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This Document Contains Chapters 9 to 10 Chapter 9 Capacity and Level of Service for Highway Segments 9-1 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. 9-2 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. Traffic composition – The effect of large or heavy vehicles in the traffic stream reduces the maximum flow on the highway because of 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 population 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. 9-3 Describe the possible situations that would require adjustments from the base conditions for freeway capacity analysis. Base conditions are the criteria that must be satisfied if a basic freeway segment is to operate at the maximum capacities. If one or more of the following conditions is not met, the capacity of the segment will be reduced. Weather and driver visibility are good and do not affect traffic flow. There are no obstructions that could impede traffic flow such as incidents, accidents or roadwork. The HCM 2010 capacity procedure does not offer a computational compensation factor for these conditions. Accordingly, the procedure is based on the assumption that each condition is met. The traffic stream is composed solely of passenger cars and there are no heavy vehicles such as trucks, buses and recreational vehicles (RVs). Traffic lanes are at least 12-ft wide and a right-side clearance exists that is at least 6-ft wide. All drivers in the traffic stream are regular users and are familiar with the freeway segment. The total ramp density is zero. Ramp density is the total number of on and off ramps per mile in one direction within the freeway segment under study. For the ramp density to be zero, there are no ramps within the segment. The HCM 2010 capacity procedure for Basic Freeway Segments applies to segments no greater than 6 miles long. A state of unsaturated flow must exist for the HCM 2010 procedures to apply. This state occurs when the traffic stream is unaffected by upstream or downstream bottlenecks. 9-4 A freeway is to be designed to provide LOS C for the following conditions: design hourly volume of 5600 veh/h; PHF: 0.92; trucks: 6%; free flow speed: 70 mi/h; no lateral obstructions; rolling terrain; total ramp density of 0.75 ramps per mile. Determine: Number of 12 ft lanes required in each direction. Step 1: Compute heavy vehicle adjustment factor Determine fp (driver population factor); assume mainly commuter traffic, fp = 1.0 Determine ET (PCE for trucks), using Table 9.3; ET = 2.5 Determine ER (PCE for RVs), using Table 9.3; ER = 2.0 Determine fHV (heavy vehicle adjustment factor), using Equation 9.5 fHV = 1 = = 0.92 1+ P ET ( T − +1) P ER( R −1) Step 2: Assume six lanes (three in each direction). Use Equation 9.4 vp = V = (5600)/(0.92)(3)(0.92)(1.0) = 2205 pc/h/ln (PHF N)( )( fHV )( fp) Step 3: Compute free flow speed using Equation 9.2 FFS = 75.4 – fLW – fLC – 3.22TRD0.84 Determine fLW using Table 9.1, fLW = 0.0 Determine fLC using Table 9.2, fLC = 0.0 TRD = 0.75 ramps/mile FFS = 75.4 – 0.0 – 0.0 – 3.22(0.75)0.84 FFS = 72.87 mi/h Step 4: Compute average passenger car speed and density to determine LOS. For FFS = 72.87 mi/h, greater than 72.5 and less than 77.5, the 75 mi/h speedflow curve will be used for this analysis. It was calculated that vp = 2205 pc/h/ln. Density = (2205 pc/h/ln) / 75.0 mi/h = 29.4 pc/mi/ln Using Figure 9.3a, this density corresponds to LOS D, therefore, a three lane section is inadequate. Repeat steps 2 and 3. Step 2: Assume eight lanes (four in each direction) vp = V = (5600)/(0.92)(4)(0.92)(1.0) = 1654 pc/h/ln (PHF N)( )( fHV )( fp) Step 3: Compute free flow speed using Equation 9.2 FFS = 72.87 mi/h Step 4: Compute average passenger car speed and density to determine LOS. For FFS = 72.87 mi/h, greater than 72.5 and less than 77.5, the 75 mi/h speedflow curve will be used for this analysis. It was calculated that vp = 1654 pc/h/ln. The expected speed of the traffic stream will be: = 75 − 0.00001107( − 1,000) = 75 − 0.00001107(1654 − 1000) = 70.26 /ℎ The density will be: Density = (1654 pc/h/ln) / 70.26 mi/h = 23.54 pc/mi/ln Using Figure 9.3a, this density corresponds to LOS C, therefore, a four lane section is adequate. 