03 public transport
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Transcript of 03 public transport
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Urban TransportUrban TransportPublic Transport
Riza Atiq bin O.K. Rahmat
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Rail based
• Mass Rapid Transit (MRT)• Light Rail Transit (LRT)• People Rapid Mover (PRT)
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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
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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
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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
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Transit Capacity
h
nSC p
3600
Cp = Theoretical passenger line capacity
n = vehicle per train
S = Maximum passenger per vehicle
h = headway in second
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Operational design
h
nSC p
3600
α = guideway utilisation factor (0.6) = load factor (0.9)
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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.
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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
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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.
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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).
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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.
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Example of Bus Service Guidelines
Service Level
•Service period : 6 am-12 pm•Headway: Peak: 5 minutes off-peak: 15 minutes
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Example of Bus Service Guidelines
Bus Stop
•City centre: 5 -7 stops / km, •sub-urban: 1 - 3 stops /km
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Example of Bus Service Guidelines
Passenger comfort
•Passenger shelter•Route and destination sign•Driver courtesy
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Bus Station
Passengers’ area (embark / disembark)
Bus holding area
Underground LRT station
Ca
r p
ark
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Bus station above underground transit stationBus station above underground transit station
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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
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Bus laneBus lane
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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)
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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.
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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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TerimakasihhTerimakasihhThank youThank you