Or, by the sliding rule. On D. is 18,94—the gaugepoint for ale or beer gallons, marked A. G: and 17,14– the gauge-point for wine gallons, marked W.G. Set the gauge-point to the length of the cask on C. and against the mean diameter, on D. you will have the answer in ale or wine gallons, accordingly as which gauge-point you make use of.* Case 2.—To gauge round tubs, 8c. RULE.—Multiply one diameter by the other, and to that product add one third of the square of their difference; multiply this sum by the length, and divide as before for beer or wine. EXAMPLE, What is the content, in beer and wine gallons, of a round tub, whose diameter at the top, within, is 40 inches, and at the bottom 34 inches, and the perpendicular height 36 inches ? 6x6 34x40+12x36 -=1371 +- beer gal. 10-34=6 3 359 Ans. And 34x40+12X36 ---=168 wine gal. Ans. 234 CASE 3.-To gauge ao square vessel. RULE.-Multiply the length by the breadth, and that product by the depth; then divide by 282 for beer or ale, (the inches in a beer or ale gallon,) and by 231 for wine, brandy, &c., (the inches contained in a wine gallon,) and the quotient will be the answer. *A rule which has been given as generally more exact, is this; multiply the product of the square of the mean diameter and the length, by 34, and point off four places from the right of the product; the figures on the left of the point will be the gallons, and those on the right decimal parts of a gallon, in wine or cider. Let the dimensions be taken exactly in inches and tenths. Take the preceding Example in Case 1. 30+25 =27,5 mean diam. and 27,5x27,5X40X34=1028500,00=102,65 gallons of wine or cider; which is 766 gal. less than by the other. EXAMPLE If a square vessel be 80 inches in length, 60 in breadth, and 40 inches deep, what is its content in beer and wine gallons ? 80 x 60 x 40 wine gal. 80x60x40 beer gal. - -=831,16+ Ans. — ---=680,85+ Ans. 231 282 Nore.—The content of any vessel, in feet, gallons, and bushels, may be thus found: Measure the inside of the vessel, according to the rule of the figure, and find the content in cubic inches; then, ( 1728 and the (Cubic feet. ) 282 l quotient Ale or beer gal. Divide by 3 031 I will be the Wine gallons. ( 2150,425 ) content in ( Bushels. Section IV.- To find the tonnage of a ship. RULE.--Multiply the length of the keel by the breadth of the beam, and that product by the depth of the hold; divide the last product by 95, and the quotient is the tonnage. EXAMPLE If a ship be 72 feet by the keel, 24 feet by the beam, and 12 feet deep, what is the tonnage ?. 72x24x12--95=218,2+tons. Ans. Or, RULE 2.-Multiply the length of the keel by the breadth of the beam, and that product again by half the breadth of the beam; divide the last product by 94, and the quotient is the tonnage.* EXAMPLE What is the tonnage of a ship that is 84 feet by the keel, and 28 feet by the beam ? 28;2=14 84x28x14:94=350,29+ lons. Ans. Note. The breadth of the beam added to two thirds the length of the keel, gives the length of a ship's main * Rule established by Congress. For double decked vessels ; length from fore part of main stem to afterpart of stern post, above upper deck ; breadth at widest part, above main wales, half of which is called the depth; déduct from length 3-5ths of breadth ; multiply the remainder by widih, . mast : Therefore the length of the mainmast of the ship last mentioned, is 84x2=3+28=84 feet. To find the thickness, the proportion is, as 84 to 28, so is the length of a mast in feet, to its thickness in inches. Consequently the thickness of the mast whose length was just found, is 28 inches. Section V.- To find the weight of anchors which cables may sustain. Rule.-As the strength of cables, and consequently the weights of their anchors, are proportioned to the cubes of their peripheries; therefore, as the cube of the periphery of any cable, is to the weight of its anchor, so is the cube of the periphery of any other cable, to the weight of its anchor. i EXAMPLES. 1. If a cable of 6 inches round require an anchor of 2 cwt., what would be the weight of an anchor, for a 12 inch cable ? cwt. 6x6x6 : 2,25 :: 12x12x12 : 18cwt. Ans. 2. If a 12 inch cable require an anchor of 18cwt. what must the circumference of a cable be, for an anchor of 2{cwt. ? cwt. cut. 18: 12x12x12::2,25:216. 3/216=6in. Ans. Section VI.-From one solid's capacity to find another's. Rule.-As the cube of any dimension is to its given weight, so is the cube of any like dimensions to its weight. EXAMPLES. 