9. The pictares of all vertical lines are vertical, First general construction. From B and P (fig. 3.) draw any two parallel lines Fig. 3. For it is evident that BA, PV, of this figure are anawhich suit particular purposes. Writers on the subject logous to BA and EV of fig, 1. and that BA : PV= have either confined themselves to one construction, 6A: 6V, from an affectation of simplicity or fondness for system; If BA' be drawn perpendicular to GL, PV will fall or have multiplied precepts, by giving every construc on PS, and need not be drawn. AV will be A'S. tion for every example, in order to make a great book, This is the most easy construction, and pearly the same and give the subject an appearance of importance and with Ferguson's, difficulty. An ingenious practitioner will avoid both Second general construction. extremes, and avail himself of the advantage of each construction as it happens to suit his purpose. We shall Draw two lines BA, BA", and two lines PV, PD, now proceed to the practical rules, which require no parallel to them, and draw AV, A"D, cutting each consideration of intersecting planes, and are all perform other in b: b is the picture of P by Cor. 2.-- This con. ed on he perspective plane by means of certain substi struction is the foundation of all the rules of perspective tations for the place of the eye and the original figure. that are to be found in the books on this subject. They The general substitution is as follows: appear in a variety of forms, owing to the ignorance or Let the plane of the paper be first supposed to be inattention of the authors to the principles. The rule the ground-plan, and the spectator to stand at F most generally adhered to is as follows: (fig. 3.). Let it be proposed that the ground-plan is Draw BA' (fig. 4.) perpendicular to the ground Fig. 4. to be represented on a plane surface, standing perpen line, and AS to the point of sight, and set off A B equal dịcularly on the line GKL of the plan, and that the to BA. Set off SD equal to the distance of the eye point K is immediately opposite to the spectator, in the opposite direction from S that B is from A, or that FK is perpendicular to GL: then FK is where B and E of fig. 1. are on opposite sides of the equal to the distance of the spectator's eye from the picture; otherwise set them the same way. D is called picture. the point of distance. Draw & D, cutting AS in b. Now suppose a piece of paper laid on the plan with This is evidently equivalent to drawing BA and PS its straight edge lying on the line GL ; draw on perpendicular to the ground-line and horizon-line, and this paper KS perpendicular to GL, and make it BA' and PD (fig. 3.) making an angle of 45° with equal to the height of the eye above the ground plan. these lines, with the additional puzzle about the way of This may be much greater than the height of a man setting off A’A" and SD, which is avoided in the con-Figo because the spectator may be standing on a place much struction here given. rained above the ground-plan. Observe also that KS This usual construction, however, by a perpendicular must be measured on the same scale on wbich the and the point of distance, is extremely simple and conground-plan and the distance FK were measured. Then venient; and two points of distance, one on each side of draw. HSO parallel to GL. This will be a horizontal $, serve for all points of the ground plan. But the first line, and (when the picture is set upright on GL) will general construction requires still fewer lines, if BA be be on a level with the spectator's eye, and the poivt S drawn perpendicular to GL, because PV will then cowill be directly opposite to bis eye. It is therefore call incide with PS. ed the principal point, or point of sight. The distance of his eye from this point will be equal to FK. Therefore Third general construction. make SP (in the line SK) equal to FK, and P is the Draw BA (fig 4.) from the given point B perpen- Fig. 4. projecting point or substitute for the place of the eye. dicular to the ground-line, and AS to the point of sight. It is sometimes convenient to place B above $, some From the point of distance D set off D d equal to BA, times to one side of it on the horizontal line, and in va op the same or the contrary sides as S, according as B is the and given poiut from the picture, and SD:Dd This construction does not naturally arise from the , called the most generally convenient, as the perpendicular distance solid figures, or sych @gures as are raised above without confusion, and their direct situations trapsferred this plan. to the ground-lipe by perpendiculars such as BA; and nothing Fig. 3. lowing Fig. 6. nothing is easier than drawing parallels, either by a pa- all the subsequent constructions with the transposed