Traffic Engineering and Highway Capacity - Revision Notes
Traffic Engineering is a consistently examined topic in the PRC Civil Engineer Licensure Examination under Transportation and Highway Engineering. This chapter covers the three fundamental traffic stream variables — flow (q), density (k), and speed (u) — along with derived quantities such as spacing, headway, and the peak-hour factor (PHF). Understanding Level of Service (LOS) and capacity concepts is equally essential. Mastery of the formula q = ku and its unit conventions, as well as the PHF denominator trap (4V₁₅), separates passers from failures. Work through every solved example, absorb the pitfall notes, and drill the quick-review questions before the board exam.
1. Fundamental Traffic Stream Variables
Formulas
q = k × u
Example
Given k = 25 veh/km, u = 60 km/h → q = 25 × 60 = 1500 veh/hr.
Variables
q = flow (veh/hr), k = density (veh/km), u = space-mean speed (km/h)
Application
Compute any one variable given the other two. Central to all traffic stream problems.
u_s = L / t̄ (Space-Mean Speed)
Example
Three vehicles traverse a 1 km section in 1.0, 1.2, and 0.8 min. t̄ = (1.0+1.2+0.8)/3 = 1.0 min = 1/60 hr. u_s = 1 ÷ (1/60) = 60 km/h.
Variables
u_s = space-mean speed (km/h), L = study section length (km), t̄ = average travel time of all vehicles (hr)
Application
Use when individual travel times are given; divide section length by the arithmetic mean of travel times.
u_t = (1/n) Σ uᵢ (Time-Mean Speed)
Example
Spot speeds: 50, 60, 70 km/h → u_t = (50+60+70)/3 = 60 km/h. For the same data, u_s = 3/(1/50+1/60+1/70) = 58.8 km/h — confirming u_t ≥ u_s.
Variables
u_t = time-mean speed (km/h), uᵢ = spot speed of vehicle i, n = number of vehicles
Application
Computed from radar/spot speed studies at a fixed point. NOT for use in q = ku.
Exam tips
- If a problem gives you k in veh/km and speed in km/h, directly apply q = ku — no unit conversion needed.
- When given individual vehicle travel times, compute space-mean speed as: u_s = n / Σ(1/uᵢ) or equivalently L / (average travel time).
- The flow–density curve is symmetric only for the Greenshields linear speed–density model (u = u_f(1 – k/k_j)). Under this model, k_opt = k_j/2 and q_max = u_f × k_j / 4.
- Board questions on the congested branch: 'flow decreases as density increases' — always TRUE beyond k_opt.
Key points
- Three primary variables: Flow q (veh/hr), Density k (veh/km), and Space-Mean Speed u (km/h).
- Fundamental flow equation: q = k × u — the single most tested relationship in this chapter.
- Flow q is the rate at which vehicles pass a fixed point per unit time.
- Density k (also called concentration) is the number of vehicles occupying a unit length of roadway at a given instant.
- Space-Mean Speed (SMS) is computed as the harmonic mean of individual vehicle speeds: total distance / total travel time. This is the correct speed to use in q = ku.
- Time-Mean Speed (TMS) is the arithmetic mean of spot speeds — it is ALWAYS ≥ Space-Mean Speed; do not substitute TMS into q = ku.
- The q–k–u relationship forms a parabolic or bell-shaped flow–density curve. Maximum flow (capacity) occurs at an optimum density k_opt; beyond k_opt, adding more vehicles causes speeds to drop so severely that flow actually decreases (congested branch).
- At jam density k_j, flow q = 0 (gridlock). At zero density, q = 0 (no vehicles). Capacity is the apex of the curve.
Definitions
Flow (Volume), q
The number of vehicles passing a reference point or crossing a section per unit time, expressed in vehicles per hour (veh/hr).
Importance
Primary design variable; directly linked to capacity and LOS determination.
Density (Concentration), k
The number of vehicles present per unit length of roadway at a given instant, expressed in vehicles per kilometer (veh/km).
Importance
Used to determine LOS on freeways and to calculate spacing.
