Question

The capacity (in PCU/hr) of the roundabout for given data is
Average entry width = 8 m
Width of weaving section = 14 m
Length of the weaving section between the ends of channelizing islands = 38 m
Proportion of the weaving traffic = 0.5

• A. 3421
• B. 3128
• C. 3755
• D. 3628

Answer 1

The correct option is C 3755

Given, e = 8 m, w = 14 m, L = 38 m, p = 0.5
Capacity$=\frac{280\times w(1+\frac{e}{w})(1-\frac{p}{3})}{(1+w/L)}$
$=\frac{280\times 14(1+\frac{8}{14})(1-\frac{0.5}{3})}{(1+14/38)}$
= 3751.3 PCU/hr