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Chapter 2

Traffic Control Methods

7 teaching hours 12 marks Transportation Engineering II

2.1 Types of Intersections

Past Exam Questions on this Topic

  1. What are the basic requirements of intersection at grade? Mention the factors affecting night visibility at road. How much is the recommended lateral offset of the lighting poles at the road? Explain with reasoning: [3+3+2] (2082 Bhadra, Q3)
    Solution

    Basic requirements of an at-grade intersection: adequate sight distance, minimum number of conflict points, channelization to guide traffic, low relative speed of crossing streams, adequate capacity, simple and clear layout, and provision for pedestrians, drainage and lighting.

    Factors affecting night visibility: brightness/luminance of source, contrast, glare, size of object, and adaptation time of the eye.

    Lateral offset of lighting poles: poles are placed clear of the carriageway - a lateral offset of about 0.3 m beyond the kerb face (outside the clear zone) is recommended so that poles do not become a collision hazard, with mounting height at least equal to the road width for full coverage.

    Exact recommended offset for local practice should be read from the design Manual.pdf (street-lighting section).

  2. What are the advantages of one way traffic movement? Draw a right angle four arm intersection of two roads and show various conflict points, if both roads with two way movements? Classify the type of conflicts [8] (2081 Baishakh, Q3)
    Solution

    Advantages of one-way traffic movement: increased capacity, fewer conflict points, higher operating speed, easier signal coordination, fewer accidents, better parking use and simpler turning movements.

    Conflict points at a right-angle four-arm intersection (both roads two-way): there are 32 conflict points = 16 crossing + 8 merging + 8 diverging. Making the roads one-way sharply reduces them. Types of conflict: crossing, merging (converging) and diverging.

  3. What are the basic requirements of intersection at grade? Write down the design steps of rotary intersection. (2075 Chaitra, Q1)
    Solution

    Basic requirements of an at-grade intersection: adequate sight distance, minimum number of conflict points, channelization to guide traffic, low relative speed of crossing streams, adequate capacity, simple and clear layout, and provision for pedestrians, drainage and lighting.

    Design steps of a rotary intersection:

    1. Collect traffic (turning-movement) and design data.
    2. Select the rotary design speed (usually 30-40 km/h).
    3. Fix the shape and size of the central island and the rotary radius.
    4. Determine the weaving length and weaving width.
    5. Fix entry and exit widths and their radii.
    6. Check the weaving capacity (Wardrop's formula) against design flow.
    7. Provide islands, markings, signs, lighting and channelization.
  4. What are the basic requirements of intersection at grade? Mention the importance of street lighting. Explain the different types of traffic islands. How is accident study carried out? Two vehicles A and B approaching at right angles, vehicle A from West and vehicle B from south, collide with each other. After the collision, vehicle A skids in a direction 49° N of W and vehicle B skids 27° E of N. The initial skid distances of vehicles A and B are 37 m and 19 m respectively before collision. If the weight of vehicle A is 4 tonne and vehicle B is 6 tonne, and the skid distances after collision for vehicle A is 15 m and for vehicle B is 36 m, calculate the initial speeds of the vehicles if the average skid resistance of the pavement is found to be 0.55. (2075 Ashwin, Q1)
    Solution

    Basic requirements of an at-grade intersection: adequate sight distance, minimum conflict points, channelization to guide traffic, adequate capacity, low relative speed, clear and simple layout, provision for pedestrians and proper drainage/lighting.

    Importance of street lighting: improves night visibility & safety, reduces night accidents, aids pedestrian security, improves traffic flow and comfort, and enhances aesthetics.

    Types of traffic islands: divisional islands, channelizing islands, pedestrian refuge islands, and rotary (central) islands.

    Accident study is carried out by collecting accident data (spot maps, collision & condition diagrams), analysing causes (road user, vehicle, road, environment), and applying remedial 3-E measures (Engineering, Enforcement, Education).

    Collision numerical

    skid coefficient f = 0.55 · vehicle A 4 t, vehicle B 6 t · post-collision skid 15 m (A) and 36 m (B) · pre-collision skid 37 m (A) and 19 m (B) · deflections A 49° N of W and B 27° E of N

    Speeds just after the collision

    v = √(2 g f d) vA' = √(2 × 9.81 × 0.55 × 15) = 12.72 m/s vB' = √(2 × 9.81 × 0.55 × 36) = 19.71 m/s

    Momentum conservation to recover the impact speeds

    The paper does not state which way each vehicle was heading. The deflections given — A to 49° N of W and B to 27° E of N — are only consistent with A travelling east and B travelling north; any other pairing makes one impact speed come out negative. Measuring all directions anticlockwise from east, A leaves at 131° and B at 63°.

    x (east): mAuA = mAvA' cos131° + mBvB' cos63° 4 uA = 50.89(−0.656) + 118.26(0.454) 4 uA = −33.38 + 53.69 = 20.30 uA = 5.08 m/s y (north): mBuB = mAvA' sin131° + mBvB' sin63° 6 uB = 50.89(0.755) + 118.26(0.891) 6 uB = 38.41 + 105.37 = 143.78 uB = 23.96 m/s

    Add the pre-collision skid to get the original speeds

    u0 = √(uimpact² + 2 g f dpre) A (dpre = 37 m): u0A = √(5.08² + 2×9.81×0.55×37) = √425.1 = 20.62 m/s = ≈ 74 km/h B (dpre = 19 m): u0B = √(23.96² + 2×9.81×0.55×19) = √779.6 = 27.92 m/s = ≈ 100 km/h
    Vehicle A was travelling at about 74 km/h and vehicle B at about 100 km/h before braking.

    The weights (A = 4 t, B = 6 t) enter only as the mass ratio in the momentum equations, so they need not be converted to newtons. Note that A leaves the impact faster than it entered it (12.72 vs 5.08 m/s) — the heavier, faster vehicle B drove it forward, which is exactly why the wreck slid 15 m.

  5. What are the basic requirements of intersection at grade? Describe grade separated intersection with its advantages and disadvantages. (2074 Ashwin, Q1)
    Solution

    Basic requirements of an at-grade intersection: adequate sight distance, minimum number of conflict points, channelization to guide traffic, low relative speed of crossing streams, adequate capacity, simple and clear layout, and provision for pedestrians, drainage and lighting.

    Grade-separated intersection - advantages: no crossing conflicts, high capacity, continuous high-speed flow, greater safety, minimal delay. Disadvantages: very high cost, large land requirement, complex/long structures and ramps, not economical for low volumes, and possible aesthetic/environmental impact.

Review

Road Intersection and Types20812076

At Grade Intersection

Basic Requirement of At Grade Intersection208220752074

Types of At Grade Intersection

Based on number of intersecting legs:

2.2 Un-Channelized and Channelized Intersections

Past Exam Questions on this Topic

  1. What are the factors affecting night visibility? What are the different types of road marking and traffic island used to regulate and control traffic flow? Explain. [2+3+3] (2076 Ashwin, Q4)
    Solution

    Factors affecting night visibility: luminance/brightness of the source, contrast between object and background, glare (disability and discomfort), size of the object, and the driver's adaptation/available time.

    Types of road markings: longitudinal (centre line, lane lines, edge lines, no-overtaking lines), transverse (stop line, give-way line, pedestrian crossing) and other markings (directional arrows, word messages, hazard/parking markings).

    Traffic islands used to regulate and control flow: divisional islands, channelizing islands, pedestrian refuge islands, and rotary (central) islands.

