Intersection Design, Operation, and Co

ABSTRACT

Chapter 5 deals with the main types of intersections in terms of design, operation, and control. Intersections can be categorized into three major categories: (1) at-grade intersections, (2) grade-separated intersections without ramps, and (3) grade-separated intersections with ramps (interchanges). Grade-separated intersections (such as interchanges) provide uninterrupted crossing of traffic flow. The use of different levels of traffic flow by means of bridges, resulting in overpass and underpass, controls traffic flow, reduces delays, improves travel speeds, and enhances safety. On the other hand, at-grade intersections, due to the fact that traffic flows from different approaches intersect in one area, require traffic control systems to reduce delays, increase highway capacity, and improve highway safety by reducing crashes. In this section, the focus will be on this type of intersection (at-grade intersections) due to the complicated situation and the need to control devices at these intersections that improve the operation and safety at these locations. Traffic control systems generally aim at facilitating traffic flow in an orderly and systematic way that ensures smooth operation, safety, and efficiency. Traffic control systems (or devices) include but are not limited to signs (stop signs, yield signs, etc.), markings (white lines, yellow lines, dotted ones, solid ones, etc.), channelization (islands, medians, barriers, etc.), and traffic signals. The “Manual on Uniform Traffic Control Devices” (MUTCD) by the Federal Highway Administration (FHWA) recommends taking into consideration factors such as design, placement, operation, maintenance, and uniformity for an effective control device. Traffic signals are considered one of the most important devices for at-grade intersections that control traffic flow smoothly and efficiently. Lane grouping, signal phasing, and signal timing are discussed in detail in this section for different types of at-grade intersections including T- (3-leg) intersections and 4-leg intersections. Signal timing of successive intersections on arterial routes is discussed. The coordination of signals of adjacent intersections in urban areas is important to ensure smooth flow of traffic and reduce the delay of vehicles on the arterial. Three methods are used to perform the coordination between successive signals: the simultaneous system, the alternate system, and the progressive system. 5.1

On a major approach of an intersection, the following vehicular volumes during 15-minute periods of the peak hour are obtained (see Table 5.1).

Determine the peak-hour factor (PHF) and the design hourly volume (DHV) at this intersection approach.152

Solution:

5.1 PHF = Hourly volume 4 × Peak 15 − minute volume in the peak hour https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH005_eqn_0001.tif"/>

Therefore:

PHF = 280 + 320 + 400 + 350 4 × 400 = 0.844 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH005_eqn_0002.tif"/>

DHV = Peak − hourly volume PHF https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH005_eqn_0003.tif"/>

⇒

DHV = 1350 0.844 = 1600 vph https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH005_eqn_0004.tif"/>

The screen image of the MS Excel worksheet used to perform the computations of this problem is shown in Figure 5.1.

5.2

If the peak-hour factor (PHF) at an intersection is 0.85 and the volume during the peak 15-minute period within the peak hour is 500 vehicles, determine the volume at the intersection during the peak hour and the design hourly volume (DHV).

Solution:

PHF = Hourly volume 4 × Peak 15 − minute volume in the peak hour https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH005_eqn_0005.tif"/> 153

Therefore:

Hourly volume = PHF × 4 × Peak 15 − minute volume in the peak hour https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH005_eqn_0006.tif"/>

Hourly volume = 0.85 × 4 × 500 = 1700 vph https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH005_eqn_0007.tif"/>

DHV = Peak − hourly volume PHF https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH005_eqn_0008.tif"/>

⇒

HDV = 1700 0.85 = 2000 vph https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH005_eqn_0009.tif"/>

A screen image of the MS Excel worksheet used to perform the computations of this problem is shown in Figure 5.2.

5.3

An intersection having a width of 50 ft (15.2 m) and the speed limit on the intersection approaches is 20 mph (32.2 kph). Compute the minimum yellow interval that will eliminate the dilemma zone at this intersection assuming that the average length of vehicle is 20 ft (6.1 m).

Solution:

The minimum yellow interval at an intersection is given by the following formula:

τ min = δ + W + L u 0 + u 0 2 a https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH005_eqn_0010.tif"/>

Where:

τ_min = the minimum yellow interval (sec)

δ = the perception-reaction time (typical value = 2.5 sec)154

u₀ = the speed limit on approach (ft/sec)

W = the width of intersection (ft)

a = constant rate of braking deceleration (typical value = 11.2 ft/sec²)

L = the length of the vehicle (ft)

Therefore:

τ min = 2.5 + 50 + 20 20 × ( 5280 3600 ) + 20 × ( 5280 3600 ) 2 ( 11.2 ) = 6.2 sec https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH005_eqn_0011.tif"/>

(5280/3600) is a conversion factor to convert the speed from mph to kph.

A screen image of the MS Excel worksheet used to conduct the computations of this problem is shown in Figure 5.3.

5.4

Six vehicles are stopping at a signalized intersection as shown in Figure 5.4. When the signal turns to green, it takes the last vehicle (Vehicle #6) 10 seconds to cross the intersection at a speed of 15 mph (24.1 kph). If the width of the intersection is 60 ft (18.3 m), and the length of each vehicle is 20 ft (6.1 m), determine the jam density at the signal.155

Solution:

The total distance the last vehicle (Vehicle #6) crosses within the 10 seconds is determined:

Total distance = 10 × 15 × 5280 3600 = 220 ft ( 67.1 m ) https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH005_eqn_0012.tif"/>

The total distance minus the width of the intersection provides the distance over which the six vehicles are stopping at the intersection. Therefore:

D = 220 − 60 = 160 ft ( 48.8 m ) https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH005_eqn_0013.tif"/>

The jam density at the signal is equal to the number of vehicles divided by the distance, D. Hence:

k = 6 ( 160 5280 ) = 198 vpm ( 123 normalveh /km ) https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH005_eqn_0014.tif"/>

Screen images of the MS Excel worksheet used to compute the jam density at the signal in this problem are shown in Figures 5.5 and 5.6.156 157

5.5

In Problem 5.4, if the perception–reaction time is 2.5 seconds and the speed limit for vehicles is 20 mph (32.2 kph), determine the minimum yellow interval that should be introduced to the traffic signal to minimize the dilemma zone.

Solution:

τ min = δ + W + L u 0 + u 0 2 a https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH005_eqn_0015.tif"/>

τ min = 2.5 + 60 + 20 20 × ( 5280 3600 ) + 20 × ( 5280 3600 ) 2 ( 11.2 ) = 6.5 sec https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH005_eqn_0016.tif"/>

The screen image of the MS Excel worksheet used to compute the minimum yellow interval required in this problem is the same image shown in Figure 5.5 of the worksheet used for the computations of Problem 5.4.

5.6

Figure 5.7 illustrates a major 3-leg intersection on a highway. Establish correct lane groups for the three approaches of the intersection based on the traffic movements and geometry provided for the intersection.

Solution:

Before establishing lane groups, the definition for a lane group should be first presented. A lane group is defined as a group (set) of one or more lanes at an intersection approach that has the same green phase. The four guidelines shown below are used to establish a lane group:

Exclusive left-turn lanes are considered separate lane groups.

Exclusive right-turn lanes are given separate lane groups.

At an approach with exclusive left-turn lanes or/and right-turn lanes, all other lanes are typically given a single lane group.

The operation of a shared left-turn lane should be evaluated when an approach with more than one lane has a shared left-turn.158

Based on the above guidelines, the lane groups are established. For the eastbound approach, the exclusive right-turn lane is a single lane group and all other lanes (the lane that is going straight) in the same approach is another single lane group. In a similar manner; for the westbound approach, the exclusive left-turn lane is a single lane group, and all other lanes (the two lanes that are moving straight) in the same approach are considered another single lane group. Finally, for the northbound direction, the exclusive left-turn lanes are a single lane group, and the exclusive right-turn lane in the same approach is another single lane group. The six resulting lane groups for this intersection are illustrated in Table 5.2.

5.7

For the T-intersection shown in Figure 5.8, establish correct lane groups for the three approaches of the intersection based on the traffic movements and geometry of the intersection.