1. If a ship of 300 tons' burthen be 75 feet by the keel, what is the burthen of one, 100 feet by the keel, of Yike form ; 251b. a qr. ? tons. . tons. cwt. t. 753 : 300 :: 1003 : 711 2 222. Ans. 2. If a brass cannon, 11 finch. diameter, weigh 100011, what will another, 20,83 inch. diameter, of like metal and shape, weigh? Ans. 5942,5697th.t , and the product by depth; divide by 95; the quotient is the true tonnage, For single-decked, take depth from under side of deck plank to ceiling in the hold; and proceed as before.. SECTION VII. Rule.—Multiply the length, breadth, and thickness together, agreeably to the rule of Duodecimals, and the last product will be the solid content of the pile, parcel, or load. EXAMPLE. · If a load of wood be 8 feet 4 inches long, 3 feet 8 inches wide, and 4 feet 6 inches high, how many cubic feet does it contain: ft. ft., ft., 84x3 8X4 6=137) sol. ft. Ans. CASE 2.--To find how many cords of wood or bark are contained in any pile, foc. RULE.--Find the solid content as before, and divide that product by 128; the quotient will be the cords, and the remainder cubic feet, or so many 128ths of a cord. Or, divide the solid content of the pile, &c. by 16, and the quotient will be cord-feet, 8 of them being 1 cord, and the remainder so many 16ths of a cord-foot. EXAMPLES 1. In a pile or load of wood 9 feet 4 inches long, 3 feet 8 inches wide, and 4 feet 9 inches high, how many cords, and how many cord-feet? ft. ft. ft. , sol.ft. ft. 94X3 8X4 9=162 6 6 And 162 68:128=1 cord, . and 34 sol. ft. 6 8. Ans. 1. ft. , Or, 1626 8:16=104cord ft. 68=1 cord 24 feet. Ans. 2. If a load of wood or bark be 8 feet long, 4 feet wide, and 2 feet 6 inches high, how many cord-feet does it contain ? Ans. 5 feet, or of a cord. MODE OF ASSESSING TAXES. It may not, perhaps, be here amiss to show the general method of Assessing Taxes. But as the quantity of new matter with which we have enlarged this edition of our Work, bus extended its pages considerably beyond the limits first intended., a brie! explanation of the general principle and rule, will, we trust, fully suffice for the purpose. ARGUMENT. There is a certain towo which contains 8 inhabitants, whom we will call A, B, C, D, E, F, G and H. The town is divideil into 2 school districts or classes, which are numbered 1 and 2. A, B, C, and D, from District No 1, and E, F, G, and H, No. 2. On these inhabitants the following taxes are to be assessed, namely: State, $14, 88cts. 6m. County, 19, 84cts. 8m. Town, 39, 69cts. 6m. School, 29 77cts. 205. And the Highway Tax is to be equal to each person's amount of inventory.* 'The first step is, to learn at what rates the various species of property are to be taxed, agreeably to the laws of the State by which they have been fixed; and for that purpose all assessors consult, of course, the latest acts that have been passed on the subject. Let a poll be taxed $1,30; an acre of orchard 25cts ; an acre of tillage 16ets ; an acre of mowing 16cts. ; an acre of pasturage 4cts. ; an ox of five years 35cts.; and a cow 20cts. id In order to find each person's proportion of the several taxes, and each school district's proportion of school money, according to the rateable estates of the members of each district or class, or accord ing to the number of scholars in each district; each man's inventory must be taken, and the amount cast by the following rule. RULE.-Multiply the value of a poll by his number of polls ; his acres of orchard by the tax-value of one ; vis number of oxen by the tax-rate of one ; and so of every other kind of property add the products, and the sum is the amount of his rateable estate; find the amount of all in the same way; add these amounts, and their sum is the value of the inventory of the town. I demand the rateable estate of A, who has 2 Polls, at $1,30 amount to $2,60 0,52 at 0,35 0,70 0,80 Amount of A's rateable estate, $4,42 Find the other amounts, in the following inventory, in the same way. To prove the Inventory. RuiE. - Add up the column of polls, and multiply the sum by the value of one : add up each of the other columns, and multiply its sum by the tax-rate of one in that column; then add the several amounts of the columns together and the sum will be equal to the total amount of the Inventory, if the work be right. EXAMPLE The total amount of rateable estates in the following inventory, is $49,62cts. And proceeding by the method given in the rule of proof, the sum of the products is $49.62cts. It is, therefore, evident the work is right. * Assess money taxes so far over the sum to be raised, as to meet abatements, |