical axis at JJ'. The intersection of this cone with the screw, 
 is easily found by joining the points of intersection of several 
 neighboring elements of the screw with the surface of the cone. 
 The intersection of the ray element with this curve, is its inter- 
 section with the screw, and therefore is the shadow of J, J'. 
 
 Fourth. Let a plane of rays be passed through 3,J', so as to 
 cut a rectilinear element from the screw. That is, as such a 
 plane contains a ray through JjJ', it is really required to pass a 
 plane through a given line, and so as to cut a rectilinear element 
 from the screw. The given ray pierces the horizontal plane, 
 W'w', at V. Any line through V may be considered as the 
 horizontal trace on 7t'"w' of a plane of rays through J,J'. 
 The upper surface of the upper thread, produced, intersects the 
 plane, TJ!"W', in the spiral ivxyz, and a line from A to any point 
 of this spiral is the horizontal projection of an element of this 
 thread. 
 
 Next, make J,J' the vertex of a cone, generated by the revo- 
 lution, about a vertical line at J, J', of a line, Jt, which makes an 
 angle with the plane Z'"w/, equal to the angle made by the ele- 
 ments of the screw with that plane. The circle, with radius Jt, 
 and centre J, is the base of this cone, in the plane Z"'/. If 
 now a trace, VS, can be drawn, such that SJ and 5A shall be 
 parallel, then these lines, being the projections of lines of equal
 
 SHADES AND SHADOWS. 63 
 
 declivity, will be parallel in space. Further, as their tracts, S, 
 and 5, are in the trace of the plane of rays, the lines themselves 
 are in that plane, and as 5 is also a point in the trace of the 
 screw, 5A is the element of the screw which is contained in a 
 plane of rays through J,J'. 
 
 Now, to find YS, draw a number of trial traces Y6, Y7, etc., 
 and note the corresponding radius vectors, 6 A, 7 A, etc. Then 
 draw J8, Ji, etc., parallel to 7 A, 6A, etc., and note their intersec- 
 tions, 8, 4, etc., with the traces V7, V6, etc. Then the curve, 
 8, 4, etc., is the locus, or place, of all intersections of radii through 
 J, parallel to radius vectors through A, with the traces through 
 V. Hence S, where this trial curve, or locus, intersects the base, 
 /IS, of the cone, is the point from which can be drawn the 
 required trace, SY, of a plane of rays through J, J', and cutting a 
 rectilinear element, 5A, from the screw. Then "W, W, where the 
 ray, JW J'W, meets this element, is the required shadow of 
 J,J' on the screw. 
 
 Remark. It is now apparent, from the preceding construction, 
 that it reduces itself to this problem of two dimensions. Through 
 a given point (V) in the plane containing a spiral of Archimedes 
 and a circle, it is required to draw a line (VS) intersecting the 
 spiral and the circle at points, from which parallels can be drawn, 
 respectively, to the pole of the spiral, and the centre of the circle. ^ 
 
 67. To find ike shadow of the curve of shade on the thread below. 
 
 ls. On any assumed element. 
 
 Principles. This shadow is found by the indirect special method 
 of (66 1st. Kem. a). Thus, the shadow of the curve of shade 
 upon the horizontal plane, will intersect the shadow, on the same 
 plane, of an assumed element, supposed to contain a point of 
 shadow, in a point, through which, if a ray be drawn, it will 
 intersect both the element and the curve of shade. The former 
 intersection will be the shadow of the latter. 
 
 Construction. PI. VII., Fig. 22. To find that point of the 
 shadow of the curve of shade, hf /?//, which falls on the element 
 G"s GV. Drawing rays, G"X and sY, not shown in the verti- 
 cal projection, through G^G', and s,s', the line YX will be found, 
 as the shadow of the element upon the horizontal plane of pro- 
 jection. Similarly, &Z is found, for the shadow of hf h'f on 
 the same plane. These shadows intersect at t. Then tu is the 
 ray which intersects G"s GV at ,w', which is the required point 
 of shadow.
 
 64: GENERAL PROBLEMS. 
 
 Remark. By producing the ray, that point of hj h'j', whose 
 shadow is tm', could be found. 
 
 2d On an assumed helix. This shadow is found in the same 
 way as was the one just explained. Thus: let the outer helix, 
 soe s'o'e', be the assumed helix. VZ is a portion of its indefinite 
 shadow, found by connecting the shadows of several of the points 
 of the helix, on the horizontal plane. This shadow intersects 
 the shadow, 7cZ, of the curve of shade, hj h'j', at Z. Then the 
 ray Zy v't' meets the helix at v,v' which is the shadow of t' 
 (between h and i, in horizontal projection) on the curve of shade 
 produced in the present case. 
 
 3d. On another branch of the curve of shade. Find, by (66 1st.) 
 a point, y", of the shadow of the curve of shade hj h'j' produced 
 upon an indefinite element, as Ae, beyond b on the supposed 
 branch, a&, of the curve of shade. Join this point with the pre- 
 viously found points of shadow, juy", of hj h'j upon the screw. 
 The indefinite curve of shadow thus found, will intersect ab in 
 the required point, d,d'. 
 
 Remark. To find how far this curve of shadow of hj h'j' is 
 real, consider that it begins aty,/, the intersection of hj h'j' with 
 the inner helix /GV, and is limited by the ray through h t h f , 
 which meets the shadow, juy"j'u'v', at p"p'". Beyond p"p'", a 
 real shadow is cast, by the outer helix through hh'. 
 
 68. To find the shadow of the outer helix on the upper surface of 
 the thread below. 
 
 1st. On any assumed element. First Method. This is the method 
 of (66 1st. Bern, a) and of (67 1st). It is shown, in PI. VIII., 
 Fig. 23, in the construction of the point //. Here AM C X/ M ; 
 is any assumed element produced, on which a point of the pro- 
 posed shadow may fall. MP is its shadow on the horizontal 
 plane, and B"F is the shadow of a portion of that outer helix, 
 whose vertical projection is b f "s". Then B'/ the horizontal 
 projection of a ray through the intersection of these shadows, 
 gives/ which, projected in C"M', at/', gives// as the point of 
 shadow required. 
 
 Second Method. This is a special method which may be called 
 the method ly helical translation of a plane of rays containing an 
 assumed element. 
 
 Principles. When a plane is perpendicular to either plane of 
 projection, all lines in it will be projected, on that plane to which 
 the given plane is perpendicular, in the trace of the given plane.
 
 SHADES AND SHADOWS. 65 
 
 Also, the intersection of the plane of rays through the element, 
 with the given helix, is the point casting a shadow on that ele- 
 ment, at its intersection with a ray through that point. 
 
 Construction. l\-i PL VIII., Fig. 23, let AT I"T' (532) be 
 any assumed element. It pierces the horizontal plane at N',N. 
 The ray AB F'B', through the upper extremity of the element, 
 pierces the horizontal plane at B',B. Then NB is the horizontal 
 trace of a plane of rays through the element AN I"N'. Let 
 this plane now be revolved till it becomes perpendicular to the 
 vertical plane of projection. Its line of declivity, AH, will then 
 be horizontally projected in AH". Producing AH, the arc CU 
 corresponds to HH". All points of the plane of rays revolve 
 equally, therefore make T3 = CU, and project 3 at 3' on the outer 
 helix. All points of the plane also rise equally, and at the rate 
 of the rise of the helix, in order that it may still continue toxion- 
 tain an element of the screw. Therefore make I"I"' equal to the 
 height of 3' above T', and draw r"3'H"'. As the plane, NB, is 
 now perpendicular to the vertical plane of projection, I'"H'" is 
 its trace, and it cuts the outer helix at u'"u'' '. In the counter 
 revolution of this plane, u'"u" traverses a helical arc whose hori- 
 zontal projection is equal to CU, hence make u"h = CU, project 
 h at h', and then draw the ray hg h'g', which intersects the 
 assumed element AT A'"!" at g,g\ the required point of 
 shadow. 
 
 2d. On any helix. 
 
 This shadow is found by the method of intersecting shadows 
 (67 1st}. Thus, let the shadow of a spire of the outer helix be 
 found on the spire below, of the same helix. F is the intersec- 
 tion of the shadows, cast on the horizontal plane, by fragments 
 of the outer helixes, near s f and u'. Then the ray Fs gives s, 
 whose vertical projection is s'. The same ray produced not 
 shown in horizontal projection gives u',u. Therefore s,s' on 
 the outer helix, is the shadow of w,w', a point on the spire 
 above. In the same way d",d'" is found. 
 
 3d. On any particular element. 
 
 The point cZ,cZ', in the meridian plane OA, is found by the 
 ordinary direct method, as the intersection of a ray, Id b"'d f , with 
 the element be b'c'. 2,2' may be found either as //' was, or as 
 g,g' was. sdd" s'd'Z'd'" is now the complete shadow of the 
 outer helix, through &,&'", on the lower thread. The visible 
 portion of an equal shadow, similarly situated, is shown on the 
 
 5
 
 66 GENERAL PROBLEMS. 
 
 thread above. On the upper thread, the dotted edge of a 
 lar shadow is shown, as it would appear if the nut were re- 
 moved. 
 
 69. The shadow of the screw and nut on the horizontal plane. 
 
 1st. The determination of the lines casting this shadow. 
 
 These lines are, so much of each of the curves of shade as is 
 real, that is, not concealed by a shadow, and on the exterior of 
 the helicoid ; so much of each outer helix as divides the illu- 
 minated part of the upper surface of a thread from the unillumiri- 
 ated part of the lower surface of the same thread ; and, lastly, 
 those edges of shade of the nut, which do not cast shadows on 
 the screw. 
 
 2d. To construct the shadow of any point in any offfie above por- 
 tions offlie total line of shade of the screw. 
 
 This construction is made by the simple method of (Prob. III.). 
 
 PROBLEM XXIII. 
 
 To determine, in general, the shadows of a vertical right conoid. 
 
 70. In considering the shadows of the conoid, PL VIII., Fig. 25, 
 it is to be remembered, that, as all the elements of the conoid 
 intersect the line of striction, AA" A', a plane can be passed 
 through this line and any two parallel elements, equidistant from 
 the meridian plane, HK. All such planes will have horizontal 
 traces, parallel to AA", and intersecting the base of the conoid. 
 A tangent plane along the element JK, is the extreme case of 
 the above group of planes. 
 
 From the preceding it follows that any plane of rays, passed 
 through the line of striction, and whose trace, parallel to AA", 
 is between AA" and K, as at ER, will intersect the conoid in two 
 rectilinear elements. Of these two elements, the foremost one, 
 as ET, will be the indefinite shadow of the portion, TA, of the 
 line of striction upon the exterior of the conoid. This shadow 
 will be limited by the ray, as A&, through A,A'. The opposite 
 parallel element, T"E", is the imnginary shadow, on the interior 
 of the conoid, of the portion TT" of the line of striction. 
 
 The line Ab is also the horizontal projection of the imaginary 
 curve of shadow, cast upon the interior of the conoid by the ver- 
 tical element at A ; it being the horizontal projection of the
 
 SHADES AND SHADOWS. 67 
 
 intersection of a vertical plane of rays, through this element, 
 with the conoid. 
 
 "When the plane of rays, through the line of striction, does not 
 otherwise intersect the conoid, the shadow of this line will fall 
 wholly on the horizontal plane. In the case just considered, only 
 the portion T^A" casts a shadow on the horizontal plane. 
 
 The remaining shadows on the horizontal plane are cast by the 
 curves of shade. They are found by the usual obvious methods, 
 
 In this article, only the lower nappe of the conoid has been 
 considered.
 
 68 GENEKAL PROBLEMS. 
 
 DIVISION II. 
 
 SHADES AND SHADOWS ON DOUBLE CURVED SURFACES. 
 
 1. Shades. 
 
 71. Either a line, or a plane, may be placed tangent to a 
 double curved surface, at a point of contact, only. Hence a point 
 in the curve of shade of such a surface, may be conceived of, 
 either as the point of contact of a tangent ray, or of a tangent plane 
 of rays. 
 
 72. A tangent plane of rays is, practically, conceived of as con- 
 taining some given ray, which, with the point of contact, deter- 
 mines that tangent plane. A tangent ray is practically conceived 
 of as a tangent to some given, or constructed, section of the given 
 double-curved surface. 
 
 73. From the two preceding articles, we have the two follow- 
 ing general methods for finding points in the curve of shade of any 
 double-curved surface. 
 
 First general method. That of tangent planes of rays. By any 
 of the solutions of the problem of orthographic projections, 
 requiring a plane to be drawn through a given line and tangent 
 to a double-curved surface, construct a tangent plane through any 
 assumed ray. Its point of contact with the. given double-curved 
 surface, will be a point in the curve of shade of that surface. 
 
 74. Second general method. That of tangent rays. Construct 
 any simple section of the given surface, whose plane shall contain a 
 ray of light. Draw a parallel ray, tangent to this section, and its 
 point of contact ivill be a point of the required curve of shade. 
 
 Remarlcs. a. The first of these methods has no formal illustra- 
 tions among the following problems, numerous special methods 
 being more convenient in the various cases. 
 
 b. The second general method is convenient, in finding certain 
 extreme points, as the highest and lowest, on double-curved sur- 
 faces of revolution. It, therefore, has no separate formal illustra- 
 tions in the succeeding problems.
 
 SHADES AND SHADOWS. 69 
 
 c. Each of the following problems is solved by a special method, 
 which is convenient for the purpose, but is not necessarily appli- 
 cable only to the problem thus solved. 
 
 75. First special method. That of normal planes of shade. The 
 essential principle of this method, which is applicable only to the 
 sphere, is, that the plane of the curve of shade briefly called the 
 plane of shade of the sphere, is perpendicular to the direction 
 of the light. 
 
 The peculiarity and convenience of this method is, that it 
 requires but one projection of the sphere, as will now be 
 shown. 
 
 PROBLEM XXIY. 
 
 To find the curve of shade of the sphere, using only one projection 
 
 of the sphere. 
 
 Principles. Let the vertical projection, only, be given, and 
 let the vertical plane of projection be taken through the centre 
 of the sphere. 
 
 Let that plane of rays, which is perpendicular to the vertical 
 plane of projection, be called the perpendicular plane of rays, 
 and let the one, whose angle with the vertical plane equals that 
 made by the light with that plane, be called the oblique plane 
 of rays. Let this oblique plane of rays contain the centre of 
 the sphere. 
 
 The oblique plane of rays, and the plane of shade, are per- 
 pendicular to each other, and to the perpendicular plane of rays, 
 hence their traces on the latter plane will be perpendicular to each 
 other at the centre of the sphere, and their common vertical 
 trace, T, will be perpendicular to the trace of the perpendicular 
 plane of rays, and will pass through the centre of the sphere. 
 
 Now let the perpendicular plane be revolved about its trace, 
 and into the vertical plane of projection. Then consider the 
 great circle, which is the vertical projection of the sphere, as the 
 circle of shade, revolved about its diameter, T, into the vertical 
 plane. Project any of its points into the trace of the perpen- 
 dicular plane. Eevolve it into the trace of the plane of shade 
 on the perpendicular plane. Counter revolve the latter plane 
 to its primitive position, carrying the revolved point. Counter
 
 70 GENERAL PROBLEMS 
 
 project the point to the parallel plane from which it originally 
 came. 
 
 Construction. PI. IX., Fig. 26. LPKE is the vertical pro 
 jection of the sphere. LK is the vertical projection of a ray, 
 and the vertical trace of the perpendicular plane of rays through 
 the centre of the sphere. Or, assumed at pleasure, is the 
 revolved position of a ray, about LK as an axis, and is also the 
 revolved position of the trace of the oblique plane of rays, on 
 the perpendicular plane of rays, LK. OS, perpendicular to rO, 
 is the revolved position, about the trace LK as an axis, of the 
 trace of the plane of shade on the plane LK. POE is the com- 
 mon vertical trace of the oblique plane of rays, rO, and the 
 plane of shade, OS, on the vertical plane of projection. It is 
 perpendicular to LK, and is the transverse axis of the ellipse 
 into which the circle of shade is projected. LRS is the circle 
 of shade, after revolution about POR, and into the vertical 
 plane of projection. Assume any point 5, project it, at &', into 
 the trace LK. Revolve it, as at b'b", about O as a centre, into 
 the trace OS of the plane of shade. Counter revolve Z>", with 
 the plane LK to V" in the primitive position of the plane LK, 
 and counter project V" to B, in the plane db whence it came. 
 Then B will be a point in the required projection of the curve 
 of shade of the given sphere. 
 
 All the points of this curve may be found as just described, 
 but it is not necessary to find more than a quadrant, REs, of the 
 curve, in this way. For, taking parallel planes, as db and CA, 
 equidistant from LK, the point V" can also be counter projected 
 to C, a point of shade opposite to B from the axis sk. Then r. 
 and B ; and h and C, etc., will be equidistant from the transverse 
 axis PR. 
 
 76. For the practical case in which the projections of a ray 
 make angles of 45 with the ground line, the angle, as rOL. 
 made by the ray with the vertical plane of projection, will be 
 35-16' (21). OS, being perpendicular to Or, KOS will be an 
 angle of 54-44:'. This angle may easily be constructed, then it 
 will not be necessary to draw Or. Thus, take on OK any dis 
 tance, Os, and find the hypothenuse of a right angled triangle, 
 each of whose other sides equals Os. On a perpendicular, a* 
 S, at s, lay off this hypothenuse, then SO, the new hypothenuse 
 thus found, will make the required angle of 54-44' with OK, 
 and will therefore be the true position, after revolution of the
 
 SHADES AND SHADOWS. 71 
 
 plane LK, of the trace of the plane of shade upon the plane 
 LK. 
 
 77. Second special method. That of the intersection of the plane 
 of shade with the given double curved surface. This method can 
 be applied to all the double curved surfaces of the second 
 order (Des. Geom.) for the curve of contact of a cylinder, or 
 cone, of rays, with such a surface, is a plane curve of shade. This 
 method consists principally in making an auxiliary rectilinear projec- 
 tion of 'the curve of shade, which may be done by talcing an auxiliary 
 vertical plane of projection perpendicular to that plane of shade. 
 
 PROBLEM XXY. 
 To find the curve of shade on an ellipsoid of revolution. 
 
 Principles. Let the ellipsoid be a prolate spheroid (Des. 
 Geom. 219) and let its principal axis be vertical. Any auxiliary 
 horizontal plane will then cut a horizontal line from the plane 
 of shade and a circle from the ellipsoid. 
 
 The two points of intersection of these two lines will be 
 two points in the curve of shade. 
 
 Construction. PL IX., Fig. 27. Only half of the ellipsoid is 
 here shown. A AT)' is its semi-principal axis. BC B'C' is 
 the diameter of its greatest horizontal section. E^F", parallel 
 to LA, the horizontal projection of a ray of light, is the ground 
 line of a new vertical plane of projection, parallel to the light. 
 The projection of the semi-ellipsoid on this plane is a semi-ellipse 
 equal to B'D'C'. Having noted L and L', two projections of 
 any point in the ray LA I/ A', project A at A", and L at L", 
 making ~L"h" equal to L7i'. Then L"A" is the projection of 
 the light on the new plane of projection. Next, draw the tan- 
 gent ray at T", and T"A" will be the auxiliary rectilinear 
 projection of the curve of shade. Now the auxiliary horizontal 
 plane, m"n", taken at pleasure, cuts from the plane of shade the 
 line a" ab, and from the ellipsoid, the circle m"n" anb. 
 These intersect at a,a" and &,cr", two points of the curve of 
 shade. On the primitive vertical projection, these points are 
 projected at a' and &', on a line a'b', as far from B'C' as a" is 
 from the ground line E"F."
 
 72 GENERAL PROBLEMS. 
 
 Similarly other points are found. A." is projected at S and V, 
 and then at S' and V. T" is likewise projected at T and T' s 
 Drawing a tangent, parallel to I/ A', we find t',t. on the meridian 
 curve parallel to the vertical plane whose ground line is B'C'. 
 
 Only the portion t'b'VC' of the shade is visible in vertical 
 projection. The boundary of the invisible portion of the shade 
 is dotted. 
 
 Remark. The point T" is exactly constructed on the princi- 
 ple that the diameter of an ellipse, through the point of tangency, 
 bisects all the chords of the ellipse parallel to the tangent. 
 Hence take any two chords, parallel to L"A", bisect them, and 
 draw a line through their middle points. Such a line will meet 
 the circumference at T", the point of tangency of a tangent 
 parallel to L"A". 
 
 78. Third special method. This is essentially the second general 
 method (74) but modified so as to be most conveniently applica- 
 ble to the construction of the curve of shade on an ellipsoid of 
 three unequal axes. 
 
