Highway Capacity and Level of Service

ABSTRACT

Chapter 4 of this book discusses the capacity and the level of service (LOS) expected on a highway based on the existing conditions of the highway. Capacity refers typically to the maximum hourly flow rate expected on a roadway or a lane of roadway under existing traffic and road conditions. In this section, the traffic-related factors as well as road conditions are studied. Computation of related factors including the peak-hour factor and heavy vehicles adjustment factor are presented. The level of service, defined as a qualitative measure for the traffic service based on some performance indicators such as speed and density, is also presented in this part. The effect of the different existing traffic and road conditions on the determination of level of service for a highway is discussed. The level of service is determined for freeway segments including general segments and segments with specific grades (downgrades or upgrades). Additionally, the level of service for two-lane two-way highways and multi-lane highways is discussed. The performance measures required to determine the level of service are computed using the given traffic and roadway conditions and the standard tables and figures in the Highway Capacity Manual (HCM). 4.1

The traffic count on a roadway segment during the peak hour is shown in Table 4.1.

Determine the peak-hour factor (PHF) for this roadway segment.

Solution:

4.1 PHF = Hourly volume Peak 15 − minute hourly flow rate https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH004_eqn_0001.tif"/>

4.2 Peak 15 − minute hourly flow rate = 4 × Traffic volume in peak 15 minutes https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH004_eqn_0002.tif"/>

Therefore:

PHF = 800 + 750 + 710 + 940 4 × 940 = 0.85 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH004_eqn_0003.tif"/>

The screen image of the MS Excel worksheet and the function (f _x ) used to compute the PHF for this problem is shown in Figure 4.1.84

4.2

If the peak-hourly volume on a highway segment is 2,700 vph and the PHF is 0.90, compute the traffic count during the peak 15-minute period.

Solution:

The two formulas shown below for PHF and peak 15-minute hourly flow rate, respectively, are used:

PHF = Hourly volume Peak 15 − minute hourly flow rate https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH004_eqn_0004.tif"/>

Peak 15 − minute hourly flow rate = 4 × Traffic volume in peak 15 minutes https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH004_eqn_0005.tif"/>

Therefore:

0.90 = 2700 4 × Traffic volume in peak 15 minutes = 0.85 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH004_eqn_0006.tif"/>

⇒

Traffic volume in peak 15 minutes = 2700 4 × 0.90 = 750 vehicles https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH004_eqn_0007.tif"/>

The screen image of the MS Excel worksheet with the function (f _x ) used to obtain the traffic volume in the peak 15 minutes in this problem is shown in Figure 4.2.85

4.3

The traffic volume for a highway segment was collected in the peak hour for 15-minute time periods. If the difference between the peak 15-minute traffic count and the average 15-minute traffic count is 25%, compute the peak-hour factor (PHF).

Solution:

The peak-hour factor is calculated using the formula below:

Hourly volume = 4 × Average 15 − minute traffic count https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH004_eqn_0009.tif"/>

Peak 15 − minute hourly flow rate = 4 × Peak 15 − minute traffic count https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH004_eqn_0010.tif"/>

Since the difference between the peak 15-minute traffic count and the average 15-minute traffic count is 25%, then:

Peak 15 − min traffic count = 1.25 × Average 15 − min traffic count https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH004_eqn_0011.tif"/>

⇒

Average 15 − min traffic count = Peak 15 − min traffic count 1.25 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH004_eqn_0012.tif"/>

Therefore:

PHF = 4 × Peak 15 − minute traffic count 1.25 4 × Peak 15 − minute traffic count = 1 1.25 = 0.85 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH004_eqn_0013.tif"/>

The screen image of the MS Excel worksheet with the function (f _x ) used to obtain the PHF based on the given data in this problem is shown in Figure 4.3.86

4.4

On a 10-lane freeway segment, the PHF and the peak-hourly volume in one direction are 0.90 and 4500 vph, respectively. If the percentage of heavy vehicles on this segment is 0% and drivers are all commuters, determine the 15-minute passenger car equivalent flow rate (pcphpl).

Solution:

The formula used to compute the 15-minute passenger car equivalent flow rate is shown in the equation below:

v p = V ( PHF ) ( N ) ( f p ) ( f HV ) https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH004_eqn_0014.tif"/>

Where:

v _p = 15-minute passenger car equivalent flow rate (pcphpl)

V = peak-hour volume in one direction (vph)

PHF = peak-hour factor

N = number of lanes in one direction

f_p = driver population factor (0.85−1.00)

f_HV = adjustment factor for heavy vehicles

The peak-hour volume, V is given as 3600 vph, PHF = 0.90, N = 5 lanes (in one direction), f_p = 1.00 since drivers are all commuters, and f_HV = 1.00 since there is no heavy vehicles using the freeway.

Therefore:

v p = 4500 ( 0.90 ) ( 5 ) ( 1.00 ) ( 1.00 ) = 1000 pcphpl https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH004_eqn_0015.tif"/>

A screen image of the MS Excel worksheet with the function (f _x ) used to obtain the 15-minute passenger car equivalent flow rate (pcphpl) of this problem is shown in Figure 4.4.87

4.5

A 6-lane general extended freeway segment is designed in a level terrain to have a maximum service flow rate of 900 pcphpl. The designer intends to determine the percentage of trucks and buses that will be allowed to utilize this segment to keep the flow rate at this limit. If the PHF is 0.95, the peak-hourly volume is 2500 vph in one direction, and the drivers are all commuters, determine the percentage of trucks and buses.

Solution:

The formula for the 15-minute passenger car equivalent flow rate is used to back-calculate the adjustment factor for heavy vehicles (f_HV) as shown below:

v p = V ( PHF ) ( N ) ( f p ) ( f HV ) https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH004_eqn_0016.tif"/>

900 = 2500 ( 0.95 ) ( 3 ) ( 1.00 ) ( f HV ) https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH004_eqn_0017.tif"/>

⇒

f HV = 2500 ( 900 ) ( 0.95 ) ( 3 ) ( 1.00 ) = 0.975 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH004_eqn_0018.tif"/>

But the adjustment factor for heavy vehicles (f_HV) is given by the following formula:

f HV = 1 1 + P T ( E T − 1 ) + P R ( E R − 1 ) https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH004_eqn_0019.tif"/>

Where:

P_T and P_R = percentages of trucks/buses and recreational vehicles (RVs) in traffic, respectively (represented as decimal numbers)

E_T and E_R = passenger car equivalent for trucks/buses and RVs, respectively

E_T = 1.5 for level terrain from Table 4.1

P_R = 0% (no recreational vehicles using the freeway) (see Table 4.2)88

Therefore:

0.975 = 1 1 + P T ( 1.5 − 1 ) https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH004_eqn_0020.tif"/>

⇒

P T = 0.052 ( 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/TNF-CH004_eqn_0021.tif"/>

A screen image of the MS Excel worksheet with the function (f _x ) used to obtain the percentage of trucks/buses (P_T) of this problem is shown in Figure 4.5. 89

4.6

If the adjustment factor for heavy vehicles on a general extended freeway segment with level terrain is 0.956, determine the percentage of trucks and buses assuming that no recreational vehicles is using the freeway.

Solution:

The adjustment factor for heavy vehicles (f_HV) is given by the formula:

E_T = 1.5 (for level terrain from Table 4.2), and P_R = 0, therefore:

0.956 = 1 1 + P T ( 1.5 − 1 ) https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH004_eqn_0023.tif"/>

⇒

P T = 1 − 1 0.956 ( 1.5 − 1 ) = 0.092 = 9.2 % https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH004_eqn_0024.tif"/>

A screen image of the MS Excel worksheet used to compute the percentage of trucks/buses (P_T) for this problem is shown in Figure 4.6.90

4.7

A highway traffic designer would like to maintain the service flow rate for a 4-lane freeway segment designed in a mountainous terrain to have a maximum service flow rate of 1200 pcphpl. The designer intends to determine the percentage of recreational vehicles that will be allowed to use this freeway to keep the flow rate at this limit as the freeway leads to national park and camping areas. If the PHF is 0.90, the peak-hourly volume is 1700 vph in one direction, and the drivers are assumed to be commuters, determine the percentage of recreational vehicles. No truck/buses are allowed to use the highway.

Solution:

To back-calculate the adjustment factor for heavy vehicles (f_HV), the formula for the 15-minute passenger car equivalent flow rate is used as shown below:

v p = V ( PHF ) ( N ) ( f p ) ( f HV ) https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH004_eqn_0025.tif"/>

1200 = 1700 ( 0.90 ) ( 2 ) ( 1.00 ) ( f HV ) https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH004_eqn_0026.tif"/>

⇒

f HV = 1700 ( 1200 ) ( 0.90 ) ( 2 ) ( 1.00 ) = 0.787 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH004_eqn_0027.tif"/>

But the adjustment factor for heavy vehicles (f_HV) is given by the following formula:

E_T = 4.0 for mountainous terrain from Table 4.2.

P_T = 0% (no trucks/buses on the freeway).

Therefore:

0.787 = 1 1 + P R ( 4.0 − 1 ) https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH004_eqn_0029.tif"/>

⇒

P R = 0.090 ( 9.0 % ) https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH004_eqn_0030.tif"/>

A screen image of the MS Excel worksheet with the function (f _x ) used to obtain the percentage of recreational vehicles (P_R) in this problem is shown in Figure 4.7.91

4.8

If a freeway segment consists of two consecutive upgrades with grade and length of 3%–1200 ft and 2%–1800 ft, respectively, estimate the average grade of this segment.

Solution:

% G Average = 100 ∑ i = 1 n G i L i ∑ i = 1 n L i https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH004_eqn_0031.tif"/>

Where:

G_Average = average grade (percent)

G_i = grade for freeway segment i

L_i= length of freeway segment i

Therefore:

% G Average = 100 G 1 L 1 + G 2 L 2 L 1 + L 2 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH004_eqn_0032.tif"/>

% G Average = 100 ( 0.03 ) ( 1200 ) + ( 0.02 ) ( 1800 ) 1200 + 1800 = 2.4 % https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH004_eqn_0033.tif"/>

A screen image of the MS Excel worksheet with the function (f _x ) used to obtain the average grade of the two freeway segments as a percentage is shown in Figure 4.8.92

4.9

If a freeway upgrade segment consists of two successive subsections of 2%, 3,000 ft long and 3%, 3,000 ft long, determine the equivalent grade for this segment (assume an entry truck speed is 55 mph).

Solution:

The equivalent grade for successive grades is determined using the performance curves for standard trucks in Figure 4.9. The procedure is described below:93

The truck speed of 55 mph (88.5 km/h) is used to enter the curve of the first grade (2% curve). Move on the curve for a distance equal to the length of the first grade (3000 ft = 914.4 m).io8

After that, the reduced speed is determined at this point by moving horizontally on the figure to the left to read the speed value (78 km/h = 48.5 mph).

From this speed value, the curve of second grade (3% curve) is entered by moving horizontally to the right until reaching the curve. From the intersection point with the 3%-curve, move on the curve for a distance equal to the length of the second grade (3000 ft = 914.4 m).

Read the new reduced speed by moving horizontally on the figure to the left. Read the new value of the speed (64 km/h = 39.8 mph).

From this speed value, make a horizontal line.

On the x-axis of the figure where the distance measurement (length of grade) is read, from the total length of the segment (3000 + 3000 = 6000 ft = 1828.8 m), make a vertical line.

The intersection point of the horizontal line and the vertical line represents the location of the equivalent grade of the freeway segment (2.8%).

The procedure is illustrated in Figure 4.10. From this figure, the equivalent grade for this segment is equal to 2.8%.

4.10

A rural 10-lane freeway in rolling terrain was designed to operate at a level of service (LOS) B with the following characteristics:

Traffic data:

PHF = 0.95

Trucks = 8%

Commuter traffic

Geometric data:

Lane width = 12 ft

Shoulder width = 10 ft

Interchange spacing = 2/3 mile

Determine the maximum peak-hourly volume that this freeway can carry to maintain this LOS.94

Solution:

The given data (inputs) are summarized in Table 4.3.

The following data (outputs) shown in Table 4.4 are determined from standard Tables 4.2 and 4.5 through 4.8, respectively.95 96

The following parameters and factors (outputs) shown in Table 4.9 are computed/determined from standard tables.97

The adjustment factor for heavy vehicles (f_HV) is computed using the following formula:

f HV = 1 1 + 0.08 ( 2.5 − 1 ) = 0.893 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH004_eqn_0035.tif"/>

The free-flow speed is computed from the base free-flow speed (BFFS) using the following formula:

FFS = BFFS − f LW − f LC − f N − f ID https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH004_eqn_0036.tif"/>

Where:

FFS = free-flow speed

BFFS = base free-flow speed

f_LW = adjustment factor for lane width

f_LC = adjustment factor for lateral clearance

f_N = adjustment factor for number of lanes

f_ID = adjustment factor for interchange spacing

⇒

FFS = 75.0 − 0.0 − 0.0 − 0.0 − 5.0 = 70 mph https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH004_eqn_0037.tif"/>

From Table 4.10, at LOS = B and FFS = 70 mph, the maximum service flow rate (v _p ) = 1210 pcphpl.

The peak-hourly volume is then calculated using the formula below:

v p = V ( PHF ) ( N ) ( f p ) ( f HV ) https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH004_eqn_0038.tif"/>

⇒

V = ( 1210 ) ( 0.95 ) ( 5 ) ( 1.00 ) ( 0.893 ) = 5132 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-CH004_eqn_0039.tif"/>

The MS Excel worksheet used to determine the maximum peak-hourly volume based on the given LOS data in this problem is shown in Figure 4.11.98

4.11

For a 6-lane highway segment with the following characteristics (see Table 4.11):

Traffic composition: Trucks = 8%, RVs = 4%

Commuter traffic

Upgrade = 4%

Length = 0.30 mile

Determine the following:

Peak-hour factor (PHF)

Passenger car equivalent values for trucks and RVs

Adjustment factor for heavy vehicles (f_HV)

15-minute passenger car equivalent flow rate (v _p)99

Solution:

The peak-hour factor (PHF) is estimated based on the following formula:

⇒

PHF = 600 + 580 + 630 + 700 4 × 700 = 0.90 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH004_eqn_0041.tif"/>

The passenger car equivalent values for trucks and RVs in Table 4.12 are determined from Tables 4.13 and 4.14. First, the freeway segment should be classified whether it is a specific upgrade or general segment. This is based on the following condition:

A freeway/highway segment having a grade ≥ 3% and length > ¼ mile or a grade < 3% and a length > ½ mile is considered a specific upgrade.

Since the given freeway segment has an upgrade of 4% ≥ 3% and length of 0.30 mile > ¼ mile ⇒ the segment is considered a specific upgrade.

Therefore, the tables for specific upgrades are used. E_T and E_R are determined as shown in Table 4.12.100 101

The adjustment factor for heavy vehicles (f_HV) is computed using the E_T and E_R values determined in part (b) above.

⇒

f HV = 1 1 + 0.08 ( 2.0 − 1 ) + 0.04 ( 2.5 − 1 ) = 0.877 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH004_eqn_0043.tif"/>

The 15-minute passenger car equivalent flow rate (v _p) is calculated using the formula shown below:

v p = V ( PHF ) ( N ) ( f p ) ( f HV ) https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH004_eqn_0044.tif"/>

⇒

v p = ( 600 + 580 + 630 + 700 ) ( 0.90 ) ( 3 ) ( 1.00 ) ( 0.877 ) = 1064 pcphpl https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH004_eqn_0045.tif"/>

A screen image of the MS Excel worksheet with the function (f _x ) used to compute the 15-minute passenger car equivalent flow rate (v _p ) for this problem is shown in Figure 4.12.

102

4.12

A freeway segment was expanded from 4 to 5 lanes in each direction. What is the effect of this lane expansion on the 15-minute hourly flow rate assuming all other factors are the same?

