Let U T V in Fig. 523 be the elevation of a cone, at right angles to the axis of which another cone or frustum of a cone, O F G P, is to miter. Let L K N M be a section of the frustum on the line F G. Let U2 W V2 be a half plan of the larger cone at the base. The first step in describing the patterns is to obtain the miter line in the elevation, as shown by the curved line from 0 to P. With this obtained the development of the pattern is a comparatively simple operation.

To obtain the miter line O P proceed as follows: Divide the profile LKNM into any convenient number of equal parts, as shown by the small figures. Inasmuch as the divisions of this profile are used in the construction of the sections - or, in other words, since sections through the cone must be constructed to correspond to certain lines through this profile - it is desirable that each half be divided into the same number of equal parts, as shown in the diagrams. Thus 2 and 2, 3 and 3, 4 and 4 of the opposite sides correspond, and sections, shown in the upper part of the diagram, are taken upon the planes which they represent. From the points in the profile LKNM draw lines parallel to B E cutting the end F G of the frustum. Produce the sides O F and P G until they meet in E, which is the apex of the cone. Through the points in F G draw lines from E, producing them until they cut the axis of the cone, as shown at A, A1, A2

Next construct sections of the cone as it would appear if cut through upon the lines A C, A1 B, A2 D. Divide the plan U2 W V2 into any convenient number of parts. From the points thus established carry lines vertically to the base line U V, and thence carry them toward the apex T, cutting the lines A C, A1 B, A2 D, all as shown. Through each of the several points of intersection in these lines draw horizontal lines from the axis of the cone to the side, all as shown. At right angles to the lines A C, A1 B, A2 D project lines to any convenient point at which to construct the required sections. Upon the lines drawn from the points

A, A1, A2 locate at convenience the points A3, A4, A5. Inasmuch as A1 B is at right angles to the axis of the cone, the section corresponding to it will be a semicircle whose radius will be equal to A1 B. Therefore, from A3 as center, with radius A1 B, describe the semicircle S Bl R. For the section corresponding to A3 D lay off from A5 the distances A5 S2 and A5 R2, in a line drawn at right angles to A3 D of the elevation, each in length equal to the horizontal line drawn through the point A3 from the axis to the side of the cone. At right angles to S2 R2 draw A5 D1, in length equal to A3 D of the elevation. Set off in it points 5, 4 and 2, corresponding to similar points in A3 D of the elevation. Through these points 5, 4 and 2, at right angles to A5 D1, draw lines indefinitely. From A' as center, with radius equal to the length of horizontal line passed through point 5 in A3 D of the elevation, describe an arc cutting line 5 of the section. From the same center, with a radius equal to the length of the horizontal line drawn through point 4 in the line A2 D of the elevation, strike an are cutting the line 4, etc. Then a line traced through these points, as shown by S2 D1 R2, will be the section of the cone as it would appear if cut on the line A3 D of the elevation. In like manner obtain the section S1 C1 R1, corresponding to A C of the elevation.

These sections may, if preferred, be obtained in the manner described in connection with Problems 159 and 160.

As these sections are obtained solely for the purpose of determining at what point in their perimeters - that is, at what distance from points C1, B1and D1 - they will be intersected by the lines representing the points 2, 3 and 4 of the profile L K M N, it is not necessary that the complete half sections should be developed. In the engraving, the small intersecting cone has so little flare that the lines A C and A3 D cross the large cone so nearly at right angles to its axis that sections 2 2 and 4 4 could be constructed with sufficient accuracy for practical purposes, as in the case of section 33, by small arcs of circles with radii respectively equal to A C and A2 D, and of only sufficient length to include the points c c and d d.

Fig. 523. - The Patterns of Two Unequal Cones Intersecting at Right Angles to their Axes.

Prolong A5 D1, as shown by E3, making A5 E3 in length equal to A1 E of the elevation. In like manner make A1 E3 and A3 E1 equal to A E and A1 E of the elevation respectively. At right angles to these lines in the sections set off F3 G3, F3 G3, F4 G4, in position coresponding to F G of the elevation. Make the length of F2 G2 equal to the length across the section of the frustum marked 2 2. In like manner make F2 G3 equal to 3 3, and F4 G4 equal to 4 4 of the section. From E1, E2 and E3 respectively, through these points in the several sections, draw lines cutting the oblique sections just obtained. From the several points of intersection between the lines drawn from E1, E2, E3 and the sections of the cone, as shown by d d, c c and b b, carry lines back to the elevation, intersecting the lines A C, A1 B, A2 D. Then a line traced through these several intersections, as shown from O to P, will be the miter line in elevation.

Having thus obtained the miter line, proceed to describe the patterns, as follows: For the envelope of the small cone, from E as center, with radius E G, describe the arc F1 G1, upon which set off the stretchout of the section L K M N. Through the points in this arc, from E, draw radial lines indefinitely. From E as center, with radii corresponding to the several points in the miter line O P, but obtained from the oblique sections above, Cut corresponding radial lines. Thus with the radius E2 d cut lines 4 and 4, with the radius E2 C cut lines 2 and 2 and with radius E1 b cut lines 3 and 3.

Then a line traced through these points of intersection, as shown by P1 O1 P1, will be the shape of the pattern of the frustum to lit against the larger cone.

For the pattern of the larger cone, from T as center,with radius T U, describe the arc V1 U1, in length equal to the circumference of the entire plan of the cone. From the points in the miter line O P carry lines parallel to the base of the cone cutting its side T U, as shown between O3and P3. also through the points in O P draw lines from the apex cutting the base and thence carry them vertically to the plan.

These points can be numbered upon the side of the cone to correspond with the plan, but entirely independent of the system of numbers employed upon the smaller cone. Upon the are V1 U1 set off points corresponding to the points just obtained in the plan from the miter line, from which draw lines toward the center T. With one foot of the compasses Bet at the point T, bring the pencil point successively to the points between O3 and P2 and cut radial lines of corresponding number in the pattern. Then a line traced through these intersections, as shown by X Y Z Y1, will be the shape of the opening to be cut in the envelope of the larger cone, over which the smaller cone will lit, and T U1 V1 will be the envelope of the entire cone.