gj^^sgse^^sss^ 
 
 i^aaEaW3«aad!i^«:xee£lfe/ 
 
The original of this book is in 
 the Cornell University Library. 
 
 There are no known copyright restrictions in 
 the United States on the use of the text. 
 
 http://www.archive.org/details/cu31924032226726 
 
Cornell University Library 
 arY676 
 A course in shades and shadows for the u 
 
 3 1924 032 226 726 
 olin,anx 
 
A COURSE 
 
 IN 
 
 SHADES AND SHADOWS 
 
 FOR THE USE OF 
 
 COLLEGES AND SCIENTIFIC SCHOOLS. 
 
 BY 
 
 WILLIAM WATSON, Ph.D., 
 
 fellow of the ambrioau academy of auts and scrbnces; membek of the rational academy of 
 
 cheeboueg; of the fbench society of civil engineeks; of the peussian 
 
 society of industkial engineers; etc, etc 
 
 BOSTON: 
 CUPPLES, UPHAM, AND COMPANY. 
 
 LONDON : 
 LONGMANS, GREEN, & CO. 
 
 ® 
 
Entered according to Act of Congress, in the year 1885, by 
 
 WILLIAM WATSON, 
 in the Office of the Librarian of Congress, Washington. 
 
TABLE OF CONTENTS. 
 
 Art. 
 
 1. The Geometrical Delineation of Shades and 
 Shadoios. Ray of liglit: — Pencil of rays: — 
 In the case of sunlight, the rays are assumed 
 to he parallel : — Plane of rays. ( Fig. 0. ) 
 2-3. Shade of a body. Line of shade : — Shadow. 
 4-8. Shadov) of a body. Line of shadow : — Shadow 
 of a curve: — The shadows of tangent curves 
 are tangent to each other: — Shadow of a 
 plane curve on a plane : — Shadow of the cir- 
 cumference of a circle upon a plane : — The 
 shadows of two diameters of a circle taken 
 at right angles to each other are conjugate 
 diameters of the ellipse of shadow. 
 9-15. Shadow of a Polyhedron. Notations: — Shadows 
 of points and lines : — Conventional direction 
 of the rays of light : — Values of the trigo- 
 nometrical functions of 9: — Advantages of 
 the conventional direction : — Abridged con- 
 structions. (Figs. 1-5.) 
 
 16-17. Problem!. To find the shadow on one of the co- 
 ordinate planes of a square parallel to the other 
 coordinate plane : — Abridged method : — 
 i?' and iJ" at 45° with GL. (Fig. 6.) 
 
 18 - 19. Problem 2. To find the shadow on one of the 
 coordinate planes of a right line perpendicu- 
 lar to the other coordinate plane : — Abridged 
 solution. (Figs. 7, 8.) 
 
 20 - 21. Problem 3. To find the shade and shadow of 
 an upright prism: — Abridged solution. 
 (Fig. 9.) 
 22. Problem 4. To find the shade and shadow of 
 an oblique cone. (Fig. 10.) 
 
 23 - 24. Problem 5. To find the shade and shadow of a 
 sphere: — Abridged solution. (Figs. 11, a.) 
 25. Theorem. Wlien we cut off by a plane, and re- 
 move, a portion of the surface of the second 
 order, such as a cylinder, a cone, an hyper- 
 boloid, an ellipsoid, etc. ; the shadow of the 
 section cast upon the interior surface so ex- 
 posed is a plane curve, and consequently one 
 of the second order. 
 
 Art. 
 
 26. Problem 6. To find the shadow of the edge of 
 
 a hollow hemispherical shell upon its inte- 
 rior surface. (Fig. 13.) 
 
 27. Problem 7. To construct the shadow of a niche 
 
 upon its interior surface. (Fig. 12. ) 
 
 28 - 29, Shadows of circles. Problem 8. To find the 
 shadow cast by a given circle parallel to one 
 coordinate plane on the other coordinate 
 plane (Fig. 14): — To find the magnitude 
 and position of the axes of the ellipse of 
 shadow. 
 Problem 9. To find the shadow of a circle' situ- 
 ated in the profile plane (Fig. 15): — Axes of 
 the ellipse of shadow (Fig. 16). 
 30. Construction of the axes of an ellipse from two 
 given conjugate diameters. (Fig. 16.) 
 
 31 - 33. Problem 10. To construct the shade of an up- 
 right cylinder with a circular base and the 
 shadow of the upper circle upon the interior 
 surface : — Abridged constructions. (Fig. 17. ) 
 
 34 - 36. Problem 11. To find the shadow of a rectangu- 
 lar abacus upon a cylindrical column, and 
 on the vertical coordinate plane: — Abridged 
 construction : — Shadow of a point on a cylin- 
 der. (Fig. 19.) 
 37. Problem 12. To find the shade of a cylindrical 
 column and its cylindrical abacus; also the 
 shadow of the abacus, both on the column and 
 on the vertical coordinate plane. (Fig. 20.) 
 
 38 - 39. Brilliant points : — General solution : — Prob- 
 lem 13. To find the brilliant point on a 
 spherical surface. (Fig. 26. ) 
 
 40. Problem 14. To find the brilliant point on a 
 
 surface of revolution. (Fig. 21.) 
 
 41. Shadows of points on curved surfaces. Prob- 
 
 lem 15. To find the shadow of a given point 
 on the surface of a cone. (Plate I. Fig. 10. ) 
 
 42. Problem 16. To find the shadow of a given 
 
 point on the surface of a sphere. (Fig. 28.) 
 
 43. Nature of the line of shade for surfaces of the 
 
 second order. (Fig. 35. ) 
 
niv 
 
 TABLE OF CONTENTS. 
 
 Art. 
 
 44. General methods of finding the line of shade:— 
 Methods of secant and tangent planes: — 
 Method of circumscribed surfaces: — Method 
 of oblique projections. (Fig. 6, Plate V.) 
 
 45 - 47. Points of the line of shade on the apparent con- 
 cour. Problem 18. To find the Une of 
 shade of a torus : — Abridged method. (Figs. 
 22, 23.) 
 
 48 - 51. Problem 19. To find the line of shade of a sur- 
 face of revolution: 1st, Method by circum- 
 scribed cones; 2d, Method by inscribed 
 spheres; 3d, Method by enveloping surfaces. 
 (Fig. 27.) 
 52. Problem 20. Having given a portion of a sur- 
 face of revolution convex toward the axis, it 
 is required to find the line of shadow cast by 
 the circumference of the upper base upon 
 the surface. (Fig. 31.) 
 
 53 - 57. Fillet's method of casting shadows by means of 
 a diagonal plane : — Shadow of a right line 
 parallel to GL: — Shadow of a horizontal 
 circle: — Shadow of a horizontal circle hav- 
 ing its centre in the diagonal plane: — Appli- 
 cations: — Expeditive solution of Problems 
 20, 4, 15. (Figs. 32-34, 45; Plate V. Fig. C.) 
 
 58 - 59. Problem 21. To find the shadow cast upon the 
 interior surface of a cone by the complete 
 circle of its base : — Expeditive method of 
 solving Problem 12. (Figs. 38 and 40.) 
 60. Problem 22. Shadow of an abacus and a quarter 
 round or ovolo. (Fig. 36.) 
 
 Art. 
 
 61. Problem 23. To find the shades and shadows 
 
 on the base of a Tuscan column. (Fig. 37.) 
 
 62. Shadows on sloping planes. Problem 24. To 
 
 find the shadow of a given point on a given 
 plane parallel to the ground line, and making 
 a given angle with the horizontal plane. 
 (Fig. 41.) 
 
 63. Problem 25. To find the shadow of a given 
 
 horizontal circle on a plane parallel to the 
 ground line, and making a given angle with 
 the horizontal plane. (Fig. 44.) 
 
 64. Problem 26. To find the shadows cast by a 
 
 chimney and dormer window upon a sloping 
 roof. (Figs. 42, 43.) 
 
 65. Problem 27. To determine the shadow of a 
 
 cornice. 
 
 66 - 67. The helicoid. Properties of the warped heli- 
 coid: — Tangent planes. Problem 28. To 
 find the line of shade on the siurface of a 
 given helicoid: — Centre of radiation. (Figs. 
 46,49.) 
 
 68 - 69. To find the point of the line of shade situated 
 upon any given helix traced upon the sur- 
 face: — A second construction. (Figs. 47, 
 50.) 
 
 70 - 73. Problem 29. To construct the shades and 
 shadows on the different parts of a screw, 
 and the shadow on the horizontal coordinate 
 plane: — Shadow of a helix, the shadow 
 shown to be a curtate cycloid : — Method of 
 description. (Figs. 48, 51-54.J 
 
SHADES AND SHADOWS. 
 
 THE GEOMETRICAL DELINEATION OF SHADES AND SHADOWS. 
 
 1. Light, whatever hypothesis we may adopt 
 respecting its nature, is invariably propagated m 
 right lines as long as it passes through the same 
 medium; the right line along which the light 
 holds its course is called a rai/ of light, and any 
 collection of such rays, of definite thickness, is 
 called a pencil. 
 
 In nature, light diverges in all possible direc- 
 tions from each luminous point, so that the pen- 
 cils are all primarily diverging; but when the 
 luminous origin is very distant, as in the case of 
 a heavenly body, the rays in every pencil we 
 consider are sensibly parallel. On account of 
 the great distance between the earth and the 
 sun, the rays may, without material error, be con- 
 sidered parallel ; draughtsmen so consider them, 
 thereby simplifying their constructions. In the 
 following problems, the rays will be so assumed. 
 
 A ray of light from any point of a luminous 
 body is represented by a right line. 
 
 A -plane, of rays is a plane passing through a 
 ray. 
 
 SHADE OB" A BODY. 
 
 2. Let Fig. 0, Plate I., B, be an opaque body 
 illuminated by a pencil of solar rays whose 
 direction is indicated by the arrow R. The 
 surface of this cylindrical pencil touches B in 
 a curve C. The portion of this cylinder from 
 which the rays are excluded by B is called the 
 
 indefinite shadow of B ; and any object situated 
 within this portion of the cylinder is in the 
 shadow of B, or has the shadow of B cast upon 
 it. 
 
 3. lAne of shade of a body. 
 
 The curve C separates B into two parts : viz., 
 that toward the source of light, called the illumi- 
 nated part ; and that opposite the source of light, 
 called the shade. The rays are excluded from the 
 shade by the body itself. 
 
 Any plane tangent to the cylindrical pencil 
 must be tangent to B at some point of C (Art. 
 130, Bes. Q-eom.') ; and, conversely, any plane tan- 
 gent to B at a point of C will be tangent to this 
 cylindrical pencil, and therefore contain a ray 
 (Art. Ill, Des. G-eom.'), and thus be a plane of 
 rays. Hence we may always determine points 
 on the line of shade of any opaque body by 
 passing planes of rays tangent to the body, and 
 finding their points of contact. 
 
 4. Shadow of a hody. 
 
 If we suppose any surface, as a screen S, to be 
 placed behind the body B, a part of S will have 
 the shadow of B cast upon it at M. M is the 
 portion of S from which light is excluded by B, 
 or M is the shadow of B on S. 
 
 Line of shadow. The periphery C, of M, or the 
 line which separates the illuminated portion of S 
 from the shadow, is called the line of shadow. 
 It is also the intersection of the surface of the 
 
SHADES AND SHADOWS. 
 
 cylindrical pencil with the surface on which the 
 shadow is cast. 
 
 5. Shadow of a curve. 
 
 The pencil of rays passing through a curve 
 forms a cylindrical surface ; the intersection of 
 this pencil with any other surface forms the 
 shadow of the curve on that surface. Thus, in 
 Fig. 0, Cj is the shadow of C ; i.e., the line of 
 shadow is always the shadow of the line of shade. 
 
 6. The two cylindrical pencils passing through 
 two tangent curves are tangent along the ray 
 passing tlirough the point of contact of the 
 curves (Art. 131, Bes. Geom.') ; hence the inter- 
 sections of the pencils by any surface wdl be 
 tangent at the point in which this ray pierces the 
 surface ; these intersections are the shadows of 
 the curves, hence the shadows of tangent curves are 
 tangent to each other ; this is also true when one 
 of the curves becomes a right line. 
 
 7. The shadow of a plane curve on a plane. 
 Designate the curve by C, and its shadow on 
 
 the plane P by Cj ; then, 1st, if the plane of C is 
 a plane of rays, Cj will be a right line ; 2d, if the 
 plane of C is parallel to P, then Cj wiU. be equal 
 and parallel to C. 
 
 8. The shadow of the circumference of a circle 
 on a plane is, in general, an ellipse' (Art. 140, 
 Des. Geom.) ; and the shadows of any two diam- 
 eters of the circle, taken at right angles to each 
 other, are conjugate diameters of the ellipse of 
 shadow.^ Denote by A and B two such diam- 
 eters of the circle, and by Aj and Bj their re- 
 spective shadows; draw tangents T and U at 
 the extremities of A : then as T, U, and B are 
 parallel, their shadows Tj, Ui, and Bj will also 
 be parallel, and Ti and Ui will be tangent to the 
 ellipse of shadow at the extremities of the diam- 
 eter Ai : hence Ai and Bj are conjugate diameters 
 of the ellipse of shadow. 
 
 9. Shadow of a polyhedron. 
 
 When the opaque body B is bounded by planes, 
 the pencil of rays touching B has the form of a 
 prism, whose exterior surface is made up of planes 
 of rays not mathematically tangent to B. In this 
 
 1 ExCBPTious. 1°. When the plane is parallel to the 
 curve. 2°. When the plane cuts a sub-contrary section from 
 the pencil of rays (Art. 170, Des. Geom.). 
 
 2 Two diameters of an ellipse are conjugate when one of 
 them is parallel to tangents drawn through the vertices of the 
 other. 
 
 case, the lines of shade and shadow are broken 
 lines. 
 
 10. Notation. As far as practicable, the 
 shadows of points a, b, c, etc., will be denoted 
 by ai, 6i, Ci, etc. ; those of lines A, B, C, etc., by 
 Ai, Bj, Ci, etc. 
 
 The direction of a ray of light R will be given 
 by its projections R'' and R", and the angle of R 
 with the horizontal plane wiU be denoted by $. 
 
 SHADOWS OF POINTS AND LINES. 
 
 11. When the opaque body is reduced to a 
 point, the pencil becomes a single ray, and the 
 shadow on a surface is the point in which this 
 ray pierces the surface. Thus, Fig. 1, Plate I., a 
 ray of light R, drawn through a given point a, 
 pierces the coordinate plane V in the point a^, 
 which is therefore the shadow of a on V. 
 
 The point Cj, Fig. 2, in which a ray R, drawn 
 through c, pierces the coordinate plane H, is the 
 shadow of c on H. 
 
 12. The shadow of a right line on a plane is 
 determined by finding the shadow cast by any 
 two of its points on the plane : thus the shadow 
 of the right luae A, Fig. 3, on V, is found by con- 
 structing the shadows of any two of its jDoints, as 
 a and h. Drawing rays through a and b, we have 
 theii- shadows «! and b^ ; and the line a^ 5j, or Aj 
 is the shadow required. 
 
 The two rays a «! and b b^ determine a plane 
 of rays. 
 
 The shadow of a right line A, on any surface 
 S, may be found by passing a plane of rays 
 through A, and finding its intersection with S. 
 
 CONVENTIONAL DIEECTION OF THE RAYS OP 
 LIGHT. 
 
 13. In delineating the shadows of structures 
 and machines, a conventional dii-ection for the 
 light has been adopted which presents great ad- 
 vantages, both in clearness of design and facility 
 of construction. 
 
 We suppose the direction of the ray of light 
 R, Fig. 4, to be the diagonal oc of a cube so 
 placed as to have two of its faces parallel to the 
 vertical, and two to the horizontal plane of pro- 
 jection. Consequently R" and R* make angles 
 of 45° with the co„; again, R', the projection of 
 R on the profile plane oo„, makes an angle of 4l^° 
 with o„o''. 
 
SHADES AND SHADOWS. 
 
 If we call the side of the cube unity, the par- 
 ticular value of 6 for this case is ^ = oco^, whence 
 
 CO* = V^ ; tan e = -i| = 1^2 = 0.707, 
 
 oc = Vs ; sin e = -;= = 1^3 = 0.577, 
 
 cos 2^ = i ; cos ^ = ^ = -1^6 = 0.816, 
 V3 
 
 sec 61 = ^ = *V6 = 1.225, 
 
 6 = 35°15' 52" ; cosec^ = ^3 = 1-732. 
 
