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.
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| 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.
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.
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.
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| 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.
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| Fig. 155. | Fig. 156. |
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.
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| 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.
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| 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.
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.
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| Fig. 159. | Fig. 160. |
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.
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.
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.
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.
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| Fig. 165. | Fig. 166. |
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.
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.
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| Fig. 169. |
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| Fig. 170. |
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| 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.
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| Fig. 172. | Fig. 173. |
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.
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.
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.
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| 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.
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.
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| 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.
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.
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.
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| 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.
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| 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.
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.
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.
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.
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.
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| 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.