Question

A roundabout is provided with an average entry width of 8.4 m, width of weaving section as 14 m and length of the weaving section between channelizing islands as 35 m. The crossing traffic and total traffic on the weaving section are 1000 and 2000 PCU/h respectively. The nearest rounded capacity of the roundabout (in PCU/h) is

• A. 3700
• B. 4500
• C. 5200
• D. 3300

Answer 1

The correct option is A 3700

Capacity of roundabout (rotary) is dependent on the minimum capacity of the individual weaving action
$Q_P=\frac{280w(1+\frac{e}{w})(1-\frac{P}{3})}{(1+\frac{w}{L})}$
where , w = Width of weaving section
= 14 m
e = average width of entry
= 8.4 m
L = length of weaving section
= 35 m
P = proportion of weaving traffic
$=\frac{1000}{2000}=0.5$
$Q_P=\frac{280\times 14(1+\frac{8.4}{14})(1-\frac{0.5}{3})}{(1+\frac{14}{35})}$
$=3733.33PCU$
$=3700PCUperhour$