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Transportation Engineering - I Traffic Flow Characteristics. Dr. Attaullah Shah Traffic Flow • Traffic flow: No of vehicle passing per unit time at a road section q = n/t n: No of vehicles t: duration of time, usually an hour • Time head way: The time between passing of successive vehicles i.e. bumpers – t = Σ hi t; duration of time interval hi; time head way of ith vehicle. – q = n/ Σ hi = Average mean time headway • The average speed is defined in two ways: – Arithmetic mean of total speeds at a particular point also called time mean speed: Mean Ut = Σui/n – Space Mean Speed: Assumed that the average speed for all vehicles is measured at the same length of roadway. – Space mean speed in unit distance per unit time = length of strip/mean time Example • The speed of five vehicles is measured with radar at the midpoint of 0.8Km strip (0.5 mi) The speeds are measured as 44,42,51,49 and 46 mi/h. Assuming the vehicles were travelling constant speed, determine the time mean speed. • Tim Mean speed Ut = Σui/n = (44+42+51+49+46)/5 = 46.4 mi/h L nL n • The space mean = t t 1 / v = 5/ [1/44+1/42+1/51+1/49+1/46] n = 46.17 mi/h Space mean speed is always smaller than time mean speed. v s i i i i i i Traffic density • K = n/l The No of vehicles per unit length. • The density is also related to individual spacing of the vehicles; The total length of roadway l = Σ si k= 1/( Mean space) Traffic flow = q = uk q: flow of traffic No of vehicles/hour Speed ( Space mean speed) k = density units of vehicle/mile. Example 5.2 Traffic Flow • Complex: between vehicles and drivers, among vehicles • Stochastic: variability in outcome, cannot predict with certainty • Theories and models – Macroscopic: aggregate, steady state – Microscopic: disaggregate, dynamics – Human factor: driver behavior Speed (v) • Rate of motion • Individual speed L L v , vavg T T • Average speed – Time mean speed Arithmetic mean – Space mean speed Harmonic mean vt v i i n L nL vs ti ti i n i vt vs Individual Speed (1) Time of Location (ft) Speed Passing (sec) 600 700 (ft/sec) 700 600 v1 50 ft/sec 2.0 0.0 50* 3600/ 5280 34.09 (mi/hr or mph) Vehicle Spot Speed 1 0.0 2.0 50.0 2 4.4 6.7 45.0 3 6.0 8.0 50.0 4 11.4 14.3 35.0 5 15.0 17.5 40.0 6 17.5 20.0 40.0 7 21.1 23.3 45.0 8 23.3 25.0 60.0 Time Mean Speed Mile post Observation Period vt 50 45 50 35 40 40 45 60 45.6 (ft/sec) 8 Observation Distance Space Mean Speed Observation Period 100* 8 vs 44.2(ft/sec) 30.1(mi/hr) 2 2.3 2 2.9 2.5 2.5 2.2 1.7 Volume (q) • Number of vehicles passing a point during a given time interval • Typically quantified by Rate of Flow (vehicles per hour) Time-Space Plot 1000 900 800 Distance (ft) 700 q 600 500 400 300 200 100 0 0 5 10 15 Time (sec) 20 25 8 1152 (veh/hr or vph) 25 / 3600 Volume (q) Density (k) • Number of vehicles occupying a given length of roadway • Typically measured as vehicles per mile (vpm), or vehicles per mile per lane (vpmpl) Density (k) Density (k) q(veh/hr) v(mi/hr) * k(veh/mi) 1152 (veh/hr) 30 .1(mi/hr) * k k 1152 / 30 .1 38 .22 (veh/mi) Spacing (s) • Front bumper to front bumper distance between successive vehicles S2-3 S1-2 Headway (h) • Time between successive vehicles passing a fixed point T=0 sec T=3sec h1-2=3sec Spacing and Headway spacing headway Spacing and Headway What are the individual headways and the average headway measured at location A during the 25 sec period? A Spacing and Headway What are the individual headways and the average headway measured at location A during the 25 sec period? Time of Location (ft) Passing (sec) 600 700 Vehicle A 1 0.0 2.0 2 4.4 6.7 3 6.0 8.0 4 11.4 14.3 5 15.0 17.5 6 17.5 20.0 7 21.1 23.3 8 23.3 25.0 h1-2 h2-3 Lane Occupancy • Ratio of roadway occupied by vehicles L2 L3 LO L i i D L1 D Clearance (c) and Gap (g) • Front bumper to back bumper distance and time Clearance (ft) or Gap (sec) g avg havg Lavg vavg cavg gavg * vavg Spacing (ft) or headway (sec) Basic Traffic Stream models 1. Speed density Model: Relationship between speed and density – Initially when there is only one vehicle, the density is very low and driver will move freely at the speed closely to the design speed of the highway. This is called free flow speed. – When more and more vehicles would come, the density of the road will increase and the speed of vehicles would reduce. The speed of the vehicle may ultimately reduce to u=0 – This high traffic density is referred as “Jam Density” – This model can be expressed as: U = Uf ( 1- k/kj) • • • • U= Space mean speed in mi/h (km/h) Uf = Free flow speed in mi/h k = density in veh/mi ( veh/km) This gives relationship between traffic flow speed and density – The speed density relationship tends to be non linear at low density and very high density. • Non linear relationship at low densities that has speed slowly declining from free flow to value • Linear relationship over the medium density region ( speed declining linearly with the density). • Non linear relationship near the traffic jam density