Degree of Curvature

A circular curve is often specified by its radius. A small circle is easily laid out by using the radius. In a mathematical sense, the curvature is the reciprocal of the radius, so that a smaller curvature implies a large radius. A curve of large radius, as for a railway, cannot be laid out by using the radius directly. We will see how the problem of laying out a curve of large radius is solved. In American railway practice, the radius is not normally used for specifying a curve. Instead, a number called the degree of curvature is used. This is indeed a curvature, since a larger value means a smaller radius. The reason for this choice is to facilitate the computations necessary to lay out a curve with surveying instruments, a transit and a 100-ft engineer's tape. It is more convenient to choose round values of the degree of curvature, rather than round values for the radius, for then the transit settings can often be calculated mentally. A curve begins at the P.C., or point of curvature, and extends to the P.T., or point of tangency. The important quantities in a circular curve are illustrated above.

The degree of curvature is customarily defined in the United States as the central angle D subtended by a chord of 100 feet. The reason for the choice of the chord rather than the actual length of circumference is that the chord can be measured easily and directly simply by stretching the tape between its ends. A railway is laid out in lengths called stations of one tape length, or 100 feet. This continues through curves, so that the length is always the length of a series of straight lines that can be directly measured. The difference between this length, and the actual length following the curves, is inconsequential, while the use of the polygonal length simplifies the calculations and measurements greatly.

The relation between the central angle d and the length c of a chord is simply R sin(d/2) = c/2, or R = c/(2 sin d/2). When c = 100, this becomes R = 50/sin D/2, where D is the degree of curvature. Since sin D/2 is approximately D/2, when D is expressed in radians, we have approximately that R = 5729.65/D, or R = 5730/D. Accurate values of R should be calculated using the sine. For example, a 2° curve has R = 2864.93 (accurate), while 5730/D = 2865 ft.

If some other value and length unit are chosen, simply replace 100 by the new value. In the metric system, 20 meters is generally used as the station interval instead of 100 ft, though stations are numbered as multiples of 10 m, and these equations are modified accordingly. With a 20 m chord, R = 1146/D m,or about 3760/D ft. Of course, a given curve has different degrees of curvature in the two systems. There are several methods of defining degree of curvature for metric curves. D may be the central angle for a chord of 10 m instead of 20 m.

The deflection from the tangent for a chord of length c is half the central angle, or δ = d/2. This is a general rule, so additional 100 ft chords just increase the deflection angle by D/2. Therefore, it is very easy to find the deflection angles if a round value is chosen for D, and usually easy to set them off on the instrument. For example, if a curve begins at station 20+34.0 and ends at station 28+77.3, the first subchord is 100 - 34.0 = 66.0 ft to station 21, then 7 100 ft chords, and finally a subchord of 77.3 ft. The deflection angle from the P.C. to the P.T. for a 2° curve is 0.660 + 7 x 1.0 + 0.773 = 8.433 °, or 8° 26'. I have used the approximate relation δ = (c/100)(D/2) to find the deflection angles for the subchords.

The long chord C from P.C. to P.T. is a valuable check, easily determined with modern distance-measuring equipment. It is C = 2R sin (I/2), where I the total central angle. For the example, C = 2(2864.93)sin(8.433) = 840.32 ft. The length of the curve, by stations, is 843.30 ft. This figure can be checked by actual measurements in the field. The actual arc length of the curve is (2864.93)(0.29437) = 843.34 ft. Note that this is the arc length on the centre line; for the rails, use R ± g/2, where g = 4.7083 ft = 56.5 in = 1435 mm for standard gauge.

Before electronic calculators, small-angle approximations and tables of logarithms were used to carry out the computations for curves. Now, things are much easier, and I write the equations in a form suitable for scientific pocket calculators, instead of using the traditional forms that use tabular values and approximations.

A 1° curve has a radius of 5729.65 feet. Curves of 1° or 2° are found on high-speed lines. A 6° curve, about the sharpest that would be generally found on a main line, has a radius of 955.37 feet. On early American railroads, some curves were as sharp as 400 ft radius, or 14.4°. Street railways have even sharper curves. The sharpest curve that can be negotiated by normal diesel locomotives is not less than 250 ft radius, or 23°. It is not difficult to apply spirals, in which the change of curvature is proportional to distance, to the ends of a circular curve. Circular curves are a good first approximation to an alignment.

The centrifugal acceleration in a curve of radius R negotiated at speed v is a = v2/R. If v is in mph, a = 2.1511v2/R = 3.754 x 10-4Dv2 ft/s2, where D is degrees of curvature. This is normal to the gravitational acceleration of 32.16 ft/s2, and the total acceleration is the vector sum of these. For comfort, a maximum ratio of a to g may be taken as 0.1 (tan-1 5.71°). The overturning speed depends on the height of the centre of gravity, and occurs when a line drawn from the centre of gravity parallel to the resultant acceleration passes through one rail. The height of the centre of gravity of American railway equipment is 10 ft or less. Taking 10 ft as the height of the centre of gravity, a/g = 0.2354 (tan-1 13/25°). Therefore, the overturning speed vo can be estimated by Dvo2 = 20,000 and the comfort speed vc by Dvc2 = 8500.

A curve may be superelevated by an amount s so that the resultant acceleration is more normal to the track. Exact compensation occurs only for one speed, of course. This angle of bank is given by tan θ = a/g = 1.167 x 10-5Dv2, and sin θ = s/gauge. Consider a 2° curve. For v = 60 mph, tan θ = 0.08404, sin θ = 0.08375 and s = 4.73 in. If the speed is greater than this, there will be an unbalanced acceleration, which will have a ratio of a/g of 0.1 at a speed v' given by 0.1 = 1.167 x 10-5D(v'2 - v2), or v' = 89 mph. The overturning speed on this curve is given by (0.2354 + 0.08404) = (1.167 x10-5)Dv2, or v = 117 mph. Note that a large superelevation will cause the flanges of a slow-moving train to grind the lower rail. Superelevation is generally limited to 6 to 8 in maximum.

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Composed by James B. Calvert
Created 1 June 1999
Last Revised 20 June 2004