  2. What are the basic requirements of intersection at grade? Mention the importance of street lighting. Explain the different types of traffic islands. How is accident study carried out? Two vehicles A and B approaching at right angles, vehicle A from West and vehicle B from south, collide with each other. After the collision, vehicle A skids in a direction 49° N of W and vehicle B skids 27° E of N. The initial skid distances of vehicles A and B are 37 m and 19 m respectively before collision. If the weight of vehicle A is 4 tonne and vehicle B is 6 tonne, and the skid distances after collision for vehicle A is 15 m and for vehicle B is 36 m, calculate the initial speeds of the vehicles if the average skid resistance of the pavement is found to be 0.55. (2075 Ashwin, Q1)
    Solution

    Basic requirements of an at-grade intersection: adequate sight distance, minimum conflict points, channelization to guide traffic, adequate capacity, low relative speed, clear and simple layout, provision for pedestrians and proper drainage/lighting.

    Importance of street lighting: improves night visibility & safety, reduces night accidents, aids pedestrian security, improves traffic flow and comfort, and enhances aesthetics.

    Types of traffic islands: divisional islands, channelizing islands, pedestrian refuge islands, and rotary (central) islands.

    Accident study is carried out by collecting accident data (spot maps, collision & condition diagrams), analysing causes (road user, vehicle, road, environment), and applying remedial 3-E measures (Engineering, Enforcement, Education).

    Collision numerical

    skid coefficient f = 0.55 · vehicle A 4 t, vehicle B 6 t · post-collision skid 15 m (A) and 36 m (B) · pre-collision skid 37 m (A) and 19 m (B) · deflections A 49° N of W and B 27° E of N

    Speeds just after the collision

    v = √(2 g f d) vA' = √(2 × 9.81 × 0.55 × 15) = 12.72 m/s vB' = √(2 × 9.81 × 0.55 × 36) = 19.71 m/s

    Momentum conservation to recover the impact speeds

    The paper does not state which way each vehicle was heading. The deflections given — A to 49° N of W and B to 27° E of N — are only consistent with A travelling east and B travelling north; any other pairing makes one impact speed come out negative. Measuring all directions anticlockwise from east, A leaves at 131° and B at 63°.

    x (east): mAuA = mAvA' cos131° + mBvB' cos63° 4 uA = 50.89(−0.656) + 118.26(0.454) 4 uA = −33.38 + 53.69 = 20.30 uA = 5.08 m/s y (north): mBuB = mAvA' sin131° + mBvB' sin63° 6 uB = 50.89(0.755) + 118.26(0.891) 6 uB = 38.41 + 105.37 = 143.78 uB = 23.96 m/s

    Add the pre-collision skid to get the original speeds

    u0 = √(uimpact² + 2 g f dpre) A (dpre = 37 m): u0A = √(5.08² + 2×9.81×0.55×37) = √425.1 = 20.62 m/s = ≈ 74 km/h B (dpre = 19 m): u0B = √(23.96² + 2×9.81×0.55×19) = √779.6 = 27.92 m/s = ≈ 100 km/h
    Vehicle A was travelling at about 74 km/h and vehicle B at about 100 km/h before braking.

    The weights (A = 4 t, B = 6 t) enter only as the mass ratio in the momentum equations, so they need not be converted to newtons. Note that A leaves the impact faster than it entered it (12.72 vs 5.08 m/s) — the heavier, faster vehicle B drove it forward, which is exactly why the wreck slid 15 m.

Unchannelized Intersection

Plain (Simple) Intersection

Flared Intersection

Channelized Intersection20762075

Advantages: Disadvantages:

Conflict Points at Intersections

Types of Conflicts

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Conflict Points

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2.3 Grade Separated Intersections (Overpass, Underpass and Interchanges)

Past Exam Questions on this Topic

  1. Discuss different types of intersection Write the advantages and limitations of grade separated intersection: [8] (2076 Chaitra, Q2)
    Solution

    Types of intersection: (i) at-grade - uncontrolled, channelized, rotary and signalized; (ii) grade-separated - overpass, underpass and interchanges (diamond, cloverleaf, trumpet, directional).

    Grade-separated intersection - advantages: no crossing conflict, high capacity, safe continuous flow, no delay. Limitations: high construction cost, large land requirement, structural complexity, and not economical for low traffic volumes.

  2. What are the basic requirements of intersection at grade? Describe grade separated intersection with its advantages and disadvantages. (2074 Ashwin, Q1)
    Solution

    Basic requirements of an at-grade intersection: adequate sight distance, minimum number of conflict points, channelization to guide traffic, low relative speed of crossing streams, adequate capacity, simple and clear layout, and provision for pedestrians, drainage and lighting.

    Grade-separated intersection - advantages: no crossing conflicts, high capacity, continuous high-speed flow, greater safety, minimal delay. Disadvantages: very high cost, large land requirement, complex/long structures and ramps, not economical for low volumes, and possible aesthetic/environmental impact.

Grade Separated Intersection20762074

Advantages: Disadvantages:

Grade Separated Intersection without Interchange

Grade Separation Structures (Overpass)

Advantages: Disadvantages:

Grade Separation Structures (Underpass)

Advantages: Disadvantages:

Grade Separated Intersection with Interchange

Types of Ramp Facilities

Semi Direct

Direct
In direct

Grade Separated Intersection with Interchange

Advantages: Disadvantages:

Types of Interchange

Trumpet Interchanges

Directional Y

Diamond Interchange

Cloverleaf Interchange

Advantages: Disadvantages:

Partial Cloverleaf

Directional Interchange

2.4 Rotary Intersection

Past Exam Questions on this Topic

  1. Define spot speed and explain the methods of conducting spot speed study. Write down the advantages and limitations of rotary intersection: [8] (2081 Bhadra, Q2)
    Solution

    Spot speed = instantaneous speed of a vehicle at a specified point on the road.

    Methods of conducting a spot-speed study:

    • Manual: stopwatch + short measured base length, or enoscope (mirror box).
    • Automatic: pneumatic road tubes, inductive loop detectors, Doppler radar gun, LIDAR speed gun, and video / image-processing systems.

    Advantages of a rotary intersection: continuous (stop-free) movement; all crossing manoeuvres converted to merging, weaving and diverging; self-regulating (no signal); eliminates severe right-angle and head-on conflicts; orderly and aesthetic.

    Limitations: requires large land area; unsuitable for very high volumes or heavy pedestrian movement; causes delay under heavy flow; not suited to high-speed traffic or steep gradients; costly where land is expensive.

  2. What are the basic requirements of intersection at grade? Write down the design steps of rotary intersection. (2075 Chaitra, Q1)
    Solution

    Basic requirements of an at-grade intersection: adequate sight distance, minimum number of conflict points, channelization to guide traffic, low relative speed of crossing streams, adequate capacity, simple and clear layout, and provision for pedestrians, drainage and lighting.

    Design steps of a rotary intersection:

    1. Collect traffic (turning-movement) and design data.
    2. Select the rotary design speed (usually 30-40 km/h).
    3. Fix the shape and size of the central island and the rotary radius.
    4. Determine the weaving length and weaving width.
    5. Fix entry and exit widths and their radii.
    6. Check the weaving capacity (Wardrop's formula) against design flow.
    7. Provide islands, markings, signs, lighting and channelization.
  3. a) The vehicle arrivals at the section of road are assumed to be Poisson distributed with an average arrival rate of 1 vehicle every 5 minutes. What is the probability of (i) Exactly 3 vehicles arrive in a 15 minute interval (ii) less than 3 vehicles arrive in a 15 minute interval? (iii) More than 3 vehicles arrive in a 15 minute interval? [4] b) Calculate the capacity of rotary with entry and exit width of 8m, the width of non-weaving section is 9m, width of rotary is 12m, length of weaving section is 60m. The ratio of weaving to total traffic in weaving section is 0.7. [4] (2076 Chaitra, Q3)
    Solution

    (a) Poisson arrivals

    Mean arrival rate = 1 vehicle per 5 min · interval of interest = 15 min

    Mean number of arrivals in the interval

    m = 15 / 5 = 3 vehicles P(x) = e−m mx / x!

