CE 543 - Traffic Engineering and Highway Safety
Traffic Control
Bachu Anilkumar
Assistant Professor
Dept. of Civil and Env. Engineering
IIT Patna
E-mail: anilkumar@iitp.ac.in
Module 5: Traffic Control
Signs, markings, islands and signals; At-grade and grade separated Intersections;
Rotaries; Basic principles of intersection signalization; Signal Design – HCM approach;
Analysis of signalized intersection, signal coordination
Traffic Control Devices
Traffic Signs - IRC 67 : 2012
Regulatory Signs
Warning Signs
Informatory Signs
Road Markings - IRC 35 : 2015
Traffic Signals
Anil (IITP) | CE 543 - Traffic Engineering and Highway Safety
Traffic Control Devices
Purpose
To promote highway safety and efficiency
Orderly movement of all road users on streets and highways
No advertising message or any other message that is not related to traffic control.
Principles
Fulfil a need;
Command attention;
Convey a clear, simple meaning;
Command respect from road users; and
Give adequate time for proper response.
Traffic Signs and Road Markings MUST BE the primary, and sole, means of
communication between Road System and Road User for efficient navigation
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Anil (IITP) | CE 543 - Traffic Engineering and Highway Safety
Traffic Signs
Regulatory Signs
Stop and Give-way
Prohibitory
No parking and No stoppage
Speed limit
Vehicle control
Restriction ends
Compulsory direction control
Violation of these signs is a legal
offence.
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Anil (IITP) | CE 543 - Traffic Engineering and Highway Safety
Traffic Signs
Purpose: Warn the road users of certain hazardous conditions
Shape - Equilateral triangle - apex upwards - White B/G - Red Boarders - symbol in
black colour
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Anil (IITP) | CE 543 - Traffic Engineering and Highway Safety
Traffic Signs
Guide the road users along routes, inform them directions and distance
Direction and Place identification, Facility information, parking and flood gauge
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Anil (IITP) | CE 543 - Traffic Engineering and Highway Safety
Traffic Signs
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Anil (IITP) | CE 543 - Traffic Engineering and Highway Safety
Traffic Signs
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Anil (IITP) | CE 543 - Traffic Engineering and Highway Safety
Road Markings
Guiding and controlling traffic on a highway
Serve as a psychological barrier
Help to signify the delineation of traffic path and its lateral clearance from traffic
hazards facilitating safe movement
Types of markings
Longitudinal markings
Transverse markings
Object markings
Word messages
Parking
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Anil (IITP) | CE 543 - Traffic Engineering and Highway Safety
Longitudinal Markings
Longitudinal Markings- Can be in White or Yellow in colour - Solid or Broken
Solid white - you cannot change the lane while driving
Broken white - you are allowed to change lane but cautiously
Yellow solid - No overtaking
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Anil (IITP) | CE 543 - Traffic Engineering and Highway Safety
Longitudinal Markings
Longitudinal Markings- Can be in White or Yellow in colour - Solid or Broken
Double yellow - Passing or changing lane is NOT allowed
Broken yellow - passing cautiously
Solid with broken - Not allowed to overtake from solid line side.
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Anil (IITP) | CE 543 - Traffic Engineering and Highway Safety
Longitudinal Markings
Longitudinal markings
Center Line
Traffic lane line
Bus lane markings
Warning lines
Edge lines
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Anil (IITP) | CE 543 - Traffic Engineering and Highway Safety
Transverse Markings
Transverse marking - marking provided across the carriageway for traffic control with
broken lines, single/ double continuous lines
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Anil (IITP) | CE 543 - Traffic Engineering and Highway Safety
Road Markings
Object marking - Marking on obstructions in a carriageway (traffic island, medians) or
obstructions near carriageway (signal posts, pier, etc.)
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Anil (IITP) | CE 543 - Traffic Engineering and Highway Safety
Word Messages
Word Messages - Information to guide, regulate, or warn the road user through word
message on road surface.
Characters for word messages are usually capital letters.
STOP, SLOW, RIGHT TURN ONLY
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Anil (IITP) | CE 543 - Traffic Engineering and Highway Safety
Parking Marking
Promotes more efficient use of the parking spaces
To prevent encroachment on places like bus stops, fire hydrant zones etc. where
parking is undesirable.
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Anil (IITP) | CE 543 - Traffic Engineering and Highway Safety
Hazardous Location Marking
Change in width of road
Hazardous location
Road-rail level crossing
Check barriers
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Anil (IITP) | CE 543 - Traffic Engineering and Highway Safety
Traffic Signals
Traffic signals must operate at all times
If properly designed signals will:
Provide for orderly flow of traffic
Reduce frequency of some crashes
Increase capacity
Provide gaps for minor movements
If NOT
Result in excessive delay
Increase frequency of some crashes
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Anil (IITP) | CE 543 - Traffic Engineering and Highway Safety
Traffic Signals
Signal Faces and Visibility
Generally 3 to 5 lenses
8 in or 12 in diameter
Minimum sight distance
Must operate continuously
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Anil (IITP) | CE 543 - Traffic Engineering and Highway Safety
Traffic Signals
Pedestrian Signals
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Anil (IITP) | CE 543 - Traffic Engineering and Highway Safety
Traffic Signals
Other Traffic Signals
Beacons
Lane-use control
Ramp meters
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Intersection Control
Anil (IITP) | CE 543 - Traffic Engineering and Highway Safety
Intersection Control
# of conflicting points of conventional 4- legged intersection
Merging - 8
Diverging - 8
Crossing - 16
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Anil (IITP) | CE 543 - Traffic Engineering and Highway Safety
Levels of Control
Level I - Basic rules of the road
Level II - YIELD or STOP control
Level III - Signalization
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Anil (IITP) | CE 543 - Traffic Engineering and Highway Safety
Basic Rules of the Road
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Anil (IITP) | CE 543 - Traffic Engineering and Highway Safety
Basic Rules of the Road
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Stopping Sight Distance
Distance travelled during perception/reaction time
Distance required to physically brake/stop the vehicle
SSD = vt +
v2
2g(f ± G)
Anil (IITP) | CE 543 - Traffic Engineering and Highway Safety
Basic Rules of the Road
Drivers on conflicting approaches must be
able to see each other in time to assess
whether an "impending hazard" is imposed
dA − b
adA
b
=
→ dB =
dB − a
a
dA − b
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Anil (IITP) | CE 543 - Traffic Engineering and Highway Safety
Basic Rules of the Road
Analysis Steps:
Calculate dA
Calculate dB,act
Calculate minimum dB
Check if dB,act < minimum dB , unsafe
and move to level II or III.
