Basic Freeway Segment Highway Capacity

Transportation · FE Reference Handbook section

Transportation

33 formulas

10 exam-style examples

~60 min

All Transportation lectures

Learning objectives

What you must be able to do before leaving this section.

This chapter section covers Basic Freeway Segment Highway Capacity within Transportation. Read it the way you would read a textbook chapter: the theory first so the relations mean something, then every equation with its use and its trap, then 10 fully worked examples with the arithmetic shown line by line, and finally a self-check you should be able to answer without notes.

Explain, in your own words, what basic freeway segment highway capacity describes physically and when it applies.
State every one of the 33 relations the handbook lists here and name each symbol with its unit.
Select the correct relation from the wording of an exam stem within 20 seconds.
Carry a complete solution from givens to a "most nearly" answer with the correct unit.
Recognise the distractors generated by the unit trap: grades as decimals in curve formulas, percent in the stem.

Lecture

Why this section exists. Basic Freeway Segment Highway Capacity is the part of Transportation that lets you connect a vertical or horizontal alignment, or a traffic stream to a number you can defend. Before any equation is useful you must be able to picture the physical situation it describes; the schematic below is that picture.

How the theory is built. The handbook prints results, not derivations. Each relation in this section comes from one governing principle applied to the idealised system: state the principle, impose the stated assumptions, and the printed equation follows. Knowing which assumption each relation rests on is what lets you reject a wrong answer choice in seconds.

How it is examined. Items from this page are written as a curve geometry element or a capacity/flow relationship. Roughly two thirds are direct substitution, one third require one intermediate quantity from a neighbouring relation, and a small number are conceptual — testing whether you know the assumption, not the arithmetic.

The habit that earns the points. Unit discipline. grades as decimals in curve formulas, percent in the stem. Every relation below is dimensionally consistent only when that rule is honoured, and the distractor set is deliberately built from candidates who ignored it. Write the unit next to every number you substitute, every time.

How to study this page. Read the theory, then cover the formula cards and try to reproduce each relation from its description. Then work the examples with the solution hidden, revealing one line at a time. Finish with the self-check questions; if you cannot answer one, return to the matching formula card.

Dense peak-hour traffic queued on an urban arterial at dusk. — Photo 1. Where this shows up in practice: basic freeway segment highway capacity.
Wikimedia Commons, CC BY 2.0

Transportation — Basic Freeway Segment Highway Capacity: reference schematic for orienting the symbols used in this section.

Theory, developed

Read this before the equations — it is what makes them memorable.

The physical situation

Every item from this section describes a vertical or horizontal alignment, or a traffic stream. Sketch it before you compute — a labelled sketch with the givens on it converts a wordy stem into a solvable problem and exposes the quantity the examiner left out on purpose.

The governing principle

The 33 relations on this page are consequences of one principle applied to that idealised system. Identify which quantity is conserved, balanced, or defined, and the correct equation follows without memorisation.

Assumptions and limits of validity

Each printed relation carries silent assumptions — linearity, steady state, uniformity, small deformation, or standard conditions, depending on the subject. Conceptual exam items are written by violating exactly one of these, so read the sentence above the equation as carefully as the equation itself.

Solution procedure you should automate

1) Read the last sentence of the stem to identify the requested quantity. 2) Locate the relation on this page whose left-hand side is that quantity. 3) Tabulate the givens with units and mark the missing symbol. 4) If a symbol is missing, find the one relation that produces it. 5) Rearrange symbolically, substitute once, evaluate, and round only at the end.

