ch 9

Chapter 9 Capacity and Level of Service for Highway Segments 9-1 Define the elements of a Class I and a Class II highway.

A Class I two-lane highway functions as an arterial or a link to an arterial, often carrying primarily commuter traffic. Motorists expect relatively high-speed high-speed travel. A Class II two-lane highway may serve as access to to a class I two-lane two-lane highway, as a scenic or recreational route, or be located in rugged terrain, thereby constraining travel speeds. Motorists expect lower speeds speeds on Class II than than on Class I roads. Average trip lengths are shorter on Class II roads than Class I. 9-2 What are the two measures used to describe service quality for a two-lane highway? Which of the measures are used to describe level-of-service for Class I and Class II highways?

Service quality on a two-lane highway is described using percent of time spent following another vehicle (PTSF) and average travel speed (ATS). (ATS). A vehicle considered to be following when the time headway between it and the preceding vehicle is less than 3 seconds. The space mean speed is used as ATS and can be found by dividing segment length by average travel time of vehicles traversing the segment in both direction during a designated time time interval. PTSF and ATS are used to determine level-of-service on Class I roads while only PTSF is used on Class II roads.

9-3 Describe the traffic characteristics associated with each of the six levels of service for two-lane highways.

In LOS A, motorists are able to travel at their desired speeds, with if any platoons and minimal passing maneuvers. With LOS B, passing demand increases significantly and approaches passing capacity at the lower end of LOS B. In LOS C, formation of platoons and platoon size size increase significantly. In LOS D, flow begins to become unstable, and passing maneuvers are extremely difficult if not possible possible to complete. In LOS E, passing is nearly impossible, and most vehicles are in long platoons; operating conditions are unstable and difficult to predict. LOS F occurs when demand exceeds capacity and volume fall back from capacity with highly variable speeds.

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(c) 2009 Cengage Learning ALL RIGHTS RESERVED.

Chapter 9: Capacity and Level of Service for Highway Segments

9-4 The following values of PTSF and ATS have been determined for three separate two-lane segments. Determine LOS if the segments are: (a) Class I, and (b) Class II.

Segment 10 11 12

PTSF (%) 25 46 67

ATS (mi/h) 52 39 39

(a) For a Class I highway, using Table 9.1, segment 10 operates at LOS A (when PTSF and ATS correspond to differing levels of service, the lower value of LOS is used), segment 11 operates at LOS B, and segment 12 operates at LOS D. (b) For a Class II highway, using Table 9.2, segment 10 operates at LOS A, segment 11 operates at LOS B, and segment 12 operates at LOS C.

9-5 Determine the PTSF for a 4.5 mile two-lane highway segment in level terrain. Traffic volumes (two-way) are 1100 veh/h. Trucks: 10%; RVs: 7%; PHF: 0.97; directional split: 60/40; no passing zones: 40%.

Step 1: Compute passenger car equivalent flow rate for peak 15-minute period, v p Determine f g (grade adjustment factor), using Table 9.4; f g = 1.00 Determine ET (PCE for trucks), using Table 9.5; E T = 1.1 Determine ER (PCE for RVs), using Table 9.5; ER = 1.0 Determine f HV (heavy vehicle adjustment factor), using Equation 9.4 f HV =

1 1 + PT ( E T − 1) + P R ( E R − 1)

=

1 1 + 0.10(1.1 − 1) + 0.07(1.0 − 1)

= 0.99

Determine v p (flow rate), using Equation 9.3 V = (1100)/(0.97)(0.99)(1.0) = 1146 pc/h v p = ( PHF )( f HV )( f G ) Step 2: Compute base percent time spent following, BPTSF, using Equation 9.2 BPTSF = 100(1 − e

−0.000879V p

) = 100(1 − e −0.000879(1146 ) ) = 63.5%

Step 3: Compute percent time spent following, PTSF, using Equation 9.1 Determine f d/np using Table 9.3; by interpolation, f d/np = 7.6% PTSF = BPTSF + f d/np = 63.5% + 7.6% = 71.1%


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9-6 Use the data provided in Problem 9-5 to estimate the ATS. Base free flow speed: 55 mi/h; lane width: 11 ft; shoulder width: 3 ft; access points per mile: 15.