9-5 An existing rural freeway in rolling terrain is to be analyzed to determine LOS using the following information: number of lanes in each direction equals 2, peak hour volume of 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 is 3 mi; all interchanges are diamond interchanges. Step 1: Compute demand flow rate under equivalent base conditions using Equation 9.4 Determine fp (driver population factor); assume mainly commuter traffic, fp = 1.0 Determine ET (PCE for trucks), using Table 9.3; ET = 2.5 Determine ER (PCE for RVs), using Table 9.3; ER = 2.0 Determine fHV (heavy vehicle adjustment factor), using Equation 9.5 fHV = 1 = = 0.775 1+ P ET ( T − +1) P ER( R −1) vp = V = (2640)/(0.91)(2)(0.775)(1.0) = 1871 pc/h/ln (PHF N)( )( fHV )( fp) Step 2: Compute free flow speed using Equation 9.2 FFS = 75.4 – fLW – fLC – 3.22TRD0.84 Determine fLW using Table 9.1, fLW = 0.0 Determine fLC using Table 9.2, fLC = 0.0 TRD = 1/3 = 0.33 ramps/mile FFS = 75.4 – 0.0 – 0.0 – 3.22(0.33)0.84 FFS = 74.13 mi/h Step 3: Compute average passenger car speed and density to determine LOS. For FFS = 74.13 mi/h, greater than 72.5 and less than 77.5, the 75 mi/h speed-flow curve will be used for this analysis. It was calculated that vp = 1871 pc/h/ln. The expected speed of the traffic stream will be: = 75 − 0.00001107( − 1,000) = 75 − 0.00001107(1871 − 1000) = 66.60 /ℎ The density will be: Density = (1871 pc/h/ln) / 66.60 mi/h = 28.09 pc/mi/ln Using Figure 9.3a, this density corresponds to LOS D. 9-6 An existing urban freeway with 4 lanes in each direction has the following characteristics: peak hour volume (in the peak direction) of 7070 veh/h, trucks are 10% of peak hour volume, PHF = 0.94, lane width = 11 ft, shoulder width = 6 ft, total ramp density = 1.8 ramps per mile, rolling terrain. 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 9.4 Determine fp (driver population factor); assume mainly commuter traffic, fp = 1.0 Determine ET (PCE for trucks), using Table 9.3; ET = 2.5 Determine ER (PCE for RVs), using Table 9.3; ER = 2.0 PR = 0.00, as nothing is mentioned about recreational vehicles Determine fHV (heavy vehicle adjustment factor), using Equation 9.5 fHV = 1 = = 0.870 1+ P ET ( T − +1) P ER( R −1) vp = V = (7070)/(0.94)(4)(0.870)(1.0) = 2163 pc/h/ln (PHF N)( )( fHV )( fp) Step 2: Compute free flow speed using Equation 9.2 FFS = 75.4 – fLW – fLC – 3.22TRD0.84 Determine fLW using Table 9.1, fLW = 1.9 Determine fLC using Table 9.2, fLC = 0.0 TRD = 1.8 ramps/mile FFS = 75.4 – 1.9 – 0.0 – 3.22(1.8)0.84 FFS = 68.22 mi/h Step 3: Compute average passenger car speed and density to determine LOS. For FFS = 68.22 mi/h, greater than 67.5 and less than 72.5, the 70 mi/h speed-flow curve will be used for this analysis. It was calculated that vp = 2163 pc/h/ln. The expected speed of the traffic stream will be: = 70 − 0.00001160( − 1,200) = 70 − 0.00001160(2163 − 1200) = 59.24 /ℎ The density will be: Density = (2163 pc/h/ln) / 59.24 mi/h = 36.51 pc/mi/ln Using Figure 9.3a, this density corresponds to LOS E in the peak hour. 9-7 An urban freeway is to be designed using the following information: AADT = 52,000 veh/day; K (proportion of AADT occurring during the peak hour) = 0.11; D (proportion of peak hour traffic traveling in the peak direction) = 0.65; trucks = 8% of peak hour volume; PHF = 0.95; lane width = 12 ft; shoulder width = 10 ft; total ramp density = 0.5 interchange/mile; all interchanges are to be cloverleaf interchanges; rolling terrain. 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) = (52000)(0.11)(0.65) = 3718 veh/h Step 2: Compute passenger car equivalent flow rate for peak 15-minute period using Equation 9.4 Assume 2 lanes in each direction will be sufficient. Determine fp (driver population factor); assume mainly commuter traffic, fp = 1.0 Determine ET (PCE for trucks), using Table 9.3; ET = 2.5 Determine ER (PCE for RVs), using Table 9.3; ER = 2.0 PR = 0.00, as nothing is mentioned about recreational vehicles Determine fHV (heavy vehicle adjustment factor), using Equation 9.5 fHV = 1 = = 0.893 1+ P ET ( T − +1) P ER( R −1) vp = V = (3718)/(0.95)(2)(0.893)(1.0) = 2191 pc/h/ln (PHF N)( )( fHV )( fp) Step 3: Compute free flow speed using Equation 9.2 FFS = 75.4 – fLW – fLC – 3.22TRD0.84 Determine fLW using Table 9.1, fLW = 0.0 Determine fLC using Table 9.2, fLC = 0.0 TRD = 0.5 ramps/mile FFS = 75.4 – 0.0 – 0.0 – 3.22(0.5)0.84 FFS = 73.60 mi/h Step 4: Compute average passenger car speed and density to determine LOS. For FFS = 73.60 mi/h, greater than 72.5 and less than 77.5, the 75 mi/h speedflow curve will be used for this analysis. It was calculated that vp = 2191 pc/h/ln. The expected speed of the traffic stream will be: = 75 − 0.00001107( − 1,000) = 75 − 0.00001107(2191 − 1000) = 59.30 /ℎ The density will be: Density = (2191 pc/h/ln) / 59.30 mi/h = 36.94 pc/mi/ln Using Figure 9.3a, this density corresponds to LOS D in the peak hour, so 2 lanes are not enough. Step 2: Try for 3 lanes vp = V = (3718)/(0.95)(3)(0.893)(1.0) = 1460 pc/h/ln (PHF N)( )( fHV )( fp) FFS = 73.60 mi/h a) Compute average passenger car speed and density to determine LOS. For FFS = 73.60 mi/h, greater than 72.5 and less than 77.5, the 75 mi/h speedflow curve will be used for this analysis. It was calculated that vp = 1460 pc/h/ln. The expected speed of the traffic stream will be: = 75 − 0.00001107( − 1,000) = 75 − 0.00001107(1460 − 1000) = 72.66 /ℎ The density will be: Density = (1460 pc/h/ln) / 72.66 mi/h = 20.09 pc/mi/ln Using Figure 9.3a, this density corresponds to LOS C, so 3 lanes are adequate. 