Prob. To put any curvilineal figure on the ground- Put a sufficient number of its points in perspective Find the pictures b, c, of its extreme points by any of by the foregoing rules, and draw a curve line through the foregoing constructions, and join them by the straight them. line bc. It is well known that the conic sections and some Perhaps the following construction will be found very other curves, when viewed obliquely, are conic sections generally convenient. or curves of the same kind with the originals, with difProduce CB till it meet the ground-line in A, and ferent positions and proportions of their principal lines, draw PV parallel to it; join AV, and draw PB, PC, and rules may be given for describing their pictures cutting AV in b, c. V is its vanishing point, by Cor. 3. founded on this property. But these rules are very vaof the fundamental theorem. rious, unconnected with the general theory of perspecIt must be left to the experience and sagacity of the tive, and more tedious in the execution, without being drawer to select such circumstances as are most suitable more accurate than the general rule now given. It to the multiplicity of the figures to be drawn. would be a useless affectation to insert them in this elePROB. 3. To put any rectilineal figure of the ground.plan mentary treatise. We come in the next place to the delineation of in perspective. figures not in a horizontal plane, and of solid figures. Put the bounding lines in perspective, and the pro- For this purpose it is necessary to demonstrate the fola blem is solved. The variety of constructions of this problem is very great, and it would fill a volume to give them all. THEOREM II. The most generally convenient is to find the vanishing points of the bounding lines, and connect these with The length of any vertical line standing on the the points of their intersection with the ground line.ground plane is to that of its picture, as the beight of the For example, to put the square ABCD (fig. 6.) into eye to the distance of the horizon line from the picture perspective. of its foot. Draw from the projecting point PV, PW, parallel to Let BC (fig. 2) be the vertical line standing on B, Fig. 2o, AB, BC, and let AB, BC, CD, DA, meet the ground- and let EF be a vertical line through the eye. Make line, in «, x, d, 8, and draw a V, V, « W, BW, cutting BD equal to EF, and draw DE, CE, BE.' It is evieach other in a bed, the picture of the square ABCD. dent that DE will cut the horizon line in some point d, The demonstration is evident. CE will cut the picture plane in c, and BE will cut it This construction, however, runs the figure to great in b, and that b c will be the picture of BC, and is vetdistances on each side of the middle line, when any of tical, and that BC is to bc as BD to bd, or as EF to the lines of the original figure are nearly parallel to the bd. ground-line. Cor. The picture of a vertical line is divided in The following construction (fig. 7.) avoids this in the same ratio as the line itself. For BC: BM= convenience. bc:bm. Let D be the point of distance. Draw the perpendi- Prob. 5. To put a vertical line of a given length in perculars A , B B, CX, D d, and the lines A e, Bj, Cg, Dh, parallel to PD. Draw Sa, SB, Sm, Sd, and De, spective standing on a given point of the picture. Df, Dg, D h, cutting the former in a, b, c, d, the angles Through the given point b (fig. 9.) of the picture, Fig. 9 of the picture. draw 8 6 A from the point of sight, and draw the verIt is not necessary that D be the point of distance, tical line AD, and make AE equal to the length oř only the lines A e, B. f, &c. must be parallel to PD. height of the given line. Join ĖS, and draw b c pa Remark. In all the foregoing constructions the ne rallel to AD, producing b c, when necessary, till it cessary lines (and even the finished picture) are frequent cut the horizontal line in d, and we have bc: 6d,= ly confounded with the original figure. To avoid this AE : AD, that is, as the length of the given line to great inconvenience, the writers on perspective direct us the height of the eye, and b d is the distance of the to transpose the figure; that is, to transfer it to the other horizon-line from the point b, which is the pieture of side of the ground.line, by producing the perpendiculars the foot of the line. Therefore (Theot. 2.) 