XXXIII

THE REDUCED DISTANCE. HOW TO PROCEED WHEN THE POINT OF DISTANCE IS INACCESSIBLE

It has already been shown that too near a point of distance is objectionable on account of the distortion and disproportion resulting from it. At the same time, the long distance-point must be some way out of the picture and therefore inconvenient. The object of the reduced distance is to bring that point within the picture.



In Fig. 85 we have made the distance nearly twice the length of the base of the picture, and consequently a long way out of it. Draw Sa, Sb, and from a draw aD to point of distance, which cuts Sb at o, and determines the depth of the square acob. But we can find that same point if we take half the base and draw a line from 1/2 base to 1/2 distance. But even this 1/2 distance-point does not come inside the picture, so we take a fourth of the base and a fourth of the distance and draw a line from 1/4 base to 1/4 distance. We shall find that it passes precisely through the same point o as the other lines aD, &c. We are thus able to find the required point o without going outside the picture.

Of course we could in the same way take an 8th or even a 16th distance, but the great use of this reduced distance, in addition to the above, is that it enables us to measure any depth into the picture with the greatest ease.

It will be seen in the next figure that without having to extend the base, as is usually done, we can multiply that base to any amount by making use of these reduced distances on the horizontal line. This is quite a new method of proceeding, and it will be seen is mathematically correct.



XXXIV

HOW TO DRAW A LONG PASSAGE OR CLOISTER BY MEANS OF THE REDUCED DISTANCE



In Fig. 86 we have divided the base of the first square into four equal parts, which may represent so many feet, so that A4 and Bd being the retreating sides of the square each represents 4 feet. But we found point 1/4 D by drawing 3D from 1/4 base to 1/4 distance, and by proceeding in the same way from each division, A, 1, 2, 3, we mark off on SB four spaces each equal to 4 feet, in all 16 feet, so that by taking the whole base and the 1/4 distance we find point O, which is distant four times the length of the base AB. We can multiply this distance to any amount by drawing other diagonals to 8th distance, &c. The same rule applies to this corridor (Fig. 87 and Fig. 88).







XXXV

HOW TO FORM A VANISHING SCALE THAT SHALL GIVE THE HEIGHT, DEPTH, AND DISTANCE OF ANY OBJECT IN THE PICTURE

If we make our scale to vanish to the point of sight, as in Fig. 89, we can make SB, the lower line thereof, a measuring line for distances. Let us first of all divide the base AB into eight parts, each part representing 5 feet. From each division draw lines to 8th distance; by their intersections with SB we obtain measurements of 40, 80, 120, 160, &c., feet. Now divide the side of the picture BE in the same manner as the base, which gives us the height of 40 feet. From the side BE draw lines 5S, 15S, &c., to point of sight, and from each division on the base line also draw lines 5S, 10S, 15S, &c., to point of sight, and from each division on SB, such as 40, 80, &c., draw horizontals parallel to base. We thus obtain squares 40 feet wide, beginning at base AB and reaching as far as required. Note how the height of the flagstaff, which is 140 feet high and 280 feet distant, is obtained. So also any buildings or other objects can be measured, such as those shown on the left of the picture.





XXXVI

MEASURING SCALE ON GROUND

A simple and very old method of drawing buildings, &c., and giving them their right width and height is by means of squares of a given size, drawn on the ground.



In the above sketch (Fig. 90) the squares on the ground represent 3 feet each way, or one square yard. Taking this as our standard measure, we find the door on the left is 10 feet high, that the archway at the end is 21 feet high and 12 feet wide, and so on.





Fig. 91 is a sketch made at Sandwich, Kent, and shows a somewhat similar subject to Fig. 84, but the irregularity and freedom of the perspective gives it a charm far beyond the rigid precision of the other, while it conforms to its main laws. This sketch, however, is the real artist's perspective, or what we might term natural perspective.



XXXVII

APPLICATION OF THE REDUCED DISTANCE AND THE VANISHING SCALE TO DRAWING A LIGHTHOUSE, &C.

[Above illustration: Perspective of a lighthouse 135 feet high at 800 feet distance.]



In the drawing of Honfleur (Fig. 92) we divide the base AB as in the previous figure, but the spaces measure 5 feet instead of 3 feet: so that taking the 8th distance, the divisions on the vanishing line BS measure 40 feet each, and at point O we have 400 feet of distance, but we require 800. So we again reduce the distance to a 16th. We thus multiply the base by 16. Now let us take a base of 50 feet at f and draw line fD to 16th distance; if we multiply 50 feet by 16 we obtain the 800 feet required.

The height of the lighthouse is found by means of the vanishing scale, which is 15 feet below and 15 feet above the horizon, or 30 feet from the sea-level. At L we raise a vertical LM, which shows the position of the lighthouse. Then on that vertical measure the height required as shown in the figure.

The 800 feet could be obtained at once by drawing line fD, or 50 feet, to 16th distance. The other measurements obtained by 8th distance serve for nearer buildings.



XXXVIII

HOW TO MEASURE LONG DISTANCES SUCH AS A MILE OR UPWARDS

The wonderful effect of distance in Turner's pictures is not to be achieved by mere measurement, and indeed can only be properly done by studying Nature and drawing her perspective as she presents it to us. At the same time it is useful to be able to test and to set out distances in arranging a composition. This latter, if neglected, often leads to great difficulties and sometimes to repainting.

To show the method of measuring very long distances we have to work with a very small scale to the foot, and in Fig. 94 I have divided the base AB into eleven parts, each part representing 10 feet. First draw AS and BS to point of sight. From A draw AD to 1/4 distance, and we obtain at 440 on line BS four times the length of AB, or 110 feet x 4 = 440 feet. Again, taking the whole base and drawing a line from S to 8th distance we obtain eight times 110 feet or 880 feet. If now we use the 16th distance we get sixteen times 110 feet, or 1,760 feet, one-third of a mile; by repeating this process, but by using the base at 1,760, which is the same length in perspective as AB, we obtain 3,520 feet, and then again using the base at 3,520 and proceeding in the same way we obtain 5,280 feet, or one mile to the archway. The flags show their heights at their respective distances from the base. By the scale at the side of the picture, BO, we can measure any height above or any depth below the perspective plane.



Note.—This figure (here much reduced) should be drawn large by the student, so that the numbering, &c., may be made more distinct. Indeed, many of the other figures should be copied large, and worked out with care, as lessons in perspective.



XXXIX

FURTHER ILLUSTRATION OF LONG DISTANCES AND EXTENDED VIEWS

An extended view is generally taken from an elevated position, so that the principal part of the landscape lies beneath the perspective plane, as already noted, and we shall presently treat of objects and figures on uneven ground. In the previous figure is shown how we can measure heights and depths to any extent. But when we turn to a drawing by Turner, such as the 'View from Richmond Hill', we feel that the only way to accomplish such perspective as this, is to go and draw it from nature, and even then to use our judgement, as he did, as to how much we may emphasize or even exaggerate certain features.



Note in this view the foreground on which the principal figures stand is on a level with the perspective plane, while the river and surrounding park and woods are hundreds of feet below us and stretch away for miles into the distance. The contrasts obtained by this arrangement increase the illusion of space, and the figures in the foreground give as it were a standard of measurement, and by their contrast to the size of the trees show us how far away those trees are.



XL

HOW TO ASCERTAIN THE RELATIVE HEIGHTS OF FIGURES ON AN INCLINED PLANE



The three figures to the right marked f, g, b (Fig. 96) are on level ground, and we measure them by the vanishing scale aS, bS. Those to the left, which are repetitions of them, are on an inclined plane, the vanishing point of which is S'; by the side of this plane we have placed another vanishing scale a'S', b'S', by which we measure the figures on that incline in the same way as on the level plane. It will be seen that if a horizontal line is drawn from the foot of one of these figures, say G, to point O on the edge of the incline, then dropped vertically to o', then again carried on to o'' where the other figure g is, we find it is the same height and also that the other vanishing scale is the same width at that distance, so that we can work from either one or the other. In the event of the rising ground being uneven we can make use of the scale on the level plane.



XLI

HOW TO FIND THE DISTANCE OF A GIVEN FIGURE OR POINT FROM THE BASE LINE



Let P be the given figure. Form scale ACS, S being the point of sight and D the distance. Draw horizontal do through P. From A draw diagonal AD to distance point, cutting do in o, through o draw SB to base, and we now have a square AdoB on the perspective plane; and as figure P is standing on the far side of that square it must be the distance AB, which is one side of it, from the base line—or picture plane. For figures very far away it might be necessary to make use of half-distance.



XLII

HOW TO MEASURE THE HEIGHT OF FIGURES ON UNEVEN GROUND

In previous problems we have drawn figures on level planes, which is easy enough. We have now to represent some above and some below the perspective plane.