Step 1: Compute free flow speed, FFS, using Equation 9.7 Determine f LS (lane and shoulder width adjustment), using Table 9.9; f LS = 3.0 Determine f A (access point density adjustment), using Table 9.10; f A =3.8 FFS = BFFS – f LS – f A = 55 – 3.0 – 3.8 = 48.2 mi/h Step 2: Compute passenger car equivalent flow rate for peak 15-minute period, v p Determine f g (grade adjustment factor), using Table 9.7; f g = 1.00 Determine ET (PCE for trucks), using Table 9.8; E T = 1.2 Determine ER (PCE for RVs), using Table 9.8; ER = 1.0 Determine f HV (heavy vehicle adjustment factor); using Equation 9.4 1 1 = 0.98 = f HV = 1 + PT ( E T − 1) + P R ( E R − 1) 1 + 0.10(1.2 − 1) + 0.07(1.0 − 1) Determine v p (flow rate), using Equation 9.3 V = (1100)/(0.97)(0.98)(1.0) = 1157 pc/h v p = ( PHF )( f HV )( f G ) Step 3: Compute average travel speed, ATS, using Equation 9.5 Determine f np (adjustment for effect of passing zones), using Table 9.6; f np = 1.5 ATS = FFS – 0.00776v p – f np = 48.2 – 0.00776(1157) – 1.5 = 37.7 mi/h

9-7 Use the results of Problems 9-5 and 9-6 to compute: LOS, v/c, and veh-mi in the peak 15 minutes and peak hour, and total travel time in the peak 15 minutes.

Results from Problems 9-5 and 9-6: PTSF = 71.1% ATS = 37.7 mi/h Level of service (if a Class I highway), using Table 9.1: LOS D Level of service (if a Class II highway), u sing Table 9.2: LOS D Using Equation 9.8, v/c = 1157/3200 = 0.36 Using Equation 9.9, V ) Lt = (0.25)(1100)(4.5)/(0.97) = 1275 veh-mi VMT 15 = 0.25( PHF Using Equation 9.10, VMT 60 = VLt = (1100)(4.5) = 4950 veh-mi Using Equation 9.11, TT 15 = VMT 15 /ATS = 1275/37.7 = 33.8 veh-h 125



9-8 Use the data provided in Examples 9-5 and 9-6 to compute to determine PTSF and ATS in the peak direction if northbound volume is 1000 veh/h and southbound volume is 600 veh/h.

Class I, 5 mile, two-lane highway segment in rolling terrain. Trucks: 14%; RVs: 4%; PHF: 0.95; no passing zones: 50%; BFFS = 60 mi/h. Lane width = 11 ft; shoulder width = 4 ft; 20 access points per mile. ATS analysis: Step 1: Compute passenger car equivalent flow rate for peak 15-minute period in the peak direction, vd Determine f g (grade adjustment factor), using Table 9.4; f g = 0.99 Determine ET (PCE for trucks), using Table 9.5; E T = 1.5 Determine ER (PCE for RVs), using Table 9.5; ER = 1.1 Determine f HV (heavy vehicle adjustment factor), using Equation 9.4 1 1 = 0.931 f HV = = 1 + PT ( E T − 1) + P R ( E R − 1) 1 + 0.14(1.5 − 1) + 0.04(1.1 − 1) Determine vd (flow rate), using Equation 9.3 V = (1000)/(0.95)(0.931)(0.99) = 1142 pc/h vd = ( PHF )( f HV )( f G ) Step 2: Compute passenger car equivalent flow rate for peak 15-minute period in the opposing direction, vo Determine f g (grade adjustment factor), using Table 9.4; f g = 0.99 Determine ET (PCE for trucks), using Table 9.5; E T = 1.5 Determine ER (PCE for RVs), using Table 9.5; ER = 1.1 Determine f HV (heavy vehicle adjustment factor), using Equation 9.4 1 1 = 0.931 = f HV = 1 + PT ( E T − 1) + P R ( E R − 1) 1 + 0.14(1.5 − 1) + 0.04(1.1 − 1) Determine vo (flow rate), using Equation 9.3 V = (600)/(0.95)(0.931)(0.99) = 685 pc/h vo = ( PHF )( f HV )( f G ) Step 3: Compute free flow speed, FFS, using Equation 9.7 Determine f LS from Table 9.9; by interpolation, f LS = 1.7 Determine f A from Table 9.10; by interpolation, f LS = 5.0 FFS = BFFS – f LS – f A = 60 – 1.7 – 5.0 = 53.3 mi/h Step 4: Compute average travel speed, ATS, using Equation 9.14 Determine f np from Table 9.13; by interpolation, f np = 1.0 ATSd = FFSd – 0.00776(vd + vo) – f np = 53.3 – 0.00776 (1142+685) – 1.1 = 38.0 mi/h (c) 2009 Cengage Learning ALL RIGHTS RESERVED.

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PTSF analysis: Step 1: Compute passenger car equivalent flow rate for peak 15-minute period in the peak direction, vd Determine f g (grade adjustment factor), using Table 9.4; f g = 1.00 Determine ET (PCE for trucks), using Table 9.5; E T = 1.0 Determine ER (PCE for RVs), using Table 9.5; ER = 1.0 Determine f HV (heavy vehicle adjustment factor), using Equation 9.4 1 1 = 1.00 f HV = = 1 + PT ( E T − 1) + P R ( E R − 1) 1 + 0.14(1.0 − 1) + 0.04(1.0 − 1) Determine vd (flow rate), using Equation 9.3 V = (1000)/(0.95)(1.0)(1.0) = 1053 pc/h vd = ( PHF )( f HV )( f G ) Step 2: Compute passenger car equivalent flow rate for peak 15-minute period in the opposing direction, vo Determine f g (grade adjustment factor), using Table 9.4; f g = 1.00 Determine ET (PCE for trucks), using Table 9.5; E T = 1.0 Determine ER (PCE for RVs), using Table 9.5; ER = 1.0 Determine f HV (heavy vehicle adjustment factor), using Equation 9.4 1 1 = 1.0 f HV = = 1 + PT ( E T − 1) + P R ( E R − 1) 1 + 0.14(1.0 − 1) + 0.04(1.0 − 1) Determine vo (flow rate), using Equation 9.3 V = (600)/(0.95)(1.0)(1.0) = 632 pc/h vo = ( PHF )( f HV )( f G ) Step 3: Compute base percent time spent following, BPTSF, using Equation 9.12 Find values of a and b from Table 9.12; by interpolation, a = -0.112 b = 0.403 b av ( −0.112 )(1053) 0.403 ) = 84.2% BPTSF d = 100(1 − e d ) = 100(1 − e Step 4: Compute percent time spent following, PTSF, using Equation 9.11 Determine f np using Table 9.11; by interpolation, f np = 8.1% PTSFd = BPTSFd + f d/np = 84.2% + 8.1% = 92.3%

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9-9 Use the data and results obtained in Problem 9-8 to determine the level of service of a two-lane section if a passing lane 1.5 mi long is added. The passing lane begins 0.75 mi from the starting point of the analysis segment.

To determine the level of service, lengths of the regions in the segment, PTSF pl and ATS pl, must first be determined. From Problem 9-8, PTSFd = 87.6% and ATSd = 38.5 mi/h Step 1: Determine region lengths. Region I: Lu =0.75 mi Region II: L pl = 1.5 mi Region III: For PTSF, from Table 9.22, Lde = 3.6 mi; from Table 9.20, f pl = 0.62 For ATS, from Table 9.22, Lde = 1.7 mi; from Table 9.20, f pl = 1.11 Region IV: For PTSF, Ld = 5 – 0.75 –1.5 – 3.6 = -0.85; use Ld = 0 and L’de = 5 – 0.75 – 1.5 = 2.75 For ATS, Ld = 5 – 0.75 – 1.5 – 1.7 = 1.05 Step 2: Compute PTSF pl using Equation 9.16 1 − f pl Lde ′2 )( )] PTSF d [ Lu + ( f pl )( L pl ) + ( 2 Lde PTSF pl = Lt 1 − 0.62 2.75 2 87.6[0.75 + (0.62)(1.5) + ( )( )] 2 3 . 6 PTSF pl = = (87.6)(2.079)/5 = 36.4% 5 Step 3: Compute ATS pl using Equation 9.18 ( ATS d )( Lt ) ATS pl = L pl 2 Lde Lu + + + Ld f pl f pl + 1 ATS pl =

0.75 +

(38.5)(5) 1.5 ( 2)(1.7) 1.11

+

1.11 + 1

= (38.5)(5)/4.7627 = 40.4 mi/h

+ 1.05

Step 4: Determine level of service from Table 9.1 (for a Class I highway) Due to PTSF = 36.4%, the level of service is LOS B.


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9-10 An existing Class I two-lane highway is to be analyzed to determine the two-way level of service, given the following information: Traffic data: PHV = 600 veh/h 60% in the peak direction 8% trucks 2% recreational vehicles PHF = 0.86 No passing zones: 40% Geometric data: Rolling terrain BFFS = 55 mi/h Lane width = 11 ft Shoulder width = 2 ft 8 access points per mile

PTSF analysis: Step 1: Compute passenger car equivalent flow rate for peak 15-minute period, v p Determine f g (grade adjustment factor), using Table 9.4; f g = 0.94 Determine ET (PCE for trucks), using Table 9.5; E T = 1.5 Determine ER (PCE for RVs), using Table 9.5; ER = 1.0 Determine f HV (heavy vehicle adjustment factor), using Equation 9.4 1 1 = 0.962 f HV = = 1 + PT ( E T − 1) + P R ( E R − 1) 1 + 0.08(1.5 − 1) + 0.02(1.0 − 1) v p =

V

( PHF )( f HV )( f G )