    Probability of exactly 3 arrivals

    P(3) = e−3 × 3³ / 3! = 0.0498 × 27/6 = 0.0498 × 4.5 = 0.224

    Probability of fewer than 3 arrivals

    P(<3) = P(0) + P(1) + P(2) = e−3(1 + 3 + 4.5) = 0.0498 × 8.5 = 0.423

    Probability of more than 3 arrivals

    P(>3) = 1 − P(≤3) = 1 − (0.423 + 0.224) = 0.353

    (b) Rotary weaving capacity

    Weaving width w = 12 m · average entry/exit width e = 8 m · proportion of weaving traffic p = 0.7 · weaving length L = 60 m

    Wardrop's weaving-capacity formula

    Qp = 280 w (1 + e/w)(1 − p/3) / (1 + w/L)

    Substitute

    Qp = 280 × 12 × (1 + 8/12) × (1 − 0.7/3) / (1 + 12/60) = 3360 × 1.667 × 0.767 / 1.2 = ≈ 3580 PCU/h
    P(exactly 3) = 0.224, P(fewer than 3) = 0.423, P(more than 3) = 0.353; the rotary's weaving capacity is ≈ 3580 PCU/h.
  4. Explain the factors to be considered in the timing design of traffic and pedestrian signals. Discuss the design of a rotary intersection. [8] (2078 Bhadra, Q9)
    Solution

    Factors in signal timing (traffic & pedestrian): approach volumes and saturation flows, pedestrian crossing time (walking speed & road width), minimum green, amber and all-red (clearance) intervals, start-up lost time, cycle-length limits, and signal coordination.

    Rotary design factors: design speed, weaving length and width, entry/exit geometry, and weaving capacity.

  5. a) Define rotary intersection. What are the advantages and disadvantages of rotary intersection? [8] b) A vehicle skids through a distance of 40m before colliding with another parked vehicle; the weight of which is 75 percent of the former After collision both the vehicles skid through 14m before stopping: Compute the initial speed of moving vehicle. Assume coefficient of friction of 0.62. [8] (2068 Baishakh, Q2)
    Solution

    Rotary intersection = a channelized at-grade intersection with a central island around which all traffic moves one-way (anticlockwise here), converting crossing conflicts into merging, weaving and diverging.

    Advantages: continuous stop-free movement, no severe right-angle/head-on conflicts, self-regulating, orderly, aesthetic. Disadvantages: large land area, poor for high volumes/heavy pedestrian flow, delay under heavy traffic, unsuitable at high speeds or steep grades.

    Numerical

    skid coefficient f = 0.62 · g = 9.81 m/s² · a moving vehicle skids 40 m, strikes a parked vehicle weighing 0.75 × its own weight, and the two then skid 14 m together

    Common speed of the pair just after impact

    v' = √(2 g f d) = √(2 × 9.81 × 0.62 × 14) = 13.06 m/s

    Speed of the moving vehicle just before impact

    plastic collision, the parked vehicle starts at rest: m · uimpact = (m + 0.75m) v' uimpact = 1.75 × 13.06 = 22.86 m/s

    Initial speed, adding the 40 m pre-impact skid

    u0 = √(uimpact² + 2 g f d) = √(22.86² + 2 × 9.81 × 0.62 × 40) = √(522.6 + 486.6) = √1009.2 = 31.77 m/s = ≈ 114 km/h
    The moving vehicle was travelling at about 114 km/h before it began to skid.

Rotary Interchange

Rotary Intersection20812068

Guideline for Selecting Rotary

Design Guidelines (IRC: 65-1976)207820762075

  1. Design Speed
  1. Shape of Central Island
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Rotary Intersection

Example: 350 630 405 At an urban intersection, two highways of 15m carriageway intersect at right angle with the traffic flow in PCU/hr for design year as mentioned in figure. Design rotary intersection making suitable assumption. 540 350 410

Tasks

2.5 Traffic Control Devices: Traffic Sign, Marking and Islands

Past Exam Questions on this Topic

  1. What are the general requirements of traffic control devices? Explain the different types of traffic sign. [8] (2082 Baishakh, Q2)
    Solution

    General requirements of a traffic-control device: it should (1) fulfil a need, (2) command attention, (3) convey a clear and simple meaning, (4) command the respect of road users, and (5) give adequate time for a proper response.

    Types: traffic signs (regulatory, warning, informatory), road markings, traffic signals and traffic islands.

  2. What are the factors affecting night visibility? What are the different types of road marking and traffic island used to regulate and control traffic flow? Explain. [2+3+3] (2076 Ashwin, Q4)
    Solution

    Factors affecting night visibility: luminance/brightness of the source, contrast between object and background, glare (disability and discomfort), size of the object, and the driver's adaptation/available time.

    Types of road markings: longitudinal (centre line, lane lines, edge lines, no-overtaking lines), transverse (stop line, give-way line, pedestrian crossing) and other markings (directional arrows, word messages, hazard/parking markings).

    Traffic islands used to regulate and control flow: divisional islands, channelizing islands, pedestrian refuge islands, and rotary (central) islands.

  3. Describe various types of traffic control devices. Write down the advantages and disadvantages of traffic signal. (2072 Chaitra, Q1)
    Solution

    Types of traffic signals: traffic-control (stop-go) signals, pedestrian signals, special/flashing signals, and vehicle-actuated signals; control signals may be fixed-time, traffic-actuated (semi- or fully) or manually operated.

    Disadvantages of traffic signals: may increase rear-end collisions, cause delay to light/side traffic, are costly to install and maintain, and create hazards on failure or when disobeyed.

Review

Traffic Control Devices2072

Traffic Signs2082

According to Traffic Sign Manual 1997 DoR “Traffic sign means any object, device, line or mark on the road whose object is to convey to road users, restrictions, prohibitions, warnings or information of any description.”

Three main types:

Importance of Traffic Sign

Principle of Traffic Signs

Basic Requirements of Traffic Signs

Limitation of Traffic Sign

Regulatory Signs

Regulatory Signs (Mandatory)

Regulatory Signs: Mandatory

STOP

Give Way

Ahead only

Restriction

Ends

Turn Left Ahead

Keep Left

Turn Left

Regulatory Signs (Prohibitory)

No Parking
No Stopping
No Overtaking
No Right Turn
No Horn
Maximum Speed

No Vehicles Over

Height Shown

No Entry

No Trucks

No Vehicles Over Maximum

Gross Weight Shown

Axle Weight Limit

Warning Signs

Pedestrian Crossing
Children
Cattle
Wild Animal
Uneven Road
Slippery Road

Informatory Signs

Informatory Signs (Direction and Place Identification Sign)

Bus Stop
One Way Street
Place Identification
Pedestrian crossing Sign

Advance Direction Sign

Route Confirmation Sign

After Junctions

Informatory Signs (Facility Information)

Overnight

Accommodation

Hospital

Refreshments

Restaurants
Picnic Site

Informatory Signs (Other Information)

Telephone

Parking

Place

Bridge Name Plate

Supplementary Signs

Road Marking or Traffic Marking2076

Road markings are lines, patterns, symbols, words, colors or reflectors on the surface of the pavement or curb or fixed objects near highway to regulate, warn, or guide traffic.