Notation used in this section

pc/h/ln	Quantity produced by "pc/h/ln = passenger cars per hour per lane" — read its definition and unit from the handbook line directly above the equation.
FFS	Quantity produced by "FFS = BFFS – fLW – fRLC – 3.22 TRD0.84" — read its definition and unit from the handbook line directly above the equation.
BFFS	Quantity produced by "BFFS = base free flow speed of basic freeway segment (mph); default is 75.4 mph" — read its definition and unit from the handbook line directly above the equation.
fLW	Quantity produced by "fLW = adjustment for lane width (mph)" — read its definition and unit from the handbook line directly above the equation.
fRLC	Quantity produced by "fRLC = adjustment for right-side lateral clearance (mph)" — read its definition and unit from the handbook line directly above the equation.
TRD	Quantity produced by "TRD = total ramp density (ramps/mi)" — read its definition and unit from the handbook line directly above the equation.
vp	Quantity produced by "vp = demand flowrate under equivalent base conditions (pc/h/ln)" — read its definition and unit from the handbook line directly above the equation.
V	Quantity produced by "V = demand volume under prevailing conditions (veh/h)" — read its definition and unit from the handbook line directly above the equation.
PHF	Quantity produced by "PHF = peak-hour factor" — read its definition and unit from the handbook line directly above the equation.
N	Quantity produced by "N = number of lanes in analysis direction" — read its definition and unit from the handbook line directly above the equation.
fHV	Quantity produced by "fHV = adjustment factor for presence of heavy vehicles in traffic stream, calculated with" — read its definition and unit from the handbook line directly above the equation.
PT	Quantity produced by "PT = proportion of single unit trucks and tractor trailers in traffic stream" — read its definition and unit from the handbook line directly above the equation.

Handbook notes for this section

Definitions and conditions exactly as the handbook states them.

Speed Flow Relationship for Basic Freeway Segments
(mph) (pc/h/ln) (pc/h/ln) (mph)
* All equations are based on Exhibit 12-6 and Equation 12-1 from the HCM 6th Edition assuming
all calibration factors (CAF and SAF) set to 1.0
where
Level of Service (LOS) Density (pc/mi/ln)
B >11 – 18
C >18 – 26
D >26 – 35
E >35 – 45
Demand exceeds capacity
F >45
where
Average Lane Width (ft) Reduction in FFS, fLW (mph)
Right-Side
Lanes in One Direction
Lateral
5 0.6 0.4 0.2 0.1
4 1.2 0.8 0.4 0.2
3 1.8 1.2 0.6 0.3
2 2.4 1.6 0.8 0.4
1 3.0 2.0 1.0 0.5
0 3.6 2.4 1.2 0.6
HCM: Highway Capacity Manual, 6th ed., A Guide for Multimodal Mobility Analysis, Transportation

Core formulas for this FE topic

Definitions, applicability, units, assumptions and worked examples for each relation.

Worked exam-style examples

The four ways this section is written on the real exam — thoughts first, then equations, then substitution.

Example 1

Greenshields flow, speed and density — Basic Freeway Segment Highway Capacity

A freeway segment follows Greenshields with free-flow speed 69 mph and jam density 205 veh/mi/ln. Find the capacity, and the speed and flow at a density of 56 veh/mi/ln.

Given

vf = 69 mph
kj = 205 veh/mi/ln
k = 56 veh/mi/ln

Find

qmax, v(k) and q(k)

Start with the thinking

Greenshields is linear: v = vf(1 − k/kj).
Capacity occurs at k = kj/2, giving qmax = vf·kj/4.

Step-by-step solution

Capacity
$q ma x = v f k j /4 = 69 (205) /4 = 3, 536 v e h / h / ln$
Optimum density
$k_{0} = k j /2 = 102.5 v e h / mi / ln$
Speed at k
$v = 69 (1 - 56/205) = 50.15 m p h$
Flow
$q = v k = 50.15 (56) = 2, 808 v e h / h / ln$
Utilization
$q / q ma x = 79.4$

Answer: qmax ≈ 3,536 veh/h/ln; at k = 56, v ≈ 50.2 mph and q ≈ 2,808 veh/h/ln

Why the other options are there

qmax = 14,145 veh/h/ln (factor of 4 omitted)
v = 18.8 mph (relation inverted)

Reference: FE Reference Handbook — Transportation → Basic Freeway Segment Highway Capacity

Example 2

Freeway density and level of service — Basic Freeway Segment Highway Capacity

A 3-lane (one direction) freeway carries 6,302 veh/h with PHF = 0.95 and an average speed of 67 mph. Find the per-lane flow rate, density and LOS.

Given

V = 6,302 veh/h
PHF = 0.95
N = 3 lanes
S = 67 mph

Find

vp, density D and LOS

Start with the thinking

Convert the hourly volume to a 15-minute peak flow rate with the PHF first.
Density = flow rate ÷ speed, in passenger cars per mile per lane.

