Railway engineering – equilibrium super elevation (cant): For a locomotive running at speed V (km/h) on a circular curve of radius R (m), with g as acceleration due to gravity (m/s^2) and G as the distance between running faces of rails (m), what is the required super elevation e?

Introduction / Context:
On curved tracks, raising the outer rail (super elevation or cant) balances the lateral acceleration so that the resultant pressure acts normal to the rail plane. The equilibrium cant depends on speed, curve radius, and track gauge. This relation is essential for safe and comfortable railway operation and for setting out permanent way on curves.

Given Data / Assumptions:

• V is train speed in km/h.
• R is curve radius in meters.
• G is the distance between running faces of the rails (m).
• g ≈ 9.81 m/s^2 is acceleration due to gravity.

Concept / Approach:
For equilibrium: e / G = v^2 / (g * R), where v is speed in m/s. Converting V (km/h) to v: v = V / 3.6. Substituting gives e = G * (V^2 / 3.6^2) / (g * R) = (G * V^2) / (g * R * 12.96). With g ≈ 9.81, the denominator becomes about 127; hence, the practical design formula is e = (G * V^2) / (127 * R), with e in meters when G is in meters and V in km/h.

Step-by-Step Solution:

Verification / Alternative check:
For standard gauge G = 1.676 m and V = 60 km/h, R = 300 m, e = 1.676 * 3600 / (127 * 300) ≈ 0.158 m (≈ 158 mm), matching handbook values.

Why Other Options Are Wrong:

• e = (V^2) / (g * R): missing G factor and wrong units.
• e = (G * g) / (R * V): incorrect dependence; cant increases with V^2, not decreases with V.
• e = (R * G) / (127 * V^2): inverted speed dependence.

Common Pitfalls:
Mixing unit systems (m/s vs km/h); forgetting that e is measured vertically at the rail; applying equilibrium cant to circumstances that require limiting cant due to low-speed freight operations.

Final Answer:
e = (G * V^2) / (127 * R)