In Fig. 553 is shown a perspective view of a soap-maker's float. In general characteristics it is very similar to piece No. 2 of the hip bath treated in the preceding problem. It also resembles the bathtub in that its bottom is bulged or raised with the hammer, and is therefore not included in the field of accurate pattern cutting. The sides are to be considered as parts of two cones having elliptical bases, the short diameters of which are alike, but the long diameters of which vary.

In Fig. 554 is shown a plan and an inverted elevation of the flaring sides, showing in dotted lines the completed cones of which the sides form a part. Thus LD1M represents one-half the base of an elliptical cone of which L M is the short diameter and D1 K1 one-half the long diameter. As all sections of a circular cone taken parallel to its base are perfect circles, so all sections of an elliptical cone parallel to its base must be ellipses of like proportions with the base. Therefore the plan of the upper base of the frustum A1 P must be so drawn that a straight line from A1 to P will be parallel to a straight line joining D1 and L.

Fig. 553.   Perspective View of Soapmaker's Float.

Fig. 553. - Perspective View of Soapmaker's Float.

For the pattern of the portion shown by D A E F of the elevation, first produce the line F E of the elevation in the direction of K indefinitely. In like manner produce D A of the elevation until it reaches F E produced in the point K. Then DKF may be regarded as the elevation of a half cone, of which that part of the vessel is a portion, and K F its perpendicular hight. Next, divide one-half the plan L D1 M into any number of equal parts, as shown by the small figures 1, 2, 3, etc. Construct the diagram of triangles, shown in Fig. 555, by drawing the line D K1 of indefinite length, and the line K1 K at right angles to it, making K1 K in length equal to F K, Fig. 554. Establish the point K2 by making the distance K1 K2 equal to E F of Fig. 554. Draw K2 A parallel to K1 D. From each of the points 2, 3, 4, etc., of the plan draw lines to the center K1, and set off distances equal to these lines upon the line K1 D of Fig. 555, measuring from K1 toward D. From each of the points thus obtained draw lines to the point K, cutting A K2 With one foot of the compasses in the point K, and the other brought successively to the points 1, 2, 3, etc., in the line D K1 and also to the points in the line A K2, describe arcs indefinitely.

Fig. 554.   Plan and Inverted Elevation of Soapmaker's Float.

Fig. 554. - Plan and Inverted Elevation of Soapmaker's Float.

Take in the dividers a space equal to the divisions in D1 L of Fig. 554, and, commencing at the point a in arc 7 (Fig. 555), step to arc 6 and thence to arc 5, and thus continue stepping from one arc to the next until the entire stretchout of the half plan has been laid off, as shown in Fig. 555. From each of the points thus obtained in a d draw lines to K, cutting arcs of corresponding number drawn from A K2 Then a line traced through the several points of intersection thus obtained, as shown by b c and a d, will be the boundary lines of the pattern.

The pattern for the other end of the article is to be, in the main, developed in the same manner as above described. One additional condition, however, exists in connection with this piece, viz.: To determine the dimensions of the cone of which this piece (E B C F, Fig. 554) is a part, since the flare at B C is much greater than A D, while the flare of both pieces at the side is the same as shown by P L of the plan. The quarter ellipse P B1 of the plan being given, and also the point C1, it becomes necessary to draw from C1 a quarter ellipse which shall be of like proportions with P B1, which, as remarked above, is a necessary condition, both being horizontal sections of the same cone. To do this proceed as follows: Connect the points P and B1 by means of a straight line. From the point C1 draw a line parallel to P B1, and produce it until it cuts the line L G1, which is a straight line drawn at right angles to L M. Then Gl becomes a point in the lower base of the cone corresponding to the point P in the upper base. Draw the line G1 P, and continue it until it intersects the long diameter in IP. Drop the point G1 vertical from the plan on to the base line D C of the elevation, as indicated by the point G. Draw a line through the points G and E, which produce indefinitely in the direction of H. In like manner produce the side C B of the cone until it intersects G E produced in the point H. Then it will be found that the point H of the elevation and the point H' of the plan coincide, as indicated by the line H H1 H G of the elevation thus represents the axis and H C one side of the cone of which the piece E B C G is a part, from which it will be seen that this cone is at once elliptical and scalene. The operation of developing the pattern from this stage forward is the same as in the previous case, all as clearly shown in Fig. 556, save only in the addition of the triangular piece indicated by G E F of the elevation. After completing the other portions of the pattern, this triangular piece is added as follows: The distance H1 L in Fig. 556 is to be set off on the line H1 C in the same manner as the distances to the other points - i.e., H1 L is equal to IP L of Fig. 554. Then L is to be treated in the same manner as the other points, an arc being struck from it, as indicated in the engraving, by which to determine the corresponding point L2 in the outline of the pattern. L2 G2 is made equal to L G1 of the plan, Fig. 554. From L2 draw a line to E. Then E L2 G2 will be the pattern of the triangular piece indicated in Fig. 554 by E F G. It is to be added upon the opposite end of the pattern in like manner, as indicated by E1 G1 L1

Fig. 555.   Diagram of Triangles and Pattern of Piece LD1 M of Fig. 554.

Fig. 555. - Diagram of Triangles and Pattern of Piece LD1 M of Fig. 554.

Fig. 556.   Diagram of Triangles and Pattern of Piece LC1 M of Fig. 554.

Fig. 556. - Diagram of Triangles and Pattern of Piece LC1 M of Fig. 554.