Let a b c d, Fig. 37., be the portion of the cone required

As it is given in magnitude, its diameters must be given at the base and summit, A B and c d; and its vertical height, c e.1

Problem XIV To Draw A Truncated Circular Cone Give Perspective Elements 93

Fig. 37.

And as it is given in position, the centre of its base must be given.

Draw in position, about this centre2, the square pillar a f d, Fig. 38., making its height, b g, equal to c e; and its side, a b, equal to A B.

In the square of its base, a b c d, inscribe a circle,

1 Or if the length of its side, A C, is given instead, take a e, Fig. 37., equal to half the excess of A B over C D; from the point e raise the perpendicular c e. With centre a, and distance A c, describe a circle cutting c e in c. Then c e is the vertical height of the portion of cone required, or c E.

2 The direction of the side of the square will of course be regulated by convenience which therefore is of the diameter of the base of the cone, A B.

In the square of its top, e f g h, inscribe concentrically a circle whose diameter shall equal c D. (Coroll. Prob. XIII.)

Problem XIV To Draw A Truncated Circular Cone Give Perspective Elements 94

Fig. 38.

Join the extremities of the circles by the right lines, k l, n m.

Then k l n m is the portion of cone required.

Corollary I

If similar polygons be inscribed in similar positions in the circles k n and l m (Coroll. Prob. XII.), and the corresponding angles of the polygons joined by right lines, the resulting figure will be a portion of a polygonal pyramid. (The dotted lines in Fig. 38., connecting the extremities of two diameters and one diagonal in the respective circles, occupy the position of the three nearest angles of a regular octagonal pyramid, having its angles set on the diagonals and diameters of the square, a d, enclosing its base.)

If the cone or polygonal pyramid is not truncated, its apex will be the centre of the upper square, as in Fig. 26.

Corollary II

If equal circles, or equal and similar polygons, be inscribed in the upper and lower squares in Fig. 38., the resulting figure will be a vertical cylinder, or a vertical polygonal pillar, of given height and diameter, drawn in position.

Corollary III

If the circles in Fig. 38., instead of being inscribed in the squares b c and f g, be inscribed in the sides of the solid figure b e and d /, those sides being made square, and the line b d of any given length, the resulting figure will be, according to the constructions employed, a cone, polygonal pyramid, cylinder, or polygonal pillar, drawn in position about a horizontal axis parallel to b d.

Similarly, if the circles are drawn in the sides g d and e c, the resulting figures will be described about a horizontal axis parallel to a b.

Problem XIV

Several Exercises will be required on this important problem.

I. It is required to draw a circular flat-bottomed dish narrower at the bottom than the top; the vertical depth being given, and the diameter at the top and bottom.

Problem XIV Perspective Elements 121

Fig. 65.

Let a b, Fig. 65., be the diameter of the bottom, a c the diameter of the top, and a d its vertical depth.

Take A D in position equal to a c.

On A D draw the square A B C D, and inscribe in it a circle.

Therefore, the circle so inscribed has the diameter of the top of the dish.

From A and D let fall verticals, A E, D H, each equal to a d.

Join E H, and describe square E F G H, which accordingly will be equal to the square A B C D, and be at the depth a d beneath it.

Within the square E F G H describe a square I K, whose diameter shall be equal to a b.

Describe a circle within the square I K. Therefore the circle so inscribed has its diameter equal to a b; and it is in the centre of the square E F G H, which is vertically beneath the square A B C D.

Problem XIV Perspective Elements 122

Fig. 66.

Therefore the circle in the square I K represents the bottom of the dish.

Now the two circles thus drawn will either intersect one another, or they will not.

If they intersect one another, as in the figure, and they are below the eye, part of the bottom of the dish is seen within it.

To avoid confusion, let us take then two intersecting circles without the enclosing squares, as in Fig. 66.

Draw right lines, a b, c d, touching both circles externally. Then the parts of these lines which connect the circles are the sides of the dish. They are drawn in Fig. 65. without any prolongations, but the best way to construct them is as in Fig. 66.

If the circles do not intersect each othei, the smaller must either be within the larger or not within it.

If within the larger, the whole of the bottom of the dish is seen from above, Fig. 67. a.

If the smaller circle is not within the larger, none of the bottom is seen inside the dish, b.

If the circles are above instead of beneath the eye, the bottom of the dish is seen beneath it, c.

If one circle is above and another beneath the eye, neither the bottom nor top of the dish is seen, d. Unless the object be very large, the circles in this case will have little apparent curvature.