= 7, dR Es = ates = The person charging this material is re- sponsible for its return to the library from which it was withdrawn on or before the Latest Date stamped below. Theft, mutilation, and underlining of books are reasons for disciplinary action and may result in dismissal from the University. UNIVERSITY OF ILLINOIS LIBRARY AT URBANA-CHAMPAIGN WAR 4 1 1976 Wits (4 ity! Y i 7 RED NOV 29 4975 aImVE a ft Rr NUY CF iweu Ain P oO g¢ ~ Ten. s Pt 4 AUS ~* ot — MLE me 7, : gi | Le L161—0O-1096 Digitized by the Internet Archive In 2022 with funding from University of Illinois Urbana-Champaign https://archive.org/details/treatiseonshadesOO0davi_1 TREATISE ON SHADES AND SHADOWS, LINEAR PERSPECTIVE. BY CHARLES DAVIES, PROFESSOR OF MATHEMATICS IN THE MILITARY ACADEMY, AND AUTHOR OF THE COMMON SCHOOL ARITHMETIC, DESCRIPTIVE GEOMETRY, AND ELEMENTS OF SURVEYING. SECOND EDITION. PUBLISHED BY WILEY & PUTNAM, COLLINS, KEESE & CO., New-Yorx,—STATIONERS COMPANY, Boston,—THOMAS, COWPERTHWAITE & CO., PHILADELPHIA,—BELKNAP & HAMERSLEY, Hartrorp,— CUSHING & SONS, BALTIMORE,—TRUMAN & SMITH, CINCINNATI. 1838. 4 : 4 bt VD PR Fre Pig nai naar spy ede By jie ey 4 * mative ki ‘ere ar cs WRC eckeictin » eter he et ni esta Hie ‘ a con : f n Bic £ si , she wy a SEP 83 ng EXCHANGE o 3 bk“wA | pty 5 LA R > ATHEMATICS ‘ : LIBRARY. PREFACE. Wiruin the last few years a great change has taken place in public sentiment, as to the importance which should be given to mathematical studies in comparison with other branches of education. Until recently it was thought that mere practical rules, unaccompanied by demonstration, were abundantly sufficient for all useful applications of mathematical science, and that the mind of the scholar could find richer nutriment in Virgil and Homer, than in the propositions of Euclid, or the sublime theories of Newton. But it is auspicious to the cause of sound learning, that these opinions have given place to more rational views of education; that we are at length convinced it is better to reason than merely to remember; and that the value of an education is to be estimated by the ability which it gives to the mind of thinking profoundly and reasoning correctly. In presenting to the public the following Treatise on Shadows and Perspective, the author cannot but flatter himself that he shall add something to the common stock of useful knowledge. The subjects treated of are certainly useful: to the architect and draftsman a | A 2 426240 ‘ IV PREFACE. knowledge of them is indispensable. To find with ma thematical accuracy the lines of shade and shadow on a complicated building,—which parts are to be darkened, and which parts are to be made light in a drawing of it, is certainly a difficult problem unless it be solved on scientific principles. The art of Perspective teaches us how to represent on a surface one or more solid bodies, in such a manner that the picture shall exhibit the same appearance as is presented by the objects themselves. It is by this art that the painter is enabled to present to the eye the almost living landscape, with its hills, its valleys, its waterfalls, and its rich foliage, varied by the beautiful tints of colour, and relieved by alternate light and shade. It is this art which has stamped the canvass with the intelligence of the human countenance, and caused it to be looked upon as the remembrancer of departed worth and the record of former times. It is this art which presents in a panorama a city in all its proportions, and causes the spectator to feel that he almost partici- pates in its bustle and business. This art also enlarges the pleasures of sight, the sense through which the mind receives the most numerous and pleasing impres- sions. Without perfect accuracy in the perspective, the pro- portions of objects cannot be preserved; and the skill of the draftsman, or the genius of the painter, is exerted in vain, if nature be not correctly copied. PREFACE. V The manner of finding the shadows of objects, and the common methods of perspective, depend on mathe- matical principles, and are susceptible of demonstration. The want of a work demonstrating those principles rigorously, has long been felt, and especially in the Military Academy, where the subjects have been taught by lecture for several years. If the one now submitted be found not to merit the approbation of the public, the author hopes it will at least be received with indulgence. The author would beg leave to express his acknow- ledgments to two or three friends, to whom he is indebted for drafts of the diagrams; and also, to repeat his ex- pressions of thankfulness to the Cadets for the interest they have taken in the work. But for their liberality it could not have appeared Military Academy, } West Point, March, 1832. Ra i Zz.) z. gh ae _ ‘aildhsehiinisaml Se fe sitheniebers ¥ nt ano ous FE atits, pri t od te ade ’ Soh etme et vr pee ae hisriy tah nial 88 oft ? 2a ti sana i ; “tRotetih ats 24 pr wie bi tt sadll Heat 1 Bet . Ye) > 4 ie 5 y, aS» vk it i r io a) a . “1 i Pi ft i” re el ee | wo = a | Qo oO» Ww faered ~J 18. 19. 20. 21, 22. SO DSH oP ww m CONTENTS. CHAPTER I. DEFINITIONS AND FIRST PRINCIPLES. CHAPTER II. APPLICATIONS AND CONSTRUCTIONS. - To-find the Shadow of a Right Line . . 2... « «1% . To find the Shadow of the Abacus and Pillar . . : . To find the Shadows of the Chimneys and House . - To find the Shade and Shadow of the Cylinder . . To find the Ellipse from its Conjugate Diameters . To find the Shadow of the Rectangular Abacus ona Golan . To find the Shadow of a Cylindrical Abacus on a Column - To find the Shadow of a Cylindrical Abacus on a Wall . To find the Shadow of the Inverted Frustum ofa Cone . . To find the Shade and Shadow of a Sphere . ° . Of Brilliant Points. . . boise . To find the Shade and Shaner of the Ellipsoid - . To find the Shadow on the Interior of the Niche . To find the Curve of Shade on the Torus . To find the Shadow on a Surface of Revolution . To find the Curve of Shade on the Surface of eosin . To find the Line separating the dark from the illuminated part of a Surface of hey and the Shadow cast by the Surface To find the Shades atid Shedawe on the ae eo Shaft of the Doric Column . To find the Shades and ee on ihe Capital of ane Doris Column sil Reh an IR pan ngs Sl Of the Helicoid . a hes were To draw a Tangent Plane to nna Helicoid “ Fea BRA ae Pgs: D GW ce ert K’ SHADES AND SHADOWS. oF the vertical plane of projection, the plane of rays through it will also be perpendicular to the vertical plane; and hence its vertical trace will be the line h’'Z'F", parallel to the vertical projection ofthe light. This plane ofrays in- tersects the horizontal plane of the eaves in a line which is perpendicular to the vertical plane at F”, and of which I’, perpendicular to the ground line, is the horizontal projection. It also cuts the line of intersection of the side roofs in a point whose vertical projection is /’ and. whose horizontal projection is £; and hence LF’, kf’ are the horizontal] projections of the intersections of the plane of rays through (hq, h’) with the side roofs; which intersections are the indefinite shadows cast by (hq, h’) on these roofs. ‘The shadows are limited in horizontal projection by the lines hn, and gm,drawn parallel to the horizontal projection of the light. Therefore kn and km are the horizontal projections of the shadows cast on the side roofs by the line (hq, h’). It may be here remarked, that all planes of rays which are perpendicular to the vertical plane of projection are parallel to each other, and will consequently intersect the side roofs in lines parallel to k¥" and kF. Drawing through m the line ms, parallel to the ground line, and through 2, 7s parallel to the horizontal projec- tion of the light, gives ms for the shadow of the lne | (iq, fh’) on the side roof. ‘The line us,is the horizontal projection*of the shadow cast by the perpendicular (2, ff’). This shadow is in the trace of a vertical plane of rays through (2, ff"), and is limited at u by the intersec- tion of a plane of rays through (/7, /”) with the side roof. The line wv, lying in this intersection, is the shadow cast bya part of the line (f7,f") on the side roof. The shadow at v is Cast by that point of the back edge of the chimney which is vertically projected at o. JA TREATISE. ON From v the shadow is cast by the back edge of the chimney. Returning to the shadow cast by the point (A, /), we find, first, the shadow ng,-which is-cast by the perpen- dicular (h, Wh"), and then draw through « a parallel to the ground line; this parallel i is the indefinite shadow cast by the line (fh, /’h’) on the side roof. We next find the point in which a ray of light through the point whose vertical projection is p, pierces the end roof, which is at (¢,/). Joining the point ¢ with the point in which the parallel through x meets the intersection of the side and end roofs, gives the shadow cast by the line (fh, fh’) on the end roof. From ¢ the shadow is cast by the edge of the chimney. We find the vertical projection of this shadow by pro- jecting its points into the vertical traces of the planes to which they respectively belong. ‘Thus m is verti- cally projected at ms ats,uatu,vatv,y at 7, k at kijnatn,and «at 2’. That part of the shadow which is on the front roof is all that can be seen in vertical projection. The boundary of that on the back roof is therefore dotted. The shadows of the other chimney are found ina manner so entirely similar as hot to require a particular explanation. Let us now find the shadows on the walls of the house. Through (BC, ac), the upper line of the convex part of the eave-trough, conceive a plane of rays to be passed. Its trace on the front wall limits the shadow cast by the eave-trough. The point (6, 6’) is in the trace of this plane, and 6’e isthe required line of shadow. The shadow b'z is cast by the corresponding line of the end eave- trough, which also casts a. shadow on the end wall that is vertically projected at z. «& SHADES“ AND SHADOWS. 25 - To find the shadow cast brine house on the horizontal Alar of projection. Through (CE, C’), the upper ‘line of the end eave- trough, conceive a plane of rays. to be passed. ‘The line LE is its trace on the horizontal plane, and consequently the shadow of the line ( The small vertical Tine (GC) casts its shadow Le’ in the line CL. » Through ¢ draw cd parallel to the “ ground line, and ities it till H meets Dd drawn par- allel to the horizontal projection of the light.” Then dc’ is the shadow cast by the le (DC,D’c), and the ray through (D, D’) touches the corner of the house at e, and pierces the horizontal plane at d. The line aE’, drawn through FE’ parallel to the ground line, and limited by Aa’ drawn parallel to the horizontal projection of the light, is the shadow cast on the hon- « zontal plane by the line (AE, A’C’). The line (A, aA’) casts the shadow a'r; a part of the line (BA, a) casts the shadow rr: from 7’ the line of shadow is cast by the corner of the house below z. The shadow on the horizontal plane which lies under the eave-trough and cornice, is not seen in Bs os projection. * PROBLEM IV. Having ‘given an oblique cylinder, and the direction of the hight, it 1s required to find the shade on the exterior surface, the shadow of the upper circle on the interior surface, and. the shadow of the cylinder on the horizontal plane. 27. Let (AB, A’B’), (PI. 3. Fig. 1), be the axis of the cylinder, and suppose the projections to be made as seen in the figure. af AM tiles 2 Qe OR ee = ‘ the : ae a "% me i nih Mp * ®. > . at bg o is * . ot og ‘ t 7 «’ * ¥s ; f ae : Mane aa = ths - 28 OY ans raga TSH ON * - ‘. If we suppose two abe planes of rays to be drawn: — pe to the surface of the eylifider, the elements of contact . . will separate the dark from the illuminated part of the. me -.» surface, and will consequently be lines of shade. yf In order to pass these planes, we first-pass aplane of. - *. rays through the axis of the cylinder, to which the ha ~- tangent planes will both be parallel. | “* Through (B,B’), a point of the axis, conceive arayof _ . light to be drawn—it pierces the horizontal plane at C. The avis of the cylinder pierces the horizontal plane at — A—therefore AC is the horizontal trace of a plane of - rays passing through the axis of the cylinder. ‘Let two'lines be now drawn tangent to the base of the cylinder, and parailel to AC. ‘These lines are De and Gf", and aré also the horizontal traces of the planes of rays tangent to, the surface of the cylinder. The - elements of contact pierce the horizontal plane at D and ~G. Hence, DE and GF, drawn parallel to AB, are the horizontal projections of the elements of contact: and by projecting D and G into the ground line at D’ andG’, and E and F into the vertical projection of the upper circle at E’ and F’, we determine D’E’ and G’F’, the vertical projections of the elements*of contact. The ~* points D, A and G,are in the same straight line perpen- dicular to AC; and DHG is a semicircle. | The light falls'on the half of the exterior surface cor-, vs responding to the semicircle DHG—the remaining half | - is inthe shade. Thelines (DE, D'E’,)\(GF, GF’) are. lines of shade, and being also elements of the surface, are generally called elements of shade. In the horizontal projection, we do not see the i & » which is on the under surface of the cylinder—therefore, we darken only that part which is included between the” elements DE and LN. In the vertical projection, we Legs oe wre ~ . iz | i - 3 * a . . ¥ z et BOGE pie & “ gh. a 5 | + ; 7 # a, ? %, x ? : Ss : ae re Rees AE we. Sep ge ; eh = - a eee 1s - Gag ee 2S aug org ; ahaa eae PRP BOR <_< = 4 - vs 2 2 r »- ean. Pies San Ue * ye ws ~ * u Re, rd we ; . * * . ” * wars pr * 2 * L? ITC Aal rome Suede, he Peart e * q ~ Sg ‘ ~~ % ; Lo * * ’ b Fra * 4 ~— * * oe 4 % % ae 4 * * 9 Wet ee ——* es oe = - a ete a GIRL NOU ite 9 Senor Ie hE Bi = aneteded ET: aay . —_ en. — os es “wes, = _- 3 Pi + \’ DR \\ \ b a 8 & ™ Cet f Ae ge ” - Ps + ¢ ae * + *, . te * . di a + ~~, + i 7 4 » , 7, ¥ 4 *» , a’ + ¢ $s. ; . ™~. © eB - « ! a a * a a . SHADES AND SHADOWS. : %. 27 see that part of the shade included between the element G’F’, and the element P’K’, only that portion of the surface, therefore, is shaded. ‘2 To find the shadow cast by the upper circle on the interior surface. . This shadow begins at the points (4, E’), EE) in which the elements of shade meet the upper circle, and is cast by the semicircle whose horizontal projection is. dak. If the surface be intersected by planes of rays parallel to the axis of the cylinder, each plane so drawn will in- - tersect the surface in two rectilinear elements. If then, through the upper extremity of the element towards the source of light, a ray be drawn, it will be contained in the plane of rays, and will, consequently, mtersect; the © other element lying in that plane—the point of inter- section is a point of shadow on the: interior surface. But the horizontal traces of all planes of rays parallel to the axis of the cylinder are parallel to AC, the trace of the’plane of rays through the axis. Therefore, * draw any line, as IP, parallel to AC, to represent the ». trace of a plane of rays parallel to the axis. ‘This plane, intersects the surface in two elements, whose horizontal projections are Id and PK. ‘Through d, the upper ex- tremity of the element towards the source of light, draw df parallel to the horizontal projection of the light —the point f, in which it meets.PK, is the horizontal pro- jection of a point. of shadow on the mner surface. The vert.cal projection of this point is found by pro- jecting P into the ground line, drawing the vertical, pro- jection of the element on which the shadow falls, and noting its intersection with the ‘perpendicular to the ground line, drawn through the point f; or,by pro- jecting the point dinto the upper base, drawing through ae « ; a . - + : a =e ” 23 ; TREATISE ON @ d' the vertical anaes of a ray of hght, and noting its intersection f’ with the vertical projection of the element on which the shadow falls. If it be required to find the shadow on any particular element, as the one, for example, whose horizontal pro- jection is LN, we have merely to pass a plane of rays through the clement, and determine the other element towards the source of light, in which it intersects the surface. ‘Then, through the upper extremity of the element casting the shadow, draw a ray of light, and where it meets the given element, is the penned point of the shadow. If we take the element whose “hori- zontal projection, is LN, Riis the horizontal trace of the plane of rays passing through it—the element towards the source of light, in which this plane inter- sects the surface, is horizontally projected*in Ré, and g is the horizontal projection of the pointof shadow cast by its upper extremity. | | At g, the horizontal projection of the curve of shadow is tangent to the line LN: for the vertical plane . of which LN is the horizontal trace, is tangent to the ‘ vertical cylinder, which projects the curve ef shadow, * along the element that pierces the horizontal plane at g. The lowest point of the curve of shadow, or that point which is farthest from the upper circle, is found by passing a plane of rays through the axis. For, since all the rays of hight make equal angles with the horizontal] plane, or with the plane of the upper base, the distance of the points of shadow below the upper base will increase, as the distance between the element casting and the element receiving the shadow, is increased. But, the plane of rays through the axis intersects the surface in elements farther from each other than the elements cut out by any other of the parallel planes of rays. Hence, ? * ° ”* 4 i ; f » &? - bral Bis SHADES AND SHADOWS. 29 it determines the lowest point’ of shadow, which is (A, h’). ig Having found as many points of the shadow as may be necessary, let the projections of the curve be accu- rately described. . In horizontal projection, the curve of shadow is seen from fF until it crosses the horizontal projection of the upper circle, thence to the point Eit is concealed by the surface of the cylinder. The part of the inner sur- face lying above the curve of shadow, receives the light ; the part lying below it is in the shadow. That part of the surface which is in the shadow, and seen, is shaded inthe drawing. With respect to the vertical projection, no part of the curve of shadow is seen, since it lies en- tirely on the inner surface of the cylinder. . To find the shadow of the cylinder on the horizontal plane. — 7 “The lines 6f the cylinder which cast shadows on the horizontal plane are, the two elements of shade, and , the half of the upper circle which is opposite the source | » of light, and of which ENF is the horizontal pro- jection. | The shadows cast by the elements of shade fall in the horizontal traces of the tangent planes of rays—hence, we may consider De and Gf”, as the indefinite shadows of the elements of shade. If through the upper eircle of the cylinder we con- ceive a cylinder of rays to be passed, its axis will be a ray of light passing through the centre’(B, B’), and will pierce the horizontal plane at C. But, since the plane of the upper circle and the horizontal plane are parallel, they will intersect the surface of the cylinder of rays in equal circles. Therefore, a circle described with C as a centre, and a radius equal to BK’, will be the ® w * zc”)!lU OS ee iy +" ai a + ™ a + + ri ei 4 =. ’ - ¥ é ¥ % a" .” re . agi . Bee, * oe ‘ a ? A é il %.. * * ‘ . 7 a b 30 TREATISE ON shadow cast by the upper circle on the horizontal plane. The shadows cast by the elements of shade will be tangent to this circle of shadow at the points f” and e; ‘which points are found by drawing a line through C per- pendicular to AC, or by drawing the horizontal projec- _tions of rays through the points F and Ey ’ ” Although we have spoken of the shadow-cast on the horizontal plane bythe whole of the upper circle, yet it is obvious that the semicircle which is towards the source of light, and which casts a shadow on the inner surface of the cylinder, cannot cast a shadow on the hori- zontal plane, unless we suppose the-surface to be trans- parent, so'that the light may not be intercepted by it. | The part of the shadow on the horizontal plane ‘which is concealed by the horizontal projection of the rface, is not shaded in the drawing. 5 28. The shadow of a circleyor iddeadl of any curve, * ona plane to which it is parallel, is an equal circle Oe curve. But*the shadow of a cirele on a piane to which it 1s it parallel, is, in general, an ellipse. The shadow ~ «« : cast by the centre of the circle isthe centre of the ellipse, »’ and the shadow cast by any diameter of thé circle is a 7 diameter of the* ellip e.» Two diameters of an ellipse ™ are said to be conjugate when either of them is parallel» to tangent lines drawn through the vertices of the other. ae If any two diameters be taken tn a circle at right angles to rat: each other, their shadows will be conjugate diameters of the € ellipse of shadob. bie 3 For, if through the vertices of either diameter, tangent _lines be drawn to the circle, they will be parallel to the other diameter; hence, their shadows on any plane will “be parallel to the shadow, of this latter diameter. But the shadows of the tangents will be tangent to the | shadow of the circle (16), that is, to the ellipse of . oe . » es. in, me *. =| etme .¥ | Fefough 0’. - ey SHADES AND» SHADOWS. 31 shadow. Therefore, the shadow of ‘one diameter is parallel to the tangents drawn through thé vertices of the shadow of dhe other. Hence the shadows cast by two diameters of a circle at right angles to each other, are conjugate diameters of the ellipse'o 8 hadow.” - 29. Itis,oftemrequired to construct ellipses when their « conjugate diameters only are known. The construction is easiest made by finding the axes Fi the curve. * We shall, therefore, give a problem, the principles of which are found inCr ouets Conic Sections, for finding the axes of an ellipse when two of its conjugate diameters are given. Let AB and DS (P1.3, Fig.2), be two conjugate diame- ters of an ellipse. Ehren either vertex, as D, of either diameter, draw a parallel EF to the other diameter. At D erect in a plane perpendicular to AOD, DO’ perpen- . dicular to EF, and make it equal to half the parallel diameter AB. With O' as a centre, and radius O'D, describe the circle CaDb. Now we may regard the ellipse whose conjugate di- _ ameters are AB and DS, as the shadow of the Byele ‘Cabo. ~ The ray of ight through the centre O’ pierces the -horizontal plane at O; the diameter ad casts the shadow » AB, and the diameter CD the shadow DS. It is now required to find two diameters of the circle CaDb at right angles to each other, whose shadows _ shall also be at. right angles ; ‘ for) 'the conjugate diameters of an ellipse, which are perpendicular to each other, are the axes. Seg To find these diameters it is necessary to construct two semicircles having a common diameter in the line | « F, one lying in the horizontal plane and passing ugh O, the other in the vertical plane and passing : ad ; * = se ~s * *, t 32 TREATISE ON Produce O'Dand make DP equal to it. Jom OP and bisect it by a perpendicular line—the point N, where the perpendicular meets EF’, 1s the common centre of the two semicircles. Let them be described with the radius NO, or NO”. Then draw the radii O’gE and OUF. These radii are at right angles to each other; and so are their shadows EO and FO. Hence,GO and LO are the semi-axes of the ellipse. ‘The extremities G and L are determined by drawing rays through g and s the ex- tremities of the perpendicular radii. ‘. PROBLEM V. Lo find the shadow of a rectangular abacus on a a cyline drical column; and also, the shade os the column. 30. Let the semicircle z SP (Pl. 3, Fig. 3) be the plan of a semi-column, and op'd¢' its elevation ; the rectangle ep the plan of the abacus, and ¢’d’ its elevation. Through the lower line (cz,c’) of the abacus, conceive a plane of rays to be passed. This plane will be per- pendicular to the vertical plane of projection, and its intersection with the surface of the column will be the. * shadow cast by the line (cz, ¢’), (15). ‘ ‘Through ¢ draw c? parallel to the vertical pro- - . jection of the light. This line is the vertical trace of the plane of rays passed through the line (cz, ¢), and it is also the vertical projection of the indefinite shadow “east on the column. (Des. Geom. 82.) But the ray of light through the point (¢, ¢’) pierces the surface of the column at ‘Cre therefore ¢2’ is the vertical projection of the shadow cast on the column by _ the line (cz, ¢’). The shadow itself 1s an ellipse, and is. PA projected in the are zz. aa SHADES AND SHADOWS. 