.16401 .317 .21387 263 .16489 .318 .21480 .264 .16578 .319 .21573 .265 .16666 .320 .16754 .321 .372 26611 .427 .31995 .373 .26707 .428 .32094 .374 .26804 .429 .32193 .21666 .375 .26901 .430 .32292 .21759 .376 .26998 .431 .32391 .267 .16843 .322 .268 .21853 .377 .27095 .16931 .323 .21946 378 .269 .17020 .324 .22040 .379 .270 .17108 .325 .22134 .271 .17197 .326 .22227 .381 27483 436 .32887 USE OF THE ABOVE TABLE. To find the Area of a Segment of a Circle. RULE.-Divide the height or versed sine by the diameter of the circle, and find the quotient in the column of versed sines. Take the area noted in the next column, and multiply it by the square of the diameter, and it will give the area required. EXAMPLE.-Required the area of a segment; its height being 10, and the diameter of the circle 50 feet. 10-50.2, and .2, per table,=.11182; then .11182X502279.55 Ans. TABLE (Continued). 325 .326 .327 1.2644 328 1.2659 .329 1.2674 .372 .373 .374 .330 .331 .332 333 .337 .338 .339 1.2689 1.2704 .376 1.3423 .420 1.2720 .377 1.3440 1.2735 .378 1.3456 .422 .334 1.2750 .379 1.3473 .423 .335 1.2766 .380 .336 1.2781 .381 1.2786 .382 1.2812 .383 1.2827 .384 Height. Length. Height. Length. Height. Length. Height. Length. 1.4870 1.4132 .461 1.4946 .462 1.4965 1.4984 .421 1.4204 .465 1.5022 1.3490 .424 1.3507 .425 1.3524 .426 To find the Length of an Arc of a Circle by the foregoing Table. RULE. Divide the height by the base, find the quotient in the column of heights, and take the length of that height from the next right-hand column. Multiply the length thus obtained by the base of the arc, and the product will be the length of the arc required. EXAMPLE.-What is the length of an arc of a circle, the span or base being 100 feet, and the height 25 feet? 25÷100.25, and .25, per table, gives 1.1591; which, being multiplied by 100 115.9100, the length. 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? 8÷35.228.228=1.1333, .229= 1.1344, 1.1333×35=39.6655, 1.1344×35 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 .6875×12, for inches=84, making the length of the arc 39 feet 8 inches. To find the length of an Elliptic Curve which is less than half of the entire Figure. b 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 ch in i, 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. G 2 |