Traffic and Transport: Flow, Geometric Design, Signals, Parking, Modes and ITS
2. Traffic survey methods
Every design and plan starts from measured traffic. The standard surveys are summarised below; know what each measures and what it is used for.
| Survey | What it gives | Method |
|---|---|---|
| Volume count | flow by class and time; peak hour, ADT, AADT | manual classified count; automatic tube or loop detector, radar, video; cordon and screen-line counts |
| Spot speed | speed distribution at a point; 85th percentile; pace | radar gun, paired detectors, video |
| Journey time and delay | travel time and delay along a route | floating-car (test-car) runs |
| Origin–destination | O–D matrix, desire lines, trip purpose | roadside interview, licence-plate matching, postcard, home interview, tag |
| Parking | accumulation, turnover, duration, occupancy | in-out count, fixed-period sampling, licence-plate survey |
Volume terms. ADT is the average daily traffic over a period shorter than a year, AADT the average over a full year; the design hourly volume is a high hour (often the 30th highest of the year) expressed as a share of AADT. The peak-hour factor is PHF = hourly volume ÷ (4 × maximum 15-minute volume), at most 1; a low PHF means a peaky flow. Example: 1,200 veh in the peak hour with a busiest quarter-hour of 360 gives PHF = 1,200/1,440 = 0.833, and the peak flow rate is 1,440 veh/h.
Spot speed. The 85th-percentile speed, below which 85% of vehicles travel, is the usual basis for a speed limit and a check on design speed; the 15th percentile marks a sensible minimum; the pace is the 10 km/h band containing the most vehicles. Spot speeds averaged arithmetically give the time-mean speed (see the next section). Origin–destination surveys differ in what they cost and give: a roadside interview stops a sample at a cordon or screen line and records origin, destination, purpose and load; licence-plate matching interrupts no one but cannot give purpose; a home interview gives full-day trips with household data and is the base survey for planning models.
Moving-observer method measures flow and travel time together. A test car runs repeatedly along the section against the stream under study (time t_a, with x vehicles of that stream met) and with it (time t_w, with y = vehicles overtaking the car minus vehicles overtaken). The flow of the stream is q = (x + y)/(t_a + t_w), and its mean journey time is t_w − y/q. Example: x = 88, y = 2, t_a = t_w = 3 min gives q = 90/6 = 15 veh/min = 900 veh/h.
3. Traffic flow analysis: q = k·v, Greenshields, headway and capacity
Three variables describe a stream: flow q (veh/h), density k (veh/km) and space-mean speed v (km/h), tied by the fundamental relation q = k·v. Their reciprocals are the time headway h = 3600/q seconds and the spacing s = 1000/k metres, and s = v × h when v is in m/s. At 1,800 veh/h on a lane the headway is 2 s, and at 54 km/h (15 m/s) the spacing is 30 m.
Two mean speeds. The time-mean speed vₜ is the arithmetic mean of spot speeds at a point; the space-mean speed vₛ is the harmonic mean, the speed that gives q = k·v, and vₜ ≥ vₛ with vₜ = vₛ + σ²/vₛ, where σ² is the variance of the space distribution of speeds. For 20, 30, 30 and 60 km/h: vₜ = 140/4 = 35, vₛ = 4/(1/20 + 1/30 + 1/30 + 1/60) = 4/(8/60) = 30 km/h (here σ² = 150, so vₜ = 30 + 150/30 = 35). Fast vehicles pass a point more often, so the time mean is biased upward.
Greenshields' linear model assumes speed falls linearly with density: v = v_f(1 − k/k_j), where v_f is the free-flow speed and k_j the jam density. Then q = v_f·k(1 − k/k_j), a parabola with its maximum, the capacity, at k = k_j/2, where v = v_f/2 and q_max = v_f·k_j/4. Any flow below capacity occurs at two densities, one uncongested (high speed, low density) and one congested (low speed, high density).
