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# tte4004%20Chapter%202%2C6 - Traffic Flow(pages...

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Unformatted text preview: Traffic Flow (pages 173-187) (pages Traffic A mathematical way to describe traffic mathematical behavior Relationships of speed-flow-density Relationships Queuing, car following, gap acceptance, Queuing, lane changing models and algorithms to help better design roads and control intersections The use of discrete event simulation to test The different scenarios and design new systems Traffic Flow Elements Traffic Time-Space Diagram: A representation of Time vehicle trajectory It is a linear representation It It is used for modeling vehicle flow on It streets and signal control on arterial streets Time-Space Diagram Time Green=Green+Clearance Green=Green+Clearance Ideal Offset=t(ideal)=L/V Ideal L=Block Length (feet) L=Block V=Vehicle Speed (fps) V=Vehicle One-Way Progression: Ideal One progression with the widest band 6000 4800 3600 2400 1200 0 30 V=60 fps 60 20 40 60 80 100 90 120 150 6000 4800 3600 2400 1200 0 0 30 V=60 fps 60 20 40 50 60 70 80 100 110 90 130 90 120 150 Traffic Flow Elements Traffic Flow: Vehicle rate passing a point during Flow: a given time period (one hour) – q = (n)(3600)/T vehicle per hour Density: Number of vehicles per unit Density: distance (vehicle per mile) Speed: Time Mean Speed is the Speed: arithmetic mean of vehicle instantaneous speed ut = (1/n) ∑ui Time Headway Time Space Headway Space u/q/k Relationships u/q/k SMS=S=(n)(d)/∑t SMS=S=(n)(d)/ q=(u)(k) q=( How do we measure density? How –Air Photo –Loop Detector (Occupancy) k={(5280)(t)/(Lv)+(LD)} k={(5280)(t)/(L t=fraction of time loop is occ. t=fraction Lv=Avg. Veh. Length LD=Loop Detector Length k=(52.8)(% OCC)/{(Lv)+(LD)} k=(52.8)(% %OCC= (∑ tOCC)(100)/T tOCC=occupancy time of individual vehicles= ton T = Selected time period 00000000 111111111 tocc Output Signals 000000 1111111111111 tocc SC DJ C DC DJ •Greenshield Model (Linear) u=u f − u f kJ k uf a = uf , b = kJ uf 2 q = uk = u f k − k kJ dq uf = 0, u f − 2 kc = 0 dk kJ kJ kc = 2 q u = u f − ( )( ) kJ u uf q = uf −u kJ u kJ kJ 2 q= ufu − u uf uf kJ 2 q = kJ u − u uf uf kJ dq = 0, 0 = k J − 2 uc du uf uc = uf 2 Capacity = C = uc kc C= u f kJ 4 •Greenberg Model (Logarithmic) kJ u = u c ln k kJ q = uk = u c kln k kJ kJ d dq = 0 = u c ln + u c k [ln ] k k dk dk kJ - k kJ = u c ln + u c k[ ] 2 k kJ k kJ = u c [ln − 1] k kJ kc = e kJ Capacity = C = u c e Numerical Example 1 Textbook 6-7 Numerical Flow-density relationship was found to be a Flow second degree quadratic Following data was observed: Following Speed (mph) Density (vpm) 50 18 45 25 40 41 34 58 22 71 13 12 12 99 Develop q-u-k relationships and determine the capacity of this section The relationship between u and k is linear The Y= a + b x where Y is u and x is k Y= Minimum of Sum of Squares of Errors Minimum a= 1/n ∑u – b/n ∑k a= b= {∑uk – 1/n(∑u ∑k)}/{∑k2 – 1/n(∑k)2} b= Speed u 50 45 40 34 22 13 12 Density k 18 25 41 58 71 88 99 u*k 900 1125 1640 1972 1562 1144 1188 k*k 324 625 1681 3364 5041 7744 9801 ∑ =216 Mean=30.86 ∑ = 400 Mean =57.14 ∑ =9531 a=58.91 ∑=28580 b=-.491 q=58.91k-.491k2 q=58.91k dq/dk=58.91-.982k=0 dq/dk Density at max flow = k = 58.91/.982 = 60.0 Density Jam Density = kj Jam Max Flow = 58.91(60.0) – 0.491 (60) Max = 1767 VPH Or: Max Flow = (58.91)*(120/4) = 1767 VPH Or: Numerical Example 2 Textbook 6-13 Numerical The capacity of a highway is suddenly The reduced to 60% Linear relationship describes speed and Linear density Jam Density = 112 vpm Jam Mean Free speed = 64.5 mph Mean What drop in space mean speed near a What work zone if the flow upstream is reduced by 80%? kj = 112, uf = 64.5 qmax = 112 X 64.5/4 = 1,806 vph Upstream flow: qu = 1806 X .8 = 1444.8 Work zone: qw = 1806 X .6 = 1083 u=64.5-k(64.5/112)=64.5-0.576k u=64.5 For upstream section: For 1444.8=64.5k-0.576k2 k=81.01 or k=30.96 u = 1444.8/81.01=17.38 For construction zone: 1083=64.5k-0.576k2 k=91.4 or k=20.57 u = 1083/91.4=11.84 Reduction in Percentage = (17.38-11.84)/17.38= 33.6% ...
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