PERSPECTIVE

P E R S P E C T I V E.

PERSPECTIVE is the art of drawing on a plane
surface true resemblances or pictures of objects, as
the objects themselves appear to the eye from any dis-
tance and situation, real or imaginary.

It was in the 16th century that Perspective was re-
vived, or rather reinvented. It owes its birth to paint-
ing, and particularly to that branch of it which was em-
ployed in the decorations of the theatre, where land-
scapes were properly introduced, and which would have
looked unnatural and horrid if the size of the objects
had not been pretty nearly proportioned to their distance
from the eye. We learn from Vitruvius, that Agathar-
chus, instructed by Eschylus, was the first who wrote
upon this subject; and that afterwards the principles
of the art were more distinctly taught by Democritus
and Anaxagoras, the disciples of Agatharchus. Of
the theory of this art, as described by them, we know
nothing; since none of their writings have escaped
the general wreck that was made of ancient literature
in the dark ages of Europe. However, the revival of

painting in Italy was accompanied with a revival of this
art.

The first person who attempted to lay down the rules
of perspective was Pietro del Borgo, an Italian. He
supposed objects to be placed beyond a transparent tab-
let, and endeavoured to trace the images which rays of
light, emitted from them, would make upon it. But
we do not know what success he had in this attempt,
because the book which he wrote upon the subject is
not now extant. It is, however, very much commended
by the famous Egrazio Dante; and, upon the prin-
ciples of Borgo, Albert Durer constructed a machine,
by which he could trace the perspective appearance of
objects.

Balthazar Perussi studied the writings of Borgo, and
endeavoured to make them more intelligible. To him
we owe the discovery of points of distance, to which
all lines that make an angle of 45 degrees with the
ground-line are drawn. A little time after, Guido Ul-
balni, another Italian, found that all the lines that are
parallel

parallel to one another, if they be inclined to the ground-line, converge to some point in the horizontal line; and that through this point also, a line drawn from the eye, parallel to them, will pass. These principles put together enabled him to make out a pretty complete theory of perspective.

Great improvements were made in the rules of perspective by subsequent geometers; particularly by Professor Gravesende, and still more by Dr Brooke Taylor, whose principles are in a great measure new, and far more general than any before him.

In order to understand the principles of perspective, it will be proper to consider the plane on which the representation is to be made as transparent, and interpolated between the eye of the spectator and the object to be represented. Thus, suppose a person at a window looks through an upright pane of glass at any object beyond it, and, keeping his head steady, draws the figure of the object upon the glass with a black lead pencil, as if the point of the pencil touched the object itself; he would then have a true representation of the object in perspective as it appears to his eye.

In order to this two things are necessary: first, that the glass be laid over with strong gum-water, which, when dry, will be fit for drawing upon, and will retain the traces of the pencil: and, secondly, that he looks through a small hole in a thin plate of metal, fixed about a foot from the glass, between it and his eye, and that he keeps his eye close to the hole; otherwise he might shift the position of his head, and consequently make a false delineation of the object.

Having traced out the figure of the object, he may go over it again with pen and ink; and when that is dry, put a sheet of paper upon it, and trace it thereon with a pencil: then taking away the paper and laying it on a table, he may finish the picture by giving it the colours, lights, and shades, as he sees them in the object itself; and then he will have a true resemblance of the object.

To every person who has a general knowledge of the principles of optics, this must be self-evident: For as vision is occasioned by pencils of rays coming in straight lines to the eye from every point of the visible object, it is plain that, by joining the points in the transparent plane, through which all those pencils respectively pass, an exact representation must be formed of the object, as it appears to the eye in that particular position, and at that determined distance: and were pictures of things to be always first drawn on transparent planes, this simple operation, with the principle on which it is founded, would comprise the whole theory and practice of perspective. As this, however, is far from being the case, rules must be deduced from the sciences of optics and geometry for drawing representations of visible objects on opaque planes; and the application of these rules constitutes what is properly called the art of perspective.

Previous to our laying down the fundamental principles of this art, it may not be improper to observe, that when a person stands right against the middle of one end of a long avenue or walk, which is straight and equally broad throughout, the sides thereof seem to approach nearer and nearer to each other as they are fur-

ther and further from his eye; or the angles, under which their different parts are seen, become less and less according as the distance from his eye increases; and if the avenue be very long, the sides of it at the farthest end will seem to meet: and there an object that would cover the whole breadth of the avenue, and be of a height equal to that breadth, would appear only to be a mere point.

Having made these preliminary observations, we now proceed to the practice of perspective, which is built upon the following

(Fundamental) THEOREM I.

Let $abcd$ (fig. 1.) represent the ground-plan of the figure to be thrown into perspective, and $efgh$ the transparent plane through which it is viewed by the eye at $E$. Let these planes intersect in the straight line $KL$. Let $B$ be any point in the ground plan, and $BE$ a straight line, the path of a ray of light from that point to the eye. This will pass through the plane $efgh$ in some point $b$; or $B$ will be seen through that point, and $b$ will be the picture, image, or representation of $B$.

If $BA$ be drawn in the ground-plan, making any angle $BAK$ with the common intersection, and $EV$ be drawn parallel to it, meeting the picture-plane or perspective-plane in $V$, and $VA$ be drawn, the point $b$ is in the line $VA$ so situated that $BA$ is to $EV$ as $bA$ is to $bV$.

For since $EV$ and $BA$ are parallel, the figure $BAbVEbB$ is in one plane, cutting the perspective-plane in the straight line $VA$; the triangles $BAb$, $EVb$, are similar, and $BA\; :\; EV\; =\; bA\; :\; bV$.

Cor. 1. If $B$ be beyond the picture, its picture $b$ is above the intersection $KL$; but if $B$ be between the eye and the picture, as at $B\text{'}$, its picture $b\text{'}$ is below $KL$.

2. If two other parallel lines $BA\text{'}$, $ES$, be drawn, and $A\text{'}$, $S$, be joined, the picture of $B$ is in the intersection of the lines $AV$ and $A\text{'}S$.

3. The line $BA$ is represented by $bA$, or $bA$ is the picture of $BA$; and if $AB$ be infinitely extended, it will be represented by $AV$. $V$ is therefore called the vanishing point of the line $AB$.

4. All lines parallel to $AB$ are represented by lines converging to $V$ from the points where these lines intersect the perspective plane; and therefore $V$ is the vanishing point of all such parallel lines.

5. The pictures of all lines parallel to the perspective plane are parallel to the lines themselves. Fig. 2.

6. If through $V$ be drawn $HVO$ parallel to $KL$, the angle $EVH$ is equal to $BAK$. Fig. 1.

Remark. The proposition now demonstrated is not limited to any inclination of the picture-plane to the ground-plane; but it is usual to consider them as perpendicular to each other, and the ground-plane as horizontal. Hence the line $KL$ is called the ground line, and $OH$ the horizon line; and $OK$, perpendicular to both, is called the height of the eye.

If $ES$ be drawn perpendicular to the picture-plane, it will cut it in a point $S$ of the horizon-line directly opposite to the eye. This is called the point of sight, or principal point.

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 affection 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:

Fig. 3. 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 eye 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 $P$ 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.

The art of perspective is conveniently divided into ICHOGRAPHY, 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 PRON. 1. 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 $BA,\; PV$, 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 $BA\; :\; PV\; =\; bA\; :\; bV$.