Worked example: v_f = 60 km/h and k_j = 120 veh/km. Capacity = 60 × 120/4 = 1,800 veh/h at k = 60 veh/km and v = 30 km/h. For q = 1,000: 60k − 0.5k² = 1,000, so k² − 120k + 2,000 = 0 and k = 20 or 100 veh/km, i.e. 50 km/h uncongested or 10 km/h congested. Capacity is the maximum hourly rate at which vehicles can be expected to traverse a section under the prevailing road, traffic and control conditions; the headway limit gives it too, since one lane at a minimum headway of 2 s carries 3,600/2 = 1,800 veh/h. Practical capacity is the flow at the chosen design LOS, below the possible capacity.
4. Geometric design of roads: sight distance and horizontal curves
Design speed is the master variable: it fixes the minimum curve radius, superelevation and sight distance, so a road's cross-section, horizontal alignment, vertical alignment and sight distances are all designed to it. Sight distance is the length of road a driver can see ahead, and the design question is always whether it is enough to act safely.
Stopping sight distance (SSD) is the lag distance plus the braking distance: SSD = v·t + v²/(2g(f ± G)), with v in m/s, t the reaction time (2.5 s in IRC practice), f the longitudinal friction coefficient and G the gradient as a fraction, plus for an upgrade and minus for a downgrade. In km/h it is 0.278·V·t + V²/(254(f ± G)). Example: V = 72 km/h = 20 m/s, t = 2.5 s, f = 0.4 and level road: lag = 50 m, braking = 400/(2 × 9.81 × 0.4) = 51 m, so SSD is about 101 m. On a 3% downgrade f − G = 0.37, braking rises to 55 m and SSD to 105 m. SSD grows with the square of speed in braking, with longer reaction time, with a wet (low-friction) surface and downhill.
Other sight distances. The intermediate sight distance is twice the SSD. The overtaking sight distance OSD = d₁ + d₂ + d₃ (distance covered in the reaction of the overtaker, the overtaking itself and the opposing vehicle's travel meanwhile) is the largest and governs two-lane, two-way roads where passing is allowed. On a valley curve the headlight beam must reveal an obstacle at least one SSD away, and a summit curve is designed so that SSD is available over the crest.
Horizontal curves and superelevation. On a curve, the centrifugal tendency is resisted by banking (superelevation e) and side friction f: e + f = V²/(127R), V in km/h and R in m. In IRC practice the design friction is f = 0.15 and the superelevation is capped at 7% in plain and rolling terrain. The procedure is to provide e for about 75% of the design speed with friction ignored, e = V²/(225R), and then check at full speed that the friction demanded does not exceed 0.15; if the e needed exceeds the cap, adopt the cap and check f = V²/(127R) − e, and if that exceeds 0.15 restrict the speed or increase the radius. The least radius is R_min = V²/(127(e_max + f)).
Example: V = 80 km/h, R = 250 m. e = V²/(225R) = 6,400/56,250 = 0.114, above the 7% cap, so adopt e = 7%. The friction demanded at full speed is V²/(127R) − e = 0.2016 − 0.07 = 0.132, below 0.15, so the curve is acceptable. For R = 200 m, e = 6,400/(225 × 200) = 0.142 is again capped at 7%, and the friction demanded is 0.252 − 0.07 = 0.182 > 0.15, so the radius must grow or the speed must be restricted. The least radius for 80 km/h is 6,400/(127 × 0.22) = 229 m. Transition (spiral) curves introduce curvature and superelevation gradually, sharp curves are widened for off-tracking, and vertical curves are parabolic.
5. Intersections, roundabouts, grade separators and signal design
An intersection is where flows cross, merge and diverge, and the number of conflict points measures its hazard. A three-arm junction has 9 conflict points (3 crossing, 3 merging, 3 diverging) and a four-arm two-way junction has 32 (16 crossing, 8 merging, 8 diverging). Improvement means removing or reducing conflicts by separating them in space or time: channelisation with islands, one-way streets, banned turns, signals, roundabouts and grade separation.
Roundabouts. Traffic circulates one way around a central island, clockwise in India because traffic keeps left, and enters and leaves by merging and diverging, so a four-arm roundabout has only 8 conflict points and no crossing conflicts. A traditional rotary relies on weaving; a modern roundabout gives priority to circulating traffic and uses entry deflection to slow vehicles, which is safer. The empirical weaving-section capacity used in IRC practice is Qₚ = 280·w·(1 + e/w)·(1 − p/3)/(1 + w/l) PCU/h, where w is the weaving width, e the mean entry width, l the weaving length and p the proportion of weaving traffic: capacity rises with a wider, longer weaving section and falls as the weaving proportion grows.
