Traffic Engineering and Management1.pptx

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1. 18CePRO :TRAFFIC ENGINEERING AND Safety Dr. R. VELKENNEDY PROFESSOR & DEAN(ECA) DEPARTMENT OF CIVIL ENGINEERING THIAGARAJAR COLLEGE OF ENGINEERING MADURAI-625015

2. Introduction Definition: Phase of engineering which deals with the planning and geometric design of streets, highways and abutting lands, and with traffic operation thereon, as their use is related to the safe, convenient and economic transportation of persons and goods. Basic Objective of Traffic Engineering: To achieve efficient, free and rapid flow of traffic with least number of traffic accidents and nominal maintenance cost. Seven Major Areas: Traffic Characteristics: Road User Characteristics and Vehicular Characteristics Traffic Studies and Analysis: Volume ,Delay ,Speed ,Parking , Pedestrian and Accident studies Planning and Analysis: Traffic Demands and Driver Behaviour Geometric Design : Design and Redesign of Intersection , channelization, and Parking facilities Traffic Operation, Regulation and Control: Traffic Signs , Signals, Marking , and Lighting Road Safety aspects : Collection Tabulation and Analysis of Accident data Administration and Management: Dealing with Education, Legislation and Enforcement measures

3. Functions of Traffic Engineering. Collection, analysis and interpretation of data pertaining to traffic. Traffic and transportation planning. Traffic design. Measures for operation of traffic and Administration.

4. Road User Characteristics Permanent physical characteristics : Vision- include the acuity of vision, peripheral vision and eye movement ,glare vision, glare recovery and depth judgment. Hearing-More important for pedestrians and cyclists. Temporary physical characteristics: Alcohol or drugs consumption .Fatigue, and illness. Mental characteristics: Experience , skill, knowledge, literacy intelligence can affect the road user characteristics. Pyschological characteristics: Attentiveness, Anger ,Fear , Tension, and superstition impatience. Environmental characteristics: Mixed traffic or heavy traffic, Climate, Traffic stream can be social, recreational, business ,routine movement or an emergency dash. Driver characteristics: Simple, reaction time, depth judgment, field of vision, vision acuity, glare recovery etc.

5. Vehicular Characteristics Static characteristics -Dimension of vehicle –Width of vehicle=2.5m, Height of vehicle-Single - decked vehicle =3.80m Double- decked vehicle =4.75m length of vehicle, Weight of vehicle, Maximum turning angle Dynamic characteristics Speed of vehicle affects design of -sight distance-super elevation- length of transition curve and limiting radius on horizontal curves -width of lane / shoulders -design gradient -capacity of traffic lane -design and control measures at intersection. Brake characteristics depends on the design and type of braking system Power of vehicle depends on the permissible and limiting values of gradient on roads. Resistance to motion of a vehicle Rolling resistance Air resistance Grade resistance Inertia force during acceleration and declaration and Transmission losses.

6. Various factors affecting friction. Type of pavement surface Condition of pavement. Type and condition of tyre . Speed of vehicle. Load and tyre pressure. Temperature of tyre and pavement. Type of skid, if any

7. PIEV Theory. Time required for response to a traffic situation such as putting on brakes, etc.depends upon the psychological process of perception, intellection, emotion and volitions. Perception time is the time required for the sensations received by the eyes or ears to be transmitted to the brain through the nervous system and spinal cord. In other words, it is the time required to perceive an object or situation. Intellection time is the time required for understanding the situation. It is also the time required for comparing the different thoughts, regrouping and registering new sensations. Emotion time is the time elapsed during emotional sensations and disturbance such as fear, anger or any other emotional feelings such as superstition etc with reference to the situation. Therefore the emotion time of a driver is likely to vary considerably depending upon the problems involved. Volition time is the time taken for final action. It is also possible that the driver may apply brakes or take any avoiding action by the reflex action, even without thinking. The total reaction time of an average driver may vary from 0.5 second for simple situations to as much as 3 to 4 seconds or even more in complex problems.