Functions:

Limitations:

Road Marking: Design

Road Marking: Types

Pavement Markings: Longitudinal Markings

Lane Line

Hazard Line

Warning

Barrier Line Do

Not Cross

Edge of Carriageway No Parking

Multi-Lane Road

Pavement Markings: Transverse Markings

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Pavement Markings: Parking Space Markings

Pavement Markings: Lane Arrows

Pavement Markings

Pavement Markings: Words

Pavement Markings: Symbol

Kerb Marking

Object Marking

Reflector Unit

Traffic Island

Types of Traffic Island

Divisional Island

Channelizing Island

Rotary Island

Pedestrian Refuse Island

Tasks

2.6 Traffic Signal2072

Why signals are used

At an intersection the conflicting streams cannot be separated in space, so a traffic signal separates them in time: each stream is given exclusive use of the conflict area for part of a repeating cycle. It is the cheapest form of at-grade control that can handle heavy crossing flows, because it needs only simple geometric design.

A signal installation must:

Definitions and notation

The whole of signal design rests on a small vocabulary, so it is worth stating it precisely:

TermMeaning
CycleOne complete rotation through all the indications provided.
Cycle length, CThe time in seconds for one full cycle — the interval from the start of green on one approach to the next start of green on that same approach.
IntervalA period during which the indications do not change. The change interval is the amber (yellow) between green and red; the clearance interval is the all-red that follows the amber and lets vehicles already inside the intersection get clear.
Green interval, GThe actual duration for which the green is displayed to a movement or set of movements.
Red interval, RThe actual duration for which the red is displayed to that movement.
PhaseA green interval plus the change and clearance intervals that follow it. Only non-conflicting movements are assigned to the same phase.
Lost timeTime during which the intersection is not effectively used by any movement — the start-up reaction of the first driver in the queue after the signal turns green, plus the amber and all-red at the end of the phase.
Saturation flow, SThe maximum discharge rate of a compact queue during a fully saturated green, expressed in PCU/h.
Normal (design) flow, qThe actual approach demand during the design hour, in PCU/h.

Classification by purpose

  1. Traffic control signals — the familiar three-colour system: red = stop, green = go, amber = clearance time.
  2. Pedestrian signals — installed so pedestrians can cross safely; they show WALK and DON'T WALK indications.
  3. Special traffic signals — installed at particular locations to warn traffic of a specific situation (flashing beacons, lane-use signals and the like).

Classification of traffic control signals by mode of operation

  1. Manually operated — worked by a traffic police officer according to the demand seen on the ground; still common at many Nepalese intersections.
  2. Fixed-time (pre-timed) automatic — the same set of phases and the same cycle time repeat continuously; the settings can only be changed at the controller.
  3. Traffic-actuated (automatic) — detectors vary the phases and the cycle with the demand. Much more efficient, but also much more costly.

Advantages and disadvantages

Advantages

  • Increases the traffic-handling capacity of an at-grade intersection.
  • Provides orderly, disciplined movement through the intersection.
  • Improves the quality of traffic flow.
  • Reduces accidents by eliminating the simultaneous crossing conflicts.
  • Gives minor-road traffic a guaranteed chance to cross the continuous major-road stream at reasonable intervals.
  • Maintains a reasonable running speed along the major road if the signals in a series are properly co-ordinated (linked / green wave).

Disadvantages

  • Failure of the signal — power cut or any other defect — causes confusion, because drivers have come to rely on it.
  • Rear-end collisions may increase.
  • Improper design or poor location invites violation of the control.
  • With fixed-time signals, the natural variation in arrivals means one approach can be queueing while the other wastes its green.

Sections 2.6–2.8 are written from P. M. Parajuli, Complete Manual on Transportation Engineering II, Ch.1 §1.3.2 (book pp.39–51) — the reference cited by the worked solutions below. Lecture 12 of the course deck, which covers this topic, was not available.

Past Exam Questions on this Topic

  1. Describe various types of traffic control devices. Write down the advantages and disadvantages of traffic signal. (2072 Chaitra, Q1)
    Solution

    Types of traffic signals: traffic-control (stop-go) signals, pedestrian signals, special/flashing signals, and vehicle-actuated signals; control signals may be fixed-time, traffic-actuated (semi- or fully) or manually operated.

    Disadvantages of traffic signals: may increase rear-end collisions, cause delay to light/side traffic, are costly to install and maintain, and create hazards on failure or when disobeyed.

2.7 Warrants for Traffic Signalization

Warrants for installation of a traffic control signal

A signal is not automatically the right answer at a busy junction: it introduces delay to the major stream and can increase rear-end collisions. A signal should therefore not be installed unless at least one of the following warrants is met, and the data supporting it must come from proper traffic engineering studies.

(a) Minimum vehicular volume warrant

If the average approach speed or the 85th-percentile speed on the major street exceeds 60 km/h, or the intersection lies inside a built-up area, these figures may be reduced to 70% of the above.

(b) Interruption of continuous traffic flow on the major road

When the major street carries 1000–1200 veh/h and the minor road, with only 100–150 veh/h in one direction, suffers undue delay or hazard during any eight hours of an average day.

(c) Minimum pedestrian volume warrant

When 150 or more pedestrians per hour cross a major street carrying more than 600 veh/h on both approaches — or 1000 veh/h where the main street has a raised median. The 70% relaxation for high approach speed applies here too.

(d) Accident experience warrant

When the accident record shows that other remedial measures have failed and five or more accidents involving injury or property damage have occurred within a 12-month period. Even then the signal must not seriously disrupt the traffic flow.

(e) Combination of warrants

When no single warrant is satisfied on its own, but two or more of warrants (a), (b) and (c) are each satisfied to the extent of 80% of their stated values.

Sections 2.6–2.8 are written from P. M. Parajuli, Complete Manual on Transportation Engineering II, Ch.1 §1.3.2 (book pp.39–51) — the reference cited by the worked solutions below. Lecture 12 of the course deck, which covers this topic, was not available.

2.8 Traffic Signal Design by Webster Method2082208120802076207520732071

General principles behind the timings

  1. The red phase of one road is the sum of the green and amber of the other road. Pedestrian crossing time is accommodated inside it when turning movements are barred.
  2. Near the end of red, amber may be shown together with red — the red–amber or initial amber — meaning "get set". Vehicles must not cross the stop line during it.
  3. Clearance amber is provided after green, before red, to satisfy two requirements:
    • a vehicle approaching the stop line can stop at it, and
    • a vehicle already too close to stop at the legal speed can clear the intersection area.
    Usually 2–4 s, longer for higher approach speeds.
  4. Green time follows from the peak-hour approach volume, and must clear the queue in most cycles.

Typical design data

One cycle of a two-phase signal (C = 45 s) Road A Road B GREENAMBREDREDGREENAMB 021.125.14245 t (s) cycle length C phase 1 phase 2
Phase diagram of a two-phase signal. The red of one road spans the green plus amber of the other, so C = GA + YA + GB + YB.

Method A — trial cycle method

The crudest of the three methods; useful only as a first estimate.

  1. Count traffic on roads A and B simultaneously for 15 minutes; call the counts N1 and N2.
  2. Assume a trial cycle length C seconds.
  3. Number of cycles in the 15-minute period = 900/C.
  4. Taking a headway of 2.5 s per vehicle, the green times are
GA = 2.5 × N1 × C / 900 GB = 2.5 × N2 × C / 900
  1. Cycle time C = GA + GB + YA + YB; iterate until the assumed and computed C agree.

Method B — Webster's method (optimum cycle)

Webster's method finds the optimum cycle length C0 — the cycle that gives the least total delay at the intersection. It is the method the examiners ask for most often. The field work is to estimate, for every approach:

In the absence of field data the saturation flow may be taken as 160 PCU/h per 0.3 m of approach width. Mixed traffic must be converted to PCU separately for the normal flow and for the saturation flow.