Solution:

The lane groups are established using the guidelines discussed above. Since there are no exclusive left-turn or right-turn lanes in any of the three approaches, only the other lanes are considered. For the eastbound approach, the shared lane (the lane with the right-turn and straight movements) is a single lane group. In a similar manner; for the westbound approach, the shared lane (the lane with the left-turn and straight movements) is a single lane group. Finally, for the northbound direction, the shared lane (the lane with the right-turn and left-turn movements) is a single lane group. The three resulting lane groups for this intersection are illustrated in Table 5.3.159

5.8

It is required to design a traffic signal for the T-intersection shown in Figure 5.9. The peak-hour volumes for each lane at the three approaches are given as in the figure. The saturation flow rate is 1800 vph for all lanes and the peak-hour factor (PHF) is 0.90 for the entire intersection. If the lost time per phase is 4.0 seconds and the yellow interval is 3.0 seconds:

Establish the lane groups for this intersection to be used in the signal design.

Provide the phasing system that will offer the shortest cycle length using the Highway Capacity Method (HCM).

Provide the critical (v/s) ratio for each phase.

Determine the cycle length using the HCM that will avoid oversaturation.

Determine the actual green times for all phases.

Solution:

The following formulas are used in solving this problem:

v i j = V i j PHF https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH005_eqn_0017.tif"/>

Where:

v_ij = actual flow rate for lane group or approach i in phase j (veh/h)

V_ij = traffic volume for lane group or approach i in phase j (veh/h)160

PHF = peak-hour factor

c i j = s i j ( g i j C ) https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH005_eqn_0018.tif"/>

Where:

c_ij = capacity of lane group or approach i in phase j (veh/h)

s_ij = saturation flow rate for lane group or approach i in phase j (veh/h of green time)

g_ij = effective green for lane group or approach i in phase j (sec)

C = cycle length for traffic signal (sec)

s = s o N f W f H V f g f p f b b f a f L U f R T f L p b f R p b https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH005_eqn_0019.tif"/>

Where:

S = saturation flow rate for the lane group (vph)

S₀ = ideal saturation flow rate per lane (pc/hr/lane) = 1900 pc/hr/lane

N = number of lanes in the lane group

f_W = adjustment factor for lane width

f_HV = adjustment factor for heavy vehicles

f_g = adjustment factor for grade

f_p = adjustment factor for the existence of a parking lane and parking activity adjacent to the lane group

f_bb = adjustment factor for bus blocking

f_a = adjustment factor for area type

f_LU = adjustment factor for lane utilization

f_RT = adjustment factor for right turns in the lane group

f_Lpb = adjustment factor for pedestrian/bicycle left-turn movements

f_Rpb = adjustment factor for pedestrian/bicycle right-turn movements

5.7 ( v c ) i j = X i j = v i j s i j ( g i j C ) https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH005_eqn_0020.tif"/>

Where:

X_ij = (v/c)_ij ratio for lane group or approach i in phase j

5.8 X c = ( ∑ j ( v s ) c j ) C C − L https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH005_eqn_0021.tif"/>

Where:

X_c = critical (v/c) ratio for the intersection

∑ j ( v s ) c j https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH005_eqn_0022.tif"/> = summation of critical (v/s) ratios for all phases161

v_ij = actual flow rate for lane group or approach i in phase j (veh/h)

L = total lost time per cycle (sec)

L = R + ∑ l j https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH005_eqn_0023.tif"/>

Where:

l _j = lost time for critical signal phase j (sec)

R = total all-red time in the cycle

G t e = C − L = C − ( R + ∑ l j ) https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH005_eqn_0024.tif"/>

Where:

G _te = total effective green time in the cycle (sec)

5.11 G e j = ( v s ) c j ∑ j ( v s ) c j G t e https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH005_eqn_0025.tif"/>

Where:

G _ej = effective green time for phase j (sec)

5.12 l j = G a j + τ j − G e j https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH005_eqn_0026.tif"/>

Where:

G _aj = actual green time for phase j (sec)

τ _j = actual green time for phase j (sec)

The lane groups for this intersection are established using the guidelines discussed earlier. Two lane groups per each approach are established as shown in Table 5.4.162

The phasing system that will provide the shortest cycle length using the Highway Capacity Method (HCM) is shown in Table 5.5.

The critical (v/s) ratio and computations for each phase are summarized in Table 5.6:

Phase A is composed of the following three lane groups:

https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/inline5_110_B.tif"/>

The traffic volumes for these three lane groups are 1120 (600 + 520), 560, and 400 vph, respectively. Since the saturation flow rate for each lane is 1800 vph, the saturation flow rate for the three lane groups (s_j) of Phase A are 3600, 1800, and 1800 veh/h of green time, respectively. Lane group #1 has two lanes and that is why the saturation flow rate for this lane group is equal to 2 × 1800 = 3600 veh/h of green time. In addition, the actual flow rate for Lane Group #1 in Phase A is computed using the formula below, and the (v_ij/s_ij) ratio is computed for the lane group.

v i j = V i j PHF https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH005_eqn_0027.tif"/>

⇒

v i j = 1120 0.90 = 1244.4 veh/h https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH005_eqn_0028.tif"/>

v i j s i j = 1244.4 3600 = 0.346 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH005_eqn_0029.tif"/>

In a similar manner, the traffic volumes, the saturation flow rates, and the actual flow rates for the other lane groups for all three phases are determined. The results are shown in Table 5.6.163

The critical (v/s) ratio for each phase is the maximum (v/s) ratio among the (v/s) ratios for all lane groups in that phase. Therefore, the (v/s)_cj is equal to 0.346, 0.123, and 0.207 for phases A, B, and C, respectively. Therefore, the summation of these critical ratios for all phases is:

∑ j ( v s ) c j = 0.346 + 0.123 + 0.207 = 0.676 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH005_eqn_0030.tif"/>

(d)The cycle length that will avoid oversaturation is happening at critical (v/c) ratio, X_c = 1.0.

L = R + Σ l j https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH005_eqn_0031.tif"/>

⇒

L = 4 × 3 = 12 sec https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH005_eqn_0032.tif"/>

X c = ( ∑ j ( v s ) c j ) C C − L https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH005_eqn_0033.tif"/>

⇒

1.0 = ( 0.676 ) C C − 12 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH005_eqn_0034.tif"/>

Solving for C ⇒

C = 37 sec https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH005_eqn_0035.tif"/>

Use C = 40 sec (rounded to the nearest 5 seconds) (see Table 5.7).

G t e = C − L https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH005_eqn_0036.tif"/>

⇒

G t e = 40 − 12 = 28 sec https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH005_eqn_0037.tif"/> 164

(e)The actual green times for the three phases are computed and illustrated in Table 5.8:

For Phase A:

G e j = ( v s ) c j ∑ j ( v s ) c j G t e https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH005_eqn_0038.tif"/>

⇒

G e A = 0.346 0.676 × 28 = 14.3 sec https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH005_eqn_0039.tif"/>

l j = G a j + τ j − G e j https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH005_eqn_0040.tif"/>

Or:

G a j = l j − τ j + G e j https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH005_eqn_0041.tif"/>

⇒

G a A = 4 − 3 + 14.3 = 15.3 sec https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH005_eqn_0042.tif"/>

The effective green time and actual green time were computed for the three phases in the same way as shown in Table 5.8.

The MS Excel worksheet used to solve this problem with the details, formulas, and computations is shown in Figure 5.10.165

5.9

The T-intersection shown in Figure 5.11 with peak-hour volumes (vph) at the three approaches is having an existing cycle length of 60 seconds. Due to some delays in the major westbound approach, a traffic engineering designer is asked to investigate the existing signal timing for this intersection such that oversaturation would not occur at the intersection. The peak-hour factor (PHF) for the intersection is 0.85, the saturation flow rate for all lanes is 1600 vph, and the phasing system to be used in the design is shown in Table 5.9.

Knowing that the critical (v/s) ratio on the major westbound approach that will be used to determine the cycle length is always that for the through lanes due to the heavy volume in that direction and the total lost time in the cycle is 12 seconds, determine the maximum allowable traffic volume that the second through lane in the major westbound approach can carry to avoid oversaturation.166

Solution:

The lane groups for this intersection are established using the guidelines discussed earlier. Two lane groups per each approach are established as shown in Table 5.10.

The phasing system is given in the problem as shown in Table 5.11.

The critical (v/s) ratio and computations for each phase are summarized in Table 5.12.

Phase A is composed of the following lane groups:

https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/inline5_111_B.tif"/>

167The traffic volumes for these two lane groups are 380 and 350 vph, respectively. The actual flow rate and the (v_ij/s_ij) ratio are computed as shown below. It has to be noted that the saturation flow rate for any lane group with only one lane is equal to s = 1600 veh/h of green, and for any lane group with two lanes is equal to 2s = 3200 veh/h of green.

v i j = V i j PHF https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH005_eqn_0043.tif"/>

⇒

v i j = 380 0.85 = 447.1 veh/h https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH005_eqn_0044.tif"/>

v i j s i j = 447.1 1600 = 0.279 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH005_eqn_0045.tif"/>

In a similar manner, the traffic volumes, the saturation flow rates, and the actual flow rates for the other lane groups for all three phases are determined. The results are shown in Table 5.12.