9-8 An existing 4-lane freeway (2 lanes in each direction) is to be expanded. The segment length is 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 9.4. Note: this segment is considered to be in mountainous terrain since the maximum sustained grade is 4%. Determine fp (driver population factor); assume mainly commuter traffic, fp = 1.0 Determine ET (PCE for trucks), using Table 9.3; ET = 4.5 Determine ER (PCE for RVs), using Table 9.3; ER = 4.0 Determine fHV (heavy vehicle adjustment factor), using Equation 9.5 fHV = 1 = = 0.66 1+ P ET ( T − +1) P ER( R −1) Step 2: Assume one additional lane in each direction vp = V = (3000)/(0.95)(3)(0.66)(1.0) = 1589 pc/h/ln (PHF N)( )( fHV )( fp) Step 3: Compute free flow speed using Equation 9.2 FFS = 75.4 – fLW – fLC – 3.22TRD0.84 Determine fLW using Table 9.1, fLW = 0.0 Determine fLC using Table 9.2, fLC = 0.4 TRD = ½ = 0.5 ramps/mile FFS = 75.4 – 0.0 – 0.4 – 3.22(0.5)0.84 FFS = 73.20 mi/h Step 4: Compute average passenger car speed and density to determine LOS. For FFS = 73.20 mi/h, greater than 72.5 and less than 77.5, the 75 mi/h speedflow curve will be used for this analysis. It was calculated that vp = 1589 pc/h/ln. The expected speed of the traffic stream will be: = 75 − 0.00001107( − 1,000) = 75 − 0.00001107(1589 − 1000) = 71.16 /ℎ The density will be: Density = (1589 pc/h/ln) / 71.16 mi/h = 22.32 pc/mi/ln Using Figure 9.3a, this density corresponds to LOS C, therefore, a three lane section is inadequate. Repeat steps 2 and 3. Step 2: Assume two additional lanes in each direction vp = V = (3000)/(0.95)(4)(0.66)(1.0) = 1192 pc/h/ln (PHF N)( )( fHV )( fp) Step 3: Compute free flow speed using Equation 9.2 FFS = 75.4 – fLW – fLC – 3.22TRD0.84 Determine fLW using Table 9.1, fLW = 0.0 Determine fLC using Table 9.2, fLC = 0.2 TRD = ½ = 0.5 ramps/mile FFS = 75.4 – 0.0 – 0.2 – 3.22(0.5)0.84 FFS = 73.40 mi/h Step 4: Compute average passenger car speed and density to determine LOS. For FFS = 73.40 mi/h, greater than 72.5 and less than 77.5, the 75 mi/h speedflow curve will be used for this analysis. It was calculated that vp = 1192 pc/h/ln. The expected speed of the traffic stream will be: = 75 − 0.00001107( − 1,000) = 75 − 0.00001107(1192 − 1000) = 74.59 /ℎ The density will be: Density = (1192 pc/h/ln) / 74.59 mi/h = 15.98 pc/mi/ln Using Figure 9.3a, this density corresponds to LOS B, therefore, a four lane section is adequate, requiring two additional lanes in each direction. 9-9 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 9.4. 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 9.4, for 8% trucks, 5% grade, and greater than 1 mile, ET = 3.5 From Table 9.5, for 4% RVs, 5% grade, and greater than 0.5 mile, ER = 3.5 9-10 Describe the situations that require adjustments from the base conditions for multilane highway capacity analysis. Base conditions for multilane highways include the following elements (other characteristics such as total lateral clearance (TLC), median type, and accesspoint density will also impact the FFS of multilane highways). If one or more of the following conditions is not met, base conditions require adjustments. Weather and driver visibility are good and do not affect traffic flow. There are no obstructions that could impede traffic flow such as incidents, accidents or roadwork. The HCM procedures are based on the assumption that each condition is met. If this is not the case, the speed, LOS and capacity is likely to be lower than estimated. The traffic stream is composed solely of passenger cars and there are no heavy vehicles such as trucks, buses and recreational vehicles (RVs). All drivers in the traffic stream are regular users and are familiar with the freeway segment. 