Step-by-step solution

Peak flow rate — vp = V/(PHF·N) = 6,302/(0.95 × 3)
Evaluate
$v p = 2, 211 p c / h / ln$
Density
$D = v p / S = 2, 211/67 = 33.00 p c / mi / ln$
LOS — density 33.0 pc/mi/ln → LOS D

Answer: vp ≈ 2,211 pc/h/ln, D ≈ 33.0 pc/mi/ln → LOS D

Why the other options are there

vp = 2,101 pc/h/ln (PHF not applied)
D = 148,152 (flow multiplied by speed instead of divided)

Reference: FE Reference Handbook — Transportation → Basic Freeway Segment Highway Capacity

Example 3

Greenshields flow, speed and density — Basic Freeway Segment Highway Capacity (2)

A freeway segment follows Greenshields with free-flow speed 65 mph and jam density 219 veh/mi/ln. Find the capacity, and the speed and flow at a density of 87 veh/mi/ln.

Given

vf = 65 mph
kj = 219 veh/mi/ln
k = 87 veh/mi/ln

Find

qmax, v(k) and q(k)

Start with the thinking

Greenshields is linear: v = vf(1 − k/kj).
Capacity occurs at k = kj/2, giving qmax = vf·kj/4.

Step-by-step solution

Capacity
$q ma x = v f k j /4 = 65 (219) /4 = 3, 559 v e h / h / ln$
Optimum density
$k_{0} = k j /2 = 109.5 v e h / mi / ln$
Speed at k
$v = 65 (1 - 87/219) = 39.18 m p h$
Flow
$q = v k = 39.18 (87) = 3, 408 v e h / h / ln$
Utilization
$q / q ma x = 95.8$

Answer: qmax ≈ 3,559 veh/h/ln; at k = 87, v ≈ 39.2 mph and q ≈ 3,408 veh/h/ln

Why the other options are there

qmax = 14,235 veh/h/ln (factor of 4 omitted)
v = 25.8 mph (relation inverted)

Reference: FE Reference Handbook — Transportation → Basic Freeway Segment Highway Capacity

Example 4

Freeway density and level of service — Basic Freeway Segment Highway Capacity (2)

A 3-lane (one direction) freeway carries 6,241 veh/h with PHF = 0.96 and an average speed of 62 mph. Find the per-lane flow rate, density and LOS.

Given

V = 6,241 veh/h
PHF = 0.96
N = 3 lanes
S = 62 mph

Find

vp, density D and LOS

Start with the thinking

Convert the hourly volume to a 15-minute peak flow rate with the PHF first.
Density = flow rate ÷ speed, in passenger cars per mile per lane.

Step-by-step solution

Peak flow rate — vp = V/(PHF·N) = 6,241/(0.96 × 3)
Evaluate
$v p = 2, 167 p c / h / ln$
Density
$D = v p / S = 2, 167/62 = 34.95 p c / mi / ln$
LOS — density 35.0 pc/mi/ln → LOS D

Answer: vp ≈ 2,167 pc/h/ln, D ≈ 35.0 pc/mi/ln → LOS D

Why the other options are there

vp = 2,080 pc/h/ln (PHF not applied)
D = 134,355 (flow multiplied by speed instead of divided)

Reference: FE Reference Handbook — Transportation → Basic Freeway Segment Highway Capacity

Example 5

Greenshields flow, speed and density — Basic Freeway Segment Highway Capacity (3)

A freeway segment follows Greenshields with free-flow speed 66 mph and jam density 219 veh/mi/ln. Find the capacity, and the speed and flow at a density of 33 veh/mi/ln.

Given

vf = 66 mph
kj = 219 veh/mi/ln
k = 33 veh/mi/ln

Find

qmax, v(k) and q(k)

Start with the thinking

Greenshields is linear: v = vf(1 − k/kj).
Capacity occurs at k = kj/2, giving qmax = vf·kj/4.

