CORNELL UNIVERSITY LIBRARY GIFT OF Ifrs. K. 1, Turner R^s Airrs impm * MMMMMiK^^^ and p 089 Cornell University Library The original of tiiis book is in tine Cornell University Library. There are no known copyright restrictions in the United States on the use of the text. http://www.archive.org/details/cu31924020549089 A TEXT-BOOK SHADES AND SHADOWS, PERSPECTIVE. PREPARED FOR THE USE OF STUDENTS IN TECHNICAL SCHOOLS JOHN E. HILL, M.S., M.C.E., Professor of Civil Engineeritig, Brown University. SECOND EDITION, REVISED AND ENLARGED. FIRST THOUSAND. l^ a NEW YORK : t ^ '< JOHN WILEY & SONS. _<^ London : CHAPMAN & HALL, LiMiTEa ^ ^ 1900. , ^^^ \ c ■ .- y ^r Copyright, 1S96, BY JOHN E. HILL, ^:[>.l N. K.^ .t i>f/^lr ROBFET DEUMMOND, ELECTROTYPER AND PRINTER, NEW YORK. PREFACE TO THE FIRST EDITION. This text-book is designed to furnish a short course in Shades and Shadows, and One-Plane Perspective, for students who have studied the elements of Descriptive Geometry. Prominence has been given to the fact that the subjects treated are branches or applications of Descriptive Geometry and follow that subject in logical sequence. The author is greatly indebted to many friends who have kindly assisted in the preparation of the work, and especially to Messrs. C. W. Comstock and C. W. Sherman, Instructors in Civil Engineering, Cornell University. The method of perspective given is that elaborated in " Modern Perspective " by Professor W. R. Ware, and the method of shading used is to be found in the works of M. Jules Pillet. To many works, but especially to those men- tioned, the author desires to express his obligation. J. E. H. Ithaca, N. Y., March i, 1894. PREFACE TO THE SECOND EDITION. For this edition the book has been throughly revised and in the main rewritten. The most notable changes are in the notation of Shades and Shadows ; the extension and addition of certain problems ; and the addition of examples of shaded objects showing the application of the principles of shading. My thanks are due Professor H. S. Jacoby, of Cornell Uni- versity, for many suggestions. J. E. H. Providence, R. I., July i, 1896. TABLE OF CONTENTS. SHADES AND SHADOWS. CHAPTER I. PAGE g§ 1-22, General Principles, 'Definitions AND Conventions., i-ii CHAPTER II. §§ 23-33. Shadows OF Right Lines ■ 12-21 CHAPTER III. §§ 34-47. Shadows OF Curves 22-4>2 PERSPECTIVE. CHAPTER IV. §§ 48-77. General Principles, Definitions and Conventions. . 43-54 CHAPTER V. §§ 78-94. Measurement of Lines in Perspective 55-6i CHAPTER VI. §§ 95-ioo«. Parallel, Angular AND Oblique Perspective 62-79 CHAPTER VII. ^§ loi-iog. Perspective of Circles 80-88 CHAPTER VIII. §§110-116. Perspective of Shadows 89-96 APPENDIX. §§ 117-121. Shading 97-101 SHADES AND SHADOWS. CHAPTER I, GENERAL PRINCIPLES, DEFINITIONS AND CONVENTIONS. Shadows. 1. Shades and Shadows is primarily an application of Descriptive Geometry. (See §§ i6 and 19.) 2. The subject defined. The subject treats of one of the means employed by which the representation of any object may be given the appearance of reality. This result is here accomplished by giving certain tints or colors to one or both of the orthographic projections of the object upon the co- ordinate planes, and by the determination of its shadow either upon those planes, upon the object itself, or upon both. Shading or tinting is treated in the Appendix. 3. Shades and Shadows in Nature. Non-luminous ob- jects are seen by the light which they reflect and refract.* It is evident that without light an object is invisible. It is also evident that if an object be illuminated by but one source of light, e. g., the sun, that the portion of the object towards the sun will appear light or bright, and the portion on the oppo- * The detailed reasons for some of the statements in this chapter may be found in any of the larger text-books on Physics. 2 SHADES AND SHADOWS. site side, or out of the light, will appear dark; also that various portions will appear to have different tints depending upon the relative positions which they bear to the sun. On a sphere the various tints blend and there are no sharp con- trasts ; on an hexagonal prism the tints on the several sides are sharply contrasted. Because of this variation in the gradation of tints or colors in the appearance of objects, we are able to distinguish one from another, and were it not for this phenomenon all objects would appear flat, i. e., like a plane. In many cases we may determine by analogy the form of an object from its shadow; e.g., one would never mistake the shadow of a tree for that of a building. If then the proper shading be given to the representation, and the shadow be determined, we have a picture of the object. 4. Sunlight The light from the sun is transmitted to the earth along slightly curved and slightly divergent lines, because the atmosphere is not homogeneous and the sun is not in- finitely distant. We may consider, however, for drafting pur- poses that the lines of light are parallel right lines, as the error caused by such assumption is not appreciable for any drafting that would be made. 5. Conventions. The majority of draftsmen desiring uni- form results have adopted the following conventional method of representation. It is unnecessary to discuss the scientific accuracy of these assumptions ; let it be taken for granted that they supply the want and give the desired results. Conventions. (yd) The source of light is considered to be a fixed point situated at an infinite distance from the object to be repre- sented. PRINCIPLES, DEFINITIONS AND CONVENTIONS. 3 (^) All lines of light emanating from the source are par- allel right lines. (c) The source of light is further limited in position by- being so placed with reference to the coordinate planes that a line of light forms one of the diagonals of a cube, two of whose faces are the coordinate planes. If the spectator were in the first dihedral angle looking at V, the source of light would be at an infinite distance above and behind his left shoulder ; the direction of a line of light would be the diago- nal of the cube above mentioned running from the near upper left vertex to the far lower right vertex. Fig. i gives a per- Fig, t. spective view of a cube, two of whose faces, gfe i and b c e i , are in H and V respectively, and the diagonal R through a and e gives the direction of a line of light to which all lines of light are parallel. [d) All lines of sight for either coordinate plane are parallel right lines perpendicular to that plane. {e) All objects situated in the second, third, or fourth di- hedral angles are considered invisible. (y) All objects are to be considered opaque. 6. Rays. Lines of light are called Rays of light, or simply Rays. (See § 5, sec. (b)) 4 SHADES AND SHADOWS. 7. Plane of Rays. Any plane containing a ray of light is a Plane of rays, for such a plane contains an infinite number of lines parallel to the ray, and consequently an infinite num- ber of rays. 8. Pencil of Rays. Any collection of rays other than a plane is called a Pencil of rays. (By some authors " cylinder of rays.") 9. Line of Shade. The Line of Shade of a surface or solid is the locus of the points of tangency of all rays which can be drawn tangent to that surface or solid. NOTE: The word tangent as used in this connection means tangent in its geo- metric sense, and also touching without immediately intersect- ing. The general idea of tangent rays is, however, always maintained, because the rays determining the line of shade do not intersect the object before touching, and do not im- mediately intersect after touching. (See Fig. i.) The line of shade of the cube is composed of the six edges fd, do, cb, bi, ig, and gf, determined by six planes of rays passed through them, and is an example of " touching without immediately intersecting." If a plane of rays, were passed through the edge ag\\. would immediately intersect the cube, consequently «^ is not a part of the line of shade. The line of shade of a sphere is a great circle determined by a tangent pencil (cylinder) of rays. The line of shade divides the object into two parts, one illuminated, or in the light, and the other unilluminated, or in the dark ; the light being excluded from the unilluminated parb by the object. The line of shade of a plane surface is the perimeter of the surface ; of a plane figure, the lines of that figure. 10. Umbra. The f/wi^ra or indefinite shadow of an object is that portion of space from which light is excluded by the PRINCIPL-ES, DEFINITIONS AND CONVENTIONS. 5 object. The umbra of a point is a line ; of a right Hne is a plane, etc. The umbra of a surface or solid is bounded by the pencil of rays tangent to the object. 11. Shadow. The Shadow of an object is the intersection of the umbra with the surface of any object. If no object is situated in the umbra of an object there is no shadow. A shadow is always a point, line, or surface. (See Plate II, Fig. II.) The shadow of the lower edge, e b, of the parallelopiped is partly on the three front faces and one side face of the tinder prism and partly on H. Care must be taken to obtain the complete shadow in every problem. 12. Primary and Secondary Shadows. A distinction is made between the primary shadow, or, briefly, the shadow, and the secondary shadow. The former is the intersection of the umbra with the first object which it meets, and the latter is a subsequent intersection. In many problems the secondary shadow is found in order to more easily determine the primary shadow. 13. Line of Shadow. The Line of Shadow is the perimeter of the shadow, or the intersection of the tangent pencil of rays Tvith the object on which the shadow is cast. 14. Shade. The Shade is that portion of the object from which light is excluded by the object itself, and which would not be illuminated were any part of the object removed. The shade is bounded by the line of shade. (See § 9.) 15. It follows from §§ 9 and 13 that the line of shadow is the shadow of the line of shade. 16. Oblique projection. Since shadows are determined by lines (rays) which are oblique to the coordinate planes, the method is one of oblique projection, and the problems pre- sented are similar to those of Descriptive Geometry involving 6 SHADES AND SHADOWS. intersections. This fact should be constantly in mind whea studying this subject. (See § i.) 17. Conventional direction of rays. Since a ray forms the diagonal of a cube as shown on Fig. i (see § 5, sec. (c)), the projections of a ray make angles of 45° with G. L. The ray itself makes angles of 35" 15' 52" with H, V, or a profile plane ; this, angle is called for convenience. Trigonometric functions of 0. (See Fig. i. Each edge of the cube is unity.) I V 2 sin ^ = -^ = 0.57735. cos ^= —= = 0.81649.. ^3 ^3 tan = —7= = 0.7071 1 = cos 45° = sin 45°. V2 V^ I sec ^ = — £ = 1.22475. cos 2 6* = - = 0,33333.. V2 3 18. Advantages of conventional direction of rays. By using the conventional direction of rays the construction of many problems may be shortened ; e. g., to find the shadow of the point a (Plate I, Fig. 3) : As a^u is greater than a^ u, from a^ lay off a^ p = cC u, draw p a^y parallel to G. L., and «°«si making 45" with G. L. The intersection of the two lines, a^i, is the shadow of a required, because a' u is the distance of the point above H, and if a right-angled triangle be constructed with aa^ = a7 u for the altitude and a ray for the hypotenuse, the base in H is a^ agy. In certain problems it may be advantageous to disregard the conventional position of the source of light and assume some other position in order to bring out more clearly, either PRINCIPLES, DEFINITIONS AND CONVENTIONS. 7 the character of the object or a particular detail. Because such cases are rare and their treatment requires mature judg- ment no departure is made in this work from the ordinary- method of procedure. Usually a change in the position of the object rather than in the position of the light will give the de- sired result. 19. The general problem. In general the shadow of an object is found by passing a pencil of rays tangent to the ob- ject and determining the intersection of this pencil with the object which receives the shadow ; this method involves, first, the determination of the line of shade, and second, its shadow. When the line of shade is a right line and the surface receiving the shadow is a plane, the problem is simple and may be stated thus : To find the intersection of a plane passed through a given line with a given plane. When the line of shade is a curve and the surface receiving the shadow is curved the problem is more difificult ; stated in Descriptive Geometry language it is : To find the intersection of any surface with a cylinder. In problems relating to curves or surfaces of the second degree, the axes or diameters of parts of the shadows may often be found and the solution greatly simplified. As a last resort a number of rays may always be passed through the line of shade and their intersections with the object receiving the shadow found. This method should not be employed how- ever, unless necessary, as the great number of construction lines required increases the chances for error and has a ten- dency to mar the appearance of the drawing. The line of shade in many cases is readily determined by inspection. When this cannot be done, or uncertainty exists in regard to lines probably composing the line of shade, the shadow of the entire contour line of the object may be found an SHADES AND SHADOWS. d the parts which cast shadows within the perimeter of the shadow rejected. The general test is, that no part of the line of shade casts a shadow within the line of shadow. 20. Notation. The following notation will be employed throughout Shades and Shadows : (a) The projections of a point on the coordinate planes will be designated by small letters with the exponents « and ^ rep- resenting the horizontal and vertical projections respectively ; e. g. , the projections of the point a are a^ and a^. (b) The projections of the revolved position of a point will be designated by the letter of the point with the subscript „ ^ or 3 and the exponents ^ and ' ; e. g., a^ and «/. (c) The projection of a point upon a new coordinate plane will be designated by the letter of the point with the exponent ^1 or ■^i, °' or ^« ; e. g., «"! or d^^. ' {d') The shadow of a point on one of the coordinate planes will be designated by the letter of the point with the subscript g ; e. g., «s. When the shadows* of a point on both coordinate planes are required they will be designated by the letter of the point with the subscripts si and si\ e. g., «si and a^z. {e) The projections of the shadow of a point on a surface other than the coordinate planes will be designated by the let- ter of the point with the exponents " and ^ and the subscripts si and s2, etc., e.g., a\, ajj. When only one projection of the shadow is required it will be designated by the letter of the point with the subscript s ; e. g., a^. {/) A line will be designated by a capital letter, and its projections, revolved position and shadows by the exponents or subscripts used for points. See {a), {p], (c), {d), (e). {g) The trace of a plane will be designated by a capital let- ter preceded by a capital H or V; e. g., HP or VP. * Primary and Secondary shadows (see § 12) usually. PRINCIPLES, DEFINITIONS AND CONVENTIONS. 9 (Ji) The trace of a plane in a revolved •^os\\\ox\ will be desig- nated by the letter of the plane with the subscript j, , or 3, and preceded by a capital H or F; e. g., HP^ or FP,. (?) The trace of a plane on a w^ze/ coordinate plane will be ■designated by the letter of the plane, preceded by a capital H ■or V with the subscript 1, j or , ; e. g., H^P or V^P. 21. Drafting Conventions. (i) Given and required lines, if visible, are represented by heavy full lines ; if invisible, by heavy long dashes. (2) Auxiliary lines, visible and invisible, are represented by light dashes. Projecting lines and lines indicating the paths of rotated points are short dashes and other auxiliary lines long dashes. (3) Traces of given and required planes, if visible, are rep- resented by heavy full lines ; if invisible, by one long dash and two short dashes, also heavy. (4) Traces of auxiliary planes — planes used as a means for solving the problem — are represented by light broken lines consisting of one long dash and two short dashes alternately. (5) Intersections of given and required planes, if visible, are represented by heavy full lines ; if invisible, by one long dash and one short dash, also heavy. (6) Intersections of auxiliary planes, or of auxiliary with given or required planes, are represented by light broken lines consisting of one long dash and one short dash alternately. ■ lO SHADES AND SHADOWS. (7) The long dashes should measure about one-eighth of aii inch and the short dashes about one twenty-fourth of an inch, t. e., about one-third as long as the long dashes. The spaces between the dashes should be very short. (8) New ground lines are represented by heavy full lines. (9) All objects of which the shadows are to be found are considered opaque. Recapitulation. Given and required lines, traces, and inter- sections, if visible, are distinguished from the invisible in form only, the weight remaining the same. When visible all of- them are represented by full lines. When given or required, but invisible, they differ from the auxiliary in weight only, the form remaining the same. Traces have one long dash and two short dashes, intersections one long dash and one short dash, and other lines are composed of either long or short dashes. All given and required lines, etc., are heavy, and all auxil- iary lines are light. 