157. PROBLEM VII.—To find the content of a frustum of a cone or pyramid, or of a prismoid or cylindroid.

I Find the volume, and multiply or divide it by the proper multiplier or divisor for the required denomination; or,

To the areas of the ends add 4 times that of the middle section, multiply the sum by the height, and divide or multiply the product by the divisor or multiplier for prismoidal vessels, or by one-sixth of those for square vessels.'

For the rule for the volume of a prismoid in article 347 (Part I.) is applicable to frustums of pyramids and cones, and of cylindroids.

Let E, e, e', be the sides of the ends and of the middle section of a frustum of a pyramid with a regular polygonal base, then, as it is a prismoid, the formula in article 347 (Part I.) is applicable to it. Let B, b, be the areas of the ends, and M that of the middle section.

}

Then (B+b+4M)=¿h { E2 + e2 + (2 e')2 } a' For e' (E+e), and (2 e')2=E2+ e2+2 Ee. Hence, v = h (E2 + e2 + Ee) a', which is the third formula in article 342 (Part. I.)

It is similarly proved that the rule is true for the frustum of a cone, or that v=h{ D2 + d2 + (2 d')2 } × ·7854.

By the Sliding Rule.

When the ends are regular polygons or circles, set the gauge point on D to the length on C; then opposite to the sides, or the diameters of the ends, and to twice the side or diameter of the middle section on D, will be found three numbers on C, the sum of which is the content."

When the ends are rectangles, find the mean proportionals between the sides of each, and use these as the sides of the ends of a frustum of a square pyramid by the rule above; or when the area of each end is the product of two factors, find their mean proportionals, and use them in the same manner. When the ends of a conic frustum are ellipses, find the mean proportional between the major and minor axes of each end, and use them as the diameters of an equivalent frustum of a cone with circular ends. If the gauge points for prismatic and cylindric vessels are used, only one-sixth of the length must be taken.

EXAMPLES.

1. Find the content in imperial gallons of a vessel, which is a frustum of a square pyramid, the sides of its ends being 78 and 42 inches, and its depth 60 inches.

and

=

v = f h { E2 + e2 + (2 e')} a', where a' 1
v = 10 (782 + 422 +1202) = 222480,

c=

v 222480

=

m 277.274

= 802.4 imperial gallons.

Or set 40.79 on D to 60 on C, then 78 on D gives 218 on C

Content in imperial gallons = 802

Instead of the prismoidal gauge point 40-79, the square prismatic gauge point 16-65 may be taken, and only of the height; thus-Set 16·65 on D to 10 on C, then 78 on D gives 218 on C, and similarly for the rest.

2. What is the content in imperial gallons of a frustum of a regular hexagonal pyramid, the sides of its ends being 72 and 48 inches, and its depth 72 inches?

It is found that v = 682400-25 cubic inches,

Or set 10.33 on D to 12 on C, then 72 on D gives 579 on C

...

...

...

48 120

258 1620

Content in imperial gallons = 2457

Here 12, which is of the height, is taken; but if a table were calculated containing prismoidal gauge points for a hexagon, that number, and the whole height, 72 inches, might then be taken.

EXERCISES.

1. What is the content in imperial gallons of a frustum of a square pyramid, the sides of its ends being 52 and 28, and its depth 36 inches? Ans. 213.97.

2. What is the content in imperial gallons of a frustum of a regular hexagonal pyramid, the sides of its ends being 54 and 36, and its depth 48 inches? Ans. 922.9.

3. Find the content in imperial gallons of a frustum of a rectangular pyramid, the sides of its greater end being 36 and 16, those of its smaller end 27 and 12, and its depth 80 inches. Ans. 131-565. 4. What is the content of a conic frustum in imperial gallons, the diameters of its ends being 44 and 16, and its depth 40 inches? Ans. 109.4.

5. What is the content in imperial gallons of a vessel of the form of an elliptic cone, the diameters of one end being 48 and 42, and those of the other 40 and 34 inches, the corresponding axes of the ends being parallel, and the depth 30 inches? Ans. 164.8.

158. The contents of other solids can be found by determining their volumes by the rules in Part I., and then dividing by the proper number for the required measure of quantity. The contents of many solids of rather irregular figures can be calculated by means of the first and second rules of article 443 (Part I.); which may also be used for regular figures, as portions of conoids and spheres.

159 PROBLEM VIII. To gauge mash-tuns, stills, and other brewing and distilling vessels.

'Divide the vessel into small portions by means of planes parallel to its base; find the areas of the middle sections of these portions, and multiply these areas by the corresponding depths of the portions to which they belong: the products are the cubic contents of the portions, and the sum of these contents is the whole content; divide the whole content by the number corresponding to the required measure or weight, and the result is the required content."

The vessel is divided into portions of 6 or of 10 inches deep, according as the sides are more or less inclined; and so that the difference of the corresponding diameters of two successive middle sections may not differ by more than one inch.

When the vessel is nearly circular at any depth, cross,

that is, perpendicular diameters are taken at the middle of any portion, and the mean of them is considered to be the diameter of a cylinder of the same depth with that portion, whose volume is nearly equal to it. In this case the volumes of the different portions are calculated as cylinders. Let d the mean diameter of a cylindric portion,

then

a = the area of the section for 1 inch deep,
m = the divisor for cylindric vessels,

Set m on D to d on C, and opposite d on D is a on C. Or set the gauge point g on D to 1 on C, and opposite to d on D is a on C (see 151).

EXAMPLE. Find the content, in imperial gallons, of an underback, the form of which is nearly the frustum of a cone, from the following dimensions, the cross diameters being measured at the middle of the several portions into which the vessel is divided, their depths being those in the first column:—

The whole depth is 38 inches. The area to 1 inch deep of the middle section of the first portion is found thus :ad2 ÷ 353 69-42 ÷ 353 = 13.64.

Or set 353 on D to 69-4 on C, and against 69-4 on D is 13.6 on C.

Or set the gauge point 18-8 on D to 1 on C, and against 69.4 on D is 13.6 on C.

In the same manner, the areas to inch deep for the other

middle sections is found to be, in order, 14-73, 15-3, 15.43; and each of these being multiplied by the depths, and the sum of the results taken, it will be the content as under :

For the 1st portion, content = 13.64 × 8 = 109·12
For the 2d portion, content = 14·73 × 10 = 147-3
For the 3d portion, content = 15·3 × 10 153-0
For the 4th portion, content = 15·43 × 10 = 154·3

Content of vessel in imperial gallons = 563-7

160. It is usual to construct a table containing the contents of a fixed vessel, as of a mash-tun or still, for every inch in depth. The contents for the first inch of depth from the bottom of the preceding vessel is 15.43; for two inches, it is the double of this, or 30-86; for three inches, it is three times this, or 46-29; and so on. For the area of each of the first ten inches, it is 15-43; for each of the next ten inches, it is 15·3; and a table is thus easily constructed.

EXERCISES.

1. Find the content in imperial gallons of a flat-bottomed copper, the mean diameters at the middle of four portions into which it is divided by horizontal sections being as under :

The bottoms of coppers are seldom flat; they are generally rising or falling, that is, convex or concave internally. The content of a vessel with a rising or falling crown, as the bottom is in this case called, is found by calculating, as in the preceding example, the content above the centre of the crown when it is rising, and then adding the content of