XLV

FURTHER ILl.u.s.tRATION OF THE DESCENDING PLANE

In the centre of this picture (Fig. 102) we suppose the road to be descending till it reaches a tunnel which goes under a road or leads to a river (like one leading out of the Strand near Somerset House). It is drawn on the same principle as the foregoing figure. Of course to see the road the spectator must get pretty near to it, otherwise it will be out of sight. Also a level plane must be shown, as by its contrast to the other we perceive that the latter is going down hill.

XLVI

FURTHER ILl.u.s.tRATION OF UNEVEN GROUND

An extended view drawn from a height of about 30 feet from a road that descends about 45 feet.

[Ill.u.s.tration: Fig. 103. Farningham.]

In drawing a landscape such as Fig. 103 we have to bear in mind the height of the horizon, which being exactly opposite the eye, shows us at once which objects are below and which are above us, and to draw them accordingly, especially roofs, buildings, walls, hedges, &c.; also it is well to sketch in the different fields figures of men and cattle, as from the size of these we can judge of the rest.

XLVII

THE PICTURE STANDING ON THE GROUND

Let _K_ represent a frame placed vertically and at a given distance in front of us. If stood on the ground our foreground will touch the base line of the picture, and we can fix up a standard of measurement both on the base and on the side as in this sketch, taking 6 feet as about the height of the figures.

[Ill.u.s.tration: Fig. 104. Toledo.]

XLVIII

THE PICTURE ON A HEIGHT

If we are looking at a scene from a height, that is from a terrace, or a window, or a cliff, then the near foreground, unless it be the terrace, window-sill, &c., would not come into the picture, and we could not see the near figures at _A_, and the nearest to come into view would be those at _B_, so that a view from a window, &c., would be as it were without a foreground. Note that the figures at _B_ would be (according to this sketch) 30 feet from the picture plane and about 18 feet below the base line.

[Ill.u.s.tration: Fig. 105.]

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.

[Ill.u.s.tration: 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 a.s.sistance, 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_.

[Ill.u.s.tration: 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

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.

[Ill.u.s.tration: Fig. 108.]

LII

HOW TO PUT A GIVEN LINE INTO PERSPECTIVE