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.
Fig. 106.
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.
Fig. 107.
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°, 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
| Fig. 108. |
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.
Fig. 109.
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.
Fig. 110.
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).
Fig. 111.
[ 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.
| Fig. 112. |
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.
Fig. 113.
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.
Fig. 114.
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.
Fig. 115.
[ 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.
Fig. 116.
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.
Fig. 117.
[ 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.
Fig. 118.
The same method applies to a given line to be divided into various proportions, as shown in this lower figure.
Fig. 119.
[ 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.
Fig. 120.
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.
Fig. 121.
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.
Fig. 122 is a front view of the portico, Fig. 123.
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.
Fig. 123.
[ 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.
Fig. 124.
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°. So that in fixing upon the angle at which the square or other figure is to be placed, we say ‘let it be 60° and 30°, or 70° and 20°’, &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° on one side, and 80° 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.
Fig. 125.
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.
Fig. 126.
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.
Fig. 127.
[ 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.
Fig. 128.
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.
Fig. 129.
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]).
Fig. 130.
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.
Fig. 131.
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].
Fig. 132.
[ 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.
Fig. 133.
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° 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.
Fig. 134.
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.
Fig. 135.
Fig. 136.
[ LXXIII]
Of Parallels and Diagonals
| Fig. 137 A. |
| Fig. 137 B. |
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 ¼, ⅓, ½. 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.
| Fig. 137 C. | Fig. 137 D. |
[ 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.
| Fig. 140. | Fig. 141. |
[ 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.
| Fig. 142. | Fig. 143. |
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.
Fig. 144.
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.
| Fig. 148 A. | Fig. 148 B. |
| Fig. 149 A. | Fig. 149 B. |
[ 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 A. | Fig. 150 B. |
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.
Fig. 151.
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.
| Fig. 152. | Fig. 153. |
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.
Fig. 154.
By a similar process we draw these two figures, one on the square, the other on the circle.
| Fig. 155. | Fig. 156. |
[ 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.
| Fig. 157. |
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.
| Fig. 158. |
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.
| Fig. 159. | Fig. 160. |
[ 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.
Fig. 161.
[ 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.
Fig. 162.
[ 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.
Fig. 163. Honfleur.
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.
Fig. 164. Toledo.
[ 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.
| Fig. 165. | Fig. 166. |
[ 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.
Fig. 167.
[ 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.
Fig. 168.
[ XCI]
How To Draw a Circle in Perspective Without a Geometrical Plan
| Fig. 169. |
| Fig. 170. |
| Fig. 171. |
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.
| Fig. 172. | Fig. 173. |
[ 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.
Fig. 174.
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.
Fig. 175.
[ 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.
| Fig. 176. | Fig. 177. |
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.
| Fig. 178. | Fig. 179. |
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.
Fig. 180.
[ 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.
Fig. 181.
Fig. 182.
[ XCVII]
How to Correct Disproportion in the Width of Columns
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.
Fig. 183.
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 ½-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.
Fig. 184.
[ XCVIII]
How to Draw a Circle over a Circle or a Cylinder
| Fig. 185. |
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
| Fig. 186. |
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.
Fig. 187.
Fig. 188.
[ 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.
Fig. 189.
[ 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.
Fig. 190.
[ 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.
Fig. 191.
[ CIV]
Section of a Dome or Niche
| Fig. 192. |
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.
Fig. 193.
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
| Fig. 194. 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.
Fig. 195.
[ CVII]
Columns and Capitals
This figure shows the application of the square and diagonal in drawing and placing columns in angular perspective.
Fig. 196.
[ 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.
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.
Fig. 197. A method of angular Perspective employed by architects.
[To face p. 171]
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
| Fig. 198. |
| Fig. 199. |
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.
Fig. 200.
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.
Fig. 201.
[ 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.
| Fig. 202 A. | Fig. 202 B. |
Fig. 203.
[ 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]).
Fig. 204.
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.
Fig. 205.
[ 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.
| Fig. 206. | Fig. 207. |
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 ½ 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.
Fig. 208.
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.
Fig. 209.