9-11 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 include directional design hourly volume = 3600 veh/h, PHF = 0.94, assumed base free flow speed = 55 mi/h. Geometric data include: urban setting, rolling terrain, lane width = 11 ft, shoulder widths = 4 ft (right side) and 1 ft (left side), and 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 9.4. Determine fp (driver population factor); assume mainly commuter traffic, fp = 1.0 Determine ET (PCE for trucks), using Table 9.3; ET = 2.5 Determine fHV (heavy vehicle adjustment factor), using Equation 9.5 fHV = 1 = = 0.930 1+ P ET ( T − +1) P ER( R −1) vp = V = (3600)/(0.94)(3)(0.930)(1.0) = 1373 pc/h/ln (PHF N)( )( fHV )( fp) Step 2: Compute free flow speed using Equation 9.7 FFS = BFFS – fLW – fLC – fM – fA Determine fLW using Table 9.1, fLW = 1.9 Determine fLC using Table 9.7, fLC = 1.5 Determine fM using Table 9.8, fM = 0.0 Determine fA using Table 9.9, fA = 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. For FFS = 48.6 mi/h, greater than 47.5 and less than 52.5, the 50 mi/h speed-flow curve will be used for this analysis. It was calculated that vp = 1373 pc/h/ln. The expected speed of the traffic stream will be 50mi/h, as vp <1,400 pc/h/ln. The density will be: Density = (1373 pc/h/ln) / 50 mi/h = 27.46 pc/mi/ln This corresponds to LOS D (Table 9.10). 9-12 Define the elements of a Class I, Class II, and Class III two-lane highway. Class I. Two-lane highways whose function it is to serve as primary arterials, daily commuter routes, and links to other state or national highway networks. Motorists’ expectations are that travel will be at relatively high speeds. Class II. Two-lane highways whose function it is to serve as access to Class I highways. They also serve as scenic byways and can be used by motorists for sightseeing; some are located in rugged terrain. Average trip lengths are shorter than on Class I highways. The expectation of motorists is that travel speeds will be lower than for Class I roads. Class III. Two-lane highways that serve moderately developed areas. They may be a portion of a Class I or Class II highway passing through a small town or a recreational area. These segments may be used by local traffic and the number of unsignalized access points is greater than in rural areas. They may contain longer segments that pass through spread-out recreational areas with increased roadside activity and reduced speed limits. 9-13 What are the three 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 Class II, and Class III highways? There are three measures used to describe the service quality of a two-lane highway. These are (1) percent time following another vehicle (PTSF) (2) average travel speed (ATS) (3) and percent of free-flow speed (PFFS). LOS for Class I highways is based on two measures: PTSF and ATS. Because speed and delay due to passing restrictions are important factors to these motorists. LOS for Class II highways is based on a single measure: PTSF. As travel speed is not a key factor but delay measures the motorist’s sense of the quality of the trip. LOS for Class III highways is based on a single measure: PFFS. In this situation where the length of the Class III segment is limited, neither high speeds nor passing restrictions are of major concerns. Motorists in a Class III situation are most concerned with the ability to maintain steady progress through these sections at or near the speed limit. 9-14 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 increase significantly. In LOS D, flow begins to become unstable, and passing maneuvers are extremely difficult if not 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. 9-15 The following values of PTSF, FFS, and ATS have been determined for three separate two-lane segments. Determine LOS if the segments are: (a) Class I, (b) Class II, and (c) Class III.

Segment PTSF (%) ATS (mi/h) FFS (mi/h)

10 25 52 60

11 46 49 55

12 67 39 60

For a Class I highway, using Table 9.11, segment 10 operates at LOS B (when PTSF and ATS correspond to differing levels of service, the lower value of LOS is used), segment 11 operates at LOS C, and segment 12 operates at LOS E. For a Class II highway, using Table 9.11, segment 10 operates at LOS A, segment 11 operates at LOS B, and segment 12 operates at LOS C. For a Class III highway, using Table 9.11: PFFS=ATS/FFS Segment 10: PFFS=52/60= 86.7% Segment 11: PFFS=49/55= 89.1% Segment 12: PFFS=39/60= 65.0% Segment 10 operates at LOS B, segment 11 operates at LOS B, and segment 12 operates at LOS E. 9-16 Determine the PTSF in each direction 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%. vvphd = V/PHF vvphd = 660/0.97 = 680.4 veh/h vvpho = V/PHF vvpho = 440/0.97 = 453.6 veh/h Step 1: Compute passenger car equivalent flow rate for peak 15-minute period, vp a) For 60% traffic (660 veh/h) Determine fg (grade adjustment factor), using Table 9.22; fg = 1.00 Determine ETd (PCE for trucks), using Table 9.23; ETd = 1.0 Determine ERd (PCE for RVs), using Table 9.23; ERd = 1.0 Determine fHVd (heavy vehicle adjustment factor), using Equation 9.5 fHVd = 1 = = 1.00 1+ P ET ( T − +1) P ER( R −1) Determine vpd (flow rate), using Equation 9.14 vpd = Vi = (660)/(0.97)(1.00)(1.0) = 680.41 pc/h (PHF)(fHV )(fG) b) For 40% traffic (440 veh/h) Determine ETo (PCE for