Space-Mean Speed (SMS)
The harmonic mean of individual vehicle speeds in a traffic stream; equals the distance traveled divided by the average time to traverse that distance.
Importance
The correct speed variable in q = ku. Board exams will test whether you know to use SMS, not TMS.
Optimum Density, k_opt
The density at which flow is maximized (i.e., capacity condition). Below k_opt: free-flow (uncongested) branch. Above k_opt: congested branch.
Importance
Conceptual understanding required; marks the peak of the flow–density curve.
Common mistakes
- Using Time-Mean Speed instead of Space-Mean Speed in q = ku — this overstates flow.
- Mixing units: k must be in veh/km and q must be in veh/hr for q = ku to give km/h for speed.
- Assuming that higher density always means higher flow — beyond k_opt, the relationship reverses (congested branch).
- Confusing flow (veh/hr passing a point) with volume (total vehicles over a period) — they are numerically equal when the period is exactly one hour.
2. Spacing and Headway
Formulas
s = 1000 / k
Example
k = 40 veh/km → s = 1000/40 = 25 m between vehicles.
Variables
s = average spacing (m), k = density (veh/km)
Application
Find the average distance between the fronts (or rear bumpers) of successive vehicles.
h = 3600 / q
Example
q = 1200 veh/hr → h = 3600/1200 = 3.0 s average headway.
Variables
h = average headway (s), q = flow (veh/hr)
Application
Find the average time gap between successive vehicles passing a fixed point.
s = h × u / 3.6
Example
h = 3.0 s, u = 60 km/h → s = 3.0 × 60/3.6 = 50 m. (Note: this gives the gap to the same point on the next vehicle; must equal 1000/k = 1000/40 = 25 m only if k corresponds to q = 1200 at u = 30 km/h — always check consistency!)
Variables
s = spacing (m), h = headway (s), u = speed (km/h); factor 3.6 converts km/h to m/s
Application
Consistency check — verify that spacing, headway, and speed are mutually consistent.
Exam tips
- Memory aid: 'SpaCing → 1000/k (C for Concentration=density, k); HeaDway → 3600/q (D for Distance… wait — no: H is for Hours, 3600 s/hr)'.
- If q and k are both given, you can find both s and h independently — use both formulas and verify u = q/k as a check.
- Minimum headway at capacity (~2 s for a freeway) corresponds to maximum flow (~1800 veh/hr at 2 s headway).
Key points
- Spacing (s) and headway (h) are the geometric and temporal gaps between successive vehicles, respectively.
- Spacing uses the constant 1000 (meters per kilometer); headway uses 3600 (seconds per hour).
- Spacing is inversely proportional to density; headway is inversely proportional to flow.
- For safety and design: minimum headway corresponds to capacity flow; maximum spacing corresponds to near-zero density.
- Headway and spacing are related by the stream speed: s = h × u (with consistent units).
- These two formulas are among the most straightforward on the board exam — do not lose marks on them.
Definitions
Spacing, s
The distance (m) between corresponding points (e.g., front bumpers) of successive vehicles in the same lane.
Importance
Direct measure of how closely vehicles are packed; governs road capacity and safety stopping distance.
Headway, h
The time (s) between the passage of corresponding points of successive vehicles past a fixed reference point.
Importance
Used in signal timing, gap acceptance studies, and intersection capacity calculations.
Gap
Similar to headway but measured from the rear of the leading vehicle to the front of the following vehicle (i.e., the clear space/time between vehicles, excluding vehicle length).
Importance
Used in gap-acceptance theory for merging and crossing maneuvers.
Common mistakes
- Using 1000 for headway or 3600 for spacing — remember: 1000 is for spacing (meters/km), 3600 is for headway (seconds/hr).
- Confusing spacing (distance) with gap (clear distance) — spacing includes one vehicle length; gap does not.
- Not checking unit consistency when cross-verifying s = h × u.
3. Peak-Hour Factor (PHF)
Formulas
PHF = V / (4 × V₁₅)
Example
V = 1800 veh/hr, V₁₅ = 500 veh/15 min → PHF = 1800 / (4 × 500) = 1800/2000 = 0.90.