If $BA\text{'}$ be drawn perpendicular to $GL$, $PV$ will fall on $PS$, and need not be drawn. $AV$ will be $AS$.—This is the most easy construction, and nearly the same with Ferguson's.

Second general construction.

Draw two lines $BA,\; BA\text{'}\text{'}$, and two lines $PV,\; PD$, parallel to them, and draw $AV,\; A\text{'}D$, cutting each other in $b$: $b$ is the picture of $B$ 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. The rule most generally adhered to is as follows:

Draw $BA$ (fig. 4.) perpendicular to the ground-line, and $AS$ to the point of sight, and set off $A\backslash beta$ equal to $BA$. Set off $SD$ equal to the distance of the eye in the opposite direction from $S$ that $\backslash beta$ 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 $\backslash beta\; D$, cutting $AS$ in $b$. This is evidently equivalent to drawing $BA\text{'}$ and $PS$ perpendicular to the ground-line and horizon-line, and $BA\text{'}\text{'}$ and $PD$ (fig. 3.) making an angle of $45^\backslash circ$ with these lines, with the additional puzzle about the way of setting off $A\text{'}A\text{'}\text{'}$ and $SD$, which is avoided in the construction 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 $B$ perpendicular to the ground-line, and $AS$ to the point of sight. From the point of distance $D$ set off $Dd$ equal to $BA$, on the same or the contrary side as $S$, according as $B$ is on the same or the contrary side of the picture as the eye. Join $d,\; A$, and draw $Dd$ parallel to $dA$. $b$ is the picture of $B$. For $SD,\; Dd$, 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 distances 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.

Fig. 5. 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 $bc$.

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 rectilinear 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.

Fig. 6. Draw from the projecting point PV, PW, parallel to AB, BC, and let AB, BC, CD, DA, meet the ground-line, in $\backslash alpha,\; \backslash beta,\; \backslash gamma,\; \backslash delta$, and draw $\backslash alpha\; V,\; \backslash beta\; V,\; \backslash gamma\; W,\; \backslash delta\; W$, cutting each other in $a,\; b,\; c,\; d$, the picture of the square ABCD. The demonstration is evident.

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.

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

Let D be the point of distance. Draw the perpendiculars $A\backslash alpha,\; B\backslash beta,\; C\backslash gamma,\; D\backslash delta$, and the lines $Ae,\; Bf,\; Cg,\; Dh$, parallel to PD. Draw $S\backslash alpha,\; S\backslash beta,\; S\backslash gamma,\; S\backslash delta$, and $D\backslash epsilon,\; D\backslash zeta,\; D\backslash eta,\; D\backslash theta$, 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 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 $A\backslash alpha,\; B\backslash beta,\; C\backslash gamma,\; D\backslash delta$, till are respectively equal to ; or, instead of the original figure, to use only its transposed substitute $A\text{'}B\text{'}C\text{'}D\text{'}$. This is an extremely proper method. But in this case the point P must also be transposed to $P\text{'}$ above S, in order to retain the first or most natural and simple construction, as in fig. 8.; where it is evident, that when $BA=AB\text{'}$, and $SP=SP\text{'}$, and $B\text{'}P\text{'}$ is drawn, cutting AS in $b$, we have $bA\; :\; bS\; =\; BA\; :\; PS$, and $b$ is the picture of B: whence follows the truth of

all the subsequent constructions with the transposed figure.

PROB. 4. To put any curvilinear figure on the ground-plan 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 kinds 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. It would be a useless affectation to insert them in this elementary treatise.

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, 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 $d\text{'}$, CE will cut the picture plane in $c\text{'}$, and BE will cut it in $b\text{'}$, and that $b\text{'}c\text{'}$ will be the picture of BC, and is vertical, and that BC is to $b\text{'}c\text{'}$ as BD to $b\text{'}d\text{'}$, or as EF to $b\text{'}d\text{'}$.

Cor. The picture of a vertical line is divided in the same ratio as the line itself. For $BC\; :\; BM\; =\; b\text{'}c\text{'}\; :\; b\text{'}m$.

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, draw $SbA$ 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 $b\text{'}c\text{'}$ parallel to AD, producing $b\text{'}c\text{'}$, when necessary, till it cut the horizon line in $d\text{'}$, and we have $b\text{'}c\text{'}\; :\; b\text{'}d\text{'}\; =\; AE\; :\; AD$, that is, as the length of the given line to the height of the eye, and $b\text{'}d\text{'}$ is the distance of the horizon-line from the point $b$, which is the picture of the foot of the line. Therefore (Theor. 2.) $b\text{'}c\text{'}$ 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 peculiarly convenient. For this purpose, draw a vertical 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 $b\text{'}r,\; r\text{'}r$, parallel to the horizon-line, and draw the verticals

verticals $xz,\; yz$: these have the lengths required, which may be transferred to $b$ and $r$. This, with the third general construction for the base points, will save all the confusion of lines which would arise from constructing each line apart.

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.

Fig. 10. 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 $AOD$ (under which he would have seen $AD$ from $o$) being 60 degrees.

Fig. 11. Make $AD$ in fig. 11. equal to $AD$ in fig. 10. and draw $SP$ and $Oo$ 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 the 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.

Fig. 12. A reticulated square is one that is divided into several 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 $Ax$ and $Cx$.

Fig. 13. Make $AD$ in fig. 13. equal to $AD$ in fig. 12. and divide it into four equal parts, as $Ae,\; eg,\; g,\; i$, and $iD$.

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 $e,\; g,\; i$, 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 $e$ and $i$, where these diagonals cut $ef$ and $ik$, draw $lm$ parallel to $AD$. Through the centre-point $x$, where the diagonals cut $gh$, draw $wo$ 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 perpendicularly 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 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 diameter as you would have the circle; and setting one foot of your 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 ellipse, 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 $ha$ and $de$ (fig. 14.) cutting one another at right angles; draw the straight lines $fg$ and $bc$ 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 $ab,\; bd,\; df$, and $cd,\; eh,\; ga$ 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 points in the circumference $C,\; F,\; G,\; H,\; I$, &c. draw to the ground line the perpendiculars $C1,\; F2,\; G3,\; H4,\; I5$, &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 $aefghibigfca$.

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 fix 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, perspective 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 perspective 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 apparent, 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 plane, and seen from one of its angles, is a square seen from one of its angles, draw first such a perspective square: then raise from any point of the ground-line DE (fig. 18) the perpendicular HI equal to the side of the square, and draw to any point V in the horizontal line HR the straight lines VI and VH. From the angles $d\_1\; b\_1$ and $c\_1$, draw the dotted lines $d\_2$ and $c\_1$ parallel to the ground line DE. Perpendicular to those dotted lines, and from the points 1 and 2, draw the straight lines L 1 and M 2. Lastly, since $H\; I$ 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 $fa$ perpendicular to $a\; E$, and from the points $b$ and $c$, $bg$ and $ce$, perpendicular to $b\; c\; 1$, and $a\; b\; d\; e$ being according to rule, make $a\; f\; =\; H\; I$, $b\; g\; =\; c\; e\; =\; L\; 1$, and $h\; d\; =\; M\; 2$. Then, if the points $g,\; h,\; e,\; f$, be joined, the whole cube will 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 toward 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 diagonal 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 finished; 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 $a$, 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.

would be a true geometrical and not a perspective 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.

Fig. 21. In fig. 21. let ABCD be a perspective square on a horizontal plane, drawn according to the foregoing rules, S being 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 $a,\; b,\; c,\; d,\; e$. Connect the upper points $f,\; g,\; h,\; i,\; k$, by straight lines; and draw $l,\; k,\; m,\; n$, 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.

