spective ratio by the dividing-point, the line E G being

used as a measuring-line.

[Obs. In drawing Fig. 31. the station-point has been taken much nearer the paper than is usually advisable, in order to show the cha racter of the curve in a very distinct form.

If the student turns the book so that E G may be vertical, Fig. 31 will represent the construction for drawing a circle in a vertical plane, the sight-line being then of course parallel to G L; and the semicircles A D B, A C B, on each side of the diameter A B, will represent ordinary semicircular arches seen in perspective. In that case, if the book be held so that the line E H is the top of the square, the upper semicircle will represent a semicircular arch, above the eye, drawn in perspective. But if the book be held so that the line G F is the top of the square, the upper semicircle will represent a semicircular arch, below the eye, drawn in perspective.

If the book be turned upside down, the figure will represent a circle drawn on the ceiling, or any other horizontal plane above the eye; and the construction is, of course, accurate in every case.]

PROBLEM XII.

TO DIVIDE A CIRCLE DRAWN IN PERSPECTIVE INTO ANY

GIVEN NUMBER OF EQUAL PARTS.

LEI A B, Fig. 32., be the circle drawn in perspective. It is required to divide it into a given number of equal parts; in this case, 20.

Let KA L be the semicircle used in the construction. Divide the semicircle K A L into half the number of parts required; in this case, 10.

Produce the line E G laterally, as far as may be

necessary.

From o, the centre of the semicircle K A L, draw radii through the points of division of the semicircle, p, q, r, &c., and produce them to cut the line E G in P, Q, R, &c.

From the points P Q R draw the lines P P', Q Q', R R', &c., through the centre of the circle A в, each cutting the circle in two points of its circumference.

Then these points divide the perspective circle as required.

K

Fig.

If from each of the points p, q, r, a vertical were raised to the line E G, as in Fig. 31. and from the point where it cut E G a line were drawn to the vanishing point, as q o' in Fig. 31., this line would also determine two of the points of division.

If it is required to divide a circle into any number of given unequal parts (as in the points A, B, and C, Fig. 33.), the shortest way is thus to raise vertical

lines from A and B to the side of the perspective square x Y, and then draw to the vanishing-point, cutting the perspective circle in a and b, the points required. Only notice that if any point, as A, is on the nearer side of the circle A B C, representative

its

point, a, must be on the nearer side of the circle a bc, and if the point в is on the farther side of the circle A B C, b must be on the farther side of a b c. If any point, as c, is so much in the lateral arc of the circle as not to be easily determinable by the vertical live, draw the horizontal c P, find the correspondent p in the side of the perspective square, and draw p c parallel to x Y, cutting the perspective circle in c.

COROLLARY.

It is obvious that if the points P', q', R, &c., by which the circle is divided in Fig. 32., be joined by right lines, the resulting figure will be a regular equilateral figure of twenty sides inscribed in the circle. And if the circle be divided into given unequal parts, and the points of division joined by right lines, the resulting figure will be an irregular polygon inscribed in the circle with sides of given length.

Thus any polygon, regular or irregular, inscribed in a circle, may be inscribed in position in a per spective circle.