COMPARISON OF LOCATED LINES.
65. In this comparison there is an element which does not enter the approximate comparison of surveyed lines, curvature. The resistance arising from this cause has never been accurately determined. Mr. McCallum estimates the resistance at one half pound per degree of curvature per one hundred feet; i. e., the resistance due to curvature on a 4° curve, would be two lbs. per ton, (see report of September 30, 1855). Mr. Clark estimates the resistance due to curves of one mile radius and under, as 6.3 lbs. per ton, or twenty per cent. of the whole resistance. The average radius encountered, therefore, by Mr. Clark, would be, at Mr. McCallum’s estimate,
6.3
0.5 = 12° nearly, or 477.5 feet.
So small a radius is by no means allowable upon English roads; thus the estimate of Mr. Clark and of Mr. McCallum differ considerably. Experiments might easily be made with the dynamometer upon different curves, by which we might find very nearly the correct resistance caused by curves.
The curvature on any road cannot be adjusted to trains moving at different speeds.
66. The tractive power acts always tangent to the curve at the point where the engine is, and thus tends to pull the cars against the inner rail. The tangential force, generated by the motion of the cars, tends to keep the flanges of the wheels against the outer rail; and only when a just balance is made between the tractive and tangential forces, the wheel will run without impinging on either rail, (the wheel being properly coned). For these forces to balance, there must be a fixed ratio between the weight of a car and the speed, (not the weight of a train, as the shackling allows the cars to act nearly independently, some indeed rubbing hard for a moment against the rail, while the next car is working at ease). Whenever the right proportion is departed from, as it nearly always is, (and perhaps necessarily in some cases,) upon railroads, the wheels will rub against one rail or the other. Thus on any road where the speed on the same curve, or the radii of curvature under the same speed, differ, there must be loss of power, and dragging or pushing against the rails.
67. We are obliged to elevate the outer rail (see chapter XIII.), for the fastest trains, and the slower trains on such roads will therefore always drag against the inner rails. Thus in practice we generally find the inside of the outer rail most worn on passenger roads, and the inside of the inner rail upon chiefly freight roads.
68. It has been the practice of some engineers in equating for curvature, to add one fourth of a mile to the measured length for each 360° of curvature, disregarding the radius, as the length of circumference increases inversely as the degree of curvature.
69. Now in equating for grades, in doubling the power we do not double the expense of working. We however increase it more by curvature than we do by grades, because besides requiring double power, the wear and tear of cars and rails and all machinery is increased upon curves, which is not the case upon grades.
70. The analysis of expense (in Appendix F.) upon the New York system of roads, gives the following:—
| Locomotives, | 40 | per cent. |
| Cars, | 20 | per cent. |
| Way and works, | 15 | per cent. |
| or in all, | 75 | per cent. |
Now each 360° will be equal to 75
100 of one quarter of a mile, or 75
400 of a mile; whence the number of degrees which shall cause an expense equal to one straight and level mile, will be 1920°.
71. The number of degrees by Mr. McCallum’s estimate would be thus:—
The resistance upon a level being ten lbs. per ton, and that due to curves one half pound per ton, per degree per one hundred feet; the length of a 2° curve to equal one mile will be
10 lbs.
1 lb.,
or ten miles. Also ten miles, or 530 hundred feet by 2° is 1060°.
72. Again, by Mr. Clark’s resistance of twenty per cent. of the level resistance, upon curves averaging 2°, we have as the length of 2° curve
10
2 = 5 miles,
or 265 hundred feet, which by 2° gives 530°.
73. Averaging the first and last, we have as the number of degrees which should be considered as causing an amount of expense equal to one straight and level mile, 1225°, which averaging with the estimated resistance by Mr. McCallum, gives finally 1142½° as causing an expense equal to one straight and level mile, or, in round numbers, 1140°.
74. Suppose now that we would know which of the lines below to choose.
| Line A. | Line B. | Description. |
|---|---|---|
| 100 miles, | 110 miles, | Actual length, |
| 5000 feet, | 3000 feet, | Rise, |
| 3500 feet, | 1500 feet, | Fall, |
| 3600° | 9000° | Degrees of curvature. |
Assuming the speed as twenty miles per hour, the number by which to equate for grades, see chapter II., is ninety-six, also the number of degrees for curvature 1140, whence,
| Line A ascending 100 + 52.1 + 3.16 = 155.26 | 147.46 |
| Line A descending 100 + 36.5 + 3.16 = 139.66 | |
| Line B ascending 110 + 31.25 + 7.89 = 149.14 | 141.31, |
| Line B descending 110 + 15.62 + 7.89 = 133.49 |
and if the cost of construction is as the actual, and the cost of maintaining and working as the mean equated length, we have, as a final comparison,
A to B as 100 + 147.46 to 110 + 141.31,
or as
247.46 to 251.31.