Grade separators remove crossing conflicts entirely: a flyover carries one road over another, an underpass under it, an interchange (diamond, trumpet for three arms, cloverleaf for four arms, directional) lets turning traffic change roads without crossing, and railway crossings are replaced by road over-bridges or under-bridges. They are justified where cross volumes are so heavy that even well-timed signals leave persistent queues, accident records are poor and land for ramps exists. In a cloverleaf, loop ramps serve the turn that would cross opposing traffic, the right turn with left-hand traffic.
Signal design (Webster). Phases serve the conflicting flows in turn. The saturation flow is the flow that would pass in a stream of continuous green; for an approach of width w metres, s = 525·w PCU/h (w from 5.5 to 18 m). Each approach has the flow ratio y = q/s; the critical y of each phase is taken, and Y = Σ y_crit. With total lost time L per cycle (start-up loss plus clearance, plus any all-red), Webster's optimum cycle is C₀ = (1.5L + 5)/(1 − Y). The effective green is shared in proportion to y: gᵢ = (yᵢ/Y)(C − L). Y must be below 1, and as it nears 1 the cycle explodes and the geometry or a grade separator must be reconsidered. Pedestrian green must allow an initial time plus crossing width divided by about 1.2 m/s.
Worked example: L = 12 s, y₁ = 0.25 and y₂ = 0.20, so Y = 0.45. C₀ = (1.5 × 12 + 5)/(1 − 0.45) = 23/0.55 = 41.8 s, say 42 s. Total effective green = 42 − 12 = 30 s, shared as g₁ = (0.25/0.45) × 30 = 16.7 s and g₂ = (0.20/0.45) × 30 = 13.3 s.
6. Parking: studies, layouts and design
Parking is a land-hungry use, and each vehicle spends most of its life parked. Supply is on-street (cheap, quick, but uses carriageway) or off-street (surface, multi-storey, underground, mechanical). Policy is demand management: price parking, limit duration, and set maximum rather than minimum standards near transit so that supply does not invite cars. In planning standards other vehicles are converted into equivalent car spaces (ECS).
The parking study terms are exact and examinable. Accumulation is the number of vehicles parked at an instant: initial count plus cumulative entries minus cumulative exits. Occupancy is accumulation ÷ capacity. Parking load is the total vehicle-hours consumed, Σ(vehicles × duration). Capacity over a period is bays × hours. Efficiency = load ÷ capacity × 100. Turnover is the number of vehicles served per bay in the period, = vehicles parked ÷ bays. Average duration = load ÷ vehicles.
Worked example: a lot of 50 bays works 10 h and serves 300 vehicles with a mean stay of 1.5 h. Load = 300 × 1.5 = 450 vehicle-hours; capacity = 50 × 10 = 500 bay-hours; efficiency = 450/500 = 90%. Turnover = 300/50 = 6 vehicles per bay per day. Accumulation example: 20 vehicles parked at the start of an hour, 120 enter and 90 leave, so accumulation = 20 + 120 − 90 = 50, which fills a 50-bay lot.
| Layout | Kerb length per car | Depth taken from the road | Remarks |
|---|---|---|---|
| Parallel | largest; fewest cars per length | least | slow to enter and leave; least width needed |
| 30°, 45°, 60° angle | intermediate | intermediate | easy entry; usually one-way aisle |
| 90° (right angle) | smallest; most cars per length | greatest | needs reversing out, a hazard to passing traffic |
Parking studies are done by in-out counts (entries and exits per interval, which give accumulation), fixed-period sampling (count parked vehicles at intervals) and licence-plate surveys (give durations, turnover and violators). Their output sizes supply: peak accumulation sets the number of bays needed, and turnover and duration show whether short-stay pricing or time limits will free space.
