Water Resources HubFE section 13 of 16 · free theory
Transportation Engineering
Sight distance, curves, traffic flow, and pavement basics — the Transportation slice of the FE Civil exam, with the equations you would actually reach for.
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Stopping sight distance
SSD is the distance a driver needs to see an obstacle, react, and stop. It is the sum of a perception–reaction distance and a braking distance, and it drives the design of crest vertical curves and intersection sight triangles:
SSD = 1.47 V t + V² / [30 (f ± G)]
| SSD | stopping sight distance, ft |
| V | design speed, mph |
| t | perception–reaction time, 2.5 s for design |
| f | coefficient of friction (0.35 typical for SSD design) |
| G | longitudinal grade as a decimal: + upgrade, − downgrade |
The 1.47 converts mph to ft/s (1 mph = 1.47 ft/s). In SI: SSD = 0.278 V t + V² / [254 (f ± G)] with V in km/h and SSD in metres.
Worked example SSD at 55 mph on level ground
Given:
- Design speed V = 55 mph, level grade (G = 0), t = 2.5 s, f = 0.35.
Solution:
- Perception–reaction distance: 1.47 × 55 × 2.5 = 202.1 ft.
- Braking distance: 55² / (30 × 0.35) = 3,025 / 10.5 = 288.1 ft.
- SSD = 202.1 + 288.1 = 490.2 ft.
Answer: SSD ≈ 490 ft.
Horizontal curves
A simple horizontal curve is defined by its radius R and deflection angle Δ. The exam expects you to move fluently between the curve components, and to tie radius to superelevation and side friction:
T = R tan(Δ/2) L = R Δ (Δ in radians) E = R [sec(Δ/2) − 1] M = R [1 − cos(Δ/2)]
| T | tangent length, ft |
| L | curve length, ft |
| E | external distance (PI to curve midpoint), ft |
| M | middle ordinate (curve midpoint to long chord), ft |
| Δ | deflection (central) angle |
Rmin = V² / [15 (e + f)] D = 5,729.58 / R
| e | superelevation rate (decimal) |
| f | side-friction factor |
| D | degree of curve (arc definition), degrees per 100-ft arc |
L = RΔ needs Δ in radians; convert with Δ(rad) = Δ(deg) × π/180. The long chord is C = 2R sin(Δ/2) — a favourite distractor when the question asks for curve length.
Worked example Curve components and minimum radius
Given:
- Curve with R = 1,000 ft and Δ = 60°.
- Design check: V = 50 mph, emax = 6%, f = 0.14.
Solution:
- T = 1,000 × tan(30°) = 577.4 ft. L = 1,000 × (60π/180) = 1,047.2 ft.
- E = 1,000 × (sec 30° − 1) = 1,000 × (1.1547 − 1) = 154.7 ft. M = 1,000 × (1 − cos 30°) = 134.0 ft.
- Minimum radius for 50 mph: Rmin = 50² / [15 × (0.06 + 0.14)] = 2,500 / 3.0 = 833 ft. The 1,000-ft curve is flatter than the minimum, so it is acceptable.
Answer: T ≈ 577 ft, L ≈ 1,047 ft, E ≈ 155 ft, M ≈ 134 ft; Rmin ≈ 833 ft.
Vertical curves
Vertical curves are parabolas joining grade g1 to grade g2 over length L. Crest curves are sized by sight distance; sag curves by headlight sight distance and rider comfort. The rate of vertical curvature K = L/A is the workhorse — it is the horizontal distance needed for a 1% change in grade:
y(x) = yPVC + g1x + (g2 − g1) x² / (2L)
| y(x) | elevation at distance x (ft) from the PVC |
| g1, g2 | approach and departure grades as decimals (3% = 0.03), signed |
| L | curve length, ft |
xhp = g1L / (g1 − g2) K = L / A
| xhp | distance from PVC to the high (crest) or low (sag) point, ft — only meaningful when it falls between 0 and L |
| A | absolute grade change |g1 − g2| in percent |
Crest (SSD): L = AS²/2158 (L ≤ S) or L = 2S − 2158/A (L > S)
Sag: L = AS²/(400 + 3.5S) (L ≤ S) or L = 2S − (400 + 3.5S)/A (L > S)
| S | sight distance (SSD), ft |
Two unit traps in one topic: g1, g2 are decimals in the elevation and high/low-point equations, but A is in percent in K = L/A and the sight-distance curve-length formulas. The crest formulas assume eye height 3.5 ft and object height 2.0 ft.
Worked example Crest curve high point and elevations
Given:
- g1 = +3%, g2 = −2%, L = 600 ft.
- PVC at station 10+00 (1,000 ft), elevation 100.00 ft.
Solution:
- A = |3 − (−2)| = 5%; K = 600/5 = 120 ft per 1% grade change.
- High-point distance from PVC: xhp = 0.03 × 600 / (0.03 − (−0.02)) = 18 / 0.05 = 360 ft → station 13+60. It lies within the curve (0 < 360 < 600), so it is real.
- Elevation there: y = 100.00 + 0.03(360) + (−0.05)(360)²/(2 × 600) = 100.00 + 10.80 − 5.40 = 105.40 ft.
- Elevation at station 12+00 (x = 200 ft): y = 100.00 + 6.00 − 0.05 × 40,000/1,200 = 104.33 ft.
Answer: High point at station 13+60, elevation 105.40 ft; elevation at station 12+00 is 104.33 ft.
Traffic flow, capacity, and level of service
Traffic stream theory rests on one relationship — flow equals density times speed — plus the Greenshields model that gives the parabolic flow–density curve the exam loves:
q = k v v = vf (1 − k/kj) qmax = vf kj / 4
| q | flow, veh/h (per lane) |
| k | density, veh/mi (per lane) |
| v | space-mean speed, mph |
| vf | free-flow speed; kj jam density |
| qmax | capacity, reached at k = kj/2 and v = vf/2 |
c = s (g / C)
| c | lane-group capacity, veh/h |
| s | saturation flow rate, ≈ 1,900 veh/h/ln |
| g | effective green time (G + Y − lost time), s |
| C | cycle length, s |
Level of service runs A (free flow) through F (forced flow / breakdown). Capacity is the maximum of the q–k curve; demand above capacity produces LOS F and growing queues. For signals, remember the concept: cycle = green + yellow + red for each phase, and capacity scales with the fraction of the cycle that is effectively green.
Pavement basics
Pavement questions on the FE are about damage and thickness concepts, not detailed design. Two ideas carry the section:
LEF ≈ (P / 18)4 SN = a1D1 + a2D2m2 + a3D3m3
| LEF | load equivalency factor — damage of one axle relative to an 18-kip single axle |
| P | axle load, kips |
| SN | structural number (flexible pavement) |
| ai | layer coefficient; Di layer thickness, in; mi drainage coefficient |
The fourth-power law means doubling an axle load multiplies pavement damage by about 16 — that is why trucks, not cars, control pavement design. ESALs (equivalent single-axle loads) accumulate LEF × axle passes over the design life. Flexible pavements distribute load through layers (SN); rigid pavements (concrete slabs) carry load by slab bending and are characterised by thickness and modulus of rupture rather than SN.