Solution:

The 15-minute passenger car equivalent flow rate (v _p ) is determined using the following formula:

v p = V ( PHF ) ( N ) ( f p ) ( f HV ) https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH004_eqn_0046.tif"/>

At number of lanes, N = 4:

v p = V 4 ( PHF ) ( f p ) ( f HV ) https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH004_eqn_0047.tif"/>

And at number of lanes, N = 5:

v p = V 5 ( PHF ) ( f p ) ( f HV ) https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH004_eqn_0048.tif"/>

Reduction in v p = 100 ( V ( PHF ) ( f p ) ( f HV ) ) ( 1 4 − 1 5 ) ( V ( PHF ) ( f p ) ( f HV ) ) ( 1 4 ) = 20.0 % https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH004_eqn_0049.tif"/>

The screen image of the MS Excel worksheet used to compute the reduction in the 15-minute hourly flow rate for this problem is shown in Figure 4.13.103

4.13

A freeway segment is designed to have a free-flow speed (FFS) of 65 mph and operate at level of service (LOS) B. If the freeway segment with a peak-hourly volume of 3600 vph and PHF of 0.90 is intended to operate at the maximum service flow rate, determine the minimum number of lanes needed for each direction (assume commuter traffic and no heavy vehicles exist).

Solution:

Since the traffic drivers are commuters and no heavy vehicles will use the freeway, the driver population factor (fp) is 1.00 and the adjustment factor for heavy vehicles (f_HV) is also 1.00.

For a level of service (LOS) B, and a free-flow speed (FFS) of 65 mph, the maximum service flow rate (v_p) is determined from Table 4.10:

v _p = 1170 pcphpl.

The required number of lanes to maintain a LOS B and this flow rate is computed using the formula:

v p = V ( PHF ) ( N ) ( f p ) ( f HV ) https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH004_eqn_0050.tif"/>

⇒

N = V ( v p ) ( PHF ) ( f p ) ( f HV ) = 3600 ( 1170 ) ( 0.90 ) ( 1.00 ) ( 1.00 ) = 3.4 ≅ 4 lanes https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH004_eqn_0051.tif"/>

Therefore, for this freeway segment to have a LOS B and operate at the maximum service flow rate, at least 4 lanes are required in each direction under the existing conditions.

The MS Excel worksheet with a function (f _x ) is used to compute the required number of lanes in this problem. A screen image of the MS Excel worksheet along with the function used is shown in Figure 4.14.104

4.14

Determine the free flow speed (FFS) of a rural 8-lane freeway segment having a lane width of 10 ft, no obstruction, and an interchange density of 1.0 interchange/mile.

Solution:

The free-flow speed is computed from the base free-flow speed (BFFS) using the following formula:

From Tables 4.5 through 4.8:

BFFS = 75 mph (for rural highways).

f_LW = 6.6 mph (using Table 4.5 at lane width = 10 ft)

f_LC = 0.0 mph (using Table 4.6 at right-shoulder lateral clearance ≥ 6 ft and number of lanes in one direction = 4 lanes)

f_N = 1.5 mph (using Table 4.7 at number of lanes in one direction = 4 lanes)

f_ID = 2.5 mph (using Table 4.8 at interchange density = 1 interchange/mile)

⇒

FFS = 75.0 − 6.6 − 0.0 − 1.5 − 2.5 = 64.4 mph ( 103.6 kph ) https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH004_eqn_0053.tif"/>

The MS Excel worksheet is used to compute the free flow speed (FFS) of the freeway in this problem. A screen image of the MS Excel worksheet along with the function used is shown in Figure 4.15.105

4.15

A highway traffic designer is conducting a study to check if the level of service (LOS) on a rural 6-lane freeway segment with the following characteristics located in a mountainous terrain is maintained at C after an increase in the number of trucks on the freeway segment from 5% to 10% due to the opening of a new nearby factory. Determine whether the LOS is maintained at C or not. If not, what measures or solutions can the designer take to keep the LOS at C?

Traffic data:

PHF = 0.92

Peak-hourly volume = 4200 vph (in one direction)

Trucks = 5% (increased to 10%)

No recreational vehicles

Commuter traffic

Geometric data:

Length = 2 miles

Upgrade = 3%

Lane width = 10 ft

Shoulder width = 5 ft

Interchange spacing = 2.0 miles

Solution:

Since the freeway segment has a grade of 3% ≥ 3% and le ngth of 2.0 miles > ¼ mile, the segment is considered a specific upgrade.

Therefore, the tables for specific upgrades are used. E_T is determined as shown in Table 4.15.106

The adjustment factor for heavy vehicles (f_HV) is computed using the E_T determined above and P_T value of 5% (before the increase).

⇒

f HV = 1 1 + 0.05 ( 2.5 − 1 ) = 0.930 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH004_eqn_0055.tif"/>

f_p = 1.00 since traffic drivers are commuters.

The 15-minute passenger car equivalent flow rate (v _p) is calculated using the following formula:

v p = V ( PHF ) ( N ) ( f p ) ( f HV ) https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH004_eqn_0056.tif"/>

⇒

v p = 4200 ( 0.92 ) ( 3 ) ( 1.00 ) ( 0.930 ) ≅ 1636 pcphpl https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH004_eqn_0057.tif"/>

From standard tables, the four adjustment factors in the formula are determined as shown below:

BFFS = 75 mph (for rural highways).

f_LW = 6.6 mph (using Table 4.5 at lane width = 10 ft)

f_LC = 0.4 mph (using Table 4.6 at right-shoulder lateral clearance = 5 ft and number of lanes in one direction = 3 lanes)

f_N = 3.0 mph (using Table 4.7 at number of lanes in one direction = 3 lanes)

f_ID = 0.0 mph (using Table 4.8 at interchange density = 0.50 ≤ 0.50 interchange/mile)

⇒

FFS = 75.0 − 6.6 − 0.4 − 3.0 − 0.0 = 65.0 mph ( 104.6 kph ) https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH004_eqn_0059.tif"/>

The average passenger car speed (S) can be computed using one of the formulas shone below:

For FFS > 70 mph:

4.7 S = FFS − [ ( FFS − 160 3 ) ( v p + 30 FFS − 3400 30 FFS − 1000 ) 2.6 ] https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH004_eqn_0060.tif"/> 107

For FFS ≤ 70 mph:

4.8 S = FFS − [ 1 9 ( 7 FFS − 340 ) ( v p + 30 FFS − 3400 40 FFS − 1700 ) 2.6 ] https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH004_eqn_0061.tif"/>

Since the FFS = 65 mph ≤ 70 mph, the following formula is used to determine the average passenger car speed (S):

S = FFS − [ 1 9 ( 7 FFS − 340 ) ( v p + 30 FFS − 3400 40 FFS − 1700 ) 2.6 ] https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH004_eqn_0062.tif"/>

S = 65 − [ 1 9 ( 7 ( 65 ) − 340 ) ( 1636 + 30 ( 65 ) − 3400 40 ( 65 ) − 1700 ) 2.6 ] = 64.2 mph ( 103.3 kph ) https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH004_eqn_0063.tif"/>

The density on the freeway segment is computed using the formula below:

4.9 D = v p 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-CH004_eqn_0064.tif"/>

Where:

D = density on the freeway segment (pc/mi/lane).

Hence:

D = 1636 64.2 ≅ 26 pc/mi/lane https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH004_eqn_0065.tif"/>

Since the density (D) is in the range of 18 ≤ D = 26 ≤ 26, where: 18 is the maximum density for LOS B and 26 is the maximum density for LOS C, the LOS is C.

After the increase when P_T = 10%, the adjustment factor for heavy vehicles (f_HV) is computed as:

f HV = 1 1 + 0.10 ( 2.5 − 1 ) = 0.870 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH004_eqn_0066.tif"/>

Therefore,

v p = 4200 ( 0.92 ) ( 3 ) ( 1.00 ) ( 0.870 ) ≅ 1750 pcphpl https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH004_eqn_0067.tif"/>

Since the FFS = 65 mph ≤ 70 mph, the same formula is used:

S = 65 − [ 1 9 ( 7 ( 65 ) − 340 ) ( 1750 + 30 ( 65 ) − 3400 40 ( 65 ) − 1700 ) 2.6 ] = 62.3 mph ( 100.3 kph ) https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH004_eqn_0068.tif"/>

The density is computed using the same formula:

D = v p 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-CH004_eqn_0069.tif"/> 108

D = 1750 62.3 ≅ 29 pc/mi/lane https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH004_eqn_0070.tif"/>

Since the density (D) is in the range of 26 ≤ D = 29 ≤ 35, where: 26 is the maximum density for LOS C and 35 is the maximum density for LOS D, the LOS is D.

After the increase in the percentage of trucks from 5% to 10%, the computed flow rate has increased from 1636 to 1750 pcphpl, the density has also increased from 26 to 29 pc/mi/lane and the LOS is reduced from C to D.

The scenarios that can improve the LOS under the existing traffic conditions should be related to the existing geometric conditions. In other words, a change in one of the existing geometric conditions would possibly improve the LOS. Two possible scenarios can be done to maintain the LOS at C:

Scenario #1: An additional lane on this freeway segment can be constructed. The additional lane reduces the maximum flow rate on the segment as shown below:

v p = 4200 ( 0.92 ) ( 4 ) ( 1.00 ) ( 0.870 ) ≅ 1312 pcphpl https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH004_eqn_0071.tif"/>

The adjustment factors f_LW, f_LC, and f_ID stay the same, but the adjustment factor f_N changes since the number of lanes is changed from 3 to 4 lanes. Therefore:

BFFS = 75 mph (for rural highways).

f_LW = 6.6 mph (using Table 4.5 at lane width = 10 ft)

f_LC = 0.4 mph (using Table 4.6 at right-shoulder lateral clearance = 5 ft and number of lanes in one direction = 3 lanes)

f_N = 1.5 mph (using Table 4.7 at number of lanes in one direction = 3 lanes)

f_ID = 0.0 mph (using Table 4.8 at interchange density = 0.50 ≤ 0.50 interchange/mile)

⇒

FFS = 75.0 − 6.6 − 0.4 − 1.5 − 0.0 = 66.5 mph ( 107.0 kph ) https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH004_eqn_0073.tif"/>

Since the FFS = 66.5 mph ≤ 70 mph, the following formula is used to determine the average passenger car speed (S):

But since v _p = 1312 < 3400−30FFS = 3400−30(66.5) = 1405 pcphpl, then S = FFS.

Therefore,

S = 66.5 mph ( 107.0 kph ) https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH004_eqn_0075.tif"/>

The density on the freeway segment is computed using the formula below:

D = v p 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-CH004_eqn_0076.tif"/>

Hence:

D = 1312 66.5 ≅ 20 pc/mi/lane https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH004_eqn_0077.tif"/> 109

Since the density (D) is in the range of 18 ≤ D = 20 ≤ 26, where: 18 is the maximum density for LOS B and 26 is the maximum density for LOS C, the LOS is C.

Scenario #2: Increase the lane width only 1 ft from 10 to 11 ft; that is a total of 3 ft for the 3 lanes (in one direction). The reduction in the service flow rate and the effect on the LOS based on the 1-ft lane increase is computed as shown below:

The increase in the lane width will affect the free-flow speed (FFS):

f_LW = 1.9 mph (using Table 4.5 at lane width = 11 ft)

All other factors are the same as before.

⇒

FFS = 75.0 − 1.9 − 0.4 − 3.0 − 0.0 = 69.7 mph ≅ 70 mph ( 112.6 kph ) https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH004_eqn_0078.tif"/>

Since the FFS = 69.7 mph ≤ 70 mph, the following formula is used to determine the average passenger car speed (S):

S = 69.7 − [ 1 9 ( 7 ( 69.7 ) − 340 ) ( 1312 + 30 ( 69.7 ) − 3400 40 ( 69.7 ) − 1700 ) 2.6 ] = 69.7 mph https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH004_eqn_0080.tif"/>

v p = 1750 pcphpl https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH004_eqn_0081.tif"/>

The density is computed using the formula:

D = v p 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-CH004_eqn_0082.tif"/>

D = 1750 69.7 ≅ 26 pc/mi/lane https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH004_eqn_0083.tif"/>

Since the density (D) is in the range of 18 ≤ D = 26 ≤ 26, where: 18 is the maximum density for LOS B and 26 is the maximum density for LOS C, the LOS is C.

The scenario of increasing the interchange spacing on the freeway would not benefit improving the LOS. In other words, looking at Table 4.8, the zero reduction in the free-flow speed (FFS) corresponds to an interchange density of ≤ 0.50 mile. In this problem, the interchange spacing of 2 miles (i.e., interchange density of 0.50 mile) provides a zero reduction in the FFS. Hence, no further increase in the spacing (reduction in density) would improve the FFS and hence the LOS on this freeway segment.

4.16

A designer would like to assess which of the following three scenarios would mostly impact the LOS of a freeway segment with the following characteristics:

Classification of freeway:

Urban

Rolling terrain110

Traffic data:

PHF = 0.90

Peak-hourly volume = 7020 vph (in one direction)

Trucks = 8%

RVs = 5%

Commuter traffic

Geometric data:

Number of lanes = 4 lanes (in one direction)

Length = 2 miles

Upgrade = 2%

Lane width = 12 ft

Shoulder width = 0 ft (no lateral clearance)

Interchange spacing = 2.0 miles

Rolling terrain

Scenario #1: increasing the number of lanes from 4 to 5 lanes.

Scenario #2: prohibiting large-size trucks with heavy loads from using the freeway during normal hours of the day and allow them to use the freeway after midnight when traffic is not heavy. This will reduce the percentage of trucks from 8% to 4%.

Scenario #3: adding a shoulder having a width of 6 ft to the freeway.

Solution:

Since the freeway segment has a grade of 2% < 3% and length of 2.0 miles > ½ mile, the segment is considered a specific upgrade.

Therefore, the tables for specific upgrades are used. E_T is determined as shown in Table 4.16.

The adjustment factor for heavy vehicles (f_HV) is computed using the E_T determined above and P_T value of 8% (before the decrease).

⇒

f HV = 1 1 + 0.08 ( 2.0 − 1 ) + 0.05 ( 1.2 − 1 ) = 0.917 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH004_eqn_0085.tif"/>

f_p = 1.00 since traffic drivers are commuters.

The 15-minute passenger car equivalent flow rate (v _p) is calculated using the following formula:

v p = V ( PHF ) ( N ) ( f p ) ( f HV ) https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH004_eqn_0086.tif"/> 111

⇒

v p = 7020 ( 0.90 ) ( 4 ) ( 1.00 ) ( 0.917 ) ≅ 2126 pcphpl https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH004_eqn_0087.tif"/>

BFFS = 70 mph (for urban highways)

f_LW = 0.0 mph (using Table 4.5 at lane width = 12 ft)

f_LC = 1.2 mph (using Table 4.6 at right-shoulder lateral clearance = 0 ft and number of lanes in one direction = 4 lanes)

f_N = 1.5 mph (using Table 4.7 at number of lanes in one direction = 4 lanes)

f_ID = 0.0 mph (using Table 4.8 at interchange density = 0.50 ≤ 0.50 interchange/mile since spacing = 2.0 miles)

⇒

FFS = 70.0 − 0.0 − 1.2 − 1.5 − 0.0 = 67.3 mph ( 108.3 kph ) https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH004_eqn_0089.tif"/>

Since the FFS = 67.3 mph ≤ 70 mph, the following formula is used to determine the average passenger car speed (S):

S = 67.3 − [ 1 9 ( 7 ( 67.3 ) − 340 ) ( 2126 + 30 ( 67.3 ) − 3400 40 ( 67.3 ) − 1700 ) 2.6 ] = 60.4 mph https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH004_eqn_0091.tif"/>

The density is computed using the formula:

D = v p 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-CH004_eqn_0092.tif"/>

D = 2126 60.4 ≅ 36 pc/mi/lane https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH004_eqn_0093.tif"/>

Since the density (D) is in the range of 35 ≤ D = 36 ≤ 45, where: 35 is the maximum density for LOS D and 45 is the maximum density for LOS E, the LOS is E.

Assessment of scenarios:

Scenario #1: increasing the number of lanes from 4 to 5 lanes.