Form scale bS, cS; mark off distances 20 feet, 40 feet, &c. Suppose figure K to be 60 feet off. From point at his feet draw horizontal to meet vertical On, which is 60 feet distant. At the point m where this line meets the vertical, measure height mn equal to width of scale at that distance, transfer this to K, and you have the required height of the figure in black.

For the figures under the cliff 20 feet below the perspective plane, form scale FS, GS, making it the same width as the other, namely 5 feet, and proceed in the usual way to find the height of the figures on the sands, which are here supposed to be nearly on a level with the sea, of course making allowance for different heights and various other things.



XLIII

FURTHER ILLUSTRATION OF THE SIZE OF FIGURES AT DIFFERENT DISTANCES AND ON UNEVEN GROUND



Let ab be the height of a figure, say 6 feet. First form scale aS, bS, the lower line of which, aS, is on a level with the base or on the perspective plane. The figure marked C is close to base, the group of three is farther off (24 feet), and 6 feet higher up, so we measure the height on the vanishing scale and also above it. The two girls carrying fish are still farther off, and about 12 feet below. To tell how far a figure is away, refer its measurements to the vanishing scale (see Fig. 96).



XLIV

FIGURES ON A DESCENDING PLANE

In this case (Fig. 100) the same rule applies as in the previous problem, but as the road on the left is going down hill, the vanishing point of the inclined plane is below the horizon at point S'; AS, BS is the vanishing scale on the level plane; and A'S', B'S', that on the incline.

Fig. 101. This is an outline of above figure to show the working more plainly.

Note the wall to the left marked W and the manner in which it appears to drop at certain intervals, its base corresponding with the inclined plane, but the upper lines of each division being made level are drawn to the point of sight, or to their vanishing point on the horizon; it is important to observe this, as it aids greatly in drawing a road going down hill.









XLV

FURTHER ILLUSTRATION OF THE DESCENDING PLANE

In the centre of this picture (Fig. 102) we suppose the road to be descending till it reaches a tunnel which goes under a road or leads to a river (like one leading out of the Strand near Somerset House). It is drawn on the same principle as the foregoing figure. Of course to see the road the spectator must get pretty near to it, otherwise it will be out of sight. Also a level plane must be shown, as by its contrast to the other we perceive that the latter is going down hill.



XLVI

FURTHER ILLUSTRATION OF UNEVEN GROUND

An extended view drawn from a height of about 30 feet from a road that descends about 45 feet.



In drawing a landscape such as Fig. 103 we have to bear in mind the height of the horizon, which being exactly opposite the eye, shows us at once which objects are below and which are above us, and to draw them accordingly, especially roofs, buildings, walls, hedges, &c.; also it is well to sketch in the different fields figures of men and cattle, as from the size of these we can judge of the rest.



XLVII

THE PICTURE STANDING ON THE GROUND

Let K represent a frame placed vertically and at a given distance in front of us. If stood on the ground our foreground will touch the base line of the picture, and we can fix up a standard of measurement both on the base and on the side as in this sketch, taking 6 feet as about the height of the figures.





XLVIII

THE PICTURE ON A HEIGHT

If we are looking at a scene from a height, that is from a terrace, or a window, or a cliff, then the near foreground, unless it be the terrace, window-sill, &c., would not come into the picture, and we could not see the near figures at A, and the nearest to come into view would be those at B, so that a view from a window, &c., would be as it were without a foreground. Note that the figures at B would be (according to this sketch) 30 feet from the picture plane and about 18 feet below the base line.





BOOK THIRD

XLIX

ANGULAR PERSPECTIVE

Hitherto we have spoken only of parallel perspective, which is comparatively easy, and in our first figure we placed the cube with one of its sides either touching or parallel to the transparent plane. We now place it so that one angle only (ab), touches the picture.



Its sides are no longer drawn to the point of sight as in Fig. 7, nor its diagonal to the point of distance, but to some other points on the horizon, although the same rule holds good as regards their parallelism; as for instance, in the case of bc and ad, which, if produced, would meet at V, a point on the horizon called a vanishing point. In this figure only one vanishing point is seen, which is to the right of the point of sight S, whilst the other is some distance to the left, and outside the picture. If the cube is correctly drawn, it will be found that the lines ae, bg, &c., if produced, will meet on the horizon at this other vanishing point. This far-away vanishing point is one of the inconveniences of oblique or angular perspective, and therefore it will be a considerable gain to the draughtsman if we can dispense with it. This can be easily done, as in the above figure, and here our geometry will come to our assistance, as I shall show presently.



L

HOW TO PUT A GIVEN POINT INTO PERSPECTIVE

Let us place the given point P on a geometrical plane, to show how far it is from the base line, and indeed in the exact position we wish it to be in the picture. The geometrical plane is supposed to face us, to hang down, as it were, from the base line AB, like the side of a table, the top of which represents the perspective plane. It is to that perspective plane that we now have to transfer the point P.



From P raise perpendicular Pm till it touches the base line at m. With centre m and radius mP describe arc Pn so that mn is now the same length as mP. As point P is opposite point m, so must it be in the perspective, therefore we draw a line at right angles to the base, that is to the point of sight, and somewhere on this line will be found the required point P'. We now have to find how far from m must that point be. It must be the length of mn, which is the same as mP. We therefore from n draw nD to the point of distance, which being at an angle of 45 deg, or half a right angle, makes mP' the perspective length of mn by its intersection with mS, and thus gives us the point P', which is the perspective of the original point.



LI

A PERSPECTIVE POINT BEING GIVEN, FIND ITS POSITION ON THE GEOMETRICAL PLANE

To do this we simply reverse the foregoing problem. Thus let P be the given perspective point. From point of sight S draw a line through P till it cuts AB at m. From distance D draw another line through P till it cuts the base at n. From m drop perpendicular, and then with centre m and radius mn describe arc, and where it cuts that perpendicular is the required point P'. We often have to make use of this problem.





LII

HOW TO PUT A GIVEN LINE INTO PERSPECTIVE

This is simply a question of putting two points into perspective, instead of one, or like doing the previous problem twice over, for the two points represent the two extremities of the line. Thus we have to find the perspective of A and B, namely a'b'. Join those points, and we have the line required.





If one end touches the base, as at A (Fig. 110), then we have but to find one point, namely b. We also find the perspective of the angle mAB, namely the shaded triangle mAb. Note also that the perspective triangle equals the geometrical triangle.



When the line required is parallel to the base line of the picture, then the perspective of it is also parallel to that base (see Rule 3).



LIII

TO FIND THE LENGTH OF A GIVEN PERSPECTIVE LINE

A perspective line AB being given, find its actual length and the angle at which it is placed.

This is simply the reverse of the previous problem. Let AB be the given line. From distance D through A draw DC, and from S, point of sight, through A draw SO. Drop OP at right angles to base, making it equal to OC. Join PB, and line PB is the actual length of AB.

This problem is useful in finding the position of any given line or point on the perspective plane.





LIV

TO FIND THESE POINTS WHEN THE DISTANCE-POINT IS INACCESSIBLE



If the distance-point is a long way out of the picture, then the same result can be obtained by using the half distance and half base, as already shown.

From a, half of mP', draw quadrant ab, from b (half base), draw line from b to half Dist., which intersects Sm at P, precisely the same point as would be obtained by using the whole distance.



LV

HOW TO PUT A GIVEN TRIANGLE OR OTHER RECTILINEAL FIGURE INTO PERSPECTIVE

Here we simply put three points into perspective to obtain the given triangle A, or five points to obtain the five-sided figure at B. So can we deal with any number of figures placed at any angle.



Both the above figures are placed in the same diagram, showing how any number can be drawn by means of the same point of sight and the same point of distance, which makes them belong to the same picture.

It is to be noted that the figures appear reversed in the perspective. That is, in the geometrical triangle the base at ab is uppermost, whereas in the perspective ab is lowermost, yet both are nearest to the ground line.



LVI

HOW TO PUT A GIVEN SQUARE INTO ANGULAR PERSPECTIVE

Let ABCD (Fig. 115) be the given square on the geometrical plane, where we can place it as near or as far from the base and at any angle that we wish. We then proceed to find its perspective on the picture by finding the perspective of the four points ABCD as already shown. Note that the two sides of the perspective square dc and ab being produced, meet at point V on the horizon, which is their vanishing point, but to find the point on the horizon where sides bc and ad meet, we should have to go a long way to the left of the figure, which by this method is not necessary.





LVII

OF MEASURING POINTS

We now have to find certain points by which to measure those vanishing or retreating lines which are no longer at right angles to the picture plane, as in parallel perspective, and have to be measured in a different way, and here geometry comes to our assistance.