For each phase take the critical (highest) ratio of normal to saturation flow:

y1 = q1/S1 , y2 = q2/S2 , … Y = y1 + y2 + … (sum of the critical ratios of all phases)

The total lost time per cycle is

L = 2n + R n = number of phases (2 s of start-up loss per phase) R = total all-red time per cycle (s)

The optimum cycle and the green split are then

C0 = (1.5 L + 5) / (1 − Y) G1 = (y1/Y)(C0 − L) , G2 = (y2/Y)(C0 − L) , …

Y must be less than 1, otherwise the intersection is over-saturated and no cycle length can clear it. The same formulation extends to any number of phases.

Method C — design on the basis of pedestrian crossing time

Used when a two-phase signal at a cross-road has to be designed together with pedestrian signals — the situation in most of the past questions.

  1. Choose the amber from the approach speed: 3 s up to 50 km/h, 4 s for 50–65 km/h, 5 s for 65–80 km/h.
  2. Pedestrian clearance time for a road = its carriageway width ÷ 1.2 m/s.
  3. Minimum red on a road = pedestrian clearance time for that road + the initial walk interval (7 s where a pedestrian signal is installed, else 5 s).
  4. Minimum green on a road = (pedestrian clearance of the cross road + initial interval) − the amber of that road. Equivalently, min green of one road = min red of the cross road − that road's amber.
  5. Increase the green of the heavier road in proportion to the approach volume per lane:
GA/GB = VA/VB
  1. Cycle length C = GA + YA + GB + YB, then round up to the next multiple of 5 s. Distribute the excess between the two greens in proportion to their volumes.
  2. Red on a road = green + amber of the other road.
  3. For the pedestrian signals: DON'T WALK on a road = the vehicular red of the other road, and
WALK time = C − (DON'T WALK + pedestrian clearance interval)

Timings are finally expressed as a percentage of the cycle, because controller settings are made in per cent, and checked on site during the peak hour.

Method D — signal timings as per IRC guidelines

IRC's guidance combines the pedestrian criterion with a queue-clearance check and with Webster's formula, and is the most complete of the three procedures.

  1. Pedestrian walking speed 1.2 m/s and initial walk time 7 s. The pedestrian green so obtained is the minimum green for the vehicular traffic on the minor and major roads.
  2. The green for the major road is increased in proportion to the traffic on the two approaches.
  3. Cycle time is then computed allowing 2 s of amber for each phase.
  4. Check the queue clearance green for each approach lane: the first vehicle takes 6 s and every subsequent PCU 2 s; the minimum green so found is limited to 16 s.
  5. Compute the optimum cycle from Webster's formula. Saturation flow may be assumed as below; above 5.5 m take 525 PCU/h per metre of width.
  6. Lost time comes from the amber, the inter-green and an initial delay of 4 s for the first vehicle on each leg.
  7. Revise the cycle and the phases in the light of steps (4) and (5), and set the cycle to a convenient multiple of 5 s.
Approach width, kerb to centre line (m) 3.03.54.04.55.05.5
Saturation flow (PCU/h) 185018901950225025502990

Sections 2.6–2.8 are written from P. M. Parajuli, Complete Manual on Transportation Engineering II, Ch.1 §1.3.2 (book pp.39–51) — the reference cited by the worked solutions below. Lecture 12 of the course deck, which covers this topic, was not available.

Past Exam Questions on this Topic

  1. Design an isolated traffic Signal at an intersection of road X (15 m width), and road Y (10.5 m width) The highest observed traffic volume per lane is 250 vehicles per hour on road X and 200 vehicles per hour on road Y. The approach speeds are 50 km/h for road X and 35 km/h for road Y. Design the traffic and pedestrian signal timings for this intersection. Also, prepare a phase diagram showing both vehicular and pedestrian signal timing. [8] (2082 Bhadra, Q4)
    Solution
    Isolated two-phase signal with pedestrian indications · Road X 15 m wide, approach speed 50 km/h, 250 veh/h per lane · Road Y 10.5 m wide, approach speed 35 km/h, 200 veh/h per lane

    Amber from the approach speed

    3 s up to 50 km/h , 4 s for 50–65 , 5 s for 65–80 both approaches ≤ 50 km/h → AX = AY = 3 s

    Pedestrian clearance time at 1.2 m/s

    CIX = 15 / 1.2 = 12.50 s CIY = 10.5 / 1.2 = 8.75 s

    Minimum red = clearance + 7 s initial walk

    RX = 12.50 + 7 = 19.50 s RY = 8.75 + 7 = 15.75 s

    Minimum green from the pedestrian criterion

    green of one road = other road's red − own amber GY = RX − AY = 19.50 − 3 = 16.50 s GX = RY − AX = 15.75 − 3 = 12.75 s

    Adjust for approach volume

    road X carries the higher volume, so anchor road Y at its pedestrian minimum and scale road X by the volume ratio: GX = GY × (nX/nY) = 16.50 × (250/200) = 20.63 s 20.63 > 12.75 → road X pedestrian minimum satisfied ✓

    Cycle length, rounded up to a multiple of 5 s

    C = GX + AX + GY + AY = 20.63 + 3 + 16.50 + 3 = 43.13 s adopt C = 45 s distribute the excess 45 − 43.13 = 1.87 s by volume: GX = 20.63 + (250/450) × 1.87 = 21.67 s GY = 16.50 + (200/450) × 1.87 = 17.33 s

    Red times

    RX = GY + AY = 17.33 + 3 = 20.33 s RY = GX + AX = 21.67 + 3 = 24.67 s

    Pedestrian signal timings

    DON'T WALK = the parallel road's red WALK = C − DON'T WALK − clearance PSX: DW = RY = 24.67 s , W = 45 − 24.67 − 12.50 = 7.83 s PSY: DW = RX = 20.33 s , W = 45 − 20.33 − 8.75 = 15.92 s
    Cycle 45 s — phase 1: road X green 21.67 s + amber 3 s; phase 2: road Y green 17.33 s + amber 3 s. Pedestrian WALK is 7.83 s across X and 15.92 s across Y.
    Phase diagram, C = 45 sGREENAMBREDREDGREENAMBDON'T WALKWALKCLEARDON'T WALKWALKCLEARRoad XRoad YPed. across XPed. across Y021.6724.674245t (s)
    Vehicular and pedestrian signal timings. A pedestrian crossing road X walks while road X is red, so DON'T WALK equals road Y's green plus amber.

    Marsani's approach-speed pedestrian-crossing procedure (pp. 73-75); the worked text uses the same 1.2 m/s walking speed, 7 s initial walk and 3/4/5 s amber rule.

  2. An isolated signal with pedestrian indication is to be installed on a right angled intersection with road A,l8m wide and road B, 12m wide. The heaviest volume per hour for each lane of road A and B are 300 and 250, respectively. The approach speeds are 55 and 40 kmph for road A and B respectively. Design the timing of traffic and pedestrian signals. [8] (2081 Baishakh, Q4)
    Solution
    Isolated two-phase signal with pedestrian indications at a right-angled intersection · Road A 18 m wide, approach speed 55 km/h, 300 veh/h per lane · Road B 12 m wide, approach speed 40 km/h, 250 veh/h per lane

    Amber from the approach speed

    Road A 55 km/h (50–65 band) → AA = 4 s Road B 40 km/h (≤ 50) → AB = 3 s

    Pedestrian clearance time at 1.2 m/s

    CIA = 18 / 1.2 = 15 s CIB = 12 / 1.2 = 10 s

    Minimum red = clearance + 7 s initial walk

    RA = 15 + 7 = 22 s RB = 10 + 7 = 17 s

    Minimum green from the pedestrian criterion

    GB = RA − AB = 22 − 3 = 19 s GA = RB − AA = 17 − 4 = 13 s

    Adjust for approach volume

    road A carries the higher volume, so anchor road B at 19 s: GA = 19 × (300/250) = 22.8 s 22.8 > 13 ✓