8. Problem A Vehicle moving at 50kmph speed, coefficient of friction between the tyre and pavement is 0.35, the highway is on a 4% down grade. Determine the braking distance. Distance covered by the vehicle after applying brake is known as Braking distance. Braking distance (l) = v 2 /2g(f ±n%) Where v is m/sec =V 2 /254(f ±n%) Where V is kmph =50 2 /254(0.35-.04) =31.75 m

9. Problem A vehicle travelling at a speed of 40kmph was stopped by applying brakes fully and the skid marks were 6m in length. Determine the average skid resistance of the pavement surface. Data: Braking distance or Skid L=6m, Speed =40kmph Braking distance or Skid L=6m= V 2 /254f Skid resistance f =V 2 /254L =40 2 /254*6=1.05 A Vehicle moving at 35kmph speed was stopped by applying the brake and the length of skid mark was 12m. If the average skid resistance of the pavement is known to be 0.62. Determine the braking efficiency. Data: Braking distance or Skid L=12m, Speed =35kmph Braking distance or Skid L=12m= V 2 /254f Skid resistance f’ =V 2 /254L =35 2 /254*12=0.40 Braking efficiency,%=100f’/f =100*0.40/0.62=64.52%

10. Problem Calculate rolling resistance, air resistance ,grade resistance ,engine horse power needed and the speed of the engine experienced by a passenger car weighing (m) 2000kg is required to accelerate at a rate of (a)3.5m/sec 2 in the first gear from a speed of (V)15kmph moving on a surface with coefficient of rolling resistance equal to (f) 0.02, coefficient of air resistance is (Ca)0.39, gradient ( i )+2.0 percent, the frontal projection area of the car is (A) 2.5m 2 and the car tyres have radius of ( r o )0.32m.The rear axle gear ratio is 3.70:1 and the first gear ratio is 2.70:1. Make suitable assumptions. Tyre deformation factor (λ) will have a value of 0.945-0.950 for high pressure air tyres and 0.930-0.935 for low pressure air tyres ,on hard surfaces. Assume Tyre deformation factor (λ)=0.935 r w = λ* r o =0.935*0.32=0.299

11. Solution: Rolling resistance ±P f =mfg=2000*0.02*9.81= 392.4 N Air resistance ±P a =C a * A*v 2 = 0.39*2.5*(15 2 /3.6)= 60.9 N Grade resistance ±P i = mig /100=2000 *2*9.81/100=392.4N Inertia forces due to acceleration and deceleration ± Pj = ma=2000 *3.5=7000N Tractive force needed P p =±P f ±P a ±P i ± Pj N P p =392.4+60.9+392.4+7000=7845.7 N Power output = P p *v =7845.7* 15/3.6=32690.4N-m/sec =32690.4/745 Watts=43.88hp Assume a transmission efficiency of 0.90, Engine Horse-Power=48.75hp v = π *D*n/(60* Gt * Ga )= π * 2 * r w * n/(60 * Gt * Ga ) 0.278 V= 3.14 * 2 r w * n/(60* Gt * Ga ) V=0.377* r w *n/ Gt * Ga , Engine Speed n=V* Gt * Ga /0.377* r w n=15*2.7*3.7/0.377*0.299=1329 rpm

12. Assignment-1 A Vehicle moving at 50kmph speed was stopped by applying the brake and the length of skid mark was 12.1m. and the average skid resistance of the pavement is known to be 0.61. Determine the braking efficiency. A passenger car weighing 2100Kg is required to accelerate at a rate of 3.5m/sec 2 in the first gear from a speed of 10kmph .The gradient is +1.5 percent and the road has a black topped surface .The frontal projection area of the car is 2.0m 2 .The car tyres have radius of 0.34m.The rear axle gear ratio is 3.58:1 and the first gear ratio is 2.48:1.Compute the engine horse power needed and the speed of the engine. Make suitable assumptions. Calculate the basic capacity of traffic lane at a speed of 50 kmph . Assume that average length of vehicles 6m and coefficient of friction is 0.30.