22. Problems in Drafting Room. The work in the draft- ing room may often be greatly facilitated and improved by assigning problems in which the positions of points, etc., are designated by means of coordinates, as in analytic geometry. The position of a point in space is known when its distance from three planes (no two of which are parallel) is known, hence to designate the position of a point in a problem in shades and shadows (descriptive geometry), we may refer it to H, V, and a profile plane, P. Following the notation of ana- lytic geometry, G. L. is the axis of x, the horizontal trace of P\s the axis of ^, and the vertical trace of Pis the axis of z. The point a with the following reference, x = 2", y = 3", z = 6", means that the point is in the first dihedral angle 2" to the right ol P, 3" in front of Tand 6" above H. The point b PRINCIPLES, DEFINITIONS AND CONVENTIONS. IT with X = 3", J = — 3", z = — 4", means that the point is in. the third dihedral angle 3" to the right of P, 3" behind V, and 4" below U. The designation of lines and planes readily fol- lows. It is preferable to assume that the reference profile plane {P) passes through the left border line of the sheet upon which the drawing is to be made. CHAPTER II. Shadows of Right Lines. 23. Problem I. To find the shadow of a line on the coordi- nate planes. Plate I, Fig. 3. Analysis. The shadow of a line is determined by the in- tersection of a plane of rays through the line with the object receiving the shadow ; or, in other words, the shadow is the locus of the intersections of all rays passed through the line with the object receiving the shadow. The object in this case is the coordinate planes, hence if the shadows of any two points of the line on H and on V be found and joined by a line, the required shadow is determined. Or the shadow of one point of the line may be found on H and on V, and joined with the H and V traces of the line respectively, for since the traces of a line are points in the coordinate planes, each is its own shadow on the plane in which it is situated, and the shadow of the line is determined as before. Construction. Let M be the given line. Take any two points of the line as a and b and through each pass a ray. The projections of the rays pass through the projections of the points and make angles of 45° with G. L. The intersection (trace) of the ray through a with H is a^l and with V is a^^. The intersection (trace) of the ray through b with H is b^ and •with V is b^Y Joining a^^ b^^ we have the shadow of M on H, and joining a^ bg^, the shadow of M on V. The parts of the shadow on H behind V and on V below H are secondary (see SHADOWS OF RIGHT LINES. 13 § 12), hence the shadow of M on the coordinate planes is the line agifisbsi or M^x- In this problem, as in many others, by finding the second- ary shadow the solution is simplified. The method of § 18 might have been employed with advantage. Another Solution. Since «gi and b^i are the H traces of rays passed through two points of the line M and a^^ and b^i are the V traces of the same rays, a^^ bg^ and a^^ b^i are the H and V traces respectively of a plane of rays passed through M, and hence are the shadow of M on H and V. The visible portion of these traces and hence the primary shadow is the broken line Mg^. 24. Problem II. To find the shadow of a plane surface on the coordinate planes. Let the given surface be a triangle. Plate I, Fig. 4. Analysis. The line of shade of the surface is the perim- eter of the surface (§ 9) and the line of shadow is the shadow of the perimeter (§ 15), hence the shadow of a triangle is determined by the shadows of its three sides. If we find the shadows of the vertices of the triangle we will have the shadows of two points in each side, hence the shadows of the vertices determine the shadow of the triangle. Construction. Let abc h^ the given triangle. Find the shadows of the vertices. The shadow of a is a^ on H, of b is bs\ on V, and of c is c^ on H, all found as in Prob. i, § 23. Join «s Cg for the shadow of ac. As a^ and c^ are on .Awhile ^si is on V, it is necessary to find the shadow (secondary) of b on H, or of a and c on V, in order to determine the direction of the shadow of ab and of cb on H. The shadow o{ b on H is b^, hence a^ m^ and c^ n^ are the primary or visible shadows of ab and cb respectively on H. To complete the shadow of the triangle draw ;«s b^^ and Wg bsu the primary shadows of ab 14 SHADES AND SHADOWS. and cb respectively on V. The required line of shadow is 25. Problem III. 7^^ find the shadow of a solid on the coordinate planes. Let the solid be a cube. (Simplest case.) Plate I, Fig. 5. Analysis. The line of shadow of a solid is the shadow of the line of shade (§ 15), hence first determine, if possible, the line of shade. In complicated problems in solids it is some- times difficult if not impossible to determine the line of shade without virtually finding the shadows of various lines of the object. The criterion to be employed is : A line of the soli(^ is a part of the line of shade when its shadow is a part of the perimeter of the shadow of the solid. A line of the solid is not a part of the line of shade when its shadow falls within the perimeter of the shadow of the solid. For rectilinear solids we may usually determine the position of the line of shade by inspection if we remember the definition of § 9. The line of shade of the cube is found by inspection to be the edges dj,jg,gf,fb, and the edges in H, ab and «d? which coincide with their shadows. The proof of this statement will be given below. Having determined the line of shade, pro- ceed as in Probs. I and II. Construction. Given the cube as shown with one face in H and the other faces perpendicular or parallel to V. The shadow of the edge ^' is d^j^, oijgisj^g^, oig/isg^f, oi fb is fs 5", found as in Prob. I by joining the shadows of the two points of each edge. The shadows of the edges ab and ad coincide with the lines. The line of shadow is, therefore, d^JsKsfs l>^ ^^ d^ on H. The visible shadow is bounded by the line d'^jsgsfs b^ ^ d^. Proof of statement in analysis: — Suppose the edge ej to be part of the line of shade, its shadow is c^j\ which falls SHADOWS OF RIGHT LINES. 1$ within the line of shadow, hence ej is not a part of the line of shade. (See § 19.) A plane of rays through ej immediately intersects the cube, hence ej is not a part of the line of shade. (See § 9.) In the same manner it may be shown that the edges ae, ef, «tc., are not parts of the line of shade. Note : From the solution of this problem we see several facts which are principles of geometry, {a) If a line is par- allel to a plane its shadow on that plane is parallel to the line. {V) Parallel lines cast parallel shadows on the same plane, etc. 26. Problem IV. To Jind the shadow of a vertical prism en the coordinate planes. Plate I, Fig. 6. Analysis. The analysis of this problem is similar to that of Prob. III. Construction. The line of shade is found by inspection to be the edges cj and af, three sides of the upper base jk, kl, and If, and the two sides of the lower base ab and be ; which statement may be proved as in Prob. III. The shadow of cj is c^y'si, oijk \sjsi n^ k^i (see Prob. I), of kl is k^i 4i ,oilfls l^ifi, of af is fi ps a^, of ab is a" b", and of be is b^ c^. The complete line of shadow is now determined. The visible portion of the shadow is bounded by the line c^jsi n^ k^ m^ d^ p^ «" e^ d^ c^ 27. The shadow of the upper base of the prism on //"and on Fis shown and may be used as the solution of the prob- lem : To find the shadow of a horizontal polygon on one or both of the coordinate planes. The shadow (visible) of the polygon on the coordinate planes is the lineygi n^ k^i ^sifsi ?s^si7si- (See Prob. II.) 28. Problem V. To find the shadow of a plane surf ace on Ml oblique plane. Let the surface be a triangle. Plate I, Fig. 7. Analysis. See analysis of Prob. II, § 24. The shadow of i6 SHADES AND SHADOWS. a point on a plane is the intersection of a ray through the point with the plane, and of a line on a plane is the intersection of a plane of rays through the line with the plane. Hence two methods of solution are available : either find the shadows of the vertices of the triangle by means of rays, or the shadows of the sides of the triangle by means of planes of rays. In general,, the first method is the simpler. Construction. Let a b c\t& the given triangle, and Q the plane on which the shadow is cast. To find the shadow of c on Q pass a vertical plane of rays Z through the point. This plane intersects the plane Q in the line M{M'', HZ). Hence ^5, the intersection of a ray through c with M, is the shadow of c required. In a like manner pass the vertical planes of rays Fand X through a and b respectively. As the vertical planes of rays Z, Y, and X are parallel, their intersections with the plane Q are parallel, hence U" and T" may be drawn through the intersection of the F traces parallel to Jkf. A ray through a intersects [/ at a^, and a ray through b intersects T at b^.. The shadow of the triangle is, therefore, a^b^c^. 29. The student should constantly have in mind the fact that the solutions of the following problems are exactly alike l " To find the shadow of a point on a plane " and " To find the intersection with a given plane of a line passed through a given point and parallel to a given hne." " To find the shadow of a right Hne on a plane " and " To find the intersection with a. given plane of a plane passed through a given line and parallel to a given line." 30. Problem VI. To find the shadow of one line upon another line. Plate I, Fig. 8. Analysis. If one line casts a shadow upon another line a plane of rays through the first line intersects the second line,, otherwise the two lines are in parallel planes of rays and the SHADOWS OF RIGHT LINES. IJ second cannot receive a shadow from the first. If a plane of rays be passed through each hne these planes intersect and the line of intersection is a ray, because the line of intersection of any two planes of rays is a ray. Hence, if the shadows of the two lines on the same object be found and a ray passed through their intersection, the required shadow and the point which casts the shadow is determined by the intersections of the ray with the two lines. A limiting case of the above is when the two lines are in the same plane of rays, in which case the sec- ond line is the shadow of the first line. Construction. Let M and N be the given lines. The shadows of tliese lines on H are M^, and N^ respectively, which intersect at the point o^^. A ray through o^ is the line of in- , tersection of the planes of rays which determined the shadows of the lines. This ray intersects i\^ at i^si and M sX o ; hence the shadow of M on N is o^^. 31. Problem VII- To find the shadow of a hexagonal' abacus on a vertical hexagonal prism. Plate I, Fig. 9. Analysis. See analysis of Prob. V, § 28. Only the shadow of the abacus on the prism is considered. It is evident from an inspection of the figure that the lower edges ef,fa, and ab of the abacus compose the line of shade for the purpose named. Having found the line of shade the problem becomes : To find the shadow of a line on a plane (§ 28). Construction. Let M, N, O, etc., be a vertical prism, and a be, etc., be the abacus. As the faces of the prism are vertical the intersection of the H projection of a ray through any point of the line of shade, as f, with the H projection (base) of the prism gives the H projection of the shadow of that point on the prism. Through / pass a ray : it intersects the prism at /s; hence /s is the shadow of /on the prism. Through k,j\ g, a, and i pass rays : they intersect the prism at k^,j\, gs, a^, and 15 SHADES AND SHADOWS. 4, respectively; hence Kfsjsgs ^s 4 's the line of shadow required of which the portion g^ a^ 4 is invisible on V projection and all is invisible on H projection. The points k,j,g, and? were determined as follows: Take the line of shade /«: it is evident tliat its shadow falls on three faces of the prism, PQ, QL, and LM; hence one point oi fa must cast a shadow on Q and one point on L. Since Q is perpendicular to H through Q^ draw the //"projection of a ray: it intersects /" a" aty"; hence /is that point of yQ the conjugate diameters of an ellipse intersecting at o, the center. With ^ as a center and radius equal to the semi-longer diameter describe a circle. Draw n in perpendicular \.o ab at o, and join inc and nd. Draw any line as wr q parallel to n m and intersecting a b and the circle, then draw t rp parallel to do c and qp and w t parallel \.o mc and n d. The points / and t are points of the ellipse. 38. To construct the axes of an ellipse having given any pair of its conjugate diameters. Plate I, Fig. lO. Analysis and Construction. Let a^^b^ and c^d^ be conjugate diameters of an ellipse in H. Draw G. L. through b^ parallel to Cs,d^; then G. L. is tangent to the ellipse. As only one ellipse can be constructed on a given pair of conjugate diameters, as a^b^, and c^d^, and as these lines may be considered the shadows of a pair of rectangular diameters in an infinite number of ellipses, we may assume for the purpose of our demonstration that these lines are the shadows of a pair of rectangular diameters of a circle situated in a plane which passes through G. L. On this assumption the diameter of the circle which casts the shadow is equal to c^d^. NOTE: It is not necessary to determine in any case the actual ellipse or circle which casts the shadow, as we wish to find only the magnitude and direction of the axes of the shadow ; also, the rays of light are not assumed in the conventional direction for the same reason. If the radii of the circle be produced they will intersect SHADOWS OF CURVES. 25 G. L. and the shadows of the produced radii will run from the intersections on G. L. to o^, the shadow of the center. A certain pair of rectangular radii cast rectangular shadows, which are the axe:; of the elliptical shadow. As each radius and its shadow intersects G. L. at the same point, the required pair of radii and their shadows may each be inscribed in a semi-circle of the same radius and having a common diameter in G. L. To find the semi-circles revolve the plane of the circle cast- ing the shadow about G. L. into H, the circle takes the position as shown with center at o^, join 0^ o^, bisect 0^ o^ and draw the perpendicular g i intersecting G. L. at i, with i as a center and radius equal to io^ = io^ describe a circle which gives the semi- circles required. Draw (3j/and 0^ e, fo^p^ and e 0^ m^, also q^ q^ and n^ n^ parallel to 0^0^, then q^pa and n^ m^ are the required axes. 39. To find the position [not magnitude) of the axes of an Mipse, having given any pair of its cotijugate diameters.* Plate II, Fig. 15. Construction. Let oa and obh^ the semi-conjugate diame- ters of an ellipse. Produce aVo c making oaXac^=ob'Xob. Draw man parallel to ob and through o and c describe a circle having its center in m a n. Draw m and o n, which are the indefinite axes of the ellipse. 40. Problem XI. To find the shadow of the edge of a hemispherical shell on the interior surface. Plate III, Fig. 16. Analysis. The line of shade or part of the edge which casts a shadow on the interior surface is determined by the diameter of the line of shade of the sphere which is parallel to H (in this case) and is the semi-circumference of the edge. The shadow is a conic section (§ 36), in this case the arc of *See Salmon's Conic Sections. 26 SHADES AND SHADOWS. a circle, because the only plane curve which can be cut from a sphere is a circle. Construction. Given the hemispherical shell ab c, with the plane of the edge parallel to H. The diameter b c perpendicu- lar to the rays and parallel to ZT is a diameter of the line of shade and also of the shadow, as it joins two points b and c on the line of shadow. (See Prob. X, § 34.) The radius o« is perpendicular to be, hence its shadow oa^, on the plane of the shadow of the edge, is a semi-diameter of the shadow conju- gate to b c; but b c and a^ are perpendicular, hence the axes of the projection of the shadow on H are determined. The V projections of the axes be and o a^ are conjugate diameters of the F projection of the shadow. To find the shadow of « on the interior, either pass a meridian plane of rays through a and revolve parallel to F about an axis perpendicular to /f through the center of the shell, or project on a F" plane of rays through G^. Z,. The V^ projection of a is «^i. Through «^i draw R'\ making an angle of 35° 15' 52" with G^. Z,.: the in- tersection tfji with the contour is the F, projection of the shadow required ; hence a^ is the shadow of a on the interior. Short Construction. Draw «s' o^^ and project flgi on a^i m at n. The angle M(7^i«Ji ^ 2 angle o^'a^iaji. Therefore o^in=: r r o'^ flji cos 2 p = -. Hence we may at once lay off c^a^ = _ 3 3 and construct an ellipse. The shadow of the edge on the in- terior is b a^c found by constructing ellipses on axes in H pro- jection and conjugate diameters in V projection. Only the shadow in //'projection is visible. 41. Problem XII. To find the shadow of an oblique cylin- drical surface on the coordinate planes and the shadow of the upper base on the interior af the surface. Plate III, Fig. 17. Analysis. The line of shade is determined by two tangent SHADOWS OF CURVES. 2/ planes of rays and two pencils (cylinders) of rays through the. bases. The cylinder, being hollow, part of the shadow of the upper base is on the interior surface and part on the coordinate planes. To determine the elements of shade, pass a plane through the axis (or element) of the cylinder and a ray ; this plane is parallel to the tangent planes, which are now readily determined. The problem "to pass a. plane of raj/s tangent to a cylinder or cone " is the same problem as " to pass a plane tangent to a cylinder or cone and parallel to a given line." From § 36 we know that the shadow on the interior is a conic section, in this case an ellipse, and hence ap/ane curve. Construction. Given the oblique cylindrical surface with one base in and the other parallel to H, as shown. Shadow on the coordinate planes. The shadow of a the cen- ter of the upper base on H is a^^ ; hence b^a^^ is the H trace of a plane of rays through the axis, because it contains the H trace of the axis and the H trace of a ray through a. The H traces of the two planes parallel to this plane and tangent to the cylinder are (^r^ and e^ x^, and the elements of shade are therefore cdan^ef, which cast shadows c^r^d^ and e^x^f^ respectively. The line of shade (edge) of the upper base is divided in two parts by the elements of shade, the shadow of fgd being on the coordinate planes and the shadow oifj'd on the interior. Draw the rectangular diameters df and J g: their shadows on V are conjugate diameters of the shadow of the upper base; hence, constructing an arc of an ellipse on d^f^ and j\g^ limited by the points ^^ and d^, we have the complete shadow of the surface on the coordinate planes. Shadow on the interior surface. Since df is a diameter of the line of shade and contains two points of the shadow it is a 28 SHADES AND SHADOWS. diameter of the shadow. The shadow oij, the extremity of t-he rectangular diameter gj, \sj\, found as follows: Through the element tj pass a plane of rays ; it intersects the cylinder in the element wg\ a ray through/ intersects w^ at /s, hence /s is the shadow of/ on the interior. A plane through fd and j^ is the plane of the line of shadow (see § 36), because the curve of shadow is a plane curve, and this plane contains three points of that curve. Produce /s a to k on the element //, then/g a = ak and df and/s/fe are conjugate diameters of the elliptical shadow. A semi-ellipse constructed on d^ f^ and cFj\ is the //projection of the shadow, and a semi-ellipse on/^ d?"' and a^'/s is the V projection of the shadow. The visible shadow on the interior is bounded by the line/^'/g u d^j'^f^, and on the coordinate, planes by the line c^ r^ y i" x^ z c^- 42. Another solution for finding shadow on interior of the surface of preceding problem. Same figure. Analysis. The intersection of the umbra of any curve with another curve is the shadow of the first curve on the second (see § 30). Hence the intersection of a pencil of rays through the upper base of the cylinder with any other section is the shadow of the base on that section. Construction. Pass any auxiliary plane through the cylinder as the horizontal plane X\ it cuts out the section B. The shadow of the upper base on the plane X is A^, the circum- ference of a circle of radius equal to that of the base, since the plane of the base and Xare parallel. A^ and B intersect at/^ and q^, hence these points are two points of the required shadow. A sufficient number of planes to accurately determine the line of shadow must be employed. 43. Problem XIII. To find the shadow of a conical surface SHADOWS OF CURVES. 29 on the coordinate planes and tJie shadow of the base on tJie interior of the surface. Plate III, Fig. 18. Analysis. This problem presents a fact concerning tangent planes to cones which is often overlooked. The elements of shade are determined by two tangent planes of rays. As these tangent planes are planes of rays each must contain the ray the which passes through the vertex of the cone, hence this ray is line of intersection of the two planes. Using the conventional direction of rays (§ 5) this line of intersection or ray cannot be parallel to the coordinate planes, but must intersect them. The ZT and F traces of the tangent planes must pass through the corresponding traces of this line, hence the //^ traces of these planes cannot be parallel, and the F traces cannot be parallel. As in the preceding problems, the shadow on the in- terior is a conic section, in this case an ellipse. Notice the fact that in this problem the line joining a and 3 is a chord and not diajtieter of the line of shadow as the line fd\n the preceding problem (§ 41). Construction. Let the conical surface be given as shown with axis perpendicular to H, and base the circumference of a circle parallel to H. To find the elements of shade pass a ray through the vertex o : it intersects the plane of the base at 7'; henceya andjb drawn tangent to the base are the inter- sections of the tangent planes of rays with the plane of the base, and a and o b are the elements of shade. The shadows oi o a and b are o^ 4 a^ and o^ k^ b^ respectively. The shadows of any two rectangular diameters of the base as y"^ and c ^ give con- jugate diameters of the shadow of the base. Hence an ellipse constructed on fg dg and c^ e^ as conjugate diameters is the shadow of the base on V. But as the line of shade A is di- vided in two parts by the elements of shade and the shadow of so SHADES AND SHADOWS. the ^xc alb is on the interior, the shadow of the base on Fis limited by the shadows of a and b. Hence Os K b^ d^ c^ «s 4 ^'s Js the shadow on the coordinate planes required. Shadow on the interior surface. The chord a b oi the line of shade is a chord of the interior line of shadow, since it joins two points a and b of that shadow. To find diameters of the line of shadow revolve the meridian plane containing a ray about the axis of the conical surface parallel to V. The point /, one extremity of the diameter of the line of shade perpen- dicular to a b, takes the position /j, and a ray through / in re- volved position is /^ /gj, hence the revolved position of the shadow of / is 4, and the true position is 4. The plane of the line of shadow, therefore, passes through the points a, b, 4 (see § 36). In revolved position (perpendicular to Fin this case) this plane contains the line a^ b^ (vertically projected at gX) and the point 4ii and intersects the revolved position of the element ^sfs of the rectangular diameters n q and rf. On H the shadow is B^^i, an arc of a circle, since B is horizontal. The shadow of the lower base C on V is C^^t ^ part of an ellipse constructed on the shadows w^x^, t^ti^ of the rectangular di- ameters w X and t u ; on H the shadow is composed of two arcs of the same circle whose center is p^^ and radius = p^ /g=/ if, since C is horizontal. The shadow on the double- curved surface is determined by the method of § 42. Let Z be any horizontal secant plane ; it cuts from the surface the circumference G. The shadow of C on this plane is found by constructing an arc with/gi, the shadow of / on the plane, as a center and radius = radius of C. This arc intersects G at ■m^ and 4 ; hence these points are two points of the required shadow. Other points were determined by the same method. The shadow of C on the surface is Qj. The line of shade of the double-curved portion intersects C^ at the points 2 and 5, at which points rays are tangent to the curve C^. Only that part of C^i which is on the left of or above the line 2, 5 is on the outside of the scotia ; the remainder of the line (on the 36 SHADES AND SHADOWS. right of or below 2, 5) can only exist as a shadow on the sup- position that the scotia is hollow and that the upper cylinder is removed ; under these conditions this part of the shadow _ of Cis on the interior. Third : Through the points /, _?, 4, and 6 of the line of shade of the double-curved portion draw tangent rays ; these rays divide the line into four parts, as follows: The part above /, 6 is on the interior or convex side of the surface, which is readily seen by drawing a ray tangent to the surface at some point above these points, e.g. in the meridian plane of rays through /, /s2 (see § 45). Such a ray is within the scotia and intersects the curve cut from the surface by the plane in a point on the side opposite the point of tangency. At the point / (and 6) the point of tangency and the point of inter- section of the ray coincide, hence at this point the line of shade changes from the convex to the concave side of the sur- face. The parts J 2 ) and 6^4 are on the concave side while the remainder (below ^, 4) is on the interior or convex side. We have shown that the shadow of C on the surface intersects the line of shade at the points 2 and 5; hence the part of the line of shade 2 i 6 ^ (i. e. all above 2,5) is in shadow and cannot cast a shadow even though the parts 2 / and 5 6 are on the concave side of the surface, hence the only portions of the entire line which cast a shadow are the parts 2 5 and 5 4. The shadows of 2 _J and 5 4 are partly on the coordinate planes and partly on the surface. The shadow on the surface is deter- mined by the method of §§ 30 and 42, thus : Find the shad- ows on H by individual rays ; they are 2s ^^ and 5^ 4^. Pass any horizontal secant plane as U below the points _J, 4, it cuts the circumference L from the surface. The shadow oiLon H is Zs, which intersects the shadow of the line of shade at ^s2 andysji a ray through each of these points intersects Z at g^i SHADOWS OF CURVES. 37 and/si, hence these points are points of the required shadow. The shadow thus determined is J^<^, only part of which is vis- ible as shown. The shadow on HisJ^i limited by Qi and A^ which is the shadow of the base A (see below.) Fourth : The shadows of the element Z and W a.ve Z^ and W^ respectively, and of the upper base A is A^, as shown. As the lower base D is in H, it is its own shadow. The complete shadow both on the surface and the coordi- nate planes is now determined. The visible shadow on the coordinate planes is bounded by the lines Z^As/siC^i Y^B^^ B^ , thence along the V contour of the surface to G. L., thence along G. L. to 4> thence CsiJs\As W^ A^. The visible shadcAV on the surface is, in V projection, bounded by the lines Z^, part of A"', Jl^, 4", ^ , Cjj, thence along the V con- tour of the surface to f, C^ Y", thence along B' to r", r^ u^ , thence along the ^contour ol the sJrrace Lo G. L., thence along G. L. to t^ In H projection the visible shadow of the surface is bounded by the lines y"2 C^J^^^- 47. Problem XVII. To find the shadotv on the siirface of a triangular-threaded screw. Axis vertical. Plate IV, Fig. 22. Analysis. The surfaces of a triangular-threaded screw are warped helicoids, and the outside and the root of the thread are helices. Planes tangent to a warped helicoid at points on the same helix make equal angles with the axis and equal angles with H. (See Des. Geom.) If a series of meridian planes of the helicoid be passed perpendicular to these tangent planes, the lines cut from the tangent planes make a constant angle with H equal to the angle which the tangent planes make with H; and with the elements passing through the points of tangency a constant angle, since the relative positions are the same in all cases. Hence the angle between the H projection of any 38 SHADES AND SHADOWS. line of intersection of the meridian and tangent planes and the ^projection of the corresponding element is constant. If now a plane of rays tangent at some point of the given helix be determined, the element containing the point of tangency and hence the point of tangency may be determined. Having determined the line of shade, construct the line of shadow by individual rays. Construction. Let A be an element of one of the helicoidal surfaces of the screw and C the helix on which a point of the line of shade is to be determined. The element A pierces H at a, and a tangent line to the helix C aX b (the intersection of A and C) pierces /f at (^, hence H T \s the horizontal trace of a tangent plane. H U is. the U trace of a meridian plane perpendicular to the tangent plane T, hence the angle x oj is the constant angle made by the H projection of the intersection of the two planes with the H projection of the element passing through the point of tan- gency. The angle which D, the line of intersection of the planes T and U, makes with H is equal to the angle which any tangent plane to the helicoid at a point of the helix C makes, with H. Revolve D about the axis X; a cone is generated with vertex at {d', o) and base Q in H. A ray M through the vertex of the cone pierces ZT at ^; hence ^Z*, through ^ and tangent to the base, is the H trace of a plane of rays tangent to the cone, and hence is parallel to the //"trace of a plane of rays tangent to the helicoid at some point of C. The H trace of a meridian plane perpendicular to the plane P is o/; hence making the angle yoin^ = angle x oy we have the element which passes through the required point of tan- gency. This element intersects the helix C at m; hence m is the point of tangency and therefore a point of the line of shade. SHADOWS OF CURVES. 39 A point of the line of shade on the helix B is determined in the same manner, thus : H Q'ls the /^ trace of a tangent plane through the element F on which is the point ti of B, and o i is the //"trace of a perpendicular meridian plane; hence ioii^ is the constant angle. The line G is the intersection of the me- ridian and tangent planes and pierces H aX i; hence C^ is the base* andy, o the vertex of a cone generated by rotating G about the axis, and H R is the H trace of a plane of rays tangent to the cone. The line o k is the H trace of a meridian plane perpendicu- lar to the plane R ; hence laying off the angle k o g^ = i ou^ we have the element which intersects the helix B 3Xg, thus deter- mining another point of the line of shade. To determine an intermediate point of the line of shade : Let K^ be the H projection of an intermediate helix. For convenience pass an auxiliary horizontal plane Z intersecting the helix at w. The intersection with the plane Z oi a. tangent plane to the helicoid, through the element A at some point of the hehx K, is L and of a perpendicular meridian plane is oz; hence zoy is the constant angle. The intersection of the me- ridian and tangent planes is H, which revolved about the axis generates a cone with base T on the auxiliary plane Z. U is the intersection with Z oi z. plane of rays tangent to the cone, and ^7- of the perpendicular meridian plane; hence laying off the angle r o t" = angle z o y the element containing the re- quired point t is determined. The Fprojection of t is t" on the V projection of the element (not shown). The line of shade on the lower thread \s gtm and part of the helix B on the right, of^. As the several threads of the screw are alike in all respects, * Almost coincides with C^. 40 SHADES AND SHADOWS. g^ t^ m^ is the H projection of all the lines of shade on the front heHcoids, hence npq is the line of shade of the second thread. The visible line of shade is composed of the lines in tg, np q and part of the helices on the right of ^ and n. The Shadow. The shadows of npq and N on H are q^a, ps2 , etc. (this line is on the right of the border, and hence is not shown). The shadow of any element P of the helicoid .Zis P^. This shadow intdrsects the shadow of the line of shade at/sj; hence passing a ray through ps2 the shadow p^i on the ele- ment P is determined. (See § 30.) In this manner the shadow on V projection q^ p^si «s 4 was determined. The shadow on the lower thread is of exactly the same form. The JI projec- tions of the shadows and the shadows on H are not shown. Problems. 1. Find the shadow of an oblique line on an oblique plane. 2. Find the shadow of a line parallel to G. L. on an oblique plane. 3. Find the shadow of a line situated in a profile plane on the coordinate planes. 4. Find the shadow of a pyramid, base on H, on the coor- dinate planes. 5. Find the shadow of an inverted pyramid, base parallel to H, on the coordinate planes. 6. Find the shadow of a pyramid, base parallel to H, on the coordinate planes. 7. Given a square pyramid on a square pedestal, find the complete visible shadow. 8. Given a square pyramid on a cylindrical pedestal, find the complete visible shadow. 9. Find the shadow of any parallelopipeji on the coordinate planes. SHADOWS OF CURVES. 4 1 10. Find the shadow of a hollow hexagonal prism on the interior surface and on the coordinate planes. 11. Find the shadow of a vertical cross. 12. Given a vertical cross on a cylindrical pedestal, find the complete visible shadow. 13. Given a square column surmounted by a square abacus, find the complete visible shadow. 14. Given a hexagonal column surmounted by a square abacus, find the complete visible shadow. 15. Find the complete shadow of a rectangular box with its lid partly raised. 16. Given a book-case containing three shelves, find the complete shadow. 17. Given a square-topped table with four square legs, find the complete shadow. 18. Find the complete shadow of any building having the plan and elevation. 19. Given a hollow cylinder, axis parallel to G. L., find the complete visible shadow. 20. Given a hollow cylinder, axis perpendicular to H or V, find the complete visible shadow. 21. Find the shadow of a cone, axis perpendicular to V, on the coordinate planes. 22. Find the shadow of any oblique cone on the coordinate planes. 23. Given a hollow cone, axis parallel to G. L., base on left, find the complete shadow. 24. Given a vertical cylinder surmounted by a square abacus, find the complete visible shadow. 25. Given a cylindrical column surmounted by a cylindrical abacus, find the complete visible shadow. 42 SHADES AND SHADOWS. 26. Given a cylindrical column surmounted by a hexagonal abacus, find the complete visible shadow. 27. Find the shadow of an ellipsoid of revolution, axis per- pendicular to H or V. 28. Find the shadow of the frustum of a cone. 29. Find the complete shadow of the frustum of a hollow cone. 30. Given a small vertical cylinder intersecting a large hori- zontal cylinder, find the complete visible shadow. 31. Find the complete visible shadow of a groined arch. 32. Find the complete visible shadow of a cylindrical niche capped by a quarter sphere. 33. Find the shadow on the interior of a recess of any shape. PERSPECTIVE. CHAPTER IV. GENERAL PRINCIPLES, DEFINITIONS AND CONVENTIONS. 48. Perspective in Nature. The phenomena of perspec- tive in nature are so familiar that they scarcely require descrip- tion. St.and at one end of a long straight street and look towards the other end ; the rows of trees on the sides, the fronts of the buildings, the sidewalks and curb-stones, seem tO' approach each other and grow gradually smaller as they recede from the spectator. Straight railroad tracks make an excel- lent example of perspective, especially if they be several miles long. If you stand on one rail and look along the tracks all the rails will apparently meet it at a distant point. At the sea shore or on shipboard the sky and water seem to meet at the horizon — the effect is the same as if the sky and water were 'two vast horizontal planes. It is this phenomenon or effect of distance which the draftsman tries to reproduce in a perspec- tive drawing. 49. Deductions. Examples of natural perspective might be multiplied indefinitely, but all teach the same facts, z. e., that any series of parallel lines if sufficiently produced appar- ently approach each other and meet in a point, and that any series of parallel planes if sufficiently extended apparently ap- proach each other and meet in a line ; also, that if a line be 43 44 PERSPECTIVE. passed through the eye parallel to any series of parallel lines, it will pass through their meeting point, and that if a plane be passed through the eye parallel to any series of parallel planes, it will pass through their' meeting line. As lines or planes may be extended in two directions an infinite distance from any given point, there are two meeting points or lines 1 80° removed from each other for each series, but as the spectator can only look in one direction at a time it is not necessary to consider the second case. In the case of parallel planes the meeting line would become the circumfer- ence of a circle because planes may be infinitely extended in all directions. 50. Methods. The method of perspective in general use by draftsmen is called One-plane Perspective, and difiersin many essentials from that used by artists, which is called Cylinder or Multiplane Perspective. The drawing produced by the One-plane method is the same as that which would be made by placing a sheet of trans- parent paper between the object to be represented and the spectator, and covering or concealing every line which is seen on the object by a line on the paper. NOTE : For purposes of representation the paper might just as well be placed be- hind the object as in front of it, although not usually so con- sidered. In the first case the drawing is smaller than the ob- ject, in the latter case larger. In Multi-plane Perspective a transparent cylindrical surface is supposed to be placed between the spectator, whose eye is in the axis of the cylinder, and the object ; lines are then drawn covering or concealing those on the object, and the cyl- inder is developed. This method is essentially that used by artists, but the difficulty of considering it geometrically is great and the practical application tiresome. GENERAL PRINCIPLES, DEFINITIONS, ETC. 45 51. Relation to Geometry. Suppose a plane placed be- tween a given point and an object, and the object projected on the plane by means of lines drawn from the point to the object ; the drawing obtained will be a duplicate of that made by the process described in § 50. In this case the natural method has been considered geometrically. The lines from the point to the object, i. e., the projecting lines, form a cone because they radiate from a point, hence the object is conically projected. If the object to be projected be a point, the pro- jecting cone becomes a line ; if the object be a line, the pro- jecting cone becomes a plane, etc. 52. Definition. One-plane * Perspective is Conical projection. Note : If the student will remember that One-plane Perspective mtSins projection from a fixed center upon a plane, much of the difificulty sometimes found with the subject may be avoided. In Descriptive Geometry the projecting lines are parallel, in Perspective the projecting lines radiate from a point. 53. Picture-Plane. The Picture Plane (also called Per- spective Plane and Plane of Projectioii) is the plane upon which the object is projected or the drawing is made. It is usually assumed vertical. 54. Perspective. The conical projection of an object on the picture plane is called the Perspective of the object. Any point or line in the picture plane may be considered to be the Perspective of some point or line in space. 55. Station Point. The Station Point (also called the Point of Sight) is the point from which the object is projected on the picture plane. It is the vertex of the projecting cone of the object. This point may, in general, be placed anywhere within a finite distance of the picture plane, but the position * Also called Linear Perspective. 46 PERSPECTIVE. once selected must remain fixed for any one drawing. In order that a perspective drawing may appear natural or give one the correct impression of the object represented, the eye of the spectator must be placed at the station point, otherwise the drawing will appear distorted. 56. Center of the Picture. The Center of the Picture is the orthographic projection of the station point on the picture plane. 57. Axis. The Axis of the picture is the perpendicular projecting the station point on the picture plane. 58. System. Any series of parallel lines is called a System of lines ; any series of parallel planes is called a System of planes. 59. Axis of a System. The Axis of a System of planes is any line perpendicular to the system. 60. Vanishing Point. Given a System of lines and a point : If a plane be passed through each line and the point the common intersection of the planes is parallel to the System of lines. Given a System of lines, the Station Point and the Picture Plane : If a plane be passed through each line of the system and the station point, the intersection of each plane with the picture plane is the Perspective of the line through which the plane is passed (§ 54). But these planes have a common intersection which pierces the picture plane in a point, and the intersection of each plane with the picture plane passes through this point ; hence the Perspective of each line passes through this point. The point of intersection of the perspec- tives of a system of lines is called the Vanishing Point of that system or of any line of that system. The Vanishing Point of a line may also be defined as the perspective (§ 54) of that point on the line which is at an in- finite distance from the station point. Since perspective is conical projection and parallel lines intersect only at infinity. GENERAL PRINCIPLES, DEFINITIONS, ETC. 47 the projection (perspective) of this point of intersection on tlie picture plane must be determined hyaline through the station point parallel to the given line. 6i. Determination of Vanishing Points. Since the common intersection of the projecting (conical projection) •planes of a system of lines is parallel to the system and passes through its vanishing point (§ 60), for any system of lines the "vanishing point is determined by the intersection with the picture plane of a line of the system through the station point. 62. Vanishing Points of Systems of Lines Parallel to the Picture Plane. The perspectives of a system of lines parallel to the picture plane are parallel to each other and to the system, for the common intersection of their projecting planes is parallel to the picture plane and therefore intersects it only at infinity. This fact limits but does not violate the preceding rule (§61), 63. Vanishing Trace. The Vanishing Trace of a plane is the locus of the vanishing points of all lines in the plane. 64. Horizon. The Horizon is the vanishing trace of hori- zontal planes or the locus of the vanishing points of horizontal lines. It is the horizontal line in the picture plane which passes through the center of the picture (§ 56). 65. Determination of Vanishing Traces. For any sys- tem of planes the vanishing trace is determined by the inter- section with the picture plane of a plane of the system through the station point. (See § 61.) 66. Relative Position of Vanishing Points and Van- ishing Traces. From §§ 6i and 63 it follows that the vanish- ing point of a line is in the vanishing trace of any plane which contains that line. Also, that if a line be the intersection of two planes its vanishing point is the intersection of the vanish- ing traces of the two planes. 48 PERSPECTIVE. 67. Initial Point. The Initial Point of a line is the point in which the line pierces the picture plane. As the picture plane is usually assumed vertical, initial point is only another name for vertical trace. Throughout this book the initial point of a line is the vertical trace of the line, and the initial line (§ 68) of a plane is the vertical trace of the plane. As the initial point of a line is in the picture plane it is one point in the perspective of the line. 68. Initial Line. The Initial Line of a plane is the line in which the plane intersects the picture plane. The initial line of a plane is parallel to the vanishing trace of the plane (§§ 63 and 65). 69. The Initial Point of a line determines the position of the perspective of the line. The Vanishing Point of a line determines the direction of the perspective of the line. 70. Principal Lines of an Object. The Principal Lines of an object are the systems of lines which most frequently appear in the representation of the object. Take, for example, a building such as a factory: the princi- pal lines shown in the drawing belong at most to but three systems. First : On the front of the building, the horizontal lines, such as the courses of brick, water-tables, window-sills, etc. Second : On the side of the building, the horizontal lines corresponding to those on the front. Third : On the front and side, the vertical lines, such as the edges of the building, window- frames, etc. Hence in order to make a perspective drawing of an object which consists essentially of systems of right lines, it is neces- sary in general to determine but three vanishing points. 71. Position of Object to be Represented. Objects which have three (or less) systems of principal lines may be placed, with reference to the picture plane, in three general GENERAL PRINCIPLES, DEFINITIONS, ETC. 49 positions. First : With the systems parallel to, and perpendic- ular to, the picture plane. Second : With one system parallel to, and two systems oblique to, the picture plane. Third : With the three systems oblique to the picture plane. 72. Parallel or One-Point Perspective. The repre- sentation of an object situated as in Case I, § 71, is called Parallel or One-Point Perspective, because but one vanishing point is required to make the drawing. 73. Angular or Two-Point Perspective. The repre- sentation of an object situated as in Case II, § 71, is called Angular or Two-Point Perspective, because two vanishing points are required to make the drawing. 74. Oblique or Three-Point Perspective. The repre- sentation of an object situated as in Case III, § 71, is called Oblique or Three-Point Perspective, because three vanishing points are required to make the drawing. 75. Notation. (i) The position of the station poiitt is denoted by e. (2) The center of the picture is denoted by e^ (see § 56). (3) I" general, when the projections of an object are attached to the drawing, the perspective of appoint is denoted by the letter of the point with the exponent ' ; e. g., a', b', etc. (4) The perspectives of a system of lines are denoted by the letter of the system with the exponents ', '', ', etc. ; e. g., R", R^ etc. (5) The vanishing point of a system of lines is denoted by a capital V with an exponent which is the letter of the system, e. g., VK, yP etc. (6) The vanishing trace oi a system of planes is denoted by a capital T succeeded by the letters of any two systems of lines in the planes, e. g., T R P., T M N, etc. 5° PERSPECTIVE. (7) The initial point of a line is denoted by a capital I with an exponent which is the letter of the line; e.g., P, P, etc. (8) The initial line of a plane is denoted by a capital I suc- ceeded by the letters of any two lines in the plane, or of any line in the plane, or by the letter of the plane ; e.g., I R B, I M, I E, etc. (9) K point of distance for a system of lines is denoted by a capital D with an exponent, which is the letter of the system ; e. g, D^ D^ etc. (10) A point of proportional distance for any line is denoted by a capital P with an exponent, which is the letter of the line; e.g., P^ P^ etc. 76. Perspective is Descriptive Geometry. Plate V, Fig. 23. This figure is the Descriptive Geometry representa- tion of three lines, Q, W, and U, and of a plane P, passed through the intersecting lines Q and W. To find the vanishing points of the lines Q, W, U, and the vanishing trace of the plane P. Plate V, Fig. 231^. Let the picture plane be the V coordinate plane of Fig. 23, and also use the H coordinate plane of the same figure. Let e^, e^ be the or- thographic projections of the station point (§ 55), which is about one and one-half inches in front of the picture plane, e' is the center of the picture (§ 56). Through the station point pass a line parallel to Q: it pierces the picture plane at V^ ; hence V*^ is the vanishing point (§§ 60 and 61) of Q and of all Hues parallel to Q, i. e., of the system Q. In the same manner the vanishing points V^ and V^ of the lines f/and W^ respectively, were determined. The line e^W^ is the horizon, since U a.r\d Q are horizontal (§ 64), and T W Q) passing through V^ and V^, is the vanishing trace of the plane P and of all planes parallel to P, i. e., of the system P, since it contains the vanishing points of two lines of GENERAL PRINCIPLES, DEFINITIONS, ETC. $1 that plane (§§ 65 and 66). TWQ is parallel to VP (Fig- 23)- It is thus evident that the determination of a vanishing point is the determination of the V trace of a line passed through a given point, and that the determination of the vanish- ing trace of a plane is the determination of the J^ trace of a plane which passes through two given lines intersecting at a given point. 