33 To find the shadow cast by the line (cd, cd’): The shadow cast by the point (¢,c’) is already found at (2,0’). Let the surface of the column be intersected by planes of rays perpendicular to the horizontal plane. These planes will cut the line (ed, cd’) of the abacus in points, and the surface of the column in right lines— then, drawing through the points of the abacus,rays of light, where they meet the elements of the column are points of the shadow. The line 64, drawn parallel to the horizontal projection of the light, is the horizontal trace of a vertical plane of rays, and this plane determines (f, k’) a point of the shadow. The points (4,7) (g, 2") (g, q) and (n, x’) are found in a similar manner. The vertical plane of rays, whose horizontal trace 1s EF, is tangent to the column. The ray, therefore, through the point (14, E’,) touches the column at (a, 7’). At this point the line of shadow terminates, and the line of shade begins and passes down the column vertically, being the element in which the plane EF is tangent to the column. The part of the surface of the column which is above the line of shadow, and the portion which is in the shade, are shade in the drawing. PROBLEM VI. To find the shadow cast by a cylindrical abacus on a eylin- drical column ; and the shade on the abacus and column. 31. Let nsr (Pl. 6, Fig. 2) be the plan of a semi- column, fr’ its elevation—cemd the plan of the abacus, and c’d its elevation. Let the surface of the column be intersected by ver- C 34 TREATISE ON tical planes ‘of rays. These planes will intersect the lower circle of the abacus, which casts the-lime of shadow on the column, in points, and the surface of the ‘column in right lines. Through the points of the abacus let rays of light be drawn—the points in which they intersect the elements of the column are points of the line of shadow. The plane of. rays, whose horizontal trace is hn, parallel to the horizontal projection of the light, determines the point of shadow (n, n); and the points of shadow (0, 0), (9,q) (8, s) (t, ) (u, w) and (¥, v’) are found in a similar manner. The vertical plane of rays, whose horizontal trace is mv, is tangent to the column along the element of shade. The shadow ends and the shade begins at the point (v,v’). The shade on the abacus is found by drawing a plane of rays which shall be tangent to it—the line p drawn tangent to cmd and parallel to the horizontal projection | of the rays of light, is the horizontal trace of the — tangent plane—the element of contact is the element of shade. The part of the surface above the line of shadow, as well as that portion of it which is in the shade, is shaded in the drawing. PROBLEM VII. Having given a cylindrical abacus and the direction of the — light, wt ws required to find the shadow of the abacus on a vertical plane, or wall. € 32. Let anbp (Pl. 6 Fig. 3) be the. plan, and a’cd'b the elevation of the abacus, the abacus touching the vertical plane on which the shadow falls,in the element (p, 70). < ‘3S . get = Plate o. ‘es, "4 | | in Hl ii * 2 : ‘ ; + + 4 : . é # . = 2 = = | : ae = ae ee eee, a. = ny © EB Prudhomme fc | OE LL : 2 - = 2 —— —_ - = Se eee, 2 me -- : t, . * = —* SHADES AND SHADOWS. 30. Suppose two squares to be constructed, the one cir- cumscribing the upper circle of the abacus, the other the lower. The square cdgl is the projection of both the squares on the horizontal plane. Let two planes of rays be drawn tangent to the abacus. The lines ¢¢” and «”, drawn parallel to the horizontal projection of the ray of light and tangent to the circle nbpa, are their traces; and (t, ¢s), (¢, yx) are the elements of contact, and consequently the elements of shade (27). Now the lines of the abacus which cast lines of shadow on the vertical plane, are, Ist, the upper sem1- circle (tb, bt'y); 2dly, the two elements of shade (/, ¢'s) (7, yx); and 3dly, the lower semicircle (cant, c'xs). Through (z,7’), the centre of the upper circle of the abacus, let a ray of light be drawn,—the point 2’, in which it pierces the vertical plane, is the centre of the ellipse of shadow cast by the upper circle. The ray through (n, 7’) pierces the vertical at p, hence n'z'p is the Neonat cast by the diameter (np, n’). Drawing rays through (6, 6’) and (a, a’) determines w and v, the extremities of the shadow cast by the diameter (ad, ab’). Hence n'p and vw are conjugate diameters of the ellipse of shadow (28). Drawing rays of hght through the points (d, 6’) and (c, a ),determines Of the shadow cast by the tangent ; (dg, 6’); also ef the shadow cast by the tangent (cd, ab’), and ae the shadow cast by the tangent (cl, a’). But since these lines are all tangent to the upper circle ° of the abacus, their shadows will be tangent to the ellipse of shadow (16); and the same may be shown of the lower circle of the abacus and its tangents. | The ellipse of shadow cast by the lower circle of the abacus is easily found. The point 2” is its centre, og, “ 36 TREATISE ON z'y are its conjugate diameters, and the lines dh, dc, cg, and gh are tangent to it. The elements of shade cast the shadows ¢’u and 7/. Hence, the lines of shadow on the vertical plane, are—the semi-ellipse dz qu, cast by the lower semicircle (cant, c'xs); the right line wt’, cast by the element (¢, ¢s); the semi-ellipse fwn'7’, cast by the upper seimicircle (2, ty), and the right line #1 cast by the element of shade (2, yz). PROBLEM VIII. Having given the frustum of an inverted cone, tt 1s re~ quired to find the shadow cast by the upper circle on the tnner surface, and the shadow cast by the frustum on the plane of the lower base. 33. Let the circle CLHD (P1.4) be the horizontal pro- jection of the upper circle of the frustum, and G’H’ its vertical projection. Let the circle described with the centre A and radius AK be the horizontal, and I’K’ the vertical projection of the lower base. The axis of the cone being supposed perpendicular to the horizontal plane, is projected on the horizontal plane at A, and on the vertical plane in the line A’A” perpendicular to the ground line, and GT'K’H' is the vertical projection of the frustum. Producing the line GY till it intersects A”A’, gives A’, the vertical projection of the vertex of the cone of which the frustum is a part. To find the shadow on the inner surface . If two tangent planes of rays be drawn to the cone, the points in which the elements of contact meet the upper circle, are the points where the shadow on the inner surface begins. Through (A, A’), the vertex of the cone, let a rav of SHADES AND SHADOWS. aay hight be drawn. Such ray pierces the plane of the upper base of the frustum in the point (B, B’). Through the point (B, B’), suppose two lines to be drawn tangent to _the upper circle of the frustum. These tangents are the traces, on the plane of the upper base, of two planes of rays drawn tangent to the cone. ‘The lines BC, BD, drawn tangent to the circle GDHL, are the horizontal projections of these traces; and (D, D’) (C, C’) are the points at which the shadow on the inner surface begins. Let the surface of the cone be now intersected by planes of rays passing through the vertex. Each se- cant plane so drawn will intersect the surface in two elements; the element towards the source of light will cast the shadow, and the other will receive it. Each plane will also contain the ray of light passing through the vertex of the cone, and consequently every trace on the plane of the upper base will pass through the point (B, Bb’). | Draw any line, as BaL, to represent the trace of such a plane. The horizontal projections of the elements in which it intersects the surface of the cone, are AL and Aa. Projecting L and a into the upper base at L’ and a’, and joining these points with A’, the vertical projection of the vertex of the cone, gives the vertical projections of these elements. Through (a, a’) the upper extremity of the element towards the source of light, conceive a ray of light to be drawn; the point (4, 4’) in which it meets the ele- ment (AL, A’L’) is a point of the curve of shadow. If we suppose the surface of the frustum to be pro- duced below the plane EF’, the shadow that would fall on the part of the surface below this plane is easily found. The lowest point is determined by passing a 38 TREATISE ON plane of rays through the axis of the cone. The trace of this plane, on the upper base of the frustum, is hori- zontally projected in the line BAd. ‘The line A’‘d’ is the vertical projection of the element which receives the shadow, and (p, p’) is the point of shadow cast by (A, /’). By passing a plane of rays through the element (AH, A‘H’) we shall find the point at which the vertical projec- tion of the shadow is tangent to the element A'H’. The horizontal projection of the trace of this plane, on the upper base of the cone, is BkH. Projecting & into the vertical plane at /’, and drawing £'l parallel to the ver- tical projection of the light, gives 7 for the point of tan- gency. ‘The horizontal projection of the point is found by projecting / into the horizontal projection of the ele- ment at /. Having found as many points of the curve as are necessary to describe it accurately, let its pro- jections be made as in the figure. It is to be observed, that the shadow on the inner surface of the frustum intersects the plane of the lower base at the points (c¢,¢’), (f,/”); and that the curve of shadow below this plane is found on the supposition that the plane does not intercept the light, and that the frus- tum is produced below it. If the plane of the lower base be supposed to inter- cept the light, there will be no shadow on the surface of the cone below it; for the plane itself will receive the shadow cast by the upper circle. Through the centre (A, A”) of the upper circle, con- ceive a ray of light to be drawn. It pierces the plane of the lower base at (I, F’). A circle described with F as a centre, and radius equal to the radius of the upper base, is the horizontal projection of the shadow cast by the upper circle on the plane EF’. The arc eqf falls within the circle of the lower base, and therefore is the SHADES AND SHADOWS. 39 line of shadow, within the surface, on the plane of that base. The circle described with the centre F will pass through ¢ and f, the horizontal projections of the points in which the curve of shadow intersects the plane E’F’. The shadow cast by the frustum on the plane of the lower base is limited by the traces of the two tangent planes of rays that determine the elements of shade. These planes contain the ray of light drawn through the vertex; hence their traces pass through the point (E, E’) in which this ray pierces the plane E’'F’; and En, Em drawn through E and tangent to the circle If Ke are their horizontal projections. The shadows cast by the elements of shade on the plane EE’ begin at the points in which they pierce it, and terminate at the points m and n, where the shadows are tangent to the shadow cast by the upper circle. The points m and x may be determined by drawing horizon- tal projections of rays through the points C and D, and noting their intersections with the shadow cast by the upper circle. That part of the shadow on the plane E’F’ which is under the surface of the cone is not seen in horizontal projection. For the purpose of showing the appearance of the shadow on the inner surface, we will suppose that part of the frustum which is in front of the vertical plane GAH to be removed, and then represent in vertical projection the remaining semi-frustum, as it appears to the eye. That part of the surface which lies between the ele- ment (AG, A’G’) and the curve of shadow, being in the shadow, is darkened in vertical projection. _ 40 | TREATISE ON PROBLEM IX. Having given a sphere in space, and the direction of the light, ut ws required to find the curve of shade, and the shadows cast on the planes of projection. ; 34. Let the centre of the sphere be taken at equal dis- . tances from the planes of projection; let A (PI. 5) be its horizontal and A’ its vertical projection; and suppose the projections of the light to make equal angles with the ground line. Suppose the sphere to be circumscribed by a tangent cylinder of rays. ‘The axis of the cylinder is the ray of light passing through the centre of the sphere; the curve of contact is a great circle whose plane is perpeéen- dicular to the axis of the cylinder, that 1s, to the direc- tion of the light in space. ‘This cirele of contact is the curve of shade. Since the axis of the tangent cylinder of rays 1s oblique to both planes of projection, the plane of the circle of shade which is perpendicular to it, is also oblique to both the planes of projection, and conse- quently its projections on these planes will be ellipses. (Des. Geom. 180.) To find the projection of the circle of shade on the horizontal! plane. The horizontal projection of that diameter of the cir cle of shade which in space is parallel to the horizontal plane, is the transverse axis of the ellipse into which the circle of shade is projected. The projection of that diameter which makes the greatest angle with the hori- zontal plane, or which is perpendicular to the parallel diameter, is the conjugate axis.—(Des. Geom. 180.) 2 % Pe eae 2 ia i: ig eas es — 2 -¥ * 3 tae ea . v . ] ; * - . > - ag ' ae * * ~ “ * er dae FiO EA rahe gt fg eas ge ; Leben age” Sap carne pple es oS ~ - a SS See So ea et nie a ae a ees a ELPrudhomme Se. \\ / \\\ Zs, AI Ly a Wr = SHADES AND SHADOWS. A} The ray of light through the centre of the sphere being perpendicular to the plane of shade, is also per- pendicular to that diameter of the circle of shade which is parallel to the horizontal plane, and therefore the pro- jections of this diameter and the ray of light,on the hori- zontal plane, are at right angles to each other.. Hence the diameter CAD, drawn through the centre A, and pet- pendicular to AB, the horizontal projection of the ray through the centre, is the transverse axis of the ellipse into which the circle of shade is projected. Through the centre of the sphere suppose a plane of rays to be drawn perpendicular to the horizontal plane. Its horizontal trace is EAB. This plane inter- seets the surface of the sphere in a great circle, the _ . cylinder of rays in two elements tangent to this circle, and the plane of the circle of shade in a diameter passing through the pots of contact. The projection of this diameter is the conjugate axis of the ellipse. To find the conjugate axis: Let this plane be revolved about a vertical axis pass- ing through the point A, till it becomes parallel to the vertical plane of projection. Inthe revolution, the point 5 describes in the horizontal plane,the arc BD’, and the point (A, A’) being in the axis, remains fixed. Project- ing B’ into the ground line, and drawing A’B", we have the vertical projection of the revolved ray. But the great circle cut out of the sphere 1s,.when revolved par- allel to the vertical plane, projected into the circle rep- resenting the vertical projection of the sphere, and the elements cut from the cylinder of rays are projected into lines parallel to A’B”. Drawing therefore the tangents IH’ and G’b’ parallel to A’B", we determine, in the revolved position of the plane of rays, the extremities of that diameter of the circle of shade whose projection is the conjugate axis. a 42 TREATISE ON When the plane of rays is parallel to the vertical plane, the point I” is horizontally projected at F, and the point G' at G. In the counter-revolution the points (1, F’) (G, G’) describe the horizontal arcs (F/, Ff), (Gg, G'g'), and gf becomes the horizontal projection of the diameter, or conjugate axis of the ellipse. Having found the two axes of the ellipse, let it be described. It is plain that (f,f’) is the highest point of the curve of shade, and (g, g’) the lowest; therefore the tangents to the curve of shade at these poits are horizontal, and their vertical projections parallel to the ground line. Since the points which are horizontally Praisetell at C and D are contained in the horizontal plane through the centre of the sphere, they will be vertically projected in its trace N’B”, at the points C’ and D’. The points which are horizontally projected at the intersections of the ellipse CgDf, with the line NB’, are vertically projected in the circumference of the circle representing the vertical projection of the sphere. We have thus found six poimts of the ellipse into which the circle of shade is projected on the vertical plane. And since the tangents passing through the highest and the lowest points f’ and g’ are parallel to the ground line, it follows that the diameter passing through these points is conjugate with the diameter D’C’; therefore let the curve be described (29). The axes of the ellipse into which the circle of shade is projected on the vertical plane can, however, be found by a construction similar to that used for the horizontal projection. Through A’ draw a diameter perpendicular to A’B, the vertical projection of the ray. This will be the transverse axis of the ellipse. Through the centre of the sphere conceive a plane of rays to be drawn per- SHADES AND SHADOWS. 43 pendicular to the vertical plane—A’B is its vertical trace. Let this plane be revolved about an axis passing through the centre of the sphere and perpendicular to the ver- tical plane, until it becomes parallel to the horizontal plane. The point B describes in the vertical plane the arc BB”; and since (A, A’) remains fixed, AB” is the horizontal projection of the revolved ray passing through the centre of the sphere. The lines cH, hd’ drawn parallel to the revolved ray and tangent to the circle NDC, determine (c, c’), (h, h’), the extremities of the diameter, in its revolved position, whose vertical projec- tion is the conjugate axis of the ellipse. In the counter revolution, the points (c, ¢’) and (h, h’) describe the arcs (cd, cd’) and (hk, Wk’): and kd’ is the conjugate axis of the ellipse into which the circle of shade is projected on the vertical plane. In horizontal projection, we see only that part of the shade which lies above the horizontal plane N’B’”’; the part of the curve of shade lying below this plane is dotted. In vertical projection, we see that part of the shade which lies in front of the plane NB’. To find the shadow on the horizontal plane: The curve of shadow on the horizontal plane is the mtersection of the horizontal plane with the surface of the cylinder of rays tangent to the sphere. This curve will be an ellipse, unless the circle of shade be parallel to the horizontal plane, or the section a sub-contrary one. When the plane of rays AB was revolved par- allel to the vertical plane, we drew F’H’ and G0’ par- allel to the revolved ray A’B”. The tangent (G’d’, Gd) pierces the horizontal plane at 6. In the counter revolution, the point 6 describes in the horizontal plane the arc 6 6’; then 6’’B is the shadow cast by the radius of the circle of shade which passes through the lowest point—hence it is the semi-transverse axis of the ellipse 44 TREATISE ON of shadow. The conjugate axis of this ellipse is the shadow cast by the horizontal diameter of the circle of shade. It passes through B,is perpendicular to 6B, and equalto CD. ‘The shadow cast on the vertical plane is found in a manner entirely similar. OF BRILLIANT POINTS. } 35. When a ray of light falls upon a surface which turns it from its course and gives it another direction, the ray is said to be reflected. ‘The ray, as it falls upon the surface, is called the incident ray, and after it leaves the surface, the reflected ray. The poimt at which the reflection takes place is called the point of incidence. it is ascertained by experiment, 1°. That the plane of the incident and reflected rays is always normal to the surface at the point of incidence. 2°, That at the point of mcidence, the incident and reflected rays make equal angles with the tangent plane or normal line to the surface. If therefore, we suppose a single luminous point, and the light emanating from it to fall upon any surface and to be reflected to the eye, the poimt at which the reflection takes place is called the brillant point. The brilliant point of a surface is, then, the point at which a ray of light and a line drawn to the eye make equal angles with the tangent plane or normal line—the plane of the two Imes being normal to the surface. 36. The rays of light being parallel, and the place of the eye at an infinite distance, the brilliant point of any surface is thus found: Through any point in space draw a ray of light and a line to the eye, and bisect the angle included between them. Thendraw a tangent plane to the surface and per- pendicular to the bisecting line—the point of contact is SHADES AND SHADOWS. 45 the brilliant point. For, let AC (Fig.n) be the line drawn to the eye, AB the ray of light, AD the bisecting line, and P the point of contact of a plane passed perpendi- cular to AD, and tangent to the surface. Through P let there be drawn a ray of light PG,which will be paral- lel to AB, the line PE to the eye,which will be parallel to AC, and PF the normal line to the surface at the point P. Since the normal PF is perpendicular to the tangent plane, it is parallel to the bisecting line AD, for AD is perpendicular to the tangent plane by construction. But the line AD is in the plane of the lmes BA, AC, and bisects the angle BAC: therefore the parallel PF lies in the plane of the lines GP, PE, and bisects the angle ‘GPE. Hence P is the brilhant point. 36. To apply these principles in finding the brilliant point on the surface of a sphere. Suppose the eye to be in a line perpendicular to the vertical plane, and at an infinite distance from it. Through (A, A’), the centre of the sphere, suppose a ray of light to be drawn, and also a line to theeye. For the purpose of bisecting the angle included between these ‘ lines, let their plane be revolved about (AP, A’) the line drawn to the eye, until it becomes parallel to the horizon- tal plane. Any point of the ray, as (E, E’), will describe an arc (Ke, He’); and Ae and AP are the horizontal pro- jections of the lines when revolved parallel to the horizon- tal plane. Bisect then, the angle PAe by the line Aq: and from any point of the axis, as P, draw the line Pge. After the counter revolution, the point e is horizontally pro- jected at E; and since P remains fixed, Pge is horizon- tally projected in PE; the point q of the bisecting line, is projected at q’, and Aq’ is the horizontal projection of the bisecting line. Its vertical projection is A’E’. The plane drawn perpendicular to this bisecting line, > ¥ 46 TREATISE ON. and tangent to the sphere, touches the surface at the point wher’ the bisecting line pierces it—that is, at (a’, a”). Consequently, (a’, a’) is the brilliant point. A similar construction would determine the brilliant point, if the eye were taken in a line perpendicular to the hori- zontal plane. % PROBLEM X. Having given an ellipsoid in space, and the direction of the hight, vt 1s required to find the curve of shade, and the shadow cast on the horizontal plane. 37. Let the horizontal plane be taken perpendicular to the axis of the surface. Let A (PI. 6. Fig. 1) be the horizontal projection of the axis, and A’B its vertical projection. Let us suppose the ellipsoid to be circum- scribed by a tangent cylinder of rays. ‘The axis of the cylinder is a ray of light passing through the centre of the ellipsoid and piercing the horizontal plane at C. Through this axis let a plane of rays be passed perpen- dicular to the horizontal plane—ACD is its trace. Since this plane is a plane of rays, and divides the ellip- soid into.two equal and symmetrical parts, the parts of the curve of shade lying on either side are equal and ~ symmetrical. Hence, both parts are projected on the meridian plane AC, into the same right line. But, since the contact of the ellipsoid and surface of the cylinder is an ellipse whose plane passes through the centre of the ellipsoid, the curve of shade is projected into a right line passing through the centre, which line is the inter section of the plane of the curve of shade with the surface of the ellipsoid in a meridian curve, and the sur- > % - meridian plane ACD. The plane ACD intersects the - s iw Pid a . . 4 ee =-----------~~—~-—~-7*-------- oe ‘ LLP rudhomme: Se} ] “’ “‘— . , eh Ne rT HAA Ae SHADES AND SHADOWS. 