SME
Transportation Engineering
Capacity analysis
Use this for “A freeway carries 4000 veh/h on 3 lanes, PHF 0.92, 10% trucks with ET = 2, at 70 mph. Find vₚ, the density and the level of service.”
Diagram
D
=
vₚ
S
=
1,590
70
=
22.8
pc/mi/ln:
LOS
C
(18
<
D
≤
26).
That is about 23 cars in 1 mi of one lane, one every 232 ft if the mile is 5280 ft.
This is an example. Type your own values to start a new problem.
fHV
=
1
÷
(1
+
PT
%
×
(
ET
−
1))
vₚ (pc/h/ln)
=
V (veh/h)
÷
(
PHF
×
N
×
fHV
)
D (pc/mi/ln)
=
vₚ (pc/h/ln)
÷
S (mph)
V is the hourly volume in one direction, N the lanes that way and PHF the peak-hour factor; fHV turns the trucks into passenger cars.
LOS
Level of service
worked out (not typed)
Formulas
fHV
=
1
1 + PT% × (ET − 1)
0.9091
=
1
1 + 10% × (2 − 1)
vₚ
=
V
PHF ×
N
× fHV
1,594.2029
=
4,000
0.92 × 3 × 0.9091
D
=
vₚ
S
22.7743
=
1,594.2029
70
LOS
from
D
:
A
up
to
11,
B
to
18,
C
to
26,
D
to
35,
E
to
45
pc/mi/ln,
F
above
22.7743 ≤ 26, so C
Assumptions
•
An HCM 7th edition basic freeway segment: LOS A to E end at 11, 18, 26, 35 and 45 pc/mi/ln.
•
The mean speed S is read from the speed–flow curve and typed here.
•
Each truck takes the room of ET passenger cars; PT is the trucks’ share in %.
Step-by-step
WE KNOW
V = 4,000 veh/h, PHF = 0.92, N = 3, PT = 10%, ET = 2, S = 70 mph
FIND
heavy-vehicle factor (fHV), flow rate (vₚ), density (D), level of service (LOS)
Step 1 · Find heavy-vehicle factor (fHV)
Use fHV = 1 ÷ (1 + PT% × (ET − 1))
Each truck takes the room of ET cars, so trucks add PT(ET − 1) cars to every car.
fHV
=
1
1 + 10% × (2 − 1)
PT = 10% = 0.1
fHV
=
1
1 + 0.1 × 1
fHV
=
1
1.1
fHV = 0.909091
Step 2 · Find flow rate (vₚ)
Use vₚ = V ÷ (PHF × N × fHV)
The peak 15 minutes as an hourly rate, shared by the lanes, in passenger cars.
vₚ
=
4,000
0.92 × 3 × 0.909091
vₚ
=
4,000
2.76 × 0.9091
vₚ
=
4,000
2.509
vₚ = 1,594.203 pc/h/ln
Step 3 · Find density (D)
Use D = vₚ ÷ S
Cars passing an hour over miles driven an hour is cars in each mile of lane.
D
=
1,594.203
70
D
=
22.7743
pc/mi/ln
Step 4 · Find level of service (LOS)
Use LOS from D: A up to 11, B to 18, C to 26, D to 35, E to 45 pc/mi/ln, F above
The
denser
the
lane,
the
less
room
each
driver
has:
the
band
of
D
is
the
level
of
service.
22.7743 ≤ 26: C
LOS
=
the
band
of
the
density
D
LOS = the band of the density 22.7743
LOS = C
Check
Put the numbers back in. Both sides should match.
0.909091
=
1
1 + 10% × (2 − 1)
✓
1,594.203
=
4,000
0.92 × 3 × 0.909091
✓
22.7743
=
1,594.203
70
✓
22.7743 ≤ 26, so C ✓
Topic check
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