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Unformatted text preview: 1 Traffic Engineering, 4th Edition Roess, R.P., Prassas, E.P., and McShane, W.R. Solution to Problems in Chapter 21 Problem 21‐1 The change (or yellow) interval is timed to allow a driver that is too close to the signal to stop safely to proceed through the intersection safely. It is computed as follows: 1.47 S 85 1.47 (35 + 5) y=t+ = 1.0 + 2a + (64.4 * 0.01G ) 2 (10) + (64.4 * 0.01 * −2) 58.8 y = 1.0 + = 1.0 + 3.14 = 4.14 s 20 − 1.288 The clearance (or all red) interval can be found using one of several equations. In this case, given a significant pedestrian crossing movement, vehicles should be cleared beyond the far crosswalk during the all red interval. Therefore: P+L (50 + 10) + 20 80 ar = = = = 1.81 s. 1.47 S15 1.47 (35 − 5) 44.1 Problem 21‐2 The solution depends upon whether the local jurisdiction allows pedestrians in the crosswalk during the yellow and all‐red intervals. Assuming that pedestrians are permitted in the crosswalk during the yellow and all –red intervals, Gp ≤ G+Y for each phas Phase A: 30.0 ≤ 18.0+4.5 = 22.5 ? NO Phase B: 15.0 ≤ 60.0+4.0 = 64.0 ? YES This is a classic situation. During the longer vehicular phase, pedestrians cross a narrower side street (Phase B). During the shorter vehicular phase, pedestrians generally cross a wider main street (Phase A). Pedestrians, therefore, need more green time during the shorter phase than they do during the longer phase. 1 2 The solution to this problem is to lengthen the cycle length so that the proportioning of green time remains the same, while satisfying the minimum pedestrian needs in both phases. To satisfy the Phase A pedestrian requirement, the green time must be increased to at least 30.0 – 4.5 = 25.5 s. However, to maintain the current balance of green time, the green time in Phase B would also have to be increased: 18 25.5 = 60 gB 25.5 * 60 gB = = 85.0 s 18 This leads to a cycle length of (25.5+4.5)+(85.0+4.0) = 119 s. If this is a pre‐timed signal, a cycle length of 120 s would be used. Keeping the split of green times constant, the green times become: ⎛ 18 ⎞ g A = (120 − 4.5 − 4.0) * ⎜ ⎟ = 25.7 s ⎝ 18 + 60 ⎠ ⎛ 60 ⎞ g B = (120 − 4.5 − 4.0) * ⎜ ⎟ = 85.8 s ⎝ 18 + 60 ⎠ All pedestrian requirements are met. Yellow and all‐red timings are unchanged. If the jurisdiction decided to allow pedestrians in the crosswalk only during the green, then the same computations would be repeated, but gA would have to be at least 30.0 s. Problem 21‐3 Step 1 – Phasing Left‐turn volumes should be checked to determine whether protected LT phasing will be necessary. These are checked using three criteria: a) minimum LT flow rate, b) cross‐product rule, and c) ITE criteria. This is done below: EB LT: vLT = 210 veh/h > 200 veh/h Protected phase needed. Protected phase needed. WB LT: vLT = 350 veh/h > 200 veh/h 2 3 NB LT: vLT = 30 veh/h < 200 veh/h Criterion not met. x‐prod = 30*(265/1) = 7,950 < 50,000 Criterion not met. ITE (Figure 18.1) Criterion not met. SB LT: vLT = 15 veh/h < 200 veh/h Criterion not met. x‐prod = 15*(250/1) = 3,500 < 50,000 Criterion not met. ITE (Figure 18.1) Criterion not met. From these results, a protected LT is needed for the E‐W arterial, while permitted phasing will be ok for the N‐S arterial. The latter is important, as the one‐lane approaches do not provide for exclusive LT lanes, which are required to implement many protected LT solutions. As there is significant imbalance between the two LT movements needing protection, a split LT phasing will be used. The NEMA phasing approach will be used: A1 A2 Phase/Ring Diagram A3 B Step 2: Convert to Through Vehicle