#### Transcript Basic principles of intersection signalization

Chapter 20: Basic principles of intersection signalization Chapter objectives: By the end of this chapter the student will be able to: Explain the meanings of the terms related to signalized intersections Explain the relationship among discharge headway, saturation flow, lost times, and capacity Explain the “critical lane” and “time budget” concepts Model left-turn vehicles in signal timing State the definitions of various delays taking place at signalized intersections Graph the relation between delay, waiting time, and queue length Explain three delay scenarios (uniform) Explain the components of Webster’s delay model and use it to estimate delay Explain the concept behind the modeling of random and overflow delay Know inconsistencies existing between stochastic and overflow delay models Chapter 20 1 Four critical aspects of signalized intersection operation discussed in this chapter 1. 2. 3. 4. Discharge headways, saturation flow rates, and lost times Allocation of time and the critical lane concept The concept of left-turn equivalency Delay as a measure of service quality Chapter 20 2 20.1.1 Components of a Signal Cycle Cycle length Phase Controller Interval Change interval All-red interval (clearance interval) Chapter 20 3 Signal timing with a pedestrian signal: Example Interval Pine St. Veh. 1 G-26 2 3 Y-3.5 4 R-25.5 5 Oak St. Ped. W-20 Veh. R-31 % Ped. DW-31 36.4 FDW-6 10.9 DW-29 6.4 AR 2.7 G-19 6 7 Y-3 8 R-2 Cycle length = 55 seconds Chapter 20 W-8 14.5 FDW-11 20.0 DW-5 5.5 AR 3.6 4 20.1.2 Signal operation modes and left-turn treatments & 20.1.3 Left-turn treatments Operation modes: Pretimed (fixed) operation Semi-actuated operation Full-actuated operation Master controller, computer control, adaptive traffic control systems for coordinated systems Left-turn treatments: Permitted left turns Protected left turns Protected/permitted (compound) or permitted/protected left turns Chapter 20 5 Factors affecting the permitted LT movement LT flow rate Opposing flow rate Number of opposing lanes Whether LTs flow from an exclusive LT lane or from a shared lane Details of the signal timing Chapter 20 6 CFI (Continuous Flow Intersection) Bangerter Highway & 3500 South Chapter 20 7 DDI (Diverging Diamond Interchange) Chapter 20 8 Four basic mechanisms for building an analytic model or description of a signalized intersection Discharge headways at a signalized intersection The “critical lane” and “time budget” concepts The effects of LT vehicles Delay and other MOEs (like queue size and the number of stops) Chapter 20 9 20.2 Discharge headways, saturation flow, lost times, and capacity Δ(i) Start-up lost time Effective green h 12 3 4 56 7 s 3600 h l1 (i ) T l1 nh Vehicles in queue Y i y i ar i Total lost time t l l L 1 2 Saturation flow rate Startup lost time Gi g i G i Yi t L Clearance lost time l y ar e 2 e Capacity yi ari Chapter 20 g i Extension ci si of green C Cycle length 10 Sample problem, p. 467 First approach: Second approach: Eq. 20-6 Chapter 20 11 20.2.6 Saturation flow rates from a nationwide survey Chapter 20 12 20.3 The “critical lane” and “time budget” concepts Each phase has one and only one critical lane (the most intense traffic demand). If you have a 2-phase signal, then you have two critical lanes. 345 L H Nt L 3600 C T G 3600 Nt L 100 Total loss in one hour 3600 C Total effective green in one hour 1 3600 Vc 3600 Nt L h h C TG 75 450 Max. sum of critical traffic demand; this is the total demand that the intersection can handle. N = No. of phases; tL = Lost time in seconds per phase; C = Cycle length, sec; h = saturation headway, sec/veh Chapter 20 13 20.3.2 Finding an Appropriate Cycle Length Desirable cycle length, incorporating PHF and the desired level of v/c C min Nt L Vc 1 3600 / h Nt L C des Eq. 20-13 Eq. 20-14 Vc 1 PHF ( v / c )( 3600 / h ) The benefit of longer cycle length tapers around 90 to 100 seconds. This is one reason why shorter cycle lengths are better. N = # of phases. Larger N, more lost time, lower Vc. Doesn’t this look like the Webster model? C0 1 .5 L 5 1 Y i i 1 Y i flow _ ratio ( v / s ) i Chapter 20 14 Webster’s optimal cycle length model C0 1 .5 L 5 C0 = optimal cycle length for minimum delay, sec 1 v s i 1 L = Total lost time per cycle, sec i Sum (v/s)i = Sum of v/s ratios for critical lanes Delay is not so sensitive for a certain range of cycle length This is the reason why we can round up the cycle length to, say, a multiple of 5 seconds. Chapter 20 15 20.3.2 Finding an Appropriate Cycle Length Desirable cycle length, Cdes Cycle length 100% increase Vc 8% increase Marginal gain in Vc decreases as the cycle length increases. (Review the sample problem on page 473) Fig. 20.4 Chapter 20 16 A sample problem, p.473 Vc TG h 1 3600 3600 Nt L h C Nt L C des 1 Vc PHF ( v / c )( 3600 / h ) Chapter 20 17 20.4 The Concept of Left-Turn (and RightTurn) Equivalency In the same amount of time, the left lane discharges 5 through vehicles and 2 left-turning vehicles, while the right lane discharges 11 through vehicles. Chapter 20 5 2 E LT 11 and : E LT 11 5 3 .0 2 18 Left-turn vehicles are affected by opposing vehicles and number of opposing lanes. 