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Anil (IITP) | CE 543 - Traffic Engineering and Highway Safety
Basic Rules of the Road
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Approach 2
To avoid collision from the point at which visibility is established, Vehicle A must travel 188
past the collision point in the same time that Vehicle B travels to a point 12ft before the
collision point.
dB − 12
dA + 18
=
1.47vA
1.47vB
dB = (dA + 18)
vB
+ 12
vA
Anil (IITP) | CE 543 - Traffic Engineering and Highway Safety
Basic Rules of the Road
Analysis Steps:
Calculate dA
Calculate dB,act
Calculate minimum dB
Check if dB,act < minimum dB , unsafe
and move to level II or III.
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Anil (IITP) | CE 543 - Traffic Engineering and Highway Safety
Basic Rules of the Road
dA = 1.47vt +
dB =
v2
= 196.5ft
30(0.348 ± 0.01G)
20 × 196.5
adA
=
= 25.4ft
dA − b
196.5 − 42
dB,min = 1.47×40×2.5t+
402
= 300.3ft
30 × 0.348
or
40
+ 12 = 298ft.
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Both are far larger than the actual distance !!
dB = (196.5 + 18)
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Anil (IITP) | CE 543 - Traffic Engineering and Highway Safety
Basic Rules of the Road
Potential Remedies??
Implement intersection control, using STOP or YIELD control or traffic signals
Lower the speed limit on the major street to a point where sight distances are
adequate
Remove or reduce sight obstructions to provide adequate sight distances
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Anil (IITP) | CE 543 - Traffic Engineering and Highway Safety
YIELD and STOP Control
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Anil (IITP) | CE 543 - Traffic Engineering and Highway Safety
YIELD and STOP Control
Based on gap acceptance behaviour
Three components
Distance from the driver’s eye to the
front of the vehicle (assumed to be 8 ft)
Distance from the front of the vehicle to
the curb line (assumed to be 10 ft)
Distance from the curb line to the center
of the right-most travel lane
approaching from the left, or from the
curb line to the left-most travel lane
approaching from the right
dA−STOP = 18 + dcl
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Anil (IITP) | CE 543 - Traffic Engineering and Highway Safety
YIELD and STOP Control
The required sight distances for Vehicle B
dB = 1.47 × Vmaj × tg
tg - average gap accepted by minor street
driver to enter the major road, seconds
Range - 6.5 to 12.5 seconds
Typical value - 7.5 seconds
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Anil (IITP) | CE 543 - Traffic Engineering and Highway Safety
YIELD and STOP Control
Consider the case of a STOP-controlled approach at an intersection with a two-lane
arterial with a design speed of 40 mi/h.
dA−STOP (from left) = 18+6 = 24 ft
dA−STOP (from right) = 18+18 = 36 ft
dB,min = 1.47 × 40 × 7.5 = 441 ft.
Calculate actual distance of Vehicle B
from the collision point when visibility is
established
dB,act,left =
adA
36 × 24
=
= 216 < 441
dA − b
24 − 20
dB,act,right =
16 × 36
= 576 > 441
36 − 35
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Anil (IITP) | CE 543 - Traffic Engineering and Highway Safety
YIELD and STOP Control
Warrants for YIELD Control
When the ability to see all potentially conflicting traffic is sufficient to allow a road use
traveling at the posted speed, 85th percentile speed, or the statutory speed to pass
through the intersection or stop in a safe manner
If controlling a merge-type movement on the entering roadway where acceleration
geometry or sight distance is not adequate for merging traffic operations.
At a second crossroad of a divided highway, where the median width is 30 ft or
greater. A STOP sign may be installed at the entrance to the first roadway of a divided
highway, and a YIELD sign may be installed at the entrance to the second roadway.
At an intersection where a special problem exists and where engineering judgment
indicates that the problem is susceptible to correction by use of a YIELD sign.
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Anil (IITP) | CE 543 - Traffic Engineering and Highway Safety
Traffic Control Signals
Ultimate form of intersection control
Alternately assigns right-of-way to specific movements
Provide for the orderly movement of traffic
Increase the traffic-handling capacity
Provide continuous movement - if coordinated !!
Interrupt heavy traffic at intervals to permit other traffic,
vehicular or pedestrian, to cross.
To increase capacity, to improve safety, and to provide for orderly movement
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Anil (IITP) | CE 543 - Traffic Engineering and Highway Safety
Traffic Control Signals
Disadvantages
Excessive delay
Excessive disobedience of the signal indications
Increased use of less adequate routes as road users attempt to avoid the traffic
control signal
Significant increases in the frequency of collisions (especially rear-end collisions)
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Anil (IITP) | CE 543 - Traffic Engineering and Highway Safety
Traffic Control Signals
Warrants for Traffic Signals
Warrant 1 : Eight-Hour Vehicular Volume
Warrant 2: Four-Hour Vehicular Volume
Warrant 3: Peak Hour
Warrant 4: Pedestrian Volume
Warrant 5: School Crossing
Warrant 6 Coordinated Signal System
Warrant 7: Crash Experience
Warrant 8: Roadway Network
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Basic Principles of Intersection Signalization
Components of Signal Cycle
Types of Signal Operation
Treatment of Left Turns
Discharge Headways, Saturation Flow, Lost Times, Capacity
Critical Lane Group Concept
Optimum Cycle Length
Left Turn Equivalency
Delay as a Measure of Effectiveness
Anil (IITP) | CE 543 - Traffic Engineering and Highway Safety
Components of Signal Cycle
Cycle - One complete set of rotation through all of the indications provided
Cycle length - Time that it takes to complete one full cycle
Interval - Period of time during which no signal indication changes
Change interval
Clearance interval
Green interval
Red Interval
Phase - consists of a green interval, plus the change and clearance intervals that
follow it.
It is a set of intervals that allows a designated movement or set of movements to flow
and to be safely halted before release of a conflicting set of movements.
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Anil (IITP) | CE 543 - Traffic Engineering and Highway Safety
Components of Signal Cycle
Change interval
Yellow
Transition from GREEN to RED
Timed to allow a vehicle that cannot safely stop when the “green” is withdrawn to
enter the intersection legally.
Clearance interval
Red, Transition from GREEN to RED
All movements will have RED
Timed to allow a vehicle that legally enters the intersection on “yellow” to safely cross
the intersection before conflicting flows are released.