13. Problem The following data were obtained from the spot speed studies Suggest i ) Upper Speed limit for regulation ii) Speed to check geometric design elements iii) Lower speed group causing congestion iv) Dispersion v) Median speed and vi) Modal speed . Mid Speed kmph 5 15 25 35 45 55 65 75 85 95 No. of vehicles observed 12 18 68 89 204 255 119 43 33 9 Spot Speed : Instaneous speed of a vehicle at a specified location Various Methods Enoscope Pressure contact Tubes Radar Speed Meters

14. Spot speed equipment ( Enoscope )

15. Frequency distribution of spot speed Data Speed range kmph Mid speed Kmph Frequency Frequency in % Cumulative frequency 0-10 5 12 1.41(12/850)x100 1.41 10-20 15 18 2.12 3.53 20-30 25 68 8 11.53 30-40 35 89 10.47 22 40-50 45 204 24 46 50-60 55 255 33 76 60-70 65 119 14 19 70-80 75 43 5.06 95.06 80-90 85 33 3.88 98.94 90-100 95 9 1.06 100 Total = 850

16. i )Upper Speed limit for regulation =85 th percentile speed = 60kmph ii) Speed to check geometric design elements = 98 th percentile speed = 84kmph iii) Lower speed group causing congestion = 15 th percentile speed = 30kmph iv) Dispersion = 85 th percentile speed - 15 th percentile speed = 30kmph v) Median speed = 50 th percentile speed = 41kmph Cumulative frequency in % Mid speed in kmph

17. Frequency distribution curve of the spot speeds Modal speed is 48 Kmph Frequency in % Mid speed in kmph

18. Various Speeds Spot Speed is the Instantaneous speed of a vehicle at a specified location. Running Speed is the average Speed maintained by a vehicle over a given course while the vehicle is in motion. Running speed= Length of course / Running time = Length of course /( Journey time- Delays) Journey Speed , also known as overall travel speed, is the effective speed of a vehicle between two points, and is the distance between two points divided by the total time taken by the vehicle to complete the journey, including all delays incurred en-route. Journey speed = Distance/ Total journey Time (including Delays). Average speed in a traffic stream can be computed in two ways: Space -mean speed represents the average speed of vehicles in a certain road length at any time. Time-mean speed represents the speed distribution of vehicles at a point on the roadway and it is the average of instantaneous speeds of observed vehicles at the spot.

19. Problem in Moving observer method The following tables give the particulars collected for a section of road 0.7km long during the course of a moving observer study: Calculate the flow in PCU per hour in both directions of traffic assuming an equivalency factor of one per car, 3 for bus and two for trucks. Calculate the journey speed and running speed.

20. Solution: Flow in North bound direction, q n = ( x s + y n ) / ( t n + t s ) = 18 +(1.2 – 0.7) (1.06+0.94) = 9.25 PCUs/min = 9.25x 60 = 555 PCUs/ hour t n = t n – ( y n / q n ) = 0.94 – [(1.2 – 0.7)/ 9.25] = 0.89 min Mean journey speed in north bound direction t n = 0.7 x 60 = 47kmph 0.89

21. t s = t s – ( y s / q s ) = 1.06 – [(1.5 – 0.5)/ 14.5] = 0.99 min Mean journey speed in south bound direction = 0.70 x 60 = 42.5kmph 0.99 Mean running time in north bound direction, t n = 0.89 – 0.10 =0.79min Running speed in north bound direction = (0.7/0.79) x 60 = 53kmph Mean running time in south bound direction, t s = 0.99 – 0.09 =0.90min Running speed in south bound direction = (0.7/0.9) x 60 = 46.7kmph Flow in South bound direction, q s = ( x n + y s ) / ( t n + t s ) = 28 +(1.5 – 0.5) (1.06+0.94) = 14.50 PCUs/min = 14.50x 60 = 870 PCUs/ hour

22. Assignment problem 1 The consolidated data collected from speed and delay studies by floating car method on a stretch of urban road of length 3 km, running North-South are given below. Determine the average values of volume, journey speed and running speed of the traffic stream along either direction.

