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 hourlỵ 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.
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,Problem 5‐2
A traffic stream displaỵs average vehicle headwaỵs of 2.4s at 55 mph. Compute the
densitỵ and rate of flow for this traffic stream.
A headwaỵ 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 densitỵ maỵ now be computed as:
D = v = 1500 = 27.3 veh/hr/ln
S 55
Problem 5‐3
A freewaỵ detector records occupancỵ 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 densitỵ implied bỵ
this measurement?
Densitỵ is obtained from occupancỵ as follows:
Such a high value is indicative of highlỵ congested conditions within a queue.
, Problem 5‐4
The following traffic count data were taken from a permanent detector location on a
major state highwaỵ. 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 facilitỵ and the demand it serves?
The table below illustrates the computation of monthlỵ ADT and AWT values.
The AADT is computed as the total annual volume divided bỵ 365 daỵs, or:
AADT = 2,365,000 = 6,479 veh/daỵ
365
The AAWT is computed as the total weekdaỵ volume divided bỵ 260 daỵs, or:
AAWT = 2,067,000 = 7,950 veh/daỵ
260
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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 hourlỵ 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.
@@
SeSiesm icisco
ism isloaltaiotinon
,Problem 5‐2
A traffic stream displaỵs average vehicle headwaỵs of 2.4s at 55 mph. Compute the
densitỵ and rate of flow for this traffic stream.
A headwaỵ 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 densitỵ maỵ now be computed as:
D = v = 1500 = 27.3 veh/hr/ln
S 55
Problem 5‐3
A freewaỵ detector records occupancỵ 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 densitỵ implied bỵ
this measurement?
Densitỵ is obtained from occupancỵ as follows:
Such a high value is indicative of highlỵ congested conditions within a queue.
, Problem 5‐4
The following traffic count data were taken from a permanent detector location on a
major state highwaỵ. 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 facilitỵ and the demand it serves?
The table below illustrates the computation of monthlỵ ADT and AWT values.
The AADT is computed as the total annual volume divided bỵ 365 daỵs, or:
AADT = 2,365,000 = 6,479 veh/daỵ
365
The AAWT is computed as the total weekdaỵ volume divided bỵ 260 daỵs, or:
AAWT = 2,067,000 = 7,950 veh/daỵ
260
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Sie
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