 The angle 9 is easily constructed. Fig. 5, by 
 assuming a point (o*, o") on the ray (R*, R"), and 
 revolTing this ray around ex untU it coincides 
 with the vertical plane V, whence we have 
 ocG = e. 
 
 14. Advantages of assuming R* and R" at an 
 angle of Ji5° with GL. 
 
 To show how the construction may be short- 
 ened by assuming R* and R" at an angle of 45° 
 with GL, we will suppose. Fig. 1, it was required 
 to find the shadow of a; we will denote the given 
 distance of the point a from the plane on which 
 its shadow is to be cast (in this case, the vertical 
 coordinate plane) by S, i^., Ga* = S. Construc- 
 tion : Lay off aPo = 8, draw through o a line par- 
 allel to GL, and set off on it a distance oai = 8 ; 
 «! is the shadow required. 
 
 15. In Fig. 2, the shadow of c on H is required. 
 In this case the distance (8) of c from the plane 
 on which the shadow is cast is &n. Construc- 
 tion : Lay off d^x = 8, draw through x a line 
 parallel to GL equal to 8, and its extremity c^ is 
 the shadow required. 
 
 16. Pboblem 1. To find the shadow on one of 
 the coordinate planes, of a square, parallel to the other 
 coordinate plane. 
 
 Let, Fig. 6, abed be a square parallel to V : it is 
 required to find its shadow on H. 
 
 1st method.'^ Rays making any angle whatever. 
 Draw through a, b, c, and d, rays, and their hori- 
 zontal traces aj, Sj, Cj, and di will be the four an- 
 gular points of the required shadow. We see 
 that the lines ab and cd parallel to H have paral- 
 lel shadows aib^ and e^d^ on H. 
 
 17. M method. R* and R" making angles of 
 45° with GL. Denote the given distance tc" by 
 8, and the length of the side a*6* by I; make 
 a*?i =zB, no = OS :=l; erect at n, o, and s, three 
 
 perpendiculars to GL, and lay off on the first the 
 distance 8, on the second the distances 8 and 
 8 -f- Z, and on the third the distance S -\-l: the 
 extremities of these perpendiculars will be the 
 angular points sought ; viz., Ci, c?j, aj, Sj, whence 
 the shadow is known. 
 
 18. Peoblem 2. To find the shadow on one of 
 the coordinate planes of a right line perpendicular 
 to either coordinate plane. 
 
 Case 1st. Let A, Fig. 7, be a vertical right line : 
 it is required to find its shadow on V. Assume 
 any point (as w". A*) on A, and draw through it a 
 ray ; this ray has its vertical trace at Wj, which is 
 one point of the required shadow ; since A is par- 
 allel to V, its shadow on V must be parallel to 
 A and to A": therefore, drawing through % a 
 right line parallel to A", we have the required 
 shadow A J. 
 
 Abridged Construction. R* and R" making an- 
 gles of 45° with GL. Let i5A* = 8. Lay off tx 
 = 8, and erect a perpendicular Aj to GL at x; 
 Ai is the shadow of the unlimited line A. 
 
 19. Case 2d. Let B, Fig. 8, be the right Hne 
 perpendicular to V. 
 
 Construction. Assume any point of B (as o*, 
 B"), and through it draw a ray ; its trace on V is 
 Oi, but B" is the trace of the line itself on V: there- 
 fore the required shadow is Bj. In this case the 
 shadow on V of an unlimited line B perpendicu- 
 lar to V is the unlimited vertical projection of 
 the ray through B". 
 
 20. Problem 3. To find the shade of an up- 
 right prism, and its shadow on the vertical coordinate 
 plane. 
 
 Let the prism be given as in Fig. 9. If we pass 
 vertical planes of rays through 6* and e*, we per- 
 ceive at once, that the faces horizontally projected 
 in 5''a'', ay*, and /*e'' are illuminated, and those 
 projected in 5*0*, c/'d'', and d''e^ are deprived of 
 light and form the shade. The upper base is illu- 
 minated. The line of shade is composed of the 
 edges (mb", 5*), be, cd, (dl'^, d'c"'), and Qfn, e"'). 
 The only portion of the shade visible is the rect- 
 angle c"nld^. 
 
 The line of shadow is the shadow of these 
 lines of shade : therefore, drawing rays through 
 b, c, d, and (e*, c"), we have the points b^, c^, d^, gj, as 
 the shadows of the angular points of the upper 
 base ; b^, however, is invisible, as it is hidden by 
 the prism itself. The shadow of the vertical edge 
 
SHADES AND SHADOWS. 
 
 through b is ijO, invisible on the vertical plane : 
 its shadow on the horizontal plane, as far as GL, 
 is 5*0 ; the shadow of the vertical edge through 
 (e*, e") is eV on the plane H, and re^ on V. Join- 
 ^S ^i^n ^'I'^D <^i^i) we have obiCidieiV the portion on 
 the plane V, and r^&'^h^or the portion on the 
 plane H. 
 
 21. Abridged construction. Suppose, as is gen- 
 erally the case in architectural drawings, it were 
 only required to find the shade and shadow on 
 the vertical coordinate plane ; R* and R" at 45° 
 with GL. 
 
 The visible shade, as before, is c'dHn. The 
 points o,p, q, and r may be found by laying off 
 on GL the following distances : mo = «j6* ; np = 
 me* ; Iq := M* ; and nr = we* ; erecting perpendic- 
 idars to GL at o, p, q, r, and drawing the vertical 
 projections of rays through 6", c", d", their inter- 
 sections with the respective perpendiculars give 
 the points bi,Ci,di,ei, which, withe^r, determine the 
 shadow on the vertical plane. 
 
 22. Peoblbm 4. To find the shade and shadow 
 of an oblique cone. 
 
 Let the cone be given as in Fig. 10. The shade 
 is found by passing planes of rays tangent to the 
 cone, and determining their elements of contact 
 (Art. 3). For this purpose, draw a ray through 
 the vertex o; o^is its horizontal trace. Drawing 
 OiW* and OjC* tangent to the base of the cone, we 
 have the horizontal traces of the two tangent 
 planes of rays ; the lines of contact are oc and on; 
 hence noc is the line of shade. The portion of 
 the shade visible on the horizontal projection is 
 oHc^ ; that visible on the vertical projection is 
 o%n''. The shadow on H is c^OiW*. 
 
 23. Pkoblem 5. To find the shade and shadow 
 of a sphere. 
 
 Let the sphere be represented as in Fig. 11, 
 Plate IL, and let the light have the direction 
 indicated by R. 
 
 The surface of the cylindrical pencil of rays is 
 tangent to the sphere, and has for its line of con- 
 tact the circumference of a great circle, the line 
 of shade, which is horizontally projected in an 
 ellipse having c* for its centre, and the horizontal 
 line l<^x drawn through c* perpendicular to R* for 
 its transverse axis. The conjugate axis passes 
 through c* parallel to R*; its length may be 
 found as follows: Imagine a vertical plane of 
 rays passed through the centre c of the sphere ; 
 
 ooi will be its horizontal trace; this plane will 
 cut from the tangent pencil two rays tangent to 
 the great circle cut from the sphere ; this great 
 circle mil intersect the circle of shade^ in a diam- 
 eter having for its horizontal projection the re- 
 quired conjugate axis. The construction may be 
 made as follows: Rotate this vertical plane of 
 rays about its horizontal trace ooj, until it coin- 
 cides vsdth H ; c falls at e2((?*(?2 = V)> ^^^ ^^^ 
 circle Ojiwj is the revolved position of the inter- 
 section of the sphere with the plane of rays ; the 
 ray through the centre c, and the tangent rays, 
 have the positions C2C1, ru^-y, and O2O1, after rota- 
 tion ; the intersection of the two circles will be 
 the line OjWj, drawn through c^ and perpendicular 
 to C2C1 : making the counter revolution, o^ and n^ 
 are found at and n, whence on is the required 
 conjugate axis. 
 
 The angle S^e-^c^ of the rays with H is equal to 
 c^n,^; hence the required semi-conjugate axis 
 is c*ra := C2W2 sin c^fi^n., i.e., r sin 6, denoting C2W2 
 by r. 
 
 The line of shadow of the sphere on H is cast by 
 the circumference of shade : it is therefore an 
 ellipse, with its centre at <?i, having for its conju- 
 gate axis l\Xi drawn through c^ perpendicular to 
 R* and equal to Ix. The transverse axis is the 
 shadow of the diameter horizontally projected at 
 on perpendicular to Ix; its revolved position is 
 O2W2 ; the rays through its extremities pierce H at 
 Oi and Jij. Hence o^ni is the transverse axis re- 
 quired. 
 
 From the figure we have n^Oi = n^^ cosec on^n^ 
 = 2r cosec 0. 
 
 The vertical projection of the line of shade, ef, 
 the vertical projection of the diameter ef parallel 
 to V, is the required transverse axis. To find the 
 conjugate axis : The ray and the circle of shade 
 are perpendicular to each other, and therefore 
 make complementary angles with V. If we 
 denote the angle of R with V by i>, found as 
 in Fig. a, we shall have 90° — <^ as the angle 
 of the circle of shade with V ; hence (Art. 169, 
 Des. Geom.') the required semi-conjugate axis e's 
 is radius X sin <^. In Fig. a, laying off oa' = oa'' 
 = c'f Fig. 11, we have cs as the required semi- 
 conjugate axis which may be laid off in Fig. 11, 
 and the ellipse of shade readily constructed. 
 
 1 For brevity we will call the circle, having the line of shade 
 for its circumference, the circle of shacle. 
 
SHADES AND SHADOWS. 
 
 This method might have been used to deter- 
 mine the horizontal projection of the line of 
 shade. 
 
 24. Abridged construction. R* and R" at 45° 
 with GL ; denote the diameter of the sphere by 
 d : then (Fig. 11) qs =i on = d sine = O.Blld ; 
 oiw, = d cosec e = LlS2d (Art. 13). 
 
 25. Theorem 1. When ive cut off hy a plane, 
 and remove a portion of a surface of the second 
 order, such as a cylinder, a cone, an hyperboloid, an 
 ellipsoid, etc., the shadow of the section cast upon 
 the interior surface so exposed is a plane curve, and 
 consequently one of the second order} 
 
 26. Pkoblem 6. To find the shadow of the 
 edge of a hollow hemispherical shell upon its in- 
 terior surface. 
 
 Let the hemispherical shell be represented as 
 in Fig. 13 ; and let the direction of the light be 
 indicated by the arrows as parallel to V, aiid 
 
 1 This theorem is usually expressed thus, the surfaces being 
 understood to mean those of the second order : — 
 
 If one surface enters \interseets\ another in a curve of tlie 
 second order, it will leave it [intersect it again} in a curve of 
 the same order. This is a very well-kn()wn theorem of ana- 
 lytic geometry. The demonstration here given is due to M. 
 Binnet. 
 
 Let us imagine two surfaces of the second degree; suppose 
 XT to be the plane of the curve in which one surface enters 
 the other. 
 
 The two equations of the surfaces may be written in their 
 most general form, thus : — 
 
 ax" + hy'^ + c%^ + te;/ + mxz + nyz + pa; + gy + >■« + cZ = (1) 
 oV-|-J)y+c'z2+rj;2/+m'j:24-?i'2/«+p'x+8'2/+'-'2+(2'=0 (2) 
 The curve of intersection being contained in the plane XT. 
 
 If we put z = in (1) and (2), we shall have 
 
 (V3? + 62/^ + fiy + pa; + gy + (2 = (3) 
 
 aV + 6 V + Vxy + p'x + g'?/ + (i' = (4) 
 
 each of which represents the known curve of intersection of 
 the two surfaces; (3) and (4) are identical, and consequently 
 the co-efficients of the corresponding terms must be equal, or 
 cannot differ except by a constant factor ^; hence, 
 a = 7ia'\ h - Vj'; I = M'; p = Ap'; q = Ag'; d = W; (5) 
 if we now multiply (2) by A, and subtract it from (1), taking 
 account of equations (5) after performing the subtractions 
 and substitutions, we have 
 
 (c — Ac')2^ + {m - hn')xz + (n — ?in')yz + (r —?i.r')z - (6) 
 which must be satisfied for all values of x, y, and z, common 
 to the two surfaces. Since z is a common factor, (6) is satis- 
 fied by placing z = 0, ov 
 
 (c — lc.')z + (m — Am')x + (n — ln')y + {r- Ir') - (7) 
 z = 0, belongs to the plane XY, in which the known curve of 
 intersection lies; (7) is the equation of a plane, which, by its 
 combination with (1) or (2), will give another curve common 
 to both surfaces, and this curve must, of course, be one of 
 the second order. 
 
 making an angle with H. This problem is 
 taken as an illustration of the foregoing theorem. 
 The two surfaces of the second degree are the 
 hemisphere, and the half-cylinder of rays enter- 
 ing the hemisphere in the semicircle, whose pro- 
 jections are a^c", ed!^f. The entering curve being 
 thus a plane curve, and one of the second order, 
 the exit curve must also be a plane curve of the 
 second order. The only curve which a plane can 
 cut from the sphere is a circle : the curve of shade 
 in space is, therefore, a semi-circumference ; it is 
 horizontally projected in a semi-ellipse, of which 
 ef is the transverse axis.* To find the semi-con- 
 jugate axis A'': Pass a vertical plane of rays 
 through c, it will cut from, the illuminating pencil 
 the ray as, whence c*s* is the semi-conjugate axis 
 of the horizontal projection of this semicircle. 
 Join (? and s' : then the angle ^'c-'a; = 2^, and c^x 
 - cV cos 2^. If 61 - 35° 15' 52", then. Art. 13, 
 cos 2^ ^ 1^ ; hence c^x =: -Jc's ^ c**''. 
 
 27. Problem 7. To construct the shadow of 
 a niche upon its interior surface. 
 
 The niche (Fig. 12) is an upright hollow semi- 
 cylinder, projected vertically in the rectangle dh^ 
 and horizontally in the semicircle d'Sih^, termi- 
 nated by a quadrant of a sphere vertically pro- 
 jected in the semicircle dsf, and horizontally on 
 the base of the semi-cylinder. R" and R* are 
 assumed at an angle of 45° with GL. The line 
 of shadow is divided into three portions : 1st, fx 
 cast by the shadow of the arc A upon the 
 spherical surface ; 2d, xd{ cast by A on the 
 cylindrical surface ; 3d, n-^d^ cast by the element 
 icCd, a'') on the cylindrical surface, fx (by Art. 
 26) is the arc- of an ellipse extending from /, the 
 point of contact of the tangent ray, to x, having 
 for its semi-axes zf, the radius, and zo = ^zf; 
 any point oi xdi" as Sj" may be found by passing 
 a ray (s'Si", s\'') through s, and finding its trace Sy" 
 on the surface of the cylinder ; di'iii is the shadow 
 of dn, a portion of (^da", a''), upon the interior 
 surface of the cylinder ; n is found by drawing 
 WiW parallel to R". 
 
 SHADOWS OF circles. 
 
 28. Since in architectural drawings it is often 
 required to find the shadows of circles in various 
 
 1 As the semi-circumference cut from the hemisphere has 
 {c/, e") for its diameter, it must be half of the circumference 
 of a great circle. 
 
SHADES AND SHADOWS. 
 
 positions, we give here two examples : R" and R* 
 at 45° with GL. 
 
 Peoblem 8. To find the shadow cast hy a given 
 circle 'parallel to one coordinate plane on the other 
 coordinate plane. 
 
 Let the circle be represented as in Fig. 14, 
 vertically projected in k'ge", and horizontally in 
 Pe". Imagine a square Ir circumscribed about 
 the circle, mr being horizontal. The shadow of 
 this square on the coordinate plane H (Art. 16) 
 is found by drawing K'mi and eVj parallel to R*, 
 e*mi perpendicular and equal to A*e^ and Wirj 
 parallel and equal to Pe* ; the middle points g, e,/, 
 k have ^i, ei, /j, ^, for their shadows. The shadow 
 of the inscribed circle is an ellipse tangent to the 
 parallelogram Prj at the points g^, ei,/i, k■^, and hav- 
 ing.^i/i and ej^ifor conjugate diameters (Art. 8). 
 
 To find the position and magnitude of the axes 
 of the ellipse of shadow. Construction : Lay off 
 mn = to the radius of the given circle, join n 
 with c, and produce me to o ; then no is equal to 
 the required transverse axis, nx the conjugate, 
 and nck^ double the angle of the transverse axis 
 with the horizontal conjugate diameter; i.e., no 
 =■ S1S2 ; nx = ZiZ.2 ; and s.^CiCi = inck"} 
 
 1 Verification of the construction given in Fig. 14. We 
 assume the three equations of Analytic Geometry relating to 
 conjugate diameters, as follows : — 
 
 ab 
 
 a'h' sin(a' — a] 
 a'l + 6'2 = a2 + 62 
 
 tan a' tan a = 5 
 
 (1) 
 (2) 
 
 (3) 
 
 in which 
 r = radius of the given circle 
 a' = r\l^; b' = r 
 a' — o = 45°: sin45° = {sfi 
 
 a'h' sin(a' 
 a' 
 
 -a] 
 
 ab ~ r'-' 
 
 "2 + b'i = 3r2 
 2ab = 2r2 
 
 + 6 
 -6 
 
 5,2 _ {,2 = ,.2^ 
 
 • _iw _^^- 3 -y^ _ 3\/"> 
 
 +V/5' 
 
 a = 2(v/5 + l); 6 = 2(V^-1); „. 
 substituting these values in (3), we have 
 
 ^ ,., , , tana(l + tano) Sv/s • 
 tan a tan(45 + a) = ■ ' — 
 
 for brevity let i/ = tan a, then 
 
 2 
 
 1 — 1 
 
 , or 2/" 
 
 sy/s -5 
 
 2 
 
 -y 
 
 whence y = tan a — 2 — V^, or i5 ; whence tan 2a = —2 
 
 or —i;2a- 116°34' or — 63026'; whence a = 58°17' or — 31°43'; 
 whence S2C,ei —a— — 31°43'; s^Cif/i = o' = 45 + a — 45° 
 - 31°43' = 13°ir; or ZzCn-, = 58°17' = a and a + 45 = a' = 
 103°17' = ZiCiffi- Using the other value, we obtain tan 2a 
 = — i; we have 2a = 153°2G' or — 26°34'; hence we have as 
 another value, a = 76°43' or — 13°17'; whence a' = 121°43'or 
 31°43'. In Fig. 15, SjCff, = - 13°17', and SjCiCi - 31°43'; 
 also, ZiCiQ, = 76°43', and XsCie, = 121°43'. 
 
 29. Problem 9. To find the shadow of a 
 circle situated in the profile plane. 
 
 Let the circle (Fig. 15) be given by its centre 
 c, and the length of its diameter /^^ = ek; R* and 
 R" being taken at 45° with GL. Imagine a square, 
 circumscribing the circle, two of its sides parallel 
 to the plane H, and the other two perpendicular 
 to it. The shadow of the square on H is formed 
 of the two lines km and er drawn through k and 
 e parallel to R* (Art. 16), and two others ek 
 and mr perpendicular to GL and separated by an 
 interval em =^fg. The shadow of the circle is 
 tangent at the points d\ C], ^1, ^1 ; and c^g^ and ejc^ 
 are its conjugate diameters. The axes SjSj, Z1Z2, 
 are found by Fig. 14. 
 
 30. Construction of the axes of an ellipse from 
 two given conjugate diameters. Let ce and en 
 (Fig. 16) be the conjugate diameters, making a 
 given angle <■> with each other. Construction : 
 From the extremity n, of one diameter, drop a 
 perpendicular nf on the other produced ; from n 
 set off distances nl = nr = ce ; join c with I and 
 r, and draw through n a parallel nd to cr, inter- 
 secting cl in ; make 02; = oc? =; ol, and join 
 with d and x ; make ca =r 7id, and cb = nx, then 
 cb and ca will be the new axes both in magnitude 
 and position.' 
 
 ^ The complete verification of this construction is made by 
 showing that the new axes in magnitude and position satisfy 
 the usual equations of Analytic Geometry relative to conju- 
 gate diameters ; viz., — 
 a'6' sin(a' — a) = ab (1) dn = a 
 a2 + 62 = a'^ + 6'2 (2) nx =b 
 62 
 
 tan a tan a' = — 
 
 (3) 
 
 = a' 
 = h' 
 
 angle nrc = r en 
 
 ncr = f ce 
 
 nca = y = den — 00° . een = a 
 Drop from d a perpendicular dz on nc produced; then rdz 
 = y = nea. With o as a centre, and a radius cl describing a 
 circumference, we have from the two secants nd and nf, ab 
 
 = a'6' sin (J (1). 
 
 In the triangle cfr, cr = yjef^ + fr^ 
 
 = Va'^ cos^u -I- (a' sin u + 6')2 = a + b. 
 In the triangle elr, since In — Ir = 6', we have cr- + ~cl- 
 = 2^2 + 2^(2, or (a -I- 6)2 -1- (a - 6)2 = 2a'2 -|- 26'2; whence 
 a2 + 62 = a'2 + b" (2). 
 
 6' sin ?■ 5' cos u 
 
 cos2(4 = 1 — sin20 = 
 
 (rt+ 6)2 - 6'2 C0S2<J 
 
 a -I- 6 ' 
 ((!' -I- 6' sin(j)2 
 
 (u + 6)2 
 
 ah' sin a 
 ez = a cos ^ — o = 
 
 -a'b 
 
 {a + 6)- 
 
 zd — o sin = 
 
 a + b 
 
 ih' COS U 
 
 a + b ' 
 
SHADES AND SHADOWS. 
 
 31. Problem 10. To construct the shade of 
 a right cylinder with a circular base, and the shadow 
 of the upper circle on the interior surface. 
 
 Let the cylinder be given as in Fig. 17. 1st, 
 To find the shade. The tangent planes of rays 
 touch the cylinder along the vertical elements 
 through ;• and t, thus giving rfbp as the shade 
 visible on V. 
 
 The shadow on the interior surface is cast 
 by the semicircle ter. The line of shadow is 
 (Art. 25) a plane curve, an ellipse, having tr 
 for a diameter and c for its centre. Other con- 
 jugate diameters of this ellipse may be found 
 (Art. 8) by selecting pairs of radii of the semi- 
 circle ter at right angles to each other and finding 
 their shadows ; besides ct we will take c (I, t") ; 
 also ce and c («/, c"). Through the extremities 
 of these radii t, e (I, ("), (y, c"), pass rays and 
 vertical planes of rays ; these planes cut from the 
 cylindrical surface elements horizontally projected 
 at t'', u, 2, and /*; and the rays intersect these 
 elements in the points ej, li, and yi, the shadows 
 of the assumed extremities ; hence joining c with 
 «!, li, and «/i, we have two pairs of semi-conjugate 
 diameters vertically projected in c"*", c'li, and 
 c^gj, (fi/j. Since tft" and c% are one pair of semi- 
 conjugate diameters, the line gli parallel to c^i" 
 must be tangent to the vertical projection of the 
 ellipse of shadow. For the same reason fg 
 parallel to (fli and a line parallel to it at r" are 
 also tangents at f and r" ; fyi parallel to c^gj is 
 
 cz _ _ ab' sinu —a'b 
 
 ^ - tan y - „(,' cos a 5 
 
 ■ y — a; u —y = a'. Substituting in (1), we have 
 
 (4) 
 
 hence 
 
 62 
 
 — tan y tan(u — y) = ~^' 
 
 tan 7 tan u — tan^y _ Ifi 
 
 1 + tan u tan y 
 Substituting, we have for the numerator, — 
 
 (a6'sin°<j-a';>sin6))n;/ a^b''^ sin^u-2cn''+ a'V _ b^a^-a'^) 
 D2 D2 - D2 
 
 T> denoting the denominator ab' cos u, and remembering that 
 a'b' sin u = ab. For the denominator 
 
 an/-^ cos^M + aV^ sin^u - aba'b' sin u a2(y2 - y) . 
 
 D2 
 
 D^ 
 
 since a' + b^ = a'^ + b'^, we have jpj-. 
 
 1; 
 
 62 
 
 62 
 
 hence tan y tan(u — y) = ^; or tan a' tan o = — — ^j which is 
 (3). Substituting in (4) the values of a', 6', a, 6, and a, given 
 in the note to Art. 28, we have tan y = 2 — 05, or tan a 
 = ^ — 2, which is the same as that given in Art. 28. 
 
 also tangent at tji : we have, therefore, five points 
 of the vertical projection of the line of shade, 
 viz., <", Ci, Zi, «/i, r", and also the directions of the five 
 corresponding tangents. Hence we have the fol- 
 lowing simple construction for the ellipse of shadow. 
 Determine as above the shadows yj and Z, of the 
 points (y, c") and (I, t"), draw ei?/i and gl^ parallel 
 to GL ; join gt" (^ being on the axis) ; then we 
 have the five points t", ej, Z,, yj and r", and the five 
 corresponding tangents ; viz., fg, e^e^, gli, fb, and 
 a parallel to gt" through r", which are entirely 
 sufficient to determine the shadow. 
 
 The shadow on any particular element, e.g., that 
 on (/"6, /*), is found by passing a plane of rays 
 through this element, and finding its intersection 
 (c", y) with the semi-circumference ter, whence, 
 drawing a ray through (c", «/), we have the re- 
 quired shadow yi. 
 
 The lowest point of the shadow. Since the dis- 
 tance c^fii of any point e^ of the shadow below the 
 upper base is proportional to the distance e^u 
 between the element casting, and that receiving 
 the shadow, it follows that this vertical distance 
 will be greatest when the distance between these 
 elements is a maximum, i.e., when the plane of 
 rays passes through the axis. 
 
 32. Abridged construction. R° and R* at 45° 
 with GL. Denote the radius of the base of the 
 cylinder by p ; then bp = p versin 45° = -^p^ ; c''ei 
 = p ; c^yi = py2 ; the angle yic°ei = 45° ; hence 
 the axes of the ellipse of shadow may be con- 
 structed as in Art. 28 : therefore making e'n = 
 2p, and with c" as a centre describing the semi- 
 circumference, xe^o, no, and nx are the lengths of 
 the axes of the ellipse of shadow, and the angle 
 nc^e' = 2y. 
 
 33. Simplified construction. Lay off c't" = e"r° 
 = J^P, and draw the elements r'p and c^s ; set off 
 c^e, = c'e", and c'g = IcH^ ; draw e^y^ and g\ par- 
 allel to GL ; their intersections with /"S and r°p 
 determine y^ and \. We then have the five 
 points t", «!, li, yi and r", and their corresponding 
 tangents t^'g, e^e^, gl^, f^y^, and a right line through 
 r" parallel to fg. 
 
 34. PnOBLEM 11. To find the shadow of a 
 rectangular abacus on a cylindrical column, and on 
 the vertical coordinate plane. 
 
 1 If the diameter is 0"'.l, tlien V\; of 0™.05 is accurate to 
 within O^.OOOS. 
 
8 
 
 shadp:s and shadows. 
 
 Let (Fig. 19) a'e" and a''d be the projections of 
 the abacus, and p'y and p'^xi-'' those of the semi- 
 cohimn. 
 
 1°. The sJunhic of the edge a''i, a" is obtained 
 by passing a plane of rays through it : this plane 
 will be perpendicular to the vertical plane, and 
 have a'ai" for its vertical trace. The ray through 
 a pierces the column at a^, which is the end of 
 this shadow ; a^l is the shadow on V ; {lai", p^a^'^ 
 is the shadow on the cylinder ; the triangle aHq is 
 the shadow cast on V by a portion of the lower 
 surface of the abacus. 
 
 2°. The shadow cast hy ah on the cylinder is the 
 curve ai'7iy ; that of ab on V is Uibi, obtained by 
 drawing rays through a, u, and b, and finding the 
 points in which they pierce the cylinder and V. 
 The curve is, in general, an ellipse. uH, the trace 
 of the plane of rays tangent to the column, de- 
 termines yz, the line of shade of the column, and 
 the shadow UiXi of a portion of (zz/, a;). The line 
 (u^y, vl^x) is in the tangent plane, and in the plane 
 of rays ; therefore it is their intersection, and 
 tangent to the curve of shade at y; y is the point 
 where the line of shadow disappears in the shade 
 of the cylinder. The shadow of (J^e", J*) is l^Cx ; 
 that of le\ dh") is eVj. 
 
 35. Abridged construction. R' and R* at 45° 
 with GL. 
 
 Pass a plane of rays through the axis of the 
 column ; c''s'' is its horizontal trace ; this plane in- 
 tersects the edge ab in s, and s^o is the vertical 
 projection of the ray through s. Denote the 
 length of the radius of the column by p. 
 
 With as a centre and p as a radius, describe 
 an arc s^ny ; this arc will be the vertical projec- 
 tion of the shadow of ab on the column ^ ; take 
 
 1 Let be the origin of a system of rectangular coordinate 
 planes, ob' the axis of Z ; the plane ZX the vertical plane hav- 
 ing ik for its horizontal trace. [If we refer to Fig. 4, Plate I., 
 we see that a plane of rays through ot makes an angle of 
 45° with H.] The equation of any plane is z = cx+ dy + g 
 ( 1 ) ; the angle ^, which this plane makes with the plane XY, is 
 
 cos ^ — ;, , „ , ,. (2). The equations of the cylinder are 
 
 a;2 + 2/2 = p2, z indeterminate (3). Let oh' =c''f—l;<^ — 45° ; 
 cos ^ = Vi . The equation of the plane of rays through ah 
 may be found as follows : Since ab is parallel to the axis of X, 
 c — ; since it makes an angle of 45° with the plane XY, 
 
 cos^^ = i ; hence \ = jqr^a or d = 1. When x = l, z = I; 
 
 hence 1 = 1 + g ; .: g = o; and the equation of the plane of 
 rays is z = y, x indeterminate (4). Substituting this value 
 of 7/ in (3), we obtain, as the equation of the vertical projec- 
 
 zr' = -^p, and zxj = c^z; erecting a perpendicular 
 at a-,, and drawing yu at 45°, Ave have m,. Draw- 
 ing rays through b and e, we complete the outline 
 by drawing m,?)] parallel, and b^ei perpendicular, 
 respectively, to GL. 
 
 36. Shadoiu of a point on a cylinder. R" and 
 R* at 45° with GL. We have seen (Art. 35) 
 that the shadow of a (Fig. 19) on the column 
 was «!, which might be found as follows : Draw 
 through a" a line parallel to GL, meeting the axis 
 of the cylinder in b'. Set off b'o = a\ the dis- 
 tance of a in front of the axis : o is the centre 
 of s^'ny., and the radius is that of the cylinder. 
 Drawing through a" a line parallel to R", we have 
 the shadow a, of a upon the cylinder, without 
 drawing any line on H. If the cj'linder were 
 hollow, and the shadow were required uj)on the 
 interior surface, b'o would be laid off above a'i', 
 instead of below it. 
 
 37. Problem 12. To find the shade of a cyl- 
 indrical column and of its cylindrical abacus ; also 
 the shadow of the abacus, both on the column and on 
 the vertical coordinate plane. 
 
 Let the abacus and column be represented as 
 in Fig. 20, the projections of the abacus being 
 y''n^ and j/'eW ; those of the column, kt and ol'hH''. 
 
 Let the column be intersected by vertical planes 
 of rays cutting the lower base of the abacus in 
 points a, b, d, etc., and the surface of the column 
 in elements horizontally projected in a'', ^8*, S*, etc. ; 
 drawing rays through a, b, d, etc., of the abacus, 
 we have their corresponding shadows a, ;8, S, etc. 
 
 The lines of shade and shadow of the column are 
 determuied by its tangent planes of rays : these 
 lines are (t\\ t") and (ciO", o") : (w^'w^", w"} is the 
 line of shade of the abacus. The line of shadow 
 of the abacus on V, cast by the arc (y^w^", y''u''), 
 is a portion of an ellipse beginning at y and ex- 
 tending to a : here it is intercepted by the column 
 and its shadow as far as ej, where it beguis again, 
 and extends to w.^ ; w^w^ is the shadow of (w^v^", 
 w*) ; Wiw" is the shadow of the upper edge of the 
 abacus wn. 
 
 BRILLIANT POIKTS. 
 
 38. When a pencil of rays falls on a polished 
 surface S, one ray, at least, is usually reflected to 
 
 tion of the intersection of the cylinder by the plane of rays, 
 x2 + 22 — p2 . tijg equation of a circumference with o as a 
 centre. 
 
SHADES AND SHADOWS. 
 
 9 
 
 the eye of the observer, who thus sees on S one 
 or more hrilliant points. 
 
 General Construction. Let us noAV suppose S 
 to be represented on the vertical plane V. Let 
 R be the incident ray ; Z, perpendicular to V, 
 the reflected ray passing through the point of 
 sight ; X the brilliant point, and N a normal to 
 S at X. Then, according to a well-known prin- 
 ciple of optics, R, N, and Z must be in the same 
 plane ; also, R and Z make equal angles with N 
 on opposite sides. 
 
 The directions of R and Z being known, that 
 of N is found by bisecting the angle of R with Z ; 
 the brilliant point x will be the point of contact 
 of a plane tangent to S and perpendicular to N. 
 
 39. Peoblem 13. To find the hrilliant point 
 on a spherical surface. 
 
 This example is taken to illustrate the general 
 method given in Art. 38. 
 
 Fig. 26 represents a quadrant of a sphere, the 
 centre c being in GL ; R a ray of light to the cen- 
 tre e; R' is the position of the ray revolved to 
 coincide with H ; en is the bisecting normal in 
 its revolved position; and d the real position of 
 the required brilliant point. 
 
 40. Problem 14. To find the hrilliant point 
 on a surface of revolution. 
 
 Let the sui-face be given as in Fig. 21, to find 
 the brilliant point on its vertical projection. 
 Through any point of GL, as o, draw a ray ao ; 
 oz perpendicular to GL is the required direction 
 of the reflected ray, to pass through the point of 
 sight. Bisect the angle of ao with oz (Art. 40, 
 Des. G-eom.~). [Rotate ao about oz till it coin- 
 cides with H; it takes the position a'd; oh' bisects 
 a'oz \ making the counter revolution, it takes the 
 position oh.'] Any plane perpendicular to oh, 
 such as the one having for its traces on and op, is 
 parallel to the plane tangent to the surface at its 
 brilliant point ; denote this unknown plane by X ; 
 then, since the meridian plane M through the 
 brilliant point is perpendicular to X (Art. 133, 
 Bes. Greom.), it must be perpendicular to the 
 plane nop parallel to X ; therefore, ce perpendicu- 
 lar to op is the horizontal trace of M, and ex its 
 vertical trace ; again, X and nop being parallel, 
 their intersections with M mvist be parallel ; but 
 the intersection of M with nop is the line having 
 n and /' for its traces, and nr' as its revolved 
 position when M is rotated about ex so as to coin- 
 
 cide with V. Draw d"g tangent to U" and par- 
 allel to nr'. d" is the revolved position of the 
 point of contact ; making the counter revolution, 
 (d", d') is found at (c^*, d"}, d"d'' being parallel to 
 GL. Hence d is the brilliant point required. 
 
 This construction gives the brilliant point only 
 on the vertical projection of the surface : a simi- 
 lar construction will give the brilliant point on 
 the horizontal projection, these points being en- 
 tirely distinct from each other. 
 
 Although two tangents can be drawn to U" 
 parallel to nr', gd" only determines the real bril- 
 liant point. 
 
 When R" and R* are at 45° with GL, the angle 
 6"o2 = ecy = 20° 6' 14"i. 
 
 SHADOWS OF POINTS ON CTJEVED STJEFACES. 
 
 41. We have given (Art. 156, Des. G-eom.') a 
 general method of finding the point in which a 
 right line pierces a surface : we shall now apply 
 it to the cone and sphere. 
 
 Pkoblem 15. To find the shadow of a given 
 point on the surface of a cone. 
 
 Let a. Fig. 10, be the given point ; pass a plane 
 of rays through a so as to cut the simplest line 
 from the cone : for this purpose draw two rays 
 through a and o ; aj and Oi are their horizontal 
 traces, and o^s the trace of the auxiliary plane : 
 this plane cuts from the cone the element os; 
 which intersects the ray through a at the point a, 
 the required shadow. 
 
 42. Problem 16. To find the shadow of a 
 given point on the surface of a sphere. 
 
 Let the sphere be given as in Fig. 28, its centre 
 in GL, and let a be" the point. Pass through a a 
 plane of rays perpendicular to V, intersecting the 
 sphere in a circle having xz for its diameter; 
 rotating this plane about its vertical trace a^z 
 until it coincides with V, xa{z is the revolved 
 position of the circle, and a'z that of the ray. 
 a^a' = a^a'' ; a/ is the revolved position of the 
 shadow of a upon the sphere ; making the coun- 
 ter revolution, a/ returns to aj, which is the 
 
 1 Denote the angle a'oz by <j>, then tan ^ = 
 
 oa" 
 oa''cos45° V2, 
 b'o cos 45° 
 
 tan 2 = cosec f - cot ^ = ^1 - V/i ; tan h''oz = -^^ ^^^-^pp 
 
 1 * 1 V^ — 1 \/3-l 
 
 = -7= tan 5 = "7= ,- - —^ — = 0.3660234 = tan of 20° 0' 
 
 14". But 6*o« = r>'oc = 20° C 14". 
 
10 
 
 SHADES AND SHADOWS. 
 
 shadow of a. If the distance «„«'' = 8 = cCa! be 
 given, no construction on tlie liorizontal coordi- 
 nate plane is required. 
 
 NATURE OF THE LINE OF SHADE FOE SIMPLE 
 SURFACES. 
 
 43. We have already seen that the line of 
 shade of a sphere is the circumference of a circle, 
 that of a polyhedron is a broken line, v^hile that 
 of a cylinder or cone is made up of two elements 
 of the surface. 
 
 Problem 17. To find the line of shade of a 
 surface of the second order. 
 
 Let S denote such a surface having a centre, 
 and for the moment suppose the source of light 
 reduced to a point; then the line of shade is a 
 plane curve. 
 
 Proof. Let s, Fig. 35, be the luminous point, 
 and c the centre of the surface ; draw through s 
 and c a secant plane ; it cuts from the surface a 
 curve mdt. We obtain two points, m and t, of 
 the line of shade, by drawing the tangents sm and 
 St. Draw the chord mt, and join s and c ; sc in- 
 tersects the curve at x and the chord at n. Then 
 we know that the chord is parallel to the tangent 
 xy of the curve, and also that en X cs := W.^ 
 
 We conclude from this, that, whatever secant 
 plane is drawn through cs, the point n is always 
 the same. Also, /or each secant plane, we draw 
 through n a right line parallel to a tangent at x ; 
 but all these tangents at x belong to the tangent 
 plane to the curved surface at x, and the locus of 
 all the parallels to this plane drawn through n is 
 a plane through n parallel to the tangent plane 
 at X, and the intersection of S by this plane is 
 the line of shade, which is a curve of the second 
 order. 
 
 When s is at an infinite distance, as in the case 
 of parallel rays, the plane of the curve of shade 
 passes through the centre ; and if the surface of 
 the second order has no centre, the plane of the 
 curve of shade is parallel to the axis. 
 
 1 The second portion becomes evident wlien we see that uy 
 must be parallel to the diameter b' conjugate to ex, which we 
 will denote by a' ; then denoting en by x', and nm by y', we 
 have the equation of the tangent 7ns, a''^yy' ± V'-xx' — ± 
 a'^b'^ ; as s is on the axis of x, for the point s we have y ~ o 
 and . •. xx* = a? or en X cs = ex^, Q. E. D. 
 
 If mM were a parabola, then any diameter so must bisect 
 a system of chords parallel to iiy ; also sn = 2xn. 
 
 44. There are three general methods of finding 
 the line of shade of a surface. 
 
 1st, The method of secant planes. This consists 
 in intersecting the surface S by a series of planes 
 of rays cutting from S curves C, C", C", etc., and 
 from the surface of the illuminating pencil of rays 
 tangents to these curves: the points of contact 
 are points of the required line of shade. 
 
 The method of tangent planes already used may 
 be cited in this connection. 
 
 2d, The method of circumscribed surfaces. For 
 example, let. Fig. 29, a cone and cylinder circum- 
 scribe a sphere having the circumferences he and 
 ef as the respective lines of contact. Then the 
 plane of rays tangent to the cone along an must 
 be tangent to the sphere at m, and, therefore, n is 
 a point of the line of shade of the sphere : for a 
 similar reason, s is also a point of the same line of 
 shade. 
 
 Remark. Fig. 29 illustrates the fact that when 
 two surfaces are tangent, the lines of shade are 
 not therefore tangent, as an and st make acute 
 angles with rsnlc. 
 
 3d, The method of oblique projections. Let, 
 Fig. C, PI. v., C and D be two curves in space. 
 It is required to find the point on C which casts 
 its shadow on D. For this purpose we find, on 
 an auxiliary plane P, the shadows Ci and Di of the 
 two given curves, which are oblique projections 
 of C and D. The point of intersection x^ of Ci 
 and Di is the trace on P of the ray which meets 
 C and D ; its point of meeting x on C and x^ on 
 D are the points required. 
 
 The principle may be thus enunciated: To 
 determine the shadow cast by one curve upon 
 another. Find the oblique projections of the given 
 curves upon the same plane, and the points of inter- 
 section of these projections are the traces of the 
 projecting lines which intersect both of the given 
 curves. 
 
 45. To determine the points of the line of shade 
 situated upon the apparent contour. 
 
 The apparent contour of a surface is the base 
 of a cylinder circumscribing the surface and per- 
 pendicular to the plane of projection. Any curve, 
 as the line of shade, traced around the surface, is 
 tangent, in projection, .to the apparent contour, 
 the points of contact being the traces on the 
 plane of projection of the elements of shade of 
 the projecting cylinder : these traces are the 
 
SHADES AND SHADOWS. 
 
 11 
 
 points of contact of the apparent contour with 
 tangents drawn parallel to the projection of the 
 light. 
 
 The foregoing remarks will now be applied in 
 the solution of the following : — 
 
 46. Problem 18. To find the line of shade of 
 a torus. 
 
 Let the torus be represented as in Fig. 22, R" 
 and R* at 45° with GL. 
 
 Points on the contour lines X" and Y* are found 
 by drawing tangents to these lines parallel re- 
 spectively to R" and R*; thus determining the 
 points a", €«, j8*, 7j*, in Avhich the projections of 
 the line of shade are tangent to the contour lines 
 of the torus : a*, £*, /3*, and rf are found on X* 
 and Y" respectively. 
 
 The highest and lowest points are in a vertical 
 plane of rays P, drawn through the centre ; for 
 the illumination is the same on each side of P, 
 which therefore divides the line of shade, as well 
 as the surface, symmetrically. Again, P cuts from 
 the torus a meridian, and, from the illuminating 
 pencil, two rays tangent to it : rotating this me- 
 ridian about the axis until its horizontal trace 
 HP becomes parallel to GL, and drawing tan- 
 gents making an angle 6 with GL, we have the 
 revolved position x and z of the highest and low- 
 est points ; making the counter revolution, x and 
 z take the positions v and S. 
 
 Points on the profile meridian. On account of 
 the symmetry of position of the principal and 
 the profile meridians with respect to the direction 
 of the light, each point of shade on the first cor- 
 responds to a point of the second upon the same 
 parallel. Hence the arcs oA and ty determine 
 the required points A. and y. 
 
 47. Abridged method for finding the visible line 
 of shade on the vertical plane of projection, R" and 
 R* still being at 45° with GL. 
 
 Let the vertical projection of the torus be 
 given, as in Fig. 23. Tangents to the vertical 
 contour line parallel to R" determine the points ". 
 and e ; ey, drawn parallel to GL, determines y at 
 its intersection with the axis on. 
 
 In Fig. 22, oh = o/B" cos 45° = O.TojSft. Hence 
 0^, Fig. 23, = oe = dm cos 45° = .Tom. Draw to 
 the contour a tangent zn, making an angle 6 with 
 om, and produce it to the axis at n. z is the re- 
 volved position of the lowest point ; making the 
 counter revolution, n remains in the axis, and nz 
 
 takes the direction wS, parallel to R" ; drawing gS 
 parallel to om, we have S as the intersection of w8 
 and 28. We have then five points and three 
 tangents to aySySe, which are abundantly sufficient 
 for its construction in practice. 
 
 48. Problem 19. To find the line of shade of 
 a surface of revolution : general method. We have 
 shown, in the previous problem, how to find the 
 highest and lowest points of the line of shade, 
 and also those on the contour lines. The object 
 in the present problem is to show how to find 
 any point whatever of the line of shade. Let the 
 surface of revolution be given as in Fig. 27 (o^a, 
 0*), a vertical axis, bVe" the vertical projection of 
 the meridian curve, ef' its horizontal projection. 
 It is required to find any point of the line of shade, 
 as that situated on the parallel np. 
 
 49. First method, hy circumscribed cones. Draw 
 at w a tangent ns to the meridian curve, and let the 
 meridian curve and the triangle sin revolve about 
 the axis si. The meridian cvirve generates the sur- 
 face of revolution ; the tangent sn, a cone tangent 
 to it, having the circumference n4p as its line of 
 contact. Then the lines of shade of this cone 
 will determine two points on the circumference 
 of contact, which will be the points required. 
 
 Take the circle of contact np as the base of this 
 cone. Through the vertex of the cone (s, 0'') draw 
 a ray ; t is its trace on B, the plane of the base of 
 the cone ; ^''4'', i''3'' are the traces on B of the 
 planes of rays tangent to the cone ; 3* and 4* are 
 traces on B of the elements of shade ; projecting 
 these traces upon VB, we have 3 and 4 as the re- 
 quired points. 
 
 If the circle of the gorge G is assumed as the 
 circle of contact, the auxiliary cone becomes a 
 vertical cylinder. If two planes of rays be drawn 
 tangent to this cylinder, the traces of the ele- 
 ments of shade on G will be A* and t'', which will 
 be horizontal projections of points of the line of 
 shade ; X" and t" will be found in VG. 
 
 50. Second method, by inscribed tangent spheres. 
 Assume the same circle of contact np, and at 
 
 n" draw a normal n'o" to the meridian curve ; it 
 meets the axis at 0, which is taken as the centre 
 of a tangent sphere. The circle of shade of this 
 sphere is perpendicular to R; its trace on the 
 plane of the principal meridian is x^o°r, drawn 
 through perpendicular to R" ; the circle of shade 
 intersects the circle of contact B in a horizontal 
 
12 
 
 SHADES AND SHADOWS. 
 
 right line perpendicular to R, and piercing the 
 meridian plane at x ; 3V'4* is its horizontal pro- 
 jection, and the points 3 and 4, in which it inter- 
 sects the circumference of contact of B, are the 
 required points of the line of shade. 
 
 Remark 1. The point a;" may be determined 
 thus : Draw viPy parallel to R", and yx^ parallel to 
 «"«; for the three perpendiculars dropped from 
 the three angular points of the triangle vU'yo'' 
 upon the opposite sides must meet in the same 
 point x". 
 
 Remark 2. When R' and R* are at 45° with 
 GL, iy = in". 
 
 51. Third method, iy enveloping surfaces. In 
 Figs. 24 and 25 we have a sphere and a surface of 
 revolution with its inscribed sphere (np being 
 the assumed circumference of contact), both 
 illuminated by the same system of parallel rays : 
 the projections of the circumferences of shade on 
 each sphere are ellipses similar both in form and 
 position. Therefore drawing OP and PN respec- 
 tively parallel to op and pn, joining or and OA, 
 we determine y and 8 by drawing oy and o8 paral- 
 lel to or and OA respectively. 
 
 A surface of revolution M may be considered 
 as the envelope of the surface of a sphere S, the 
 centre moving on the axis of M, and the radius 
 varying according to a fixed law ; the consecutive 
 intersections of S are circumferences C, and the 
 point in which the line of shade of S intersects 
 the corresponding circumference C is a point of 
 the line of shade of M. 
 
 52. Problem 20. Having given a portion of 
 a surface of revolution convex toward the axis, it 
 is required to find the line of shadow cast hy the 
 circumference of the upper base upon the surface. 
 
 This problem is the continuation of Problem 
 19. Let the surface be given as in Fig. 31. To 
 find the highest point, draw through b a line bp, 
 making, with GL, the angle 0, which R makes 
 with H, and meeting the contour line in r and 
 the axis in p ; draw through r a horizontal line, 
 and through p a line parallel to R"; the intersec- 
 tion 3" is the vertical projection of the required 
 point ; its horizontal projection is 3* upon kz\ the 
 horizontal trace of the vertical meridian plane of 
 rays ; for bp is the revolved position of the ray 
 Sp. To find other points of the curve, pass hori- 
 zontal secant planes cutting the surface in cir- 
 cumferences W, X, Y, Z. The shadows of x on 
 
 these planes are s, Sj, Sj, s^ ; and the shadows of the 
 circumference of the upper base (bd, a^n^') upon 
 these planes, are the circumferences W2, Xj, Y2, 
 Z2, whence we have, as their intersections, the 
 points 12456789 as the points required. 
 
 53. Fillet'' s method of casting shadows by means 
 of a diagonal plane : R" and R* at 45° with GL.i 
 
 For brevity, we will designate the plane P, 
 Fig. 32, perpendicular to H, and making an angle 
 of 45° with V as a diagonal plane. 
 
 The shadow on the diagonal plane P of a line 
 ab, parallel to GL, is vertically projected in a2''b2^ 
 at 45° with GL, found by drawing rays through 
 a and b, and determining their traces a"2, a^, bi, 
 Sj* on P. 
 
 Let (Z, ffj) be the projections of an axis, and 
 the distance of a point b in front of that axis be 
 denoted by 8 = aj*^' ; then the point b^ may be 
 found without using the horizontal plane, in the 
 following manner : Draw a line through 5" paral- 
 lel to GL, intersecting Z at a;" ; lay off x"a2 = 8, 
 and draw Ui^cf at 45° parallel to R'' ; draw b^" at 
 45°, and their point of meeting V is the required 
 shadow: for the quadrilateral JVaj'Sj' equals 
 b^a^'a^'b^. If b is behind the axis, then 8 should 
 be laid off above x", instead of below it, as in the 
 figure. 
 
 54. Shadow of a circle on the diagonal plane. 
 Let the circle be parallel to H, Fig. 33, and let 
 
 it be circumscribed by the square ad. Then we 
 have the square a^d^ as its projection upon the 
 diagonal plane and any side as a2b2 = ab cos 45° 
 = a6 \/^ : hence its surface is one-half of tliat of 
 the original square, and the circle inscribed will 
 be one-half of the original circle. 
 
 If the circle had the position indicated in Fig. 
 34, having the diagonal plane passing through its 
 centre, the vertical projection of the shadow on 
 the diagonal plane is a circle whose radius is the 
 line V2 = ab cos 45°. 
 
 55. We shall apply this device in obtaining 
 rapid solutions of a number of problems. 
 
 Application 1st. Let it be required, Fig. 45, to 
 find the shadow of the circumference projected in 
 bd on the surface of revolution. Imagine a diag- 
 onal plane drawn through the axis of the surface. 
 Tlien the oblique projection of the circumference 
 bd is obtained as follows : DraAv bt at 45°, and zt 
 
 1 This method is due to M. Pillet, Professor at the J^coie 
 des Beaux Arts, Paris. 
 
SHADES AND SHADOWS. 
 
 13 
 
 perpendicular to it ; with s as a centre, describe 
 the arc tef: this is the required projection. In 
 the same way the oblique projection of np is xys; 
 they intersect at Ij and £2, which are the oblique 
 projections of points of the shadow. Drawing 
 lines I2I and 222 at 45° with GL, we have their 
 intersections 1 and 2 with jip as the points re- 
 quired. 
 
 66. Application M. To find the line of shade 
 of a cone. 
 
 Let it be required to find the line of shade of 
 the cone vertically projected in ade, Fig. 39, with- 
 out using the horizontal projection. Project the 
 cone on the diagonal plane P through its axis. 
 Draw dd^ at 45°, and o(4 perpendicular to it; 
 with as a centre and od^ as a radius, describe a 
 circle ; this is the projection of the shadow of 
 the base on P. The vertex a being in P is its 
 own projection; therefore, drawing as^ and ar^ 
 tangent to this circle, Ve have the vertical pro- 
 jection of the shadow of the cone on P. But the 
 projections of lines of shadow must be the shad- 
 ows of the lines of shade : hence, drawing the 
 projecting lines S2S and r^r at 45°, we have ar and 
 as as the required lines of shade of the cone. 
 
 67. Application 3d. To find the shadow of a 
 given point on the surface of a cone by the method 
 of the diagonal plane. 
 
 Let, Fig. b, PI. V., abe be a cone, and n the 
 projection .of the given point situated at a given 
 distance S in front of the axis ao. The projection 
 of the cone on the diagonal plane is aJa'^a ; that of 
 the point n is n^ [found thus: draw nt perpen- 
 dicular to the axis ; it meets it at ^ ; set off td = 
 8, and draw dn^ at 45° parallel to R* ; draw nn^ 
 parallel to R"] ; join a with n^, and produce it to 
 €2 on the circumference of shadow ; the point e 
 of the base, which casts the shadow e^, is found by 
 drawing eej at 45°. Now join a with e, and ae is 
 the element upon which the shadow of n is cast 
 at nj. 
 
 58. Problem 21. To find the shadow cast 
 upon a hollow cone by the complete circle of its base. 
 
 Let the cone be given as in Fig. 38. For brev- 
 ity, instead of saying the vertical projection of 
 the shadow of a point a upon the diagonal plane, 
 we shall say simply the projection of a upon P, 
 or the oblique projection of a. Take the diagonal 
 plane passing through its axis. Any plane of 
 rays through cuts two elements from the cone, 
 
 one of which is the shadow of the other, and pro- 
 jects both on the diagonal plane in the same line : 
 thus 0^2 and 0Z2 are the vertical projections of an 
 element and its shadow. Therefore, drawing a 
 ray through a, we have X2 as its oblique projec- 
 tion on the base of the cone ; 0x2 is the oblique 
 projection of oa, and ofej that of its shadow. Draw- 
 ing ^2^ at 45°, we have ot the shadow of oa ; produ- 
 cing ax2 to r, we have r as the shadow of the point 
 a. Draw oy2, 0%, and oy; then, drawing rays 
 through s and ^3, we have n and k as points of 
 the line of shadow. 
 
 Remark. The lowest point n may also be found 
 by drawing a line af making an angle with az, 
 and through its intersection s with oc a ray sn 
 intersecting the horizontal through / at n. The 
 point I projected on the axis is symmetrical with 
 k ; hence it is found at the intersection of a hori- 
 zontal line through k with oe (Art. 47). 
 
 59. An abridged solution of Problem 12 is 
 shown in Fig. 40. The highest point b is found 
 by drawing a line mo, making an angle with xf; 
 it is the revolved position of a ray in the meridian 
 plane of rays ; making the counter revolution, 
 the ray is vertically projected in ob ; drawing rb 
 parallel to xf., we have the highest point b. The 
 diagonal plane cuts the column along ed., its line 
 of shade, and the abacus in yh ; dk is the oblique 
 projection of the lower circle of the abacus, and 
 d is the point common to this shadow and the 
 line of shade ; here the ray becomes tangent to 
 the curve of shadow. The curve of shadow is 
 tangent to sx., and falls between s and r; c is at 
 the same height as a. Any intermediate points, 
 if desired, may be found by the method explained 
 in Art. 36. 
 
 60. Peoblem 22. Shadow of an abacus and a 
 quarter-round, or ovolo. 
 
 As an example of the use of the diagonal plane, 
 let it be required to determine the shadows in 
 Fig. 36. This consists of an abacus pV, whose 
 horizontal sections are squares, a quarter-round 
 or semi-torus lO'e, a fillet/^, and a cylinder vn. 
 
 1st, For the line of shade of the ovolo, the con- 
 struction is the same as that of Fig. 23. 
 
 2d, The shadow of this line aj8 45 on the diag- 
 onal plane through the axis ^s", is the curve 
 aySj?/, constructed by points as in Art. 53. a is in 
 the diagonal plane ; (the distance of /3 from the 
 axis is /85 ; laying off pz = jS5, and drawing a line 
 
14 
 
 SHADES AND SHADOWS. 
 
 zPi parallel to R* and a ray fiP^^ we haye /Sj at 
 their point of intersection; or drawing 6^3, we 
 have also /SySj = PP^. Again, the tangent at P2 
 makes an angle of 45° with mn : hence we have 
 the directions of three tangents to the curve ; viz., 
 at u, 90°, at ySj, 45°, and at y, 0° with mn. 
 
 3d, The shadow of the edge of the abacus pq on 
 the ovolo. This line casts a shadow qw on the 
 diagonal plane. The shadow of pq on the ovolo 
 is the section by a plane parallel to mn and 
 inclined to V at an angle of 45°. If this plane 
 rotates 45° about the axis of the capital, it be- 
 comes perpendicular to V, and will be entirely 
 projected along its trace pw, which will be the 
 shadow of the edge vertically projected at p. 
 (To see how this plane revolves, imagine (Fig. '4, 
 PL I.) R to revolve 45° about a vertical line 
 passing through its centre: would pass to <?", 
 c to i*, and the plane oo^ct would become o"ct''o, 
 having R" as its vertical trace. The shadow of 
 00" is o")?.) 
 
 We have the highest point, 10, of this shadow 
 by projecting 10' on zs". 
 
 The points 2 and 3 are obtained by taking the 
 intersections £3 and 32 of the auxiliary shadows 
 wq and ajB^, and projecting back by reversed rays 
 upon the line of shade of the ovolo. These two 
 extreme rays, since they are tangent to the ovolo 
 and lie in the intersecting plane of the curve 2, 
 10, 3, must also be tangent to the curve at the 
 points 2 and 3 respectively. 
 
 Bemarh 1. By symmetry, the point 1 is an- 
 other point of the line of shadow. 
 
 Remark 2. The edge of the abacus projected 
 at p, having pw for its shadow, meets the line of 
 shade at 16 ; hence 16. and 3 are at the same 
 height, for both are symmetrically situated with 
 respect to 4y. 
 
 4th. The shadow of the abacus on the column. 
 With w as a centre and a radius equal to zm, 
 '■"escribe an indefinite arc 67 ; it is the indefinite 
 shadow of the edge pq on the column. By Art. 
 59 we determine the shadow of /8345 on the col- 
 umn. 12y makes an angle 6, and determines 13 ; 
 14 and 15 are at the same height ; 8 is determined 
 by the intersection of the line of shade ?-8 with 
 -the curve '^ifi^iy; the intersection of the two 
 shadows at 7 gives the final shadow on the col- 
 umn, .', 6 7, 15, 8. 
 
 Remark 3. We have found the shadow by the 
 
 method of the diagonal plane ; but as a verifica- 
 tion, and to show the accuracy of the method, we 
 have drawn the plan, and passed a vertical secant 
 plane of rays through any point, as b, taken at 
 random on the edge of the abacus : this plane 
 has 6*0)* as its horizontal trace. It cuts from the 
 abacus the line (Ifc, &*) ; from the ovolo a curve 
 di (e, 0*) ; from the fillet (eo, 0*) ; and from the 
 column («<"", 0)*) ; (the point i is found by inter- 
 secting the surface by a horizontal plane ; X" and 
 X* are the horizontal circumferences it cuts from 
 the surface, and i is the point in which X meets 
 the vertical plane of rays). It is seen that c gives 
 the point q, and t the point 9. 
 
 61. Problem 23. To find the shades and 
 shadows on the base of a Tuscan column. 
 
 The base of such a column, represented in 
 elevation in Fig. 37, consists of three parts : viz., 
 1°, a rectangular prism gk, called a plinth, whose 
 horizontal sections are squares ; 2°, a torus mfx 
 3°, a cylindrical fillet rn. The shaft is concave 
 outward as far up as J: it then becomes nearly 
 cylindrical ; and, in casting the shadows, it is 
 considered as a cylinder above b. 
 
 The line of shade px of the shaft casts a shadow 
 upon the surface formed by the revolution of bsd 
 called the conge, and also on the torus. To find 
 any point of this shadow, we make use of the 
 diagonal plane. This plane passes through so 
 and px. Any horizontal plane ts cuts a circum- 
 ference from the conge, a part of which is pro- 
 jected on the diagonal plane in the arc Yy, inter- 
 secting pxiny; projecting y on ts by the ray «/£, 
 we have e as a point of the shadow of px on the 
 conge ; the shadow on the torus can be deter- 
 mined in the same manner. These shadows con- 
 sist, 1°, of the shadow of the line of shade 1 2 of 
 the fillet ; 2°, of that of the arc 1 4 ; and, 3°, of a 
 portion of py ; and are represented by the curve 
 li4,5. 
 
 SHADOWS ON SLOPING PLANES. 
 
 62. Problem 24. To find the shadow of a 
 given point on a given plane parallel to the ground 
 line. 
 
 Let, Fig. 41, a be the given point, and let the 
 given plane M pass through GL, and make a 
 given angle u. with the horizontal plane II. Let 
 R" and R* make 45° with GL. Pass a profile 
 plane through a, and revolve it about its vertical 
 
SHADES AND SHADOWS. 
 
 15 
 
 trace aVj until it coincides with V ; a is found at 
 a' ; a^m is the revolved position of the trace of M 
 on the profile plane P. Since the projection of 
 R on P makes 45° with c^a^^ it has aV (making 
 45° with GL) for its revolved position ; making 
 the counter revolution, a/ returns to a;, which is 
 the projection on P of the shadow of a on M : this 
 shadow therefore lies in a perpendicular xy to 
 a*ao, and also in cPy, the vertical projection of the 
 ray through a : hence it lies at their intersection 
 y. Through y draw hn parallel to a^m ; then, since 
 AqJ = «„«* = 8, we have the following rule for de- 
 termining immediately y, knowing a", u, and 8. 
 
 Rule. From a^ set off a^b = B; draw bn, 
 making the given angle a with GL, and through 
 a' a right line at 45° with GL : its intersection y 
 with bn is the shadow required. 
 
 63. Peoblbm 25. To find the shadow of a 
 given horizontal circle on a plane parallel to the 
 ground line, and making a given angle with the hori- 
 zontal coordinate plane. R" and R* at 45° with GL. 
 
 Let (Fig. 44) ad be a square circumscribing the 
 given circle, and let a be the angle of the plane 
 with H. The shadow of the centre o is found at 
 Oi by taking Oqo' = OqO* ; c/bi making an angle a 
 with aV, and o's at 45° with GL ; r^ti = ab ; cfii 
 =. b'y = ab cos a ; whence we have the parallelo- 
 gram (XiSicZ 1^1, and su ■=■ ab sin a ; the inscribed 
 ellipse is the shadow of the given circle. 
 
 64. PiEOBLEM 26. To find the shadows cast by 
 the chimneys and dormer window upon a sloping roof. 
 
 Let the roof be given as in Fig. 42 ; qrp =i a, 
 the angle of the roof with H. 
 
 1st, The shadow of the chimney. Fig. 43 shows 
 the plan of the chimney. Make l^m^ = twice the 
 distance of m in front of I ; lay off 0(,e and o^u = 
 Oqo', and Oqo", Fig. 43, and through y draw a ray 
 yyi, and through e and u lines parallel to rq, meet- 
 ing yyi in yi and y^ ; «/i and y2 are the shadows of 
 the vertices of the conical chimney-pots, and Oi 
 and Oj the centres of the shadows of their upper 
 bases: the shadows of these bases are ellipses 
 found by Problem 25. Through Iq and m^ draw 
 lines parallel to rq ; lj,i is the shadow of IJ, ; m^ is 
 the shadow of m ; m^^ is the shadow of mn, and 
 «i02 is the shadow of the edge vertically projected 
 at n. Lay off vx := ms. Fig. 43, and draw xz^ par- 
 allel to qr ; drawing a ray through t, we have t-^ as 
 the shadow of t, and t^o^ parallel to ab, as the 
 shadow of the edge us, Fig. 43. 
 
 2d, To find the shadow of the dormer window, 
 draw g'V; then 1 is situated at a distance A^^ in 
 front of ^f, and also the same distance to the 
 right of 4(1 ; hence, setting off 4J = twice 4olo, 
 and drawing 4o4i and Jlj, and also rays through 
 1, 2, 3, 4, we have A^A^, the shadow of 4o4, and 
 1 li the shadow of the edge vertically projected 
 at 1 ; in the same way the points 2i and 3i are 
 determined. 
 
 65. Peoblem 27. To determine the shadow of 
 a cornice. 
 
 Fig. 30, PI. IV., is the elevation of a cornice. 
 To find the shadows cast, draw ca' = ga' ; em' = 
 ym' ; tvf = zif ; then lines drawn through a', m', 
 v', parallel to ac, are the shadows of ac, me, and 
 uv respectively. 
 
16 
 
 SHADES AND SHADOWS. 
 
 THE HELICOID. 
 
 66. Properties of the warped Jielicoid. 
 
 We have seen (Arts. 234, 235, Des. Cieom.') how 
 to assume a point on the surface, and pass a plane 
 tangent to the surface at that point. Fig. 49 is 
 reproduced from Fig. 137 QBes. Greom.y, with a few 
 omissions, oep is the directing helix, (AA;*,/P) 
 an element making a constant angle a with the 
 axis, a the point of contact of a plane tangent 
 to the helicoid through this element, and dV the 
 horizontal trace of this tangent plane. Denote 
 the pitch of the helix by p. Draw through a* the 
 right line ty perpendicular to A;*c?*, and erect at 
 A a perpendicular At to AP- The two triangles 
 dVF' and a''Ai are similar, since their sides are 
 perpendicular: hence, 
 
 cW ^ Ag* . 
 a"k" At ' 
 
 0) 
 
 cZ* is the horizontal trace of the tangent to the 
 helix passing through a ; hence 
 
 a''d'' 
 
 P 
 
 2ir X Aa" ' 
 
 (2) 
 
 both expressions being the tangent of the angle 
 made by the helix with H. Multiplying (1) by 
 
 = cot a, whence At 
 
 (2), we have ||- 
 
 27rX At 
 
 = ^ tan a = c. If we put ^ =: h the reduced 
 
 pitch, we have c = A tan a.i 
 
 Hence, to draw a plane tangent to a warped 
 helicoid at any point a of a given element fk, it 
 is sufficient to draw At ^ <? perpendicular to 
 A/c*, to join t and a* ; then the perpendicular A*?/ 
 dropped from P upon to* produced, is the hori- 
 zontal trace of the required tangent plane. 
 
 Again, if the trace A*d* of the tangent plane 
 containing the element fk be given, to find the 
 point of contact of this plane, it is sufficient to 
 erect the perpendicular A* = c to Ak'', and drop 
 ti/ upon AW ; tt/ intersects AP at a*, the horizon- 
 tal projection of the required point of contact. 
 
 If the point of contact a moves along the ele- 
 ment fk, dV and At rotate around P and A 
 
 1 This relation is immediately obtained from the equation 
 of tlie base of the helicoid ; viz., r — K = c^ given in the note 
 
 to Art. 234, Des. Oeom. ; for differentiating — = subnormal 
 
 - At-c. 
 
 respectively, and when a reaches k, dV becomes 
 tangent to the spiral (j^k\ and the normal at P 
 will pass through the revolved position of t. 
 
 67. Peoblem 28. To find the line of shade on 
 the surface of a given helicoid. 
 
 Let aye, Fig. 46, be the directing helix, D a 
 given element making a given angle a. with the 
 axis (sf, o). The direction of the light indicated 
 by the arrows R", R*, is assumed at an angle 9 
 with the horizontal, and parallel to the vertical 
 coordinate plane. The shadow of D is Dj, found 
 by drawing a ray of light through any point, as 
 (o, /) ; this ray pierces H at /i", which, joined 
 with a*, gives Dj. 
 
 From s lay off on the axis, sa; = A =: ^, and 
 
 2ir 
 
 make sxt' equal to the given angle a, which every 
 element makes with the axis. Draw xr" parallel 
 to R", so that the angle xr"s = ; then h tan a =: 
 c ; denote sr" =: h cot by m. 
 
 Now, the plane of rays passing through D and 
 Di is tangent to the helicoid at some point of D. 
 To find this point, erect at o a perpendicular od to 
 D*, and make od =: c; then from d drop a per- 
 pendicular on Di, and the point s", in which it 
 meets D* produced, is the horizontal projection 
 of the point of contact z of the plane of rays with 
 the helicoid, i.e., a point of its line of shade. 
 Draw through s* and d lines parallel to GL ; pro- 
 duce Di to Zi and pd to r. Denote the angle 
 k'^oa'' by <^, and the distance oz* by p. The tri- 
 angles rod and a'^of' have their sides respectively 
 perpendicular : they are therefore similar, and we 
 
 have -^ = ~; but o/i* 
 oji" oa" 
 
 sf tan a 
 c cotO 
 
 sf{' = sf cote; oa^ = 
 od=. c\ hence ro =■ sf cot B ■ 
 
 sf tan a 
 
 tan 
 
 = h cot 6 = m. 
 
 The point r is called the centre of radiation, 
 and is a fixed point for the same values of h and 
 6. With as a centre and oc^ = c as a radius, 
 describe the circumference t^d: then, with r and 
 the distance c, we can at once find the horizontal 
 projection of the point of contact s* for any ele- 
 ment whatever, as D*, by erecting a perpendicular 
 od joining d, its intersection with the circumfer- 
 
SHADES AND SHADOWS. 
 
 17 
 
 ence M, with r ; the point s*, ia which dr crosses 
 D*, is the point required. In this way the curve 
 1032r03*4 is obtained.! 
 
 68. To find the point of the line of shade situated 
 upon any given helix traced upon the surface. 
 
 Given c, m, the position of r, and the base aeh. 
 Fig. 50, of a helix, it is required to find that point 
 of the line of shade which touches this helix. 
 From the centre o, with c as a radius, describe a 
 circle d'd"d"'. Let r be the centre of radiation, 
 ro being taken equal to m; draw roa; draw he 
 perpendicular to it ; join ad', and with o as a cen- 
 tre and TO as a radius, describe an arc r/, meeting 
 ad' produced in /. Now, if the whole system 
 revolve about a vertical axis at o until / coin- 
 cides with the fixed point r, ad'r' will take the 
 direction 1 dr 2, and the point 1 is the horizontal 
 projection of the shade on the element ol, making 
 with oh an angle <i> : 2, situated on another ele- 
 ment 2, is a second point. In the same way, by 
 drawing ad" and revolving it in the contrary 
 direction, we obtain the points 3 and 4, all, how- 
 ever, on different elements. Comparing Figs. 46 
 and 50, we see that the points of shade are situ- 
 ated on lines passing through r symmetrically 
 disposed with respect to ao, so that when one 
 point has been found the others can be immedi- 
 ately determined. 
 
 Join with /, and produce it to Z; then evi- 
 dently loh = hoi. = <^,2 which, with the assigned 
 value of p, determines the point 1. 
 
 1 The implicit polar equation of this curve is at once 
 obtained: thus, draw dl parallel to GL ; then by the similar 
 
 triangles j-ng* and rU, we have ??£_ z= ™ or P£2ii = 
 
 Id rl c sin ^ 
 m p sm <i> ^ j^ Cartesian coordinates, 
 TO + c cos 
 
 x — p cos (p p = v'x^ + 2/2 sm(j) = 
 
 y — p siu 
 
 cos (jl- 
 
 X 
 
 Substitutmg these values in the equation, we have "^^^l^ — -^ 
 
 cy 
 
 (m-y)\/x^ + y^ 
 
 — nr-, — 5 , ! whence mx^ifl + y^ = c [my — (x^ + y^)], 
 m\x^ + y^ + ex 
 
 or nAfi (x^ + y^) = c^ [{my - (a;^ + y'')^. 
 
 2 Put oa = p ; dlo - y ; ta,n y = - ; (1) 
 
 P 
 
 oa=c; lcol = <l>;^^^=sme; (2) 
 
 m 
 
 or = m;orl — e; ^ = 90° — y — e. (3) 
 
 The equations (1),- (2), (3) must be fulfilled for every value of 
 p, i.e., for every helix traced on the surface. 
 
 69. Fig. 47 shows this construction with the 
 values of c, m, and p, taken from Fig. 46. The 
 triangle txr = fxr" ; sa = p = oh''; join a and t, 
 and with s as a centre and a radius sr = to, de- 
 scribe the arc ro meeting at produced in ; join s 
 and 0, and with a radius =. sl=: oP describe the 
 arc lu ; then lu = W*, and determines the point 1. 
 It is evident that lu, in Fig. 47, is constructed in 
 exactly the same manner as Ih in Fig. 50.^ 
 
 70. Problem 29. To construct the shades and 
 shadows on the different parts of a screw, and the 
 shadow cast on the horizontal coordinate plane. 
 
 If the vertices of an isosceles triangle move 
 around an axis situated in the plane of the trian- 
 gle, and parallel to its unequal side so as to de- 
 scribe three helices of the same pitch,^ the equal 
 sides of the triangle will describe the surface of 
 the triangular threaded screw. 
 
 This surface, represented in Fig. 51, is gener- 
 ated by the isosceles triangle (V'nH^', ^V) moving 
 around the axis (Z", 0) so that the vertices de- 
 scribe the three helices Ip, ns, and l^pi. The 
 direction of the light is denoted by R. 
 
 If we attempt to apply directly the method 
 explained in Art. 69, we shall find the values of 
 c, h, and to to be so small as to preclude their use 
 in the construction : to obviate this difficulty, we 
 shall use these constants multiplied by ir. This 
 
 gives 
 
 E^m^ = ^j, = ^, Therefore, draw- 
 
 ing an indefinite horizontal line ur. Fig. 52, and 
 erecting a perpendicular sx = ^ ^ , and drawing 
 
 st parallel to Z^w", and xr parallel to R', we have 
 ts = TTC and rs = wro ; take saz=.-n- y, oV^, and draw 
 ato ; with s as a centre and a radius sr, draw the 
 arc ro ; then with s as a centre and a radius si = 
 oV', describe the arc lu. Draw, in Fig. 51, ou 
 parallel to R*, and from u lay off the arc u^ = 
 lu. Fig. 52, and we have the point 4* of the line 
 of shade, whence we obtain the vertical projec- 
 tions 1", 4" on the vertical projection of the helix. 
 The points 2, 3, and 5 are obtained in the same 
 manner. Drawing oX perpendicular to ou, the 
 points 7* and 8* will be symmetric with 4* and 5*. 
 
 1 Dropping the perpendicular sz upon at, we have tan y = 
 
 c ^)_ psiny _ ^^ ^ (2), and ^ = 90° - y - e (3). 
 p m 
 
 2 The two vertices adjacent to the unequal side describe 
 different spires of the same helix. 
 
18 
 
 SHADES AND SHADOWS. 
 
 SHADOWS. 
 
 The shadow of the screw on the coordinate 
 plane H is limited by that of the outer helix, 
 which is determined by points ; e.g., the points 
 4i, 7], and 9i are the shadows of 4, 7, and 9. 
 
 Shadow cast on the surface hy the curve of shade. 
 
 1st, To find the shadow cast upon the surface 
 by the curve 4, 5, we use the method of oblique 
 projections (Art. 44) ; the shadows upon H of 
 the curves 4 5 and 7 8 are 4i 5i and 7i 8i ; these, 
 as shown in Fig. 54, intersect in a point yi; there- 
 fore, drawing a ray through y-i, we determine the 
 point yoi which casts its shadow y upon 7 8. The 
 curves 4 yj and y 7 have real shadows yi^i and 
 
 The curve 7 8 is not illuminated. The curve 
 5 z/o casts the shadow 5" d" y on the upper fillet. 
 To find any intermediate point as d" : Draw an ele- 
 ment ae ; its shadow is e^^iflj, intersecting 4i 6i at 
 di, which, projected back by a reversed ray on to 
 ae, gives the required point d", the shadow on the 
 upper fillet of a point of the curve 4 5. 
 
 71. Shadow of a helix. Fig. 48 shows the 
 oblique projection or shadow of a helix. Ima- 
 gine the helix to be illuminated by rays whose 
 direction is indicated by the arrows ; the shadow 
 
 . of any point nis n^^; draw a^'s = E.^ — ; 
 
 ^ The equation of the curve may be deduced thus : Assume 
 a* as the origin, o*A; and a*6 as the axes of X and T respec- 
 tively ; put r = 0*6 ; A = %r^ = tlie reduced pitch ; 6 = 
 
 02"fcoa'', the angle of the light ; ^ = a^ohiK Then, for the 
 shadow of any point n, we have n\, whose coordinates are x — 
 mt = [nooo = n 0" = n*i] -|- ooa", or a; = 0ft cot 9 — r sin ^ ; 
 y — r + (^t=zr — r Qos^. This is the equation of a curtate 
 
 02*^0 parallel to R" ; sq parallel to h^a" ; hd = h 
 arc 6wV; join o^d and project q upon it at e; 
 draw ef then the curve aSi^k is a curtate cycloid 
 described by the point a\ carried around by the 
 circle iqf rolling along the line ii'.'^ 
 
 72. In Fig. 53, the form of several spires of 
 the shadow of the outer helix of the triangular 
 threaded screw is shown. By Art. 71, os =. h 
 cot 6 ; but. Fig. 52, sr =■ irh cot 6 ; and sa = Trp; 
 
 if we take ai = p = ou, Fig. 51, we have 
 
 TTp 
 
 = —; ox ij =: h cot 0. 
 
 73. To show the form (not the position) of the 
 shadow of the outer helix phi, Fig. 51, we choose 
 at random a centre o, Fig. 53, and with a radius 
 ij. Fig. 52, describe the circle smz ; draw ol and 
 OS parallel and perpendicular respectively to R* ; 
 draw sy parallel to ol, and ov parallel and equal 
 to yv, Fig. 51^; erect at " an indefinite perpen- 
 dicular ; with as a centre and a radius ou, de- 
 scribe an arc meeting vn in n; n is the initial 
 point of the displaced shadow (its real position 
 being obtained by moving the figure parallel to 
 itself until o coincides with o. Fig. 51) ; the 
 curve is described by the extremity of the radius 
 vector on when the circumference zms rolls along 
 the line sg. 
 
 cycloid, in which r is the describing radius, and h cot B that 
 of the rolling circle. 
 
 1 For by construction oa'-s = £ = M ; e/ = — cot 9 ; (i6 = 
 
 '!^ : since ^ = ^, we have fo* = h cot B. The angle 6o*(Z 
 2 ^ /o* 
 
 = tan !I = 57° 31' 06". 
 2 
 
 2 V is the vertical projection of the horizontal trace of the 
 outer helix pU, found by producing p^bH" until it meets GL. 
 
SHADES AND SHADOWS. 
 
 19 
 
 ABEIDGED CONSTRUCTIONS 
 
 OF THE PROBLEMS 
 SHADOWS. 
 
 OF SHADES AND 
 
 R" AND R* AT 45° WITH THE GeOTJND LiNE. 
 [The theoretical explanations of the constructions, having been already fully given, are omitted.] 
 
 Fig. 0. B represents an opaque body sepa- 
 rated by the curve C into the illuminated por- 
 tion and the shade ; C is the line of shade, M 
 the shadow on S, and Cj the line of shadow. 
 
 Fig. 1. To find the shadow of a on V. De- 
 note Ga* by S. Lay off a^o = 8, draw through o 
 a line parallel to GL, and set off on it a distance 
 oai = 8 : Ui is the shadow required. 
 
 Fig. 2. To find the shadow of c on H. De- 
 note (?"« by 8. Lay off (^'x = 8, draw through x 
 a line parallel to GL equal to 8, and its extremity 
 Ci is the shadow required. 
 
 Fig. 3. To find the shadow of A on V. aj and 
 &i are the shadows of a and b, and the line ai 6i, 
 or Aj, is the shadow required. 
 
 Fig. 4 shows the conventional direction of the 
 ray of light R to be the diagonal oc of a cube so 
 placed as to have two of its faces parallel to the 
 vertical, and two to the horizontal plane of pro- 
 jection. Consequently R" and R* make angles 
 of 45° with the co„; again, R', the projection of 
 R on the profile plane oo„, makes an angle of 45° 
 with o„o''. 
 
 Denoting oco'' by 0, we have approximately 
 tan e = 0.3; cosec 6 = 1.73 ; sin 5 = 0.577 ; and 
 cos 26 = ^ exactly. 
 
 Fig. 5 shows that the angle is easily con- 
 structed by assuming a point (o*, o") on the ray 
 (R*, R"), and revolving this ray around ex until 
 it coincides with the vertical plane V, whence we 
 have ocG = 0. 
 
 Fig. 6. Let abed be a square parallel to V : it 
 is required to find its shadow on H. Denote the 
 given distance te" by 8, and the length of the side 
 flAj" tiy I ; make a*w =l8, no ^ os ^l; erect at n, 
 0, and s, three perpendiculars to GL, and lay off 
 on the first the distance 8, on the second the dis- 
 tances 8 and 8 -|- Z, and on the third the distance 
 8 -\- 1: the extremities of these perpendiculars will 
 be the angular points sought ; viz., Ci, d^, a^ &i, 
 whence the shadow is known. 
 