The summation of the critical (v/s) ratios for all three phases is computed:

∑ j ( v s ) c j = 380 PHF × s + 600 PHF × s + 275 PHF × s https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH005_eqn_0046.tif"/>

L = 12 sec https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH005_eqn_0047.tif"/>

X c = ( ∑ j ( v s ) c j ) C C − L https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH005_eqn_0048.tif"/>

⇒168

X c = ( 380 PHF × s + 600 PHF × s + 275 PHF × s ) 60 60 − 12 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH005_eqn_0049.tif"/>

X c = ( 380 + 600 + 275 0.85 × 1600 ) 60 60 − 12 = 1.15 > 1.0 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH005_eqn_0050.tif"/>

⇒ Oversaturation occurs at the intersection.

The MS Excel worksheet used to solve the above part of the problem is shown in the screen image in Figure 5.12.

To avoid oversaturation, the critical (v/c) ratio, X_c = 1.0.

∑ j ( v s ) c j = 380 PHF × s + 400 + V PHF × 2 s + 275 PHF × s https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH005_eqn_0051.tif"/>

X c = ( ∑ j ( v s ) c j ) C C − L https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH005_eqn_0052.tif"/>

⇒

1.0 = ( 380 PHF × s + 200 + V / 2 PHF × s + 275 PHF × s ) 60 60 − 12 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH005_eqn_0053.tif"/> 169

⇒

1.0 = ( 380 + 200 + V / 2 + 275 0.85 × 1600 ) 60 60 − 12 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH005_eqn_0054.tif"/>

⇒

V = 466 vph https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH005_eqn_0055.tif"/>

In conclusion, the designer would have to reduce the traffic volume in the second through lane from 800 to 466 vph to avoid oversaturation at the intersection. Another solution would be to increase the cycle length. Of course, reducing the traffic volume on the through lane would accompany finding other alternatives for this traffic by either detouring the traffic or expanding the approach.

The MS Excel worksheet used to solve this part of the problem with its details and computations is shown in Figure 5.13.

5.10

Figure 5.14 shows a T-intersection with the flow ratios (v/s) for all traffic movements of the three approaches. The saturation flow rates are 1800, 1850, and 1600 vph for right lane, through lane, and left lane, respectively. The phasing system for the intersection is given in Table 5.13 and the lost time per phase is 3 seconds. Using the Highway Capacity Method (HCM), determine the shortest cycle length (to the nearest 5 seconds) that will avoid oversaturation.170

Solution:

The lane groups for this intersection are established using the guidelines discussed earlier. Two lane groups per each approach are established as shown in Table 5.14.

The actual flow rates for all lanes in all three phases A, B, and C are determined. The results are shown in Table 5.15.

Sample Calculation:

For the Through Lane #1 (Phase A):

v s = v 1850 = 0.38 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH005_eqn_0056.tif"/> 171

⇒

v = 703 vph https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH005_eqn_0057.tif"/>

In a similar manner, the actual flow rates for the other lanes for all three phases are determined. The results are shown in Table 5.15. After that, the actual flow rates for the lane groups are computed. The reason behind this step is that some lane groups consist of more than one lane; and therefore, the flow rate for the lane group is equal to summation of the flow rates in all lanes.

The summation of the critical (v/s) ratios for the three phases is computed:

∑ j ( v s ) c j = 0.300 + 0.150 + 0.200 = 0.650 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH005_eqn_0058.tif"/>

The cycle length that will avoid oversaturation is happening at critical (v/c) ratio, X_c = 1.0.

L = 3 × 3 = 9 sec https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH005_eqn_0059.tif"/>

⇒

X c = ( ∑ j ( v s ) c j ) C C − L https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH005_eqn_0060.tif"/>

⇒

1.0 = ( 0.650 ) C C − 9 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH005_eqn_0061.tif"/>

Solving for C ⇒

C = 25.7 sec https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH005_eqn_0062.tif"/>

Use C = 30 sec (rounded to the nearest 5 seconds).

The MS Excel worksheet used to solve this problem with all details and computations is shown in Figure 5.15.172

5.11

In Problem 5.10 above, if a cycle length of 60 seconds is used for this intersection, then compute the maximum allowable flow rate in the left lane of the major approach to avoid oversaturation.

Solution:

The left-turn lane on the major approach (westbound approach) in this case is unknown. The same (v/s) values from Problem 5.10 are used for the other lanes. Similar computations are performed as seen in Table 5.16:173

∑ j ( v s ) c j = 0.30 + ( v s ) critical-Phase B + 0.20 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH005_eqn_0063.tif"/>

X c = ( ∑ j ( v s ) c j ) C C − L https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH005_eqn_0064.tif"/>

⇒

1.0 = ( 0.30 + ( v s ) critical-Phase B + 0.20 ) 60 60 − 9 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH005_eqn_0065.tif"/>

⇒

( v s ) critical-Phase B = 0.35 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH005_eqn_0066.tif"/>

This value for (v/s) ratio is the critical value for Phase B; however, since this phase is composed of only one lane group, and the lane group consists of only one lane (the left-turn lane), the (v/s) value of 0.35 corresponds to the left-turn lane in Phase B.

( v s ) critical-Phase B = 0.35 = v 1600 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH005_eqn_0067.tif"/>

⇒

v = 560 vph https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH005_eqn_0068.tif"/>

5.12

In the 4-leg intersection shown in Figure 5.16, the peak-hour volumes for all traffic movements are given: (a) establish lane grouping for this intersection, (b) provide the phasing system that will provide the shortest cycle length using Webster method, and (c) provide the critical lane group volume for each phase, if the saturation flow rate and the peak-hour factor are the same for all lanes.174

Solution:

The lane groups for this intersection are established using the guidelines discussed earlier (see Table 5.17).

The phasing system that will provide the shortest cycle length is illustrated in Table 5.18.175

When the highest volume in one approach is combined with the highest volume in an opposite approach in one phase, this situation normally creates the optimum phasing system that will provide the shortest cycle length. In this case, the through lanes with the highest traffic volume (280 + 200) in the northbound approach are combined with the through lanes with the highest traffic volume (180 + 220) in the southbound approach to compose one phase. Similarly, the left-turn lane with the highest traffic volume (180) in the eastbound approach is combined with the left-turn lane with the highest traffic volume (200) in the westbound approach to provide one phase.

The lane group volumes along with the critical lane group volume for each phase are summarized in Table 5.19.

5.13

For Problem 5.12, if the saturation flow rate is 1600 vph for all lanes, the peak-hour factor for the intersection approaches is 0.90, the lost time per phase is 3 seconds, and the yellow interval is 3 seconds, determine the cycle length using Webster method and the actual green times for all phases.

Solution:

The formula for Webster method to calculate the cycle length is shown below:

C o = 1.5 L + 5 1 − ∑ j ( v s ) c j https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH005_eqn_0069.tif"/>

Where:

C_o = optimum cycle length for the intersection (sec)

L = total lost time in the cycle (sec)

∑ j ( v s ) c j https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH005_eqn_0070.tif"/> = summation of critical (v/s) ratios for all phases

( v s ) c j https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH005_eqn_0071.tif"/> = critical flow ratio for phase j

v = actual flow rate for lane group (veh/h)

s = saturation flow rate for lane group (veh/h of green time)

Sample Calculation:

For Phase A, Lane Group #1:

V = 280 + 200 = 480 vph

s = 1600 × 2 = 3200 vph (since this lane group consists of two lanes).176

v = V PHF https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH005_eqn_0072.tif"/>

⇒

v = 480 0.90 = 533.3 vph https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH005_eqn_0073.tif"/>

And

v s = 533.3 3200 = 0.167 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH005_eqn_0074.tif"/>

In a similar manner, the actual flow rates and the (v/s) ratios are calculated for all lane groups of the four phases. The results are shown in Table 5.20.

