the picture: the completion of this problem will be very easy, if the drawing of the squares is well understood.

In such simple objects as these it will not be necessary to draw a plan; when one side is parallel to the picture, and dimensions are known. In fig. 2, the same objects as those in fig. 1 are drawn without a plan thus:

Draw the ground line M M, then the vertical line S' V, and the horizontal line D D, at the height of the eye; making DD the same distance on each side of V, that the eye is from the transparent plane; for drawing the squares mark off from S' to b', on the ground line, the distance that the square is on one side of the observer; let b'a' be the length of one side of the square; from b' and a' draw lines to V, which represent the sides of the square carried on indefinitely; to cut off the required perspective width of the side b'e' of the square, lay off the width, a' b', from b' to p, then draw from p to D on the left, and the point e' where the line Dp intersects b' V, will give the apparent width required; then draw ƒ" e' parallel to a'b', and the square is complete: this may be proved in the same way as in fig. 1. The further square may be obtained in a similar manner, setting off the distance between the squares from p to q, and the width of the square beyond that, and drawing lines to D as before: some of the lines in this plate are not continued to the ground line, in order to avoid confusion. Proceed with the cubes by the same rule. Let 1, 2, 3, 4, be the size of one side of the cube if continued until touching the picture; from these points draw rays to V: from 3 to t set off the distance the cube is from the picture, and from t to r, the width of the cube; draw from these points to D on the right, and their intersection of the line 3 V in m, o, will give the perspective width and position of that side of the cube: draw lines perpendicular to the ground line from m and o, and lines parallel to 4-2 from the angles of the cube, l', g', m; then draw the side n h', and the cube is complete. The operation of drawing the other cube is similar, and easy to be understood.

From the drawing of a square in parallel perspective, we deduce rules for the construction of a scale in perspective. Let DMM D, (fig 6,) be the plane of the picture, the same letters of reference being used as in preceding figures. From S' lay off the distance o S' equal to some unit of measure, as may be most convenient; from o draw the diagonal to D the point of distance; now draw 1 1' parallel to the ground line M M, again draw from 1' the diagonal 1' D, and lay off the parallel 22′, proceed in the same way with the diagonal 2' D and the parallel 3 3', and extend the construction as far as may be necessary. It is evident o S' 11', 1'122', 2'233' are the perspective projectors of equal squares, and

therefore o S', 11', 22' 3 3', etc., and S' 1, 1 2, 2 3, etc., are equal to each other, and that if o S' is set off to represent any unit of measure, as one foot, one yard, or ten feet, &c., each of these lines represents the same dis

tance, the one being measures parallel to the base line, the others perpendicular to it. In making a perspective drawing a scale thus drawn will be found very convenient; but as in the centre of the picture it might interfere with the construction lines of the object to be put in perspective, it is better that the scale be transferred to the side of the picture a Mo, the diagonals to be laid off to a point to the right of D equal to the point of distance.

The scales thus projected are for lines in the base or ground plane; for lines perpendicular to this plane the following construction is to be adopted; upon any point of the base line removed from S', as a for instance, erect a perpendicular, a d; on this line, lay off as many of the units o S' as may be necessary; in this example three have been laid off, that is, a d

30 S'. From a and d draw lines to the centre of view, and extend the parallels 11', 2 2', 33'; at the intersection of these lines with a V erect perpendiculars. The portions comprehended between the lines a V and d V will be the perspective representations of the line a d, in planes at distances of 1, 2, 3, o S' from the base line, and as b, c, d are laid off at intervals equal to o S', by drawing the lines c V and b V six equal squares are constructed, of which the sides correspond to the unit of measure, o S'.

To determine the Perspective Position of any point in the Ground Plane. Thus (fig. 7), to determine the position of the point p, which in plane would be six feet distant from the plane of the picture, M M, and ten feet from the central plane, to the left.

Lay off from S', to the left, the distance a S', equal to six feet on the

scale adopted; draw the diagonal to the point of distance D, on the right; at its intersection a' with the vertical line V S', draw a parallel to M M the base line; lay off from S', S'b equal to ten feet, draw 6 V; the intersection of this line p, with the parallel previously drawn, will be the position of the point required.

By a similar construction the position of any point in the ground plan may be determined. It is not necessary that the distances should be expressed numerically; they may be shown on the plan and thence be transferred to the base line, and thrown into perspective by the diagonals and parallels. As the intersections of the various lines of the outlines of objects are points, by projecting perspectively these points, and afterwards. connecting by lines, the perspective of any plane surface, on the ground plane, may be shown.

If the point p were not in the ground plane, but in a position directly above that already assumed, that is, the distances from the plane of the picture, and the central plane being the same, but its distances above the ground plane were, say, five feet; then at b erect a perpendicular, and lay off bb' equal to five feet, connect b' V, at p erect another perpendicular, and its intersection p' with the line b' V will be the position of the point required.

Or the plane of the point p' might be assumed as the position of the ground plane, M' M' becoming the base line, and laying off from S", S" a" and S" b' — equal respectively to six and ten feet; drawing the diagonal a" D and b' V and the parallel as before, the point p' will be determined.

To draw an Octagon in Parallel Perspective.-Let A (fig. 8) represent the plan of an octagon. Draw M M, S' V, and DD, as before; from the points M, a, b, c, draw rays to V. Set off on M M from c to the right the distances ce, cd, cf, from which draw diagonals to D on the left, and at their intersection with the ray c V, draw parallels e' g', d' h', k' l', to the base line; these points will correspond to the angles on the plan. Now