Note that the perspective square P equals the geometrical square K, so that side AB of the one equals side ab of the other. With centre A and radius AB describe arc Bm' till it cuts the base line at m'. Now AB = Am', and if we join bm' then triangle BAm' is an isosceles triangle. So likewise if we join m'b in the perspective figure will m'Ab be the same isosceles triangle in perspective. Continue line m'b till it cuts the horizon in m, which point will be the measuring point for the vanishing line AbV. For if in an isosceles triangle we draw lines across it, parallel to its base from one side to the other, we divide both sides in exactly the same quantities and proportions, so that if we measure on the base line of the picture the spaces we require, such as 1, 2, 3, on the length Am', and then from these divisions draw lines to the measuring point, these lines will intersect the vanishing line AbV in the lengths and proportions required. To find a measuring point for the lines that go to the other vanishing point, we proceed in the same way. Of course great accuracy is necessary.

Note that the dotted lines 1,1, 2,2, &c., are parallel in the perspective, as in the geometrical figure. In the former the lines are drawn to the same point m on the horizon.



LVIII

HOW TO DIVIDE ANY GIVEN STRAIGHT LINE INTO EQUAL OR PROPORTIONATE PARTS



Let AB (Fig. 117) be the given straight line that we wish to divide into five equal parts. Draw AC at any convenient angle, and measure off five equal parts with the compasses thereon, as 1, 2, 3, 4, 5. From 5C draw line to 5B. Now from each division on AC draw lines 4,4, 3,3, &c., parallel to 5,5. Then AB will be divided into the required number of equal parts.



LIX

HOW TO DIVIDE A DIAGONAL VANISHING LINE INTO ANY NUMBER OF EQUAL OR PROPORTIONAL PARTS

In a previous figure (Fig. 116) we have shown how to find a measuring point when the exact measure of a vanishing line is required, but if it suffices merely to divide a line into a given number of equal parts, then the following simple method can be adopted.

We wish to divide ab into five equal parts. From a, measure off on the ground line the five equal spaces required. From 5, the point to which these measures extend (as they are taken at random), draw a line through b till it cuts the horizon at O. Then proceed to draw lines from each division on the base to point O, and they will intersect and divide ab into the required number of equal parts.





The same method applies to a given line to be divided into various proportions, as shown in this lower figure.







LX

FURTHER USE OF THE MEASURING POINT O

One square in oblique or angular perspective being given, draw any number of other squares equal to it by means of this point O and the diagonals.

Let ABCD (Fig. 120) be the given square; produce its sides AB, DC till they meet at point V. From D measure off on base any number of equal spaces of any convenient length, as 1, 2, 3, &c.; from 1, through corner of square C, draw a line to meet the horizon at O, and from O draw lines to the several divisions on base line. These lines will divide the vanishing line DV into the required number of parts equal to DC, the side of the square. Produce the diagonal of the square DB till it cuts the horizon at G. From the divisions on line DV draw diagonals to point G: their intersections with the other vanishing line AV will determine the direction of the cross-lines which form the bases of other squares without the necessity of drawing them to the other vanishing point, which in this case is some distance to the left of the picture. If we produce these cross-lines to the horizon we shall find that they all meet at the other vanishing point, to which of course it is easy to draw them when that point is accessible, as in Fig. 121; but if it is too far out of the picture, then this method enables us to do without it.

Figure 121 corroborates the above by showing the two vanishing points and additional squares. Note the working of the diagonals drawn to point G, in both figures.



LXI

FURTHER USE OF THE MEASURING POINT O

Suppose we wish to divide the side of a building, as in Fig. 123, or to draw a balcony, a series of windows, or columns, or what not, or, in other words, any line above the horizon, as AB. Then from A we draw AC parallel to the horizon, and mark thereon the required divisions 5, 10, 15, &c.: in this case twenty-five (Fig. 122). From C draw a line through B till it cuts the horizon at O. Then proceed to draw the other lines from each division to O, and thus divide the vanishing line AB as required.





In this portico there are thirteen triglyphs with twelve spaces between them, making twenty-five divisions. The required number of parts to draw the columns can be obtained in the same way.



LXII

ANOTHER METHOD OF ANGULAR PERSPECTIVE, BEING THAT ADOPTED IN OUR ART SCHOOLS

In the previous method we have drawn our squares by means of a geometrical plan, putting each point into perspective as required, and then by means of the perspective drawing thus obtained, finding our vanishing and measuring points. In this method we proceed in exactly the opposite way, setting out our points first, and drawing the square (or other figure) afterwards.



Having drawn the horizontal and base lines, and fixed upon the position of the point of sight, we next mark the position of the spectator by dropping a perpendicular, S ST, from that point of sight, making it the same length as the distance we suppose the spectator to be from the picture, and thus we make ST the station-point.

To understand this figure we must first look upon it as a ground-plan or bird's-eye view, the line V2V1 or horizon line representing the picture seen edgeways, because of course the station-point cannot be in the picture itself, but a certain distance in front of it. The angle at ST, that is the angle which decides the positions of the two vanishing points V1, V2, is always a right angle, and the two remaining angles on that side of the line, called the directing line, are together equal to a right angle or 90 deg. So that in fixing upon the angle at which the square or other figure is to be placed, we say 'let it be 60 deg and 30 deg, or 70 deg and 20 deg', &c. Having decided upon the station-point and the angle at which the square is to be placed, draw TV1 and TV2, till they cut the horizon at V1 and V2. These are the two vanishing points to which the sides of the figure are respectively drawn. But we still want the measuring points for these two vanishing lines. We therefore take first, V1 as centre and V1T as radius, and describe arc of circle till it cuts the horizon in M1, which is the measuring point for all lines drawn to V1. Then with radius V2T describe arc from centre V2 till it cuts the horizon in M2, which is the measuring point for all vanishing lines drawn to V2. We have now set out our points. Let us proceed to draw the square Abcd. From A, the nearest angle (in this instance touching the base line), measure on each side of it the equal lengths AB and AE, which represent the width or side of the square. Draw EM2 and BM1 from the two measuring points, which give us, by their intersections with the vanishing lines AV1 and AV2, the perspective lengths of the sides of the square Abcd. Join b and V1 and dV2, which intersect each other at C, then Adcb is the square required.

This method, which is easy when you know it, has certain drawbacks, the chief one being that if we require a long-distance point, and a small angle, such as 10 deg on one side, and 80 deg on the other, then the size of the diagram becomes so large that it has to be carried out on the floor of the studio with long strings, &c., which is a very clumsy and unscientific way of setting to work. The architects in such cases make use of the centrolinead, a clever mechanical contrivance for getting over the difficulty of the far-off vanishing point, but by the method I have shown you, and shall further illustrate, you will find that you can dispense with all this trouble, and do all your perspective either inside the picture or on a very small margin outside it.

Perhaps another drawback to this method is that it is not self-evident, as in the former one, and being rather difficult to explain, the student is apt to take it on trust, and not to trouble about the reasons for its construction: but to show that it is equally correct, I will draw the two methods in one figure.



LXIII

TWO METHODS OF ANGULAR PERSPECTIVE IN ONE FIGURE



It matters little whether the station-point is placed above or below the horizon, as the result is the same. In Fig. 125 it is placed above, as the lower part of the figure is occupied with the geometrical plan of the other method.

In each case we make the square K the same size and at the same angle, its near corner being at A. It must be seen that by whichever method we work out this perspective, the result is the same, so that both are correct: the great advantage of the first or geometrical system being, that we can place the square at any angle, as it is drawn without reference to vanishing points.

We will, however, work out a few figures by the second method.



LXIV

TO DRAW A CUBE, THE POINTS BEING GIVEN

As in a previous figure (124) we found the various working points of angular perspective, we need now merely transfer them to the horizontal line in this figure, as in this case they will answer our purpose perfectly well.



Let A be the nearest angle touching the base. Draw AV1, AV2. From A, raise vertical Ae, the height of the cube. From e draw eV1, eV2, from the other angles raise verticals bf, dh, cg, to meet eV1, eV2, fV2, &c., and the cube is complete.



LXV

AMPLIFICATION OF THE CUBE APPLIED TO DRAWING A COTTAGE



Note that we have started this figure with the cube Adhefb. We have taken three times AB, its width, for the front of our house, and twice AB for the side, and have made it two cubes high, not counting the roof. Note also the use of the measuring-points in connexion with the measurements on the base line, and the upper measuring line TPK.



LXVI

HOW TO DRAW AN INTERIOR AT AN ANGLE

Here we make use of the same points as in a previous figure, with the addition of the point G, which is the vanishing point of the diagonals of the squares on the floor.



From A draw square Abcd, and produce its sides in all directions; again from A, through the opposite angle of the square C, draw a diagonal till it cuts the horizon at G. From G draw diagonals through b and d, cutting the base at o, o, make spaces o, o, equal to Ao all along the base, and from them draw diagonals to G; through the points where these diagonals intersect the vanishing lines drawn in the direction of Ab, dc and Ad, bc, draw lines to the other vanishing point V1, thus completing the squares, and so cover the floor with them; they will then serve to measure width of door, windows, &c. Of course horizontal lines on wall 1 are drawn to V1, and those on wall 2 to V2.

