Solutions HW 9 with 2 ft - 1 Traffic Engineering, 4th...

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Unformatted text preview: 1 Traffic Engineering, 4th Edition Roess, R.P., Prassas, E.S., and McShane, W.R. FALL 2010 Solutions to Homework 9 Problem 22‐1 Detector Placement Detectors are to be placed to produce a minimum green of 6 s. Then: ⎛d⎞ Gmin = l 1 + 2.0 Int ⎜ ⎟ ⎝ 25 ⎠ ⎛d⎞ 6 = 2.0 + 2.0 Int ⎜ ⎟ ⎝ 25 ⎠ ⎛ d ⎞ 6−2 Int ⎜ ⎟ = = 2.0 2 ⎝ 25 ⎠ d = 25.01 − 50.00 ft Passage Time As a point detector is in use, the passage time (PT) should be set to the minimum allowable headway (MAH), which is generally set at 3.0 s. The passage time must be at least long enough to allow a vehicle to traverse the distance between the front edge of the detector and the intersection line: 40 PT = 3.0 ≤ = 1.09 s OK 1.47 * 25 Yellow and All‐Red Times All semi‐actuated signals are two‐phase. Thus, there are two sets of yellow and all red times, one for each street. For the minor street: 1.47 S 85 1.47 * (25 + 5) y=t+ = 1.0 + = 1.0 + 2.2 = 3.2 s 2a + (64.4 * 0.01G ) 2 * 10 (12 * 6) + 20 w+ L ar = = = 3 .1 s 1.47 S15 1.47 * (25 − 5) 1 2 For the major street: 1.47 * (45 + 5) y = 1.0 + = 1.0 + 3.7 = 4.7 s 2 * 10 (12 * 2) + 20 ar = = 0.7 s 1.47 * (45 − 5) Note that because pedestrian flows are “low,” the all red is timed to allow vehicles to clear the width of the intersection, w, NOT the distance to the far edge of the crosswalk. Assuming that the standard default values for start‐up lost time and encroachment are in use, the total lost time in the cycle is equal to the sum of the yellow and all‐red intervals, or (3.2+3.1) + (4.7+0.7) = 6.3 + 5.4 = 11.7 s. Maximum Green Time (Minor) and Minimum Green Time (Major) To time the critical cycle for a semi‐actuated signal, the Maximum Green Time for the minor street and the Minimum Green Time for the major street must be determined. To do this, all volumes must be converted to tvu’s/ln using the equivalents shown in Tables 18.1 and 18.2 of the textbook. All left and right turns are permitted. Approach Mvt Volume Equiv Volume Volume (veh/h) (tvu/h) (tvu/h/ln) EB L 2 10.00 20 1226/3 = 409 T 1200 1.00 1200 R 5 1.21 6 WB L 3 15.00 45 1052/3 = T 1000 1.00 1000 351 R 6 1.21 7 NB L 10 1.94* 20 130/1 = T 100 1.00 100 130 R 8 1.21 10 SB L 14 1.80* 25 160/1 = 160 T 120 1.00 120 R 12 1.21 15 *interpolated; boldface = critical lane volume. 2 3 Because this is a simple 2‐phase signal, the critical lane volumes are easily identified as the larger of the two lane volumes moving in each phase. The sum of critical lane volumes is: 409 + 160 = 569 tvu/h. Then: L 11.7 11.7 C= = = = 19.9 s 569 Vc ⎛ ⎞ ⎛ ⎞ 1 − 0.412 1− ⎜ ⎟ 1− ⎜ ⎟ ⎝ 1615 * 0.9 * 0.95 ⎠ ⎝ 1615 * PHF * v / c ⎠ This would be a very short cycle length indeed, particularly when it is noted that there are only 19.9‐11.7 = 8.2 s of effective green time to allocate. Nevertheless, the computation of green‐time allocations can be worked out: g EW ( Minimum ) = 8.2 * 409 = 5.9 s x 1.5 = 8.9 s 569 160 g NS ( Maximum ) = 8.2 * = 2.3 s x 1.5 = 3.5 s 569 This is not viable, as the maximum green time for the minor street (3.5 s) is less than the minimum green time of 6.0 s. A reasonable maximum green time for the side