ART. 54. To ullage a Cask, lying on one side, by the Gauging Rod, when the Bung Diameter, and the Content, one, or both are greater or less than the Table on the Rod is made for. RULE. As the bung diameter of the cask to be measured, is to the bung diameter that the table is made for; so are the dry inches of the cask, to a fourth number, which find in the table on the rod, and note the number of gallons answering to it. Then as the content of the cask that the table is made for, is to the content of the cask to be measured; so is the number of gallons answering to the aforesaid fourth number, to the number of gallons your cask wants of being full. ART. 55. To find a Ship's Burthen, or to Gauge a Ship. There is such a diversity in the forms of ships, that no general rule can be applied to answer all varieties; however, the following rules are practised. RULE 1. Multiply the breadth at the main beam, half the breadth, and length together; divide the product by 94, and the quotient is the tons. RULE 2. Divide the continued product of the length, breadth, and depth, in feet, by 100, for ships of war, and 95 for merchant ships, in which nothing is allowed for guns, &c. and the quotient is the tons. RULE 3. Take the length from the stern post to the upper part of the stem; subtract two thirds of her breadth from that length; multiply the remainder by the whole breadth, and that product by half the breadth, in feet, and divide by 100 for war, and 94 for merchant tonnage. RULE 4. The weight of a ship's burthen is half the weight of water she can hold. What is the tonnage of a ship, whose length is 97 feet, breadth 31 feet, and depth 15 feet. By Rule 1st. breadth 15.5. Breadth 31 155 465 480:5 Length 97 33635 43245 94)46609-5(495 C3 tons. By Rule 2d. 97 291 3007 Depth 15.5 15035 15035 3007 95)46608-5(490-61 tons. Carried over. Allowing the Cubit, as it is found by modern travellers, to be 22 inches, the content of Noah's Ark is as follows, viz. QUESTIONS IN MENSURATION. 1. THE largest of the Egyptian pyramids is square at the base, and measures 693 feet on a side: how much ground does it cover? 696 X393 272.25 1764 =1764 poles, and =11 acres and 4 poles, Ans. 2. What difference is there between a floor 20 feet square, and two others, each 10 feet square? 20×20-10×10+10×10=200 feet, Ans. 3. There is a square of 2500 yards in area: what is each side of the square, and the breadth of a walk along one side and one end, which may take up just one half of the square? √2500=50 yards, each side. 2 =14.65 yards, breadth of the walk, Ans. 4. A pine plank is 16 feet and 5 inches long, and I would have just a square yard slit off: at what distance from the edge must the line be drawn? A square yard=1296 inches, and 16 feet 5 inches=197 inches. 1296 Therefore, 197611 inches, Ans. 5. If the area of a triangle be 900 yards, and the perpendicular 40 yards required the length of the base? 900×2 45 yards, Ans. 6. If the three sides of a plane triangle be 24, 16, and 12 perches required its area? 24+16+12 2 -=26; 26—242; 26-16=10; 26-12=14, and 26×14×10×2=85.32 perches, area. Again, as 24: 16+12 :: 16-12: 4.6+, the difference of the segments of the base; then, 4.6+ 12- 29.6, and 12×12 —9·6×9.67.11 the perpendicular on the longest side; whence 24-2×7·11=85.32, area as abovė. 7. Required the area of a circular garden, whose diameter is 12 rods? 12×12×·7854-113 0976 poles, Ans. 8. The wheel of a perambulator turns just once and a half in a rod what is its diameter ? : 16.5x11 circumference, and 11X-31831-3 feet, Ans. 9. Agreed for a platform to the curb of a round well, at 74d. per square foot: the inward part, round the mouth of the well, is 36 inches diameter, and the breadth of the platform was to be 151 inches what will it come to? : 36+15.5×2=67the greatest diam.; 67×67×·7854-36 × 36 ×·7854 17.4157 square feet, at 74d. per foot,=10s. 10, d. [Ans. 10. Required the difference between the area of a circle, whose radius (or semidiameter) is 50 yards, and its greatest inscribed square? 50x2100 the diameter, and 100X100X 7854-7854 the area of the circle; then, 50×50×2=5000 the area of the greatest inscribed square, and 7854-5000=2854, Ans. 11. There is a section of a tree 25 inches over; I demand the difference of the areas of the inscribed and circumscribed squares, and how far they differ from the area of the section? 25×25-12.5×12.5×2=312·5 the difference of the squares. 25×25 —25×25×·7854=134·125 the circumscribed square, more than the section, and 25X25X 7854-12·5×12·5×2=178.375 inscribed square, less than the area of the section. 12. Four men bought a grindstone of 60 inches diameter: bow much of its diameter must each grind off, to have an equal share of the stone, if one first grind his share, and then another, till the stone is ground away, making no allowance for the eye? RULE. Divide the square of the diameter by the number of men, subtract the quotient from the square, and extract the square root of the remainder, which is the length of the diameter after the first man has ground his share; this work being repeated by subtracting the same quotient from the remainder, for every man, to the last; extract the square root of the remainders, and subtract those roots from the diameters, one after another; the several remainders will be the answers. 13. If a bick foot of iron were hammered, or drawn, into a square bar, an inch about, that is, of an inch square its length, supposing there is no waste of metal? 12×12×12 required •25×·25×4 =6912 inches,=576 feet, Ans. 14. Required the axis of a globe, whose solidity may be just equal to the area of its surface? •7854×4 ⚫5236 6 inches, Aus. 15. A joist is 7 inches wide, and 24 thick; but I want one just twice as large, which shall be 33 inches thick: what will be the breadth ? 7.5×2.25×2 3.75 =9 inches, Ans. 16. I have a square stick of timber 18 inches by 14; but one of a third part of the timber in it, provided it be 8 inches deep, will serve how wide will it be? 18×14 3 8-10 inches, Ans. 17. A had a beam of oak timber, 18 inches square throughout, and 25 feet long, which he bartered with B, for an equilateral triangular beam of the same length, each side 24 inches: required the balance at 1s. 4d. per foot? 18x18x25 144 =56.95, solidity of the square beam. The perpendicular let fall on one of the sides of the triangu lar beam is 20-7846 inches, and the half perp.=10-3923; then 10.3923 x 24 144 =1.732 foot, area at the end, and 1·732×25=43·3 feet, solidity of the triangular beam; therefore 56-25-43-3-12-95 feet, at 1s. 4d. per foot=17s. 3 2d. balance due to A, Ans. 18. What is the difference between a solid half foot, and half a foot solid? 12x12x6 6×6×6 4, therefore, one is but of the other. 19. A lent B a solid stack of hay, measuring 20 feet every way; sometime afterward, B returned a quantity measuring every way 10 feet: what proportion of the hay remains due? 20×20×20-10×10×10=7000 feet, Ans. 20. A ship's hold is 75 feet long, 184 wide, and 71 deep: how many bales of goods 34 feet long, 24 deep, and 23 wide, may be stowed therein, leaving a gang way the whole length, of 31 feet wide? 75.5X18.5×7·25—75·5×7·25×3·25 =385 44 bales, Ans. 21. If a stick of timber be 20 feet long, 16 inches broad, and 8 inches thick, and 31 solid feet be sawed off one end: how long will the stick then be? 1728×3.5 |