= (600)/(0.86)(0.962)(0.94) = 772 pc/h

Step 2: Compute base percent time spent following, BPTSF, using Equation 9.2 BPTSF = 100(1 − e

−0.000879V p

) = 100(1 − e −0.000879( 772 ) ) = 49.3%

Step 3: Compute percent time spent following, PTSF, using Equation 9.1 Determine f d/np using Table 9.3; by interpolation, f d/np = 11.0% PTSF = BPTSF + f d/np = 49.3% + 11.0% = 60.3% This corresponds to LOS C (Table 9.1) based solely on PTSF. ATS analysis: Step 1: Compute free flow speed, FFS, using Equation 9.7 Determine f LS (lane and shoulder width adjustment), using Table 9.9; f LS = 3.0 Determine f A (access point density adjustment), using Table 9.10; f A = 2.0 FFS = BFFS – f LS – f A = 55 – 3.0 – 2.0 = 50.0 mi/h

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Step 2: Compute passenger car equivalent flow rate for peak 15-minute period, v p Determine f g (grade adjustment factor), using Table 9.7; f g = 0.930 Determine ET (PCE for trucks), using Table 9.8; E T = 1.9 Determine ER (PCE for RVs), using Table 9.8; ER = 1.1 Determine f HV (heavy vehicle adjustment factor); using Equation 9.4 1 1 = 0.931 f HV = = 1 + PT ( E T − 1) + P R ( E R − 1) 1 + 0.10(1.2 − 1) + 0.07(1.0 − 1) V p =

V

( PHF )( f HV )( f G )

= (1100)/(0.86)(0.931)(0.93) = 806 pc/h

Step 3: Compute average travel speed, ATS, using Equation 9.5 Determine f np (adjustment for effect of passing zones), using Table 9.6; f np = 2.0 ATS = FFS – 0.00776v p – f np = 50.0 – 0.00776(806) – 2.0 = 41.9 mi/h This corresponds to LOS D (Table 9.1) based solely on ATS. The overall LOS is taken as the poorer of the two (for PTSF and ATS), LOS D.

9-11 An existing Class II two-lane highway is to be analyzed to determine LOS in the peak direction given the following information: Peak hourly volume in the analysis direction: 900 veh/h Peak hourly volume in the opposing direction: 400 veh/h Trucks: 12% of total volume Recreational vehicles: 2% of total volume PHF: 0.95 Lane width: 12 ft Shoulder width: 10 ft Access points per mile: 20 Terrain: rolling Base free flow speed: 60 mi/h No passing zones: 40% of analysis segment length

PTSF analysis: Step 1: Compute passenger car equivalent flow rate for peak 15-minute period in the peak direction, vd Determine f g (grade adjustment factor), using Table 9.4; f g = 0.94 Determine ET (PCE for trucks), using Table 9.5; E T = 1.0 Determine ER (PCE for RVs), using Table 9.5; ER = 1.0 Determine f HV (heavy vehicle adjustment factor), using Equation 9.4 f HV =

1 1 + PT ( E T − 1) + P R ( E R − 1)


=

1 1 + 0.12(1.0 − 1) + 0.02(1.0 − 1)

130

= 1.00


vd =

V

( PHF )( f HV )( f G )

= (900)/(0.95)(1.0)(1.0) = 947 pc/h

Step 2: Compute passenger car equivalent flow rate for peak 15-minute period in the opposing direction, vo Determine f g (grade adjustment factor), using Table 9.4; f g = 0.94 Determine ET (PCE for trucks), using Table 9.5; E T = 1.5 Determine ER (PCE for RVs), using Table 9.5; ER = 1.0 Determine f HV (heavy vehicle adjustment factor), using Equation 9.4 1 1 = 0.943 f HV = = 1 + PT ( E T − 1) + P R ( E R − 1) 1 + 0.12(1.5 − 1) + 0.02(1.0 − 1) vo =

V

( PHF )( f HV )( f G )

= (400)/(0.95)(0.943)(0.94) = 475 pc/h

Step 3: Compute base percent time spent following, BPTSF, using Equation 9.13 Find values of a and b from Table 9.12; by interpolation, a = -0.073 b = 0.454 b av ( −0.073 )( 947 ) 0.454 ) = 80.7% BPTSF d = 100(1 − e d ) = 100(1 − e Step 4: Compute free flow speed, FFS, using Equation 9.6 Determine f LS (lane and shoulder width adjustment), using Table 9.9; f LS = 0 Determine f A (access point density adjustment), using Table 9.10; f A = 5.0 FFS = BFFS – f LS – f A = 60 – 0 – 5.0 = 55.0 mi/h Step 5: Compute percent time spent following, PTSF, using Equation 9.12 Determine f np using Table 9.11; by interpolation, f np = 10.1% PTSFd = BPTSFd + f d/np = 80.7% + 10.1% = 90.8% This corresponds to LOS E (Table 9.2)