In order to see this drawing properly, the eye should be placed about 3 inches from it, and opposite the point of sight; it will then stand out like a stereoscopic picture, and appear as actual space, but otherwise the perspective seems deformed, and the angles exaggerated. To make this drawing look right from a reasonable distance, the point of distance should be at least twice as far off as it is here, and this would mean altering all the other points and sending them a long way out of the picture; this is why artists use those long strings referred to above. I would however, advise them to make their perspective drawing on a small scale, and then square it up to the size of the canvas.



LXVII

HOW TO CORRECT DISTORTED PERSPECTIVE BY DOUBLING THE LINE OF DISTANCE

Here we have the same interior as the foregoing, but drawn with double the distance, so that the perspective is not so violent and the objects are truer in proportion to each other.



To redraw the whole figure double the size, including the station-point, would require a very large diagram, that we could not get into this book without a folding plate, but it comes to the same thing if we double the distances between the various points. Thus, if from S to G in the small diagram is 1 inch, in the larger one make it 2 inches. If from S to M2 is 2 inches, in the larger make it 4, and so on.

Or this form may be used: make AB twice the length of AC (Fig. 130), or in any other proportion required. On AC mark the points as in the drawing you wish to enlarge. Make AB the length that you wish to enlarge to, draw CB, and then from each division on AC draw lines parallel to CB, and AB will be divided in the same proportions, as I have already shown (Fig. 117).

There is no doubt that it is easier to work direct from the vanishing points themselves, especially in complicated architectural work, but at the same time I will now show you how we can dispense with, at all events, one of them, and that the farthest away.





LXVIII

HOW TO DRAW A CUBE ON A GIVEN SQUARE, USING ONLY ONE VANISHING POINT

ABCD is the given square (Fig. 131). At A raise vertical Aa equal to side of square AB', from a draw ab to the vanishing point. Raise Bb. Produce VD to E to touch the base line. From E raise vertical EF, making it equal to Aa. From F draw FV. Raise Dd and Cc, their heights being determined by the line FV. Join da and the cube is complete. It will be seen that the verticals raised at each corner of the square are equal perspectively, as they are drawn between parallels which start from equal heights, namely, from EF and Aa to the same point V, the vanishing point. Any other line, such as OO', can be directed to the inaccessible vanishing point in the same way as ad, &c.

Note. This is only one of many original figures and problems in this book which have been called up by the wish to facilitate the work of the artist, and as it were by necessity.





LXIX

A COURTYARD OR CLOISTER DRAWN WITH ONE VANISHING POINT



In this figure I have first drawn the pavement by means of the diagonals GA, Go, Go, &c., and the vanishing point V, the square at A being given. From A draw diagonal through opposite corner till it cuts the horizon at G. From this same point G draw lines through the other corners of the square till they cut the ground line at o, o. Take this measurement Ao and mark it along the base right and left of A, and the lines drawn from these points o to point G will give the diagonals of all the squares on the pavement. Produce sides of square A, and where these lines are intersected by the diagonals Go draw lines from the vanishing point V to base. These will give us the outlines of the squares lying between them and also guiding points that will enable us to draw as many more as we please. These again will give us our measurements for the widths of the arches, &c., or between the columns. Having fixed the height of wall or dado, we make use of V point to draw the sides of the building, and by means of proportionate measurement complete the rest, as in Fig. 128.



LXX

HOW TO DRAW LINES WHICH SHALL MEET AT A DISTANT POINT, BY MEANS OF DIAGONALS

This is in a great measure a repetition of the foregoing figure, and therefore needs no further explanation.



I must, however, point out the importance of the point G. In angular perspective it in a measure takes the place of the point of distance in parallel perspective, since it is the vanishing point of diagonals at 45 deg drawn between parallels such as AV, DV, drawn to a vanishing point V. The method of dividing line AV into a number of parts equal to AB, the side of the square, is also shown in a previous figure (Fig. 120).



LXXI

HOW TO DIVIDE A SQUARE PLACED AT AN ANGLE INTO A GIVEN NUMBER OF SMALL SQUARES

ABCD is the given square, and only one vanishing point is accessible. Let us divide it into sixteen small squares. Produce side CD to base at E. Divide EA into four equal parts. From each division draw lines to vanishing point V. Draw diagonals BD and AC, and produce the latter till it cuts the horizon in G. Draw the three cross-lines through the intersections made by the diagonals and the lines drawn to V, and thus divide the square into sixteen.



This is to some extent the reverse of the previous problem. It also shows how the long vanishing point can be dispensed with, and the perspective drawing brought within the picture.



LXXII

FURTHER EXAMPLE OF HOW TO DIVIDE A GIVEN OBLIQUE SQUARE INTO A GIVEN NUMBER OF EQUAL SQUARES, SAY TWENTY-FIVE

Having drawn the square ABCD, which is enclosed, as will be seen, in a dotted square in parallel perspective, I divide the line EA into five equal parts instead of four (Fig. 135), and have made use of the device for that purpose by measuring off the required number on line EF, &c. Fig. 136 is introduced here simply to show that the square can be divided into any number of smaller squares. Nor need the figure be necessarily a square; it is just as easy to make it an oblong, as ABEF (Fig. 136); for although we begin with a square we can extend it in any direction we please, as here shown.







LXXIII

OF PARALLELS AND DIAGONALS







To find the centre of a square or other rectangular figure we have but to draw its two diagonals, and their intersection will give us the centre of the figure (see 137 A). We do the same with perspective figures, as at B. In Fig. C is shown how a diagonal, drawn from one angle of a square B through the centre O of the opposite side of the square, will enable us to find a second square lying between the same parallels, then a third, a fourth, and so on. At figure K lying on the ground, I have divided the farther side of the square mn into 1/4, 1/3, 1/2. If I draw a diagonal from G (at the base) through the half of this line I cut off on FS the lengths or sides of two squares; if through the quarter I cut off the length of four squares on the vanishing line FS, and so on. In Fig. 137 D is shown how easily any number of objects at any equal distances apart, such as posts, trees, columns, &c., can be drawn by means of diagonals between parallels, guided by a central line GS.





LXXIV

THE SQUARE, THE OBLONG, AND THEIR DIAGONALS





Having found the centre of a square or oblong, such as Figs. 138 and 139, if we draw a third line through that centre at a given angle and then at each of its extremities draw perpendiculars AB, DC, we divide that square or oblong into three parts, the two outer portions being equal to each other, and the centre one either larger or smaller as desired; as, for instance, in the triumphal arch we make the centre portion larger than the two outer sides. When certain architectural details and spaces are to be put into perspective, a scale such as that in Fig. 123 will be found of great convenience; but if only a ready division of the principal proportions is required, then these diagonals will be found of the greatest use.



LXXV

SHOWING THE USE OF THE SQUARE AND DIAGONALS IN DRAWING DOORWAYS, WINDOWS, AND OTHER ARCHITECTURAL FEATURES

This example is from Serlio's Architecture (1663), showing what excellent proportion can be obtained by the square and diagonals. The width of the door is one-third of the base of square, the height two-thirds. As a further illustration we have drawn the same figure in perspective.







LXXVI

HOW TO MEASURE DEPTHS BY DIAGONALS

If we take any length on the base of a square, say from A to g, and from g raise a perpendicular till it cuts the diagonal AB in O, then from O draw horizontal Og', we form a square AgOg', and thus measure on one side of the square the distance or depth Ag'. So can we measure any other length, such as fg, in like manner.





To do this in perspective we pursue precisely the same method, as shown in this figure (143).

To measure a length Ag on the side of square AC, we draw a line from g to the point of sight S, and where it crosses diagonal AB at O we draw horizontal Og, and thus find the required depth Ag in the picture.



LXXVII

HOW TO MEASURE DISTANCES BY THE SQUARE AND DIAGONAL

It may sometimes be convenient to have a ready method by which to measure the width and length of objects standing against the wall of a gallery, without referring to distance-points, &c.



In Fig. 144 the floor is divided into two large squares with their diagonals. Suppose we wish to draw a fireplace or a piece of furniture K, we measure its base ef on AB, as far from B as we wish it to be in the picture; draw eo and fo to point of sight, and proceed as in the previous figure by drawing parallels from Oo, &c.

Let it be observed that the great advantage of this method is, that we can use it to measure such distant objects as XY just as easily as those near to us.

There is, however, a still further advantage arising from it, and that is that it introduces us to a new and simpler method of perspective, to which I have already referred, and it will, I hope, be found of infinite use to the artist.

Note.—As we have founded many of these figures on a given square in angular perspective, it is as well to have a ready and certain means of drawing that square without the elaborate setting out of a geometrical plan, as in the first method, or the more cumbersome and extended system of the second method. I shall therefore show you another method equally correct, but much simpler than either, which I have invented for our use, and which indeed forms one of the chief features of this book.



