The Theory and Practice of Perspective Part 22

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 pa.s.sing through their intersections. Then mark with a strong line the outlines of the hexagonals, shading some of them; but the figure explains itself.

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

First draw the hexagon on _AD_ as in the previous figure, dividing _AD_ into four, &c., set off right and left s.p.a.ces 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.

[Ill.u.s.tration: Fig. 208.]

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.

[Ill.u.s.tration: Fig. 209.]

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.

[Ill.u.s.tration: Fig. 210.]

CXVII

FURTHER ILl.u.s.tRATION OF THE HEXAGON

[Ill.u.s.tration: Fig. 211 A.]

[Ill.u.s.tration: Fig. 211 B.]

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 s.p.a.ces 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.

[Ill.u.s.tration: Fig. 212.]

CXVIII

ANOTHER VIEW OF THE HEXAGON IN ANGULAR PERSPECTIVE

[Ill.u.s.tration: 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.

[Ill.u.s.tration: 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

[Ill.u.s.tration: Fig. 215.]

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

CXX

THE PENTAGON

[Ill.u.s.tration: Fig. 216.]

The pentagon is a figure with five equal sides, and if inscribed in a circle will touch its circ.u.mference 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 circ.u.mference 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.

[Ill.u.s.tration: Fig. 217.]

[Ill.u.s.tration: 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.

[Ill.u.s.tration: Fig. 219.]

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.

[Ill.u.s.tration: Fig. 220.]