23. Moving observer method If the mean journey time in the North bound direction is ( t n )132 secs , the mean journey time in the South bound direction is ( t s ) 138 secs , the opposing traffic count of vehicles met when the test car was travelling South is(Xs) 28 and the number of vehicles overtaking the test car minus the number of overtaken by the test car when the test car is travelling North is( Yn ) 1, the flow in North bound direction q n is q n = ( x s + y n )/ ( t n + t s ) = [(28 +1)/ (132+138)] * 3600 veh /hr = 387 veh /hr

24. Problem If the mean journey time in the North bound direction t n is 152 secs , the number of vehicles overtaking the test car minus the number of overtaken by the test car when the test car is travelling North y n is 1, the flow in North bound direction q n is 557 veh /hour ,the running time in north bound direction is Mean journey time in north bound direction t n = t n – ( y n / q n ) = 152 – (1*3600/557) = 152 – 6.5sec = 145.5 sec

25. Traffic Volume One of the fundamental measures of traffic on a road system is the volume of traffic using the road in a given interval of time. It is also termed as flow and it is expressed in vehicle per hour or vehicles per day. When the traffic is composed of a number of types of vehicle , it is the normal practice to convert the flow into equivalent passenger car unit(PCUs), by using certain equivalency factors. The flow is then expressed as PCUs per hour or PCUs per day. Annual Average Daily Traffic(AADT) is 1/365 th of the total annual flow, is a common measure of flow utilized in geometric standards for highways, improvement of existing facilities and standards for pavement design and maintenance. If the flow is not measured for all the 365 days ,but only for a few days the average flow is called Average Daily Traffic(ADT) . Method available for Traffic counts: i ) Manual methods ii)Combination of Manual and mechanical methods iii) Automatic devices iv) Moving observer method v) Photographic methods

26. Basic manoeuvres in traffic stream.    

27. Nature of collision Head on collision Rear end collision Brush/side swipe Right angled collision Right turn collision

28. Collision diagram

29. Condition diagram

30. TYPES OF ACCIDENTS Fatal accident Major injury Minor injury Property damage

31. Factors cause accidents The road The vehicle The driver The road user other than the motorist Environmental factors Cause of Accidents Fault of driver of motor vehicle Fault of driver of other motor vehicle Fault of cyclist ,Fault of pedestrian and Fault of passenger Defect in mechanical condition of motor vehicle Poor light condition (including street light) Defect in road condition Result of weather conditions and Stray animal

32. Accident analysis Methods of Accident Analysis Regression Method Poisson Distribution Use of Chi-Square test for comparing accident data and Quality Control Method From Poisson Distribution method Probability of occurrence of ‘r’ event P(r)= e - m x m r r! m = average rate of occurrence of event e = base of Naperian Logarithms Applying the above formula to determine the probability of a driver causing an accident. N = Number of drivers M = kilometers driven by each driver p = probability of having an accident per kilometre travelled m be the average rate of occurrence of accidents in a length of travel of M kilometres (m= pM ) or (m= pN )

33. Problem in Accident analysis The accident records for three consecutive years at an uncontrolled junction indicate the following number of accidents: Calculate the probability of 4 accidents occurring per year at the site. Average rate of occurrence of events (m)=(3+6+9)/3=6 Probability of occurrence of r events P(r)= e - m x m r r! Probability of occurrence of r events P(r)= e -6 x 6 4 4! P (r=4 accidents) = 0.1339 Year No. of accidents 2018 3 2019 6 2020 9

34. Problem in Accident analysis It has been found that on an average 1 in150 drivers in a bus company are involved in accident every year. If there are 600 drivers in the company, what is the probability that there are exactly 5 drivers who are involved in an accident during a year? Solution: Probability of occurrence of r events P(r)= e - m x m r r! p = 1/150= 0.007 N = 600 m = pN = 0.007x 600=4.2 P(r= 5)= e -4.2 x 4.2r 5! = 0.1633

