Transportation Learning HubTransportation Learning Hub · PE Civil Transportation · Area 2 of 10
Traffic Engineering — PE Civil Transportation
The largest area of the PE Civil Transportation exam, and the most varied: how traffic streams behave, how much a road or junction can carry and how well it performs, how demand is measured and forecast, and how safety is analysed. The calculations rest on a handful of identities — the art is knowing which quantity the question is really asking for.
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Traffic flow fundamentals
Three stream quantities describe traffic at a point or along a length: flow \(q\) (vehicles passing per hour), density \(k\) (vehicles per mile of lane), and space-mean speed \(v\). They are bound by one identity, \(q = k v\), and by their mirror-image pairs — headway is the time between vehicles (the inverse of flow) and spacing is the distance between vehicles (the inverse of density). As density rises from free flow, speed falls; flow climbs to a maximum — the capacity — at an intermediate density, then collapses as the stream approaches jam density.
\[q = k\,v \qquad h = \frac{1}{q} \qquad s = \frac{1}{k}\]
| q | flow rate, veh/h |
|---|---|
| k | density, veh/mi |
| v | space-mean speed, mph (the speed consistent with \(q = kv\); not the time-mean spot speed) |
| h, s | average headway (time per vehicle) and average spacing (distance per vehicle) |
Keep the pairs straight: headway goes with flow (time), spacing with density (distance). Swapping them inverts the relationship and produces the planted wrong answer.
Worked example Flow from density and speed
Given: A lane carries traffic at a density of 28 veh/mi and a space-mean speed of 45 mph. Find the flow rate.
Solution:
- Apply the stream identity: \(q = k v = 28 \times 45\).
- \(q = 1{,}260\) veh/h.
Answer: \(q = 1{,}260\) veh/h per lane.
Capacity and level of service
Capacity is the maximum flow a facility can serve under stated conditions; level of service (LOS) grades how it feels to use, from A (free flow) to F (forced, breakdown conditions). The analysis splits by facility type. Uninterrupted flow — freeways and multilane highways — is judged mainly on density and speed. Interrupted flow — street segments and intersections, where signals, signs and turning traffic break the stream — is judged on control delay and travel speed: running time between stops plus stopped delay gives the segment travel time, and travel speed is length divided by that total. At intersections, capacity is assessed movement by movement — signalised approaches, roundabout entries and interchange ramps each have their own method in the Highway Capacity Manual, but the exam logic is uniform: estimate the demand flow rate, estimate the capacity under the given conditions, and compare the two.
\[\text{Travel speed} = \frac{\text{segment length}}{\text{running time} + \text{control delay}}\]
| LOS A–F | operating-quality grades; the measure behind the grade depends on the facility (density, delay, or speed) |
|---|---|
| Demand vs capacity | demand at or above capacity produces LOS F behaviour — queues grow rather than stabilise |
An LOS answer is meaningless without its measure: density for freeways, delay for signalised intersections, travel speed for urban streets. Questions that mix measures are testing exactly that.
Traffic studies, PHF and forecasting
Design almost never uses a raw hourly count, because demand inside the hour is uneven. The peak hour factor measures that unevenness, and the peak flow rate \(V/\text{PHF}\) is the rate the design must serve. Around that core, the analysis subtopic covers volume and speed studies (counts, spot speeds, travel-time runs), trip generation (trips produced by and attracted to a land use), modal split (how trips divide among modes), traffic impact studies for new development, and forecasting — projecting counted demand forward with growth and development assumptions, then monitoring whether reality tracks the forecast.
\[\text{PHF} = \frac{V}{4\,V_{15}} \qquad \text{peak flow rate} = \frac{V}{\text{PHF}} = 4\,V_{15}\]
| V | hourly volume, veh/h |
|---|---|
| V15 | volume in the peak 15 minutes of that hour |
| PHF | 1.00 if flow is perfectly uniform across the hour; lower as the peak sharpens |
Worked example Peak hour factor and peak flow rate
Given: An approach carries 1,840 veh in the peak hour; the heaviest 15 minutes inside that hour carries 540 veh. Find the PHF and the peak flow rate.
Solution:
- \(\text{PHF} = \dfrac{1{,}840}{4 \times 540} = \dfrac{1{,}840}{2{,}160} = 0.852\).
- Peak flow rate: \(\dfrac{V}{\text{PHF}} = \dfrac{1{,}840}{0.852} = 2{,}160\) veh/h — the same as \(4 \times 540\), as it must be.
Answer: PHF \(\approx 0.852\); peak flow rate \(\approx 2{,}160\) veh/h.
Safety analysis
Safety work compares locations on exposure, not raw crash counts: a long, busy segment should record more crashes than a short, quiet one without being more dangerous. The standard yardstick is the crash rate per 100 million vehicle-miles, which divides crashes by the travel that produced them. Around it sit the qualitative tools — collision diagrams that plot crash types at a junction to expose patterns, conflict analysis that watches near-misses where crash data are thin — and the Highway Safety Manual framework, in which a crash modification factor (CMF) scales the expected crash frequency when a treatment is applied: a CMF below 1.0 predicts fewer crashes, above 1.0 predicts more. Pedestrian and bicycle facilities are analysed the same way — exposure, conflict points, and the quality of the facility provided.
\[R = \frac{A \times 10^{8}}{365 \times \text{ADT} \times N \times L}\]
| R | crash rate, crashes per 100 million vehicle-miles |
|---|---|
| A | crashes in the study period |
| ADT, N, L | average daily traffic (veh/day), study period (years), segment length (mi) |
Worked example Crash rate for a highway segment
Given: A 2.5-mi segment with ADT 12,000 records 18 crashes over 3 years. Find the crash rate per 100 million vehicle-miles.
Solution:
- Exposure: \(365 \times 12{,}000 \times 3 \times 2.5 = 32{,}850{,}000\) vehicle-miles.
- \(R = \dfrac{18 \times 10^{8}}{32{,}850{,}000} = 54.8\).
Answer: \(R \approx 54.8\) crashes per 100 million vehicle-miles.