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Introduction to Transportation Engineering Instructor Dr. Norman Garrick Hamed Ahangari 27th March 2014 1 Traffic Stream Analysis 2 Key equations Flow headway (h) - (s/veh) Flow rate (q) - (veh/h) q=3600/(h) (1) T=3sec T=0 sec h1-2=3sec 3 Key equations Density Spacing (s) -(ft/veh) density-concentration(k) S2-3 k=5280/(s) - S(veh/mi) 1-2 (2) 4 Key equations 1 1 N N 1 1 vi 5 Key equations q=u.k (4) flow=u(SMS) * density q=ku (veh/hr) = (veh/mi) (mi/hr) h=1/q (sec/veh) = 1 / (veh/hr) (3600) s=1/k (ft/veh) = 1 / (veh/mi) (5280) 6 Example 1 Data obtained from aerial photography showed six vehicles on a 600 ft-long section of road. Traffic data collected at the same time indicated an average time headway of 4 sec. Determine (a) the density on the highway, (b) the flow on the road, (c) the space mean speed. 7 Solution • Given: • h=4 sec, l=600 ft, n=6 • Part (a): – Density (k): K= (n)/(l)= 6/600= 0.01 veh/ft k=0.01*5280= 52.8 veh/mile • Part (b): – flow (q): q= 1/h= ¼= 0.25 veh/sec q = 0.25*3600= 900 veh/hour 8 Solution • Part (c): – Space Mean Speed (U(sms)): U(sms)= q/k =900/52.8 U(sms)= 17 miles/hour 9 Traffic Flow Curves Maximum Flow, Jam Concentration, Freeflow Speed u u uf q kj k qmax k qmax q qmax - maximum flow kj - jam concentration u = 0, k = kj uf - free flow speed k = 0, u = uf 10 Example 2 Assume that : u=57.5*(1-0.008 k) Find: a) uf free flow speed b) kj jam concentration c) relationships q-u, d) relationships q-k, e) qmax capacity 11 Solution • Part a):free flow speed? i) when k=0 uf ii) u=57.5*(1-0.008 k) i+ii) uf kj u=57.5*(1-0.008 k)= 57.5*(1-0.008*0) uf =57.5 miles/hour 12 Solution • Part b): kj jam concentration? i) when u=0 kj ii) u=57.5*(1-0.008 k) i+ii) uf kj 0=57.5*(1-0.008 k)---0.008k=1 kj = 125 veh/miles 13 Solution • Part c): relationships q-u? i) q=u.k ii) u=57.5*(1-0.008 k)---u/57.5=1-0.008k 1-u/57.5=0.008K-----K=125-2.17u) (iii) i+iii) q= u.k= u.(125-2.17u) q= 125u-2.17u^2 q u 14 Solution • Part d): relationships q-k? i) q=u.k u=q/k ii) u=57.5*(1-0.008 k) i+ii) q/k=57.5*(1-0.008k) q=57.5k -0.46k^2 q k 15 Solution • Part e): qmax capacity ? q qmax dq i) To find qmax , set 0: dk ii) q=57.5k -0.46k^2 k d(q)/d(k)=0 57.5-0.92k=0 km=62.5 qmax =1796 veh/hour 16 Example 3 • The data shown below were obtained on a highway. Use linear regression analysis to fit these data and determine Speed (mi/h) Density (veh/mile) – (a) the free speed, – (b) the jam density, – (c) the capacity, 14.2 24.1 30.3 36.8 40.1 50.6 55.0 85 70 55 47 41 20 15 – d) the speed at maximum flow. 17 Plot data 60 y = -0.57x + 62.92 50 Speed 40 30 20 10 0 0 20 40 Density 60 80 100 18 Solution • from plot: u = -0.57k + 62.92 • Part a) uf=62.92 miles/hour • Part b) kj = 110.8 veh/mi • Part c) qmax = 1736 veh/hr • @ qmax, u = 31.5 mph and k = 55.2 veh/mi 19 Shockwave Analysis 20 Example 4Length of Queue Due to a Speed Reduction • The volume at a section of a two-lane highway is 1500 veh/h in each direction and the density is about 25 veh/mi. • A large dump truck from an adjacent construction site joins the traffic stream and travels at a speed of 10 mi/h for a length of 2.5 mi. • Vehicles just behind the truck have to travel at the speed of the truck which results in the formation of a platoon having a density of 100 veh/mi and a flow of 1000 veh/h. • Determine how many vehicles will be in the platoon by the time the truck leaves the highway. 21 Solution Approach Conditions(Case1) Platoon Conditions (Case2) q1=1500 veh/hr K1=25 veh/mi q2=1000 veh/hr K2=100 veh/mi 22 Solution • 1: = 1000−1500 = 100−25 -6.7 mile/hour • Step 2: Growth rate of platoon: = 10+6.7=16.7 mile/hour • Step 3: Time(truck)= distance/u= 2.5/10= 0.25 hour • Step 4: Length of platoon= time* = 0.25*16.7= 4.2 mile • Step 5: Queue length= density*distance =100*4.2= 420 vehicle 23 Distance Time 24 Example 5Length of Queue Due to Stop • A vehicle stream is interrupted and stopped by policeman. • The traffic volume for the vehicle stream before the interruption is 1500 veh/hr and the density is 50 veh/mi. • Assume that the jam density is 250 veh/mi. After four minutes the policeman releases the traffic. • The flow condition for the release is a traffic volume of 1800 veh/hr and a speed of 18mph. Determine : • the length of the queue • and the number of vehicles in the queue after five minutes. • how long it will take for the queue to dissipate after the 25 policeman releases the traffic. Solution Approach conditions Shockwave 1 State 1 q = 1500 veh/hr k = 50 veh/mi Platoon Conditions Release Conditions Shockwave 2 State 2 q = 0 veh/hr k = 250 veh/mi State 3 q = 1800 veh/hr u = 18 mi/hr 26 Solution • 12 = 0−1500 = 250−50 -7.5 mi/hr • Shockwave 1 is moving upstream at -7.5 mph • Length of the queue after 4 minutes Length = u*t = 7.5 mph * 4/60 hr = 0.5 mile • Vehicles are in the queue after 4 minutes # of vehicles = k * L = 250 *0.5 = 125 vehicles 27 Solution 23 = 1800−0 = 250−100 -12 mi/hr • Shockwave 2 is moving upstream at 12 mph • usw1 = - 7.5 mph usw2 = - 12 mph • The queue will dissipate at rate of 4.5 mph Time to dissipate a 0.5 mile queue is L/speed =0.5 mile / 4.5 mph = 0.012 hr = 6.6 minutes 28