Units Demand volumes must now be converted to “through vehicle units” using the equivalents of Tables 18.1 and 18.2. This is done below: 3 4 Mvt EB L EB TH EB R WB L WB TH WB R NB L NB TH NB R SB L SB TH SB R Vol (veh/h) 210 775 45 350 750 50 30 250 25 15 265 10 Equiv (T 18.1/18.2) 1.05 1.00 1.32* 1.05 1.00 1.32* 3.31** 1.00 1.32* 3.13** 1.00 1.32* Vol. (tvu/h) 221 775 59 368 750 66 99 250 33 47 265 13 No. of Lanes 1 2 1 2 1 1 Vol/Lane (tvu/h) 221 417 368 408 382 325 * for modern pedestrian activity ** interpolated Step 3 – Determine the Sum of Critical Lane Volumes The volume per lane (in tvu/h) for each portion of the phase diagram is entered. The critical path through the signal is the one that produces the highest sum of critical lane volumes. Critical Lanes ? 221 368 A1 221 + 407 = 628 tvu/h or 407 A2 368 + 417 = 785 tvu/h√ A3 417 382 tvu/h√ or 325 tvu/h B ∑ = 785+382 = 1,167 382 325 tvu/h 4 5 The sum of the critical lane volumes is 1,167 tvu/h. Step 4: Determine Yellow and All‐Red Times Yellow times are computed using Equation 18‐2. The ring diagram shows that this is a 3‐phase signal phase plan. Thus, there are 3 yellow times, two for the E‐ W arterial, one for the N‐S arterial. The two E‐W yellow times are based upon an 85th percentile approach speed of 35+5 = 40 mi/h. The N‐S yellow is based upon an 85th percentile approach speed of 30+5 = 35 mi/h: 1.47 S 85 y=t+ 2a + (64.4 * 0.01G ) 1.47 * 40 y EW = 1.0 + = 3.9 s 2(10) 1.47 * 35 y NS = 1.0 + = 3.6 s 2(10) Because there is moderate pedestrian presence, the all‐red phases should be timed to clear vehicles beyond the far pedestrian crosswalk. Again, there are 3 phases and 3 all‐red times, two for the E‐W arterial, and one for the N‐S arterial. The E‐W all‐reds are based upon clearing a pedestrian distance of 24 ft (2 lanes) plus 10 ft = 34 ft. The N‐S all‐red is based upon clearing a pedestrian distance of 24+16+24 plus 10 ft = 74 ft. The E‐W street has a 15th percentile speed of 35 – 5 = 30 mi/h; the N‐S street has a 15th percentile speed of 30‐5 = 25 mi/h. Then: P+L ar = 1.47 S15 arEW = arNS 34 + 20 = 1.0 s 1.47 * 35 74 + 20 = = 2.6 s 1.47 * 25 A yellow and all‐red transition occurs twice in Phase A (for the E‐W street). Both rings have a transition at the end of Phase A3. One ring has a transition at the end of Phase A1, the other ring at the end of Phase A2. Combined these two count as one full transition. 5 6
Step 5: Determine Lost Time Per Cycle Assuming that standard values of start‐up lost time (2 s) and encroachment time (2.0 s) are used, the lost time is the sum of the yellow and all‐red phases, or (3.9+1.0) + (3.9+1.0) + (3.6+2.6) = 16.0 s Step 6: Determine the Appropriate Cycle Length An appropriate cycle length is estimated using Equation 18‐11: L C des = Vc ⎤ ⎡ 1− ⎢ ⎥ ⎣1615 * PHF * (v / c ⎦
C des = 16.0 1,167 ⎡ ⎤ 1− ⎢ ⎥ ⎣1615 * 0.98 * 0.95 ⎦ 16.0 = = 71.4 s ⇒ 75.0 s 1 − 0.776 C des Step 7: Spitting the Green The 75‐s cycle length includes 16 s of lost time, leaving 75 – 16 = 59 s of effective green time to allocate in proportion to the critical lane volumes in the critical path of the signal. The critical path and resulting critical lane volumes are as follows: Phases A1+A2 368 tvu/h Phase A3 417 tvu/h Phase B 382 tvu/h Total 1,167 tvu/h Then: ⎛V ⎞ g i = g tot * ⎜ ci ⎟ ⎜ ∑V ⎟ ci ⎠ ⎝ g A1+ A2 = 59 * (368 / 1167) = 18.6 s g A3 = 59 * (417 / 1167) = 21.1 s g B = 59 * (382 / 1167) = 19.3 s 6 7