 A plane of rays passed through the mean axis, will be per- 
 pendicular to the plane of the longest and shortest axes. Make 
 the ellipse, E, cut from the ellipsoid by this plane, the base of a cylin- 
 der whose right section shall be a circle. Such a cylinder may be 
 considered, as the projecting cylinder of the ellipse upon the plane of 
 right section of the cylinder, taken as a new plane of projection. 
 Other parallel planes of rays will cut ellipses, similar to ~E,from the 
 ellipsoid. Their projections on the new plane of projection will 
 therefore be circles, to which tangent rays may be drawn, which will 
 give projections of points of the curve of shade. 
 
 PROBLEM XXVI. 
 
 To find die projections of the curve of shade on an ellipsoid of fliree 
 
 unequal axes. 
 
 Principles. Let the longest axis be vertical, and the plane of 
 the longest and shortest axes, parallel to the vertical plane of 
 projection. To locate the projecting cylinder described in (78) 
 make the centre of the ellipsoid the centre of an auxiliary 
 sphere, whose radius shall be equal to the semi-mean axis of the
 
 SHADES AND SHADOWS. 73 
 
 ellipsoid ; then the cylinder will be tangent to this sphere .in a 
 circle, which may ba considered as a circular projection of the 
 elliptical base of the cylinder. 
 
 Any plane, parallel to the circle just described, may be taken 
 as a new horizontal plane of projection, on which a new hori- 
 zontal projection of a ray of light must be constructed. 
 
 Construction. PI. IX., Fig. 28. 0,O' is the centre of the 
 ellipsoid. Let O'D' be the length of the semi-mean axis, whose 
 vertical projection is 0'. A circle, with radius O'D' and centre 
 0', is the vertical projection, partially shown, of the auxiliary 
 sphere above described. 
 
 LO and I/O' are the projections of a ray of light. Then 
 L'F' is the vertical trace of a plane of rays, perpendicular, here, 
 to the vertical plane of projection, in being perpendicular to the 
 plane of the longest and shortest axes. J'F' is the vertical pro- 
 jection of the ellipse contained in this plane of rays, and tan- 
 gents, from J' and F', to the vertical projection of the auxiliary 
 sphere, as at e', are the extreme elements of its projecting cylinder. 
 Take, now, any new ground line, MN, parallel to OV, and make 
 O'O" perpendicular to it, and make 0" and O equidistant from 
 their respective ground lines, MN and MP. Likewise make L" 
 in a perpendicular from L' to MN, and as far from MN as L is 
 from MP. Then L^O" is the new horizontal projection of a ray 
 of light. Now the circle with centre O" and radius 0"F (=OY} 
 is the horizontal projection of the section J'F', and its diameter 
 /h, perpendicular to L"O", determines precisely the points of 
 contact, f and n, of rays tangent to the section J'F' JnYf. 
 These points being projected at/' and n', give f,f and n,n f for 
 two points of the required curve of shade. 
 
 By drawing the tangent rays at t' and/I", we find the line t'T', 
 which bisects all the chords, as K'B', parallel to the tangents at 
 t' and T'. These chords are the vertical projections of ellipses, 
 whose centres, as ra', are in the line 2'T', and whose horizontal 
 projections are circles having their centres in T, the horizontal 
 projection of t'T'. 
 
 Thus, m is the centre, and mK the radius, of the horizontal 
 projection of K'B', and kb, its diameter perpendicular to L"O", 
 determines the points of shade, k and 5, vertically projected at 
 k' and V. In the same way g,g' and hji' are found. 
 
 Through the points now found, the projections of the required 
 curve can be sketched ; showing that it will be included between
 
 74: GENERAL PROBLEMS. 
 
 tangents (not shown) as at u and y, tangent also to the vertical 
 projection, t'f'T'u, and parallel to O'O". 
 
 The horizontal projection of the curve of shade is also tangent, 
 at u and y, to the ellipse, notshown, whose minor axis is Oa 0', 
 and whose longer axis is the distance, on KF, between two 
 lines, perpendicular to it, and tangent to the ellipse T'C'A'J' ; 
 and the former ellipse is the new horizontal projection of the 
 ellipsoid. 
 
 The visible portion of the shade is shaded in vertical pro- 
 jection. 
 
 79. fourth special method. This is essentially the special 
 method of (51), but modified so as to be also applicable to the 
 ellipsoid of three unequal axes. It is a property of these ellip- 
 soids employed in the preceding article that a set of parallel 
 sections of any one of them, is composed of similar ellipses. 
 Add to this, as just previously shown, that a diameter can 
 obviously be found, in the plane of the longest and shortest 
 axes, which shall be equal to the mean axis. The plane of this 
 diameter and mean axis will evidently intersect the ellipsoid in a 
 circle, and so will all parallel planes. Now by taking a plane 
 of projection treated most conveniently as a new horizontal 
 plane and made parallel to these circles, we can readily con- 
 struct the elements of shade on a series of auxiliary tangent 
 cones having these circles for their bases. The intersection of 
 the elements of shade with the circles will be points in the curve 
 of shade of the ellipsoid. 
 
 The construction is left for the student to make. 
 
 80. fifth special method. This is the method of projections of 
 rays on meridian planes. Its essential principle is, that the point 
 of contact of a tangent line to any meridian curve of a surface 
 of revolution, and parallel to the projection of a ray on the 
 plane of that curve, is a point of the curve of shade of that sur- 
 face. For, a point of shade is the point of contact of a tangent 
 plane of rays (73). But such a plane is perpendicular to the 
 meridian plane through its point of contact. Hence its trace on 
 the meridian plane is the projection of a ray upon that plane, and 
 is also a tangent to the meridian curve, at the point of contact 
 of the tangent plane ; that is, at a point of shade ; which proves 
 the principle above stated.
 
 SHADES AND SHADOWS. 75 
 
 The general operations of solution under this method are 
 exemplified as follows : 
 
 PROBLEM XXVII. 
 To construct the curve of shade upon a torus. 
 
 Principles. Conceive of a rectangle, with one of its shorter 
 sides made the diameter of a semi-circle. If the entire plane 
 figure, thus formed, be revolved about the opposite side of the 
 rectangle, the solid generated will be a torus. 
 
 If the axis of the torus be supposed to be vertical, the points 
 of shade on its greatest horizontal section will be the points of 
 contact of two parallel and vertical tangent planes of rays. The 
 points of shade on the meridian curve parallel to the vertical 
 plane, will be the points of contact of two planes of rajs per- 
 pendicular to the vertical plane of projection. 
 
 Hence, in finding the four points just described, we make a 
 rudimentary application of the first general method (73). Other 
 points of shade are mostly found by the fifth special method (80). 
 
 Constructions. 1. To find the four points on the visible bounda- 
 ries of the torus. PI. IX., Fig. 29. Let ATB* and A'L'B' be 
 the projections of the torus, and RL R'L' a ray of light. Then, 
 by drawing T/, perpendicular to RL, T and t' will be determined 
 as the horizontal projections of the points of contact of vertical 
 planes of rays, tangent to the greatest circle of the torus at T,T' 
 and t.t '. 
 
 Likewise, at C' and D', the centres of the semi-circular ends 
 of the elevation, draw C'n' and D'c'. perpendicular to R'L', and 
 project n' at n, and c' at c, then n,n' and c,<?' will be the points 
 of shade on that meridian section of the torus, which is parallel 
 to the vertical plane of projection. 
 
 2. The highest and lowest points. These are in the meridian 
 plane of rays KL. Revolve this plane, with the ray RL R'L', 
 about a vertical axis at L, till parallel to the vertical plane of 
 projection. Then r"L R"L' is the revolved position of the 
 ray RL R'L'. Through C' and D', draw CV and DV per- 
 pendicular to R"L", and project u' at u and a! at a. Then a,a' 
 and u,u f are the revolved positions of the highest and lowest 
 points. In the counter revolution, a,a' returns to a'V, and uju!
 
 76 GENERAL PROBLEMS. 
 
 to u'V", vhich are the highest and lowest points of the required 
 carve of shade. These points are thus found by a simple appli- 
 cation of the second general method (74). 
 
 3. To find intermediate points. Assume any meridian plane, 
 as d"Le", project the ray RL R'L' upon it by projecting lines, 
 as Rr RV, perpendicular to d"~Le,". L,I/, being in the assumed 
 plane, is its own projection on that plane, and rL and r'~L' are 
 the projections of the projection of the ray, RL R'L', upon the 
 plane d"Le". 
 
 If, now, the plane d""Le" be revolved about a vertical axis at 
 L, till it is parallel to the vertical plane of projection, the meri- 
 dian section contained in it will coincide with the vertical pro- 
 jection of the torus, and the projected ray rL r'\J will appear 
 at R""L R'"L'. From C' and D', draw lines, Q'd' and DV, 
 perpendicular to the revolved ray R ///7 L R //X I/, and project d' 
 and e' at d and e. Then, in the counter revolution of the meri- 
 dian plane, d,d' will return in a horizontal arc to d",d'", and e,e' 
 to e",e'", which are points of the required curve of shade. 
 
 Other points can be similarly found. 
 
 81. The special method of (51) may be applied to any double 
 curved surface of revolution, but for the sake of completeness 
 of illustration, and to show how it would be applied in finding 
 the curve of shade on a warped hyperboloid, it is here applied 
 to a concave double curved surface of revolution, analogous in 
 form to that warped surface. This method will therefore be next 
 described in immediate connexion with the " construction " of the 
 following example : 
 
 PROBLEM XXVIII. 
 To find the curve of shade on a piedouche. 
 
 Principles. A Piedouche, PI. X., Figs. 30, 31, is a little orna- 
 mental pedestal, used for the support of a bust, a statuette, etc. 
 
 Conceive of an ellipse, whose axes are in a vertical plane, but 
 oblique to the horizontal plane. Suppose a vertical line, L, in 
 the plane of the ellipse, and so placed as to converge downwards 
 towards its minor axis. Then let that arc of the ellipse which is 
 included between horizontal tangents, and convex towards L, be 
 revolved about L as an axis, and a concave double curved sur-
 
 &HADES AND SHADOWS. 77 
 
 face of revolution will be formed, which will be the principal 
 surface of the piedouche. The lower base will evidently be 
 larger than the upper, and both may be made the bases of short- 
 cylinders having L for their axis. See PI. X., Fig. 30, which 
 shows how the generating semi-ellipse, AEB, may be formed 
 from a semi-circle, as ACB, where DE, etc. DC, etc. 
 
 The elliptical generatrix may be replaced by a compound 
 curve, composed of circular arcs, tangent to each other, and with 
 different radii. 
 
 Remark. The piedouche formed by the second method will 
 answer every graphical purpose, but is less ornamental, since a 
 circle is a monotonous curve, its points being equidistant from 
 the centre; while the gradually varying distance of the points 
 of an ellipse from its centre gratifies the natural love of variety. 
 While dwelling on a point of taste, it may be added that the 
 generatrix should begin and end on horizontal tangents, because 
 it thus suggests completeness and this, because horizontal boun- 
 daries imply stability ; and this, again, because gravity, in refer- 
 ence to whose action the piedouche is to be stable, acts vertically. 
 
 Constructions. 1. Of the piedouche, and of some preliminary 
 points of shade. PI. X., Fig. 31. Assume the bases, H'C' and 
 F'G", of the double curved portion of the piedouche. At F', 
 take any vertical height F'E', greater than half the distance 
 between the assumed bases. Through E', draw the line E'l', 
 parallel to the bases, and, from C', let fall C'D', perpendicular to 
 E'l'. Then, with E'F' as a radius, describe the quadrant from 
 F' to E'l', and with D'C' as a radius, describe the quadrant from 
 ET to C'. 
 
 The compound curve, thus formed, will be the meridian curve, 
 whose revolution about a vertical axis, A T'A', will generate 
 the double curved portion of the piedouche. Its cylindrical 
 bases are, as shown in vertical projection, above and below the 
 double curved portion. These bases are horizontally projected 
 in the concentric circles, A IIYB, and A GWF. 
 
 In applying the method of (51) some of the assumed circles 
 of contact of the auxiliary surface may prove to be beyond the 
 limits of the curve of shade, so that it is best to construct the 
 highest and lowest points of that curve first, as in (Prob. 
 XXVII.). 
 
 Let AJ A'J' be a projection of a ray of light. Eevolve 
 this ray about the axis A T'A' to the position AJ" A'J'"
 
 78 GENERAL PROBLEMS. 
 
 \ 
 
 Draw IV and EV'" perpendicular to A'J"', then u' and r'", 
 horizontally projected at u and r", will be revolved positions of 
 the highest and lowest points. In the counter-revolution, these 
 points return to IT,!!' and r,r', which are their real positions. 
 
 By drawing IV and EV, perpendicular to A'J', we find a',a 
 and n',71, the points of tangency of planes of rays perpendicular 
 to the vertical plane of projection. That is, aa! and nn' are the 
 points of shade on the meridian curves forming the vertical pro- 
 jection of the piedouche, and contained in the meridian plane 
 HAF, parallel to the vertical plane of projection. See the first 
 general method (73). 
 
 2. Of intermediate points of shade, first illustration of (81) 
 Principles. Assume any horizontal section of the piedouche 
 such a section being circular because the axis is vertical and 
 make it the circle of contact, C, of a tangent cone, whose ele- 
 ments will therefore be tangent to the meridian curves of the 
 piedouche. Drawing planes of rays, tangent to this cone, we 
 find its elements of shade. The two points of intersection of 
 these elements with the circle of contact, C, will be points of 
 shade on the piedouche (51). At the circle of the gorge, the 
 cone will become a tangent cylinder. 
 
 Construction. Assume any horizontal plane, K'L', PI. X, Fig. 
 31, between the highest and lowest points. At either point, as 
 I/, of its intersection with the meridian boundary of the vertical 
 projection of the piedouche, draw the tangent L'S'. This tangent 
 will be an element of the auxiliary tangent cone, whose base is 
 Z'L'. 
 
 The circle, with radius A7c, equal to X'L', is the horizontal 
 projection of the circle of contact, or base, of the cone, and A,S' 
 is the vertex of this cone. The ray AK S'K', through the 
 vertex A,S', pierces the plane, KT/, of the base at K',K. Hence 
 K& is the horizontal projection of the trace on the plane K'L', 
 of a plane of rays, tangent to the auxiliary cone. Project k at k' ; 
 then, as k,k' is the intersection of the element of shade, Ak 
 S'&', of the cone, with the circle of contact, Z'L', it is a point of 
 tangency of the plane of rays, K&, with the piedouche ; that is, 
 k,k' is a point in the curve of shade of the piedouche. But as 
 two planes of rays can be drawn, tangent to the cone, each cone 
 will give two points of shade. Thus, by drawing the chord kt, 
 perpendicular to KA, we find the point of shade, t } t' } both of 
 whose projections are invisible.
 
 SHADES AND SHADOWS. 79 
 
 Likewise, by assuming a circle of contact 5V, whose radius, in 
 horizontal projection, is AQ, equal to half of Z/c', and which is 
 the base of a tangent cone whose vertex is W, we find Q,Q' and 
 R,R' for two more points of the curve of shade. 
 
 Finally, at the circle of the gorge, /'m', horizontally projected 
 with a radius A/z, equal to half of f'm', the auxiliary cone becomes 
 a cylinder. Hence, by drawing the diameter gh, perpendicular 
 to AK, the direction of the light, we find the two points of shade, 
 g,g f and h,h'. 
 
 Remark. The elements of shade, kA and A, produced in 
 the direction kA and tA, will meet the base of a cone, tangent 
 to the lower nappe of the piedouche, and having the same inclina- 
 tion that the element I/S' has, and will therefore give two more 
 points of shade, which can be easily found. 
 
 3. Second illustration of the method of (51). 
 
 Principles. Make any circle, between the highest and lowest 
 points of shade, the circle of contact of an auxiliary sphere, tan- 
 gent to the piedouche internally. The meridian plane, parallel 
 to the vertical plane of projection, and the plane of the assumed 
 circle of contact, which is horizontal, may be taken in this con- 
 struction as the planes of projection. But the plane of the curve 
 of shade of the sphere is perpendicular to the direction of the 
 light, hence its traces will be perpendicular to the projections of 
 a ray. The -vertical trace will pass through the centre of the 
 sphere, and its intersection with that diameter of the circle of 
 contact which lies in the meridian plane just mentioned, is a point 
 of the horizontal trace of the plane of shade of the sphere. The 
 intersections of this horizontal trace with the circumference of 
 the assumed circle of contact, will be two points of the curve of 
 shade of the piedouche, being the points in which the curve 
 of shade of the sphere intersects the circle of contact of the sphere 
 and piedouche (51). 
 
 Construction. PI. X, Fig. 31. Let M'N' dNe be the assumed 
 circle of contact. Produce E'N' to T' in the axis of the piedouche ; 
 then T'N' is the radius of the auxiliary sphere, whose*circle of 
 contact with the piedouche is M'N'. Then T'O', perpendicular 
 to AM', is the vertical trace of the plane of shade of the sphere, 
 and O',O is a point of its horizontal trace, eOd. The points e,e' 
 and d,d', as explained above, are therefore points of the curve of 
 shade of the piedouche. Through the points now found, the 
 curve can be sketched. It is wholly invisible in horizontal pro-
 
 80 GENERAL PROBLEMS. 
 
 jection. That part is visible in vertical projection which is in 
 front of the meridian plane GAF. 
 
 Remarks. a. The curve of shade on a sphere may be found as 
 in Prob. XXIV., in Prob. XXV., in Prob. XXVfl. Its points 
 may also all be found as are the highest and lowest on the 
 piedouche or torus, since a set of vertical planes of rays would 
 cut the sphere in circles ; or they may be found by auxiliary 
 tangent cones, as in (2.) 
 
 b. Since the piedouche is a surface of revolution, its curve of 
 shade may be found as in the problem of the torus. Likewise the 
 curve of shade of the torus may be found by the methods just 
 employed. In fact, these methods are especially applicable to the 
 interior half of the annular torus, a surface generated by the 
 revolution of a circle about a line exterior to it, but in its plane. 
 
 c. By inspection of the figure of the piedouche, the following 
 curious property may be observed. Near the points R,R', Q,Q', 
 1,1', and 2,2', tangent rays may be drawn to the projections of 
 the curve of shade at the two projections of the same point. 
 Hence such tangent rays are really tangent to the curve of shade 
 in space. These tangents evidently include the greatest hori- 
 zontal chords of the curve of shade, one on each nappe of the 
 piedouche. 
 
 d. It may be shown in several ways, that there must be such 
 points of maximum width of the curve, at which tangent rays 
 may be drawn to the curve. This may be done, in an elementary 
 manner, as follows. Suppose all points of the curve of shade to 
 be found by the general method of (74) employed in finding the 
 highest and lowest points of shade. A vertical plane of rays 
 near the gorge, will cut a curve similar to A"B", PI. X., Fig. 32, 
 to which two tangent ra} r s can be drawn, giving two points of 
 shade, T and T'. Consideration of the form of the piedouche 
 will show, that, as the vertical secant plane recedes from the 
 gorge, the curve, A"B", will become flatter, the points T and T' 
 will approach each other, and will finally unite, as at T, on A'B'; 
 T being a point of inflexion, or change of curvature. 
 
 If the cutting plane recedes still further from the gorge, the 
 curve will be, as at AB, so flat that no tangent ray can be drawn 
 to it ; hence the point T marks the position of the greatest width 
 of the curve. 
 
 e. Moreover, the ray is tangent, at T, to the curve of shade 
 itself, for, on that position of A"B" which is consecutive with
 
 SHADES AND SHADOWS. 81 
 
 A'B', T and T' will be consecutive points of shade, and their 
 tangents consecutive parallels. On A'B', therefore, T and T' 
 unite, and their separate tangents coalesce, forming a tangent to 
 the curve of shade at T. 
 
 f. Those of the secant planes which intersect the gorge, cut 
 the piedouche in curves analogous to the meridian curve. 
 
 g. Furthermore, the points, as T, Fig. 32, near 1, 2, E, etc., 
 on PL X., Fig. 31, are the limits between that part of the curve 
 of shade which really exists on the exterior of a solid opaque 
 piedouche, and on a transparent one, open at the top, and con- 
 sisting merely of a surface without thickness. All points, as T, 
 on the upper nappe, and similar ones at such points at T', on the 
 curves like A X/ B 7/ cut from the lower nappe, are imaginary points, 
 on the interior side of the hollow surface. Only such points as 
 T', on curves shaped like A^B" and cut from the upper nappe, 
 and such as T, situated on curves cut from the lower nappe, are 
 points in which rays in space are truly tangent to the external 
 surface of the piedouche. The construction of a few intersec- 
 tions of the piedouche by vertical planes of rays will make these 
 statements clear. 
 
 h. There is a direct construction, given by M. LEROY, of the 
 points, as T, of maximum width of the curve of shade, but it is 
 extremely tedious and comparatively unimportant, being, in the 
 graphical point of view, only approximate. Hence it is not 
 here given. Moreover, it is unimportant, for the further reason 
 that these points are found with sufficient accuracy in sketching 
 the curve by hand through a considerable number of other con- 
 structed points. 
 
