Super-elevation formula for hill road curves: If V is the design speed in km/h and R is the radius of a horizontal curve (m), what relationship correctly expresses the required super-elevation when combined with side friction?

Introduction / Context:On horizontal curves, the centripetal demand V^2/R is met by a combination of super-elevation (e) and side friction (f). For highway design in metric units with speed in km/h and radius in metres, a standard constant appears in the relationship used to size super-elevation and check comfort and safety.

Given Data / Assumptions:

• V in km/h, R in m.
• e is super-elevation (dimensionless; e = tan θ).
• f is side friction factor (dimensionless).
• Standard highway design conditions (not rail).

Concept / Approach:

The governing equilibrium is e + f = V^2/(225 R) in highway practice with the 225 constant valid for km/h and metres. Designers choose e up to an adopted maximum (often around 7–10% depending on terrain/urban constraints) and use a limiting f for comfort and safety, then solve for R or e as necessary.

Step-by-Step Solution:

Verification / Alternative check:

Design charts and IRC formulae consistently employ this relationship for highway curves. Values with 127 are associated with railway practice (speed in km/h), not standard highway equations with f included.

Why Other Options Are Wrong:

• e = V^2/(225 R): Ignores side friction contribution.
• Expressions with 127: Not the conventional constant for highway super-elevation with friction.
• e = f = V/(15 R): Dimensionally inconsistent with the standard equation.

Common Pitfalls:

• Mixing units and constants from rail formulas or from m/s forms without proper conversion.

Final Answer:

e + f = V^2 / (225 R).