6c is the A a, B e, Cx, Dd, till * A', &B', &c. are respectively required picture of the vertical line. equal to A a, B B, &c.; or, instead of the original fi This problem occurs frequently in views of archigure, to use only its transposed substitute A'B'C'D'. tecture; and a compendious method of solving it would This is an extremely proper method. But in this case be particularly convenient. For this purpose, draw a verthe point P must also be transposed to P' above S, in tical line XZ at the margin of the picture, or on a sepaorder to retain the first or most natural and simple con tate paper, and through any point V of the horizon-line struction, as in fig. 8 ; where it is evident, that when draw vx. Set off XY, the height of the vertical-line, BA=AB', and SP=SP', and B'P' is drawn, cutting and draw VY. Then from any points by r, on which it AS in b, we have b A : 6 S=B'A: PS,=BA : PS, is required to have the pietores of lines equal to XY, and b is the picture of B: whence fellonus the truth of draw b Sort, parablel to the horizon-line, and draw the verticals Fig. 7. Fig. 8. corners. Fig. 10. Fig. 11. verticals su, tv: these have the lengths required, which ef, gh, i k, tending towards the point of sight S; and cut ef and ik, draw l m parallel to AD. Through the PROB. 6. To put any sloping line in perspective. centre-point x, where the diagonals cut g h, draw no parallel to AD.-Lastly, through the points v and u, From the extremities of this line, suppose perpendi- where the diagonals cut e f and i k, draw p q parallel to culars meeting the ground plane in two points, which AD; and the reticulated perspective square will be we shall call the base points of the sloping line. Put finished. these base points in perspective, and draw, by last pro This square is truly represented, as if seen by an obblem, the perpendiculars from the extremities. Join server standing at 0, and having his eye above the hothese by a straight line. It will be the picture re rizontal plane ABCD on which it is drawn ; as if OS quired. was the height of his eye above that plane : and the PROB. 7. To put a square in perspective, as seen by a lines which form the small squares within it have the person not standing right against the middle of either drawn as it would appear to an eye placed perpendicu same letters of reference with those in fig. 12. which is of its sides, but rather nearly even with one of its larly above its centre x. In fig. 10. let ABCD be a true square, viewed by an PROB. 9. To put a circle in perspective. observer, not standing at o, directly against the middle If a circle be viewed by an eye placed directly over of its sides AD, but at almost even with its corner its centre, it appears perfectly round, but if it be obD, and viewing the side AD under the angle AOD; liquely viewed, it appears of an elliptical shape. Tbiš the angle A o D (under which he would have seen AD is plain by looking at a common wine-glass set upright from being 60 degrees. on a table. Make AD in fig. 11. equal to AD in fig. 10. and Make a true reticulated square, as fig. 12. of the same Fig. 12. draw $P and O parallel to AD. Then, in fig. 11. diameter as you would have the circle ; and setting one let O be the place of the observer's eye, and SO be foot of the compasses in the centre X, describe as large perpendicular to SP; then S shall be the point of sight a circle as the sides of the square will contain. Then, in the horizon SP. having put this reticulated square into perspective, as in Take SO in your compasses, and set that extent from 'fig. 13. observe through what points of the cross lines S to P; then P shall be the true point of distance, taken and diagonals of fig. 12. the circle passes ; and through according to the foregoing rules. the like points in fig. 13. draw the ellipsis, which will From A and D draw the straight lines AS and be as true a perspective representation of the circle, as DS; draw also the straight line AP, intersecting DS the square in fig. 13. is of the square in fig. 12. in C. This is Mr Ferguson's rule for putting a circle in M is the centre of each square, and scribe a square about it. · Draw first the diagonals of the square, and then the diameters h a and de (fig. 14.) Fig. 14. PROB. 8. To put a reticulated square in perspective, as cutting one another at right angles ; draw the straight seen by a person standing opposite to the middle of one lines fg and b c parallel to the diameter de. Through b and f, and likewise c and draw straight lines meetof its sides. ing DE, the ground line of the picture in the points 3 gec a will be the circle in perspective. scribing, from the point of bisection as a centre, the Make AD in fig. 13. equal to AD in fig. 12. and di- semicircle AGB (fig. 15.), and from any number of Fig. 15 vide it into four equal parts, as A e, e gigi, and i D. points in the circumference C, F, G, H, I, &c. draw to Draw SP for the horizon, parallel to AD, and, the ground line the perpendiculars C1, F2, G 3, H4, through the middle point g of AD, draw OS perpendi- I 5, &c. From the points A, 1, 2, 3, 4, 5, B, draw cular to AD and SP.