 Fig. 7. To find the shadow of A on V. Let 
 ^A* = S. Lay off tx = 8, and erect a perpen- 
 dicular Ai to GL at x; Aj is the shadow of the 
 unlimited line A. 
 
 Fig. 8. To find the shadow of B on V. The 
 shadow of any point of B as (o*, B") is Oi, which 
 joined with B* gives the required shadow Bj. 
 
 Fig. 9. To find the shade and shadow of the 
 prism on V. The visible shade is (fdHn. The 
 points 0, p, q, and r are found by laying off on 
 GL the following distances: mo = mb''; np = 
 we* ; Iq = Id'' ; and nr = we* ; erecting perpendic- 
 ulars to GL at 0, p, q, r, and drawing the vertical 
 projections of rays through b", e", cZ", their inter- 
 sections with the respective perpendiculars give 
 the points b^eidiBi, which, with SiV, determine the 
 shadow on the vertical plane. 
 
 Fig. 10. To find the shade and shadow of the 
 cone. Draw a ray through the vertex o; Oj is its 
 horizontal trace. Drawing OiW* and OjC* tangent 
 to the base of the cone, we have the horizontal 
 traces of the two tangent planes of rays; the 
 lines of contact are oe and on ; hence noe is the 
 line of shade. The portion of the shade visible 
 on the horizontal projection is oHc^ ; that visible on 
 the vertical projection is oH^n". The shadow on 
 H is c^OiJi*. 
 
 Fig. 11. To find the shade and shadow of a 
 sphere. Denote the diameter of the sphere by d : 
 then qs = on z= d sin 6 = 0.577 d; OiUi = d co- 
 sec 6 = 1.732c?. Fig. a shows that es = oe sin 9 
 
 = e\ Fig. 11 ; oc = ^ = oa' = oa". <j> = 6. 
 
 Fig. 12. To find the shadow of a niche upon 
 its interior surface. The line of shadow is partly 
 composed of the arc of an ellipse extending from 
 /, the point of contact of the tangent ray, to x, 
 having for its semi-axes zf, the radius, and zo = 
 ^zf; any point of xdi" as Si" may be found by 
 passing a ray (^s''s{', s*8i*) through s, and finding its 
 trace (si", s*) on the surface of the cylinder ; Sj^w, 
 is the shadow of dn, a portion of (^a", a*), upon 
 the interior surface of the cylinder ; n is found 
 by drawing w,w parallel to R". 
 
 Fig. 13 shows the shadow of the edge of a 
 hemispherical shell upon its interior surface, to 
 
 be a semi-ellipse having ec* and c*** =: -— for its 
 axes when 6 — 35° 14' 52". 
 
20 
 
 SHADES AND SHADOWS. 
 
 Fig. 14. To find the shadow of the circle hfo. 
 The shadow of e is Cj ; the shadows of the diam- 
 eters he and (^gf, /,) are kye^ and figi, two conju- 
 gate diameters of the ellipse of shadow. Taking 
 fn = gf, joining n with c and producing it to o, 
 no and nx are equal to the axes of the ellipse of 
 shadow SjSi and Z2^\ \ filso SiC^Ci = y = ^nck^. 
 
 Fig. 15. To find the shadow of the circle in 
 the profile plane given by its centre c, and its 
 diameter fg = ek. The shadow of e is Cj, that 
 of the horizontal diameter (^ek, c") is eiki, that of 
 the vertical diameter (c*, fg} is e*(/i ; S1S2 and ZiZj 
 are found from Fig. 14. 
 
 Fig. 16. To find the axes a and b of an ellipse 
 when a', b' and o are given. From the extremity 
 n, of one diameter, drop a perpendicular nf on 
 the other produced ; from n set off distances nl = 
 nr = ce; join with I and r, and draw through n 
 a parallel nd to cr, intersecting cl in ; make ox = 
 0(^ = ol, and join c with d and a; ; make ca = «c?, 
 and e6 = nx, then cJ and ca will be the new axes 
 both in magnitude and position. 
 
 Fig. 17. To find the shade of the cylinder, 
 and the shadow of its upper base on the interior 
 surface. Denote the radius of the base of the 
 cylinder by p; then_6^ = p versin 45° =: ^p'; 
 0% = p; c^gi = p\/2; the angle j/jcVi =: 45° ; 
 making e'n = 2p, and with c" as a centre describ- 
 ing the semi-circumference, xe^o, no, and nx are 
 the lengths of the axes of the ellipse of shadow, 
 and the angle wcV = 2y. 
 
 Simplified construction. Lay off c^t" = cV = 
 ■j^P, and draw the elements r'p and c's ; set off 
 eV, = eV, and c^g = 2cV ; draw eji/i and gli par- 
 allel to GL; their intersections with f^b and r^p 
 determine «/i and Z,. We then have the five 
 points i", e , Zj, «/], and r", and their corresponding 
 tangents t^g, e%, gli, fy^, and a right line through 
 r" parallel to t^g. 
 
 Fig. 19. To find the shadow of a rectangular 
 abacus on a cylindrical column, and on V. Pass 
 a plane of rays through the axis of the column ; 
 0*8* is its horizontal trace ; this plane intersects 
 the edge ab in s, and fl'o is the vertical projection 
 of the ray through s. Denote the length of the 
 radius of the column by p. 
 
 With as a centre and p as a radius, describe 
 an arc s^ny ; this arc will be the vertica,l projec- 
 tion of the shadow of ab on the column ; take 
 zr° =. -^-gp, and 2% =: c^z ; erecting a perpendicular 
 at Xi, and drawing yu at 45°, we have Mj. Draw- 
 ing rays through b and e, we complete the outline 
 by drawing Uibi parallel, and Jie, perpendicular, 
 respectively, to GL. 
 
 To find the shadow of a given point a on a 
 cylinder. Draw through a" a line parallel to GL, 
 meeting the axis of the cylinder in b'. Set off 
 
 ^ If the diameter is O"".!, then ^ of O^.OS is accurate to 
 within O^.OOOS. 
 
 b'o = aH, the distance of a in front of the axis : 
 is the centre of Sy'ny, and the radius is that of 
 the cylinder. Drawing through a" a line parallel 
 to R", we have the shadow aj of a upon the cyl- 
 inder, without drawing any line on H. If the 
 cylinder were hollow, and the shadow were re- 
 quired upon the interior surface, b'o would be 
 laid off above «'i', instead of below it. 
 
 Fig. 20. To find the shades and shadows of 
 the abacus and column in Fig. 20. 
 
 Let the column be intersected by vertical planes 
 of rays cutting the lower base of the abacus in 
 points a, b, d, etc., and the surface of the column 
 in elements horizontally projected in a*, /S*, S*, etc. ; 
 drawing rays through a, b, d, etc., of the abacus, 
 we have their corresponding shadows a, /8, S, 
 etc. 
 
 The lines of shade and shadow of the column are 
 determined by its tangent planes of rays : these 
 lines are (e"«', t*) and (ejo", 0'') : {jifw^, w*) is the 
 line of shade of the abacus. The line of shadow 
 of the abacus on V, cast by the arc (jfw{, y^uF), 
 is a portion of an ellipse beginning at y and ex- 
 tending to a : here it is intercepted by the column 
 and its shadow as far as «!, where it begins again, 
 and extends to w^ ; w^uo^ is the shadow of (w^w^, 
 w'*) ; w iw" is the shadow of the upper edge of the 
 abacus wn. 
 
 Fig. 21. To find the brilliant point d on a 
 surface of revolution, ao is the incident, oz the 
 reflected ray, and ob the line bisecting the angle 
 aoz. on and op are respectively perpendicular to 
 a'o and oJ*. ce, perpendicular to op, gives r*, 
 which revolves to / ; w/ is. parallel to the tan- 
 gent to U", passing through the revolved position 
 d" of the brilliant point d. 
 
 Fig. 26. To find the brilliant point on a spher- 
 ical surface. 
 
 Fig. 26 represents a quadrant of a sphere, the 
 centre c being in GL ; R a ray of light to the cen- 
 tre c; R' is the position of the ray revolved to 
 coincide with H ; en is the bisecting normal in 
 its revolved position ; and d the real position of 
 the required brilliant point. 
 
 Fig. 22. To find the line of shade of a torus. 
 
 Points on the contour lines X" and Y'' are found 
 by drawing tangents to these lines parallel re- 
 spectively to R' and R*; thus determining the 
 points a", €", li\ rf, in which the projections of 
 the line of shade are tangent to the contour lines 
 of the torus : a*, «*, ^, and tf are found on X* 
 and Y" respectively; drawing tangents making 
 an angle with GL, we have the revolved posi- 
 tion X and z of the highest and lowest points; 
 making the counter revolution, x and z take the 
 positions v and 8. 
 
 The arcs oA and ey determine the points A and 
 y on the profile meridian. 
 _ Fig. 23. Abridged method for finding the visible 
 line of shade on the vertical plane of projection. 
 
SHADES AND SHADOWS. 
 
 21 
 
 Tangents to the vertical contour line parallel 
 to R" determine the points <* and c; £7, drawn 
 parallel to GL, determines 7 at its intersection 
 with the axis on. Draw to the contour a tangent 
 zn, making an angle with om, and produce it to 
 the axis at n. z is the revolved position of the 
 lowest point ; making the counter revolution, n 
 remains in the axis, and nz takes the direction n8, 
 parallel to R" ; drawmg zS parallel to om, we have 
 S as the intersection of n8 and 28. We have then 
 five points and three tangents to a/^y^^, which are 
 abundantly sufficient for its construction in prac- 
 tice. 
 
 Fig. 27. To find the line of shade of any sur- 
 face of revolution. Assume np as the circle of 
 contact of the surface with that of an inscribed 
 sphere, and at n' draw a normal rn'o" ; lay off iy 
 ;= in", and draw yx" perpendicular to n^o^. a^ is 
 found on ey*. Through a/' draw a line perpen- 
 dicular to R*, intersecting the circumference 
 ^AgApA in the points 3* and 4*, which are the hori- 
 zontal projections of 3 and 4, the points required. 
 
 In Figs. 24 and 25 we have a sphere and a sur- 
 face of revolution with its inscribed sphere («p 
 being the assumed circumference of contact), both 
 illuminated by the same system of parallel rays : 
 the projections of the circumferences of shade on 
 each sphere are ellipses similar both in form and 
 position. Therefore drawing OP and PN respec- 
 tively parallel to op and pn, joining or and OA, 
 we determine 7 and S by drawing 07 and oS paral- 
 lel to or and OA respectively. 
 
 Fig. 81. To find the shadow of the circle 
 (^bd, al'nd^) on the surface of revolution. To 
 find the highest point, draw through h a line bp, 
 making, with GL, the angle 0, which R makes 
 with H, and meeting the contour line in r and 
 the axis in p ; draw through r a horizontal line, 
 and through p a line parallel to R" ; the intersec- 
 tion 3" is the vertical projection of the required 
 point ; its horizontal projection is 3* upon kz^, the 
 horizontal trace of the vertical meridian plane of 
 rays. To find other points of the curve, pass hori- 
 zontal secant planes cutting the surface in cir- 
 cumferences W, X, Y, Z. The shadows of x on 
 these planes are «, Sj, S3, 84 ; and the shadows of the 
 circumference of the upper base (bd, a'^nd'') upon 
 these planes, are the circumferences W2, Xj, Yj, 
 Z2, whence we have, as their intersections, the 
 points 1 2 4 5 6 7 8 9 as the points required. 
 
 Shadows of Points on Curved Surfaces. 
 
 Fig. 10. To find the shadow of the point a on 
 the cone. Draw rays through a and ; a^ and Oi 
 are their horizontal traces, and o^s the trace of 
 their plane : this plane cuts from the cone the 
 element os ; which intersects the ray through a 
 at the point <i, the required shadow. 
 
 Fig. 28. To find the shadow of a point, a, on 
 the surface of a sphere. 
 
 Pass through a a plane of rays perpendicular 
 to V, intersecting the sphere in a circle having xz 
 for its diameter; rotating this plane about its 
 vertical trace a"z until it coincides with V, xa{z 
 is the revolved position of the circle, and a'z that 
 of the ray. a"a' = «„a* ; a{ is the revolved posi- 
 tion of the shadow of a upon the sphere ; making 
 the counter revolution, a/ returns to aj, which is 
 the shadow of a. 
 
 Fig. 29. The line of shade of an inscribed and 
 circumscribed surface. For example, let, Fig. 29, 
 a cone and cylinder circumscribe a sphere having 
 the circumferences be and ef as the respective 
 lines of contact. Then the plane of rays tangent 
 to the cone along an must be tangent to the 
 sphere at n, and, therefore, w is a point of the 
 line of shade of the sphere : for a similar reason, 
 s is also a point of the same line of shade. 
 
 Remark. Fig. 29 illustrates the fact that when 
 two surfaces are tangent, the lines of shade are 
 not therefore tangent, as an and st make acute 
 angles with rsnk. 
 
 Fig. C The method of oblique projections. Let, 
 Fig. C, PI. v., C and D be two curves in space. 
 It is required to find the point on C which casts 
 its shadow on D. For this purpose we find, on 
 an auxiliary plane P, the shadows Cj and Dj of the 
 two given curves, which are oblique projections 
 of C and D. The point of intersection x^ of Ci 
 and Di is the trace on P of the ray which meets 
 C and D ; its point of meeting 2; on C and x^ on 
 D are the points required. 
 
 Fig. 30, PI. IV., is the elevation of a cornice. 
 To find the shadows cast, draw ca' = go! ; erd = 
 ywl ; ti/ = zv' ; then lines drawn through a', m', 
 v', parallel to ac, are the shadows of ao, me, and 
 uv respectively. 
 
 PilleVs method of casting shadows by means of a 
 diagonal plane. 
 
 The shadow on the diagonal plane P of a line 
 ab, parallel to GL, is vertically projected in a^h^ 
 at 45° with GL, found by drawing rays through 
 a and b, and determining their traces a°2, a^, b^, 
 
 V on P. 
 
 Let (Z, Oj) be the projections of an axis, and 
 the distance of a point h in front of that axis be 
 denoted by 8 = aaV' ; then the point b^ may be 
 found in the following manner: Draw a line 
 through 6" parallel to GL(, intersecting Z at a;" ; 
 lay off x^a^ = 8, and draw a^c" at 45° parallel to 
 R* ; draw h'b^ at 45°, and their point of meeting 
 
 V is the required shadow. If b is behind the 
 axis, then 8 should be laid off above a;", instead of 
 below it, as in the figure. 
 
 Fig. 33. Shadow of a circle on the diagonal 
 plane. 
 
 Let the circle be parallel to H, and let it be 
 circumscribed by the square ad. Then we have 
 
22 
 
 SHADES AND SHADOWS. 
 
 the square a^d^ as its projection upon the diag- 
 onal plane and any side as a^h^ = ah cos 45° 
 = a6 V'i = hence its surface is one-half of that of 
 the original square, and the circle inscribed will 
 be one-half of the original circle. 
 
 If the circle had the position indicated in Fig. 
 34, having the diagonal plane passing through its 
 centre, the vertical projection of the shadow on 
 the diagonal plane is a circle whose radius is the 
 
 line 
 
 ah cos 45°. 
 
 Fig. 45. Application 1st. Let it be required to 
 find the shadow of the circumference projected in 
 hd on the surface of revolution. Imagine a diag- 
 onal plane drawn through the axis of the surface. 
 Then the oblique projection of the circumference 
 hd is obtained as follows : DraAV ht at 45°, and zt 
 perpendicular to it ; with s as a centre, describe 
 the arc tef: this is the required projection. In the 
 same way the oblique projection of np is xi/s; 
 they intersect at Ij and £3, which are the oblique 
 projections of points of the shadow. Drawing 
 lines I2I and 222 at 45° with GL, we have their 
 intersections 1 and 2 with np as the points re- 
 quired. 
 
 Fig. 39. Application 2d. Let it be required 
 to find the line of shade of the cone vertically 
 projected in ade. Draw ddi at 45°, and od^ per- 
 pendicular to it , with as a centre and od^ as a 
 radius, describe a circle ; drawing as2 and arj tan- 
 gent to this circle, we have the vertical projection 
 of the shadow of the cone on P ; drawing the 
 projecting lines SjS and r.^r at 45°, we have ar and 
 as as the required lines of shade of the cone. 
 
