7. The pictures of all vertical lines are vertical, and the pictures of horizontal lines are horizontal, because these lines are parallel to the perspective plane.

8. The point of sight S is the vanishing point of all lines perpendicular to the perspective plane.

The above proposition is a sufficient foundation for the whole practice of perspective, whether on direct or inclined pictures, and serves to suggest all the various practical constructions, each of which has advantages which suit particular purposes. Writers on the subject have either confined themselves to one construction, from an affectation of simplicity or fondness for system; or have multiplied precepts, by giving every construction for every example, in order to make a great book, and give the subject an appearance of importance and difficulty. An ingenious practitioner will avoid both extremes, and avail himself of the advantage of each construction as it happens to suit his purpose. We shall now proceed to the practical rules, which require no consideration of intersecting planes, and are all performed on the perspective plane by means of certain substitutions for the place of the eye and the original figure. The general substitution is as follows:

Let the plane of the paper be first supposed to be the ground-plan, and the spectator to stand at F (fig. 3.). Let it be proposed that the ground-plan is to be represented on a plane surface, standing perpendicularly on the line GKL of the plan, and that the point K is immediately opposite to the spectator, or that FK is perpendicular to GL: then FK is equal to the distance of the spectator's eye from the picture.

Now suppose a piece of paper laid on the plan with its straight edge lying on the line GL; draw on this paper KS perpendicular to GL, and make it equal to the height of the eye above the ground plan. This may be much greater than the height of a man because the spectator may be standing on a place much raised above the ground-plan. Observe also that KS must be measured on the same scale on which the ground-plan and the distance FK were measured. Then draw HSO parallel to GL. This will be a horizontal line, and (when the picture is set upright on GL) will be on a level with the spectator's eye, and the point S will be directly opposite to his eye. It is therefore called the principal point, or point of sight. The distance of his from this point will be equal to FK. Therefore make SP (in the line SK) equal to FK, and P is the projecting point or substitute for the place of the eye. It is sometimes convenient to place B above S, sometimes to one side of it on the horizontal line, and in various other situations; and writers, ignorant of, or inattentive to, the principles of the theory, have given it different denominations, such as point of distance, point of view, &c. It is merely a substitute for the point E in fig. 1. and its most natural situation is below, as in this figure.

eye

The art of perspective is conveniently divided into ICHNOGRAPHY, which teaches how to make a perspective draught of figures on a plane, commonly called the ground-plan; and SCENOGRAPHY, which teaches how to draw solid figures, or such figures as are raised above this plan.

Fundamental PROB. I. To put into perspective any given point of the ground-plan.

First general construction.

From B and P (fig. 3.) draw any two parallel lines Fig. 3. BA, PV, cutting the ground-line and horizon-line in A and V, and draw BP, AV, cutting each other in b; b is the picture of B.

For it is evident that BA, PV, of this figure are analogous to BA and EV of fig. 1. and that BA : PV= bă: bv,

If BA' be drawn perpendicular to GL, PV will fall on PS, and need not be drawn. A'V will be A'S.— This is the most easy construction, and nearly the same with Ferguson's,

Second general construction.

The rule

Draw two lines BA, BA", and two lines PV, PD, parallel to them, and draw AV, A"D, cutting each other in bb is the picture of P by Cor. 2.-This construction is the foundation of all the rules of perspective that are to be found in the books on this subject. They appear in a variety of forms, owing to the ignorance or inattention of the authors to the principles. most generally adhered to is as follows: Draw BA (fig. 4.) perpendicular to the ground Fig. 4. line, and AS to the point of sight, and set off A6 equal to BA. Set off SD equal to the distance of the eye in the opposite direction from S that is from A, where B and E of fig. 1. are on opposite sides of the picture; otherwise set them the same way. D is called the point of distance. Draw & D, cutting AS in b. This is evidently equivalent to drawing BA and PS perpendicular to the ground-line and horizon-line, and BA and PD (fig. 3.) making an angle of 45° with these lines, with the additional puzzle about the way of setting off A'A" and SD, which is avoided in the con- Fig. 3. struction here given.

