Download - 03 public transport
Urban TransportUrban TransportPublic Transport
Riza Atiq bin O.K. Rahmat
Rail based
• Mass Rapid Transit (MRT)• Light Rail Transit (LRT)• People Rapid Mover (PRT)
MRT• Speed – up to 100 km/hr• 4 – 12 couches per train• Couches 22m x 3.1 m • Capacity – up to 80,000 passengers / hr /
direction• Acceleration / deceleration ≈ 1.2 m/s2
• Rail – 1435mm gauge• Headway ≥ 120 s• Suitable for radial movement• For high density and high plot ratio area.• Feeder bus service is required• Power supply: 750 V dc• Sub-station: 3 – 5 km spacing
LRT• Speed – up to 40 km/hr• 2 – 6 couches per train• Capacity – up to 40,000 passengers / hr / direction• Acceleration / deceleration ≈ 1.2 m/s2
• Rail – 1000 or 1435mm gauge• Headway ≥ 120 s• Suitable for radial movement• For high density and high plot ratio area.• Feeder bus service is required• Power supply: 750 V dc• Sub-station: 3 – 5 km spacing
PRT• Speed – up to 30 km/hr• 2 – 4 couches per train• Capacity – up to 10,000 passengers / hr / direction• Rail – 1000 gauge or monorail• Headway ≥ 90 s• Suitable for intra-city travel
Transit Capacity
h
nSC p
3600
Cp = Theoretical passenger line capacity
n = vehicle per train
S = Maximum passenger per vehicle
h = headway in second
Operational design
h
nSC p
3600
α = guideway utilisation factor (0.6) = load factor (0.9)
ExampleExample
•City Hall of KL intends to provide a transit line to meet peak hour demand of 12,000 passengers/hr. Required speed = 35 - 40 km/hr. Minimum headway = 120s maximum headway = 240s. Guideway utilisation factor = 0.6 and load factor = 0.9s. Station platform limit = 10 vehicles. Vehicle capacity = 130 passengers.
h
nSC p
3600
12,000 = 3600 x 0.6 x 0.9 x n x 130 / h n = 0.04748 h
n (veh/train) h (headway (s) 1 21.06 2 42.12 3 63.18 4 84.24 5 105.30 6 126.36
Possible range 120 ≤ h ≤ 240
7 147.428 168.489 189.54
10 210.60
Bus Service• Capacity = 12 - 240 passengers.• Flexible - Expansions and extensions
can be introduced easily• Transit systems using buses are
capable of carrying 2,400 to 15,000 passengers per hr per direction.
• Volume of up to 30,000 passengers per hr per direction can be achieved with special bus lane, off-line stations and multiple boarding platform.
Bus travel pattern• Radial service - sub-urban to CBD (Shuttle buses)• Ring road service to link up various sub-centres (Stage
buses)• Local travel service (Mini bus)• Special travel in the CBD (eg. tourism)• Travel service between activity centres (Shuttle of mini
bus).
Example of Bus Service Guidelines
Service pattern•Service major activity centres such as office buildings, school and hospital.•Provide 300 meters coverage where population density > 30. Serve at least 90% of the residents.•Space routes at about 0.75 km in urban area and 1.5 in sub-urban area.
Example of Bus Service Guidelines
Service Level
•Service period : 6 am-12 pm•Headway: Peak: 5 minutes off-peak: 15 minutes
Example of Bus Service Guidelines
Bus Stop
•City centre: 5 -7 stops / km, •sub-urban: 1 - 3 stops /km
Example of Bus Service Guidelines
Passenger comfort
•Passenger shelter•Route and destination sign•Driver courtesy
Bus Station
Passengers’ area (embark / disembark)
Bus holding area
Underground LRT station
Ca
r p
ark
Bus station above underground transit stationBus station above underground transit station
Bus Priority lane
The first stop line for regular traffic
The second stop line for buses
Special treatment for buses at intersection of Jalan Raja Laut and Jalan Sultan Ismail
Bus laneBus lane
Bus Operation DesignFrequency, f = n / N
•n = Demand for service (passengers / hr)•N = Maximum number of passengers per bus
•Usually minimum headway is set in multiples of 7.5 or 10 minutes for the shake of coordination (or service frequency of 8, 6, 4 …. per hr)
ExampleA bus service is planned between Bangi and Putrjaya, a distance of 20 km. The operating time is 45 minutes. Estimated demand is 500 passengers/hr. 45-seater buses will be used, which can accommodate 20 standees. Design basic system and determine the fleet size. Maximum headway is 30 minutes and the minimum terminal time is 5 minutes.
Solution:• Operating speed v = 60L/t
= 60 x 20 / 45 = 26.67km/hr • headway, h = 60(45 + 20)/500 = 7.8 min (adopt 7.5 min.)• Cycle time, T = 2 (45 + 5) = 100 • Fleet size, N = T / h = 100 / 7.5 = 14 vehicles• Revised cycle time, T'= N x h = 14 x 7.5 = 105 min• Revised terminal time = (T' - 2 travel time)/2
= (105 - 2 x 45)/2 = 7.5 minutes• Commercial speed = 20 / (105 / 60 / 2) • = 22.86 km/hr
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