EXAMPLE. Required the area of the figure ABCDE, in which CE = 33 feet, BE 22 feet, and the perpendicular AF = 13 feet, BG =14 feet, and DH = 12 feet.

The diagonal BE, 22 feet, multiplied by half the perpendicular AF, 6.5 feet, gives the area of the triangle ABE, 143 square feet; and the diagonal CE, 33 feet, multiplied by half A the sum of the perpendiculars BG, DH, 13 feet, gives the area of the figure BCDE, 429 feet; this, added to the triangle ABE, 143 feet, gives the whole area 572 square feet.

PROBLEM IV.

To find the area of a circle.

E

B

RULE. Multiply the square of the diameter of the circle by the quantity 0.7854, and you will have the sought area.

Note. Instead of multiplying by 0.7854, you may multiply by 11 and divide by 14; the quotient will be the area nearly. This quantity, 0.7854, represents the area of a circle whose diameter is 1; the circumference of the same circle being 3.1416 nearly. The proportion of the diameter to the circumference is expressed in whole numbers by the ratio of 7 to 22 nearly, or more exactly by 113 to 355.*

EXAMPLE. Required the area of a circle ABCD, whose diameter BD is 10.6 feet.

The diameter 10.6 multiplied by itself and by 0.7854 gives the sought area, 88.247544 square feet.

C

PROBLEM V.

To find the area of an ellipsis.

RULE. Multiply the longest diameter by the least, and the product by 0.7854; this last product will be the area required.

EXAMPLE. Required the area of an ellipsis ABCD, whose longest diameter AC is 12 feet, and the shortest diameter BD 10 feet.

The product of the two diameters is 12 × 10=120; this, multiplied by 0.7854, gives the sought area, 94.2480 square feet.

D

B

The area of a sector of a circle may be found by means of the whole area of the circle obtained in Problem IV., by saying, As 360 degrees is to the angle contained between the two legs of the sector, so is the whole area of the circle to the area of the

sector.

There are various regular solids. The most noted are the following:-(1.) A Cube, which is a figure bounded by six equal squares. (2.) A Parallelopiped, which is a solid terminated by six quadrilateral figures, of which the opposite ones are equal and parallel. (3.) A Cylinder, which is a figure formed by the revolution of a rectangular parallelogram about one of its sides. (4.) A Pyramid, which is a solid decreasing gradually from the base till it comes to a point. There are various kinds of pyramids, according to the figure of their bases. Thus, the base be a triangle, the solid is called a triangular pyramid; if a parallelogram, a parallelogramic pyramid; and if a circle, a circular pyramid, or simply a cone. The point in which the pyramid ends is called the vertex, and a line drawn from the vertex perpendicular to the base is called the height of the pyramid.

*This ratio may be easily remembered by observing that, if the first three odd numbers, 1, 3, 5, are repeated twice, they will produce the quantity 113355; the three first figures of which make the first

PROBLEM VI.

To find the solidity of a cube.

RULE. Multiplying the length of a side of the cube by itself, and the product by the same length, gives the solidity required; which will be expressed in cubic feet if the dimensions be given in feet, but in cubic inches if the dimensions be given in inches, &c.

EXAMPLE. If the side AB of the cube be 6.3 feet, it is required to determine the solidity.

The product of 6.3 by 6.3 is 39.69; this, multiplied again by 6.3, gives the solidity 250.047 cubic feet.

D

To find the solidity of a rectangular parallelopiped.

RULE. Multiply the length, breadth, and depth, into each other; the product will be the solidity required.

RULE. Multiply the square of the diameter of the base by the length, and this product by the constant quantity 0.7854; the last product will be the solidity required.

EXAMPLE. Required the solidity of a cylinder ADHF,

whose length DH is 13 feet, and diameter of the base

AD 11 feet.

The diameter 11, multiplied by itself and by the length 13, gives 1573, which, being multiplied by 0.7854, gives the solidity in cubic feet 1235.4342.

PROBLEM IX.

To find the solidity of a grindstone.

H

Grindstones, in the form of cylinders, are sold by the stone of 24 inches diameter, and 4 inches thick. The number of stones that any one contains, may be obtained by the following rule.

RULE. Multiply the square of the diameter in inches by the thickness in inches, and divide the product by 2304, and you will have the number of stones required.

EXAMPLE. Required the number of stones in a grindstone whose diameter is 36 inches, and thickness 8 inches.

The square of the diameter 36 is 1296, which, being multiplied by the thickness 8, gives 10368. This, divided by 2304, gives 4.5, or 44 stones, the solidity required.