5 1000 1500 The LT equivalent increases as the opposing flow increases. For any given opposing flow, however, the equivalent decreases as the number of opposing lanes is increased. Chapter 20 19 Left-turn consideration: 2 methods Given conditions: Solution 1: Each LT consumes 5 times more effective green time. 2-lane approach Permitted LT 10% LT, TVE (ELT) =5 h = 2 sec for through h prev ( 0 . 1)( 5 2 . 00 ) ( 0 . 9 )( 2 . 00 ) 2 . 80 sec/ h s 3600 h prev 3600 1286 vphgpl 2 . 80 Solution 2: Calibrate a factor that would multiply the saturation flow rate for through vehicles to produce the actual saturation flow rate. s 3600 f LT h ideal h prev 2 s 1800 ( 0 . 714 ) 1286 vphgpl 1800 vphgpl or h ideal PLT E LT hideal (1 PLT )(1 . 0 ) hideal 1 1 PLT ( E LT 1) 1 1 0 . 10 ( 5 1) s 1800 ( 2 . 0 / 2 . 8 ) 1800 ( 0 . 714 ) 1286 0 . 714 Chapter 20 20 20.5 Delay as an MOE Stopped time delay: The time a vehicle is stopped while waiting to pass through the intersection Approach delay: Includes stopped time, time lost for acceleration and deceleration from/to a stop Travel time delay: the difference between the driver’s desired total time to traverse the intersection and the actual time required to traverse it. Common MOEs: • Delay • Queuing • No. of stops (or percent stops) Time-in-queue delay: the total time from a vehicle joining an intersection queue to its discharge across the stopline or curb-line. Control delay: time-in-queue delay + acceleration/deceleration delay) Chapter 20 21 20.5.2 Basic theoretical models of delay Uniform arrival rate assumed, v Here we assume queued vehicles are completely released during the green. Note that W(i) is approach delay in this model. At saturation flow rate, s The area of the triangle is the aggregate delay. Figure 20.10 Chapter 20 22 Three delay scenarios This is acceptable. This is great. UD = uniform delay OD = overflow delay due to prolonged demand > supply (Overall v/c > 1.0) OD = overflow delay due to randomness (“random delay”). Overall v/c < 1.0 If this is the case, we have to do something about this signal. Chapter 20 A(t) = arrival function D(t) = discharge function 23 Arrival patterns compared Isolated intersections Signalized arterials HCM uses the Arrival Type factor to adjust the delay computed as an isolated intersection to reflect the platoon effect on delay. Chapter 20 24 Webster’s uniform delay model, p480 g R C 1 C V v R t c vR vt c st c UDa tc vR sv g C 1 sv C v g vs V st c C 1 C s v 2 g vs 2 UD a ( base : R )( height : V ) C 1 2 2 C s v 1 1 Total approach delay The area of the triangle is the aggregated delay, “Uniform Delay (UD)”. 2 UD To get average approach delay/vehicle, divide this by vC C 1 g C 2 Chapter 20 1 v s 25 Modeling for random delay, p.481 C 1 g C 2 D 2 1 v s 0 . 65 c v v c 2 1 3 UD = uniform delay Adjustment term for overestimation (between 5% and 15%) OD = overflow delay due to randomness (in reality “random delay”). Overall v/c < 1.0 v c 2 2 v 1 v / c 2g C Analytical model for random delay D = 0.90[UD + RD] Chapter 20 26 Random delay derivation Chapter 20. Chapter 20 27 Modeling overflow delay C 1 g C 2 UD o 1 v s C 1 ( g C ) 2 1 g / C 2 1 g / C v / c C 2 2 because c = s (g/C), divide both sides by v and you get (g/C)(v/c) = (v/s). And v/c = 1.0. The aggregate overflow delay is: OD a 1 2 T vT cT T 2 2 v c Because the total vehicle discharged during T is cT, OD T 2 v c 1 T 2 X 1 See the right column of p.482 for the of this model. 28 Chaptercharacteristics 20 Average overflow delay between T1 and T2 OD T1 T 2 v c 1 2 Average delay/vehicle = (Area of trapezoid)/(No. vehicles within T2-T1). Derive it by yourself. Hint: the denominator is c(T2-T1). Chapter 20 29 20.5.3 Inconsistencies in random and overflow delay C 1 g C 2 D 2 1 v s 0 . 65 c v v c 2 2 v 1 v / c v c 2 1 3 2g C OD T 2 v c 1 The stochastic model’s overflow delay is asymptotic to v/c = 1.0 and the overflow model’s delay is 0 at v/c =1.0. The real overflow delay is somewhere between these two models. Chapter 20 30 Comparison of various overflow delay model 20.5.4 Delay model in the HCM 2000 The 4th edition dropped the HCM 2000 model (I don’t know why…). It looks like Akcelik’s model that you see in p. 484 (eq. 20-26). These models try to address delays for 0.85<v/c<1.15 cases. Chapter 20 31 20.5.5 Sample delay computations We will walk through sample problems (pages 484-485). This will review all delay models we studied in this chapter. Start reading Synchro 9.0 User Manual and SimTraffic 9.0 User Manual. We will use these software programs starting Mon, October 20, 2014. Chapter 20 32