 IL Shadows. 
 
 82. The direct method of finding shadows on double curved 
 surfaces, is essentially the same as for rinding shadows on other 
 surfaces. Pass a plane of rays through any point, P, of the line 
 casting ike shadow, and construct the intersection, I, of this plane 
 with the given double curved surface. The point, where a ray through 
 P meets the line, I, is the shadow of P on the double curved surface. 
 
 It being desirable, for graphical convenience, that the line I 
 should be a circle, it is evident that the foregoing direct method 
 is always conveniently applicable only to spheres ; and to other 
 
 6
 
 82 GENERAL PROBLEMS. 
 
 surfaces of revolution, only when so situated in reference to the 
 light, that planes of rays may be made to cut them in circles. 
 As these conditions do not generally occur, the shadows on 
 double curved surfaces are mostly found by a variety of indirect 
 and special methods, which will be sufficiently understood from 
 the following problems. 
 
 PROBLEM XXIX. 
 
 To construct tJie shadow of the front circle of a niche upon its spheri- 
 cal part. 
 
 Principles. This problem, enunciated more as an abstract one, 
 is this : To construct the shadow of a vertical circle upon the 
 interior of a hemispherical surface of which the given circle is 
 the front edge. 
 
 83. If a straight line be moved, so as to remain constantly 
 parallel to its first position, and to rest on the circumferences of 
 two great circles of a sphere, it will generate a cylinder, of which 
 those circles will be equal oblique sections. All the elements of 
 this cylinder will intersect the sphere, except the two which are 
 tangent to it at the extremities of the common diameter of the 
 two great circles. Hence we have the principle, that a cylinder 
 which intersects a sphere in a great circle, intersects it also in a second 
 great circle. The right section of this cylinder will be an ellipse, 
 equally inclined to these great circles, and whose transverse axis 
 is their common diameter. 
 
 It now follows, that the cylinder of rays passing through the 
 front edge of the quarter sphere, will intersect its interior in an 
 arc of a great circle of that sphere, which arc will be the shadow 
 required. 
 
 1. Tlie first solution, based on the foregoing general princi- 
 ples, is made by the direct general method (82), and depends imme- 
 diately on the following 
 
 Principles. A plane of rays, perpendicular to the vertical plane 
 of projection, will cut a circular arc from the spherical part of the 
 niche, and a ray from the cylinder of rays, whose intersection will 
 be a point of the required shadow. 
 
 Construction. PI. X., Fig. 33. In this construction, three 
 planes of projection are used.
 
 SHADES AND SHADOWS. 83 
 
 The plane of the front of the niche is taken as the vertical plane 
 of projection. The plane of the upper base of the niche is taken 
 as the horizontal plane of projection, AC is therefore the ground 
 line for these planes, and the spherical quadrant of the niche is 
 in their second angle. D'H', parallel to the vertical projection, 
 DO, of a ray of light, is the trace, or ground line, of the auxili- 
 ary plane of projection. D'H' is in the vertical plane, and the 
 auxiliary plane is revolved about it, into the plane of the paper, 
 so that the part in front of the vertical plane falls to the right 
 of D'H'. 
 
 DO D"p is a ray of light. When a point is projected on 
 two planes, each perpendicular to a third, its projections on those 
 planes are at equal distances from their respective ground lines, 
 hence, making O'p' equal to Op, p f is the auxiliary projection 
 of the point O^p of the ray DO D"p. D is projected at D', 
 therefore D'p' is the projection of the ray, DO D'^9, on the 
 auxiliary plane. 
 
 The plane of rays, whose vertical trace is DO, cuts from the 
 spherical part of the niche, produced, the semicircle DOc 
 D'T'H' and from the cylinder of rays (83), the ray, DO D'/, 
 which pierces the semicircle at c', which, being projected back 
 at c, gives c,c f as one point of the required shadow produced. 
 
 Likewise, any other parallel plane of rays, as ak, cuts from 
 the niche the semicircle ak a'h'f", and from the cylinder of 
 rays, the ray ah a'h', giving h,h f for another point of shadow. 
 
 As N'T' is the ray which is the tangent element (83) of the 
 cylinder of rays, T' is where the shadow begins, and T7ic is the 
 quadrant of shadow, on the spherical part of the niche, pro- 
 duced. 
 
 2. The second Solution is made by the special method of (42). 
 
 Construction. PI. X., Fig. 34, CHB C'A'B' is the upper 
 base of the semi-cylindrical part of the niche. CAB C'A^B' 
 is the front edge of its spherical part. 
 
 By the first solution, T,T r , where a plane of rays, N'T', perpen- 
 dicular to the vertical plane of projection, is tangent to the 
 spherical part, on its front edge, is the point where the required 
 shadow on the spherical interior begins. 
 
 To find another point, take any auxiliary plane, parallel to the 
 front face of the niche. EH may be assumed as the trace of 
 such a plane, on the plane C'A'B'. The ray AE A'E', 
 through the centre of the front semicircle, pierces the vertical
 
 84 GENERAL PROBLEMS. 
 
 plane EH, at E,E', which is therefore the centre of the shadow 
 of that semicircle on the plane EH. With E' as a centre, and 
 a radius equal to A'C' (33), describe the arc which intersects the 
 semicircle Wh', cut from the quarter sphere by the plane EH, 
 at h' ; whose horizontal projection is 7i, on the trace EH. hfi is 
 therefore a required point of shadow. 
 
 The point y,y f is similarly found, on a similar vertical secant 
 plane, aY. 
 
 3. Construction of the point where the shadow leaves the sphwical 
 part of the niche. 
 
 This point, Z>,5', may here be found, without reference to the 
 cylindrical part of the niche. 
 
 Remembering that the plane of shadow contains a great circle 
 of the spherical part, the diameter A'T' is its trace on the front 
 face of the niche, and therefore A' A is one point of its trace on 
 the base, CHB, of the spherical part. That trace intersects the 
 semicircle CHB at the point required. Here then is given one 
 trace, A'T', of a plane, and a point, as y,y', in that plane, to find 
 its other trace. Now ya y'a' is a line through y,y', and paral- 
 lel to the trace A'T', and therefore is a line of the plane of 
 shadow. This line pierces the plane C'A'B' at a',a, giving Aab 
 for the required trace on C'A'B', and b,b' for the required point 
 at which the shadow leaves the spherical part of the niche ; it 
 being the intersection of the plane of shadow with the base 
 line CHB. 
 
 4. Third Solution. This is made by the special method of 
 (43) which, in its present application, depends immediately on 
 the following 
 
 Principles. Knowing, in advance, from (83) that the curve 
 of shadow is a plane curve, we can find its rectilinear projection 
 on a plane of projection, taken perpendicular to the plane of 
 shadow. Then, having this rectilinear projection, both projec- 
 tions of particular points in it can be found as in (43). 
 
 Construction. PI. X., Fig. 33. D'H' is the trace, on the plane 
 of the face of the niche, of a plane perpendicular to the plane 
 of shadow. All the planes of projection are, moreover, the 
 same here, as in the first solution. 
 
 Having found, as before, c', we have OT'Y for the vertical, 
 and O'c' for the auxiliary, trace of the plane of the shadow, the 
 latter trace being also the projection of the shadow on the auxili- 
 ary plane. Any semicircle, as a'h'f", lying in the spherical
 
 SHADES AND SHADOWS. 85 
 
 surface, or any ray, as a/A', will cut from the given projection of 
 the shadow a point, as A', whose other projection, A, is found by 
 drawing h'h, perpendicular to the ground line D'H', and noting 
 its intersection, A, with the vertical projection, ok, of the same 
 semicircle, a'A/", or ray a'h'. 
 
 Discussion. 
 
 FIRST. If it were not known already that the shadow on the 
 spherical part of the niche is a plane curve, it could be proved by 
 reference to PI. X., Fig. 33. For, O'D' and O'c' are equal, hence 
 O'c'D' is an isosceles triangle. But a'h' being a chord, parallel 
 to D'c', in a semicircle a'h'f", concentric with D'c'H', it follows 
 that OVA' is an isosceles triangle also, and is similar to O'D'c'. 
 Then, as 0V and O'D' coincide in direction, O'A' and O'c' like- 
 wise coincide, and O'A'c', the projection of the shadow on the 
 auxiliary plane, is a straight line, which shows the shadow itself 
 to be a plane curve. 
 
 SECOND. Another construction of the point, e, where the sha- 
 dow leaves the spherical part of the niche, may here be given. 
 C'B" is the auxiliary projection of a line, perpendicular to the ver- 
 tical plane of projection at C. Its horizontal projection would be 
 a perpendicular to the ground line, AC, at C ; but, to avoid con- 
 fusion of the figure, a similar line is shown at AB. Producing 
 O'c' to B", gives B" as the point in which the plane of shadow 
 intersects C'B". Now makeAB equal to C'B", and OB will be 
 the horizontal trace of the plane of shadow, after revolving 180 
 about OU, in the horizontal plane, as an axis. The point m is 
 then the revolved position of the intersection of the horizontal 
 trace of the plane of shadow, with the horizontal semicircle, 
 AUC, of the niche. This intersection is the point sought. Then 
 the horizontal . projection, ran, of the vertical arc of counter- 
 revolution, gives n for the primitive horizontal projection of this 
 point, and ne, perpendicular to the ground line, AC, gives e, the 
 vertical projection of the same required point of shadow. 
 
 THIRD. The point e can again be found, but considered still 
 as the intersection of the horizontal trace of the plane of shadow 
 with the horizontal semicircle, AUC, of the niche. This is done 
 by viewing this trace as the intersection of the plane of shadow 
 with the plane of the horizontal circle AUC. Then, after show- 
 ing the traces of both of these planes on the planes of projection
 
 86 GENEEAL PEOBLEMS. 
 
 whose ground line is D'H', their line of intersection can be 
 found. 
 
 The traces of the plane of shadow are OT'Y, on the vertical 
 plane, and O'B", on the auxiliary plane D'H'. The traces of the 
 plane of the base of the niche are T'O' Y, on the auxiliary plane, 
 and O'e", on the vertical plane ; for, thus considered, D'H' is 
 supposed to be the trace of the plane DO itself, after being trans- 
 lated, parallel to itself, to the position D'H', and then revolved, 
 as before described. O'e" is thus parallel to OC, being only a 
 transferred position of OC. 
 
 The intersection of the plane of shadow and the plane of the 
 base is now projected in O'c' and O'e". The arc/"/" represents 
 a revolution of the plane of the base, together with the intersec- 
 tion just noted, about O' T'O'Y as an axis, and into the trans- 
 lated position, D'H', of the plane of rays DO. The base then 
 appears in D'c'II', and its intersection, O'e" O'c', with the 
 plane of shadow, at O'f. At e',e f " is therefore the revolved 
 position of the point of shadow on the base of the niche. In 
 counter-revolution, e'e" returns to e"E, from which is projected, 
 at e, the required projection of the required point of shadow. Or, 
 e' may be first counter-projected at e'"' ', and then counter-revolved, 
 as shown by the arc e""e, which gives, again, the same required 
 point, e. 
 
 FOURTH. Considering D'H', again, as the ground line of the 
 original position of the auxiliary plane of projection, e may again 
 be found by the intersection of the projection, O'c', of the shadow 
 on the spherical quadrant, with the auxiliary projection, C'ET', 
 of the base, AUC, of the niche. C'ET' is a quarter ellipse on 
 the semi-axes O'T' and O'C', C' being the projection of C. 
 
 FIFTH. A little consideration will show that C'O' : C'H':: 
 Er I Ee'. 
 
 Hence O'T', C'E, and H'e' all meet at I. Hence, from any 
 point, I, on O'O, draw lines to C' and H'. From the intersec- 
 tion, as e', of IH' with the semicircle D'c'IF, draw a perpendicu- 
 lar to O'O ; its intersection, as E, with 1C' will be a point of the 
 ellipse. 
 
 Thus, after finding e', as above, we may find E, and hence e, 
 by drawing H'e'I, and then 1C' which will intersect O'c' at E ; 
 which is projected by E<?, perpendicular to D'H' at e. 
 
 Moreover, we have here a construction of the ellipse by homo- 
 logous secants, analogous to the homologous tangents which may
 
 SHADES AND SHADOWS. 87 
 
 be drawn at e' and E, and which will meet O'O at a common 
 point. 
 
 SIXTH. Finally, e may be approximately found as the point 
 where the entire quadrant of shadow, The, on the spherical part 
 produced, crosses AC. (See 42.) 
 
 SEVENTH. To construct the tangent line at the point, e. 
 
 Principles. At the point e, the required line is tangent both to 
 the spherical branch, ehT', and to the cylindrical branch of the 
 shadow. For, the tangent, L, to the spherical branch, is the 
 intersection of a tangent plane, P, to the sphere of the niche at <?, 
 with the tangent plane, P', to the cylinder of rays, along the ele- 
 ment eW. The tangent, L', to the cylindrical branch, is the 
 intersection of the same tangent plane, P', to the cylinder of rays, 
 along the element eE', with the tangent plane, P", along the 
 element, eB", of the cylinder of the niche. Now, since this 
 cylinder and the sphere are tangent to each other on AUC, the 
 planes, P and P", coincide, hence the tangent lines, L and L', 
 coincide also, forming a common tangent to the two curves of 
 shadow which coalesce at e. 
 
 Again : The curve ehT' is a plane curve, and the tangent to 
 a plane curve of intersection is the intersection of the plane of 
 that curve, with a tangent plane to the surface on which it. lies. 
 Hence, finally, the required tangent line is best found as the 
 intersection of the tangent plane to the cylinder of rays, along the 
 element eE, with the plane of the curve of shadow. 
 
 Construction. PI. X., Fig. 33. E'Y, tangent to the front of 
 the niche, or base of the cylinder of rays, at E', is the vertical 
 trace of the tangent plane to the cylinder of rays, along the ele- 
 ment eFj'. OT'Y is the vertical trace of the plane of shadow. 
 Then Ye, joining the intersection of these traces with the given 
 point e, which is common to both planes, is their intersection, or, 
 the required tangent line. 
 
 EIGHTH. Having now, by previous constructions, the tangents 
 at T', b', and c, PI. X., Fig. 34, where c represents the upper end 
 of the vertical shadow of D'C', it is easy to construct, according 
 to elementary plane geometry, the curved shadow by approxi- 
 mate circular arcs. Thus, from T' to V may be made part of an 
 oval of three centres, having AT' and A'E' for its axes ; and b'c 
 may be a single arc, tangent to the tangents at b' and c, or, if 
 necessary, it may be an arc of a compound curve having the same 
 tangents, and a horizontal line at c for one axis. When the
 
 88 GENERAL PROBLEMS. 
 
 projections of the light make an angle of 45 with the ground 
 line, a single arc can be drawn through b' and tangent at T', and 
 another, through b' and tangent at c; which will nearly coincide 
 with the shadow. 
 
 PROBLEM XXX. 
 
 {To find the shadow of the upper circle of a piedouche upon its 
 
 concave surface. 
 
 Principles. The highest and lowest points of this shadow are 
 found by the direct general method (41). Other points are found 
 by the special method of (42). 
 
 Construction. PL XL, Fig. 35, shows the surface represented 
 as in PI. X.., Fig. 31. Its projections need not again be described. 
 
 1. The highest and lowest points of shadow. 
 
 AK is the horizontal trace of a meridian plane of rays. Since 
 the piedouche is a surface of revolution, if we revolve AK about 
 the vertical axis at A, till it becomes parallel to the vertical plane 
 of projection, the meridian curves contained in it will coincide 
 with. E'c^" and B'D'C', which are seen in the vertical projection, 
 of the piedouche. 
 
 F,F' is the revolved position of the point whose horizontal 
 projection is F", and which is cut by the plane of rays from the 
 upper circle casting the shadow. Then the revolved position of 
 a ray through F,F', will intersect F'cT'E' and B'D'C', respec- 
 tively, in the revolved positions of the highest and lowest points 
 of shadow, viz. at d'",d' r , and r"V", in the meridian plane FAC. 
 
 The revolved ray, F W'V", is parallel to AV", which is found 
 by revolving the ray Ac AV to the position Ac" AV", 
 parallel to the vertical plane of projection. 
 
 In the counter-revolution, d",d"' and r",r'" return, respectively, 
 in the horizontal arcs, d"dd'"d', and r'"r r"'r' t to their true 
 positions, at d,d' and r"V, as the highest and lowest points of 
 the required shadow. 
 
 2. To find any intermediate points. 
 
 Any horizontal plane, as j'D', cuts from the surface of the 
 piedouche a circle, q'D' qsf; and from the cylinder of rays, 
 whose given base is the circle, BNF F'A'B', a circle equal to 
 the latter circle, and which is the shadow of BNF F'A'B' on
 
 SHADES AND SHADOWS. 89 
 
 ihe plane ql)'. The intersection of this auxiliary shadow with 
 the circle qsfq'D', gives two points of shadow on the given 
 surface (42). 
 
 The centre of the auxiliary shadow is b\b- where the ray Ab 
 A'6', through the centre of the upper circle of the concave sur- 
 face, pierces the plane g-T)'. Then, drawing an arc, fq, with a 
 centre, &, and a radius equal to A'B', we find /and q, horizontal 
 projections of two points of the required shadow. Projecting 
 these points on q'T>', the vertical trace of the horizontal plane 
 containing them, gives q' and/ for their vertical projections. 
 
 Two points can be similarly found, on any other auxiliary 
 horizontal plane between the highest and lowest points, as shown 
 at e,e', on the plane e'a', and &tp,p f on the plane p'c' . 
 
 Discussion. 
 
 FIRST. The whole of the shadow just found, would be real, 
 only in a geometrical sense ; or upon a hollow and transparent 
 piedouche. Observing that the highest point of the curve of shade 
 would be found by drawing a tangent ray, parallel to A'c'", and 
 counter- revolving as before, it is plain that the highest point of 
 shadow is below the highest point of shade. Hence the shadow 
 on an opaque solid piedouche will be real, from its highest 
 point to its intersection with the curve of shade. The real part 
 of the curve of shade was described in Prob. XXVIII. (Eem.^). 
 This real part will cast a shadow on the lower part of the 
 piedouche. 
 
 Thus the complete line of separation, between the illuminated 
 and the dark portions of the piedouche, consists of the real por- 
 tion of the shadow of the upper circle, the real part of the curve 
 of shade, and the shadow of this real shade. 
 
 SECOND. The latter shadow is thus found. See PI. XI., Fig. 
 36, which represents, in horizontal projection, and sufficiently for 
 purposes of explanation, the shadow just mentioned, together 
 with the entire shadow of the piedouche on the horizontal 
 plane. 
 
 The shadow of the curve of shade upon the lower part of the 
 piedouche, is found by the special method of two auxiliary 
 intersecting shadows (66-ls/. Eem. a). Thus, qkt and pmu are 
 shadows of the real parts of the curve of shade, on the horizontal 
 plane. They are found from the projections of those parts,
 
 90 GENERAL PEOBLEMS. 
 
 shown in PI. X., Fig. 31, and just as any shadow is found on a 
 plane of projection (30). 
 
 The circle sr is the shadow of an equal circle, ITT, near the 
 foot of the piedouche ; r and s, the intersections of these sha- 
 dows, are the shadows of points of shadow cast by the curve 
 of shade on the circle UT. That is, rays, as sU and rT, through 
 5 and r, will meet both UT, and the curves of shade. U and T 
 are the shadows, on UT, of the points cut from the curve of 
 shade by these rays. 
 
 THIRD. pq, and ut, are the shadows of the unreal portions, 
 1U2, and QrE, PL X.. Fig. 31, of the curve of shade. 
 
 These unreal shadows join the real ones at p, q, u, and t, 
 forming cusps of the first order. 
 
 This result being apparent, it enables us to learn, as we could 
 not easily do from an inspection of the curve of shade only, in 
 PI. X., Fig. 31, the real nature of the tangent cylinder of rays, 
 whose curve of contact with the piedouche is the curve of 
 shade. 
 
 Considering putq as its base, and that its elements are parallel, 
 it is now evident, that above the point 2,2', for example, its sur- 
 face curves upward, and backward towards the vertical plane ; 
 and that below the same point, its surface bends downward, and 
 backward towards the vertical plane ; so that it has an edge on 
 the tangent ray at 2,2'. A plane through this ray, or element, 
 and tangent on the same, to both the branches, j ust described, 
 of the cylinder, is an osculatory plane to the curve of shade at 
 2,2' ; and its horizontal trace is a tangent to the base of the 
 cylinder at q, PL XL, Fig. 36, between hq and pq. But the 
 tangent ray at 2,2', being such a salient edge as just described, a 
 plane section, as the base of the cylinder, will present a cusp as 
 at q, where that edge pierces the plane of the section. 
 