trucks), using Table 9.23; ETo = 1.1 Determine ERo (PCE for RVs), using Table 9.23; ERo = 1.0 Determine fHVo (heavy vehicle adjustment factor), using Equation 9.5 fHVo = 1 = = 0.99 1+ P ET ( T − +1) P ER( R −1) Determine vpo (flow rate), using Equation 9.14 vpo = Vi = (440)/(0.97)(0.99)(1.0) = 458.19 pc/h (PHF)( fHV )( fG ) Step 2: Compute base percent time spent following, BPTSF, using Equation 9.17. coefficients a and b from Table 9.26. a) For 60% traffic (660 veh/h) = 100 1 − = 100 1 − (−0.0025 × 680.41.) =60.68% b) For 40% traffic (440 veh/h) = 100 1 − = 100 1 − (−0.00378 × 458.19.) =50.95% Step 3: Compute percent time spent following, PTSF, using Equation 9.16 a) For 60% traffic (660 veh/h) = + , , , + , Determine , using Table 9.25 v = 680.41 + 458.19 = 1138.6 pc/h By interpolation, ,= 27.02% = 60.68 + 27.02 = 76.83% b) For 40% traffic (440 veh/h) = + , , , + , = 50.95 + 27.02 = 61.82% 9-17 Use the data provided in Problem 9-16 to estimate the ATS in each direction. Base free flow speed: 55 mi/h; lane width: 11 ft; shoulder width: 3 ft; access points per mile: 15. Step 1: Equation 9.9, estimate free flow speed. Adjustment coefficients fLS and fA from Tables 9.12 and 9.13, respectively. FFS = BFFS – fLS – fA FFS = 55 – 3.0 – 3.75 FFS =48.25 mi/h Step 2: Estimate heavy vehicle adjustment factor, use Equation 9.11: , ( ) ( ) For 60% traffic (660veh/h) PT = 0.10 ET = 1.1 (Table 9.16) PR = 0.07 ER = 1.0 (Table 9.16) , .(.) . ( . ) , . For 40% traffic (440 veh/h) PT = 0.10 ET = 1.3 (Table 9.16) PR = 0.07 ER = 1.0 (Table 9.16) , . Step 3: Estimate demand volume for each direction. Use Equation 9.10. For 60% traffic (660 veh/h) Vd vd = = (660)/(0.97)(1.00)(0.99) = 687.28 pc/h (PHF)( fg ATS, )(fHV ATSd, ) For 40% traffic (440 veh/h) v = Vo = (440)/(0.97)(1.0)(0.97) = 467.64 pc/h o (PHF)( fg ATS, )( fHV ATSo, ) Step 4: Estimate ATS for each direction. Use Equation 9.13. Adjustment factors for no-passing zones interpolated from Table 9.20. = − 0.00776 , + ,−, For 60% traffic (660 veh/h) = 48.25 − 0.00776(687.28 + 467.64) − 1.4 = 37.88 /ℎ For 40% traffic (440 veh/h) = 48.25 − 0.00776(687.28 + 467.64) − 0.8 = 38.49 /ℎ 9-18 Use the results of Problems 9-16 and 9-17 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. For the main direction (660 veh/h): Step 1: LOS determination. See Table 9.11. If Class I Highway: ATS = 37.88 mi/h PTSF = 76.83% LOS = E If Class II Highway: PTSF = 76.83% LOS = D If Class III Highway: ATS = 37.88 mi/h FFS = 48.25 mi/h PFFS = 78.51% LOS = C Step 2: Ratio v/c. Use vd from ATS estimation, as it provides a higher v/c ratio. / = 1,700 / = / = 0.40 Step 3: Estimate vehicle miles traveled. Use Equation 9.26 VMT15 = 0.25( Vi )Lt = (0.25)(660)(4.5)/(0.97) = 765.46 veh-mi PHF Step 4: Estimate total travel time. Use Equation 9.27. TT15 = VMT15/ATSi = 765.46/37.88 = 20.21 veh-h 9-19 Use the data and results obtained in Problems 9-16 and 17 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, PTSFpl and ATSpl, must first be determined. From Problem 9-16, PTSFd = 76.83% and ATSd = 37.88 mi/h Step 1: Determine region lengths. Region I: Lu = 0.75 mi Region II: Lpl = 1.5 mi Region III: For PTSF, from Table 9.26, Lde = 5.8 mi; from Table 9.29, fpl = 0.62 For ATS, from Table 9.26, Lde = 1.7 mi; from Table 9.30, fpl = 1.11 Region IV:Use Equation 9.21. Ld = Lt - (Lu +Lpl + Lde ) For PTSF, Ld = 4.5 – 0.75 –1.5 – 5.8 = –3.55; use Ld = 0 and L’de = 4.5 – 0.75 – 1.5 = 2.25 For ATS, Ld = 4.5 – 0.75 – 1.5 – 1.7 = 0.55 Step 2: Compute PTSFpl using Equation 9.23 1− f pl Lde′ 2 + ( f ) + ( PTSFpl = Lt 76.83[0.75 + (0.62)(2.25) + (1− 0.62)(2.252 )] PTSFpl = 2 5.8 = 40.48% 4.5 Step 3: Compute ATSpl using Equation 9.24 ATS pl = (ATSd )( )Lt L 2L Lu + d f pl f pl +1 ATS pl = (37.88)(4.5) = 39.99 mi/h 0.75 + 1.5 + 0.55 1.11 1.11 1 Step 4: Determine level of service from Table 9.11 (for Class I highway): LOS E. 9-20 An existing Class I two-lane highway is to be analyzed to determine the level of service in each direction, 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 Analysis for the peak direction: 360 veh/h vvphd = V/PHF vvphd = 360/0.86 = 418.6 veh/h vvpho = V/PHF vvpho = 240/0.86 = 279.1 veh/h Step 1: Compute passenger car equivalent flow rate for peak 15-minute period, vp Main direction: Determine fg (grade adjustment factor), using Table 9.22; fg = 0.90 Determine ETd (PCE for trucks), using Table 9.23; ETd = 1.6 Determine ERd (PCE for RVs), using Table 9.23; ERd = 1.0 Determine fHVd (heavy vehicle adjustment factor), using Equation 9.5 fHVd = 1 = = 0.95 1+ P ET ( T − +1) P ER( R −1) Opposite direction: Determine fg (grade adjustment factor), using Table 9.22; fg = 0.81 Determine ETo (PCE for trucks), using Table 9.23; ETo = 1.7 Determine