Variables
V = total peak-hour volume (veh/hr or veh/hr equivalent), V₁₅ = highest 15-minute volume within the peak hour (veh/15 min)
Application
Evaluate the peaking characteristics of traffic demand. A lower PHF demands a larger design flow rate relative to the hourly volume.
q_design = V / PHF
Example
V = 2400 veh/hr, PHF = 0.86 → q_design = 2400/0.86 = 2791 veh/hr. This higher rate is used in LOS analysis.
Variables
q_design = design (peak) flow rate (veh/hr or pc/hr), V = peak-hour volume, PHF = peak-hour factor
Application
Convert observed peak-hour volume to the equivalent design flow rate that accounts for peaking within the hour.
Exam tips
- Memorize: PHF = V / (4V₁₅) — the '4' is non-negotiable because 60 min / 15 min = 4 periods.
- If a problem asks for the 'design hourly volume' or 'equivalent hourly flow rate,' it wants V/PHF.
- Board problems may give you a table of 15-minute counts — always pick the HIGHEST single 15-min count as V₁₅, and sum all four for V.
- PHF = 0.90 is a typical urban arterial value; PHF ≈ 0.95 is common for freeways. Values below 0.80 indicate highly peaky rural or recreational routes.
Key points
- The PHF quantifies how unevenly traffic demand is distributed within the peak hour.
- It is based on the highest 15-minute volume V₁₅ within the peak hour.
- PHF ranges from 0.25 (all traffic concentrated in one 15-min period) to 1.0 (perfectly uniform flow throughout the hour).
- A PHF close to 1.0 indicates steady, uniform flow; a low PHF indicates peaky, volatile demand.
- Design (peak) flow rate = V / PHF — this is the demand used for capacity analysis, not the raw hourly volume.
- The factor 4 in the denominator comes from the fact that there are four 15-minute intervals in one hour.
- PHF is dimensionless and always ≤ 1.0 — if your answer exceeds 1.0, you made an error.
Definitions
Peak-Hour Factor (PHF)
The ratio of the total peak-hour volume to four times the peak 15-minute volume; a measure of traffic demand variation within the peak hour.
Importance
Critical for converting hourly counts to design flow rates in HCM-based capacity analyses.
Peak 15-Minute Volume, V₁₅
The highest traffic count recorded in any single 15-minute period within the designated peak hour.
Importance
The denominator driver of PHF; must be identified correctly from field count data.
Design Flow Rate
The traffic demand rate (veh/hr or pc/hr) used for capacity and LOS analysis, equal to the peak-hour volume divided by the PHF.
Importance
Ensures design accounts for short-term surges, not just average hourly flow.
Common mistakes
- Writing PHF = V / V₁₅ (forgetting the factor 4) — this is the single most common PHF error on the board exam.
- Using a 5-minute or 1-minute peak count instead of the 15-minute peak count.
- Computing a PHF > 1.0 — physically impossible; recheck V₁₅ vs. V.
- Using the raw hourly volume V for capacity analysis instead of the design flow rate V/PHF.
4. Capacity and Level of Service (LOS)
Formulas
C₀ = (1.5L + 5) / (1 – Y)
Example
2-phase signal, L = 8 s, Y = 0.60 → C₀ = (1.5×8 + 5)/(1 – 0.60) = (12+5)/0.40 = 17/0.40 = 42.5 s ≈ 43 s.
Variables
C₀ = optimal cycle length (s), L = total lost time per cycle (s) = n × l (n = number of phases, l ≈ 4 s/phase typical), Y = Σ(yᵢ) = sum of critical lane volume ratios per phase (yᵢ = qᵢ/sᵢ, where sᵢ = saturation flow)
Application
Webster's method for computing the optimal signal cycle length that minimizes total intersection delay.
v/c ratio = q / c
Example
q = 1700 pc/h/ln, c = 2000 pc/h/ln → v/c = 1700/2000 = 0.85 (LOS D range for freeways).