35. Problem in Accident analysis It is observed that on an average a vehicle driver drives 4500km during the course of a year. The probability of having an accident is 90 per 200 million vehicle kilometers. What is the probability of a driver having at least three accidents during his driving career extending to 20 years? Solution: No. of accidents = 90 per 200,000,000 vehicle kilometers= 0.0000045 Probability of having an accidents per km traveled (p) = 0.0000045 accidents per veh . Km Km driven by each driver(M) = 4500 x 20 = 90,000 km Ave. rate of occurrence of accidents in a length of travel of M km(m) = pM = 90,000 x 0.0000045 = 0.405 Probability of occurrence of r events P(r)= e - m x m r r! Probability of a driver having r accidents P(r)= e -0.405 x 0.405 r r! P(r= atleast 3) = 1 – P(r=0) - P(r=1) -P(r=2) = 1- e -0.405 x 0.405 - e -0.405 x 0.405 1 - e -0.405 x 0.405 2 0! 1! 2! = 1 – 0.667 – 0.270 – 0.054 = 0.009

36. Parking survey Ill effects of parking: Congestion: Considerable street space-lowering of the road capacity-speed will be reduced-journey time and delay will increase-operational cost of the vehicle increases-economical loss. Accidents: Careless monitoring-parking and un parking leads to accidents-accidents occur driving out a car from the parking area. Careless opening of the doors of parked cars. Environmental pollution: Stopping and starting of vehicles-while parking and un parking-results in noise and fumes-affects the aesthetic beauty of the buildings. Obstruction to firefighting operations: Parked vehicles-obstruct the movement of firefighting vehicles-block access to hydrants-access to buildings.

37. Definition of common terms Parking accumulation: The total number of vehicles parked in an area at a specified moment. The curve of parking accumulation for a typical day is given below

38. Parking load: The area under the parking accumulation curve during a specified period. For example, in above figure, the hatched area represents the parking load in vehicle – hour for a period of 4 hours from 6am to 10am. Parking index: Percentage of parking bays actually occupied by parked vehicle as compared to the theoretical number available. Parking index = Number of bays occupied x 100 Theoretical number of bays available

39. Parking Turn-over: Rate of the usage of the available parking space. Thus if there were 10 parking spaces used by 100 vehicles in a period of, say 12 hours, Then the Parking Turn-over =100/10 vehicles per space in a period of 12 hours. Parking Volume: The number of vehicles parking in a particular area over a given period of time. It is usually measured in vehicles per day. Parking duration: The length of time spent in a parking space.

40. Types of parking surveys Parking space inventory Parking usage survey by patrol Questionnaire type parking usage survey Cardon count

41. Parking Space inventory

42. Traffic Capacity Traffic Capacity is defined as the maximum hourly rate at which persons or vehicle can reasonably be expected to traverse a point or uniform section of a lane or roadway during a given time period under prevailing roadway, traffic and control conditions. Expressed in so many Vehicles or Persons per hour. If V is the speed in kmph and S is the average spacing in metres of moving vehicle, the capacity in vehicles per hour per lane is 1000V/S Average spacing in metres of moving vehicle (S)=L+ vt+v 2 /2gf Where v is m/Sec and Road is level = L+ vt+v 2 /2g(f ±n% ) When road is in Grade. =L+0.278Vt+V 2 /254f Where V is kmph and Road is level. If L is the length of vehicles in metres ,C is the clear distance between two consecutive vehicles (stopping sight distance), V is the speed of vehicles in kmph , the capacity N of vehicles per hour per lane is 1000V/(L +C) Stopping sight distance(C)= vt+v 2 /2gf Where v is m/sec =0.278Vt+V 2 /254f Where V is kmph

43. Problem: Calculate the basic capacity of traffic lane at a speed of 50 kmph . Assume that average length of vehicles(L) is 6m and coefficient of friction(f) is 0.30. Assume Reaction time of the driver (t) is 2.5sec When the road is level, n becomes zero . Average spacing in metres of moving vehicle (S)=L+vt+v 2 /2g(f ±n%) Where v is m/sec = L +0.278Vt+ V 2 /254(f ±n%) Where V is kmph = 6+ 0.278x50x2.5 +50 2 /254x(0.30 ± 0) =6+34.75+32.81 =73.56 m Basic capacity = 1000V/S = 1000x50/73.56 = 679.72 Vehicles per hour

44. Basic Capacity The maximum number of passenger cars that can pass a point on a lane or roadway during one hour under the most nearly ideal roadway and traffic conditions which can possibly be attained. Possible capacity The maximum number of vehicles that can pass a given point on a lane or roadway during one hour, under prevailing roadway and traffic conditions.