ERo (PCE for RVs), using Table 9.23; ERo = 1.0 Determine fHVo (heavy vehicle adjustment factor), using Equation 9.5 fHVo = 1 = = 0.95 1+ P ET ( T − +1) P ER( R −1) Determine vpd (flow rate), using Equation 9.14 vpd = Vi = (360)/(0.86)(0.95)(0.90) = 489.60 pc/h (PHF)(fHV )(fG) vpo = Vi = (240)/(0.86)(0.95)(0.81) = 362.66 pc/h (PHF)( fHV )( fG ) Step 2: Compute base percent time spent following, BPTSF, using Equation 9.17. coefficients a and b from Table 9.26. = 100 1 − = 100 1 − (−0.0015 × 489.60.) =44.43% Step 3: Compute percent time spent following, PTSF, using Equation 9.16 = + , , , + , Determine , using Table 9.25 v = 489.60 + 362.66 = 852.26 pc/h By interpolation, ,= 32.4% = 44.43 + 32.4 = 63.86% Step 4: Equation 9.9, estimate free flow speed. Adjustment coefficients fLS and fA from Tables 9.12 and 9.13, respectively. FFS = BFFS – fLS – fA FFS = 55 – 3.0 – 2.0 FFS =50 mi/h Step 5: Estimate heavy vehicle adjustment factor, use Equation 9.11: , ( ) ( ) a) Main direction: PT = 0.08 ET = 2.0 (Table 9.16) PR = 0.02 ER = 1.1 (Table 9.16) , .(.) . ( . ) , .

b) Opposite direction: PT = 0.08 ET = 2.2 (Table 9.16) PR = 0.02 ER = 1.1 (Table 9.16)

,

.(.) . ( . ) , . Step 6: Estimate demand volume for each direction. Use Equation 9.10. vd = Vd = (360)/(0.86)(0.87)(0.92) = 522.99 pc/h (PHF)( fg ATS, )( fHV ATSd, ) vo = Vo = (240)/(0.86)(0.81)(0.91) = 378.60 pc/h (PHF)(fg ATS, )(fHV ATSo, ) Step 4: Estimate ATS. Use Equation 9.13. Adjustment factors for no-passing zones interpolated from Table 9.20. = − 0.00776 , + ,−, = 50 − 0.00776(522.99 + 378.60) − 1.7 = 41.30 /ℎ Step 5: LOS. Table 9.11 ATS = 41.30 mi/h PTSF = 63.86% LOS = D 9-21 An existing Class II two-lane highway is to be analyzed to determine the level of service in each 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 Analysis for the peak direction: 900 veh/h vvphd = V/PHF vvphd = 900/0.95 = 947.4 veh/h vvpho = V/PHF vvpho = 400/0.95 = 421.1 veh/h Step 1: Compute passenger car equivalent flow rate for peak 15-minute period, vp a) Main direction: Determine fg (grade adjustment factor), using Table 9.22; fg = 1.00 Determine ETd (PCE for trucks), using Table 9.23; ETd = 1.0 Determine ERd (PCE for RVs), using Table 9.23; ERd = 1.0 Determine fHVd (heavy vehicle adjustment factor), using Equation 9.5 fHVd = 1 = = 1.00 1+ P ET ( T − +1) P ER( R −1) b) Opposite direction: Determine fg (grade adjustment factor), using Table 9.22; fg = 0.91 Determine ETo (PCE for trucks), using Table 9.23; ETo = 1.6 Determine ERo (PCE for RVs), using Table 9.23; ERo = 1.0 Determine fHVo (heavy vehicle adjustment factor), using Equation 9.5 fHVo = 1 = = 0.93 1+ P ET ( T − +1) P ER( R −1) Determine vpd (flow rate), using Equation 9.14 vpd = Vi = (900)/(0.95)(1.00)(1.00) = 947.4 pc/h (PHF)( fHV )( fG) vpo = Vi = (400)/(0.95)(0.93)(0.91) = 497.5 pc/h (PHF)(fHV )(fG) Step 2: Compute base percent time spent following, BPTSF, using Equation 9.17. coefficients a and b from Table 9.26. = 100 1 − = 100 1 − (−0.0027 × 947.4.) = 71.71% Step 3: Compute percent time spent following, PTSF, using Equation 9.16 = + , , , + , Determine , using Table 9.25 v = 947.4 + 497.5 = 1444.9 pc/h Assume directional split 70/30. By interpolation, ,= 17.2% = 71.71 + 17.2 = 82.99% Step 4: LOS. Table 9.11 PTSF = 82.99% LOS = E 9-22 An existing Class III two-lane highway is to be analyzed to determine the level of service in each direction given the following information: Peak hourly volume in the analysis direction: 900 veh/h Peak hourly volume in the opposing direction: 720 veh/h Trucks: 10% of total volume Recreational vehicles: 2% of total volume PHF: 0.94 Lane width: 12 ft Shoulder width: 2 ft Access points per mile: 30 Terrain: level Measured free flow speed: 45 mi/h No passing zones: 60% of analysis segment length Analysis for the peak direction: 900 veh/h vvphd = V/PHF vvphd = 900/0.94 = 957.4 veh/h vvpho = V/PHF vvpho = 720/0.94 = 766.0 veh/h Step 1: Estimate heavy vehicle adjustment factor, use Equation 9.11: , ( ) ( ) Main direction: PT = 0.10 ET = 1.0 (Table 9.16) PR = 0.02 ER = 1.0 (Table 9.16) , .(.) . ( . ) , . Opposite direction: PT = 0.10 ET = 1.1 (Table 9.16) PR = 0.02 ER = 1.0 (Table 9.16) , .