6210 78 Then the depths a and b per General Table give 2a+b 10 =16.1, or its nearest whole No. 16 and (a') .6 per Table No. 2. 2b+a=13.9, or its nearest whole No. 14 and (b) .4 per 10 Table No. 2. 46 By Table No. 1., for bottom width 36 feet, and ratio of slopes } 2 to 1, No. of cubic yds. to be deducted Content in cubic yards 12668 396 12272 NOTE 1. O Care must be taken to use the decimal a' with in finding the quantities in Table No. 2., as a mistake might easily be made in this matter, which would lead to an erroneous result. 2. If any two succeeding depths be nearly equal, and the sum of their decimal parts be together equal to 1 foot, or nearly so, by adding 1 foot to the lesser depth, and rejecting the decimal in the larger depth, and using the depths thus altered as whole numbers, a result sufficiently correct will be obtained; as in the following example. Ex. 2. Let the depths of a cutting be 50.29 and 48.7 feet, their distance 1 chain, the bottom width 33 feet, and the ratio of the slopes 1 to 1; required the content. By adding the decimal .29, in the larger depth, to 48.7, the depths may be called 50 and 49, for which, by the General Table, the content is 5990 for 1 to 1 From which deduct, from Table No. 1. NOTE 1. The results of all the examples in the two preceding cases only differ by a very small fraction of a cubic yard from the true contents obtained by calculation from formula (1.), page 51. The method of finding the contents to decimals, or tenths of a foot in the depths having been particularly discussed, on account of its utility in finding the contents for actual contract-work from the working drawings; in which, as great accuracy is required, the contents should be found to two places of decimals, or to ths of a foot in the depths, as in Case II. of the following Problem. 2. When one or both of the given depths exceed the limits of the table, find the content corresponding to half the two depths, and four times the result will be the content required. PROBLEM II. CASE I. To find the Content of a Cutting between two Cross Sections the Areas of which, to the Intersection of the Slopes, the Length, the Bottom Width, and the Ratio of the Slopes, being given. RULE. Find the square roots of the given areas either by a table of square roots*, or by actual extraction; with these roots, as depths, proceed to find the content from the General Table, as in Prob. I., from which deduct the quantity corresponding to the given bottom width and ratio of slopes from Table No. 1., and multiply the remainder by the length for the content. NOTE. If the length be given in feet, multiply the content found by the above rule by the feet, and divide by 66 for the content. Ex. 1. Let the areas of the two ends of a cutting be 5141 and 1444 square feet, the bottom width 30 feet, the length 1.60 chains, and the ratio of the slopes 11⁄2 to 1; required the content of the cutting in cubic yards. = 1444; and since the bottom-width A B = a b, and the ratio of the slopes are given, the solidity of the prism, the ends of which are ABN = a bn, is given in Table No. 1., and is to be deducted from the content found by the General Table, as in Prob. I. The square roots of 5141 and 1444 are 71.7 and 38 7483 By Table No. 1., for bottom width 30, and slopes 1 to 1 366.67 CASE II. When great accuracy is required, especially in the mea- RULE. Find the content by the Tables, as in the two preceding Problems, multiply the result by half the sum of the side slopes, and from the product deduct half the sum of the cubic yards, correspond- ing to the given bottom width and the ratios of the slopes of the two sides, multiplied by the whole length of the cutting, and the re- mainder is the content in cubic yards. Ex. Let the depths and their dis- tances be as in the annexed table, the bottom width 36 feet, and the ratios of the slopes of the two sides 1 to 1, and 1 to 1; required the content of the 44.46 1.00 7820 (528+792) × 2 and ratios of the two slopes, is multiplied by the whole length of the cutting, and deducted, agreeable to the rule. CASE II. When the Sectional Areas are given. RULE. Find the content from the Tables as in Prob. II., and deduct for the quantity below the cutting, as in Case I. of this Problem. NOTE. This Rule seems too obvious to require an example. PROBLEM IV. To apply the General Table and Table No. 2., to such Bottom Widths and Ratios of Slopes as are not found in Table No. 1. Put w = bottom width, and r to 1, the ratio of slopes. feet to be added to depth of cutting, or distance from bottom of cutting to the intersection of slopes. And 11 w2 18 r = cubic yards to be deducted for each chain in length. Ex. If the bottom width be 28 feet, and the ratio of slopes 14 to 1; then w÷2 r = 28 ÷ 2 × 14 = 11} = 11.2 feet = distance from bottom of cutting to intersection of slopes, and 11 w2 ÷ 18 r = 11 × 28218 × 14 = 383.29 cubic yards to be deducted for each chain in length. PROBLEM V. To find the Content of the Cutting of a Tunnel. 1. When the length is given in yards, and the width and height in feet. RULE. Multiply continually together the length, width, and mean height, and divide the product by 9. 2. When the length is given in chains, and the width and height in feet. RULE. Multiply the continued product of the length, width and mean height by 22, and divide by 9. Ex. The length of cutting of a tunnel is 1053 yards, its width 30 feet, and mean height 32 feet; required the content. The following Example shows the Magnitude of the Errors of several Methods, practically used, for calculating the Contents of Cuttings. Ex. The areas of the two cross sections of a cutting, to the intersection of the slopes, are 10324 and 400 square feet, their distance 4 chains, the bottom width 36 feet, and the ratio of the slopes 1 to 1; required the content of the cutting by the General Table, &c., and by the erroneous' methods practically used. NOTE. The great difference of the sectional areas in this example, shows very prominently the errors of several methods of finding the contents of cuttings : at the same time, it is proper to remark, that similar differences in the sectional areas very frequently occur in practice, and that the erroneous methods give the contents very near the truth only when these areas are nearly equal. (1.) By Table No. 1., the depth below the formation-level to the meeting of the slopes is 18 ft.; hence the area of the triangle below it is (36 × 18) = 324 sq. ft., which, deducted from the given sectional areas, gives 10000 and 76 sq. ft. for the sectional areas, as used by Mr. Bashforth. Whence 10000 = 100 76= 8.718} Content by Gen. Tab., &c. 8920.4 Content for 4 chains in length Error in defect, compared with the true content given above, being above 71 per cent., by Mr. B.'s method. 4 35681.6 c. y. 2726.4 (2.) By taking a mean of the areas for a mean Section. 22 (10000+76) ×· ×4 = 49260.4 cubic yards; which exceeds the true content by 10852.4 cubic yards, being above 22 per cent. in (3.) By taking a mean depth the error in defect is just half of the preceding error, or nearly 11 per cent. |