SOLUTIONS TO PROBLEMS ON DESIGN OF TRAFFIC SIGNALS
- Given:
Cycle time at an intersection = 60 s
Green time for a phase = 27 s
Yellow time = 4 s
Saturation headway = 2.4 s/vehicle
Start-up time lost = 2 s/phase
Clearance lost time = 1 s/phase
Calculate the capacity of movement per lane
Solution:
Total time lost = tL = 2 + 1 = 3 seconds
Effective green time = gi= 27 + 4 - 3 = 28 seconds
As per the equation for saturation flow rate = Si = 3600 / h = 3600/2.4
= 1500 vehicles / hour
Capacity of the given phase is found by the equation Ci = 1500 * (28/60)
= 700 vehicles/hour/lane
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2. In a right angled intersection of two roads, one road has four lanes with a total width of 12 m. The other road has two lanes with a total width of 6.6 m. The traffic volume of two approaching roads is 900 and 743 PCU per hour. on the two approaches of road-1 and 278 and 189 PCU/hour on the two approaches of road-1 and 278 and 180 PCU.hour on the two approaches of road-2. Design the signal timing as per IRC guidelines
Solution
Width of road-1 = 12.0m, 4 lanes = 2 lanes in each direction
Width of road-2 = 6.6m, 2 lanes = 1 lane in each direction
Approach volume on road-1 = 90 & 743 PCU/hr
Approach volume on road-2 = 278 & 180 PCU/hr
Pedestrian walking speed = 1.2m/s
Design traffic on road-1 = higher of the two approach volume per lane
= 900/2 = 450PCU/hr
Design traffic on road-2 = 278 PCU/hr
STEP-1
Pedestrian crossing time
Pedestrian green time for road-1 = (12/1.2) + 7.0 = 17 seconds
Pedestrian green time for road-2 = (6.6/1.2) + 7.0 = 12.5 seconds
STEP-2
Minimum green time for traffic
Minimum green time for vehicles on road-1 = G(i) = 17 seconds
Minimum green time for vehicle on road-1 = G1 = 17 * (450/278) = 27.5 seconds
STEP 3
Revised green time for traffic signals
Adding 2.0 seconds each towards clearance amber and inter-green period for each phase, total cycle time required = (2 + 17 + 2) +(2 + 27.5 + 2) = 52.5 seconds
Approximating this to the next multiple of 5, cycle time = 55 seconds
Signal time is set conveniently in multiples of 5
The extra time of (55 - 52.5) may be apportioned to green times of road-1 and road-2 as 1.5 and 1 seconds respectively. Adopting G1 = 27.5 + 1.5 = 29.0 seconds and G2 = 17.0 + 1.0 = 18 seconds
STEP 4
Checking for clearing the vehicles during the green phase
Vehicle arrival per lane per cycle on road-1 = (450/55) = 8.2 PCU/cycle
Minimum green time required per cycle to vehicles on road-1 =
= 6 + (8.2 - 1.0) * 2 = 20.4 seconds (This is less than 29 seconds, hence accepted)
Vehicle arrival per lane per cycle on road-2 = (278/55) = 5.1 PCU/cycle
Minimum green time required per cycle to vehicles on road-2 =
= 6 + (5.1 - 1.0) * 2 = 14.2 seconds (This is less than 18 seconds, hence acceptable)
Since the green time provided by the road by pedestrian crossing criteria is higher than the values calculated, the design values are correct
STEP 5
Check for optimum signal cycle by webster's equation
Lost time per cycle = (amber time + inter-green period + time lost for initial delay of first vehicle) for two phases. = (2 + 2 + 4) *2 = 16 seconds
Saturation flow for road-1 of width 6 m = 525 * 6 = 3150 PCU/hour
Saturation flow for road-2 of width 3.3 m = 180 PCU 3.0 wide road + (40 * 3/5)
= 1874 PCU/hour
y1 = 900 / 3150 = 0.286 & y2 = 278/1874 = 0.148
Optimum cycle time C0 = (1.5L + 5)/(1 - y) = 51.2 seconds
Hence, cycle time of 55 seconds as designed earlier is acceptable. The details of signal timing are tabulated as follows.
Road Green phase Amber time Red phase Cycle time
Road-1 29 s 2 s (22 + 2) 55
Road-2 18 s 2 s (33 + 2) 55
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3. Average speed on a roadway = 80 kmph
Average spacing between vehicles (under stopped conditions) = 6.9m
Maximum flow of vehicles (C) = 100 * Average Velocity / Stopping distance
= 100 * (80/6.9)
= 1159.42 vehicles/hour
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4. 15-minute traffic counts n1= 178 and n2= 142
A1= 3 sec, A2= 2 sec, Ht= 2.5 sec
Trial (i) Assume a trial cycle C1= 50 sec
Number of cycles in 15 min = 15 * 60/50 = 18 sec
Green time for Road-1, allowing average time headway of 2.5 sec per vehicle,G1= 178×2.518= 24.7 sec
Similarly for Road-2, G2= 142×2.518= 19.7 sec
Amber times A1 and A2 are 3 and 2 sec (given)
Total cycle length, C = (G1+ G2+ A1+ A2)= 24.7 + 19.7 + 3.0 + 2.0
= 49.4 sec
As this is lower than the assumed trial cycle of 50 sec, another lower cycle length may be tried.