Fig. 23. Let ABDEC (fig. 23.) be a section of such a prism. Draw HI parallel to AB, and distant from it the breadth of the side on which the prism rests; and from each angle internal and external of the prism let fall perpendiculars to HI. The parallelogram will be thus divided by the scenographical process below the ground-line, so as that the side AB of the real prism will be parallel to the corresponding side of the scenographic view of it.—To determine the altitude of the internal and external angles. From H (fig. 24.) raise HI perpendicular to the ground-line, and on it mark off the true altitudes $H\_1,\; H\_2,\; H\_3,\; H\_4$, and $H\_5$. Then if

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 process similar to that of the preceding problem, will be determined the height of the internal angles, viz. $1=a,\; 2=b,\; 4=d$; and of the external angles, $3=c,\; 5=e$; 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 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 $i$ and $k$; the other two ($l$ 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 $Aa$ and $Dd$ equal to the intended thickness of the legs, and $ab$ and $dc$ also equal thereto. Draw the diagonals 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 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 table-board, 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 free 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 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 and from these points of division, draw lines toward the point of sight S, terminating at the furthest 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 $Aa$.—This done, draw $ah,\; bm,\; cn,\; do,\; ep,\; fq,\; gr$, and $hz$, all parallel to AK; and the space ADLK will be subdivided into

Fig.

Fig. 24.

into nine equal squares, which are the outer upright surfaces of the nine cubes in the side AD of the square ABCD.

From the points where the lines, which are parallel to AD in this square, meet the side CD thereof, draw short lines to LN, all parallel to DL, and they will divide that side into the outer upright surfaces of the nine cubes which compose it: and then the outfides of all the cubes that can be visible to an observer, placed at a proper distance from the corner D of the square, will be finished.

As taught in Prob. 17. place the pyramid AE upright on its square base Atv a, making it as high as you please; and the pyramid DH on its square base hwt D, of equal height with AE.

Draw EH from the top of one of these pyramids to the top of the other; and EH will be parallel to AD.

Draw ES and HS to the point of sight S, and HP to the point of distance P, intersecting ES in F.

From the point F, draw FG parallel to EH; then draw EG, and you will have a perspective square EFGH (parallel to ABCD) with its two diagonals EG and FH, intersecting one another in the centre of the square at I. The four corners of this square, E, F, G, H, give the perspective heights of the four pyramids AE, BF, CG, and DH; and the intersection I of the diagonals gives the height of the pyramid MI, the centre of whose base is the centre of the perspective square ABCD.

Lastly, place the three pyramids BF, CG, MI, upright on their respective bases at B, C, and M; and the required perspective representation will be finished, as in the figure.

PROB. 23. To put upright pyramids in perspective, on the sides of an oblong square or parallelogram; so that their distances from one another shall be equal to the breadth of the parallelogram.

Fig. 26. In most of the foregoing operations we have considered the observer to be so placed, as to have an oblique view of the perspective objects: in this, we shall suppose him to have a direct view of fig. 26. that is, standing right against the middle of the end AD which is nearest to his eye, and viewing AD under an angle of 60 degrees.

Having cut AD in the middle, by the perpendicular line St, take S therein at pleasure for the point of sight, and draw ES for the horizon, parallel to AD. Here St must be supposed to be produced downward, below the limits of the plate, to the place of the observer; and SE to be produced towards the left hand beyond E, far enough to take a proper point of distance therein, according to the foregoing rules.

Take Ad at pleasure, and Dg equal to Ad, for the breadths of the square bases of the two pyramids AE and DF next the eye: then draw AS and d'S, and likewise DS and g'S, to the point of sight S; and DG on to the point of distance, intersecting AS in G: then, from G draw GI parallel to AD, you will have the first perspective square AGID of the parallelogram ABCD.

From I draw IH to (or toward) the point of distance, intersecting AS in H: then, from H draw

HK parallel to AD, and you will have the second perspective square GHKI of the parallelogram. Go on in this manner till you have drawn as many perspective squares up toward S as you please.

Through the point e, where DG intersect g'S, draw hf parallel to AD; and you will have formed the two perspective square bases Abcd and efDg of the two pyramids at A and D.

From the point f (the upper outward corner of efDg) draw fh toward the point of distance, till it meets AS in h; then, from this point of meeting, draw hm parallel to GI, and you will have formed the two perspective squares Ghit and ImIn, for the square bases of the two pyramids at G and I.

Proceed in the same manner to find the bases of all the other pyramids, at the corners of the rest of the perspective squares in the parallelogram ABCD, as shown by the figure. Then,

Having placed the first two pyramids at A and D upright on their square bases, as shown in Prob. 9. and made them of any equal heights at pleasure, draw ES and FS from the tops of these pyramids to the point of sight S: place all the rest of the pyramids upright on their respective bases, making their tops touch the straight lines ES and FS; and all the work, except the shading part, will be finished.

PROB. 24. To put a square pyramid of equal fixed cubes in perspective.

Fig. 27. represents a pyramid of this kind; consist-
ing as it were of square tables of cubes, one table above another; 81 in the lowest, 49 in the next, 25 in the third, 9 in the fourth, and 1 in the fifth or uppermost. These are the square numbers of 9, 7, 5, 3, and 1.

If the artist is already master of all the preceding operations, he will find less difficulty in this than in attending to the following description of it: for it cannot be described in a few words, but may be executed in a very short time.

In fig. 28. having drawn PS for the horizon, and ken S for the point of sight therein (the observer being at O) draw AD parallel to PS for the side (next the eye) of the first or lowermost table of cubes. Draw AS and DS to the point of sight S, and DP to the point of distance P, intersecting AS in the point B. Then, from B, draw BC parallel to AD, and you will have the surface ABCD of the first table.

Divide AD into nine equal parts, as Aa, ab, bc, cd, &c. then make AK and DL equal to Aa, and perpendicular to AD. Draw KL parallel to AD, and from the points of equal division at a, b, c, &c. draw lines to KL, all parallel to AK. Then draw hS to the point of sight S, and from the division points a, b, c, &c. draw lines with a black lead pencil, all tending towards the point of sight, till they meet the diagonal BD of the square.

From these points of meeting draw black lead lines to DC, all parallel to AD; then draw the parts of these lines with black ink which are marked 1, 2, 3, 4, &c. between hE and DC.

Having drawn the first of these lines hq with black ink, draw the parts ai, bk, cl, &c. (of the former lines which met the diagonal BD) with black ink also; and rub out the rest of the black lead lines, which would

would otherwise confuse the following part of the work. Then, draw $LF$ toward the point of sight $S$; and, from the points where the lines 1, 2, 3, 4, &c. meet the line $DC$, draw lines down to $LF$, all parallel to $DL$; and all the visible lines between the cubes in the first table will be finished.

Make $iG$ equal and perpendicular to $\backslash beta\; i$, and $qM$ equal and parallel to $iG$: then draw $GM$, which will be equal and parallel to $iq$. From the points $k,\; l,\; m,\; n$, &c. draw $kn,\; lo,\; mp$, &c. all parallel to $iG$, and the outfaces of the seven cubes in the side $Gq$ of the second table will be finished.

Draw $GS$ and $MS$ to the point of sight $S$, and $MP$ to the point of distance $P$, intersecting $GS$ in $H$; then, from the point of intersection $H$, draw $HI$ parallel to $AD$; and you will have the surface $GHIM$ of the second table of cubes.