Here the extra grades on the one hand nearly equal the curvature and the extra length on the other hand.
75. As a further example in the comparison of competing lines, let us take the actual case of the location of the eastern part of the New York and Erie Railroad.
It was questioned which of the two lines between Binghampton and Deposit should be adopted, and also between the mouth of Callicoon Creek and Port Jervis.
Fig. 30.
Between A and c, fig. 30, were located the lines shown in the sketch, one following the Susquehanna river from A to B, thence crossing the dividing ridge between that river and the Delaware to Deposit (c). The other passing up the Chenango river to a, thence crossing first the summit M to the Susquehanna at L, and second the summit K, to Deposit (c). The elements of the two lines are as follows:—
| A route, A B c. | B route, A M K c. | |
|---|---|---|
| Length, | 39.29 | 43.58 |
| Rise A to c, | 540.00 | 1087.00 |
| Rise c to A, | 395.00 | 936.00 |
| Whole rise and fall, | 935.00 | 2023.00 |
| Degrees of curvature, | 2371°.00 | 3253°.00 |
| Estimated cost, | $746,900.00 | $628,600.00 |
Assuming the number by which to equate for grades, as 96, and the equating number of degrees of curvature as 1140°; equating for grades and curvature in both directions, we have,
| Route A. A to c. | Mean, 46.25. | ||
| 39.29 + | 540 96 + 2371 1140 = 39.29 + | 5.63 + 2.08 = 47.00 | |
| Route A. c to A. | |||
| 39.29 + | 395 96 + 2371 1140 = 39.29 + | 4.12 + 2.08 = 45.49 | |
| Route B. A to c. | Mean, 56.96. | ||
| 43.58 + | 1087 96 + 3253 1140 = 43.58 + | 11.32 + 2.85 = 57.75 | |
| Route B. c to A. | |||
| 43.58 + | 936 96 + 3253 1140 = 43.58 + | 9.75 + 2.85 = 56.18 | |
Assuming the cost of working and of maintaining as $4,000 per mile, we have
The cost of building A to B as $746,900 to $628,600
The cost of operating A to B as (46.25 × 4000) × 100
6 to (56.96 × 4000) × 100
6,
| or as | $3,083,334 | to | $3,797,334 |
| and the sum as | $3,830,234 | $4,425,934 |
giving the preference of $595,700 to the route A B c, notwithstanding that the estimate thereon exceeds that on B by $118,300. The route A B c was adopted.
Again, it was doubtful whether to adopt the route E F, in going from D to G, or the line I H. The following are the elements of the two lines:—
| I H. | E F. | |
|---|---|---|
| Measured length, | 61.14 | 58.53 |
| Rise D to G, | 1187 | 454 |
| Rise G to D, | 1049 | 316 |
| Degrees curve, | 7609° | 4588° |
| Estimated cost, | $1,094,950 | $1,496,430 |
The mean equated lengths are as follows:—
| Line I H. D to G. | Mean, 79.46, | ||
| 61.14 + | 1187 96 + 7609 1140 = 61.14 + | 12.36 + 6.68 = 80.18 | |
| Line I H. G to D. | |||
| 61.14 + | 1049 96 + 7609 1140 = 61.14 + | 10.93 + 6.68 = 78.75 | |
| Line E F. D to G. | Mean, 66.56. | ||
| 58.53 + | 454 96 + 4588 1140 = 58.53 + | 4.73 + 4.02 = 67.28 | |
| Line E F. G to D. | |||
| 58.53 + | 316 96 + 4588 1140 = 58.53 + | 3.29 + 4.02 = 65.84 | |
The comparison as to cost is
I H to E F as $1,094,950 to $1,496,430,
and as to working,
I H to E F as (79.46 × 4000) × 100
6 to (66.56 × 4000) × 100
6,
and the sum as
| 1,094,950 | to | 1,496,430 | |
| + 5,297,334 | + 4,437,334 | ||
| or | $6,392,284 | to | $5,933,764 |
Although the cost of E F is $401,480 more than that of I H, the line E F was adopted.