Roundabout capacity by weaving formula (IRC concept) Given: average entry width e = 8.4 m, weaving width w = 14 m, weaving length l = 35 m, crossing traffic = 1000 PCU/h, total traffic on weaving section = 2000 PCU/h. Estimate the capacity of the roundabout (nearest PCU/h).

Introduction / Context:
At unsignalized roundabouts, the practical capacity of the weaving section can be estimated using the IRC-style empirical weaving formula. It accounts for geometry (entry width, weaving width, length) and traffic composition (weaving proportion).

Given Data / Assumptions:

• Average entry width e = 8.4 m.
• Weaving width w = 14 m.
• Weaving length l = 35 m.
• Total traffic on weaving section = 2000 PCU/h.
• Crossing (weaving) traffic = 1000 PCU/h → proportion p = 1000/2000 = 0.5.

Concept / Approach:
IRC weaving capacity formula (one common form): Q = 280 * w * (1 + e/w) * (1 - p/3) / (1 + w/l), where Q is in PCU/h, w and e in metres, l in metres, and p is the proportion of weaving traffic.

Step-by-Step Solution:
1) Compute e/w = 8.4 / 14 = 0.6 → (1 + e/w) = 1.6.2) Compute w/l = 14 / 35 = 0.4 → denominator (1 + w/l) = 1.4.3) Compute (1 - p/3) with p = 0.5 → 1 - 0.5/3 = 1 - 0.1667 = 0.8333.4) Q = 280 * 14 * 1.6 * 0.8333 / 1.4 ≈ 3730 PCU/h → nearest option 3700.

Verification / Alternative check:
Magnitude consistency: larger w, longer l, and lower p increase capacity; here values yield capacity in the mid-3,000s PCU/h, which is reasonable for the geometry.

Why Other Options Are Wrong:
3300 underestimates the computed value; 4500 and 5200 are too high for the given geometry and weaving proportion.

Common Pitfalls:
Using wrong p (must be weaving proportion); mixing up units; rounding too early.

Final Answer:
3700