TUNNELS.
103. The maintaining a correct centre line through tunnels is generally considered difficult. The fixing of the line in deep shafts requires great care, owing to the short distance between the only two fixed points, that can be transferred from the surface to the bottom of the pit. This is a matter of manual skill and of instrumental manipulation. There is no difficulty in aligning the upper ends of two plumb-lines; and the lower ones will certainly be governed by their position. The following method has been found to answer every purpose.
Let the opening of the shaft be ten feet in diameter. Place two horizontal bars at right angles to the road across the opening, upon which slide blocks holding the upper end of the plumb-lines. Adjust these lines, at the surface, with a transit; and when fixed, place iron pins at the point marked by the plumbs at the bottom of the shaft. Upon these pins fix the exact centres. For keeping the line in the shaft headings, a straight rod, with steel points at each end, should be used, which being placed upon the iron centre pins, fixes the centre line of the tunnel. When the tunnel is curved, the line should be laid off by offsets from the tangent to the curve at the shaft.
By this method points at ten feet distance may be fixed within 1
100 of an inch, a difference of which would cause an error of ⅒ of an inch per one hundred, or an inch per thousand feet.
CHAPTER VI.
EARTHWORK.
FORM OF RAILROAD SECTIONS.
104. The reader is presumed to be acquainted with the manner of finding the areas and cubes of simple geometric figures and bodies. The following fifteen figures show the forms which may be taken by the cross section of a railroad in cutting; for embankment invert the same. They are easily separable into simple figures.
Fig. 39.
Fig. 40.
Fig. 41.
Fig. 42.
Fig. 43.
Fig. 44.
Fig. 45.
Fig. 46.
Fig. 47.
Fig. 48.
Fig. 49.
Fig. 50.
Fig. 51.
Fig. 52.
Fig. 53.
Fig. 54.
105. The formation of tables for the amount of earth in level cutting is very simple. The area of the following section, where B is the base, and R the horizontal dimension of the slope, is
B + B + 2R
2 × h, or 2B + 2R
2 × h,
or finally
B + R × h,
i. e., the base of a rectangle by its height. Multiply this by 100 and divide the product by 27; or divide by 27
100, and we have the cubic amount in a prism one hundred feet long. The road-bed being nineteen feet wide, and slopes one and a half to one, the formula for the amount of a prism one hundred feet long is
(19 + 1½h)h
0.27,
and assuming the base of rock cutting as eighteen feet, and slope one quarter to one, and embankment eighteen feet at subgrade, we have, rock,
(72 + h)h
1.08,
and embankment,
(18 + 1½h)h
0.27,
the figure being inverted for embankment. For a prism of ten or of one thousand feet in length, we have only to move the decimal point. In forming a table, proceed as follows:—
| h | B + 1½h | B + 1½h × h | (B + 1½h) × h 0.27 |
| a | b | c | d |
| a′ | b′ | c′ | d′ |
| an | bn | cn | dn. |
Fig. 55.
It is evident from inspection of fig. 55, that c exceeds co by h × 2r; and that c″ exceeds c′ by h′ × 2r′; and so on as far as we go; this increase being constant, we have then to find the area of c, and for the area c + c′ double c, and add the increment; whence the rule:—
Having found the increase (which varies with the angle of the slope) for the second section, add the increase to twice the first. For the third, add twice the increase to three times the first; and for the nth, add n – 1 times the increment to n times the first area, or algebraically calling a the first area, a′ the second, a″ the third, an the nth area, and we have
| The first area | a | = a; |
| The second area | 2a + i | = a′; |
| The third area | 3a + 2i | = a″; |
| The nth area | na + (n – 1)i | = an. |
We might operate at once upon the cubic contents, but for the length to which some decimals run; some indeed circulating.
106. The table thus made may be of the following form:—
| Cut (or fill), in feet. | Cubic yards Earth. Slopes 1½ to 1. | Cubic yards Rock. Slopes ¼ to 1. |
|---|---|---|
| 1 | 76 | 68 |
| 2 | 163 | 137 |
| 3 | 261 | 208 |
| 4 | 371 | 282 |
| 5 | 491 | 356 |
| 6 | 622 | 433 |
| 7 | 802 | 512 |
| 8 | 919 | 593 |
| 9 | 1083 | 675 |
| 10 | 1260 | 759 |
i. e., cut being eight feet, each one hundred feet length gives nine hundred and nineteen cubic yards; one thousand feet, 9190 yards, and ten feet of length 91.9 cubic yards.
107. The preceding system is intended only for approximate estimates. Let one person read off the cuts or fills from the profile, a second give the corresponding number of yards by the table made as above, while a third sets the figures down; being careful to separate the cuts from the fills.
For final measurements, none but the prismoidal formula should be used; the length of the prismoids being taken at each one hundred feet, and nearer when the ground is rough.
108. As an example of the comparative amounts given by the above formula, and by the common method of averaging end areas, take the following, the slopes being 1½ to 1.
| Base. | Distance. | Cut. | End Area. | Mean Area. | Middle Area. |
|---|---|---|---|---|---|
| 20 | 0 | 0 | 000 | 000 | 000 |
| 20 | 50 | 5 | 137 | 069 | 059 |
| 20 | 50 | 10 | 350 | 244 | 236 |
| 20 | 50 | 15 | 637 | 493 | 483 |
| 20 | 50 | 00 | 000 | 318 | 236 |
By averaging end areas we have
| 50 × | 69 = | 3,450 | |
| 50 × | 244 = | 12,200 | |
| 50 × | 493 = | 24,650 | |
| 50 × | 318 = | 15,900 | Sum, 56,200. |
And by the prismoidal formula,
| 50 × | 305 | |
| 50 × | 1,257 | |
| 50 × | 2,669 | |
| 50 × | 1,755 | Sum 299,300 ÷ 6 = 49,000, |
| and 56,200 – 49,000 = 7,200 | ||
cubic feet in favor of the method of end areas.
109. The prismoidal formula is algebraically
a + a′ + 4a″
6L = c,
when L = length,
c = cubic contents,
a = area of one end,
a′ = area of other end,
a″ = middle area;
or, verbally, to the sum of the end areas add four times the middle area, and multiply the result by one sixth of the length; the middle area being the area made upon the mean height of the two ends. Thus if the length is one hundred feet, and one end ten feet high, the other twenty feet high, and slopes one and a half to one, the cubic amount is, (the base being twenty-two feet,)
[(22 + 22 + 30
2 × 10) + (22 + 22 + 60
2 × 20) + (22 + 22 + 45
2 × 15 × 4)] × 100./6