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.
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,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 ffollowing ftraffic fcount fdata fwere ftaken ffrom fa fpermanent fdetector flocation fon fa
fmajor fstate fhighway. fFrom fthis fdata, fdetermine f(a) fthe fAADT, f(b) fthe fADT ffor feach
fmonth. f (c) fthe fAAWT, fand f(d) fthe fAWT ffor feach fmonth. f From fthis finformation, fwhat
fcan fbe fdiscerned fabout fthe fcharacter fof fthe ffacility fand fthe fdemand fit fserves?
The ftable fbelow fillustrates fthe fcomputation fof fmonthly fADT fand fAWT fvalues.
The fAADT fis fcomputed fas fthe ftotal fannual fvolume fdivided fby f365 fdays, for:
AADT f= f 2,365,000 f = f 6,479 fveh/day
365
The fAAWT fis fcomputed fas fthe ftotal fweekday fvolume fdivided fby f260 fdays, for:
AAWT f= f 2,067,000 f = f 7,950 fveh/day
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 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.
@@
SeSiesim
smiciicsiosloaltaiotinon
,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 ffollowing ftraffic fcount fdata fwere ftaken ffrom fa fpermanent fdetector flocation fon fa
fmajor fstate fhighway. fFrom fthis fdata, fdetermine f(a) fthe fAADT, f(b) fthe fADT ffor feach
fmonth. f (c) fthe fAAWT, fand f(d) fthe fAWT ffor feach fmonth. f From fthis finformation, fwhat
fcan fbe fdiscerned fabout fthe fcharacter fof fthe ffacility fand fthe fdemand fit fserves?
The ftable fbelow fillustrates fthe fcomputation fof fmonthly fADT fand fAWT fvalues.
The fAADT fis fcomputed fas fthe ftotal fannual fvolume fdivided fby f365 fdays, for:
AADT f= f 2,365,000 f = f 6,479 fveh/day
365
The fAAWT fis fcomputed fas fthe ftotal fweekday fvolume fdivided fby f260 fdays, for:
AAWT f= f 2,067,000 f = f 7,950 fveh/day
260
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