HIGHWAY CAPACITY
Capacity
• Capacity: It is the ability of a road to accommodate
traffic volume. It is the maximum hourly flow rate at
which the maximum number of vehicles per unit
time can reasonably be expected to cross a point on
a roadway during a given time period under
prevailing traffic roadway and control condition.
• Highways capacity depends on
– Road way characteristics-Geometric design elements
– Traffic conditions-type of vehicle in traffic streams
– Control conditions-TCDs
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Capacity Unit
• Passenger car unit: It is a vehicle unit used for
expressing highway capacity. One car is considered as a
single unit, cycle, motorcycle is considered as half car
unit.
• Bus , truck causes a lot of inconvenience because of its
large size and is considered equivalent to 3 cars or 3
PCU.
• For multi lane highway it is 2000pcphpl, and for 2-lane
highway is 2800pcph (passenger car per hour).
• Units : vph (vehicle per hour or vphpl (vehicle Per hour
per lane)
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PCU
Type of vehicle PCU
Car, taxi, pick up 1.0
Cycle, motor cycle 0.5
Bus, truck, 3.0 (4.o in some cases)
Horse drawn cart 4.0
Bullock cart 6.0
Bullock cart (Large) 8.0
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Capacity
• Basic capacity: the maximum volume of vehicles per
hour that can pass a certain point or section of a road
in a given time under the ideal condition (most ideal
road way, traffic and control conditions that can
possibly be attained). It assume that all the vehicles are
travelling at the same speed and minimum spacing is
allowed.
• Possible capacity: the maximum number of vehicles
that can pass a given section during a given period of
time under prevailing (most frequent/usual) roadway,
traffic and control condition.
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Capacity
• Practical/design capacity: Capacity without the
traffic density being so great as to cause
unreasonable delays, hazard or restriction to the
drivers freedom under the prevailing condition of
road way, traffic and control.
• Basic capacity>possible capacity>design capacity
• It is carried out for design and extension of existence
roads. relate demand or existing flow levels,
geometric characteristics, and controls to measures
of the resulting quality of operations.
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Factors affecting capacity
• Lane width
• Width of shoulder
• Lateral clearance
• Commercial vehicles
• Road alignment and geometry (curves,
Superelevation etc)
• Existence of intersections.
• One way or two way traffic and number of lanes
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Factors affecting capacity
• Drivers and vehicular characteristics
• Single type or mixed traffic
• Flow speed
• Weather condition
• Parking
• Presence of pedestrians
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Level of service
Level of service (LOS) is a quality measure describing
operational conditions within a traffic stream, generally
in terms of such service measures
as speed and travel time, driver freedom, traffic
interruptions, comfort and convenience.
Highway capacity manual (HCM) divides the quality of
traffic into six levels ranging from level A to level F.
Level A represents the best quality of traffic where the
driver has the freedom to drive with free flow speed
and level F represents the worst quality of traffic.
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Multilane Highway LOS
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LOS
• LOS A: density is low enough that closeness of vehicle do not
effect vehicle movement
• LOS E: No usable gap between vehicles, speed is slow,,
condition can easily cross over into LOS F region.
• LOS F: Breakdown condition, number of vehicles arriving >
number of vehicles leaving, speed is zero
LOS Max. Density (PC/min/lane)
A 12
B 20
C 30
D 42
E 67
F >67 13
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Factors affecting LOS
• Speed and Travel time
• Traffic interruptions/restrictions
• Freedom to travel with desired speed
• Driver comfort and convenience
• Lane width
• Lateral obstruction
• Traffic composition
• Grade and
• Driver Population
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LOS F
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LOS
• Elements to evaluate LOS:
• Travel speed and travel time
• Volume to capacity (V/C) ratio
• Density (number of vehicles per unit length on
highway, D=F/S, where S is speed and F is flow in pcph)
• Delay (at intersection, signals)
If V/C =1 (level E) , V/C >1 level of service is low
• Service flow =SF=V/PHF=Peak hour volume/PHF
• Peak hour factor (PHF): The ratio of Hourly volume to
max flow rate or service flow i.e. V15 X 4
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Example
• Find service flow rate (capacity), level of
service (LOS), for an existing facility of 6 lanes,
peak hr traffic=3000vph (10% trucks, 3%
buses), PHF=0.9
• Solution: As PHF=peak hr volume/service flow
• SF=peak hr volume/PHF=3000/0.9=3333vph
• Cj=2000pcphpl, fw=0.93,fp=1
SF C J (V / C ) * N * fW * f HV * f P
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Example
f HV
1 /((1 Pt ( Et 1) Pb ( Eb 1) PR ( E R 1))
Et=4, Eb=4, Pt=0.1, Pb=0.03
fHV=1/(1+0.1(4-1)+0.03(4-1))=0.74
V/C=3333/(2000*3*0.93*1.0*0.74)=0.81
From table, B=0.54,C=0.77, D=0.93, so LOS is D
Density =31 pc/mile/lane, speed is 52mph
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Example
• 31=0.1*4*D+0.03*3*D+0.87*1*D
• D=23 veh/mile/lane
• F=S*D
• S=3333/(23*3)=48mph
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Example 2 (Design)
• A rural freeway is being designed, the vertical
profile design indicate a 3 mile segment of level
terrain followed by a continuous 2 mile at 4%
upgrade. The peak hr demand is 2500vph in one
direction. There are 15% trucks and 10% buses.