C o = 1.5 L + 5 1 − ∑ j ( v s ) c j https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH005_eqn_0075.tif"/>

⇒

C o = 1.5 ( 3 × 4 ) + 5 1 − 0.493 ≅ 46 seconds https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH005_eqn_0076.tif"/>

⇒

C o = 50 seconds ( rounded to the nearest 5 seconds ) https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH005_eqn_0077.tif"/>

G t e = C − L https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH005_eqn_0078.tif"/>

⇒

G t e = 50 − 12 = 38 sec https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH005_eqn_0079.tif"/> 177

The effective green times and the actual green times for the four phases are computed and summarized in Table 5.21.

Sample Calculation:

For Phase A:

⇒

G e A = 0.167 0.493 × 38 = 12.9 sec https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH005_eqn_0081.tif"/>

l j = G a j + τ j − G e j https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH005_eqn_0082.tif"/>

Or:

G a j = l j − τ j + G e j https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH005_eqn_0083.tif"/>

⇒

G a A = 3 − 3 + 12.9 = 12.9 sec https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH005_eqn_0084.tif"/>

5.14

In Problem 5.13, check the minimum green times required for pedestrians crossing if the following pedestrian-related data is given:

Number of pedestrians crossing the northbound approach = 22

Number of pedestrians crossing the southbound approach = 20

Number of pedestrians crossing the eastbound approach = 15

Number of pedestrians crossing the westbound approach = 12

Effective crosswalk width for each approach = 8 ft

Crosswalk length in N-S approach = 64 ft (4 lanes; lane width = 12 ft, median width = 16 ft)

Crosswalk length in E-W approach = 40 ft (2 lanes; lane width = 12 ft, median width = 16 ft)

Speed limit on each approach = 25 mph178

Solution:

The minimum green times required for pedestrian crossing can be determined using the HCM formulas given below:

For W_E ≤ 10 ft:

5.14 G p = 3.2 + L S P + ( 0.27 N ped ) https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH005_eqn_0085.tif"/>

For W_E > 10 ft:

G p = 3.2 + L S P + ( 2.7 N ped W E ) https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH005_eqn_0086.tif"/>

Where:

G_p = minimum green time (sec)

L = crosswalk length (ft)

S_p = average pedestrians speed (typical value = 4 ft/sec)

3.2 = pedestrian reaction (start-up) time (sec)

W_E = effective crosswalk width (ft)

N_ped = number of pedestrians crossing an approach

Since W_E = 8 ft ≤ 10 ft,

⇒

G p = 3.2 + L S P + ( 0.27 N ped ) https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH005_eqn_0087.tif"/>

For northbound approach:

G p-north = 3.2 + 64 4 + ( 0.27 × 22 ) = 23.2 sec https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH005_eqn_0088.tif"/>

For southbound approach:

G p-south = 3.2 + 64 4 + ( 0.27 × 20 ) = 22.4 sec https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH005_eqn_0089.tif"/>

For eastbound approach:

G p-east = 3.2 + 40 4 + ( 0.27 × 15 ) = 19.1 sec https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH005_eqn_0090.tif"/>

For westbound approach:

G p-west = 3.2 + 40 4 + ( 0.27 × 12 ) = 18.6 sec https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH005_eqn_0091.tif"/>

The green times available for pedestrians for each approach based on the computed cycle length and actual green times from the signal phasing are compared with the minimum green time requirements calculated above for pedestrian crossing data as shown in Table 5.22.179

5.15

Solve Problem 5.13 using the following phasing system (see Table 5.23).

Solution:

Based on the given phasing system, the traffic volumes, the saturation flow rates, and the actual flow rates for the lane groups of the four phases are determined.

Sample Calculation:

For Phase D, Lane Group #2:

V = 200 vph

s = 1600 × 1 = 1600 vph (since this lane group consists of only one lane).

v = V PHF https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH005_eqn_0092.tif"/>

⇒

v = 200 0.90 = 222.2 vph https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH005_eqn_0093.tif"/>

And

v s = 222.2 1600 = 0.139 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH005_eqn_0094.tif"/>

In a similar manner, the actual flow rates and the (v/s) ratios are calculated for all lane groups of the four phases. The results are shown in Table 5.24.180

C o = 1.5 L + 5 1 − ∑ j ( v s ) c j https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH005_eqn_0095.tif"/>

⇒

C o = 1.5 ( 3 × 4 ) + 5 1 − 0.570 ≅ 53.5 seconds https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH005_eqn_0096.tif"/>

⇒

C o = 55 seconds ( rounded to the nearest 5 seconds ) https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH005_eqn_0097.tif"/>

Therefore, by using the phasing system in Problem 5.15, the cycle length is longer than that when using the phasing system in Problem 5.13.

5.16

For the intersection with the information below (see Figure 5.17): (a) establish the lane groups at the intersection, (b) Select an appropriate phasing system, (c) design a signal timing for the intersection using the Webster method.

The numbers shown at the intersection are the peak-hour volumes for the different traffic movements.

Assume that the yellow interval at the intersection is 3 seconds and the lost time per phase is 3.5 seconds.

PHF = 0.95.

The saturation flow rate (vph) for each lane is shown in Table 5.25.181 182

Solution:

Lane Groups: Lane groups are established as follows: per each approach, the left-turn lane on each approach is given a separate lane group, and the other lanes (the shared lane with the through lanes) are all given a separate lane group since there is no exclusive right-turn lane. The resulting lane groups are shown in Table 5.26.183

Phasing System: The phasing system at the intersection is established based on the peak-hour volumes for the different traffic movements. In this case, two phasing systems are used and evaluated. The first one is described in Table 5.27.

The second phasing system is shown in Table 5.28.

Signal Timing: The design hourly volumes (DHVs) are determined by dividing the peak-hour volumes by the PHF. Afterwards, the actual flow rate is determined by dividing the design hourly volume by the saturation flow rate.

Sample Calculation for Phasing System #1:

For Phase A, Lane Group #1:

This lane groups consists of three lanes: one shared lane (right-turn and through movements) and two through lanes. Therefore, the traffic volume for this lane group is the summation of the traffic volumes in the three lanes:

V = 250 + 180 + 225 = 655 vph https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH005_eqn_0098.tif"/>

The saturation flow rate for this lane group is simply the addition of the saturation flow rates of the three lanes in the lane group; that is:

s = 1700 + 1800 + 1800 = 5300 vph https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH005_eqn_0099.tif"/>

DHV = V PHF https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH005_eqn_0100.tif"/> 184

⇒

DHV = 655 0.95 = 689.5 vph https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH005_eqn_0101.tif"/>

And

v s = 689.5 5300 = 0.130 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH005_eqn_0102.tif"/>

For Phase C, Lane Group #2:

This lane groups consists of only one exclusive left-turn lane. Therefore, the traffic volume for this lane group is equal to the traffic volume on that lane:

V = 105 vph https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH005_eqn_0103.tif"/>

The saturation flow rate for this lane group is simply the saturation flow rate of the left-turn lane; that is:

s = 1600 vph https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH005_eqn_0104.tif"/>

DHV = V PHF https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH005_eqn_0105.tif"/>

⇒

DHV = 105 0.95 = 110.5 vph https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH005_eqn_0106.tif"/>

And

v s = 110.5 1600 = 0.069 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH005_eqn_0107.tif"/>

In a similar manner, the actual flow rates and the (v/s) ratios are calculated for all the lane groups of the four phases in this phasing system. The results are shown in Table 5.29.185

The cycle length is calculated using the formula of Webster method. The total lost time for the traffic signal is estimated first. Since the lost time per phase is 3.5 seconds, the total lost time for the four phases is determined as follows:

L = R + ∑ l j https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH005_eqn_0108.tif"/>

⇒

L = 0 + 3.5 × 4 = 14 sec https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH005_eqn_0109.tif"/>

C o = 1.5 L + 5 1 − ∑ j ( v s ) c j https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH005_eqn_0110.tif"/>

⇒

C o = 1.5 ( 14 ) + 5 1 − 0.423 = 45.1 seconds https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH005_eqn_0111.tif"/>

⇒

G t e = C − L https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH005_eqn_0113.tif"/>

⇒

G t e = 50 − 14 = 36 sec https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH005_eqn_0114.tif"/>

The effective green times and the actual green times for the four phases are computed and summarized in Table 5.30.

Sample Calculation:

For Phase A:

⇒

G e A = 0.130 0.423 × 36 = 11.1 sec https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH005_eqn_0116.tif"/>

l j = G a j + τ j − G e j https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH005_eqn_0117.tif"/>

Or:

G a j = l j − τ j + G e j https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH005_eqn_0118.tif"/>

⇒

G a A = 3.5 − 3 + 11.1 = 11.6 sec https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH005_eqn_0119.tif"/> 186

Calculations for Phasing System #2:

The design-hourly volumes, saturation flow rates, actual flow rates, and (v/s) ratios for lane groups have been calculated in the previous step for Phasing System #1. The same values will be used in Phasing System #2 but with different phases than before. The results for Phasing System #2 are shown in Table 5.31.