45. Practical Capacity or Design capacity The maximum number of vehicles that can pass a given point on a lane or a roadway during one hour , without the traffic density being so great as to cause unreasonable delay, hazard or restriction to the driver's freedom to manoeuvre under prevailing roadway and traffic conditions. Factors affecting capacity and Level of Service: Lane width Lateral clearance Shoulders Surface condition Alignment and Grades

46. Level of service concept in the HCM Manual As per HCM Manual Free flow ,with low volumes ,low density and high speeds is called Level of service A   Types of Level of service Level of service A Level of service B Level of service C Level of service D Level of service E Level of service F: Forced flow operations at low speeds, where volumes are below capacity.

47. Intersection Intersection is an area where two or more highways join or cross, within which are included the roadway and road-side facilities for traffic movements in that area. Classification of intersections: At-grade intersection Grade Separated intersection

48. Basic forms of at-grade intersection

49. Grade separators Types of Grade-separated intersections Grade separated intersections without interchange. Grade-separated intersections with interchange Interchange is a system where by facility is provided for movement of traffic between two or more roadways at different levels in the grade-separated junction. A structure without interchange is on over-bridge or underpass or flyover where by the traffic at different levels moves separately without a provision for an interchange between them.

50. Classification of Interchange based on numbers of legs. A. Three-leg interchange T-interchange Y – interchange Trumpet interchange A partial rotary interchange B. Four-leg interchange Diamond interchange Half clover leaf interchange Clover leaf interchange Rotary interchange Directional interchange C. Multi-leg interchange Rotary interchange

51. Grade separated intersections without interchange :Bangalore - Mysore Expressway

52. Yamuna Expressway- Trumpet interchange

53. Cloverleaf Interchange (Two-tier)

54. Directional Interchange

55. Rotary Intersection A specialized form of at-grade intersection laid out for movement of traffic in one direction round a central island. The vehicles from the converging areas are forced to move around the central island in a clockwise direction in an orderly manner and weave out of the rotary movement into their desired direction. Advantages of rotary intersection: A orderly and regimented traffic flow is provided by rotary one way movement. Normally, all traffic proceeds simultaneously and continuously at fairly uniform, through low speed. Frequent stopping and starting are avoided. Weaving movements replace the usual angular crossing of typical at-grade intersection. Direct conflict is eliminated. Suited for intersections with five or more intersection legs and/or heavy right turning movements. All turns can be made easy, although little extra travel distance is required for all movements except left turns.

56. Classification based on shape of Rotary island

57. Rotary Elements

58. Rotary design Design Speed: The design speed of a rotary governs the various elements such as radii and weaving lengths. To design rotaries in rural areas for a speed of 40 K.P.H and those in urban areas to a speed of 30 K.P.H. Radius at entry: The radius at entry is determined by the design speed, super elevation and coefficient of friction. A range of 20-35m is found to be suitable for rural designs and a range of 15-20m in suitable for urban designs. Radius at exit: The exit radius should be higher than the radius of the rotary island so that it favors a higher speed by drivers. The radius of the exit curves 1.5 to 2 times the radius of the entry curves. Radius of the central island: The radius of the central island is governed by the rotary design speed and theoretically it should be equal to the radius at entry. A value of 1.33 times the radius of entry curve is probably adequate for this purpose. Weaving lengths (l): The weaving length determines the ease with which the traffic can merge and diverge. A ratio of 4:1 is regarded as a minimum current Indian practice Design Speed (K.P.H) Minimum weaving length (l) m 40 45 30 30