 Observing, further, that the cylinder of rays, to which the 
 plane of rays is tangent, is, itself, tangent to the piedouche at R, 
 the plane is also tangent to the piedouche at R. Hence its hori- 
 zontal trace can readily be found. 
 
 FOURTH. The rest of the shadow, PL XI., Fig. 36, is thus 
 composed : ahc is the shadow of that lower semicircle of the 
 upper base of the piedouche which is towards the light, deb is 
 the shadow of the opposite upper semicircle, KSB. ab and cd 
 are the shadows of the elements of shade of the upper base at K 
 and B. Likewise, nmy is the shadow of the upper semicircle,
 
 SHADES AND SHADOWS. 91 
 
 NHY, of the lower base of the piedouche, and ~Nn and Yy are 
 the shadows of the elements of shade of the lower base, at N 
 and Y 
 
 III. General Problem in Review of Shades and 
 Shadows, determined by Parallel Rays. 
 
 84. No better problem can perhaps be found for this review 
 than that of the shades and shadows on the Roman Doric column, 
 embracing, as it does, the shadows both of straight and of curved 
 lines ; on planes, and on a variety of single curved, and double 
 curved surfaces. 
 
 Description of the Column. PI. XL, Fig. 37, represents, with- 
 out regard to precise architectural proportions, 'a fragment of the 
 shaft and base of the column. 
 
 Indicating its horizontal circles by their radii, the circle Ca is 
 the plan of the shaft a'SK. Next, Cb is the plan of the smallest 
 section of the scotia e'b'c". Then Cc is the plan of the upper 
 and middle fillets, c'g' and c"h'. And Cd is the largest section of 
 the upper torus, c"d'h'g'. Ce e'n' is the lower fillet. CF -f'nf is 
 the lower torus. FGH F'H/ is a square member, called the 
 plinth, a! eg' is the foot of the shaft, and is a concave double 
 curved surface. The fillets are short vertical cylinders. 
 
 PI. XL, Fig. 38, similarly represents the principal parts of the 
 capital and shaft of the column. a"ahw a'b'io'g"h' is the 
 abacus, whose horizontal sections are squares. The portion, 
 on o'n'n", of the abacus, is the cyma reversa, and is cylindrical. 
 Cg and C/are the greatest and least circular sections of the half 
 torus g'g'f'i called the echinus. Cff'f" is a fillet. Between 
 C/and Ge is e'f'f ', the cavetto. Cc is the neck of the column, 
 limited by a small torus, not shown, and called the astragal, just 
 under which is the lower fillet and then a double curved concave 
 surface similar to the foot of the column. 
 
 Thus the whole column, between the plinth and the abacus, is 
 a surface of revolution having a vertical axis.
 
 92 GENERAL PROBLEMS. 
 
 PROBLEM XXXI. 
 
 To find the Shades and Shadows of the Shaft and Base of a RomLn 
 Doric Column. 
 
 Such parts of the solution as present peculiar features, will be 
 figured. Others will be merely referred to previous problems 
 containing similar constructions. 
 
 1. The shaft is strictly a double curved surface, its upper 
 diameter being, in practice, a little less than its lower, and its 
 generatrix slightly curved. The fragments shown in the figures 
 may, however, be treated as vertical cylinders, in finding their 
 elements of shade. 
 
 2. The shadow of the shaft on the foot of the column, is 
 found by the special method of (43) (Prob. XVL, 3), its hori- 
 zontal projection being known as a straight line, tangent to the 
 plan of the shaft. 
 
 3. The element of shade, and the upper circle of the fillet, cast 
 shadows on the upper torus. See PL XII., Fig. 39. 
 
 The element of shade is the vertical line at NN 7 , and its shadow 
 is found as in (2). Thus, NK is the indefinite horizontal pro- 
 jection of the shadow, NK being the intersection of a vertical 
 tangent plane of rays, at N", with the torus, and therefore paral- 
 lel to the horizontal projections of rays, da d'a! is any assumed 
 horizontal circle of the torus, intersecting the indefinite shadow 
 NK at a! whose vertical projection, a', is on d'a', the vertical 
 projection of the assumed circle. 
 
 This shadow begins at N', the foot of the element of shade. 
 
 The upper circle casts a shadow which is found by the method 
 of (42), see also (Prob. XXX 2). Thus, the ray, G/iG'h' t 
 through the centre of the upper circle, pierces an auxiliary hori- 
 zontal plane, h'e', cutting the torus in a circle, at h,h f . The cir- 
 cle, ybj with centre h and radius hy, equal to GV, is the auxiliary 
 shadow of the upper circle, GV, on the plane h'e'. This shadow 
 meets the circle eb, cut from the torus by the plane h'e', at two 
 points, one of which is b. Then b is vertically projected at &', 
 on the vertical projection, e'k', of the circle cut from the torus. 
 
 On a complete construction, the latter shadow will intersect 
 the shadow of the element of shade, just before found, and the
 
 SHADES AND SHADOWS. 93 
 
 curve of shade of the upper torus, in points which the student 
 may construct. 
 
 4. The curve of shade on each torus in found as in Prob. 
 XXVII. 
 
 5. The curve of shade of the upper torus casts a shadow on 
 the middle fillet, which is found as in Prob. IV. But PI. XIL, 
 Fig. 39, serves, however, to show how this shadow can be found, 
 having given but one projection of the curve of shade of the 
 torus. Let EV be a fragment of the vertical projection of the 
 curve of shade of the torus. Let o' be a point on this curve, 
 which is supposed to cast a shadow on the fillet. This point is 
 transferred, on a circular section of the torus, to a point whose 
 revolved position is D',D", found by revolving the semicircular 
 meridian curve, in the plane DGGr', about its vertical diameter 
 atS, 
 
 In the counter revolution about S, D'D" proceeds to D, in 
 horizontal projection ; and in the counter revolution about the 
 vertical axis at G, it returns to o, the desired horizontal projec- 
 tion of o'. 
 
 This done, the shadow of 0,0' is found by the simple direct 
 method of (40). Thus, op, the horizontal projection of the ray 
 through 0,0', pierces the fillet at a point of shadow whose hori- 
 zontal projection is^>. The vertical projection, p', of this point, 
 is at the intersection of the projecting line, pp', with the vertical 
 projection, o'p', of the same ray. 
 
 6. Part of the same curve of shade on the upper torus, casts 
 a shadow on the scotia. PI. XIL, Fig. 40. This curve of shade 
 being of double curvature, its shadow on an auxiliary plane, P 
 as in the special method of (42), will be an irregular curve, a num- 
 ber of whose points must be found, and joined together, to give 
 the auxiliary shadow, whose intersection with the circle cut from 
 the scotia by the plane P, would be a point of the required 
 shadow. 
 
 Hence the special method of (42), which, as seen in Prob. 
 XXX., is so convenient when the shadow is cast by a circle, is 
 no more convenient than the direct method (82) when the curve 
 casting the shadow is of double curvature, as in this case. 
 
 For this reason the direct method (S2) as applied in problems 
 like the present, is here used. Let b'c' be a fragment of the ver- 
 tical projection of the curve of shade of the torus. Let V be 
 any point on this curve, whose shadow on the scotia is to be
 
 94 GENERAL PROBLEMS. 
 
 found. It is transferred, on a circular section of the torus, to a', 
 whose horizontal projection is a. In counterrevolution about a 
 vertical axis at D ; a',a returns to b',b. Now, b'r f is the vertical 
 trace of a plane of rays, taken through &,&', and perpendicular 
 to the vertical plane of projection. It cuts from the scotia the 
 curve efr e'f'r'] which is found, as fully shown, by auxiliary 
 horizontal planes, as G'f. This plane cuts from the scotia the 
 circle G'f C/J, and from the curve c'r' the point/', whose hori- 
 zontal projection, f, is on the horizontal projection, C/ of the 
 circle containing it. Having thus constructed the intersection 
 of the plane of rays with the scotia, by means of a system of 
 auxiliary planes, draw the ray bs b's', which pierces the curve 
 efr e'r f at 5,5', the required shadow of b,b'. Other points of 
 shadow on the scotia may be similarly found. 
 
 The plane of rays may also be taken vertical. 
 
 7. There will be two points of the curve of shade b'c', the 
 rays through which will pass through the points, as e,e', on the 
 lower edge of the fillet. That part of the curve b'c', lying 
 between these points, casts a shadow on the fillet. These points 
 themselves cast shadows on the lower edge of the fillet, and the 
 part of this edge between these points, will cast a shadow on the 
 scotia. The latter shadow will be found as in Prob. XXX. " 
 
 8. The curve of shade of the scotia will be found as in Prob. 
 XXVIIL 
 
 9. If, on account of a certain direction of the light, the 
 curve of shade of the upper torus casts a shadow on the lower 
 torus, it will be indicated by the passing of some of the rays, 
 similar to bs, in front of r,r', so as not to cut the corresponding 
 curve similar to efr. This shadow will be found as in (6). 
 
 10. The shadows of the lower fillet on the lower torus are 
 found as in (3). 
 
 11. After finding the curve of shade on the lower torus as in 
 Prob. XXVII., its shadow on the top of the plinth will be found, 
 a point at a time, as in Prob. III.
 
 SHADES AND SHADOWS. 95 
 
 PROBLEM XXXII. 
 
 To find Hie Shades and Shadows of the Capital and tihaft of a Roman 
 Doric Column. 
 
 In making the solution of this problem, we have 
 
 1. The shadow of the upper member of the abacus on the 
 cyma reversa ; the element of shade on the latter ; its shadow 
 on itself, and on the lower part of the abacus. These shadows 
 form one topic, being all found in the same general way, as seen 
 in PI. XIL, Fig. 41. 
 
 The cyma reversa, being bounded by cylindrical surfaces, it is 
 necessary to have an auxiliary elevation of it, as it appears when 
 seen in the direction of the arrow. The point T" then is the 
 auxiliary vertical projection of the lower front edge, NY T'A'. 
 Seen in the direction of the arrow, the point, B,B', of the ray 
 AB A'B', appears at a distance, A's, below T" T'A', and at 
 a distance, rB, to the left of NY r" . Therefore, make TV 
 equal to AY, and r"B" equal to rB, then T"B" will be the 
 auxiliary vertical projection of a ray, or the trace of a plane of 
 rays through NT T'A' T'. This plane cuts from the front 
 of the cyma reversa the line u" uu', the upper shadow on the 
 cyma reversa. A parallel, and tangent, plane of rays, La", 
 gives the element of shade, n" nn', and its shadow, a" aa'. 
 Another plane of rays, through the edge below a" aa', would 
 give the shadow on the lower part of the abacus, not shown. 
 
 2. The shadow of the edge h'g", PL XI., Fig. 38, on all the 
 cylindrical parts below, is found as in Prob. IV. 
 
 The shadow of the edge which is perpendicular to the paper 
 at A', on all parts below, is bounded by the intersection of a 
 plane of rays through this edge with those lower surfaces. This 
 intersection will appear in vertical projection as the vertical pro- 
 jection of a ray at h' (43). 
 
 3. The shadow of the lower front edge of the abacus, RT 
 K'T', PL XII., Fig. 42, on the echinus is found thus : Let N'K' 
 ~Ney be a circle, cut from the echinus by a horizontal plane, 
 N'K'. A ray, cd c'o", through any point, as c,c', of the edge 
 casting the shadow, will meet this plane in dd', which determines 
 dg, the shadow of RT R'T' on the parallel plane, N'K' (5).
 
 96 GENERAL PROBLEMS. 
 
 This auxiliary shadow meets the circle, Neg N'K', at e,e' and 
 g,g', two points of the required shadow, which is thus found by 
 (42) see also Prob. XXX. 
 
 Observing that un is equal to Sf, the highest point, /', of this 
 shadow, may be thus found. The plane of rays, through ET 
 R'T', intersects the meridian plane, Dn, in aline, nb, which is the 
 projection of the light on the plane Dn. The intersection of 
 this trace nb with the echinus is the point/ 7 . Revolve bn to ba", 
 whose vertical projection is a'b'. Draw R'L', parallel to a'b', 
 and revolve L to /', and f is the required highest point of the 
 shadow. By drawing a ray through /', we could find the point 
 whose shadow is f. 
 
 THe shadow of RT R'T' is real, above its intersection with 
 the curve of shade of the echinus, which is found as in Prob. 
 XXVII. 
 
 The remaining shadows involve no new operations.
 
 SHADES AND SHADOWS. 97 
 
 SERIES II. 
 
 SHADOWS DETERMINED BY DIVERGING RAYS. 
 
 SECTION I. 
 G-eneral Principles. 
 
 85. Having seen, in the preceding general problems, how 
 shades and shadows are found, when the rays of light are parallel, 
 it now remains to examine Arts. (2) and (16) somewhat in detail. 
 "When a body, B, is illuminated by a single luminous point, that 
 point may be considered as the vertex of a cone of rays, C, tan- 
 gent to the body, B. The line of contact of C and B, is the line 
 of shade of B. The interior of the cone, beyond B, is the 
 shadow, in space, of B, and that portion of any secant surface, 
 S, which is within the cone, and beyond B, is the shadow of B 
 on S. If, now, B be illuminated simultaneously by two points, 
 P and P', a portion, both of B and S, will receive light from 
 neither point. This is the total shade on B, or shadow on S, 
 respectively. Another portion of B and of S will receive light 
 from both points. This will be their completely illumined por- 
 tion. A third portion of B, and of S, will receive light from 
 one or the other of the points, P and P', but not from both. 
 This is their penumbra, or partial shade, or shadow. 
 
 86. Let this case be extended to an assemblage of luminous 
 points, forming a luminous body, L. In general, L and B will 
 be of different sizes, hence they may have two common tangent 
 cones of rays inclosing them ; one with its vertex beyond both 
 bodies, the other with its vertex between the two bodies. The 
 area, or zone, on B, between the curves of contact of the two 
 cones, will be the partial shade of B. The annular space on S, 
 between the intersections of the two cones with S, will be the 
 penumbra or partial shadow of B on S the part which is reached 
 by only a part of the rays from L. 
 
 7
 
 08 GENERAL PROBLEMS. 
 
 87. See, now, PI. XIII., Fig. 44, where AC represents a 
 luminous body ; BD, an opaque body, and PQ an opaque sur- 
 face which receives the shadow of BD. The dark space, 6c?, is 
 the total shadow, and is bounded by the cone of rays, tangent to 
 both bodies, and having its vertex on the same side of both. 
 The lighter space, ca, is the penumbra, and is bounded by the 
 tangent cone whose vertex is L. 
 
 88. Eegarding L at first as the original source of light, and the 
 vertex of the single cone Lea, it may then be regarded as the 
 vertex merely of a complete cone of two nappes, in the outer 
 nappe of which is inscribed the extended source of light, AC, 
 around which, and the body BD, may be inscribed the second 
 cone which determines the shadow bd. 
 
 89. It should be noted, that the volumes of rays will only be 
 cones, when the bodies, AC or BD, are similar, or have similar 
 curves of contact ; or when, at least, they can both be inscribed 
 in one given cone, so as to have a continuous curve of contact 
 with that cone. 
 
 In other cases, the volumes of rays will be bounded by warped 
 surfaces, but these cases are unimportant, and will not be further 
 considered. In every case whatever, the surface of rays will be 
 a ruled surface, since rays of light are straight lines. 
 
 90. For the practical case, let L be the sun, and B a body 
 near the earth, S. Here the angular breadth of the penumbra 
 of shade upon a spherical body, B, would be, as found by a 
 simple calculation, only T ^j- of the radius of B. 
 
 See PL XIIL, Fig. 43, where O is the sun's centre, a, the 
 centre of a small spherical body near the earth, and V, and v, 
 the vertices of the two tangent cones of rays already described. 
 Then, on account of the relatively great distance of the sun, 
 AB, in the triangle ABC, becomes sensibly equal to his diame- 
 ter, and aC and ab, radii of the small body, being perpendicular 
 to BC and AC, respectively, the chord 6C sensibly reduces to 
 the segment, 6C, of the side AC, and ABC and o&C are similar 
 triangles, and AB : BC::Z>C : aC : or 
 
 885 000 : 95 000 000::iC : aC ; or 
 ftC = aC very nearly. 
 Tu7 
 
 This penumbra of shade, whose breadth is Z>C, may therefore 
 be disregarded on terrestrial objects, and the solar rays ma}', 
 accordingly, be considered parallel.
 
 SHADES AND SHADOWS. 99 
 
 Remark. The three following problems, embracing both 
 shades and shadows upon a variety of surfaces, plane, single- 
 curved, and double-curved, are meant to be sufficient to initiate 
 the student in the solution of problems in which the source of 
 light is supposed to be a near luminous point. 
 
 After the explanations already made, separate statements of 
 "principles" will not precede the "constructions" of these 
 problems. 
 
 SECTION II. 
 
 !Pro"blems, involving divergent Rays. 
 
 PROBLEM XXXIII. 
 
 To find the shadow of a semi-cylindrical abacus, upon a vertical 
 plane through its axis. 
 
 PL XIII., Fig. 45, FcE F'c'E' is the half abacus, with hori- 
 zontal semi-circular bases, whose diameters are in the vertical 
 plane of projection. 
 
 L,I/ is the source of light. A line from this point to any 
 point of an edge, or element, of shade of the abacus, is a ray of 
 light, whose intersection with the vertical plane of projection, 
 will be a point of the required shadow. 
 
 LD is the horizontal trace of a vertical tangent plane of rays, 
 which determines the element of shade d d'd" . Then from 
 d,d" to E,E', is the edge of shade of the upper base, and from 
 djd' to F,F', is the edge of shade of the lower base. 
 
 These edges, and the element of shade, cast the line of shadow 
 which is required. 
 
 The shadow then begins at F,F'. Any point, as a,a', casts a 
 point of shadow A, A', which is found where the ray La LV 
 pierces the vertical plane of projection. 
 
 The element of shade, d d'd", casts the shadow D'D", and 
 the curve D"E' is the shadow of dEd"W. The shadow of 
 c,c', the foremost point of the lower edge of shade, is C', the 
 lowest point of the shadow ; which is now fully determined. 
 
 The shadow of a tangent at c,c' will be a tangent at C', parallel 
 to the ground line. Other obvious tangents, useful in sketching 
 the curve, can readily be determined.
 
 100 DIVERGENT RAYS. 
 
 PROBLEM XXXIV. 
 
 To find the curve of shade on a sphere, the light proceeding from an 
 
 adjacent point. 
 
 PI. XIIL, Fig. 46. Let the centre of the sphere, Sa'cfc, be in 
 the ground line, and let L,L' be the luminous point. 
 
 1. To find the highest and loivest points of shade. LO is the 
 horizontal trace of a vertical plane of rays containing these 
 points. After revolving this plane about the vertical diameter 
 of the sphere, till it coincides with the vertical plane of projec- 
 tion, the luminous point will appear at I/',!/", and the great 
 circle cut from the sphere, at Sa'dc. The revolved rays, I/"A'", 
 ~L'"l'", tangent to this circle, then determine h"',h", and I'", the 
 revolved positions of the required points. In the counter revo- 
 lution, h'",h" proceeds in the horizontal arc, h"h h'"h' to its 
 true position h,h'. The construction of 1,1', the lowest point, 
 will be given presently. 
 
 2. To find the foremost and hindmost points. Go through a 
 series of operations, similar to the foregoing, upon a plane of 
 rays, I/O, perpendicular to the vertical plane of projection, 
 beginning by revolving it into the horizontal plane of projection, 
 about the horizontal diameter, whose vertical projection is 0. 
 This will give the points/,/ 7 , and e,e'. The middle point, n, of 
 f'e'j is the vertical projection of the centre of the circle of shade, 
 then I', the vertical projection of the lowest point, is at the inter- 
 section of Z"T, the vertical projection of an arc of counter revo- 
 lution, with the diameter, h'n'l'. Then I' is horizontally projected 
 in LO, at I. 
 
 3. The points on tfie circles which are in the planes of projection, 
 are found by drawing tangent planes of rays, perpendicular to 
 the planes of projection. As the centre of the sphere is in the 
 ground line, one circle of the diagram represents both of its pro- 
 jections. Then IS a' and L'i' are the vertical traces of tangent 
 planes of rays perpendicular to the vertical plane. They give 
 a',a and b',b, as the points of shade, on the circle, SZ> Sa'c?c ? 
 which is in the vertical plane of projection. 
 
 Likewise, Lc and ~Ld, the horizontal traces of vertical tangent 
 planes of rays, give the points c,c' and d,d f on the circle Sa'dc 
 Si, which is in the horizontal plane of projection.
 
 SHADES AND SHADOWS. 101 
 
 The construction of other points, as, for instance, those on the 
 great circle, perpendicular to the ground line, is left for the 
 student. 
 
 Having now eight points of shade, the curve of shade may 
 be sketched. To avoid confusion, only the vertical projection 
 of the shade is shown, the horizontal projection of the points 
 of the curve of shade being left unconnected. Also, the whole 
 shade on the front hemisphere, is represented as visible, for the 
 sake of clearness, though only a quadrant of the sphere is in the 
 first angle. 
 
