If the staves of the cask be much curved or bulging - between the bung and the head, multiply the difference by 7; if not quite so curve, by 65; if they bulge yet less, by 6; and if they are almost or quite straight, by 55, and add the product to the head diameter; the sum will be a mean diameter, by which the cask is reduced to a cylinder. Square the mean diameter, thus found, then multiply it by the length; divide the product by 359 for ale or beer gallons, and by 294 for wine gallons. ; Note 1. The length is most conveniently taken by callipers, allowing, for the thickness of both heads, 1 inch, 14 inch, or 2 inches, according to the size of the cask; but if you have no callipers, do thus measure the length of the stave, then take the depth of the chimes, which, with the thickness of the head, being subtracted from the length of the stave, leaves the length within. Note 2. You must take the head diameter, close to its outside, and, for small casks, add three tenths of an inch; for casks of 30, 40, or 50 gallons, 4 tenths, and for larger casks, 5 or 6 tenths, and the sum will be very nearly the head diameter within. In taking the bung diameter, observe, by moving the rod backward and forward, wheth er the stave, opposite the bung, be thicker or thinner than the rest, and if it be, make allowance accordingly. By the Sliding Rule. On D is 18.94, the gauge point for ale or beer gallons, marked AG, and 17.14, the gauge point for wine gallons, marked WG: 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. By the Scale. Take the extent from the gauge point to the mean diameter, set one foot of the dividers in the length, and turning them twice over, they will point out the content. ART. 51. Required the content in ale and wine gallons, of a cask, whose bung diameter is 35 inches, head diameter, 27 inches, and length 45 inches? Square of the diameter = 1062.76 Bung diameter = 35 ART. 52 A round mash tub is 42 inches diameter at the top, within, and 36 inches at the bottom, and the perpendicular height 48 inches; required the content in beer and wine gallons? This being the lower frustum of a cone, to the product of the diameters add of the square of their difference; multiply this sum® by the length, and it will give the solidity in such parts as the dimen sions are taken in. If they be taken in inches, divide by 359 for beer, and 294 for wine gallons. ART. 53. Let the difference of diameters of this tub be 6 inches, the height 48 inches, and the content 2033 gallons, to find the diameters? Multiply the content, if beer measure, by 359; if wine measure, by 294, and divide the product by the length: from the quotient subtract of the square of the difference of the diameters; to this remainder add the square of the difference of the diameters, and extract the square root of the sum; from the square root subtract the difference of the diameters, and it will give the least diameter to great exactness, to which add the difference of the diameters, and the sum is the greatest diameter. The content of any vessel, in gallons, &c. may be thus found: measure the inside of the vessel, according to the rule of the figure, and find the content in cubick inches; then, 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 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 By Rule 2d. 97 291 3007 Depth = 15.5 15035 7633 22899 2366.23 Multiply by breadth 15.5 1183115 1183115 236623 94)36676-565(390-176 tons. Allowing the Cubit, as it is found by modern travellers, to be 22 inches, the content of Noah's Ark is as follows, viz. Cubits. 300 Length of the keel, Breadth by the midship beam 50 30 } Its burthen as a man of war 27729 ton's. As a merchant ship, 29188-6 ts. Depth in the hold QUESTIONS IN MENSURATION. 1. THE largest of the Egytian pyramids is square at the base, and measures 693 feet on a side: how much ground does it cover? 696x393 1764 1764 poles, and 11 acres and 4 poles, Ans. 160 272.25 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. = 2500 = 35-35, and 50 35.35 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 fine be drawn? = A square yard = 1296 inches, and 16 feet 5 inches 197 inches. 1296 5. If the area of a triangle be 900 yards, and the perpendicular 40 yards required the length of the base? 900X2 45 yards, Ans. 40 6. If the three sides of a plain triangle be 24, 16 and 12 perches: required its area? 24+16+12 2 26; 26-24-2; 26-16=10; 26-12=14, and 26×14×10x2 = 85.32 perches, = area. Again, as 24: 16+124 16-12:46+, the difference of the segments of the base; then, 4.6+ 12-- -= 2 9.6, and 12x12—9·6x9•6 = 7·11 the perpendicular on the longest side; whence 242x7·11=85·32, area as above. 7. Required the area of a circular garden, whose diameter is 12 rods? 12x12x·7854" 113.0976 poles, Ans. 8. The wheel of a perambulator turns just once and an half in a rod: what is its diameter ? 16.5x 11, circumference, and 11x-318313 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 15 inches: what will it come to? 36+15.5×2=67 the greatest diameter; 67x67x-7854-36x36x 7854 2507.8722 17.4157 square feet, at 7d. per foot, = 10s. 10, d. [Ans. 10. Required the difference between the area of a circle, whose ra. dius (or semidiameter) is 50 yards, and its greatest inscribed square? 50x2100 the diameter, and 100x100x78547854 the area of the circle; then, 50x50x2 = 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 —25x25x-7854 = 134-125 the circumscribed square, more than the section, and 25x25x•7854—12.5x12·5x2 = 178.375 inscribed square, less than the area of the section. 12. Four men bought a grindstone of 60 inches diameter: how 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? man 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 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 60' answers. |