Question

For the intersection of problem 20-5, consider a case in which the e-w arterial has two lanes in each direction and the n-s arterial has only one lane in each direction. for this case: a. what is the absolute minimum cycle length that could be used? b. what cycle length would be required to provide for a ratio of 0.90 during the worst 15 minutes of the hour if the phf is 0.92?

Answer 1

The absolute minimum cycle length is 3 seconds, and the cycle length required to provide for a ratio of 0.90 during the worst 15 minutes of the hour is 0.90V / 0.92 seconds.

Step-by-step explanation:

a. The absolute minimum cycle length that could be used is determined by the sum of the time it takes for vehicles on the e-w arterial to clear the intersection and the time it takes for vehicles on the n-s arterial to clear the intersection. Since the e-w arterial has two lanes in each direction, it takes 2 seconds for the vehicles to clear the intersection. The n-s arterial has only one lane, so it takes 1 second for the vehicles to clear the intersection. Therefore, the absolute minimum cycle length is 2 seconds + 1 second = 3 seconds.

b. To calculate the cycle length required to provide for a ratio of 0.90 during the worst 15 minutes of the hour, we need to first calculate the cycle length:

Cycle Length = (Worst 15 minutes volume) / (PHF) = (0.90 x Total volume in the worst 15 minutes) / (0.92)

Let's assume the total volume in the worst 15 minutes is V.

Taking the ratio of 0.90 and PHF of 0.92, the cycle length required is 0.90V / 0.92.

Therefore, the cycle length required is 0.90V / 0.92 seconds.