    Cycle length, rounded up to a multiple of 5 s

    C = 22.8 + 4 + 19 + 3 = 48.8 s adopt C = 50 s distribute the excess 1.2 s by volume: GA = 22.8 + (300/550) × 1.2 = 23.45 s GB = 19.0 + (250/550) × 1.2 = 19.55 s

    Red times

    RA = GB + AB = 19.55 + 3 = 22.55 s RB = GA + AA = 23.45 + 4 = 27.45 s

    Pedestrian signal timings

    PSA: DW = RB = 27.45 s , W = 50 − 27.45 − 15 = 7.55 s PSB: DW = RA = 22.55 s , W = 50 − 22.55 − 10 = 17.45 s
    Cycle 50 s — phase 1: road A green 23.45 s + amber 4 s; phase 2: road B green 19.55 s + amber 3 s. Pedestrian WALK is 7.55 s across A and 17.45 s across B.
    Phase diagram, C = 50 sGREENAMBREDREDGREENAMBDON'T WALKWALKCLEARDON'T WALKWALKCLEARRoad ARoad BPed. across APed. across B023.4527.454750t (s)
    Road A takes the longer amber (4 s) because its approach speed of 55 km/h falls in the 50–65 km/h band.

    Marsani's approach-speed pedestrian-crossing procedure (pp. 73-75): amber 4 s for 50–65 km/h and 3 s for ≤ 50 km/h, 1.2 m/s walk, 7 s initial walk.

  3. Using the data in the table below, come up with the cycle length and actual green for a 2-phase signal. Assume a lost time per phase of 3 seconds, yellow time of 3 seconds and all-red time of 4 sec. [8] (2080 Bhadra, Q3)
    Approach flows & saturation flows
    ApproachEastNorthWestSouth
    Flow (veh/hr)488338115217
    Saturation flow (veh/hr)1725172517251725
    Solution
    Two-phase signal · phase 1 = North–South, phase 2 = East–West · saturation flow S = 1725 veh/h on every approach · approach flows 338 and 217 (N–S), 488 and 115 (E–W) veh/h · lost time 3 s per phase · amber 3 s · all-red 4 s

    Critical flow ratio of each phase

    take the highest q/S of the approaches in that phase: y1 = max(338, 217)/1725 = 338/1725 = 0.196 y2 = max(488, 115)/1725 = 488/1725 = 0.283 Y = 0.196 + 0.283 = 0.479

    Total lost time per cycle

    L = 2 × 3 (start-up) + 4 (all-red) = 10 s the 3 s amber is counted as effective green

    Optimum cycle length

    C0 = (1.5 L + 5) / (1 − Y) = (1.5 × 10 + 5) / (1 − 0.479) = 20 / 0.521 = 38.4 s → adopt ≈ 40 s

    Green split

    effective green = C0 − L = 40 − 10 = 30 s g1 = (0.196/0.479) × 30 = 12.3 s g2 = (0.283/0.479) × 30 = 17.7 s
    Cycle 40 s with effective greens of 12.3 s (N–S) and 17.7 s (E–W). The displayed green is the effective green plus the start-up loss minus the amber.
    Phase diagram, C = 40 sGREENAMBREDREDGREENAMBNorth&#8211;SouthEast&#8211;WestALL-RED012.315.319.33740t (s)
    East–West gets the longer green because its critical flow ratio (488/1725 = 0.283) exceeds that of North–South (338/1725 = 0.196).

    The amber is taken as effective green and one all-red is counted per cycle. If both all-red intervals and the amber losses are counted instead, L and hence C0 increase accordingly.

  4. The average normal flow of traffic on the cross roads 1 and 2 during design period are 450 and 350 PCU/hr. The saturation headway on these roads are estimated as 2.5 sec and 3.75 sec respectively. The all red time required for pedestrian crossing is 15 sec. Design two phase signal by Webster's method with neat phase diagram. Take amber time of 2 sec on each phase for clearance and start-up loss time of 2 sec and 3 sec for roads 1 & 2 respectively. [8] (2076 Chaitra, Q4)
    Solution
    Two-phase signal · design flows 450 and 350 PCU/h · saturation headways 2.5 s and 3.75 s · start-up losses 2 s and 3 s · ambers 2 s and 2 s · all-red 15 s

    Saturation flow of each approach

    S = 3600 / saturation headway S1 = 3600 / 2.50 = 1440 PCU/h S2 = 3600 / 3.75 = 960 PCU/h

    Critical flow ratios

    y1 = 450 / 1440 = 0.313 y2 = 350 / 960 = 0.365 Y = Σy = 0.313 + 0.365 = 0.678

    Total lost time per cycle

    L = Σ(start-up) + Σ(amber) + all-red = (2 + 3) + (2 + 2) + 15 = 24 s

    Optimum cycle length

    C0 = (1.5 L + 5) / (1 − Y) = (1.5 × 24 + 5) / (1 − 0.678) = 41 / 0.322 = ≈ 127 s

    Green split

    effective green = C0 − L = 127 − 24 = 103 s g1 = (y1/Y)(C0 − L) = (0.313/0.678) × 103 = ≈ 47.5 s g2 = (y2/Y)(C0 − L) = (0.365/0.678) × 103 = ≈ 55.5 s

    Check

    g1 + g2 + ambers + start-up + all-red = 47.5 + 55.5 + (2+2) + (2+3) + 15 = 127 s = C0 ✓
    Cycle 127 s with greens of 47.5 s (road 1) and 55.5 s (road 2). Finish with a neat phase diagram.

    Lost-time accounting per Marsani (L = Σstart-up + Σamber + all-red).

  5. The average normal flow of traffic on cross roads A and B, of width 7m both, during the design period are 400 and 250 PCU/hr. The saturation flow values on these roads are estimated as 1250 and 1000 PCU per hour respectively. The all-red time is provided for pedestrian crossing with walking speed 1 m/sec and initial walk time 6 sec. Design a two phase traffic signal by Webster's method. [8] (2075 Chaitra, Q4)
    Solution
    Roads A and B, each 7 m wide · design flows 400 and 250 PCU/h · saturation flows 1250 and 1000 PCU/h · amber 2 s per phase · pedestrian walking speed 1 m/s with a 6 s initial walk interval

    Flow ratios

    yA = qA/SA = 400 / 1250 = 0.32 yB = qB/SB = 250 / 1000 = 0.25 Y = Σy = 0.57

    All-red time from the pedestrian crossing requirement

    crossing time = width / walking speed = 7 / 1 = 7 s initial walk interval = 6 s R = 6 + 7 = 13 s

    Total lost time

    L = nℓ + R (ℓ = 2 s amber, n = 2 phases) = 2 × 2 + 13 = 17 s

    Optimum cycle length

    C0 = (1.5 L + 5) / (1 − Y) = (1.5 × 17 + 5) / (1 − 0.57) = 30.5 / 0.43 = ≈ 71 s

    Green split

    effective green = C0 − L = 71 − 17 = 53.9 s gA = (0.32/0.57) × 53.9 = ≈ 30.3 s gB = (0.25/0.57) × 53.9 = ≈ 23.7 s

    Check

    gA + gB + 2 × amber + all-red = 30.3 + 23.7 + 4 + 13 ≈ 71 s = C0 ✓
    Cycle 71 s with greens of 30.3 s (road A) and 23.7 s (road B), separated by 2 s ambers and a 13 s pedestrian all-red.
    Phase diagram, C = 71 sGREENREDREDGREENRoad ARoad BALL-RED030.332.345.36971t (s)
    The 13 s all-red is the pedestrian interval (6 s initial walk + 7 s to cross 7 m at 1 m/s); it is inserted once per cycle.