47 face of the cylinder of rays in two elements that are tangent to it. The points of tangency are the highest and lowest points of the curve of shade. The line in which the plane of shade intersects the meridian plane AC, also passes through these points. Let the ellipsoid be now projected on a vertical plane, parallel to the meridian plane ACD. Assuming E’D’ for the new ground line, the centre of the ellipsoid will be projected at I’, in the perpendicular AF to the ground line E’D’, and at a distance from it, equal to the distance of the centre of the ellipsoid from the horizontal plane. The ray of light through the centre will be vertically projectedin FC’. The ellipse described about the cen- tre F, and equal to a meridian curve of the surface, represents the vertical projection of the ellipsoid. Having described this ellipse, draw the two tangents d’D’ and e’E’ parallel to FC’, the vertical projection of the ray. The points of contact d”,¢’ are the vertical pro- jections of the highest and lowest points of the curve of shade; and d”Fe’ is the vertical projection of the curve of shade, since the plane of shade is perpendicu- lar to the new vertical plane. The horizontal projec- tions of the highest and lowest points are d and e. Let us suppose a system of horizontal planes to inter- sect the ellipsoid between the highest and lowest points of the curve of shade. Each of such secant planes, being perpendicular to the axis, will intersect the sur- face of the ellipsoid in a horizontal circle, and the plane of shade in a horizontal le perpendicular to the meri- dian plane ACD, and consequently, to the new vertical plane. Let f’n be the vertical trace, on the new vertical plane, , of one of the secant planes. The line f'n is the vertical projection of the circle in which the plane intersects the 48 TREATISE ON surface of the ellipsoid, and h” of the line in which it intersects the plane of shade. Projecting.the circle and line on the horizontal plane, the points fh and &, in which they intersect, are two points in the horizontal projection of the curve of shade. In a similar manne any number of points may be determined. The hori zonta] plane through the centre of the ellipsoid, inter sects the plane of shade in a line whose horizontal pro- jection is st, and this line is the transverse axis of the ellipse into which the curve of shade is projected; and ed, the projection of the diameter joining the highest and lowest points, is the conjugate axis. The projection of the curve of shade on the primitive vertical plane, is found by projecting the horizontal cir- cles, and drawing perpendiculars to the ground line through points of t the horizontal projection of the curve. Thus, projecting the horizontal circles passing through the highest and lowest pomts, and drawing perpendicu- lars to the ground line from d and e, determines d’ and eé, the vertical projections of the highest and lowest points. The vertical projection of the circle f’h'n, is Wk’, and h’,k’ are the vertical projections of the points of shade that are horizontally projected ath and k. The diameters d’e and /s’ are conjugate. The part of the surface which is in the shade, and in front of the meridian plane gAp, is shaded in vertical projection. ‘The part in the shade and above the horizontal plane passing through the centre of the ellip- soid, is shaded in horizontal projection. To find the shadow on the horizontal plane: If we suppose the lines in which the horizontal secant planes intersect the plane of shade, to cast shadows on the horizontal plane, the shadows cast will be parallel to the lines themselves. The line in SHADES AND SHADOWS. AY which the horizontal plane /’n intersects the plane of Mw shade, casts the shadowh’"/”, and the points h” and ”” in which it is intersected by hh” and £k”, drawn parallel to the horizontal projection of the light, are pomts in the shadow on the horizontal plane. The highest and lowest points cast their shadows at D and E. If through the highest and lowest points of the curve of shade, two planes be drawn tangent to the cylinder of rays circumscribing the ellipsoid, their traces on the horizontal plane will be tangent to the ellipse of shadow; but since the planes are perpendicular to the vertical plane EAD, their traces will be perpendicular to ED, at the points E and D. Hence, ED is an axis of the ellipse of shadow. The other axis passes through the middle point C, and is the shadow cast by the diameter (st, s‘7’). PROBLEM XI. Having given the plan and elevation of a niche, and the direction of the light, it is required to find the shadow which the niche casts upon tiself. 38. A niche is a recess in the wall of a building. It is generally composed of a semi-cylinder, and the quad- rant of a sphere having the same radius as the base of the cylinder. ‘The quadrant of the sphere rests on the upper base of the semi-cylinder, forming the upper part of the niche. ‘The quadrant and semi-cylinder are tan- gent to each other in the semicircle which separates the cylindrical from the spherical surface. Let AB (Pl. 7. Fig. 1) be the intersection of the face of a vertical wall with a horizontal plane, and the semi- circle ACB the plan of the niche. The rectangle A’B’ is the elevation of the cylindrical part, and the semi- D 50 TREATISE ON © circle A”F’B” of the spherical part of the niche—A’B’ is the vertical projection of the semicircle that separates the cylindrical from the spherical surface. The lines of the niche which cast lines of shadow on - the surface, are, 1°. ‘The element (A, A’A”). 2°. The semicircle A’ FB”. . A part of the shadow cast falls on the base of the niche, a part on the cylindrical, and a part on the spheri cal surface. To find the shadow on the base and cylindrical surface. Through the element (A, A’A”) conceive a plane of rays to be passed. Its horizontal trace AC is parallel to the horizontal projection of the rays of light, and the plane intersects the cylindrical surface in a second ele- ment (C, C’a), opposite the source of light, and on this element the shadow falls. The line A’a, drawn through A” parallel to the vertical projection of the rays of light, limits the shadow on the element: and drawing through C’ the line C’d, parallel to the vertical projection of the ray of light, determines the point 6, which casts a shadow atC. ‘The point C is common both to the cylindrical surface and to the base of the niche. Hence, the part A’b of the element (A, A’A”) casts the shadow AC on the base of the niche, and the part 6A”, the shadow. C’a on the cylindrical surface. Above the point a, the shadow on the cylindrical surface is cast by the semi- circle (AB, A’F’B”). ' Through any point of the semicircle, as (I, F’), con- ceive a vertical plane of rays to be passed. Its hori- | zontal trace Ff is parallel to the horizontal projection of the rays of light, and the plane intersects the surface of the cylinder in the element (f, ff’). The point f’, in which the vertical projection of the ray drawn through ~* 7 + ¥2 eS , | \ P a | Hil il | ii wee = Sy PE, ATR il d ' x MoO < 58 wn as “ K\ mS ea : 4 m ner x \ | beak \ \\ SN vat ‘ \ ur ' oth ‘ Be FE Pridhomme Se. SHADES AND SHADOWS. 51 F’ meets (f'f”), is the vertical projection of the shadow cast by the point (1*°,F”). In a similar manner, we may determine other points of shadow on the cylindrical surface. Above the line A”B” the shadow will fall on the spheri- cal surface. Before finding this shadow we will demon- strate, that when a cylinder intersects a sphere, the curve in which it enters on the one side, is equal to the curve in which it leaves the sphere on the other. For, the parts of the elements of the cylinder inter- cepted between the points at which they enter, and the points at which they leave the sphere, are parallel chords of the sphere. Conceive a plane to be passed through the middle point of one of these chords and perpendicular to it. Such plane passes through the centre of the sphere and bisects all the other parallel chords. Hence, the curves in which the cylinder enters and leaves the sphere, are symmetrical with respect to this plane, and are consequently, equal. Therefore, if one of themis a circle, the other will be an equal circle; and hence, * when the surface of a cylinder intersects that of a sphere in a great circle, it will at the same time intersect it in a second great circle, and the planes of these circles will intersect each other in a diameter of the sphere. The two elements of the cylinder, passing through the extremities of this common diameter, are perpendicular to it, since they are tangent to the sphere. Let us suppose an entire hemisphere to be described on the diameter (AB, A’B”), having the plane of its great circle vertical. Through this circle suppose a | cylinder of rays to be passed. Half the surface of this cylinder intersects the surface of the hemisphere in a semicircle, which is the shadow that the semicircle (AB, A’F’B") would cast on the hemisphere... . 2 a , 52 TREATISE ON The plane of the circle casting the shadow, and the plane of the circle of shadow, intersect each other in a line passing through the centre of the hemisphere and perpendicular to the elements of the cylinder of rays, that is, perpendicular to the direction of the light. But, since this diameter is a line of the vertical plane AB, it is parallel to the vertical plane of projection—hence, its vertical projection is at right angles to the vertical pro- jection of the rays of light (Des. Geom. 49). ‘There- fore, if through D’, we draw the line D//'D” perpendicular to the projection of the ray, we determine Ih’, the ver- tical projection of the diameter in which the plane of the circle casting the shadow intersects the plane of the circle of shadow. The semicircle IA’) casts the semicircle of shadow on the hemisphere, the vertical projection of which shadow is an ellipse, whose trans- verse axis is Ih’ (Des. Geom. 180). Through (D, D’) let a plane of rays be passed per- pendicular to the face of the niche, and let us suppose for a moment the vertical plane of projection to be moved forward to coincide with this face. The point D’ will then be directly over D, and n'D'p will be the trace of the plane of rays. Let us now assume an auxiliary plane of projection, parallel to the plane whose trace is np, and at a given distance from it. Draw n’y’ parallel to n'p, to represent the trace of the auxiliary plane, and let it be borne in mind that this trace, as well as n’p, is in the plane of the face of the niche. The semicircle in which the plane of rays n'p intersects the hemisphere, is projected on the auxiliary plane in the semicircle n’hp’. The pro- jections of the ray of light passing through the point (n,n), are nk and nD’. This ray is projected on the auxiliary plane by projecting 7’ into the trace at n”, and SHADES AND SHADOWS. 53 projecting any other point of the ray, as (4, D’) (Des. Geom. 14): this is done by laying off Dk’ equal to Dk The line n"&l is the projection of the ray on the aux iliary plane, and the point J, in which it meets the semi circle n/p’, is the projection of a point of the curve of shadow. But smce D‘h’, a line in the plane of the cir- cle of shadow, is perpendicular to the plane of rays whose trace is np, and consequently to the auxiliary plane, it follows, that the plane of shadow 1s _ per- pendicular to these planes; hence, the projection of the shadow on either of them is a right line (Des. Geom. 82). But D” is the projection of h', a point of the curve of shadow, and a second point is projected at J; hence, Dl is the projection of the shadow on the auxiliary plane. This line is also the projection of the intersec- tion of the plane of rays, whose trace is n'p, with the plane of the circle of shadow. In space, therefore, the line whose projection on the auxiliary plane is D’/, is perpendicular to the diameter Ih’: hence, D7?’ its projection on the vertical plane, is the semi-conjugate axis of the ellipse into which the circle of shadow is projected (Des. Geom. 180). Let the ellipse be described. [rom h’ to the point q’, where the ellipse intersects at A"B", the shadow falls on the sur- face of the sphere. From q' the shadow falls on the cylindrical surface. The arc /’E casts the shadow on the spherical, and the are A”E the shadow on the cylin- \ drical surface. Had the quadrant of the sphere below the plane A” B’ been permitted to intercept the light, the shadow af’q, instead of being on the cylinder, would have been the shadow q/l'I on the surface of the sphere. We have also dotted the line of shadow ae'vg’, which the front circle 54 TREATISE ON of the hemisphere would cast upon the vertical cylinder, if the sphere did not intercept the rays of light. The part of the surface of the niche lying between the lines A’A” Ef’ and the line of shadow, is shaded in vertical projection. It is not necessary to find the semi-conjugate axis of the ellipse into which the circle of shadow is projected. We may find points of the curve, and describe it through them. Let z's’ be the vertical trace of a plane of rays per- pendicular to the face of the niche. This plane inter- sects the hemisphere ina semicircle which is projected on the auxiliary plane in the semicircle described with the radius D’?”.. Draw2"s” parallel to nl. The point s”, _ where it meets the semi-circle and the line DJ, is the pro- jection on the auxiliary plane of a point of the circle of shadow. This point is projected on the vertical plane of projection at s’, and on the horizontal plane at s, by making ms equal to m’s”. In a similar manner other points of the curve may be found. The point A’ is projected on the horizontal plane at h, and the point g at qg: therefore, hsq is the horizontal pro- jection of the shadow which falls on the spherical part of the niche. The shadow on the cylindrical part is horizontally projected in the are Ceg. The point (q, 9), at which the shadow passes from the cylindrical to the spherical part of the niche, can be found by a direct construction. The point in space, is the one in which the intersec- tion of the upper base of the cylinder with the plane of shadow, meets the circle of the upper base. But h'1 is the trace of the plane of shadow on the face of the niche, and (s, s') is a pot of this plane. Therefore, we have one trace of an oblique plane, and a point of the plane, SHADES AND SHADOWS. 55 to find its other trace (Des. Geom. 43). Draw su parallel to h’D’; its horizontal projection is su, and (u, wv’) is the point in which it pierces the horizontal plane A’ B": therefore, (uw, u’) is a point in the trace of the plane of shadow. But the trace passes through (D, D’); hence Dug is the horizontal projection of the required trace. But q' is the vertical projection of g; hence (q, q’) 1s the point at which the shadow passes from the cylin- drical to the spherical surface. PROBLEM XI. To find the curve of shade on the surface of a torus. 39. Let abcg (PI. 7, Fig. 2) be a rectangle, having the semicircles aA’é and cB'g described on its vertical sides. If the figure aA’bcB'g be revolved about a vertical axis (C, C”O), it will generate a solid, called a torus. This solid is used in some of the orders of architecture in forming the bases and capitals of the columns. Before finding the curve of shade, we will demonstrate, that if a surface of revolution be intersected by a me- ridian plane, and a line be drawn tangent to the meridian curve and parallel to the projection of the rays of light on the meridian plane, the point of contact will be a point of the curve of shade. For, every plane of rays, tangent to a surface, touches it ina point of the curve of shade. But since the surface is one of revolution, such plane is perpendicular to,the meridian plane passing through the point of contact (Des. Geom.105). ‘Therefore, its trace will not only be tangent to the meridian curve, but also parallel to the projection of the rays of light, since the rays are projected by per- ned 56 TREATISE ON pendicular planes. Hence, the tangent to a meridian curve, drawn parallel to the projection of the rays of light on ets plane, determines a point of shade. Let the circle ATBH represent the horizontal pro- jection of the surface, ¢B’cbA’a its vertical projection, and (A, A’) the ray of light. Through any point of the axis of the surface, as (C, ©’), suppose a ray of light to be drawn, CD, C’D’ are its projections. ‘Through this ray let a meridian plane be passed, and then revolved about the axis of the surface until it becomes parallel to the vertical plane of projection. The point (C, C’) being in the axis, remains fixed, and any point of the ray, as (D, D’), describes a horizontal arc (Dd, D'd’); and the vertical projection of the ray, from its revolved position, is C’d’. The section of the surface, when revolved parallel to the vertical plane, is vertically projected in the meridian line gB’cbA’a. If now, two lines be drawn tangent to the semicircles and parallel to C’d’, the revolved ray, they will determine the points (£, £’) and (¢, 2’): these are the highest and low- est points of the curve of shade, in their revolved position. In the counter revolution of the meridian plane, they describe horizontal arcs, and when the counter revolu- tion is completed, are horizontally projected at h and J, and vertically at A’ andl’. The lines drawn parallel to C’D’, the vertical projec- tion of the ray, and tangent to the semicircles, determine the points (q,9') and(p,p') at which the curve of shade ’ intersects the meridian plane AB. The points (T, ¢’) and (H, 7”) of the curve of shade are found by passing two planes of rays tangent to the surface and perpendicular to the horizontal plane. The planes will touch the surface in the circumference of the circle whose vertical projection is AB’. The points of SHADES AND SHADOWS. 57 tangency are also found in a meridian plane passed per- pendicular to the plane DCA. To find other points of the curve of shade: Draw any meridian plane, as nCm, and project the ray of light upon it. The point (C, C’) is its own projec- tion, and (D, D’) is projected at (F, fF’); therefore, (CF, C’F’) is the projection of the ray on the meridian plane nCm. Let the meridian plane be revolved till it becomes parallel to the vertical plane of projection. The line (CF, C’F’) will then be vertically projected inthe lineC/’. Draw two lines parallel to Cf’ and tangent to the semicircles; their points of contact (7, 7’) and (v, v’) are points of the curve of shade in their revolved position. In the counter revolution these points describe horizontal arcs, and when. it is completed are horizontally projected at n and m, and vertically atm and m. Any number of points of the curve of shade may be found in a similar manner. The curve HqlTphmis the horizontal,and ql'n (pm the vertical projection of the curve of shade. PROBLEM XUIL Having given a surface of revolution, conver towards the axis, and the direction of the light, it 1s required to find the shadow which the upper circle casts upon the surface, and also the brilliant point. 40. Let DRN (PI. 8, Fig. 1) be the horizontal, and D'L the vertical projection of a cylinder or pedestal, on which the surface rests. Take two-thirds of the radius AD, and lay it off from A’ to v, on the vertical line A’B. Through v draw the horizontal line vu, which meets the vertical line D’u at #«. With was a centre, and radius wD’, describe the SOF . (PREATISE ON quadrant D’H’X’. Then ‘lay off vB equal to one-third of the radius A’D’, and draw Bl’ parallel to A’D’, and equal to two-thirds of it. Withy asa centre, and a radius equal to yX’ or yl’; describe the quadrant X’dl’. The two quadrants will have a common tangent line at the point X’, which will be vertical. Now, ifthe curve D'H’X’ be revolved about the vertical axis (A, A’B) it will gene- rate a surface of revolution, convex towards the axis, and concave outwards. The circle whose radius is X’v is called the circle of the gorge. Let us also suppose a SringeR Hayne the radius of its base equal to the radius of the upper circle of the sur- face, to be placed on the surface. It is required to find the curve of shadow which the upper circle (daQS, /'z’) casts upon the surface, under the supposition that the light is not intercepted by the sur- face. Admitting this supposition, it follows, that each point of the upper circle casting a shadow will, in gene- ral, cast two points of shadow; one where the ray through the point enters the surface, and the other where it leaves the surface. Through the upper circle of the surface suppose a cylinder of rays to be passed. The ray (AC, BC’) is the axis of this cylinder, and the curve in which it intersects the surface of revolution is the curve of shadow r equired. Through the axis of the cylinder of rays and the axis of the surface, suppose a meridian plane to be passed. Its horizontal trace is AC, and it cuts the upper circle of the surface in two points, one of which (a, a’) casts shadows on the meridian curves. Let this plane be revolved parallel to the vertical ,plane of projection. After it is so revolved, the ray of light will be vertically projected in Be’, the meridian curves in the curves repre- ‘senting the vertical projection of the surface, and the point (a, a’) at 7’. o as ks E i Tl N — =n SSS ————— | | | | WIN ——— —- --__ eee? LE. Prudhomme St, pe ts SHADES AND SHADOWS. 59 Through /’ draw /'éd parallel to the revolved ray Be’; the points 6 and d in which it intersects the meridian curves, are the highest and lowest points of the curve of shadow, in their revolved position. In the counter revolution the points 6 and d describe arcs of horizontal circles, and when the revolution is completed, are hori- zontally projected at and g, and vertically at h’ and ¢’. These points may also be found by drawing through », the point in which the revolved ray l’bd meets the axis, the line g’vh' parallel to BC’, the vertical projection of the ray, and noting its intersection with the horizontal lines dg’ and 6h’; and then projecting the points g’ and h’ into the horizontal plane in the line aAC. To find points of the curve between the highest and lowest points, we intersect by horizontal planes. Each horizontal plane will intersect the surface of revolution in a circle, and the surface of the cylinder of rays in a~ circle equal to the upper circle of the surface : the points in which these circles intersect are points of the curve of shadow. Let H'C’ be the trace of an auxiliary horizontal plane. This plane cuts the axis of the cylinder of rays in the point (C, C’). With C as a centre, and a radius equal to Bz’, describe the arc pm; this will be the horizontal projection of an arc of the circle in which the auxiliary horizontal plane intersects the surface of the cylinder of rays. Projecting H’ into the line DA, we have AH for the radius of the circle in which the same plane inter- sects the surface of revolution. Describing that circle, and noting the points in which it intersects the circle described witl the centre C, we find m and p, the hori- zontal projections of two points of the curve of shadow. These points are vertically projected in the vertical trace of the auxiliary plane, at m’ and »’. Similar con- ee ae “ ' 60 TREATISE ON structions determine the points (2, 2’), (e, €), (0, 0’), and (n,n). ‘The points (2, 2’) and (e, e’) are im the circle of the gorge. That part of the curve is made full, in vertical pro- jection, which is in front of the meridian plane DAz; and the part of the surface which is above the curve of shadow is shaded. The elements of shade on the pedestal are (N, N’) and (R, R’); and on the upper cylinder (Q, Q’) and (8, 8’). 41. To find the brilliant point : Suppose the eye to be situated in a perpendicular to the vertical plane, and at an infinite distance from it. Through any point of the axis, as (A, A’), suppose a ray of light to be drawn, and also a line to the eye, and the angle contained between them to be bisected as in Art. 36. The bisecting line is (AK, AK’). It is now required to pass a plane perpendicular to this line and tangent to the surface; the point of contact will be the brilliant point. If we suppose the tangent plane to be drawn, its trace on the meridian plane passing through the bisecting line (AK, A’K’) will be perpendicular to the bisecting line, and tangent to the meridian curve. Let this meridian plane be revolved parallel to the vertical plane of pro- jection. The bisecting line will then be vertically pro- jected in the line A’é’, and the meridian curve im the curve DHT’. Let H'l be drawn perpendicular to A’k’, and tangent to the curve D'H’; the point of contact (H, H’) is the revolved position of the brilliant point. Draw the normal H’V perpendicular to the tangent, or parallel to A‘k’. In the counter revolution, V being in the axis remains fixed, and H’ describes the arc of a horizontal circle. After the counter revolution, the bisecting line is vertically projected in A’K’, and the normal VH’ in SHADES AND SHADOWS. 61 VP’, parallel to A’K’. Hence P’, where the horizontal line H'P’ intersects VP’, is the vertical projection of the brilliant point.. its horizontal projection is at P, in the horizontal trace of the meridian plane AK. The construction here given, is general for all sur- faces of revolution. If the eye is supposed to be in a line perpendicular to the horizontal plane, the brilliant point is easily found; for we bisect the angle as before, and draw a tangent plane perpendicular to the bisect- ing line. | It is plain that a second line can be drawn perpen- dicular to £'A’, produced on the other side of A’, which shall be tangent to the meridian curve 2d. A second tangent plane can therefore be drawn perpendicular to the bisecting line, and the point of contact will answer the mathematical conditions of a brillant point. ‘The point, however, will be on thas’ part of the surface which is not seen by the eye. PROBLEM XIV. Having given a surface of revolution and the direction of the hght, tt 1s required to find the curve of shade. 