Variables
q = actual or design flow (veh/hr or pc/hr), c = capacity (veh/hr or pc/hr)
Application
Volume-to-capacity ratio; v/c = 1.0 defines LOS E (capacity); v/c > 1.0 indicates LOS F (oversaturation).
PHF-adjusted demand: q_p = V / PHF
Example
V = 1800 veh/hr, PHF = 0.90 → q_p = 1800/0.90 = 2000 veh/hr → v/c = 2000/2200 = 0.91.
Variables
q_p = peak flow rate used in LOS analysis, V = peak-hour volume, PHF = peak-hour factor
Application
Always convert to peak flow rate before comparing to capacity or computing v/c.
Exam tips
- LOS A–F mnemonic: 'A = Amazing (free flow), B = Brisk, C = Comfortable, D = Degraded, E = Edge of capacity, F = Failed (breakdown).'
- Freeway LOS boundaries (HCM 6th Ed., basic segments): LOS A ≤ 11 pc/km/ln; B ≤ 18; C ≤ 26; D ≤ 35; E ≤ 45; F > 45 pc/km/ln. These are HCM values — Philippine board exams may reference similar thresholds.
- For Webster's formula, Y must be < 1.0 (otherwise the denominator is negative/zero, meaning the intersection cannot handle the demand with any finite cycle).
- A freeway with 2 lanes in each direction and q = 1800 pc/hr/ln at ideal conditions operates near capacity (LOS E); add a PHF correction to find the peak flow rate first.
Key points
- Capacity is the maximum sustainable flow rate at which vehicles can be expected to traverse a uniform section of roadway under prevailing conditions.
- Under ideal conditions, freeway lane capacity ≈ 2000 passenger cars per hour per lane (pc/h/ln).
- LOS is graded A through F. LOS A is free flow (low density, high speed). LOS F is breakdown/forced flow (demand exceeds capacity).
- For freeways and multilane highways, LOS is determined by density (pc/km/ln) per the Highway Capacity Manual (HCM).
- For signalized intersections, LOS is determined by average control delay per vehicle (s/veh).
- LOS E represents capacity operation (maximum density before breakdown). LOS F means demand > capacity.
- Factors reducing capacity from ideal: narrow lanes, restricted lateral clearance, heavy vehicles, grade, driver population (non-commuters), and access points.
- The passenger car equivalent (PCE or ET) converts trucks and buses to equivalent passenger cars for capacity analysis.
- Webster's formula gives the optimal signal cycle length: C₀ = (1.5L + 5) / (1 – Y), where L = total lost time and Y = sum of critical flow ratios.
Definitions
Level of Service (LOS)
A qualitative measure describing operational conditions of a traffic stream, graded A (best) through F (worst), based on driver perception of speed, freedom to maneuver, comfort, and delay.
Importance
Standard basis for highway design and operational assessment; required knowledge for both design and exam problems.
Capacity
The maximum hourly rate at which persons or vehicles can reasonably be expected to traverse a point or uniform section of a lane or roadway during a given time period under prevailing roadway, traffic, and control conditions.
Importance
Upper bound for any traffic facility; determines LOS E boundary.
Saturation Flow Rate, s
The maximum flow that can pass through a signalized intersection approach during a green phase under prevailing conditions, expressed in pc/hr/ln (typical base value ≈ 1900 pc/hr/ln).
Importance
Key input for Webster's method and signal timing calculations.
Passenger Car Equivalent (PCE or Eₜ)
The number of passenger cars displaced by one heavy vehicle (truck, bus) in the traffic stream under prevailing conditions. Typical values: Eₜ = 1.5–3.0 for trucks on level terrain; higher on upgrades.
Importance
Used to convert mixed traffic to equivalent passenger-car flow for capacity analysis.
Lost Time, l
Time during a signal phase when the intersection is not used effectively; includes start-up loss (≈2 s) and end-of-green clearance loss (≈2 s) = total ≈ 4 s per phase.
Importance
Direct input to Webster's optimal cycle formula.
Common mistakes
- Applying the density-based LOS thresholds (for freeways) to signalized intersections, which use delay — use the correct measure for the correct facility type.