    Matches Marsani's worked signal-design example exactly (400/250 flows, S = 1250/1000, 7 m width, 1 m/s walk, 6 s initial walk): Y = 0.57, C0 = 70.9 s, gA = 30.3 s, gB = 23.7 s.

  6. A four-legged right angled intersection is to be signalized with a fixed time 2-phase signal. The design hour flow and saturation flow are given in the table below. The lost time is 2 seconds per phase due to starting delays and the amber times for north-south and east-west are 3 seconds and 4 seconds respectively. Determine the optimum cycle time and allocate the green times to the two phases. (2075 Ashwin, Q4)
    Design-hour flow & saturation flow
    ApproachNorth (N)South (S)East (E)West (W)
    Design hour flow900500800700
    Saturation flow2500200032003000
    Solution
    Four-legged intersection, two phases · N–S flows 900 and 500 PCU/h with saturation flows 2500 and 2000 · E–W flows 800 and 700 PCU/h with saturation flows 3200 and 3000 · start-up losses 2 s and 2 s · ambers 3 s and 4 s

    Each phase is governed by its critical approach — the one with the highest q/S ratio, not the highest flow.

    Critical flow ratio of each phase

    phase 1 (N–S): max(900/2500, 500/2000) = max(0.360, 0.250) = 0.360 phase 2 (E–W): max(800/3200, 700/3000) = max(0.250, 0.233) = 0.250 Y = 0.360 + 0.250 = 0.610

    Total lost time

    L = Σ(start-up) + Σ(amber) = (2 + 2) + (3 + 4) = 11 s

    Optimum cycle length

    C0 = (1.5 × 11 + 5) / (1 − 0.610) = 21.5 / 0.390 = ≈ 55 s

    Green split

    effective green = C0 − L = 55 − 11 = 44 s g1 = (0.360/0.610) × 44 = ≈ 26 s g2 = (0.250/0.610) × 44 = ≈ 18 s
    Cycle 55 s with effective greens of 26 s (N–S) and 18 s (E–W). The displayed green is G = g − amber + start-up loss.
  7. The average normal flow of traffic on cross roads A and B are 400 and 250 PCU per hour. The saturation headway on these roads are 3 sec and 4 sec respectively. The all-red time required for pedestrian crossing is 15 sec. Design a two phase traffic signal by Webster's method. [8] (2073 Shrawan, Q2)
    Solution
    Two-phase signal · design flows 400 and 250 PCU/h · saturation headways 3 s and 4 s · amber 2 s per phase · all-red 15 s for pedestrian crossing

    Saturation flow of each approach

    S = 3600 / h S1 = 3600 / 3 = 1200 PCU/h S2 = 3600 / 4 = 900 PCU/h

    Flow ratios

    y1 = 400 / 1200 = 0.333 y2 = 250 / 900 = 0.278 Y = Σy = 0.611

    Total lost time

    L = 2n + R (n = 2 phases, R = all-red) = (2 + 2) + 15 = 19 s

    Optimum cycle length

    C0 = (1.5 L + 5) / (1 − Y) = (1.5 × 19 + 5) / (1 − 0.611) = 33.5 / 0.389 = ≈ 86 s

    Green split

    effective green = C0 − L = 86 − 19 = 67 s g1 = (0.333/0.611) × 67 = ≈ 36.5 s g2 = (0.278/0.611) × 67 = ≈ 30.5 s

    Check

    g1 + g2 + ambers + all-red = 36.5 + 30.5 + 4 + 15 = 86 s = C0 ✓
    Cycle 86 s with greens of 36.5 s and 30.5 s, 2 s ambers and a 15 s all-red pedestrian interval.
    Phase diagram, C = 86 sGREENREDREDGREENRoad 1Road 2ALL-RED036.538.553.58486t (s)
    The 15 s all-red lets pedestrians cross; it is counted once per cycle in L = 2n + R.

    Lost-time convention L = 2n + R (amber + all-red) and the whole procedure follow Marsani's worked Webster example (q = 375/225, S = 1050/850, R = 10 s → Y = 0.62, C0 = 68.5 s).

  8. An isolated signal with pedestrian indication is to be installed on a right-angled intersection with road A 15 m wide and road B 12 m wide. The heaviest volume per hour for each lane of road A and road B are 300 and 250 respectively. The amber times for roads A and B are 3 and 2 seconds respectively. Design the timings of the traffic and pedestrian signals. [8] (2071 Chaitra, Q4)
    Solution
    Isolated two-phase signal with pedestrian indications · Road A 15 m wide, 300 veh/h per lane, amber 3 s · Road B 12 m wide, 250 veh/h per lane, amber 2 s

    Pedestrian clearance time at 1.2 m/s

    CIA = 15 / 1.2 = 12.5 s CIB = 12 / 1.2 = 10.0 s

    Minimum red = clearance + 7 s initial walk

    RA = 12.5 + 7 = 19.5 s RB = 10.0 + 7 = 17.0 s

    Minimum green from the pedestrian criterion

    GB = RA − AB = 19.5 − 2 = 17.5 s GA = RB − AA = 17.0 − 3 = 14.0 s

    Adjust for approach volume

    road A carries the higher volume, so anchor road B at 17.5 s: GA = 17.5 × (300/250) = 21 s 21 > 14 ✓

    Cycle length, rounded up to a multiple of 5 s

    C = 21 + 3 + 17.5 + 2 = 43.5 s adopt C = 45 s distribute the excess 1.5 s by volume: GA = 21.0 + (300/550) × 1.5 = 21.82 s GB = 17.5 + (250/550) × 1.5 = 18.18 s

    Red times

    RA = GB + AB = 18.18 + 2 = 20.18 s RB = GA + AA = 21.82 + 3 = 24.82 s

    Pedestrian signal timings

    PSA: DW = RB = 24.82 s , W = 45 − 24.82 − 12.5 = 7.68 s PSB: DW = RA = 20.18 s , W = 45 − 20.18 − 10.0 = 14.82 s
    Cycle 45 s — phase 1: road A green 21.82 s + amber 3 s; phase 2: road B green 18.18 s + amber 2 s. Pedestrian WALK is 7.68 s across A and 14.82 s across B.
    Phase diagram, C = 45 sGREENAMBREDREDGREENDON'T WALKWALKCLEARDON'T WALKWALKCLEARRoad ARoad BPed. across APed. across B021.8224.824345t (s)
    The ambers are the values given in the question; every other interval follows from the pedestrian criterion and the volume split.

    Marsani's approach-speed pedestrian-crossing procedure (pp. 73-75); the amber values are those given in the question, with 1.2 m/s walking speed and a 7 s initial walk.

2.9 Traffic Calming Measures

Review

Traffic Calming

Traffic Calming Measures

Changes in Vertical Alignment

Rumble strips

Speed hump

Uneven road surface

Raised intersection

Raised pedestrian crossings

Changes in Horizontal Alignment

Chicane

Lane narrowing

Mini roundabout
Junction area reduction

Other measures

Driver’s feedback sign

Gates

2.10 Importance and Design of Street Lighting

Past Exam Questions on this Topic

  1. What are the basic requirements of intersection at grade? Mention the factors affecting night visibility at road. How much is the recommended lateral offset of the lighting poles at the road? Explain with reasoning: [3+3+2] (2082 Bhadra, Q3)
    Solution

    Basic requirements of an at-grade intersection: adequate sight distance, minimum number of conflict points, channelization to guide traffic, low relative speed of crossing streams, adequate capacity, simple and clear layout, and provision for pedestrians, drainage and lighting.

    Factors affecting night visibility: brightness/luminance of source, contrast, glare, size of object, and adaptation time of the eye.