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].
Fig. 210.
[ 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.
| Fig. 211 A. | Fig. 211 B. |
Fig. 212.
[ CXVIII]
Another View of the Hexagon in Angular Perspective
| Fig. 213. |
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.
Fig. 214.
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.
Fig. 215.
[ CXX]
The Pentagon
| Fig. 216. |
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.
| Fig. 217. | 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.
| Fig. 219. | Fig. 220. |
Another way of drawing a pentagon (Fig. 220) is to draw an isosceles triangle with an angle of 36° 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°, double that at A, which is 36°.
[ 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.
![[Larger View]](#3252062959678807837_fig197large.png.id-2965045035112079104.wrap-0.html.html){kind=link}
[ 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.
Fig. 224.
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 ½ 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.
Note the way in which we find the measurements on base of pyramid and on line MN. By drawing AS and BS to point of sight we find Te, which measures 100 feet at a distance of 1,600 feet. We mark off seven of these lengths, and an additional 64 feet by the scale, and so obtain the required length. The position of the third corner of the base is found by dropping a perpendicular from K, till it meets the line eS.
Another thing to note is that the side of the pyramid that faces us, although an equilateral triangle, does not appear so, as its top angle is 382 feet farther off than its base owing to its leaning position.
[ CXXIII]
The Pyramid in Angular Perspective
In order to show the working of this proposition I have taken a much higher horizon, which immediately detracts from the impression of the bigness of the pyramid.
Fig. 225.
We proceed to make our ground-plan abcd high above the horizon instead of below it, drawing first the parallel square and then the oblique one. From all the principal points drop perpendiculars to the ground and thus find the points through which to draw the base of the pyramid. Find centres OO· and decide upon the height OP. Draw the sloping lines from P to the corners of the base, and the figure is complete.
[ CXXIV]
To Divide the Sides of the Pyramid Horizontally
Having raised the pyramid on a given oblique square, divide the vertical line OP into the required number of parts. From
A through C draw AG to horizon, which gives us G, the vanishing point of all the diagonals of squares parallel to and at the same angle as ABCD. From G draw lines through the divisions 2, 3, &c., on OP cutting the lines PA and PC, thus dividing them into the required parts. Through the points thus found draw from V all those sides of the squares that have V for their vanishing point, as ab, cd, &c. Then join bd, ac, and the rest, and thus make the horizontal divisions required.
| Fig. 226. | Fig. 227. |
The same method will apply to drawing steps, square blocks, &c., as shown in Fig. 227, which is at the same angle as the above.
[ CXXV]
Of Roofs
The pyramidal roof (Fig. 228) is so simple that it explains itself. The chief thing to be noted is the way in which the diagonals are produced beyond the square of the walls, to give the width of the eaves, according to their position.
| Fig. 228. | Fig. 229. |
Another form of the pyramidal roof is here given (Fig. 229). First draw the cube edcba at the required height, and on the side facing us, adcb, draw triangle K, which represents the end of a gable roof. Then draw similar triangles on the other sides of the cube (see [Fig. 159], LXXXIV). Join the opposite triangles
at the apex, and thus form two gable roofs crossing each other at right angles. From o, centre of base of cube, raise vertical OP, and then from P draw sloping lines to each corner of base a, b, &c., and by means of central lines drawn from P to half base, find the points where the gable roofs intersect the central spire or pyramid. Any other proportions can be obtained by adding to or altering the cube.
Fig. 230.
To draw a sloping or hip-roof which falls back at each end we must first draw its base, CBDA (Fig. 230). Having found the centre O and central line SP, and how far the roof is to fall back at each end, namely the distance Pm, draw horizontal line RB through m. Then from B through O draw diagonal BA, and from
A draw horizontal AD, which gives us point n. From these two points m and n raise perpendiculars the height required for the roof, and from these draw sloping lines to the corners of the base. Join ef, that is, draw the top line of the roof, which completes it. Fig. 231 shows a plan or bird's-eye view of the roof and the diagonal AB passing through centre O. But there are so many varieties of roofs they would take almost a book to themselves to illustrate them, especially the cottages and farm-buildings, barns, &c., besides churches, old mansions, and others. There is also such irregularity about some of them that perspective rules, beyond those few here given, are of very little use. So that the best thing for an artist to do is to sketch them from the real whenever he has an opportunity.