From the points $n,\; o,\; p,\; q$, &c. draw black lead lines toward the point of sight $S$, till they meet the diagonal $MH$ of the perspective square surface $GHIM$; and draw $SM$, with black ink, toward the point of sight.

From those points where the lines drawn from $n,\; o,\; p,\; q$, &c. meet the diagonal $MH$, draw black lead lines to $MI$, all parallel to $AD$; only draw the whole first line $\backslash gamma\; i$ with black ink, and the parts 2, 3, 4, &c. and $\backslash eta\; i,\; \backslash theta\; u,\; p\; v$, &c. of the other lines between $\backslash gamma\; N$ and $MI$, and $GM$ and $\backslash gamma\; i$, with the same; and rub out all the rest of the black lead lines, to avoid further confusion. Then, from the points where the short lines 1, 2, 3, &c. meet the line $MI$, draw lines down to $qE$, all parallel to $Mq$, and the outer surfaces of the seven cubes in the side $ME$ will be finished; and all these last lines will meet the former parallels 2, 3, 4, &c. in the line $qE$.

Make $iO$ equal and perpendicular to $\backslash gamma\; i$, and $yP$ equal and parallel to $iO$; then draw $OP$, which will be equal and parallel to $iy$.—This done, draw $OS$ and $PS$ to the point of sight $S$, and $PP$ to the point of distance $P$ in the horizon. Lastly, from the point $Q$, where $PP$ intersects $OS$, draw $QR$ parallel to $OP$; and you will have the outlines $OQRP$ of the surface of the third perspective table of cubes.

From the points $u,\; v,\; w,\; x$, draw upright lines to $OP$, all parallel to $iO$, and you will have the outer surfaces of the five cubes in the side $Oy$ of this third table.

From the points where these upright lines meet $OP$, draw lines toward the point of sight $S$, till they meet the diagonal $PQ$; and from these points of meeting draw lines to $PR$, all parallel to $OP$, making the parts 2, 3, 4, 5, of these lines with black ink which lie between $ZY$ and $PR$. Then, from the points where these lines meet $PR$, draw lines down to $\backslash gamma\; N$; which will bound the outer surfaces of the five cubes in the side $PN$ of the third table.

Draw the line $\backslash delta\; i$ with black ink; and, at a fourth part of its length between $\backslash delta$ and $Z$, draw an upright line to $S$, equal in length to that fourth part, and another equal and parallel thereto from $Z$ to $V$: then draw $SV$ parallel to $\backslash delta\; Z$, and draw the two upright and equidistant lines between $\backslash delta\; Z$ and $SV$, and you will have the outer surfaces of the three cubes in the side $SZ$ of the fourth table.

Draw $SS$ and $VS$ to the point of sight $S$ in the ho-

rizon, and $VP$ to the point of distance therein, intersecting $SS$ in $T$; then draw $TU$ parallel to $SV$, and you have $STUV$, the surface of the fourth table, which being reticulated or divided into 9 perspective small squares, and the uppermost cube $W$ placed on the middlemost of the squares, all the outlines will be finished; and when the whole is properly shaded, as in fig. 27, the work will be done.

PROB. 25. To represent a double cross in perspective.

In fig. 29. let $ABCD$ and $EFGH$ be the two perspective squares, equal and parallel to one another, the uppermost directly above the lowermost, drawn by the rules already laid down, and as far asunder as is equal to the given height of the upright part of the cross; $S$ being the point of sight, and $P$ the point of distance, in the horizon $PS$ taken parallel to $AD$.

Draw $AE$, $DH$, and $CG$; then $AEHD$ and $DHGC$ shall be the two visible sides of the upright part of the cross; of which, the length $AE$ is here made equal to three times the breadth $EH$.

Divide $DH$ into three equal parts, $HI$, $IK$, and $KD$. Through these points of division, at $I$ and $K$, draw $MO$ and $PR$ parallel to $AD$; and make the parts $MN$, $IO$, $PQ$, $KR$, each equal to $HI$; then draw $MP$ and $OR$ parallel to $DH$.

From $M$ and $O$, draw $MS$ and $OS$ to the point of sight $S$; and from the point of distance $P$ draw $PN$ cutting $MS$ in $T$; from $T$ draw $TU$ parallel to $MO$, and meeting $OS$ in $U$; and you will have the uppermost surface $MTUO$ of one of the cross pieces of the figure. —From $R$, draw $RS$ to the point of sight $S$; and from $U$ draw $UV$ parallel to $OR$; and $OUVR$ shall be the perspective square end next the eye of that cross part.

Draw $PMX$ (as long as you please) from the point of distance $P$, through the corner $M$; lay a ruler to $N$ and $S$, and draw $XN$ from the line $PX$:—then lay the ruler to $I$ and $S$, and draw $YZS$.—Draw $XY$ parallel to $MO$, and make $XW$ and $YB$ equal and perpendicular to $XY$: then draw $WB$ parallel to $XY$, and $WXYB$ shall be the square visible end of the other cross part of the figure.

Draw $BK$ toward the point of sight $S$; and from $U$ draw $UP$ to the point of distance $P$, intersecting $YS$ in $Z$: then, from the intersection $Z$, draw $Za$ parallel to $MO$, and $Zb$ parallel to $DH$, and the whole delineation will be finished.

This done, shade the whole, as in fig. 30. and you will have a true perspective representation of a double cross.

PROB. 26. To put three rows of upright square objects in perspective, equal in size, and at equal distances from each other, on an oblong square plane, the breadth of which shall be of any assigned proportion to the length thereof.

Fig. 31. is a perspective representation of an oblong square plane, three times as long as it is broad, having a row of nine upright square objects on each side, and one of the same number in the middle; all equally high, and at equal distances from one another, both long-wise and cross-wise, on the same plane.

In fig. 32. $PS$ is the horizon, $S$ the point of sight, $P$ the

the point of distance, and AD (parallel to PS) the breadth of the plane.

Draw AS, NS, and DS, to the point of sight S; the point N being in the middle of the line AD; and draw DP to the point of distance P, intersecting AS in the point B; then, from B draw BC parallel to AD, and you have the perspective square ABCD.

Through the point i, where DB intersects NS, draw ae parallel to AD; and you will have subdivided the perspective square ABCD into four lesser squares, as A a i N, N i e D, a B k i, and i k C e.

From the point C (at the top of the perspective square ABCD) draw CP to the point of distance P, intersecting AS in E; then, from the point E draw EF parallel to AD; and you will have the second perspective square BEFC.

Through the point l, where CE intersects NS, draw bf parallel to AD; and you will have subdivided the square BEFC into the four squares B b l k, k l f C, b E m l, and l m F f.

From the point F (at the top of the perspective square BEFC) draw FP to the point of distance P, intersecting AS in I; then from the point I draw IK parallel to AD; and you will have the third perspective square EIKF.

Through the point n, where FI intersects NS, draw cg parallel to AD; and you will have subdivided the square EIKF into four lesser squares, E e n m, m n g F, e I o n, and n o K g.

From the point K (at the top of the third perspective square EIKF) draw KP to the point of distance P, intersecting AS in L; then from the point L draw LM parallel to AD; and you will have the fourth perspective square ILMK.

Through the point p, where KL intersects NS, draw dh parallel to AD; and you will have subdivided the square ILMK into the four lesser squares I d p o, o p h K, d L q p, and p q M h.

