Highway Engineering

Basic Elements of Highway Traffic Analysis 89

Example 4.4 Contd

Solution

Since the road is operating at capacity, the level of service is assumed to be
E.

Therefore:

fw=0.76 (2.75 m wide lanes, no roadside obstructions)

fd=0.89 (70/30 distributional split)

Rolling terrain, level of service E, therefore ET= 5.0, EB= 2.9 and

ER=3.3

Therefore:

f

v

c

ff f

HV

E

E

wHVd

E

=

+{}()+ ()+ ()

=

=¥ÊË ˆ ̄ ¥¥ ¥

=¥¥¥¥=

1

10104004 19 00223

0 657

2800

2800 0 92 0 76 0 657 0 89 1145

.....

.

....

SF

SF v h

Example 4.5
Determine the level of service provided by a 2-lane highway with a peak-
hour volume (V) of 1200 and a peak-hour factor of 0.8. No passing is per-
mitted on the highway. The directional split is 60/40 in favour of the peak
direction. Both lanes are 12 ft (3.65 m) wide. There is a 1.22 m (4 ft) clearance
on both hard shoulders.
The percentages for the various heavy vehicle types are:

PT– 10%

PB– 4%

PR– 2%

The terrain is level.

Solution

SF =V∏PHF

= 1200 ∏0.8

= 1500

Correction factors:

fd=0.94 (80/20 distributional split)

Contd