For better understanding of topic, we cover theoretical portion and a numerical step by step simultaneously.
Numerical: At right angle intersection of two roads- one road of four lane of total width 12m and another one road with two lane of total width of 6.6m. Volume of traffic approaching intersection during design hour are 900 and 743 PCU/hr on two approaches of road 1 and 278 and 180 PCU/hr on two approaches of road 2. Design signal time by IRC method.
Method
Step 1
Theory: IRC suggested
Walking speed= 1.2m/sec
Intial walking time = 7sec
These are minimum green time for any vehicle traffic on any road. We calculate green pedestrian time required for major and minor roads on the above basis.
Numerical:
Design traffic on road 1= highest of two approaches= 900
Design traffic on road 2= highest of two approaches = 278
Volume/lane on road 1= 900/2= 450PCU/hr
Volume/lane on road 2= 278= 278PCU/hr
( As road 1 has total four lane, two lanes on each side of road while road 2 has total two lanes, one on each side)
Pedestrian green time for any road=
(total road width/ walking speed)+ 7 sec.
Pedestrian green time for road 1=
(12/1.2)+7 = 17sec
Pedestrian green time for road 2= (6.6/1.2)+7 = 12.5 sec
Step 2
Theory:
Numerical:
Minimum green time for vehicle on Road 2 = G2= 17 sec
Minimum green time for vehicle on Road 1 =G1= 17× ( 450/278)= 27.5 sec
Adding 2 sec for clearing camber and 2 sec for inter green period.
Total cycle= ( 2+17+2) + ( 2+27.5+2)= 52.5
For convenience, take total cycle time= 55 sec ( mutliple of 5).
Now extra 2.5 (55-52.5) sec is added to each green time according to volume/per hour.
Now G1= 27.5+1.5=29 sec
G2= 17+1= 18 sec.
Step 3
Theory: For calculating the minimum green time required for clearing vehicles for each lane of approach road, we assume that first vehicle take 6 sec and other queue of PCU will clear at rate of 1PCU in 2 sec. The minimum green time for vehicle traffic on any approach is limited to 16 sec.
Numerical: vehicle arriving per lane per cycle= 450/55= 8.2 PCU/ cycle
Minimum green time per cycle=
6+ (8.2-1)×2= 20.4 sec ( less than 29 sec)
Vehicle arriving per lane per cycle for road 2= 278/55= 5.1 PCU/cycle.
Minimum green time per cycle=
6+ (5.1-1)×2= 14.2sec ( less than 19 sec)
Hence our design Green time is correct.
Step 4
Theory: The optimum signal time is calculated using webster formula. The saturation flow values may be assumed
PCU/hr Road Width
1850 3.0
1890 3.5
1950 4.0
2250 4.5
2550 5.0
2990 5.5
For width above 5.5m, the saturation flow may be assumed as 525 PCU/hr per metre width.
The lost time is calculated from amber time, inter green time and Intial delay of 4 sec for first vehicle on each leg.
Numerical:
Lost time per cycle=( amber+inter green+time lost for intial delay for two phrases=(2+2+4)×2= 16 sec.
Saturation flow for road 1 of width 6m=
525×6=3150 PCU/hr
Saturation flow for road 2 of width 3.3m= 1850 PCU/hr for 3.0 m wide+ 40×3/5= 1874 PCU/hr
y1=900/3150=0.286 y2= 278/1874= 0.434
Y=0.286+ 0.148= 0.434
Optimum signal cycle time= 51.2 sec
Therefore cycle time of 55 sec is acceptable.
Road Green Amber Red Cycle
Road 1 29 2 22+2 55
Road 2 18 2 33+2 55
All units are in seconds.
Numerical: At right angle intersection of two roads- one road of four lane of total width 12m and another one road with two lane of total width of 6.6m. Volume of traffic approaching intersection during design hour are 900 and 743 PCU/hr on two approaches of road 1 and 278 and 180 PCU/hr on two approaches of road 2. Design signal time by IRC method.
Method
Step 1
Theory: IRC suggested
Walking speed= 1.2m/sec
Intial walking time = 7sec
These are minimum green time for any vehicle traffic on any road. We calculate green pedestrian time required for major and minor roads on the above basis.
Numerical:
Design traffic on road 1= highest of two approaches= 900
Design traffic on road 2= highest of two approaches = 278
Volume/lane on road 1= 900/2= 450PCU/hr
Volume/lane on road 2= 278= 278PCU/hr
( As road 1 has total four lane, two lanes on each side of road while road 2 has total two lanes, one on each side)
Pedestrian green time for any road=
(total road width/ walking speed)+ 7 sec.
Pedestrian green time for road 1=
(12/1.2)+7 = 17sec
Pedestrian green time for road 2= (6.6/1.2)+7 = 12.5 sec
Step 2
Theory:
- For major road, green time required for vehicle traffic is increased according to the ratio of approaches on two roads.
- Cycle time is calculated after allowing amber of 2 sec each and 2 sec for intergreen period.
Numerical:
Minimum green time for vehicle on Road 2 = G2= 17 sec
Minimum green time for vehicle on Road 1 =G1= 17× ( 450/278)= 27.5 sec
Adding 2 sec for clearing camber and 2 sec for inter green period.
Total cycle= ( 2+17+2) + ( 2+27.5+2)= 52.5
For convenience, take total cycle time= 55 sec ( mutliple of 5).
Now extra 2.5 (55-52.5) sec is added to each green time according to volume/per hour.
Now G1= 27.5+1.5=29 sec
G2= 17+1= 18 sec.
Step 3
Theory: For calculating the minimum green time required for clearing vehicles for each lane of approach road, we assume that first vehicle take 6 sec and other queue of PCU will clear at rate of 1PCU in 2 sec. The minimum green time for vehicle traffic on any approach is limited to 16 sec.
Numerical: vehicle arriving per lane per cycle= 450/55= 8.2 PCU/ cycle
Minimum green time per cycle=
6+ (8.2-1)×2= 20.4 sec ( less than 29 sec)
Vehicle arriving per lane per cycle for road 2= 278/55= 5.1 PCU/cycle.
Minimum green time per cycle=
6+ (5.1-1)×2= 14.2sec ( less than 19 sec)
Hence our design Green time is correct.
Step 4
Theory: The optimum signal time is calculated using webster formula. The saturation flow values may be assumed
PCU/hr Road Width
1850 3.0
1890 3.5
1950 4.0
2250 4.5
2550 5.0
2990 5.5
For width above 5.5m, the saturation flow may be assumed as 525 PCU/hr per metre width.
The lost time is calculated from amber time, inter green time and Intial delay of 4 sec for first vehicle on each leg.
Numerical:
Lost time per cycle=( amber+inter green+time lost for intial delay for two phrases=(2+2+4)×2= 16 sec.
Saturation flow for road 1 of width 6m=
525×6=3150 PCU/hr
Saturation flow for road 2 of width 3.3m= 1850 PCU/hr for 3.0 m wide+ 40×3/5= 1874 PCU/hr
y1=900/3150=0.286 y2= 278/1874= 0.434
Y=0.286+ 0.148= 0.434
Optimum signal cycle time= 51.2 sec
Therefore cycle time of 55 sec is acceptable.
Road Green Amber Red Cycle
Road 1 29 2 22+2 55
Road 2 18 2 33+2 55
All units are in seconds.
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