The PHF is 0.85. level of service C is desired, how
many lanes will be needed to provide for this.
• Solution::Assume lane width is 12 feet,
fw=1.0,fp=1.0, fHV=?
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Example 2
• PB=0.1, PT=0.15
• Part (1) zero grade:
• ET=1.7, EB=1.5
• FHV=1/(1+0.15(1.7-1)+0.1(1.5-1))=0.87
• SF=2500/0.85=2000*0.65*1.0*1.0*N*0.87
• V/C for 70mph is 0.54-0.77 for LOS C, let take
V/C=0.65
• N=2.6, say 3.0, and with N=3, V/C=0.56
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Example 2
• Part (2), 4% up grade:
• For up grade, trucks are extremely heavy so
Et=8, Eb=1.6,
• FHV=1/((1+0.15(8-1)+0.10(1.6-1))
• =0.47 (up grade), Now N=?
• SF=2500/0.85=2000*0.65*1.0*1.0*N*0.47
• Or N=4.0 Lanes
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Problem 3
• A four lanes freeway runs through an area of
level terrain. The freeway has 12 feet lanes, 8
foot clear shoulders and a 70 mph design
speed. Traffic consists of 10% trucks and has
PHF of 0.91. Compute the service flow rates
for each level of service for this facility.
• Solution:
SF C J (V / C ) * N * fW * f HV * f P
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Solution
• Cj=2000pcphpl (for design speed of 70 mph)
• N=2 lanes per direction
• Fw=1.00 (ideal condition)
• Fp=1.00 (regular user)
• V/C ratio different for each level of service
• LOS A=0.35
• LOS B=0.54
• LOS C=0.77
• LOS D=0.93
• LOS E=1.00
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Solution
• The heavy vehicles factor is determined from
f HV
1 /((1 Pt ( Et 1) Pb ( Eb 1) PR ( E R 1))
• =0.93
• Service flow rate for each level of service may
now be computed as:
• SF(A)=2000*2*0.35*1.0*1.0*0.93=1308 vph
• SF (B)=2000*2*0.54*1.0*1.0*0.93=2019 vph
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Solution
• SF (C)=2000*2*0.77*1.0*1.0*0.93=2879 vph
• SF (D)=2000*2*0.93*1.0*1.0*0.93=3477 vph
• SF (E)=2000*2*1.0*1.0*1.0*0.93=3738 vph
• these are service flow values during 15 minutes of peak hour.
• The service volumes during peak hour will be:
• SV(A)=1308*0.91 (PHF)=1191 vph
• SV (B)=2019*0.91=1837 vph
• SV (C)=2879*0.91=2619 vph
• SV (D)=3477*0.91=3164 vph
• SV (E)=3738*0.91=3402 vph
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Problem 4
• A long viaduct on a freeway is currently
operating with three 12 feet lanes and 6 ft
lateral clearances on both sides. Due to some
congestion, two alternatives are suggested,
four 10 ft lanes with 4 ft clearance on the left
and right and four 12 ft lanes with no lateral
clearance on either side. Compare the
capacity of existing facility and the two
proposed improvements. f (HV)= 0.74
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Problem 4
• Solution: Capacity is the service flow rate at LOS E, by
definition. The capacity for each of three cases can be
determined from:
SF C J * N * fW * f HV * f P
• V/C is 1.0 for LOS E, Cj is 2000 pcphpl, Fp=1.0, N=3 or
4 depend on case. The lateral width factor fw for 12
ft lanes, 6 ft clearances= 1.0
• for 10 ft lanes, 4 ft clearances=0.87
• For 12 ft lanes, 0 ft clearances =0.91
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Problem 4
• SF(E) for 3 12 ft lanes=2000*3*1.0*1.0*0.74
• =4412 vph
• SF (E) for 4 10 ft lanes and 4 ft clearance on both
sides=2000*4*0.87*1.0*0.74
• =5118 vph
• SF (E) for 4 12 ft lanes and no
clearance=2000*4*0.91*1.0*0.74
• =5353 vph
• The 4 lanes increase the capacity , though the
clearance is not ideal
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