Traffic Capacity & LOS

9% of exam

Uninterrupted and interrupted flow: peak-hour factor, flow rate, density and speed relationships, freeway/segment level of service, and signalized-intersection and roundabout capacity.

5 concepts

Exam areas

A. Uninterrupted flow

Flow, Speed, Density & the Peak-Hour Factor

The fundamental q = k·v relation, the Greenshields speed-density line, jam density and capacity, and the peak-hour factor that scales an hourly volume to a 15-minute flow rate.

Three numbers describe any stream of traffic at a point: how many vehicles pass per hour (flow

q

), how fast they move (speed

v

), and how tightly they are packed (density

k

). They are not independent — they are locked together by one identity,

q = k\,v

, and everything downstream in capacity and level-of-service analysis is built on it. The PE Civil Transportation exam tests this relentlessly in the uninterrupted-flow area, and the points that get lost are almost always unit slips (vehicles per hour versus per 15 minutes) or a confusion between the two kinds of average speed. In the NCEES PE Civil Reference Handbook the fundamental relation and the speed-density/flow-density material sit under §5.1.1 'Uninterrupted Flow' (incl. §5.1.1.2 Space Mean Speed and §5.1.1.4 Greenshields Maximum Flow Rate) and §5.1.2 'Street Segment Interrupted Flow' (§5.1.2.1 Speed-Density Model, §5.1.2.2 Flow-Density Model), with the peak-hour factor in §5.1.3 'Traffic Analysis'; this concept makes you fluent in all of it.

The fundamental relation q = k·u

Flow is the product of density and space-mean speed. Read the equation dimensionally and it is obvious:

\left[\tfrac{\text{veh}}{\text{mi}}\right]\times\left[\tfrac{\text{mi}}{\text{h}}\right]=\left[\tfrac{\text{veh}}{\text{h}}\right]

. Density is the snapshot quantity — vehicles per mile at one instant — while flow is the temporal quantity — vehicles per hour past one point. The single most important habit on this topic is to keep these two views separate: a detector measures flow and speed directly, but it must infer density from

k = q/u_s

. The speed in this identity must be the space-mean speed

u_s

, never the time-mean speed.

q = k\,u_s

Space-mean versus time-mean speed

Average speed is ambiguous until you say which average. The time-mean speed

u_t

is the arithmetic mean of the spot speeds of individual vehicles passing a point; the space-mean speed

u_s

is the harmonic mean — the length of a section divided by the average travel time across it. Because slower vehicles spend more time in the section, they are weighted more heavily in the space average, so

u_s \le u_t

always, with the gap growing as speeds scatter. Only

u_s

is consistent with

q = k\,u_s

; using

u_t

there is a classic exam trap.

u_s = \dfrac{n L}{\displaystyle\sum_{i=1}^{n} t_i}, \qquad u_t = \dfrac{1}{n}\sum_{i=1}^{n} u_i, \qquad u_s \le u_t

The Greenshields linear model

Greenshields proposed the simplest realistic closure: speed falls linearly as density rises, from the free-flow speed

u_f

at near-zero density to zero at the jam density

k_j

where vehicles are bumper to bumper. With this one straight line the whole stream is determined. Substituting the line into

q=k\,u_s

gives a parabola in density — flow climbs from zero, peaks, and falls back to zero at jam — which is the shape every fundamental diagram takes.

u_s = u_f\left(1 - \dfrac{k}{k_j}\right)

Capacity from the parabola

Put the Greenshields line into the flow identity,

q = u_f\,k\left(1 - k/k_j\right)

, and maximize: the peak sits exactly halfway along, at the optimum density

k_{cap}=k_j/2

and the optimum speed

u_{cap}=u_f/2

. The peak value is the capacity of the Greenshields stream, and the handbook gives it directly as

q_{max}=u_f k_j/4

. The lesson examinees must internalize is that maximum flow does NOT happen at maximum speed — it happens at half the free-flow speed, where speed and density are balanced.

q_{max} = \dfrac{u_f\,k_j}{4}, \qquad k_{cap} = \dfrac{k_j}{2}, \qquad u_{cap} = \dfrac{u_f}{2}

Two regimes, one diagram

Every flow below capacity can be served at two different densities — one on the free-flow (uncongested) branch to the left of the peak and one on the congested branch to the right. The left branch is the stable operating side: low density, high speed. The right branch is breakdown: high density, crawling speed, the same vehicles per hour but moving in a queue. Knowing which branch you are on is what distinguishes a healthy facility from a jammed one carrying identical flow.