This problem may be solved by means of the line of numbers on Gunter's Scale, in a very expeditious manner, by the following rule.

RULE. Extend from 48 to the diameter; that extent, turned over twice the same way, from the thickness, will reach to the number of stones required.

Thus, in the preceding example, the extent from 48 to the diameter 36, turned over twice, from the thickness 8, will reach to 4.5, or 44, which is the number of stones sought.

PROBLEM X.

To find the solidity of any pyramid or cone.

RULE. Multiply the area of the base by one third of the perpendicular height of the pyramid or cone; the product will be the solidity required.

EXAMPLE I. If the pyramid have a square base, the side of which is 4 feet, and the perpendicular height 6 feet, it is required to determine the solidity.

The area of the base is 4 X 4 = 16 square feet; this, being multiplied by one third of the height, or 2 feet, gives 32 feet, the solidity required.

EXAMPLE II. If the diameter of the base of a cone be 10.6 feet, and the perpendicular height 30 feet, it is required to find the solidity.

The area of this base was found in Problem IV. equal to 88.247544; this, multiplied by one third of the height, or 10 feet, gives the solidity required, equal to 882.47544 cubic feet.

Having obtained, by the foregoing rules, the number of cubic feet in any body, you may find the corresponding number of tons by dividing the number of cubic feet by 40, which is the number of cubic feet contained in one ton. Thus, the solidity of the abovementioned cone, 882.47544, being divided by 40, gives 22.061886, which is the number of tons in that cone.

PROBLEM XI.

To find the tonnage of a ship.

By a law of the Congress of the United States of America, the tonnage of a ship is to be found in the following manner:

If the vessel be double-decked, take the length thereof from the fore part of the main stem to the after part of the stern-post above the upper deck; the breadth thereof at the broadest part above the main wales; half of this breadth shall be accounted the depth of such vessel; then deduct from the length three fifths of the breadth, multiply the remainder by the breadth, and the product by the depth; divide this last product by ninety-five, and the quotient will be the true content or tonnage of such vessel.

If the vessel be single-decked, take the length and breadth as above directed in respect to a double-decked vessel, and deduct from the length three fifths of the breadth, and taking the depth from the under side of the deck-plank to the ceiling of the hold, multiply and divide as aforesaid; the quotient will be the true content or tonnage of such vessel.

EXAMPLE. Suppose the length of a double-decked vessel is 80 feet, and the breadth 24 feet, what is her tonnage?

Three fifths of the breadth, 24 feet, is 14.4 feet, which, being subtracted from the length, 80 feet, leaves 65.6. This, multiplied by the breadth, 24 feet, gives 1574.4; this multiplied by the depth, 12 feet (half of 24), gives 18892.8, which, being divided by 95, gives the tonnage 198.9.

Carpenters, in finding the tonnage, multiply the length of the keel by the breadth of the main beam and the depth of the hold in feet, and divide the product by 95; the quotient is the number of tons. In double-decked vessels, half the breadth is taken

GAUGING.

HAVING found the number of cubic inches in any body, by the preceding rules, you may thence determine the content in gallons, bushels, &c., by dividing that number of cubic inches by the number of cubic inches in a gallon, bushel, &c., respectively. A wine gallon, by which most liquors are measured, contains 231 cubic inches. A beer gallon, by which beer, ale, and a few other liquors, are measured, contains 282 cubic inches. A bushel of corn, malt, &c., contains 2150.4 cubic inches; this measure is subdivided into 8 gallons, each of which contains 268.8 cubic inches.

In all the following rules, it will be supposed that the dimensions of the body are given in inches, and decimal parts of an inch.

PROBLEM I.

To find the number of gallons or bushels in a body of a cubic form.

RULE. Divide the cube of the sides by 231, the quotient will be the answer in wine gallons; or by 282, and the quotient will be the answer in beer gallons; or by 2150.4, and the quotient will be the number of bushels.

EXAMPLE. Required the number of wine gallons contained in a cubic cistern, the length of whose side is 62 inches..

Multiplying 62 by itself, and the product again by 62, gives the solidity 238328 which, being divided by 231, gives the content 10313 wine gallons.

PROBLEM II.

To find the number of gallons or bushels contained in a body of the form of a rectangular parallelopiped. (See figure of Problem VII. of Mensuration.)

RULE. Multiply the length, breadth, and depth, together; divide this last product by 231 for wine gallons, by 282 for beer gallons, or by 2150.4 for bushels.

