To find the solidity of a Grindstone. 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 multiplied by the thickness 8 gives 10368. This divided by 2304 gives 4.5 or 4 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

D

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 41⁄2, 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 has 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 4x4-16 square feet, this multiplied by one third of the height or 2 feet gives 32 feet the solidity required.

G

K

E

F

A

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, quotes 2,061886, 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 which 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 threefifths 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, 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 divided by 95 quotes the tonnage 198.8.

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 for the depth.

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 divided by 231 gives the content 1031 wine gallons.

PROBLEM II.

To find the number of gallons or bushels contained in a body of the form of a rectangular Parallelepiped. See the 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; and 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 parallelepiped, whose length EF is 66 inches, its breadth GF 35 inches, and its depth FD 24 inches. Multiplying the length 66 by the breadth 35 gives 2310, this multiplied by the depth 24, gives the solidity 55440; which divided by 231 quotes 240 wine gallons.

PROBLEM III.

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

.7854.

EXAMPLE. Required the number of wine gallons contained in the cylinder AFHD (See the fig. of Prob. VIII. of Mensuration) the diameter AD of its base being 26 inches, and length HD 18 inches? The diameter 26 multiplied by itself gives 676, this multiplied by the length 18, gives the solidity 12168, which divided by 294.12, gives the answer 41 wine gallons.

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 onethird of its perpendicular height; the product divided by 231 will give the answer in wine gallons; if divided by 282, the quotient will be the number of beer gallons; and if divided 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 multiplied by one-third of the altitude 20, gives the solidity 18000, which 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 frustrum of a cone. (See the figure below.)

A

B

F

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, and by 2738 for bushels. EXAMPLE. Given the diameter DC of the bottom of a frustrum of a cone 36 inches, the top diameter AB=27 inches, and the perpendicular height, FE 50 inches. Required the content in wine gallons? The product of the two diameters 36 and 27 is 972; their difference is 9, which squared and divided by 3 gives 27; this added to 972 gives 999, which D multiplied by the height 50 gives the solidity

EC

49950; this divided by 294.12 quotes the content in wine gallons 169.

PROBLEM VI.

To gauge a cask.

To gauge a cask, you must measure the head diameters FA, DC, and take the mean of them when they differ; measure also the diameter EB 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 calculate the content in gallons or bushels, by the following rule.

RULE. Take the difference between the head and bung diameters, multiply this by .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, and by 2738 for bushels.

The quantity .62 is generally used by gaugers in finding the mean diameter of a cask; but if the staves are nearly straight, it would be more accurate to take 55 or less; if, on the contrary, the cask is full on the quarter, it would be best to take .64 or .65.

EXAMPLE. Given the bung diameter EB-34.5 inches, the head diameter FA=DC=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 multiplied by .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, this multiplied by the length 59.3 gives the solidity 64969.673, which, divided by 294.12, quotes the content in wine gallons 220.8. To gauge a cask by means of the line of numbers on Gunter's Scale, or on the calipers used by gaugers.

Make marks on the scale at the points 17.15, 18.95, and 52.33, which numbers are the square roots of 294.12, 359.05, and 2738, respectively. A brass pin is generally fixed on the calipers at each of these points, which are called the gauge points. Having prepared the scale in this manner, you may calculate the number of gallons or bushels by the following

RULE. Extend from 1 towards the left hand to .62, (or less if the staves be nearly straight) that extent will reach from the difference between the head and bung diameters, to a number to the left hand, which added to the head diameter will give the mean diameter; then put one foot of the compasses upon the gauge point-which is 17.15 for wine gallons, 18.95 for ale gallons, and 52.33 for bushels-and extend the other to the mean diameter; this extent turned over twice the same way, from the length of the cask, will give the number of gallons or bushels respectively.

In the preceding example the extent from 1 to .62 will reach from 3.8 to 2.4 nearly, which added to 30.7 gives the mean diameter 33.1.

Then the extent from the gauge point 17.15 to 33.1, turned over twice from the length 59.3, will reach to 220.8 wine gallons.

