Understanding Equilibrium Superelevation in Railways

Equilibrium superelevation, also known as equilibrium cant, is the amount by which the outer rail of a railway track is raised above the inner rail on a curve. This elevation is provided to counteract the centrifugal force experienced by a train moving on the curve, ensuring that the resultant of the centrifugal force and the weight of the train is perpendicular to the plane of the rails. This ideally eliminates the lateral pressure on the rails at a specific speed, known as the equilibrium speed.

Factors Affecting Superelevation

The required superelevation depends on several factors:

• Gauge (G): The distance between the inner faces of the rails.
• Speed (V): The speed of the train on the curve.
• Radius (R): The radius of the curve.
• Acceleration due to gravity (g): A constant value.

Deriving the Equilibrium Superelevation Formula

The theoretical formula for equilibrium superelevation (\(e\)) can be derived by considering the forces acting on a train on a curve. For equilibrium conditions, the ratio of the centrifugal force (\(F_c\)) to the weight of the train (\(W\)) is equal to the ratio of the superelevation (\(e\)) to the gauge (\(G\)) for small angles. Assuming a train mass \(m\), speed \(v\) (in m/s), radius \(R\) (in meters), and gauge \(G\) (in meters):

Centrifugal force, \(F_c = \frac{mv^2}{R}\)

Weight, \(W = mg\)

For equilibrium, \(\frac{F_c}{W} = \frac{e}{G}\) (approximately for small e/G)

So, \(\frac{mv^2/R}{mg} = \frac{e}{G}\)

\(\frac{v^2}{gR} = \frac{e}{G}\)

Rearranging for \(e\): \(e = \frac{Gv^2}{gR}\)

This formula requires all units to be consistent (e.g., meters for G, R, e; m/s for v; m/s\(\text{\textsuperscript{2}}\) for g).

Equilibrium Superelevation Formula with Mixed Units

In railway engineering practice, speed is usually given in kilometers per hour (km/hour). To use the formula with V in km/hour, we need to convert it to m/s:

\(v \, (\text{m/s}) = V \, (\text{km/hour}) \times \frac{1000 \, \text{m}}{3600 \, \text{s}} = V \times \frac{5}{18} \, (\text{m/s})\)

Substitute this into the formula \(e = \frac{Gv^2}{gR}\):

\(e = \frac{G \left( V \times \frac{5}{18} \right)^2}{gR} = \frac{G \times \frac{25V^2}{324}}{gR} = \frac{25GV^2}{324gR}\)

Using the standard value of acceleration due to gravity \(g \approx 9.81 \, \text{m/s}\text{\textsuperscript{2}}\):

\(e \approx \frac{25GV^2}{324 \times 9.81 R} \approx \frac{25GV^2}{3179.64R}\)

To simplify this for practical use, the constant is approximated. The commonly accepted formula for equilibrium superelevation (\(e\)) in meters when G is in meters, V is in km/hour, and R is in meters is:

\(e = \frac{GV^2}{127R}\)

This formula accounts for the unit conversions and the value of \(g\).

Applying the Formula to the Question

The question asks for the equilibrium superelevation given G in meters, V in km/hour, and R in meters. Based on the standard formula used in railway engineering with these units, the equilibrium superelevation is given by:

\(e = \frac{GV^2}{127R}\)

Comparing this with the given options, we find that the formula matches one of the options.

Summary of Formula Components

Therefore, the correct formula for equilibrium superelevation is \(\rm \frac{GV^2}{127R}\).