ART. 14. To find the tonnage of a ship. RULE. Multiply the tength of the keel by the breadth of the beam, and that product by the depth of the hold, and divide the last product by 95, and that quotient by the tonnage. EXAMPLE. EXAMPLE. Suppose a ship 72 feet by the keel, and 24 feet by the beam, and 12 feet deep; what is the tonnage ? 72x24x12+95=218,2+tons. Ans. RULE IL. Multiply the length of the keel by the breadth of the beam, and that product by half the breadth of the beam, and divide by 95. A ship 84 feet by the keel, 28 feet by the beam; what is the tonnage 2 84 x 28 x1495x350,29 tons. Ans. ART. 15. From the proof of any cable, to find the strength of another. RULE. Is to the weight of its anchor; EXAMPLES 1. If a cable 6 inches about, require an anchor of 2 cwt. of what weight must an anchor be for a 12 inch cable ? A3 6x6x6 : 21cwt. ; : 12x12x12 : 18cwt. Ans. 2. If a 12 inch cable require an anchor of 18 cwt. what must the circumference of a cable be, for an anchor of 21 cwt, ? cwt. cut. in. As 18 : 12x12x12 : : 2,25 : 216 216=6 Ans. Art. 16. Having the dimensions of two similar built ships of a different capacity, with the burthen of one of them, to find the burthen of the other. a RULE. The burthens of similar built ships are to each other, as the cubes of their like dimensions. EXAMPLE. If a ship of 300 tons burthen be 75 feet long in the keel, I demand the burthen of another ship, whose keel is 100 feet long? T.cwt.qrs.lb. As 75x75x75 : 300 :: 100x100x100 : 711 % 24+ DUODECIMALS, CROSS MULTIPLICATION, Is a rule made use of by workmen and artificers in casting up the contents of their work. RULE. 1. Under the multiplicand write the corresponding doo nominations of the multiplier. 2. Multiply each term into the multiplicand, beginning at the lowest, by the highest denomination in the multiplier, and write the result of each under its respective term; observing to carry an unit for cvery 12, from each Jower denomination to its next superior. 3. In the same manner multiply all the niultiplicand by the inches, or second denomination, in the multiplier, and set the result of each term one place removed to the right hand of those in the multiplicand. 4. Do the same with the seconds in the multiplier, set: ting the result of each term two places to the right hand of those in the multipicard, &c. EXAMPLZS. 4 6 9 7 9 7 87 99 225 6 91 10 1 29 0 Product, 39 29 FEET, INCHES AND SECONDS. F. I. Multiply 9 8 6 Ву 7 9 3 [tiplier. 67 11 6 =prod. by the feet in the nrúl. 7 3 4 6 =ditto by the inches. 2 5 1 6=ditto by the seconds. How many square feet in a board 16 feet 9 inches long, and 2 feet 3 inches wide ? By Duodecimals. By Decimals. F. I. F. 1. 16 9 16 9=16,75 feet. 2.3 2 3= 2,25 a TO MEASURE LOADS OF WOOD. RULE. Multiply the length by the breadth, and the product by the depth or height, which will give the content in solid feet; of which 64 make half a cord, and 128 a cord. EXAMPLE. How many solid feet are contained in a load of wood, 7 feet 6 inches long, 4 feet 2 inches wide, and 2 feet s inches high? 7 ft. 6 in.=7,5 and 4 ft. 2 in.=4,167 and 2 ft. 3 in= 2,25; then, 7,5X4,167 = $1,2525 X2,25=70,318125 solid feet, Ans. But loads of wood are commonly estimated by the foot, allowing the load to be 8 feet long, 4 feet wide, and then feet high will make half a cord, which is called 4 feet of wood; but if the breadth of the load be less than 4 feet, its height must be increased so as to make half a cord, which is still called 4 feet of wood. By measuring the breadth and heighth of the load, the content may be found by the following RULE, Multiply the breadth by the height, and half the product will be the content in feet and inches. EXAMPLE. Required the content of a load of wood which is s feet 9 inches wide and 2 feet 6 inches high. By Duodecimals. By Decimals. F. in. F. 3 9 3,75 2 6 2,5 7 6 1875 9 4 6 9,375 F. in. Ans. 4 8 S 4,687554 81, or half a cord and 84 inches over. The foregoing unethod is concise and easy wo those who are well acquainted with Duodecimals, but the following Table will give the content of any load of wood, by inspection only, sufficiently exact for common practice ; wbich will be found very convenient, Breadth. || Height A TALLE of Breadth, Height, and Content. Inches. ft. in. 1|2|3|4|1|23| 4 | 5 | 6171819110111 2 6 115 30 45 601) 1 2145 6 7 9 10 11 12 14 7 16|31|17|62| 1 6 8 9.10 12 131141 8 | 16/3214864|| 1 3 4 5 7 8 9 11 12 13 15 9 17 33 49/66|| 1 314 6 7 8 9 11 12 14/15 10 17|94|51|68|1 2 3 4 5 6 7 9 10 11 13114/16 11 18/35/55 701 2346 7 9 10 12 13 15 16 3 0 18 65472|| 2 3 5 6 8 9111|12/14/15117 1 19|37|56741|2|3 5 6 8 9|11|12|14|16|17 2 ll19 38|57|76|| 21 3 5 6 8 10 11 13 14 16/17 3 19|39|59|78!) 2 3 5 7 8 10 11 13 15 16 18 4 20/40/60/8023517 8 10 12|19|1517|18 5 21 41/6282 3 5 7 8 110 112 14 16/1719 6 21 42|63|84|2415 7 911 |12|14|16|18/19 7 2243/64/861 24 5 1 79 911 13 14|16|1890 8 | 22 44.66.88|| 2 4 6 en han 9 11 13 15 17 18 20 9 |23 4568|90|| 2 467 9 11 13 15 17 19 21 10 1123 46 69 921 2 467 9 12 13 15 17 19:21 11 23 4770194! 2141618 10 12 14 16 18 20 22 4 0 24/48172|96| 214168 10712 14 16 18 20 22 Aeo TO USE THE FOREGOING TABLE. First measure the breadth and height of your load to the nearest average inch ; then find the breadth in the left hand column of the table; then move to the right on the same line till you come under the height in feet, and you will have the content in inches, answering the feet, to which add the content of the inches on the right and divide the sum by 12, and you will have the true content of the load in feet and inches. NOTE.The contents answering the inches being always small, may be added by inspection. 1. Admit a load of wood is 3 feet 4 inches wide, and 2 fect 10 inches high ; required the content. Thus, against 8 ft. 4 inches, and under 2 feet, stands 40 inchc9; and under 10 inches at top, stands 17 inches then 40+ 17=57 true content in inches, which divide by 12 gives 4 feet 'Q inches, the answer. 2. The breadth being 3 feet, and height 2 feet i inches, required the content.- Thus, with breadth feet o inches and under feet EXAMPLES. a |