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Anil (IITP) | CE 543 - Traffic Engineering and Highway Safety
Components of Signal Cycle
Green interval
Movements permitted have a “green” light
Other movements will have RED light
Denoted with Gi
Red interval
All movements NOT permitted will have a “RED” light
Denoted with Ri
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Anil (IITP) | CE 543 - Traffic Engineering and Highway Safety
Types of Operation
Pre-timed Operation
Semi actuated operation
Fully actuated operation
Computer Control
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Anil (IITP) | CE 543 - Traffic Engineering and Highway Safety
Components of Signal Cycle
Pre-Timed Operation
Cycle length, phase sequence, and timing of each interval are constant
Follows a pre-determined plan
Multi-dial controllers: One for peak and off-peak
Semi actuated operation
Detectors placed in minor street
Always green for major road
Minor will get green when there is a demand
Green returns to Major road (i) after reaching maximum green on minor or (ii) no
further demand
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Anil (IITP) | CE 543 - Traffic Engineering and Highway Safety
Components of Signal Cycle
Fully-Actuated Operation
All approaches will be monitored by detectors
Green allocation - information from detectors and programmed rules
Cycle length, phase split, sequence of operation MAY CHANGE !!
Computer Control
Computer acts as a master controller - coordinating lot of signals
All signals in the network must use the same cycle length (or an even multiple
thereof).
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Anil (IITP) | CE 543 - Traffic Engineering and Highway Safety
Treatment of Left Turns
Permitted left turn - Left turn is allowed along with opposing through movement
Must select an appropriate gap in the opposing traffic stream.
left-turn volumes are reasonable and where gaps in the opposing flow are adequate
to accommodate left turns safely.
Protected left turn - Left turn is allowed when opposing through movement is stopped
Protects left-turning vehicles by stopping the opposing through movement.
Can lead to multi-phase signalization
Can be a combination of permitted and protected.
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Anil (IITP) | CE 543 - Traffic Engineering and Highway Safety
Discharge Headway
First headway
Second
First headway is relatively long !!
Must go through the full
perception-reaction sequence
Second is shorter - PRT overlaps
Constant headway achieved is referred to as
the saturation headway (h).
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Anil (IITP) | CE 543 - Traffic Engineering and Highway Safety
Saturation Flow
s=
3600
h
Remarks
Similar to capacity of approach - if the
signal were always green
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Anil (IITP) | CE 543 - Traffic Engineering and Highway Safety
Lost Times
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Start-up Lost Time
The time lost in the starting of the green
time interval when a traffic signal phase
changes from red to green
Headway vary for first few vehicles and
then remain almost constant
Sum of additional time involved in initial
headway
X
l1 =
∆i
i
Anil (IITP) | CE 543 - Traffic Engineering and Highway Safety
Lost Times
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Start-up Lost Time
Possible to model the amount of green time required to discharge "n" number of
vehicles !!
Tn = l1 + nh
Anil (IITP) | CE 543 - Traffic Engineering and Highway Safety
Lost Times
Clearance Lost Time
Time interval between the last vehicle’s front wheels crossing the stop line, and the
initiation of the GREEN for the next phase.
Denoted by l2
Un-utilized time.
Occurs at the end of the GREEN signal.
Start-up lost time: beginning of the GREEN signal.
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Anil (IITP) | CE 543 - Traffic Engineering and Highway Safety
Lost Times
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Total Lost Time
Start-up lost time: beginning of the GREEN signal.
Clearance: End of the GREEN signal
Total lost time = Startup + clearance
tL = l1 + l2
NOT ABLE TO USE THE TOTAL GREEN TIME !! - Effective Green Time
Anil (IITP) | CE 543 - Traffic Engineering and Highway Safety
Effective Green Time
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The actual signal goes through a sequnce of intervals:
Green - All-Red - RED.
Yellow and All-Red should be provided
One set - RED
Another set - GREEN, YELLOW and ALL-RED
Modelling perspective
Effective GREEN - Time that vehicles are moving
Effective RED - Time vehicles are NOT moving
Anil (IITP) | CE 543 - Traffic Engineering and Highway Safety
Effective Green Time
Effective Green, g = Gi + Yi − tL,i
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Anil (IITP) | CE 543 - Traffic Engineering and Highway Safety
Capacity of an Intersection Lane
Saturation flow rate(s) - Capacity of an intersection lane - assuming that the light is
always GREEN.
g i
ci = si
C
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Anil (IITP) | CE 543 - Traffic Engineering and Highway Safety
Sample Problem
Consider a given movement at a signalized intersection with the following known
characteristics:
Cycle length, C = 60 s
Green time, G = 27 s
Yellow plus all-red time, Y = 3 s
Saturation headway, h = 2.4 s/veh
Start-up lost time = 2.0 s
Clearance lost time = 1.0 s
For these characteristics, what is the capacity (per lane) for this movement?
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Anil (IITP) | CE 543 - Traffic Engineering and Highway Safety
Critical Lane and Time-Budget Concepts
Time-Budget: Allocation of time to
various vehicular and pedestrian
movements at an intersection through
signal control
Critical-lane: Identification of specific
lane movements that will control the
timing of a given signal phase
Critical lane - Lane with the most intense traffic demand, not the lane with the highest
volume.
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Anil (IITP) | CE 543 - Traffic Engineering and Highway Safety
Critical Lane and Time-Budget Concepts
Establishment of Time Budget
Movement of vehicles in critical lane 1
Movement of vehicles in critical lane 2
Start-up and clearance lost times for
vehicles in critical lane 1
Start-up and clearance lost times for
vehicles in critical lane 2
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Anil (IITP) | CE 543 - Traffic Engineering and Highway Safety
Signalized Intersection Capacity
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Assume total lost time is constant for all phases. Total lost time per signal cycle
L = N × tL
Total lost time in an hour
LH = L
3600
C
Time devoted for critical lane movements
TG = 3600 − LH
Maximum sum of critical lane volumes
TG
1
3600
Vc =
=
3600 − NtL
h
h
C
Anil (IITP) | CE 543 - Traffic Engineering and Highway Safety
Example
Example: If the signal at this location has
two phases, a cycle length of 60 seconds,
total lost times of 4 s/phase, and a saturation
headway of 2.5 s/veh. Calculate maximum
sum of critical lane volumes
3600
1
3600 − NtL
Vc =
h
C
Vc =
1
3600
3600 − 2 × 4
= 1248veh/h
2.5
60
Total available time = 3600-(60 × 8) = 3120 sec. i.e. 2.5 seconds per vehicle.
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Anil (IITP) | CE 543 - Traffic Engineering and Highway Safety
Observations
Cycle length ↑ - Capacity ↑
Longer cycle lengths - cycles ↓ - lost
times↓ - effective green time ↑ - sum of
critical lane volumes ↑
Relationship gets flatter - cycle length ↑
Capacity ↓ - Phases ↑
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Anil (IITP) | CE 543 - Traffic Engineering and Highway Safety
Minimum Cycle Length
60
Minimum Acceptable Cycle length
3600
1
3600 − NtL
Vc =
h
C
C=
1−
NtL
Vc
3600/h
Example: Vc =100 veh/h; h=2.5 s; tL = 4 s and N=2
C=
1−
2×4
= 26.2s
1000
3600/2.5
Anil (IITP) | CE 543 - Traffic Engineering and Highway Safety
Minimum Cycle Length
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Assumptions
Demand is uniformly distributed
Every second of effective green time is utilized
NOT POSSIBLE PRACTICALLY !!