street will have to be determined, and a related minimum green time for the major street estimated. Given that the high‐volume SB minor street approach has a demand flow rate of 146 veh/h, the average arrival rate of vehicles would be 146/60 = 2.4 veh/min. The minimum green of 6 s allows two vehicles to be services (2.0 s start‐up lost time + 2 vehicles at 2.0 s). If the maximum green is set at the minimum green of 6 s plus one passage time (6 + 3 = 9 s), a third vehicle could be serviced on one green. With the low demand, anything more than this would be excessive. The minimum main street green, however, should remain at the same multiple as indicated by the original timing computation. Thus: g EW (min imum ) 8.9 s = 9s 3.5 s 8.9 * 9 g EW (min imum ) = = 22.9 s 3.5 Critical Cycle Length The critical cycle length includes the minimum major street green, the maximum side‐street green, and the yellow and all red intervals for each street, or: C cr = (9.0 + 3.2 + 3.1) + (22.9 + 4.7 + 0.7) = 43.6 s. ( ( ) ) 3 4 Pedestrian Safety Pedestrians will cross the minor street during the major street green, which has 22.9‐second minimum. For low pedestrian flows, Nped = 50/(3600/43.6) = 0.6 peds/cycle. Then: 24 G pEW = 3.2 + (0.27 * 0.6) + = 3.2 + 0.2 + 6.9 = 10.3 s < 22.9 s OK 3 .5 No pedestrian push‐buttons are needed for the EW phase. Pedestrians will cross the major street during the NS green, which has a minimum of 6.0 s. This will obviously be too little to accommodate safe crossings: 72 G pNS = 3.2 + (0.27 * 0.6) + = 3.2 + 0.2 + 20.6 = 24.0 s >>> 6.0 s (or even 6.0 + 3.2 + 3.1 = 12.3 s ) 3 .5 In addition, the side‐street will never get a green without a vehicle actuation. Thus, a pedestrian present without any vehicles would not get a green indication under any circumstances. Thus, a pedestrian push‐button must be provided for pedestrians crossing the major street. When it is pushed, the following times will be allocated: Walking Man = 3.2 + (0.27 * 0.6) = 3.6 s 72 Flashing Upraised Hand = = 20.6 s 3.5 Problem 22‐2 Phase Plan There are no left‐turn lanes provided, so it would be very difficult to provide for protected left turns with this geometry. All left‐turn volumes are well under 200 veh/h, none of the cross‐products are anywhere near 50,000, and none of the ITE criteria are met. A two‐phase full actuated signal timing will be used. Passage Time Once again, with point detectors, the passage time (PT) is set equal to the maximum allowable gap (MAH), which will be set at 3.0 s. This must be at least sufficient to allow a vehicle to move from the detector to the intersection line, or: 4 5 PT = 3 ≤ 40 = 0.8 s OK 1.47 * 35 Minimum Green Times Because all detector locations are fixed, the minimum green times are: ⎛d⎞ ⎛ 40 ⎞ g min NS = l 1 + 2.0 Int ⎜ ⎟ = 2.0 + 2.0 Int ⎜ ⎟ = 6.0 s ⎝ 25 ⎠ ⎝ 25 ⎠ ⎛ 60 ⎞ g min EW = 2.0 + 2.0 Int ⎜ ⎟ = 8.0 s ⎝ 25 ⎠ Yellow and All Red Intervals Because both streets are 48 ft wide, and have the same approach speeds (ave = 35 mi/h), both phases will have the same yellow and all‐red times: 1.47 * S 85 1.47 * (35 + 5) y = 1 .0 + = 1.0 + = 1 .0 + 2 .9 + 3 .9 s 2a + (64.4 * 0.01G ) 2 * 10 48 + 20 w+ L ar = = = 1 .5 s 1.47 * S15 1.47 * (35 − 5) Since pedestrian