CASE II.-To find the content when the depths are given în RULE.-Let a and b be the feet in any two succeeding depths, 2a, rejecting the last figure, and a will shew the number After which deduct for the quantity below the formation level EXAMPLE. Let the sectional areas be 1406 and 2560 square feet, the ..... 4660 Here 1406 = 37.5 and 2540 50.4. Put 37 = a, For bottom width 36 and ratio of slopes 1 to 39 57 4746 528 4218 4 16872 cubic yds. CASE III.-In measuring contract work, where great ac- curacy is required, theths of a foot, or second decimals, EXAMPLE. The areas of seven cross sections of a Ans. The content, per General Table, and given bottom width and ratio of slopes x by the whole length, viz. 275 x 18 = = 4950 cubic yards, must be deducted, which leaves 167568 cubic yards, the content required. NOTE 1.-When the distances of the sectional areas are given in feet, the quantities of the General Table must be multiplied by their respective distances, and the final result divided by 66, as in Example 3, Prob. I. NOTE 2.-When the surface lines of the sectional areas are either level or are readily reducible to that position, the decimals, if any, in depths must be taken into the calculation, as in Cases II. and III. PROBLEM III. To adapt the General Table to such widths of the formation level and ratios of slope as are not found in Table No. 1. Put w = width of formation level, and r : 1 the ratio of slope. Then = feet to be added to the depth of cutting below w r = 10.4 ft. distance Let the width of formation level be 26 feet, and the ratio of slopes 1 to 1; then w÷r = 13 ÷ 11 below formation level to meeting of slopes. And 22 x 132 ÷ 9 × 11 = 330.5 cubic yards to be deducted for each chain in length from the contents of the General Table. PROBLEM IV. To find the content of a cutting, when each of the sides have two different ratios of slope. RULE.-When the cutting ABCD has two different slopes, as Aa or Bb and a C or bD; it must be divided into two parts by the line a b, and the quantities of the parts ABba, abDC of the cutting must be found separately by Prob. I. If the surface line CD be sloping or curved, Prob. II. will also be required; and, if the depths m'n, m N and their corresponding quantities of cubic yards are not found in M m B Table No. 1, they must be found by Prob. III., the sum of the contents of the two parts being the required content of the cutting. NOTE. Cuttings of this kind are often advantageously adopted, where their upper parts are of loose or springy earth, and their lower parts strong clay or rocky. PROBLEM V. To find the content of a cutting when the ratio of the slopes of the two sides are different. RULE.-Find the central depths to the meeting of each of the side slopes, and take their corresponding contents by Prob. I. for the whole length of the cutting, in the same manner as if it were for two cuttings, and from the sum, subtract the sum of the cubic yards corresponding to the given bottom width and ratios of slope multiplied by the whole length of the cutting, and half the sum will be the content required. CASE II. When sectional areas are given. deduct RULE. Find the contents corresponding to the sectional areas to the meeting of the side slopes by Prob. II., and from their sum 44 w2 l cubic yards for the content. In this for9(r+r') mula w = bottom width, 1 = whole length of cutting, and r and r' the first terms of the ratios of the side slopes. NOTE. Cuttings with slopes of this kind are frequently adopted in practice, where the ground is springy on one side of them, the greater ratio of slope being on the springy side. PROBLEM VI. To find the quantity of the cutting of a tunnel. RULE.-Multiply continually together the width, mean height and length; divide the product by 9, if the length be given in yards, but, if the length be given in chains, multiply the product by 22 and divide by 9, the width and height, in both cases, being given in feet. NOTE. Examples are not given in the three last Problems, the methods of solving which being sufficiently obvious from the Rules and the first three Problems. THE FOLLOWING EXAMPLES SHOW THE ERRORS OF METHODS PUT FORTH BY MR. BASHFORTH AND OTHERS, TO FIND THE CONTENT OF CUTTINGS. EXAMPLES. 1. The areas of two cross sections of a cutting to the meeting of the intersection of the side slopes, are 1296 and 361 square feet, their distance one chain, and bottom width 36 feet, and the ratio of the slopes 1 to 1; required the content of the cutting by the true and the erroneous methods practically used. True content for 1 chain in length......... 1115 cubic yds. (2.) By Mr. Bashforth's method:- By Table No. 1, the depth of the meeting of the side slopes below formation level is 18 feet; hence the area of the triangle below it is 36 × 9 = 324 square feet, which, being taken from the given sectional areas, will give 972 and 37 square feet for Whence 975 140 :: 100: 14·36, or above 14 per cent. the error in defect by Mr. Bashforth's method. 2. Let the areas of the two cross sections to the intersection of the slopes be 324 and 2916, and the other things as in the last example, required the content of the cutting by the correct By taking a mean of the areas in example 2. (2592 +0) × 22 3168 cubic yards, which exceeds the 9 true content by 528 cubic yards, being 163 per cent, in excess. NOTE. By taking a mean depth the error in defect is just half the preceding one, or 8 per cent. These methods only give a near approximation to the true content, when the sectional areas of a cutting are nearly equal. See the results of the following formula and remarks. The above examples shew with sufficient clearness the inaccuracy of Mr. Bashforth's method; but since it has been defended, as sufficiently correct for practical purposes, the following formula will more rigidly prove its failure. Let A' and B' be the sectional areas of a cutting, as used by Mr. Bashforth, and a the area between the formation level and the meeting of the slopes; then by Baker's Railway Engineering, page 55, s2ì (√ A ́ + a × B' + a — √ A' X B' = his error in de fect, the content by his method being (A' + B'x √ A'+B'), the length being unity; whence 81 · a) = A' + B' + √ A' × B' cent. by Mr. Bashforth's method. 2. When A' 100 (2-0-1) 3 = 10 1) a = 10/1/ 2 1200 = 91 = 0 and B' = 3 a, the above formula gives 331 per cent. the error in defect. 3. When A' = 0 and B' = a, the error in defect is 413 per cent. Errors of this kind will repeatedly occur in practice, where cuttings and embankments are made, as in the following figure; wherein AB is the formation level, and the corresponding curved line is a section of the earth's surface; the small figures below being cross sections taken to the intersection of the slopes, in which MN, m n, represent the formation level, the sectional areas, used by Mr. Bashforth, being those above the lines MN, mn, the cross section c d mn may either be very |