How to overcome??
Take PHF into account - estimate the flow rate in worst 15-minute period
Capacity utilization - 80 % to 90%.
Divide the Vc with PHF and expected utilization of capacity
C=
NtL
Vc
1−
(3600/h) × PHF × (v /c)
Anil (IITP) | CE 543 - Traffic Engineering and Highway Safety
Minimum Cycle Length
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Example: Vc =100 veh/h; h=2.5 s; tL = 4 s and N=2
C=
1−
2×4
= 26.2s
1000
3600/2.5
Example: Vc =1000 veh/h; h=2.5 s; tL = 4 s and N=2, PHF - 0.95 and v/c ratio - 0.9
C=
2×4
= 42.6 sec
1000
1−
(3600/2.5) × 0.95 × 0.90
Anil (IITP) | CE 543 - Traffic Engineering and Highway Safety
Minimum Cycle Length
Impact of Utilization of Capacity
Example: Vc =1200 veh/h; h=2.2 s; tL = 4 s and N=3, PHF - 0.90 and v/c ratios from 1.00
to 0.80
C=
3×4
= 64.8 ≈ 65
1200
1−
(3600/2.2) × 0.90 × 1.00
v/c ratio
0.95 → 85 seconds
0.90 → 130 seconds
0.85 → 290 seconds
0.80 → -648.6 seconds
Practical limit: 120 seconds
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Anil (IITP) | CE 543 - Traffic Engineering and Highway Safety
Example
Using the time-budget and critical-lane concepts, determine the number of lanes required
for each of the critical movements and the minimum desirable cycle length.
3600
1
3600 − 2 × 4
= 1357veh/h
Vc =
2.3
60
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Anil (IITP) | CE 543 - Traffic Engineering and Highway Safety
Example
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Vc =
1
3600
3600 − 2 × 4
= 1357veh/h
2.3
60
N-S artery: Six lanes
E-W artery: Four lanes
PHF: 0.95 and v/c - 0.9
Desirable cycle length
Cdes =
1−
2×4
= 77.7 ≈ 80
1200
(3600/2.3) × 0.95 × 0.90
Anil (IITP) | CE 543 - Traffic Engineering and Highway Safety
Concept of Left-Turn Equivalency
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Anil (IITP) | CE 543 - Traffic Engineering and Highway Safety
Concept of Left-Turn Equivalency
How to consider the effect of left-Turn vehicles??
Through vehicle equivalents
How many through vehicles would consume the same amount of effective green time
traversing the stop-line as one left-turning vehicle ??
In the same amount of time, the left lane discharges five through vehicles and two
left-turning vehicles, while the right lane discharges eleven through vehicles.
11 = 5 + 2ELT → ELT = 3.0
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Anil (IITP) | CE 543 - Traffic Engineering and Highway Safety
Concept of Left-Turn Equivalency
Relationship Among Left-Turn Equivalents, Opposing Flow, and Number of Opposing
Lanes
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Anil (IITP) | CE 543 - Traffic Engineering and Highway Safety
Concept of Let-Turn Equivalency
An approach to a signalized intersection has two lanes, permitted left-turn phasing, 10%
left-turning vehicles, and a left-turn equivalent of 5.0. The saturation headway for through
vehicles is 2.0 s/veh. Determine the equivalent saturation flow rate and headway for all
vehicles on this approach.
One left turn vehicle is equivalent to 5 through vehicles
One left turn vehicle consumes 5.0 times the effective green time as a through vehicle
10% of the traffic stream will have 2 × 5 = 10 sec of saturation headway
90% of vehicles will have 10 seconds
Average sat headway = (0.9 x 2.0)+(0.1 x 10.0) = 2.8
Saturation flow rate = 3600/2.8 = 1286 veh/hg/ln.
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Anil (IITP) | CE 543 - Traffic Engineering and Highway Safety
Concept of Left-Turn Equivalency
Relationship between Ideal (only through) and Prevailing Conditions
sprev = sideal × fLT
Left turn adjustment factor, fLT =
3600/hprev
sprev
hideal
=
=
sideal
3600/hideal
hprev
hprev = (PLT ELT hideal ) + ((1 − PLT )hideal )
fLT =
hideal
1
=
(PLT ELT hideal ) + ((1 − PLT )hideal )
1 + PLT (ELT − 1)
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Anil (IITP) | CE 543 - Traffic Engineering and Highway Safety
Delay as Measure of Effectiveness
Point locations within a network
Measures of effectiveness - Highway - not relevant
Speed ??
Density ??
Delay, Queuing and Stops
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Anil (IITP) | CE 543 - Traffic Engineering and Highway Safety
Tyes of Delay
Most common measure - delay
Stopped-time delay
The time a vehicle is stopped in queue while waiting to pass through the intersection
Approach delay
Includes stopped-time delay
Also adds time loss due to deceleration from the approach speed to a stop and the time
loss due to reacceleration back to the desired speed.
Time-in-queue delay
Time from a vehicle joining an intersection queue to its discharge across the STOP line on
departure.
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Anil (IITP) | CE 543 - Traffic Engineering and Highway Safety
Tyes of Delay
Travel time delay
Difference between the driver’s expected travel time through the intersection (or any
roadway segment) and the actual time taken.
Difficult to establishing a “desired” travel time
Control delay
Delay caused by a control device, either a traffic signal or a STOP-sign.
Approximately equal to the time-in-queue delay
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Anil (IITP) | CE 543 - Traffic Engineering and Highway Safety
Types of Delay
Delay types for a single vehicle approaching a RED signal
Units: Seconds per vehicle, Vehicle-seconds, Vehicle-minutes and Vehicle-hours
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Anil (IITP) | CE 543 - Traffic Engineering and Highway Safety
Theoretical Models of Delay
N-T Plot
Number of vehicles arriving and departing
vs. time
X-axis: Time
Divided into effective Red and Green
Y-axis: No. of vehicles
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Anil (IITP) | CE 543 - Traffic Engineering and Highway Safety
Theoretical Models of Delay
Waiting time in queue, W (i)
Number of vehicles queued at any time, Q(t)
Aggregate delay for all vehicles (vehicles x
time)
Major Assumptions
Uniform arrival rate
Queue is building at a point
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Anil (IITP) | CE 543 - Traffic Engineering and Highway Safety
Theoretical Models of Delay
Case 1: Stable Flow
No signal cycle fails
Departure function catches up with arrival
function
Sum of triangular areas
Uniform Delay
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Anil (IITP) | CE 543 - Traffic Engineering and Highway Safety
Theoretical Models of Delay
Case 2: Individual cycle failures within a stable
operation
Few signal cycles will fail
Some vehicles may not get served
Departure function catches up with arrival
function - at the end of time period.
