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.
A perspective line AB being given, find its actual length and the angle at which it is placed.
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| 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.
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.
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.
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.
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.
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.
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.
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.
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.
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.
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.
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.
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.
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.
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.
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.
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.
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).
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.
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.
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| Fig. 137 A. |
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| 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.
| figure | figure |
| Fig. 137 C. | Fig. 137 D. |
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.
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| Fig. 138. | Fig. 139. |
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.
| figure | figure |
| Fig. 140. | Fig. 141. |
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.
| figure | figure |
| 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.
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.
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.
| figure | figure |
| Fig. 145. | |
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.
| figure | figure |
| Fig. 146. | |
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.
| figure | figure |
| Fig. 147. | |
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.
| figure | figure |
| Fig. 148 A. | Fig. 148 B. |
| figure | figure |
| Fig. 149 A. | Fig. 149 B. |