In Fig. 587, the elevation of the article is shown by A B D C, below which is drawn the plan of the same, corresponding parts in each being connected by the vertical dotted lines. There are two methods of triangulation available in the solution of this problem only one of which is given in this connection.

Fig. 687.   Plan and Elevation of an Irregular Flaring Article, with Lines of Triangulation Shown in the Plan.

Fig. 687. - Plan and Elevation of an Irregular Flaring Article, with Lines of Triangulation Shown in the Plan.

Divide one-half of the circle representing the base of the article into any convenient number of spaces, as indicated by the small figures, 1, 2, 3, etc. In like manner divide the inner circle, which represents the top, into the same number of spaces, as indicated by 1', 2', 3', etc. Between the points of like numbers in these two circles, as for example between 2 and 2', 3 and 3', etc., draw lines, as shown; also connect the points in the inner circle with points in the outer circle of the next higher number, as indicated by the dotted lines. Thus, connect 1' with 2, and 2' with 3, and so on, as shown. These lines just drawn across the plan are the bases of a number of right angled triangles whose altitudes are equal to the hight of the article, and whose hypothenuses, when drawn, will give the correct distances across the pattern, or envelope of the article, between the points in the top and those in the bottom in the direction indicated by the lines of the plan. The triangles having the solid lines of the plan as their bases arc shown in Fig. 588, while those constructed upon the dotted lines are shown in Fig. 589 and are obtained in the following manner:

At any convenient point erect a perpendicular, E F, Fig. 588, which in length make equal to the straight hight of the article, as shown in the elevation. From F at right angles set off a base line of indefinite length. On this line, measuring from F, set off lengths equal to the several solid lines in the plan. For example, make the space F 10 equal to the length 10

10' in the plan, and the space F 9 equal 9 9' in the plan, and so on, until F 1 is set off equal to 1 1' in the plan. From the points thus established in the base line draw lines to the point E. The triangles thus constructed will represent sections through the article on the solid lines in the plan. In other words, the several hypo then uses of the triangles shown in Fig. 588 are equal in length to lines measured at corresponding points on the surface of the completed article.

In like manner construct the triangles shown in Fig. 589, representing measurements taken on the dotted lines shown in the plan. Draw the perpendicular K G, equal in length to the straight hight of the article. From K lay off a horizontal base line indefinite in length, drawing it at right angles to K G. From K set off lengths equal to the dotted lines in the plan - that is, making the distance K 10 equal to the distance 9' 10 in the plan, and K 9 equal to the distance 8' 9 in the plan, and so on until K 2 is made equal to the distance 1' 2 in the plan. From the points thus established in the base line draw lines to the point G. Then the hypothenuses of the triangles thus constructed will equal measurements along the surface of the completed article at points corresponding to the dotted lines in the plan. With distances thus established upon the surface of the article, and with the stretchout of the required pattern determined at both top and bottom, it is easy to lay out the pattern upon the general plan of constructing a triangle when the three sides are given.

Fig. 588.   Diagram of Triangles Based upon the Solid Lines of the Plan in Fig. 587.

Fig. 588. - Diagram of Triangles Based upon the Solid Lines of the Plan in Fig. 587.

Fig. 589.   D'agram of Triangles Based upon the Dotted Lines of the Plan in Fig. 587.

Fig. 589. - D'agram of Triangles Based upon the Dotted Lines of the Plan in Fig. 587.

The development of the pattern can be begun at either end according to convenience, and the operation is conducted as follows: Assume any line, as 1 1' of Fig. 590, which make equal in length to A B of the elevation, or, what is the same thing, equal to E 1 of

Fig. 588, which is one side of the first triangle. The other two sides are respectively the distance 1 2 of the plan and the bypothenuse of the triangle shown in Fig. 589 corresponding to the line 2 1' of the plan. Accordingly, take the distance 1 2 of the plan in the dividers, and from 1 as center describe a short arc. Then, taking the distance G 2 of Fig. 589 in the dividers, and with 1' as a center, intersect the arc already struck, thus establishing the point 2 of the pattern.

Fig. 590.   The Pattern of Article Shown in Fig. 587.

Fig. 590. - The Pattern of Article Shown in Fig. 587.

The elements of the second triangle arc the distance 1' 2' on the inner circle of the plan, the hypothennse of the triangle in Fig. 588 corresponding to the line 2 2' in the plan and the side just established in the pattern 2 1'. From 1' as center, with 1' 2' of the plan as radius, describe a short arc. Then from 2 as center, with E 2 of Fig. 588 as radius, describe a second arc, intersecting the one already made, thus establishing the point 2'. Proceed in this manner until triangles have been described adjacent to each other, corresponding to the divisions first established in the plan. Then lines traced through the points thus established, as shown from 1 to 10 and from 1' to 10', will constitute the pattern of half the article. The other half may be obtained by any' convenient means of duplication and may be added on to cither end of the half already obtained, according as it is desired to make the joint at the widest or narrowest part of the pattern.

Fig. 591   Model of One Half the Article Shown in Fig. 587, Illustrating the Construction and Use of the Triangles.

Fig. 591 - Model of One-Half the Article Shown in Fig. 587, Illustrating the Construction and Use of the Triangles.

In Fig. 591 is shown a model which may be constructed of thin metal and wires, or of cardboard and threads, according to convenience, which will assist the student in forming a correct idea of the relationship existing between the various lines drawn upon the plan and the lines of which the pattern is constructed. The top and bottom of the model are duplicates of the inner and outer circles of the plan. The piece forming the bottom should have the solid lines and the inner circle of the plan drawn upon it as a means of placing in position the several triangular pieces shown, which are duplicates of the several triangles shown in Fig. 588. These triangular pieces having been placed in position according to their numbers, and fastened at the top and bottom, their outer edges, or hypothenuses, will then represent the solid lines drawn across the pattern, and will bear the same relation or angle to the edges of the top and bottom pieces of the model that the solid lines of the pattern bear to the top and bottom outlines of the pattern. Finally, threads or wires having been attached, as shown, will represent the dotted lines drawn across the pattern, and will bear the same angle to the edges of the solid triangles, as measured upon the model, that the dotted lines of the pattern bear to the solid lines, as measured upon the pattern.

Since the top and the bottom of the shape here shown are both round and horizontal, the figure becomes that of the frustum of a scalene cone; and, therefore, its sides, if continued upward, would terminate in an apex which can be made the common apex of a number of triangles whose bases are the spaces upon the outer line of the plan. This method of solving the problem as applied to a full scalene cone is given in Problem 163, which see.