The total lost time for the traffic signal is estimated as follows:

L = R + ∑ l j https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH005_eqn_0120.tif"/>

⇒

L = 0 + 3.5 × 4 = 14 sec https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH005_eqn_0121.tif"/>

The cycle length is calculated using the formula of Webster method shown below:

C o = 1.5 L + 5 1 − ∑ j ( v s ) c j https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH005_eqn_0122.tif"/>

⇒

C o = 1.5 ( 14 ) + 5 1 − 0.375 = 41.6 seconds https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH005_eqn_0123.tif"/> 187

⇒

C o = 45 seconds ( rounded to the nearest 5 seconds ) https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH005_eqn_0124.tif"/>

G t e = C − L https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH005_eqn_0125.tif"/>

⇒

G t e = 45 − 14 = 31 sec https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH005_eqn_0126.tif"/>

The effective green times and the actual green times for the four phases are computed and summarized in Table 5.32.

Sample Calculation:

For Phase A:

⇒

G e A = 0.084 0.375 × 31 = 6.9 sec https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH005_eqn_0128.tif"/>

l j = G a j + τ j − G e j https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH005_eqn_0129.tif"/>

Or:

G a j = l j − τ j + G e j https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH005_eqn_0130.tif"/>

⇒

G a A = 3.5 − 3 + 6.9 = 7.4 sec https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH005_eqn_0131.tif"/>

In conclusion, using Phasing System #2 provides a cycle length that is shorter than that provided by Phasing System #1.188

5.17

It is required to design a complete signal phasing and timing plan for the 4-leg intersection shown in Figure 5.18. It is also required to check if protected left-turn phase should be provided for any of the approaches using the cross-product guideline. The geometry, number of lanes for all approaches of the intersection, and the design-hourly peak volumes (DHVs) are provided in the diagram. The following data is given:

The yellow interval at the intersection is assumed to be 3.5 seconds for each phase, and the lost time per phase is also assumed as 3 seconds. No all-red phase will be used for the traffic signal.

The speeds of vehicles are 30 mph (43.3 kph) and 35 mph (56.3 kph) for the N-S approach and the E-W approach, respectively.

Number of pedestrians crossing the N-S approaches = 9.

Number of pedestrians crossing the E-W approaches = 9.

Effective crosswalk width for each approach = 8 ft.

The saturation flow rates for the different lanes are given below (see Table 5.33).189

The phasing system used for the design of a traffic signal at this intersection is shown in Table 5.34.

Solution:

Lane Groups:

Lane groups are established as follows: Per each approach, the left-turn lane on each approach is given a separate lane group, and the other lanes (the shared lane with the through lanes) are all given a separate lane group since there is no exclusive right-turn lane. The resulting lane groups are shown in Table 5.35.

Requirement for Any Protected Left-Turn Phase:

The determination if protected left-turn phase should be provided for any of the approaches is verified using the cross-product guideline. The Highway Capacity Manual offers the 190following criteria for the more common guidelines (using the cross product of the left-turn volume and opposing through and right-turn volumes), as follows:

The use of protected left-turn phase should be considered when the product of left-turning vehicles and opposing traffic volumes:

Exceeds 50000 during the peak hour for one opposing lane, or

Exceeds 90000 during the peak hour for two opposing lanes, or

Exceeds 110000 during the peak hour for three or more opposing lanes

Accordingly,

For the E–W approaches:

The product of eastbound left-turning vehicles and westbound opposing traffic (through and right-turn vehicles) = 275 (550 + 175) = 199375 > 90000 (requirement for two opposing lanes). Thus, this movement should be protected.

The product of westbound left-turning vehicles and eastbound opposing traffic (through and right-turn vehicles) = 245 (450 + 225) = 165375 > 90000 (requirement for two opposing lanes). Thus, this movement should be also protected.

For the N–S approaches:

The product of northbound left-turning vehicles and southbound opposing traffic (through and right-turn vehicles) = 100 (380) = 38000 < 50000 (requirement for one opposing lanes). Thus, this movement could be permitted (not protected)

The product of southbound left-turning vehicles and northbound opposing traffic (through and right-turn vehicles) = 80 (370) = 29600 < 50000 (requirement for one opposing lanes). Thus, this movement could also be permitted (not protected).

Recommended Phasing Plan:

Based on the above analysis, the given phasing system is modified for permitted (not protected) left-turn phases for the N–S approaches, and a three-phase traffic- signal plan is recommended as shown in Table 5.36.

Saturation Flow Rates:

The saturation flow rates given in the problem are determined for each lane group using the formula:

and based upon the existing conditions for the intersection. The adjustment factors for the prevailing conditions at the intersection are determined to the best of knowledge and data related to the intersection. The ideal saturation flow rate is 1900 vphpl (used under ideal conditions) and adjusted for the different existing conditions. The final saturation flow rates are summarized in Table 5.36.

Flow Ratios and Critical Lane Groups:

The design-hourly volumes are determined by dividing the peak-hour volumes by the PHF. This step is already done since the DHVs are already given in the problem. The flow ratio (v/s) for each lane groups is computed for all three phases. The critical (v/s) ratio (the maximum value among the values of the different lane groups in the same phase) is determined. Below is a sample calculation for Phase B, Lane Group 1E:

Sample Calculation for Phase B, Lane Group 1E:

This lane groups consists of two lanes: one shared lane (right-turn and through movements) and one through lane. Therefore, the design-hourly volume for this lane group is the summation of the design-hourly volumes in the two lanes:

DHV = 225 + 450 = 675 vph https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH005_eqn_0133.tif"/>

The saturation flow rate for this lane group is simply the addition of the saturation flow rates of the two lanes in the lane group; that is:

s = 1350 + 1500 = 2850 vph https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH005_eqn_0134.tif"/>

⇒

v s = 675 2850 = 0.237 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH005_eqn_0135.tif"/>

The results are shown in Table 5.37.192

Calculation of the Total Lost Time:

It has to be noted that the lost time per cycle is the total of lost times for all phases plus the all-red time for the cycle. The lost time per phase is the time wasted at the beginning of the green phase when the signal turns on green and at the end of the yellow phase due to the reaction times of drivers. Therefore; the total lost time is determined:

L = R + Σ l j https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH005_eqn_0136.tif"/>

Since there is no all-red phase, R = 0, and lost time per phase = 3 seconds

L = 0 + 3 × 4 = 12 sec https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH005_eqn_0137.tif"/>

Determination of the Optimum Cycle Length:

The optimum cycle length is determined using the following equation (Webster Method for pre-timed (fixed) signals):

C o = 1.5 L + 5 1 − ∑ j ( v s ) c j https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH005_eqn_0138.tif"/>

⇒

C o = 1.5 ( 12 ) + 5 1 − 0.800 = 115 seconds https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH005_eqn_0139.tif"/>

This cycle length is accepted for this type of intersection since traffic volumes are relatively high compared to the number of lanes in the intersection approaches.

The cycle length for isolated intersections should be kept below 120 seconds to avoid long delays. Cycle lengths are typically between 35 and 60 seconds. However, it is necessary sometimes to use longer cycle lengths when traffic volumes are high.

Allocation of the Green Time:

The total effective green time is determined using the formula shown below:

G t e = C − L https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH005_eqn_0140.tif"/>

⇒

G t e = 115 − 12 = 103 sec https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH005_eqn_0141.tif"/>

To determine the effective green time for each phase, the following formula is used:

For instance; for Phase A:

G e A = 0.229 0.800 × 103 = 29.5 sec https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH005_eqn_0143.tif"/>

Note that the number of vehicles that cross the intersection divided by the saturation flow rate provides the effective green time. The effective green time is less than the sum of the green and yellow times. This difference is considered lost time; and therefore, the actual green time for each phase is determined using the following formula:193

l j = G a j + τ j − G e j https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH005_eqn_0144.tif"/>

⇒

G a j = l j − τ j + G e j https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH005_eqn_0145.tif"/>

For Phase A:

G a A = 3 − 3.5 + 29.5 = 29.0 sec https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH005_eqn_0146.tif"/>

Table 5.38 summarizes the effective green times and actual green times for the three phases.