The peak-hour factor

Demand is never uniform within an hour. The peak-hour factor (PHF) measures how peaked it is: it is the hourly volume divided by four times the peak 15-minute count,

PHF = V/(4 V_{15})

. A perfectly uniform hour gives

PHF = 1.00

; real urban peaks run

0.85

0.95

. Design and capacity work uses the flow RATE within the peak 15 minutes, not the hourly average, so you convert by dividing the hourly volume by the PHF:

v = V/PHF

. Forgetting this step under-sizes a facility for the surge it must actually handle.

PHF = \dfrac{V}{4\,V_{15}}, \qquad v = \dfrac{V}{PHF}

Exam strategy

Anchor on

q = k\,u_s

and always write the units beside each symbol before substituting; a detector gives you flow and speed, so density is

k=q/u_s

. For any Greenshields question, immediately note the three anchors

q_{max}=u_f k_j/4

k_{cap}=k_j/2

u_{cap}=u_f/2

, and remember capacity is at half the free-flow speed. When a problem gives a 15-minute count, decide at once whether it wants the hourly volume (

\times 4

) or the equivalent flow rate (

V/PHF

). If a speed is described as a spot-speed average, it is time-mean and must NOT be put into

q=k\,u_s

Key equations

Fundamental relation

q = k\,u_s

Flow $q$ (veh/h) equals density $k$ (veh/mi) times space-mean speed $u_s$ (mph). The backbone of all uninterrupted-flow analysis.

Density from a detector

k = \dfrac{q}{u_s}

Density inferred from measured flow and space-mean speed; vehicles per mile per lane (or per roadway).

Equivalent hourly flow

q = \dfrac{3600\,n}{T}

Converts $n$ vehicles counted in $T$ seconds to an equivalent hourly rate (vph). Same identity scales a 15-min count by 4.

Space-mean speed

u_s = \dfrac{nL}{\sum t_i}

Harmonic-type average: section length $L$ over mean travel time. The only speed valid in $q=k\,u_s$ .

Greenshields speed-density

u_s = u_f\left(1 - \dfrac{k}{k_j}\right)

Linear model; $u_f$ = free-flow speed (mph), $k_j$ = jam density (veh/mi). Speed is zero at jam.

Greenshields flow-density

q = u_f\,k\left(1 - \dfrac{k}{k_j}\right)

Parabola obtained by substituting the speed-density line into $q=k\,u_s$ ; peak gives capacity.

Greenshields capacity

q_{max} = \dfrac{u_f\,k_j}{4}

Maximum flow of the Greenshields stream (vph), occurring at $k_{cap}=k_j/2$ and $u_{cap}=u_f/2$ .

Optimum density and speed

k_{cap} = \dfrac{k_j}{2}, \qquad u_{cap} = \dfrac{u_f}{2}

Density and speed at capacity. Capacity is reached at half the free-flow speed, not at maximum speed.

Peak-hour factor

PHF = \dfrac{V}{4\,V_{15}}

Hourly volume $V$ over four times the peak 15-min count $V_{15}$ . Range 0.25–1.0; uniform hour = 1.0.

Peak flow rate

v = \dfrac{V}{PHF}

Equivalent hourly flow rate during the peak 15 minutes; the value used in capacity and LOS work.

Worked examples

Capacity of a Greenshields stream

Problem. A freeway lane has a free-flow speed of $60\ \text{mph}$ and a jam density of $240\ \text{veh/mi}$ . Using the Greenshields model, find the capacity, the density and speed at capacity, and the flow when the density is $80\ \text{veh/mi}$ .

Solution. Capacity:

q_{max}=u_f k_j/4 = (60)(240)/4 = 3600\ \text{vph}

. At capacity:

k_{cap}=k_j/2 = 120\ \text{veh/mi}

and

u_{cap}=u_f/2 = 30\ \text{mph}

(check:

120\times 30 = 3600\ \text{vph}

, consistent). At

k=80\ \text{veh/mi}

u_s = 60\left(1-\tfrac{80}{240}\right)=60(0.667)=40.0\ \text{mph}

, so

q = k\,u_s = 80\times 40.0 = 3200\ \text{vph}

. Sanity check:

80\ \text{veh/mi}<k_{cap}

, so we are on the uncongested branch and the flow (

3200

) is below capacity (

3600

) — consistent.

q_{max}=\dfrac{(60)(240)}{4}=3600\ \text{vph}

Density from detector data

Problem. A loop detector reports a flow rate of $1500\ \text{vph}$ in a single lane at a space-mean speed of $50\ \text{mph}$ . Find the density, and the average spacing between vehicles.