LXXVIII

HOW BY MEANS OF THE SQUARE AND DIAGONAL WE CAN DETERMINE THE POSITION OF POINTS IN SPACE

Apart from the aid that perspective affords the draughtsman, there is a further value in it, in that it teaches us almost a new science, which we might call the mystery of aspect, and how it is that the objects around us take so many different forms, or rather appearances, although they themselves remain the same. And also that it enables us, with, I think, great pleasure to ourselves, to fathom space, to work out difficult problems by simple reasoning, and to exercise those inventive and critical faculties which give strength and enjoyment to mental life.

And now, after this brief excursion into philosophy, let us come down to the simple question of the perspective of a point.





Here, for instance, are two aspects of the same thing: the geometrical square A, which is facing us, and the perspective square B, which we suppose to lie flat on the table, or rather on the perspective plane. Line A'C' is the perspective of line AC. On the geometrical square we can make what measurements we please with the compasses, but on the perspective square B' the only line we can actually measure is the base line. In both figures this base line is the same length. Suppose we want to find the perspective of point P (Fig. 146), we make use of the diagonal CA. From P in the geometrical square draw PO to meet the diagonal in O; through O draw perpendicular fe; transfer length fB, so found, to the base of the perspective square; from f draw fS to point of sight; where it cuts the diagonal in O, draw horizontal OP', which gives us the point required. In the same way we can find the perspective of any number of points on any side of the square.



LXXIX

PERSPECTIVE OF A POINT PLACED IN ANY POSITION WITHIN THE SQUARE

Let the point P be the one we wish to put into perspective. We have but to repeat the process of the previous problem, making use of our measurements on the base, the diagonals, &c.



Indeed these figures are so plain and evident that further description of them is hardly necessary, so I will here give two drawings of triangles which explain themselves. To put a triangle into perspective we have but to find three points, such as fEP, Fig. 148 A, and then transfer these points to the perspective square 148 B, as there shown, and form the perspective triangle; but these figures explain themselves. Any other triangle or rectilineal figure can be worked out in the same way, which is not only the simplest method, but it carries its mathematical proof with it.











LXXX

PERSPECTIVE OF A SQUARE PLACED AT AN ANGLE NEW METHOD

As we have drawn a triangle in a square so can we draw an oblique square in a parallel square. In Figure 150 A we have drawn the oblique square GEPn. We find the points on the base Am, as in the previous figures, which enable us to construct the oblique perspective square n'G'E'P' in the parallel perspective square Fig. 150 B. But it is not necessary to construct the geometrical figure, as I will show presently. It is here introduced to explain the method.





Fig. 150 B. To test the accuracy of the above, produce sides G'E' and n'P' of perspective square till they touch the horizon, where they will meet at V, their vanishing point, and again produce the other sides n'G' and P'E' till they meet on the horizon at the other vanishing point, which they must do if the figure is correctly drawn.

In any parallel square construct an oblique square from a given point—given the parallel square at Fig. 150 B, and given point n' on base. Make A'f' equal to n'm', draw f'S and n'S to point of sight. Where these lines cut the diagonal AC draw horizontals to P' and G', and so find the four points G'E'P'n' through which to draw the square.



LXXXI

ON A GIVEN LINE PLACED AT AN ANGLE TO THE BASE DRAW A SQUARE IN ANGULAR PERSPECTIVE, THE POINT OF SIGHT, AND DISTANCE, BEING GIVEN.



Let AB be the given line, S the point of sight, and D the distance (Fig. 151, 1). Through A draw SC from point of sight to base (Fig. 151, 2 and 3). From C draw CD to point of distance. Draw Ao parallel to base till it cuts CD at o, through O draw SP, from B mark off BE equal to CP. From E draw ES intersecting CD at K, from K draw KM, thus completing the outer parallel square. Through F, where PS intersects MK, draw AV till it cuts the horizon in V, its vanishing point. From V draw VB cutting side KE of outer square in G, and we have the four points AFGB, which are the four angles of the square required. Join FG, and the figure is complete.

Any other side of the square might be given, such as AF. First through A and F draw SC, SP, then draw Ao, then through o draw CD. From C draw base of parallel square CE, and at M through F draw MK cutting diagonal at K, which gives top of square. Now through K draw SE, giving KE the remaining side thereof, produce AF to V, from V draw VB. Join FG, GB, and BA, and the square required is complete.

The student can try the remaining two sides, and he will find they work out in a similar way.



LXXXII

HOW TO DRAW SOLID FIGURES AT ANY ANGLE BY THE NEW METHOD

As we can draw planes by this method so can we draw solids, as shown in these figures. The heights of the corners of the triangles are obtained by means of the vanishing scales AS, OS, which have already been explained.





In the same manner we can draw a cubic figure (Fig. 154)—a box, for instance—at any required angle. In this case, besides the scale AS, OS, we have made use of the vanishing lines DV, BV, to corroborate the scale, but they can be dispensed with in these simple objects, or we can use a scale on each side of the figure as a'o'S, should both vanishing points be inaccessible. Let it be noted that in the scale AOS, AO is made equal to BC, the height of the box.



By a similar process we draw these two figures, one on the square, the other on the circle.







LXXXIII

POINTS IN SPACE

The chief use of these figures is to show how by means of diagonals, horizontals, and perpendiculars almost any figure in space can be set down. Lines at any slope and at any angle can be drawn by this descriptive geometry.

The student can examine these figures for himself, and will understand their working from what has gone before. Here (Fig. 157) in the geometrical square we have a vertical plane AabB standing on its base AB. We wish to place a projection of this figure at a certain distance and at a given angle in space. First of all we transfer it to the side of the cube, where it is seen in perspective, whilst at its side is another perspective square lying flat, on which we have to stand our figure. By means of the diagonal of this flat square, horizontals from figure on side of cube, and lines drawn from point of sight (as already explained), we obtain the direction of base line AB, and also by means of lines aa' and bb' we obtain the two points in space a'b'. Join Aa', a'b' and Bb', and we have the projection required, and which may be said to possess the third dimension.



In this other case (Fig. 158) we have a wedge-shaped figure standing on a triangle placed on the ground, as in the previous figure, its three corners being the same height. In the vertical geometrical square we have a ground-plan of the figure, from which we draw lines to diagonal and to base, and notify by numerals 1, 3, 2, 1, 3; these we transfer to base of the horizontal perspective square, and then construct shaded triangle 1, 2, 3, and raise to the height required as shown at 1', 2', 3'. Although we may not want to make use of these special figures, they show us how we could work out almost any form or object suspended in space.





LXXXIV

THE SQUARE AND DIAGONAL APPLIED TO CUBES AND SOLIDS DRAWN THEREIN



As we have made use of the square and diagonal to draw figures at various angles so can we make use of cubes either in parallel or angular perspective to draw other solid figures within them, as shown in these drawings, for this is simply an amplification of that method. Indeed we might invent many more such things. But subjects for perspective treatment will constantly present themselves to the artist or draughtsman in the course of his experience, and while I endeavour to show him how to grapple with any new difficulty or subject that may arise, it is impossible to set down all of them in this book.





LXXXV

TO DRAW AN OBLIQUE SQUARE IN ANOTHER OBLIQUE SQUARE WITHOUT USING VANISHING POINTS

It is not often that both vanishing points are inaccessible, still it is well to know how to proceed when this is the case. We first draw the square ABCD inside the parallel square, as in previous figures. To draw the smaller square K we simply draw a smaller parallel square h h h h, and within that, guided by the intersections of the diagonals therewith, we obtain the four points through which to draw square K. To raise a solid figure on these squares we can make use of the vanishing scales as shown on each side of the figure, thus obtaining the upper square 1 2 3 4, then by means of the diagonal 1 3 and 2 4 and verticals raised from each corner of square K to meet them we obtain the smaller upper square corresponding to K.

It might be said that all this can be done by using the two vanishing points in the usual way. In the first place, if they were as far off as required for this figure we could not get them into a page unless it were three or four times the width of this one, and to use shorter distances results in distortion, so that the real use of this system is that we can make our figures look quite natural and with much less trouble than by the other method.





LXXXVI

SHOWING HOW A PEDESTAL CAN BE DRAWN BY THE NEW METHOD

This is a repetition of the previous problem, or rather the application of it to architecture, although when there are many details it may be more convenient to use vanishing points or the centrolinead.







LXXXVII

SCALE ON EACH SIDE OF THE PICTURE

As one of my objects in writing this book is to facilitate the working of our perspective, partly for the comfort of the artist, and partly that he may have no excuse for neglecting it, I will here show you how you may, by a very simple means, secure the general correctness of your perspective when sketching or painting out of doors.