Overall period is stable, with individual cycle
failures
Overflow delay.
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Anil (IITP) | CE 543 - Traffic Engineering and Highway Safety
Theoretical Models of Delay
Case 3: Demand exceeds capacity for a significant period
Worst case scenario - All cycles will fail
Vehicles may not get served at the end of
each cycle - accumulate
Typically happens when v/c > 1
Overflow delay
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Anil (IITP) | CE 543 - Traffic Engineering and Highway Safety
Components of Delay
Components of delay - Analytical models
Uniform delay
Assumption of uniform arrivals
Stable flow with no cycle failures
Random delay
Additional delay beyond uniform delay
Random flow
With few cycle failures
Overflow delay
Capacity is less than the demand
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Anil (IITP) | CE 543 - Traffic Engineering and Highway Safety
Webster’s Uniform Delay Model
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Anil (IITP) | CE 543 - Traffic Engineering and Highway Safety
Webster’s Uniform Delay Model
Aggregate uniform delay, UD =
1
RV
2
h
g i
Length of RED phase, R = C 1 −
C
V = v (R + tc )
v (R + tc ) = stc → R + tc =
R = tc
s
s
v
tc
R
− 1 → tc = s
v
−1
v
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Anil (IITP) | CE 543 - Traffic Engineering and Highway Safety
Webster’s Uniform Delay Model

83

R



V = v (R + tc ) = v R + s
−1
v
V =R
vs
s−v
h
g i vs =C 1−
C
s−v
Aggregate uniform delay, UD =
g i2 vs 1 2h
C 1−
2
C
s−v
Anil (IITP) | CE 543 - Traffic Engineering and Highway Safety
Webster’s Uniform Delay Model
Aggregate uniform delay, UD =
84
g i2 vs 1 2h
C 1−
2
C
s−v
Average uniform delay per vehicle? - divide with number of vehicles arriving during the
cycle = arrival rate x cycle length
2 h vs i
g 2
C 2 1 − Cg
s−v
1 1− C
UD =
= C
2vC
2
1 − vs
c
g/C
g 2
1 1− C
1 1−
= C UD = C 2
2
1 − vc Cg
1−
Another form in terms of capacity: s =
g 2
C g
CX
Anil (IITP) | CE 543 - Traffic Engineering and Highway Safety
Random Delay
85
Vehicle arrivals are random
Inter-vehicle arrivals - possion distribution
v/c < 1.00
Webster’s random delay formulation:
Random delay per vehicle, RD =
X2
2v (1 − X )
Total delay, D = 0.90(UD + RD)
Anil (IITP) | CE 543 - Traffic Engineering and Highway Safety
Overflow Delay
Arriving vehicles exceeds capacity
Queue grows - overflow delay, in
addition to UD
v/c > 1.00
Two components: UD and OD
2
h
1 − Cg
1
gi
= 0.5C 1 −
UD = C
g
2 1 − C 1.00
C
86
Anil (IITP) | CE 543 - Traffic Engineering and Highway Safety
Overflow Delay
87
Aggregate overflow delay?
Agg OD, ODa =
1
T2
T (vT − cT ) =
(v − c)
2
2
Average overflow delay? - Divide with cT
Avg OD, OD =
T
(X − 1)
2
Anil (IITP) | CE 543 - Traffic Engineering and Highway Safety
Overflow Delay
88
Average delay per vehicle during time
period from T1 to T2
Avg OD, OD =
T1 + T2
(X − 1)
2
Anil (IITP) | CE 543 - Traffic Engineering and Highway Safety
Inconsistencies in Random and Overflow Delay
Random delay per vehicle, RD =
Avg OD, OD =
X2
2v (1 − X )
T
(X − 1)
2
What happens when X → 1 ??
RD - Infinity
OD - ZERO ??
89
Anil (IITP) | CE 543 - Traffic Engineering and Highway Safety
Inconsistencies in Random and Overflow Delay
Uniform delay - v/c less than 0.85
Overflow delay - v/c more than 1.5
Most frequent cases - 0.85 to 1.15
How to bridge this gap ?? - Akcelik
Model
90
Anil (IITP) | CE 543 - Traffic Engineering and Highway Safety
Inconsistencies in Random and Overflow Delay
cT
OD =
4
s
"
(X − 1) +
(X −
1)2
Xo = 0.67 +
+
12(X − Xo )
cT
sg
600
T - Analysis period. h
X - v/c ratio
C - Capacity, veh/h
S - Saturation flow rate, veh/sg
g - Effective green time, s
#
91
Anil (IITP) | CE 543 - Traffic Engineering and Highway Safety
Delay Estimation - Example
Example: An intersection approach has an approach flow rate of 1,000 veh/h, a saturation
flow rate of 2,800 veh/hg, a cycle length of 90 s, and a g/C ratio of 0.55. What average
delay per vehicle is expected under these conditions?
Calculate v/c ratio
c = s(g/C) = 2800(0.55) = 1540 veh/h
v/c = 1000/1540 = 0.55
Decide what delay equation to use ?
Uniform delay equation may be applied
d=
C [1 − (g/C)]2
90
[1 − 0.55]2
=
= 14.2sec/veh
2 1 − (v /s)
2 1 − (1000/2800)
92
Anil (IITP) | CE 543 - Traffic Engineering and Highway Safety
Delay Estimation - Example
93
Example: What will happen if the demand flow rate increased to 1,600 veh/h?
Calculate v/c ratio - v/c = 1600/1540 = 1.039
Decide what delay equation to use ? - UD and OD: Akcelik Equation for OD
UD = 0.5C[1 − (g/C)] = 0.5 × 90[1 − 0.55] = 20.3s/veh
OD =
cT
4
s
"
(X − 1) +
(X − 1)2 +
12(X − Xo )
cT
#
s
"
#
12(1.039 − 0.734)
1540 × 1
2
(1.039 − 1) + (1.039 − 1) +
= 39.1s/veh
OD =
4
1540 × 1
Xo = 0.67 +
sg
→ 0.67 +
600
Total Delay = UD+OD = 20.3+39.1 = 59.4 s/veh
2800
3600
× (0.55 × 90)
600
= 0.734
Anil (IITP) | CE 543 - Traffic Engineering and Highway Safety
Signal Design
HCM Method
Steps involved:
Development of a safe and effective phase plan and sequence
Determination of vehicular signal needs
Yellow and All-RED intervals
Critical lane volumes
Lost times per phase
Cycle length
Allocation of effective green time - Splitting the green
Determination of pedestrian signal needs
Minimum pedestrian time
Check whether vehicular green meet minimum pedestrian needs
If not, adjust timings to ensure pedestrian safety
94
Anil (IITP) | CE 543 - Traffic Engineering and Highway Safety
Development of a Signal Phase Plan
Provision for left turn
Left-turn volume
Conflicting volume
vLT ≥ 200 veh/h
xprod = vLT ×
vo
≥ 50, 000
No
vLT = left-turn flow rate, veh/h
vo = opposing through movement flow rate, veh/h
No = number of lanes for opposing through movement
95
Anil (IITP) | CE 543 - Traffic Engineering and Highway Safety
Phase and Ring Diagrams
Signal Phase Arrows
96
Anil (IITP) | CE 543 - Traffic Engineering and Highway Safety
Phase and Ring Diagrams
Used to illustrate signal phase plans phase plans
Phase diagram - shows all movements being made in a given phase within a single
block of the diagram.