Step-by-step solution

Capacity
$q ma x = v f k j /4 = 66 (219) /4 = 3, 614 v e h / h / ln$
Optimum density
$k_{0} = k j /2 = 109.5 v e h / mi / ln$
Speed at k
$v = 66 (1 - 33/219) = 56.05 m p h$
Flow
$q = v k = 56.05 (33) = 1, 850 v e h / h / ln$
Utilization
$q / q ma x = 51.2$

Answer: qmax ≈ 3,614 veh/h/ln; at k = 33, v ≈ 56.1 mph and q ≈ 1,850 veh/h/ln

Why the other options are there

qmax = 14,454 veh/h/ln (factor of 4 omitted)
v = 9.9 mph (relation inverted)

Reference: FE Reference Handbook — Transportation → Basic Freeway Segment Highway Capacity

Example 6

Freeway density and level of service — Basic Freeway Segment Highway Capacity (3)

A 3-lane (one direction) freeway carries 4,058 veh/h with PHF = 0.95 and an average speed of 56 mph. Find the per-lane flow rate, density and LOS.

Given

V = 4,058 veh/h
PHF = 0.95
N = 3 lanes
S = 56 mph

Find

vp, density D and LOS

Start with the thinking

Convert the hourly volume to a 15-minute peak flow rate with the PHF first.
Density = flow rate ÷ speed, in passenger cars per mile per lane.

Step-by-step solution

Peak flow rate — vp = V/(PHF·N) = 4,058/(0.95 × 3)
Evaluate
$v p = 1, 424 p c / h / ln$
Density
$D = v p / S = 1, 424/56 = 25.43 p c / mi / ln$
LOS — density 25.4 pc/mi/ln → LOS C

Answer: vp ≈ 1,424 pc/h/ln, D ≈ 25.4 pc/mi/ln → LOS C

Why the other options are there

vp = 1,353 pc/h/ln (PHF not applied)
D = 79,736 (flow multiplied by speed instead of divided)

Reference: FE Reference Handbook — Transportation → Basic Freeway Segment Highway Capacity

Example 7

Greenshields flow, speed and density — Basic Freeway Segment Highway Capacity (4)

A freeway segment follows Greenshields with free-flow speed 55 mph and jam density 178 veh/mi/ln. Find the capacity, and the speed and flow at a density of 46 veh/mi/ln.

Given

vf = 55 mph
kj = 178 veh/mi/ln
k = 46 veh/mi/ln

Find

qmax, v(k) and q(k)

Start with the thinking

Greenshields is linear: v = vf(1 − k/kj).
Capacity occurs at k = kj/2, giving qmax = vf·kj/4.

Step-by-step solution

Capacity
$q ma x = v f k j /4 = 55 (178) /4 = 2, 448 v e h / h / ln$
Optimum density
$k_{0} = k j /2 = 89 v e h / mi / ln$
Speed at k
$v = 55 (1 - 46/178) = 40.79 m p h$
Flow
$q = v k = 40.79 (46) = 1, 876 v e h / h / ln$
Utilization
$q / q ma x = 76.7$

Answer: qmax ≈ 2,448 veh/h/ln; at k = 46, v ≈ 40.8 mph and q ≈ 1,876 veh/h/ln

Why the other options are there

qmax = 9,790 veh/h/ln (factor of 4 omitted)
v = 14.2 mph (relation inverted)

Reference: FE Reference Handbook — Transportation → Basic Freeway Segment Highway Capacity

Example 8

Freeway density and level of service — Basic Freeway Segment Highway Capacity (4)

A 3-lane (one direction) freeway carries 5,159 veh/h with PHF = 0.87 and an average speed of 68 mph. Find the per-lane flow rate, density and LOS.

Given

V = 5,159 veh/h
PHF = 0.87
N = 3 lanes
S = 68 mph

Find

vp, density D and LOS

Start with the thinking

Convert the hourly volume to a 15-minute peak flow rate with the PHF first.
Density = flow rate ÷ speed, in passenger cars per mile per lane.

