If a curve setup on a highway has a deflection angle of 120° between its tangents and a versine distance of 225 m, then the sharpness of the curve subtended by an arc of length 30 m will be
- A5°
- B4°Correct
- C3°
- D6°
Solution
Concept
The versine of a circular curve gives the radius. The degree of curve is the angle subtended at the centre by a standard 30 m arc.
Formula
V = R(1 - cos(Δ/2)); D = (30/R) × (180/π) ≈ 1718.9/R
Step-by-step solution
1. Deflection angle Δ = 120°, so Δ/2 = 60°. 2. Versine V = R(1 - cos 60°) = R(1 - 0.5) = R/2. 3. 225 = R/2, so R = 450 m. 4. Degree of curve for a 30 m arc: D = (30/450) × (180/π) = 3.82°. 5. The nearest option is 4°.
Explanation
The versine is the distance between the mid-point of the curve and the mid-point of the long chord. From it we get R = 450 m. A 30 m arc on this radius subtends about 3.82°, which is nearly 4°.
Why the other options are wrong
A (5°) and D (6°) would need a much smaller radius, about 344 m and 286 m. C (3°) would need a radius of about 573 m.
Shortcut
When Δ = 120°, the versine is R/2. Then use D ≈ 1719/R.
Common mistake
Using Δ instead of Δ/2 in the versine formula, or using the 20 m chord formula D = 1146/R.
Correct answer: B) 4°