Fig. 231.
[ CXXVI]
Of Arches, Arcades, Bridges, &c.
For an arcade or cloister (Fig. 232) first set up the outer frame ABCD according to the proportions required. For round arches the height may be twice that of the base, varying to one and a half. In Gothic arches the height may be about three times the width, all of which proportions are chosen to suit the different purposes and effects required. Divide the base AB into the desired number of parts, 8, 10, 12, &c., each part representing 1 foot. (In this case the base is 10 feet and the horizon 5 feet.) Set out floor by means of ¼ distance. Divide it into squares of 1 foot, so that there will be 8 feet between each column or pilaster, supposing we make them to stand on a square foot. Draw the first archway EKF facing us, and its inner semicircle gh, with also its thickness or depth of 1 foot. Draw the span of the archway EF, then central line PO to point of sight. Proceed to raise as many other arches as required at the given distances. The intersections of the central line with the chords mn, &c., will give the centres from which to describe the semicircles.
Fig. 232.
[ CXXVII]
Outline of an Arcade with Semicircular Arches
This is to show the method of drawing a long passage, corridor, or cloister with arches and columns at equal distances, and is worked in the same way as the previous figure, using ¼ distance and ¼ base. The floor consists of five squares; the semicircles of the arches are described from the numbered points on the central line OS, where it intersects the chords of the arches.
Fig. 233.
[ CXXVIII]
Semicircular Arches on a Retreating Plane
First draw perspective square abcd. Let ae· be the height of the figure. Draw ae·f·b and proceed with the rest of the outline. To draw the arches begin with the one facing us, Eo·F enclosed in the quadrangle Ee·f·F. With centre O describe the semicircle and across it draw the diagonals e·F, Ef·, and through nn, where these lines intersect the semicircle, draw horizontal KK and also KS to point of sight. It will be seen that the half-squares at the side are the same size in perspective as the one facing us, and we carry out in them much the same operation; that is, we draw the diagonals, find the point O, and the points nn, &c., through which to draw our arches. See perspective of the circle (Fig. 165).
Fig. 234.
If more points are required an additional diagonal from O to
K may be used, as shown in the figure, which perhaps explains itself. The method is very old and very simple, and of course can be applied to any kind of arch, pointed or stunted, as in this drawing of a pointed arch (Fig. 235).
Fig. 235.
[ CXXIX]
An Arcade in Angular Perspective
First draw the perspective square ABCD at the angle required, by new method. Produce sides AD and BC to V. Draw diagonal BD and produce to point G, from whence we draw the other diagonals to cfh. Make spaces 1, 2, 3, &c., on base line equal to B 1 to obtain sides of squares. Raise vertical BM the height required. Produce DA to O on base line, and from O raise vertical OP equal to BM. This line enables us to dispense with the long vanishing point to the left; its working has been explained at Fig. 131. From P draw PRV to vanishing point V, which will intersect vertical AR at R. Join MR, and this line, if produced, would meet the horizon at the other vanishing point.
In like manner make O2 equal to B2·. From 2 draw line to V, and at 2, its intersection with AR, draw line 2 2, which will also meet the horizon at the other vanishing point. By means of the quarter-circle A we can obtain the points through which to draw the semicircular arches in the same way as in the previous figure.
Fig. 236.
[ CXXX]
A Vaulted Ceiling
From the square ceiling ABCD we have, as it were, suspended two arches from the two diagonals DB, AC, which spring from the four corners of the square EFGH, just underneath it. The curves of these arches, which are not semicircular but elongated, are obtained by means of the vanishing scales mS, nS. Take any two convenient points P, R, on each side of the semicircle, and
raise verticals Pm, Rn to AB, and on these verticals form the scales. Where mS and nS cut the diagonal AC drop perpendiculars to meet the lower line of the scale at points 1, 2. On the other side, using the other scales, we have dropped perpendiculars in the same way from the diagonal to 3, 4. These points, together
with EOG, enable us to trace the curve E 1 2 O 3 4 G. We draw the arch under the other diagonal in precisely the same way.
| Fig. 237. | Fig. 238. |
The reason for thus proceeding is that the cross arches, although elongated, hang from their diagonals just as the semicircular arch EKF hangs from AB, and the lines mn, touching the circle at PR, are represented by 1, 2, hanging from the diagonal AC.