Thus we have formed an oblong square ALMD, whose perspective length is equal to four times its breadth, and it contains 16 equal perspective squares.—If greater length was still wanted, we might proceed further on toward S.

Take A 3, equal to the intended breadth of the side of the upright square object AQ (all the other sides being of the same breadth), and AO for the intended height. Draw O 18 parallel to AD, and make D 8 and 4 7 equal to A 3; then draw 3 S, 4 S, 7 S, and 8 S to the point of sight S; and among them we shall have the perspective square bases of all the 27 upright objects on the plane.

Through the point 9, where DB intersects 8 S, draw 1 10 parallel to AD, and you have the three perspective square bases A 1 2 3, 4 5 6 7, 8 9 10 D, of the three upright square objects at A, N, and D.

Through the point 21, where eb intersects 8 S, draw 14, 11 parallel to AD; and you will have the three perspective squares a 14 15 16 17 18 19 20, and 21 11 e 22, for the bases of the second cross row of objects; namely, the next beyond the first three at A, N, and D.

Through the point w, where CE intersects 8 S, draw a line parallel to BC; and you will have three perspective squares, at B, z, and C, for the bases of the third row of objects; one of which is set up at B.

Through the point x, where fc intersects 8 S, draw a line parallel to bf; and you will have three perspective squares, at b, l, and x, for the bases of the fourth cross row of objects.

Go on in this manner, as you see in the figure, to find the rest of the square bases, up to LM; and you will have 27 upon the whole oblong square plane, on which you are to place the like number of objects, as in fig. 31.

Having assumed AO for the perspective height of the three objects at A, N, and D (fig. 32.) next the observer's eye, and drawn O 18 parallel to AD, in order to make the objects at N and D of the same height as that at O; and having drawn the upright lines 4 15, 7 W, 8 X, and D 22, for the heights at N and D; draw OS and RS, 15 S and WS, XS and 22 S, all to the point of sight S; and these lines will determine the perspective equal heights of all the rest of the upright objects, as shown by the two placed at a and B.

To draw the square tops of these objects, equal and parallel to their bases, we need only give one example, which will serve for all.

Draw 3 R and 2 Q parallel to AO, and up to the line RS; then draw PQ parallel to OR, and OPQR shall be the top of the object at A, equal and parallel to its square base A 1 2 3.—In the same easy way the tops of all the other objects are formed.

When all the rest of the objects are delineated, shade them properly, and the whole perspective scheme will have the appearance of fig. 31.

PROB. 27. To put a square box in perspective, containing a given number of lesser square boxes of a depth equal to their width.

Let the given number of little square boxes or cells Fig. 33. be 16, then 4 of them make the length of each side of the four outer sides ab, bc, cd, da, as in fig. 33. and the depth af is equal to the width ae. Whoever can draw the reticulated square, by the rules laid down towards the beginning of this article, will be at no loss about putting this perspective scheme in practice.

PROB. 28. To put stairs with equal and parallel steps in perspective.

In fig. 34. let ab be the given breadth of each step, Fig. 34. and ai the height thereof. Make bc, cd, de, &c. each equal to ab; and draw all the upright lines ai, bl, cn, dp, &c. perpendicular to ah (to which the horizon rs is parallel) and from the points i, l, n, p, r, &c. draw the equidistant lines iB, lC, nD, &c. parallel to ah; these distances being equal to that of iB from ah.

Draw xi touching all the corner-points i, n, p, r, t, v, and draw 2 16 parallel to xi, as far from it as you want the length of the steps to be.

Toward the point of sight S draw the lines a 1, i 2, k 3, l 4, &c. and draw 16 15, 14 13, 12 11, 10 9, 8 7, 6 5, 4 3, and 2 1, all parallel to ah, and meeting the lines w 15, u 13, t 11, &c. in the points 15, 13, 11, 9, 7, 5, 3, and 1; then from these points draw 15 14, 13 12, 11 10, 9 8, 7 6, 5 4, and 3 2, all parallel to ha; and the outlines of the steps will be finished. From the point 16 draw 16 A parallel to ha, and A x 16 will be part of the flat at the top of the uppermost step.

Fig. 35. This done, shade the work as in fig. 35. and the whole will be finished.

PROB. 29. To put stairs with flats and opening in perspective, standing on a horizontal pavement of squares.

Fig. 36. In fig. 36. having made S the point of sight, and drawn a reticulated pavement AB with black lead lines, which may be rubbed out again; at any distance from the side AB of the pavement which is nearest to the eye, and at any point where you choose to begin the stair at that distance, as a, draw Ga parallel to BA, and take ab at pleasure for the height of each step.

Take ab in your compasses, and set that extent as many times upward from F to E as is equal to the first required number of steps O, N, M, L, K; and from these points of division in EF draw 1b, 2d, 3f, 4h, and Ek, all equidistant from one another, and parallel to Fa: then draw the equidistant upright lines ab, td, uf, vh, wk, and lm, all perpendicular to Fa: then draw mb, touching the outer corners of these steps at m, k, h, f, d, and b; and draw ns parallel to mb, as far from it as you want the length of the steps K, L, M, N, O to be.

Towards the point of sight S drawn mn, l5, ko, i6, hp, fq, dr, and br. Then (parallel to the bottom-line BA) through the points n, o, p, q, r, s, draw n8; 5, 14; 6, 15; 7, 16; 1, 17; and 2s: which done, draw n5 and o6 parallel to lm, and the outlines of the steps K, L, M, N, O will be finished.

At equal distances with that between the lines marked 8 and 14, draw the parallel lines above marked 9 to 11, 12 and 13; and draw perpendicular lines upwards from the points n, o, p, q, r, s, as in the figure.

Make Hm equal to the intended breadth of the flat above the square opening at the left hand, and draw HW toward the point of sight S, equal to the intended length of the flat; then draw WP parallel to Hm, and the outlines of the flat will be finished.

Take the width of the opening at pleasure, as from F to C, and draw CD equal and parallel to FE. Draw GH parallel to CD, and the short lines marked 33, 34, &c. just even with the parallel lines 1, 2, &c. From the points where these short lines meet CD draw lines toward the point of sight S till they meet DE; then from the points where the lines 38, 39, 40, &c. of the pavement meet Cy, draw upright lines parallel to CD; and the lines which form the opening will be finished.

The steps P, Q, R, S, T, and the flat U above the arch V, are done in the same manner with those in fig. 34. as taught in Prob. 28. and the equidistant parallel lines marked 18, 19, &c. are directly even with those on the left-hand side of the arch V, and the upright lines on the right-hand side are equidistant with those on the left.

From the points where the lines 18, 19, 20, &c. meet the right-hand side of the arch, draw lines toward the point of sight S; and from the points where the pavement lines 29, 30, 31, 32, meet the line drawn from A towards the point of sight, draw upright lines toward the top of the arch.

Having done the top of the arch, as in the figure, and the few steps to the right hand thereof, shade the whole as in fig. 37. and the work will be finished.

Fig. 37.

PROB. 30. To put upright conical objects in perspective, as if standing on the sides of an oblong square, at distances from one another equal to the breadth of the oblong.

In fig. 38. the bases of the upright cones are perspective circles, inscribed in squares of the same diameter; and the cones are set upright on their bases by the same rules as are given for pyramids, which we need not repeat here.

In most of the foregoing operations we have considered the observer's eye to be above the level of the tops of all the objects, as if he viewed them when standing on high ground. In this figure, and in fig. 41. and fig. 42. we shall suppose him to be standing on low ground, and the tops of the objects to be above the level of his eye.

