On a highway with ruling gradient 3%, what gradient should be provided on a horizontal curve of radius 300 m after applying grade compensation?

Introduction / Context:Grade compensation reduces the algebraic gradient on horizontal curves to offset the additional tractive resistance that vehicles experience while negotiating curvature. Without compensation, the effective effort demanded on a curve would be higher than on a tangent with the same gradient, penalizing heavy vehicles.

Given Data / Assumptions:

• Ruling gradient on tangent, G_ruling = 3.00%.
• Curve radius, R = 300 m.
• Use the standard compensation formula for curves (subject to a cap).

Concept / Approach:IRC practice uses grade compensation (%) ≈ 75 / R for R in metres, limited to a maximum (often 0.8%). For R = 300 m, compensation = 75 / 300 = 0.25%. The compensated gradient on the curve = ruling gradient − compensation.

Step-by-Step Solution:Compute compensation: C = 75 / 300 = 0.25%.Apply to ruling gradient: G_curve = 3.00% − 0.25% = 2.75%.Check cap: 0.25% < 0.8%, so no capping required.

Verification / Alternative check:As R increases, compensation decreases towards zero; at small R (sharp curves), compensation approaches the cap, aligning with practical experience.

Why Other Options Are Wrong:

• 2.00%, 2.25%, 2.50%: over-compensate for R = 300 m.
• 3.00%: ignores compensation and would demand unnecessary tractive effort on the curve.

Common Pitfalls:

• Using the compensation formula with R in km instead of metres.
• Forgetting to enforce the maximum cap when R is very small.

Final Answer:2.75%