0 1.00 in length, is multiplied by 4 chains the whole length of the cutting, thus giving the whole quantity to be deducted, the remainder being the true content in cubic yards of the cutting. 2. The several depths of a railway cutting to the meeting of the side slopes are as in the annexed table, the bottom width being 30 feet, and the ratio of the slopes 1 to 1; required the content of the cutting. NOTE. When any of the distances between two succeeding depths is greater or less than 1 chain, the corresponding quantity from the General Table must be multiplied by that particular distance; as the distances between the depths 20 and 25, and between 32 and 39, &c., the distance being 2 chains. The last distance, viz., that between 30 and 10, is 1.46, For side slopes For side slopes 1 to 1 366-67 x 14.46 in this case 2 figures must be cut off for decimals, after multiplying. 3. Let the depths of a meeting of the side slopes, and their distances in feet be as in the annexed table, the bottom width 30 feet, and the ratio of the slopes 1 to 1; required the content in cubic yards. railway cutting to the Dist. in Depths Quantities multi feet. in feet. plied by length. Total quantities. For slopes 1 to 1 2557728 366-67 × 278 = 101933 66)2455795 Content in cubic yards...... 37209 NOTE. When the distances are in feet the quantities from General Table must be respectively multiplied by their distances, the quantity from Table No. 1, by the whole distance, and the result divided by 66, the feet in 1 chain, for the content in cubic yards. PROBLEM II. CASE I.—The areas of two cross sections of a railway cutting to the intersection of the side slopes, its length in chains, bottom width, and the ratio of the slopes are given; required the content of the cutting in cubic yards. RULE.-With the square roots of the given areas as depths, find the content from the General Table, as in the last Problem, from which subtract the quantity answering to the given width, and the ratio of side slopes from Table No. 1, and the remainder, being multiplied by the length, will be the content required. NOTE. If the length be given in feet, proceed as in Example 3, last Problem. EXAMPLE. 1. Let the two sectional areas of a cutting be 4761 and 1296 square feet, the bottom width 36 feet, the length 3.25 chains, and the ratio of the side slopes 2 to 1; required the content in cubic yards. ✓4761 = 69 69} content per General Table 6959 ✓1296 = 36 For bottom width 36 and slopes 2 to 1, per 396 Table No. 1 Content for 1 chain in length 6563 31 19689 1641 Content for 3.15 chains 21330 cubic yds. CASE II.-In measuring contract work, where great accuracy is required, the oths of a fot, or second decimals, must be used in the calculation, by taking for them oth of their respective quantities in Table No. 2. EXAMPLE. The areas of seven cross sections of a railway cutting to the meeting of the side slopes and their distances are as in the annexed table; the bottom width is 30 feet, and the ratio of the slopes 1 to 1; required the cubic yards in the cutting. 9:00 4100 14:00 5141 Ans. The content, per General Table, and Table No. 2, is 172318 cubic yards, from which the quantity corresponding to the 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. For further explanations and numerous examples of the methods of finding the contents of earthwork, see Baker's Land and Engineering Surveying. See, also, Tate's Geometry, page 252. GENERAL RULE FOR FINDING THE CONTENTS OF SOLIDS. The wedge, the prismoid, the pyramids, and their frustums; the whole or a segment, or any portion of the whole, contained between two parallel planes perpendicular to the axis of a sphere, of an ellipsoid, of a paraboloid, of an hyperboloid, may be found by the following general formula. Let A and B be the areas of the ends of the solid, C the area of a section parallel to and equidistant from the ends, and L the distance between the ends; then The following investigation of this very general Rule was given by B. Gompertz, Esq., F.R.S., &c., in the Gentlemen's Mathematical Companion for 1822. Let x be the variable distance from a given point of another section parallel to the two ends, and a, b, c be given quantities, the area of the said section will be a + bx + c x2, as this will contain the cases of the sections of the solids, enumerated in the Rule; for instance, in the pyramid or cone, the area of the section may be expressed by c x2, in the wedge the areas may be expressed by a + b x + c x2, a, b, c being constant for the same point and wedge for any parallel section to a plane given in position, as long as the section has the same number of sides. In the paraboloid all planes perpendicular to the axis, if the given point be the vertex, will have the areas of their sections expressed by b x. In the ellipsoid or hyperboloid, the point being in that vertex of the axis about which it is revolved, the area of the sections may be expressed by bx + c x2, c being negative in the ellipsoid and positive in the hyperboloid. And I observe from the method of equidistant ordinates in curves, or of sections in solids, see the method of differences, if A, B represent the areas of the two ends, C the area of the section in the middle between them, and L the length; then The solidity= A+B+4 C X L Q. E. I. EXAMPLES ON THE FOREGOING RULE. 1. The length of a railway cutting is 5 chains or 110 yards, the top width and depth at one end are respectively 120 and 30 feet, the top width and depth at the other end are respectively 90 and 20 feet, and the bottom width 30 feet; required the content of the cutting in cubic yards. Ans. 20777 cubic yards. 2. Required the content of a sphere, the diameter of which is Ans. 33.5104 cubic yards. 12 feet. NOTE. Here the areas of the extreme sections are each = 0. 3. A cask is in the form of the middle zone of a sphere, its top and bottom diameters being 34 inches, and its height 30 inches, inside measure; how many gallons will it hold? Ans. 149.22. 4. How many gallons are contained in a cask, in the form of the middle zone of a spheroid, the bung and head diameter being 40 and 32 inches, and the length 36 inches, all internal measures? Ans. 143 imperial gallons. 5. How many cubic feet are there in a parabolic conoid, the height of which is 42, and the diameter of its base 24 inches? Ans. 5.4978 cubic feet. 6. A 5 inch cube of ivory is turned into a sphere of the same diameter; what weight of ivory will be lost, its weight being 1820 ounces (Av.) per cubic foot? Ans. 61.44 ounces. TABLE No. 1.-THE AREAS OF SEGMENTS OF CIRCLES, DIAMETER UNITY. NOTE. When the tabular height exceeds the heights given in this Table the segment must be divided into two equal parts and their common versed sine found by Prob. VI., page 26. The tabular heights will then fall within this Table, whence the area of the whole segment may be found. |