7. Modes of transportation, pedestrians and non-motorised transport
No single mode suits every corridor. Modes differ in right-of-way, capacity, speed, flexibility and capital cost, and the planner matches them to the peak-hour, peak-direction demand of a corridor, to the cost the city can bear and to how well modes connect. Capacity of a line mode is simple arithmetic: passengers per hour per direction = vehicles (trains) per hour × cars per train × passengers per car.
| Mode | Right-of-way | Capacity class | Typical role |
|---|---|---|---|
| Walking, cycling, cycle-rickshaw (NMT) | footpaths, cycle tracks, shared streets | low per vehicle, high per metre of width | short trips and the first and last mile |
| Para-transit (auto-rickshaw, e-rickshaw, shared taxi) | mixed traffic | low | flexible door-to-door feeder where buses are thin; adds to congestion |
| City bus | mixed traffic or priority lane | low to medium | backbone of most Indian cities; low capital cost, slowed by congestion |
| BRT (bus rapid transit) | segregated bus lanes with junction priority | medium to high | metro-like service at much lower capital cost on medium-demand corridors |
| LRT, tram (light rail) | dedicated or semi-segregated, at grade or elevated | medium | medium-demand corridors and city centres |
| Metro (heavy rail) | fully grade-separated, elevated or underground | high | high-demand trunk corridors; highest capital cost |
| Suburban (commuter) rail | railway tracks, often shared with long-distance trains | high | long commutes from the periphery into the core |
The ranges are orders of magnitude, not norms: buses in mixed traffic carry a few thousand passengers per hour per direction, while heavy metro and suburban rail can exceed 40,000. Worked example: a 2-minute headway is 30 trains per hour; with 6 cars of 250 passengers each, capacity = 30 × 6 × 250 = 45,000 passengers per hour per direction. Mode choice follows demand: a metro on a thin corridor wastes capital, and buses in mixed traffic on a corridor of tens of thousands collapse.
Pedestrian and non-motorised transport (NMT) planning. Every trip begins and ends on foot, and in Indian cities walking, cycling and cycle-rickshaws carry a large share of trips, with pedestrians the most vulnerable road users. The principles: a continuous and connected network with direct routes and short blocks; safety through separation or calming of traffic and good lighting; comfort (clear, level, shaded, encroachment-free footpaths wide enough for the flow, with a separate zone for vendors and street furniture); accessibility for all (kerb ramps no steeper than 1:12, tactile paving, no steps on the route); and directness, since walkers make little detour at a typical speed of about 1.2 m/s, so at-grade crossings on desire lines are used and foot over-bridges and subways often are not.
Cyclists need segregated tracks on fast arterials and can mix on calm streets, safe parking and links to stations. India's National Urban Transport Policy (2006) asks for road space to be shared equitably by people moved, not vehicles, and for NMT to be given priority. A complete street serves pedestrians, cyclists, transit and vehicles within one design, and pedestrian flow per metre of effective footpath width, per minute, grades the pedestrian LOS.
8. Demand management and control, terminals and allied facilities, and ITS
Two families of measures manage congestion without simply adding lanes. Traffic system management (TSM) is supply-side and makes the existing system work better; travel demand management (TDM) is demand-side and reduces vehicle trips or moves them in space, time or mode. A sound strategy follows the sequence avoid (reduce the need to travel, through land use), shift (to walking, cycling, transit) and improve (efficiency of what remains).
| Supply-side (TSM) | Demand-side (TDM) |
|---|---|
| signal coordination and green waves; one-way streets; reversible lanes; turn bans | congestion or road pricing; parking pricing and limits; low-emission zones |
| junction improvement and channelisation; bus priority lanes and signals | staggered or flexible work hours; telework; car-pooling; employer travel plans |
| parking management; enforcement and incident management | better transit and NMT; TOD and mixed use that shorten trips |
Control devices regulate and guide. Regulatory signs are circular, warning signs triangular and informatory signs rectangular; markings and signals complete the set. Signals along a corridor can be simultaneous, alternate, progressive (offsets set so that a platoon meets green after green) or flexible progressive; area traffic control with adaptive timing responds to measured flows across a network.