If you had used the gauge point 18.95, the answer would have been in ale gallons; and if you had used 52.33, the answer would have been in bushels.

SURVEYING.

LAND is generally measured by a chain of 66 feet in length, divided into 100 equal parts called links, each link being 7.92 inches.

A pole or rod is 164 feet, or 25 links, in length; hence a square pole contains 2724 square feet, or 625 square links.

An acre of land is equal to 160 square poles, and therefore contains 43560 square feet, or 100,000 square links.

To find the number of square poles in any piece of land, you may take the dimensions of it in feet, and find the area in square feet, as in the preceding Problems; divide this area by 43560, the quotient will be the number of acres: or by 272.25, and the quotient will be the number of square poles. If the dimensions be taken in links, and the area be found in square links, you may obtain the number of acres by dividing by 100000 (that is, by crossing off

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 .51. For this multiplied by 9, the difference between AB and DC, produces 4.59 or 4.6 nearly, which, added to 27, and the sum 31.6 squared, multiplied by 50 and divided by 294.12 quotes 169 gallons nearly..

the five right hand figures;) and the number of square poles may be obtained by dividing by 625.

PROBLEM I.

To find the number of acres and poles in a piece of land in the form of a rectangular parallelogram.

RULE. Multiply the base by the perpendicular height, and divide by 625, if the dimensions were taken in links, but by 272.25, if they were taken in feet; the quotient will be the number of poles, which, divided by 160, will give the number of acres.

EXAMPLE I. Suppose the base DB (see the figure of Ex. I. Prob. I. of Mensuration) of the rectangular parallelogram ACBD is 60 feet, and the perpendicular BC 25 feet; required the area in poles?

The product of the base 60 by the perpendicular 25 gives the content 1500 square feet, and by dividing by 272,25, we obtain the answer in square poles 5.5.

PROBLEM II.

To find the number of acres and poles in a piece of land in the form of an oblique-angular parallelogram. (See the figure of Prob. I. Ex. IV. of Mensuration.)

RULE. This area may be found in exactly the same manner as in the preceding Problem, by multiplying the base AD by the perpendicular height BE, and dividing by 625, when the dimensions are taken in links, but by 272.25 when taken in feet; the quotient will be the answer in poles, which divided by 160 will give the answer in acres.

EXAMPLE. Suppose the base AD is 632 links, and the perpendicular BE 326 links; required the number of poles?

Multiply the base 632 links by the perpendicular 326 links, the product 206032 divided by 625, gives the answer in poles 329.

PROBLEM III.

To find the number of acres and poles in a piece of land of a triangular form.

RULE. Multiply the base by the perpendicular height, and divide the product by 1250 when the dimensions are given in links, but by 544.5 when they are given in feet; the quotient will be the answer in poles.

NOTE. Instead of dividing by 1250, you may multiply by 8, and cross off the four right hand figures.

EXAMPLE. Given the base AC (see fig. of Prob. II. of Mensuration) equal to 300 feet, and the perpendicular BD 150 feet; required the area in poles? Multiply the base 300 by the perpendicular 150, the product 45000, divided by 544.5, quotes the answer in poles 82.6.

PROBLEM IV.

To find the number of acres and poles in a piece of land of any irregular right-lined figure.

RULE. Find the area as in Problem III. of Mensuration, by drawing diagonals, and reducing the figure to triangles: the base of each triangle being multiplied by the perpendicular, (or by the sum of the perpendiculars falling on it) and the sum of all these products divided by 1250 when the dimensions are given in links, but by 544.5 when in feet, will give the area of the figure in poles.

EXAMPLE. Suppose that the piece of land is of the same form as the figure in Prob. III. of Mensuration, and that EB=22 feet, EC=33 feet, AF=18 feet, BG=14 feet, and DH=12 feet: it is required to find the area in poles?

The product of EB 22 feet, by AF 13 feet, gives double the triangle EAB 286 square feet; and the diagonal EC 33 feet, multiplied by the sum of the perpendiculars BG, DH, 26 feet, gives double the figure BCDE, 858 square feet; the sum of this and 286, divided by 544.5 gives the area 2,1 or 2 poles.