Question

(15 points) Assume that all volumes shown have been converted to compatible through-car equivalent" values for the conditions shown. The phase plan in given in the following. Assume that critical volumes reverse in the other daily peak hour. The Saturation headway is 2 s/veh, start- up lost time is 1.5 s, clearance lost time is 2 s, Cycle length is 90s, only critical volumes shown and all volumes in tvu/h. Please determine: a. Find the appropriate number of lanes for each lane group needed. (5 points) b. Consider a case in which the E-W arterial has four lanes in each direction and the N-S arterial also has three lanes in each direction. What is the absolute minimum cycle length that could be used? (2 points) c. What cycle length would be required to provide for a v/c ratio of 0.90 during the worst 15- minutes of the hour if the PHF is 0.95 if the E-W arterial has four lanes in each direction and the N-S arterial also has four lanes in each direction? (3 points) d. What is the lane capacity of Westbound approach if the actual green time for phase 1 is 20s, Yellow and all-red interval is 5 s. (5 points) 1200 tvu/h Phase 1 Phase 2 2000 tvu/h

Answer 1

So we have a two phase signal.

saturation headway = 2 sec/vehicle . Therfore saturation flow = 3600/2 = 1800 pcu/rhr/ln

Since saturation flow is 1800 at minimum for EB/WB we need two lanes, and we will start with one lane on NB/SB

I am assumign that in all my scenarios the # of lanes on EB=WB and # of lanes on NB =SB

Assume that PHF = 1 to start with.

Also we are given total lost time = start up lost time + clearance lost time = 3.5 seconds/phase.

So total lost time per cycle = 3.5 *2 = 7 seconds

Cycle Length = 90 seconds

Assume that the intersection operates

We know from formulas that C_minimum is given by

LEX Cmin LEX Χ. - Υ. Χ. - Σ. (3).

So assume that at worst intersection is at full capacity and Xc = 1, so Cmin = L/(1- sum of critical flow ratios) = 90

so 90 = 7(1-sum of critical flow ratios)

Therefore at worst case sum of critical flow ratios should be equal to 1-7/90 = 0.9222

So we can set up our phase, lane group and flow ratio table as follows

# of Saturation lanes Flow/lane Total Saturation Flow Given Peak Hour Flow Design Flow tvu = peak hour flow/PHF Flow Ratio = Design Critical Flow/Saturati Flow Ratio PHE on Flow Phase 1 0.556 EB WB NB SB 2 2 1 1 1800 1800 1800 1800 3600 3600 1800 1800 2000 2000 1000 1000 1.00 1.00 1.00 1.00 2000 2000 1000 1000 0.556 0.556 0.556 0.556 Phase 2 0.556 1.111

So this WILL NOT WORK, since sum of critical flow ratios = 1.111

The sum of critical flow RATIOS must be less than 1, and at max it should be 0.9222

So try another combination by increasing the number of lanes on NB and SB to 2 lanes each

# of Saturation lanes Flow/lane Total Saturation Flow Given Peak Hour Flow PHF Design Flow tvu = peak hour flow/PHF Flow Ratio = Design Critical Flow/Saturati Flow Ratio on Flow 0.556 | ЕВ Phase 1 WB NB Phase 2 SB 2 2 2 2 1800 1800 1800 1800 3600 3600 3600 3600 2000 2000 1000 1000 1.00 1.00 1.00 1.00 2000 2000 1000 1000 0.556 0.556 0.278 0.278 0.278 0.833

The above combination works

Sum of critical flow ratios = 0.833 which is less than .922

THEFORE 2 LANES ARE REQUIRED IN EACH DIRECTION IN EACH APPROACH.

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B) Now we have a scenario with 4 lanes each direcion in E-W and 3 lanes each direction in N-S

In this case we have the following table

# of Saturation lanes Flow/lane Total Saturation Flow Given Peak Hour Flow PHE Design Flow tvu = peak hour flow/PHE Flow Ratio = Design Critical Flow/Saturati Flow Ratio on Flow Phase 1 0.278 EB WB NB SB 4 4 3 3 1800 1800 1800 1800 7200 7200 5400 5400 2000 2000 1000 1000 1.00 1.00 1.00 1.00 2000 2000 1000 1000 0.278 0.278 0.185 0.185 Phase 2 0.185 0.463

So sum of critical flow ratios = 0.463

So absolute minimum cycle length is Cmin = L/(1- sum of critical flow ratios) = 7(1-.463) = 13.03 seconds.

Maybe you can round it up to 15 seconds

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c) Now we have 4 lanes each in E-W and 4 lanes in N-S

Also we are given PHF = 0.90

# of Saturation lanes Flow/lane Total Saturation Flow Given Peak Hour Flow PHE Design Flow tvu = peak hour flow/PHF Flow Ratio = Design Critical Flow/Saturati Flow Ratio on Flow Phase 1 0.292 EB WB NB SB 4 4 4 4 1800 1800 1800 1800 7200 7200 7200 7200 2000 2000 1000 1000 0.95 0.95 0.95 0.95 2105 2105 1053 1053 0.292 0.292 0.146 0.146 Phase 2 0.146 0.439

We are given V/C ratio = 0.90 Therefore that means X_c = 0.90

Cycle Length according to HCM formula and based on X_c is given by

LEX Cmin LEX Χ. - Υ. Χ. - Σ. (3).

So we have minimum cycle length for getting v/c = 0.90

= 7 * 0.90/(0.90- .439) = 13.66 seconds

So round it up to 15 seconds

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d) For this we go back to our original problem, where cycle length = 90 seconds

Effective Green = Green time + yellow time - lost time = 20 + 5 - 7 = 18 seconds

so effective green for Phase 1 (i.e EB/WB approach) = critical flow ratio/sum of critical flow ratios * 18

= 0.292/.439 * 18 = 11.972 seconds

so we have saturation flow = 1800 pcu/hr/lane

Cycle Length = 90 seconds

Effective green per cycle = 11.972 seconds

so total effective green time which WB gets in one hour = 3600/90 * 11.972 = 478.88 seconds

So lane capacity = 1800 pcu/hr/lane 478.88/3600 = 239.44 vehicles/hour/lane

So capacity of WB approach assumeing green time is 20 second and cycle length is 90 seconds is 240 vehicles/hour

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