-Make S the point of sight, and straight lines to the principal point or point of sight V, O the place of the observer's eye. likewise straight lines from B and A to the points of diTake SP equal to So, and P shall be the true point stance L and K. Through the common intersections of distance.-Draw AS and DS to the point of sight, draw straight lines as in the preceding case; and you and AP to the point of distance, intersecting DS in will have the points a, c, f, g, h, i, b, representatives of C: then draw BC parallel to AD, and the outlines of A, C, F, G, H, I, B.' Then join the points a, c, f, the reticulated square ABCD will be finished. &c. as formerly directed, and you have the perspective From the division points e,8,i, draw the straight lines circle a cfgh ibih gf ca. Hence 10. Fig. 12. Fig. 13. Hence it is apparent how we may put not only a cir PROB. 15. To put any solid in perspective. cle but also a pavement laid with stones of any form in perspective. It is likewise apparent how useful the Put the base of the solid, whatever it be, in perspecs square is in perspective; for, as in the second case, a true tive by the preceding rules. From each bounding square was described round the circle to be put in point of the base, raise lines representing in perspective perspective, and divided into several smaller squares, the altitude of the object; by joining these lines and so in this third case we make use of the semicircle shading the figure according to the directions in the only for the sake of brevity instead of that square and preceding problem, you will have a scenographic recircle. presentation of the object. This rule is general; but as its application to particular cases may not be appa? PROB. 10. To put a reticuluted square in perspective as rent, it will be proper to give the following example: seen by a person not standing right against the middle of it! PROB. 16. To put a cube in perspective as seen from: one of its angles. Fig. 15. In fig. 16. let O be the place of an observer, viewing Since the base of a cube standing on a geometrical the ABCD almost even with its corner D. square Draw at pleasure SP for the horizon, parallel to AD, from one of its angles, draw first such a perspective plane, and seen from one of its angles, is a square seen and make SO perpendicular to SP: then S shall be the point of sight, and P the true point of distance, if SP DE (fig. 18.) the perpendicular HI equal to the side of Fig. 18. square : then raise from any point of the ground-line be made equal to SO. Draw AS and DS to the point of sight, and AP to line HR the straight lines VI and VH. From the the square, and draw to any point V in the horizontal the point of distance, intersecting DS in the point C; then draw BC parallel to AD, and the outlines of the angles d, b, and c, draw the dotted lines d 2 and c 1 pa. perspective square will be finished. This done, draw dotted lines, and from the points 1 and 2; draw the rallel to the ground line DE. Perpendicular to those the lines which form the lesser squares, as taught in straight lines L 1 and M 2. Lastly, since HI is the alProb. 8. and the work will be completed. You may put a perspective circle in this square by the same rule in d, draw from the point a the straight line f a perpen titude of the intended cube in a, Li in c and b, M 2 as it was done in fig. 13. dicular to a E, and from the points b and c, bg and ce, PROB. 14. To put a cube in perspective, as if viewed perpendicular to bc1, and abdc being according to by a person standing almost even with one of its rule, make af=HI, bg=e=L 1, and h d=M 2. edges, and seeing three of its sides. Then, if the points g, h, e, f, be joined, the whole cube will be in perspective. In fig. 17. let AB be the breadth of either of the six equal square sides of the cube AG; O the place of the Prob. 17. To put a square pyramid in perspective, as . observer, almost even with the edge CD of the cube, S standing upright on its base, and viewed obliqucly. the point of sight, SP the horizon parallel to AD, and In fig. 19. let AD be the breadth of either of the Fig. 19. P the point of distance taken as before. four sides of the pyramid ATCD at its base ABCD ;. Make ABCD a true square; draw BS and CS to and MT its perpendicular height. Let O be the place the point of sight, and BP to the point of distance, in- of the observer, 's his point of sight, SE his horizon, patersecting CS in G.