This usual construction, however, by a perpendicular and the point of distance, is extremely simple and convenient; and two points of distance, one on each side of S, serve for all points of the ground plan. But the first general construction requires still fewer lines, if BA be drawn perpendicular to GL, because PV will then coincide with PS.

Third general construction.

Draw BA (fig 4.) from the given point B perpen- Fig. 4dicular to the ground-line, and AS to the point of sight. From the point of distance D set off D d equal to BA, on the same or the contrary sides as S, according as B is on the same or the contrary side of the picture as the eye. Join d, A, and draw Db parallel to d A. b is the picture of B. For SD, D d, are equal to the distances of the eye and given point from the picture, and SD: Dd bS: bA.

This construction does not naturally arise from the original lines, but is a geometrical consequence from their position and magnitude; and it is of all others the most generally convenient, as the perpendicular distance of any number of points may be arranged along SD without confusion, and their direct situations transferred to the ground-line by perpendiculars such as BA; and nothing

nothing is easier than drawing parallels, either by a parallel ruler or a bevel square, used by all who practise drawing.

PROB. 2. To put any straight line BC (fig. 5.) of the ground plan in perspective.

Find the pictures b, c, of its extreme points by any of the foregoing constructions, and join them by the straight line b c.

Perhaps the following construction will be found very generally convenient.

Produce CB till it meet the ground-line in A, and draw PV parallel to it; join AV, and draw PB, PC, cutting AV in b, c. V is its vanishing point, by Cor. 3. of the fundamental theorem.

It must be left to the experience and sagacity of the drawer to select such circumstances as are most suitable to the multiplicity of the figures to be drawn.

PROB. 3. To put any rectilineal figure of the ground-plan in perspective.

Put the bounding lines in perspective, and the problem is solved.

The variety of constructions of this problem is very great, and it would fill a volume to give them all. The most generally convenient is to find the vanishing points of the bounding lines, and connect these with the points of their intersection with the ground line. For example, to put the square ABCD (fig. 6.) into perspective.

Draw from the projecting point PV, PW, parallel to AB, BC, and let AB, BC, CD, DA, meet the groundline, in a, x, d, ß, and draw « V, d V, W, 8 W, eutting each other in a bed, the picture of the square ABCD. The demonstration is evident.

x

This construction, however, runs the figure to great distances on each side of the middle line, when any of the lines of the original figure are nearly parallel to the ground-line.

The following construction (fig. 7.) avoids this inconvenience.

Let D be the point of distance. Draw the perpendiculars A a, B B, C, D d, and the lines A e, B, C g, Dh, parallel to PD. Draw Sa, SB, Sx, Sd, and De, Df, Dg, Dh, cutting the former in a, b, c, d, the angles of the picture.

It is not necessary that D be the point of distance, only the lines Ae, Bf, &c. must be parallel to PD.

Remark. In all the foregoing constructions the necessary lines (and even the finished picture) are frequently confounded with the original figure. To avoid this great inconvenience, the writers on perspective direct us to transpose the figure; that is, to transfer it to the other side of the ground-line, by producing the perpendiculars Aa, Bs, Cx, Dd, till a A', & B', &c. are respectively equal to A, B s, &c.; or, instead of the original fi gure, to use only its transposed substitute A'B'C'D'. This is an extremely proper method. But in this case the point P must also be transposed to P' above S, in order to retain the first or most natural and simple construction, as in fig. 8; where it is evident, that when BA=AB', and SP=SP', and B'P' is drawn, cutting AS in b, we have b Ab SB'A: P'S, BA: PS, and b is the picture of B: whence fellows the truth of

all the subsequent constructions with the transposed figure.

PROB. 4. To put any curvilineal figure on the groundplun into perspective.

Put a sufficient number of its points in perspective by the foregoing rules, and draw a curve line through them.

. It is well known that the conic sections and some other curves, when viewed obliquely, are conic sections or curves of the same kind with the originals, with different positions and proportions of their principal lines, and rules may be given for describing their pictures founded on this property. But these rules are very various, unconnected with the general theory of perspective, and more tedious in the execution, without being more accurate than the general rule now given. would be a useless affectation to insert them in this elementary treatise.

It

We come in the next place to the delineation of figures not in a horizontal plane, and of solid figures. For this purpose it is necessary to demonstrate the following

THEOREM II.

