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| 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.
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.
This figure shows the application of the square and diagonal in drawing and placing columns in angular perspective.
Fig. 196.
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.
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| Fig. 198. |
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| 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.
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.
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.
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.
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| Fig. 202 A. | Fig. 202 B. |
Fig. 203.
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.
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.
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| 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.
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.
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.
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.
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| Fig. 211 A. | Fig. 211 B. |
Fig. 212.
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| 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.
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.
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| 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.
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| 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.
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| 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°.
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.
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| Fig. 221. | Fig. 222. | Fig. 223. |
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.
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.
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.
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| 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.
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.
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| 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.
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.
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.