Solution. Density:

k = q/u_s = 1500/50 = 30.0\ \text{veh/mi}

. Average spacing is the inverse of density:

s = 1/k = 1/30\ \text{mi} = 5280/30 = 176\ \text{ft per vehicle}

. Sanity check:

30\ \text{veh/mi}

50\ \text{mph}

is light, free-flowing traffic, and

176\ \text{ft}

headway is a comfortable gap — consistent with uncongested operation.

k = \dfrac{1500}{50} = 30.0\ \text{veh/mi}

Peak-hour factor and design flow rate

Problem. During the analysis hour a count station records $2900\ \text{veh}$ , with the peak 15-minute count being $820\ \text{veh}$ . Find the peak-hour factor and the equivalent peak flow rate.

Solution. Peak-hour factor:

PHF = V/(4V_{15}) = 2900/(4\times 820) = 2900/3280 = 0.884

. Peak flow rate:

v = V/PHF = 2900/0.884 = 3280\ \text{vph}

(equivalently

4\times 820

, the 15-min count scaled to an hourly rate). Sanity check:

PHF<1

as required, and the flow rate (

3280

) exceeds the hourly volume (

2900

) because demand surges within the hour — consistent.

PHF = \dfrac{2900}{4(820)} = 0.884

Common pitfalls

•Putting time-mean speed into $q = k\,u_s$ . Only space-mean (harmonic) speed is consistent with the fundamental relation; a spot-speed average will understate density (u_t ≥ u_s, so k = q/u_t comes out low).
•Assuming capacity occurs at the free-flow speed. In Greenshields capacity is at HALF the free-flow speed ( $u_{cap}=u_f/2$ ), where the flow-density parabola peaks.
•Confusing volume with flow rate. A 15-min count $\times 4$ gives an hourly rate; the hourly volume must be divided by PHF to recover the peak flow rate — they are not the same number.
•Mixing per-lane and per-roadway density. Greenshields $k_j$ and the resulting capacity are per lane unless stated; multiply by the number of lanes only at the end.
•Reading off the wrong branch of the fundamental diagram. A given sub-capacity flow corresponds to two densities; high density at low speed means you are in the congested (queued) regime, not free flow.
•Forgetting that $PHF\le 1$ always. A computed PHF above 1.0 means the 15-min count or the hourly volume was transcribed wrong.

References

NCEES PE Civil Reference Handbook — §5.1.1 Uninterrupted Flow (§5.1.1.2 Space Mean Speed, §5.1.1.4 Greenshields Maximum Flow Rate, $q_{max}=u_f k_j/4$ )
NCEES PE Civil Reference Handbook — §5.1.2 Street Segment Interrupted Flow (§5.1.2.1 Speed-Density Model, §5.1.2.2 Flow-Density Model)
NCEES PE Civil Reference Handbook — §5.1.3 Traffic Analysis (§5.1.3.3 Peak-Hour Factor)
Garber & Hoel, Traffic and Highway Engineering, 5th ed. — Source of the Greenshields maximum-flow relationship cited by the handbook.

Uninterrupted-Flow Level of Service

Convert a measured volume to an equivalent passenger-car flow rate, divide by speed to get density, and read the A-F level of service for a basic freeway segment.

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B. Street segment interrupted flow

Urban Street Segment Running Time & LOS

Segment running time and travel speed, how signal spacing and control delay erode it, and the travel-speed level of service for an urban street segment.

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C. Intersection capacity

Signalized Intersection Capacity & Delay

Lane-group capacity from saturation flow and green ratio, the volume-to-capacity ratio, and HCM/Webster control delay with its A-F level-of-service grade.

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Roundabout Capacity & Control Delay

The gap-acceptance exponential capacity model driven by circulating flow, the entry volume-to-capacity ratio, and roundabout control delay with its level-of-service grade.

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