Trial (ii)
Assume a trial cycle C2= 40 sec
Number of cycles in 15 min = 15 * 60/40 = 22.5sec
Green time for Road-1, allowing average time headway of 2.5 sec per vehicle, G1= 178×2.522.5= 19.8 sec
Similarly for Road-2, G2= 142×2.522.5= 15.8 sec
Amber times A1and A2are 3 and 2 sec (given)
Total cycle length, C = (G1+ G2+ A1+ A2)= 19.8 + 15.8 + 3.0 + 2.0= 40.6 sec
As this is lower than the assumed trial cycle of 50 sec, another higher cycle length may be tried.
Trial (iii)Assume a trial cycle C3= 45 sec
Number of cycles in 15 min = 15*60 / 45= 20sec
Green time for Road-1, allowing average time headway of 2.5 sec per vehicle,G1= 178×2.5/20= 22.25 sec
Similarly for Road-2, G2= 142×2.5/20= 17.75 sec
Amber times A1and A2are 3 and 2 sec (given)
Total cycle length, C = (G1+ G2+ A1+ A2)
= 22.25+17.75+3.0+2.0= 45.0 sec
Therefore, the trial cycle of 45 sec may be adopted with the following signal phases:
G1= 22.25, say adopt G1= 22 sec
G2= 17.75, say adopt G2= 18 sec
Adopt A1= 3 sec, A2= 2 sec
Total cycle length, C = (G1+ G2+ A1+ A2)= 22.0 + 18.0 + 3.0 + 2.0= 45.0 sec
Since this is greater than the assumed cycle of 40 seconds, the value of 45 seconds of cycle time is adopted.
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5. Flow of traffic on road A = 400 PCU/hour
Flow of traffic on road B = 250 PCU/hour
Saturation flow value on road A =1250 PCU/hour
Saturation flow value on road B =1000 PCU/hour
Pedestrian crossing time = 12 seconds
Using WEBSTERs method to design traffic signal
Normal flow on roads A & B in PCU/hr
Saturation flow
All – red time, R=12 sec
Number of phase, n = 2
Total lost time in sec
Optimum cycle time say (~ 67.5 sec)
Providing an all-red time, R for pedestrian crossing = 12 sec
Providing Amber times of 2.0 sec each for clearance
Total cycle time = 29 + 22.5 + 12 + 2 + 2 = 67.5 sec.
Width of road – 1 = 12.0 m or total 4 lanes, with 2 lanes in each direction; Width of road - 2 = 6.6 m or total 2 lanes, with one lane in each direction. Approach volumes on road – 1 = 900 & 743 PCU/hr On road - 2 = 278 & 180 PCU/hr Pedestrian walking speed = 1.2 m/sec. Design traffic on road - 1= higher of the two approach volume per lane = 900/2 = 450 PCU/hr Design traffic on road – 2 = 278 PCU/hr Step – 1. Pedestrian crossing time Pedestrian green time for road – 1 = sec Pedestrian green time for road – 2 = sec Step – 2, Minimum green time for traffic Minimum green time for vehicles on Road – 1, G (1) = 17 sec Minimum green time for Road – 1, sec =
revised green time for traffic signals Adding 2.0 sec each towards clearance amber and 2.0 sec inter-green period for each phase, total cycle time required = (2 + 17 + 2) + (2 + 27.5 + 2) = 52.5 sec. Signal cycle time may be conveniently set in multiples of five sec and so the cycle time = 55 sec. The extra time of 55.0 – 52.5 = 2.5 sec per cycle may be apportioned to the green times of Road – 1 and Road – 2, as 1.5 and 1.0 sec respectively. Therefore adopt sec and sec Step – 4, check for clearing the vehicles arrived during the green phase Vehicle arrivals per lane per cycle on Road – 1 = 450/55=8.2 PCU/cycle Minimum green time required per cycle to clear vehicles on Road – 1 = 6 + (8.2 – 1.0)2 = 20.4 sec (less than 29.0 sec and therefore accepted) Vehicle arrivals per lane per cycle on Road – 2 = 278/55 = 5.1 PCU/cycle Minimum green time for clearing vehicles on Road – 2 = 6 + (5.1 - 1.0) 2 = 14.2 sec(less than 18.0 sec) As the green time already provided for the two roads by pedestrian crossing criteria in Step (2) above are higher than these values (29.0 and 18.0 sec), the above design values are alright.