77. To find the perspective of a rectangular prism by direct projection. Plate V, Fig. 26. Analysis. Since perspective is conical projection or projec- tion from a given point upon a plane (§§ 51, 52, 54), it follows that if a line be drawn from the given point (station point, § 5S) to each point of the figure, that the intersections of these lines with the plane (picture plane, § 53) give the perspective of the figure. This is the same as finding the V traces of a number of lines which radiate from a point. In this method both pro- jections (//' and F) of the object are required, and are usually placed so that the object is in either the second or third dihedral angle. A much simpler solution than the above, which may be cahed direct projection, will be explained in the next chapter. Construction. Let ab dcik gfa be a rectangular prism situated in the second dihedral angle. The top and bottom faces are parallel to H and the other four faces are vertical planes inclined at angles of 60° and 30° to V as shown. The station point (§ 5S) is ^ and the picture plane (§ 53) is V. From e draw a line to a : it intersects the picture plane ( V") at d\ hence d is th.& perspective of the point a. From e draw a line to /fe: it intersects the picture plane (F) at ^'; hence k" is the prospective of k. In like manner draw lines to all the corners of the prism and find their intersections with the picture plane (F). 52 PERSPECTIVE. These intersections or perspectives joined in the proper manner give the perspective d ¥ d^ c" i' k' g^ f^ d of the prism. Note the following facts : The perspective lines «"/', c' i\ d' U,3.x\A F g' converge and meet at V^', which is the vanishing point of these lines (§ 60). This is as it should be, because the corresponding Hnes on the prism are parallel, and hence must have a common vanishing point. Through e (station point) draw a line parallel to one of these lines on the prism, e. g., af: it intersects the picture plane at V^, which is the same vanish- ing point as before. Hence we reach the same result whether we begin by determining the vanishing point of a system of lines (§ 61), or whether we find the vanishing point from the per- spectives of the lines. Similarly the vanishing point of the hnes inclined to the right is V^ (not shown, about one inch to right of border line on V^^'*'). The perspective lines d i>\ c' d\ t' k\ a.ndf'g' are parallel, which is correct, because the correspond- ing lines on the prism are parallel to the picture plane (V); hence their vanishing point is at infinity, for if a line be drawn through the station point (e) parallel to these lines it will inter- sect the picture plane only at infinity. The systems af, c i, etc., and ac,bd, etc., are horizontal, hence the line V*^ e^ is the hori- zon (§ 64) or the vanishing trace (§ 63) of the planes afica.nd bgk d, because it contains the vanishing points (V^ and V^) of two or more lines in each plane. This is also true from the fact that the two planes afic and bgkd are horizontal ; hence if a plane be passed through the station point parallel to these planes it will intersect the picture plane in the line V'^ e', or the horizon. As the planes abgf a.nA cdki are vertical their vanishing trace must be perpendicular to G. L. And as V is the vanish- ing point of a line in each plane, Y^ z is the vanishing trace of GENERAL PRINCIPLES, DEFINITIONS, ETC. 53 these planes. This fact might also have been established by finding the vanishing point of any other line in either plane, £. g., one of the diagonals of either of the two faces. The V trace of the line af, or the point in which the line pierces the picture plane, is q^; hence cf is the initial point .(§ 6"]) of the line af. The perspective of this line, «'/"', passes through q^, because q"' is a point in the line and also a point in the picture plane, in other words, if and the perspective of (f •coincide. Similarly m' is the initial point of /? and the perspective /' «' passes through it, also li' is the initial point of gk and the perspective g^ ¥ passes through it. Since iif and tC are the initial points (F traces) of two lines in the plane y"_^/^2, the line in^ vf iC must be the initial line (§ 68) or f^ trace of the plane fgki. Also m^ q' is the initial line of the plane a fie. The line in^ (f is horizontal, which is correct, since the vanish- ing trace V'^^'^ is horizontal, and hence parallel to in'' q'^ (§ 69). We thus establish the fact that the vanishing trace and initial line of a plane are parallel, which must be true because the plane which determines the vanishing trace is parallel to the plane which has the initial line and the intersections of parallel planes with the same plane are parallel. The line in^ ri' which is the initial line of the plane f gk i is also parallel to the vanishing trace (not shown) of the plane. The line if cf is the initial line of the plane afgb be- cause it contains the initial point f of a line in the plane and is parallel to N^ z which is the vanishing trace of the plane. We might also have determined this initial line by drawing a line through if parallel to a b, since the initial point o{ ab is at infinity. Note also the fact that the initial point of a line is at the intersection of the initial lines of any two planes which con 54 PERSPECTIVE. tain the line, e.g., nf , and that the vanishing point of a line is at the intersection of the vanishing traces of any two planes which contain the line ; e.g., V"^ is the vanishing point of af and the intersection of if^V'^ and ^•V'^. For the duplication of this problem by the method of measurement see Prob. I, § 95. CHAPTER V. THE MEASUREMENT OF LINES IN PERSPECTIVE. 78. The perspectives of lines which lie in the picture plane coincide with the lines, and hence are equal to them in length. In general, however, a line and its perspective are unequal in length. If by any means we may transfer distances from the per- spectives of lines which are actually in the picture plane to the perspectives of other lines which are not in the picture plane, a method of measuring or laying off distances in perspective is effected. 79. Proposition I. If parallels to the base of a triangle be drawn, the corresponding segments on the two sides are in a constant ratio. (See Plate VII, Fig. 34.) If the triangle be isosceles the ratio is unity, hence the cor- responding segments on the two sides are equal. (See Plate VII, Fig. 35)- 80. Proposition II. A line drawn through the intersection of the diagonals of a parallelogram parallel to two of its sides bisects the other two sides and the parallelogram. (See Plate VII, Fig. 36.) This method of subdivision may be continued as shown in the figure. 81. Proposition III. If one side of a parallelogram be di- vided in any way at one end, equal divisions may be laid off at the other end by means of two diagonals. (See Plate VII, Fig- 37-) 82. To duplicate Proposition I in perspective. Plate V, Fig. 24. Isosceles triangle. 55 56 PERSPECTIVE. Analysis and Construction. Let a z x \)& an isosceles tri, angle situated in the third dihedral angle and having one of its equal sides, a x,in V (or picture plane.) Assume G. L. to be the horizon, e the station point, and e'' the center of the picture (§ 56). Find the perspective of the triangle. The vanishing point oi a z {/) is V'' found by passing a line through the station point e parallel to / (see § "jj). Since a' is in the picture plane it is the initial point of y, and hence one point in the per- spective ofy; hence d^'V'' is the indefinite perspective of J. The vanishing point o{ x z {F) is D'', found by passing a line' through the station point parallel to F, and x^ is the initial point of F; hence, ;i:''D'' is the indefinite perspective of the line. The hnes a'V'' and x^ T)^ intersect at z\ which is therefore the perspective of the point z. Since a^f is in the picture plane, a^ x^ is the perspective, or in other words the line is its own perspective. Hence a^ z^ x^ is the perspective of the triangle az x. But the triangle a z x '\% isosceles, hence d^ x^ = d" s' in perspective, that is, the per- spective length d' z^ = the true length d' x''. Parallels to the base (i^) of the triangle vanish at D'^; hence if we lay off any distance on d" x^, as for example a^ c =: 1/2", and draw c m D'', we will again have an isosceles triangle in which a^ c = 1/2" = a^ m in perspective. Also lay off any other distance as a^ d= i^', draw ^D^ intersecting J' at/; then a' d— ij" = d^ p in perspective. Notice that T J F is the vanishing trace of the plane of the triangle because it contains the vanishing points of two sides of the triangle, and that d' x^ is the initial line of the plane of the triangle, and also that T J F and cC x"^ are parallel (see §§ 68 and Tj). As the chief requirement is to find a means to lay off dis- THE MEASUREMENT OF LINES IN PERSPECTIVE. 57 tances on the perspective /', any line in the picture plane other than a x which would complete an isosceles triangle having a z for one of its equal sides may be used. Let ay = ax = as he such a line; then we have the isosceles tri- angle a 2 J. The vanishing point of the base /^ is D'''; hence n' z'y" is the perspective of the triangle azy, and we may use the side a"'y"', which is in the picture plane, to lay off distances to be transferred (by parallels to the base) to the perspective side/'. 82a. Usually in a problem where measurements are nec- essary we have the following data (Fig. 24) : The vanishing point of the line as V-', the indefinite perspective of the hne as y ', the vanishing trace of a plane containing the line as T J F, the horizon as G. L. {G. L. is the horizon only for convenience), and the station point e. It is required to transfer measure- ments to the line y in perspective by means of an isosceles triangle. We know that the line T J F is parallel to the in- itial line of the plane of the triangle, and hence is parallel to one of the equal sides ; we also know that the line joining the station point with the vanishing point V''^ of the line is parallel to the line and hence parallel to the other equal side ; hence all we are required to find is the base or vanishing point of the base of the triangle having for its equal sides e N^ , and part of T J F equal in length to e V'', to form a triangle whose sides are re- spectively parallel to the triangle in space which has the side /. Revolve the plane containing T J F and e about T J F (which is the F trace of such plane) into F (picture plane): the station point e revolves to e^, and the line eN^ to ^, V''. Now lay off V'^ D'' = V'' ^,, and join D-^ e^, and we have an isosceles triangle V''D''^£j. But D''^ is a point in the picture plane, and hence the vanishing point of the base e D'' of the 58 PERSPECTIVE. isosceles triangle e V'' D'^. We have thus determined the same point, D'', for the vanishing point of the base of the isosceles triangle that we had in § 82. The point D-'' or D'' is called a Point of Distance i^ 83), and the line (f x" or a"" y is called a Line of Measures (§ 87). 83. Point of Distance. A Point of Distance for a given oblique line is the vanishing point of the base of an isosceles triangle whose equal sides are the given line and the initial line of any plane which contains the given line. From § 'is2a we may derive the following definition : A Point of Distance for a given oblique line is the vanishing point of the base of an isos- celes triangle whose equal sides are, the line from the station point to the vanishing point of the given line, and the vanish- ing trace of any plane containing the given line. 84. Since a line may be contained in an infinite number of planes having an infinite number of vanishing traces, it follows that a Point of Distance for a line may be any point in the picture plane situated as far from the vanishing point of the line as the vanishing point of the line is from the station point. 85. When- the given line \% parallel to the picture plane its vanishing point is at infinity; hence the equal sides of the isosceles triangle are parallel, and the point of distance may be any point on the vanishing trace of a plane containing the line (see §§ 95 and 97). 86. Point of Half Distance. Plate V, Fig. 24. Suppose the lines revolved, etc., as in § 82«. Bisect V D"^, and draw e^ D^ ; a triangle e, D^'' V-^ is formed, of which the side V'^D^'^ is one-half e^ V-'. Di-^ is the vanishing point of the side e Di'' of the triangle. Hence if we lay off on a^ x" distances to half scale and draw lines to Di"^ intersecting /', the same subdi- visions are effected as in § 82 ; e.g., lay off a' f= \ inch, draw fiW intersecting /' at m, then a' m in perspective = 2^"'/, THE MEASUSEMENT OF LINES IN PERSPECTIVE. 59 because the sides of the triangle a7 fm are parallel respectively to the sides of the triangle e D^'^ V'^. The point D^'^ is called a Point of Half Distance. If one-quarter of V'' e^ were laid off on V'' D'' from V'', a Point of Quarter Distance would be ob- tained, and the distances ox\ ax would be laid off to quarter scale. This process may, of course, be continued indefinitely, and Points of Three-Quarter Distance, One Tenth Distance, etc.,, obtained. 87. Line of Measures. A Line of Measures for a given line is the initial line of the plane containing the given line and its point of distance. Hence a line of measures for an oblique line is a line in the picture plane which passes through, the initial point of the line and xs parallel to the line joining the vanishing point of the given line with its point of distance (see § 8212). 88. Plane of Measures. A Plane of Measures is a plane in which the measurements which are to be transferred to the- perspective are made. Throughout this book the Plane of Measures is supposed to- coincide with the picture plane. 89. Scale. The scale to which the drawing is made may be anything convenient, such as i inch to 10 feet, i inch to i foot, etc., or even i inch to i inch, and depends entirely upon the purposes for which the drawing is required. A decimal or chain scale is, however, preferable to any other although not used in practice. 90. Proportional Measures. It is often necessary to di- vide a given line into a certain number of equal or propor- tional parts irrespective of the actual length of any part. For this purpose a method somewhat shorter than the method of equal measures (§ 82 et seg.) is given. 6o PERSPECTIVE. To duplicate Proposition i in Perspective. Plate V, Fig. 25. Any triangle. See § %2a. Analysis and Construction. Let V"^ be the vanishing point of a line, and T J the vanishing trace of any plane passed through the line/. Revolve the plane of T J and e about TJ as an axis into the picture plane : the point e revolves to e^, and the Hne e V-' to e^ V-'. Draw any line as e^ P^ thus cojnpleting a triangle e^ V' P-^. Revolve the triangle back to its true position, and P''^ is the vanishing point of the base e '?^ . Let V b be the perspective of a line of the system /which is to be divided into three (or any number) of equal (or un- equal) parts. The point V is the initial point of/', but as far as the solu- tion of the problem is concerned may be anywhere on the perspective line /'. Through P draw P^ parallel to T J, draw Y^ b d: then the perspective triangle V b d'vs, similar to the tri- angle ^ V'' P''. Divide P(3?into three equal parts (the method of geometric division is shown), and from the points of division draw / P''^ and ^ P'' intersecting/' at m and n respectively: then V p : p q : etc., : : V nt : m n : etc., in perspective. The point P-"^ is called a Point of Proportional Distance, and the line F d is called a Line of Proportional Measures. Any other point as P''' might have been taken on T J, and the same subdivisions affected, as shown in the figure. Instead of using the line F d through the initial point of /, we may use any other line parallel to T J, as in g, and obtain the same result. 91. Point of Proportional Distance. A Point of Propor- tional Distance for an oblique line is the vanishing point of the base of a triangle, the sides of which are the given line, and any line parallel to the vanishing trace of a plane which con- tains the given line. THE MEASUREMENT OF LINES IN PERSPECTIVE. 