Ring diagram - shows which movements are controlled by which “ring” on a signal
controller
97
Anil (IITP) | CE 543 - Traffic Engineering and Highway Safety
Phase and Ring Diagrams
Common phase plans
Simple two-phase signalization
Exclusive left-turn phasing
Leading and lagging green phases
Exclusive left tun with leading green
Eight-phase actuated control
Exclusive pedestrian phase
98
Anil (IITP) | CE 543 - Traffic Engineering and Highway Safety
Phase and Ring Diagrams
Two-Phase Signal
99
Anil (IITP) | CE 543 - Traffic Engineering and Highway Safety
Phase and Ring Diagrams
Exclusive left-turn phasing
100
Anil (IITP) | CE 543 - Traffic Engineering and Highway Safety
Phase and Ring Diagrams
Leading and Lagging Green Phases
101
Anil (IITP) | CE 543 - Traffic Engineering and Highway Safety
Phase and Ring Diagrams
Exclusive left turn with leading green
102
Anil (IITP) | CE 543 - Traffic Engineering and Highway Safety
Phase and Ring Diagrams
The Exclusive Pedestrian Phase
103
Anil (IITP) | CE 543 - Traffic Engineering and Highway Safety
Signal Design
Determination of vehicular signal needs
Determination of vehicular signal needs
Yellow and All-RED intervals
Critical lane volumes
Lost times per phase
Cycle length
Allocation of effective green time - Splitting the green
104
Anil (IITP) | CE 543 - Traffic Engineering and Highway Safety
Signal Design
105
Determination of vehicular signal needs
Determination of Change and Clearance Intervals
Change Interval (Yellow): Allows a vehicle that is one safe stopping distance away
from the STOP line.
y =t+
1.47S85
2a + (64.4 × 0.01G)
y =t+
S85
, SI Units
2a
y= length of yellow interval
t = driver reaction time
S85 = 85th percentile speed of approaching vehicles
a = deceleration rate of vehicles
G = Grade of approach
Anil (IITP) | CE 543 - Traffic Engineering and Highway Safety
Signal Design
Determination of vehicular signal needs
Determination of Change and Clearance Intervals
Clearance Interval (All-Red): Vehicle enters intersection legally - Provide sufficient
time for the vehicle to cross the intersection
106
Anil (IITP) | CE 543 - Traffic Engineering and Highway Safety
Signal Design
107
Determination of vehicular signal needs
No pedestrian traffic
ar =
w +L
1.47S15
Significant pedestrian traffic exists
ar =
P +L
1.47S15
Some pedestrian traffic exists
w +L
P +L
ar = max
,
1.47S15 1.47S15
Anil (IITP) | CE 543 - Traffic Engineering and Highway Safety
Signal Design
108
Determination of vehicular signal needs
Points to Remember
Yellow time - 85th percentile speeds
All-red time - 15th percentile speed
S15 = S − 5
S85 = S + 5
S = average speed in mi/h.
Anil (IITP) | CE 543 - Traffic Engineering and Highway Safety
Determination of Lost Times
109
Total lost time = Start-up lost time + clearance lost time
Start-up lost time, l1i = 2 s/phase
Clearance lost time, l2i = y + ar − e
e=motorist use of yellow and all-red, approximately 2 s/phase.
Lost time per phase, tLi = l1i + l2i = l1i + yi + ari − e
tLi = yi + ari → Total Lost time, L =
n
X
i
tLi
Anil (IITP) | CE 543 - Traffic Engineering and Highway Safety
Critical Lane Volumes
Calculate critical lane volumes under each phase
Finally sum up all critical volumes of each phase
Proceed signal design with Vc
How to calculate critical volume ??
Convert demand volumes to through vehicle equivalents
Left turning vehicles
No. of left-turning vehicles
No. of lanes in opposite direction
Opposing flow
Right turning vehicles
Pedestrian volume
110
Anil (IITP) | CE 543 - Traffic Engineering and Highway Safety
Critical Lane Volumes
VEQ = VLTE + VTH + VRTE = VLT ELT + VTH + VRT ERT
111
Anil (IITP) | CE 543 - Traffic Engineering and Highway Safety
Determine Cycle Length
112
Inputs
Number of phases
Total lost time
Peak hour factor
volume to capacity ratio
Saturation headway (or) saturation flow rate
Cdes =
NtL


1−

Vc
3600
× PHF × (v /c)
h


Anil (IITP) | CE 543 - Traffic Engineering and Highway Safety
Splitting the Green
113
Split the total available effective green time to various phases
Total effective green time in the cycle
gToT = C − L
gi = gTOT ×
Vci
Vc
C - Cycle length
L - Total lost time
gTOT - Total effective green time in cycle
Vci - Critical lane volume for phase or subphase, in tvu/h
Vc - Sum of the critical lane volumes
Anil (IITP) | CE 543 - Traffic Engineering and Highway Safety
Check for Pedestrian Requirements
Minimum pedestrian green requirement
GP = 3.2 +
Nped
L
+ (2.7 ×
), for WE > 10ft
Sp
WE
GP = 3.2 +
L
+ (0.27 × Nped ), for WE ≤ 10ft
Sp
G - Minimum pedestrian crossing time
L - Length of crosswalk
Sp - Average walking speed of pedestrians
Nped - Number of pedestrians crossing per phase in a single crosswalk
WE - Width of crosswalk
If GP < G + Y OK, otherwise re-design.