Figure 238, which is practically the same as the preceding only differently shaded, is drawn in the following manner. Draw arch EGF facing us, and proceed with the rest of the corridor, but first finding the flat ceiling above the square on the ground ABcd. Draw diagonals ac, bd, and the curves pending from them. But we no longer see the clear arch as in the other drawing, for the spaces between the curves are filled in and arched across.
[ CXXXI]
A Cloister, from a Photograph
This drawing of a cloister from a photograph shows the correctness of our perspective, and the manner of applying it to practical work.
Fig. 239.
[ CXXXII]
The Low or Elliptical Arch
Let AB be the span of the arch and Oh its height. From centre O, with OA, or half the span, for radius, describe outer semicircle. From same centre and oh for radius describe the inner semicircle. Divide outer circle into a convenient number of parts, 1, 2, 3, &c., to which draw radii from centre O. From each division drop perpendiculars. Where the radii intersect the inner circle, as at gkmo, draw horizontals op, mn, kj, &c., and
through their intersections with the perpendiculars f, j, n, p, draw the curve of the flattened arch. Transfer this to the lower figure, and proceed to draw the tunnel. Note how the vanishing scale is formed on either side by horizontals ba, fe, &c., which enable us to make the distant arches similar to the near ones.
Fig. 240.
Fig. 241.
[ CXXXIII]
Opening or Arched Window in a Vault
First draw the vault AEB. To introduce the window K, the upper part of which follows the form of the vault, we first decide on its width, which is mn, and its height from floor Ba. On line Ba at the side of the arch form scales aa·S, bb·S, &c. Raise the semicircular arch K, shown by a dotted line. The scale at the side will give the lengths aa·, bb·, &c., from different parts of this dotted arch to corresponding points in the curved archway or window required.
Fig. 242.
Note that to obtain the width of the window K we have used
the diagonals on the floor and width m n on base. This method of measurement is explained at Fig. 144, and is of ready application in a case of this kind.
[ CXXXIV]
Stairs, Steps, &c.
Having decided upon the incline or angle, such as CBA, at which the steps are to be placed, and the height Bm of each step, draw mn to CB, which will give the width. Then measure along base AB this width equal to DB, which will give that for all the other steps. Obtain length BF of steps, and draw EF parallel to CB. These lines will aid in securing the exactness of the figure.
| Fig. 243. | Fig. 244. |
[ CXXXV]
Steps, Front View
In this figure the height of each step is measured on the vertical line AB (this line is sometimes called the line of heights), and their depth is found by diagonals drawn to the point of distance D. The rest of the figure explains itself.
Fig. 245.
[ CXXXVI]
Square Steps
Draw first step ABEF and its two diagonals. Raise vertical AH, and measure thereon the required height of each step, and thus form scale. Let the second step CD be less all round than the first by Ao or Bo. Draw oC till it cuts the diagonal, and proceed to draw the second step, guided by the diagonals and taking its height from the scale as shown. Draw the third step in the same way.
Fig. 246.
[ CXXXVII]
To Divide an Inclined Plane into Equal Parts—such as a Ladder Placed against a Wall
Divide the vertical EC into the required number of parts, and draw lines from point of sight S through these divisions 1, 2, 3, &c., cutting the line AC at 1, 2, 3, &c. Draw parallels to AB, such as mn, from AC to BD, which will represent the steps of the ladder.
Fig. 247.