6r 92. Line of Proportional Measures, A Line of Propor- tional Measures for an oblique line is a line drawn through any point of the given line parallel to the vanishing trace of any- plane containing the given line. 93. Of any two perspective lines having the same vanishing point, one may be taken as the trace of a plane passing through the other ; and if a third line be drawn parallel to the first through any point of the second, any parts taken upon this third line may be transferred to the second in their true propor- tions by means of a point of proportional distance taken upon the first.* 94. It is evident from the preceding paragraphs that lines which are parallel to the picture plane may be sub-divided directly. * Modern Perspective, by W. R. Ware. CHAPTER VI. PARALLEL, ANGULAR, AND OBLIQUE PERSPECTIVE. 95. Problem I. To find the perspective of a rectangular prism. Angular perspective. Plate V, Fig. 27. Analysis. This problem is a duplication of § "jy (Fig. 26) by the method of measurements described in the last chapter. The prism is supposed to rest on a horizontal plane through G. L. which corresponds to the line vi" (f of Fig. 26. All dimensions are given in Fig. 26. The principal (and only) lines of the prism are the edges which belong to three systems (§ 70). Four of the edges are horizontal and inclined to the right [K lines), four are horizontal and inclined to the left (iV lines), and four are parallel to the picture plane and perpen- dicular to H {A lines). Hence the drawing is to be made in angular perspective (§ 73), and only the two vanishing points of the horizontal edges are required. Construction. Let V V^ be the horizon, assume the station point 2y (= e^ X, Fig. 26) in front of the picture plane, and let e^ be the center of the picture (§ 56). Consider a horizontal plane passed through V^ e^ N^ ; then e^ is the projection of the station point on this plane, and also its true position, and e^ e^ = 2|-". This horizontal plane through the horizon is shown in a number of the following problems and is useful in determining vanishing points, etc. Since K and N are horizontal, their vanishing points are on the horizon (§ 64) ; hence by drawing through e^ a line K making an angle of 60° with V V^ the vanishing point V^ of 62 PARALLEL, ANGULAR, AND OBLIQUE PERSPECTIVE. 63 the edges inclined to the right is determined, and through i\ a line TV perpendicular to ^the vanishing point V^ of the edges inclined to the left is determined. A point of distance for ^is D'^, found by laying off V^ D^ = V^ e^ on the horizon (§§ 82-84), and a point of distance for N is D'', found by laying off V^ D'^ In practical work it is usually much easier to compute the positions of vanishing points and points of distance than to determine them by geometric methods. In the present problem the angles which the edges make with the picture plane are given ; hence ^'^ V^ = 2.5 tan 30" = 1.44", e^Y^ = 2.5 tan 60° = 4.33", V D^' = .^'^ ,=: 5.00", and V^ D^ =^^^=2.89". ^ ^^ sm 30 sm 60 ^ These distances may be laid off at once. Since the prism rests on a plane through G. L., this line must be the initial line (§ 68) of the lower face of the prism ; and as it is parallel to the horizon which contains the vanishing points and points of distance of the horizontal lines of this face, it is also a line of measures (§ 87) for this face. The point a (Fig. 26) is i|-" to the left of the station point and 3/4" behind the picture plane. Hence lay off ji/ P (Fig. 27) = i-|". The perspective of the point « is on a perpendicular to the picture plane through P, and as all perpendiculars vanish at e^ (V^), 1^ e^ must contain the perspective of the point. A point of distance for 1^ e"' is V''^"^, found by laying off ^vY45''L_ g^g'^^ since the perpendicular is in the plane of the (lower) face. Lay off l^J — 3/4" and draw/ V*'°-^ intersecting I^ e^ at rt' ; we thus have an isosceles triangle formed in which I^y= P«' in perspective; hence «' is the perspective of the point required (§ 82). Draw the indefinite lines K\ N\ and A^ for the indefinite perspectives of the edges of the prism ■which intersect at a (Fig. 26). 64 PERSPECTIVE. We know, that the initial points of the lower edges are on C. Z., because G.L. is the initial line of the lower face ; also that this line is a line of measures and we have determined the points of distances for TV' and K\ The initial point of N' is F', found by producing N' to G. L. Draw D*' a^ p intersecting N^ at d ; then P' dp is an isosceles triangle in which P'/ = P' a} in perspective. Lay off / m = i^" (= a/, Fig. 26) and draw m D^ intersecting yV' at/'; then/ m — «'/' in perspective, being corresponding segments on the equal sides of an isosceles tri- angle (§§ 79 and 82), and the perspective of the lower left edge is determined. It was of course unnecessary to find the initial point P' except to show how the isosceles triangle was formed as we wished to lay off a distance (rom a\ To lay off the length of K' draw D^ a' n, lay ofi ng— 1/2" (=: ac, Fig. 26),. draw ^D"^ intersecting K' at <:', then «^= «'c' in perspective. We may now draw N' K", which intersect at i\ and thus com- plete the perspective of the lower face. Draw the indefinite lines/' ^', t ' k\ and c' d^ perpendicular to G. L. {i. e., parallel to the corresponding lines of Fig. 26) for the perspectives of the edges which are parallel to the picture plane. To lay off /'^' : The point in is the initial point of the line /«/' D^ (see before), and this line passes through the point /' of /'_^'. The initial poifit of /'^' is at infinity; hence mr parallel to/'^' is the initial line of a plane containing/'^', and also a line of meas- ures for/ g^. Lay off m r = 3/4" {—fg, Fig. 26), draw D" r intersecting/'^' at ^', then ;« r = /' _§-' in perspective. This is true, because mr and/'^' are actually parallel, and mD^ and r D^ are parallel in perspective. Note : The measurement of a line parallel to the picture is effected by laying off distances on the initial line of any plane containing the line (§ 85), and transferring them by means of PARALLEL, ANGULAR, AND OBLIQUE PERSPECTIVE. 65 parallels to the perspective of the line. In the case of lines parallel to the picture plane the initial line {?, parallel to the perspective line. Similarly we may lay off' c';^' as follows : P'^ is the initial point of N", which passes through the point c' of c^ (P; hence JN2 iN4jg ^ ]if,g Qf measures fortV, to which it is parallel. Lay off F^F^ = 3/4" (= cd, Fig. 26), draw piV intersecting c^ d^ at d\ then c' ^"^ is the perspective required. Draw N^ {e-'d') intersecting A' at 5', also K' {d' d'), N' and /^ thus completing the perspective of the prism, which is a duplicate of Fig. 26. Instead of erecting the perpendiculars to G.L., a' b\ f g^, i' k\ and c' d', and laying off the height of the prism upon one of them, we may find the perspective of the upper face directly as follows : Since the prism is 3/4" high and the plane of the up- per face is horizontal, lay off m r = 3/4" and draw r F^ par- allel to G. L.; then r 1™ is the initial line of the upper face, and also its line of measures, and we may duplicate in this plane the work of finding the perspective of the lower face. After the perspective of the upper face is determined by this method join the corresponding points in the two faces to complete the perspective. 96. Object at 45°. When an object is so placed that a number of its principal lines (§ 70) are horizontal and inclined at an angle of 45° to the picture plane, the determination of the vanishing points and points of distance is much simplified, as these points are symmetrically placed with reference to the center of the picture. The vanishing points of horizontal lines making angles of 45" with the picture plane are on the horizon at a distance from the center of the picture equal to the length of the axis (§ 57) ! hence, if the center of the picture and one of these ^ PERSPECTIVE. vanishing points be given or assumed, the position of the sta- tion point is known. Plate VI, Fig. 28. The horizon is the horizontal hne passing through e", the center of the picture, and V^°^ and V*""^ are the vanishing points of horizontal lines making angles of 45° to the right and left. These points are usu- ally denoted V''^°» and V*=°^ or simply V^=°. V' '^ e" — V«°i-^^ = length of axis = 3 inches (in this case) ; hence the station point is 3 inches in front of the picture plane. T W J and T F O are the vanishing traces of diagonal planes (§ 44), and T Y Z is the vanishing trace of profile planes. Lines in pro- file planes and making angles of 45° with the picture plane vanish at V^ and V^; hence V«°« e' = e" N^ = e' V^ = V^=°^^\ and V*=° V^ = V"° V^ = etc. = the distance that the station point is from V*°°^, or V*^°^ = the distance that a point of distance (§§ 83 and 84) for R, L, Z, or Y lines is from V*=°^, V*^°^, V^, or V^ respectively. With V*'°^ as a center and radius = V^^°^ V^ describe a circle (only an arc is shown); this circle is the locus of the points of distance for the system of lines vanishing at V*^°^. Similarly with V'''°^ as a center and radius = V^'° V"*^ de- scribe a circle which is the locus of the points of distance for the system vanishing at V*'""^- The symmetry of the 45° posi- tion is thus shown. The vanishing point of any line in a diagonal plane may be readily determined if the angle which it makes with a horizon- tal plane be known, as follows : The line if inclined to the right vanishes in T W J, if inclined to the left it vanishes in T F O (§ 63). If the station point be revolved about T W J into the pic- ture plane, D^^°^ is its revolved position, since V*^°^ qw-r _ V^'^ e ; hence through D*^°'* draw any Hne, as D*^°^ V'', making PARALLEL, ANGULAR, AND OBLIQUE PERSPECTIVE. 6/ the given angle (30° in this case) with the horizon. V'^, the intersection of D*'°^ V'' and T W J, is the vanishing point required. Similarly V^, V°, and V'^ are the vanishing points of lines in diagonal planes inclined down to the right and up and down to the left respectively, and making angles of 30° with a horizontal plane. We might at once have laid off the distances Y45°K v"^ = V*^°^ V-' = V^"^ V° = V*'°'^ V^ = 3 V2 tan 30" = 2.45". V'' D*^°^ = V'^f; hence, if with V'^ as a center and ra- dius V'' D^°^ we describe a circle, the locus of the points of distance for the system vanishing at V''^ is obtained. Notice that this circle passes through V^. Why ? T F R, T L W, TOR, and T L J are the vanishing traces ■of planes containing horizontal lines inclined at angles of 45° to the picture plane, and lines in diagonal planes inclined at an- gles of 30° to the horizontal plane. Another advantage in placing an object at 45° is that in this position the station point is at a maximum distance from the picture plane, and the drawing is concentrated within nar- rower limits; e.g., Plate V, Fig. 27, the object is placed at 30° and 60°, and the station point is at the distance e^ e^ from the picture plane ; if now the station point be moved to ^,, its dis- tance from the picture plane is ^^ D^ > e^ e^, and the vanishing points, V^ and V*^, used for the 30°, 60° position, are now used for the 45° position, since they are equally distant from ^j, and V^ e^ V^ is a right angle, being inscribed in a semicircle. 97. Different Ways of Measuring the Same Line. Plate VI, Fig. 28. Analysis and Construction. Let J be the indefinite perspec- tive of a line which vanishes at V''^, and let R be the perspec- tive of its horizontal projection on a plane through I^; then P is the initial point of R, and I W J perpendicular to the hori- 68 PERSPECTIVE. zon (i. e., parallel to T W J) is the initial line of a plane contain- ing R and J. Note the fact that the vanishing point "of a line and the vanishing point of its horizontal projection are in the same perpendicular to the horizon. Produce J until it intersects I W J at F, which is the in- itial point oij. It is required to lay off 2.5" on y from the point/. Since V'' is the vanishing point of the line, TWJ, T L J, and T Q J, all passing through V, may be considered the vanishing traces of planes containing the line, and V^, D''^, and V on the " locus of D''" are the corresponding points of distance (see § 96). {a) We have already found the line of measures I W J parallel to T W J; hence draw N^ p i inter- secting I W J at i, from i lay off ik ^ 2.5", draw k V^ inter- secting y at q; then/^ in perspective = 2.5". [B) Through F draw I L J parallel to T L J for a line of measures, draw D''//, from / lay off /^= 2.5" and draw^^^D""; then /) ^ in perspective = 2.5" as before, (c) Through F draw I Q J par- allel to T Q J, draw V/ r, from r lay off r ^ = 2.5" and draw tqY'^; then/ ^ in perspective = 2.5". Three ways of obtaining the same result have thus been shown. The measurement of the line L, which is in a horizontal plane through F, is also shown. V^ is a point of distance, be- ing on " locus of D^ " (see § 96), and F 3, parallel to T L U and passing through F, the initial point of the line, is a line of measures ; then w x in perspective = j/ ^. Let A be the perspective of a hne parallel to the picture plane and B the perspective of its projection on a horizontal plane, of which G. L. is the initial line. It is required to lay off a certain distance on A from n to the left. Suppose we measure the hne by parallels (§ 85) situated in profile planes (compare with § 95). [NOTE: We make this as- PARALLEL, ANGULAR, AND OBLIQUE PERSPECTIVE. 69 sumption only for convenience in drawing; any other assumed position of the parallels would have answered the purpose.] As the vanishing trace of profile planes is T Y Z (a line through e^ perpendicular to the horizon), the point of distance or vanishing point of the parallels to be used must be on T Y Z. Through n, I pass a profile plane, its initial line is I N M; hence a line of measures for A must have one extrem- ity in I N M, because this line is the initial line of a profile plane through one extremity of A, and hence must contain the parallel which passes through n. Assume V**^, or any point on T Y Z, as the vanishing point of the parallels, draw V**^ n d in- tersecting I N M at of; then ^ is the initial point of this line. Through d draw dj parallel to A for a line of measures, lay off the required distance dj and drawy'^^V"^; then e ^ 3.5" (= 7' to scale). Draw P' e^ , the indefinite perspective of one of the edges of the base ; on G. L., the line of measures corresponding to V*'°^ lay off F' I" = 0.5" (= i' to scale), draw I" Y'^"^ in- tersecting P^^^ at b; then b is i' behind the wall or picture- plane. Lay off F' F* = 2" (= 4' to scale), draw p-* e', draw A^ parallel to G.L. intersecting P* e^ at a; then A^ is the per- spective of the near edge in the floor. The diagonal of the base inclined to the left vanishes at V^^"-"^, because it is a hori- zontal line inclined at an angle of 45° to the picture plane ; hence draw b V^'""^ intersecting P^ e^ at k, and the perspective of the edge P^ is determined. Draw A* parallel to A^ and in- tersecting /" at/", thus completing the perspective of the base. We might have laid oft piy= 2" and drawn/ V^^"'^ intersecting- P' aty, etc. The line \^^ m perpendicular to G.L. is the in- itial line of the face /", B\ and hence a line of measures for £^. -Lay off Pi l^ = 2" (= 4' to scale), draw P^ e' and £\. which intersect each other at c; then 3 c is the perspective of the edge £\ Draw B', B', B* perpendicular to G. L. and par- allel to B^ ; also draw P\ A\ P\ and A^ to complete the figure (see § 95). The diagonals- of the upper and lower faces vanish at V-^' and V^ (see before). The diagonals Z of the side faces vanish at V^, since e^ V^ = e^ Y^ = 3.5", and the other two diagonals (not shown) vanish at V^. Since Z^ and Z' vanish at V^, and PARALLEL, ANGULAR, AND OBLIQUE PERSPECTIVE. ",'1 A' and A' are parallel to the picture plane and horizontal, the vanishing trace of the plane ^4' Z'' ^^ is TQZ; and since D vanishes at N^, and B^ and B'' are vertical, the vanishing trace of the plane D B^ is T L Q; hence V^, the intersection of these two vanishingtraces, is the vanishing point of the diagonal Q of the cube (§ 66). Similarly the diagonal W^ vanishes at V^, etc. (c) Measurement of the Diagonals. The vanishing trace of a plane containing Z'' is T Y Z ; hence D^, found by laying off V^ D^ =. V^ Vi- = 4.9s" (§ 84), is a point of distance for Z\ Since I"»? is the initial line of the plane of the face /-" B^ (see before), it is the line of measures (§ 87) corresponding to D^ ; hence draw D^ bn and D^ I in intersecting I^' in at n and in re- spectively ; then n m = 2.83" is the true length of the diagonal Z ^. The vanishing trace of a plane containing R} is the horizon, since R} is horizontal; hence the required point of distance is D«°^ found by laying off V^ D« = V^ D^ = 4.95". G. L. is the line of measures; hence draw 'D^fp and Ti^aq intersecting G.L. at p and q respectively; then /^ = »;;/ = 2.83" is the length of R required. Since the diagonals W^ and Q^ inter- sect, T W Q is the vanishing trace of a plane containincr W^, and D^, found by laying off V^ D^ = the hypotenuse of a triangle of which V"^ ^"^ and e e" are the sides = 6.06", is the point of distance. Since P^ is the initial point of P", it is one point of the initial line of the plane W^ , Q\ which contains P' ; hence I W" C parallel to T W Q is the line of measures. Draw D^/r and D^/z z intersecting I W C at r and 2 respect- ively ; then rt — 3.46" is the length of W\ T L Q is the vanish- ing trace of a plane containing Q\ and D<5 is a point of distance since V intersecting I C at/, then np ^= r i (Fig. 30) = 3-46". The diagonal C is horizontal and vanishes at V ; hence the vanishing trace of a plane containing G^ is the horizon, and D"*, found by laying off V« D» = V* V^ = -^^ = 4.20", is the cos d "f ^ ' point of distance. A^ is the line of measures, hence draw T}^ gq intersecting I G' M' (or A-) at q ; then b q— pn = 3.46" is the length required. The Jf diagonals of the two faces in- clined down from A"- and A^ vanish at V'", found by laying off yY yM _ ^v yw _ 4.