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9-12 A new section of Richmond Highway is being designed as a six-lane facility (three in each direction) with a two-way left-turn lane. Determine the peak hour LOS. Traffic data: • Directional design hourly volume = 3600 veh/h • PHF = 0.94 • Assumed base free flow speed = 55 mi/h Geometric data: • Urban setting • Rolling terrain • Lane width = 11 ft • Shoulder widths = 4 ft (right side) and 1 ft (left side) • Average access point spacing = 12 points per mile on each side

Step 1: Compute passenger car equivalent flow rate for peak 15-minute period using Equation 9.21. Determine f p (driver population factor); assume mainly commuter traffic, f p = 1.0 Determine ET (PCE for trucks), using Table 9.25; ET = 2.5 Determine f HV (heavy vehicle adjustment factor), using Equation 9.4 1 1 = 0.930 f HV = = 1 + PT ( E T − 1) + P R ( E R − 1) 1 + 0.05( 2.5 − 1) + 0.00 v p =

V

( PHF )( N )( f HV )( f p )

= (3600)/(0.94)(3)(0.930)(1.0) = 1373 pc/h/ln

Step 2: Compute free flow speed using Equation 9.25 FFS = BFFS – f LW – f LC – f M – f A Determine f LW using Table 9.29, f LW = 1.9 Determine f LC using Table 9.34, f LC = 1.5 Determine f M using Table 9.35, f M = 0.0 Determine f A using Table 9.36, f A = 3.0 FFS = 55 – 1.9 – 1.5 – 0.0 – 3.0 = 48.6 mi/h Step 3: Compute average passenger car speed and density to determine LOS. Since FFS = 48.6 mi/h and v p = 1373 pc/h/ln, S = FFS = 48.6 mi/h. Density = (1373 pc/h/ln) / (48.6 mi/h) = 28.3 pc/mi/ln This corresponds to LOS D (Table 9.33).


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9-13 Briefly describe the traffic characteristics associated with each of the levels of service for basic freeway sections.

At LOS A free flow conditions prevail. Vehicles are not impeded by other vehicles. Effects of minor incidents or breakdowns are easily absorbed. At LOS B, reasonably free flow conditions still exist and vehicles continue to travel at free flow speeds. Ability to maneuver within the traffic stream is slightly restricted. Effects of minor incidents or breakdowns are easily absorbed. At LOS C, speeds are at or near the free flow speed, but freedom to maneuver is noticeably restricted. Lane changes are more difficult. Minor incidents result in significant deterioration in local level of service. At LOS D, speeds begin to decline slightly with increasing flows. Freedom to maneuver is noticeably limited and drivers experience reduced physical and psychological comfort. Minor incidents will result in queuing as little space is left in the traffic stream to absorb disruptions. At LOS E, operations are volatile because there are virtually no gaps; volume is at or near capacity. Maneuvers such as lane changing or merging traffic at entrance ramps will result in a disturbance of the traffic flow. Any incident can be expected to cause extensive queues as the traffic stream has no ability to dissipate its effects. At LOS F, breakdown conditions exist and uniform moving flow cannot be maintained. The flow conditions are such that the number of vehicles that can pass a point is less than the number of vehicles arriving at that point.

9-14 Describe the factors that affect the level of service of a freeway section and the impact each has on flow. Lane width - Traffic is restricted when lane widths are narrower than 12 ft. Motorists tend to travel more cautiously because of the reduced lateral distance between vehicles by reducing their speeds. Lateral clearance - Lateral obstructions tend to have an effect similar to reduced lane width. Drivers in the lane adjacent to the obstruction will tend to shy away from the obstruction, moving them closer to vehicles in the adjacent lane and resulting in a reduction in speeds. Lateral obstructions more than 6 ft from the edge of the traveled lane have no significant effect of traffic flow.

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Traffic composition - The effect of large or heavy vehicles in the traffic stream reduces the maximum flow on the highway because of their operating characteristics; a heavy vehicle occupies more space in the traffic stream than does a passenger car. Grade - The effect of a grade depends on the length and slope of the grade. Traffic operations are significantly affected when grades of 3% or greater are 0.25 mile or greater in length and when grades of 3% or less are greater than 0.5 mile in length. The effect of grades on heavy vehicles is much greater than on passenger cars. Driver population - A driver population consisting primarily of weekday commuters will have significantly different behavior than a driver po pulation consisting of drivers unfamiliar with the roadway. For example, recreational traffic capacities can be as much as 20% lower than commuter traffic capacities. Interchange spacing – As interchanges are more closely spaced, the lengths of basic freeway segments unaffected by interchanges decreases as the weaving movements at interchanges have an increasing impact on traffic flow resulting in a reduction of speeds.