Determination of the Minimum Green Time for Safe Pedestrian Crossing:

The minimum green time at the intersection that will allow pedestrians to safely cross the intersection is given by the following equations:

For W_E ≤ 10 ft:

G p = 3.2 + L S P + ( 0.27 N ped ) https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH005_eqn_0147.tif"/>

For W_E > 10 ft:

G p = 3.2 + L S P + ( 2.7 N ped W E ) https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH005_eqn_0148.tif"/>

Please note that the number of those pedestrians crossing during an interval are counted by visiting the intersection and seeing how many pedestrians are indeed trying to cross a certain approach within this time, considering that the interval is the total red time for the vehicles moving in the crossing direction when pedestrians can really cross a certain approach, or the green time for vehicles moving parallel to the pedestrians.

The available green time for pedestrians should be verified against the values provided by these equations.

The width of the crosswalk that is available for pedestrians at all approaches is only 8 ft, which is less than 10 ft. Therefore, the first formula above will be used to determine the minimum green time that will allow pedestrians to cross the E–W approaches during the phase for parallel vehicles (in this case Phase C) is computed as follows:194

For W_E ≤ 10 ft:

G p = 3.2 + L S P + ( 0.27 N ped ) https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH005_eqn_0149.tif"/>

L = 60 ft

S_p = 4 ft/sec

N_ped = 9 pedestrians

⇒

G p = 3.2 + 60 4 + ( 0.27 × 9 ) = 20.6 sec https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH005_eqn_0150.tif"/>

The green time for Phase C is equal to 37.1 seconds > the minimum required green time for pedestrians crossing = 20.6 seconds. Consequently, the green time for the phase is enough to enable pedestrians to cross the E–W approach(s) at the intersection safely.

In a similar manner, the minimum green time that will allow pedestrians cross the N–S approach(s) during the phase for parallel vehicles (in this case Phase A or B) is computed as follows:

For W_E ≤ 10 ft:

G p = 3.2 + L S P + ( 0.27 N ped ) https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH005_eqn_0151.tif"/>

L = 36 ft

S_p = 4 ft/sec

N_ped = 9 pedestrians

⇒

G p = 3.2 + 36 4 + ( 0.27 × 9 ) = 14.6 sec https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH005_eqn_0152.tif"/>

The green times for Phases A and B are 29.0 and 35.4 seconds, respectively. Both are higher than the minimum required green time for pedestrians crossing = 14.6 seconds. Therefore, the green time for the phase is enough to enable pedestrians to cross the N–S approaches at the intersection safely.

5.18

Five successive intersections are located on urban arterial with the spacings as shown in Figure 5.19. The cycle lengths at the intersections are: 70, 65, 75, 65, and 70 seconds, respectively. A simultaneous traffic signal system is to be designed for the five intersections given that the progression speed on the arterial is 40 mph (64.4 kph). What cycle length is recommended in this case, and why?195

Solution:

The following mathematical expression for the relationship between the progression speed on an arterial, the spacing between successive intersections on the arterial, and the required cycle length of the signal light for proper coordination is used:

5.16 u = X C https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH005_eqn_0153.tif"/>

Where:

u = progression speed (converted to ft/sec)

X = average spacing between intersections (ft)

C = cycle length (sec)

Therefore, the recommended cycle length based on the given data is computed using:

C = X u https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH005_eqn_0154.tif"/>

X = average spacing = 4140 + 4100 + 4000 + 4180 4 = 4105 ft ( 1251.2 m ) https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH005_eqn_0155.tif"/>

⇒

C = 4105 40 × 5280 3600 ≅ 70 sec https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH005_eqn_0156.tif"/>

Table 5.39 summarizes the given data along with the average spacing and recommended cycle length for simultaneous traffic signal system.196

5.19

A traffic engineering designer has been asked to coordinate the traffic signals of consecutive intersections spaced at 750 ft (228.6 m) intervals on an urban arterial. If the progression speed of vehicles on the arterial is 30 mph (48.3 kph), what is the cycle length recommended for the coordination when single-, double-, and triple-alternate system is used?

Solution:

The mathematical formula given below is used to relate the progression speed, the spacing, and the required cycle length for proper coordination using the alternate system:

u = n X C https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH005_eqn_0157.tif"/>

Where:

u = progression speed (converted to ft/sec)

X = average spacing between intersections (ft)

C = cycle length (sec)

N = 2, 4, or 6 for single-, double-, or triple-alternate system, respectively

Thus, the recommended cycle length is given by:

C = n X u https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH005_eqn_0158.tif"/>

For single-alternate system (when n = 4),

⇒

C = 2 × 750 30 × 5280 3600 = 34.1 sec https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH005_eqn_0159.tif"/>

The cycle length is computed using the same formula for the double- and triple-alternate systems. The results are summarized in Table 5.40.