 Fig. b. Application 3d. To find the shadow of 
 a given point on the surface of a cone hy the method 
 of the diagonal plane. 
 
 Let, Fig. b, PI. V., aho be a cone, and n the 
 projection of the given point situated at a given 
 distance S in front of the axis ao. The projection 
 of the cone on the diagonal plane is aJj^z ; that of 
 the point nis, n^; join a with n^, and produce it to 
 eg on the circumference of shadow ; the point e 
 of the base, which casts the shadow ej, is found by 
 drawing ee^ at 45°. Now join a with e, and ae is 
 the element upon which the shadow of n is cast 
 at «i. 
 
 Fig. 38. Application Ijih. To find the shadow 
 cast on the surface of a hollow cone hy the complete 
 circle of its hase. 
 
 Drawing a ray through a, we have X2 as its 
 oblique projection on the base of the cone ; ox^ 
 is the oblique projection of oa, and 0^2 th^t of its 
 shadow. Drawing tjt at 45°, we have ot the 
 shadow of oa ; producing ax^ to r, we have r as 
 the shadow of the point a. Draw oy^, oz^, and 
 oy ; then, drawing rays through s and k^, we have 
 n and k as points of the line of shadow. 
 
 Fig. 35 is a diagram used to demonstrate the 
 fact that the line of shade of any surface of the 
 second order is a curve of the same order. 
 
 Fig. 40 shows an abridged construction of the 
 shadow of the cylindrical abacus on its column. 
 The highest point h is found by drawing a line 
 mo, making an angle ^with xf; it is the revolved 
 position of a ray in the meridian plane of rays ; 
 making the counter revolution, the ray is ver- 
 tically projected in oh ; drawing rh parallel to xf, 
 we have the highest point h. The diagonal plane 
 cuts the column along ed, its line of shade, and 
 the abacus in yk ; dk is the oblique projection of 
 the lower circle of the abacus, and d is the point 
 common to this shadow and the line of shade ; 
 here the ray becomes tangent to the curve of 
 shadow. The curve of shadow is tangent to sx, 
 and falls between s and r; c is at the same height 
 as a. Any intermediate points, if desired, may 
 be found by the method previously explained. 
 
 Fig. 36. As an example of the use of the diag- 
 onal plane, let it be required to determine the 
 shadows in Fig. 36. 
 
 1st, For the line of shade of the ovolo, the con- 
 struction is the same as that of Fig. 23. 
 
 2d, The shadow of this line aj8 46 on the diag- 
 onal plane through the axis as", is the curve 
 a^22/' Again, the tangent at 1^2 makes an angle 
 of 45° with mn : hence we have the directions of 
 three tangents to the curve ; viz., at a, 90°, at ySj, 
 45°, and at y, 0° with mn. 
 
 3d, The shadow of the edge of the abacus pq on 
 the ovolo. This line casts a shadow qw on the 
 diagonal plane. Qpw is the shadow on V of the 
 edge vertically projected at p.') 
 
 We have the highest point, 10, of this shadow 
 by projecting 10' on zs°. 
 
 The points 2 and 3 are obtained by taking the 
 intersections 22 and 32 of the auxiliary shadows 
 wq and a^^y, and projecting back by reversed rays 
 upon the line of shade of the ovolo. These two 
 extreme rays are also tangent to the curve at the 
 points 2 and 3 respectively. 
 
 Remark 1. By symmetry, the point 1 is an- 
 other point of the line of shadow. 
 
 Remark 2. The edge of the abacus projected 
 at p, having pw for its shadow, meets the line of 
 shade at 16 ; hence 16 and 3 are at the same 
 height, for both are symmetrically situated with 
 respect to ^y. 
 
 4th. The shadow of the abacus on the column. 
 With w as a centre and a radius equal to zm, 
 describe an indefinite arc 67 ; it is the indefinite 
 shadow of the edge pq on the column. By Fig. 
 40 we determine the shadow of y3345 on the col- 
 umn. 12y makes an angle 6, and determines 13 ; 
 14 and 15 are at the same height ; 8 is determined 
 by the intersection of the line of shade ?-8 with 
 the curve %ifii2>.iy; the intersection of the two 
 shadows at 7 gives the final shadow on the col- 
 umn, V, 6, 7, 15, 8. 
 
 Remark 3. We liave found the shadow by the 
 method of the diagonal plane ; but as a vcrifica- 
 
SHADES AND SHADOWS. 
 
 23 
 
 tion, and to show the accuracy of the method, we 
 have drawn the plan, and passed a vertical secant 
 plane of rays through any point, as b, taken at 
 random on the edge of the abacus : this plane 
 has 6*0)* as its horizontal trace. It cuts from the 
 abacus the line (6"c, J'') ; from the ovolo a curve 
 di (e, o*) ; from the fillet (eo, o'') ; and from the 
 column (xo)", to*). It is seen that c gives the 
 point Cj, and t the point 9. 
 
 Fig. 37. To find the shades and shadows on 
 the base of a Tuscan column. 
 
 The line of shade px of the shaft casts a shadow 
 upon the surface formed by the revolution of bsd 
 called the conge, and also on the torus. To find 
 any point of this shadow, we make use of the 
 diagonal plane. This plane passes through zo 
 and px. Any horizontal plane ts cuts a circum- 
 ference from the conge, a part of which is pro- 
 jected on the diagonal plane in the arc V?/, inter- 
 secting pxin y ; projecting y on ts by the ray ye, 
 we have e as a point of the shadow of px on the 
 conge ; the shadow on the torus can be deter- 
 mined in the same manner. These shadows con- 
 sist, 1°, of the shadow of the line of shade 1 2 of 
 the fillet : 2°, of that of the arc 1 4 ; and, 3°, of a 
 portion of py ; and are represented by the curve 
 li4.5. 
 
 Shadows on Sloping Planes. 
 
 Fig. 41. The shadow of a, on the plane M 
 passing through GL, and inclined to H at an 
 angle a, is thus found : Denote «o** by 8. From 
 «(, set off a|,6 = 8 ; draw bn, making the given 
 angle a with GL, and through a" a right line at 
 45° with GL: its intersection y with bn is the 
 shadow required. 
 
 Fig. 44. To find the shadow of the circle 
 inscribed in the square ad, on a plane passing 
 through GL, and inclined to H at an angle «. 
 
 The shadow of the centre o is found at Oj by 
 taking o„(j' = 0(,o* ; </bi making an angle a with 
 a'c', and o's at 45° with GL ; rit^ = ab ; Ciby = b'y 
 := ab cos a ; whence we have the parallelogram 
 afiidiCi, and su =: ab sin a; the inscribed ellipse 
 is the shadow of the given circle. 
 
 Fig. 42. To find the shadows cast by the chim- 
 neys and dormer window upon a sloping roof. 
 
 Let the roof be given as in the figure ; qrp = a, 
 the angle of the roof with H. 
 
 1st, The shadow of the chimney. Fig. 43 shows 
 the plan of the chimney. Make l^ms = twice the 
 distance of m in front of I ; lay off o^e and o„u = 
 0^0, and o^p". Fig. 48, and through y draw a ray 
 yy-i, and through e and u lines parallel to rq, meet- 
 ing y^/i in yi and y^ ; y^ and y^ are the shadows of 
 the vertices of the conical chimney-pots, and Oy 
 and Oj the centres of the shadows of their upper 
 bases : the shadows of these bases are ellipses 
 found by Fig. 44. Through ^o and wij draw lines 
 
 parallel to rq ; l^^ is the shadow of l^ ; m^ is the 
 shadow of m ; «i,Wi is the shadow of mn, and n^o^ 
 is the shadow of the edge vertically projected at 
 n. Lay off vx = ms. Fig. 43, and draw xz^ par- 
 allel to qr ; drawing a ray through t, we have t^ as 
 the shadow of t, and t-fi-^ parallel to ab, as the 
 shadow of the edge ms. Fig. 43. 
 
 2d, To find the shadow of the dormer window, 
 draw ^V ; then 1 is situated at a distance 4olo in 
 front of ^V, and also the same distance to the 
 right of 4o; hence, setting off 4J ^ twice 4olo, 
 and drawing 4o4i and Jlj, and also rays through 
 1, 2, 3, 4, we have 4o4i, the shadow of 4o4, and 
 1 li the shadow of the edge vertically projected 
 at 1 ; in the same way the points 2, and 3i are 
 determined. 
 
 THE HELICOID. 
 
 Fig. 49. Let p = the pitch of the helix de- 
 scribed by every point of om ; a, its angle with 
 the axis. 
 
 If we put ^ = A the reduced pitch, we have 
 
 G ■=■ h tan a. 
 
 To draw a plane tangent to a warped helicoid 
 at any point a of a given element fk, it is suf- 
 ficient to draw Ai = e perpendicular to AA;*, to 
 join t and a* ; then the perpendicular W'y dropped 
 from ^*.upon ta^ produced,. is the horizontal trace 
 of the required tangent plane. 
 
 Again, if the trace PcZ* of the tangent plane 
 containing the element fk be given, to find the 
 point of contact of this plane, it is sufficient to 
 erect the perpendicular At =zc to Ak\ and drop 
 ty iipon P5* ; ty intersects AP at a\ the horizon- 
 tal projection of the required point of contact. 
 
 Fig. 46. Or — m = h cot 6, being the angle 
 of the rays with H. The point r is called the 
 centre of radiation, and is a fixed point for the 
 same values of h and 0. With o as a centre and 
 o(? = c as a radius, describe the circumference 
 t'^d: then, with r and the distance e, we can at 
 once find the horizontal projection of the point 
 of shade s'' for any element whatever, as D'', by 
 erecting a perpendicular od joining d, its inter- 
 section with the circumference t''d, with r; the 
 point g*, in which dr crosses D*, is the point 
 required. In this way the curve 10322rO''4 is 
 obtained. 
 
 Fig. 60. Given e, m, the position of r, and the 
 base aek, of a helix, it is required to find that point 
 of the line of shade which touches this helix. 
 From the centre o, with c as a radius, describe a 
 circle d!d"d!". Let r be the centre of radiation, 
 ro being taken equal to m ; draw roa ; draw ke 
 perpendicular to it ; join ad', and with o as a cen- 
 tre and TO as a radius, describe an arc r/, meeting 
 ad' produced in /. Now, if the whole system 
 revolve about a vertical axis at o until /coin- 
 
24 
 
 SHADES AND SHADOWS. 
 
 cides with the fixed point r, ac?V' will take the 
 direction 1 dr 2, and the point 1 is the horizontal 
 projection of the shade on tlie element ol, making 
 with ok an angle <^ : 2, situated on another ele- 
 ment o2, is a second point. 
 
 Fig. 47 shows this construction with the values 
 of c, m, and p, taken from Fig. 46. The triangle 
 ixr = fxr" ; sa = p ^ o/c* ; join a and t, and with 
 s as a centre and a radius sr = m, describe the 
 arc ro meeting at produced in o ; join s and o, and 
 with a radius := si := ok'* describe the arc lu; 
 then lu = Ai*l*, and determines the point 1. 
 
 Fig. 51. The surface here represented is gener- 
 ated by tlie isosceles triangle (hifU^, V'n''') moving 
 around the axis (Z", o) so that the vertices de- 
 scribe the three helices Ip, ns, and 4^2- The 
 direction of the liglit is denoted by R. 
 
 Drawing an indefinite horizontal line ur, Fig. 52, 
 
 and erecting a perpendicular sx = '' , and draw- 
 
 ing it parallel to ?"»", and xr parallel to R', we have 
 tH := ire and rs ^ -nm ; take sa = t x o?*, and draw , 
 ato ; with s as a centre and a radius sr, draw the 
 arc ro ; then with s as a centre and a radius si = 
 oZ*, describe the arc lu. Draw, in Fig. 51, ou 
 parallel to R'', and from u lay off the arc m4* =^ 
 lu., Fig. 52, and we have the point 4'' of the line 
 of shade, whence we obtain the vertical projec- 
 tions 1", 4", on the vertical projection of the helix. 
 The jDoints 2, 3, and 6 are obtained in the same 
 manner. Drawing oX perpendicular to o?(, the 
 points 7* and 8'' will be symmetric with 4* and 5*. 
 
 Shadows. The shadow of the screw on the 
 coordinate plane H is limited by that of the 
 outer helix, which is determined by points ; e.g., 
 the jjoints 4i, 7], and 9, are the shadows of 4, 7, 
 and 9. 
 
 Shadow cast on the surface hy the curve of shade. 
 
 1st, To find the shadow cast upon the surface 
 by the curve 4, 5, we use the method of oblique 
 
 projections (Art. 44) ; the shadows upon H of 
 the curves 4 5 and 7 8 are 4, 5, and 7i 8i ; these, 
 as shown in Fig. 54, intersect in a point yi ; there- 
 fore, drawing a ray through i/,, we determine the 
 point 2/oi which casts its shadow y upon 7 8. The 
 curves 4 2/o and yl have real shadows z/i4i and 
 z/i7,. 
 
 The curve 7 8 is not illuminated. The curve 
 5,yo casts the shadow 5° d^ y on the upper fillet. 
 To find any intermediate point as d' : Draw an ele- 
 ment ae ; its shadow is e'^h^ai, intersecting 4i 5i at 
 di, which, projected back by a reversed ray on to 
 ae, gives the required point J^ the shadow on the 
 upper fillet of a point of the curve 4 5. 
 
 Fig. 48. Shadow of a helix. Imagine the helix 
 to be illuminated by rays whose direction is indi- 
 cated by the arrows ; the shadow of any point n 
 is Wi, and the curve of shadow ahijc is a curtate 
 cycloid described by the point a*, carried around 
 by the circle iqf rolling along the line ii'. 
 
 If, in Fig. 52, we take ai = p = ou, Fig. 51, 
 we have ij = the radius of the rolling circle. 
 
 Fig. 53. To show the form (not the position) 
 of the shadow of the outer helix pbl. Fig. 61, we 
 choose at random a centre o, Fig. 53, and with a 
 radius iJ, Fig. 52, describe the circle S7nz ; draw ol 
 and OS parallel and perpendicular lespectively to 
 R*; draw sy parallel to ol, and ov parallel and 
 equal to yv. Fig. 51 ; (v is the vertical projection 
 of the horizontal trace of the outer helix pbl, 
 found by producing p'b'l" until it meets GL;) 
 erect at v an indefinite perpendicular ; with o as 
 a centre and a radius ou, describe an arc meeting 
 vn in w; n is the initial point of the displaced 
 shadow (its real position being obtained by mov- 
 ing the figure parallel to itself until o coincides 
 with 0, Fig. 51) ; the curve is described by the 
 extremity of the radius vector on when the cir- 
 cumference 2ms rolls along the line sg. 
 
Win. Walsoji. 
 
 SHADES AND SHADOWS. 
 
 PI. I. 
 
Wm. Watson. 
 
 SHADES AND SHADOWS. 
 
 PLU. 
 
 .A.N/' 
 
 Fig. 11 
 
 Fi^.l2. 
 
 Ii6. 13. 
 
AVm-Watson. 
 
 SHADES AND SHADOWS. 
 
 PLm 
 
 Yi6. 19. 
 
 -^ kC 
 
 Fi^. 20 .'^^^„, 
 
^'in. Wat son. 
 
 SHADES AJND SHADOWS 
 
 pliy; 
 
Wm. Walson. 
 
 SHADES AND SHADOWS 
 
 P1.V. 
 
 Fig.b. 
 
 Fig. 32. 
 
 Fi^. 31. 
 
Wm. Wat soil. 
 
 SHADES AND SHADOWS. 
 
 PL VI 
 
 FiA. 36 
 
Will .Wat son. 
 
 SHADES AND SHADOWS. 
 
 PLvn. 
 
 .F^ 
 
 FiA.43. 
 
 u s 
 
 ri6.4'4' 
 
Wm.Watson. 
 
 SHADES AND SHADOWS. 
 
 PI. Yin. 
 
 Fid. /fc6. 
 
Wm.Watsor 
 
 SHADfJS AND SHADOWS. 
 
 PI. IX. 
 
 Fig. 51 
 
 \ \ 
 
 V"-- 
 
 '<'jn' 
 
 V^ 
 
 '-'-\i 
 
 
 
 1 
 
 / 
 
^;««TWy^''=^ffl«MtaGSK »f^^