Highway design – superelevation formula If V is the design speed in km/h and R is the radius of a horizontal circular curve in metres, the required superelevation e (as a fraction) is given by which expression?

Introduction / Context:Superelevation is the transverse upward tilt of the pavement on horizontal curves to counteract the lateral acceleration experienced by vehicles. In highway engineering, it is expressed as a fraction e (rise per unit width) related to speed and curvature. Choosing the correct formula is essential for comfort and safety.

Given Data / Assumptions:

• V = design speed in km/h.
• R = radius of the horizontal curve in metres.
• Standard highway relationship that neglects lateral friction at the chosen design speed (all centripetal demand provided by superelevation).

Concept / Approach:For equilibrium on a banked curve, the required centripetal acceleration V^2/(225R) (with V in km/h) is balanced by the component of gravity along the road surface. This yields e = V^2/(225R) when lateral friction is not relied upon at the design speed. When friction is also considered, the combined relation becomes e + f = V^2/(127R), where f is the side-friction factor.

Step-by-Step Solution:

Verification / Alternative check:Compare with the combined design relation e + f = V^2/(127R). Setting f = 0 for the design (conservative) case gives e = V^2/(127R). But standard IRC-type practice commonly uses e = V^2/(225R) for the equilibrium-only expression and keeps a separate check with friction for higher speeds.

Why Other Options Are Wrong:Options C and D invert the relationship. Option B corresponds to the combined e + f form. Option E has incorrect dimensions.

Common Pitfalls:Mixing the two constants 127 and 225; using metres per second for V without unit conversion; exceeding maximum permissible e in practice (typically capped).

Final Answer:e = V^2 / (225 * R)