Terminals and allied facilities are planned from the peak-hour load. Typical elements: inter-city bus terminals at the city edge on an arterial and near rail; local bus terminals and depots for parking, maintenance and fuelling of the fleet; interchange hubs where metro, rail and bus meet, with short, level, covered, signed transfers and integrated ticketing; truck terminals on the periphery to keep heavy goods vehicles out of the core; and park-and-ride at the edge of the built-up area. Sizing: bays needed = arrival rate × bay occupancy time, plus an allowance. Example: 90 buses per hour, each occupying a bay 8 minutes, need 90 × 8/60 = 12 bays, and with a 25% allowance 15 bays. Keep passenger and vehicle flows separate and circulation clear of the public road.
Intelligent transportation systems (ITS) apply information, communication and control technology to make transport safer and more efficient, using what exists better rather than building more. Its parts: advanced traffic management (adaptive signals, variable message signs, incident detection, enforcement cameras); traveller information (real-time apps and signs); public transport systems (vehicle location, passenger information, smart cards such as the National Common Mobility Card); electronic toll collection (FASTag); emergency management; and commercial vehicle operations. Data come from loop detectors, cameras with number-plate recognition, GPS probes and signal controllers, and integrated command and control centres bring them together.
ITS improves the use of capacity but does not replace it, and in Indian conditions its limits are mixed traffic, weak lane discipline, patchy data, maintenance and interoperability, and privacy. It works best on top of sound basics: good geometry, enforced regulation, priority for transit and a walkable street.
Key takeaways
- Hierarchy is the mobility–access trade-off: arterials for through movement, local streets for access. LOS A to F grades operation, with E at capacity; PCU puts mixed traffic on a common scale, PCU/h = Σ N × factor.
- Flow q = k·v with the space-mean (harmonic) speed. Greenshields: v = v_f(1 − k/k_j), capacity v_f·k_j/4 at half the jam density and half the free-flow speed. Headway 3600/q seconds, spacing 1000/k metres.
- Surveys: volume (PHF = V/(4 × V₁₅)), spot speed (85th percentile), O–D (roadside, licence plate, home), parking, and the moving-observer flow q = (x + y)/(t_a + t_w).
- SSD = vt + v²/(2g(f ± G)); superelevation e + f = V²/(127R), capped, with friction checked at full speed. A four-arm junction has 32 conflict points, a roundabout 8; grade separation is a last resort, not a free cure.
- Webster: C₀ = (1.5L + 5)/(1 − Y), with y = q/s, s = 525w, and green shared in proportion to y. Parking: efficiency = load ÷ (bays × hours), turnover = vehicles ÷ bays, accumulation = initial + entries − exits.
- Match the mode to corridor demand: capacity = trains per hour × cars × passengers per car; BRT, LRT, metro and suburban rail differ in right-of-way and cost. Plan for pedestrians and NMT first, manage demand (avoid, shift, improve), size terminals by arrival rate × occupancy time, and use ITS to improve, not replace, capacity.
Practice questions (20)
Attempt each one before opening the answer. Every explanation names the tempting wrong option as well as the right one, because that is where marks are lost.
In the functional classification of urban roads, which class best serves long through trips with limited, controlled access to adjoining property?
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Answer: C — Arterial road, which gives priority to through movement and limits or controls access to property
Arterials sit at the mobility end of the mobility–access trade-off. Collectors balance the two, and local streets and service roads sit at the access end.Under Greenshields' linear speed–density model, a road has free-flow speed 80 km/h and jam density 120 veh/km. Its capacity, in veh/h, is ____.
Numerical answer — type the value.
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Answer: 2400
q_max = v_f × k_j/4 = 80 × 120/4 = 2,400 veh/h. Check: capacity occurs at k = 60 veh/km and v = 40 km/h, so q = 60 × 40 = 2,400.In Greenshields' model v_f = 60 km/h and k_j = 120 veh/km. A flow of 1,000 veh/h occurs at two densities, one uncongested and one congested. The density of the uncongested state, in veh/km, is ____.
Numerical answer — type the value.
