Highway Engineering – Ideal Banking Angle on Curves On a frictionless ideal banked road curve, the required banking angle depends primarily on which factor?

Difficulty: Easy

Correct Answer: (velocity)^2 of the vehicle

Explanation:


Introduction / Context:
Banking of roads allows vehicles to negotiate curves safely at a design speed without relying solely on tire-road friction. The ideal banking angle comes from balancing components of the normal reaction against centripetal requirements in a frictionless design scenario.



Given Data / Assumptions:

  • Frictionless idealization (design speed condition).
  • Vehicle travels on a circular arc of radius R at speed v.
  • Gravity g acts vertically downward.



Concept / Approach:
The standard relation for ideal banking is tan(θ) = v^2 / (g R). Thus θ depends on the square of speed and inversely on radius, independent of vehicle weight. Friction, surface type, and power are not part of the ideal balance condition (though they matter in real-world safety margins).



Step-by-Step Solution:
Equate horizontal component of normal reaction to centripetal requirement: N sinθ = m v^2 / R. Vertical equilibrium: N cosθ = m g. Divide: tanθ = (m v^2 / R) / (m g) = v^2 / (g R). Hence θ is determined by v^2 (and R), not by weight or engine power.



Verification / Alternative check:
Doubling speed quadruples v^2 and significantly increases θ; doubling vehicle mass cancels out of the ratio and does not change θ.



Why Other Options Are Wrong:
Weight: cancels from the equations. Nature of surface and coefficient of friction: relevant for non-ideal (with friction) conditions, not for ideal banking angle. Engine power: unrelated to the geometric equilibrium for banking.



Common Pitfalls:
Assuming heavier vehicles need a different banking angle; under ideal theory, they do not.



Final Answer:
(velocity)^2 of the vehicle

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