The √h1 + ( D - d 2 / 2) will give the slant height S. The full equation will therefore be

Area = π {D+d/2 √h1+( D - d / 2)2}+d2π /4 = 3.1416 { 4+2 /2 √6.25+1}+ 3.1416 = 28.519 square inches.

Fig. 271.

Diameter of blank = √28.5l9 x 1.12838 = 6. 026 inches.

The above method is approximately correct.

Another method of dealing with fig. 271 is to carefully set out the figure on paper, when the mean diameter at M M will be found to be 3 in., and the slant height to be 2.75 in., and the bottom of the vessel d being 2 in. diameter; then

Area = (3 x π x 2.75) + d2π /4

= 25.9182 + 3.1416 = 2906 square inches.

Diameter of blank = √29.06 x 1.12838

= 6.083 inches diameter.

Fig. 272, Elliptical flat-bottomed vessel:

Area - (circumference x h) + (area of bottom)

= 10.79925 + 5.4978 = 16.297 square inches.

Then an elliptical blank 3.45 in. x 6.037 in. will have its diameters in the same ratio as those in the vessel, fig. 272, and will contain 16.359 square inches, which is slightly larger than area actually required to be in the blank. Fig. 273, Rectangular vessel:

Fig. 272.

Note.-The length of sides of the rectangular blank must be in the same ratio as the sides of the vessel, fig. 273; so that, since the sides of the vessel are 2.5 in. and 3.5 in., the sides of the rectangular blank will be 3.86 to 5.404.

Then 3.86 in. x 5.404 in. = 20.859 square inch.

It will here be noticed that the area of the blank is slightly in excess of that of the example; this size blank will, however, be sufficiently near for all practical purposes.

Fig. 273.

Fig. 274, Semi-oblate spheroid: A formula which gives the approximate area of a whole spheroid is as follows:

Area = π D √ D2+d2 / 2 = 44.430 square inch.

Therefore 22.215 square inch = area of semi-oblate spheroid.

Fig. 274.

Fig. 275, Cylindrical conical-topped vessel:

Area = { π d h + d2 π / 4 } + π D + d /2 S

S = slant height, and may be obtained in the same manner as was done in the case of fig. 271.

The full equation for fig. 275 will therefore be

Fig. 275.

The diameter of required blank =√12.268 x 1.12838

= 3.952 inches.

Note.-The 1.12838 is a constant used to find diameters of blanks for all cylindrical vessels.

Another method by which the area of fig. 275 may be found is-

Area = πdh + (mean dia. of conical top x π x S) + -when πdh = belt of shell and d2π/4= bottom of shell, and mean diameter of conical top x π x s = area of conical belt.

Set out the conical top on paper, when the mean diameter at M M, fig. 275, will be found equal 1.625 in., and the slant height S equal to .625 in.

The equation will be-

Area = π d h + d 2π /4 + (1.625 x 3.1416 x .625)

= 12.271 square inch area.

Fig. 276, Example of cylindrical stamping: The simplest way to measure this is to first lay out the outline of the figure on paper as shown, then treat each section of the figure separately. It will be seen that the figure has been divided into six sections, and numbered 1st, 2nd, 3rd, 4th, 5th, and 6th. These divisions are bisected to obtain their mean diameters (see each section, d1, d2, d3, d4, d5, and d6,).

Fig. 276.

The area of each section should be now found by separate working, and all the six areas added finally together. The area of the bottom of the vessel is found, and adding this to the areas of the six sections, previously found, will give the approximate total area of the vessel.

The semi-oblate spheroid may be treated in two sections, as will be seen at figs. 278 and 279. This enables the area to be obtained more accurately than by the formula) shows the method by which the figure can be traced upon paper to enable the figure to be measured up in sections. This is done by an approximate method used for constructing an ellipse by means of arcs of circles. Draw the major and minor arcs, bisecting each other at right angles; draw the rectangle OBEC; bisect BE at F; draw C F and E D intersecting each other in G; bisect C G by a line at right angles to it; the bisecting line meets the line CD in J, the centre from which the arc L C G is described. Complete the quadrantal arc C L K, join K A, and produce to L, and join L J; the point M is the centre from which the arc L A is described; the ellipse can be completed by symmetry shows the one section, which is a slice cut off the top, and it will be seen that the ellipse or figure has been cut through at L G, thereby forming the spherical segment LCG.

Area = πD

Fig. 277.

Fig. 277.

Fig. 278.

Area = 2 π R h e = length of the curve LA.

= 2 x 3.1416 x 2.4375 x .825 = 9.572 square inches.

Fig. 279 gives the remaining section or bottom half of the ellipse, in the form of a band or belt. The area of this band = e x 2πa. First find e.

Fig. 278.

Fig. 279.

Then area of band = 1.015 x 6.2832 x 1.875 = 11-957 square inches.

Therefore the area of semi-oblate spheroid = 9.572 + 11.957 = 21.529 square inches.

When stamping articles of irregular shape from sheet metals, such, for instance, as fluted cake pans, the work is usually stamped from a piece of sheet metal larger in area than what is actually required to form the finished article, the surplus metal being afterwards clipped off by special clipping tools made to suit the particular shape of the work. Further, as the work varies in shape and depth, so does the number of stamp blows and required number of annealings vary accordingly. But when raising articles in the power press, it is usual to make the blank the necessary area to simply form the required article. In such cases there will necessarily be provision made upon the top of the bottom die to ensure that an operator shall place the blank perfectly central upon the die, so that when the article is raised the raising shall be done evenly all round the circumference of the finished article. Such a pair of dies for raising are seen at fig. 280, where it will be noticed that three bits of metal have been screwed down upon the top of the bottom die for the purpose of evenly adjusting the blank, as before explained.

Fig. 280.