In fig. 38. let AD be the perspective breadth of the oblong square ABCD; and let Aa and Dd (equal to Aa) be taken for the diameters of the circular bases of the two cones next the eye, whose intended equal heights shall be AE and DF.

Having made S the point of sight in the horizon parallel to AD, and found the proper point of distance therein, draw AS and aS to contain the bases of the cones on the left-hand side, and DS and dS for those on the right.

Having made the two first cones at A and D of equal height at pleasure, draw ES and FS from their tops to the point of sight, for limiting the perspective heights of all the rest of the cones. Then divide the parallelogram ABCD into as many equal perspective squares as you please; find the bases of the cones at the corners of these squares, and make the cones thereon, as in the figure.

If you would represent a ceiling equal and parallel to ABCD, supported on the tops of these cones, draw EF, then EFGH shall be the ceiling; and by drawing ef parallel to EF, you will have the thickness of the floor-boards and beams, which may be what you please.

This shows how any number of equidistant pillars may be drawn of equal heights to support the ceiling of a long room, and how the walls of such a room may be represented in perspective at the backs of these pillars. It also shows how a street of houses may be drawn in perspective.

PROB. 31. To put a square hollow in perspective, the depth of which shall bear any assigned proportion to its width.

Fig. 41. is the representation of a square hollow, of which the depth AG is equal to three times its width AD; and S is the point of sight over which the observer's eye is supposed to be placed, looking perpendicularly down into it, but not directly over the middle.

Draw AS and DS to the point of sight S; make ST the horizon parallel to AD, and produce it to such a length beyond T that you may find a point of distance therein not nearer S than if AD was seen under an angle of 60 degrees.

Draw DU to the point of distance, intersecting AS in B; then from the point B draw BC parallel to AD; and

and you will have the first perspective square ABCD, equal to a third part of the intended depth.

Draw CV to the point of distance, intersecting AS in E; then from the point E draw EF parallel to AD; and you will have the second perspective square BEFC, which, added to the former one, makes two-thirds of the intended depth.

Draw FW to the point of distance, intersecting AS in G; then from the point G draw GH parallel to AD; and you will have the third perspective square EGHF, which, with the former two, makes the whole depth AGHD three times as great as the width AD, in a perspective view.

Divide AD into any number of equal parts, as suppose 8; and from the division-points $a,\; b,\; c,\; d,\; \backslash dots$ draw lines toward the point of sight S, and ending at GH; then through the points where the diagonals BD, EC, GF, cut these lines, draw lines parallel to AD; and you will have the parallelogram AGHD reticulated, or divided into 192 small and equal perspective squares.

Make AI and DM equal and perpendicular to AD; then draw IM, which will be equal and parallel to AD; and draw IS and MS to the point of sight S.

Divide AI, IM, and MD, into the same number of equal parts as AD is divided; and from these points of division draw lines toward the point of sight S, ending respectively at GK, KL, and LH.

From those points where the lines parallel to AD meet AG and DH draw upright lines parallel to AI and DM; and from the points where these lines meet IK and LM draw lines parallel to IM; then shade the work, as in the figure.

PROB. 32. To represent a semicircular arch in perspective, as if it were standing on two upright walls, equal in height to the height of the observer's eye.

After having gone through the preceding operation, this will be more easy by a bare view of fig. 42. than it could be made by any description; the method being so much like that of drawing and shading the square hollow.—We need only mention, that $aTbEA$ and $DFcd$ are the upright walls on which the semicircular arch is built; that S is the point of sight in the horizon $Tt$, taken in the centre of the arch; and $d$ in fig. 41. is the point of distance; and that the two perspective squares ABCD and BEFC make the parallelogram AEFD of a length equal to twice its breadth AD.

PROB. 33. To represent a square in perspective, as viewed by an observer standing directly even with one of its corners.

In fig. 43. let $A9BC$ be a true square, viewed by an observer standing at some distance from the corner C, and just even with the diagonal C9.

Let $\backslash rho\; SP$ be the horizon, parallel to the diagonal AB; and S the point of sight, even with the diagonal C9. Here it will be proper to have two points of distance $\backslash rho$ and P, equidistant from the point of sight S.

Draw the straight line 1 17 parallel to AB, and draw A 8 and B 10 parallel to CS. Take the distance between 8 and 9 in your compasses, and set it off all the way in equal parts from 8 to 1, and from 10 to 17.—The line 1 17 should be produced a good way further,

both to right and left hand from 9, and divided all the way in the same manner.

From these points of equal division, 8, 9, 10, &c. draw lines to the point of sight S, and also to the two points of distance $\backslash rho$ and P, as in the figure.

Now it is plain, that $acbg$ is the perspective representation of $A9BC$, viewed by an observer even with the corner C and diagonal C9.—But if there are other such squares lying even with this, and having the same position with respect to the line 1 17, it is evident that the observer, who stands directly even with the corner C of the first square, will not be even with the like corners G and K of the others; but will have an oblique view of them, over the sides FG and IK, which are nearest his eye: and their perspective representations will be $egf6$ and $hki3$, drawn among the lines in the figure: of which the spaces taken up by each side lie between three of the lines drawn toward the point of distance $\backslash rho$, and three drawn to the other point of distance P.

PROB. 34. To represent a common chair, in an oblique perspective view.

The original lines to the point of sight S, and points of distance $\backslash rho$ and P, being drawn as in the preceding operation, choose any part of the plane, as $lmn13$, on which you would have the chair I to stand.—There are just as many lines (namely two) between $l$ and $m$ or $13$ and $n$, drawn toward the point of distance $\backslash rho$, at the left hand, as between $l$ and $13$, or $m$ and $n$, drawn to the point of distance P on the right: so that $lm,\; mn,\; n13$, and $13l$, form a perspective square.

From the four corners $l,\; m,\; n,\; 13$, of this square raise the four legs of the chair to the perspective perpendicular height you would have them: then make the seat of the chair a square equal and parallel to $lmn13$, as taught in Prob. 18. which will make the two sides of the seat in the direction of the lines drawn toward the point of distance $\backslash rho$, and the fore and back part of the seat in direction of the lines drawn to the other point of distance P. This done, draw the back of the chair leaning a little backward, and the cross bars therein tending toward the point of distance P. Then shade the work as in the figure; and the perspective chair will be finished.

PROB. 35. To present an oblong square table in an oblique perspective view.

In fig. 43. M is an oblong square table, as seen by Fig. 43. an observer standing directly even with C9 (see Prob. 33.), the side next the eye being perspective parallel to the side $ac$ of the square $acbg$.—The forementioned lines drawn from the line 1 17 to the two points of distance $\backslash rho$ and P, form equal perspective squares on the ground plane.

Choose any part of this plane of squares for the feet of the table to stand upon; as at $\backslash rho,\; q,\; r,\; s$, in direction of the lines $\backslash rho\; p$ and $rs$ for the two long sides, and $ts$ and $qr$ for the two ends; and you will have the oblong square or parallelogram $qrst$ for the part of the floor or ground-plane whereon the table is to stand: and the breadth of this plane is here taken in proportion to the length as 6 to 10; so that, if the length of the table be ten feet, its breadth will be six.

On the four little perspective squares at $q,\; r,\; s$, and

place the four upright legs of the table, of what height you please, so that the height of the two next the eye, at $o$ and $p$, shall be terminated by a straight line $uv$ drawn to the point of distance $P$. This done, make the leaf $M$ of the table an oblong square, perspective equal and parallel to the oblong square $qrst$ on which the feet of the table stand. Then shade the whole, as in the figure, and the work will be finished.