Let us take this example from a sketch made at Honfleur (Fig. 163), and in which my eye was my only guide, but it stands the test of the rule. First of all note that line HH, drawn from one side of the picture to the other, is the horizontal line; below that is a wall and a pavement marked aV, also going from one side of the picture to the other, and being lower down at a than at V it runs up as it were to meet the horizon at some distant point. In order to form our scale I take first the length of Ha, and measure it above and below the horizon, along the side to our left as many times as required, in this case four or five. I now take the length HV on the right side of the picture and measure it above and below the horizon, as in the other case; and then from these divisions obtain dotted lines crossing the picture from one side to the other which must all meet at some distant point on the horizon. These act as guiding lines, and are sufficient to give us the direction of any vanishing lines going to the same point. For those that go in the opposite direction we proceed in the same way, as from b on the right to V' on the left. They are here put in faintly, so as not to interfere with the drawing. In the sketch of Toledo (Fig. 164) the same thing is shown by double lines on each side to separate the two sets of lines, and to make the principle more evident.





LXXXVIII

THE CIRCLE

If we inscribe a circle in a square we find that it touches that square at four points which are in the middle of each side, as at a b c d. It will also intersect the two diagonals at the four points o (Fig. 165). If, then, we put this square and its diagonals, &c., into perspective we shall have eight guiding points through which to trace the required circle, as shown in Fig. 166, which has the same base as Fig. 165.







LXXXIX

THE CIRCLE IN PERSPECTIVE A TRUE ELLIPSE

Although the circle drawn through certain points must be a freehand drawing, which requires a little practice to make it true, it is sufficient for ordinary purposes and on a small scale, but to be mathematically true it must be an ellipse. We will first draw an ellipse (Fig. 167). Let ee be its long, or transverse, diameter, and db its short or conjugate diameter. Now take half of the long diameter eE, and from point d with cE for radius mark on ee the two points ff, which are the foci of the ellipse. At each focus fix a pin, then make a loop of fine string that does not stretch and of such a length that when drawn out the double thread will reach from f to e. Now place this double thread round the two pins at the foci ff' and distend it with the pencil point until it forms triangle fdf', then push the pencil along and right round the two foci, which being guided by the thread will draw the curve, which is a true ellipse, and will pass through the eight points indicated in our first figure. This will be a sufficient proof that the circle in perspective and the ellipse are identical curves. We must also remember that the ellipse is an oblique projection of a circle, or an oblique section of a cone. The difference between the two figures consists in their centres not being in the same place, that of the perspective circle being at c, higher up than e the centre of the ellipse. The latter being a geometrical figure, its long diameter is exactly in the centre of the figure, whereas the centre c and the diameter of the perspective are at the intersection of the diagonals of the perspective square in which it is inscribed.





XC

FURTHER ILLUSTRATION OF THE ELLIPSE

In order to show that the ellipse drawn by a loop as in the previous figure is also a circle in perspective we must reconstruct around it the square and its eight points by means of which it was drawn in the first instance. We start with nothing but the ellipse itself. We have to find the points of sight and distance, the base, &c. Let us start with base AB, a horizontal tangent to the curve extending beyond it on either side. From A and B draw two other tangents so that they shall touch the curve at points such as TT' a little above the transverse diameter and on a level with each other. Produce these tangents till they meet at point S, which will be the point of sight. Through this point draw horizontal line H. Now draw tangent CD parallel to AB. Draw diagonal AD till it cuts the horizon at the point of distance, this will cut through diameter of circle at its centre, and so proceed to find the eight points through which the perspective circle passes, when it will be found that they all lie on the ellipse we have drawn with the loop, showing that the two curves are identical although their centres are distinct.





XCI

HOW TO DRAW A CIRCLE IN PERSPECTIVE WITHOUT A GEOMETRICAL PLAN

Divide base AB into four equal parts. At B drop perpendicular Bn, making Bn equal to Bm, or one-fourth of base. Join mn and transfer this measurement to each side of d on base line; that is, make df and df' equal to mn. Draw fS and f'S, and the intersections of these lines with the diagonals of square will give us the four points o o o o.



The reason of this is that ff' is the measurement on the base AB of another square o o o o which is exactly half of the outer square. For if we inscribe a circle in a square and then inscribe a second square in that circle, this second square will be exactly half the area of the larger one; for its side will be equal to half the diagonal of the larger square, as can be seen by studying the following figures. In Fig. 170, for instance, the side of small square K is half the diagonal of large square o.





In Fig. 171, CB represents half of diagonal EB of the outer square in which the circle is inscribed. By taking a fourth of the base mB and drawing perpendicular mh we cut CB at h in two equal parts, Ch, hB. It will be seen that hB is equal to mn, one-quarter of the diagonal, so if we measure mn on each side of D we get ff' equal to CB, or half the diagonal. By drawing ff, f'f passing through the diagonals we get the four points o o o o through which to draw the smaller square. Without referring to geometry we can see at a glance by Fig. 172, where we have simply turned the square o o o o on its centre so that its angles touch the sides of the outer square, that it is exactly half of square ABEF, since each quarter of it, such as EoCo, is bisected by its diagonal oo.







XCII

HOW TO DRAW A CIRCLE IN ANGULAR PERSPECTIVE

Let ABCD be the oblique square. Produce VA till it cuts the base line at G.



Take mD, the fourth of the base. Find mn as in Fig. 171, measure it on each side of E, and so obtain Ef and Ef', and proceed to draw fV, EV, f'V and the diagonals, whose intersections with these lines will give us the eight points through which to draw the circle. In fact the process is the same as in parallel perspective, only instead of making our divisions on the actual base AD of the square, we make them on GD, the base line.

To obtain the central line hh passing through O, we can make use of diagonals of the half squares; that is, if the other vanishing point is inaccessible, as in this case.



XCIII

HOW TO DRAW A CIRCLE IN PERSPECTIVE MORE CORRECTLY, BY USING SIXTEEN GUIDING POINTS

First draw square ABCD. From O, the middle of the base, draw semicircle AKB, and divide it into eight equal parts. From each division raise perpendiculars to the base, such as 2 O, 3 O, 5 O, &c., and from divisions O, O, O draw lines to point of sight, and where these lines cut the diagonals AC, DB, draw horizontals parallel to base AB. Then through the points thus obtained draw the circle as shown in this figure, which also shows us how the circumference of a circle in perspective may be divided into any number of equal parts.





XCIV

HOW TO DIVIDE A PERSPECTIVE CIRCLE INTO ANY NUMBER OF EQUAL PARTS

This is simply a repetition of the previous figure as far as its construction is concerned, only in this case we have divided the semicircle into twelve parts and the perspective into twenty-four.



We have raised perpendiculars from the divisions on the semicircle, and proceeded as before to draw lines to the point of sight, and have thus by their intersections with the circumference already drawn in perspective divided it into the required number of equal parts, to which from the centre we have drawn the radii. This will show us how to draw traceries in Gothic windows, columns in a circle, cart-wheels, &c.

The geometrical figure (177) will explain the construction of the perspective one by showing how the divisions are obtained on the line AB, which represents base of square, from the divisions on the semicircle AKB.



XCV

HOW TO DRAW CONCENTRIC CIRCLES



First draw a square with its diagonals (Fig. 178), and from its centre O inscribe a circle; in this circle inscribe a square, and in this again inscribe a second circle, and so on. Through their intersections with the diagonals draw lines to base, and number them 1, 2, 3, 4, &c.; transfer these measurements to the base of the perspective square (Fig. 179), and proceed to construct the circles as before, drawing lines from each point on the base to the point of sight, and drawing the curves through the inter-sections of these lines with the diagonals.



Should it be required to make the circles at equal distances, as for steps for instance, then the geometrical plan should be made accordingly.

Or we may adopt the method shown at Fig. 180, by taking quarter base of both outer and inner square, and finding the measurement mn on each side of C, &c.





XCVI

THE ANGLE OF THE DIAMETER OF THE CIRCLE IN ANGULAR AND PARALLEL PERSPECTIVE

The circle, whether in angular or parallel perspective, is always an ellipse. In angular perspective the angle of the circle's diameter varies in accordance with the angle of the square in which it is placed, as in Fig. 181, cc is the diameter of the circle and ee the diameter of the ellipse. In parallel perspective the diameter of the circle always remains horizontal, although the long diameter of the ellipse varies in inclination according to the distance it is from the point of sight, as shown in Fig. 182, in which the third circle is much elongated and distorted, owing to its being outside the angle of vision.







XCVII

HOW TO CORRECT DISPROPORTION IN THE WIDTH OF COLUMNS

[Transcriber's Note: The column referred to as "1" in the text is marked "S" in both Figures.]

The disproportion in the width of columns in Fig. 183 arises from the point of distance being too near the point of sight, or, in other words, taking too wide an angle of vision. It will be seen that column 3 is much wider than column 1.





In our second figure (184) is shown how this defect is remedied, by doubling the distance, or by counting the same distance as half, which is easily effected by drawing the diagonal from O to 1/2-D, instead of from A, as in the other figure, O being at half base. Here the squares lie much more level, and the columns are nearly the same width, showing the advantage of a long distance.