check for optimum signal cycle by Webster’s equation Lost time per cycle = (amber time + inter – green time + time lost for initial delay of first vehicle) for two phases = (2 + 2 + 4) x 2 = 16 sec. Saturation flow for Road – 1 of width 6 m = 525 x 6 = 3150 PCU/hr Saturation flow for Road – 2 of width 3.3 m =1850 PCU for 3.0 m wide road + ( 40 * 3/5) = 1874 PCU/hr Y = 0.286 + 0.148 = 0.434 Optimum signal cycle time, sec Therefore the cycle time of 55 sec designed earlier is acceptable. The details of the signal timings are given below. These may also be shown in the form of phase diagram as in Fig. 5.30. Road Green phase, G sec Amber time, sec Red phase, R sec Cycle time, C sec Road 1 29 2 (22 + 2) 55 Road 2 18 2 (33 + 2) 55
APPROXIMATE METHOD BASED ON PEDESTRIAN CROSSING REQUIREMENT
The following design procedure is suggested for the approximate design of a two phase traffic signal unit at cross roads, along with pedestrian signals:
Based on pedestrian walking speed of 1.2 m per second and the roadway width of each approach road, the minimum time for the pedestrian to cross each road is also calculated
Total pedestrian crossing time is taken as minimum pedestrian crossing time plus initial interval for pedestrians to start crossing, which should not be less than 7 sec and during this period when the pedestrian will be crossing the road, the traffic signal shall indicate red or ‘stop’.
The red signal time is also equal to the minimum green time plus amber time for the traffic of the cross road.
The actual green time needed for the road with higher traffic is then increased in proportion to the ratio of approach volumes of the two roads in vehicles per hour per lane.
Based on approach speeds of the vehicles, the suitable clearance interval between green and red period i.e., clearance amber periods are selected.
The amber periods may be taken as 2, 3 or 4 seconds for low, medium and fast approach speeds
The cycle length so obtained is adjusted for the next higher 5 – sec interval; the extra time is then distributed to green timings in proportion to the traffic volumes The timings so obtained are installed in the controller and the operations are then observed at the site during peak traffic hours; modification in signal timings are carried out if needed
The design of a simple two-phase signal is given below.
- Example - 4 An isolated traffic signal with pedestrian indication is to be installed on a right angled intersection with road A, 18 m wide and road B, 12 m wide. During the peak our, traffic volume per hour per lane of road A and road B are 275 and 225 respectively. The approach speeds are 55 and 40 kmph, on roads A and road B respectively. Assume pedestrian crossing speed as 1.2 m per sec. Design the timings two-phase traffic and pedestrian by the approximate method. Solution Given: Widths of road A = 18 m and of road B = 12 m Traffic volumes on road A = 275 and on road B = 225 vehicles/lane/hour Approach speeds on road A = 55 and on road B = 40 kmph Pedestrian crossing speed = 1.2 m/sec Design of two-phase traffic control signals Pedestrian crossing/clearance time for Road A = 18/1.2 =15 sec Pedestrian crossing/clearance time for Road B = 12/1.2 = 10 sec Adding 7 sec initial walk period, minimum red time for traffic of road A, is (15 + 7) = 22 sec and that for road B is (10 + 7) = 17 sec. Minimum green time, for traffic of road B, based on pedestrian crossing requirement = 22-3 = 19 sec. Minimum green time, for traffic of road A, based on pedestrian crossing requirement = 17- 4= 13 sec.
- The minimum green time calculated for road A is with respect to pedestrian crossing time required for the narrower road B. As road A has higher traffic volume per lane than road B, the green time of road A has to be higher than that of road B; the increase may be proportion to the approach volume of road A with respect to that of road B. Let and be the green times and be the approach volume per lane Using the relation, Green time for traffic is taken as the minimum value = 19 sec as obtained from pedestrian crossing criterion for the wider road A. Green time for traffic of road A may be increased in proportion to higher traffic volume Using relation sec Based on the approach speed of 55 kmph for road A, amber period, sec For road B with 40 kmph, amber period, sec Total cycle length Therefore adopt signal cycle length of 50 sec. The additional period of 50 – 49.2 = 0.8 sec is distributed to green timings in proportion to the approach traffic volume. Therefore the revised signal phase are: sec, adopt 23.5 sec sec, adopt 19.5 sec Therefore cycle time, C = 23.5 + 19.5 + 4 + 3 = 50 sec
- Design of pedestrian signals: Do not Walk (DW) period of pedestrian signal at road A (is red period of traffic signal at B). Pedestrian clearance intervals (CI) are of 15 and 10 sec respectively, for roads A and B for crossing. The walk time (W) is calculated from total cycle length.
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