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SOLUTIONS
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Traffic Engineering, 5 Edition W W
Roess, R.P., Prassas, E.S., and McShane, W.R.
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Solutions to Homework No. 2
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Problem 5‐1 W
A volume of 1,200 veh/h is observed at an intersection approach. Find the peak flow rate
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within the hour for the following peak-
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hour factors: 1.00, 0.90., 0.80, 0.70. Plot and comment on the results.
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The peak flow rate of flow is computed as v = V/PHF. The table below summarizes the resul
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ts for the information given. A plot follows.
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Even with the same hourly volume,
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a small difference in PHF leads to a
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n enormous difference in peak flo
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w rates. Traffic engineers must be
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able to deal with this peaking char
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acteristic on a regular basis. W W W W
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,Problem 5‐2 W
A traffic stream displays average vehicle headways of 2.4s at 55 mph. Compute the de
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nsity and rate of flow for this traffic stream.
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A headway can be converted to a flow rate as follows:
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v = 3600 = 3600 = 1,500 veh/hr/ln
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h 2.4
Knowing both flow rate and speed (given), the density may now be computed as:
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D = v = 1500 = 27.3 veh/hr/ln
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S 55
Problem 5‐3 W
A freeway detector records occupancy of 0.26 for a 15- minute period. If the detector is
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3.5 ft long, and the average vehicle has a length of 18 ft. what is the density implied by this
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measurement?
Density is obtained from occupancy as follows:
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Such a high value is indicative of highly congested conditions within a queue.
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, Problem 5‐4 W
The following traffic count data were taken from a permanent detector location on a maj
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or state highway. From this data, determine (a) the AADT, (b) the ADT for each month. (c)
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the AAWT, and (d) the AWT for each month. From this information, what can be discerned
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about the character of the facility and the demand it serves?
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The table below illustrates the computation of monthly ADT and AWT values.
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The AADT is computed as the total annual volume divided by 365 days, or:
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AADT = 2,365,000 = 6,479 veh/day
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365
The AAWT is computed as the total weekday volume divided by 260 days, or:
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AAWT = 2,067,000 = 7,950 veh/day
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260
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