The increase of the number of lanes from 4 to 5 lanes affects the following:

v p = 7020 ( 0.90 ) ( 5 ) ( 1.00 ) ( 0.917 ) ≅ 1701 pcphpl https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH004_eqn_0094.tif"/>

f_LC = 0.6 mph (using Table 4.6 at right-shoulder lateral clearance = 0 ft and number of lanes in one direction = 5 lanes)

f_N = 0.0 mph (using Table 4.7 at number of lanes in one direction = 5 lanes)

⇒112

FFS = 70.0 − 0.0 − 0.6 − 0.0 − 0.0 = 69.4 mph ( 111.7 kph ) https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH004_eqn_0095.tif"/>

Since the FFS = 69.4 mph ≤ 70 mph, the following formula is used to determine the average passenger car speed (S):

S = 69.4 − [ 1 9 ( 7 ( 69.4 ) − 340 ) ( 1700 + 30 ( 69.4 ) − 3400 40 ( 69.4 ) − 1700 ) 2.6 ] = 68.3 mph https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH004_eqn_0097.tif"/>

The density is computed using the formula:

D = v p 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-CH004_eqn_0098.tif"/>

D = 1701 68.3 ≅ 25 pc/mi/lane https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH004_eqn_0099.tif"/>

Since the density (D) is in the range of 18 ≤ D = 25 ≤ 26, where: 18 is the maximum density for LOS B and 26 is the maximum density for LOS C, the LOS is C.

In this case, the E_T value is changed (see Table 4.17).

The adjustment factor for heavy vehicles is computed accordingly as shown below:

f HV = 1 1 + 0.04 ( 3.0 − 1 ) + 0.05 ( 1.2 − 1 ) = 0.917 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH004_eqn_0100.tif"/>

v p = V ( PHF ) ( N ) ( f p ) ( f HV ) https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH004_eqn_0101.tif"/>

⇒

The free-flow speed is the same as before applying any scenario since the change in the percentage in trucks does not affect the FFS adjustment factors:

Since the FFS = 67.3 mph ≤ 70 mph, the following formula is used to determine the average passenger car speed (S):

The density is computed using the formula:

D = v p 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-CH004_eqn_0106.tif"/>

D = 2126 60.4 ≅ 36 pc/mi/lane https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH004_eqn_0107.tif"/>

Since the density (D) is in the range of 35 ≤ D = 25 ≤ 45, where: 35 is the maximum density for LOS D and 45 is the maximum density for LOS E, the LOS is E.

Scenario #3: adding a shoulder having a width of 6 ft to the freeway.

The service flow rate is the same as before applying any scenario since the change in the shoulder width does not affect the flow rate:

f_LC = 0.0 mph (using Table 4.6 at right-shoulder lateral clearance = 6 ft and number of lanes in one direction = 4 lanes)

⇒

FFS = 70.0 − 0.0 − 0.0 − 1.5 − 0.0 = 68.5 mph ( 110.2 kph ) https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH004_eqn_0109.tif"/>

Since the FFS = 68.5 mph ≤ 70 mph, the following formula is used to determine the average passenger car speed (S):

S = 68.5 − [ 1 9 ( 7 ( 68.5 ) − 340 ) ( 2126 + 30 ( 68.5 ) − 3400 40 ( 68.5 ) − 1700 ) 2.6 ] = 61.1 mph https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH004_eqn_0111.tif"/> 114

The density is computed using the formula:

D = v p 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-CH004_eqn_0112.tif"/>

D = 2126 61.1 ≅ 35 pc/mi/lane https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH004_eqn_0113.tif"/>

Since the density (D) is in the range of 35 ≤ D = 35 ≤ 45, where: 35 is the maximum density for LOS D and 45 is the maximum density for LOS E, the LOS is D.

Investigating the three scenarios, Scenarios #1 (increasing the number of lanes from 4 to 5 lanes) improved the LOS from E to C and Scenario #3 (adding a shoulder having a width of 6 ft to the freeway) improved the LOS from E to D. On the other hand, Scenario #2 (reducing the percentage of trucks from 8% to 4%) did not improve the LOS of E. In conclusion, the improvement in Scenario #1 by having an additional lane is more significant (improvement by two levels) than the improvement by adding a 6-ft shoulder (improvement by one level). Yet, adding a shoulder of 6-ft width is less costly than adding another 12-ft lane to the freeway segment of 2.0 miles long.

4.17

Determine the LOS for the peak direction of a Class I two-lane highway with the following characteristics:

Traffic data:

PHF = 0.85

Hourly volume = 850 vph (in the peak direction)

Opposing volume = 650 vph

Trucks = 6%

RVs = 3%

BFFS = 55 mph

Geometric data:

Lane width = 12 ft

Shoulder width = 6 ft

20 access points per mile

No-passing zones = 20%

Rolling terrain

Solution:

To determine the level of service of the analysis direction on Class I two-lane highways, the percent time-spent following (PTSF) and the average travel speed (ATS) in the analysis direction should be computed.

PTSF Computation:

A trial initial value for the flow rate in the peak direction can be computed using the following formula:

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-CH004_eqn_0114.tif"/>

v = 850 0.85 = 1000 pcph https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH004_eqn_0115.tif"/>

The values of f_G (adjustment factor for grade), E_T, and E_R are determined from Table 4.18. The value of f_G is determined from Table 4.19 and the values of E_T and E_R are determined from Table 4.20.115

⇒

f HV = 1 1 + 0.06 ( 1.0 − 1 ) + 0.03 ( 1.0 − 1 ) = 1.00 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH004_eqn_0117.tif"/>

v = V ( PHF ) ( f G ) ( f HV ) https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH004_eqn_0118.tif"/> 116

⇒

v = 850 ( 0.85 ) ( 1.00 ) ( 1.00 ) = 1000 pcph https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH004_eqn_0119.tif"/>

The opposing flow rate is calculated using the procedure shown below:

A trial initial value for the opposing flow rate can be computed using the following formula:

v o = 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-CH004_eqn_0120.tif"/>

v o = 650 0.85 ≅ 765 pcph https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH004_eqn_0121.tif"/>

The values of f_G (adjustment factor for grade), E_T, and E_R are determined as shown in Table 4.21. The value of f_G is determined from Table 4.19 and the values of E_T and E_R are determined from Table 4.20.

⇒

v o = V ( PHF ) ( f G ) ( f HV ) https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH004_eqn_0124.tif"/>

⇒

v o = 650 ( 0.85 ) ( 1.00 ) ( 1.00 ) = 765 pcph https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH004_eqn_0125.tif"/>

The percent time-spent following (PTSF) is calculated using the following formula:

PTSF = BPTSF + f np https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH004_eqn_0126.tif"/>

Where:

BPTSF = base percent time-spent following.

f_np = adjustment factor for percentage of no-passing zones in the analysis direction.117

The base percent time-spent-following (BPTSF) is computed using the following equation:

BPTSF = 100 [ 1 − e a v 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-CH004_eqn_0127.tif"/>

Where:

v = passenger car equivalent flow rate for the peak 15-minute period in the analysis direction.

a and b = coefficients based on peak 15-minute passenger car equivalent opposing flow rate.

The coefficients a and b for an opposing demand flow rate = 765 pcph are calculated using interpolation between the values of a and b at opposing flow rates of 600 and 800 pcph as shown in Table 4.22. The values of the coefficients a and b are determined from Table 4.23.118

⇒ at 765 pcph,

a = − 0.173 − { ( 800 − 765 ) × ( − 0.173 − ( − 0.100 ) ) ( 800 − 600 ) } = − 0.160 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH004_eqn_0128.tif"/>

and

b = 0.349 − { ( 800 − 765 ) × ( 0.349 − 0.413 ) ( 800 − 600 ) } = − 0.160 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH004_eqn_0129.tif"/>

Therefore,

BPTSF = 100 [ 1 − e − 0.160 ( 1000 ) 0.360 ] = 85.4 % https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH004_eqn_0130.tif"/>

4.13 FFS = BFFS − f LS − f 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-CH004_eqn_0131.tif"/>

Where:

FFS = free-flow speed (mph)

BFFS = base free-flow speed (mph)

f_LS = adjustment factor for lane and shoulder widths

f_A = adjustment factor for number of access points

f_LS = 0.0 (from Table 4.24 for lane width = 12 ft ≥ 12 ft and shoulder width = 6 ft ≥ 6 ft)

f_A = 5.0 (from Table 4.25 for number of access point = 20 per mile)119

Hence,

FFS = 55 − 0.0 − 5.0 = 50 mph https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH004_eqn_0132.tif"/>

From Table 4.26 by interpolation between the values at opposing flow rates of 600 and 800 pcph, and using the data: no-passing zones = 20% ≤ 20%, opposing demand flow rate = 765 pcph and FFS = 50 mph, the adjustment factor (f_np) is calculated as shown below (see Table 4.27):120 121

⇒

f np = 2.5 − { ( 800 − 765 ) × ( 2.5 − 4.7 ) ( 800 − 600 ) } = 2.885 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH004_eqn_0133.tif"/>

Therefore:

PTSF = 85.4 + 2.885 ≅ 88.3 % https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH004_eqn_0134.tif"/>

ATS Computation:

A trial initial value for the flow rate in the peak direction can be computed using the following formula:

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-CH004_eqn_0135.tif"/>

v = 850 0.85 = 1000 pcph https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH004_eqn_0136.tif"/>

The values of f_G (adjustment factor for grade), E_T, and E_R are determined as shown in Table 4.28. The value of f_G is determined from Table 4.29 and the values of E_T and E_R are determined from Table 4.30.122

⇒

f HV = 1 1 + 0.06 ( 1.5 − 1 ) + 0.03 ( 1.1 − 1 ) = 0.968 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH004_eqn_0138.tif"/>

v = V ( PHF ) ( f G ) ( f HV ) https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH004_eqn_0139.tif"/>

⇒

v = 850 ( 0.85 ) ( 0.99 ) ( 0.968 ) ≅ 1044 pcph https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH004_eqn_0140.tif"/>

The opposing flow rate is computed using the procedure shown below:

A trial initial value for the opposing flow rate can be computed using the following formula:

v o = 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-CH004_eqn_0141.tif"/>

v o = 650 0.85 ≅ 765 pcph https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH004_eqn_0142.tif"/>

The values of f_G (adjustment factor for grade), E_T, and E_R are determined as shown in Table 4.31. The value of f_G is determined from Table 4.29 and the values of E_T and E_R are determined from Table 4.30.123

⇒

v o = V ( PHF ) ( f G ) ( f HV ) https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH004_eqn_0145.tif"/>

⇒

v o = 650 ( 0.85 ) ( 0.99 ) ( 0.968 ) ≅ 798 pcph https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH004_eqn_0146.tif"/>

ATS = FFS − 0.00776 ( v + v o ) − f np https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH004_eqn_0147.tif"/>

Where:

ATS = average travel speed in the analysis direction

FFS = free-flow speed in the analysis direction

v = passenger car equivalent flow rate for the peak 15-minute period in the analysis direction

v _o = passenger car equivalent flow rate for the peak 15-minute period in the opposing direction

f_np = adjustment factor for the percentage of no-passing zones in the analysis direction

From Table 4.32 by interpolation between the values at opposing flow rates of 600 and 800 pcph, and using the data: no-passing zones=20%≤20%, opposing demand flow rate=798 pcph and FFS=50 mph, the adjustment factor (f_np) is calculated as shown below (see Table 4.33):124 125

⇒

f np = 0.4 − { ( 800 − 765 ) × ( 0.4 − 0.6 ) ( 800 − 600 ) } = 0.402 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH004_eqn_0148.tif"/>

⇒

ATS = 50 − 0.00776 ( 1044 + 798 ) − 0.402 = 35.3 mph https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH004_eqn_0149.tif"/>

From Table 4.34 for Class I two-lane highways and using the PTSF value of 88.3% > 80%, and the ATS value of 35.3 mph ≤ 40 mph, the level of service (LOS) for the analysis direction of this highway is E.126

4.18

Solve Problem 4.17 but using a Class II two-lane highway with the same characteristics.

Solution:

For a Class II two-lane highway, only the PTSF is needed to determine the LOS.

Therefore, from Table 4.35 and using the PTSF value (that was computed in Problem 4.17) of 88.3% > 85%, the LOS for the peak direction of this highway is E.127

4.19

Determine the LOS of service for a Class I two-way two-lane highway with the following characteristics:

Traffic data:

PHF = 0.90

Volume = 1500 vph (two ways)

Directional split = 50/50

Trucks = 12%

RVs = 2%

BFFS = 60 mph

Geometric data:

Length of segment = 5 miles

Lane width = 12 ft

Shoulder width = 8 ft

40 access points per mile

No-passing zones = 40%

Level terrain

Solution:

To determine the level of service of the two-way two-lane highway segment classified as Class I, the percent time-spent following (PTSF) and the average travel speed (ATS) should be computed.

PTSF Computation:

A trial value for the 15-minute passenger car equivalent flow rate (v _p ) can be computed using the following formula:

v p = 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-CH004_eqn_0150.tif"/>

v p = 1500 0.90 ≅ 1667 pcph https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH004_eqn_0151.tif"/>

The values of f_G (adjustment factor for grade), E_T, and E_R are determined as shown in Table 4.36. The value of f_G is determined from Table 4.19 and the values of E_T and E_R are determined from Table 4.20.128

⇒

f HV = 1 1 + 0.12 ( 1.0 − 1 ) + 0.02 ( 1.0 − 1 ) = 1.00 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH004_eqn_0153.tif"/>

v p = V ( PHF ) ( f G ) ( f HV ) https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH004_eqn_0154.tif"/>

⇒

v p = 1500 ( 0.90 ) ( 1.00 ) ( 1.00 ) ≅ 1667 pcph https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH004_eqn_0155.tif"/>

The percent time-spent following (PTSF) is calculated using the following formula:

PTSF = BPTSF + f d / np https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH004_eqn_0156.tif"/>

Where:

f_d/np = adjustment factor for the combined effect of directional split and percentage of no-passing zones

The base percent time-spent-following (BPTSF) is computed using the following equation:

4.17 BPTSF = 100 [ 1 − e − 0.000879 v p ] https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH004_eqn_0157.tif"/>

Where:

v _p = the 15-minute passenger car equivalent flow rate

Therefore,

BPTSF = 100 [ 1 − e − 0.000879 ( 1667 ) ] = 76.9 % https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH004_eqn_0158.tif"/>

From Table 4.37 by interpolation between the values at two-way flow rates of 1400 and 2000 pcph, and using the data: no-passing zones = 40%, and the directional split of 50/50, the adjustment factor (f_d/np) is calculated as shown below (see Table 4.38):129 130

⇒

f d /np = 2.9 − { ( 2000 − 1667 ) × ( 2.9 − 5.5 ) ( 2000 − 1400 ) } = 4.343 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH004_eqn_0159.tif"/>

Therefore:

PTSF = 76.9 + 4.343 ≅ 81.2 % https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH004_eqn_0160.tif"/>

ATS Computation:

FFS = BFFS − f LS − f 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-CH004_eqn_0161.tif"/>

Where:

FFS = free-flow speed (mph)

BFFS = base free-flow speed (mph)

f_LS = adjustment factor for lane and shoulder widths

f_A = adjustment factor for number of access points

f_LS = 0.0 (from Table 4.24 for lane width = 12 ft ≥ 12 ft and shoulder width = 8 ft ≥ 6 ft)

f_A = 10.0 (from Table 4.25 for number of access point = 40 per mile)

Hence,

FFS = 60 − 0.0 − 10.0 = 50 mph ( 80.5 kph ) https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH004_eqn_0162.tif"/>

4.19 ATS = FFS − 0.00776 v p − f np https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH004_eqn_0163.tif"/>

Where:

ATS = average travel speed for both directions combined (mph)

FFS = free-flow speed (when volume < 200 pcph)

v _p = passenger car equivalent flow rate for the peak 15-minute period

f_np = adjustment factor for the percentage of no-passing zones

From Table 4.39 by interpolation between the values at two-way flow rates of 1400 and 2000 pcph, and using the data: no-passing zones = 40%, the adjustment factor (f_np) is calculated as shown below (see Table 4.40).131

⇒

f np = 0.7 − { ( 1800 − 1600 ) × ( 0.7 − 0.8 ) ( 1800 − 1667 ) } = 0.85 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH004_eqn_0164.tif"/>

⇒

ATS = 50 − 0.00776 ( 1667 ) − 0.85 = 36.2 mph ( 58.3 kph ) https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH004_eqn_0165.tif"/>

From Table 4.34 for Class I two-lane highways and using the PTSF value of 81.2% > 80%, and the ATS value of 36.2 mph ≤ 40 mph, the level of service (LOS) for the analysis direction of this highway is E.132

4.20

Determine the LOS for the peak direction of a Class I two-lane highway with the following characteristics:

Traffic data:

PHF = 0.90

Hourly volume = 900 vph (in the peak direction)

Opposing volume = 500 vph

Trucks = 4% (of which 10% are semi-trailers)

RVs = 2%

BFFS = 60 mph

Difference between FFS and truck crawl speed = 40 mph

Geometric data:

Length of segment = 2 miles

Upgrade = 4%

Lane width = 11 ft

Shoulder width = 5 ft

40 access points per mile

No-passing zones = 10%

Solution:

To determine the level of service of the peak direction on Class I two-lane highways, the percent time-spent following (PTSF) and the average travel speed (ATS) in the analysis direction should be computed.