114
Anil (IITP) | CE 543 - Traffic Engineering and Highway Safety
Signal Design
Steps - HCM Approach
Develop a phase plan
Convert volumes to Through-Vehicle Equivalents
Determine critical lane volumes
Determine Yellow and All-Red intervals
Determination of lost times
Determine desirable cycle length
Allocate effective green to each phase
Check pedestrian requirements
115
Anil (IITP) | CE 543 - Traffic Engineering and Highway Safety
Design of Two-Phase Signal
Moderate pedestrian volumes
116
Anil (IITP) | CE 543 - Traffic Engineering and Highway Safety
Design of Two-Phase Signal
Step 1: Develop a phase plan - Only one lane i each
approach - not possible to protect the left turn
xprod
vLT ≥ 200 veh/h
vo
= vLT ×
≥ 50, 000
No
xprodEB =10x(315/1)=3150 < 50000
xprodWB =12x(420/1)=5040 < 50000
xprodNB =10x(400/1)=4000 < 50000
xprodSB =10x(375/1)=3750 < 50000
No criteria met !! - No need to protect the left turn movements.
117
Anil (IITP) | CE 543 - Traffic Engineering and Highway Safety
Design of Two-Phase Signal
Step 2: Convert Volumes to Through-Vehicle Equivalents
118
Anil (IITP) | CE 543 - Traffic Engineering and Highway Safety
Design of Two-Phase Signal
Step 3: Determine Critical-Lane Volumes
119
Anil (IITP) | CE 543 - Traffic Engineering and Highway Safety
Design of Two-Phase Signal
Step 4: Determine Yellow and All-Red Intervals
Average approach speed - 30 mi/h, S85 = Sav + 5 = 35mi/h, S15 = Sav − 5 = 25mi/h
y =t+
1.47S85
1.47 × 35
= 1.0 +
= 3.6s
2a + (64.4 × 0.01G)
2 × 10 + 0
P = Sum of two-15 ft lanes and 10-ft crosswalk: 15+15+10 = 40 ft.
ar =
P +L
40 + 20
=
= 1.6s.
1.47S15
1.47 × 25
Both streets have the same width, crosswalk width, and approach speed, the values of y
and ar will be same for both Phases A and B.
120
Anil (IITP) | CE 543 - Traffic Engineering and Highway Safety
Design of Two-Phase Signal
Step 5: Determination of Lost Times
Start-up lost time, l1 = 2.0 sec
Yellow time, y = 3.6 sec
All-red time, ar = 1.6 sec
Clearance lost time
l2 = y + ar − e = 3.6 + 1.6 − 2 = 3.2sec
Total lost time per phase, tL = l1 + l2 = 2 + 3.2 = 5.2sec
Number of phases = 2; Total lost time, L = 5.2+5.2 = 10.4 seconds
121
Anil (IITP) | CE 543 - Traffic Engineering and Highway Safety
Design of Two-Phase Signal
122
Step 6: Determine the Desirable Cycle Length
NtL
Cdes =
Cdes =



1−
Vc


3600
× PHF × (v /c)
h
1−
10.4
= 33.5sec
924
1615 × 0.92 × 0.90
Round off to highest 5 seconds → Cycle length = 35 seconds
Anil (IITP) | CE 543 - Traffic Engineering and Highway Safety
Design of Two-Phase Signal
Step 7: Allocate Effective Green to Each Phase
Cycle length = 35 seconds
Available green time, effective green time = 35-10.4 = 24.6 seconds
gA = gTOT
470
VcA
= 24.6 ×
= 12.5sec.
Vc
924
gB = gTOT
VcB
454
= 24.6 ×
= 12.1sec.
Vc
924
Actual green times:
GA = gA − YA + tLA = 12.5 − 5.2 + 5.2 = 12.5 s
GB = gB − YB + tLB = 12.1 − 5.2 + 5.2 = 12.1 s
123
Anil (IITP) | CE 543 - Traffic Engineering and Highway Safety
Design of Two-Phase Signal
Step 8: Check Pedestrian Requirements
Moderate pedestrian - 200 per hour
How many per cycle? - 3600/35 = 103 cycles per hour
200/103 = 1.94. Approximately 2 pedestrians per hour.
GpA,B = 3.2 +
L
30
+ 0.27Nped = 3.2 +
+ 0.27 × 2 = 11.2s
Sp
4.0
Gp < G + Y
GpA = 11.2 < 12.5 + 5.2 = 17.7s OK
GpB = 11.2 < 12.1 + 5.2 = 17.3s OK
The signal safely accommodates all pedestrians. No changes in the signal timing for
vehicular needs is required.
124
Anil (IITP) | CE 543 - Traffic Engineering and Highway Safety
Case 2: Intersection of Major Arterials
Exclusive Left Turn
Moderate pedestrian volumes
125
Anil (IITP) | CE 543 - Traffic Engineering and Highway Safety
Case 2: Intersection of Major Arterials
Exclusive Left Turn
Step 1: Develop a phase plan
xprod
vLT ≥ 200 veh/h
vo
= vLT ×
≥ 50, 000
No
EB: VLT = 35 < 200
xprodEB =35x(500/2)=8750 < 50000, Not needed
WB: VLT = 25 < 200
xprodEB =25x(610/2)=22875 < 50000, Not needed
NB: VLT = 250 > 200; Protection required
SB: VLT = 220 > 200; Protection required
Need to protect the left turn movements of NB and SB.
126
Anil (IITP) | CE 543 - Traffic Engineering and Highway Safety
Case 2: Intersection of Major Arterials
Exclusive Left Turn
Step 2: Convert Volumes to Through-Vehicle Equivalents
127
Anil (IITP) | CE 543 - Traffic Engineering and Highway Safety
Case 2: Intersection of Major Arterials
Exclusive Left Turn
Step 3: Determine Critical-Lane Volumes
128
Anil (IITP) | CE 543 - Traffic Engineering and Highway Safety
Case 2: Intersection of Major Arterials
Exclusive Left Turn
Step 4: Determine Yellow and All-Red Intervals
Speed limit - 45 mi/h : No differentiation between S85 and S15 .
yA,B,C = t +
1.47S85
1.47 × 45
= 1.0 +
= 4.3s
2a + (64.4 × 0.01G)
2 × 10 + 0
Width of N-S street - 55 ft. → P= 55+10 = 65 ft.
arA,B =
P +L
65 + 20
=
= 1.4s.
1.47S15
1.47 × 45
Width of E-W street - 60 ft. → P= 60+10 = 70 ft.
arC =
P +L
70 + 20
=
= 1.3s.
1.47S15
1.47 × 45
129
Anil (IITP) | CE 543 - Traffic Engineering and Highway Safety
Case 2: Intersection of Major Arterials
Exclusive Left Turn
Step 5: Determination of Lost Times
Assuming start-up lost time and e as 2.0 s
Lost time per phase, tL = Y = y + ar = Yellow time + All-Red
YA,B = 4.3 + 1.4 = 5.7s
YC = 4.3 + 1.3 = 5.6s
Total lost time, L = 5.7+5.7+5.6 = 17 sec.
