CE 361 Introduction to Transportation Engineering / Out: Fri. 5 September 2008
Homework 2 (HW 2) Solutions / Due: Mon. 15 September 2008

TRAFFIC FLOW DATA

CE361 08HW2 Solutions- 1 -Fall 2008

  • 1-3 CE361 students. If 2-3, signatures required.
  • ID each problem by its number and name. State your methodology.
  1. (10 points) Mr. Posey’s letter. The largest standard deviation in FTE Table 6.15 is S = 4.378. Use z = 1.96 unless otherwise told. In FTE Table 2.7, U = 1.04 for the 85th percentile speed.

FTE (2.4) == 113.45 = 114 vehicles.

  1. Occupancy and density from loop detectors. Effective length of 7.0 feet, actuated for 4.8 percent of one-hour. 632 vehicles detected, mean speed 63.2 mph, mean length 18.24 feet.

A. (10 points) Convert the apparent occupancy rate to the actual occupancy rate.

(2.6) = 0.2732 second

(2.7) = 0.1977 second

(2.8) Actual Occupancy Oact = q * t’(P) = (632 * 0.1977)/3600 = 0.035 or 3.5%.

B. (10 points) What was the density (vpm) on the roadway during the occupancy measurements?

Use any one of three equations:

(2.9) = 9.99 veh/mi

Calculate Oapp as in FTE Example 2.10A, then use (2.10) = 10.03 veh/mi

(2.12) = 10.00 veh/mi

  1. Speed-Flow-Density relationships. Do FTE Exercise 2.40, using the data emailed to you.
  2. (15 points) Speed-density plot.

Obsvn / Flow (vph) / Speed (kph) / Occup (%) / Density (vpk)
1 / 1260 / 62 / 10 / 20.3
2 / 1620 / 59 / 13 / 27.5
3 / 1440 / 55 / 14 / 26.2
4 / 1620 / 56 / 15 / 28.9
5 / 1440 / 59 / 12 / 24.4
6 / 1800 / 54 / 18 / 33.3
7 / 1620 / 43 / 19 / 37.7
8 / 2160 / 36 / 33 / 60.0
9 / 1260 / 18 / 37 / 70.0
10 / 1800 / 32 / 27 / 56.3
11 / 1800 / 35 / 26 / 51.4
12 / 2160 / 33 / 33 / 65.5
13 / 1440 / 20 / 38 / 72.0
14 / 1800 / 25 / 35 / 72.0
15 / 2160 / 36 / 28 / 60.0
16 / 1800 / 22 / 39 / 81.8
17 / 1800 / 33 / 28 / 54.5
18 / 1800 / 37 / 24 / 48.6
19 / 1980 / 35 / 27 / 56.6
20 / 1260 / 29 / 24 / 43.4
21 / 1800 / 20 / 41 / 90.0
22 / 1440 / 25 / 29 / 57.6
23 / 1800 / 36 / 23 / 50.0
24 / 1620 / 24 / 33 / 67.5
25 / 1260 / 18 / 32 / 70.0
26 / 900 / 12 / 47 / 75.0
27 / 1260 / 17 / 32 / 74.1
28 / 1440 / 25 / 30 / 57.6
29 / 1800 / 33 / 27 / 54.5
30 / 1980 / 35 / 28 / 56.6
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  • Using Equation 2.12, calculate D = q/S values for the Density column. Figure 3A is the plot of 30 data points of S vs. D.

  • Fit a straight line through the data points.
Free-flow speed Sf = 73.362 km/hr.
Jam density Dj = 73.362/0.7161 = 102.45 veh/km.
  • The fitted line is consistent with the plotted data points.

  1. (15 points) Flow-density plot.
  2. See Figure 3B.
  3. Use Equation 2.19 to convert D = 0, 5, 10, …, 95, 100, 102.45 veh/km into corresponding “model” values of q.
  4. Sample calculation for D = 30 veh/km:


q = 1556 veh/hr
  • Plot the model values as a smooth parabolic curve.
  • The smooth curve is not inconsistent with the plotted points.
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  1. (15 points) Speed-flow plot.
  2. See Figure 3C.
  3. Use Equation 2.18 to convert S = 0, 10, 20, …, 70, 73.36 kph into corresponding “model” values of q.
  4. Sample calculation for S = 30 km/hr:


q = 1817 veh/hr
  • The smooth curve is not inconsistent with the plotted points.
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  1. The road to Shreveport. 343 miles = 552 km) from New Orleans to Shreveport. Normally 5 hours and 21 minutes; before Hurricane Gustav, 25 hours. 2 lanes each direction on a 10-kilometer segment. Metric distances.

(2.13) ; (2.19) ; (2.18)

A. (10 points) Normal mean speed of the trip to Shreveport? What is the corresponding traffic density and flow rate on the 10-kilometer segment, according to the Greenshields equations?

= 103.18 km/hr.

FTE (2.13) ;; D = 15.56 veh/km.

FTE (2.18) = 1606 veh/hr

B. (10 points) Mean speed of the pre-Gustav trip to Shreveport? Corresponding traffic density and flow rate on the 10-kilometer segment, according to the Greenshields equations?

= 22.1 km/hr.

FTE (2.13) ;; D = 68.0 veh/km.

FTE (2.18) = 1503 veh/hr

C. (5 points) Do all the values you calculated make sense? (Are any suspicious?)

  • 103.18 km/hr = 64.11 mph, which is a little low for an Interstate segment under normal conditions, but it is not unreasonable.
  • A density 68.0 veh/km. equates to a vehicle every 1000 m/68 = 14.7 meters or 48.2 feet. If an average vehicle is about 15 feet long, the D=68 veh/km value represents a situation in which each vehicle (on the average) has a little more than two vehicle lengths behind it until the next vehicle. This is possible in an evacuation situation.
  • The flow rates of 1606 veh/hr and 1503 veh/hr are reasonable.