It is commonly known in your area that the heaviest traffic flow rates occur between 4:00 PM and 6:30 PM. Your assignment for the day is to find the peak hour volume, peak hour factor (PHF), and the actual or design flow rate for an existing one-lane approach. To do this, you obtain a click-counter and position yourself at the intersection. For each fifteen-minute interval, you record the numbers of right-turns, left-turns, straight-through trucks, and straight-through passenger cars. Your tabulated values are as shown below.
If a truck is equal to 1.5 passenger cars and a right-turn is as well, and if a left-turn is equal to 2.5 passenger cars, then calculate the peak hour volume, peak hour factor (PHF), and the actual (design) flow rate for this approach.
[Solution Shown Below]
The first step in this solution is to find the total traffic volume for each 15 minute period in terms of passenger car units. This is done by multiplying the number of trucks by 1.5, the number of right turns by 1.5, and the number of left turns by 2.5. We then add these three numbers and the volume of straight-through cars together to get the total volume of traffic serviced in each interval. Once we have this, we can locate the hour with the highest volume and the 15 minute interval with the highest volume. The peak hour is shown in blue below with the peak 15 minute period shown in a darker shade of blue.
The peak hour volume is just the sum of the volumes of the four 15 minute intervals within the peak hour (464 pcu). The peak 15 minute volume is 135 pcu in this case. The peak hour factor (PHF) is found by dividing the peak hour volume by four times the peak 15 minute volume.
PHF = 464 /(4 * 135) = 0.86
The actual (design) flow rate can be calculated by dividing the peak hour volume by the PHF, 464/0.86 = 540 pcu/hr, or by multiplying the peak 15 minute volume by four, 4 * 135 = 540 pcu/hr.