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.
@@
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 sfollowing straffic scount sdata swere staken sfrom sa spermanent sdetector slocation son sa
smajor sstate shighway. sFrom sthis sdata, sdetermine s(a) sthe sAADT, s(b) sthe sADT sfor seach
smonth. s (c) sthe sAAWT, sand s(d) sthe sAWT sfor seach smonth. s From sthis sinformation, swhat
scan sbe sdiscerned sabout sthe scharacter sof sthe sfacility sand sthe sdemand sit sserves?
The stable sbelow sillustrates sthe scomputation sof smonthly sADT sand sAWT svalues.
The sAADT sis scomputed sas sthe stotal sannual svolume sdivided sby s365 sdays, sor:
AADT s= s 2,365,000 s = s 6,479 sveh/day
365
The sAAWT sis scomputed sas sthe stotal sweekday svolume sdivided sby s260 sdays, sor:
AAWT s= s 2,067,000 s = s 7,950 sveh/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.
@@
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 sfollowing straffic scount sdata swere staken sfrom sa spermanent sdetector slocation son sa
smajor sstate shighway. sFrom sthis sdata, sdetermine s(a) sthe sAADT, s(b) sthe sADT sfor seach
smonth. s (c) sthe sAAWT, sand s(d) sthe sAWT sfor seach smonth. s From sthis sinformation, swhat
scan sbe sdiscerned sabout sthe scharacter sof sthe sfacility sand sthe sdemand sit sserves?
The stable sbelow sillustrates sthe scomputation sof smonthly sADT sand sAWT svalues.
The sAADT sis scomputed sas sthe stotal sannual svolume sdivided sby s365 sdays, sor:
AADT s= s 2,365,000 s = s 6,479 sveh/day
365
The sAAWT sis scomputed sas sthe stotal sweekday svolume sdivided sby s260 sdays, sor:
AAWT s= s 2,067,000 s = s 7,950 sveh/day
260
@@
Se
Sie
sm icic
ism sioslo
altaiotinon