(.) . ( . ) , . Step 2: Estimate demand volume for each direction. Use Equation 9.10. vd = Vd = (900)/(0.94)(1.00)(1.00) = 957.4 pc/h (PHF)( fg ATS, )( fHV ATSd, ) vo = Vo = (720)/(0.94)(1.00)(0.99) = 773.7 pc/h (PHF)(fg ATS, )(fHV ATSo, ) Step 3: Estimate ATS. Use Equation 9.13. Adjustment factors for no-passing zones interpolated from Table 9.20. = − 0.00776 , + ,−, = 45 − 0.00776(957.4 + 773.7) − 0.9 = 30.66 /ℎ Step 4: LOS. Table 9.11 ATS = 30.66 mi/h FFS = 45 mi/h PFFS = ATS/FFS = 30.66/45 = 68.1% LOS = D Chapter 10 Capacity and Level of Service at Signalized Intersections 10-1 Explain fundamental differences between how LOS is determined for the automobile mode compared with human-powered modes. The procedures for the automobile LOS deal with the computation of the level of service at each lane group, intersection approach, and at the intersection as a whole. The LOS at each intersection approach and the entire intersection is represented by only the control delay, while the LOS for a lane group is represented by both the control delay and the volume to capacity ratio. Six levels of service are prescribed. The criteria for each are described below and are shown in Table 10.1. The criteria for the non-automobile modes are based on scores that are computed from factors that can be described as either performance measures such as pedestrian delay or are indicators of the intersection characteristics such as pedestrian corner circulation area. The relationship between the computed scores and the LOS was obtained from the results of a survey in which travelers rated their perceived quality of service for a specific crossing of a signalized intersection. Table 10.2 gives the relationship between the scores and the LOS for the non-automobile mode. 10-2 What factors influence the volume of vehicles turning right on a red indication and why is this important? What does the Highway Capacity Manual recommend if field data on rights turns on red cannot be collected? Many factors influence the RTOR flow rate, which makes it difficult to predict. These include whether right turns are on shared or exclusive lane, the right turn flow rate, v/c ratio for conflicting movements, volume of conflicting pedestrians, etc. The HCM suggests that actual data should be obtained in the field for this traffic element. The HCM also recommends the conservative approach of using a value of 0 veh/h when field data are not available, or when future conditions are being considered. This is important because the delay incurred by right-turning traffic is affected by its ability to turn on red. 10-3 Name and briefly describe the five flow or movement rates that are inputs to the automobile LOS procedure for signalized intersections. The Base Saturation Flow Rate (so) is the maximum rate of flow across the stop line that can occur in a traffic lane. Default values given by the HCM are1900 pc/ln/h for metropolitan areas having a population of 250,000 and 1,700 pc/ln/h for all other areas. The pedestrian flow rate is computed from the number of pedestrians in the crosswalk that interrupt the flow of right-turning vehicles from the approach under consideration, during the analysis period. Pedestrian counts should be taken for each walking direction of the pedestrians. Pedestrian flow rate is given as pedestrians per hour. The bicycle flow rate is computed from the number of bicycles with a travel path that results in an interruption of the flow of right-turning vehicles from the approach under consideration, during the analysis period. Bicycle flow rate is given as bicycles per hour. The On-Street Parking Maneuver Rate is the number of parking maneuvers per hour during the analysis period that have an impact on an intersection leg. The HCM suggests that a practical limit of 180maneuvers/h should be used. It should be measured on one side of the road on two-way legs and on both sides on a one-way leg. The Local Bus Stopping Rate is given as the number of buses per hour during the analysis period that block traffic by stopping for embarking or disembarking passengers within 250 ft upstream of the stop line. The HCM suggests that a practical upper limit of 250 buses/ h should be used. 10-4 Determine the adjusted saturation flow rate for a lane group comprised of one lane for the through movement, under the following conditions: Base rate = 1900 pc/h/ln Lane width = 11 ft Heavy vehicles = 4% of the traffic stream Approach grade = +3% No on-street parking No bus stops Bicycle and pedestrian traffic conflicting with this lane group is negligible Intersection is in a central business district Use Equation 10.8 to determine saturation flow rate: s =( )(so fw)( fHV )( fg)( fp)( fa)( fbb)( fLu)( fRT )( fLT )( fLPb)( fRPb) so = 1900 (pc/h/ln) N = 1 fw = 1.00 (Table 10.7) fHV = use Equation 10.9: = = 100 + 4(2.0 − 1) = 0.962 fg = use Equation 10.10: = 1 − 200 = 1 − = 0.985 fp = no on-street parking, 1.00 fa = 0.90 fbb = no bus stops, 1.00 fLu = one lane, 1.00 fRT = 1.0 fLT = 1.0 fLpb = 1.0 fRpb = 1.0 = 1900 × 1.00 × 0.962 × 0.985 × 1.00 × 0.90 × 1.00 × 1.00 × 1.00 × 1.00 × 1.00 × 1.00 = 1620 /ℎ/ 10-5 Determine the capacity of the lane group described in Problem 10-4 if the effective green time for this