NOTE. When great accuracy is required, if, in the division of a height by the base, there should be a remainder. Find the lengths of the curves from the two nearest tabular heights, and subtract the one length from the other. Then, as the base of the arc of which the length is required is to the remainder in the operation of division, so is the difference of the lengths of the curves to the complement required, to be added to the length. EXAMPLE. What is the length of an arc of a circle, the base of which is 35 feet, and the height or versed sine 8 feet? 39.6655, 1.1344X35 8-35.2283, 228=1.1333, 229=1.1344, 1.1333x35 39.7040, 39.7040-39.6655.0385, difference of lengths. Hence, as 35: 20:0385 0220, the length for the remainder, and .0220+ 39.665539.6875, and .6875X12, for inches=84, making the length of the arc 39 feet 84 inches. To find the length of an Elliptic Curve which is less than half of the entire Figure... k a GEOMETRICALLY.-Let the curve of which the length is required be abc, Draw the line ce, and from e, with the distance eb, describe bh; bisect chini, and from e, with the radius ei, describe ki, and it is equal half the arc abc. To find the length when the Curve is greater than half the entire Figure. RULE. Find by the above problem the curve of the less portion of the figure, and subtract it from the circumference of the ellipse, and the remainder will be the length of the curve required. G2 285 1.2567 .520 1.5941 .745 1.9675 .975 2.3810 1.6019 .750 1.9760 .980 2.3906 .290 1.2634 .525 1.6097 295 1.2700 .530 1.6175 .300 1.2767 .535 1.6253 .765 2.0016 .995 2.4194 .305 1.2834 .540 1.6331 .770 2.0102.1000 .310 1.2901 .755 1.9845 .985 2.4002 1.9931 .990 2.4098 2.4291 To find the Length of the Curve of a Right Semi-Ellipse. Proceed with the foregoing table by the rule for ascertaining the lengths of circular arcs, page 76. EXAMPLE.-What is the length of the curve of the arch of a bridge, the span being 70 feet, and the height 30.10 feet? 30.10÷70.430 per table, 1.4627, and 1.4627X70=102.3890, the length re quired. When the Curve is not that of a Right Semi-Ellipse, the height being half of the Transverse Diameter. RULE.-Divide half the base by twice the height; then proceed as in the foregoing example, and multiply the tabular length by twice the height, and the product will be the length required. EXAMPLE.-What is the length of the profile of arch (it being that of a semi-ellipse), the height measuring 35 feet and the base 60 feet? 60÷2=30÷35×2.428, the tabular length of which is 1.4597. Then, 1.4597x35×2=102.1790, the length required.. NOTE-When the quotient is not given in the column of heights, divide the difference between the two nearest heights by .5; multiply the quotient by the excess of he height given and the height in the table first above it, and add this sum to the tabular area of the least height. Thus, if the height is 118, To find the Area of a Zone by the above Table. RULE 1.-When the zone is greater than a part of a semicircle, take the height on each side of the diameter of the circle, of which it is a part; divide the heights by the diameter; find the respective quotients in the column of heights, and take out the areas opposite to them, multiplying the areas thus found by the square of the diameter or chord, and the products, added together, will be the area required. NOTE.-When the quotient is not given in the column of heights, divide the differ ence between the two nearest heights by 5, and multiply the quotient by the excess between the height given and the height in the table first above it, and add this sum to the tabular area of the least height. Thus, if the height is .333, .30416-30790.00374÷5.000748×3 (excess of 333 over 330) = .002244+.30416 .306404, the area for 333. EXAMPLE.-What is the area of zone, the diameter of the circle being 100, and the heights respectively 20 and 10, upon each side of it? 20+100.200, and 200, per table,=.19453X1002 = 1945.3. RULE. When the zone is less than a semicircle, proceed as in rule 1 for one height. EXAMPLE.-What is the area of a zone, the longest chord being 10, and the height 4? 4÷10.400.35182×102 = 35.182 Ans. MENSURATION OF SOLIDS. 23. OF CUBES AND PARALLELOPIPEDONS. C 24. To find the Solidity of a Cube-fig. 23. RULE.-Multiply the side of the cube by itself, and that product again by the side, and this last product will be the solidity. To find the Solidity of a Parallelopipedon—fig. 24. RULE.-Multiply the length by the breadth, and that product by the depth, and this product is the solidity. OF REGULAR BODIES. To find the Solidity of any Regular Body. RULE.--Multiply the tabular solidity in the following table by the cabe of the linear edge, and the product is the solidity. TABLE of the Solidities of the Regular Bodies when the Linear Edge is To find the Solidity of Cylinders, Prisms, and Ungulas-figs. 25, 26, and 27. RULE.-Multiply the area of the base by the height, and the product is the solidity. |