The totals should add up to 59 s. 18.6+21.1+19.3 = 59.0 s. Adding the 16 s of yellow and all red brings us to the 75‐s cycle length. Because of the phase plan used, timing the critical path does not fully specify the signal timing. Phases A1 and A2 must be specified to complete the process. These phases, however, exist only on the non‐critical path through the signal. The sum of Phases A1, A2, and A3 is known: 18.6+21.1 = 39.7 s. On the non‐critical path, Phase 1A has a lane volume of 221 tvu/h, while the sum of Phases A2 and A3 has a lane volume of 407 tvu/h. Therefore: ⎛ 221 ⎞ g A1 = 39.7 * ⎜ ⎟ = 14.0 s ⎝ 221 + 407 ⎠ The timing of Phase A2 cannot be computed directly, as no one movement uses this phase exclusively. This is, in effect, the overlap phase. However, the sum of Phases A1, A2, and A3 is known, and the length of Phases A1 and A3 is known. By inference: g A2 = 39.7 − 14.0 − 21.1 = 4.6 s Step 8: Pedestrian Requirements Pedestrian requirements are estimated using Equation 18‐15: ⎛L⎞ G p = 3.2 + (0.27 * N ped ) + ⎜ ⎟ ⎜S ⎟ ⎝ p⎠ The default value for “moderate” pedestrian activity (Table 18.2) is 200 peds/h in each crosswalk. With a cycle length of 75 s, the number of pedestrian per cycle in each crosswalk is: 200 N peds = = 4.2 peds / cycle (3600 / 75) Pedestrians cross the 24‐ft N‐S street during Phase A3 only. The minimum pedestrian green time is: ⎛ 24 ⎞ G pA3 = 3.2 + (0.27 * 4.2 ) + ⎜ ⎟ = 3.2 + 1.1 + 6.0 = 10.3 s ⎝4⎠ 7 8
The green for Phase A3 is 21.1 s, which is more than the required 10.3 s. This is safe for pedestrians. Pedestrians cross the 64‐ft E‐W street during Phase B. The minimum pedestrian green time is: 64 G pB = 3.2 + (0.27 * 4.2 ) + = 3.2 + 1.1 + 16.0 = 20.3 s 4 Phase B is 19.3 s, which is less than that needed. If, however, pedestrians are permitted to be in the crosswalk during yellow and/or all‐red phases, the situation may be deemed safe. G+y = 19.2+3.6 = 22.8 s; G+y+ar = 22.8+2.6 = 25.4 s, both of which are more than the required 20.3 s. Discussion Technically, the signal timing as proposed is acceptable and safe for pedestrians. It should be noted, however, that the overlap Phase A2 is quite short – 4.6 s. Given the greater complexity that this type of phasing presents to drivers, it is wise to question whether or not it is justified, given the short overlap provided. If the proposed ring diagram is examined, a simple 3‐phase signal with a simultaneous left‐turn phase for the E‐W street results in the same sum of critical lane volumes as the original timing. Thus, the cycle length and splits would be unaffected (except that both LT phases would be 18.6 s, and there would be no Phase A2. In this case, there is no advantage to the overlapping phase plan. This is because the higher EW LT movement and the higher EW TH movement on are opposite approaches, which places them on the same ring of the signal. The advantage of using a split LT phasing is significant only when the higher LT and higher TH volume are on the same approach. In the end, a simple 3‐phase signal would be recommended. 8 9
Problem 21‐4 Check Left Turns This problem features an intersection between a major artery and a significant one‐way street or artery. Because of the one‐way street, there is only one potentially‐opposed left turn in the EB direction. This left turn should be protected because: v LT = 275 veh / h > 200 veh / h There is not WB left turn, and there is no flow of any kind in the SB direction. The left‐turn from the one‐way street is not opposed by a vehicular flow, but is opposed by a pedestrian flow, much as a right turn is. The value of ELT for this movement can be treated in any one of three ways, as a (a) protected LT (ELT = 1.05), (b) a permitted LT with no opposing flow (ELT = 1.1), or (c) a right turn against moderate pedestrian flow (ELT = 1.32). While these will yield slightly different signal timings, any of the three are logically acceptable. For this solution, a value of 1.1 will be used. Conversion of Demand Flows to tvu’s The table below illustrates the computation of through‐vehicle‐equivalents and their assignment to lane groups and lane volumes. Lane Approach Mvt Vol Equiv. Vol Group Vol/Ln Vol (veh/h) (T18.1,2) (tvu/h) (tvu/h) (tvu/h/ln) EB LT 275 1.05 289 289 289 TH 1,100 1.00 1,100 1,100 550 WB TH 730 1.00 730 862 431 RT 100 1.32 132 NB LT 45 1.10 50 TH 900 1.00 900 983 328 RT 25 1.32 33 9 10
Ring Diagram and Sum of Critical Lane Volumes With only one protected left turn in the EB direction, a leading EB green will be used with no corresponding WB lag. The ring diagram is shown below: 289 A1 289+431 = 720 550 Or 550 A2 431 B 328 328 Vc = 720+328 = 1,048 tvu/h Yellow and All Red Times and Lost Times With an average speed of 40 mi/h on all approaches, the 85th percentile speed is estimated to be 40+5 = 45 mi/h, while the 15th percentile speed is estimated to be 40‐5 = 35 mi/h. Then: 1.47 S 85 y=t+ 2a + (64.4 * 0.01G ) 1.47 * 45 y A1, A 2, B = 1.0 + = 4.3 s 2 * 10 P+L ar = 1.47 S15 (36 + 10) + 18 = 1.2 s 1.47 * 35 (76 + 10) + 18 = 2.0 s arB = 1.47 * 35 arA1, A 2 = 10 11
Because there are three phases on the critical path, there are three sets of lost times in this cycle. As the standard default values for start‐up lost time and encroachment time (2 s ea) are being used, lost time is equal to the sum of the yellow and all red intervals, and effective green is equal to actual green. Then: L = ( 4.3 + 1.2) + ( 4.3 + 1.2) + ( 4.3 + 2.0) = 17.3 s Cycle Length The cycle length is estimated as: L C des = Vc ⎛ ⎞ 1− ⎜ ⎟ ⎝ 1615 * PHF * v / c ⎠ 17.3 17.3 = = 71.8 s ⇒ 75 s C des = 1,048 ⎛ ⎞ 1 − 0.759 1− ⎜ ⎟ ⎝ 1615 * 0.95 * 0.90 ⎠ Splitting the Green In a 75‐s cycle, with 17.3 s of lost time, the amount of effective green time to be allocated is: g tot = 75.0 − 17.3 = 57.7 s Then: ⎛ 289 ⎞ g A1 = 57.7 * ⎜ ⎟ = 15.9 s ⎝ 1,048 ⎠ ⎛ 431 ⎞ g A 2 = 57.7 * ⎜ ⎟ = 23.7 s ⎝ 1,048 ⎠ ⎛ 328 ⎞ g B = 57.7 * ⎜ ⎟ =18.1 s ⎝ 1,048 ⎠ 11 12
Checking Pedestrian Safety The minimum green needed for pedestrians to safely cross the street is given by: ⎛L⎞ G p = 3.2 + (0.27 N peds ) + ⎜ ⎟ ⎜S ⎟ ⎝ p⎠ The default volume of pedestrians for “moderate” activity is 200 peds/h per crosswalk. Therefore: 200 = 3.9 peds / cycle ⇒ 4 peds / cycle N peds = 3600 70 To safely cross the E‐W artery, pedestrians need: ⎛ 76 ⎞ G pEW = 3.2 + (0.27 * 4) + ⎜ ⎟ = 3.2 + 1.1 + 19.0 = 23.3 s ⎝4⎠ These pedestrians cross during Phase B, which has a green time of 18.1 s. The yellow is 4.3 s, and all red is 2.0 s. If pedestrians are permitted to cross during the yellow and all red in addition to the green, the total available crossing time is 18.1+4.3+2.0 = 24.4 s, which is sufficient. If pedestrians are permitted to cross only during green, or during green and yellow, a deficiency would exist, and some adjustment to the signal timing would be needed. To safely cross the NS artery, pedestrians need: ⎛ 36 ⎞ G pEW = 3.2 + (0.27 * 4) + ⎜ ⎟ = 3.2 + 1.1 + 9.0 = 13.3 s ⎝4⎠ These pedestrians cross the street during Phase A2, which has a green time of 22.1 s, which is sufficient. ( ) 12 13
Problem 21‐5 The unique feature of this intersection is the offset. The offset will cause two specific recommendations: 1. Despite the results of any left‐turn checks, the N‐S street must get a fully protected LT phase. This will avoid potential accidents between left‐turn vehicles and opposing through vehicles, which are in much more dangerous positions than in a standard 90o intersection. 