 PROBLEM XXXV. 
 
 Having a niche, whose base is produced, forming a full circle; and 
 a right cone, the centre of whose circular base coincides with the 
 centre of the base of the niche, it is required to find the shades and 
 shadows of this system, when illumined by an adjacent point. 
 
 1. The shadows on the base and cylindrical part. 
 
 PI. XIII., Fig. 47. The given magnitudes being familiar 
 ones, their projections, in the position described, may be under- 
 stood from the figure. L,L' is the luminous point. 
 
 La is the horizontal trace of a vertical plane of rays, through, 
 the edge A A' A" of the niche, and limits the shadow, Aa, of 
 the niche upon its base. From a,a", the shadow of the same 
 edge is the element, a a"a', of the cylindrical part of the niche, 
 limited by the ray La L'a". 
 
 The shadow, e,e', of a point, F,F', of the front circle of the 
 spherical part, upon the cylindrical part, is seen to be found as 
 in Prob. XVIIL, the ray being drawn through L,L'. 
 
 2. The shadow of the spherical part on itself, is found essentially 
 as in Prob. XXIX., only each point of this shadow requires a 
 separate auxiliary plane, whose vertical trace will pass through 
 L'. Thus, L'N' is the trace, on the front face of the niche, of an 
 auxiliary plane perpendicular to the vertical plane of projection. 
 The ray, in this plane, meets the semicircle, cut by it from the 
 niche, in a point of shadow. Wl"N', described on M'N' as a 
 diameter, is the semicircle just named, after revolution about 
 M'N' into the front face of the niche. The point L,L' is at a 
 distance, LJ, from the front of the niche, hence, making L'L" 
 equal to LJ, and perpendicular to L'N', it results that L"MT' 
 is the revolved position of a ray, giving I" for the revolved
 
 102 DIVERGENT RAYS. 
 
 position of a point of shadow. By counter-revolution about 
 L'N', the point I" returns to ?', its true position as the shadow 
 ofM'. 
 
 3. The method of Prob. XXIX. (2d solution) is here shown, 
 also. Thus, let DE be the horizontal trace of a vertical plane, 
 parallel to the front of the niche. It cuts from the spherical 
 part, the semicircle DE D/'E'. The ray, LV I/O', through 
 the centre of the front semicircle, AYB, pierces this plane at 0,0'. 
 The horizontal line, o'a'", of this plane, limited, at a'", by the 
 ray L' A'V", is the shadow of the radius O'A" on the plane DE. 
 Hence a'"f, with centre o', and radius, o'a'", is an arc of the 
 shadow of the front circle, on DE, and /', its intersection with 
 the semicircle D/'E', cut from the niche by the plane DE, is a 
 point of shadow on the spherical part. 
 
 It will now be easy to find the point n' ; the point of which 
 f is the shadow ; and the horizontal projection of the curve 
 T7V, remembering that it is a plane curve (88). 
 
 4. To find the elements of shade of the cone, and its shadows. 
 Draw the ray LV LV'' and find U, where it pierces the plarut) 
 of the base of the niche. U and ~Uq are the horizontal traces 
 of planes of rays, tangent to the cone, and tV t'V, and qV q f 
 V, are its elements of shade. 
 
 The traces Utf and U^ bound the shadow on the base of the 
 niche. At ni,m' and p,p' the shadow on the cylindrical part of 
 the niche begins. The ray LV L'V pierces this part at v,v', the 
 shadow of the vertex on the interior of the niche. The shadows 
 of the elements of shade being curves, other points besides m,m' 
 and p,p' muse be found. To avoid the acute intersections of pro- 
 jecting lines with q'V and t'V, find, by division, their middle 
 points, giving <7,#' and h,h'. Kays, L^ L'#' and Ui L'/i', through 
 these points, pierce the cylinder of the niche, produced, at s,s' 
 and r,r', respectively. Through these, and the previously found 
 points, the indefinite shadows, v'm's' and v'p'r', of the elements of 
 shade, can be drawn. The points, as u, where the horizontal pro- 
 jections of the rays just drawn meet the horizontal traces of the 
 planes of shade, are the points in which those rays pierce the hori- 
 zontal plane, and should therefore appear in perpendiculars to the 
 ground line through u', etc., which are their vertical projections. 
 
 The niche conceals the horizontal projections of the shades 
 and shadows except the portion of shade V/K on the cone, and 
 tlcK, a portion of the shadow of the cone, on the base of the niche.
 
 SHADES AND SHADOWS. 103 
 
 PART II. 
 SHADES AND SHADOWS IN ISOMETRICAL PROJECTION. 
 
 > 
 
 SECTION I. 
 General Principles. 
 
 91. Isometrical projections of shadows may be found by two 
 quite different methods. 
 
 First They may be found directly on the isometrical projec- 
 tion of the object which receives them. Second. They may be 
 first constructed in ordinary orthographic projection, and then, 
 from such projections, their isometrical projections may be found. 
 Examples of both of these modes of procedure will be given 
 among the following problems. 
 
 92. Since the essential feature of isometrical drawing is, that 
 it shows three dimensions, at right angles to each other, in their 
 real size, it follows, that this kind of drawing is chiefly useful in 
 its applications to rectangular bodies. This fact, also, will be 
 illustrated in the following constructions. 
 
 93. In isometrical projection, only one principal plane of pro- 
 jection is used ; but, according to first principles, two projections 
 of a line, i. e , its position with respect to at least two planes, are 
 necessary to determine its position in space ; hence, in isometrical 
 projection, the rays of light are more conveniently determined by 
 referring them to isometrical surfaces, as those of the body illu- 
 minated, than to planes of projection. Hence, again, isometrical 
 drawing is most useful in connection with rectangular objects ; 
 since, in representing other objects, auxiliary projections of rays 
 must be employed, as will soon be seen. 
 
 In the following problems only parallel rays are employed. 
 In PI. XIV., Fig. 48, the light is supposed to enter the cube at 
 the upper left hand corner, E, and to leave it at the lower right 
 hand corner, J, and this is the conventional direction of the light, 
 generally, in isometrical drawing.
 
 101 ISOMETRICAL SHADOWS. 
 
 The ray, being thus a diagonal of the cube, makes an angle of 
 35 16' with each of its faces. Its projections, as EK and ET, 
 on those faces, are diagonals of the faces, and therefore make 
 angles of 45 with the edges of the cube. In the absence of 
 these planes and edges, as when a curved surface is the subject 
 of a problem, it may be convenient to know the angle made by 
 the ray with the plane of projection. This we next turn to find. 
 
 PROBLEM XXXVI. 
 
 To find the angle made by the isometrical ray of light with, the 
 isometrical plane of projection. 
 
 The point, C, Fig. 48, is the projection of the' diagonal from 
 the foremost to the hindmost points of the cube, hence the plane 
 of projection is perpendicular to this diagonal. Then, in Fig. 49, 
 let ACBD be the plan of a cube, and D'C'E'G' its elevation ; 
 the cube being in the fourth angle, with its vertical faces making 
 equal angles with the vertical plane of projection, parallel to the 
 paper. Then PQ, perpendicular to the diagonal D'F', is the 
 trace, on the plane of the paper, of the plane of isometrical pro- 
 jection, and AB and B'E' are the two projections of the ray seen 
 at EJ, in Fig. 48. Observe that EK, El, and KI, Fig. 48, are 
 parallel to the isometrical plane of projection. Then B'G', Fig. 
 49, is the trace of a plane, parallel to the isometrical plane PQ, 
 and the point A,B' is its own projection on this plane, and the 
 point B,E' is projected upon it by the perpendicular E'<7, which 
 is seen in its true size. But the diagonals of the cube being 
 equal, D'F'= the line AB B'E', also E'# = i of D'F', and hence 
 of the ray AB B'E', and if then the ray, considered as radius, 
 be called 1, E'<7 = will represent the sine of the angle made by 
 the ray with the parallel plane B'G' and hence with the isome- 
 trical plane PQ. But 1 is the Nat. sine of 19 28', very nearly, 
 which is therefore the desired value of the angle made by a ray 
 of light with the plane of isometrical projection.
 
 -.n- 
 SHADES AXD SHADOWS. lOo 
 
 SECTION II. 
 Shad.es and. Shadows on Isornetrical Planes. 
 
 94. Shadows on isometrics! planes will be the intersections of 
 rays, through points casting shadows, with their projections on 
 such planes, or with the traces, on the same planes, of any planes 
 of rays containing those points. 
 
 PROBLEM XXXVII. 
 
 Having given a cube, with thin plates projecting vertically and for- 
 ward, in Hie plane of its left-hand back face, to find the shadows 
 of the edges of these plates upon the cube and its base. 
 
 PI. XIV., Fig. 48. To find the shadow of AB upon the top of 
 the cube, to which it is parallel. Acr and B6 represent the rays 
 themselves, through A and B ; hence, as AE and Be are project- 
 ing linen, perpendicular to the top of the cube, E# and c&, on, 
 and parallel to EK, are the projections of these rays, upon the 
 face ECKL. These rays meet their projections at a and 5, the 
 shadows of A and B, hence ab is the shadow of AB. EaJc is 
 now evidently the shadow of the rectangle ABcE. 
 
 To find the shadow o/"EDFG. Here DE, being perpendicular to 
 the face ECIGr, its shadow on that face is in the trace, El, of a 
 plane of rays, through DE, upon that face, and is limited at c?, by 
 the ray T)d. From d, dh, parallel to DF, is the shadow of the 
 portion, DH, of the edge DF. Then JiF is the shadow of HF 
 on the plane of the lower base of the cube. 
 
 We see that PI. XIV., Fig. 50, shows the case in which the 
 line, AB, casting the shadow, coincides, in projection, with a ray. 
 Either point, as 6, of the shadow, may be found, either by pass- 
 ing a plane of rays through B, and perpendicular to the top sur- 
 face FHK, or to the face FKI. The line BD, perpendicular, 
 and the trace Db parallel to FI, determine a plane in the former 
 position, and 6, the intersection of the trace D6, and ray BZ>, is 
 the shadow of B. BE, parallel to KH, and the trace E6 in the 
 direction of El, Fig. 48, determine a plane in the latter position, 
 above named ; and b is seen to be, as before, the shadow of B 
 on FKI.
 
 106 ISOMETRICAL SHADOWS. 
 
 Observe that Eb, and El, Fig. 48, make angles of 60 with a 
 horizontal line. 
 
 SECTION III. 
 Sb.ad.es and. Shadows on. N"on-Isometrical 3?lanes.' 
 
 PROBLEM XXXVIII. 
 
 To find the shadow of a hexagonal cupola on a coupled roof, one 
 face of the cupola making equal angles with two adjacent walls of 
 the house. 
 
 PL XIV., Fig. 51. To locate the walls and roof Let A'"A"Y 
 and YA'V be portions of the adjacent walls, which are at right 
 angles to each other. 
 
 Suppose the inclination of the roof to the horizontal plane, 
 A'"A'W, of the eaves, to be 30. Find 0, the middle point of 
 A'" A". Then make OA', horizontal, and equal to OA'", and 
 make OA / B / =30, then B', the intersection of the vertical OB', 
 with A'B', is the extremity of the summit of the roof; through 
 it B'A"', B'A", and B'M, may be drawn, and through M, the 
 ends of the roof, parallel to A"'B' and A"B'. 
 
 1. To locate the cupola. Let n be the point at which its axis, 
 riN, pierces the centre line, On, of the plane of the eaves, and 
 let nv be the half width of the square vgde, in which the circum- 
 scribing circle, acbe, of the base of the cupola is inscribed. The 
 semicircle, abb, represents the semicircle, ac5, after revolution 
 about the axis ai, till it is parallel to the plane of the paper, i. e., 
 to the isometrical plane of projection. Hence trisect akb, and 
 at k and h draw Ice and Ac?, the projections of the arcs of coun- 
 ter-revolution described by k and h; and ac, cd, and db, will be 
 three sides of the base of the cupola. The remaining sides are 
 parallel to these three. At the corners a, c, etc., of the base, 
 draw the equal vertical edges aC, cl, etc., and join their upper 
 extremities, which will complete the cupola. 
 
 2. To construct the intersection of the cupola with the roof. The 
 plane of the face led gives the trace, urn, on the plane of the 
 eaves, and the trace uU, on the vertical plane through the ridge 
 B'M. Then Urn is its trace on the front roof, and CD, the defi- 
 nite intersection of the face led with the front roof. Similar
 
 SHADES AND SHADOWS. 107 
 
 planes, through Hi. and Gcr, give the traces Nn, on the front 
 roof, and NP, on the back roof, and give B and A, as the inter- 
 sections of these edges with those roofs. 
 
 The vertical plane through the ridge, cuts the edges, ac and 
 be, of the base, at s and t. Vertical lines, sS, and iT, from these 
 points, meet the ridge at S and T, where it meets the walls of 
 the cupola. From S and T, draw S A and SO ; TB and TE, 
 which, with AF and BD, complete the required intersection. 
 
 3. To find the shades and shadows. The auxiliary planes, just 
 used, being planes of rays, determine the faces whose upper 
 edges are IJ, JH, HK, and KL, as in the shade. Kays, as Ir 
 and Hp, through the extremities of these edges, meet the traces, 
 O and Bp, of the planes of rays on the roof, produced, at r, p, 
 etc. Then Drq'poQE is the required shadow, of which only 
 the part on the actual roof is real. 
 
 The student may reconstruct the figure, with the face ID 
 parallel to the wall YA"?i. 
 
 SECTION IV. 
 
 Shades and. Shadows on. Single Curved Surfaces. 
 
 PROBLEM XXXIX. 
 
 To find the elements of shade on an inverted hollow right cone, the 
 shadow on its interior, and the shadow of one of its elements of 
 shade on an oblique plane. 
 
 1. To find the elements of shade. PI. XIY., Fig. 52. Let the 
 isometrical ellipse, SQTJST, be the upper base, 0V the axis, and 
 the tangents, VQ and VN, the extreme elements of the cone. 
 YOR is a plane of rays, and its trace, OR, on the base, meets 
 the ray YR, parallel to E J, Fig. 48, at R, the intersection of a 
 ray through the vertex, with the plane of the base. Then RS 
 and RT are the traces, on the plane of the base, of tangent planes 
 of rays, which determine TV and SV as the elements of shade. 
 
 2. To find the shadow on the interior. R6 is the trace of a 
 secant plane of rays, which cuts from the base the point a, cast- 
 ing a shadow, and from the cone, the element, 6Y, receiving the 
 shadow. Then the ray, a A, determines A, the shadow of a,
 
 108 ISOMETRICAL SHADOWS. 
 
 In the same way, other points, C, etc., may be found. The 
 shadow ends at S and T. The shadow of e is the lowest point 
 of shadow, e being the point furthest from the element which 
 receives its shadow. The shadow of e is E, on the element /V 
 (undistinguishable in the figure from NV). The shadow on 
 the element NV, is found by drawing a trace EN. 
 
 3. To find the shadow of the front element of shade on an oblique 
 plane, BLFL DKFI is in an assumed horizontal plane through 
 FI ; and as its position, relative to the cone, is undetermined by 
 the projection, it is, when produced, assumed as cutting the axis 
 0V at m. Then make Tt equal to Om, and t is the projection 
 of T on the plane DKFI. Hence tu is the trace, on this plane, 
 of a vertical plane of rays through T ; wU is its trace on the 
 vertical surface BDF, and Uy, on the oblique plane. The ray, 
 Ty, therefore gives y, as the shadow of T. 
 
 Another point of the shadow of TV, is where it pierces the 
 oblique plane. Now tm, parallel by construction to TO, is the 
 trace of a meridian plane, OTtm, on the horizontal plane DFI. 
 This meridian plane being vertical, it intersects the vertical sur- 
 faces, BDF and BDK, in the vertical lines AH and gGc, giving 
 GH for its trace on the oblique plane. But this meridian plane 
 contains, by construction, the element of shade, TV. Hence r, 
 where TV meets the trace GHr, is the intersection of TV with 
 the oblique plane, produced, and is therefore a point of its sha- 
 dow on that plane. Hence ry is the required shadow of TV. 
 
 The shadow of SV may be similarly found. 
 
 The distance equal to Om, might first be assumed as at pP, 
 then Pin, parallel to pO, will locate m, and t will be the intersec- 
 tion of Tt with P*, parallel to^T. 
 
 SECTION V. 
 
 Sh.ad.es and. Shadows on Double-Curved Surfaces. 
 
 PKOBLEM XL. 
 To find the curve of shade on a sphere.. 
 
 The projection of the sphere on the isometrical plane of pro- 
 jection is the circle CBDA, PI. XIV., Fig 53. Make the angle 
 KF'E' equal to PF'C', Fig. 49, then KF' will be the intersec-
 
 SHADES AND SHADOWS. 109 
 
 tion of the plane of the paper, which is the isometrical plane of 
 projection, with an auxiliary plane, taken as a vertical plane of 
 projection, and on which E'F' is the trace of a plane of rays. 
 Project back O at 0', and make O'E^O'F', and each equal to 
 OA. Then E'F' is the auxiliary projection of that great circle 
 which is contained in a plane of rays. Project E' and F' at E 
 and F ; then the ellipse on AB and EF as axes, will be the iso- 
 metrical projection of the great circle just named. Planes of 
 rays, parallel to E'F', will cut parallel small circles from the 
 sphere, whose diameters will be chords of the horizontal great 
 circle, whose isometrical projection is AHBG. 
 
 As these parallel circles will be projected in similar ellipses, 
 assume gh, kn, etc., as their diameters, and draw hs and nr parallel 
 to BE, which will give the conjugate axes, es and fr, of these 
 ellipses. Constructing these ellipses, and drawing rays tangent 
 to them, gives points of shade, d, c, &, etc. 
 
 The tangent at <z, parallel to the rays, is the trace of a plane 
 of rays perpendicular to the plane of isometrical projection, and 
 gives a as the point of shade on the great circle, parallel to the 
 paper, which forms the isometrical projection of the sphere. 
 
 H and Gr are points of shade, and a line aOp is the trace of the 
 plane of shade on a meridian plane parallel to the paper, giving 
 p for another point of shade. Other points can be found as first 
 described. 
 
 JZemark. For an approximate construction, circles with radii, 
 eh, etc., may be employed instead of ellipses, in ordinary con- 
 structions. 
 
 PROBLEM XLI. 
 To construct the curve of shade on a torus. 
 
 PI. XIY., Fig. 54. KT is the intersection of the isometrical 
 plane, or plane of the paper, with the auxiliary vertical plane. 
 FI/ is the ground line of the latter plane, on that auxiliary plane 
 which is used as a horizontal plane of projection, and is found 
 by making the angle I/FT=E'F'Q, Fig. 49. 
 
 ~Lda" and a'e'h"u' are the auxiliary projections of the torus. 
 The auxiliary projections of the light, in the position which it is 
 desired to have on the isometrical projection, are 0"L K'L'.
 
 110 ISOMETRICAL SHADOWS. 
 
 We now, according to (91 Second), construct the auxiliary pro- 
 jections of the curve of shade, and then the isometrical projection 
 of the torus and its shade. 
 
 1. To find four points of the curve of shade. From Fig. 48, it 
 appears that the light makes an angle of 35 16' with the plane 
 of the base of the cube, i. e., with the horizontal plane of projec- 
 tion. Then, knowing that the horizontal plane (so called in 
 Fig. 54) and seen edgewise before revolution at FL/, makes this 
 same angle with the horizon, it follows that O"L" R'L'" is a 
 ray, after being revolved about the (relatively) vertical axis 
 0" I/CX, till parallel to the auxiliary vertical plane ; and R'L'" 
 is found horizontal. Then draw rays, tangent at e' and u', and 
 parallel at R'L"', and e' and u' will be the revolved positions of 
 the highest and lowest points of the curve of shade, whose true 
 positions are h' and R'. 
 
 The points of shade on the greatest horizontal section of the 
 torus, are a", a! and b",b'. 
 
 2. To find intermediate points. Let the points in the meridian 
 plane, cd, be found. According to (80) O"l and RT represent 
 the ray 0"L RT/ after being projected on this plane, cd. 
 Then 0"l" and RT" are the revolved positions, in the parallel 
 meridian plane, a"b", of the projected ray. Next, draw from r, 
 the centre of the semicircular part of the meridian curve, rd" f , 
 perpendicular to RT", then d'",d" is the revolved position, and 
 d,d' the primitive position of a point of shade ; O" LVi' being 
 the axis of revolution. For variety of construction, revolve Yc 
 to the position Yc", parallel to the vertical plane of projection. 
 This semicircle will then be vertically projected, with the radius 
 p'm' ; and by drawing p'c'", perpendicular to RT", we find the 
 point of tangency, c'",c", f a revolved ray, parallel to RT", and 
 from this, the same point in its true position, c,c'. 
 