42. Fig. 2, Pl. 8 represents the projections of a sur- face of revolution generated as in the last problem. Every ray of light that is tangent to the surface of revolution is an element of the tangent cylinder of rays which determines the curve of shade; therefore, every point at which a ray of light is tangent to the surface is a point of the curve of shade. Through the axis of the surface let a meridian plane of rays be drawn—PB is its horizontal trace. Let this plane be revolved until it becomes parallel to the verti- cal plane of projection. The ray of light through (A, A’) 62 TREATISE ON will then be vertically projected in the line A’b’, and the meridian curves, in the curves which represent the ver- tical projection of the surface. Let the two tangents ad and ge be drawn parallel to the revolved ray A’b’; the points of contact (a, a’) and (e,¢) are the highest and lowest points of the curve of shade in their revolved position. After the counter revolution, these points are — horizontally projected at c and f, and vertically at ¢ and f. The lines gf’ and dé are parallel to A’B’ the verti- cal projection of the ray. To find other points of the curve of shade, we use auxiliary tangent surfaces. Draw any line, as E’Z’, between the highest and lowest points. At E’ or #’, draw a tangent, as EC, to the meridian curve. Conceive the meridian plane and the right-angled triangle E’CO to be revolved about the axis of the sur- face. The meridian curve generates the surface of revo- lution, and the line EC the surface of a right cone tan- gent to it ina circle whose vertical projection is E’k’, and horizontal projection Emn. If now, two tangent planes of rays be passed to this cone, they will be also tangent to the surface of revolution at two points in the circum- ference of the circle (Emn, E’k’): these points are points of the curve of shade. Through (A, C), the vertex of the cone, let a ray of light be drawn. This ray pierces the plane of the cone’s base at (D, D’). Through (D, D’) let two lines be drawn tangent to the base of the cone—these tangents are the traces of the two planes of rays that are tangent to the cone, and Dm and Dn are their horizontal projections. Projecting the points m and n into the vertical plane at m andn', we determine two points - (m,m’) and (n,7’) of the curve of shade. In a similar manner, other points of the curve of shade may be found. - SHADES AND SHADOWS. 63 If the circle of contact (Emn, E’k’) be taken nearer the circle of the gorge, the vertex of the cone will be farther from the horizontal plane, and the elements will be nearer vertical. And when the circle of the gorge is assumed for the circle of contact, the auxiliary tan- gent surface will be a vertical cylinder, having a common axis with the surface of revolution. If two planes of rays be drawn tangent to this cylinder, the elements of contact will pierce the horizontal plane at z and /, the opposite extremities of a diameter perpendicular to the horizontal projection of the ray of light. This cylinder determines the two points (J, /’) and (2, z’) of the curve of shade. There is yet a third method of finding points of the curve of shade: it is by means of auxiliary tangent spheres. Assume any line, as p’r, for the vertical projection of the circle of contact of a sphere and the surface of revolution. At p’ draw a tangent to the meridian curve, and pg perpendicular to it. ‘The point g, where the perpendicular meets the axis, is the vertical projection of the centre of the sphere, and qp’ is its radius. Let us now suppose this sphere to be circumscribed by a tangent cylinder of rays. This cylinder will touch the sphere in a great circle, the plane of which will be per- pendicular to the direction of the ight. Therefore, the trace of the plane of the circle of contact of the sphere and cylinder of rays, on the meridian plane EAQ, or on the vertical plane of projection, will be perpendicular to the projection of the ray of light on either of these planes (Des. Geom. 49). Hence, qs’, drawn through gq perpendicular to A’B,, is the vertical projection of the line in which the plane of the circle of contact of the sphere and cylinder intersects the meridian plane EAS. 64 TREATISE ON But the plane of this circle of contact intersects the plane of the circle of contact of the sphere and surface of revolution in a horizontal line perpendicular to the direction of the light. This horizontal line must pierce the meridian plane EAé in the trace qs’, and also in the trace p’r; therefore it pierces it at (s,s°). Hence, the line tsv, drawn through s perpendicular to the horizontal projection of the ray, is the horizontal projection of the intersection sought. Projecting p' mto the horizontal plane at p, and describing a circle with the radius Ap, we have the horizontal projection of the circle of con- tact of the sphere and surface of revolution. Projecting the points ¢ and v, in which the line ésy intersects this circle, into the vertical plane at ¢’ and v’, and we have the points (¢, ¢’) and (v, v') which are common to the surface of revolution, the tangent sphere, and the cylinder of rays. If through these points planes be drawn tangent to the cylinder of rays, they will be planes of rays, and tangent both to the sphere and surface of revolution. Hence, the points (/,¢’) and (v, v’) are points of the curve of shade. If we suppose the space within the surface of revo- lution to be unoccupied, a part of the curve of shade will cast a shadow that will fall onthe convex side of the surface. The ray of. light which determines (c, ¢’), the highest point of shade, being produced, will intersect the oppo- site meridian curve: the poit of intersection is a point of shadow. Thereare points of the curve of shade, on both sides of the meridian plane PAD, which also cast sha- dows on the surface. As we recede from the meridian plane PAD, on either side, the part of the ray inter- cepted between the point of shade and the correspond- ing point of shadow, continually diminishes, and finallv SHADES AND SHADOWS. 65 becomes nothing, or the point of shadow unites with the point of shade casting it. At these two points, one on each side of the meridian plane PAD, the ray of light will be tangent to the curve of shade. For, when the point of shadow unites with the point of shade, they become consecutive points of the curve of shade; hence, the ray passing through them is tangent to it (Des. Geom. 65). But if the rays of light at these two points are tangent to the curve of shade, their projections will be tangent to the projec- tions of ‘the curve (Des. Geom. 90). The horizontal projection of the curve of shade being constructed, draw two tangents to it parallel to PAD, the horizontal pro- jection of the ray of light. The points of contact are X and y, and Xcy is the horizontal projection of that part of the curve of shade which casts a shadow on the interior of the surface. Projecting the pomts X and y into the vertica! projection of the curve, at X’ and 7’, and drawing lines parallel to the vertical projection of the rays, the lines so drawn will be tangent to the vertical projection of the curve of shade. Descending along the curve of shade, from the points (X, X’), (y, 7’), the rays of light touch the surface on the concave side, and the points of the curve of shade still cast shadows upon the surface. When we reach those points, one on each side of the meridian plane PAD, at which the point of the curve of shade unites with the point of the curve of shadow, the rays become tangent to the curve. Therefore, drawing two other tangents parallel to PD, their points of contact w and z are the horizontal projections of two points at which the rays of light are tangent to the curve of shade. At these points the curve of shade returns to the convex side of the surface. Projecting these points into the vertical 66 TREATISE ON projection of the curve of shade, we find the points Ww and z’, through which, if the projections of raysbe drawn, .— they will be tangent to the vertical projection of the: ~ curve of shade. That part of the curve of shade whose beaant projection is yeX lies on the convex side of the surface ; the part Xv/mz lies on the concave side of the surface ; the part zfw lies on the convex side, and the part wnity on the concave side. The curve is symmetrical with respect to the meridian plane PAD. : The part of the curve of shade which is in front, cof the meridian plane EAd is made full in vertical projec- tion, and the part of the surface lying above the curve of shade,and seen,is darkened. PROBLEM XV. Having given the position of a surface of revolution and the direction of the hght, tt ts required to find the line which separates the dark from the wluminated part of the surface, and the shadow which is cast on the horizontal plane of pro- jection. a - 43. Let (A, A’B), (PL 9) be the axis of the surface, and let the projections of the surface be made as In the figure. | Find the shadow which the upper horizontal circle CD casts upon the surface, as in Prob. 13, and Hien find the curve of shade, as in Prob. 14. | The highest and lowest points of the curves of mids and shadow, are in the meridian plane of rays EAB’: By considering the form of the meridian curves, it j 1s plain, that the highest point of the curve of shade is above mos highest point of the curve of shadow, and the ™ « = + ra i? . & ; be r ie ype “ a x. ‘ * z 7 , .. > : * . . = no : , ’ - - , ~ a . @ . - . “ + = Ss 2 - - ~ : 7 ~ a © v . . - e 4 ‘ i 7? . ? ¢ by - * - . x 2 . 1 ‘ > . = . . ‘ = *. oy * @ : : * ~ * - * ® - : ~ = P ~ + e 3 - oe et + & w — ¥; 5, > ~ _ . wr) ated s « * & . = - ; “ 2 ¥ oS is ~. = . ; o" & » ~ “. *. Ar} 35. : : : F ae, 4 i = + BA ~ . . . ~ ; . * ¢ > : ana: a - 9 Gs ~ * ' = « — prem A i> = ™ i + . —~ ‘ 7 > in . zs - q 4 : * Cc* 5 >. = . - = ad sme y 4 » ; . ; . ine 2 > ¥ ¥ Semin A = + ' & ~ J - . - 2 “ oo ¥ = fj : > a — ~ re = r 7 i : r e— = . ¢ . - é — *. $ ; - - -_ 2 : z he : 7 7 S ‘ = o ? r- - ? ~ . . ' * f e ; . % uf e 4 Ke ; ,) - RRS oe RED ie 2a fink fy’ Nate tia mg > : / met wee am I ot) wih (ing OE ES, ) Sen!) ee Whe ut PEE) NARA aR Cae lap beac aame hans: st Testy aa i SS nye Cra ils t a a fu rea ( y e* nisl ay ‘ / a é or ¥ ; apie Re. ve | ul inal / id | . § ba f . if \ i I F A / \ \ | / Fi / / ‘ | nsf Pe / a \\ AKG \\\ . | ve vf H vi \ AI \\ ‘s c / ri fo \ AN AIS \ 1 : " yr / i / ‘ a\ \\ \\\ | vii / : hk f sys : \\ \ \ \\ pm . Y i \ \ + f / / / Shy i \ WN \\ | * iy yf J va \ ah \ \\ | + rd f / / yy = ‘ \ wy AK if ne ve cs \ \ \ ‘ \\ 2 { ad / f / / Ze Z7___= vi \ \\ \ We f / j ye = / ; Ps ie Less se : ¥ / 4 } Zé th if : | | Ze Z Z gE TW tjjj- a \ \ \ \ \ \ i \ \ \ \ & | SSS == 1/11) ER ey en ey oe ne 76 ce meme {11 (11) CCE | / 4 ta gb i H = \ If, a ew IZ) i if / ; ant %) ly / i Pt hi we | oad / q i fi feo WY WAN Han)? ie hat ie eee OS) \\\) {62 Fath / KW. yr AVS \ aia . AS : SHADES AND SHADOWS. 67 lowest point of the curve of shade below the lowest point of the curve of shadow. ‘These curves will therefore intersect each other. This they do at the points (a, a’) and (6, 6’). Above these points the curve of shadow being below the curve of shade, and on the exterior sur- face, separates the dark from the illuminated part of the surface. Below these points, the curve of shade is below the curve of shadow, and separates the dark from the illu- minated part of the surface, until it returns to the con- vex side of the surface at the points (c, c’) and (d, d’). It has already been observed, in Prob. 14, that the parts of the curve of shade (fac, f'ac’) and (gbd, g‘b'd’), which are on the concave side of the surface, cast shadows upon it. These shadows begin at the points (c, c’) and (d, a’). To find these shadows, we will, in the first place, find the shadow which the entire curve of shade would cast on the horizontal plane, under the supposition that the surface offers no obstruction to the light. This is done by finding where rays of light, drawn through the several points of the curve of shade, pierce the horizontal plane. The upper part of the curve, which is on the convex side of the surface, casts the shadow /”h’g’—the lower part of the curve, on the convex side of the surface, casts the shadow c’l”d"—the parts of the curve on the concave side of the surface, cast the shadows c’t’f” and "g'g". Assume now any horizontal circle below the points (c, ¢’), (d, d’); the one, for example, whose ver- tical projection is &'mn, and whose horizontal projection is the circle described with the centre A and radius Ak © equal to mk’. The centre of this circle casts a shadow on the horizontal plane at m’—the circumference de- scribed with m’ as a centre, and radius mp, equal to mk’, — 4% 68 TREATISE ON will be the shadow cast on the horizontal plane by the circumference of the assumed circle. If through the points p and q, in which this shadow intersects the shadow cast by the curve of shade, rays of light be drawn, they will intersect, in space, the cir- cumference of the horizontal circle and the curve of shade. The pomts in which these rays mtersect the circumference of the horizontal circle, are points of shadow on the surface which are cast by the pomts in which the rays intersect the curve of shade. Therefore, draw the horizontal projections of rays through the points p and g, and the points p’ and q’, in which they intersect the circumference kp‘q, described with the centre A and radius As, are the horizontal projections of two points of the shadow on the sur- face. ‘These pomts are vertically projected at p” and gq’. By similar constructions, we may find any num- ber of points in the shadow which the curve of shade casts on the surface. If we take the lower horizontal circle FG, we shall find the points (7, /) and (x, x’) where the shadows on the surface terminate. The curve (cpt, cpt) and (dqz, d'qx°) may now be described. ‘The curve (cp’t, cp?) intersects the meridian plane HA at the point (s, s’). We have thus found the lines on the exterior of this surface, which separate the dark from the illuminated part. They are the curve of shadow until it intersects the curve of shade, then the curve of shade until it returns to the interior surface, and then the curve of shadow cast on the surface by the curve of shade. The light does not fall on that portion of the surface which is above and within the parts of these curves. Let us now find the shadow which the surface casts on the horizontal plane. SHADES AND SHADOWS. 69 The elements of shade (I, I’) and (K, K’) cast lines of shadow onthe horizontal plane, which are found by drawing rays of light through their upper extremities. Then find the shadow cast by the circumference of the circle whose vertical projection is FG. This shadow intersects the shadow cast by the curve of shade in the points ¢” and 2”. ‘These points of shadow are cast by the points (¢, t’) and (a, 2’). » I’ind next, the shadow cast by the circumference of the upper circle of the surface. The centre of this cir- cle casts a shadow af B’.. With B’ as a centre, and a radius equal to BD, let the circumference of a circle be described—this circumference is the shadow sought. If through the points y and z, in which this shadow intersects the shadow cast by the curve of shade, rays of light be drawn, they will intersect, in space, both the curve of shade and the upper circle of the surface. | These points of shadow are therefore cast by the points (a, a’) and (6, 6’) in which the curves of shade and shadow intersect upon the surface. The shadows ¢’y and xz are cast by parts of the curve of shade. The elements of shade (P, P’), and (L, L’) of the upper cylinder, cast shadows which are tangent to the circumference described with the centre B’. The cir- cumference of the upper circle of the cylinder casts a shadow on the horizontal plane, which is also tangent to the shadows cast by the elements of shade. OF THE SHADES AND SHADOWS OF THE ROMAN DORIC COLUMN. 44, This column is composed of three principal parts :—Ist. The base; 2d. the shaft; and 3d. the capital. ‘That the figure may not be too complicated, we shall 70 TREATISE ON first find the shades and shadows on the base and shaft of the column, and then use a separate figure to deter- mine those of the capital. 45. The base of the column (PI. 10. Fig. 1) is com- posed of seven parts. 1. A rectangular prism, called a plinth, whose horizontal sections are squares: 2d. A solid of revolution, convex outward, called the lower torus: 3d.A cylindrical fillet: 4th. A solid of revolution, concave outward, called a scotia: 5th. A cylindrical fillet: 6th. The upper torus: 7th. A cylindrical fillet. The shaft of the column rises from the upper fillet. For a short distance it is concave outward, then it be- comes nearly cylindrical, and continues so to near its upper extremity, where it again becomes concave out- ward. ‘The drawing is made on the supposition that the shaft is cylindrical between the parts of it which are concave outward. 46. The capital is composed of two distinct parts— the one a member whose horizontal sections are squares, having their centres in the axis of the column, the other a solid of revolution. The projections of the capital are shown in Fig. 2. The part of the capi- tal whose vertical projection is ww’o’s's, is the portion whose horizontal sections are squares—this part is called the abacus. ‘he part of the abacus between ry and nv is called the cyma-reversa, or talon. The part of the capital whose vertical projection is ll’mm is called the echinus. The part whose vertical projection is k&T7/ is called the upper fillet. The part whose vertical projection is 7k’% is called the cavetto. The part whose vertical projection is ge"/’h is called the neck. The part ff’ is called the astragal, or colarino. The part whose projection is dd’c’e is called the lower SHADES AND SHADOWS, 71 fillet. ‘The entire column below the abacus and above the plinth is a solid of revolution, and the vertical projec- tions, in the two figures, are meridian sections of its surface. PROBLEM XVI. To find the shades and shadows on the shaft and base of the Roman Dorie column. 47, A semi-column will illustrate all the cases. Let the semicircle AMB (PI. 10. Fig. 1) be the horizontal projection of the cylindrical part of the column, CND of the upper and middle fillets, EOF of the upper torus, GPH of the lower fillet, LQS of the lower torus, and the rectangle LR of the plinth. The space included between the semicircle GPH, the horizontal projection of the lower fillet, and a semi- circle described at equal distances from AMB and CND, iimits the horizontal projection of the scotia. Let the vertical projections be made as in the figure. Through the axis of the column, let a meridian plane of rays be drawn—VI isits horizontal trace. Through I draw la at right angles tolV. ‘The plane of rays, tangent to the cylindrical part of the shaft, touches it in an element, which pierces the horizontal plane at a; this element, which is the element of shade, can there- fore be drawn. The element of shade casts a shadow on the concave part of the shaft, beginning atthe point (a, a’). This shadow is in the vertical plane of rays passing through the element of shade, and is therefore horizontally pro- jected in the right line drawn through a perpendicular to la. The vertical projection of this shadow is found 72 » °° * PREATISE ON by intersecting the shaft below a’ by horizontal planes. Each plane will intersect the shaft in a circle, which being projected on the horizontal plane, the circum- ference will intersect the tangent through a. ‘Then, pro- jecting the point of intersection into the vertical plane, we determine a point of the vertical projection of the shadow. The plane of rays tangent to the upper fillet, touches it in the element which is projected on the horizontal plane at 6. This element of shade, and the circum- ference of the upper circle of the fillet, cast shadows on the upper torus. The shadow of the element is found by a construction similar to that used in finding the shadow on the foot of the shaft. Points of the shadow cast by the circle are found by first finding the shadow _of the circle on a horizontal plane, and then finding the shadow which‘a horizontal section of the torus would cast on the same plane, and drawing rays through their points of intersection. ‘The shadow cast by the circum- ference of the upper circle of the fillet on the torus, is * made full, until it intersects at (A, h’) the curve of shade, determined as in Prob. 12. Passing the curve of shade on the upper torus, the construction for which is given in Prob. 12, we come next to the shadow which this curve casts upon the middle fillet. : To find this shadow, we intersect the torus and the fillet by a plane of rays cp’, perpendicular to the vertical plane of projection. Through the point ¢, in which the trace of this plane intersects the vertical projection of the curve of shade, let a horizontal plane be passed— this plane will intersect the torus in a horizontal circle, and let this circle be projected on the horizontal plane. The horizontal projection of ¢ is at c, and (¢,c’) is a . , ' * ® 4 ‘ ‘ : A er] . * Ps Pe + Mae a 3 Fs EC. ee ¢ > E %*« . > * o ia Le) Ti Ni ‘th | ] | wn IN l i . a | : » > » ad y i - a ae a a a n = FAS ae ee 5. eer ie LPrudhomme Se. . ; = ; ; ~— a - ay 7] . > nd 4 SHADES AND SHADOWS. 73 point of the curve of shade. The ray of light through this point pierces the fillet in the point (p, p’), which is a point of the curve of shadow. This shadow passes down. upon the fillet obliquely, until it intersects the lower circle of the fillet or upper circle of the scotia, at (X, X’). That part of the upper circle of the scotia, from h” to (X, X’), on which the light falls, casts a shadow on the scotia, which is found as in Prob. 13. The ray of light passing through the point (X, X’), in which the shadow on the fillet intersects the upper circle of the scotia, determines the first point of shadow which the curve of shade on the torus casts on the scotia. To find points of this shadow, intersect the torus and scotia vy a plane of rays ¢/’ perpendicular to the verti-. cal plane; find the point (e, e’) in which this plane inter- sects the curve of shade, and construct the curve in which it intersects the scotia. Through e draw the hori- zontal projection of a ray of light—the point f, in which it intersects the curve of the scotia, is the horizontal, and f’ is the vertical projection of a point of shadow. The shadow cast on the scotia by the curve of shade on the torus, terminates in the upper circle of the lower fillet. A part of this shadow, beginning at g, is seen m horizontal projection. The element of shade on the lower fillet is (g, 2’); and this element casts a shadow, beginning at its foot, on the lower torus. Then, that part of the upper cir- cle of the fillet, ntercepted between the element of shade and the point in which it is met by the shadow on the scotia, casts a shadow on the lower torus. ‘Then the shadow on the lower torus is. cast by the curve of shade on the upper torus; and this shadow continues until it intersects the curve of shade determined as in Prob. 12. , if £ 74 TREATISE ON PROBLEM XVII. To find the shades and shadows on the capital and shaft of the Roman Dorie column. 48. The part of the abacus, whose vertical projec- tion is s 7’, casts a shadow on the cyma-reversa. The semi-reversa is composed of parts of cylinders whose elements are respectively parallel and perpendi- cular to the vertical plane of projection. ‘Two of the cylinders intersect each other in a curve, whose hori- zontal projection is gn”, and vertical projection gpn. The line (rr, 77") of the abacus casts a line of shadow on the cyma-reversa parallel to itself. Through (7’, r) draw a ray of light, and find the point in which it pierces the surface of the cylinder whose elements are parallel to the vertical plane: gq‘ is the horizontal Ime drawn through the point, and is the lne of shadow required. The light falls on a part of a cyma-reversa below this line. Let a tangent plane of rays be drawn to the part of the cyma-reversa, whose elements are parallel to the vertical plane. It touches the cyma-reversa in the ele- i - ment pp’, which is therefore an element of shade. This ‘element of shade casts a shadow on the part of the cyma-reversa below it. This shadow is determined by finding the intersection of the tangent plane of rays with _» the cyma-reversa. It is the horizontal line near zz’. The lower element x7’ of the cyma-reversa casts a shadow oo’ parallel to itself, onthe part of the abacus below it. We come next to the echinus, which is the half of a torus. SHADES AND SHADOWS. 