=

Here 16 x 8 + 122 x 3 + 16 x 12 x 4 = 256 × 8 + 144 × 3 + 192 × 4 = 2048 + 432 + 768 3248; and 3248 × 20 × .05236 = 64960 × .05236 =3401.3056, the solidity required.

2. What is the solidity of the frustum of a parabolic spindle whose length is 25, middle diameter 20, and end diameter 15? Ans. 6643.175.

3. The length of a cask, in the form of the middle frustum of a parabolic spindle, is 30, the bung diameter 24, and the head diameter 18 inches; what is its content in ale and wine gallons?

Ans. 40.7071 ale, and 49.6944 wine gallons.

PROBLEM XIX.

To find the solidity of a hyperboloid.

RULE.

To the square of the radius of the base, add the square of the diameter in the middle between the base and the vertex; then this sum being multiplied by the altitude, and the product thus obtained by .5236, will give the solidity.

EXAMPLES.

1. What is the solidity of the hyperboloid ACB, the altitude Cr being 25, the radius Ar of the base 26, and the middle diameter mn 34?

m

Here 262 +342 = 6761156

1832; and 1832

× 25 ×.5286 = 45800 × .5236 = 23980.88, the solidity

required.

2. The altitude of a hyperboloid is 75, the radius of the base 78, and the middle diameter 102; what is the solidity? Ans. 647483.76.

PROBLEM XX.

To find the solidity of the frustum of a hyperboloid.

RULE.

To the sum of the squares of the greatest and least semi-diameters, add the square of the middle diameter ; then this sum being multiplied by the altitude, and the product thence arising by .5236, will give the solidity.

EXAMPLES.

1. What is the solidity of the hyperbolic frustum ABCD, the diameter AB of the greater end being 20, the diameter CD of the less end 12, the middle dia meter mn 17, and the altitude rs 24 ?

Here 10% +62 +172 = 100 + 36 +289 = 425; then 425 x 24 x .5236 = 10200 x .5236=5340.72, the solidity of the frustum.

2. The diameter of the greater end of a hyperbolic frustum is 40, the diameter of the less end 24, the middle diameter 34, and the altitude 48; what is the solidity? Ans. 42725.76.

PROBLEM XXI.

To find the solidity of a hyperbolic spindle.
RULE.

To the square of the greatest diameter, add the square of double the diameter taken at of the length; then this sum being multiplied by the length, and the product thus obtained by .1309, will give the solidity, nearly.

Note. When great accuracy is not required, this Rule may be used for any spindle formed by the revolution of a conic section, or part of a conic section, about its axis, as it will always give nearly the true solidity.

EXAMPLES.

1. What is the solidity of a hyperbolic spindle, whose length is 36, the greater diameter 24, and the diameter at of the length 16.73338?

Here 242 +16.73338 × 2;2 =242 + 33.466762 =376 +112.00240249 1696.0240249; then 1696.0240249 X 36 x .1309 61056.8648964

=

7992.34361493876, the answer required.

1309

==

2. The length of a hyperbolic spindle is 30, the greatest diameter 20, and the diameter at of the length 13.94449; what is the solidity?

PROBLEM XXII.

Ans. 4625.20177.

To find the solidity of the middle frustum of an elliptic or a hyperbolic spindle.

RULE.

To the sum of the squares of the greatest and least diameters, add the square of double the diameter taken exactly in the middle between them; and this sum being multiplied by the length, and the product again. by .1309, will give the solidity.

Note. This Rule will also give nearly the content of any frustum or segment formed by the revolution of a conic section, or part of a conic section, either about the axis of the section, or about any other line.

EXAMPLES.

1. The length of the middle frustum of an elliptic spindle is 60, the greatest diameter 48, the least diameter 36, and the diameter in the middle between them 45.23635; what is the solidity?

=

Here 482 + 362 + 45.23635 × 2o| = 482 + 362 + 90.47272 = 2304 + 1296 +8185.30944529 11785.30944529; then 11785.30944529 × 60 ×.1309 = 707118.5667174 × .1309 = 92561.82038330766, the solidity required.

2. The length of the middle frustum of a hyperbolic spindle is 40, the greatest diameter 32, the least diameter 24, and the diameter in the middle between them 29; what is the solidity? Ans. 25991.504.

3. What is the content of the middle frustum of any spindle; the length being 50, the greatest diameter 40, the least diameter 30, and the diameter in the middle between them 37.69696 ? Ans. 53565.871567. 4. What is the content of the segment of any spindle; the length being 15, the greatest diameter 12, and the middle diameter 9? Ans. 918.918.

PART VIII.

GAUGING.

GAUGING is the art of finding the capacities or contents of all sorts of vessels used by Maltsters, Brewers, Distillers, Wine Merchants, Victuallers, &c. &c. ; such as cisterns, couches, coppers, coolers, tuns, vats, stills, casks, &c. &c.

Note 1. By this measure, wines, spirits, cyder, mead, perry, vinegar, and oil are bought and sold; and the computations in gauging these liquors are made by it.

2. One gallon, wine measure, contains 231 cubic inches; consequently a pint is 28.875, and a quart 57.75 cubic inches.

2 pints

A Table of Ale and Beer Measure.

4 quarts, or 8 pints

9 gallons, or 36 quarts

2 firkins, or 18 gallons

4 firkins, 2 kilderkins, or 36 gallons 1 barrel, bar.

6 firkins, 3 kilderkins, or 54 gallons 1 hogshead, hhd. 2 hogsheads, 12 firkins, or 108

gallons

}

1 butt, butt.

2 butts, 4 hogsheads, or 216 gallons 1 tun, tun.

Note. By this measure, ale, beer, porter, and water are measured.

2. One gallon, ale measure, contains 282 cubic inches; consequently a pint is 35.25, and a quart 70.5 cubic inches.

Note 1. By this measure, corn, seeds, roots, fruits, sand, and salt are mea

sured.

1