XCVIII

HOW TO DRAW A CIRCLE OVER A CIRCLE OR A CYLINDER

First construct square and circle ABE, then draw square CDF with its diagonals. Then find the various points O, and from these raise perpendiculars to meet the diagonals of the upper square at points P, which, with the other points will be sufficient guides to draw the circle required. This can be applied to towers, columns, &c. The size of the circles can be varied so that the upper portion of a cylinder or column shall be smaller than the lower.





XCIX

TO DRAW A CIRCLE BELOW A GIVEN CIRCLE

Construct the upper square and circle as before, then by means of the vanishing scale POV, which should be made the depth required, drop perpendiculars from the various points marked O, obtained by the diagonals, making them the right depth by referring them to the vanishing scale, as shown in this figure. This can be used for drawing garden fountains, basins, and various architectural objects.





C

APPLICATION OF PREVIOUS PROBLEM

That is, to draw a circle above a circle. In Fig. 187 can be seen how by means of the vanishing scale at the side we obtain the height of the verticals 1, 2, 3, 4, &c., which determine the direction of the upper circle; and in this second figure, how we resort to the same means to draw circular steps.







CI

DORIC COLUMNS

It is as well for the art student to study the different orders of architecture, whether architect or not, as he frequently has to introduce them into his pictures, and at least must know their proportions, and how columns diminish from base to capital, as shown in this illustration.





CII

TO DRAW SEMICIRCLES STANDING UPON A CIRCLE AT ANY ANGLE



Given the circle ACBH, on diagonal AB draw semicircle AKB, and on the same line AB draw rectangle AEFB, its height being determined by radius OK of semicircle. From centre O draw OF to corner of rectangle. Through f', where that line intersects the semicircle, draw mn parallel to AB. This will give intersection O' on the vertical OK, through which all such horizontals as m'n', level with mn, must pass. Now take any other diameter, such as GH, and thereon raise rectangle GghH, the same height as the other. The manner of doing this is to produce diameter GH to the horizon till it finds its vanishing point at V. From V through K draw hg, and through O' draw n'm'. From O draw the two diagonals og and oh, intersecting m'n' at O, O, and thus we have the five points GOKOH through which to draw the required semicircle.



CIII

A DOME STANDING ON A CYLINDER



This figure is a combination of the two preceding it. A cylinder is first raised on the circle, and on the top of that we draw semicircles from the different divisions on the circumference of the upper circle. This, however, only represents a small half-globular object. To draw the dome of a cathedral, or other building high above us, is another matter. From outside, where we can get to a distance, it is not difficult, but from within it will tax all our knowledge of perspective to give it effect.

We shall go more into this subject when we come to archways and vaulted roofs, &c.



CIV

SECTION OF A DOME OR NICHE



First draw outline of the niche GFDBA (Fig. 193), then at its base draw square and circle GOA, S being the point of sight, and divide the circumference of the circle into the required number of parts. Then draw semicircle FOB, and over that another semicircle EOC. The manner of drawing them is shown in Fig. 192. From the divisions on the circle GOA raise verticals to semicircle FOB, which will divide it in the same way. Divide the smaller semicircle EOC into the same number of parts as the others, which divisions will serve as guiding points in drawing the curves of the dome that are drawn towards D, but the shading must assist greatly in giving the effect of the recess.



In Fig. 192 will be seen how to draw semicircles in perspective. We first draw the half squares by drawing from centres O of their diameters diagonals to distance-point, as OD, which cuts the vanishing line BS at m, and gives us the depth of the square, and in this we draw the semicircle in the usual way.





CV

A DOME

First draw a section of the dome ACEDB (Fig. 194) the shape required. Draw AB at its base and CD at some distance above it. Keeping these as central lines, form squares thereon by drawing SA, SB, SC, SD, &c., from point of sight, and determining their lengths by diagonals fh, f'h' from point of distance, passing through O. Having formed the two squares, draw perspective circles in each, and divide their circumferences into twelve or whatever number of parts are needed. To complete the figure draw from each division in the lower circle curves passing through the corresponding divisions in the upper one, to the apex. But as these are freehand lines, it requires some taste and knowledge to draw them properly, and of course in a large drawing several more squares and circles might be added to aid the draughtsman. The interior of the dome can be drawn in the same way.







CVI

HOW TO DRAW COLUMNS STANDING IN A CIRCLE

In Fig. 195 are sixteen cylinders or columns standing in a circle. First draw the circle on the ground, then divide it into sixteen equal parts, and let each division be the centre of the circle on which to raise the column. The question is how to make each one the right width in accordance with its position, for it is evident that a near column must appear wider than the opposite one. On the right of the figure is the vertical scale A, which gives the heights of the columns, and at its foot is a horizontal scale, or a scale of widths B. Now, according to the line on which the column stands, we find its apparent width marked on the scale. Thus take the small square and circle at 15, without its column, or the broken column at 16; and note that on each side of its centre O I have measured oa, ob, equal to spaces marked 3 on the same horizontal in the scale B. Through these points a and b I have drawn lines towards point of sight S. Through their intersections with diagonal e, which is directed to point of distance, draw the farther and nearer sides of the square in which to describe the circle and the cylinder or column thereon. I have made all the squares thus obtained in parallel perspective, but they do not represent the bases of columns arranged in circles, which should converge towards the centre, and I believe in some cases are modified in form to suit that design.



CVII

COLUMNS AND CAPITALS

This figure shows the application of the square and diagonal in drawing and placing columns in angular perspective.





CVIII

METHOD OF PERSPECTIVE EMPLOYED BY ARCHITECTS

The architects first draw a plan and elevation of the building to be put into perspective. Having placed the plan at the required angle to the picture plane, they fix upon the point of sight, and the distance from which the drawing is to be viewed. They then draw a line SP at right angles to the picture plane VV', which represents that distance so that P is the station-point. The eye is generally considered to be the station-point, but when lines are drawn to that point from the ground-plan, the station-point is placed on the ground, and is in fact the trace or projection exactly under the point at which the eye is placed. From this station-point P, draw lines PV and PV' parallel to the two sides of the plan ba and ad (which will be at right angles to each other), and produce them to the horizon, which they will touch at points V and V'. These points thus obtained will be the two vanishing points.

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The next operation is to draw lines from the principal points of the plan to the station-point P, such as bP, cP, dP, &c., and where these lines intersect the picture plane (VV' here represents it as well as the horizon), drop perpendiculars b'B, aA, d'D, &c., to meet the vanishing lines AV, AV', which will determine the points A, B, C, D, 1, 2, 3, &c., and also the perspective lengths of the sides of the figure AB, AD, and the divisions B, 1, 2, &c. Taking the height of the figure AE from the elevation, we measure it on Aa; as in this instance A touches the ground line, it may be used as a line of heights.

I have here placed the perspective drawing under the ground plan to show the relation between the two, and how the perspective is worked out, but the general practice is to find the required measurements as here shown, to mark them on a straight edge of card or paper, and transfer them to the paper on which the drawing is to be made.

This of course is the simplest form of a plan and elevation. It is easy to see, however, that we could set out an elaborate building in the same way as this figure, but in that case we should not place the drawing underneath the ground-plan, but transfer the measurements to another sheet of paper as mentioned above.



CIX

THE OCTAGON

To draw the geometrical figure of an octagon contained in a square, take half of the diagonal of that square as radius, and from each corner describe a quarter circle. At the eight points where they touch the sides of the square, draw the eight sides of the octagon.





To put this into perspective take the base of the square AB and thereon form the perspective square ABCD. From either extremity of that base (say B) drop perpendicular BF, draw diagonal AF, and then from B with radius BO, half that diagonal, describe arc EOE. This will give us the measurement AE. Make GB equal to AE. Then draw lines from G and E towards S, and by means of the diagonals find the transverse lines KK, hh, which will give us the eight points through which to draw the octagon.



CX

HOW TO DRAW THE OCTAGON IN ANGULAR PERSPECTIVE

Form square ABCD (new method), produce sides BC and AD to the horizon at V, and produce VA to a' on base. Drop perpendicular from B to F the same length as a'B, and proceed as in the previous figure to find the eight points on the oblique square through which to draw the octagon.



It will be seen that this operation is very much the same as in parallel perspective, only we make our measurements on the base line a'B as we cannot measure the vanishing line BA otherwise.



CXI

HOW TO DRAW AN OCTAGONAL FIGURE IN ANGULAR PERSPECTIVE

In this figure in angular perspective we do precisely the same thing as in the previous problem, taking our measurements on the base line EB instead of on the vanishing line BA. If we wish to raise a figure on this octagon the height of EG we form the vanishing scale EGO, and from the eight points on the ground draw horizontals to EO and thus find all the points that give us the perspective height of each angle of the octagonal figure.





CXII

HOW TO DRAW CONCENTRIC OCTAGONS, WITH ILLUSTRATION OF A WELL

The geometrical figure 202 A shows how by means of diagonals AC and BD and the radii 1 2 3, &c., we can obtain smaller octagons inside the larger ones. Note how these are carried out in the second figure (202 B), and their application to this drawing of an octagonal well on an octagonal base.