relationship Speed vs. Density k u u f 1 k j Speed (mph) uf Free Flow Speed kj Density (veh/mile) Jam Density Basic Traffic Stream models • Flow Density Model: – – – – As we know that q = uk and U = Uf ( 1- k/kj) Parabolic density model q = Uf ( k – k2/Kj ) The max flow rate qcap: The highest flow rate The traffic density that corresponds to this capacity flow rate is kcap and the corresponding values qcap ,Kcap ,Ucap • dq/dk = Uf (1-2k/kf )= 0 Since the free flow speed cannot be zero. • Kcap = Kj / 2 • Substituting the value: Ucap = ( k – Kj /2kj ) = Uf /2 q c ap u c ap k c ap u k f j /4 Flow vs. Density 2 k q uf k k j Congested Flow FLow (veh/hr) Optimal flow, capacity, qm Uncongested Flow km Optimal density Density (veh/mile) kj Jam Density Speed Flow Model k u u 1 k u k k 1 u u q uk k u u f j j f 2 j f • This is parabolic relationship • Two speeds are possible as it is quadratic equation, up to the highway capacity q • Similarly two densities are possible for given flow cap Speed vs. Flow Speed (mph) uf Free Flow Speed 2 u q k j u u f Uncongested Flow um Congested Flow Optimal flow, Flow (veh/hr) capacity, qm Example At a section of highway in Islamabad the free flow speed is 90km/h and capacity of 5000 veh/h. During traffic census at a particular section 2500 vehicles were counted. Determine the space mean speed of the vehicle. Solution: The traffic flow is related to the space mean speed and jam density as The jam 2 k q uf k k j k 4q / u 4 5000 / 90 222veh / km To determine the jam density j f From above equation of traffic flow we can rearrange the equation as k u k u q 0 u j 2 j f 222 u 222u 2500 0 90 2.47u 222u 2500 0 u 80.53km / h 2 2 28.36km / h Both the speeds are feasible as shown on previous slide Traffic Flow Models-Poisson Model • The Poisson distribution: – Is a discrete (as opposed to continuous) distribution – Is commonly referred to as a ‘counting distribution’ – Represents the count distribution of random events Poisson Distribution ( t ) e P ( n) n! n t P(n) = probability of having n vehicles arrive in time t λ = average vehicle arrival rate in vehicles per unit time t= duration of time interval over which vehicles are counted e= base of the natural logarithm (e=2.718) Example • An observer counts 360veh/h at a specific highway location. Assuming the arrival of vehicles at this point follows Poisson’s distribution. Estimate the probabilities of 0,1,2,3,4 and 5 vehicles arriving in 20 sec time interval. • Solution: – The average arrival rate =λ= 360 veh/h = 360/60x60=0.1 veh/sec ( t ) n e t P ( n) n! – t=20 sec ( 20 0.1) e P ( n) n! – For probability of no vehicle n 0.120 ( 20 0.1) e P (0) 0! – P(1) = 0.271 P(2) =0.271 P(3) = 0.180 0 0.120 0.135 P(4) = 0.090 – For 5 or more vehicles: – P(n> 5) =1-P(n<5)= 1-(0.135+0.271+0.271+0.18+0.090 = 0.053 – Draw the histogram of the distribution Poisson Example • Example: – Consider a 1-hour traffic volume of 120 vehicles, during which the analyst is interested in obtaining the distribution of 1minute volume counts Poisson Example What is the probability of more than 6 cars arriving (in 1-min interval)? Pn 6 1 Pn 6 6 1 Pn i i 0 Pn 6 1 (0.135 0.271 0.271 0.180 0.090 0.036 0.012) 1 0.995 0.005or (0.5%) Poisson Example What is the probability of between 1 and 3 cars arriving (in 1-min interval)? P1 n 3 Pn 1 Pn 2 Pn 3 P1 n 3 27.1% 27.1% 18.0% 72.2% Poisson distribution • The assumption of Poisson distributed vehicle arrivals also implies a distribution of the time intervals between the arrivals of successive vehicles (i.e., time headway) Negative Exponential • To demonstrate this, let the average arrival rate, , be in units of vehicles per second, so that q veh h veh sec h sec 3600 Substituting into Poisson equation yields ( t ) e P ( n) n! n t n qt 3600 qt e 3600 P ( n) n! (Eq. 5.25) Negative Exponential • Note that the probability of having no vehicles arrive in a time interval of length t [i.e., P (0)] is equivalent to the probability of a vehicle headway, h, being greater than or equal to the time interval t. Negative Exponential • So from Eq. 5.25, P(0) P(h t ) 1e qt 3600 1 e (Eq. 5.26) qt 3600 This distribution of vehicle headways is known as the negative exponential distribution. Note: x0 1 0! 1 Negative Exponential Example • Assume vehicle arrivals are Poisson distributed with an hourly traffic flow of 360 veh/h. Determine the probability that the headway between successive vehicles will be less than 8 seconds. Determine the probability that the headway between successive vehicles will be between 8 and 11 seconds. Negative Exponential Example • By definition, Ph t 1 Ph t Ph 8 1 Ph 8 Ph 8 1 e 1 e qt 3600 360 ( 8 ) 3600 1 0.4493 0.551 Negative Exponential Example P8 h 11 Ph 11 Ph 8 1 Ph 11 Ph 8 360 (11) 3600 1 e 0.551 1 0.3329 0.551 0.1161 Negative Exponential 1.0 Prob (h >= t) 0.8 e^(-qt/3600) 0.6 0.4 0.2 0.0 0 5 10 15 20 Time (sec) For q = 360 veh/hr 25 30 35 Negative Exponential c.d.f. 1.0 Probability (h < t) 0.8 1 - e^(-qt/3600) 0.6 0.551 0.4 0.2 0.0 0 5 8 10 15 20 Time (sec) 25 30 35