The following rules apply:

Any segment having a grade ≥ 3% and a length ≥ 0.6 mile is considered a specific upgrade.

Segments in mountainous terrain are considered specific upgrades.

When a segment has multiple grades, a composite grade as a percentage is computed as the total change in elevation divided by the total length.

In this case, since the segment has a length of 2 miles ≥ 0.6 mile and a grade of 4% ≥ 3%, the segment is analyzed as a specific upgrade.

PTSF Computation:

A trial initial value for the flow rate in the peak direction can be computed using the following formula:

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-CH004_eqn_0166.tif"/>

v = 900 0.90 = 1000 pcph https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH004_eqn_0167.tif"/>

The values of f_G (adjustment factor for grade), E_T, and E_R are determined from Tables 4.41 and 4.42, respectively as shown in Table 4.43:133 134 135

⇒

f HV = 1 1 + 0.04 ( 1.0 − 1 ) + 0.02 ( 1.0 − 1 ) = 1.00 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH004_eqn_0169.tif"/>

v = V ( PHF ) ( f G ) ( f HV ) https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH004_eqn_0170.tif"/>

⇒

v = 900 ( 0.90 ) ( 0.98 ) ( 1.00 ) ≅ 1021 pcph https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH004_eqn_0171.tif"/>

The opposing flow rate is calculated using the procedure shown below:

A trial initial value for the opposing flow rate can be computed using the following formula:

v o = 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-CH004_eqn_0172.tif"/>

v o = 500 0.90 ≅ 556 pcph https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH004_eqn_0173.tif"/>

The values of f_G (adjustment factor for grade) for downgrades = 1.00

E_T and E_R for the downgrade direction is determined as if it is a directional segment in level terrain from Table 4.20, and E_TC is determined from Table 4.44 as shown in Table 4.45:136

4.20 f HV = 1 1 + P TC P T ( E TC − 1 ) + ( 1 − P TC ) P T ( E T − 1 ) + P R ( E R − 1 ) https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH004_eqn_0174.tif"/>

Where:

P_TC = percentage of trucks that travel at crawl speed from the total percentage of trucks

E_TC = passenger car equivalent for trucks traveling at crawl speed

⇒

f HV = 1 1 + 0.10 ( 0.04 ) ( 23.1 − 1 ) + ( 1 − 0.10 ) ( 0.04 ) ( 1.1 − 1 ) + 0.02 ( 1.0 − 1 ) https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH004_eqn_0175.tif"/>

f HV = 0.916 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH004_eqn_0176.tif"/>

v o = V ( PHF ) ( f G ) ( f HV ) https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH004_eqn_0177.tif"/> 137

⇒

v o = 500 ( 0.90 ) ( 1.00 ) ( 0.916 ) ≅ 607 pcph https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH004_eqn_0178.tif"/>

The percent time-spent following (PTSF) is calculated using the following formula:

PTSF = BPTSF + f np https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH004_eqn_0179.tif"/>

The base percent time-spent-following (BPTSF) is computed using the following equation:

BPTSF = 100 [ 1 − e a v 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-CH004_eqn_0180.tif"/>

From Table 4.23, the coefficients a and b for an opposing demand flow rate = 607 pcph are calculated using interpolation between the values of a and b at opposing flow rates of 600 and 800 pcph as shown below (see Table 4.46).

⇒ at 607 pcph,

a = − 0.173 − { ( 800 − 607 ) × ( − 0.173 − ( − 0.100 ) ) ( 800 − 600 ) } = − 0.103 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH004_eqn_0181.tif"/>

and

b = − 0.349 − { ( 800 − 607 ) × ( 0.349 − 0.413 ) ( 800 − 600 ) } = 0.411 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH004_eqn_0182.tif"/>

Therefore,

BPTSF = 100 [ 1 − e − 0.103 ( 1021 ) 0.411 ] = 83.1 % https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH004_eqn_0183.tif"/>

FFS = BFFS − f LS − f 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-CH004_eqn_0184.tif"/>

f_LS = 1.7 (from Table 4.24 for lane width = 11 ft > 11 ≤ 12 ft and shoulder width = 5 ft ≥ 4 < 6 ft)

f_A = 10.0 (from Table 4.25 for number of access point = 40 per mile)

Hence,

FFS = 60 − 1.7 − 10.0 = 48.3 mph ( 77.7 kph ) https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH004_eqn_0185.tif"/> 138

The adjustment factor (f_np) is determined using interpolation between opposing flow rates of 600 and 800 pcph for FFS = 50 mph, and again another interpolation between opposing flow rates 600 and 800 pcph for FFS = 45 mph. After that, a third interpolation is performed between FFS of 45 and 50 mph to calculate the value of at FFS = 48.3 mph.

From Table 4.26 by interpolation between the values at opposing flow rates of 600 and 800 pcph, and using the data: no-passing zones = 10% ≤ 20%, opposing demand flow rate = 607 pcph and FFS = 50 mph, the adjustment factor (f_np) is calculated as shown below (see Table 4.47):

⇒

f np = 2.5 − { ( 800 − 607 ) × ( 2.5 − 4.7 ) ( 800 − 600 ) } = 4.623 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH004_eqn_0186.tif"/>

And again, from Table 4.26 by interpolation between the values at opposing flow rates of 600 and 800 pcph and using the data: no-passing zones =10%≤20%, opposing demand flow rate =607 pcph and FFS =45 mph, the adjustment factor (f_np) is calculated as shown below (see Table 4.48). Then, by interpolation between the values at FFS of 45 mph and 50 mph, the adjustment factor (f_np) is calculated (see Table 4.49):139

⇒

f np = 2.3 − { ( 800 − 607 ) × ( 2.3 − 4.5 ) ( 800 − 600 ) } = 4.423 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH004_eqn_0187.tif"/>

⇒

f np = 4.623 − { ( 50 − 48.3 ) × ( 4.623 − 4.423 ) ( 50 − 45 ) } = 4.555 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH004_eqn_0188.tif"/>

Therefore:

PTSF = 83.1 + 4.555 ≅ 87.7 % https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH004_eqn_0189.tif"/>

ATS Computation:

A trial initial value for the flow rate in the peak direction can be computed using the following formula:

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-CH004_eqn_0190.tif"/>

v = 900 0.90 = 1000 pcph https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH004_eqn_0191.tif"/>

The values of f_G (adjustment factor for grade), E_T, and E_R are determined from Tables 4.50 through 4.52, respectively as shown in Table 4.53:140 141 142 143

⇒

f HV = 1 1 + 0.04 ( 8.1 − 1 ) + 0.02 ( 1.0 − 1 ) = 0.779 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH004_eqn_0193.tif"/>

v = V ( PHF ) ( f G ) ( f HV ) https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH004_eqn_0194.tif"/>

⇒

v = 900 ( 0.90 ) ( 1.00 ) ( 0.779 ) ≅ 1284 pcph https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH004_eqn_0195.tif"/>

The opposing flow rate is computed using the procedure shown below:

A trial initial value for the opposing flow rate can be computed using the following formula:

v o = 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-CH004_eqn_0196.tif"/>

v o = 500 0.90 ≅ 556 pcph https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH004_eqn_0197.tif"/>

The values of f_G (adjustment factor for grade) for downgrades = 1.00

E_T and E_R for the downgrade direction is determined as if it is a directional segment in level terrain from Table 4.20, and E_TC is determined from Table 4.44 as shown in Table 4.54:144

f HV = 1 1 + P TC P T ( E TC − 1 ) + ( 1 − P TC ) P T ( E T − 1 ) + P R ( E R − 1 ) https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH004_eqn_0198.tif"/>

⇒

f HV = 1 1 + 0.10 ( 0.04 ) ( 23.1 − 1 ) + ( 1 − 0.10 ) ( 0.04 ) ( 1.2 − 1 ) + 0.02 ( 1.0 − 1 ) https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH004_eqn_0199.tif"/>

f HV = 0.913 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH004_eqn_0200.tif"/>

v o = V ( PHF ) ( f G ) ( f HV ) https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH004_eqn_0201.tif"/>

⇒

v o = 500 ( 0.90 ) ( 1.00 ) ( 0.913 ) ≅ 609 pcph https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH004_eqn_0202.tif"/>

The free-flow speed (FFS) was computed above as follows:

From Table 4.39 by interpolation between the values at opposing flow rates of 600 and 800 pcph, and using the data: no-passing zones = 10% ≤ 20%, opposing demand flow rate = 609 pcph and FFS = 50 mph, the adjustment factor (f_np) is calculated as shown below (see Table 4.55):

⇒

f np = 0.4 − { ( 800 − 609 ) × ( 0.4 − 0.6 ) ( 800 − 600 ) } = 0.591 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH004_eqn_0205.tif"/> 145

And again, from Table 4.39 by interpolation between the values at opposing flow rates of 600 and 800 pcph and using the data: no-passing zones =10%≤20%, opposing demand flow rate =607 pcph and FFS =45 mph, the adjustment factor (f_np) is calculated as shown below (see Table 4.56). Then, by interpolation between the values at FFS of 45 mph and 50 mph, the adjustment factor (f_np) is calculated (see Table 4.57).

⇒

f np = 0.3 − { ( 800 − 609 ) × ( 0.3 − 0.4 ) ( 800 − 600 ) } = 0.396 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH004_eqn_0206.tif"/>

⇒

f np = 0.591 − { ( 50 − 48.3 ) × ( 0.591 − 0.396 ) ( 50 − 45 ) } = 0.525 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH004_eqn_0207.tif"/>

⇒

ATS = 48.3 − 0.00776 ( 1284 + 609 ) − 0.525 = 33.1 mph ( 53.3 kph ) https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH004_eqn_0209.tif"/> 146

From Table 4.34 for Class I two-lane highways and using the PTSF value of 87.7% > 80%, and the ATS value of 33.1 mph ≤ 40 mph, the level of service (LOS) for the analysis direction of this highway is E.

4.21

An urban 6-lane divided highway segment in level terrain has the following characteristics:

Traffic data:

Peak-hourly volume = 1800 vph

PHF = 0.95

Trucks = 12%

RVs = 2%

Free-flow speed (FFS) = 50 mph

Commuter traffic

Geometric data:

Length = 0.75 mile

Grade = 2%

Lane width = 11 ft

Shoulder width = 4 ft

Determine the level of service (LOS) for the upgrade direction.

Solution:

A highway segment having a grade ≥ 3% and length > 0.5 mile or a grade < 3% and a length > 1.0 mile is considered a specific grade.

Since the given highway segment has an upgrade of 2% < 3% and length of 0.75 mile < 1.0 mile ⇒ the segment is not considered a specific grade, but rather it is an extended general highway segment.

Therefore, Table 4.2 for extended general highway segments is used. E_T and E_R are determined as shown in Table 4.58:

⇒

f HV = 1 1 + 0.12 ( 1.5 − 1 ) + 0.02 ( 1.2 − 1 ) = 0.940 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH004_eqn_0211.tif"/>

The adjustment factor for drivers, f_p = 1.00 since traffic drivers are commuters.

v p = V ( PHF ) ( N ) ( f p ) ( f HV ) https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH004_eqn_0212.tif"/>

⇒

v p = 1800 ( 0.95 ) ( 2 ) ( 1.00 ) ( 0.940 ) https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH004_eqn_0213.tif"/>

v p ≅ 1008 pcphpl https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH004_eqn_0214.tif"/>

The average passenger car speed (S) can be determined using speed-flow curves with LOS criteria for multilane highways.

When the flow rate ≤ 1400 pcphpl, the average passenger car speed (S) is equal to the FFS; and when the flow rate > 1400 pcphpl, S is determined from the curves.

In this case; since v _p = 1008 pcphpl ≤ 1400 pcphpl,

S = FFS = 50 mph.

D = v p 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-CH004_eqn_0215.tif"/>

Where:

D = density (pc/mile/lane)

S = average passenger car speed (mph)

v_p = 15-minute service flow rate (pcphpl)

⇒

D = 1008 50 ≅ 21 pc/mi/lane https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH004_eqn_0216.tif"/>

From Table 4.59, for 18 ≤ D ≤ 26 pc/mi/lane, the LOS is C.148

4.22

Determine the level of service (LOS) for the upgrade direction (peak direction) of a rural 4-lane divided highway segment with the following characteristics:

Traffic data:

Average Annual Daily Traffic (AADT) = 22000 veh/day

K (proportion of the AADT in the peak hour) = 0.15

D (proportion of the peak-hourly traffic in the peak direction) = 0.60

PHF = 0.90

Trucks = 5%

RVs = 4%

Free-flow speed (FFS) = 55 mph

Commuter traffic149

Geometric data:

Rolling terrain

Length = 1.00 mile

Grade = 4%

Lane width = 12 ft

Shoulder width = 6 ft

Solution:

A highway segment having a grade ≥ 3% and length > 0.5 mile or a grade < 3% and a length > 1.0 mile is considered a specific grade.

Since the given highway segment has an upgrade of 4% ≥ 3% and length of 1.0 mile > 0.5 mile ⇒ the segment is considered a specific grade.

Therefore, Tables 4.13 and 4.14 for upgrades are used to determine E_T and E_R, respectively as shown in Table 4.60.

⇒

f HV = 1 1 + 0.05 ( 2.5 − 1 ) + 0.04 ( 2.5 − 1 ) = 0.881 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH004_eqn_0218.tif"/>

The adjustment factor for drivers, f_p = 1.00 since traffic drivers are commuters.

DDHV = AADT × K × D https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH004_eqn_0219.tif"/>

Where:

DDHV = directional design hourly volume

K = proportion of the AADT in the peak hour

D = proportion of the peak-hourly traffic in the peak direction

⇒

DDHV = 22000 × 0.15 × 0.60 = 1980 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-CH004_eqn_0220.tif"/> 150

v p = V ( PHF ) ( N ) ( f p ) ( f HV ) https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH004_eqn_0221.tif"/>

⇒

v p = 1980 ( 0.90 ) ( 2 ) ( 1.00 ) ( 0.881 ) https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH004_eqn_0222.tif"/>

v p ≅ 1249 pcphpl https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH004_eqn_0223.tif"/>

Since the flow rate = 1249 pcphpl ≤ 1400 pcphpl, the average passenger car speed (S) is equal to the FFS.

S = FFS = 55 mph.

D = v p 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-CH004_eqn_0224.tif"/>

⇒

D = 1249 55 ≅ 23 pc/mi/lane https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH004_eqn_0225.tif"/>

From Table 4.59, for 18 ≤ D ≤ 26 pc/mi/lane, the LOS is C.

