Question

Consider the following parameters related to a rotary intersection:
1. Width of the weaving section.
2. Length of the weaving section.
3. Proportion of weaving traffic.
4. Weaving angle.
5. Width of the carriageway at entry.
Capacity is generally expressed in terms of

• A. 1, 2, 3 and 5
• B. 4 and 5
• C. 1, 2, 3 and 4
• D. 1, 2 and 3

Answer 1

The correct option is A 1, 2, 3 and 5

The practical capacity $\left(Q_p\right)$ of the rotary is dependent on the minimum capacity of the individual weaving section.

$Q_p=\frac{280W(1+e/W)(1-p/3)}{(1+W/L)}$

where,
W is width of weaving section
e is average width of entry $e_1$ and width of non weaving section $e_2$ for the range e/W = 0.4 to 1.0
L is length of the weaving section
p is proportion of weaving traffic