Highway Engineering

Basic Elements of Highway Traffic Analysis 85

Example 4.3
A suburban undivided 4-lane highway on rolling terrain has a peak hour
volume (V) in one direction of 1500 vehicles per hour, with a peak hour factor
estimated at 0.85. All lanes are 3.05 m (10 ft) wide. There are no obstructions
within 1.83 m (6 ft) of the kerb.
The percentages for the various heavy vehicle types are:

PT– 12%

PB– 6%

PR– 2%

Determine the level of service of this section of highway.

Solution

The service flow is again calculated knowing the hourly volume during the
peak hour and the peak hour factor:

SF=V∏PHF

= 1500 ∏0.85 =1764.71 vehicles per hour

C 60 =2000 passenger cars per hour per lane N(the number of lanes in each

direction) = 2

fw =0.91 (3.05 m wide lanes, no roadside obstructions)

fP =1.0 (ideal driver population)

fE =0.80 (suburban undivided)

Since

Therefore

(4.20)

Using the data from Table 4.2, the highway operates at level of service E.

v

c

CNf f f f

v

c

i

ij wHVpE

i

Ê

Ë

ˆ

̄ =∏¥¥¥ ¥¥()

Ê

Ë

ˆ

̄ =∏¥¥¥¥¥()=

SF

1764 71. 2000 2 0 91 0 66 1 0 0 8.....0 92

SFij

i

C whvpE

v

c

()=¥ÊË ˆ ̄ ¥¥ ¥ ¥ ¥Nf f f f

f

HV PETT PEBB PERR

=

+-{}()+-()+-()

+{}()+ ()+ ()

=

1

1 111

1

101230 06 2 0 02 2

066

...

.