CE 372 - Fundamentals of Transportation Engineering
30 March
2007 - Introduction to Traffic Operations Analysis/Signalized Intersections
2 April 2007 - Introduction to Traffic Operations
Analysis/Two-Way Stop-Controlled Intersections
Introduction to Traffic Operations - 1998 notes (but still worth reviewing)
Written notes
Questions from class (30 March 2007)
Questions from class (2 April 2007)
Assignment
As the traffic engineer for the City of Moscow, you have been asked to
determine whether or not a traffic signal should be installed at the
intersection of US 95/Styner/Lauder. The signal has been requested by
residents of the area who claim that delays on the minor streets are high. Your
task is to recommend to the city council whether you support this request or
whether you think that the current stop control should remain.
Here are some data that might need for your analysis (note that I'm providing you with a much simplified intersection, one that doesn't quite look like today's intersection!).
For signal control, try a range of cycle lengths from 60 to 90 seconds (in increments of 10 seconds). Compute the green time based on the relative volumes on each approach: consider the ratio of the maximum of the NS movement volumes and the minor street volume.
Assume a critical gap of 7.1 seconds and a follow up time of 3.5 seconds for the minor street left turn movement.
Summarize your recommendation and your analysis in a two page report, clearly presenting the justification for your recommendation. Your report (written together as a two-person team) is due on Friday, April 4, 2007. Please send the Word document electronically to me at mkyte@uidaho.edu.
| Given data | ||
| Saturation flow rate | 1900 | |
| Cycle | 60-90 | |
| Flow rate | 2007 (today) | 2017 (forecast) |
| NB-TH | 400 | 488 |
| SB-TH | 500 | 609 |
| EB-LT | 300 | 366 |
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Equations of interest:
capacity = saturation flow rate (green time/cycle length), or c = s(g/C)
delay = (cycle length/2)(1-green time/cycle length)2/(1-volume/saturation flow rate), or d=(C/2)(1-g/C)2/(1-v/s)
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delay = 3600/capacity