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times the area of a section parallel to, equally distant from both ends; the last sum, multiplied into one-sixth the length, will give the solidity.

To find the solidity of a pyramid.

Rule.-Multiply the area of the base by one-third the height for the solidity. If you meet with a wedge, divide it into two pyramids, and find the area of both: or, find the area according to the rule in mensuration.

Examples. 1.-Suppose a section of a railway cutting to have 26 feet wide at bottom, 52 feet wide at top, and 8 feet deep throughout, and the length 850 feet; require the number of solid yards in the section.

= 78, 782


39 mean breadth.


26 × 52 = 850 X 8 = 6800; then, 6800 × 39 265200 cubic feet, 265200 ÷ 3 = 88400, and 88400 ÷ 9 = 9822 cubic yards, which is the whole cutting.

Note. The foregoing examples are very simple, but will seldom answer in practice, particularly on a railway. But the following methods of calculation will answer in general.

2. Suppose the cutting for a section of a railway to have the following dimensions, viz: perpendicular height at one end 19 feet, and at the the other end 8 feet, length 100 feet, and breadth 50 feet; required number of solid yards, when the ratio of the slope is 2 to 1.

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Note. To find the area of the end, multiply the ratio of the slope by the height, and to the product add the breadth of the base, and the sum multiplied by the height gives the area of the end.

To find the area of the middle section. Multiply the ratio of the slope by half the sum of the perpendicular heights of both ends, and to the product add the breadth. of the base, the sum multiplied by the height in the middle gives the area of the middle section.


Demonstration of the last example, which represents a prismoid, (Fig. 27.) Let m = height of the greater end, n height of the lesser end, b = the breadth, t = the ratio of the slope to unity, 1 = the length of the section. It is evident if you multiply the ratio of the slope by the altitude m, that it will give Cx, to this add DC the breadth of the base, or roadway, and multiply the sum by the height m, which will give the area of the cross-section or end DA BC. This is not the case, unless the perpendicular Bx, is equal the perpendicular Ay; if not, you must find the area of each separately, and m, that is, the height by the breadth. find the area of the end OHSR. tudes, m and n, half the sum will give the middle altitude o; having the altitude in the middle, you can find the area of the cross-section, d an m, as before. Now conceive the two side wedges cut off by a plane perpendicular to the base and another plane through the end of the lesser altitude n, parallel to the base DS.

add them to DC X In a similar manner Then add both alti

The area of the smaller end is equal n × (b × tn) equal nb x tn2. Therefore the area of the lower prismoid (nb + 1. (1)


(1.) The contents of the upper prismoid equal (bm+ 2 tmn bn 2 tn') X (2)

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(2.)—The base of the two pyramids equal (m — n) × t × (mn) equal tm2 2tmn + tn': Hence the area of the pyramids is equal (t m2 - 2 tmnt n') X onethird.

6 t

(3.)-Now if we unite (1,) (2) (3,) and reduce them, we get (6bn + 6t n2 + 3'bm + 6 tmn 3 b n n2tm2 - 4 tmn+ 2 tn') X one-sixth, equal (3 b n + 3 bm + 2tmn + 2t m2 + 2t n2) × one-sixth.

Now it is evident, that bn + tn2 + bmt m2, the first 4 terms are equal the area of both ends of the pris


moid; the other terms may be thus expressed, 4 < 2

× (b + tx (b + tx "+m) > equal four, the area of the


middle section. Hence the rule is evident.

Another method.-Let A B C D PQOS, (Fig. 27,) represent the prismoid, produce off AP and DS to meet in n; also Co and BQ to meet in m; then will n m be the common section of the two planes, A BQ P and D SOC.


Now allow AC and Co at right angles to DC and SO. (4.6) we get the altitude at A, to the altitude at P, so is Dn to ns; and per division we get s n. In a similar manner we may find mo, if not equal s n: We may also find the perpendiculars from m and n, on DC, when m C and n D are not so. Join D m, Am, S m, and Pm. It is evident the whole figure is divided into two pyramids by the lines Dm and A m: also the figure PQ n mos, into two other pyramids, by the lines Sm and Pm. The area of the pyramid A B D Cm is the area of A B C D multiplied by one-third the altitude Cm, and the area of the pyramid D Á n m equal the area of the base D m n by onethird the altitude from A, upon the base DC; add both, and you get the area of the whole figure. Again, find the area of the pyramid Pm Qos, by multiplying the area of the base PQos by one-third the altitude mo, and the pyramid Pnm S, by multiplying the base Snm by onethird the altitude from P, on the base o S. The area of the two last pyramids taken from sum of the areas of the two first ones, will give the area of the prismoid APQ C DSO.

1st Note. Together with finding the area of the prismoid by taking the difference between the above expressions, you may reduce the expression down to the common


2nd. It has before been observed, where the slopes

stand at different angles of inclination, owing to the vari ation of the different strata, each slope must be calculated separately.

3d. When a change of slope takes place, a hedge or benching is left, so.when the angle of slope changes all these should be taken into account.

Equalizing the Excavation and Embankments.

After having determined the direction of the line, from a careful comparison of the several trial lines, the next point of importance to be attended to, is the balancing of the excavations and embankments; that is, to regulate the slopes in such a way, that the cuttings and filling may be equal or nearly so-so as to bring the surface of the road to the required inclination.

In order to effect this, it is manifest a variety of slopes must be assumed; and the calculations of all the cuttings and embankments compared. The result of this comparison will help to point out the most advantageous slopes. Few surfaces will admit of one unvaried inclination for any considerable distance; therefore, the only way would be, to divide the whole line into several parts, and then equalize these; taking care that the rise or fall, at the connecting points, be not too abrupt, avoiding, if possible, such ascents and descents as would require the use of a stationary engine. But after a consideration, if it be found that the expense of cutting and embanking should exceed the sum that would compensate for the delay and expense of the engine, in this case, of course, a stationary engine should be employed.

On a Curved and Straight Railway.

A carriage on going with considerable velocity on a curved track, exerts a certain influence in the direction of

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