9-15 Given: Freeway; design volume of 5000 veh/h; PHF: 0.9; trucks: 10%; design LOS: C; free flow speed: 70 mi/h; no lateral obstructions; rolling terrain; interchange spacing: 3 mi. Determine: Number of 12 ft lanes required in each direction.

Step 1: Compute passenger car equivalent flow rate for peak 15-minute period using Equation 9.21. Determine f p (driver population factor); assume mainly commuter traffic, f p = 1.0 Determine ET (PCE for trucks), using Table 9.25; ET = 2.5 Determine ER (PCE for RVs), using Table 9.25; ER = 2.0 Determine f HV (heavy vehicle adjustment factor), using Equation 9.4 f HV =

1 1 + PT ( E T − 1) + P R ( E R − 1)

=

1 1 + 0.10(2.5 − 1) + 0.00( 2.0 − 1)

= 0.87

Step 2: Assume six lanes (three in each direction) V = (5000)/(0.90)(3)(0.87)(1.0) = 2128 pc/h/ln v p = ( PHF )( N )( f HV )( f p ) Step 3: Compute free flow speed using Equation 9.23 FFS = BFFS – f LW – f LC – f N - f ID Determine f LW using Table 9.29, f LW = 0.0 (c) 2009 Cengage Learning ALL RIGHTS RESERVED.

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Determine f LC using Table 9.30, f LC = 0.0 Determine f N using Table 9.31, f N = 3.0 Determine f ID using Table 9.32, f ID = 0.0 FFS = 70 – 0.0 – 0.0 – 3.0 – 0.0 = 67.0 mi/h Step 4: Compute average passenger car speed and density to determine LOS. For FFS = 67 mi/h and v p = 2128 pc/h/ln, using Figure 9.9, the average passenger car speed, S = 60.0 mi/h Density = (2128 pc/h/ln) / 60.0 mi/h = 35.4 pc/mi/ln Using Table 9.24, this density corresponds to L OS E, therefore, a three lane section is inadequate. Repeat steps 2 and 3. Step 2: Assume eight lanes (four in each direction) V v p = = (5000)/(0.90)(4)(0.87)(1.0) = 1597 pc/h/ln ( PHF )( N )( f HV )( f p ) Step 3: Compute free flow speed using Equation 9.23 FFS = BFFS – f LW – f LC – f N - f ID Determine f LW using Table 9.29 f LW = 0.0 Determine f LC using Table 9.30, f LC = 0.0 Determine f N using Table 9.31, f N = 1.5 Determine f ID using Table 9.32, f ID = 0.0 FFS = 70 – 0.0 – 0.0 – 1.5 – 0.0 = 68.5 mi/h Step 4: Compute average passenger car speed and density to determine LOS. For FFS = 68.5 mi/h and v p = 1597 pc/h/ln, using Figure 9.9, the average passenger car speed, S = 68.1 mi/h Density = (1597 pc/h/ln) / 68.1 mi/h = 23.4 pc/mi/ln Using Table 9.24, this density corresponds to L OS C, therefore, a four lane section is adequate.

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9-16 An existing rural freeway in rolling terrain is to be analyzed to determine LOS using the following information: Number of lanes in each direction: 2 Peak hour volume: 2640 veh/h (in the peak direction) 18% trucks 2% recreational vehicles PHF = 0.91 Lane width: 12 ft Lateral clearance: 10 ft Average interchange spacing: 3 mi

Step 1: Compute passenger car equivalent flow rate for peak 15-minute period using Equation 9.21. Determine f p (driver population factor); assume mainly commuter traffic, f p = 1.0 Determine ET (PCE for trucks), using Table 9.25; ET = 2.5 Determine ER (PCE for RVs), using Table 9.25; ER = 2.0 Determine f HV (heavy vehicle adjustment factor), using Equation 9.4 f HV = v p =

1 1 + PT ( E T − 1) + P R ( E R − 1) V

( PHF )( N )( f HV )( f p )

=

1 1 + 0.18( 2.5 − 1) + 0.02( 2.0 − 1)

= 0.775

= (2640)/(0.91)(2)(0.775)(1.0) = 1871 pc/h/ln

Step 2: Compute free flow speed using Equation 9.23 FFS = BFFS – f LW – f LC – f N - f ID Determine f LW using Table 9.29, f LW = 0.0 Determine f LC using Table 9.30, f LC = 0.0 Determine f N using Table 9.31, f N = 0.0 Determine f ID using Table 9.32, f ID = 0.0 FFS = 75 – 0.0 – 0.0 – 0.0 – 0.0 = 75.0 mi/h Step 3: Compute average passenger car speed and density to determine LOS. For FFS = 75.0 mi/h and v p = 1871 pc/h/ln, 2 .6 ⎡⎛ 160 ⎞⎛ 1871 + (30)(75) − 3400 ⎞ ⎤ ⎟⎟ ⎥ S = 75 − ⎢⎜ 75 − ⎟⎜⎜ 3 ( 30 )( 75 ) 1000 − ⎠⎝ ⎢⎣⎝ ⎠ ⎥⎦

The average passenger car speed, S = 69.8 mi/h Density = (1871 pc/h/ln) / (69.8 mi/h) = 26.8 pc/mi/ln This corresponds to LOS D.


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9-17 An existing urban freeway with 4 lanes in each direction has the following characteristics: Traffic data: Peak hour volume (in the peak direction): 7070 veh/h Trucks: 10% of peak hour volume PHF = 0.94 Geometric data: Lane width: 11 ft Shoulder width: 6 ft Interchange spacing (average): 1.4 mile Terrain: rolling Determine the LOS in the peak hour. Clearly state assumptions used for any values not given. Show the demand flow rate, mean speed, and density for the given conditions.

Step 1: Compute passenger car equivalent flow rate for peak 15-minute period using Equation 9.21. Determine f p (driver population factor); assume mainly commuter traffic, f p = 1.0 Determine ET (PCE for trucks), using Table 9.25; ET = 2.5 Determine f HV (heavy vehicle adjustment factor), using Equation 9.4 f HV = v p =

1 1 + PT ( E T − 1) + P R ( E R − 1) V

( PHF )( N )( f HV )( f p )

=

1 1 + 0.10(2.5 − 1) + 0.00

= 0.870

= (7070)/(0.94)(4)(0.870)(1.0) = 2163 pc/h/ln

Step 2: Compute free flow speed using Equation 9.23 FFS = BFFS – f LW – f LC – f N - f ID Determine f LW using Table 9.29, f LW = 1.9 Determine f LC using Table 9.30, f LC = 0.0 Determine f N using Table 9.31, f N = 1.5 Determine f ID using Table 9.32, f ID = 1.1 FFS = 70 – 1.9 – 0.0 – 1.5 – 1.1 = 65.5 mi/h Step 3: Compute average passenger car speed and density to determine LOS. For FFS = 65.5 mi/h and v p = 2163 pc/h/ln, 2 .6 ⎡1 ⎛ 2163 + (30)(65.5) − 3400 ⎞ ⎤ ⎟⎟ ⎥ S = 65.5 − ⎢ ((7)(65.5) − 340 )⎜⎜ ( 40 )( 65 . 5 ) 1700 − ⎢⎣ 9 ⎝ ⎠ ⎥⎦

The average passenger car speed, S = 58.3 mi/h Density = (2163 pc/h/ln) / (58.3 mi/h) = 37.1 pc/mi/ln This corresponds to LOS E. 137



9-18 An urban freeway is to be designed using the following information: AADT = 44,000 veh/day K (proportion of AADT occurring during the peak hour): 0.10 D (proportion of peak hour traffic traveling in the peak direction): 0.60 Trucks: 10% of peak hour volume PHF = 0.94 Lane width: 12 ft Shoulder width: 10 ft Interchange density: 0.5 interchange/mile Terrain: rolling Determine the number of lanes required to provide LOS C. Clearly state assumptions used for any values not given, and show all calculations required.

Step 1: Compute directional design hourly volume (DDHV). DDHV = (AADT)(K)(D) = (44000)(0.10)(0.60) = 2640 veh/h Step 2: Compute passenger car equivalent flow rate for peak 15-minute period using Equation 9.21. Assume 2 lanes in each direction will be sufficient. Determine f p (driver population factor); assume mainly commuter traffic, f p = 1.0 Determine ET (PCE for trucks), using Table 9.25; ET = 2.5 Determine ER (PCE for RVs), using Table 9.25; ER = 2.0 Determine f HV (heavy vehicle adjustment factor), using Equation 9.4 f HV =

v p =

1 1 + PT ( E T − 1) + P R ( E R − 1) V

( PHF )( N )( f HV )( f p )