Table 4.1 Traffic Count During the Peak Hour for Problem 4.1 Time Period Traffic Count 7:00–7:15am 800 7:15–7:30am 750 7:30–7:45am 710 7:45–8:00am 940 Figure 4.1 A screen image of the MS Excel worksheet used for the computations of Problem 4.1. https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/fig4_1_B.tif"/> Figure 4.2 A screen image of the MS Excel worksheet used for the computations of Problem 4.2. https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/fig4_2_B.tif"/> Figure 4.3 A screen image of the MS Excel worksheet used for the computations of Problem 4.3. https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/fig4_3_B.tif"/> Figure 4.4 A screen image of the MS Excel worksheet used for the computations of Problem 4.4. https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/fig4_4_B.tif"/> Table 4.2 Passenger Car Equivalents for Trucks (ET) and Recreational Vehicles (ER) on Extended General Multilane Highway Segments and Extended Freeway Segments Factor Type of Terrain Level Rolling Mountainous E_T (trucks and buses) 1.5 2.5 4.5 E_R (RVs) 1.2 2.0 4.0 Reproduced with permission from Highway Capacity Manual, 2000, Transportation Research Board, National Academy of Sciences, Courtesy of the National Academies Press, Washington, DC, USA. Figure 4.5 A screen image of the MS Excel worksheet used for the computation of the percentage of trucks/buses for Problem 4.5. https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/fig4_5_B.tif"/> Figure 4.6 A screen image of the MS Excel worksheet used for the computation of the percentage of trucks/buses for Problem 4.6. https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/fig4_6_B.tif"/> Figure 4.7 A screen image of the MS Excel worksheet used for the computation of the percentage of recreational vehicles for Problem 4.7. https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/fig4_7_B.tif"/> Figure 4.8 A screen image of the MS Excel worksheet used for determining the average grade of the two freeway segments for Problem 4.8. https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/fig4_8_B.tif"/> Figure 4.9 Performance curves for trucks (120 kg/kW = 200 lb/hp). Used with permission from <italic>Highway Capacity Manual</italic>, 2000, Transportation Research Board, National Academy of Sciences, Courtesy of the National Academies Press, Washington, DC, USA. https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/fig4_9_B.tif"/> Figure 4.10 Determining the equivalent grade from performance curves for trucks (120 kg/kW = 200 lb/hp). Used with permission from <italic>Highway Capacity Manual</italic>, 2000, Transportation Research Board, National Academy of Sciences, Courtesy of the National Academies Press, Washington, DC, USA. https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/fig4_10_B.tif"/> Table 4.3 Level of Service (LOS) Data for Problem 4.10 LOS B No. of Lanes, N 5 PHF 0.95 Trucks 8% Drivers Commuters Lane Width (ft) 12 Shoulder Width (10 ft) 0.0 Interchange Spacing (mi) 2/3 Table 4.4 Determining the Values of ET, ER, fLW, fLC, fN, and fID for Problem 4.10 Item Value Table Comment E_T 2.5 Table 4.2 Using “rolling terrain” E_R --- No RVs f_LW 0.0 Table 4.5 Using “lane width of 12 ft” f_LC 0.0 Table 4.6 Using “lanes in one direction ≥ 5 and right-shoulder lateral clearance of ≥ 6 ft (shoulder width 10 ft)” f_N 0.0 Table 4.7 Using “lanes in one direction ≥ 5” f_ID 5.0 Table 4.8 Using “interchanges per mile of 1.50” since the interchange spacing is 2/3 mile BFFS (mph) 75.0 For rural freeways Table 4.5 Adjustment for Lane Width (fLW) Lane Width (ft) Lane Width (m) Reduction in FFS, f_LW (mi/h) Reduction in FFS, f_LW (km/h) 12.0 3.6 0.0 0.0 11.5 3.5 0.6 1.0 11.2 3.4 1.3 2.1 11.0 3.3 1.9 3.1 10.5 3.2 3.5 5.6 10.2 3.1 5.0 8.1 10.0 3.0 6.6 10.6 Reproduced with permission from Highway Capacity Manual, 2000, Transportation Research Board, National Academy of Sciences, Courtesy of the National Academies Press, Washington, DC, USA. Table 4.6 Adjustment for Right-Shoulder Lateral Clearance (fLC) Right-Shoulder Lateral Clearance (ft) Right-Shoulder Lateral Clearance (m) Reduction in FFS, f_LC (mi/h) Reduction in FFS, f_LC (km/h) Lanes in One Direction Lanes in One Direction 2 3 4 ≥ 5 2 3 4 ≥ 5 ≥ 6 ≥ 1.8 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 5 1.5 0.6 0.4 0.2 0.1 1.0 0.7 0.3 0.2 4 1.2 1.2 0.8 0.4 0.2 1.9 1.3 0.7 0.4 3 0.9 1.8 1.2 0.6 0.3 2.9 1.9 1.0 0.6 2 0.6 2.4 1.6 0.8 0.4 3.9 2.6 1.3 0.8 1 0.3 3.0 2.0 1.0 0.5 4.8 3.2 1.6 1.1 0 0.0 3.6 2.4 1.2 0.6 5.8 3.9 1.9 1.3 Reproduced with permission from Highway Capacity Manual, 2000, Transportation Research Board, National Academy of Sciences, Courtesy of the National Academies Press, Washington, DC, USA. Table 4.7 Adjustment for Number of Lanes (fN) Number of Lanes Reduction in FFS, f_N (mi/h) Reduction in FFS, f_N (km/h) ≥ 5 0.0 0.0 4 1.5 2.4 3 3.0 4.8 2 4.5 7.3 Reproduced with permission from Highway Capacity Manual, 2000, Transportation Research Board, National Academy of Sciences, Courtesy of the National Academies Press, Washington, DC, USA. Table 4.8 Adjustment for Interchange Density (fID) Interchanges per Mile Interchanges per Kilometer Reduction in FFS, f_N (mi/h) Reduction in FFS, f_N (km/h) ≤ 0.50 ≤ 0.3 0.0 0.0 0.65 0.4 0.7 1.1 0.75 0.5 1.3 2.1 1.00 0.6 2.5 3.9 1.15 0.7 3.1 5.0 1.25 0.8 3.7 6.0 1.50 0.9 5.0 8.1 1.60 1.0 5.7 9.2 1.75 1.1 6.3 10.2 2.00 1.2 7.5 12.1 Reproduced with permission from Highway Capacity Manual, 2000, Transportation Research Board, National Academy of Sciences, Courtesy of the National Academies Press, Washington, DC, USA. Table 4.9 Results for Problem 4.10 Item Value Table f_HV 0.893 Computed f_p 1.00 Commuters FFS (mph) 75.0 Computed v _p (pcphpl) 1,210 Table 4.10 V (vph) 5,344 Computed Table 4.10 Level-of-Service (LOS) Criteria for Basic Freeway Segments Criteria LOS A B C D E FFS = 75 mi/h (120 km/h)* Maximum density (pc/mi/ln, pc/km/ln) 11 (7) 18 (11) 26 (16) 35 (22) 45 (28) Maximum speed (mi/h, km/h) 75 (120.0) 75 (120.0) 68.2 (114.6) 61.5 (99.6) 53.3 (85.7) Maximum v/c 0.35 0.55 0.77 0.92 1 Maximum service flow rate (pc/h/ln) 840 1320 1840 2200 2400 FFS = 70 mi/h (110 km/h) Maximum density (pc/mi/ln, pc/km/ln) 11 (7) 18 (11) 26 (16) 35 (22) 45 (28) Maximum speed (mi/h, km/h) 70.0 (110.0) 70.0 (110.0) 68.2 (108.5) 61.5 (97.2) 53.3 (83.9) Maximum v/c 0.33 0.51 0.74 0.91 1 Maximum service flow rate (pc/h/ln) 770 1210 1740 2135 2350 FFS = 65 mi/h (105 km/h) Maximum density (pc/mi/ln, pc/km/ln) 11 (7) 18 (11) 26 (16) 35 (22) 45 (28) Maximum speed (mi/h, km/h) 65.0 (105.0) 65.0 (105.0) 64.6 (104.0) 59.7 (96.1) 52.2 (84.0) Maximum v/c 0.30 0.50 0.71 0.89 1.00 Maximum service flow rate (pc/h/ln) 710 1170 1680 2090 2350 FFS = 60 mi/h (100 km/h) Maximum density (pc/mi/ln, pc/km/ln) 11 (7) 18 (11) 26 (16) 35 (22) 45 (28) Maximum speed (mi/h, km/h) 60.0 (100.0) 60.0 (100.0) 60.0 (100.0) 57.6 (93.8) 51.1 (82.1) Maximum v/c 0.30 0.48 0.70 0.90 1.00 Maximum service flow rate (pc/h/ln) 700 1100 1600 2065 2300 FFS = 55 mi/h (90 km/h) Maximum density (pc/mi/ln, pc/km/ln) 11 (7) 18 (11) 26 (16) 35 (22) 45 (28) Maximum speed (mi/h, km/h) 55.0 (90.0) 55.0 (90.0) 55.0 (90.0) 54.7 (89.1) 50.0 (80.4) Maximum v/c 0.28 0.44 0.64 0.87 1.00 Maximum service flow rate (pc/h/ln) 630 990 1440 1955 2250 *Number between brackets are in SI units. Reproduced with permission from Highway Capacity Manual, 2000, Transportation Research Board, National Academy of Sciences, Courtesy of the National Academies Press, Washington, DC, USA. Figure 4.11 A screen image of the MS Excel worksheet used to compute the maximum peak-hourly volume in Problem 4.10. https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/fig4_11_B.tif"/> Table 4.11 Traffic Count During the Peak Hour for Problem 4.11 Time Period Traffic Count 7:00–7:15am 600 7:15–7:30am 580 7:30–7:45am 630 7:45–8:00am 700 Table 4.12 Determining the ET and ER Values for a Specific Upgrade for Problem 4.11 Item Value Table Comment E_T 2.0 Table 4.13 Using 8% Trucks and G of 4% > 3–4%, and L = 0.30 mile > 0.25–0.50 mile E_R 2.5 Table 4.14 Using 4% RVs and G of 4% > 3–4%, and L = 0.30 mile > 0.25–0.50 mile Table 4.13 Passenger-Car Equivalents for Trucks and Buses (ET) on Upgrades, Multilane Highways, and Basic Freeway Segments Upgrade (%) Length (mi) Length (km) E_T Percentage of Trucks and Buses 2 4 5 6 8 10 15 20 25 < 2 All All 1.5 1.5 1.5 1.5 1.5 1.5 1.5 1.5 1.5 ≥ 2–3 > 0.00–0.25 > 0.0–0.4 1.5 1.5 1.5 1.5 1.5 1.5 1.5 1.5 1.5 > 0.25–0.50 > 0.4–0.8 1.5 1.5 1.5 1.5 1.5 1.5 1.5 1.5 1.5 > 0.50–0.75 > 0.8–1.2 1.5 1.5 1.5 1.5 1.5 1.5 1.5 1.5 1.5 > 0.75–1.00 > 1.2–1.6 2.0 2.0 2.0 2.0 1.5 1.5 1.5 1.5 1.5 > 1.00–1.50 > 1.6–2.4 2.5 2.5 2.5 2.5 2.0 2.0 2.0 2.0 2.0 > 1.50 > 2.4 3.0 3.0 2.5 2.5 2.0 2.0 2.0 2.0 2.0 > 3–4 > 0.00–0.25 > 0.0–0.4 1.5 1.5 1.5 1.5 1.5 1.5 1.5 1.5 1.5 > 0.25–0.50 > 0.4–0.8 2.0 2.0 2.0 2.0 2.0 2.0 1.5 1.5 1.5 > 0.50–0.75 > 0.8–1.2 2.5 2.5 2.0 2.0 2.0 2.0 2.0 2.0 2.0 > 0.75–1.00 > 1.2–1.6 3.0 3.0 2.5 2.5 2.5 2.5 2.0 2.0 2.0 > 1.00–1.50 > 1.6–2.4 3.5 3.5 3.0 3.0 3.0 3.0 2.5 2.5 2.5 > 1.50 > 2.4 4.0 3.5 3.0 3.0 3.0 3.0 2.5 2.5 2.5 > 4–5 > 0.00–0.25 > 0.0–0.4 1.5 1.5 1.5 1.5 1.5 1.5 1.5 1.5 1.5 > 0.25–0.50 > 0.4–0.8 3.0 2.5 2.5 2.5 2.0 2.0 2.0 2.0 2.0 > 0.50–0.75 > 0.8–1.2 3.5 3.0 3.0 3.0 2.5 2.5 2.5 2.5 2.5 > 0.75–1.00 > 1.2–1.6 4.0 3.5 3.5 3.5 3.0 3.0 3.0 3.0 3.0 > 1.00 > 1.6 5.0 4.0 4.0 4.0 3.5 3.5 3.0 3.0 3.0 > 5–6 > 0.00–0.25 > 0.0–0.4 2.0 2.0 1.5 1.5 1.5 1.5 1.5 1.5 1.5 > 0.25–0.30 > 0.4–0.5 4.0 3.0 2.5 2.5 2.0 2.0 2.0 2.0 2.0 > 0.30–0.50 > 0.5–0.8 4.5 4.0 3.5 3.0 2.5 2.5 2.5 2.5 2.5 > 0.50–0.75 > 0.8–1.2 5.0 4.5 4.0 3.5 3.0 3.0 3.0 3.0 3.0 > 0.75–1.00 > 1.2–1.6 5.5 5.0 4.5 4.0 3.0 3.0 3.0 3.0 3.0 > 1.00 > 1.6 6.0 5.0 5.0 4.5 3.5 3.5 3.5 3.5 3.5 > 6 > 0.00–0.25 > 0.0–0.4 4.0 3.0 2.5 2.5 2.5 2.5 2.0 2.0 2.0 > 0.25–0.30 > 0.4–0.5 4.5 4.0 3.5 3.5 3.5 3.0 2.5 2.5 2.5 > 0.30–0.50 > 0.5–0.8 5.0 4.5 4.0 4.0 3.5 3.0 2.5 2.5 2.5 > 0.50–0.75 > 0.8–1.2 5.5 5.0 4.5 4.5 4.0 3.5 3.0 3.0 3.0 > 0.75–1.00 > 1.2–1.6 6.0 5.5 5.0 5.0 4.5 4.0 3.5 3.5 3.5 > 1.00 > 1.6 7.0 6.0 5.5 5.5 5.0 4.5 4.0 4.0 4.0 Reproduced with permission from Highway Capacity Manual, 2000, Transportation Research Board, National Academy of Sciences, Courtesy of the National Academies Press, Washington, DC, USA. Table 4.14 Passenger-Car Equivalents for Recreational Vehicles (ER) on Uniform Upgrades, Multilane Highways, and Basic Freeway Segments Upgrade (%) Length (mi) Length (km) E_R Percentage of RVs 2 4 5 6 8 10 15 20 25 ≤ 2 All All 1.2 1.2 1.2 1.2 1.2 1.2 1.2 1.2 1.2 > 2–3 > 0.00–0.50 0.0–0.8 1.2 1.2 1.2 1.2 1.2 1.2 1.2 1.2 1.2 > 0.50 > 0.8 3.0 1.5 1.5 1.5 1.5 1.5 1.2 1.2 1.2 > 3–4 > 0.00–0.25 0.0–0.4 1.2 1.2 1.2 1.2 1.2 1.2 1.2 1.2 1.2 > 0.25–0.50 > 0.4–0.8 2.5 2.5 2.0 2.0 2.0 2.0 1.5 1.5 1.5 > 0.50 > 0.8 3.0 2.5 2.5 2.5 2.0 2.0 2.0 1.5 1.5 > 4–5 > 0.00–0.25 > 0.0–0.4 2.5 2.0 2.0 2.0 1.5 1.5 1.5 1.5 1.5 > 0.25–0.50 > 0.4–0.8 4.0 3.0 3.0 3.0 2.5 2.5 2.0 2.0 2.0 > 0.50 > 0.8 4.5 3.5 3.0 3.0 3.0 2.5 2.5 2.0 2.0 > 5 > 0.00–0.25 > 0.0–0.4 4.0 3.0 2.5 2.5 2.5 2.0 2.0 2.0 1.5 > 0.25–0.50 > 0.4–0.8 6.0 4.0 4.0 3.5 3.0 3.0 2.5 2.5 2.0 > 0.50 > 0.8 6.0 4.5 4.0 4.5 3.5 3.0 3.0 2.5 2.0 Reproduced with permission from Highway Capacity Manual, 2000, Transportation Research Board, National Academy of Sciences, Courtesy of the National Academies Press, Washington, DC, USA. Figure 4.12 A screen image of the MS Excel worksheet used to compute the 15-minute passenger car equivalent flow rate for Problem 4.11. https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/fig4_12_B.tif"/> Figure 4.13 A screen image of the MS Excel worksheet used to compute the reduction in the 15-minute hourly flow rate for Problem 4.12. https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/fig4_13_B.tif"/> Figure 4.14 A screen image of the MS Excel worksheet used to compute the required number of lanes for Problem 4.13. https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/fig4_14_B.tif"/> Figure 4.15 A screen image of the MS Excel worksheet used to compute the free flow speed on the freeway for Problem 4.14. https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/fig4_15_B.tif"/> Table 4.15 Determining the ET Value for a Specific Upgrade for Problem 4.15 Item Value Table Comment E_T 2.5 Table 4.13 Using 5% Trucks and G of 3% ≥ 2–3%, and L of 2.0 miles > 1.50 mile Table 4.16 Determining the ET and ER Values for a Specific Upgrade for Problem 4.16 Item Value Table Comment E_T 2.0 Table 4.13 Using 8% Trucks and G of 2% ≥ 2–3%, and L of 2.0 miles > 1.50 mile E_R 1.2 Table 4.14 Using 5% RVs and G of 2% ≤ 2%, and L of 2.0 miles (“All” in the table) Table 4.17 Determining the ET Value for a Specific Upgrade for Problem 4.16 Item Value Table Comment E_T 3.0 Table 4.13 Using 4% Trucks and G of 2% ≥ 2–3%, and L of 2.0 miles > 1.50 mile Table 4.18 Determining fG, ET, and ER for the Peak Direction (PTSF Method) for Problem 4.17 Item Value Table Comment f_G 1.00 Table 4.19 Using directional flow rate = 1000 > 600 pcph, and rolling terrain E_T 1.0 Table 4.20 Using directional flow rate = 1000 > 600 pcph, and rolling terrain E_R 1.0 Table 4.20 Using directional flow rate = 1000 > 600 pcph, and rolling terrain Table 4.19 Grade Adjustment Factor (fG) to Determine Percent Time-Spent-Following on Two-Way and Directional Segments Range of Two-Way Flow Rates (pc/h) Range of Directional Flow Rates (pc/h) Type of Terrain Level Rolling 0–600 0–300 1.00 0.77 > 600–1200 > 300–600 1.00 0.94 > 1200 > 600 1.00 1.00 Reproduced with permission from Highway Capacity Manual, 2000, Transportation Research Board, National Academy of Sciences, Courtesy of the National Academies Press, Washington, DC, USA. Table 4.20 Passenger-Car Equivalents for Trucks (ET) and Recreational Vehicles (ER) to Determine Percent Time-Spent Following on Two-Way and Directional Segments Vehicle Type Range of Two-Way Flow Rates (pc/h) Range of Directional Flow Rates (pc/h) Type of Terrain Level Rolling Trucks, E_T 0–600 0–300 1.1 1.8 > 600–1200 > 300–600 1.1 1.5 > 1200 > 600 1.0 1.0 RVs, E_R 0–600 0–300 1.0 1.0 > 600–1200 > 300–600 1.0 1.0 > 1200 > 600 1.0 1.0 Reproduced with permission from Highway Capacity Manual, 2000, Transportation Research Board, National Academy of Sciences, Courtesy of the National Academies Press, Washington, DC, USA. Table 4.21 Determining fG, ET, and ER for the Opposing Direction (PTSF Method) for Problem 4.17 Item Value Table Comment f_G 1.00 Table 4.18 Using directional opposing flow rate = 765 > 600 pcph, and rolling terrain E_T 1.0 Table 4.19 Using directional opposing flow rate = 765 > 600 pcph, and rolling terrain E_R 1.0 Table 4.19 Using directional opposing flow rate = 765 > 600 pcph, and rolling terrain Table 4.22 Tabulating the Values of a and b for the Opposing Direction for Problem 4.17 Opposing Flow Rate (pcph) a b 600 −0.100 0.413 765 ? ? 800 −0.173 0.349 Table 4.23 Values of Coefficients (a, b) Used in Estimating Percent Time-Spent-Following for Directional Segments Opposing Demand Flow Rate (pc/h) a b ≤ 200 −0.013 0.668 400 −0.057 0.479 600 −0.100 0.413 800 −0.173 0.349 1000 −0.320 0.276 1200 −0.430 0.242 1400 −0.522 0.225 ≥ 1600 −0.665 0.199 Reproduced with permission from Highway Capacity Manual, 2000, Transportation Research Board, National Academy of Sciences, Courtesy of the National Academies Press, Washington, DC, USA. Table 4.24 Adjustment for Lane Width and Shoulder Width (fLS) Lane Width (ft) Lane Width (m) Reduction in FFS (mi/h, km/h)* Shoulder Width ≥ 0 – < 2 (ft) ≥ 2 − < 4 (ft) ≥ 4 − < 6 (ft) ≥ 6 (ft) ≥ 0.0 − < 0.6 (m) ≥ 0.6 − < 1.2 (m) ≥ 1.2 − < 1.8 (m) ≥ 1.8 (m) 9 − < 10 2.7 − < 3.0 6.4 (10.3) 4.8 (7.7) 3.5 (5.6) 2.2 (3.5) ≥ 10 − < 11 ≥ 3.0 − < 3.3 5.3 (8.5) 3.7 (5.9) 2.4 (3.8) 1.1 (1.7) ≥ 11 − < 12 ≥ 3.3 − < 3.6 4.7 (7.5) 3.0 (4.9) 1.7 (2.8) 0.4 (0.7) ≥ 12 ≥ 3.6 4.2 (6.8) 2.6 (4.2) 1.3 (2.1) 0.0 (0.0) * Number between brackets are in SI units. Reproduced with permission from Highway Capacity Manual, 2000, Transportation Research Board, National Academy of Sciences, Courtesy of the National Academies Press, Washington, DC, USA. Table 4.25 Adjustment for Access-Point Density (fA) Access Points per Mile Access Points per Kilometer Reduction in FFS (mi/h) Reduction in FFS (km/h) 0 0 0.0 0.0 10 6 2.5 4.0 20 12 5.0 8.0 30 18 7.5 12.0 ≥ 40 ≥ 24 10.0 16.0 Reproduced with permission from Highway Capacity Manual, 2000, Transportation Research Board, National Academy of Sciences, Courtesy of the National Academies Press, Washington, DC, USA. Table 4.26 Adjustment for No-Passing Zones (fnp) to Percent Time-Spent Following in Directional Segments Opposing Demand Flow Rate (pc/h) No Passing Zones (%) ≤ 20 40 60 80 100 FFS = 65 mi/h (110 km/h) ≤ 100 10.1 17.2 20.2 21.0 21.8 200 12.4 19.0 22.7 23.8 24.8 400 9.0 12.3 14.1 14.4 15.4 600 5.3 7.7 9.2 9.7 10.4 800 3.0 4.6 5.7 6.2 6.7 1000 1.8 2.9 3.7 4.1 4.4 1200 1.3 2.0 2.6 2.9 3.1 1400 0.9 1.4 1.7 1.9 2.1 ≥ 1600 0.7 0.9 1.1 1.2 1.4 FFS = 60 mi/h (100 km/h) ≤ 100 8.4 14.9 20.9 22.8 26.6 200 11.5 18.2 24.1 26.2 29.7 400 8.6 12.1 14.8 15.9 18.1 600 5.1 7.5 9.6 10.6 12.1 800 2.8 4.5 5.9 6.7 7.7 1000 1.6 2.8 3.7 4.3 4.9 1200 1.2 1.9 2.6 3.0 3.4 1400 0.8 1.3 1.7 2.0 2.3 ≥ 1600 0.6 0.9 1.1 1.2 1.5 FFS = 55 mi/h (90 km/h) ≤ 100 6.7 12.7 21.7 24.5 31.3 200 10.5 17.5 25.4 28.6 34.7 400 8.3 11.8 15.5 17.5 20.7 600 4.9 7.3 10.0 11.5 13.9 800 2.7 4.3 6.1 7.2 8.8 1000 1.5 2.7 3.8 4.5 5.4 1200 1.0 1.8 2.6 3.1 3.8 1400 0.7 1.2 1.7 2.0 2.4 ≥ 1600 0.6 0.9 1.2 1.3 1.5 FFS = 50 mi/h (80 km/h) ≤ 100 5.0 10.4 22.4 26.3 36.1 200 9.6 16.7 26.8 31.0 39.6 400 7.9 11.6 16.2 19.0 23.4 600 4.7 7.1 10.4 12.4 15.6 800 2.5 4.2 6.3 7.7 9.8 1000 1.3 2.6 3.8 4.7 5.9 1200 0.9 1.7 2.6 3.2 4.1 1400 0.6 1.1 1.7 2.1 2.6 ≥ 1600 0.5 0.9 1.2 1.3 1.6 FFS = 45 mi/h (70 km/h) ≤ 100 3.7 8.5 23.2 28.2 41.6 200 8.7 16.0 28.2 33.6 45.2 400 7.5 11.4 16.9 20.7 26.4 600 4.5 6.9 10.8 13.4 17.6 800 2.3 4.1 6.5 8.2 11.0 1000 1.2 2.5 3.8 4.9 6.4 1200 0.8 1.6 2.6 3.3 4.5 1400 0.5 1.0 1.7 2.2 2.8 ≥ 1600 0.4 0.9 1.2 1.3 1.7 Reproduced with permission from Highway Capacity Manual, 2000, Transportation Research Board, National Academy of Sciences, Courtesy of the National Academies Press, Washington, DC, USA. Table 4.27 Tabulating the Values of fnp for the Opposing Direction (PTSF Method) for Problem 4.17 Opposing Flow Rate (pcph) f_np 600 4.7 765 ? 800 2.5 Table 4.28 Determining the Values of fG, ET, and ER for the Peak Direction (ATS Method) for Problem 4.17 Item Value Table Comment f_G 0.99 Table 4.29 Using directional flow rate = 1000 > 600 pcph, and rolling terrain E_T 1.5 Table 4.30 Using directional flow rate = 1000 > 600 pcph, and rolling terrain E_R 1.1 Table 4.30 Using directional flow rate = 1000 > 600 pcph, and rolling terrain Table 4.29 Grade Adjustment Factor (fG) to Determine Speeds on Two-Way and Directional Segments Range of Two-Way Flow Rates (pc/h) Range of Directional Flow Rates (pc/h) Type of Terrain Level Rolling 0–600 0–300 1.00 0.71 > 600–1200 > 300–600 1.00 0.93 > 1200 > 600 1.00 0.99 Reproduced with permission from Highway Capacity Manual, 2000, Transportation Research Board, National Academy of Sciences, Courtesy of the National Academies Press, Washington, DC, USA. Table 4.30 Passenger-Car Equivalents for Trucks (ET) and Recreational Vehicles (ER) to Determine Speeds on Two-Way and Directional Segments Vehicle Type Range of Two-Way Flow Rates (pc/h) Range of Directional Flow Rates (pc/h) Type of Terrain Level Rolling Trucks, E_T 0–600 0–300 1.7 2.5 > 600–1200 > 300–600 1.2 1.9 > 1200 > 600 1.1 1.5 RVs, E_R 0–600 0–300 1.0 1.1 > 600–1200 > 300–600 1.0 1.1 > 1200 > 600 1.0 1.1 Reproduced with permission from Highway Capacity Manual, 2000, Transportation Research Board, National Academy of Sciences, Courtesy of the National Academies Press, Washington, DC, USA. Table 4.31 Determining the Values of fG, ET, and ER for the Opposing Direction (ATS Method) for Problem 4.17 Item Value Table Comment f_G 0.99 Table 4.29 Using directional opposing flow rate = 765 > 600 pcph, and rolling terrain E_T 1.5 Table 4.30 Using directional opposing flow rate = 765 > 600 pcph, and rolling terrain E_R 1.1 Table 4.30 Using directional flow rate = 1000 > 600 pcph, and rolling terrain Table 4.32 Adjustment for No-Passing Zones (fnp) to Average Travel Speed in Directional Segments Opposing Demand Flow Rate (pc/h) No Passing Zones (%) ≤ 20 40 60 80 100 FFS = 65 mi/h (110 km/h)* ≤ 100 1.1 (1.7) 2.2 (3.5) 2.8 (4.5) 3.0 (4.8) 3.1 (5.0) 200 2.2 (3.5) 3.3 (5.3) 3.9 (6.2) 4.0 (6.5) 4.2 (6.8) 400 1.6 (2.6) 2.3 (3.7) 2.7 (4.4) 2.8 (4.5) 2.9 (4.7) 600 1.4 (2.2) 1.5 (2.4) 1.7 (2.8) 1.9 (3.1) 2.0 (3.3) 800 0.7 (1.1) 1.0 (1.6) 1.2 (2.0) 1.4 (2.2) 1.5 (2.4) 1000 0.6 (1.0) 0.8 (1.3) 1.1 (1.7) 1.1 (1.8) 1.2 (1.9) 1200 0.6 (0.9) 0.8 (1.3) 0.9 (1.5) 1.0 (1.6) 1.1 (1.7) 1400 0.6 (0.9) 0.7 (1.2) 0.9 (1.4) 0.9 (1.4) 0.9 (1.5) ≥ 1600 0.6 (0.9) 0.7 (1.1) 0.7 (1.2) 0.7 (1.2) 0.8 (1.3) FFS = 60 mi/h (100 km/h) ≤ 100 0.7 (1.2) 1.7 (2.7) 2.5 (4.0) 2.8 (4.5) 2.9 (4.7) 200 1.9 (3.0) 2.9 (4.6) 3.7 (5.9) 4.0 (6.4) 4.2 (6.7) 400 1.4 (2.3) 2.0 (3.3) 2.5 (4.1) 2.7 (4.4) 2.9 (4.6) 600 1.1 (1.8) 1.3 (2.1) 1.6 (2.6) 1.9 (3.0) 2.0 (3.2) 800 0.6 (0.9) 0.9 (1.4) 1.1 (1.8) 1.3 (2.1) 1.4 (2.3) 1000 0.6 (0.9) 0.7 (1.1) 0.9 (1.5) 1.1 (1.7) 1.2 (1.9) 1200 0.5 (0.8) 0.7 (1.1) 0.9 (1.4) 0.9 (1.5) 1.1 (1.7) 1400 0.5 (0.8) 0.6 (1.0) 0.8 (1.3) 0.8 (1.3) 0.9 (1.4) ≥ 1600 0.5 (0.8) 0.6 (1.0) 0.7 (1.1) 0.7 (1.1) 0.7 (1.2) FFS = 55 mi/h (90 km/h) ≤ 100 0.5 (0.8) 1.2 (1.9) 2.2 (3.6) 2.6 (4.2) 2.7 (4.4) 200 1.5 (2.4) 2.4 (3.9) 3.5 (5.6) 3.9 (6.3) 4.1 (6.6) 400 1.3 (2.1) 1.9 (3.0) 2.4 (3.8) 2.7 (4.3) 2.8 (4.5) 600 0.9 (1.4) 1.1 (1.8) 1.6 (2.5) 1.8 (2.9) 1.9 (3.1) 800 0.5 (0.8) 0.7 (1.1) 1.1 (1.7) 1.2 (2.0) 1.4 (2.2) 1000 0.5 (0.8) 0.6 (0.9) 0.8 (1.3) 0.9 (1.5) 1.1 (1.8) 1200 0.5 (0.8) 0.6 (0.9) 0.7 (1.2) 0.9 (1.4) 1.0 (1.6) 1400 0.5 (0.8) 0.6 (0.9) 0.7 (1.1) 0.7 (1.2) 0.9 (1.4) ≥ 1600 0.5 (0.8) 0.5 (0.8) 0.6 (0.9) 0.6 (0.9) 0.7 (1.1) FFS = 50 mi/h (80 km/h) ≤ 100 0.2 (0.3) 0.7 (1.1) 1.9 (3.1) 2.4 (3.9) 2.5 (4.1) 200 1.2 (1.9) 2.0 (3.2) 3.3 (5.3) 3.9 (6.2) 4.0 (6.5) 400 1.1 (1.8) 1.6 (2.6) 2.2 (3.5) 2.6 (4.2) 2.7 (4.4) 600 0.6 (1.0) 0.9 (1.5) 1.4 (2.3) 1.7 (2.8) 1.9 (3.0) 800 0.4 (0.6) 0.6 (0.9) 0.9 (1.5) 1.2 (1.9) 1.3 (2.1) 1000 0.4 (0.6) 0.4 (0.7) 0.7 (1.1) 0.9 (1.4) 1.1 (1.8) 1200 0.4 (0.6) 0.4 (0.7) 0.7 (1.1) 0.8 (1.3) 1.0 (1.6) 1400 0.4 (0.6) 0.4 (0.7) 0.6 (1.0) 0.7 (1.1) 0.8 (1.3) ≥ 1600 0.4 (0.6) 0.4 (0.7) 0.5 (0.8) 0.5 (0.8) 0.6 (1.0) FFS = 45 mi/h (70 km/h) ≤ 100 0.1 (0.1) 0.4 (0.6) 1.7 (2.7) 2.2 (3.6) 2.4 (3.8) 200 0.9 (1.5) 1.6 (2.6) 3.1 (5.0) 3.8 (6.1) 4.0 (6.4) 400 0.9 (1.5) 0.5 (0.8) 2.0 (3.2) 2.5 (4.1) 2.7 (4.3) 600 0.4 (0.7) 0.3 (0.5) 1.3 (2.1) 1.7 (2.7) 1.8 (2.9) 800 0.3 (0.5) 0.3 (0.5) 0.8 (1.3) 1.1 (1.8) 1.2 (2.0) 1000 0.3 (0.5) 0.3 (0.5) 0.6 (1.0) 0.8 (1.3) 1.1 (1.8) 1200 0.3 (0.5) 0.3 (0.5) 0.6 (1.0) 0.7 (1.2) 1.0 (1.6) 1400 0.3 (0.5) 0.3 (1.5) 0.6 (1.0) 0.6 (1.0) 0.7 (1.2) ≥ 1600 0.3 (0.5) 0.3 (0.5) 0.4 (0.7) 0.4 (0.7) 0.6 (0.9) *Number between brackets are in SI units. Reproduced with permission from Highway Capacity Manual, 2000, Transportation Research Board, National Academy of Sciences, Courtesy of the National Academies Press, Washington, DC, USA. Table 4.33 Tabulating the Values of fnp for the Opposing Direction (ATS Method) for Problem 4.17 Opposing Flow Rate (pcph) f_np 600 0.6 798 ? 800 0.4 Table 4.34 Level-of-Service (LOS) Criteria for Two-Lane Highways in Class I LOS Percent Time-Spent-Following Average Travel Speed (mi/h) Average Travel Speed (km/h) A ≤ 35 > 55 > 90 B > 35–50 > 50–55 > 80–90 C > 50–65 > 45–50 > 70–80 D > 65–80 > 40–45 > 60–70 E > 80 ≤ 40 ≤ 60 Note:LOS F applies whenever the flow rate exceeds the segment capacity. Reproduced with permission from Highway Capacity Manual, 2000, Transportation Research Board, National Academy of Sciences, Courtesy of the National Academies Press, Washington, DC, USA. Table 4.35 Level-of-Service (LOS) Criteria for Two-Lane Highways in Class II LOS Percent Time-Spent-Following A ≤ 40 B > 40–55 C > 55–70 D > 70–85 E > 85 Note:LOS F applies whenever the flow rate exceeds the segment capacity. Reproduced with permission from Highway Capacity Manual, 2000, Transportation Research Board, National Academy of Sciences, Courtesy of the National Academies Press, Washington, DC, USA. Table 4.36 Determining fG, ET, and ER for the Peak Direction (PTSF Method) for Problem 4.19 Item Value Table Comment f_G 1.00 Table 4.19 Using two-way flow rate = 1667 > 1200 pcph, and rolling terrain E_T 1.0 Table 4.20 Using two-way flow rate = 1667 > 1200 pcph, and rolling terrain E_R 1.0 Table 4.20 Using two-way flow rate = 1667 > 1200 pcph, and rolling terrain Table 4.37 Adjustment for Combined Effect of Directional Distribution of Traffic and Percentage of No-Passing Zones (fd/np) on Percent Time-Spent Following on Two-Way Segments Two-Way Flow Rate, v _p (pc/h) Increase in Percent Time-Spent-Following (%) No Passing Zones (%) 0 20 40 60 80 100 Directional Split = 50/50 ≤ 200 0.0 10.1 17.2 20.2 21.0 21.8 400 0.0 12.4 19.0 22.7 23.8 24.8 600 0.0 11.2 16.0 18.7 19.7 20.5 800 0.0 9.0 12.3 14.1 14.5 15.4 1400 0.0 3.6 5.5 6.7 7.3 7.9 2000 0.0 1.8 2.9 3.7 4.1 4.4 2600 0.0 1.1 1.6 2.0 2.3 2.4 3200 0.0 0.7 0.9 1.1 1.2 1.4 Directional Split = 60/40 ≤ 200 1.6 11.8 17.2 22.5 23.1 23.7 400 0.5 11.7 16.2 20.7 21.5 22.2 600 0.0 11.5 15.2 18.9 19.8 20.7 800 0.0 7.6 10.3 13.0 13.7 14.4 1400 0.0 3.7 5.4 7.1 7.6 8.1 2000 0.0 2.3 3.4 3.6 4.0 4.3 ≥ 2600 0.0 0.9 1.4 1.9 2.1 2.2 Directional Split = 70/30 ≤ 200 2.8 13.4 19.1 24.8 25.2 25.5 400 1.1 12.5 17.3 22.0 22.6 23.2 600 0.0 11.6 15.4 19.1 20.0 20.9 800 0.0 7.7 10.5 13.3 14.0 14.6 1400 0.0 3.8 5.6 7.4 7.9 8.3 ≥ 2000 0.0 1.4 4.9 3.5 3.9 4.2 Directional Split = 80/20 ≤ 200 5.1 17.5 24.3 31.0 31.3 31.6 400 2.5 15.8 21.5 27.1 27.6 28.0 600 0.0 14.0 18.6 23.2 23.9 24.5 800 0.0 9.3 12.7 16.0 16.5 17.0 1400 0.0 4.6 6.7 8.7 9.1 9.5 ≥ 2000 0.0 2.4 3.4 4.5 4.7 4.9 Directional Split = 90/10 ≤ 200 5.6 21.6 29.4 37.2 37.4 37.6 400 2.4 19.0 25.6 32.2 32.5 32.8 600 0.0 16.3 21.8 27.2 27.6 28.0 800 0.0 10.9 14.8 18.6 19.0 19.4 ≥ 1400 0.0 5.5 7.8 10.0 10.4 10.7 Reproduced with permission from Highway Capacity Manual, 2000, Transportation Research Board, National Academy of Sciences, Courtesy of the National Academies Press, Washington, DC, USA. Table 4.38 Tabulating the Values of fd/np for Two-Way Flow Rates (PTSF Method) for Problem 4.19 Opposing Flow Rate (pcph) f_d/np 1400 5.5 1667 ? 2000 2.9 Table 4.39 Adjustment for Effect of No-Passing Zones (fnp) on Average Travel Speed on Two-Way Segments Two-Way Demand Flow Rate, v _p (pc/h) Reduction in Average Travel Speed (mi/h, km/h) No Passing Zones (%) 0 20 40 60 80 100 0 0.0 (0.0)* 0.0 (0.0) 0.0 (0.0) 0.0 (0.0) 0.0 (0.0) 0.0 (0.0) 200 0.0 (0.0) 0.6 (1.0) 1.4 (2.3) 2.4 (3.8) 2.6 (4.2) 3.5 (5.6) 400 0.0 (0.0) 1.7 (2.7) 2.7 (4.3) 3.5 (5.7) 3.9 (6.3) 4.5 (7.3) 600 0.0 (0.0) 1.6 (2.5) 2.4 (3.8) 3.0 (4.9) 3.4 (5.5) 3.9 (6.2) 800 0.0 (0.0) 1.4 (2.2) 1.9 (3.1) 2.4 (3.9) 2.7 (4.3) 3.0 (4.9) 1000 0.0 (0.0) 1.1 (1.8) 1.6 (2.5) 2.0 (3.2) 2.2 (3.6) 2.6 (4.2) 1200 0.0 (0.0) 0.8 (1.3) 1.2 (2.0) 1.6 (2.6) 1.9 (3.0) 2.1 (3.4) 1400 0.0 (0.0) 0.6 (0.9) 0.9 (1.4) 1.2 (1.9) 1.4 (2.3) 1.7 (2.7) 1600 0.0 (0.0) 0.6 (0.9) 0.8 (1.3) 1.1 (1.7) 1.3 (2.1) 1.5 (2.4) 1800 0.0 (0.0) 0.5 (0.8) 0.7 (1.1) 1.0 (1.6) 1.1 (1.8) 1.3 (2.1) 2000 0.0 (0.0) 0.5 (0.8) 0.6 (1.0) 0.9 (1.4) 1.0 (1.6) 1.1 (1.8) 2200 0.0 (0.0) 0.5 (0.8) 0.6 (1.0) 0.9 (1.4) 0.9 (1.5) 1.1 (1.7) 2400 0.0 (0.0) 0.5 (0.8) 0.6 (1.0) 0.8 (1.3) 0.9 (1.5) 1.1 (1.7) 2600 0.0 (0.0) 0.5 (0.8) 0.6 (1.0) 0.8 (1.3) 0.9 (1.4) 1.0 (1.6) 2800 0.0 (0.0) 0.5 (0.8) 0.6 (1.0) 0.7 (1.2) 0.8 (1.3) 0.9 (1.4) 3000 0.0 (0.0) 0.5 (0.8) 0.6 (0.9) 0.7 (1.1) 0.7 (1.1) 0.8 (1.3) 3200 0.0 (0.0) 0.5 (0.8) 0.6 (0.9) 0.6 (1.0) 0.6 (1.0) 0.7 (1.1) *Numbers between brackets are in SI units. Reproduced with permission from Highway Capacity Manual, 2000, Transportation Research Board, National Academy of Sciences, Courtesy of the National Academies Press, Washington, DC, USA. Table 4.40 Tabulating the Values of fd/np for Two-Way Flow Rates (ATS Method) for Problem 4.19 Two-Way Flow Rate (pcph) f_np 1600 0.8 1667 ? 1800 0.7 Table 4.41 Grade Adjustment Factor (fG) for Estimating Percent Time-Spent-Following on Specific Upgrades Grade (%) Length of Grade (mi) Length of Grade (km) Grade Adjustment Factor (f_G) Range of Directional Flow Rates (pc/h) 0–300 > 300–600 > 600 ≥ 3 < 3.5 0.25 0.4 1.00 0.92 0.92 0.50 0.8 1.00 0.93 0.93 0.75 1.2 1.00 0.93 0.93 1.00 1.6 1.00 0.93 0.93 1.50 2.4 1.00 0.94 0.94 2.00 3.2 1.00 0.95 0.95 3.00 4.8 1.00 0.97 0.96 ≥ 4.00 ≥ 6.4 1.00 1.00 0.97 ≥ 3.5 < 4.5 0.25 0.4 1.00 0.94 0.92 0.50 0.8 1.00 0.97 0.96 0.75 1.2 1.00 0.97 0.96 1.00 1.6 1.00 0.97 0.97 1.50 2.4 1.00 0.97 0.97 2.00 3.2 1.00 0.98 0.98 3.00 4.8 1.00 1.00 1.00 ≥ 4.00 ≥ 6.4 1.00 1.00 1.00 ≥ 4.5 < 5.5 0.25 0.4 1.00 1.00 0.97 0.50 0.8 1.00 1.00 1.00 0.75 1.2 1.00 1.00 1.00 1.00 1.6 1.00 1.00 1.00 1.50 2.4 1.00 1.00 1.00 2.00 3.2 1.00 1.00 1.00 3.00 4.8 1.00 1.00 1.00 ≥ 4.00 ≥ 6.4 1.00 1.00 1.00 ≥ 5.5 < 6.5 0.25 0.4 1.00 1.00 1.00 0.50 0.8 1.00 1.00 1.00 0.75 1.2 1.00 1.00 1.00 1.00 1.6 1.00 1.00 1.00 1.50 2.4 1.00 1.00 1.00 2.00 3.2 1.00 1.00 1.00 3.00 4.8 1.00 1.00 1.00 ≥ 4.00 ≥ 6.4 1.00 1.00 1.00 ≥ 6.5 0.25 0.4 1.00 1.00 1.00 0.50 0.8 1.00 1.00 1.00 0.75 1.2 1.00 1.00 1.00 1.00 1.6 1.00 1.00 1.00 1.50 2.4 1.00 1.00 1.00 2.00 3.2 1.00 1.00 1.00 3.00 4.8 1.00 1.00 1.00 ≥ 4.00 ≥ 6.4 1.00 1.00 1.00 Reproduced with permission from Highway Capacity Manual, 2000, Transportation Research Board, National Academy of Sciences, Courtesy of the National Academies Press, Washington, DC, USA. Table 4.42 Passenger-Car Equivalents for Trucks (ET) and RVs (ER) for Estimating Percent Time-Spent-Following on Specific Upgrades Grade (%) Length of Grade (mi) Length of Grade (km) Passenger-Car Equivalent for Trucks (E_T) RVs, E_R Range of Directional Flow Rates (pc/h) 0–300 > 300–600 > 600 ≥ 3 < 3.5 0.25 0.4 1.0 1.0 1.0 1.0 0.50 0.8 1.0 1.0 1.0 1.0 0.75 1.2 1.0 1.0 1.0 1.0 1.00 1.6 1.0 1.0 1.0 1.0 1.50 2.4 1.0 1.0 1.0 1.0 2.00 3.2 1.0 1.0 1.0 1.0 3.00 4.8 1.4 1.0 1.0 1.0 ≥ 4.00 ≥ 6.4 1.5 1.0 1.0 1.0 ≥ 3.5 < 4.5 0.25 0.4 1.0 1.0 1.0 1.0 0.50 0.8 1.0 1.0 1.0 1.0 0.75 1.2 1.0 1.0 1.0 1.0 1.00 1.6 1.0 1.0 1.0 1.0 1.50 2.4 1.1 1.0 1.0 1.0 2.00 3.2 1.4 1.0 1.0 1.0 3.00 4.8 1.7 1.1 1.2 1.0 ≥ 4.00 ≥ 6.4 2.0 1.5 1.4 1.0 ≥ 4.5 < 5.5 0.25 0.4 1.0 1.0 1.0 1.0 0.50 0.8 1.0 1.0 1.0 1.0 0.75 1.2 1.0 1.0 1.0 1.0 1.00 1.6 1.0 1.0 1.0 1.0 1.50 2.4 1.1 1.2 1.2 1.0 2.00 3.2 1.6 1.3 1.5 1.0 3.00 4.8 2.3 1.9 1.7 1.0 ≥ 4.00 ≥ 6.4 3.3 2.1 1.8 1.0 ≥ 5.5 < 6.5 0.25 0.4 1.0 1.0 1.0 1.0 0.50 0.8 1.0 1.0 1.0 