 3. To construct the isometrical projection of the fonts, and 
 of its shade. Assume eu for the trace of the meridian plane, 
 a"b", upon the isometrical plane. 
 
 The highest points, e and u, are projected from e' and u' upon 
 this trace. The right and left hand points, E and G, are pro- 
 jected from 0', and EG- is the projection of the diameter E"G". 
 To find intermediate points, as those in the plane cd, project 
 the direction of vision, sO" s'O', upon this plane as at SO" 
 S'O'. Then revolve this projected ray, together with the meridian 
 plane, about a vertical axis at 0" till parallel to the vertical
 
 SHADES AND SHADOWS. Ill 
 
 plane of projection, and then by (80) the points of contact of 
 projecting lines, parallel to s'"O', will be points of apparent 
 contour in the isometrical projection. These points are best 
 found by drawing radii, as rn'", perpendicular to s'"O', which 
 gives n'"n" , and after counter-revolution, n,n'. There being 
 evidently four such points, make bN=ns, and then making 
 Oa= Ob, make al, aB, and b A, each equal to bN. The oval figure 
 passing through the points now found, will be the isometrical 
 projection of the torus. 
 
 For the curve of shade ; h' is projected at H, tR being equal 
 to h'h". Oq=0t, and then, qR=tK. The points c' and d' are 
 then projected at C and D, at distances from eu equal to the 
 distance of d from a"~b". Finally, a' and b' are projected in the 
 meridian curve, a"b"j whose isometrical projection is eu, at a ana 
 b. Through the six points now found, with f and &, the points 
 of tangency of tangent planes of rays perpendicular to the isome- 
 trical plane, the curve of shade can be" 1 drawn. 
 
 H, the highest point, determines the visible portion of the 
 curve of shade to be/*HD&. 
 
 Remarks. a. After all the labor of making the foregoing 
 construction, it is now more fully evident, according to (92), that 
 while this, and the three preceding constructions, afford good 
 examples for study, they, and the present problem particularly, 
 show that isometrical projection has little or no advantage in 
 respect to clearness of representation, except as applied to plane- 
 sided bodies having solid right angles, whose sides are equally 
 inclined to the plane of projection. 
 
 b. On account of the superior pictorial character of isometrical 
 projections, and still more, of the oblique projections explained 
 in a previous volume (Elementary Projection Drawing) under 
 the name of " Cabinet Projections," it is of less importance to 
 represent shades and shadows upon them. The cabinet projec- 
 tions of shades and shadows can readily be made from their com- 
 mon projections.
 
 112 BRILLIANT POINTS. 
 
 2. 
 
 THE FINISHED EXECUTION OF SHADES AND 
 SHADOWS. 
 
 CHAPTER I. 
 
 THEORY AND CONSTRUCTION OF BRILLIANT POINTS ; 
 AND OF GRADATIONS OF SHADE. 
 
 SECTION I. 
 Preliminary General Principles. 
 
 1. Geometrical Conditions for the adequate graphical repre 
 sentation of Form. 
 
 95. The obviously essential geometrical feature of a surface is 
 its continuity. But the bounding surface of a volume is repre- 
 sented, geometrically, as seen from a given point, by its apparent 
 contour ; which . is only that line of the surface which is its 
 visible boundary, as seen from that single point. 
 
 Geometrically, therefore, an infinite number of consecutive 
 contours, seen from as many points of view, on each of the two 
 opposite sides of a body, would be necessary to completely 
 represent its continuity and conformation. 
 
 But the method of projections usually gives but two apparent 
 contours ; viz., those which constitute the two projections of a 
 body. Furthermore, it would be graphically impossible to 
 represent consecutive apparent contours ; hence, purely geo- 
 metrical diagrams, though enabling us to represent any required 
 point of a surface, can never, of themselves, adequately repre- 
 sent the continuity of the inclosing surface of a body. 
 
 96. It is therefore our previous and definite conception of 
 the surfaces to be represented, which enables the projections of
 
 SHADES AND SHADOWS. 113 
 
 surfaces practically to express the forms of those surfaces intelligi- 
 bly. This may be illustrated by the difficulty usually experi- 
 enced at first, in comprehending projections of new surfaces, as 
 warped surfaces, before acquiring, from models or other sources, 
 some idea of their form. On the other hand, the circular and 
 rectangular projections of the familiar cylinder of revolution, 
 perfectly represent it to the mind, because we so well know that 
 all its right sections are equal circles, and all its meridian sections 
 are equal rectangles. 
 
 97. By calling in physical considerations to the aid of geo- 
 metrical ones, and confining ourselves to the sense of sight as a 
 means of judging of the configuration of bodies, the action of 
 light upon their surfaces would be observed at once. This leads 
 to the following considerations : 
 
 2. Of the physical conditions for the visibility of Bodies ', 
 and an adequate representation of their Forms. 
 
 98. The most immediately obvious result of the exposure of 
 a body to the light, is the existence, upon the body, of a line of 
 shade separating the illuminated from the shaded portion of the 
 body; and the production of a shadow upon any adjacent sur- 
 face from which light is excluded by toe given body. The con- 
 struction of the lines of shade and of shadow have, accordingly, 
 claimed attention in all the previous problems. 
 
 But from (95) the curve of shade, alone, is insufficient, together 
 with the apparent contour, to determine, unequivocally, the form 
 of a body. This becomes further apparent, as follows: Having 
 a given cylinder of rays, any curve traced upon its surface, may 
 be the curve of contact of an infinite number of different bodies, 
 all of which shall have the same projecting cylinder, and all of 
 which will have this one assumed curve for their curve of shade. 
 All these bodies, being inscribed tangentically in the same cylin- 
 der of rays, will, moreover, cast identical shadows on any other 
 given surface. 
 
 99. See, at this point, PI. XV., Fig. 55. NBEC is any opaque 
 body. ABKD is a tangent cylinder of luminous rays, giving 
 the curve of shade BHDG. Next, IFCJ represents a tangent 
 projecting cylinder, or cylinder of visual rays reflected from the 
 object, and gives the apparent contour FGrCH. Hence the por- 
 tion, E FGCH, of the body, bounded by that contour, and the 
 
 8
 
 BRILLIANT POINTS. 
 
 portion, GDH, of its curve of shade, are visible. But besides 
 the curve and that contour, there is only our imagination to sug- 
 gest the real configuration of the visible portions of the surface 
 on which no contour lines are shown. These portions may 
 indeed have any sinuous configuration, though experience and 
 association lead us to assume that the figure represents an ellip- 
 soidal body, or something like one. 
 
 100. We therefore continue the examination of the action of 
 light upon surfaces, in order to discover how their forms may be 
 distinguished, and we find that a surface is illuminated in pro- 
 portion to the directness with which the light falls upon it. See 
 PI. XV., Fig. 56. Here, let AB be the trace of a plane, tangent 
 at T, to any curved surface. The space ab is illuminated by the 
 beam of rays, M. The equal space cd, struck more obliquely 
 by the light, is illuminated by the thinner beam N, while 
 another equal'space, ef, which is struck perpendicularly, receives 
 all the light of the thickest beam, O. Hence, in this case, the 
 curved surface is most highly illuminated at the point of tan- 
 gency, T. Along the curve of shade, tangent planes coincide 
 with the direction of the light, and therefore the given surface 
 there receives no light. The curve of shade therefore is the 
 darkest part of the surface. 
 
 101. Between the curve of shade and the brightest point, 
 curves may be conceived lying on the surface, at all points "of 
 any one of which, tangent planes will make equal angles with the 
 rays of light. 
 
 But these angles represent the angles made by the surface 
 itself with the light along the supposed curve. Therefore the 
 curves just supposed will be curves of equal illumination, and 
 the illumination of the surface will vary in intensity from the 
 maximum darkness at the curve of shade, where the surface 
 makes an angle of with the light, to the maximum brightness, 
 where the sifrface is normal to the light. 
 
 Moreover, drawing aS, for example, the perpendicular width 
 of the beam of light, it represents the sine of the angle SJcz, made 
 by the light with the tangent plane AB. Hence, in this view, 
 the intensity with which a surface is illuminated is directly as the 
 sine of the angle made by the light with that surface. 
 
 102. But again : Bodies are not seen directly by the light 
 thrown upon them, but by such portion of that light as is 
 returned from them.
 
 SHADES AND SHADOWS. 115 
 
 Hence, in general terms, the comparative amounts of light re- 
 mitted to the eye, from the different points of a body, are, with 
 its apparent contour, the means of judging of its form. 
 
 103. Now, optical researches show, that, when light falls upon. 
 a body, a part is extinguished, or destroyed (absorbed according 
 to the material theory), a part is reflected, and a third portion is 
 polarised, by refraction among the molecules of the surface. 
 
 104. Strictly speaking, a body is visible, only as it shows its 
 own proper color, which it does by means of the refracted rays 
 which it remits. 
 
 Mirror-like surfaces reflect the rays just as they are received, 
 and present images of all objects, from which light proceeds to 
 fall upon them. They thus indicate their presence by their 
 effects, but are strictly invisible, except that, in practice, there 
 are no absolutely perfect mirrors ; hence they are faintly visible 
 by the few refracted rays coming from the molecules of their 
 surfaces. 
 
 ] 05. The law of reflection is, that the incident and reflected 
 rays, at the same point of a surface, make equal angles with that 
 surface. 
 
 Hence, as there will, taking the general case of a double-curved 
 surface, be but one such point, for a given fixed position of the 
 luminous point, and point of sight, a perfectly polished double- 
 curved body, exposed to a single source of light, would have but 
 a single visible point ; and that, the one at which a normal line, 
 or a tangent line or plane to the surface, would make equal angles 
 with the incident ray, and the ray reflected to the eye. 
 
 The condition for the complete visibility of a perfectly polished 
 body, therefore, is, that it shall be exposed to light from all 
 sources. 
 
 106. The majority of the objects which present themselves to 
 our observation, as a stone or piece of wood, are dull, or partially 
 polished bodies, that is to say, their exterior, without failing to 
 present an appearance of continuity, is nevertheless, by their 
 porosity, broken up into minute asperities and cavities, as 
 shown, greatly magnified, in PI. XV., Fig. 57. These irregulari- 
 ties, though separately invisible, from their minuteness as com- 
 pared with the size of the body, yet, under the action of the light, 
 produce an aggregate effect, which is appreciable, since they are 
 of considerable magnitude, as compared with the extreme tenuity 
 of the molecules of light. Each asperity, therefore, is considered
 
 116 BRILLIANT POINTS. 
 
 as having a number of indefinitely small facets, one or more of 
 which is in a position to return light to the eye. 
 
 But the exterior facets, lying in the perfectly continuous ima- 
 ginary surface of the body, are larger than interior ones, which 
 are withdrawn from polishing agencies ; and are nearer each other, 
 as seen in projection, in the vicinity of the brilliant point, L, than 
 in the more directly viewed portion of the continuous ideal sur- 
 face containing them ; and, besides, the properly disposed interior 
 facets may often, in obliquely viewed regions, be hidden by 
 asperities in front of them, so that the point on a dull body, 
 which corresponds to the only visible point of a perfectly 
 polished body, is the brightest point of that dull body. 
 
 107. Taking into the account the secondary remissions to the 
 eye, from facets which are illuminated by reflection from other 
 facets, it appears that the whole illuminated part of a dull body 
 is rendered visible by a single source of light. 
 
 Its shade, however, can become visible only by sending to the 
 eye the light received from a secondary source, as from the 
 atmosphere, or from surrounding bodies. 
 
 Hence, if PI. XV., Fig. 55, represents a dull body, illuminated 
 by a single source of light, all that part, B FGrDH, of the illu- 
 minated part, which is in the field of vision, will be actually 
 visible. The completely unilluminated shade, GCDH, though 
 in the range of vision, will be invisible, since it remits no rays to 
 the eye. 
 
 108. But the theory of minute reflecting facets alone, is insuf- 
 ficient, in not accounting for the colors of bodies ; since light that 
 is merely reflected, retains the color which it had when coming 
 in incident rays. It has therefore been presumed, and is con- 
 firmed by experiments, that bodies are visible, mainly in conse- 
 quence of light remitted by them after polarization by refraction, 
 due to vibrations in their superficial molecules. These vibrations 
 are supposed to be in unison so to speak with those of light 
 of the color presented by those molecules ; and they make those 
 molecules act as self-luminous points while illuminated. 
 
 109. Hence, besides the contours furnished by their projec- 
 tions, the conditions for the full representation of the continuity 
 and forms of bodies, are, that their surfaces, being dull or porous, 
 shall be illuminated from a primary and strong, and a lesser or 
 reflected light ; that their numberless facets shall be so distributed 
 as to reflect increasing quantities of light, in successive rings of
 
 SHADES AND SHADOWS. 117 
 
 equal reflection, from the curve of shade to the brilliant point, 
 analogous to the curves of equal illumination (101) and that the 
 vibrations of their superficial molecules, under the action of light, 
 shall cause them to act, while exposed to the light, as self-lumi- 
 nous bodies. 
 
 Finally : a true representation of these apparent varying 
 intensities of light and shade, exhibits graphically to the eye the 
 continuity and the consecutive changes of form of surfaces, hence, 
 in connection with the apparent contours of bodies, it adequately 
 represents them to the eye. 
 
 3. Of Brilliant Points and Lines. 
 
 110. Most conspicuous and important, next to the line of 
 shade of a body, is its brilliant point, or the element of greatest 
 apparent illumination in its illuminated part. As these bril- 
 liant points can, moreover, usually be constructed without diffi- 
 culty, they will now be more fully considered. 
 
 In treating of brilliant points thus in detail, the relative posi- 
 tions of the luminous point, and the point of sight, with respect 
 to a given body, are first to be noted. Since either of these 
 points may be either at a finite or an infinite distance from the 
 given body, the four following combinations may arise : 
 
 Distance of the 
 
 Luminous point. Point of sight. 
 
 Infinite. Finite. 
 
 Finite. " 
 
 Infinite. Infinite. 
 
 Finite. " 
 
 111. The two former cases are properly treated under the 
 head of "scenographic projections" or natural perspective, since 
 the point of sight is there supposed to be at a finite distance from 
 the object viewed. 
 
 The two latter cases may here be discussed, since, whatever 
 the distance of the source of light, the point of sight is at an 
 infinite distance from the object, which is characteristic of ortho- 
 graphic projections. 
 
 112. In speaking now of the position of the brilliant point 
 upon any surface, the distinction between the real, and the
 
 118 BRILLIANT POINTS. 
 
 apparent brilliant point must be observed. The real brilliant 
 point on a given surface, is the point at which the tangent plane 
 to that surface is perpendicular to the direction of the light, for 
 this point is in a position to receive the greatest amount of light 
 per unit of surface. 
 
 Hence, if T, PL XV., Fig. 56, be the point of contact of a 
 surface with a tangent plane, AB, to which the light is normal, 
 T will be the real brilliant point of the given surface. 
 
 113. The apparent brilliant point of a surface, is that point at 
 which a normal to the surface bisects the angle between the 
 incident and the reflected rays at the same point (105). The 
 apparent brilliant point is the only one which it is necessary to 
 consider, in treating of the finished execution of shading. 
 
 114. On developable single curved surfaces, and with the 
 luminous point and point of sight both at an infinite distance, 
 the brilliant point expands into a brilliant line, whose location 
 will be explained in detail, in connection with the problems of 
 the next section. 
 
 SECTION II. 
 The Construction of Brilliant 3?oints and. Lines. 
 
 1. Brilliant elements ff Planes. 
 
 115. A plane illuminated by parallel rays would be equally 
 light in every part, and, neglecting the effect of the different 
 depths of atmosphere, through which its different parts would 
 be seen were it viewed obliquely, every part of it would appear 
 equally bright. 
 
 This result would follow from the theory of the constitution 
 of surfaces, explained in connection with PI. XV., Fig. 57, for 
 then a material plane surface of uniform texture would present 
 a uniform distribution of facets, so disposed as to reflect rays to 
 the eye. 
 
 A plane, illuminated by diverging rays, will have a brilliant 
 point.
 
 SHADES AND SHADOWS. 119 
 
 PROBLEM XLII. 
 
 To find the brilliant point of a plane which receives light from 
 a near luminous point. 
 
 PL XV., Fig. 58. Let the plane be vertical, and let TP be 
 its horizontal trace ; and let S be the luminous point. Also let 
 TE be the direction of the reflected rays, -which are parallel (111). 
 
 At any point, T, construct a line, TN, perpendicular to the 
 given plane, and make the angle KTN=RTN. Then, through 
 the luminous point, S, draw a ray, SB, parallel to KT, and B, 
 its intersection with the plane TP, will evidently be the point at 
 which the incident and reflected rays make equal angles with the 
 normal to the plane. Hence B is the brilliant point required. 
 
 Remarks. a. Observe, that as NT is normal to the given 
 plane, the plane of the incident and reflected rays is perpendicu- 
 lar to the given plane. Hence, in this case, the ray SB is hori- 
 zontal in space. 
 
 b. The construction of the vertical projection of the figure, 
 and the solution of the problem when the given plane is oblique 
 to the planes of projection, may now be left to the student. 
 
 2. Brilliant elements on Developable Surfaces. 
 
 Passing to single curved surfaces, we shall first consider deve- 
 lopable surfaces and these, at first, as illuminated by parallel 
 rays. 
 
 PROBLEM XLIII. 
 
 To find the brilliant element on a cylinder, illuminated by 
 parallel rays. 
 
 First Solution. PL XV., Fig. 59. Let ANB A'B' be a 
 vertical right cylinder, and RO R'O' the projections of a ray 
 of light. OE represents the direction of those reflected rays 
 which reach the eye. Now revolve the plane, R'O'E, of the 
 incident and reflected ray?, RO R'O' and O' OE, about its 
 horizontal trace, O'E, into the horizontal plane of projection. 
 The incident ray will then appear at R"0' R'"O, and L'"O is
 
 120 BRILLIANT POINTS. 
 
 the revolved position of the bisecting line of the angle EOB 
 (105). Then, by making the counter revolution, this bisecting 
 line will appear at LO I/O', since it is in a plane B/O'E which 
 is perpendicular to the vertical plane of projection. 
 
 Now we may suppose that a row of asperities is ranged along 
 the element N N'N" in which the bisecting line pierces the 
 cylinder. If then each of these asperities has one or more 
 minute facets which are perpendicular to LO I/O', they will 
 collectively reflect the rays contained in a vertical plane of rays 
 through EO R'O', and thus form a brilliant line N N'N". 
 
 Second Solution. In this solution, some of the facets of the 
 little asperities are supposed to be arranged in parallel elliptical 
 bands, indefinitely narrow, and contained in the successive paral- 
 lel planes of incident and reflected rays. 
 
 See PL XV., Fig. 60, where the light is taken in the same 
 direction as in the last figure, for the sake of easy comparison 
 of the two. Those features of the construction, which are the 
 same in both figures, are here omitted. 
 
 HO is the revolved position of an incident ray, OE is a 
 reflected ray, and DO is the bisecting line of their included angle. 
 A'B' is the vertical trace of a plane of rays, perpendicular to the 
 vertical plane of projection, and determining the narrow elliptical 
 band A'B', along which facets are supposed to be disposed so aa 
 to reflect light to the eye. A"CB"S is the revolved position of 
 this elliptical band. On, the bisecting line of parallel chords, aa 
 C<? and #5, which are perpendicular to DO, determines N'", the 
 point at which a normal, parallel to DO, can be drawn. 
 
 In the counter revolution, N'" returns to N; hence, if the con- 
 vex surface of the cylinder be supposed to be formed of consecu- 
 tive bands of reflecting facets parallel to A'B, a vertical row of 
 them will ba found on an element N N'N". 
 
 Remarks. a. It should be remembered, that the plane of the 
 incident and reflected ra} 7 s is normal to the reflecting surface. 
 
 Otherwise, a ray striking a plane, at a point A, for example, 
 might be reflected in a cone of reflected rays, generated by the 
 revolution of the incident ray about a perpendicular to the plane 
 at the point A. 
 
 But in the case of an absolutely nnporous and polished cylin- 
 der, or cone, which is struck obliquely by the light, a plane of 
 incident and reflected rays, situated as in the present example, 
 cannot be made normal to the convex surface at N. Hence such a
 
 SHADES AND SHADOWS. 121 
 
 cylinder, when illuminated from a single source of light, should 
 be absolutely invisible throughout. 
 
 The existence of asperities of surface, presenting reflecting 
 facets, is, therefore, a necessary condition for the visibility of the 
 surface thus illuminated. Hence reasoning or experiment must 
 decide which of the theories of the location of the reflecting 
 facets, given* in the two solutions of this problem, is true, since 
 these solutions give quite different locations to the brilliant line. 
 
