Example.—How far from the toe of the switch is the point of the frog, the gauge being 4′ 8½″, the rail twenty feet long, and moving five inches; the frog being six feet long, six inches wide across the head, and three inches at the mouth?
We have
In laying the rails, the distance from the point to the end of the frog (towards the switch) is to be taken from the above.
Table showing the distance between the frog and switch, gauge 4′ 8½″, movement of switch-rail five inches. Frog six inches across head, and three inches at mouth. Main track being straight.
| Length of frog. | LENGTH OF SWITCH RAIL. | |||||
|---|---|---|---|---|---|---|
| 12 | 14 | 16 | 18 | 20 | 22 | |
| 3 | 29.1 | 29.7 | 30.1 | 30.4 | 30.7 | 30.9 |
| 3½ | 33.3 | 34.0 | 34.5 | 35.0 | 35.3 | 35.6 |
| 4 | 37.3 | 38.2 | 38.8 | 39.4 | 39.8 | 40.2 |
| 4½ | 41.1 | 42.2 | 43.0 | 43.7 | 44.3 | 44.7 |
| 5 | 44.8 | 46.1 | 47.1 | 47.9 | 48.5 | 49.1 |
| 5½ | 48.3 | 49.8 | 51.0 | 51.9 | 52.7 | 53.2 |
| 6 | 51.7 | 53.4 | 54.8 | 55.9 | 56.8 | 57.6 |
| 6½ | 55.0 | 56.9 | 58.5 | 59.8 | 60.8 | 61.7 |
| 7 | 58.1 | 60.3 | 62.1 | 63.4 | 64.7 | 65.7 |
| 7½ | 61.2 | 63.6 | 65.6 | 67.2 | 68.5 | 69.6 |
292. When the switch rail is short, the angle between the main line and the switch rail, when switched, is considerable; and causes quite a shock to the passing engine. The switch shown in fig. 145 remedies the evil, makes the machinery compact, and the calculation simple. The tangent point of the turnout curve is at n n (the usual heel). In place of adjusting the single to the double line of rails, the double is adjusted to the single line. The data given are the gauge and radius of curve; and as before, the elements required the frog angle and distance from switch to frog point.
Fig. 145.
Now
The angle of frog is also
The length of this switch rail depends upon the radius of curvature. The distance between the two rails at S must be enough to admit the wheel flange, that is, at least two inches.
Let A B, fig. 146, be the straight rail; E D the curved one. Draw G H parallel with and two inches distant from the inner edge of A B. No point of the curved rail must fall within G H; whence E is the turning-point, and E D the length, found as follows.
Let R equal the radius of curve to outside of outer rail; d equal two inches plus width of rail, or i e, and D equal D E.
Then
Fig. 146.
Example.—Let the radius of outer rail be five hundred feet, and the gauge five feet. We have, then, the distance
and the length of switch
Five hundred feet is, therefore, about the longest radius for which such switches should be used.
Crossings occur where two tracks cross, and consist of four frogs, with the corresponding guard rails, as in fig. 147.
Fig. 147.
293. The motion of a train of cars around a curve is accompanied by a tangential force, depending in amount upon the velocity of the train and the radius of curvature. This force tends to throw the cars from the track; and is counteracted by elevating the exterior rail.
The centrifugal force of any body in motion in a curved line is shown by the formula
The force tending to throw the car from the rail is not centrifugal but tangential, but it matters not whether the body is kept in position by tension upon the inside or by compression on the outside; the amount of the force is the same.
Fig. 148.
The horizontal projection of the centre of gravity of the car, when at rest, is at c, fig. 148, and when in motion the direction of the weight should be a b; and the inclination, c′ a′ b′, must be such that a b will be perpendicular to c′ a′; to effect which, c′ b′ should be to a′ b′ as the weight to the tangential force; or E being the elevation of the rail, g the gauge, W the weight, and c the tangential force; we have
g and R are the only fixed quantities in the formula; and the average weight and speed of a car must be assumed.
Examination of the formula shows how important it is that all trains should run at such a velocity as to demand the same elevation of rail. The absolute elevation must be arranged to meet the requirement of the fastest trains; and other trains must conform, even at a disadvantage.
Note.—The subject of the mechanics of traversing railroad curves, is yet quite in the dark. The action of the train, as caused by its own momentum, is tangential; while the action of the engine tends to pull the cars against the inner rail, being opposed to the first motion. This might require a reduction of the elevation given by the formula when the engine is exerting a strong tractive power, but when running without steam the full elevation is needed, (see chapter III.)
Fig. 149.
In laying and maintaining the rails to the proper elevation, a clinometer attached to a rail gauge, as in fig. 149, answers a good end: the small arc being graduated according to the different elevations required by curves of different radii. Thus the index of the level being placed at 2°, when the rails are fitted to A and B, the elevation is correct for a 2° curve; or for a curve of 2,865 feet radius.
The difference in gauge of one foot makes a difference in the elevation of but 0.009 feet, or about ⅒ of an inch.
The following table is calculated for the average of the different gauges in use, thus,—
| 4.7 | ||
| 5.0 | ||
| 5.5 | ||
| 6.0 | ||
| 4)21.2 | ||
| Average gauge, | 5.3 | feet. |
| TABLE OF ELEVATION OF OUTER RAIL. | ||||||
| Radius of curve in feet, being | ELEVATION OF OUTER RAIL IN FEET AND DECIMALS, THE VELOCITY IN MILES PER HOUR BEING | |||||
|---|---|---|---|---|---|---|
| 10. | 15. | 20. | 25. | 30. | 40. | |
| 250 | .130 | |||||
| 500 | .070 | |||||
| 1,000 | .037 | .079 | ||||
| 2,000 | .018 | .040 | .074 | .111 | ||
| 3,000 | .013 | .026 | .048 | .074 | .106 | |
| 4,000 | .009 | .020 | .037 | .058 | .079 | .154 |
| 5,000 | .007 | .016 | .031 | .045 | .065 | .119 |
| 6,000 | .006 | .013 | .024 | .037 | .053 | .095 |
| 7,000 | .005 | .011 | .021 | .033 | .046 | .086 |
| 8,000 | .004 | .010 | .018 | .029 | .039 | .077 |
| 10,000 | .003 | .008 | .010 | .022 | .032 | .059 |