    Lateral offset of lighting poles: poles are placed clear of the carriageway - a lateral offset of about 0.3 m beyond the kerb face (outside the clear zone) is recommended so that poles do not become a collision hazard, with mounting height at least equal to the road width for full coverage.

    Exact recommended offset for local practice should be read from the design Manual.pdf (street-lighting section).

  2. Explain the method to determine the spacing of street lights. Explain about the different types of parking. [8] (2082 Baishakh, Q1)
    Solution

    Spacing of street lights. From the required average illumination E (lux): E = (lamp lumens × CU × MF)/(spacing s × road width w), so s = (lumens × CU × MF)/(E × w). Spacing also depends on mounting height (spacing/height ratio ≈ 3-4) and the luminaire's light distribution.

    Layout systems: single-side, staggered, central (median) and opposite (twin) arrangements.

  3. What are the factors affecting night visibility? What are the different types of road marking and traffic island used to regulate and control traffic flow? Explain. [2+3+3] (2076 Ashwin, Q4)
    Solution

    Factors affecting night visibility: luminance/brightness of the source, contrast between object and background, glare (disability and discomfort), size of the object, and the driver's adaptation/available time.

    Types of road markings: longitudinal (centre line, lane lines, edge lines, no-overtaking lines), transverse (stop line, give-way line, pedestrian crossing) and other markings (directional arrows, word messages, hazard/parking markings).

    Traffic islands used to regulate and control flow: divisional islands, channelizing islands, pedestrian refuge islands, and rotary (central) islands.

  4. What are the causes of accident and how can accidents be prevented? Describe briefly the factors influencing street light design. (2075 Chaitra, Q3)
    Solution

    Causes of accidents: road-user (over-speeding, drunk/fatigued driving, wrong overtaking, rule violation), vehicle (brake/tyre/steering/light defects, overloading), road (poor geometry, inadequate sight distance, slippery surface, poor markings) and environment (rain, fog, glare, poor light).

    Prevention - the 3 E's: Engineering (geometry, signs, lighting, road furniture), Enforcement (speed and traffic-law enforcement) and Education (road-safety awareness and training).

  5. What are the basic requirements of intersection at grade? Mention the importance of street lighting. Explain the different types of traffic islands. How is accident study carried out? Two vehicles A and B approaching at right angles, vehicle A from West and vehicle B from south, collide with each other. After the collision, vehicle A skids in a direction 49° N of W and vehicle B skids 27° E of N. The initial skid distances of vehicles A and B are 37 m and 19 m respectively before collision. If the weight of vehicle A is 4 tonne and vehicle B is 6 tonne, and the skid distances after collision for vehicle A is 15 m and for vehicle B is 36 m, calculate the initial speeds of the vehicles if the average skid resistance of the pavement is found to be 0.55. (2075 Ashwin, Q1)
    Solution

    Basic requirements of an at-grade intersection: adequate sight distance, minimum conflict points, channelization to guide traffic, adequate capacity, low relative speed, clear and simple layout, provision for pedestrians and proper drainage/lighting.

    Importance of street lighting: improves night visibility & safety, reduces night accidents, aids pedestrian security, improves traffic flow and comfort, and enhances aesthetics.

    Types of traffic islands: divisional islands, channelizing islands, pedestrian refuge islands, and rotary (central) islands.

    Accident study is carried out by collecting accident data (spot maps, collision & condition diagrams), analysing causes (road user, vehicle, road, environment), and applying remedial 3-E measures (Engineering, Enforcement, Education).

    Collision numerical

    skid coefficient f = 0.55 · vehicle A 4 t, vehicle B 6 t · post-collision skid 15 m (A) and 36 m (B) · pre-collision skid 37 m (A) and 19 m (B) · deflections A 49° N of W and B 27° E of N

    Speeds just after the collision

    v = √(2 g f d) vA' = √(2 × 9.81 × 0.55 × 15) = 12.72 m/s vB' = √(2 × 9.81 × 0.55 × 36) = 19.71 m/s

    Momentum conservation to recover the impact speeds

    The paper does not state which way each vehicle was heading. The deflections given — A to 49° N of W and B to 27° E of N — are only consistent with A travelling east and B travelling north; any other pairing makes one impact speed come out negative. Measuring all directions anticlockwise from east, A leaves at 131° and B at 63°.

    x (east): mAuA = mAvA' cos131° + mBvB' cos63° 4 uA = 50.89(−0.656) + 118.26(0.454) 4 uA = −33.38 + 53.69 = 20.30 uA = 5.08 m/s y (north): mBuB = mAvA' sin131° + mBvB' sin63° 6 uB = 50.89(0.755) + 118.26(0.891) 6 uB = 38.41 + 105.37 = 143.78 uB = 23.96 m/s

    Add the pre-collision skid to get the original speeds

    u0 = √(uimpact² + 2 g f dpre) A (dpre = 37 m): u0A = √(5.08² + 2×9.81×0.55×37) = √425.1 = 20.62 m/s = ≈ 74 km/h B (dpre = 19 m): u0B = √(23.96² + 2×9.81×0.55×19) = √779.6 = 27.92 m/s = ≈ 100 km/h
    Vehicle A was travelling at about 74 km/h and vehicle B at about 100 km/h before braking.

    The weights (A = 4 t, B = 6 t) enter only as the mass ratio in the momentum equations, so they need not be converted to newtons. Note that A leaves the impact faster than it entered it (12.72 vs 5.08 m/s) — the heavier, faster vehicle B drove it forward, which is exactly why the wreck slid 15 m.

  6. What are the importances of street lighting? Describe the factors affecting street light design. [8] (2072 Chaitra, Q2)
    Solution

    Importance of street lighting: improves night visibility and safety, reduces night accidents, aids pedestrian movement and security, improves traffic flow and comfort, and enhances aesthetics.

    Factors affecting street-light design: required illumination level (lux), mounting height, spacing, lateral overhang, carriageway width, lamp type/lumen output, coefficient of utilization, maintenance factor, and uniformity of illumination.

Street Lighting

Importance of Street Lighting20752072

Factors Influencing Night Visibility20822076

Requirement of Level of Illumination of Roads

Design of Lighting System20802075

Design: Determination of Level of Illumination Required

Design: Choice of Light Source (Lamp)

Filament lamp
Fluorescent lamp

Mercury vapor lamp

LED lamp

Design: Luminaire Distribution of Light

Design: Lateral Light Distribution

Type I (Two way lateral distribution)

Type I lateral light distribution diagram

Type II (Narrow asymmetrical lateral distribution)

Type II lateral light distribution diagram

Type III (Medium asymmetrical lateral distribution)

Type III lateral light distribution diagram

Type IV (Wide asymmetrical lateral distribution)

Type IV lateral light distribution diagram

Type V (Normal symmetrical distribution)

Type V lateral light distribution diagram

Design: Longitudinal Light Distribution

Short distribution: The maximum candlepower beam strikes the roadway surface between 1.0 and 2.25 mounting heights from the luminaire.

Medium distribution: The maximum candlepower beam strikes the roadway surface between 2.25 and 3.75 mounting heights from the luminaire.

Long distribution: The maximum candlepower beam strikes the roadway surface between 3.75 and 6.0 mounting heights from the luminaire.

Design: Spacing of Lights2082

Lecture slide

Utilization Factor:

Lecture slide

Example: Calculate the spacing between the lighting units to produce a lux equal to 7

𝜇 from the following data: Width of road 14 m Mounting height 8 m Lamp size 7000 lumen Luminare type II.

Design: Height and Overhang of Mounting

Design: Lateral Placement

Design: Lighting Layout

Single side Pole Arrangements
Both Side Staggered Pole Arrangements
Twin Central Pole Arrangements

Tasks