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