 The first construction is, I believe, the only one hitherto used, 
 excepting the approximate one given in. my " Elementary Pro- 
 jection Drawing," and it agrees with the formation of the artifi- 
 cial grain of the surface by the operation of turning. 
 
 b. Either of the solutions of this problem may be applied, by 
 the student, in finding the brilliant element of a cone. 
 
 PROBLEM XLIV. 
 
 To find the brilliant point of a cylinder which is illuminated 
 by diverging rays. 
 
 PI. XV., Fig. 61. Let L be the luminous point, and BDK the 
 horizontal projection of the cylinder. The reflected rays being 
 parallel, because the eye is at an infinite distance from the 
 cylinder, let LG be a line in the direction of the reflected rays. 
 
 All lines through O, and in a plane perpendicular to the axis 
 of the cylinder, will be normal to its surface. The cylinder being 
 vertical, in the present case, this plane will be horizontal, and 
 EO, FO, GO, etc., are normals to the cylinder. 
 
 Now, making La=LE ; LJ=LF, etc., the curve abc is the line 
 at each of whose points incident rays, as L, and reflected rays 
 from a, etc., parallel to LG, make equal angles with the normals, 
 as E#O, etc., through the same points. 
 
 Hence B, where this curve meets the convex surface of the 
 cylinder, is the point of that surface at which the incident and 
 reflected rays, LB and BY, make equal angles with the normal 
 BO. That is, B is the horizontal projection of the required bril- 
 liant point. Its vertical projection is not shown, it being found 
 by merely projecting B into the circle cut from the cylinder by a 
 horizontal plane through the luminous point.
 
 122 BRILLIANT POINTS. 
 
 PROBLEM XLY. 
 
 To find the brilliant point on a cone which is illuminated 
 from, a near luminous point. 
 
 PI. XY., Fig. 62. In the last problem, the curve in which the 
 normals pierced the cylinder, coincided, in horizontal projection, 
 with the projection BDK of the cylinder, and was, therefore, not 
 constructed. The cone, not having a vertical surface, this curve 
 of intersection of the normals would appear as a distinct curve, 
 which must be constructed. 
 
 Let VDTH Y'D'H be a cone of revolution, having its axis 
 vertical; and let S,A' be the luminous point. 
 
 Draw SAi", in the direction of the reflected rays. At S, any 
 line will make equal angles with S5", and some incident ray, 
 hence S is a point of the trial curve, containing the intersection 
 of normals with incident rays. A is another point of the same 
 curve, since SA, regarded first as an incident, and then as a 
 reflected ray, makes the same angle with a normal YA at right 
 angles to it. 
 
 To find other points of the trial curve SAA. At any point, as 
 5", on the line S5", draw the horizontal projection, b"V, of a 
 normal. Revolve this normal to the position ZY, parallel to the 
 vertical plane of projection. N'J', perpendicular to the element 
 V'D' taken as the revolved position of the element Yy will 
 then be its vertical projection, and A'N' is the vertical projection 
 (not drawn) of its primitive position. Now revolve this normal 
 and the line SJ", about an axis perpendicular to the vertical plane 
 at N', and into the horizontal plane N'A". It will then appear 
 at A"N' BY, and S&" will appear at A" S'"B. Now make 
 S'"&" not drawn equal to S"'B, and h" will be the revolved 
 position of another point of the trial curve. In the counter- 
 revolution, h" returns in a short arc, h"h, parallel to Bi", and ^, 
 on the primitive projection, b"V, of the normal, is another point 
 of the curve S AA. 
 
 Having now found one point of this curve, in the manner just 
 described, others are easily found as follows, Revolve assumed 
 points, asp and #, top"' and #, and join p" and k with Y. Then 
 arcs, with radii S'"p" and S'"&, will locate points on p"V and 
 &Y not shown at distances from S"', equal to S'"p r/ and S"'&.
 
 SHADES AND SHADOWS. 123 
 
 Then, by counter-revolution, as before, find c and y. The trial 
 curve SAcA can now be drawn. 
 
 It is necessary, in the next place, to construct the curve in 
 which all normals, which intersect SJ", pierce the conic surface. 
 Take, for example, the normal Y5". After revolution into the 
 meridian plane AY, parallel to the vertical plane, this normal 
 appears at Vb JV", normal to the revolved position, Y'D', of 
 the element Vr, which is in the meridian plane through Y&". 
 Hence r'",r" is the revolved, and r,/ the primitive position of 
 the intersection of this normal with the conic surface. 
 
 Other similar points being found in like manner, the curve 
 uer e'n'r' may be sketched as the locus of the intersections of all 
 normals, through points of A' S6", with the conic surface. This 
 curve intersects SA^, the locus of the intersections of normals 
 with incident rays, at n,n'. This point is therefore that point, on 
 the conic surface, at which the incident and reflected rays, Sn 
 and wE, make equal angles with the normal, wY, at that point. 
 Note, however, that these angles are not equal in projection, since 
 their plane is oblique. 
 
 Remark, "We may construct the point in which the normal at 
 any point will pierce the line Sb". Thus, let F be the revolved 
 position of any point at which a normal is to be drawn. FK, 
 perpendicular to Y'H', is the revolved position of this normal. 
 The revolution of FK, about Y'K as an axis, will generate a 
 cone, whose surface will be normal to that of the given cone ; 
 then the intersection of this auxiliary cone with the line Sb" A', 
 will give the point, which, when joined with Y,K, will be the 
 normal required. 
 
 3. Brilliant Points on Warped Surfaces. 
 PROBLEM XLYI. 
 
 To find the brilliant point of a warped surface, when illuminated ~by 
 
 parallel rays. 
 
 The solution of this problem involves the construction of a 
 normal, parallel to a given line, L, viz., to the bisecting line of 
 the angle included by an incident and a reflected ray. 
 
 But since these rays make equal angles with a tangent plane
 
 124 BRILLIANT POINTS. 
 
 which is perpendicular to the supposed normal, the construction 
 of the normal may be effected by the construction of a tangent 
 plane, perpendicular to a given line. But, again, since all planes 
 which are perpendicular to the same straight line, are parallel, 
 the operation, just proposed, is equivalent to the construction of 
 a tangent plane parallel to a given plane. 
 
 To construct the tangent plane, draw any two lines, A and B, 
 not parallel, but each perpendicular to the given line, L. A 
 system of tangent lines, parallel to A, will constitute a cylinder, 
 tangent to the warped surface. Another such system, parallel to 
 B, will form another tangent cylinder. The intersection of the 
 curves of contact of these cylinders will be the point of taiigency, 
 on the warped surface, of a tangent plane perpendicular to L, and 
 therefore the point of intersection, I, of the required normal, 
 parallel to L. Hence I will be the required brilliant point. 
 
 PROBLEM XLYII. 
 
 To construct the brilliant point on a screw, when illuminated by 
 
 parallel rays. 
 
 Produce, each way, several elements on each side of the meri- 
 dian plane parallel to the vertical plane, the axis of the screw 
 being vertical, so as to draw, tangent to them, a sufficient arc of 
 that meridian curve of the screw, which lies in this meridian 
 plane. Then, as in Prob. XLL, find the bisecting line, N, of the 
 angle included between an incident and a reflected ray. Revolve 
 the line, N, till parallel to the vertical plane of projection. Draw 
 a line, parallel to this revolved position, and normal to the meri- 
 dian curve, just named, at a point, n. This point, n, will be the 
 revolved position of the brilliant point. In counter-revolution, 
 it will return in a helical arc, and its projections will appear on 
 the projections of N. 
 
 This construction can be wrought out by the student.
 
 SHADES AND SHADOWS. 125 
 
 4. Brilliant Points on Double- Curved Surfaces. 
 PROBLEM XLVIII. 
 
 To find tfie brilliant point on any double-curved surface, when 
 illuminated by diverging rays. 
 
 Let PI. XV., Fig. 63, represent any double-curved surface, 
 not of revolution. Let S be the luminous point, and SB/ a line 
 parallel to the reflected rays. Let n'Nn" be the curve in which 
 all normals, as RW, RN, and R'W, intersect the given surface. 
 Also, let N'NN" be the locus of intersections of these normals 
 with incident rays from S. That is, SN'=SR', SN=SR, and 
 SN"=SR", etc., so that if the rays, SN', etc., were drawn, the 
 angles at N', etc., made by the incident ray SN', etc., and normals 
 R'N', etc., would be equal to the angles at R', etc., made by the 
 reflected ray SR' with the same normals. Then N, the intersec- 
 tion of the two curves just described, is that point on the given 
 surface, where the incident and reflected rays make equal angles 
 with the normal RN. Hence N is the brilliant point. 
 
 The practical difficulty, in case of the class of surfaces con- 
 sidered in this article, and also in case of warped surfaces, is, that, 
 unless they are surfaces of revolution, there is no convenient 
 direct construction of the required normals. Hence the remain- 
 ing problems embrace only the construction of the brilliant points 
 of double-curved surfaces of revolution, when illuminated by 
 parallel rays. 
 
 "When such surfaces are exposed to diverging rays, their 
 brilliant points are found as in Prob. XLIY. 
 
 PKOBLEM XLIX. 
 To find the brilliant point on a sphere, illuminated by parallel rays. 
 
 PI. IX., Fig. 26. A vertical plane, through the centre of the 
 sphere, is here regarded as the vertical plane of projection. OL 
 is the vertical projection of a ray through the centre of the sphere. 
 As no horizontal projection of the sphere is shown, Or may be 
 assumed as the ray, after being revolved into the vertical plane
 
 126 BRILLIANT POINTS. 
 
 of projection, about LK, a line of that plane, as an axis. Then, 
 also, OP will represent the direction of the reflected rays. Hence 
 N'O, the revolved position of the bisecting normal of the angle 
 rOP, gives N' as the revolved position of the brilliant point. 
 By a counter-revolution about LK, its primitive position, N, ia 
 found. 
 
 PROBLEM L. 
 To find the brilliant point on a piedouche. 
 
 PI. X., Fig. 31. A^ SY is any incident ray intersecting the 
 axis A T'A'. Let S' AW represent the direction of the 
 reflected rays, as the piedouche is seen in vertical projection. 
 Revolving the assumed ray around AW, as an axis, till it is 
 parallel to the horizontal plane of projection, it appears at SY" 
 Ai". Then draw Ag-, the bisecting line of the angle t'"AW, 
 and draw an auxiliary line, i"W. In the counter-revolution, 
 i"j!" returns to ij!, the line i"W returns to i W, and q to p, 
 whence it is projected to p', because the plane WAi is perpen- 
 dicular to the vertical plane of projection. Now pA and p'S' 
 are the projections of the bisecting line of the angle iAW. The 
 brilliant point is the point of contact of a tangent plane which 
 is perpendicular to this bisecting line. Then revolve Ap S'jp', 
 till parallel to the vertical plane, taking A T'A' for an axis, 
 when it will appear at Ap" S'p"'. Next, draw the normal fl, 
 parallel to ^'''S', and I will be the revolved position of the bril- 
 liant point. Construct the projections of the horizontal circle 
 through this point, and V,V, its intersection \\ith the meridian 
 plane- through the bisecting line pA p'A, is the primitive 
 position of the required brilliant point. 
 
 Remark. The student may now construct the brilliant point 
 on any of the double curved surfaces of revolution shown in the 
 preceding plates, or upon a warped hyperboloid of revolution.
 
 SHADES AND SHADOWS. 12 T 
 
 CHAPTER H. 
 
 OP THE REPRESENTATION OF THE GRADATIONS 
 OF LIGHT AND SHADE. 
 
 116. In observing a shade, or a shadow, two things arrest 
 the attention, its position, and its intensity. 
 
 To find the position of shades and of shadows upon surfaces, 
 was the object sought in " BOOK I." Now, we are to determine 
 their varying intensities, as affected by the several circumstances 
 presently to be enumerated. Furthermore, the varying apparent 
 intensity of the illumination of those parts of surfaces which 
 are exposed to the light, will be affected by the same circum- 
 stances. Hence the gradations of the lights, and of the shades 
 and shadows of surfaces, will be discussed together. 
 
 117. The particulars affecting apparent intensities, and the 
 gradations of lights and shades, and which will next be dis- 
 cussed in detail, are arranged under the following heads : 
 
 1. Effects due to the laws of illumination and vision. 
 
 2. Those due to the supposed infinite distances of objects 
 from the eye. 
 
 3. Those due to secondary sources of light; including the 
 air as an illuminating medium. 
 
 4. Those due to the nature of different surfaces. 
 
 5. Those due to the atmosphere as an absorbing medium. 
 
 6. Those due to variations in the intensity of the light. 
 
 7. Those due to the forms of bodies. 
 
 8. Those due (on the drawing) to drawing materials and 
 processes of manipulation. 
 
 118. Effects due to the laws of illumination and vision. 
 
 a. Law of illumination. The intensity of light, or of the 
 degree of actual illumination of the same surface, at different 
 distances from the source of light, varies inversely as the square 
 of the distance from the luminous source.
 
 128 THEORY OF SHADING. 
 
 b. Case where, the light comes from an infinite, distance. The 
 principle just stated applies, sensibly, whenever the source of 
 light is at a finite distance. But when the light proceeds from 
 an infinite distance, terrestrial bodies are all at substantially the 
 same distance from it, and when similarly situated with respect 
 to it, may be considered as equally illuminated. 
 
 c. Law of apparent brilliancy. From remark a it follows, 
 that if a surface be illuminated in a given degree, its apparent 
 brilliancy should be inversely as its distance from the eye, but 
 its image being reduced in the same proportion, the retina re- 
 ceives light at the same rate or amount per superficial unit of its 
 surface. Hence, practically, the apparent brilliancy of a surface 
 of given brightness, and in a given position, will be the same at 
 whatever distance it is seen ; if we disregard atmospheric 
 effects. 
 
 d. Illustration. The last remark is easily illustrated. See 
 PI. XV., Fig. 64 
 
 Let A be a visible point of an illuminated surface, and let PQ 
 be the pupil of the eye. The point A remits to the eye the cone 
 of rays APQ, whose base is the pupil PQ. These rays converge 
 in the eye, forming, on the retina at R, the image of the point 
 A. If now the object be removed to A', double the distance 
 AC, the base, PQ, remaining of constant size, the section of the 
 new cone, at the distance equal to AC from its vertex, will evi- 
 dently have an area equal to one-fourth of PQ. But sections at 
 a constant distance from their vertices measure the relative angu- 
 lar capacities of the two cones. Hence the eye at double the dis- 
 tance AC receives one-fourth as much light from the point A as at 
 the distance AC. 
 
 But to apply this result practically, we must consider more 
 than a solitary point. Two points will be sufficient. Then 
 let B be a second point on the body AB. The axis, Br, of its 
 cone of rays gives r as the focus of those rays on the retina, and 
 Rr as the space on the retina, illuminated by rays from the space 
 AB on the object. Now remove AB to double the distance AC 
 viz. to A'B', and the point B' will be imaged at r', one-fourth as 
 brightly as at r. But Rr' is half of Rr, and hence the area, of 
 which Rr' is the diameter, is one-fourth as large as that whose 
 width is Rr. While, therefore, A'B' remits to the eye one-fourth 
 of the light that AB remits, that light is received by the area, 
 Rr', of the retina, which is one-fourth of the area Rr. Hence the
 
 SHADES AND SHADOWS. 429 
 
 rate of illumination of the retina, for a unit of surface, is the 
 same in both cases. 
 
 e. In discussing this nice point, on which learned authors 
 differ, care must be taken not to confound aggregate amount, 
 and consequent dazzling effects of brilliancy, with its rate or inten- 
 sity. A square foot of sun-lightened snow is as intensely bright, 
 as to its degree of illumination, as a whole snow-bank, but 'the 
 total amount of light from it, and extent of retina affected by it, 
 is less than in case of the whole snow drift. Hence the amount 
 of dazzling effect is less. So a star, as brilliant as the sun, can 
 easily be gazed upon, though its sensibly equal intensity of bright- 
 ness with the sun is shown by its twinkling, which is its daz- 
 zling effect upon a mere point of the retina ; an effect which has 
 only to be repeated in the same degree on many points at once 
 of the retina, in order to become blinding. 
 
 /. /Shading of the penumbra. When a penumbra (85) is occa- 
 sioned, as it is, by a near radiant body of sensible size, it receives 
 light from a larger and larger portion of the luminous body, as 
 it approaches its own outer boundary. Hence its shaded repre- 
 sentation would be lighter and lighter in the same direction, 
 unless prevented by the form of the surface containing it, or by 
 other special circumstances. 
 
 119. Effects due to the infinite distance of objects from the eye. 
 
 a. Objects at an infinite distance seen as points. At the outset, 
 and as a preliminary remark, it should be noted that supernatu- 
 rally acute powers of vision are required to perceive an object 
 at all, at an infinite distance. This may be proved by a refer- 
 ence to the laws of vision, as follows : 
 
 Each molecule of the surface of an illuminated object, having 
 facets so disposed as to diffuse light in all directions, remits to 
 the eye a cone of rays whose base is the pupil of the eye (11 Sd). 
 This cone of rays, in passing through the lenses of the eye, con- 
 verges to a point in its axis, and on the retina, which is the 
 image of the point on the object from which the rays proceeded. 
 Each point of the object thus produces its image, and these 
 image points, collectively, form the superficial image of the 
 object. 
 
 But now conceive the object viewed to be at an infinite dis- 
 tance from the eye. Frusta, of finite length, of all the cones of 
 rays remitted to the eye, and having the pupil for their common 
 base, will sensibly coincide in a single surface, which will be 
 
 9
 
 130 THEORY OF SHADING. 
 
 sensibly cylindrical. These frusta of rays, having thus a com- 
 mon axis, converge to a single common point on the retina, which 
 is the image of the object. Now in order to be sensible of this 
 image, consisting of but a single point, the retina must be of 
 extraordinary sensibility unless the light be of extreme intensity 
 also, its nervous texture must possess absolute continuity, in 
 order to be impressible, strictly, at every point. 
 
 b. Why represented as having sensible magnitude. In the two 
 respects just named, then, the eye must possess extraordinary 
 power, in order to perceive infinitely remote objects ; but, since 
 they should appear as points, why are they represented as of 
 sensible magnitude ? For the same reason that applies to ordinary 
 vision. A giant nine feet high, and ninety feet distant, appears 
 of the same size, nominally, or in a merely geometrical sense, as 
 a boy three feet high, and thirty feet distant. But, practically, 
 we say that each appears as we realize that it appears, and the 
 appearance that we realize, i. e., the one that seems real to us, is 
 determined by our knowledge, derived from other sources, of the 
 real sizes of objects. Knowledge, therefore, of the actual sizes 
 of objects, may make us believe that an infinitely remote object, 
 seen with the organs above described, does appear of sensible 
 size, according to that knowledge ; rather than as a point, accord- 
 ing to its minute image on the retina. 
 
 c. Theory of exaggeration of effects in shaded projections. But 
 again, eyes of such acute sensibility, would also indicate with 
 corresponding vividness the modifications of light and shade due 
 to the causes already stated (117), hence the usual practice of 
 greatly exaggerating effects, is quite rational in shading objects 
 when shown in projection. 
 
 This exaggeration appears principally in two particulars: 
 First, in the extremely vivid contrast, between the element of 
 shade and the brilliant point, which does not appear in common 
 experience, but which appears natural, and agreeably suggestive 
 of the forms of objects when represented in shaded projections. 
 Second, in the effect allowed in representing surfaces at different 
 distances. This is shown, first, in the manner of shading single 
 surfaces seen obliquely ; and, second, in the tinting of a series of 
 parallel surfaces, at slightly increasing distances, with tints of 
 increasing darkness. 
 
 120. Effects due to secondary sources of light; including the 
 air as an illuminating medium.
 
 SHADES AND SHADOWS. 131 
 
 a. Direction of the strongest secondary light. A surface will 
 diffuse ligbt in quantities proportioned to the amount received. 
 It will receive most when struck perpendicularly. Therefore 
 the greatest intensity of light from the particles of air, and of 
 surrounding surfaces, will be in a direction opposite to that of 
 the primitive rays. 
 
 b. Of the relative darkness of different portions of the shade on a 
 given surface. From (a) it follows that the shade of a body will 
 have its faint brilliant element, due to the secondary lights, just 
 as the illuminated part has its own brilliant element, due to the 
 primary light. Therefore the shade is shaded lighter, in reced- 
 ing from the line of shade, which is the darkest line of the body 
 in reference to both lights. 
 
 c. Of the relative darkness of different parts of a shadow. Any 
 point in a shadow receives diffused light from surrounding 
 objects, except within a cone whose vertex is the point, and 
 whose base is the curve of contact of the cone with the body 
 casting the shadow. Therefore, generally, the further a point 
 of shadow is from the object casting it, the more diffused light 
 it receives, and the lighter it is. 
 