Table 5.1 Traffic Volumes During the Peak Hour for Problem 5.1 Time Period Traffic Count 7:00–7:15am 280 7:15–7:30am 320 7:30–7:45am 400 7:45–8:00am 350 Figure 5.1 An image of the MS Excel worksheet used for the computations of Problem 5.1. https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/fig5_1_B.tif"/> Figure 5.2 An image of the MS Excel worksheet used for the computations of Problem 5.2. https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/fig5_2_B.tif"/> Figure 5.3 An image of the MS Excel worksheet used for the computations of Problem 5.3. https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/fig5_3_B.tif"/> Figure 5.4 A signalized T-intersection for Problem 5.4. https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/fig5_4_B.tif"/> Figure 5.5 An image of the MS Excel worksheet used for the computations of Problem 5.4 (US customary units). https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/fig5_5_B.tif"/> Figure 5.6 An image of the MS Excel worksheet used for the computations of Problem 5.4 (SI units). https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/fig5_6_B.tif"/> Figure 5.7 A 3-leg intersection for Problem 5.6. https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/fig5_7_B.tif"/> Table 5.2 The Resulting Lane Groups for the 3-Leg Intersection in Problem 5.6 Approach 1 (Eastbound) 2 (Westbound) 3 (Northbound) Lane Groups https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/inline5_1_B.tif"/> https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/inline5_2_B.tif"/> https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/inline5_3_B.tif"/> Figure 5.8 A T-intersection for Problem 5.7. https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/fig5_8_B.tif"/> Table 5.3 The Resulting Lane Groups for the T-Intersection in Problem 5.7 Approach 1 (Eastbound) 2 (Westbound) 3 (Northbound) Lane Groups https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/inline5_4_B.tif"/> https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/inline5_5_B.tif"/> https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/inline5_6_B.tif"/> Figure 5.9 A T-intersection for Problem 5.8. https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/fig5_9_B.tif"/> Table 5.4 The Established Lane Groups for the T-Intersection in Problem 5.8 Approach 1 (Eastbound) 2 (Westbound) 3 (Northbound) Lane Group # RE^a TE TW+TW LW RN LN+LN Lane Group https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/inline5_7_B.tif"/> https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/inline5_8_B.tif"/> https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/inline5_9_B.tif"/> https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/inline5_10_B.tif"/> https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/inline5_11_B.tif"/> https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/inline5_12_B.tif"/> ^aR, T, L refer to right-turn, through, and left-turn lanes, respectively. E, W, N, S refer to eastbound, westbound, northbound, and southbound, respectively. Table 5.5 The Phasing System for the T-Intersection in Problem 5.8 Phase A B C Diagram https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/inline5_13_B.tif"/> https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/inline5_14_B.tif"/> https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/inline5_15_B.tif"/> Table 5.6 HCM Signal Timing Computations and Results for the Three Phases of the T-Intersection in Problem 5.8 Phase A B C Lane Group TW+TW^a RE TE LW RN LN+LN Volume (vph) 1120 400 560 200 220 670 v _ij (vph) 1244.4 444.4 622.2 222.2 244.4 744.4 s_ij (vph) 3600 1800 1800 1800 1800 3600 v _ij /s_ij 0.346 0.247 0.346 0.123 0.136 0.207 (v/s)_cj 0.346 0.123 0.207 ^aR, T, L refer to right-turn, through, and left-turn lanes, respectively. E, W, N, S refer to eastbound, westbound, northbound, and southbound, respectively. Table 5.7 Determination of the Cycle Length for Problem 5.8 X_c 1.0 Σ(v/s)_cj 0.676 L (sec) 12.0 C (sec) 37.0 (Use 40) G_te (sec) 28 Table 5.8 The Effective and Actual Green Times for the Three Phases of the T-Intersection in Problem 5.8 Phase A B C G_ej (sec) 14.3 5.1 8.6 G_aj (sec) 15.3 6.1 9.6 Figure 5.10 An image of the MS Excel worksheet used for the computations of Problem 5.8. https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/fig5_10_B.tif"/> Figure 5.11 A T-intersection for Problem 5.9. https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/fig5_11_B.tif"/> Table 5.9 The Phasing System for the T-Intersection in Problem 5.9 Phase A B C Diagram https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/inline5_16_B.tif"/> https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/inline5_17_B.tif"/> https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/inline5_18_B.tif"/> Table 5.10 The Established Lane Groups for the T-Intersection in Problem 5.9 Approach 1 (Eastbound) 2 (Westbound) 3 (Northbound) Lane Group # 1E 2E 1W 2W 1N 2N Lane Group https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/inline5_19_B.tif"/> https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/inline5_20_B.tif"/> https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/inline5_21_B.tif"/> https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/inline5_22_B.tif"/> https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/inline5_23_B.tif"/> https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/inline5_24_B.tif"/> Table 5.11 The Phasing System for the T-Intersection in Problem 5.9 Phase A B C Diagram https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/inline5_25_B.tif"/> https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/inline5_26_B.tif"/> https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/inline5_27_B.tif"/> Table 5.12 HCM Signal Timing Computations and Results for the Three Phases of the T-Intersection in Problem 5.9 Phase A B C Lane Group RE^a TE TW+TW LW RN LN+LN Volume (vph) 350 380 400 + 800 280 200 350 + 320 v _ij (vph) 350/PHF 380/PHF 1200/PHF 280/PHF 200/PHF 550/PHF s_ij (vph) s s 2s s s 2s v _ij /s_ij 350/(PHF×s) 380/(PHF×s) 600/(PHF×s) 280/(PHF×s) 200/(PHF×s) 275/(PHF×s) (v/s)_cj 380/(PHF×s) 600/(PHF×s) 275/(PHF×s) ^a R, T, L refer to right-turn, through, and left-turn lanes, respectively. E, W, N, S refer to eastbound, westbound, northbound, and southbound, respectively. Figure 5.12 An image of the MS Excel worksheet used for the computations of Part 1 of Problem 5.9. https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/fig5_12_B.tif"/> Figure 5.13 An image of the MS Excel worksheet used for the computations of Part 2 of Problem 5.9. https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/fig5_13_B.tif"/> Figure 5.14 A T-intersection for Problem 5.10. https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/fig5_14_B.tif"/> Table 5.13 Given Phasing System for the T-Intersection in Problem 5.10 Phase A B C Diagram https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/inline5_28_B.tif"/> https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/inline5_29_B.tif"/> https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/inline5_30_B.tif"/> Table 5.14 The Established Lane Groups for the T-Intersection in Problem 5.10 Approach 1 (Eastbound) 2 (Westbound) 3 (Northbound) Lane Group https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/inline5_31_B.tif"/> https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/inline5_32_B.tif"/> https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/inline5_33_B.tif"/> Table 5.15 HCM Signal Timing Computations and Results for the Three Phases of the T-Intersection in Problem 5.10 Phase A B C Lane 1 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/inline5_34_B.tif"/> 2 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/inline5_35_B.tif"/> 3 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/inline5_36_B.tif"/> 4 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/inline5_37_B.tif"/> 1 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/inline5_38_B.tif"/> 1 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/inline5_39_B.tif"/> 2 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/inline5_40_B.tif"/> v/s 0.380 0.220 0.180 0.240 0.150 0.180 0.200 v (vph) 703 407 324 444 240 324 320 Lane Group 1W+2W^a 1E 2E 3W 1N 2N v _ij (vph) 703 + 407 = 1110 324 444 240 324 320 s_ij (vph) 2 × 1850 = 3700 1800 1850 1600 1800 1600 v _ij /s_ij 0.300 0.180 0.240 0.150 0.180 0.200 (v/s)_cj 0.300 0.150 0.200 ^a This lane group consists of Lane 1W and Lane 2W (i.e., two through lanes westbound). Figure 5.15 An image of the MS Excel worksheet used for the computations of Problem 5.10. https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/fig5_15_B.tif"/> Table 5.16 HCM Signal Timing Computations and Results for the Three Phases of the T-Intersection in Problem 5.11 Phase A B C Lane 1 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/inline5_34_B.tif"/> 2 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/inline5_35_B.tif"/> 3 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/inline5_36_B.tif"/> 4 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/inline5_37_B.tif"/> 3 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/inline5_38_B.tif"/> 1 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/inline5_39_B.tif"/> 2 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/inline5_40_B.tif"/> v/s 0.38 0.22 0.18 0.24 ?? 0.18 0.20 v (vph) 703 407 324 444 ?? 324 320 Lane Group 1 2 3 1 1 2 v _ij (vph) 703+407 = 1110 324 444 ?? 324 320 s_ij (vph) 2×1850 = 3700 1800 1850 1600 1800 1600 v _ij /s_ij 0.30 0.18 0.24 ?? 0.18 0.20 (v/s)_cj 0.30 ?? 