If the line 17 were prolonged to the right and left hand, and equally divided throughout (as it is from 1 to 17), and if the lines which are drawn from $p$ and $P$ to the right and left hand sides of the plate were prolonged till they came to the extended line 17, they would meet it in the equal points of division. In forming large plans of this sort, the ends of slips of paper may be pasted to the right and left edges of the sheet on which the plan is to be formed.

Of the Anamorphosis, or reformation of distorted images.

By this means pictures that are so mishapen, as to exhibit no regular appearance of any thing to the naked eye, shall, when viewed by reflection, present a regular and beautiful image. The inventor of this ingenious device is not known. Simon Stevinus, who was the first that wrote upon it, does not inform us from whom he learned it. The principles of it are laid down by S. Vauzelard in his Perspective Conique et Cylindrique; and Gaspar Schott professes to copy Marius Bettinus in his description of this piece of artificial magic.

It will be sufficient for our purpose to copy one of the simplest figures of this writer, as by this means the mystery of this art will be sufficiently unfolded. Upon the cylinder of paper, or pasteboard, $ABCD$, fig. 44, draw whatever is intended to be exhibited, as the letters $IHS$. Then with a needle make perforations along the whole outline; and placing a candle, $G$, behind this cylinder, mark upon the ground plane the shadow of them, which will be distorted more or less, according to the position of the candle or the plane, &c. This being done, let the picture be an exact copy of this distorted image, let a metallic speculum be substituted in the place of the cylinder, and let the eye of the spectator have the same position before the cylinder that the candle had behind it. Then looking upon the speculum, he will see the distorted image restored to its proper shape. The reformation of the image, he says, will not easily be made exact in this method, but it will be sufficiently so to answer the purpose.

Other methods, more exact and geometrical than this, were found out afterwards: so that these pictures could be drawn by certain rules, without the use of a candle. Schott quotes one of these methods from Bettinus, another from Herigonius, and another from Kircher, which may be seen in his Magia, vol. i. p. 162, &c. He also gives an account of the methods of re-forming pictures by speculums of conical and other figures.

Instead of copying any of these methods from Schott or Bettinus, we shall present our readers with that which Dr Smith hath given us in his Optics, vol. i. p. 250, as, no doubt, the best, and from which any person may easily make a drawing of this kind. The same description answers to two mirrors, one of which, fig. 39, is convex, and the other, fig. 40, is concave.

In order to paint upon a plane a deformed copy $ABCDEKHGF$ of an original picture, which shall appear regular, when seen from a given point $O$, elevated above the plane, by rays reflected from a polished cylinder, placed upon the circle $lnp$, equal to its given base; from the point $R$, which must be supposed to lie perpendicularly under $O$, the place of the eye, draw two lines $Ra,\; Re$; which shall either touch the base of the cylinder, or else cut off two small equal segments from the sides of it, according as the copy is intended to be more or less deformed. Then, taking the eye, raised above $R$, to the given height $RO$, somewhat greater than that of the cylinder, for a luminous point, describe the shadow $ackf$ (of a square, fig. 39, or parallelogram standing upright upon $ackf$ as a base, and containing the picture required) anywhere behind the arch $lnp$. Let the lines drawn from $R$ to the extremities and divisions of the base $a,\; b,\; c,\; d,\; e$, cut the remotest part of the shadow in the points $f,\; g,\; h,\; i,\; k$, and the arch of the base in $l,\; m,\; n,\; o,\; p$; from which points draw the lines $lAf,\; mBg,\; nCh,\; oDi,\; pEk$, as if they were rays of light that came from the focus $R$, and were reflected from the base $lnp$; so that each couple, $lA,\; lR$, produced, may cut off equal segments from the circle. Lastly, Transfer the lines and all their parts in the same order, upon the respective lines and having drawn regular curves, by estimation, through the points $A,\; B,\; C,\; D,\; E$, through $F,\; G,\; H,\; I,\; K$, and through every intermediate order of points; the figure $ACEKHGF$, so divided, will be the deformed copy of the square, drawn and divided upon the original picture, and will appear similar to it, when seen in the polished cylinder, placed upon the base $lnp$, by the eye in its given place $O$.

The practical methods of drawing these images seem to have been carried to the greatest perfection by J. Leopold, who, in the Acta Liphenfis for the year 1712, has described two machines, one for the images to be viewed with a cylindrical, and the other with a conical mirror. The person possessed of this instrument has nothing to do but to take any print he pleases, and while he goes over the outlines of it with one pen, another traces the anamorphosis.

By methods of this kind, groves of trees may be cut, so as to represent the appearance of men, horses, and other objects from some one point of view, which are not at all discernible in any other. This might easily be effected by one person placing himself in any particular situation, and giving directions to other persons what trees to lop, and in what manner. In the same method it has been contrived, that buildings of circular and other forms, and also whole groups of buildings consisting of walls at different distances and with different positions to one another, should be painted so as to exhibit the exact representation of particular objects, which could only be perceived in one situation. Bettinus has illustrated this method by drawings in his Apianis.

It may appear a bold assertion to say, that the very short sketch now given of the art of perspective is a sufficient foundation for the whole practice, and includes all the expeditions rules peculiar to the problems which most generally occur. It is, however, true, and the intelligent

telligent reader will see, that the two theorems on which the whole rests, include every possible case, and apply with equal facility to pictures and originals in any position, although the examples are selected of perpendicular pictures, and of originals referred to horizontal planes, as being the most frequent. The scientific foundation being so simple, the structure need not be complex, nor swell into such volumes as have been published on the subject: volumes which by their size deter from the perusal, and give the simple art the appearance of intricate mystery; and by their prices, defeat the design of their authors, viz. the dissemination of knowledge among the practitioners. The treatises on perspective acquire their bulk by long and tedious discourses, minute explanations of common things, or by great numbers of examples; which indeed do make some of these books valuable by the variety of curious cuts, but do not at all instruct the reader by any improvements made in the art itself. For it is evident that most of those who have treated this subject have been more conversant in the practice of designing than in the principles of geometry; and therefore when, in their practice, the cases which have offered have put them on trying particular expedients, they have thought them worth communicating to the public as improvements in the art; and each author, fond of his own little expedient, (which a scientific person would have known for an easy corollary from the general theorem), has made it the principle of a practical system—in this manner narrowing instead of enlarging the knowledge of the art; and the practitioner tired of the bulk of the volume, in which a single maxim is tediously spread out, and the principle on which it is founded kept out of his sight, contents himself with a remembrance of the maxim (not understood), and keeps it slightly in his eye to avoid gross errors. We can appeal to the whole body of painters and draughtsmen for the truth of this assertion; and it must not be considered as an imputation on them of remissness or negligence, but as a necessary consequence of the ignorance of the authors from whom they have taken their information. This is a strong term, but it is not the least just. Several mathematicians of eminence have written on perspective, treating it as the subject of pure geometry, as it really is; and the performances of Dr Brook Taylor, Gravefande, Wolf, De la Caille, Emerson, are truly valuable, by presenting the art in all its perspicuous simplicity and universality. The works of Taylor and Emerson are more valuable, on account of the very ingenious and expeditious constructions which they have given, suited to every possible case. The merit of the first author has been universally acknowledged by all the British writers on the subject, who never fail to declare that their own works are composed on the principle of Dr Brook Taylor; but any man of science will see that these authors have either not understood them, or aimed at pleasing the public by fine cuts and uncommon cases; for without exception, they have omitted his favourite constructions, which had gained his predilection by their universality, and attached themselves to inferior methods, more usually expedient perhaps, or inventions (as they thought) of their own. What has been given in this article is not professed to be according to the principles of Dr Brook Taylor, because the principles are not peculiar to him, but the necessary results of the theory itself, and incal-

cated by every mathematician who had taken the trouble to consider the subject. They are sufficient not only for directing the ordinary practice but also for suggesting modes of construction for every case out of the common track. And a person of ingenuity will have a laudable enjoyment in this, without much stretch of thought, inventing rules for himself; and will be better pleased with such fruits of his own ingenuity, than in reading the tedious explanation of examples devised by another. And for this purpose we would, with Dr Taylor, "advise all our readers not to be contented with the scheme they find here; but, on every occasion, to draw new ones of their own, in all the variety of circumstances they can think of. This will take up more time at first, but they will find the vast benefit and pleasure of it by the extensive notions it will give them of the nature of the principles."