15 First, find the curve of shade on the echinus as in Prob. 12. Then, through the horizontal line of the abacus whose vertical projection is w draw a plane of rays: we is its vertical trace. Since this plane is per- pendicular to the vertical plane of projection, the shadow cast by the line of the abacus is vertically projected in the trace we’. The ray of light through the point (w”, w) pierces the neck of the column at the point (ce, ¢’). Through the lower line of the abacus (w’n”, ww’) let a plane of rays be drawn. ‘This plane intersects the echinus in a curve which is the shadow cast by the line (w'n, ww’) on the echinus ; this curve of shadow meets the curve of shade in the points x and y; leaving a part of the echinus in the light. The curve of shadow is found by intersecting the echinus by horizontal planes —each plane will intersect the echinus in a horizontal circle and the plane of rays in a right line: the point in which the right line intersects the circle is a point of the curve of shadow The plane of rays passed through the line (w"n”, w w’) intersects the neck of the column in the curve ct, the cavetto in the curve u, and the fillet directly above, in the curve v. The curve of shade on the echinus casts a shadow on the fillet below, which begins at /, and intersects at v the shadow cast by the lower line (w’n”, ww’) of the abacus. The lower circle ££’ of the fillet, casts a shadow on the echinus, which begins at a, and intersects the shadows cast. by the abacus in two points, one near a, the other at w. From uto z the shadow is cast by the small part of the arc of the circle Zk’, whichis in the light near v. The ray through z passes through the last point of the circle kk which is in the light, and therefore passes through a point of the curve of shade onthe echinus. From z to : * * ry . ees * J ® z cory 76 sie" ATISE ON Fy the line 77’, the shadow is 5 ee by the curve of shade on the echinus. The lower circle ¢7’ of the cavetto, casts a cHadows ! on the neck of the column which intersects at ¢ the” shadow cast by the lower line (w"n’”, w w’) of the abacus, and at ¢ the shadow, cast by the curve of shade on the «+ echinus; from / to?” the shadow is cast by the curve of shade on the echinus. - We come next to the astragal ff’. Find the curve -. ,of shade as i ‘Prob:12. | A as ' The curve of shade on the echinus casts a shadow on the astragal which intersects the curve of shade near og’. Thecurve of shade on the astragal casts a © shadow on the fillet dd’e’e. The lower circle ee’ of the fillet, casts ashadowon | the shaft of the colimn, which intersects the element of shade at 6. : The shadow cast by a curve of shade on asurface of revolution, may be found by drawing rays throughits different points, and findmg where they pierce the sur- » face; but it is generally better to pursue the method adopted in Prob. 14. ae ~ : a + * y* - OF THE HELICOID.* 49. The helicoid is a surface generated by arightline —, moving uniformly in the direction of another right line which it intersects, and having at the same time a uni- form angular motion around it. . Therfixed line is called the axis of the helicoid, and “" « the moving line the generatrix. ) a “it te * The properties of the helicoid are used in the next problem, and © | although their discussion belongs rather to warped surfaces, than toa treatise on shades and shadows, yet it was thought best to give the ~ properties here, as the student may not meet with them elsewhere. - news ° “ye % lig/. | _ | ae | | | / " y wg” nd eis eT A \ ; ea . eee Mtoe | G J ae i if t a es | ) |: | : | \ 3 \ |: ¥ \ aN \ Ki: \ \ \ : \ Es PD ae er Ps “Se ee x Bg *| , \7 \ x 7 } “7 x | a x Sy | “ . — 7% * ms a 4 & wo ? ? " bad 54, o ef s : eee stp re ’ ae Pine | ¥ > ¥ ee » ot 2 : a a” ? opaque body will be the line of shade. ee 4 of light, wall be the indefinite shadow. When the opaque .. body is the largest, the luminous body and the vertex. “. gays of light may be considered as convarayas to the vw t ne ° des » TREATISE ON “ luminous point at d finite distance; and to fall upon an opaque body, the rays will be,divergent. _ Let us suppose'a cone to be drawn tangent to the — “opaque body, of which the Hane point shall be the - vertex. It is plain, 1°. That the line of contact of fae cone with the eat alk r ‘ oe a ¢ we 2°. That the part of Space included within the sur- facevof this cone, and lying on that side o “the opaque body opposite to the source of light, will be the indefi-_ nite shadow of the opaque body. a Tae 3°. That the shadow of the opaque body on any sur-) .* “ig ~ face, will be the intersection of that surface with the = # tangent cone of rays. 56. If we suppose the light to emanate froma lumin- ous body, we may Agee a cone drawn tangent to the luminous body, and to any opaque body whose shade ~~ and shadow are to be determined. The curve of con- > | _tact on the opaque body willbe the line of shade, and the part. of space within the surface of the tangent cone ad on the side of the opaque body opposite the source of the tangent cone will be on the same side of the opaque - body, and the rays of light will be divergent; but when it is the smallest, the opaque body will be between | ‘the luminous body’and the vertex of the cone, and the’. . * os 2 fh : 4 4 “?, shila of the tangent,cone. * In all the cases in which the rays F light are not... T pBeattel, the. problems in Shades and Shadows will be ‘ solved by finding the contact of a cone with an opaque body, and the intersection of this cone with any surface - on parcnich the shadow falls. er" . * * - a” CUP e3 — ae ‘ ve a. .. * « : feng ¢ -¥ ® | <— . “; - a a : , eS Se oS od E ‘ a wa 3 — . . ‘Se ‘. : * : 2 = : p * * * ; a * . ~~ ee a ‘ S ~~ \ 28 % = 4 i —————— —$—— . E.Prudhomme Sc. CHAPTER L. arate Wish 7. Hap we no knowledge’ of abe other than what is derived through the medium of sight, we should sup- pose them to differ from each other in two respects only —form and colour. Objects having different forms, or different colours, produce different effects upon the eye; but objects of the same form and colour cannot be distinguished | from each other without ‘the yaid of the other senses. « 58. When we view an uate’ all the matt fa it which | are seen are supposed either to emit or reflect rays. of light which fall upon the eye; and it is through the medium of these rays that we derive the idea both of its form and colour. . ® 59. Perspective is the art of representing objects on a surface, in such a manner that the representations shall present to the eye, situated at a particular point, the same appearance as is presented: by the objects themselves. The representation of an object so made - is called its perspective. a _*60. Let us now suppose that we are viewing an i Etiect in space, and that a transparent pic) is tic between. us and the object. * Every point of the object which is seen, is supposed: _ to emit a ray of light that falls upon) the eye, and each . »_ ad “. » ; © i fe £ oe. ° * 8 ae oil # oye a ¢ ae e $ ¥ 7 pt 96 TREATISE ON ray pierces the transparent plane ina point. If to each point so determined, a proper colouring be. given, the representation or picture, on the transparent plane, will present to the eye the same appearance as the object itself. Such representation is, therefore, the perspective of the object. j 61. The rays of ight coming from the different points of the object to the eye, are called vesual rays ; “and the plane on which the representation is made, is called the Perspective Plane. 62. The art of Perspective is therefore divided into two parts. ' 1°. To find the points in which the visual rays pierce the perspective plane, which determines the general outline of the perspective. 2°, So to shade and colour this outline, that it shall appear in every respect like the object itself. The first part is called Linear Perspective. It em- braces that portion of perspective that is strictly mathe- matical, and which will form the selltthy of the following treatise. The second part is called Merial Perspective. This branch of the art belongs to the draftsman and the painter, and is to be learned by a careful study of the © objects of nature, under the guidance of an improved and cultivated taste. 63. As it is the end of perspective to represent objects as they appear in nature, such a position ought to be given to the perspective plane as will enable us to con- ceive, most easily, of the positions of objects from viewing their perspectives. This we can do with the least difficulty, when the per- spective plane is taken parallel to the principal lines of the eis . ob LINEAR PERSPECTIVE. 97 Of the objects in nature, the larger portion of lines *4 are vertical; therefore, in Tost perspective drawings the perspective plane has a vertical position. 64. Before an object can be put in perspective three things must be given, or known. 1°. The place of the eye. 2°, The position of the perspective plane. . The position of the object to be put in perspec- aa | 65. By considering what has already been said, we may deduce the following principles : 1°. If through the eye and any point in space, a visual ray be drawn, the place at which it pierces the perspec- tive plane is the perspective of the point. 2°, If through the eye and all the poits of a right line, a system of visual rays be drawn, they will form a plane passing through the eye and the right line; this plane is called a ‘Sel plane, and its trace on the per- spective plane is the perspective of the right line. 3°. If through the eye and all the points of a curve, a system of visual rays be drawn, they will, in general, form the surface of a cone, the vertex being at the eye; this cone is called a visual cone, and its intersec- tion with the perspective plane is the perspective of the curve. Oot 3 66. In determining the perspective of an object, it is unnecessary, and indeed impracticable, to draw visual rays through all its points that are seen, and to con- struct the intersections of these rays with the perspective »plane. We therefore select the prominent points and lines only, such as the vertex and edges of a pyramid, the vertex of a cone, the edges of a prism, &c.: and having put these lines in perspective, we have, in fact, determined the pepepectrmer the body. % 98 TREATISE ON 67. If through the eye a system of visual rays be drawn tangent to the object to be put in perspective, they will, in general, form the surface of a visual cone tan- gent to the object; the line of contact is called the ap- parent contour of the object; and the intersection of the surface of this cone with the perspective plane, is the boundary of the perspective of the object. We shall now apply these principles in finding the perspectives of objects. PROBLEM I. Having given a cube and tts shadow on the horizontal plane, it ts required to find the perspective of the cube and the perspective of ats shadow. 68. Let DEGF (PI. 12) be the horizontal projection, and D'E'G’E” the vertical projection of the cube. And let DehgG be the shadow cast on the horizontal plane. Let (A, A’) be the place of the eye, and BC, BC’ the traces of the perspective plane. The perspective of the cube and its shadow, after they shall have been found, will be projected on the planes of projection in the traces of the perspective plane (Des. Geom. 24); but in order to exhibit their per- spective truly to the eye, it must be presented as it appears on the perspective plane. For this purpose we remove the perspective plane parallel to itself, any convenient distance, as BB’, and then revolve it about its vertical trace B’E’ until it coin- cides with the vertical plane of projection. The place of the eye is supposed to be moved with the perspective plane, and to have the same relative position with it after it has been revolved. LINEAR PERSPECTIVE. 99 Through the angle (D,D’) of the cube, draw the visual ray (AD, A’D’); this ray pierces the perspective plane at the point (d, a’), and after the plane has been moyed and revolved, the perspective of the point is at d”: on Through (E, D’) draw the visualray (AE, A’D’) ; nis ray pierces the perspective plane at (a, a’), and deter- mines a” the perspective of (E, D’). Hence d”a” is the » perspective of DE. The*visual ray through the point (F, F’) pierces the perspective plane at the point (/", /’), and determines e, the perspective of the point (EE). The visual ray through the point CE, E’) pierces the perspective plane at (a, is and determines £’, the perspective of the point (E, E’). Hence ak’ is the’ perspective of the edge (E, D’E’) of the cube. The visual ray through (D, E’) pierces the perspective plane at (d, £), and determines p, the perspective of the _ point (D, E’). Hence pk’ is the perspective of the edge (DE, E’) of the cube, d"“p of the edge (D, D’'E’), and the square a’k'pd” of the front face of the cube, which is parallel to the perspective plane. : Having determined, by similar constructions, the per- _ spectives of the other angles of the cube, we see that the trapezoid d"pq'f is the perspective of the face which is horizontally projected in the line DI’; the trapezoid pkn'd the perspective of the upper face of the cube; the trapezoid d"a"n'f the perspective of the base of the ~ cube; the square fn'n’d the perspective of the back face of the cube which is parallel to the perspective plane; and the trapezoid a’k'n'n' the perspective of the face of the cube which is horizontally projected in the line EG. There are but three faces of the cube which are seen, viz. the upper face, the front face parallel to the per- spective plane, and the face which is horizontally pro- G2 . ty ‘tele ing : x Po 4 sy, i * is — se. e+e = 33 we age : a . ‘ m+ °3, . g .? .? a 4 . 4 * : RE. a a Bee . 4 : aur ‘ ra ’ eT ® Ke = a ‘im ¥ % 5 a ~ ye . $r - : " : ‘ w . yr , % be 1 4 - r * Facet a Tae @ 3 at a ¥ * os had > » .#* » #4 @ a — MS ilies a J es 3 hy. *, .* ¥, . ee 5 ov oe ss ee a ee ” + t re « er” 100 , TREATISE ON eet jected in the line FD. ©The perspectives of the ines — ee ee faces are made full in the perspective ne. e perspectives of the other edges are dotted, in order to show how the cube would appear if it were a transparent body. The perspective of the shadow on the horizontal - plane is found by finding the perspectives of the elas e,h, and g, and tins the lines d"e, eh, hg and gn. 2 The jie hg, which is the perspective of hg, inter- sects at c the line fg,which is the perspective of the edge (F, FG’) of the cube. The part eg of the line hg not seen. [f through the vertical edge (1°, F’G’) of the cube, a visual plane be passed, it will cut the line hg in the point ~ whose perspective isc. ‘That part of the line hg lymg. % i between the visual plane and the vertical plane of pro- . es jection will not be seen, because the cube intervenes. The visual ray passing through the poimt whose per- spective is ce, will intersect the line hg, and the ose i (F, I"G") of the cube; and generally the visual ray pass- » ©. ang through the point inwhich the perspectives of two lines. ee will intersect both the lines in. space. 69. This method of perspective is often used advan- ~ _ tageously in finding the perspectives of bodies bounded by curved foriees. If through the eye a plane be. passed tangent to the object to be put in perspective, the point of contact will be a point of the apparent con- ©. ~ tour of the object’ (67), and the visual ray drawn * through the point of tangency will determine a point in ~ the boundary of the perspective. For example, if it +” pee ee a ghee 3 2 Es ¥ g © ~ “ were required to find the perspective of a sphere, the : ¢ « position of the body, the perspective plane, and the 4 plage ofthe eye being given, we should first pass thr ough | , the eye a system of planes tangent to the sphere, ‘and : al ’ é + +s 7 al a “ee | ~*~ Ce eae, ee tad * ee ast - . 7 = A * a . * nts 4 . * J s mee ee ae zal. Te ae ee ee ee eee ered - ail * 4 ; *, ? . » @ of oe wy ¢ os . Cru. a 4 i NR oat i: na a ae | We ey: e: ' 4S a , , ] G— AIM 4 ris J / , | == / / / / te / | 4 if ' | | | | pate torees | | ——————————— af ——— S| Cid ———SS==—S—SSSSSSSII 55 \ eS \ ——— % | —————— \ = | | ll : : | yee / , | aa ‘H | ah j * y . . @ a oe, : * a : ¥ * ? ~ a My Sa , aie +, & Pay ’ *s 7 a LINEAR PERSPECTIVES 101 ry then draw visual rays through the points of _contact ; pee the points in which these visual rays, pierce the "per- a ps spective plane are points in the boundary of the per- 7 * wi spective of the sphere, . a e ee eae a ¥ vq Fs A how “weg “ * ‘ > - ££ +i as" . .? i " .. & ee ‘ete, : . ** ky. ee CHAPTER HI. pis ty > - . ee : a ” dh *. #9 | +7 i ae “ Z . , . ie 4 » _. OF THE METHOD OF PERSPECTIVE BY MEANS OF DikGowhith ’ ra Ps pe AND PERPENDICULARS. * * Pee a » # ™ . : ae * at 70. The point from which the eye is supposed to view . ear * an object put in perspective, 1s called the point of si sight; and the projection of this ; point on the perspective plane. ca 4 is called the centre of the picture. © —_s ere d P 71. If through the point of sight a right line be drawn aM > parallel to any right line in space, the point in which it en * .« pierces the:perspective plane is — the Ciheolh & point , of that line. # Hence, all parallel right lines née the same vanishing < point : for a right line parallel to one of them will be ad parallel to all the others. . . The % Hence also, all lines perpendicular to the perspective, plane have their vanishing pone at the. centre*of-the#,’.. .,7¢ picture. wl is cabs a NEED ic Regarding a right line as indefinite i in length, i its\van Ne ga ishing point is a point of its perspectives; For, the par- S ae allel through the point of sight is contained in the visual ee plane passing through the given line; therefore, the point . at which it pierces the perspective plane is in the trace of. of the visual plane. But the trate of the visa pane a es ~ #, - = r< 2 jg ete Re a au * e* * * ‘ ral e . # ri ~w! th La wy. = » iy te a, ‘ _ ‘ot a - % e . '% me ¢ ww? & . , . ¥ ti. . ¥) ” «= @ ™ \ * > %g oe x * oe “ - : * * ‘ > , e % »® * - 2 102 -«s “TREATISE ON is the perspective of the given line (65); therefore, the vanishing point of a line is a point of its perspective ; and is the perspective of the point at an infinite distance from the perspective plane. _ When fy ' Jen ij a 2 ee ——s eee ee a eee B r | == : Hh Z = : : | 3 ae | = igen. aN Kes a as ae a CO | we = SS R : ; ai fb ie eae B Fie im | : oO l — # S R | a See Aesresi nnn c C a Cc Prudhomme Sc. | # LINEAR PERSPECTIVE. 113 it describe a rectangle equal to the front face of the pedestal. From 6 lay off dc equal to the distance be- tween the pedestals, and make cd equal toad; and oncd describe a rectangle equal to the front faces of the pedestals. ‘The perspectives of the four pedestals are then determined by constructions entirely sna to those of the last problem. To find the perspectives of the vertices of fre pyra- mids : Let S be the centre of the picture, and D and D the vanishing points of diagonals. {f a line be. drawn through the vertices of the two pyramids on the left, it will be perpendicular to the per- spective plane, and will pierce it in the vertical line AX, drawn through the middle point of a. On this line, therefore, lay off hh’ equal to the height of the vertices of the pyramids above the base of the pedestals, and /’ will be the point in which the line through the vertices of the pyramids pierces the perspective plane. The line /'S is the perspective of this perpendicular. The diagonal through the vertex of the front pyramid pierces the perspective plane in the horizontal line h’a, and also in the vertical line aa’; hence aD" is the per- spective of the diagonal, and v the perspective of the ‘ertex of the front pyramid. The perspectives of the vertices of the other pyramids are easily found. From e lay off ef equal to the distance which the pedestal projects beyond the pyramid, draw eD’ to the vanishing point of diagonals, and from f draw /S to the vanishing point of perpendiculars ; ; the point g, in which these lines intersect, is the perspective of the point in which an edge of the pyramid pierces the upper face of the pedestal, and the horizontal line gp 1s the indefinite PIO ere om 114 TREATISE ON e perspective of the line in which the front face of the pyramid intersects the upper face of the pedestal. From z lay off on ce a distance equal to e f, and through the point so determined draw a line to the centre of the picture : the point p, in which it meets gy, is the -perspect awe of the point in which a second ae of the pyramid pierces the upper face of the Redentall The diagonals through p and ¢ determine, by their intersec- tions with the eden gS and pS, the points in which the two remaining edges pierce the upper face of the pedestal. Joining these points with v, the perspec- tive of the vertex, we have the perspective of the pyramid. Only a part of the edge of the pedestal which is per- pendicular to the perspective plane at e,is seen. ‘The perspective of the edge intersects the perspective of the edge of the pyramid at / If through the poimt/a visual ray be drawn, it will intersect, in space, both the edge of the pyramid and the edge of the pedestal. At the point in which it intersects the edge of the pedestal, the edge of the pedestal passes behind the pyramid, and is not seen; and the same may be said of the edge of the pedestal parallel to ez. The perspectives of the other pedestals and pyramids are found by constructions entirely similar. To find the perspectives of the shadows cast on the horizontal plane: Let R be the vanishing point of rays, and P the vanishing point of horizontal projections. The point v’, in which the diagonals pD and gD’ intersect, is the perspective of the projection of the vertex of the pyramid on the plane of the upper base of the pedestal; hence, v’P is the perspective of the pro jection of the ray through the vertex of the pyramid, on that plane. But vR is the perspective of the ray; LINEAR PERSPECTIVE. 115 nence ¥" is the perspective of the shadow cast by the vertex of the pyramid on the plane of the upper face of the pedestal. It is plain that the edges of the pyramid which pierce the upper face of the pedestal in the points whose per- spective are p and &, will cast shadows on the pedestal and on the horizontal plane. Therefore, pv’ and kv’ are the indefinite shadows on the upper face of the pedestal. It is evident, that only the parts pm and kn fall on the pedestal, and that the points m and 2 cast shadows on the horizontal plane. The point 7, in which the diagonals aD’ and 6D inter- sect, is the perspective of the projection of the vertex of the pyramid on the plane of the base of the pedestal. From r, draw rP to the vanishing point of projections, and the point v”, in which it intersects vR, is the per- spective of the shadow cast by the vertex of the pyramid on the horizontal plane. The line 62’ is the perspective of the shadow cast on the horizontal plane by the edge 52 of the pedestal, and aS is the indefinite perspective of the shadow cast by zs. Therefore, drawing through m, mR to the vanishing point of rays, determines m’, the shadow cast on the horizontal plane by the point m: the shadow z’m‘ is cast by ¢ m. The part ms, of the edge zs, will not cast a shadow on the horizontal plane, being itself in the shadow of the pyramid. If, however, we draw from s a line to the vanishing point of rays, the point in which it mtersects 2'S, limits the shadow which would fall on the horizontal plane if the edge were in the light. The line drawn through s’, the point so determined, and parallel to the ground line, is the indefinite perspective of the shadow | cast by the edge Xs. Through » draw »R, and we de- termine n’, the shadow cast on the horizontal plane by H 2 a #. . @ > ee, " oe i et, Pa ’ % ‘ z » we — » come ee 4 ; ~ t * r vs , . > ) a « We. * “2 « =e . .. * .% #; 116 TREATISE. ON a . the point n; and n'v” is the shadow cast by the edge of the pyramid. The part ns, of the edge of the pedestal, being in the. shadow of the pyramid, cannot cast a shadow on the horizontal plane ; ang the perspective of — .. . . the shadow cast by Xn begins at n’,and is paralleltothe - * ground line. * ' ‘The perspectives of the shadows of the other pyra- | ‘ mids are found by similar constructions. ‘The faces of ¥* the.pyramids and pedestals which are in the shade and ; : seen, are shaded in the drawing. + PROPOSITION IV. THEOREM. . If aright line be tangent to a curve in space, the perspective E;’. i . ‘of the right line will be 1 tangent to the perspective of the curve. 88. For let AFCG (PI. 15, Fig.1) bea curve, to which a right he is dra awn tangent at any point as F. and a visual Satiy through the right line, the visual — = plane will be tangent to es visual cone. ‘T’he perspec- _ tive plane will intersect the visual plane in a right line a and the visual cone in a curve, and the right line and | ; curve will be tangent to each other (Des. Geom. 84). , But the right line in which the perspective plane inter- » sects the visual plane is the perspective of the tangent ~ line,.and the curve in which the perspective plane inter- sects the visual cone is the perspective of the given curve AF CG: hence, when a right line and curve are tangent * ‘In space, their perspectives are also tangent. . If two curves are tangent to each other in space, their . perspectives are also tangent. For the two visual — a cones which determine their perspectives are tangent . y to each other, and therefore the curves in which they ~ Bes intersect the perspective plane are likewise tangest. - bt a wt * REE Ge ne SE SP Py Se é = , ” a >= : ° See a = = a 2 : * 7 a ‘ . ;: 7 > a > * ~ * _ * : : Yee : * *¢ »* - #2 | : ? . * _aenee —s SS SS Plate 14: See B re Ud ¥ E-Prud homie Je | $$! \LINEAR PERSPECTIVE. ’ MWh ee 2 * * PROBLEM V. To find the perspecige of a aie. , 89. Let B (PII arty: be the centre of the circle which is to be put active and RT the trace of the perspective plane; the perspective plane being per- pendicular to the plane of the circle AFCG. Althotgh the horizontal projection of the circle is made in front of the perspective plane, all the points of it are, in fact, as far behind it as they are now projected ? in front of the trace RT. : Let the point of sight be taken in a plane passing through the centre of the circle, and perpendicular to the perspective and horizontal planes. . Let S be the centre of the picture, and P and P x the vanishing points of diagonals. Through B draw, the diagonal BN, and ae a perpendicular to the | ground line. Oe If the circle were in front of the perspective plane, 7 x the diagonal BN would haye its vanishing point at P; ..’ but since it is behind it, ‘the vanishing point. of the ~ diagonal is at P’ (79). Hence NP’ is the perspective. of the diagonal: The perpendicular through B piérces . a the perspective plane at n, and has its vanishing point » at S; therefore the point 6 where Sn intersects the per- > me spective of the diagonal, is the perspective of the iat centre B. 5 a Through 6 draw the horizontal line «4 c} this line ys is the urate perspective of the diameter ABC. | © Through A and C draw tangent lines. These tangents are perpendicular to the ground line, and their perspec- tives pass through 8. . me pone a and c in which a » 4 * _ a > s co, * ~~ . a « * ” ‘© ” - a ? > : , s -— _ ¥ 2 ve I in : ee 138 . TREATISE ON they intersect the perspective of AC) are the perspectives of the points A and C. The lines Sa and Sc are tan- gent to the ellipse, which is the perspective of the circle AFCG. Let the perspectives of the points F and G be next found; they are / and g. fpr a If through the points F and G tangent lines be drawn to the circle AFCG, their perspectives will be tangent to the perspective of the circle (88). But since the tangents are parallel to the ground line, their perspec- tives will also be parallel to the ground line; hence gf is perpendicular to the tangents drawn through its ex- tremities f and g; it is, therefore, an axis of the ellipse, which is the perspective of the circle. Bisect gf atd. ‘Throughd draw Pde, and from e draw the diagonal eD. It is plain that Pde is the perspective of the diagonal De, and that dis the perspective of D. Through D draw KDH parallel to the ground line, and find the perspectives of the points K and H, which are & andh; kh is the: perspective of KH, and is the other axis of the ellipse. ‘The ellipse therefore can be described. ~ 90. When the point of sight is not in the plane pass- ing through the centre of the given circle and perpen- dicular to the perspective and horizontal planes, we are unable to find the axes of the ellipse by a direct construc- tion. We then find the perspectives of several points of the circumference of the circle, and describe the ellipse through them. In Pl. 15, Fig. 2, the perspective of the circle is found by points. The perspectives of the tangents at, the points a, d, 6, g,c and e are tangent to the perspective of the circle at the points a’, d’, 6, g’, ¢ and é. 91. Having given a circle in space, and the point of “ es LINEAR PERSPECTIVE. 119 sight, we may so place the perspective plane that the perspective of the circle shall be any.one of the conic sections. For, when the circle and point of sight are given, the visual cone circumscribing the circle is also given, and if the position ofthe perspective plane be undetermined, it may be so chosen as to intersect the cone in any one of the conic sections. When the perspective plane is parallel to the base of the visual cone, or when it cuts the cone in a sub-con- trary section, the curve of intersection is a circle. Me PROBLEM VI. To find the perspective of a cylinder, the perspective of the shadow cast by the upper circle on the interior surface, and the perspective of the shadow on the horizontal plane. re 92. Let the circle described in the horizontal plane with the centre C (PI. 15, Fig. 3), and radius CB, be the lower base of the cylinder; the centre C being ata dis- tance behind the perspective plane equaltoCC”. Let A’B’ be the projection on the perspective plane of the upper base of the cylinder, the plane of this base intersecting the perspective plane in the horizontal line A’B’. Let the point of sight be taken in the plane through the axis of the cylinder and perpendicular to the per- spective plane. Let S be the centre of the picture, and P’ and P the vanishing points of diagonals. Find now the perspective of the lower base of the cylinder as in Prob. 5. In finding the perspective of the upper base we have merely to regard the perpendiculars and diagonals already drawn, as the projections on the horizontal plane » 4 120 TREATISE ON of corresponding perpendiculars and diagonals drawn in the upper base of the cylinder. For example, the diagonal aBé, being considered in the upper base of the cylinder, would pierce the perspec- tive plane at 0’, its perspective would be a’P’, and its intersection with C’S determines a’, a point in the per- spective of the upper base; the diagonal Be determines the point c’. If through the point of sight two tangent planes were drawn to the cylinder, they would touch it in the two elements which pierce the horizontal plane at fand g. Having found g’ and/’, the perspectives of the points g and f, in the lower base of the cylinder, and g” and f” the perspectives of the corresponding points of the upper base, draw the lines g’g" and f’f"; these lines are the per- spectives of the elements which pierce the horizontal plane at g andf- The part of the cylinder convex towards the point of sight, and limited by these elements, is seen; the other part is not seen. Therefore, the semi-ellipses g of and g"c'f", which are seen, are made full, and the semi- ellipses g’af’ and g’a'f", which are not seen, are dotted. To find the Den Kosai of the shadow cast on the interior of the cylinder by the circumference of the upper base: Let R be the vanishing point of rays, and H the vanishing point of horizontal projections. If two tangent planes of rays be drawn to the cylin- der in space, their horizontal traces will be tangent to the base of the cylinder, and the elements of contact will be the elements of shade. But the horizontal traces of these planes will be parallel to the horizontal projec- tion of the rays of light; hence, their vanishing point is at H. The horizontal traces are also tangent to the am," LINEAR PERSPECTIVE. 121 base of the cylinder, therefore their perspectives will be tangent to the perspective of the base (88). Through H draw the tangents Hk and Hh; the points of contact &# and A are the perspectives of the two points in which the elements of shade pierce the hori- zontal plane. But since the elements of shade are ver- tical lines, kX’ and hh, drawn perpendicular to the ground line, are their perspectives, and #’ and fh’ are the perspectives of the points at which the shadow on the interior of the cylinder begins. if we suppose the cylinder in space to be intersected by a plane of rays parallel to its axis, the horizontal trace of the plane will be parallel to the horizontal pro- jection of the rays of light, and consequently, will have its vanishing point at H. Every plane so drawn will in- tersect the cylinder in two elements, and the one towards the source of light will cast a shadow on the other. hrough H draw any line, as Haz, to represent the perspective of the horizontal trace of a secant plane of rays. Through the points z and m, in which it itersects the perspective of the base, draw the elements 77’ and nn. Through? draw? R to the vanishing point of rays ; the point m in which it intersects nn’ is the perspective of a point of shadow on the interior of the cylinder. To find the shadow cast on any particular element, as ff’, draw from the vanishing point of horizontal pro- jections a line, as Hf’, through its foot, and through the upper extremity of the element passing through the other point in which Hf’ intersects the perspective of the base, let a line be drawn to the vanishing of rays; the point p, in which it intersects the element /"/", 1s the point of shadow required. To find the perspective of the shadow cast on the horizontal plane : 4? » & z ’ ~ : *% | = > 4 : * ** ee 3 rs . *. s :” ‘ ‘te . mm r : | oo ”~ , > « t 222 . TREATISE ON The traces of the tangent planes of rays are tangent to the shadow cast by the upper circle of the cylinder on the horizontal plane (27); therefore, their perspec- tives are tangent to the perspective of that shadow (88). But the rays of light passing through the upper extremi- ties of the elements of shade, intersect the traces of the tangent planes at the points of tangency (27); therefore, /” and h’, where the perspectives of the rays intersect the perspectives of the traces, are the perspec- tives of the points of tangency, and hh’, kk” are the per- spectives of the shadows cast by the elements of shade. That part of the upper base whose perspective is kan fh’, casts the curve of shadow on the horizontal *- of the cylinder. | Any line, as Hz, drawn through H, may be considered as the perspective of the horizontal trace of a plane of rays; the point n”, in which the perspective of the ray » through n’ intersects Ha, is the perspective of a point of shadow on the horizontal plane. A line drawn through R, tangent to the perspective of the upper» base, will also be tangent to the curve ey aa + so leigh. te Me “s The part. of the surface of the cylinder which is in al , "the shade, and seen, is shaded in the drawing. The perspective Of the shadow on the horizontal plane is _ also shaded. : aa * . % . a t a 4 % +. r Bye & . ¥ . , =a oe bal - r" = s > al plane ; the remaining part casts a shadow on the interior Lee a DS ne ~ | | fink ftg fn —>7CL b | { f } e reir LE Privd homme ° oe * ; . + aA ; ” co a * : ‘ ~ 418 ‘ [ ¥ f $)% > >i P he LINEAR PERSPECTIVE. @ 123 e PROBLEM VIL » oo” It ts required to find the perspective of the frustum of an inverted cone ; also the perspective of the shadow on the ante- rior of the frustum, and the perspective of the shadow ow the horizontal plane. * 93. Let the circle described with the centre A and tadius AB (Pl. 16), be the horizontal projection of the upper base of the frustum, and B’CG’ the intersection of its plane with the perspective plane. Let the circle described with the centre A and radius AH, be the lower base of the frustum, and HT its vertical projection. Lhe horizontal projection of the vertex of the cone is at A, and its vertical projection, which is at L, is found » by joining B’ and H’-and producing the line until it in- tersects CAE drawn perpendicular to the ground dine, Let S be the centre of the Shes and D the vanish- ing point of diagonals. ‘ ; ‘Through (E, C), agpomt of the ee base, draw a diagonal and perpendicular. The diagonalpierces the perspective plane at G’, and the perpendicular pierces it at C. The diagonal’ has its vanishing pointrat*D, and the perpendiculaz its vanishing. point at S. e is the perspective of the point (E,C). By similar con- structions we find ‘, the perspective of. (H,C), ¢ the perspective of the point whose horizontal projec- - tion is c, and 6 the perspective of (B,B'). The perspec- tive of any point may be found by determining the per- spectives of the diagonal and perpendicular passing through it. Before describing the ellipse bec’, it will be. well to remember that ¢S and B'S, being the perspectives of Therefore, - ie es 2 a "s eg bans Phe? i ¥-¢€ * a : > ” : 7 ~ lied si "ae we a ade | e * * * -- r he ? % + & e« « . 7 ar : t oo - 4 ee: he ¢* «ty ne &* » i ” ® # 124 - TREATISE ON - - tangent lines to the upper base of the frustum, are tan- gent to the ellipse deceé : and also, that the tangent lines In space drawn through the points (E,C) and (E’,C) are parallel to the perspective plane; hence, their perspec- tives are the herizontal lines drawn through e and é, tangent to the ellipse deceé. The perspective of the lower base of the frustum is determined by constructions entirely similar. Let the perspective of the vertex of the cone be next found. ‘i The ornate through the vertex of the cone pierces the perspective plane at L, the vertical projec- tion of the vertex; and the diagonal through the vertex pierces the perspective plane at N’; therefore the point L’, where LS intersects N’D, is the perspective of the - vertex. if through the point of sight we suppose two planes to be drawn tangent to the frustum of the cone, the traces of these planes, on the perspective plane, will pass through the perspective of the vertex, and will limit the perspective of the cones hence, the perspec- tives of the upper and lower circle will be tangent to these traces. Let these tangents be then drawn through | the point'L’. The point of sight being above the upper base of the - frustum, the whole of that circle will be seen, and there- fore its perspective is made full. A part only of the ‘ lower circle is seen, the perspective of this part is made full, and is limited by the tangent lmes drawn, through L’. To find the shadow which the upper ce casts on the interior of the frustum : Let R be the vanishing point of rays, and P the vanishing point of horizontal projections. _ LINEAR PERSPECTIVE. 125 Since the ray of light through the vertex of the cone is a line of every plane of rays which intersects the cone in right-lined elements, the point in which this -ray pierces the horizontal plane is common to the horizon- tal traces of all such secant planes; hence, the perspec- tive of this point is common to the. perspectives of all the traces. But the perspective of this point is found in the per- spective of the ray through the vertex of the cone, and in the perspective of the horizontal projection of this ray (82). Through R draw RU’; this line is the indefi- nite perspective of the ray. Through a’, the. perspec- tive of A, draw Pa’; this is the indefinite perspective of the horizontal projection of the ray; the point K, in -which they intersect, 1s the perspective of the point in which the ray through the vertex of the cone pierces . the horizontal plane. Through K, draw Kf and Kd tangent to the perspec- tive of the lower base of the frustum. These tangents are the perspectives of the horizontal traces of the two planes of rays which are tangent to the frustum in space. Hence, L/h and Lidg are the perspectives of the elements _of shade, and g and / the perspectives of the points at which the shadow on the interior of the frustum begins. To find points of this shadow, draw any line through K, as Kpqs', which will be the perspective of the hori- zontal trace of a secant plane of-rays. Through the points p and gq draw the elements L’pk and L’gs. From k draw /R to the vanishing point of rays; the point # in which it intersects L’s, is the perspective of a point . of shadow. The shadow on any particular element is found by drawing a line from K through its foot; and then drawing the perspective of aray through the upper a” . ts 3 ‘ ey TREATISE. on Ly od mata ee + “He ofthe chy ti ht, as 1 efor i va ra ats The part, of F the curve. whose Pe is sabhdh bY Ms , casts. a shadow. on the interior of the frustum ; and the .: at part. whose perspective. jae seg casts a hago g on aA how alplave, <¢.!* osha - Ttis is now oe ind tg shadow on the ho horizontal 3 arte ate | ae ; | ‘A ) A Bice Pod ‘point | whose pbaspeatine is h, casts a tate: in x A ca thé trace ‘of the tangent lane of rays; but the per- Ae. .. spective of the shadow i 18 so in the perspective of the a me ray through h; hence it is at h’.. Fora similar reason. bP ae Soe ey g’ is the perspective of a point of the shadow, on ‘the ae horizontal plane.» * ae ae Ye x os. ‘The line Kpgs', as’ has already péeys srermirgads is the “0 ° perspective of the trace of a secant plane of rays. If “4 3 _ through s, the upper extremity of the “element opposite 4s it; Abe source of hight,» sR be drawn'to the vanishing point SN ie «ya * of rays, , the point s’, in which it meets Kypgs'’, is another ae . +» point of the pempecg of the shadow. By similar con- # : structions any | numb toric thes found. “3 4 s The perspective ot the shadow is tangent to the lines es : Kh and Kg’, at the points h’ and ge “Aline drawn through ab ws eR: fangent to perspective of the upper | base of the oh or frustamn vill also be tangent to the curve g's. ae i : - The part of the exterior surface of the frustum which od iat ¥ _ isin the shade, and” seen, is'shade in the drawing. West ate £8 { also shade that part-of the interior” of the frustum on. re which the shadow falls, ‘and which is seen. Of the ~ i fa * shadow which falls on the horizontal plane, all that is”. io he , seenisshaded. = at * al ee neg \e ¥ a : tay de a” *- + 4 A ~ 2 Pu A ' ‘at & ee _ a mee na Cee ae: 2 ¥ “a 7) a ad te Ge FS ty » 7] are Oa te nr Ws Pye We ~ 4 : 1 hey al Fae 2 .) a ; * * @ 4% _ *;, * Ya * : Rs Mel, os det. $ gs ’ aoe « ot * we REET Se a Soey a Pa LAr . es - * . of ? " * ra “f x ahh Ng’ > . ie 4 2 2 . * ™ ro i i* ae tat oe. > - a” “ o Pa ees i } q cas “ ae ee a . « ok a : > > > * s , “ 7 * a P. B . wae La i. ae * : ; ° oe — 2 A ¥ as a ae 2 oa _ - - >. : ’ & ees >. =— 2 ¥ > dp “> > s * : > 2 ee ? «}. Ss 5 z < { = & = . yo . — - <, * ~a + : . + * ® rf oa : ss . 1% * *. . , : +2 = = P £ a 4 pes r ” + + 52 : -* : P 3 4%, al #, = ." -* . ” > a oe a eee - 5 ail E | 9 M ro ‘ . “ i] + 1 Pred homme Se. | ES DEER | a m. LINEAR PERSPECTIVE. 127 ° £ . : , PROBLEM VIII. To find the perspective of aniche, and the per eee ms the shadow cast on tts interior surface. 94. Let the perspective plane be taken through the front face of the niche. Let AB (PI. 17) be the ground line, and the semicircle 6ar the horizontal projection of the niche. * _ Draw 4c and re, in the perspective plane, perpen dicular to the ground line, and make them equal to the height of the cylindrical part of the niche; and on ce describe the semicircle cle. ‘These are the lines of the _ niche which are in the perspective plane. _ The perspective of the lower base of the niche is - rahb, the segment of an ellipse: the perspective of the semicircle of contact of the cylindrical and spherical parts, is the elliptical segment extpe; both of these curves are found by methods already explained. The lines 6S “and 7S, are the perspectives of the two lines drawn tan- gent to the base of the niche at the points 6 and 7; hence they are tangent to the elliptical segment ra‘hd. (88.) To find the shadow on the interior of the niche. The first line which casts a shadow on the interior of the niche, is the element c 6. If through this element a plane of rays be passed, it will intersect the surface of the niche, in space, in a second element on which the shadow will fall. But, since the plane of rays is vertical, its horizontal trace is parallel to the horizontal projection of the rays of light; hence its vanishing point is at H, the vanishing point of horizontal projections of rays, and 6H is its perspective.’ 125 TREATISE ON Through d, the point in which 6H intersects ra’b, draw dd'paralleltodc; dd’ isthe indefinite perspective of the element that receives the shadow. ‘Through ce, the upper extremity of the element casting the shadow, draw cR to the vanishing point of rays; the point d’, in whichat. intersects dd’, is the perspective of the shadow cast by the point c, and dd’ is the shadow cast on the cylindrical part of the niche, by the element bc. The line éd is the perspective of the shadow which a part of the same element casts on the horizontal plane. ‘Through H draw any line, as Hf, near to Hd. This line may be regarded as the perspective of the horizontal trace of a vertical plane of rays; andAf'is the per- spective of the element in which it intersects the cylin- drical part of the niche. The plane intersects the per- spective plane in the line ff’; therefore the point A’, in which /’R intersects hi’, is the perspective of the shadow cast by the point (7, f’) onthe cylindrical part of the niche To find the perspective of the shadow which falls on the spherical part of the niche: If we suppose the quadrant of the sphere, which forms the spherical part of the niche, to be intersected by a plane parallel to the front face of the niche, the section will be a semicircle whose diameter will be a chord of the semicircle of contact of the cylindrical and spherical parts of the niche. The front circle of the niche will cast a shadow on this plane equal to itself (28); and the point where this shadow intersects the circle cut from the sphere is a point of the required shadow in space. If then, we find the perspectives of these two circles, the point in which they intersect will be the perspective of a point of the curve of shadow. Since both the circles are parallel to the perspective plane, their perspectives will be circles (91). Draw any od LINEAR PERSPECTIVE. 4 129 line, as np, to represent the perspective of the diameter of the semicircle cut from the spherical part of the niche by the parallel plane, and on it describe the semicircle nkp. ‘he perspective of the shadow cast on the plane __ of this circle by the centre g, is found in gR, and also in the perspective of the projection of the ray through g on that plane. But, since the plane is parallel to the perspective plane, the projection of the ray upon it is parallel to SR. But g’ is the perspective of one point of its projection; hence g‘g”, drawn parallel to SR, is the perspective of the projection of the ray on the parallel plane, and g” is the perspective of the shadow cast by the centre g. The shadow cast by cg is parallel to itself and to the perspective plane; hence ¢”g, drawn parallel to cg, 1s the perspective of the indefinite shadow cast by eg on the parallel plane. But this shadow is limited by the ray cR; hence g’g is the perspective of the radius of the circle of shadow. With yg” as a centre, and the radius gg, describe the are gk ; the pomt £, in which it intersects xkp, is the perspective of a point of the curve | of shadow. If through the centre g, gt be drawn in the perspec- tive plane, and perpendicular to SR, it will be a line of the plane of the circle of shadow; therefore the point 2, in which it meets the circumference c / e, is a point of the perspective of the curve of shadow. We can easily find the perspective of the point at which the shadow passes from the spherical to the cylindrical part of the niche. For, if through the point in space of which & is the perspective, a line be sup- posed drawn parallel to gz, it will be contained in the plane of shadow, and will pierce the upper base of the cylinder in the trace of the plane of shadow. But ks, drawn parallel to gz, is ng perspective of this parallel, 130 TREATISE ON and s is the perspective of the point in which it pierces © the upper base of the cylinder. Hence, gst is the per- spective of the trace of the plane of shadow, and ¢ the perspective of the point at which the shadow passes — from the spherical to the cylindrical surface. 3 There are other constructions for finding the shadow on the spherical part of the niche. Revolve the plane of rays. passing through the point of sight and perpendicular to the perspective plane, about SR, until it coincides with the perspective plane. The point of sight falls in SD”, drawn perpen- dicular to SR, and at a distance from S equal to SD (76). The ray of light through the point of sight takes the position RD”. Intersect the spherical part of the niche by a plane of rays perpendicular to the vertical plane. Let Im’z, drawn perpendicular to gz, or parallel to SR, be the trace of such a plane. Revolve this plane until it coin- cides with the perspective plane. The semicircle cut out of the sphere, when revolved, is the semicircle lmz, and the ray through / takes the position /m, parallel to RD”. Let the counter revolution be now made, and draw mS, which is the perspective of mm’, and JR, which is the perspective of the ray; the point m’, in which they intersect, is the perspective of a point of the curve of shadow. Other points may be found by similar con- _ structions. (ithe » ‘ PY a fe * . a &- <* = Se a ea Pie i Se ee ey ED eae lac all ates . ihc RSA ig FE Mua Pom ST, _B rs LINEAR PERSPECTIVE. P 131 PROBLEM IX. To jind the perspective of a sphere, the perspective of tts shade, and the perspective 8 tts shadow on a horizontal plane. 