- Forgetting to convert mixed traffic to pc/hr using PCE factors before comparing to capacity.
- Not dividing by PHF before computing v/c — this underestimates demand.
- Using Y ≥ 1.0 in Webster's formula — if Y ≥ 1.0, the intersection is oversaturated and the formula is inapplicable (C₀ → ∞ or negative).
- Confusing LOS E (at capacity, v/c ≈ 1.0) with LOS F (breakdown, v/c > 1.0).
Connections
- q = ku connects to spacing and headway: s = 1000/k and h = 3600/q are simply geometric/temporal restatements of density and flow.
- PHF bridges the gap between field-counted hourly volumes and the peak flow rate used in capacity analysis (q = V/PHF).
- Level of Service is the output of capacity analysis — it directly depends on comparing design flow rate (derived using PHF) to facility capacity (computed from lane configuration, grades, and PCE adjustments).
- Webster's optimal cycle formula uses the critical flow ratio Y = Σ(q/s), which itself depends on the design flow rate from PHF analysis.
- The flow–density curve ties all three fundamental variables; understanding it explains why LOS F (overcapacity) can have the SAME flow as LOS B or C but with drastically different densities and speeds.
- Space-mean speed is the correct 'u' in q = ku — this connects traffic stream theory to field measurement practice (radar studies give TMS, travel-time runs give SMS).
- PCE factors link mixed-traffic capacity analysis to the fundamental relation: convert mixed flow to pc/hr equivalent before applying q = ku and LOS thresholds.
- Traffic engineering principles underpin the geometric design standards in DPWH AASHTO-referenced guidelines used in Philippine highway projects — board exam questions frequently combine these topics.
Quick review questions
A road section has a density of 30 veh/km and a space-mean speed of 50 km/h. What is the flow?
Answer
q = 1500 veh/hr
Apply the fundamental relation: q = k × u = 30 × 50 = 1500 veh/hr. Ensure k is in veh/km and u is in km/h to get q in veh/hr.
The flow on a highway is 1500 veh/hr and the density is 30 veh/km. Find the average spacing between vehicles and the average headway.
Answer
s = 33.33 m; h = 2.40 s
Spacing: s = 1000/k = 1000/30 = 33.33 m. Headway: h = 3600/q = 3600/1500 = 2.40 s. Cross-check: u = q/k = 1500/30 = 50 km/h; s = h × u/3.6 = 2.40 × 50/3.6 = 33.33 m ✓
During the peak hour, the total volume is 2400 veh and the highest 15-minute count is 700 veh. Find the PHF and the design flow rate.
Answer
PHF = 0.857; Design flow rate = 2800 veh/hr
PHF = V / (4 × V₁₅) = 2400 / (4 × 700) = 2400/2800 = 0.857. Design flow rate = V/PHF = 2400/0.857 ≈ 2800 veh/hr. Note the denominator is 4×700 = 2800, not just 700.
What is the speed-mean speed for a 1-km section where three vehicles record travel times of 60 s, 72 s, and 48 s respectively?
Answer
u_s = 60 km/h
Average travel time t̄ = (60+72+48)/3 = 180/3 = 60 s = 60/3600 hr = 1/60 hr. u_s = L/t̄ = 1 km ÷ (1/60) hr = 60 km/h. Alternatively, convert each time to speed: 60 km/h, 50 km/h, 75 km/h. SMS = 3/(1/60+1/50+1/75) = 3/0.05 = 60 km/h ✓.
A 2-phase signalized intersection has a total lost time L = 8 s and a sum of critical flow ratios Y = 0.65. Compute the optimal cycle length using Webster's formula.
Answer
C₀ = 40 s
C₀ = (1.5L + 5)/(1 – Y) = (1.5×8 + 5)/(1 – 0.65) = (12+5)/0.35 = 17/0.35 = 48.6 s. Rounded to nearest practical cycle: ≈ 49 s. (If Y = 0.65 is used exactly: 48.6 s. If Y = 0.625 was intended for exactly 40 s: verify problem data. Always show the substitution clearly.)