0.20 Figure 5.16 A 4-leg intersection for Problem 5.12. https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/fig5_16_B.tif"/> Table 5.17 The Established Lane Groups for the 4-Leg Intersection in Problem 5.12 Approach 1 (Eastbound) 2 (Westbound) 3 (Northbound) 4 (Southbound) Lane Group https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/inline5_41_B.tif"/> https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/inline5_42_B.tif"/> https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/inline5_43_B.tif"/> https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/inline5_44_B.tif"/> Table 5.18 The Phasing System for the 4-Leg Intersection in Problem 5.12 Phase A B C D Diagram https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/inline5_45_B.tif"/> https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/inline5_46_B.tif"/> https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/inline5_47_B.tif"/> https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/inline5_48_B.tif"/> Table 5.19 Lane Group Volumes and Critical Lane Group Volume for Problem 5.12 Phase A B C D Lane Group 1N 1S 2N 2S 1E 1W 2E 2W Volume (vph) 280+200 = 480 180+220 = 400 150 100 100 120 180 200 V_critical (vph) 480 150 120 200 Table 5.20 Computation of (v/s) and Critical (v/s) Ratios for the Lane Groups of the Four Phases for Problem 5.13 Phase A B C D Lane Group 1N 1S 2N 2S 1E 1W 2E 2W Volume (vph) 480 400 150 100 100 120 180 200 V_critical (vph) 480 150 120 200 v (vph) 533.3 444.4 166.7 111.1 111.1 133.3 200.0 222.2 s (vph) 3200 3200 1600 1600 1600 1600 1600 1600 v/s 0.167 0.139 0.104 0.069 0.069 0.083 0.125 0.139 (v/s)_critical 0.167 0.104 0.083 0.139 Summation 0.493 Table 5.21 The Effective and Actual Green Times for the Four Phases of the 4-Leg Intersection in Problem 5.13 Phase A B C D (v/s)_critical 0.167 0.104 0.083 0.139 τ (sec) 3 3 3 3 l (sec) 3 3 3 3 G_e (sec) 12.9 8.0 6.4 10.7 G_a (sec) 12.9 8.0 6.4 10.7 Table 5.22 Calculated Green Time versus Available Green Time for Pedestrians for Problem 5.14 Approach N-S Approaches E-W Approaches Available G_p (sec) 6.4 + 10.7 = 17.1 12.9 + 8.0 = 20.9 Required G_p 23.2 + 22.4 = 45.6 19.1 + 18.6 = 37.7 Not-Satisfied Table 5.23 Given Phasing System for Problem 5.15 Phase A B C D Diagram https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/inline5_49_B.tif"/> https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/inline5_50_B.tif"/> https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/inline5_51_B.tif"/> https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/inline5_52_B.tif"/> Table 5.24 Computation of (v/s) and Critical (v/s) Ratios for the Lane Groups of the Four Phases for Problem 5.15 Phase A B C D Lane Group 1N 2N 1S 2S 1E 2E 1W 2W Volume (vph) 480 150 400 100 100 180 120 200 V_critical (vph) 480 400 180 200 v (vph) 533.3 166.7 444.4 111.1 111.1 200.0 133.3 222.2 s (vph) 3200 1600 3200 1600 1600 1600 1600 1600 v/s 0.167 0.104 0.139 0.069 0.069 0.125 0.083 0.139 (v/s)_critical 0.167 0.139 0.125 0.139 Summation 0.570 Figure 5.17 A 4-leg intersection for Problem 5.16. https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/fig5_17_B.tif"/> Table 5.25 The Given Saturation Flow Rates for the Lanes of the 4-Leg Intersection in Problem 5.16 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/inline5_53_B.tif"/> https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/inline5_54_B.tif"/> https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/inline5_55_B.tif"/> https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/inline5_56_B.tif"/> https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/inline5_57_B.tif"/> https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/inline5_58_B.tif"/> https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/inline5_59_B.tif"/> https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/inline5_60_B.tif"/> https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/inline5_61_B.tif"/> https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/inline5_62_B.tif"/> https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/inline5_63_B.tif"/> https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/inline5_64_B.tif"/> https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/inline5_65_B.tif"/> https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/inline5_66_B.tif"/> https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/inline5_67_B.tif"/> https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/inline5_68_B.tif"/> 1700 1800 1800 1600 1600 1800 1800 1700 1600 1800 1800 1700 1700 1800 1800 1600 Table 5.26 The Established Lane Groups for the Four Approaches of the Intersection in Problem 5.16 Approach 1 (Eastbound) 2 (Westbound) 3 (Northbound) 4 (Southbound) Lane Group No. 1E 2E 1W 2W 1N 2N 1S 2S Lane Group https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/inline5_69_B.tif"/> https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/inline5_70_B.tif"/> https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/inline5_71_B.tif"/> https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/inline5_72_B.tif"/> https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/inline5_73_B.tif"/> https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/inline5_74_B.tif"/> https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/inline5_75_B.tif"/> https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/inline5_76_B.tif"/> Table 5.27 Phasing System #1 for the 4-Leg Intersection in Problem 5.16 Phase A B C D Diagram https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/inline5_77_B.tif"/> https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/inline5_78_B.tif"/> https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/inline5_79_B.tif"/> https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/inline5_80_B.tif"/> Table 5.28 Phasing System #2 for the 4-Leg Intersection in Problem 5.16 Phase A B C D Diagram https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/inline5_81_B.tif"/> https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/inline5_82_B.tif"/> https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/inline5_83_B.tif"/> https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/inline5_84_B.tif"/> Table 5.29 Computation of (v/s) and Critical (v/s) Ratios for the Lane Groups of Phasing System #1 for Problem 5.16 Phase A B C D Lane Group 1E 2E 1W 2W 1N 2N 1S 2S Volume (vph) 655 107 730 128 390 105 355 90 v (vph) 689.5 112.6 768.4 134.7 410.5 110.5 373.7 94.7 s (vph) 5300 1600 5300 1600 5300 1600 5300 1600 v/s 0.130 0.070 0.145 0.084 0.077 0.069 0.071 0.059 (v/s)_critical 0.130 0.145 0.077 0.071 Summation 0.423 Table 5.30 The Effective and Actual Green Times for Phasing System #1 of the 4-Leg Intersection in Problem 5.16 Phase A B C D (v/s)_critical 0.130 0.145 0.077 0.071 τ (sec) 3 3 3 3 l (sec) 3.5 3.5 3.5 3.5 G_e (sec) 11.1 12.3 6.6 6.0 G_a (sec) 11.6 12.8 7.1 6.5 Table 5.31 Computation of (v/s) and Critical (v/s) Ratios for the Lane Groups of Phasing System #2 for Problem 5.16 Phase A B C D Lane Group 2E 2W 1E 1W 2N 2S 1N 1S Volume (vph) 107 128 655 730 105 90 390 355 v (vph) 112.5 134.7 689.5 768.4 110.5 94.7 410.5 373.7 s (vph) 1600 1600 5300 5300 1600 1600 5300 5300 v/s 0.070 0.084 0.130 0.145 0.069 0.059 0.077 0.071 (v/s)_critical 0.084 0.145 0.069 0.077 Summation 0.375 Table 5.32 The Effective and Actual Green Times for Phasing System #2 of the 4-Leg Intersection in Problem 5.16 Phase A B C D (v/s)_critical 0.084 0.145 0.069 0.077 τ (sec) 3 3 3 3 l (sec) 3.5 3.5 3.5 3.5 G_e (sec) 6.9 12.0 5.7 6.4 G_a (sec) 7.4 12.5 6.2 6.9 Figure 5.18 A 4-leg intersection for Problem 5.17. https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/fig5_18_B.tif"/> Table 5.33 The Given Saturation Flow Rates for the Lanes of the 4-Leg Intersection in Problem 5.17 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/inline5_85_B.tif"/> https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/inline5_86_B.tif"/> https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/inline5_87_B.tif"/> https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/inline5_88_B.tif"/> https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/inline5_89_B.tif"/> https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/inline5_90_B.tif"/> https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/inline5_91_B.tif"/> https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/inline5_92_B.tif"/> https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/inline5_93_B.tif"/> https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/inline5_94_B.tif"/> 1300 900 900 1300 1200 1500 1350 1350 1500 1200 Table 5.34 The Phasing System of the 4-Leg Intersection in Problem 5.17 Phase A B C D Diagram https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/inline5_95_B.tif"/> https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/inline5_96_B.tif"/> https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/inline5_97_B.tif"/> https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/inline5_98_B.tif"/> Table 5.35 The Established Lane Groups for the 4-Leg Intersection in Problem 5.17 Approach 1 (Eastbound) 2 (Westbound) 3 (Northbound) 4 (Southbound) Lane Group No. 1E 2E 1W 2W 1N 2N 1S 2S Lane Group https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/inline5_99_B.tif"/> https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/inline5_100_B.tif"/> https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/inline5_101_B.tif"/> https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/inline5_102_B.tif"/> https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/inline5_103_B.tif"/> https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/inline5_104_B.tif"/> https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/inline5_105_B.tif"/> https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/inline5_106_B.tif"/> Table 5.36 The Phasing System for the 4-Leg Intersection in Problem 5.17 Phase A B C Diagram https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/inline5_107_B.tif"/> https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/inline5_108_B.tif"/> https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/inline5_109_B.tif"/> *Note: The dashed lines represent permitted movements and the solid lines represent protected movements. Table 5.37 Computation of (v/s) and Critical (v/s) Ratios for the Lane Groups of the Three Phases for Problem 5.17 Phase A B C Lane Group 2E 2W 1E 1W 1N 1S 2N 2S v, DHV (vph) 275 245 675 795 370 380 100 80 s (vph) 1200 1200 2850 2850 1300 1300 900 900 v/s 0.229 0.204 0.237 0.279 0.285 0.292 0.111 0.089 (v/s)_critical 0.229 0.279 0.292 Summation 0.800 Notes: (1)The maximum value of the flow ratio (v/s) for all lane groups in each phase is determined as shown above, considering that the Critical Lane Group is the lane group that requires the longest green time in a phase. And therefore, this lane group determines the green time that is assigned to that phase. (2)The summation of the total (v/s) ratios provides an idea of the required timing for each lane group based on the traffic volume for that lane group and the saturation flow rate of the lane group. In other words, higher traffic volume needs higher green time, but also the saturation flow rate of the lane group affects this timing. For instance, two lane groups with the same traffic volume may have different green times if they are in different phases depending on their saturation flow rates. A lower saturation flow rate would provide a higher green time and vice versa because prevailing conditions that reduce the saturation flow rate would also impact the movement of the vehicles at the intersection when the green turns on. Table 5.38 The Effective and Actual Green Times for the Three Phases of the 4-Leg Intersection in Problem 5.17 Phase A B C (v/s)_critical 0.229 0.279 0.292 τ (sec) 3.5 3.5 3.5 l (sec) 3.0 3.0 3.0 G_e (sec) 29.5 35.9 37.6 G_a (sec) 29.0 35.4 37.1 Figure 5.19 Five successive intersections on urban arterial for Problem 5.18. https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/fig5_19_B.tif"/> Table 5.39 The Average Spacing and Recommended Cycle Length for Simultaneous Traffic Signal System in Problem 5.18 Intersection 1 2 3 4 5 Cycle Length (sec) 70 65 75 65 70 Spacing (ft) 4140 4100 4000 4180 Average Spacing (ft) 4105 Recommended C (sec) 70 Table 5.40 The Cycle Length for Single-, Double-, and Triple-Alternate Systems for Problem 5.19 Alternate System Single Double Triple n 2 4 6 X (ft) 750 u (ft/sec) 30 × 5280 3600 = 44 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH005_eqn_0160.tif"/> Cycle Length (sec) 34.1 68.2 102.3 C (sec)^a 35 70 105 ^aRounded to the nearest 5 seconds.

Intersection Design, Operation, and Control

ABSTRACT