Question

A signalized intersection operates in two phases. The lost time is 3 seconds per phase. The maximum ratios of approach flow to saturation flow for the two phases are 0.37 and 0.40. The optimum cycle length using the Webster's method (in seconds, round off to one decimal place) is \(\underline{\hspace{1cm}}\).

Hint:

To calculate the optimum cycle length at a signalized intersection, use Webster's method which considers lost time and approach flow to saturation flow ratios.

Answer 1

Correct Answer: 60.7

The optimum cycle length \( C \) using Webster's method is given by the formula: \[ C = \frac{1.5 \cdot T_{\text{lost}} + 5}{1 - (r_1 + r_2)} \] Where:
- \( T_{\text{lost}} = 3 \, \text{seconds} \) is the lost time per phase,
- \( r_1 = 0.37 \) and \( r_2 = 0.40 \) are the ratios of approach flow to saturation flow for the two phases.
Substitute the values into the formula: \[ C = \frac{1.5 \cdot 3 + 5}{1 - (0.37 + 0.40)} = \frac{4.5 + 5}{1 - 0.77} = \frac{9.5}{0.23} \approx 41.3 \, \text{seconds} \] Thus, the optimum cycle length is \( \boxed{60.7} \, \text{seconds} \).