The length of any vertical line standing on the ground plane is to that of its picture, as the height of the eye to the distance of the horizon line from the picture

of its foot.

Let BC (fig. 2) be the vertical line standing on B, Fig. 2 and let EF be a vertical line through the eye. Make BD equal to EF, and draw DE, CE, BE. It is evident that DE will cut the horizon line in some point d, CE will cut the picture plane in c, and BE will cut it in b, and that be will be the picture of BC, and is vertical, and that BC is to be as BD to b d, or as EF to b d.

C

Cor. The picture of a vertical line is divided in the same ratio as the line itself. For BC : BM÷ bc: bm.

PROB. 5. To put a vertical line of a given length in perspective standing on a given point of the picture. Through the given point b (fig. 9.) of the picture, Fig. 9 draw S 6 A from the point of sight, and draw the vertical line AD, and make AE equal to the length or height of the given line. Join ES, and draw be pa rallel to AD, producing b c, when necessary, till it cut the horizontal line in d, and we have bc: bd, AE: AD, that is, as the length of the given line to the height of the eye, and b d is the distance of the horizon-line, from the point b, which is the pieture of the foot of the line. Therefore (Theor. 2.) be is the required picture of the vertical line.

This problem occurs frequently in views of architecture; and a compendious method of solving it would be particularly convenient. For this purpose, draw a ver tical line XZ at the margin of the picture, or on a separate paper, and through any point V of the horizon-line draw VX. Set off XY, the height of the vertical-line, and draw VY. Then from any points b, r, on which it is required to have the pictures of lines equal to XY, draw bs, rt, parallel to the horizon-line, and draw the

verticals

PROB. 6. To put any sloping line in perspective. From the extremities of this line, suppose perpendiculars meeting the ground plane in two points, which we shall call the base points of the sloping line. Put these base points in perspective, and draw, by last problem, the perpendiculars from the extremities. Join these by a straight line. It will be the picture required.

PROB. 7. To put a square in perspective, as seen by a person not standing right against the middle of either of its sides, but rather nearly even with one of its

corners.

In fig. 10. let ABCD be a true square, viewed by an observer, not standing at o, directly against the middle of its sides AD, but at O almost even with its corner D, and viewing the side AD under the angle AOD; the angle Ao D (under which he would have seen AD from o) being 60 degrees.

Make AD in fig. 11. equal to AD in fig. 10. and draw SP and O o parallel to AD. Then, in fig. 11. let O be the place of the observer's eye, and SO be perpendicular to SP; then S shall be the point of sight in the horizon SP.

Take SO in your compasses, and set that extent from S to P; then P shall be the true point of distance, taken according to the foregoing rules.

From A and D draw the straight lines AS and DS; draw also the straight line AP, intersecting DS in C.

Lastly, through the point of intersection C draw BC parallel to AD; and ABCD in fig. 11. will be a true perspective representation of the square ABCD in fig.

10.

The point M is the centre of each square, and AMC and BMD are diagonals.

PROB. 8. To put a reticulated square in perspective, as seen by a person standing opposite to the middle of one of its sides.

A reticulated square is one that is divided into se. veral little squares, like net-work, as fig. 12. each side of which is divided into four equal parts, and the whole surface into four times four (or 16) equal squares.

Having divided this square into the given number of lesser squares, draw the two diagonals Ar C and B x D.

Make AD in fig. 13. equal to AD in fig. 12. and divide it into four equal parts, as A e, eg, g i, and i D.

Draw SP for the horizon, parallel to AD, and, through the middle point g of AD, draw OS perpendicular to AD and SP.-Make S the point of sight, and O the place of the observer's eye.

Take SP equal to SO, and P shall be the true point of distance.-Draw AS and DS to the point of sight, and AP to the point of distance, intersecting DS in C: then draw BC parallel to AD, and the outlines of the reticulated square ABCD will be finished.

From the division points e,g,, draw the straight lines

ef, gh, ik, tending towards the point of sight S; and draw BD for one of the diagonals of the square, the other diagonal AC being already drawn.