95. If we suppose the sphere to be circumscribed by a visual cone, the curve in which the perspective plane intersects the cone will be the perspective of the sphere. The point of sight and the place of the sphere being given, the perspective plane may be taken in such a position as to intersect the visual cone in any of the conic sections. Although it is easy to understand why the perspective of a sphere may be.an ellipse, a parabola, or an hyper- bola; yet neither of these curves seems to be a proper representation of a body perfectly round. It is for this reason that the perspective of a sphere is generally drawn on a plane perpendicular to the line joining the point of sight and its centre; when the perspective plane has this position, it intersects the visual cone in a circle. Let AB (PI. 18), be the ground line, S the centre of the picture, D the vanishing point of diagonals, R the vanishing point of rays, and H the vanishing point of horizontal projections of rays. Suppose the centre of the sphere to be at a distance behind the perspective plane, equal to its radius. Let 5S be the projection of its centre, and the circle described with the radius SC the projection of the sphere on the perspective plane. If through the point of sight we suppose a plane to be passed perpendicular to the horizontal and perspec- tive planes, LSa' will be its trace on the perspective 12 132 TREATISE ON plane. This plane will intersect the sphere in a great circle, and the visual cone in two elements which will be tangent to it. Let this plane be revolved about its trace LSa’, until it coincides with the perspective plane. The point of sight will fall at D (76), and the centre of the circle cut out of the sphere at C. With Cas a centre, and radius CS, describe the semicircle, as in the figure; and then draw from D the tangent line Da‘a, which will be, in its revolved position, an element of the visualcone. ‘This element pierces the perspective plane at a’; hence, the circle described with the radius Sa’, is the perspective of the sphere. It is now required to find the perspective of the curve of shade. It will first be necessary to find the vanishing line of the plane of the circle of shade (80). Since the plane of shade is perpendicular to the direc- tion of the light (34), the required line is the trace of a plane passing through the point of sight and perpen- dicular to the rays. Through the point of sight let a plane of rays be passed perpendicular to the perspective plane; RSE is its trace. Let this plane be revolved about RE, to coin- cide with the perspective plane. ‘The point of sight falls at F, in a perpendicular to RS, and at a distance from 5 equal to SD. Since the ray through the point of sight pierces the perspective plane at R, it will, after the revolution, take the position RF. If then FE be drawn perpendicular to RF, it will be a line of the plane through the point of sight, and perpendicular to the direction of the ray, and E will be a point of its trace. But RE is the projection, on the perspective plane, of the ray passing through the point of sight (85): and since the plane through the LINEAR PERSPECTIVE. 133 point of sight is perpendicular to this ray, the trace will be perpendicular to its projection. Therefore, GEI, drawn perpendicular to RE, is the vanishing line of the plane of shade. If we suppose two planes of rays to be drawn tangent to the visual cone which determines the perspective of the sphere, they will also be tangent to the sphere in space, and the points of contact will be points of the curve of shade. ‘The perspective of these points of contact will be found in the perspective of thé sphere, and in the traces of the tangent planes. Since the tangent planes are planes of rays, they will contain the ray through the point of sight; hence their traces will pass through the point R. But these traces must also be tangent to the perspective of the sphere; hence Ré and Re, drawn tangent to the perspective of the sphere, are the traces of the tangent planes of rays, and 6 and e are two points of the perspective of the circle of shade. And since S is the perspective of the centre of the sphere, the lines 6Sd and cSe are the indefinite perspec- tives of two diameters of the circle of shade. Let us suppose, for a moment, the circle of shade to ‘be represented by the circle dcebg, Fig. n, and let dd and ce be the diameters already referred to. The traces of the tangent planes of rays are the tangents nd and ne; and since the tangent planes of rays are perpendicular to the plane of shade, they will intersect in a ray of light perpendicular to the plane of the circle poge at n. If through this ray and the centre a, a plane be passed, its trace ng will bisect the angle cab and be parallel to the lines 6 e and cd, joining the corresponding extremities of the diameters dd and ce. Hence, the plane of rays whose trace is RE, intersects the plane of the circle of shade in a line making equal angles with the diameters 134 TREATISE ON whose perspectives are éd and ce. But this line is parallel to the chords joining the corresponding ex- tremities of these diameters, and also to the line FE, in which the plane of rays RE intersects the paralle! plane through the point of sight. Therefore, E is the common vanishing point of the trace on the plane of shade, and of the parallel chords joing the corresponding ex- tremities of the diameters. Draw 4E and cK. The points e andd in which they intersect cSe and 65d, are the perspectives of the extremities of these diameters, and consequently, of two more points of the curve of shade. To find other points of the curve of shade: Let the plane passing through the point of sight and parallel to the plane of shade be revolved about its trace GI, to coincide with the perspective plane. The point of sight falls at F’, a distance from E equal to EF. Through this point let any two lines, as F’G and F', be drawn at right angles to each other, and note the points G and lin which they meet the trace GI. Let us now consider the plane to be revolved back to its position in space. If through the two extremities of any diameter of the circle of shade, two lines be drawn parallel to F’G and F'l, they will be contained in the plane of shade, and their point of intersection on the surface of the sphere will be a point of the circle of shade. But these two lines will have G and I for their vanishing points (71). If therefore, through the points d and 6, we draw the lines dG and Ol, they will be the perspectives of two lines drawn through the extremities of a diameter at right angles to each other, and the point g,in which they intersect, is the perspective of a point of the circle of shade. The lines through the points ¢ and e determme the point f. LINEAR PERSPECTIVE 135 if through /, we draw Sh, it will be the indefinite per- spective of a diameter of the circle of shade. Through e draw cl. This line is the perspective of the chord parallel to the chord whose perspective is fe; therefore, his the perspective of the other extremity of the diam- eter, and consequently, the perspective of a point of the circle of shade. Having found a sufficient number of points of the curve of shade, let it be described. Only the part cfgé, which is in front of the circle of contact of the visual cone and sphere, is seen. It is now required to find the perspective of the shadow cast on the horizontal plane. It will first be necessary to find the perspective of the horizontal trace of the plane of shade. Since this trace is a horizontal line, its vanishing point is in the line ‘TSH (74), and since it is a line of the plane of shade, its vanishing point is in the line GI; hence it is at ‘T’. It is necessary in the next place to find the perspective of the horizontal projection of the centre of the sphere. To do this, lay off from L to P the radius SC of the sphere; P is the point at which the diagonal through the horizontal projection of the centre of the sphere pierces the perspective plane, and the perpendicular through the same point pierces it at L; hence zis the perspective of the horizontal projection of the centre of the sphere. We will here premise, in order to illustrate what fol- lows, that when a line intersects the perpendicular from the point of sight to the perspective plane, its perspec- tive and its projection on the perspective plane are the same line: for, the visual plane which determines its perspective is then perpendicular to the perspective plane. 136 TREATISE ON A second point in the perspective of the horizonta trace of the plane of shade is found, by finding the per- spective of the point in which any diameter of the circle of shade pierces the horizontal plane. ‘To simplify the construction, we will take that diameter which is parallel to the perspective plane. Since the diameter is perpendicular to the ray through the centre of the sphere, and since the perspectives of the two lines are the same as their projections, it fol- lows, that their perspectives will be at right angles to each other (Des. Geom. 51). But SR is the perspective of the ray; hence, SF drawn perpendicular to SR, is the indefinite perspective of the diameter. The projection of this diameter on the horizontal plane, passes through the horizontal projection of the centre of the sphere, and is parallel to the ground line; hence, its perspective passes through z and is parallel to AB (72). Therefore, k is the perspective of the point in which the diameter pierces the horizontal plane, and consequently, a point in the perspective of the horizontal trace of the plane of shade; and Té is the perspective of that trace. The line 7H is the perspective of the horizontal pro- jection of the ray through the centre of the sphere, and SR is the perspective of the ray; hence p, their point of intersection, is the perspective of the shadow cast on the horizontal plane by the centre of the sphere. The shadow cast on the horizontal plane by any dia- meter of the circle of shade, will pass through the point in which the diameter pierces the horizontal plane, and also through the point of which p is the perspective. Therefore, produce the diameter 6d till it meets Tg, the perspective of the horizontal trace of the plane of shade, and from r draw rd‘pb’; this line is the perspective of the indefinite shadow cast by the diameter on the LINEAR PERSPECTIVE. 137 horizontal plane. Through the extremities 6 and d of the diameter, draw lines to R; the points 4’ and d’ are points of the perspective of the shadow. The diameter ec being produced to gq, gives the points ¢ and¢. The perspective of the shadow will be tangent to the lines Ré and Re at the points 6’ and ¢. 96. We can find, by a direct construction, the axes of the ellipse, which is the perspective of the circle of shade. That the figure may not become too complicated, we will make the construction in Fig. m, in which the pro- jection and perspective of the sphere are both repre- sented, and in which R is the vanishing point of rays. If through the axis of a scalene cone, having a cir- cular base, a plane be passed perpendicular to the base, it wil divide the cone into two symmetrical parts. If then, a plane be passed perpendicular to the plane throvgh the axis, it will intersect the cone m a curve whose axis is the intersection of the two planes. The second axis of the curve 1s the line drawn through the middle point of the first, and perpendicular to 1t. The visual cone, which is formed by drawing visual rays to all the points of the circle of shade, is a scalene cone with a circular base. ‘The plane of rays, whose trace is SR, passes through the axis, is perpendicu- lar to the plane of shade, and also to the perspective plane. Hence, the line RS, in which it intersects the perspective plane, contains an axis of the ellipse in which the perspective plane intersects the visual cone, or an axis of the ellipse which is the perspective of the circle of shade. ‘The plane, whose trace is RS, also intersects the plane of the circle of shade, in a diameter perpendicular to the direction ofthe light. Let this plane be revolved about RS to coincide with the perspective | ~ 138 TREATISE ON plane. The point of sight falls at F; RF is the revolved position of the ray, the centre of the sphere falls at s, and psq, drawn perpendicular to RE’, is the diameter of the circle of shade. Through p and q draw Fp and Fq; Pq is evidently the perspective of pg, and is an axis of the ellipse. Through a, the middle point of p‘g, draw Faa’. The chord of the circle of shade, whose perspec- _ tive is the other axis of the ellipse, passes through the point a’, in its true position in space, and is perpendicular ‘to pag. But since the circle of shade is a great circle of the sphere, the length of the chord is equal to cab. Let ca'b be revolved about the middle point @’, till it has the position cab’, parallel to fR. Through F draw Fe’ and I'd’; the points f and d, in which they intersect fR, limit the other axis of the ellipse. But this axis must _pass through a, and be perpendicular to p’q’. There- fore, drawing the perpendicular fd’, and making ad’ and af’ equal to af, or ad, we have the second axis of the curve. PROBLEM X. Lo find the perspective of the groined arch and the perspec- tive of wuts shadows. 97. ‘The arch is supported by four pillars, capped by cornices, and standing on pedestals; the pedestals are placed in the four angles of a square. The arch itself is formed by portions of two equal cylinders. Fig. n (PI. 19) represents the two projections of the pedestals, the pillars, the cornices, and the arch. On the line (ab, a’b’), joining the points in which the inner front edges of the pillars, produced, pierce the upper plane of the cornices, let the semicircle a'sb’ be described, its plane Gi BE Prudhom. 9 : ” Re SS ' - : ee ¢ . . ~ : An tne lacs ar arcs SON ge: tp cette Ps, Py aM ied wean be i : , by) Way ‘ id, * Xi} ‘yn LINEAR PERSPECTIVE. 139 coinciding with that of the front faces of the pillars; and on de let there be described an equal and parallel semicircle. [If now aright line be moved from the position (ad, ’), parallel to itself, and touching the two semicircles, it will generate a cylinder whose axis is (cm, ¢’), and whose elements are perpendicular to the front face of the arch. Onthe two lines gf and rv, joining the points in which the edges of the pillars pierce the upper plane of the cornices, let two equal and parallel semicircles be described, and let a right line be moved along them, parallel to itself, generating a semi-cylinder whose axis zx is parallel to the front face of the arch, or perpendi- cular to its side faces. Ml The two cylinders which have been generated are equal; their axes are at right angles to each other, and the surfaces of the cylinders intersect in two equal semi- ellipses, called the groins of the arch. These groins spring from the upper plane of the cor- nices, the one from n to gq, the other from/to ¢. They intersect each other in a point of which o is. the hori- zontal projection, and which is at a distance above the plane of the cornices equal to the radius ¢’s. . In the construction of the arch, only a part of each cylinder is used. The parts alond and dgote, are formed by the cylinder whose axisis em; andthe parts rlogv and gnotf, by the cylinder whose axis is zx; the other parts of both cylinders are supposed to be removed. If we suppose the outer planes of the pillars to be produced above the upper plane of the cornices, they will form vertical faces of a prism, whose horizontal sections are squares. We shall call this prism the solid part of the arch. ; Let the perspective plane be taken to coincide with {40 TREATISE ON the front faces of the pedestals and cornices, and let AB be the ground line. Let S be the centre of the picture, D’ and D the vanishing points of diagonals, R the vanishing point of rays, and H the vanishing point of horizontal projections. From A and B lay off dis- tances equal to a side of the squares which form the bases of the pedestals, and on them describe rectangles equal to the front faces of the pedestals. ‘Then, let the perspectives of the pedestals be found as in Prob. 2. From a lay off ab equal to the distance which the pedestal projects beyond the pillar. From a draw aD to the vanishing point of diagonals, and from 6 draw 6S to the vanishing pomt of perpendiculars; d is the per- spective of the point in which an edge of the pillar pierces the pedestal: and the horizontal line through d is the indefinite perspective of the line in which the front face of the pillar intersects the pedestal. From the other extremity of the line aé lay off a dis- tance equal to ab, and draw a Ime to S; the point e, in which it intersects de, is the perspective of the point in which a second edge of the pillar pierces the pedestal, and the point in which it intersects the diagonal aD, is the perspective of the point in which a third edge pierces the pedestal. ‘Through this latter point draw a parallel to de; the point in which it intersects 45 is the perspec- tive of the point in which the fourth edge of the pillar pierces the pedestal ‘The four vertical lines drawn through these points are the indefinite perspectives of the vertical edges of the pillar. The front and inner side faces only are seen. From a draw a vertical line in the perspective plane, and on it lay off ac equal to the distance between the upper plane of the pedestals and the lower plane of the cornices, and through ¢ draw an indefinite horizontal LINEAR PERSPECTIVE. 14] line cc’. Since the front faces of the cornices are in the perspective plane, we can lay off the thickness of the cor- nices cc’, and their width cf, equal to the width of the pedestals. Project the point 4 into the lower plane of the cornices at 6’,and through J draw 8S. The point in which this line intersects the edge of the column through 1,1s the perspective of the point in which that edge pierces the lower plane of the cornice; and the hori- zontal line through this point is the perspective of the line in which the front face of the column intersects the lower plane of the cornice. From the point in which this horizontal line intersects the edge of the column through e, draw a line to S; this line is the perspective of the intersection of the inner side face of the column with the cornice. The line drawn from ¢ to S is the indefinite perspective of the edge of the cornice which is perpendicular to the perspective plane at c; the part which is seen is limited by the edge of the pillar. From f, draw fS, and from c, a diagonal cD; their point of intersection /’, is the perspective of the angle of the cornice, diagonally opposite to c. Through f draw a vertical line, and from the upper extremity of the perpendicular through f, draw aline toS; their point of intersection is the perspective of an angular point of the cornice. ‘The horizontal line through /’ is the perspec- tive of the edge parallel to cf’ We have now found the perspective of one pillar and one cornice. ‘The perspec- tives of the others are found by similar constructions. We will next find the perspectives of the front and back circles of the arch. Find ¢ and e’, the perspectives of the points in which the front and inner edges of the pillars pierce the upper planes of the cornices; draw ¢e’, and on it describe a semicircle, making that part full which is above the 142 TREATISE ON uppet lines of the cornices; this circle is the perspective of the front circle of the arch. Find the perspectives of the points in which the edges of the pillars that are projected at d and e (Fig.n), pierce the upper plane of the cornices ; the semicircle described on the line joining these points, is the perspective of the back circle. We next find the perspectives of the groins, and the perspectives of the side circles. First, find the perspectives of the points », gq, / and ¢ (Fig. x), in the upper plane of the cornices, from which the groins spring; and also the perspectives of the points g, f,r and v, in the same plane, from which the side circles spring. Let us now suppose the solid part of the arch to be intersected by a horizontal plane, and let hi (Fig. ), be its trace on the front face of the arch. ‘This plane will intersect the solid part of the arch ina square. It will intersect the cylinder whose axis is cm, in two elements perpendicular to the front face of the arch at the points kand p. If through the points in which the horizontal projection of either element intersects the lines ng and /t, limes be drawn parallel to the ground line, they will be the horizontal projections of the elements in which the secant plane intersects the other cylinder. The points p, p,# and &”, are in the groins, and the square k'p'p’k’ has the same diagonals with the square in which the secant plane intersects the solid part of the arch. Draw any horizontal line, as hz, for the perspective of the trace of such a secant plane. From Ah draw AS and hD; from 7? draw 7S and7zD’; and from the points & and p, draw kS and pS. The points # and p” are the per- spectives of the points & and p” (Fig. n), of the groin LINEAR PERSPECTIVE. 143 /t; and the points p’ and &”, the perspectives of the points p’ and #” of the groim nq. The line fp’ is the perspective of the element /’p’ (Fig. x), and therefore the points 4 and 7’, in which it in- tersects AS and 7S, are the perspectives of points /’ and # (Fig. 7), of the side circles. If we draw a line through the points k” and p’, the points h” and 2’, in which it in- tersects AS and 2S, are the perspectives of the two points h’ and 2” (Fig. n), of the side circles. : If we use a second secant plane, we determine two other points in the perspective of each groim, and two points in the perspective of each of the side circles. If through the point of sight a plane be passed tan gent to the cylinder whose elements are parallel to the perspective plane, it is evident that the perspective of the element of contact will pass through the highest point of each of the curves. But the perspective of this element is a horizontal line (72); hence it will be tangent to each of the curves. To find this element, and its perspective, suppose a vertical plane to be passed through the point of sight, and perpendicular to the perspective plane. This plane, whose trace is R’g, will intersect the cylinder in a circle whose centre is in the upper plane of the cornices, and equally distant from the front and back faces of the arch. Let this plane be revolved about R’g, to coincide with the perspective plane. The point of sight falls at D, and the centre of the circle at c’. With c’ asa centre, and c’a’, equal to a line corresponding to ca (Fig. 2), describe the arc of a circle, and from D draw a tangent to this arc. This tangent is a line of the tan- gent plane, and the point in which it meets Sg, produced, is the point at which it pierces the perspective plane. The horizontal line drawn through this point is the perspec- 144 TREATISE ON tive of the element to which the curves are tangent. Let the curves be now described. The points at which the perspectives of the groins intersect the perspectives of the side circles, will limit the parts of the side circles which are seen. t is now required to find the ‘perspective of the shadows cast by the different parts of the arch. The front circle of the arch casts a shadow on that part of the arch formed by the cylinder whose elements are perpendicular to the perspective plane. In finding the perspective of this shadow, we have merely to find the perspective of the shadow cast by the base of a cylinder on the interior surface. Throuhg g, draw gl perpendicular to Sik; J 1s the per- spective of the pomt at which the shadow on the interior of the cylinder begins; for the tangent line at /1is parallel to SR, and is the perspective of the trace of a plane of rays tangent to the cylinder. ‘To find other points of the shadow, draw any line, as ma, parallel to the tangent ati. This line is the perspective of the trace of a plane of rays on the front face of the arch. This plane inter- sects the cylinder in two elements; one casts, and the other receives the shadow. ‘The line nS is the perspec- tive of the element which receives the shadow, and m’, where it is intersected by mR, is the point of shadow. Let the curve of shadow from / be then described. The rays of light may have such a direction, that the shadow of the front circle will intersect the groin k‘p” When this occurs, one curve of shadow will pass from the point of intersection along the cylinder whose elements are parallel to the perspective plane, and will pass off on the side circle 2”. Another curve will pass from the point of intersection, along the cylinder whose LINEAR PERSPECTIVE. 145 elements are perpendicular to the perspective plane, and will pass off on the back circle of the arch. The side circle of the arch, on the left, casts a shadow on the interior of the arch which falls on the cylinder whose elements are parallel to the perspective plane. If we suppose a tangent plane of rays to be drawn to this cylinder, the point in which the element of contact meets the end circle towards the source of light, will be the pomt at which the shadow on the interior of the cylinder begins. But this tangent plane will be perpen- dicular to the side face of the arch; hence, its trace will be parallel to the projections of rays on the side planes. If we suppose a plane to be passed through R parallel to the side planes of the arch, and the ray through the point of sight to be projected on this plane, the projection will be parallel to the projection of rays on side planes, and consequently, to the trace of the tangent plane. [f a line be drawn through the point of sight parallel to the projection of the ray, it will pierce the perspective plane at R’, since RR’ 1s the trace, on the perspective plane, of the plane which projects the ray on the plane through R; hence R’ is the vanishing point of the projections of rays on the side planes. If, therefore, through R’ we draw Rq, tangent to the curve h’gh’, it will be the perspective of the trace of the tangent plane of rays, and gis the perspective of the point at which the shadow on the interior of the arch begins. if through R’ we draw a secant line, we may regard it as the perspective of the trace of a plane of rays, paral- lel to the tangent plane, and intersecting the cylinder in two elements. If through the lower point, in which this line cuts the curve A’gh’, we draw a horizontal line, it will be the per- spective of the element which receives the shadow; and K 146 TREATISE ON drawing through the upper point a line to the vanishing point of rays, determines o, the perspective of a point of the curve of shadow. Let the curve of shadow be then drawn. We are next to find the shadows of the cornices on the front faces of the pillars. Through ¢ draw a line to S, and another to R. The point where the line to S meets the line in which the front face of the pillar intersects the lower plane of the cor- nice, is the perspective of one point in the projection of the ray through ¢, on the front face of the pillar, and the line drawn through this point, parallel to SR, is the per- spective of the projection: the point in which this perspective intersects cR is the perspective of one point of the line of shadow: but the line of shadow is hori- zontal both in space and in perspective. To find the shadow cast by the column. The edge er casts a line of shadow on the pedestal and on the hori- zontal plane, both of which are parallel to the horizontal! projection of rays. Hence, their perspectives are easily found. At s the shadow falls on the inner side face of the back pedestal. It passes up the face in a vertical line; reaching the upper face it becomes parallel to the horizontal projections of rays, and when it reaches the face of the pillar, it ascends in a vertical line along the face. If through,, the highest point of the edge of the pillar which is in the light, a line be drawn to R, it will limit the shadow cast by the edge of the pillar. From the point in which this line intersects the vertical line before drawn, on the face of the pillar, the shadow will be cast by the lower line of the cornice. This shadow will be parallel to the projections of rays on the side planes, will therefore have its vanishing point at R’, and will be limited by the ray through f/ The vertical line LINEAR PERSPECTIVE. 