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 |