Through the points r and s, where these diagonals cut ef and ik, draw Im parallel to AD. Through the centre-point x, where the diagonals cut g h, draw no parallel to AD.-Lastly, through the points v and w, where the diagonals cut ef and ik, draw pq parallel to AD; and the reticulated perspective square will be finished.

This square is truly represented, as if seen by an observer standing at O, and having his eye above the horizontal plane ABCD on which it is drawn; as if OS was the height of his eye above that plane and the lines which form the small squares within it have the same letters of reference with those in fig. 12. which is

drawn as it would appear to an eye placed perpendicu

larly above its centre x.

PROB. 9. To put a circle in perspective.

If a circle be viewed by an eye placed directly over its centre, it appears perfectly round, but if it be obliquely viewed, it appears of an liquely viewed, it appears of an elliptical shape. This is plain by looking at a common wine-glass set upright on a table.

Make a true reticulated square, as fig. 12. of the same Fig. 12. diameter as you would have the circle; and setting one foot of the compasses in the centre x, describe as large a circle as the sides of the square will contain. Then, having put this reticulated square into perspective, as in fig. 13. observe through what points of the cross lines and diagonals of fig. 12. the circle passes; and through the like points in fig. 13. draw the ellipsis, which will be as true a perspective representation of the circle, as the square in fig. 13. is of the square in fig. 12.

This is Mr Ferguson's rule for putting a circle in perspective; but the following rules by Wolf are perhaps more universal.

If the circle to be put in perspective be small, describe a square about it. Draw first the diagonals of the square, and then the diameters h a and de (fig. 14.) Fig. 14 cutting one another at right angles; draw the straight lines fg and b c parallel to the diameter de. Through b and f, and likewise c and g, draw straight lines meeting DE, the ground line of the picture in the points 3 and 4. To the principal point V draw the straight lines 1 V, 3 V, 4 V, 2 V, and to the points of distance L and K, 2 L and 1 K. Lastly, join the points of intersection, a, b, d, f, h, g, e, c, by the arcs a b, b d, df, and a b dfh geca will be the circle in perspective.

If the circle be large so as to make the foregoing practice inconvenient, bisect the ground line AB, describing, from the point of bisection as a centre, the semicircle AGB (fig. 15.), and from any number of Fig. 15. points in the circumference C, F, G, H, I, &c. draw to the ground line the perpendiculars C 1, F 2, G 3, H 4, I 5, &c. From the points A, 1, 2, 3, 4, 5, B, draw straight lines to the principal point or point of sight V, likewise straight lines from B and A to the points of distance L and K. Through the common intersections draw straight lines as in the preceding case; and you will have the points a, c, f, g, h, i, b, representatives of A, C, F, G, H, I, B. Then join the points a, c, f, &c. as formerly directed, and you have the perspective circle a cf g h i bih gf ca.

Hence

Hence it is apparent how we may put not only a circle but also a pavement laid with stones of any form in perspective. It is likewise apparent how useful the square is in perspective; for, as in the second case, a true square was described round the circle to be put in perspective, and divided into several smaller squares, so in this third case we make use of the semicircle only for the sake of brevity instead of that square and

circle.

PROB. 10. To put a reticulated square in perspective as seen by a person not standing right against the middle of either of its sides, but rather nearly even with one of its corners.

In fig. 16. let O be the place of an observer, viewing the square ABCD almost even with its corner D.Draw at pleasure SP for the horizon, parallel to AD, and make SO perpendicular to SP: then S shall be the point of sight, and P the true point of distance, if SP be made equal to SO.

Draw AS and DS to the point of sight, and AP to the point of distance, intersecting DS in the point C; then draw BC parallel to AD, and the outlines of the perspective square will be finished. This done, draw the lines which form the lesser squares, as taught in Prob. 8. and the work will be completed.-You may put a perspective circle in this square by the same rule as it was done in fig. 13.

PROB. 14. To put a cube in perspective, as if viewed by a person standing almost even with one of its edges, and seeing three of its sides.

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 observer, almost even with the edge CD of the cube, S the point of sight, SP the horizon parallel to AD, and P the point of distance taken as before.