–Then draw FG parallel to BC, rallel to AD and perpendicular to os; and let the and the uppermost perspective square side BFGC of the proper point of distance be taken in SE produced to-cube will be finished. ward the left hand, as far from S as O is from S. Draw DS to the point of sight, and AP to the Draw AS and DS to the point of sight, and DL point of distance, intersecting DS in the point I: then to the point of distance, intersecting AS in the point. draw GI parallel to CD; and, if the cube be an o B. Then, from B, draw BC parallel to AD, and paque one, as of wood or metal, all the outlines of it ABCD shall be the perspective square base of the pywill be finished; and then it may be shaded as in the ramid, figure. Draw the diagonal AC, intersecting the other dia.. But if you want a perspective view of a transparent gonal BD at M, and this point of intersection shall be: glass cube, all the sides of which will be seen, draw the centre of the square base. ÅH toward the point of sight, FH parallel to BA, Draw MT perpendicular to AD, and of a length and HI parallel to AD: then AHID will be the equal to the intended height of the pyramid : then square base of the cube, perspectively parallel to the draw the straight outlines AT, CT, and DT; and the top BFGC; ABFH will be the square side of the outlines of the pyramid (as viewed from 0) will be fi cube, parallel to CGID, and FGIH will be the square nished; which being done, the whole may be so sbaded: side parallel to ABCD. as to give it the appearance of a solid body. As to the shading part of the work, it is such mere If the observer had stood at o, he could have only children's play, in comparison of drawing the lines seen the side ATD of the pyramid ; and two is the which form the shape of any object, that po rules need greatest number of sides that he could see from any be given for it. Let a person sit with his left side to other place of the ground. But if he were at any ward a window, and he knows full well, that if any height above the pyramid, and had his eye directly solid body be placed on a table before him, the light over its top, it would then appear as in fig. 20. and he Fig. 20. will fall on the left-hand side of the body, and the would see all its four sides E, F, G, H, with its top t, right-band, side will be in the shade. just over the centre of its square base ABCD; which, would. Fig. 17. Fig. 21. would be a true geometrical and not a perspective from any point V in the horizon be drawn the straight lines VH, V 1, V 2, V 3, V. 4; V 5 or VI ; by a pro cess similar to that of the preceding problem, will be in their proper places, the scenograph of the prism is ing on four upright square legs of any given length with respect to the breadth of the table. of the square table, parallel to the floor. from its edges equal to their thickness. Take A a and and BD, and draw straight lines from the points a, b, c, d, towards the point of sight S, and terminating at the PROB. 19. To put a truncated pyramid in perspectives side BC. Then, through the points where these lines Let the pyramid to be put in perspective be quin cut the diagonals, draw the straight lines n and o, P quaogular. If from each angle of the surface whence and q, parallel to AD; and you will have formed the top is cut off, a perpendicular be supposed to fall four perspective squares (like ABCD in fig. 19.) for Fig. 19. upon the base, these perpendiculars will mark the the bases of the four legs of the table: and then it is bounding-points of a pentagon, of which the sides will easy to draw the four upright legs by parallel lines, be parallel to the sides of the base of the pyramid, all perpendicular to AD; and to shade them as in the within which it is inscribed. Join these points, and the figure. interior pentagon will be formed with its longest side To represent the intended thickness of the tableparallel to the longest side of the base of the pyramid. board, draw eh parallel to EH, and HG toward the From the ground line EH (fig. 22.) raise the perpen- point of sight S: then shade-the spaces between these dicular IH, and make it equal to the altitude of the in- lines, and the perspective figure of the table will be tended pyramid. To any point V draw the straight finished. lines, IV and HV, and by a process similar to that in Prob. 16. determine the scenographical altitudes a, b, c, PROB. 22. To put five square pyramids in perspective, doc. Connect the upper points f, g, h, i, k, by straight standing upright on a square puvement composed of lines, and draw lk, fm, gn, and the perspective of the surfaces of 81 cubes. the truncated pyramid.will be completed. In fig. 25. let ABCD be a perspective square drawn Fig. 29 Cor. If in a geometrical plane two concentric cir- according to the foregoing rules; S the point of sight, cles be described, a truncated cone may be pet in P the point of distance in the horizon PS, and AC and perspective in the same manner as. a truncated pyra BD the two diagonals of the square. mid. Divide the side AD into 9 equal parts (because 9 times 9 is 81) as A a, a b, bc, &c. and from these point of sight S, terminating at the furthermost side BC 1 Fig. 22. Fig. 23. Fig. 24 |