movement is 35 seconds and the total cycle length is 60 seconds. Use Equation 10.16: = = 1 × 1620 × = 945 ℎ/ℎ 10-6 Describe the three components of total control delay for a given lane group. Uniform Delay. The uniform delay is that which will occur in a lane group if vehicles arrive uniformly at the intersection, and saturation does not occur during any cycle. Also, only one effective green time for the cycle and one saturation flow rate for the analysis period are assumed. Incremental Delay. Incremental delay takes into consideration that the arrivals are not uniform but random, with the number of arrivals varying from cycle to cycle. It also takes into consideration the delay that occurs when demand is higher than the capacity during the analysis period. Initial Queue Delay. This delay occurs as a result of an initial unmet demand (Ob) vehicles at the start of the analysis period T. That is, a residual event of length Qb exists at the start of the analysis period. This does not include vehicles in queue due to the random variation in demand from cycle to cycle. 10-7 Explain how the total control delay for each lane group at an intersection is used to determine the total control delay for the entire intersection. Having determined the average stopped delay for each lane group, the average stopped delay for any approach can be determined as the weighted average of the stopped delays of all lane groups on that approach. The average intersection control delay is found in a manner similar to the approach delay. In this case, the weighted average of the delays at all approaches is the average stopped delay at the intersection. 10-8 Determine the saturation flow rate for a left-turn lane that operates in a permittedonly mode, under the following conditions: Opposing demand flow rate = 1100 veh/h Lane width = 11 ft Heavy vehicles = 3% of the traffic stream Approach grade = –2% No on-street parking No bus stops Bicycle and pedestrian traffic conflicting with this lane group is negligible Intersection is not in a central business district Step 1: Determine sp,lf. Use Equation 10.54: / , 1100 × × . / , , 521 /ℎ/ Step 2: Determine the saturation flow rate for the left-turn lane. Use Equation 10.53 , , = , , = 521 /ℎ/ fw = 1.00 (Table 10.7) fHV = use Equation 10.9: = = 100 + 3(2.0 − 1) = 0.971 fg = use Equation 10.10: = 1 − 200 = 1 − = 1.01 fp = no on-street parking, 1.00 fa = 1.00 fbb = no bus stops, 1.00 fLu = one lane, 1.00 fLpb = 1.0 , , = 521 × 1.00 × 0.971 × 1.01 × 1.0 × 1.0 × 1.0 × 1.0 × 1.0 , , = 511 ℎ/ℎ/ 10-9 Determine the pedestrian LOS for pedestrians using a crosswalk under the following peak hour conditions: Number of traffic lanes crossed = 3 Conflicting right-turn-on-red flow rate = 25 veh/h Conflicting right turns are not channelized Conflicting permitted left turn flow rate = 35 veh/h Total vehicular movements conflicting with the subject crosswalk = 960 veh/h 85th percentile speed of street being crossed = 40 mi/h Additionally, a previous analysis found pedestrian delay to be 24 seconds per pedestrian. Use Equation 10.88 to calculate pedestrian LOS score for intersection. Step 1: Cross section adjustment factor. Equation 10.89: = 0.681( ). = 0.681(3). = 1.1978 Step 2: Motorized vehicle volume adjusted factor. Equation 10.90: + , = 0.00569 − ,0.0027, − 0.1946 4 = 0.00569 − 0 = 0.08535 Step 3: Motorized vehicle speed adjustment factor. Equation 10.91: = 0.00013 , , Equation 10.93 can be used to calculate n15,mj: 0.25 , = 0.25 , = × 960 , = 80 ℎ/ = 0.00013 × 80 × 40 = 0.416 Step 4: Pedestrian delay adjusted factor. Equation 10.92: = 0.0401 ( ,) = 0.0401 (24) = 0.1274 The pedestrian LOS score for intersection will be: , = 0.5997 + + + + , = 0.5997 + 1.1978 + 0.08535 + 0.416 + 0.1274 , = 2.4262 The level of service is LOS B (Table 10.2) 10-10 Determine the bicycle LOS for bicyclists under the following peak hour conditions: The street consists of three 12-ft travel lanes (one in each direction and a twoway left turn lane) and a 5-ft bicycle lane on each side (for a total pavement width of 46 ft The street has a total approach volume of 650 veh/h Use Equation 10.96 to calculate bicycle LOS score for intersection. Step 1: Cross section adjustment factor. Equation 10.97: = 0.0153 − 0.2144 Equation 10.99 can be used to calculate Wt: = + + + ∗ = 12 + 5 + 0 = 17 = 0.0153 × 46 − 0.2144 × 17 = −2.941 Step 2: Motorized vehicle volume adjusted factor. Equation 10.98: + + = 0.0066 4 = 0.0066 = 1.0725 Step 4: The bicycle LOS score for intersection will be: , = 4.1324 + + , = 4.1324 − 2.941 + 1.0725 , = 2.2639 The level of service is LOS B (Table 10.2) Solution Manual for Traffic and Highway Engineering Nicholas J. Garber, Lester A. Hoel 9781133605157