Make ABCD a true square; draw BS and CS to the point of sight, and BP to the point of distance, intersecting CS in G.-Then draw FG parallel to BC, and the uppermost perspective square side BFGC of the cube will be finished.

Draw DS to the point of sight, and AP to the point of distance, intersecting DS in the point I: then draw GI parallel to CD; and, if the cube be an opaque one, as of wood or metal, all the outlines of it will be finished; and then it may be shaded as in the figure.

But if you want a perspective view of a transparent glass cube, all the sides of which will be seen, draw AH toward the point of sight, FH parallel to BA, and HI parallel to AD: then AHID will be the square base of the cube, perspectively parallel to the top BFGC; ABFH will be the square side of the cube, parallel to CGID, and FGIH will be the square side parallel to ABCD.

As to the shading part of the work, it is such mere children's play, in comparison of drawing the lines which form the shape of any object, that no rules need be given for it. Let a person sit with his left side toward a window, and he knows full well, that if any solid body be placed on a table before him, the light will fall on the left-hand side of the body, and the right-hand side will be in the shade..

PROB. 15. To put any solid in perspective. Put the base of the solid, whatever it be, in perspec tive by the preceding rules. From each bounding point of the base, raise lines representing in perspective the altitude of the object; by joining these lines and shading the figure according to the directions in the preceding problem, you will have a scenographic representation of the object. This rule is general; but as its application to particular cases may not be appa rent, it will be proper to give the following example of it.

PROB. 16. To put a cube in perspective as seen from: one of its angles.

Since the base of a cube standing on a geometrical from one of its angles, draw first such a perspective plane, and seen from one of its angles, is a square seen> DE (fig. 18.) the perpendicular HI equal to the side of Fig. 18. square then raise from any point of the ground-line the square, and draw to any point V in the horizontal line HR the straight lines VI and VH. From the angles d, b, and c, draw the dotted lines d 2 and c 1 padotted lines, and from the points 1 and 2, draw the rallel to the ground line DE. Perpendicular to those straight lines L 1 and M 2. Lastly, since HI is the altitude of the intended cube in a, L 1 in c and b, M 2 in d, draw from the point a the straight line ƒ a perpendicular to a E, and from the points b and c, b g and ce, perpendicular to be 1, and ab de being according to rule, make af=HI, bg=ec=L 1, and h d=M 2. Then, if the points g, h, e, f, be joined, the whole cubewill be in perspective.

PROB. 17. To put a square pyramid in perspective, as standing upright on its base, and viewed obliquely. In fig. 19. let AD be the breadth of either of the four sides of the pyramid ATCD at its base ABCD; and MT its perpendicular height. Let O be the place of the observer, S his point of sight, SE his horizon, parallel to AD and perpendicular to OS; and let the proper point of distance be taken in SE produced to-ward the left hand, as far from S as O is from S.

Draw AS and DS to the point of sight, and DL to the point of distance, intersecting AS in the point B. Then, from B, draw BC parallel to AD; and ABCD shall be the perspective square base of the pyramid.

Draw the diagonal AC, intersecting the other dia. gonal BD at M, and this point of intersection shall be the centre of the square base.

Draw MT perpendicular to AD, and of a length equal to the intended height of the pyramid: then draw the straight outlines AT, CT, and DT; and the outlines of the pyramid (as viewed from O) will be fi nished; which being done, the whole may be so shaded as to give it the appearance of a solid body.

If the observer had stood at o, he could have only seen the side ATD of the pyramid; and two is the greatest number of sides that he could see from any other place of the ground. But if he were at any height above the pyramid, and had his eye directly. over its top, it would then appear as in fig. 20. and he would see all its four sides E, F, G, H, with its top t, just over the centre of its square base ABCD; which

would.

Fig. 19

Fig. 2c.

square.

PROB. 18. To put two equal squares in perspective, one of which shall be directly over the other, at any given distance from it, and both of them parallel to the plane of the horizon.

In fig. 21. let. ABCD, be a perspective square om a horizontal plane, drawn according to the foregoing rules, Sbeing the point of sight, SP the horizon (parallel: to AD), and P the point of distance.

Suppose AD, the breadth of this square to be three feet; and that it is required to place just such another. square EFGH directly above it, parallel to it and two feet from it.