activity is low, the use of “w + L” in determining the all‐red intervals is acceptable. This is a two‐phase signal. There will be two sets of yellow and all‐red intervals, one for the EW Street and one for the NS Street. Assuming standard default values for start‐up lost time and encroachment (both 2.0 s), the sum of the yellow and all‐red intervals is equal to the lost time in the cycle: L = (3.9 + 1.5) + (3.9 + 1.5) = 10.8 s Maximum Green Times To find maximum green times, we need a cycle length, which requires the sum of critical lane volumes. All volumes must be converted to tvu’s/h/ln. All left‐and right turns are permitted. Equivalents are drawn from textbook Tables 21.1 and 21.2. 5 6 Volume Equiv Volume Volume (veh/h) (tvu/h) (tvu/h/ln) EB L 7 13.00 91 1053/2 = T 950 1.00 950 527 R 10 1.21 12 WB L 5 11.75* 59 1077/2 = 539 T 1000 1.00 1000 R 15 1.21 18 NB L 10 7.25* 73 888/2 = 444 T 800 1.00 800 R 12 1.21 14.5 SB L 8 8.00* 64 832/2 = T 750 1.00 750 416 R 15 1.21 18 *interpolated; boldface = critical lane volume. The sum of the critical lane volumes is 539 + 444 = 983 tvu’s/h/ln. The beginning cycle length may now be estimated as: L 10.8 10.8 C= = = 37.5 s = Vc 983 ⎛ ⎞ ⎛ ⎞ 1 − 0.7119 1− ⎜ ⎟ ⎟ 1− ⎜ ⎝ 1615 * 0.9 * 0.95 ⎠ ⎝ 1615 * PHF * v / c ⎠ Because we are timing an actuated signal, this value need not be rounded up. With a 37.5 s cycle length, including 10.8 s of lost time, there are 37.5 – 10.8 = 26.7 s of effective green time to allocate, as follows: ⎛ 539 ⎞ g EW = 26.7 * ⎜ ⎟ = 14.6 * 1.5 = 21.9 s ⎝ 983 ⎠ ⎛ 444 ⎞ g NS = 26.7 * ⎜ ⎟ = 12.1 * 1.5 = 18.2 s ⎝ 983 ⎠ Thus, the EW Phase has a minimum green of 8.0 s and a maximum green of 21.9 s, while the NS Phase has a minimum green of 6.0 s and a maximum green of 18.2 s. Approach Mvt 6 7 Critical Cycle Length The critical cycle length includes the two maximum green times and all of the lost time in the cycle, or: C cr = 21.9 + 18.2 + 10.8 = 50.9 s Pedestrian Safety Both streets are 48 ft wide. The minimum green time for the EW Street is 8.0 s, while the minimum green time for the NS Street is 6.0 s. In both cases: ⎛L⎞ G p min = 3.2 + (0.27 * N peds ) + ⎜ ⎟ ⎜S ⎟ ⎝ p⎠ 50 N peds = = 0.7 peds / cycle 3600 50.9 ⎛ 48 ⎞ G p min = 3.2 + (0.27 * 0.7) + ⎜ ⎟ = 3.2 + 0.2 + 13.7 = 17.1 s ⎝ 3.5 ⎠ Neither minimum green time provides this. Even with 3.9 s of yellow and 1.5 s of all‐red added, the EW Phase provides only 8.0+3.9+1.5 = 13.4 s while the NS Phase provides only 6.0+3.9+1.5 = 11.4 s. Both streets require pedestrian push‐ buttons and pedestrian signals for safety. When activated by a pedestrian call, each phase will provide: Walking Man = 3.2 + (0.27 * 0.7) = 3.4 s ⎛ 48 ⎞ Flashing Upraised Hand = ⎜ ⎟ = 13.7 s ⎝ 3 .5 ⎠ Problem 22‐3 Minimum Green Times/Detector Location Detector locations are specified. The detector locations are somewhat unique. Sixty‐foot area detectors, with the front edge located 2 ft from the STOP line are being used. The minimum green will, therefore, be variable, based upon the number of vehicles detected in the detection zone. The minimum greens will vary as follows: g min ALL = l 1 + 2.0 N ( ) 7 8 where “N” is the number of vehicles detected within the 60‐ft detection zone of each phase when the green is