EXAMPLE. Required the number of wine gallons contained in a cistern ABCDFGHE (see fig. Prob. VII. of Mensuration) of the form of a parallelopiped, whose length EF is 66 inches, its breadth FG 35 inches, and its depth DF 24 inches.

Multiplying the length 66 by the breadth 35 gives 2310; multiplying this by the depth 24 gives the solidity 55440, which, being divided by 231, gives 240 wine gallons.

PROBLEM III.

To find the number of gallons or bushels contained in a body of cylindrical form.

RULE. Multiply the square of the diameter by the height of the cylinder, and divide the product by 294.12; the quotient will be the number of wine gallons. If you divide by 359.05, the quotient will be the number of ale gallons; and if you divide by 2738, the quotient will be the number of bushels.

Note. These divisors are found by dividing 231, 282, and 2150.4, by 0.7854 respectively.

EXAMPLE. Required the number of wine gallons contained in the cylinder AFHD (see the fig. of Problem VIII. of Mensuration), the diameter AD of its base being 26 inches, and length DH 18 inches.

The diameter 26 multiplied by itself gives 676; multiplying this by the length 18 gives the solidity 12168, which, being divided by 294.12, gives the answer, 41 wine gallons nearly.

PROBLEM IV.

To find the number of gallons or bushels contained in a body of the form of a pyramid or cone. (See figures of Problem X. of Mensuration.)

RULE. Multiply the area of the base of the pyramid or cone by one third of its perpendicular height; the product, divided by 231, will give the answer in wine gallons. If it be divided by 282, the quotient will be the number of beer gallons; or by 2150.4, the quotient will be the number of bushels.

EXAMPLE. Required the number of beer gallons contained in a pyramid DEFGK (see fig. Prob. X. Example I.), whose base is a square EFGK, a side of which, as EF, is equal to 30 inches, and the perpendicular height of the pyramid is 60 inches.

The square of 30 is the area of the base 900; this, being multiplied by one third of the altitude 20, gives the solidity 18000, which, being divided by 282, gives the answer in beer gallons 63.8.

PROBLEM V.

To find the number of gallons or bushels contained in a body of the form of a frustum of a cone. (See the figure below.)

RULE. Multiply the top and bottom diameters together, and to the product add one third of the square of the difference of the same diameters; multiply this sum by the perpendicular height, and divide the product by 294.12 for wine gallons, by 359.05 for ale gallons, or by 2738 for bushels.

EXAMPLE. Given the diameter CD of the bottom of a frustum of a cone 36 inches, the top diameter AB 27 inches, and the perpendicular height EF 50 inches; required the contents in wine gallons.

The product of the two diameters, 36 and 27, is 972; their difference is 9, which, being squared and divided by 3, gives 27; adding this to 972 gives 999, which, being multiplied by the height 50, gives the solidity 49950; dividing this by 294.12 gives the content in wine gallons 169.8.

PROBLEM VI.

To gauge a cask.

To gauge a cask, you must measure the head diameters, AF, CD, and take the mean of them when they differ; measure also the diameter BE at the bung (taking the measure within the cask); then measure the length of the cask, making due allowance for the thickness of the heads. Having these dimensions, you may calcu late the content, in gallons or bushels, by the following rule:

RULE. Take the difference between the head and bung diameters; multiply this by 0.62, and add the product to the head diameter; the sum will be the mean diameter; multiply the square of this by the length of the cask, and divide the product by 294.12 for wine gallons, by 359.05 for beer gallons, or by 2738 for bushels.

The quantity 0.62 is generally used by gaugers in finding the mean diameter of a cask. But if the staves are nearly straight, it will be more accurate to take 0.55, or less;* if, on the contrary, the cask is full on the quarter, it will be best to take 0.64 or 0.65.

F

E

D

EXAMPLE. Given the bung diameter EB 34.5 inches, the head diameter AF CD 30.7 inches, and the length 59,3 inches; required the number of wine gallons this cask will hold. The difference of the two diameters, 34.5 and 30.7, is 3.8; this being multiplied by 0.62, gives 2.4 nearly, to be added to the head diameter 30.7 to obtain the mean diameter 33.1. The square of 33.1 is 1095.61; multiplying this by the length 59.3, gives the solidity 64969.673; dividing this by 294.12, gives the content in wine gallons 220.9.

B

*In the example to Problem V. preceding (which may be esteemed as the half of a hogshead with staves perfectly straight), the multiplier is only 0.51. For this, being multiplied by 9 (the difference between AB and CD), produces 4.59 or 4.6 nearly; adding this to 27 gives 31.6, whose square, being