03 public transport
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03 public transport Presentation Transcript

  • 1. Urban Transport Public Transport Riza Atiq bin O.K. Rahmat
  • 2. Rail based• Mass Rapid Transit (MRT)• Light Rail Transit (LRT)• People Rapid Mover (PRT)
  • 3. 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
  • 4. 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
  • 5. 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
  • 6. Transit Capacity 3600nSCp = h Cp = Theoretical passenger line capacity n = vehicle per train S = Maximum passenger per vehicle h = headway in second
  • 7. Operational design 3600ατnSCp = h α = guideway utilisation factor (0.6) τ = load factor (0.9)
  • 8. Example•City Hall of KL intends to provide a transit line tomeet peak hour demand of 12,000 passengers/hr.Required speed = 35 - 40 km/hr. Minimumheadway = 120s maximum headway = 240s.Guideway utilisation factor = 0.6 and load factor =0.9s. Station platform limit = 10 vehicles. Vehiclecapacity = 130 passengers.
  • 9. 3600ατnSCp = h 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 7 147.42 Possible range 8 168.48 9 189.54 120 ≤ h ≤ 240 10 210.60
  • 10. 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.
  • 11. 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).
  • 12. Example of Bus Service GuidelinesService 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 insub-urban area.
  • 13. Example of Bus Service GuidelinesService Level•Service period : 6 am-12 pm•Headway: Peak: 5 minutes off-peak: 15 minutes
  • 14. Example of Bus Service GuidelinesBus Stop•City centre: 5 -7 stops / km,•sub-urban: 1 - 3 stops /km
  • 15. Example of Bus Service GuidelinesPassenger comfort•Passenger shelter•Route and destination sign•Driver courtesy
  • 16. LRT station UndergroundCar park Passengers’ area (embark / disembark) Bus holding area Bus Station
  • 17. Bus station above underground transit station
  • 18. Bus Priority lane The second stop line for buses The first stop line for regular trafficSpecial treatment for buses at intersection of Jalan Raja Laut and Jalan Sultan Ismail
  • 19. Bus lane
  • 20. Bus Operation DesignFrequency, f = n / N•n = Demand for service(passengers / hr)•N = Maximum number ofpassengers per bus•Usually minimum headway is set inmultiples of 7.5 or 10 minutes forthe shake of coordination (orservice frequency of 8, 6, 4 …. perhr)
  • 21. ExampleA bus service is planned between Bangi and Putrjaya, adistance of 20 km. The operating time is 45 minutes.Estimated demand is 500 passengers/hr. 45-seater buseswill be used, which can accommodate 20 standees. Designbasic system and determine the fleet size. Maximumheadway is 30 minutes and the minimum terminal time is 5minutes.
  • 22. 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
  • 23. Terimakasihh Thank you