initiated. Passage Time Passage time for an area detector is given as: L + Ld PT = MAH − v 1.47 S a The maximum allowable headway (MAH) will be set at the standard 3.0 s. The passage time is then: 20 + 60 = 3.0 − 1.36 = 1.64 s ⇒ 1.6 s PT = 3.0 − 1.47 * 40 This must be sufficient to allow a vehicle to traverse the distance between the front end of the detector and the intersection line, or: 30 = 0.51 s OK PT = 1.6 ≥ 1.47 * 40 Phase Plan Given that there are large LT demands on the EW artery, and LT lanes are provided, LT phases will be provided for this street. LT demands on the NS artery are small, and there are no LT lanes, so only one phase will be provided for this direction. A NEMA‐sequence will be used, as the variable left turn phases can be followed by a leading green phase in the direction requiring the larger green time on a phase‐by‐phase basis. In terms of timing, this will be a three‐phase signal. For the EW street, the phasing will be in a “Quad 8” configuration, illustrated in Figure 21.7 of the text. There will be a single NS street phase. In terms of implementation, several details need to be understood: • Because the left‐turn phase can be followed by a lead in either direction, or can be skipped entirely, the “Dual Entry” switch for the LT phases would be set to “off.” As they can end at different times, the ”Simultaneous Force‐Off” switch would be off. 8 9 • The EW through phase can start at different times for each direction. The “Dual Entry” switch would be set to “off.” The “Simultaneous Force Off” switch would be set to “on,” as both directions must end simultaneously. For the NS Phase, the “Dual Entry” switch would be set to “on,” as would the “Simultaneous Force Off” switch. • In terms of timing, we will treat this as if it were a simple 3‐phase signal, using the larger demand volume (in tvu/h/ln) for the ES LT Phase, the EW TH/RT Phase, and the NS Phase. Yellow and All Red Intervals There are three phases, and three sets of yellow and all‐red signal intervals. Two of these are on the EW Street, while one is on the NS Street. As all of the approach speeds are the same, the yellows will be the same for each phase: 1.47 S 85 1.47 * (40 + 5) = 1.0 + = 1.0 + 3.3 = 4.3 s y=t+ 2 *10 2a + (64.4 * 0.01G ) EW vehicles must cross a four‐lane (48 ft) street, while NS vehicles must cross a five‐lane (60 ft) street. Because there are low pedestrian flows, we will use the “w+L” criteria to determine all‐red phases: w+ L ar = 1.47 S15 arEW Phases = arNS 48 + 20 = 1.3 s 1.47 * (40 − 5) 60 + 20 = = 1.6 s 1.47 * (40 − 5) Because the standard default values of start‐up lost time and encroachment are assumed, the total lost time in the cycle is equal to the sum of the yellow and all‐ red intervals, or: L = ( 4.3 + 1.3) + ( 4.3 + 1.3) + ( 4.3 + 1.6) = 17.1 s Maximum Green Times To obtain maximum green times, demand volumes must be adjusted to reflect tvu/h/ln, and the sum of critical lane volumes determined. 