=

1 1 + 0.10(2.5 − 1) + 0.00

= 0.870

= (2640)/(0.94)(2)(0.870)(1.0) = 1614 pc/h/ln

Step 3: Compute free flow speed using Equation 9.23 FFS = BFFS – f LW – f LC – f N - f ID Determine f LW using Table 9.29, f LW = 0.0 Determine f LC using Table 9.30, f LC = 0.0 Determine f N using Table 9.31, f N = 4.5 Determine f ID using Table 9.32, f ID = 0.0 FFS = 70 – 0.0 – 0.0 – 4.5 – 0.0 = 65.5 mi/h Step 4: Compute average passenger car speed and density to determine LOS. For FFS = 65.5 mi/h and v p = 1614 pc/h/ln, 2 .6 ⎡1 ⎛ 1614 + (30)(65.5) − 3400 ⎞ ⎤ ⎟⎟ ⎥ S = 65.5 − ⎢ ((7)(65.5) − 340)⎜⎜ ( 40 )( 65 . 5 ) 1700 − ⎢⎣ 9 ⎝ ⎠ ⎥⎦


138


The average passenger car speed, S = 65.3 mi/h Density = (1614 pc/h/ln) / (65.3 mi/h) = 24.7 pc/mi/ln This corresponds to LOS C; N = 2 (two lanes in each direction) are required.

9-19 Given: 4-lane freeway (2 lanes in each direction); segment length: 2 mi; sustained grade: 4%; Design volume of 3000 veh/h; trucks: 10%; buses: 2%; RVs: 3%; PHF: 0.95; free flow speed: 70 mi/h; right side lateral obstruction: 5 ft; design LOS: B. Determine: number of additional lanes required in each direction.

Step 1: Compute passenger car equivalent flow rate for peak 15-minute period using Equation 9.21. Note: this segment is considered to be in mountainous terrain since the maximum sustained grade is 4%. Determine f p (driver population factor); assume mainly commuter traffic, f p = 1.0 Determine ET (PCE for trucks), using Table 9.25; ET = 4.5 Determine ER (PCE for RVs), using Table 9.25; ER = 4.0 Determine f HV (heavy vehicle adjustment factor), using Equation 9.4 f HV =

1 1 + PT ( E T − 1) + P R ( E R − 1)

=

1 1 + 0.12(4.5 − 1) + 0.03( 4.0 − 1)

= 0.66

Step 2: Assume one additional lane in each direction V v p = = (3000)/(0.95)(3)(0.66)(1.0) = 1589 pc/h/ln ( PHF )( N )( f HV )( f p ) Step 3: Compute free flow speed using Equation 9.23 FFS = BFFS – f LW – f LC – f N - f ID Determine f LW using Table 9.29, f LW = 0.0 Determine f LC using Table 9.30, f LC = 0.4 Determine f N using Table 9.31, f N = 3.0 Determine f ID using Table 9.32, f ID = 0.0 FFS = 70 – 0.0 – 0.4 – 3.0 – 0.0 = 66.6 mi/h Step 4: Compute average passenger car speed and density to determine LOS. For FFS = 66.6 mi/h and v p = 1589 pc/h/ln, using Figure 9.9, the average passenger car speed, S = 66.4 mi/h Density = (1589 pc/h/ln) / 66.4 mi/h = 23.9 pc/mi/ln Using Table 9.24, this density corresponds to L OS C, therefore, a three lane section is inadequate. Repeat steps 2 and 3. Step 2: Assume two additional lanes in each direction

139



v p =

V

( PHF )( N )( f HV )( f p )

= (3000)/(0.95)(4)(0.66)(1.0) = 1192 pc/h/ln

Step 3: Compute free flow speed using Equation 9.23 FFS = BFFS – f LW – f LC – f N - f ID Determine f LW using Table 9.29, f LW = 0.0 Determine f LC using Table 9.30, f LC = 0.2 Determine f N using Table 9.31, f N = 1.5 Determine f ID using Table 9.32, f ID = 0.0 FFS = 70 – 0.0 – 0.2 – 1.5 – 0.0 = 68.3 mi/h Step 4: Compute average passenger car speed and density to determine LOS. For FFS = 68.3 mi/h and v p = 1192 pc/h/ln, using Figure 9.9, the average passenger car speed, S = 68.3 mi/h Density = (1192 pc/h/ln) / 68.3 mi/h = 17.5 pc/mi/ln Using Table 9.24, this density corresponds to L OS B, therefore, a four lane section is adequate, requiring two additional lanes in each direction.

9-20 Given: Roadway segment with 6000 ft of 3% upgrade, followed by 5000 ft of 5% upgrade; trucks: 8%; RVs: 4%. Determine: number of PCEs.

Use performance curves provided in Figure 9.13. Assuming an entry speed of 55 mi/h, at the end of the first grade speed is 38 mi/h. At end of the second grade, speed is 27 mi/h, which is also the crawl speed for that grade. Therefore, the effective grade is 5%. From Table 9.26, for 8% trucks, 5% grade, and greater than 2 miles, ET = 3.5 From Table 9.27, for 4% RVs, 5% grade, and greater than 2 miles, ER = 4.5


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