How is the peak hour found from 15-minute counts?

Sum every four consecutive intervals and choose the largest total

Pick the hour starting at the first count

Take the largest single 15-minute count times 4

PHF is calculated as:

V / (4 x V15,max) = 1500/(4 x 490) = 0.765

Peak volume divided by PHF, about 1960 PCU/h

Peak volume minus 15-min maximum

Short answer

Peak hour is 9:30-10:30 AM with peak hourly volume 1500 PCU. PHF = 1500/(4x490) = 0.765, and design flow rate = 1500/0.765 = 1960 PCU/h.

Table below shows volume counts on an approach of an urban intersection from 8.30 AM to 10.30 AM.

Determine (i) peak hour (ii) peak hourly volume (iii) peak hour factor, and (iv) the design flow rate for the approach.

The peak hour is found by adding the 15-minute counts over every four consecutive intervals (a sliding one-hour total) and choosing the largest.

The largest one-hour total is 1500 PCU, so the peak hour is 9:30 - 10:30 AM.

Peak hour volume = 1500 PCU/h

The highest 15-minute count within the peak hour is 490 PCU (10:15 - 10:30).

PHF = \frac{V_{peak\ hour}}{4 \times V_{15,max}} = \frac{1500}{4 \times 490} = 0.765

v = \frac{V}{PHF} = \frac{1500}{0.765} = 1960\ \text{PCU/h}\quad (= 4 \times 490)

Answer: peak hour 9:30-10:30 AM, peak hour volume 1500 PCU, PHF = 0.765, design flow rate = 1960 PCU/h.

✓ Answer checked by Claude Opus 5.5 (high effort)