^5"^ since these diagonals are in the faces of the cube inclined 45° to the picture plane. Similarly the iV diagonals vanish at V^. The diagonal J/' is in the plane A" A''; hence I G' M' {A) is the initial line and line of meas- PARALLEL, ANGULAR, AND OBLIQUE PERSPECTIVE- 75 ures. A point of distance corresponding to this line is on G. L. at a distance from V™ = 7", hence draw D** a r ; then b r — 2.83" = 3 « in perspective. The diagonal M'' is in the plane A^ A'. The initial point of a line {¥') in this plane is P', hence I M'' Y* parallel to A' is a line of measures; then by us- ing D'" we obtain xj = 2.83" for the length of M\ A point of distance for iV' is D'* (on G. L.), found by laying off VN D" = 7" ; then dt =. 2.83" is the length of N\ {/) Figs. 32 and 33. The two upper cubes are symmetric- ally placed with reference to T Y Z, and similiarly placed with reference to the picture plane and ceiling; hence the explana- tion for one suffices for the other. One edge of the cube is vertical and 1.5' behind the picture plane. To find this edge draw the perpendicular i c^, lay off k z -^ 0.75", and draw ^r V*°^ intersecting ke' -aX a; then a is- 0.75' behind the picture plane, and the vertical line E" is the indefinite perspective of the edge. The L and R edges are horizontal and inclined at an angle of 45° to the picture plane; hence they vanish at V*^°^ and V*^°^ respectively. Draw U and R\ A point of distance for Z' is D*^"^ and T Q Z is the line of measures, since the face L' R' is in the ceiling. Draw D'^ a m, from m lay oS m n = 2", draw n D^ intersecting L' at d; then ad is the perspective sought. A line of measures for i?' is I U' B', hence lay off zj> = 1" and draw/ V^ intersecting B^ at b; then ab m perspective = 2". A line of measures for 72' is T Q Z and the point of distance is D^, hence draw D^ a r and from r lay off rj = 2", draw j D^ intersecting i?' at c\ then a c\s the perspective of the edge required. The cube is completed by drawing i?^ D ; then E', R', etc. {g) Measurement of the Diagonals. The C/ diagonals van- ish on T W R at V^ (not shown), 4.95" below V^- A point of distance, D^, is found by laying off V^ D^ = 7". The line •j6 PERSPECTIVE. of measures for Z7, \s,zq, hence draw VP aw and W fq inter- secting ^■^ at w and q respectively; then wq =/^ (Fig. 30) = 2.83" = the length of the diagonal U ^. The diagonal /' van- ' ishes at V'^ (not shown, 4.9S" above V^), a point of distance is D-f on T W R, and the line of measures is ^r ^ (I U' B'). It is evident that D-' c intersects zq beyond the limits of the drawing, hence we employ a point of half distance Vi^l^ , found by laying off D^D^/^^ = 1/2 D^ V = 3.5". Draw Yy'ftUs and D1/2J h X (the point x is just above D) intersecting z q ^\. s and TV Q X respectively; then .y;f = — - =1.42" = one half the length of y. The diagonal (9' vanishes at V^, and a point of distance is found by laying off V^ DO = V° V«°. Since C, is in a profile plane which contains the edges B^ and E", a perpendicular to the picture plane through a determines the line of measures I O' B' parallel to T Y Z. Draw D° ay and D° Ih intersecting I O' B' at y and h respectively; then y h = ri (Fig. 30) = 3.46". The length of the diagonal K^ isgi (on T W R) = 2.83". (See § 105 for construction of the circles inscribed on the faces of the cube in Fig. 33.) 99. Successive Perspective Plans. It is often necessary in order to save confusion in the drawing and render the construction more simple to make a perspective plan of the object at some distance either below or above the desired posi- tion. The plan of a large building, for example, is frequently quite complicated, and as at least half is invisible from a given position, it is unnecessary to show that portion in the finished drawing. The complete plan, however, is needed in order to obtain the details of the roof, the chimneys, etc. Plate VI, Fig. 29. Angular perspective. Given the vanish- ing points, etc., as shown, also given a plan. The perspective PARALLEL, ANGULAR, AND OBLIQUE PERSPECTIVE. 77 drawing is made to the same scale as the plan in this particular case. The plane of the plan is horizontal and the point a is in the picture plane. The points a, b, c, etc., of the plan correspond to the points a', (5\r', etc., of the perspective on the horizontal plane through G. L., and the method of measurement is readily traced. Take any new ground line, as G^. Z,., parallel to G. L.\ from d erect a perpendicular to G. L. (and H) intersecting G^.L^. at «^ From a' lay out the perspective of the plan by measure- ments as before. Since a' was determined by a perpendicular to G. L. through a^, all the other points, as U', c\ d'^, etc., might have been similarly determined; e.g., draw i?', erect the per- pendicular b'' b^ intersecting 72' at b'' ; draw D and a perpen- dicular through c' intersecting U at /, etc. Hence it is evident that measurements were unnecessary to determiine the second perspective plan. The geometric statement of the preceding is : If a cylinder (general sense) be perpendicular to H, all hori- zontal sections are equal, and the horizontal projections of these sections coincide with the base of the cylinder on H. 100. Problem III. To find the perspective of a cube. Ob- lique perspective. Plate VIII, Fig. 38. Analysis and Construction. All the faces and edges are ob- lique to the picture plane (§ 74). Each edge is 2" long. Let F, R, Q be three edges of the cube intersecting at the point a which is in the picture plane. Assume the vanishing points V"^ and V^ of the edges F^ and Q' respectively; then T F Q is the vanishing trace of the plane of those edges. Assume V'^, the center of V^ V^, to be the vanishing point of the diagonal N'' which bisects the angle F^ Q. From these assumptions it follows that the center of the picture is on V^ V-'' drawn through V perpendicular to T F Q. Assume e^ as the center of the picture; hence the axis (§ 57) is one side of a right- 78 PERSPECTIVE. angled triangle of which the other side and hypotenuse are e"" V^ and V^ Y^ respectively. With V^ as a center and radius V'^ V^ describe an arc of a circle V^ e^ V^, draw e' e^ perpen- dicular to V^V^ intersecting the arc at f,, then ^"^^, is the length of the axis. The edgei?' is perpendicular to the face F^ Q and to the diagonal iV'; hence i?' vanishes on V*^V^ De- scribe an arc of a circle through V*^ e^ with center in V^ V^; then V^ is the vanishing point of R', since V'^ e^ V^ is a right angle, being inscribed in a semicircle. e^ V^ is the distance of the station point from V^; hence with V^ as a center and V^ e^ as radius describe a circle which is the locus of the points of distance, D^, for the system R. With V^ as a center and radius V^ e^ describe a circle, thus ob- taining the locus of D'5; and with V^ as a center and radius V^ e^ describe a circle, thus determining the locus of D^. We may now proceed to draw the cube. Through a, the vertex in the picture plane, draw Q, Ji\ F\ the indefinite perspectives of the edges which intersect at a. Draw the line of measures I Q' R' through a parallel to T Q R, lay off a n = 2" and draw n D^ intersecting i?' at 3 ; also lay off « / = 2" and draw t D*^ intersecting Q' at d; the perspec- tives of two edges are now determined. Through a draw the line of measures I F' R' parallel to T F R, lay off a m = 2" and draw ;« D^ intersecting F^ aty, thus determining the perspective of the third edge. The drawing is finished as in the preceding problems, as all the necessary measurements have been made. Instead of assuming e'' we might have assumed V^ ; then ^^ may be found as follows: Describe semicircles through V^, V, V« ; y% V^, V« and V«, e„ V^. From the points of intersection o and/ of these semicircles draw oV^ and/ V*^ intersecting at e'', the center of picture required. Let the student prove this statement. PARALLEL, ANGULAR, AND OBLIQUE PERSPECTIVE. 79 lOOa. Second Solution. Fig. 3812:. The essential points in either solution are the determination of the position of the station point (having assumed the vanishing points of two edges which are not parallel), and the position of the third vanishing point. Suppose the vanishing point of the diagonal iV' does not bisect V V^ (Fig. 38), but has the position shown in Fig. 38a. To find the position of the station point: On V^ V^ de- scribe a semicircle which must contain the revolved position (i?j) of the station point, since F'^ and Q are perpendicular. A line from e^ to V must bisect the right angle between F^ and Q ; hence the angle V^ e^ V^ = the angle V^ «■, V = 45°. Hence in the triangles b e^r and r e^x we have the following pro- portion : zr:r X =^ sin § : sin a = cos a : sin a = cot a. Through r draw re = rx and perpendicular to zx, draw^rc; then cot a = — . The line eic intersects the semicircle at e,, which is there- r c fore the revolved position of the station point in the picture plane ; hence the center of the picture may be anywhere on the line fj (f drawn perpendicular to V^V^. The position of the third vanishing point readily follows. CHAPTER VII. PERSPECTIVE OF CIRCLES. loi. The problem, to find the perspective of a circle, is the Descriptive Geometry problem, to find the intersection of a cone with a plane (§ 51). In perspective the intersecting plane is the picture plane, and the cone is given by its vertex (the station point) and circular base in a plane which is usually oblique to the picture plane. 102. In general the perspective of the center of a circle is not the center of the perspective of the circle, {qx the same reason that in general the axis of a cone does not pass through the center of an oblique section. 103. Proposition IV. If any variable tangent to a central conic section meet two fixed parallel tangents, it will intercept portions on them, whose rectangle is constant and equal ' to the square of the semi-diameter parallel to them.* Plate VIII, Fig. 39- Construction. Let <« t be a diameter of the ellipse abed, M and Q tangents at the extremities of the diameter a c, and iV any other tangent intersecting Jf and Q at _/ and j' respect- ively; then afy. cg=^ {oby. Hence, if we have the tangents to any diameter of an ellipse and any third tangent, the length of the semi-conjugate diam- eter may be readily found (Fig. 39a:). af— af (Fig. 39), fc=gc (Fig. 39). On ac asa diameter describe a circle, draw * Salmon's Conic Sections. 80 PERSPECTIVE OF CIRCLES. 81 fm perpendicular to ac, then {fmy = {ob)" (Fig. 39) = afx. fc. 104. If we have six or more tangents to a small ellipse and the corresponding points of tangency, the ellipse maybe drawn accurately without finding diameters. 105. Problem IV. Given a cube in perspective to describe circles on the visible faces. Plate VII, Figs. 33 and 33(3;. Method of tangents (§ 104). Analysis and Construction. For the construction of the cube see § 98/. It is required to inscribe circles (in perspec- tive) in R'R\ R^ R\ and E> B\ the visible faces of the cube. Fig. 33«. To find the tangents and points of tangency (§ 104) inscribe a circle in a square (2" sides) of the same size as that of the face of the cube. Draw the diagonals n a and m h and the four tangents to the circle which are perpendicular to these diagonals, and we have eight tangents and eight points of tangency, y, c, d, i, 0, k,g, and /, which are to be transferred to the perspective. Fig. 33- The lengths of D and R^ were laid off by means of the line of measures T Q Z ; in nla the length of L\ and r g- is the length of i?'. Hence lay off np — np (Fig. S3a),po = po, q =^ q and qm =^ q m. From the points p, 0, q draw lines to D'^ intersecting L\ and thus transfer the distances laid off on m n to U. Similarly for R^. The same subdivision may be effected by proportional measures (§ 90) as follow^ : Through a draw at, a. line of pro- portional measures, parallel to the horizon, on a t lay off «;r„ xt, etc., = ax,xt, etc. (Fig. 33a) respectively, draw t c F^ ; then P^ is a point of proportion distance. From the points of subdivision on a t draw lines to P^ intersecting i?' a.tj,j, and s,. thus determining the same points as before. To divide B', which is parallel to the picture plane (§ 94), 82 PERSPECTIVE. proceed as follows : Through a (Fig. no) draw ab = ab '(Fig. 33), join bh, from the points of division draw parallels to b h intersecting ab zs shown. Transfer the distances thus obtained directly to 5' (Fig. 33), and the required subdivision is effected. Complete the octagon on each face and draw the diagonals to determine the points of tangency not on the edges of the cube. Note the fact that the tangents (Fig. 33«)/ /^ and/t are parallel to nih, and hence have the same vanishing point, and that q i and sf are parallel to n a, and hence have the same vanishing point. On each face draw a smooth curve (ellipse) tangent to the octagon at the determined points of tangency, and thus obtain the required solution. 106. Problem V. To find the perspective of a circle. Plate VIII, Figs. 40-43« inclusive. General Analysis. The same horizon, vanishing points, etc., are used for all the figures. The center of the picture is e^, and the station point is 4" in front of the picture plane i. e., ^'^ V^° = 4". The diameter of each circle is 1.5". («) First Case. When the circle is in a vertical plane, and the perspective of the center is on the principal horizon. Fig. 40. Analysis. Circumscribe a square about the circle with its sides vertical and horizontal. Assume that the horizontal sides vanish at V*^°^, thus placing the circle in a vertical plane inclined 45° to the picture plane, and assume that one of the vertical sides is in the picture plane. The circumscribed square will give two parallel tangents to the perspective of the circle, and two tangents which are not parallel ; hence diameters of the elliptical perspective are readily found by § 103. Construction. Let ab= 1.5" be the side of the square in the picture plane, and R' and R^ the indefinite perspectives of the horizontal sides. Since the perspective of the center is to PERSPECTIVE OF CIRCLES. 83 be on the horizon, ap = bp. Through b draw a Hne of meas- ures, b n, parallel to the horizon, lay o'S. bin ■= m n = 0.75", and draw m Yi^ and n D^ intersecting i?' at f and c respectively. Draw //^ and c(/ perpendicular to the horizon; then abcd\s the perspective of the circumscribed square. Since /^ is per- pendicular to the tangents a b and c d, it is an axis of the ellipse. Through 0, the center oi p q, draw go r parallel to a b, and lay ofi or = (7^ = Vdg X <^p for the major axis. An ellipse con- structed oh p g aiid^r as axes is the perspective of the circle required, e^ in this case coincides with the perspective of the center of the circle; hence the axis of the picture (§ 57) is the axis of the projecting cone of the circle. (a) Second Case. Fig. 40a. Analysis. The position of the circle is so chosen that the picture plane cuts a sub-contrary section from the projecting cone. One point of the circle is in the picture plane. A circumscribed square is used as before. Construction. Let the plane of the circle be inclined at an angle of 60° to the picture plane towards the right. The per- spective of the center is 2.31" (4 tan 30°) to the left of e^ at D^, since the axis of the projecting cone must be inclined at an angle of 60° (in this case) to the picture plane in order that the picture plane cut a sub-contrary section. The triangle lying in a horizontal plane through the horizon and formed by the axis of the projecting cone, the horizon, and a diameter of the given circle, is equilateral (in this case) ; hence to find the point of the circle in the picture plane lay off D*^ r = 0.75", and r is the point desired. Through r draw arb=^ 1.5" for the side of the circumscribed square which is in the picture plane. Draw iV' and iV". Through b draw b m parallel to the horizon, draw m D*' intersecting N at c, and draw c d parallel to « (5 to complete the square. Bisect r/ at 0; with (7 as a center and radius o r describe a circle, which is the perspective required. 