CE 361 Introduction to Transportation Engineering



|CE 361 Introduction to Transportation Engineering |Posted: Wed 13 October 2010 |

|Homework 6 (HW 6) Solutions |Due: Fri. 22 October 2010 |

EVALUATION OF PROJECTS FOR SAFETY AND EFFICIENCY

Dear Consultant:

The County wishes to determine your ability to evaluate projects as it begins to undertake a concerted effort to improve the safety and efficiency of local transportation systems. Part of your qualification process to bid on county traffic safety projects involves your ability to provide clear and correct solutions to problems related to recent and upcoming projects. In this assignment, as in all previous assignments, you are expected to:

□ Work alone or as part of a group of up to three CE361 students. Group members must sign the front page of the work submitted.

□ Label each problem with its name.

Failure to comply with these requirements will result in penalties.

|Project evaluation. A private health care provider builds an Automated People Mover |[pic] |

|(pictured at right) to connect its two hospitals that are one mile apart. Before the APM | |

|was built, doctors, nurses, staff, visitors, and even patients went from one hospital to | |

|another by going to their parked cars, driving to the other hospital’s parking garage, then | |

|walking to their destinations. This would take about 30 minutes. The same trip by APM | |

|would take ten minutes. 1000 passengers ride the APM on an average day, including weekends.| |

|The APM cost $40 million to build and costs $1.12 million per year to operate. | |

|(5 points) If the value of travel time is $25/hour, what is the dollar value of travel time| |

|saved each year? | |

|(15 points) How many years (to the nearest 0.1 year) will it take for the benefits of the | |

|APM to justify the costs? Use a discount rate of 2.9 percent per year. Explain your | |

|procedures. | |

A. A3 = Annual time savings = 365 days/yr * 1000 pax/day * (1/3 hr saved per pax) * $25/hr VoT = $3.042*106.

|P1 = |4.00E+07 |cost to build |Set up a worksheet like the one at left. A2 = annual cost to |

| | | |operate. |

| | | |Find P2 -= PW of A2 after n in [P|A] factor can be determined. In|

| | | |the worksheet, “sum P” = P3 – (P1 + P2) because P3 is the PW of |

| | | |annual benefits of time saved, P1 is cost to build and P2 is PW of|

| | | |annual costs to |

|A2 = |1.12E+06 |ann cost to operate | |

|A3 = |3.04E+06 |ann time savings | |

|i = |0.029 |discount rate | |

|n = |32.372209 |years | |

|P|A = |20.8152645 |factor | |

|P2 = |2.33E+07 |PW ann op costs | |

|P3 = |6.33E+07 |PW ann time savings | |

|sum P = |9.69E-07 | | |

operate. Calc [P|A] factor using eqn in FTE Fig 5.4: [P|A,i,n] = [pic]= [pic]with n to be determined. The value of n affects the values of [P|A], P2, and P3. Find n such that sum P = 0 by trial and error or by Solver feature of worksheet. The value of n is 33.4years.

1. Crash data. To demonstrate your understanding of Processes 1 and 2 in FTE Figure 6.4, click on Tippecanoe County’s “2008 Vehicle Crash Report”. In Table 60, find the intersection of SR 26 E/South St and 3rd St. (It is in downtown Lafayette, one block south of the courthouse.) Also read the footnotes below Table 60. Traffic count maps are available at . Show your calculations and compare the results of your calculations with the entries in Table 60.

A. (10 points) South Street is one-way EB; 3rd Street is one-way SB. What are the approach ADT values for the NB, SB, EB, and WB approaches? What value of MEV should be used in the MEVrate equation in the Crash report? How does it compare with the MEV value in Table 60?

The excerpt of the traffic count map that includes the intersection of South Street and 3rd Street is shown below. The SB approach ADT is 5482 (2007 count); the EB approach ADT is 12,750 (2008 count).

[pic]

Equation 1 on page 25 of the Crash Report is [pic]. V1 is the north (SB) approach. V1 = 5482. V4 is the west (EB) approach, V4 = 12,750. MEV = 365*(V1+V2+V3+V4) = 365*(5482+0+0+12,750) = 365*18,232 = 6,654,680 = 6.655*106. In Table 60, MEV = 6.655. The values match.

B. (20 points) For the 3-year period 2006-2008, what is the value of A (crash frequency) for use in Equation 1 for the intersection of South and 3rd Streets? Use Equation 1 to calculate the MEV Rate for this intersection.

A = [pic]=18.67 crashes per year.

MEVrate = [pic]= 2.805. Equals Table 60 entry.

C. (10 points) Use Equation 2 to calculate the CRF for this intersection. What value of Ra did you use and how did you determine it?

Equation 2 on page 25 is [pic]. Ra = 1.6066 from the “***” footnote under Table 60 for Urban Principal Arterials.

CRF =[pic] = 1.6066 + 0.808 + 0.075 = 2.490 vs. 2.098 in Table 60 for the intersection of South and 3rd Streets. Values do not match.

2. FTE Exercise 6.19 Racing the Train.

a. (5 points) Car’s x1 = 800 – (0.6 sec * 1.47 fps/mph * 55 mph) = 800 – (0.6 * 80.85) = 800 – 48.51 = 751.49 ft from tracks

b. (15 points) Find time for car to reach 85 mph.

v1 = v0 + at; t = [pic]= 1.575 sec.

Find distance for car to reach 85 mph.

x =[pic]= 127.34 + 34.73 = 162.07 ft, which is less than the distance to the tracks. The car will be traveling at 85 mph for the rest of the distance to the tracks, which is: 800 – 48.51 – 162.07 = 589.42 ft.

| |Mode |t sec |V fps |ft traveled |dist to Xing |

| |reaction |0.6 |80.85 |48.51 |751.49 |

|Car |accelerate |1.575 |up to 124.95 |162.07 |589.42 |

| |constant V |4.72 |124.95 |589.42 |0 |

| | |6.895 | | | |

|Train |constant V |6.895 |58.80 |405.43 |594.57 |

c. (10 points) Time car will travel at 85 mph to tracks = [pic]= 4.72 sec. Time for car to reach tracks = 0.6 + 1.575 + 4.72 = 6.895 sec. At this time, where will train be? 1000 ft – (1.47 * 40 * 6.895) = 1000 – 405.43 = 594.57 feet from the crossing.

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