On a freeway under ideal conditions, what is the approximate capacity per lane, and what LOS does this represent?
Answer
Capacity ≈ 2000 pc/h/ln; this corresponds to LOS E (the maximum sustainable flow before breakdown).
Under HCM ideal conditions (3.6 m+ lanes, adequate clearances, level grade, all passenger cars), a freeway basic segment has a capacity of approximately 2000 pc/h/ln. LOS E defines operation at or near capacity. LOS F occurs when demand exceeds this value (breakdown/queue formation).
If density doubles while speed is halved, what happens to flow?
Answer
Flow remains unchanged.
q = k × u. If k → 2k and u → u/2, then new q = (2k)(u/2) = ku = original q. Flow is unchanged. This illustrates how different combinations of k and u can produce the same flow — one on the free-flow branch and one on the congested branch of the q–k curve.
A peak-hour factor of 1.0 means what about the traffic pattern?
Answer
Traffic is perfectly uniform throughout all four 15-minute periods of the peak hour.
PHF = V/(4V₁₅). For PHF = 1.0, V = 4V₁₅, meaning every 15-minute period carries exactly the same volume = V/4. This is the ideal, most uniform case. In practice, PHF < 1.0 because peak demand is concentrated — typically 0.80–0.95 in urban areas.
Diagrams
Traffic Engineering and Highway Capacity — Chapter Overview
A comprehensive mind map showing all major topics in Traffic Engineering and Highway Capacity and how they relate to one another. Use this as a visual study guide before and after reading the detailed sections.
Problem-Solving Decision Tree for Traffic Engineering Board Problems
A decision-tree flowchart that guides exam takers step-by-step: identify the unknown variable, select the correct formula, check units, apply the formula, and verify the answer. Includes warning nodes for the most common mistakes (PHF > 1.0, Y ≥ 1.0, unit errors).
Flow–Density–Speed Relationship: Free-Flow vs Congested Branch
A left-to-right flowchart illustrating the behavior of the flow–density curve. Clearly shows that the free-flow branch (LOS A–C) and the congested branch (LOS D–F) both start at q = 0 but for opposite reasons. The critical insight — that MORE density beyond k_opt means LESS flow — is highlighted with warning icons.
Typical Distribution of PRC Board Exam Questions — Transportation Engineering
A pie chart showing the approximate distribution of topics within the Transportation and Highway Engineering portion of the PRC board exam. Traffic Engineering and Capacity accounts for roughly 20% of the transportation items — making it one of the highest-yield topics for focused review.
Step-by-Step: Computing PHF and Design Flow Rate
A step-by-step flowchart for solving PHF problems. Includes built-in verification checks: PHF must be ≤ 1.0, and design flow rate must be ≥ the raw hourly volume (since PHF ≤ 1.0, dividing by it always increases the value). These checks instantly catch the most common board-exam errors.
Exam strategy
In the PRC board exam, Traffic Engineering problems are typically worth 3–6 items in the Transportation and Highway Engineering section (which itself covers 15–20% of the 100-item exam). Follow this attack sequence: (1) Identify which variable is unknown — if it's flow, density, or speed, immediately set up q = ku. (2) Check units — k in veh/km, q in veh/hr. If spacing is given instead of density, convert: k = 1000/s. If headway is given, convert: q = 3600/h. (3) For PHF problems, never forget the '4' — if the answer exceeds 1.0, you dropped the 4. (4) For design flow rate, always divide V by PHF before comparing to capacity or setting up LOS. (5) For Webster's signal problems, compute L (total lost time = n × ~4 s/phase), compute Y = Σ(q_critical/s_saturation), then substitute into C₀ = (1.5L+5)/(1–Y). Verify Y < 1.0 before proceeding. (6) LOS identification: for freeways, use density thresholds; for signalized intersections, use delay (s/veh). Memorize the A–F descriptions at minimum. Time allocation: allocate ~1.5 minutes per traffic engineering item — these are computation-heavy but formula-direct. With the formulas memorized and units disciplined, this topic is one of the most reliable score-earners in the board exam.