[ CXXXVIII]
Steps and the Inclined Plane
In Fig. 248 we treat a flight of steps as if it were an inclined plane. Draw the first and second steps as in Fig. 245. Then through 1, 2, draw 1V, AV to V, the vanishing point on the vertical line SV. These two lines and the corresponding ones at BV will form a kind of vanishing scale, giving the height of each step as we ascend. It is especially useful when we pass the horizontal line and we no longer see the upper surface of the step, the scale on the right showing us how to proceed in that case.
Fig. 248.
In Fig. 249 we have an example of steps ascending and descending. First set out the ground-plan, and find its vanishing point S (point of sight). Through S draw vertical BA, and make SA equal to SB. Set out the first step CD. Draw EA, CA, DA, and GA, for the ascending guiding lines. Complete the steps facing us, at central line OO. Then draw guiding line FB for the descending steps (see Rule 8).
Fig. 249.
[ CXXXIX]
Steps in Angular Perspective
First draw the base ABCD (Fig. 251) at the required angle by the new method (Fig. 250). Produce BC to the horizon, and thus find vanishing point V. At this point raise vertical VV·. Construct
first step AB, refer its height at B to line of heights hI on left, and thus obtain height of step at A. Draw lines from A and F to V·. From n draw diagonal through O to G. Raise vertical at O to represent the height of the next step, its height being determined by the scale of heights at the side. From A and F draw lines to V·, and also similar lines from B, which will serve as guiding lines to determine the height of the steps at either end as we raise them to the required number.
Fig. 250.
Fig. 251.
[ CXL]
A Step Ladder at an Angle
First draw the ground-plan G at the required angle, using vanishing and measuring points. Find the height hH, and width at top HH·, and draw the sides HA and H·E. Note that AE is wider than HH·, and also that the back legs are not at the same angle as the front ones, and that they overlap them. From E raise vertical EF, and divide into as many parts as you require rounds to the ladder. From these divisions draw lines 1 1, 2 2, &c., towards the other vanishing point (not in the picture), but
having obtained their direction from the ground-plan in perspective at line Ee, you may set up a second vertical ef at any point on Ee and divide it into the same number of parts, which will be in proportion to those on EF, and you will obtain the same result by drawing lines from the divisions on EF to those on ef as in drawing them to the vanishing point.
Fig. 252.
[ CXLI]
Square Steps Placed over each Other
This figure shows the other method of drawing steps, which is simple enough if we have sufficient room for our vanishing points.
Fig. 253.
The manner of working it is shown at [Fig. 124].
[ CXLII]
Steps and a Double Cross Drawn by Means of Diagonals and one Vanishing Point
Although in this figure we have taken a longer distance-point than in the previous one, we are able to draw it all within the page.
Fig. 254.
Begin by setting out the square base at the angle required. Find point G by means of diagonals, and produce AB to V, &c. Mark height of step Ao, and proceed to draw the steps as already shown. Then by the diagonals and measurements on base draw the second step and the square inside it on which to stand the foot of the cross. To draw the cross, raise verticals from the four corners of its base, and a line K from its centre. Through any
point on this central line, if we draw a diagonal from point G we cut the two opposite verticals of the shaft at mn (see Fig. 255), and by means of the vanishing point V we cut the other two verticals at the opposite corners and thus obtain the four points through which to draw the other sides of the square, which go to the distant or inaccessible vanishing point. It will be seen by carefully examining the figure that by this means we are enabled to draw the double cross standing on its steps.
Fig. 255.
[ CXLIII]
A Staircase Leading to a Gallery
In this figure we have made use of the devices already set forth in the foregoing figures of steps, &c., such as the side scale on the left of the figure to ascertain the height of the steps, the double lines drawn to the high vanishing point of the inclined plane, and so on; but the principal use of this diagram is to show on the perspective plane, which as it were runs under the stairs, the trace or projection of the flights of steps, the landings and positions of other objects, which will be found very useful in placing figures in a composition of this kind. It will be seen that these underneath measurements, so to speak, are obtained by the half-distance.
Fig. 256.
