Meet Our Team FUNDAMENTA L CONCEPT OF TRAFFIC SIGNAL DESIGN AND PHASES Instructor Instructor Ts. NOR AZIAH FATMA BINTI ABDUL AYAH @ ABDUL AZIZ Civil Egineering Department Politeknik Sultan Mizan Zainal Abidin Pn. SUHAILA AZURA BINTI ABD SALAM Civil Egineering Department Politeknik Sultan Mizan Zainal Abidin Meet Our Team TRAFFIC SIGNAL DESIGN AND PHASES
DESIGN PRINCIPLE Factors in design: Traffic Flow, q (pcu/hr) Actual flow on a traffic-signal approach converted in pcu/hr. Saturated Flow, S (pcu/hr) > 5.5 , = 525 Y value (occupancy of the intersection) = =1 Total Lost Time, L (s) = =1 − + =1 Optimum Cycle Time, Co (s) = 1.5 + 5 1 −
Actual flow on a traffic-signal approach converted in pcu/hr. EXAMPLE: The volume of a road is 600 vehicles/hr. The volume includes 150 motorcycles, 50 buses, 40 medium lorries, 60 light van and the rest are cars and taxis. Convert the vehicle volume per hour to passenger car unit per hour, pcu/hr. Type of Vehicle Equivalent Value in p.c.u’s Motorcycle 0.33 Passenger car 1.00 Light Vans 1.75 Medium Lorries 2.00 Heavy Lorries 2.25 Buses 2.25 Type Of Vehicle Total Vehicle / hour Equivalent Factor PCU/Hr Motorcycle 150 0.33 49.50 Buses 50 2.25 112.50 Light Vans 60 1.75 105.00 Medium Lorries 40 2.00 80.00 Passenger car 300 1.00 300.00 TOTAL 647.00 ANSWER : A volume of 600 vehicles/hr is equivalent to 647 pcu/hr. So, this means, the traffic flow on the road with 600 different types of vehicles same as the condition of 647 pcu/hr just being there. TRAFFIC FLOW, q
Where: • effective approach width, W > 5.5 m, S = 525W (pcu/hr); • W < 5.5 m, S is refer the following table; SATURATED FLOW, S Assume that an intersection’s approach signal was to stay green for an entire hour, and the traffic was as dense as could reasonably be expected. The number of vehicles that would pass through the intersection during that hour is the saturation flow rate. W (m) 3.0 3.25 3.5 3.75 4.0 4.25 4.5 4.75 5.0 5.25 S (pcu/h r) 1845 1860 1885 1915 1965 2075 2210 2375 2560 2760 Table : Relationship Between Effective Lane Width And Saturation Flow Saturation Flow Rate can be defined with the following scenario:
= =1 where; yi = highest y value from the approach within that phase i. (where y = ) q = actual flow on a traffic-signal approach converted in pcu/hr. S = saturation flow for the approach in pcu/hr. OCCUPANCY OF THE INTERSECTION, Y To determine the occupancy of the entire intersection, the lane occupancies from all approaches or lanes are combined Y should not be higher than 0.85. If the value found is higher than 0.85, it is recommended that the geometrics of the intersection be upgraded to increase the capacity.
= =1 − + =1 Where: I = the inter-green time between the phases, usually 5 sec a = amber time, usually 3 sec l i = drivers reaction time at begin of green per phase, usually 2 sec TOTAL LOSS TIME, L The total loss time typically includes the red clearance interval, which is the time provided for vehicles already in the intersection to clear it before the opposing traffic begins to move. The red clearance interval ensures that the intersection is clear of conflicting vehicles before allowing the next phase to start. Total loss time can be defined :
= 1.5 + 5 1 − where; L = Total loss time (sec) Y = Occupancy of intersection OPTIMUM CYCLE TIME, Co This optimum cycle time, Co , gives the minimum average delay for the intersection. But this delay is not greatly increased if the cycle time varies within the range of 0.75 to 1.50 of the calculated Co . For practical purposes, cycle time should be between 45 seconds to 120 seconds, although an absolute minimum of 25 seconds can be used.
GREEN TIME EFFECTIVE GREEN TIME, g Green time plus the change interval minus the lost time for a designated phase where; gi = Effective green time within that phase i Co = Optimum cycle time (sec) L = Total loss time (sec) yi = highest y value from the approach within that phase i. Y = Occupancy of the intersection ACTUAL GREEN TIME, G It represents the time allocated for vehicles in that phase to proceed through the intersection without encountering a red signal. where; Gi = Actual green time within that phase i gi = Effective green time within that phase i a = amber time, usually 3 sec l i = Loss time within that phase i (sec) = 0 − × = + −
EXAMPLE 1 A 2-phase traffic signal system will be installed at the following intersection. Both traffic flow, q and Saturated Flow, S are shown in the figure. Based on the data, determine; a) Total loss time, L b) Optimal cycle time, Co c) Actual green time, G d) Sketch the time phasing diagrams for this 2-phase traffic signal. N W E S q = 416 pcu/hr S = 1970 pcu/hr q = 780 pcu/hr S = 3160 pcu/hr q = 350 pcu/hr S = 1970 pcu/hr q = 1450 pcu/hr S = 3160 pcu/hr Given: Inter-green time = 4 sec Amber time = 3 sec Loss time = 2 sec
ANSWER 1 Approach Phase 1 Phase 2 N S E W q (pcu/hr) 416 350 780 1450 S (pcu/hr) 1970 1970 3160 3160 = q 0.21 0.18 0.25 0.46 0.21 0.46 Y value. = =1 = =1 2 = 0.21 + 0.46 = 0.67 ≤ 0.85 a) Total loss time, L. = =1 − + =1 = =1 2 − + =1 2 = 4 − 3 + 4 − 3 + 2 + 2 = 6 sec Given: Inter-green time, I = 4 sec Amber time, a = 3 sec Loss time, l = 2 sec
ANSWER 1 b) Optimal cycle time, Co = 1.5 + 5 1 − = 1.5 6 + 5 1 − 0.67 = 42 sec < 120 sec. From previous solutions: Total Loss Time, L = 6 sec Int. occupancy, Y = 0.67 ➢ Effective green time, g Effective green time, gi of each phase must be determined first. From previous solutions: Optimal Cycle , Co = 42 sec Total Loss Time, L = 6 sec Int. occupancy, Y = 0.67 Phase 1 Int. occupancy, y1 = 0.21 Phase 2 Int. occupancy, y2 = 0.46 PHASE 1 (N-S) PHASE 2 (E-W) 1 = 42 − 6 × 0.21 0.67 = 11 2 = 42 − 6 × 0.46 0.67 = 25 1 = 11 + 2 − 3 = 10 2 = 25 + 2 − 3 = 24 = 0 − × c) Actual green time, G From previous solutions: Loss time, li = 2 sec Amber time, a = 3 sec Phase 1 effective green, g1 = 11 sec Phase 2 effective green, g2 = 25 sec = + −
ANSWER 1 d) Sketch the time phasing diagrams for this 2-phase traffic signal. From previous solutions: Inter-green time, I = 4 sec Amber time, a = 3 sec Phase 1 actual green, G1 = 10 sec Phase 2 actual green, G2 = 24 sec 0 10 13 14 38 41 42 0 Phase 1 (N-S) Phase 2 (W-E) 3s 24 sec 3s 10 sec 10 sec I=4 sec I=4 sec