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Answer: 20
q = v_f·k(1 − k/k_j) = 60k − 0.5k². Setting 60k − 0.5k² = 1,000 gives k² − 120k + 2,000 = 0, so k = (120 ± √(14,400 − 8,000))/2 = (120 ± 80)/2 = 20 or 100. The smaller root is uncongested: v = 60 × (1 − 20/120) = 50 km/h and q = 20 × 50 = 1,000.Spot speeds of four vehicles at a point are 20, 30, 30 and 60 km/h. The space-mean speed is
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Answer: A — 30 km/h
The space-mean speed is the harmonic mean: 4/(1/20 + 1/30 + 1/30 + 1/60) = 4/(8/60) = 30 km/h. The tempting 35 km/h is the arithmetic (time-mean) speed, 140/4, which is biased toward fast vehicles.In the peak hour a road carries 600 cars, 400 two-wheelers, 100 buses and 50 three-wheelers. Take PCU factors of 1.0 for a car, 0.5 for a two-wheeler, 3.0 for a bus and 0.8 for a three-wheeler. The volume in PCU/h is
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Answer: D — 1,140
Volume = 600 × 1.0 + 400 × 0.5 + 100 × 3.0 + 50 × 0.8 = 600 + 200 + 300 + 40 = 1,140 PCU/h. The figure 1,150 is the raw vehicle count, 1,100 leaves out the three-wheelers, and 1,340 forgets to halve the two-wheelers.A vehicle travels at 72 km/h on a level road. Take the reaction time as 2.5 s, the coefficient of longitudinal friction as 0.4 and g = 10 m/s², and use SSD = vt + v²/(2gf) with v in m/s. The stopping sight distance, in metres, is ____.
Numerical answer — type the value.
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Answer: 100
v = 72 km/h = 20 m/s. Lag distance = 20 × 2.5 = 50 m. Braking distance = v²/(2gf) = 400/(2 × 10 × 0.4) = 400/8 = 50 m. SSD = 50 + 50 = 100 m.A curve of 200 m radius carries a design speed of 80 km/h. Take e + f = V²/(127R), a maximum permitted superelevation of 7% and a maximum design side friction of 0.15. What should the designer conclude?
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Answer: B — Provide e = 7%; the friction then needed is about 0.18, above 0.15, so increase the radius or restrict the speed
V²/(127R) = 6,400/(127 × 200) = 0.252. With f = 0.15 the superelevation would be 10.2%, above the 7% cap. At e = 0.07 the friction demanded is 0.252 − 0.07 = 0.182, which exceeds 0.15, so the design is unsafe as it stands and the radius must be increased or the speed restricted. The sum e + f at the limits is only 0.22, not 0.252.A two-phase signal has these critical approaches. Phase 1: width 8 m, flow 1,260 PCU/h. Phase 2: width 6 m, flow 945 PCU/h. Take saturation flow s = 525w PCU/h (w in metres) and a total lost time of 10 s per cycle. Webster's optimum cycle C₀ = (1.5L + 5)/(1 − Y), in seconds, is ____.
Numerical answer — type the value.
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Answer: 50
s₁ = 525 × 8 = 4,200, so y₁ = 1,260/4,200 = 0.30. s₂ = 525 × 6 = 3,150, so y₂ = 945/3,150 = 0.30. Y = 0.60. C₀ = (1.5 × 10 + 5)/(1 − 0.60) = 20/0.40 = 50 s.A four-arm intersection with two-way traffic on every arm has 32 conflict points when uncontrolled. Converting it to a roundabout with the same arms leaves how many conflict points?
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Answer: D — 8
A roundabout removes all crossing conflicts and leaves only the merges and diverges at the arms: 4 merging and 4 diverging, a total of 8. The figure 16 is the number of crossing conflicts at the uncontrolled intersection, which is exactly what the roundabout eliminates.Which situation most strongly justifies building a grade separator at an urban junction?
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Answer: A — Heavy crossing flows leave persistent queues despite optimised signals, crashes are frequent and ramp land is available
Grade separation is costly, takes land and shifts bottlenecks, so it is justified only when volumes, delay and conflict record cannot be handled at grade even after signal and geometric improvements. The other situations call for cheaper measures first.Compared with angle and right-angle layouts, kerb parking parallel to the carriageway
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Answer: B — accommodates the fewest cars per unit kerb length but takes the least depth from the road
Parallel bays use the most kerb length per car but intrude least into the carriageway, and entering them needs manoeuvring. Right-angle parking fits the most cars per kerb length but takes the greatest depth and needs reversing out.A 50-bay car park operates for 10 hours and serves 300 vehicles with an average stay of 1.5 hours. Its parking efficiency (parking load ÷ capacity), in per cent, is ____.
Numerical answer — type the value.
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Answer: 90
Parking load = 300 × 1.5 = 450 vehicle-hours. Capacity = 50 bays × 10 h = 500 bay-hours. Efficiency = 450/500 × 100 = 90%. (Turnover would be 300/50 = 6 vehicles per bay, a different quantity.)In a moving-observer survey the test car, driven against the stream under study, meets 88 vehicles of that stream per run. In the runs with the stream, vehicles overtaking it exceed vehicles it overtakes by 2 per run. Each run takes 3 minutes. The flow of the stream, in veh/h, is
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Answer: C — 900
q = (x + y)/(t_a + t_w) = (88 + 2)/(3 + 3) = 15 veh/min = 900 veh/h. Dividing by a single run time gives 1,800, the usual slip; leaving out y gives 880.Which of the following are core features of a bus rapid transit (BRT) system?
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Answer: A — Segregated bus lanes, with priority or signal preference at junctions; B — Off-board fare collection and level boarding at stations; D — High-frequency service and stations designed for metro-like speed and reliability at lower capital cost
BRT gets its performance from segregation, priority, fast boarding and frequent service, at far lower capital cost than rail. A fully grade-separated electric rail alignment describes a metro, not BRT.Which of the following are sound principles of pedestrian and non-motorised transport planning?
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Answer: B — Make footpaths continuous, level and free of encroachment, with kerb ramps no steeper than about 1:12; D — Allocate road space by the number of people moved per hour rather than by the number of vehicles
Continuous, accessible footpaths and allocation by people moved are core NMT principles. Pedestrians tolerate little detour, so over-bridges that force one are often ignored, and pedestrians can of course be counted and planned for directly.Which of the following statements about traffic surveys are correct?
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Answer: A — The 85th-percentile speed is commonly used as a basis for fixing a speed limit; C — Peak-hour factor is the hourly volume divided by four times the highest 15-minute volume; D — A home interview survey can give full-day trips together with household characteristics
The 85th percentile, PHF = V/(4 × V₁₅) and home interviews are all correctly stated. Licence-plate matching works without stopping anyone, by noting plates at stations and matching them, which is exactly why it cannot give trip purpose.Which of the following changes increase the stopping sight distance required on a road?
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Answer: B — A higher design speed; C — A longer driver reaction time; D — A wet pavement that lowers the friction coefficient
SSD = vt + v²/(2g(f ± G)): a higher speed lengthens both terms, a longer reaction time lengthens the lag, and lower friction lengthens braking. An upgrade helps braking (f + G), so it shortens SSD, whereas a downgrade lengthens it.Which of the following is a travel demand management measure rather than a supply-side measure?
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Answer: C — Pricing kerb and off-street parking in the central area to rise with demand
Parking pricing changes whether, when and how people travel, which is demand management. A new lane, signal coordination and a flyover all make the supply of road capacity work harder or larger.A bus terminal receives 90 buses per hour at the peak and each bus occupies a bay for 8 minutes including entry and exit. Allowing 25% extra for peaks and breakdowns, the number of bays to provide is
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Answer: D — 15
Bays = arrival rate × occupancy time = 90 × 8/60 = 12; with a 25% allowance, 12 × 1.25 = 15. The figure 12 forgets the allowance, and 18 and 24 come from using longer occupancy times than the 8 minutes given.Detector-based signal control that adjusts splits and cycle length in real time to the measured flows belongs to which area of intelligent transportation systems?
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Answer: A — Advanced traffic management systems
Adaptive signal control is a core function of advanced traffic management. Traveller information tells users about conditions, commercial vehicle operations serve freight fleets, and electronic toll collection handles payment.