Step-by-step solution

Peak flow rate — vp = V/(PHF·N) = 5,159/(0.87 × 3)
Evaluate
$v p = 1, 977 p c / h / ln$
Density
$D = v p / S = 1, 977/68 = 29.07 p c / mi / ln$
LOS — density 29.1 pc/mi/ln → LOS D

Answer: vp ≈ 1,977 pc/h/ln, D ≈ 29.1 pc/mi/ln → LOS D

Why the other options are there

vp = 1,720 pc/h/ln (PHF not applied)
D = 134,411 (flow multiplied by speed instead of divided)

Reference: FE Reference Handbook — Transportation → Basic Freeway Segment Highway Capacity

Example 9

Greenshields flow, speed and density — Basic Freeway Segment Highway Capacity (5)

A freeway segment follows Greenshields with free-flow speed 62 mph and jam density 201 veh/mi/ln. Find the capacity, and the speed and flow at a density of 85 veh/mi/ln.

Given

vf = 62 mph
kj = 201 veh/mi/ln
k = 85 veh/mi/ln

Find

qmax, v(k) and q(k)

Start with the thinking

Greenshields is linear: v = vf(1 − k/kj).
Capacity occurs at k = kj/2, giving qmax = vf·kj/4.

Step-by-step solution

Capacity
$q ma x = v f k j /4 = 62 (201) /4 = 3, 116 v e h / h / ln$
Optimum density
$k_{0} = k j /2 = 100.5 v e h / mi / ln$
Speed at k
$v = 62 (1 - 85/201) = 35.78 m p h$
Flow
$q = v k = 35.78 (85) = 3, 041 v e h / h / ln$
Utilization
$q / q ma x = 97.6$

Answer: qmax ≈ 3,116 veh/h/ln; at k = 85, v ≈ 35.8 mph and q ≈ 3,041 veh/h/ln

Why the other options are there

qmax = 12,462 veh/h/ln (factor of 4 omitted)
v = 26.2 mph (relation inverted)

Reference: FE Reference Handbook — Transportation → Basic Freeway Segment Highway Capacity

Example 10

Freeway density and level of service — Basic Freeway Segment Highway Capacity (5)

A 2-lane (one direction) freeway carries 3,034 veh/h with PHF = 0.93 and an average speed of 59 mph. Find the per-lane flow rate, density and LOS.

Given

V = 3,034 veh/h
PHF = 0.93
N = 2 lanes
S = 59 mph

Find

vp, density D and LOS

Start with the thinking

Convert the hourly volume to a 15-minute peak flow rate with the PHF first.
Density = flow rate ÷ speed, in passenger cars per mile per lane.

Step-by-step solution

Peak flow rate — vp = V/(PHF·N) = 3,034/(0.93 × 2)
Evaluate
$v p = 1, 631 p c / h / ln$
Density
$D = v p / S = 1, 631/59 = 27.65 p c / mi / ln$
LOS — density 27.6 pc/mi/ln → LOS D

Answer: vp ≈ 1,631 pc/h/ln, D ≈ 27.6 pc/mi/ln → LOS D

Why the other options are there

vp = 1,517 pc/h/ln (PHF not applied)
D = 96,240 (flow multiplied by speed instead of divided)

Reference: FE Reference Handbook — Transportation → Basic Freeway Segment Highway Capacity

Self-check

Answer these without notes before moving on.

Without looking, state the relation on this page whose left-hand side is the quantity most often requested, and name every symbol in it.
Which assumption, if violated, makes the main relation of this section invalid?
Given a vertical or horizontal alignment, or a traffic stream, what is the first quantity you would compute, and why that one first?
Which unit conversion in this subject most often produces a wrong answer choice, and what is its numerical factor?
Rework Example 1 above from the givens alone, without reading the solution lines.

Chapter summary

Basic Freeway Segment Highway Capacity contains 33 relations; you must be able to find this page in under 15 seconds.
Exam style: a curve geometry element or a capacity/flow relationship.
Unit rule: grades as decimals in curve formulas, percent in the stem.
Work the 10 examples until the solution path, not the answer, is automatic.

Common traps in this section

grades as decimals in curve formulas, percent in the stem
Answering the intermediate quantity instead of the quantity requested.
Rounding intermediate values before the final step.
Using a relation from an adjacent handbook section that shares a symbol.
Skipping the sketch — most lost points on this page start with a misread geometry.