2. The clearance time for the E‐W street will be quite long due to the position of the crosswalks. Vehicles will have to clear 120 ft. Left Turns As the geometry already requires a protected LT phase for the N‐S street, this need not be further compared to criteria. The E‐W LTs, however, must be checked to see whether they will also require protection: WB: 180 veh/h < 200 veh/h No Xprod = 180*(700/2) = 63,000 > 50,000 Yes EB: 80 veh/h < 200 veh/h No Xprod = 80*(825/2) = 33,000 No ITE Criteria: Yes (barely) The N‐S street has reasonable balance between the two left turns (90, 120), so we will use an exclusive LT lane. Also, with a confusing geometry, and simple signalization plan would be preferred. The E‐W left turns are quite different, so a NEMA phasing plan with an exclusive LT phase followed by a leading green will be used. 13 14
Converting Volumes to tvu’s Approach Mvt Volume (veh/h) EB L T R L T R L T R L T R 80 700 65 180 825 95 120 750 40 90 680 75 Equiv (T18.1,2) 1.05 1.00 1.21 1.05 1.00 1.21 1.05 1.00 1.21 1.05 1.00 1.21 Vol (tvu/h) 84 700 79 189 825 115 126 750 48 95 680 91 Lane Grp Vol/Ln Vol (tvu/h/ln) (tvu/h) 84 84 779 390 189 940 126 798 95 771 189 470 126 399 95 386 WB NB SB Ring Diagram and Sum of Critical Lane Volumes The ring diagram is shown on the next page, with appropriate demand volumes shown for each portion of each ring. 95 126 95 or 126* A B 399* or 386 399 386 C1 84 84 + 470 = 554 189 Or C2 189 + 390 = 579* C3 470 390 14 15 Vc = 126+399+579 = 1,104 tvu/h Yellow and All Red Intervals, Lost Times The average speed on both streets is 35 mi/h. All yellow times will, therefore, be the same: 1.47 * S 85 y=t+ 2a + (64.4 * 0.01G ) 1.47 * (35 + 5) y = 1.0 + = 3.9 s 2 *10 Even though there are only a few pedestrians, the awkward geometry gives adequate reason to time the all red intervals to allow vehicles to clear the far crosswalks, i.e., using P instead of w. N‐S vehicles would then have to clear 50 + 10 = 60 ft. E‐W vehicles would have to clear 120 – 10 = 110 ft (note that the 120 ft on the diagram goes from the far edges of both crosswalks). Then: P+L ar = 1.47 * S15
arEW = arNS 110 + 20 = 2.9 s 1.47 * (35 − 5) 60 + 20 = = 1.8 s 1.47 * (35 − 5) This is a four‐phase signal, and there will be four sets of yellow and all red intervals, two for each street (as each street has protected LT phases). The total of all yellow and all‐red phases is, therefore: (3.9+2.9)+(3.9+2.9)+(3.9+1.8)+(3.9+1.8) = 25.0 s As we are using the standard default values for start‐up lost time and encroachment (each 2.0 s), this is also equal to the total lost time in the cycle. Cycle Length The cycle length is computed as: 15 16 L Vc ⎛ ⎞ 1− ⎜ ⎟ ⎝ 1615 * PHF * v / c ⎠ 25.0 25.0 C= = = 103 ⇒ 110 s 1104 ⎛ ⎞ 1 − 0.7574 1− ⎜ ⎟ ⎝ 1615 * 0.95 * 0.95 ⎠ Splitting the Green The 100 s cycle length has 25 s of lost time, leaving 110 – 25 = 85 s of effective green time to assign to the four phases on the critical path, as follows: ⎛ 126 ⎞ g A = 85 * ⎜ ⎟ = 9.7 s ⎝ 1104 ⎠ ⎛ 399 ⎞ g B = 85 * ⎜ ⎟ = 30.7 s ⎝ 1104 ⎠ ⎛ 189 ⎞ g C1+C 2 = 85 * ⎜ ⎟ = 14.6 s ⎝ 1104 ⎠ ⎛ 390 ⎞ g C 3 = 85 * ⎜ ⎟ = 30.0 s ⎝ 1104 ⎠ We must also establish the boundary between Phases C1 and C2, which occurs only on the non‐critical path. The total length of Phases C1, C2, and C3 is 14.6+30.0 = 44.6 s. Then: C=
⎛ 84 ⎞ g C1 = 44.6 * ⎜ ⎟ = 6 .7 s ⎝ 84 + 470 ⎠ g C 2 = 44.6 − 30.0 − 6.7 = 7.9 s Checking Pedestrian Safety Pedestrians crossing the E‐W street do so during Phase B. The crosswalk length is 50 ft. Then: N peds = 50 /(3600 / 110) = 1.5 peds / cycle ⎛ 50 ⎞ G pB = 3.2 + (0.27 *1.5) + ⎜ ⎟ = 16.1 s < 30.7 s OK ⎝4⎠ 16 17
Pedestrians crossing the N‐S street do so during Phase C3. The crosswalk length is 60 ft. Then: ⎛ 60 ⎞ G pC 3 = 3.2 + (0.27 * 1.5) + ⎜ ⎟ = 18.6 s < 30.0 s OK ⎝4⎠ 17 ...
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