[ CXLIV]
Winding Stairs in a Square Shaft
Draw square ABCD in parallel perspective. Divide each side into four, and raise verticals from each division. These verticals will mark the positions of the steps on each wall, four in number. From centre O raise vertical OP, around which the steps are to wind. Let AF be the height of each step. Form scale AB, which will give the height of each step according to its position. Thus at mn we find the height at the centre of the square, so if we transfer this measurement to the central line OP and repeat it upwards, say to fourteen, then we have the height of each step on the line where they all meet. Starting then with the first on the right, draw the rectangle gD1f, the height of AF, then draw to the central line go, f1, and 1 1, and thus complete the first step. On DE, measure heights equal to D 1. Draw 2 2 towards central line, and 2n towards point of sight till it meets the second vertical nK. Then draw n2 to centre, and so complete the second step. From 3 draw 3a to third vertical, from 4 to fourth, and so on, thus obtaining the height of each ascending step on the wall to the right, completing them in the same way as numbers 1 and 2, when we come to the sixth step, the other end of which is against the wall opposite to us. Steps 6, 7, 8, 9 are all on this wall, and are therefore equal in height all along, as they are equally distant. Step 10 is turned towards us, and abuts on the wall to our left; its measurement is taken on the scale AB just underneath it, and on the same line to which it is drawn. Step 11 is just over the centre of base mo, and is therefore parallel to it, and its height is mn. The widths of steps 12 and 13 seem gradually to increase as they come towards us, and as they rise above the horizon we begin to see underneath them. Steps 13, 14, 15, 16 are against the wall on this side of the picture, which we may suppose has been removed to show the working of the drawing, or they might be an open flight as we sometimes see in shops and galleries, although in that case they are generally enclosed in a cylindrical shaft.
Fig. 257.
[ CXLV]
Winding Stairs in a Cylindrical Shaft
First draw the circular base CD. Divide the circumference into equal parts, according to the number of steps in a complete round, say twelve. Form scale ASF and the larger scale ASB, on which is shown the perspective measurements of the steps according to their positions; raise verticals such as ef, Gh, &c. From divisions on circumference measure out the central line OP, as in the other figure, and find the heights of the steps 1, 2, 3, 4, &c., by the corresponding numbers in the large scale to the left; then proceed in much the same way as in the previous figure. Note the central column OP cuts off a small portion of the steps at that end.
Fig. 258.
In ordinary cases only a small portion of a winding staircase is actually seen, as in this sketch.
Fig. 259. Sketch of Courtyard in Toledo.
[ CXLVI]
Of the Cylindrical Picture or Diorama
Although illusion is by no means the highest form of art, there is no picture painted on a flat surface that gives such a wonderful appearance of truth as that painted on a cylindrical canvas, such as those panoramas of ‘Paris during the Siege’, exhibited some years ago; ‘The Battle of Trafalgar’, only lately shown at Earl's Court; and many others. In these pictures the spectator is in the centre of a cylinder, and although he turns round to look at the scene the point of sight is always in front of him, or nearly so. I believe on the canvas these points are from 12 to 16 feet apart.
Fig. 260.
The reason of this look of truth may be explained thus. If we place three globes of equal size in a straight line, and trace their apparent widths on to a straight transparent plane, those at the sides, as a and b, will appear much wider than the centre one at c. Whereas, if we trace them on a semicircular glass they will appear very nearly equal and, of the three, the central one c will be rather the largest, as may be seen by this figure.
We must remember that, in the first case, when we are looking at a globe or a circle, the visual rays form a cone, with a globe at its base. If these three cones are intersected by a straight glass GG, and looked at from point S, the intersection of C will be a circle, as the cone is cut straight across. The other two being intersected at an angle, will each be an ellipse. At the same time, if we look at them from the station point, with one eye only, then the three globes (or tracings of them) will appear equal and perfectly round.
Of course the cylindrical canvas is necessary for panoramas; but we have, as a rule, to paint our pictures and wall-decorations on flat surfaces, and therefore must adapt our work to these conditions.
In all cases the artist must exercise his own judgement both in the arrangement of his design and the execution of the work, for there is perspective even in the touch—a painting to be looked at from a distance requires a bold and broad handling; in small cabinet pictures that we live with in our own rooms we look for the exquisite workmanship of the best masters.