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

2. Traffic Stream Analysis 2

3. Key equations 3 Flow headway (h) - (s/veh) Flow rate (q) - (veh/h) q=3600/(h) (1) T=3sec T=0 sec h 1-2 =3sec

4. Key equations 4 Density Spacing (s)-(ft/veh) density-concentration(k) - (veh/mi) k=5280/(s) (2) S 1-2 S 2-3

5. Key equations 5 Speed U(TMS)= 1/n ∑ vi (3) U(SMS)=

6. Key equations 6 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. 7

8. Solution Given: h=4 sec, l=600 ft, n=6 8 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. 9 Solution Part (c): – Space Mean Speed (U(sms)): U(sms)= q/k =900/52.8 U(sms)= 17 miles/hour

10. Traffic Flow Curves Maximum Flow, Jam Concentration, Freeflow Speed 10 u k u q k q q max kjkj ufuf k j - jam concentration u = 0, k = k j q max - maximum flow u f - free flow speed k = 0, u = u f

11. Example 2 Assume that : u=57.5*(1-0.008 k) Find: a) u f free flow speed b) k j jam concentration c) relationships q-u, d) relationships q-k, e) q max capacity 11

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

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

14. 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 u q

15. 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 k q

16. 16 Solution Part e): q max capacity ? i) ii)q=57.5k -0.46k^2 d(q)/d(k)=0 57.5-0.92k=0 k m =62.5 q max =1796 veh/hour k q q max

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. 17 Speed (mi/h)Density (veh/mile) 14.285 24.170 30.355 36.847 40.141 50.620 55.015

18. Plot data 18

19. from plot: u = -0.57k + 62.92 Part a) u f =62.92 miles/hour Part b)k j = 110.8 veh/mi Part c) q max = 1736 veh/hr @ q max, u = 31.5 mphandk = 55.2 veh/mi 19 Solution

20. Shockwave Analysis 20

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. 21

22. 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. 23 Solution

24. 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 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 policeman releases the traffic. 25

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

27. 27 Solution

28. 28 Solution