 The general principle just before named, must, however, be 
 applied by inspection to each particular case of shadows on 
 plane, convex, and concave surfaces. In case of a vertical prism 
 casting its shadow on the horizontal plane, the shadow would 
 be darkest at its junction with the base of the prism; a result 
 which is confirmed by observation. 
 
 d. Optical, and actual effects of near surrounding objects. When 
 an illuminated surface is placed quite near a shade, it actually 
 illuminates a small neighboring region of that shade to a notice- 
 able extent, and that region should therefore receive a lighter 
 tint, in a drawing showing both bodies. 
 
 Besides this actual effect, there are optical effects, due merely to 
 the mutual influence of contrasts, as when a shadow falls on an 
 illuminated surface, and when two tints of widely different in- 
 tensity are brought together. Here, the light surface appears 
 lighter, and the dark surface darker, by contrast, as when black 
 velvet laid upon black cloth makes the latter look lighter. 
 Likewise, complementary colors heighten the effect of each other, 
 when brought together. Thus, purple and gold, or blue and 
 orange, each look more brilliant when side by side, than when 
 seen separately. Such contrasts are presented in the shaded
 
 132 THEORY OF SHADING. 
 
 drawing of an assemblage of objects, as well as by the original 
 objects, hence the shading need not be exaggerated in order to 
 produce the proper effect. Thus, if a shadow on a brilliant 
 surface be of nearly uniform intensity, as it would be on a 
 plane perpendicular to the light, its edges would, as they should 
 do, appear darker than the interior portions, by contrast with the 
 high light around them. 
 
 e. Of the relative darkness of a shade and a shadow. Some 
 interesting principles are suggested by considering the relative 
 darkness of a shade. and of a shadow, under various circum- 
 stances. For example, let a shadow fall on the surface of a 
 sphere. Any point in this shadow, will receive scattering rays 
 of reflected and refracted light from all illuminated bodies and 
 particles, in the hemisphere of space bounded by a tangent plane 
 at the given point, and exterior to the cone, whose vertex is the 
 same point, and whose base is its curve of contact with the body 
 casting the shadow. A point in the shade of the sphere, near to 
 which no other surface receives a shadow, will be illuminated by 
 reflections from the entire hemisphere of space, bounded by the 
 tangent plane at that point. Hence the latter point would be 
 somewhat lighter than the former ; as also from the fact of its 
 receiving the strongest atmospheric reflection (a). 
 
 f. In the case of the line of shade, as compared with a shadow 
 on the illuminated part of a body, the shadow is the darker of 
 the two. For, as the atmospheric reflections are strongest in a 
 direction opposite to the light, they are weakest or nothing in 
 the same direction as the light. Hence shadows cast upon sur- 
 faces in the vicinity of their brilliant points are darker than the 
 lines of shade of those surfaces. The shadow of the abacus 
 upon the cylinder, for example, is darkest at the brilliant ele- 
 ment of the cylinder, and fades away to the element of shade, 
 where it disappears. 
 
 g. When the primary and secondary lights are of nearly equal 
 intensity, as seen in the comparatively uniform diffusion of light 
 in a cloudy day, the contrast between the brilliant point, and the 
 curve of shade, and the shadows, would be diminished. The 
 latter would appear lighter, and the former darker, than when 
 exposed to a single strong light. 
 
 121. Effects due to the nature of surfaces. 
 
 a. The nature of surfaces may vary, by being dull or polished, 
 or by being differently impressed, in their molecules, by different
 
 SHADES AND SHADOWS. 133 
 
 colored rays. While no surfaces are of absolutely perfect polish, 
 owing to their porosity, yet, as they approximate to such a 
 polish, it may be possible to express that fact graphically by the 
 manner of shading them. 
 
 A perfectly polished body, we have seen, has a single, well de- 
 fined brilliant point. Hence, other things remaining the same, 
 the more polished the surface, the smaller may be the region of lifjld 
 shading containing the brilliant point ; and the more dull the surface, 
 the broader may be the area of light shading as if each large 
 asperity on a dull body, which is pretty directly illuminated, 
 had its own somewhat conspicuous brilliant point. 
 
 b. As already stated (104) bodies are, strictly speaking, not 
 seen by their reflected light, but by that which they refract. Re- 
 flected light merely furnishes images of the objects which send 
 rays to the reflecting surface, just as a surface echoes sound 
 without imparting to it any new quality of tone derived from 
 the echoing surface. 
 
 In accounting for the colors of bodies, an analogy is supposed 
 to exist between them and musical instruments, or resonant 
 volumes of air. Many persons may have observed that upon 
 sounding various notes, in a clear and strong humming manner, 
 in a closed closet or small room, some of them will, without 
 extra effort, sound much louder than others, as if the whole 
 body of air in the room were excited to musical vibrations in 
 unison with the loud sounding note. So surfaces, when exposed 
 to the pulsations of light, seem to respond, in the vibrations of 
 their molecules, only to the vibrations of rays of a certain color. 
 By thus responding, these surfaces become, for the time, lumi- 
 nous bodies, emitting, however, only such colored rays as their 
 nature causes them to be excited by. 
 
 Now, as bodies, in proportion to the perfection of their polish, 
 reflect more and more light, less will remain to express their 
 color by the refracting process just explained. Hence the higher 
 the polish, the less clearly will the proper color of a surface be 
 revealed. 
 
 c. It may be added, as a matter of interest, to the illustration 
 of the closed room, that such inclosed spaces are resonant, 
 though in a less degree, to notes which differ but little from the 
 principal note. So it is found, also, that the colors of bodies are 
 not absolutely pure, but are always mixed with small portions 
 of the neighboring colors of the spectrum. Thus the light from
 
 134 THEORY OF SHADING. 
 
 a clear yellow flower, when dispersed by a prism, shows some 
 orange and green rays, mixed with the pure yellow ones. 
 122. Effects due to the atmosphere as an absorbing medium. 
 
 a. Appearance of an illuminated surface seen obliquely. The 
 atmosphere, not being perfectly transparent, extinguishes a por- 
 tion of the light coming through it, and this portion is greater in 
 proportion to the extent of air traversed. But we attribute to 
 a surface the amount of light entering the eye in the direction 
 from it to the eye. Hence the most remote portions of an illu- 
 minated surface are the darkest in appearance. 
 
 b. Appearance of a surface of shade seen obliquely. A surface of 
 shade only remits to the eye tertiary rays, as they may be called, 
 that is, rays received by reflection from surrounding objects. 
 Hence the atmospheric reflections coming to the eye in the Direc- 
 tion of the shaded object, may be supposed to be stronger than 
 the light from the shaded object itself. But, as before, we attri- 
 bute to the object the light coming to the eye in the direction of 
 it, while the more remote it is, the greater is the body of inter- 
 vening air which sends light from its own molecules to the eye. 
 Hence the more remote a surface of shade is, the lighter it 
 appears, or the nearer it appears to be of the same brilliancy of 
 the atmosphere generally. This is very evident in nature, when 
 we compare the lights and shades of distant buildings, or hills, 
 with those of near ones. 
 
 c. ^Results of the two preceding principles. 1. A series of illu- 
 minated parallel surfaces, seen perpendicularly, will appear 
 darker and darker, as they become more distant. 2. The 
 reverse will happen, if these surfaces are in shade. 3. Near 
 shadows will appear intense, and remote ones faint. 4. Portions 
 of illuminated curved surfaces, which are more distant than the 
 brilliant point, or line, will appear darker than that point or line. 
 5. Portions of the shade of curved surfaces, which are more 
 distant than the line of shade, will appear lighter than the line 
 of shade. 6. Curved surfaces will appear of a more uniform 
 tint, the more distant they are, the lights being darkened, accord- 
 ing to (a), and the shades dimmed, according to (b). 
 
 d. Limitation of the last principle. The last principle applies, 
 however, to the relative depths of a body of air of finite thickness, 
 through which bodies are seen from an infinite distance. See 
 (119). 
 
 e. Effect of the air on apparent color. The blue color of the air,
 
 SHADES AND SHADOWS. 133 
 
 due to the rays remitted from it by refraction, as explained already, 
 is attributed to objects seen through it, as well as the intensity o 
 the atmospheric rays. Hence distant objects appear of a bluish 
 cast, which increases in clearness with their distance. It is thus 
 that, in a colored drawing, a distant object may be distinguished 
 from the dull shading due to uniformly diffused light (120*7). 
 
 123. Effects due to variations in the intensity of the light. Inte- 
 resting results follow an examination of the effect of lights, each 
 received mainly from one direction, but varying in intensity. 
 All so-called transparent media absorb some light, and, for the 
 present purpose, it is sufficiently correct to assume that, for light 
 passing through any given medium, and in greater quantity than 
 can be absorbed by the medium, the absolute amount absorbed 
 will be equal for all degrees of intensity. It is important here 
 to observe that this absorbed light is, as it were, latent, and does 
 not at all render the absorbing body visible. 
 
 It follows that in case of a feeble light, where the total amount 
 absorbed by the air and by surrounding objects is large in com- 
 parison to the whole amount of light emitted, the diffused light 
 caused by reflections and refractions among the particles of air, 
 and the light remitted from surrounding objects, will both be 
 small, but very little light will be thrown upon the shade, or 
 into the indefinite shadow in space of a body, under these cir- 
 cumstances, and hence this shade and shadow will approach in 
 darkness to those formed under the supposition that they received 
 no light, and were absolutely black (105). 
 
 Moreover, the diffused light spoken of, partially illumines 
 objects within the indefinite shadow ; but if it be feeble, it will 
 all be absorbed by those objects, and they will therefore, as 
 shown above, not throw any light into the indefinite shadow, or 
 upon the shade of the body. For this additional reason, there- 
 fore, will the shades and shadows due to a feeble light be 
 intensely dark, as compared with the illumined part of the body. 
 This conclusion corresponds to the great blackness of the shades 
 and shadows observable on and about buildings by moonlight. 
 To return, now, to the case of clear sunlight, and then to apply 
 these principles, it appears that, while the absolute amount of 
 absorption remains the same, there is a large excess of light, 
 available for producing diffused and reflected light, of sufficient 
 brightness to produce reflection from bodies in the shadow upon 
 the shade of an opaque body, so that enough light will reach
 
 136 THEORY OF SHADING. 
 
 this shadow and shade, to mitigate their blackness considera- 
 bly. 
 
 124. Effects due to the forms of bodies. 
 
 a. Form of the curves of equal shade. The lines of equal tint 
 on different bodies, will be either similar or dissimilar. They 
 will be similar on developable surfaces, where they are simply 
 rectilinear elements, and on spheres, ellipsoids, etc., where they 
 will generally be circles, ellipses, etc. But they will be dissimi- 
 lar on the more complex surfaces, such as the torus and pie- 
 douche. In the latter case, they will be gradually transformed, 
 from a ring surrounding the brilliant point, to curves more and 
 more nearly similar to the curve of shade. This should be 
 remembered in distributing the tints upon such surfaces. 
 
 b. Distribution of points of equal shade. Points of equal ap- 
 parent brilliancy are not, however, symmetrically placed around 
 the brilliant point as a centre. That is. on a sphere, for example, 
 the curves of equal illumination are not circles, having their 
 centres in the diameter through the brilliant point. See PI. 
 XV., Fig. 65, which represents a sphere, with the plane of the 
 paper taken as a plane of incident and reflected rays, as LO and 
 OR/. Then B, the middle point of the arc AC, is the brilliant 
 point. 
 
 Now from (120a) the small superficial element, shown as 
 having A for its centre, is the element having the greatest actual 
 illumination. This element returns the broad beam of rays, 
 LA, partly by reflection (106) in the condensed beam AR. An 
 equal element, equidistant from the brilliant point, as at C, re- 
 ceives only the narrow beam of rays L/C, which, furthermore, 
 it remits in the diffuse beam CR'. It is now quite evident, from 
 a comparison of the breadth of the two reflected beams, that, if 
 they diffused equal amounts of received light, the intensity of 
 this diffused light, and consequently the apparent brilliancy of 
 the elements A and C, would be inversely as the sine of the angle 
 made by the reflected ray with a tangent plane at the point from which 
 it proceeds. 
 
 But not only is the beam CR' more diluted, so to speak, than 
 AR, so far as the space filled by a given amount of light is con- 
 cerned, but it actually contains fewer rays, viz., those of the 
 narrow incident beam L'G. Hence, much more, is the element 
 at A, brighter than the element C, which is at the same distance 
 from the brilliant point, B.
 
 SHADES AND SHADOWS. 137 
 
 In fact, the superior brilliancy of B over all other points, can 
 be fully accounted for, in this view, only by taking account of 
 the effect of distance (122a), and by supposing that some of the 
 reflecting facets of molecules in the element A, which is seen 
 obliquely, are hidden by others in advance of them, so that all 
 the rays of the beam LA are not visibly reflected. 
 
 Therefore, finally, from the above, we conclude, that on cylin- 
 ders, cones, spheres, etc., points which are equidistant from the 
 brilliant line, or point, should not receive equal tints; and ob- 
 serving, in Fig. 63, that tangent rays parallel to LA determine 
 the curve of shade, that the darkest of two such points should 
 be the one which is between the brilliant element and the Visible 
 projection of the element, or curve, of shade. 
 
 c. Relative illumination of apparent contours. It may be sepa- 
 rately noted that, as a sort of balancing between the results of 
 the light received by a surface, the light reflected and refracted 
 from it, and the obstruction of some of these remissions, there 
 will be a narrow space along the apparent contour of curved 
 surfaces, which will appear lighter than the portions a trifle 
 further from that contour; for this space, which is very narrow 
 in projection, is wide in reality, and so receives a broad beam 
 of light ; which, however, it reflects within a narrow compass. 
 
 d. Rate of variation of shades on different surfaces. A consi- 
 deration of the forms of bodies shows that there must be varia- 
 tions in the rate of gradation from light to dark, on different 
 surfaces. For example, see PI. XV., Figs. 59 and 60, equal 
 vertical strips of shade on the cylindrical surface will not appear 
 equal in vertical projection, and the shading, used to express the 
 curvature of its convex surface, would therefore proceed rapidly 
 from dark to medium shades, and then, more gradually, from 
 medium to light shades. But on the vertical plane seen oblique- 
 ly, Fig. 58, similar equal strips would appear equal in projection, 
 and the shading used to express the different distances of these 
 strips from the eye (122), would proceed at a perfectly uniform- 
 rate from dark to light. 
 
 e. The proper shading of bounding edges. The theory of the 
 form of bounding edges, should be observed in the execution of 
 shading. These edges are supposed not to be mathematical 
 lines, but to be rounded, through the ultimate imperfections of 
 workmanship, and the attrition of flying particles. Thus, an 
 edge of a prism is to be regarded as minutely cylindrical ; and
 
 138 THEORY OF SHADING. 
 
 the circumference of the base of a cylinder is a portion of a 
 torus, whose meridian section is a very minute quadrant. 
 
 These edges, even when in shade, will have their own brilliant 
 points, due to the scattering light which they may receive from 
 surrounding objects, and to their superior smoothness, consequent 
 on greater exposure to polishing agencies. Accordingly, very 
 pleasing effects are produced, especially in an assemblage of 
 shaded figures, as in a machine drawing, by shading the edges 
 of each form, according to the principles here stated. 
 
 /. Illustrations. First. On a shaded drawing of a vertical 
 cylinder of revolution, the upper edge should everywhere be 
 lighter than the bordering portions of the convex surface. The 
 lower edge of its illuminated portion would be darker than the 
 convex surface, but the lower edge of its shade should be lighter 
 than the contiguous shade, though rather darker than the upper 
 edge of the shade. 
 
 Second. The successive parallel plane surfaces, forming the 
 front of an edifice, are distinctly shown in projection, by thus 
 treating their edges, which are mainly horizontal and vertical 
 quarter cylinders. Upper and left hand edges, when exposed to 
 the light, are left blank, or of a light shade, for a very narrow 
 space. Lower and right hand edges, are quarter cylinders con- 
 taining elements of shade, and may be ruled with a tint a little 
 darker than that of the flat surface bounded by them. 
 
 The light edges may be ruled with white, if not left blank in 
 tinting. This course would produce the most striking effect in 
 drawings on tinted paper. By thus distinguishing parallel sur- 
 faces by the treatment of their edges, we avoid the necessity of 
 an undue exaggeration of the difference between their shades, as 
 explained in (122 c). 
 
 125. Effects (on the drawing) due to drawing materials and 
 processes of manipulation. 
 
 a. Since the amount of remitted light received by the eye, 
 both white, or reflected, and colored, or refracted, decreases in 
 receding from the brilliant point, while the number of colored 
 rays determines the apparent clearness and strength of color, it 
 would seem that the projection of a body would be more accu- 
 rately shaded by a mixture of India ink with its conventional tint, 
 than by shading it with ink alone, to express its gradations of 
 shades, and then covering it with a flat tint of the conventional 
 color. For the latter method would give so little color compara-
 
 SHADES ANE SHADOWS. 139 
 
 lively to the very dark shades, that it would be imperceptible, 
 while in fact every part of the body does reveal its proper color 
 to some extent. Otherwise, if the conventional flat tint were 
 made strong enough to show, at the line of shade, it would obli- 
 terate the light ink-shading beneath it, in the vicinity of the bril- 
 liant point. In geometrical drawing, however, one of the exag- 
 gerations, which compensate for the artificial character of all 
 projections, is secured by the second method of treating colored 
 objects. Hence the view presented in the " Elementary Projec- 
 tion Drawing" (182). 
 
 b. There are three simple methods which may be used, sepa- 
 rately or together, in executing a shaded drawing. These 
 methods are : 1. By softened tints. 2. By superposed flat tints. 
 3. By touches. 
 
 c. Shading in softened tints. This is difficult to execute, but 
 would give, when perfectly done, the most perfectly smooth and 
 continuous gradation of shade. Moreover, it requires to be done 
 at one operation, with a mixture of ink and the appropriate con- 
 ventional tint, as explained in (a). For if the two shades be 
 made separately, though both by this same method of softened 
 tints, discrepancies between their gradations will be apt to spoil 
 the intended effect of each. 
 
 d. Shading by superposed flat tints. This consists in placing 
 a narrow dark tint on the line of shade ; then, when dry, an- 
 other, covering the first, and extending a little further, and so 
 on. In this method, the drier the brush, the less will the edges 
 of the successive bands show. 
 
 In the last method, we rely somewhat on a free conformity 
 of the softened shades to their theoretically true distribution, as 
 determined by observation merely. 
 
 In this method, we aim to have each of the well marked suc- 
 cessive bands of tint exactly located so as to represent the bands 
 of equal apparent brilliancy on the body represented. They 
 should therefore, on account of their conspicuousness, be located 
 according to precise knowledge of their proper position. 
 
 But this location it is difficult, if not impossible, to effect, by 
 any simple or generally available construction, owing to the 
 complication of the problem, even in the simplest cases, as indi- 
 cated in (124 I). 
 
 e. Shading by touches. This method consists in placing 
 directly upon each part of the drawing, a small quantity of ink,
 
 140 THEORY OF SHADING. 
 
 with a tolerably dry brush, and of a darkness equal to that of 
 the corresponding point of the surface represented. This method 
 is quite similar to that of shading with a lead pencil, and, 
 guided by practised judgment, will secure an orderly gradation 
 of shades, without involving any of the raggedness and blotted 
 appearance which are apt to appear in unskilful use of the two 
 former methods. 
 
 /. The first method is modified, by the size of the surface of 
 paper to be shaded, into what may be called the methods by 
 wet softened tints, and by dry softened tints. The former is 
 applicable to quite large surfaces, and requires a pretty large 
 and wet brush. The latter is merely the second method with 
 the edge of each tint softened by a brush damp with clear water. 
 
 Neither the first nor the second method, alone, will ever pro- 
 duce perfectly smooth shading. The third must be combined 
 with them, so far as is necessary to fill up the irregularities of 
 shade produced by those preliminary methods. The method by 
 touches alone is very tedious, as applied to large surfaces. Hence 
 it should then be combined with the second method. It is, 
 however, conveniently applied, alone, to small surfaces. 
 
 g. The great advantage of the method ty touches, is, that the 
 minute light spots, finally left among the fine strokes of the 
 brush, reproduce, in the drawing, the effect of the minute bril- 
 liant points of the asperities of the object, and thus give the 
 drawing an appearance of distinctness and solidity, which is 
 very desirable. On the other hand, the uniform saturation of 
 the paper with wet tints, used in the method of softened tints, 
 produces a cloudy effect like that of the shade of polished 
 bodies. Yet it does not, on the whole, adequately represent 
 such bodies, because its gradation of shade is gradual, while 
 theirs is apparently abrupt, on account of the dazzling bright- 
 ness of their brilliant points. 
 
 h. The effect of the minute brilliant points is also produced, 
 in part, by shading on rough paper, on whose actual asperities 
 the light may act. 
 
 i. In the use of tinted paper, brilliant effects may be produced 
 by using white tints for the brilliant points. 
 
 Also, the effect of colored drawings will be heightened, by 
 employing tinted paper of a color complementary to that which 
 prevails on the drawing.
 
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