130
Anil (IITP) | CE 543 - Traffic Engineering and Highway Safety
Case 2: Intersection of Major Arterials
131
Exclusive Left Turn
Step 6: Determine the Desirable Cycle Length
L
Cdes =
Cdes =



1−
Vc


3600
× PHF × (v /c)
h
1−
17
= 109.7sec
1130
1615 × 0.92 × 0.90
Round off to highest 5 seconds → Cycle length = 110 seconds
Anil (IITP) | CE 543 - Traffic Engineering and Highway Safety
Case 2: Intersection of Major Arterials
Exclusive Left Turn
Step 7: Allocate Effective Green to Each Phase
Cycle length = 110 seconds
Available green time, effective green time = 110-17 = 93 seconds
gA = gTOT
263
VcA
= 93 ×
= 21.6sec.
Vc
1130
gB = gTOT
VcB
516
= 93 ×
= 42.5sec.
Vc
1130
gC = gTOT
VcC
351
= 93 ×
= 28.9sec.
Vc
1130
Check for total cycle length: 21.6+42.5+28.9+17 = 110 → OK
132
Anil (IITP) | CE 543 - Traffic Engineering and Highway Safety
Case 2: Intersection of Major Arterials
133
Exclusive Left Turn
Step 8: Check Pedestrian Requirements
Moderate pedestrian - 200 per hour
How many per cycle? - 3600/110 = 32.7 cycles per hour → 200/32.7 = 6.1 ped
GpB = 3.2+
L
60
+0.27×6.1 = 19.8s < GB +YB = 42.5+5.7 = 48.2 OK
+0.27Nped = 3.2+
Sp
4.0
GpC = 3.2+
L
55
+0.27Nped = 3.2+
+0.27×6.1 = 18.6s < GC +YC = 28.9+5.6 = 34.5 OK
Sp
4.0
The signal safely accommodates all pedestrians. No changes in the signal timing for
vehicular needs is required.
Anil (IITP) | CE 543 - Traffic Engineering and Highway Safety
Case 3: Intersection of Major Arterials
Exclusive Left-Turn Phase Plus Leading Green
Moderate pedestrian volumes
134
Anil (IITP) | CE 543 - Traffic Engineering and Highway Safety
Case 3: Intersection of Major Arterials
Exclusive Left-Turn Phase Plus Leading Green
Step 1: Develop a phase plan
xprod
vLT ≥ 200 veh/h
vo
= vLT ×
≥ 50, 000
No
EB: VLT = 300 > 200, Protection required
WB: VLT = 150 < 200
xprodWB =150x(1200/3)=60000 > 50000,
Protection needed
NB: VLT = 50 < 200; Not required
SB: VLT = 300 < 200; Not required
Need to protect the left turn movements of EB and WB.
135
Anil (IITP) | CE 543 - Traffic Engineering and Highway Safety
Case 3: Intersection of Major Arterials
Exclusive Left-Turn Phase Plus Leading Green
Step 2: Convert Volumes to Through-Vehicle Equivalents
136
Anil (IITP) | CE 543 - Traffic Engineering and Highway Safety
Case 3: Intersection of Major Arterials
Exclusive Left-Turn Phase Plus Leading Green
Step 3: Determine Critical-Lane Volumes
137
Anil (IITP) | CE 543 - Traffic Engineering and Highway Safety
Case 3: Intersection of Major Arterials
Exclusive Left-Turn Phase Plus Leading Green
Step 4: Determine Yellow and All-Red Intervals
EWAvg - 50 mi/h; NSAvg - 35 mi/h.
yA1,A2,A3 = t +
1.47 × 55
1.47S85
= 1.0 +
= 5.0s
2a + (64.4 × 0.01G)
2 × 10 + 0
1.47S85
1.47 × 40
= 1.0 +
= 3.9s
2a + (64.4 × 0.01G)
2 × 10 + 0
Width of N-S street - 40 ft.
P +L
40 + 20
arA1,A2,A3 =
=
= 0.9s.
1.47S15
1.47 × 45
yB = t +
Width of E-W street - 96 ft.
arC =
P +L
90 + 20
=
= 2.6s.
1.47S15
1.47 × 30
138
Anil (IITP) | CE 543 - Traffic Engineering and Highway Safety
Case 3: Intersection of Major Arterials
Exclusive Left-Turn Phase Plus Leading Green
Step 5: Determination of Lost Times
Assuming start-up lost time and e as 2.0 s
Lost time per phase, tL = Y = y + ar = Yellow time + All-Red
YA1,A2 = 5.0 + 0.9 = 5.9s
YA3 = 5.0 + 0.9 = 5.9s
YB = 3.9 + 2.6 = 6.5s
Total lost time, L = 5.9+5.9+6.5 = 18.3 sec.
139
Anil (IITP) | CE 543 - Traffic Engineering and Highway Safety
Case 3: Intersection of Major Arterials
140
Exclusive Left-Turn Phase Plus Leading Green
Step 6: Determine the Desirable Cycle Length
L
Cdes =
Cdes =



1−
Vc


3600
× PHF × (v /c)
h
1−
18.3
= 95.3sec
998
1615 × 0.85 × 0.90
Round off to highest 5 seconds → Cycle length = 100 seconds
Anil (IITP) | CE 543 - Traffic Engineering and Highway Safety
Case 3: Intersection of Major Arterials
Exclusive Left-Turn Phase Plus Leading Green
Step 7: Allocate Effective Green to Each Phase
Cycle length = 100 seconds
Available green time, effective green time = 100-18.3 = 81.7 seconds
gA1,A2 = gTOT
VcA1,A2
315
= 81.7 ×
= 25.8sec.
Vc
998
gA3 = gTOT
VcA3
334
= 81.7 ×
= 27.3sec.
Vc
998
gB = gTOT
VcB
349
= 81.7 ×
= 28.6sec.
Vc
998
141
Anil (IITP) | CE 543 - Traffic Engineering and Highway Safety
Case 3: Intersection of Major Arterials
Exclusive Left-Turn Phase Plus Leading Green
Specific lengths of all phases
Total length of combined Phase A1 and A2 = 25.8+27.3 = 53.1 seconds
gA1 = 53.1 ×
158
= 15sec
158 + 400
gA2 = 53.1 − 15.0 − 27.3 = 10.8sec
GA1 = 15.0 sec
YA3 = 5.9 sec
GA2 = 10.8 sec
GB = 28.6 sec
YA1,A2 = 5.9 sec
YB = 6.5 sec
GA3 = 27.3 sec
C = 100 sec
No pedestrians in this intersection. No Step 8 !!
142
Bachu Anilkumar
IIT Patna
anilkumar@iitp.ac.in
Thank You !!