9 10 Approach EB Mvt L T R Volume (veh/h) 220 400 50 Equiv 1.05 1.00 1.21 Volume (tvu/h) 231 400 61 158 500 36 50 700 12 52 600 12 Volume (tvu/h/ln) 231 461/2= 232 158 536/2 = 268 762/2 = 381 664/2 = 332 L 150 1.05 T 500 1.00 R 30 1.21 NB L 10 5.00 T 700 1.00 R 10 1.21 SB L 8 6.50* T 600 1.00 R 10 1.21 *interpolated; boldface = critical lane volume. WB The sum of critical lane volumes derives from the EB LT, the WB TH/RT and the NB phases. Thus, Vc =231 + 268 + 381 = 880. The beginning cycle length may now be computed as: L 17.1 17.1 C= = = 47.1 s = Vc 880 ⎛ ⎞ ⎛ ⎞ 1 − 0.637 1− ⎜ ⎟ ⎟ 1− ⎜ ⎝ 1615 * 0.9 * 0.95 ⎠ ⎝ 1615 * PHF * v / c ⎠ In a 47.1 s cycle, with 17.1 s of lost time, the amount of effective green time to allocate is 47.1 – 17.1 = 30.0 s. Then: g EB LT = 30 * 231 = 7.9 *1.5 = 11`.9 s 880 g EB TH / RT = 30 * 268 = 9.1 * 1.5 = 13.7 s. 880 g NB = 30 * 381 = 13.0 * 1.5 = 19.5 s 880 Remember that the minimum green for each phase can vary between 6 and 12 s. This leaves very little flexibility, particularly for the EW phases. It is possible to arbitrarily increase the maximum green time for one of the phases, and re‐set the others to keep the same ratio of green time. For example, let us increase the EB ( ( ( ) ) ) 10 11 LT phase to allow a maximum of four additional vehicles (beyond the maximum 12‐s minimum green) to clear the intersection in one cycle. Then, the maximum green would be set to 12 s + 2.0*4 = 20.0 s. Then: 11.9 20.0 = 13.7 g EB TH / RT g EB TH / RT = 11.9 20.0 = 19.7 g NS g NS = 20.0 * 13.7 = 23.0 s 11.9 20.0 * 19.7 = 33.1 s 11.9 Critical Cycle Length The critical cycle length includes the maximum green for each phase, and all of the lost times in the cycle, or: C cr = 20.0 + 23.0 + 33.1 + 17.1 = 93.2 s Pedestrian Safety Pedestrians cross the 60‐ft EW street during the NS phase. ⎛L⎞ G pNS = 3.2 + (0.27 N peds ) + ⎜ ⎟ ⎜S ⎟ ⎝ p⎠ N peds = 50 /(3600 / 93.2) = 1.3 peds / cycle ⎛ 60 ⎞ G pNS = 3.2 + (0.27 *1.3) + ⎜ ⎟ = 3.2 + 0.4 + 17.1 = 20.7 s ⎝ 3.5 ⎠ The minimum green can be as little as 6 s, which clearly does not meet this criteria even if the yellow (4.3 s) and all‐red (1.6 s) are added (6.0+4.3+1.6=11.9 s < 20.7 s). Push buttons and pedestrian signals are needed for those crossing the EW street during the NS phase. Pedestrians cross the 48‐ft NS Street during the EW TH/RT phase. 11 12 ⎛ 48 ⎞ G pEW TH / RT = 3.2 + (0.27 * 1.3) + ⎜ ⎟ = 3.2 + 0.4 + 13.7 = 17.3 s ⎝ 3 .5 ⎠ Once again, the minimum green time for the EW TH/RT phase is only 6.0 s. Even with the yellow (4.3 s) and the all‐red (1.3 s) thrown in, this is not sufficient for pedestrians to cross safely: 6.0+4.3+1.3 = 11.5 s < 17.3 s. Pedestrian push‐buttons and signals would be needed for those crossing the NS street as well. It should be noted that even if minimum green times were sufficient to allow safe crossings, the pedestrian push‐buttons would still be needed to accommodate pedestrians arriving when there are no vehicles present to initiate the appropriate phase. When activated, the pedestrian push‐buttons would provide: Walking Man = 3.2 + (0.27 * 1.4) = 3.6 s Flashing Upraised Hand ( Xing EW Street ) = 60 / 3.5 = 17.1 s Flashing Upraised Hand ( Xing NS Street ) = 48 / 3.5 = 13.7 s. 12 ...
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This note was uploaded on 01/18/2011 for the course PROJECT MA PM 587 taught by Professor Lee during the Spring '10 term at Keller Graduate School of Management.

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