CXIII

A PAVEMENT COMPOSED OF OCTAGONS AND SMALL SQUARES

To draw a pavement with octagonal tiles we will begin with an octagon contained in a square abcd. Produce diagonal ac to V. This will be the vanishing point for the sides of the small squares directed towards it. The other sides are directed to an inaccessible point out of the picture, but their directions are determined by the lines drawn from divisions on base to V2 (see back, Fig. 133).





I have drawn the lower figure to show how the squares which contain the octagons are obtained by means of the diagonals, BD, AC, and the central line OV2. Given the square ABCD. From D draw diagonal to G, then from C through centre o draw CE, and so on all the way up the floor until sufficient are obtained. It is easy to see how other squares on each side of these can be produced.



CXIV

THE HEXAGON

The hexagon is a six-sided figure which, if inscribed in a circle, will have each of its sides equal to the radius of that circle (Fig. 206). If inscribed in a rectangle ABCD, that rectangle will be equal in length to two sides of the hexagon or two radii of the circle, as EF, and its width will be twice the height of an equilateral triangle mon.



To put the hexagon into perspective, draw base of quadrilateral AD, divide it into four equal parts, and from each division draw lines to point of sight. From h drop perpendicular ho, and form equilateral triangle mno. Take the height ho and measure it twice along the base from A to 2. From 2 draw line to point of distance, or from 1 to 1/2 distance, and so find length of side AB equal to A2. Draw BC, and EF through centre o', and thus we have the six points through which to draw the hexagon.





CXV

A PAVEMENT COMPOSED OF HEXAGONAL TILES

In drawing pavements, except in the cases of square tiles, it is necessary to make a plan of the required design, as in this figure composed of hexagons. First set out the hexagon as at A, then draw parallels 1 1, 2 2, &c., to mark the horizontal ends of the tiles and the intermediate lines oo. Divide the base into the required number of parts, each equal to one side of the hexagon, as 1, 2, 3, 4, &c.; from these draw perpendiculars as shown in the figure, and also the diagonals passing through their intersections. Then mark with a strong line the outlines of the hexagonals, shading some of them; but the figure explains itself.

It is easy to put all these parallels, perpendiculars, and diagonals into perspective, and then to draw the hexagons.

First draw the hexagon on AD as in the previous figure, dividing AD into four, &c., set off right and left spaces equal to these fourths, and from each division draw lines to point of sight. Produce sides me, nf till they touch the horizon in points V, V'; these will be the two vanishing points for all the sides of the tiles that are receding from us. From each division on base draw lines to each of these vanishing points, then draw parallels through their intersections as shown on the figure. Having all these guiding lines it will not be difficult to draw as many hexagons as you please.



Note that the vanishing points should be at equal distances from S, also that the parallelogram in which each tile is contained is oblong, and not square, as already pointed out.

We have also made use of the triangle omn to ascertain the length and width of that oblong. Another thing to note is that we have made use of the half distance, which enables us to make our pavement look flat without spreading our lines outside the picture.





CXVI

A PAVEMENT OF HEXAGONAL TILES IN ANGULAR PERSPECTIVE

This is more difficult than the previous figure, as we only make use of one vanishing point; but it shows how much can be done by diagonals, as nearly all this pavement is drawn by their aid. First make a geometrical plan A at the angle required. Then draw its perspective K. Divide line 4b into four equal parts, and continue these measurements all along the base: from each division draw lines to V, and draw the hexagon K. Having this one to start with we produce its sides right and left, but first to the left to find point G, the vanishing point of the diagonals. Those to the right, if produced far enough, would meet at a distant vanishing point not in the picture. But the student should study this figure for himself, and refer back to Figs. 204 and 205.





CXVII

FURTHER ILLUSTRATION OF THE HEXAGON





To draw the hexagon in perspective we must first find the rectangle in which it is inscribed, according to the view we take of it. That at A we have already drawn. We will now work out that at B. Divide the base AD into four equal parts and transfer those measurements to the perspective figure C, as at AD, measuring other equal spaces along the base. To find the depth An of the rectangle, make DK equal to base of square. Draw KO to distance-point, cutting DO at O, and thus find line LO. Draw diagonal Dn, and through its intersections with the lines 1, 2, 3, 4 draw lines parallel to the base, and we shall thus have the framework, as it were, by which to draw the pavement.





CXVIII

ANOTHER VIEW OF THE HEXAGON IN ANGULAR PERSPECTIVE



Given the rectangle ABCD in angular perspective, produce side DA to E on base line. Divide EB into four equal parts, and from each division draw lines to vanishing point, then by means of diagonals, &c., draw the hexagon.

In Fig. 214 we have first drawn a geometrical plan, G, for the sake of clearness, but the one above shows that this is not necessary.



To raise the hexagonal figure K we have made use of the vanishing scale O and the vanishing point V. Another method could be used by drawing two hexagons one over the other at the required height.



CXIX

APPLICATION OF THE HEXAGON TO DRAWING A KIOSK



This figure is built up from the hexagon standing on a rectangular base, from which we have raised verticals, &c. Note how the jutting portions of the roof are drawn from o'. But the figure explains itself, so there is no necessity to repeat descriptions already given in the foregoing problems.



CXX

THE PENTAGON



The pentagon is a figure with five equal sides, and if inscribed in a circle will touch its circumference at five equidistant points. With any convenient radius describe circle. From half this radius, marked 1, draw a line to apex, marked 2. Again, with 1 as centre and 1 2 as radius, describe arc 2 3. Now with 2 as centre and 2 3 as radius describe arc 3 4, which will cut the circumference at point 4. Then line 2 4 will be one of the sides of the pentagon, which we can measure round the circle and so produce the required figure.

To put this pentagon into parallel perspective inscribe the circle in which it is drawn in a square, and from its five angles 4, 2, 4, &c., drop perpendiculars to base and number them as in the figure. Then draw the perspective square (Fig. 217) and transfer these measurements to its base. From these draw lines to point of sight, then by their aid and the two diagonals proceed to construct the pentagon in the same way that we did the triangles and other figures. Should it be required to place this pentagon in the opposite position, then we can transfer our measurements to the far side of the square, as in Fig. 218.





Or if we wish to put it into angular perspective we adopt the same method as with the hexagon, as shown at Fig. 219.



Another way of drawing a pentagon (Fig. 220) is to draw an isosceles triangle with an angle of 36 deg at its apex, and from centre of each side of the triangle draw perpendiculars to meet at o, which will be the centre of the circle in which it is inscribed. From this centre and with radius OA describe circle A 3 2, &c. Take base of triangle 1 2, measure it round the circle, and so find the five points through which to draw the pentagon. The angles at 1 2 will each be 72 deg, double that at A, which is 36 deg.





CXXI

THE PYRAMID

Nothing can be more simple than to put a pyramid into perspective. Given the base (abc), raise from its centre a perpendicular (OP) of the required height, then draw lines from the corners of that base to a point P on the vertical line, and the thing is done. These pyramids can be used in drawing roofs, steeples, &c. The cone is drawn in the same way, so also is any other figure, whether octagonal, hexangular, triangular, &c.











CXXII

THE GREAT PYRAMID

This enormous structure stands on a square base of over thirteen acres, each side of which measures, or did measure, 764 feet. Its original height was 480 feet, each side being an equilateral triangle. Let us see how we can draw this gigantic mass on our little sheet of paper.

In the first place, to take it all in at one view we must put it very far back, and in the second the horizon must be so low down that we cannot draw the square base of thirteen acres on the perspective plane, that is on the ground, so we must draw it in the air, and also to a very small scale.

Divide the base AB into ten equal parts, and suppose each of these parts to measure 10 feet, S, the point of sight, is placed on the left of the picture near the side, in order that we may get a long line of distance, S 1/2 D; but even this line is only half the distance we require. Let us therefore take the 16th distance, as shown in our previous illustration of the lighthouse (Fig. 92), which enables us to measure sixteen times the length of base AB, or 1,600 feet. The base ef of the pyramid is 1,600 feet from the base line of the picture, and is, according to our 10-foot scale, 764 feet long.

The next thing to consider is the height of the pyramid. We make a scale to the right of the picture measuring 50 feet from B to 50 at point where BP intersects base of pyramid, raise perpendicular CG and thereon measure 480 feet. As we cannot obtain a palpable square on the ground, let us draw one 480 feet above the ground. From e and f raise verticals eM and fN, making them equal to perpendicular G, and draw line MN, which will be the same length as base, or 764 feet. On this line form square MNK parallel to the perspective plane, find its centre O' by means of diagonals, and O' will be the central height of the pyramid and exactly over the centre of the base. From this point O' draw sloping lines O'f, O'e, O'Y, &c., and the figure is complete.