1.0 0.75 1.2 1.0 1.0 1.0 1.0 1.00 1.6 1.0 1.2 1.2 1.0 1.50 2.4 1.5 1.6 1.6 1.0 2.00 3.2 1.9 1.9 1.8 1.0 3.00 4.8 3.3 2.5 2.0 1.0 ≥ 4.00 ≥ 6.4 4.3 3.1 2.0 1.0 ≥ 6.5 0.25 0.4 1.0 1.0 1.0 1.0 0.50 0.8 1.0 1.0 1.0 1.0 0.75 1.2 1.0 1.0 1.3 1.0 1.00 1.6 1.3 1.4 1.6 1.0 1.50 2.4 2.1 2.0 2.0 1.0 2.00 3.2 2.8 2.5 2.1 1.0 3.00 4.8 4.0 3.1 2.2 1.0 ≥ 4.00 ≥ 6.4 4.8 3.5 2.3 1.0 Reproduced with permission from Highway Capacity Manual, 2000, Transportation Research Board, National Academy of Sciences, Courtesy of the National Academies Press, Washington, DC, USA. Table 4.43 Determining fG, ET, and ER for a Specific Upgrade on Two-Lane Highway (PTSF Method) for Problem 4.20 Item Value Table Comment f_G 0.98 Table 4.41 Using directional flow rate = 1000 > 600 pcph, and grade = 4% ≥ 3.5 < 4.5, and length of grade = 2 miles E_T 1.0 Table 4.42 Using directional flow rate = 1000 > 600 pcph, and grade = 4% ≥ 3.5 < 4.5, and length of grade = 2 miles E_R 1.0 Table 4.42 For specific upgrades Table 4.44 Passenger-Car Equivalents (ETC) for Estimating the Effect on Average Travel Speed of Trucks that Operate at Crawl Speeds on Long Steep Downgrades Difference Between FFS and Truck Crawl Speeds (mi/h) Difference Between FFS and Truck Crawl Speeds (km/h) Passenger-Car Equivalents for Trucks at Crawl Speeds (E_TC) Range of Directional Flow Rates (pc/h) 0–300 > 300–600 > 600 ≤ 15 ≤ 20 4.4 2.8 1.4 25 40 14.3 9.6 5.7 ≥ 40 ≥ 60 34.1 23.1 13.0 Reproduced with permission from Highway Capacity Manual, 2000, Transportation Research Board, National Academy of Sciences, Courtesy of the National Academies Press, Washington, DC, USA. Table 4.45 Determining ET, ER, and ETC for Downgrade Direction on Two-Lane Highway (PTSF Method) for Problem 4.20 Item Value Table Comment E_T 1.1 Table 4.20 Using directional opposing flow rate = 556 > 300–600 pcph, and level terrain E_R 1.0 Table 4.20 Using directional opposing flow rate = 556 > 300–600 pcph, and level terrain E_TC 23.1 Table 4.44 Using directional opposing flow rate = 556 > 300–600 pcph, and difference between FFS and truck crawl speed = 40 mph ≥ 40 mph Table 4.46 Tabulating the Values of a and b for the Opposing Direction for Problem 4.20 Opposing Flow Rate (pcph) a b 600 −0.100 0.413 607 ? ? 800 −0.173 0.349 Table 4.47 Tabulating the Values of fnp for the Opposing Direction (PTSF Method) for Problem 4.20 Opposing Flow Rate (pcph) f_np 600 4.7 607 ? 800 2.5 Table 4.48 Tabulating the Values of fnp for the Opposing Direction (PTSF Method) for Problem 4.20 Opposing Flow Rate (pcph) f_np 600 4.5 607 ? 800 2.3 Table 4.49 Tabulating the Values of fnp for the Opposing Direction (PTSF Method) for Problem 4.20 FFS (mph f_np 45 4.423 48.3 ? 50 4.623 Table 4.50 Grade Adjustment Factor (fG) for Estimating Average Travel Speed on Specific Upgrades Grade (%) Length of Grade (mi) Length of Grade (km) Grade Adjustment Factor (f_G) Range of Directional Flow Rates (pc/h) 0–300 > 300–600 > 600 ≥ 3 < 3.5 0.25 0.4 0.81 1.00 1.00 0.50 0.8 0.79 1.00 1.00 0.75 1.2 0.77 1.00 1.00 1.00 1.6 0.76 1.00 1.00 1.50 2.4 0.75 0.99 1.00 2.00 3.2 0.75 0.97 1.00 3.00 4.8 0.75 0.95 0.97 ≥ 4.00 ≥ 6.4 0.75 0.94 0.95 ≥ 3.5 < 4.5 0.25 0.4 0.79 1.00 1.00 0.50 0.8 0.76 1.00 1.00 0.75 1.2 0.72 1.00 1.00 1.00 1.6 0.69 0.93 1.00 1.50 2.4 0.68 0.92 1.00 2.00 3.2 0.66 0.91 1.00 3.00 4.8 0.65 0.91 0.96 ≥ 4.00 ≥ 6.4 0.65 0.90 0.96 ≥ 4.5 < 5.5 0.25 0.4 0.75 1.00 1.00 0.50 0.8 0.65 0.93 1.00 0.75 1.2 0.60 0.89 1.00 1.00 1.6 0.59 0.89 1.00 1.50 2.4 0.57 0.86 0.99 2.00 3.2 0.56 0.85 0.98 3.00 4.8 0.56 0.84 0.97 ≥ 4.00 ≥ 6.4 0.55 0.82 0.93 ≥ 5.5 < 6.5 0.25 0.4 0.63 0.91 1.00 0.50 0.8 0.57 0.85 0.99 0.75 1.2 0.52 0.83 0.97 1.00 1.6 0.51 0.79 0.97 1.50 2.4 0.49 0.78 0.95 2.00 3.2 0.48 0.78 0.94 3.00 4.8 0.46 0.76 0.93 ≥ 4.00 ≥ 6.4 0.45 0.76 0.93 ≥ 6.5 0.25 0.4 0.59 0.86 0.98 0.50 0.8 0.48 0.76 0.94 0.75 1.2 0.44 0.74 0.91 1.00 1.6 0.41 0.70 0.91 1.50 2.4 0.40 0.67 0.91 2.00 3.2 0.39 0.67 0.89 3.00 4.8 0.39 0.66 0.88 ≥ 4.00 ≥ 6.4 0.38 0.66 0.87 Reproduced with permission from Highway Capacity Manual, 2000, Transportation Research Board, National Academy of Sciences, Courtesy of the National Academies Press, Washington, DC, USA. Table 4.51 Passenger-Car Equivalents for Trucks (ET) for Estimating Average Travel Speed on Specific Upgrades Grade (%) Length of Grade (mi) Length of Grade (km) Passenger-Car Equivalent for Trucks (E_T) Range of Directional Flow Rates (pc/h) 0–300 > 300–600 > 600 ≥ 3 < 3.5 0.25 0.4 2.5 1.9 1.5 0.50 0.8 3.5 2.8 2.3 0.75 1.2 4.5 3.9 2.9 1.00 1.6 5.1 4.6 3.5 1.50 2.4 6.1 5.5 4.1 2.00 3.2 7.1 5.9 4.7 3.00 4.8 8.2 6.7 5.3 ≥ 4.00 ≥ 6.4 9.1 7.5 5.7 ≥ 3.5 < 4.5 0.25 0.4 3.6 2.4 1.9 0.50 0.8 5.4 4.6 3.4 0.75 1.2 6.4 6.6 4.6 1.00 1.6 7.7 6.9 5.9 1.50 2.4 9.4 8.3 7.1 2.00 3.2 10.2 9.6 8.1 3.00 4.8 11.3 11.0 8.9 ≥ 4.00 ≥ 6.4 12.3 11.9 9.7 ≥ 4.5 < 5.5 0.25 0.4 4.2 3.7 2.6 0.50 0.8 6.0 6.0 5.1 0.75 1.2 7.5 7.5 7.5 1.00 1.6 9.2 9.0 8.9 1.50 2.4 10.6 10.5 10.3 2.00 3.2 11.8 11.7 11.3 3.00 4.8 13.7 13.5 12.4 ≥ 4.00 ≥ 6.4 15.3 15.0 12.5 ≥ 5.5 < 6.5 0.25 0.4 4.7 4.1 3.5 0.50 0.8 7.2 7.2 7.2 0.75 1.2 9.1 9.1 9.1 1.00 1.6 10.3 10.3 10.2 1.50 2.4 11.9 11.8 11.7 2.00 3.2 12.8 12.7 12.6 3.00 4.8 14.4 14.3 14.2 ≥ 4.00 ≥ 6.4 15.4 15.2 15.0 ≥ 6.5 0.25 0.4 5.1 4.8 4.6 0.50 0.8 7.8 7.8 7.8 0.75 1.2 9.8 9.8 9.8 1.00 1.6 10.4 10.4 10.3 1.50 2.4 12.0 11.9 11.8 2.00 3.2 12.9 12.8 12.7 3.00 4.8 14.5 14.4 14.3 ≥ 4.00 ≥ 6.4 15.4 15.3 15.2 Reproduced with permission from Highway Capacity Manual, 2000, Transportation Research Board, National Academy of Sciences, Courtesy of the National Academies Press, Washington, DC, USA. Table 4.52 Passenger-Car Equivalents for RVs (ER) for Estimating Average Travel Speed on Specific Upgrades Grade (%) Length of Grade (mi) Length of Grade (km) Passenger-Car Equivalent for RVs (E_R) Range of Directional Flow Rates (pc/h) 0–300 > 300–600 > 600 ≥ 3 < 3.5 0.25 0.4 1.1 1.0 1.0 0.50 0.8 1.2 1.0 1.0 0.75 1.2 1.2 1.0 1.0 1.00 1.6 1.3 1.0 1.0 1.50 2.4 1.4 1.0 1.0 2.00 3.2 1.4 1.0 1.0 3.00 4.8 1.5 1.0 1.0 ≥ 4.00 ≥ 6.4 1.5 1.0 1.0 ≥ 3.5 < 4.5 0.25 0.4 1.3 1.0 1.0 0.50 0.8 1.3 1.0 1.0 0.75 1.2 1.3 1.0 1.0 1.00 1.6 1.4 1.0 1.0 1.50 2.4 1.4 1.0 1.0 2.00 3.2 1.4 1.0 1.0 3.00 4.8 1.4 1.0 1.0 ≥ 4.00 ≥ 6.4 1.5 1.0 1.0 ≥ 4.5 < 5.5 0.25 0.4 1.5 1.0 1.0 0.50 0.8 1.5 1.0 1.0 0.75 1.2 1.5 1.0 1.0 1.00 1.6 1.5 1.0 1.0 1.50 2.4 1.5 1.0 1.0 2.00 3.2 1.5 1.0 1.0 3.00 4.8 1.6 1.0 1.0 ≥ 4.00 ≥ 6.4 1.6 1.0 1.0 ≥ 5.5 < 6.5 0.25 0.4 1.5 1.0 1.0 0.50 0.8 1.5 1.0 1.0 0.75 1.2 1.5 1.0 1.0 1.00 1.6 1.6 1.0 1.0 1.50 2.4 1.6 1.0 1.0 2.00 3.2 1.6 1.0 1.0 3.00 4.8 1.6 1.2 1.0 ≥ 4.00 ≥ 6.4 1.6 1.5 1.2 ≥ 6.5 0.25 0.4 1.6 1.0 1.0 0.50 0.8 1.6 1.0 1.0 0.75 1.2 1.6 1.0 1.0 1.00 1.6 1.6 1.0 1.0 1.50 2.4 1.6 1.0 1.0 2.00 3.2 1.6 1.0 1.0 3.00 4.8 1.6 1.3 1.3 ≥ 4.00 ≥ 6.4 1.6 1.5 1.4 Reproduced with permission from Highway Capacity Manual, 2000, Transportation Research Board, National Academy of Sciences, Courtesy of the National Academies Press, Washington, DC, USA. Table 4.53 Determining fG, ET, and ER for the Peak Direction (ATS Method) for Problem 4.20 Item Value Table Comment f_G 1.00 Table 4.50 Using directional flow rate = 1000 > 600 pcph, and grade = 4% ≥ 3.5 < 4.5, and length of grade = 2 miles E_T 8.1 Table 4.51 Using directional flow rate = 1000 > 600 pcph, and grade = 4% ≥ 3.5 < 4.5, and length of grade = 2 miles E_R 1.0 Table 4.52 Using directional flow rate = 1000 > 600 pcph, and grade = 4% ≥ 3.5 < 4.5, and length of grade = 2 miles Table 4.54 Determining ET, ER, and ETC for Downgrade Direction on Two-Lane Highway (ATS Method) for Problem 4.20 Item Value Table Comment E_T 1.2 Table 4.20 Using directional flow rate = 556 > 300–600 pcph, and level terrain E_R 1.0 Table 4.20 Using directional flow rate = 556 > 300–600 pcph, and level terrain E_TC 23.1 Table 4.44 Using directional flow rate = 556 > 300–600 pcph, and difference between FFS and truck crawl speed = 40 mph ≥ 40 mph Table 4.55 Tabulating the Values of fnp for the Opposing Direction (ATS Method) for Problem 4.20 Opposing Flow Rate (pcph) f_np 600 0.6 609 ? 800 0.4 Table 4.56 Tabulating the Values of fnp for the Opposing Direction (ATS Method) for Problem 4.20 Opposing Flow Rate (pcph) f_np 600 0.4 609 ? 800 0.3 Table 4.57 Tabulating the Values of fnp for the Opposing Direction (ATS Method) for Problem 4.20 FFS (mph f_np 45 0.396 48.3 ? 50 0.591 Table 4.58 Determining ET and ER Values for Multilane Highway for Problem 4.21 Item Value Table Comment E_T 1.5 Table 4.2 Using level terrain E_R 1.2 Table 4.2 Using level terrain Table 4.59 Level-of-Service (LOS) Criteria for Multilane Highways Criteria LOS A B C D E FFS = 60 mi/h (100 km/h) Maximum density (pc/mi/ln) 11 (7) 18 (11) 26 (16) 35 (22) 40 (25) Maximum speed (mi/h) 60.0 (100.0) 60.0 (100.0) 59.4 (98.4) 56.7 (91.5) 55.0 (88.0) Maximum v/c 0.32 0.50 0.72 0.92 1.00 Maximum service flow rate (pc/h/ln) 700 1100 1575 2015 2200 FFS = 55 mi/h (90 km/h) Maximum density (pc/mi/ln) 11 (7) 18 (11) 26 (16) 35 (22) 40 (25) Maximum speed (mi/h) 55.0 (90.0) 55.0 (90.0) 54.9 (89.8) 52.9 (84.7) 51.2 (80.8) Maximum v/c 0.30 0.47 0.68 0.89 1.00 Maximum service flow rate (pc/h/ln) 630 990 1435 1860 2100 FFS = 50 mi/h (80 km/h) Maximum density (pc/mi/ln) 11 (7) 18 (11) 26 (16) 35 (22) 40 (25) Maximum speed (mi/h) 50.0 (80.0) 50.0 (80.0) 50.0 (80.0) 48.9 (77.6) 47.5 (74.1) Maximum v/c 0.28 0.44 0.64 0.85 1.00 Maximum service flow rate (pc/h/ln) 560 880 1280 1705 2000 FFS = 45 mi/h (70 km/h) Maximum density (pc/mi/ln) 11 (7) 18 (11) 26 (16) 35 (22) 40 (25) Maximum speed (mi/h) 45.0 (70.0) 45.0 (70.0) 45.0 (70.0) 44.4 (69.6) 42.2 (67.9) Maximum v/c 0.26 0.41 0.59 0.81 1.00 Maximum service flow rate (pc/h/ln) 490 770 1120 1530 1900 Reproduced with permission from Highway Capacity Manual, 2000, Transportation Research Board, National Academy of Sciences, Courtesy of the National Academies Press, Washington, DC, USA. Table 4.60 Determining Values of ET and ER for Upgrade Direction on Multilane Highway for Problem 4.21 Item Value Table Comment E_T 2.5 Table 4.13 Using a percentage of trucks of 5%, upgrade of 4% > 3–4%, and length of grade of 1.00 mile > 0.75–1.00 mile E_R 2.5 Table 4.14 Using a percentage of RVs of 4%, upgrade of 4% > 3–4%, and length of grade of 1.00 mile > 0.50 mile