147 through f will then cast its shadow, which will be a vertical line, and will be hmited by the ray drawn through the upper extremity of the vertical edge of the cornice through f. There is a small part of the edge of the cornice, parallel to ff’, which is in the light, and which will cast a shadow on the face of the pillar. Its shadow will be perpendicular to the perspective piane, and consequently have its vanishing point at 5. The front circle of the arch will next cast a shadow on the side face. To find this shadow we will observe, that the shadow of the diameter ee’, on the side face, is parallel to the projections of rays on the side face, and since e” is the perspective of the point in which the diameter pierces the plane of the side face, e’H’ wii be the indefinite per- spective of its shadow. Draw any ordinate of the cir- cle, as ¢u, which it is supposed will cast a shadow. Through ¢ draw ¢R; the poimt ¢ is the shadow cast on the side plane by the foot of the ordinate. But the ordi- nate being parallel to the side plane, its shadow will be the vertical line through ¢. The point in which this vertical line intersects the ray through wis a point of shadow on the side plane. When this shadow reaches the front face of the pillar, it then falls on that face, and is a circle both in space and in perspective. Through g draw gR; and through pg’, the perspective of the point at which the centre of the front circle is projected on the plane of the front faces of the back pillars, draw ge” parallel to SR; g” is the perspective of the shadow cast by the centre of the front circle on the plane of the front faces of’ the back pillars But the shadow cast by the radius ge” is parallel to itself; “vow therefore, draw ge” parallel to the ground line, and e’R ” wt to the vanishing point of rays; g” e” is the perspective 148 TREATISE ON of the radius of the circle of shadow. With g” as a_ centre, and g” e” as-a radius, describe the arc of acircle, _ and we have the shadow cast on the pillar. The parts of the arch which are in the shade, or on which shadows fall, are shaded. PROBLEM XI. It is required to find the perspective of a house ana the perspec- . tive of its shadows. 98. Having measured all the lines of the house which are to be put in perspective, make its horizontal pro- jection to any convenient scale, as in Pl. 20. The outer lines are the horizontal projections of the outer lines of the eave-trough, and the adjacent inner ones are the lines in which the outer faces of the walls intersect the horizontal plane. ‘The lines in which the roofs intersect are also made, as well as the projections of the chimneys and steps. In selecting the position of the perspective plane, reference should be had to the part of the building which is to appear most prominent in the picture. Were it only required to represent the front of the building, we should take the perspective plane parallel to it; but when the front and an end are to be represented, it is most convenient to take the perspective plane oblique to them both; though it may be taken parallel to the one and perpendicular to the other. It simplifies the con- struction, without affecting the generality of the method, to assume the perspective plane through one or two of the prominent vertical lines. Such lines being in the perspective plane will be their own perspectives, and all the vertical distances may be laid off on them. In the construction here given, the perspective plane is passed through the vertical corners of the house oo v ™ *? of se Te ee ee Se ee ' ee 4a he + 3 i : — a foe # 3 SS S=== : | x \ \N | ; | ee : = ws = hs “4 7 ‘ é- hele ee il | i | ——— R Ih : l Ms + # L\ J. c »\ pn fi Yh | x =i = lex NS Z a = See ee ie e x 4 : BS TES ey ety Bly \ Wx N \. aa ts SST AN se \\ ‘E oie Bi MN = \\' \ e | ss -\\ : - a — | =< SE \ \ \, =e * . oe : os tk . | " | tH | | 5a ee | | | sg ‘= Sere aes \ | ee = = by of : { Nt! | {|| ; i‘ | FS x ne oe = as Saye eins | : = Sy & ek ae &. = “ \ ~ a < . i f 7 pon | \ = : | oy mes ; , . - ; Fs Soy | | \ = es oe Son haf \ Y cones | ee Hil \ | \ is ae Hl} * ‘ \ y \ a i \ otee. : ——— eee ies ae Hes ar i = aT < = | : ’ 7 Ht Sar te, Ripe de \ p —_———__,™_!'- ( i \ N Ss Mid a ee l [> \ \ —— eae =—— t | i \\ Se iN Ss < 4 = = sah est = I | K eet ? | Mi \\ = N < : er | 2 ES | j ill \ 771] -: SS | ; | : : — : mee Bee ee ap a Paes Pe | Hi \ 3 ; ae Oe & Mii] Si eda | i : = ! i \\ OY py cate Seacie \ / conn Ae \ ewan S| | A} Neos nay, : “ \ “ é = Dich ed | ae ee 77 fT RRS eo eae | , | rr £ i | Fe A Wy We : | 2 SS x a N : Sa6 | eet ~~ ee i . We IN \ \ \ = je H ~] | a \ : a eS Xe x ke: | = i /- \ \ \ \ 3 -~—---)\” H | | | ae ge 7 x 4 ‘ a8 < ae * : i Pee / | : | | a a - [ i Ty J \ % * = ‘ | ' es A= \. XX SS \ < | “ys \t : ‘ <—T | H ! rad Sox ae a 28 a = NX x | ‘ 1 : ie I! Yt . \ the ground, and draw KSH, which will be the horizon of the picture (74). On the line FG, perpendicular to DE, lay off the distance of the point of sight from the perspective plane. The horizontal projection of the point of sight does not fall on the paper, the eye being at about one and a half times the length of the building from the perspective plane. In making the drawing, a paper must be taken of such dimensions as will enable ., us to represent upon it the horizontal projection of the » point of sight, and the vanishing poate of the principal lines. Although the horizontal Prema of the house is made in front of the ground line, yet the building is in fact behind the perspective plane, with its front towards the eye. To represent to the mind the horizontal pro- jection of the house in its true position, we must con- ceive the horizontal plane to be revolved 180° around the ground line DE, which will bring every pomt now in front of the ground line at an ie distance be- hind it. At the points a and d draw add Art ded’ perpendi- | ~ cular to the ground line DE, and make 6 a’ equal to a 8, and cd’ equaltocd. The lines Aa and Adjare the lines Aa and Ad,in the positions from which their perspectives are taken. Through the point of sight Jet a line be drawn paral- lel to Ad’: its horizontal projection will pass through the horizontal projection of the point of sight, and the line will pierce the perspective plane 1 in a point of the horizontal line KSH. This point is the vanishing point of all lines parailel to Ad. The point is not on the paper, but as we have frequent occasion to refer to it, v 4 “e , 150 TREATISE ON we will designate it by the letter K. Through the point of sight let a line be drawn parallel to Aa’: this line pierces the perspective plane at H, which is, therefore, the vanishing point of all lines parallel to Aa. From A draw a line to the vanishing point K; it will be the indefinite perspective of the line Ad. From e draw a perpendicular to the ground line, and from the foot of the perpendicular draw a line to S; the point é, in which it intersects the line AK, is one point in the per- spective of the corner of the house, which pierces the horizontal plane at e; the vertical line drawn through “the point ¢ is the indefinite perspective of the corner, and the line drawn to B is the perspective of the line Be. The line drawn from B to the vanishing point K, 1s the indefinite perspective of the line Bf From f, draw a perpendicular to the ground line, and from the foot of the perpendicular draw a line to5; the point’ in which it meets the line BK, before drawn, is the perspective of the point f ‘Through the point /’, draw a vertical line, and it will be the indefinite perspective of the corner of the house which pierces the horizontal plane at f/ The perspective of the point d is in the line AK, and alsoin the line joining ¢c and 8; hence it is atd’ their point of intersection. ‘The vertical line drawn through this point is the indefinite perspective of the corner of the build- ig which meets the horizontal plane at d. On the vertical lines passing through the points A and B, lay off the distance from the ground .to the eave- trough. From the upper extremity of the vertical line through A, draw a line to the vanishing point K, and .- note the point in which it intersects the vertical line through é: the part intercepted between the vertical Jines through A and ¢,is the perspective of the lower ~ “7 AS {> a | Octery $ PAUL LINEAR ‘deg sate 151 line of the eave-trough that is horizontally projected in the line Ae. ‘The line joining the upper extremities of the vertical lines through b and ¢, is the perspective of the lower line of the eave-trough of which Be is the projection, and the perspective would, if produced, pass through H. From the upper extremity of the vertical line through B, draw a line to K. The part cut off by the vertical line through /’ is the perspective of the lower line of the eave-trough, of which Bf is the horizontal projection. ‘Through the upper extremity of the vertical line through A, a line has been drawn to K; the part of. this line intercepted between the vertical lines through ji and d’, is the perspective of all that can be seen a the lower line of the eave-trough, whose horizontal projec- tion is gd. From A draw a line to H, and ane 6 a line to S. The vertical line drawn through a’, their point of inter- section, 1s the indefinite perspective of the corner of the house which meets the horizontal plane at a. From the upper extremity of the vertical line through A, draw a line to H. The part cut off by the vertical line through a’ is the perspective of the lower line of the eave-trough of which Aais the horizontal projection. On the vertical lines through A and B, lay off the dis- tauce to the lower line of the water-table; then the width of the water-table; then find the perspectives of its upper and lower lines, in the same manner as we have already found the perspectives of the lines of the eave- trough. On the vertical line through A, lay off from the line © DE, the distance to the upper face of the lower window- sill, and draw through the point a horizontal line, and also a line to K. The horizontal line is the projection ° on the perspective plane of the lower line of the windows of the first story, and the line to K is its perspective. | a» Jo2 TREATISE ON From the point in which the perspective meets the ver tical line through ¢’, draw a line to H, and it will be the perspective of the lower line of the window, in that part of the building whose horizontal projection is Be: this line intersects the vertical line through B, at the same point in which itis intersected by the projection of the lower line of the windows. ‘The line drawn from this point to K is the indefinite perspective ‘of the lower line of the windows, corresponding to the part of the house of which Bf is the horizontal projection. By similar constructions we can find the perspectives of the upper horizontal lines of the windows, and also the perspectives of the hori- zontal lines which are tangent to their semicircular arches. In finding the perspectives of the vertical lines of the windows, it should not be forgotten that they are paral- lel to the perspective plane, and consequently, their perspectives will be parallel to the lines themselves. Through the lower extremity of the vertical line whose horizontal projection is f, draw a line perpendicular to the perspective plane. It will pierce the perspective plane in the projection of the lower line of the windows. From the point at which it pierces, draw a line to 8: its intersection with the perspective of the horizontal line before referred to, is the perspective of one point of the vertical] bounding line of the window; and the inde- finite line can be drawn, since it is vertical. The other vertical lines are found by constructions entirely similar, and all of them are limited by the perspectives, which have’ already been found, of the horizontal lines of the building. | To find the perspective of the roofs of the house. Produce nm, the horizontal projection of the line in which the side roofs intersect, until it meets the ground line at E. Draw EE’ perpendicular to the ground line, + LINEAR PERSPECTIVE. 153 and make it equal to the height of the intersection of the roofs above the ground. Then E’ will be the point in which the intersection of the roofs pierces the per- spective plane—the horizontal line through E’ is the projection, on the perspective plane, of the intersection of the roofs, and the line drawn from E’ to the vanish- ing point K is its indefinite perspective. ‘The point of which £ is the horizontal projection is vertically pro- jected atk’. From /' draw alinetoS: the point k’, in which it intersects the line E’KX, is the perspective of the point (4,£'). ‘The perspectives of the points (m, m’), (n,n'), and (p, p’), are found by similar constructions. If through the point in which the vertical line through A meets the horizontal plane of the eaves of the house, a line be drawn to #’, it will be the projection, on the perspective piane,of the line in which the side and front end roofs, intersect. The horizontal lines of the water-table, ofthe windows and of the eave-trough, corresponding to the end of the house, have a common vanishing point H; and the per- spectives of the vertical lines of the windows are found in the same manner as those in the front of the house. To find the perspectives of the chimneys. [From the horizontal line n’E’ lay off a distance above it equal to the height of the tops of the chimneys above the inter- section of the side roofs, and through the point go de- termined, draw the horizontal line xs. Produce the hori- zontal lines of the cornices which intersect at cither of the angles that are seen, until they meet the perspective | plane, which they will do inthe lmers. Since H and K - are the vanishing points of these lines, their perspectives can be drawn, and the point in which any two of them, that pass through the same angular point of the cornice, intersect, is the perspective of an angular point of the 4 154 TREATISE ON cornice. The perspectives of the lines in which the faces of the chimneys intersect the roof of the house,and the perspectives of their vertical edges, are easily found. 99. Although the perspective of every point can be found rigorously, yet the smaller parts of the building can generally be made with sufficient accuracy, without making a separate construction for each of them. Thus, after we have found the bounding lines of the windows, the window-sills, the caps and casings may be made very accurately without the aid of a geometrical construction. Having found the perspectives of the doors and steps, we may make the railing without a par- ticular construction for each vertical bar. After having found the lower line of the eave-trough, we may draw the outer upper line, always observing the rules of per- spective, and the general symmetry of the picture. In looking obliquely upon a building, the casings of the doors and windows on the side nearest the eye are not seen, while those on the other side are very distinct. To find the perspective of the shadows cast upon the house. Let R be the projection of a ray of light on the horizontal plane, ft’ its projection on the perspec- tive plane, and R” the position of its horizontal projec- tion, after it has been revolved to correspond with the horizontal projection of the building. ‘The rays of light being nearly parallel to the perspective plane, their vanishing point will be so far distant from the centre of the picture that we cannot use it conveniently in finding _the shadows. We therefore adopt other methods. In ail the constructions, we must bear in mind that the house is behind the perspective plane, and there- fore, when a ray of light is drawn through any point, its horizontal projection must not be drawn parallel to R, but to R’, which makes the same angle with the 3 . © OY CIETY OF oT pA \. LINEAR PERSPECTIVE. 159 ground line, but is differently inclined. Let us first find the shadow cast by the upper line of the end eave- trough, on the end of the house. From ¢ draw a line perpendicular to the ground line, and’make vf equal to the distance from the ground to the upper line of the eave-trough. Through (¢, #) draw aray of light; it pierces the end wall of the house in the point (u, wv’), which is a point in the line of shadow. From the point in which wu’ meets the ground line, draw a line to 8; the point in which it intersects AH is the perspective of the pointu. Through this pomt draw a vertical line, and it will be the perspective of the vertical line passing through (u,v). From uw’, draw a line to S; the poimt in which it intersects the vertical line before found, is the perspective of one point of the required line of shadow. But since the shadow is parallel to the line itself, its vanishing point is at H; therefore, the line drawn through H, and the point found, is the indefinite perspective of the line of shadow. Through the highest point, which is in the light, of the corner of the house which meets the horizontal plane at A, draw a-ray of light. Such ray will pierce the face Be of the building in a point which is horizon- tally projected at 7. The vertical line drawn through ¢ is the shadow cast on the wall by the corner of the house which meets the horizontal plane at A. The’ perspective of this vertical line of shadow, which is easily found, is the perspective of the shadow cast by the corner of the house towards the source of light. From the upper extremity of this shadow, the shadow is cast for a short distance by the upper line of the end eave-trough; then by the upper line of the front eave- trough; and the shadow terminates in the line of shadow cast by the upper line of the eave-trough belong- 156 TREATISE ON ing to the part of the house Be ;.the latter shadow being parallel in space to the shadow cast by the end eave- trough on the end of the house. The shadow of one of the chimneys only is seen. This shadow is found by drawing rays of hght thrdugh the extremities of the lines casting it; finding the points in which they pierce the roof, and determining the per- spectives of those points GENERAL REMARKS. | 100. It is, perhaps, too obvious to require illustration, that the projections in Descriptive Geometry, viz. the Orthographic and Stereographic projections of the sphere, are but particular methods of perspective. If the point of sight be at an infinite distance from the perspective plane, the perspectives of objects are the same as their orthographic projections. 101. If the plane of any circle of the sphere be as- sumed for the perspective plane, the point of sight being at the pole, the perspective representation of the sphere . upon this plane willin nowise differ from its stereographic projection. . 102. We have thus far represented objects on plane surfaces only. ‘Their perspectives may however be made on other surfaces, and the principles which have been explained will require but a slight modification to render them applicable to the case in which the per- spective is made on any surface whatever. P Suppose, for example, it were required to construct the perspectives of the different circles of the earth on the surface of a tangent cylinder, the point of sight . being at the centre.. If we suppose the equator to, be the circle of contact, the perspectives of the meridians will be elements of the cylinder, and the perspective t ef » 7 eee ee ee ae a . os em ras ae 7 ‘4 “ s ?. »* © a . te bai ’ ”, = ‘5 % a < = a : — a = ——_ : a = r at. ‘ ; ; . . apn ay 3 SS neh |D G . ¥ - ’ ' E.Prad homme Se. be? > = Pa a = atl iia nee ees . ’ ss 4 +—— = ——————__——______ a ____________ of . LINEAR PERSPECTIVE. pT ee of any other circle is found by constructing the inter- section of the cone, ef which the circle is the base, and the point of sight the vertex, with the surface of the . . tangent cylinder. Having thus found the perspectives) ._ of all the circles of the sphere, if we develop the surface of the tangent cylinder on a plane, its development will be a map of the earth. Mercator’s chart is constructed on these principles. 103. Panoramic views, which often. exhibit entire cities, are generally constructed on the surface of a- vertical cylinder, the eye of the spectator being in the _ axis. When the perspective is accurately made, and viewed from the right point, the deception is perfect. The houses seem to stand out from the canvass on which they are drawn; the streets have the aspect of bustle and business, and one feels himself transported into a populous city, and mingling in its affairs. 104. It is of great importance in linear perspective, that the distance of the eye from the plane of the picture should be judiciously chosen. If that distance is not sufficiently great to enable the eye to command the whole surface with perfect convenience, the represen- tation, though strictly according to the mathematical rules of perspective, will be deceitful. A rule to deter- mine this distance is therefore required, and it will be found, that-the eye should never approach nearer the perspective plane than to a distance equal to the diagonal of the picture. For example, let CDBA (Pi. 21, Fig. 1), represent the picture, then the distance of the point of sight from its surface should not be less than AD. I[t may be greater, and this is left to the taste of the artist. 105. Let it be required to put a pavement in perspec- tive, composed of alternate squares of black and white 158 TREATISE ON marble—the squares of equal size. Let ABDC (PI. 21, Fig. 2), represent the picture; EF its horizon, and G the centre of the picture; and let the distance of the eye from the perspective plane be equal only to GH. Let us now take the same example (PI. 21, Fig. 3), and let the distance from the point of sight to the perspec- tive plane be equal to the diagonal of the prcture. It is evident that, in the first example, the pieces of which the pavement is composed neither appear to be squares, nor of equal size, while in the second figure they do. 106. The vanishing point of perpendiculars, or the point directly opposite the eye; has been called the centre of the prcture. ‘This term has been used merely to designate the point, and does not imply that the point must be situated in the centre of the canvass in which the picture is drawn. Indeed, it is not absolutely neces- sary that it should be situated within the limits of the canvass. It always, however, determines the horizon of the picture, which, by a law of nature, rises and falls with the eye. In the example (PI. 21, Fig. 4), in which AB is the horizon, the centre C, of the ae: les without the limits of the canvass. In the example (PI. 21, Fig. 5), it lies without, and below the base line of the picture. 107. The situation of the horizontal line is altogether left to the taste of the artist, and must depend on the nature of the subject to be represented. If the appear- ance of magnitude in a building, or of majesty in a human or ideal figure is required, the choice of a very low horizon is the best. It has been remarked that the “ Apollo Belvidere was not made to muse in a valley.” In a picture, he can only be placed on an LINEAR PERSPECTIVE. 159 eminence, by allowing the horizon to be at his feet or below them. On the contrary, where a rich back-ground to figures is required, it may often be best obtained by placing the horizon very high in the picture. In landscape painting, the most usual practice is to place the horizon rather below than above the middle line of the canvass. 108. Artists are accustomed to divide linear perspec- tive into two kinds, parallel and oblique, and to consider the choice of either as entirely a matter of taste. 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