59. Width of carriageway at entry and exit: The carriage way width of the intersection legs is governed by the design year traffic entering and leaving the intersection. A minimum width of carriage way of 5 m.   Carriageway width of approach road Radius at entry (m) Width of carriageway at entry and exit (m) 7m (2 lanes) 6.5 10.5 m (3 lanes) 25-35 7.0 14 m (4 lanes) 8.0 21 m (6 lanes) 13.0 7m (2 lanes) 7.0 10.5 m (3 lanes) 15-25 7.5 14 m (4 lanes) 10.0

60. Width of rotary carriageway: The width of the non-weaving section should be equal to be widest single entry into the rotary, and should generally be less than the width of the weaving section. The width of the weaving section(w) should be one traffic lane (3.5m) wider than the mean entry width. Thus, w =[(e 1 +e 2 ) /2]+ 3.5 where, w = width of the weaving section e 1 = average entry width e 2 = width of the non weaving section Weaving length ( min. should be 30m)

61. Proportion of weaving traffic (P) = Practical Capacity Q p = This formula is valid under the following conditions: There are no standing vehicles on the approaches to the rotary. The site of the rotary is level and approach gradients do not exceed 1 in 25. e/w should be between 0.4 and 1.0. w/l should be between 0.12 and 0.40. p should be between 0.4 and 1.0. l should be between 18 and 90 m.

62. Entry and exit angles: Entry angles should be larger than exit angles and it is desirable that the entry angles should be about 60º if possible; the exit angles should be small, even tangential. A design with an entry angle of 60º and exit angle of 30º. Other miscellaneous features: Stopping sight distance should be provided for the design speed adopted. The rotary should preferably be located on a ground having a slope flatter than 1 in 50. Capacity: The capacity of a rotary is directly determined by the capacity of a weaving section. The capacity of a weaving section is determined by the geometric layout, including entrances and exits, and the percentage of weaving traffic.

63. Problem: Traffic flow in an urban section at the intersection of two highways in the design year is given below. The highways intersect at right angles and have a carriage way width of 14m.Design the rotary intersection using PCU value of car =1, commercial vehicle ( com.v ) =2.8 and scooter (SC)=0.75. Approach Left turning Straight Ahead Right Turning Car Com.V SC Car Com.V SC Car Com.V SC N 200 50 100 250 100 150 150 50 80 E 180 60 80 220 50 120 200 40 120 S 250 80 100 150 50 90 160 70 90 W 220 50 120 180 60 100 250 60 100

64. Approach Left turning Straight Ahead Right Turning Car Com.V SC Car Com.V SC Car Com.V SC North 200 50 100 250 100 150 150 50 80 PCU 200 140 75 250 280 113 150 140 60 Total PCU 415 643 350 East 180 60 80 220 50 120 200 40 120 PCU 180 168 60 220 140 90 200 112 90 Total PCU 408 450 402 South 250 8 100 150 50 9 160 70 9 PCU 250 224 75 150 140 68 160 196 68 Total PCU 549 358 424 West 220 5 120 180 60 100 250 60 100 PCU 220 140 90 180 168 75 250 168 75 Total PCU 450 423 493

65. Entry width (e 1 ) = 10 m Width of non weaving section (e 2 ) = 10 m Average entry width of rotary in m (e) = (e 1 +e 2 ) /2 = (10+10)/2 = 10 m Width of weaving section (w) =((e 1 +e 2 ) /2) + 3.5 = [(10+10)/2]+3.5 = 13.5 m Proportion of weaving traffic (P s-e) = P = (1136+852)/(350+1136+852+408) = 1988/2746 = 0.72 Length of weaving section (l) = 50 m Practical Capacity Q p = = [280x13.5(1+(10/13.5))x(1-(0.72/3))]/(1+(13.5/50)) = 4000 PCUs/hr This is very much higher than the traffic flow of 2746 PCUs/hr and the design is acceptable.

66. Sketch of the rotary design

67. Motor vehicle Act It was formed in 1939 To regulate the road traffic in the form of traffic laws, ordinances and regulations. Three phases primarily covered are Control of driver, Vehicle ownership and Vehicle operation It was revised on 1988 came into force from July 1989 , 2017 and 2019

68. Thank You….