84 PERSPECTIVE. (a) Third Case. When the circle is in a vertical plane and the perspective of the center is not on the horizon. Fig. 41. Analysis and Construction. Let the perspective be a du- plicate of Fig. 40, except that the center is below the horizon. Let a b , on a b (Fig. 40) produced, be the side of the circumscribed square in the picture plane, draw R^ and R' lirB- ited by dc (Fig. 40) produced. Then a b c d is the perspective of the circumscribed square. Through/, the center of ab, draw R". Produce ab to t, making b t =^ c g, on p t describe a semi- circle, draw b r perpendicular to p t\ then b r \s the length of the semi-diameter conjugate to p g (§ 103). Through i?, the center of / ^, drawee / parallel to a (5, and lay off ^, and M"-, which is its own shadow. The per- spective of the visible shadow on the upper base of the paral- lelepiped is bu and a small part of K!'. 115. Problem VIII. Given a vertical rectangular column near a small building having a projecting roof : To find the per- spective and the perspective of the shadow. Angular perspective. Plate X, Fig. 48. Analysis. The vanishing points, etc., are as shown, and the station point is 4" in front of the picture plane. One edge of the column is in the picture plane, and from the center of the rear edge of the upper base a guy-line {F) extends to the ground. The column is in 30°, 60° perspective, and the build- is in 45"^ perspective as shown. The special problem is to de- termine the perspectives of the shadows of the column and guy-line on the building. Construction. Let abed, f B\ B' be the perspective of the column with the guy F^ running from /to the ground, which is a horizontal plane through G. L., at g. Also, let the perspective of the building be as shown, with m^ j^ P k^ m^ for the perspective plan (§ 99) of the roof. The vanishing trace of the plane of the left side of the roof is T L J, and of the right side is T L W. V'^ is on T W R above the horizon. The shadow of the building. The line of shade is composed of the lines /', /^ L\ L\ A", A\ and g v^ 2 l\ The indefinite perspective of the shadow of A' is v^ V^- The perspective of the shadow of m (the corner of the roof) is Wg, found by § 113, i. e., the intersection of the perspective of a ray through m with the perspective of the ff projection of a ray through m^. Since L^ is horizontal and vanishes at V", its shadow on H vanishes at V", hence through m^ draw m^ V"^ intersecting v^ V at Ps, then v^ ps is the perspective of the shadow of A^. The vanishing trace of a plane of rays through y is T W R, hence PERSPECTIVE OF SHADOWS. 93 the shadow of /" on H vanishes at V^. Draw R\ through n draw the perspective of a ray intersecting E^ at n^, then m^ n^ is the perspective of the shadow of /^ V vanishes at V^ hence fig V^ is the indefinite perspective of its shadow. The per- spective of a ray through o intersects «s V^ (Z's) at Og. Through Os draw R* for the perspective of the shadow of y. The re- mainder of the shadow on H is invisible. The vanishing point of the shadow of V on the right side of the building is V^, the intersection of T W R and T L X, hence through p, the inter- section of the plane A^ A'' with IJ, draw X' for the perspective of the required shadow on the right side of the building. Through v draw v V^ for the perspective of the shadow of U on the left side. Shadow of the Column. Since E" and B' are vertical their shadows on //vanish at V^, hence through a and b draw /?' and K respectively. The perspectives of rays through x and c intersect i2' and E^ at x^ and c^ respectively, thus determining the perspective's of the shadows of E" and B' on H. Since x d vanishes at V^, its shadow on H vanishes at V^, hence draw x^ V^ ; and since d c vanishes at V*, its shadow on H vanishes at V*, hence draw V* ^s d^ intersecting x^ V^ at d^. We now have the complete shadow of the column on H in perspective. The intersections of planes of rays through B" and B'' with the side A^ A^ of the building are parallel to B^ and B^ and hence are parallel in perspective, therefore draw B\ and B\ for the perspectives of the shadows of B^ and B^ on the side of the building. Since itf- k^ is the perspective plan of L\ the plane m^ m k k^ is parallel to A" A', hence erecting perpendiculars from 5° and r", the shadows s and r of B^ and B' respectively on D are determined. The vanishing trace of a plane of rays through B^ or 5' is T W R and the vanishing trace of the left side of the roof is T L J, hence the vanishing point of the 94 PERSPECTIVE. shadows of E" and B" on this side of the roof is V-' (not shown), the intersection of T W R and TL J. Through s and r draw y andy* respectively for the required perspectives. Similarly the perspectives of the shadows of B\ and B' on the right side of the roof vanish at V^, the intersection of T W R with T L W, hence through i and h draw iN^ and h V^ respect- ively for the required perspectives. These two lines are limited by;7. Shadow of the Guy. /^vanishes at V^, hence the vanishing trace of a plane of rays through F \% T F N. The vanishing point of the shadow of ^ on H is V^ (not shown), the intersection of T F N and the horizon, hence draw N^ for the perspective of the shadow on H. Or, find the perspective of the shadow of /on H and join with g. The shadow of F on the H projecting plane of D, or the plane in^in k/^, van- ishes at V^, hence t u is the perspective of the shadow of F on that plane. The vanishing point of the shadow of F on the left side of the roof is the intersection of T F N and T L J (not shown, beyond the left border), hence through u draw E\ The vanishing point of the shadow of F on the right side of the roof is V^ the intersection of TFN and TLW, hence from the point where £■' intersects Z" draw a hne to V^ and limit it by the intersection with //. Or, to find the shadow of the guy on the roof, produce W to y and draw yw\^, which is the perspective of the intersection of the plane of the right side of the roof with H. N^ intersects yY^ at w, hence through w draw V^w intersecting D, etc. Il6. Illustration. Plate XI shows the perspective of a C. B. & Q. R. R. standard furniture car drawn by the methods out- lined in the preceding paragraphs. The scale is f" = i'. The principal lines are in 30°, 60° angular perspective ; the 30° van- ishing point is about 23" to the right of e^, and the 60° vanishing PERSPECTIVE OF SHADOWS. 95 point is about 7" to the left of e'. e' is 2.5" above O (see drawing), and the station point is 13" (nearly) in front of the picture plane. No part of the car or track is in the picture plane. Problems. Each problem may be stated in six different ways, depending upon the position of the object relative to the picture plane. {a) Parallel perspective, {b) Angular perspective, (c) Oblique perspective; (i) Part of the object is in the picture plane, (2) No part of the object is in the picture plane. When the state- ment of the problem is succeeded by the letters {a 1), etc., it means that the drawing is to be made in parallel perspective {a), and that a point or line of the object is in the picture plane (i), etc. Each problem is to be solved in the positions (« i), {a 2), (b i), and [pi), unless otherwise stated. All dimensions are to be given and the perspectives are to be drawn to scale. 1. Find the perspective of a pyramid, base on H. 2. Find the perspective of an inverted pyramid, base par- allel to H. 3. Find the perspective of a vertical hexagonal prism. 4. Find the perspective of a hexagonal prism resting on one face in H. 5. Find the perspective of a vertical cross. 6. Find the perspective of a square column surmounted by a square abacus. 7. Find the perspective of a rectangular box with its lid partly raised, {c i), {c 2), etc. 8. Find the perspective of a bookcase containing three shelves. 9. Find the perspective of a square-topped table with four square legs. 96 PERSPECTIVE. 10. Find the perspective of a flight of steps. 11. Find the perspective of a triangle situated in an oblique plane, {c i), {c 2), etc. 12. Find the perspective of a vertical circular cylinder. 13. Find the perspective of a circular cone, axis vertical. 14. Find the perspective of an ellipsoid of revolution, axis vertical. 15. Find the perspective of any building. 16. Find the perspective of a groined arch, (a i), (« 2). 17. Find the perspective and the perspective of the shadow in any one of the preceding problems. APPENDIX. Shading. 117. Hypothesis.* («) All objects to be shaded are to be considered unpolished ; the character of the surface being simi- lar to that of a plaster cast. {b) The sun is the only source of direct light and is fixed in position (§ 5). (c) Objects are also illuminated by reflected light from the minute particles which are everywhere present in the atmos- phere. Were it not for reflected light the shade of an object would be black ; hence these particles illuminate the shade or portions in shadow and cause them to be visible. Reflected light, being much less intense than direct light {U), does not cause shadows and does not affect the line of shade. (d) For the illumination of the shhde the minute reflecting particles (c) may be considered, with but slight error, to be con. centrated in a point diametrically opposite the sun or source of light. (See f and g^ (e) The only kinds of light which are supposed to reach an object are from the source {b) and from atmospheric particles {c). All surrounding objects which might reflect light and thus * This method of shading is due M. Jules Fillet, Ecole Nationale des Fonts et Chauss6es, to whose works the student is referred for a more extended treatment embracing the effect of distance, color, etc. ^7 98 APPENDIX. alter the appearance of the given object are considered to be removedi (/) The apparent illumination of an object depends only on the intensity of the source of light and the angle which the rays make with a normal to the surface. ig) The illumination of any element (any very small area) of a surface is proportional to the cosine of the angle which the incident ray makes with a normal to the surface at that element. Ii8. Deductions. Using the sphere as a typical surface, we make the following deductions from § 117. (See Plate XII, Fig. 49.) {a) The apparent darkest part of the sphere is the line of shade o, because the cosine of the angle between an incident ray and normal is a minimum, i. e., = O, at any point of the line ; the apparent lightest part is the point 8, because the cosine of the angle between the incident ray and a normal at this point is a maximum, i. e., — i. From 8 to o the shading gradually increases in depth, and from o to — 5 it gradually decreases in depth ; the part O to — 5, however, is much darker than the part o to 5, since it is unilluminated by direct light. {b) If any object be placed so as to exclude light from the sphere, the shadow is as much darker as that portion would be lighter without the interposed object, for the shadow is illuminated only by reflected light from atmospheric particles. 119. Method of Shading. Plate XII, Figs. 49 and 49a. Let the diameter of the sphere which coincides with a ray be divided into sixteen equal parts, and let planes be passed per- pendicular to the diameter (ray) through the points of division. Fig. 49« shows the lengths of the axes of the projections of the circles cut by the secant planes. Call the line of shade o, the ellipses (in projection) which SHADING. 99 are illuminated by direct light i, 2, 3, 4, etc., and the ellipses (in projection) which are illuminated only by reflected light — I, — 2, — 3, — 4, etc. We are now ready to apply the tints of india ink to produce the proper shading. {a) To obtain the tint to be applied to the shade rule a small sheet of white paper with equidistant lines, thus : so that the width of each line is one half the width of the white space between any two lines. In a small beaker place six portions of water (a brushful may serve as a measure) and enough india ink to obtain a shade equivalent in depth to that produced by the equidistant lines on the ruled sheet. Call this tint No. o and apply to the shade of the object, i. e., be- tween the o line and the right contour. Add one portion of water to tint No. o, call this tint No. i, and apply between the lines I and — i after the o tint is dry. Add two portions of water to No. i, call this tint No. 2, and apply between the lines 2 and — 2 after No. i is dry. Add three portions of water to No. 2, call this tint No. 3, and apply between the lines 3 and - 3 after No. 2 is dry, etc. This method gives a banded appearance to the drawing. In order to give a uniformly decreasing depth of tint proceed thus ; First place the No. o tint over the portion in the shade as before and allow this tint to dry. Then between the lines I and —1 apply tint No. i, but instead of allowing it to dry "draw out" the edges with a brush moistened with water, and thus eliminate the sharp lines which would otherwise appear at I and — i; after the "drawing out " is effected and the lOO APPENDIX. tint is allowed to dry lay on tint No .2, "draw out," etc. Either method is readily acquired after a few preliminary trials. (^) Suppose an object placed between the sphere and the sun so as to cut off direct light from a small portion of the area included between the lines 5 and 7. This portion is now in shadow and must be shaded according to the statement of § \\U. First, over the area inclosed by the line of shadow (or sim- ply shadow) lay on tint No. o. Second, lay on tints Nos. I, 2, 3, 4, 5, 6, 7, as if the lines 5, 6, 7 were in the shade; /. e., over 7 to 6 lay on tints Nos. i, 2, 3, 4, 5, 6, 7, arid over 6 to 5 lay on tints Nos. i, 2, 3, 4, 5, 6. Third, over the shadow lay on a tint stronger than No. o, call this tint No. S. Fourth, to tint No. S add one portion of water, call this tint No. 6', and Jay on between 7 and 6 ; to tint No. 6' add two portions of water, call this tint No. 5', and lay on between 7 and 5, etc. 120. Shading of Any Surface. The proper shading for any surface may be determined from the sphere type (Plate XII, Fig. 49) by inscribing a tangent sphere in the surface and finding the intersections of the secant planes passed through '"the sphere with the surface. Plate XII, Fig. 50, gives the shading lines on a vertical cir- cular cylinder. A tangent sphere is inscribed in the cylinder, with center at the center of the base (on i7) of the cylinder. Fifteen equidistant planes are passed perpendicular to a ray and their intersections with the base determined, thus giving the line 5, 6, 6, 5, 4, 3, 2, i, o, — i,— 2, — 3, and — 4. It is at once evident that the lightest part of the cylinder is between 6 and 6 corresponding to the zone 6 on the sphere (§ 119). PIat,e XII, Fig. 50a, gives the proper shading lines on a circular cylinder, axis parallel to <9. Z. C . . . . D, with corre- SHADING. lOI spending shading lines, is for H projection and E . . . . F, wifh corresponding shading lines, is for V projection. 121. Plate XII, Fig. 51, shows the shading on a sphere in F" projection by the "drawing out" method; and Fig. 52 shows the shading on a vertical cylinder by the " banded " method. PLATE I G— G^ L G- J— L G- -L V-L \ \\ -0-' / ^|- L \ \ o;t / \ / Fiff. 10. PLATE II. Ft (J. i:$. HW PLATE in •L IMATE IV. PLATE V. / / \\ I I \ \ //I / ~1^ F'uf. 24. \ \ \ Fuj. 25. » \ ,\ \X-^:Ae>- - ./ W /^ \/ / \ \ \i w \ \p^' v>; / i An V\ .>^ \ \ i \ \ I j i Fig, 26. / • / n^,^ / I / I / j / \ \, •i T- \ / \ ./ M./ Fig. 23 Fig. 23a. e" 9c. 1 1* Horizon \J \ \ ■^nr9- .xr-; 1^ ! A i \/N\ \ \, \ ! "\ jq" x; \ \ \ \ c. / / ' — z^ !/ / •L \ \ A \ \ \ .-i|^-: / IN'K' e' D" / 1 -^y \ \ y I \ / 1 1 .^f\\\ \ p ,p \ \ Fig.2r. \ j q |N' r \ \. '^\\ ^^^ I I I J, I / 'y PLATE VI. y->50L Fi(j. 29. j i k 1 f d 9 m n • c -J a R b PLATE VII. PLATE VIII. Fi(j. 43a. Fifj. 40a. "i — — _ Fif/.40. G — L ^ ^^ y ^ V- ' — ^ ^ ^ \ \, \ \ . ■^ V Fifj. 45 \ / / / Fig. 45(f. ^'.-"T'^^"/^ X \ \ \ X / / \N \ \. \ /\ X "% / ^J/ \, \ v" -e — \ V* ■f \ / / / / / / .r ,r// N PLATE IX. Fig. 46. :\\ .^ + -V ^ i- \ 4- + -v \ \ ^ \^ -V Xi xX\ \ \ 1^ e" PLATE X. -1^ -. O ^ <^ y V- ^/ \ / \ — r y45»R I e' ■9- D450L -9 w / \ ./ /V//../r. is-- ./ / \l ! / I /^ \ \. I N Fig. 48. /> ^ v^ ^ // <^.^ -/ y45<>R c! •\ TF.y,- :^ \ \ \ \ --^^. \ N. N. ■■-^^ PLATE XI. PLATE ATI. 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