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Introduction to Transportation Engineering

Introduction to Transportation Engineering. Instructor Dr. Norman Garrick Hamed Ahangari 27th March 2014. Traffic Stream Analysis. Key equations. T=3sec. T=0 sec. h 1-2 =3sec. Key equations. S 2-3. S 1-2. Key equations. Key equations. q= u.k (4) flow=u(SMS) * density.

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Introduction to Transportation Engineering

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  1. Introduction to Transportation Engineering Instructor Dr. Norman Garrick Hamed Ahangari 27th March 2014

  2. Traffic Stream Analysis

  3. Key equations T=3sec T=0 sec h1-2=3sec

  4. Key equations S2-3 S1-2

  5. Key equations

  6. Key equations q=u.k (4) flow=u(SMS) * density • q = k u • (veh/hr) = (veh/mi)  (mi/hr) • h= 1 /q • (sec/veh) = 1 / (veh/hr)  (3600) • s = 1 /k • (ft/veh) = 1 / (veh/mi)  (5280)

  7. 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.

  8. 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

  9. Solution • Part (c): • Space Mean Speed (U(sms)): U(sms)= q/k =900/52.8 U(sms)= 17 miles/hour

  10. Traffic Flow CurvesMaximum Flow, Jam Concentration, Freeflow Speed u u uf qmax kj k q q qmax qmax - maximum flow kj - jam concentration u = 0, k = kj k uf - free flow speed k = 0, u = uf

  11. Example 2 Assume that : u=57.5*(1-0.008 k) Find: a) uffree flow speed b) kjjam concentration c) relationshipsq-u, d) relationships q-k, e)qmaxcapacity

  12. Solution • Part a):free flow speed? i)when k=0 uf ii) u=57.5*(1-0.008 k) i+ii) u=57.5*(1-0.008 k)= 57.5*(1-0.008*0) uf=57.5 miles/hour uf kj

  13. Solution • Part b): kj jam concentration? i)when u=0 kj ii) u=57.5*(1-0.008 k) i+ii) 0=57.5*(1-0.008 k)---- 0.008k=1 kj= 125 veh/miles uf kj

  14. 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

  15. 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

  16. Solution • Part e): qmaxcapacity ? i) • q=57.5k -0.46k^2 d(q)/d(k)=0 57.5-0.92k=0 km=62.5 qmax=1796 veh/hour q qmax k

  17. Example 3 • The data shown below were obtained on a highway. Use linear regression analysis to fit these data and determine • (a) the free speed, • (b) the jam density, • (c) the capacity, • d) the speed at maximum flow.

  18. Plot data

  19. 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

  20. Shockwave Analysis

  21. Example 4-Length 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.

  22. Solution Approach Conditions(Case1) q1=1500 veh/hr K1=25 veh/mi Platoon Conditions (Case2) q2=1000 veh/hr K2=100 veh/mi

  23. Solution • -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

  24. Distance Time

  25. Example 5- Length 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 policemanreleases 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 policeman releases the traffic.

  26. Solution Approach conditions Platoon Conditions Release Conditions Shockwave 2 Shockwave 1 State 1 q = 1500 veh/hr k = 50 veh/mi State 2 q = 0 veh/hr k = 250 veh/mi State 3 q = 1800 veh/hr u= 18 mi/hr

  27. Solution • -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

  28. Solution -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

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