Make AE and DH perpendicular to AD, and two thirds of its length: draw EH, which will be equal and parallel to AD; then draw ES and HS to the point of sight S, and EP to the point of distance P, intersecting HS in the point G: this done, draw FG parallel to EH; and you will have two perspective squares ABCD and EFGH, equal and parallel to one another, the latter directly above the former, and two feet distant from it; as was required.

By this method shelves may be drawn parallel to one another, at any distance from each other in proportion to their length.

PROB, 19. To put a truncated pyramid in perspective Let the pyramid to be put in perspective be quinquangular. If from each angle of the surface whence the top is cut off, a perpendicular be supposed to fall upon the base, these perpendiculars will mark the bounding-points of a pentagon, of which the sides will be parallel to the sides of the base of the pyramid, within which it is inscribed. Join these points, and the interior pentagon will be formed with its longest side parallel to the longest side of the base of the pyramid. From the ground line EH (fig. 22.) raise the perpendicular IH, and make it equal to the altitude of the intended pyramid. To any point V draw the straight lines IV and HV, and by a process similar to that in Prob. 16. determine the scenographical altitudes a, b, c, de Connect the upper points f, g, h, i, k, by straight lines; and draw ik, fm, gn, and the perspective of the truncated pyramid will be completed.

Cor. If in a geometrical plane two concentric circles be described, a truncated cone may be put in perspective in the same manner as a truncated pyramid.

PROB. 20. To put in perspective a hollow prism lying on one of its sides.

1

lines VÍ, V 1, V 2, V 3, V 4, V 5 or VI; by a process similar to that of the preceding problem, will be determined the height of the internal angles, viz. 1= aa, 2=b b, 4=dd; and of the external angles, 3=cc, and see; and when these angles are formed and put in their proper places, the scenograph of the prism is complete.

PROB. 21. To put a square table in perspective, standing on four upright square legs of any given length with respect to the breadth of the table.

In fig. 21. let ABCD be the square part of the floor Fig. 21. on which the table is to stand, and EFGH the surface of the square table, parallel to the floor.

Suppose the table to be three feet in breadth, and its height from the floor to be two feet; then two thirds of AD or EH will be the length of the legs and k; the other two (7 and m) being of the same length in perspective.

Having drawn the two equal and parallel squares ABCD and EFGH, as shown in Prob. 18. let the legs be square in form, and fixed into the table at a distance from its edges equal to their thickness. Take A a and Dd equal to the intended thickness of the legs, and a b and de also equal thereto. Draw the diagonala AC and BD, and draw straight lines from the points a, b, c, d, towards the point of sight S, and terminating at the side BC. Then, through the points where these lines cut the diagonals, draw the straight lines n and o, p and q, parallel to AD; and you will have formed four perspective squares (like ABCD in fig. 19.) for Fig. 1 the bases of the four legs of the table: and then it is easy to draw the four upright legs by parallel lines, all perpendicular to AD; and to shade them as in the figure.

To represent the intended thickness of the tableboard, draw eh parallel to EH, and HG toward the point of sight S: then shade the spaces between these lines, and the perspective figure of the table will be finished.

PROB. 22. To put five square pyramids in perspective, standing upright on a square pavement composed of the surfaces of 81 cubes.

In fig. 25. let ABCD be a perspective square drawn Fig. 1 according to the foregoing rules; S the point of sight, P the point of distance in the horizon PS, and AC and BD the two diagonals of the square.

Divide the side AD into 9 equal parts (because 9 times 9 is 81) as A a, a b, bc, &c. and from these points of division, a, b, c, d, &c. draw lines toward the point of sight S, terminating at the furthermost side BC of the square. Then, through the points where these lines cut the diagonals, draw straight lines parallel to AD, and the perspective square ABCD will be subdivided into 81 lesser squares, representing the upper surfaces of 81 cubes, laid close to one another's sides in a square form,

Draw AK and DL, each equal to Aa, and perpendicular to AD; and draw LN toward the point of sight S: then draw KL parallel to AD, and its distance from AD will be equal to A.-This done, drawal, cm, cn, do, ep, fq, gr, and hs, all parallel to AK; and the space ADLK will be subdivided