The art of perspective is necessary to all arts where there is any occasion for designing; as architecture, fortification, carving, and generally all the mechanical arts; but it is more particularly necessary to the art of painting, which can do nothing without it. A figure in a picture, which is not drawn according to the rules of perspective, does not represent what is intended, but something else. Indeed we hesitate not to say, that a picture which is faulty in this particular, is as blameable, or more so, than any composition in writing which is faulty in point of orthography, or grammar. It is generally thought very ridiculous to pretend to write a heroic poem, or a fine discourse, upon any subject, without understanding the propriety of the language in which we write; and to us it seems no less ridiculous for one to pretend to make a good picture without understanding perspective: Yet how many pictures are there to be seen, that are highly valuable in other respects, and yet are entirely faulty in this point? Indeed this fault is so very general, that we cannot remember that we ever have seen a picture that has been entirely without it; and what is the more to be lamented, the greatest masters have been the most guilty of it. Those examples make it to be the less regarded; but the fault is not the less, but the more to be lamented, and deserves the more care in avoiding it for the future. The great occasion of this fault, is certainly the wrong method that is generally used in educating of persons in this art: for the young people are generally put immediately to drawing; and when they have acquired a facility in that, they are put to colouring. And these things they learn by rote, and by practice only; but are not at all instructed in any rules of art. By which means, when they come to make any designs of their own, though they are very expert at drawing out and colouring every thing that offers itself to their fancy; yet for want of being instructed in the strict rules of art, they do not know how to govern their inventions with judgement, and become guilty of so many gross mistakes; which prevent themselves, as well as others, from finding that satisfaction they otherwise would do in their performances. To correct this for the future, we would recommend it to the masters of the art of painting, to consider if it would not be necessary to establish a better method for the education of their scholars, and to begin their instructions with the technical parts of painting, before they let them loose to follow the inventions of their own uncultivated imaginations.

The art of painting, taken in its full extent, consists of two parts; the inventive, and the executive. The inventive part is common with poetry, and belongs more properly and immediately to the original design (which it invents and disposes in the most proper and agreeable manner) than to the picture, which is only a copy of that design already formed in the imagination of the artist. The perfection of this art of painting depends upon the thorough knowledge the artist has of all the parts of his subject; and the beauty of it consists in the happy choice and disposition that he makes of it: And it is in this that the genius of the artist discovers and shows itself, while he indulges and humours his fancy, which here is not confined. But the other, the executive part of painting, is wholly confined and strictly tied to the rules of art, which cannot be dispensed with upon any account; and therefore in this the artist ought to govern himself entirely by the rules of art, and not to take any liberties whatsoever. For any thing that is not truly drawn according to the rules of perspective, or not truly coloured or truly shaded, does not appear to be what the artist intended, but something else. Wherefore, if at any time the artist happens to imagine that his picture would look the better, if he should swerve a little from these rules, he may assure himself, that the fault belongs to his original design, and not to the strictness of the rules; for what is perfectly agreeable and just in the real original objects themselves, can never

appear defective in a picture where those objects are exactly copied.

Therefore to offer a short hint of thoughts we have some time had upon the method which ought to be followed in instructing a scholar in the executive part of painting: we would first have him learn the most common effects of practical geometry, and the first elements of plain geometry and common arithmetic. When he is sufficiently perfect in these, we would have him learn perspective. And when he has made some progress in this, so as to have prepared his judgement with the right notions of the alterations that figures must undergo, when they come to be drawn on a flat, he may then be put to drawing by view, and be exercised in this along with perspective, till he comes to be sufficiently perfect in both. Nothing ought to be more familiar to a painter than perspective; for it is the only thing that can make the judgement correct, and will help the fancy to invent with ten times the ease that it could do without it.

We earnestly recommend to our readers the careful perusal of Dr Taylor's Treatise, as published by Colson in 1749, and Emerson's published along with his Optics. They will be surprised and delighted with the instruction they will receive; and will then truly estimate the splendid volumes of other authors and see their frivolity.

PERSPECTIVE is also used for a kind of picture, or painting, frequently seen in gardens, and at the ends of galleries; designed expressly to deceive the sight by representing the continuation of an alley, a building, landscape, or the like.

Aerial PERSPECTIVE, is sometimes used as a general denomination for that which more restrictedly is called aerial perspective, or the art of giving a due diminution or degradation to the strength of light, shade, and colours of objects, according to their different distances, the quantity of light which falls upon them, and the medium through which they are seen; the chiaro oscuro, or clair obscur, which consists in expressing the different degrees of light, shade, and colour of bodies, arising from their own shape, and the position of their parts with respect to the eye and neighbouring objects, whereby their light or colours are affected; and keeping, which is the observance of a due proportion in the general light and colouring of the whole picture, so that no light or colour in one part may be too bright or strong for another. A painter, who would succeed in aerial perspective, ought carefully to study the effects which distance, or different degrees or colours of light, have on each particular original colour, to know how its hue or strength is changed in the several circumstances that occur, and to represent it accordingly. As all objects in a picture take their measures in proportion to those placed in the front, so, in aerial perspective, the strength of light, and the brightness of the colours of objects close to the picture, must serve as a measure, with respect to which all the same colours at several dis-

stances must have a proportional degradation in like circumstances.

Bird's eye view in PERSPECTIVE, is that which supposes the eye to be placed above any building, &c. as in the air at a considerable distance from it. This is applied in drawing the representations of fortifications, when it is necessary not only to exhibit one view as seen from the ground, but so much of the several buildings as the eye can possibly take in at one time from any situation. In order to this, we must suppose the eye to be removed a considerable height above the ground, and to be placed as it were in the air, so as to look down into the building like a bird that is flying. In representations of this kind, the higher the horizontal line is placed, the more of the fortification will be seen, and vice versa.

PERSPECTIVE Machine, is an instrument by which any person, without the help of the rules of art, may delineate the true perspective figures of objects. Mr Ferguson has described a machine of this sort of which he ascribes the invention to Dr Bevis.

Fig. 45. is a plan of this machine, and fig. 46. is a representation of it when made use of in drawing distant objects in perspective.

In fig. 45. abef is an oblong square board, represented by ABEF in fig. 46. x and y (X and Y) are two hinges on which the part clid (CLID) is moveable. This part consists of two arches or portions of circles cml (CML) and dnl (DNL) joined together at the top (L), and at bottom to the cross bar dc (DC), to which one part of each hinge is fixed, and the other part