Transportation Learning HubTransportation Learning Hub · PE Civil Transportation · Area 4 of 10
Horizontal Design — PE Civil Transportation
Horizontal design is the geometry of the road in plan view: the circular curves that join the tangents, the stations that locate them, the superelevation that banks them, and the sight distance that must survive inside them.
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Circular curve elements and stationing
A simple curve is fixed by its radius R and central (deflection) angle Δ. The PC is where the curve leaves the back tangent, the PT where it rejoins the forward tangent, and the PI is the intersection of the two tangents. Everything else is trigonometry on that one triangle-and-sector figure:
\[T = R \tan\left(\frac{\Delta}{2}\right) \qquad L = R\Delta \qquad C = 2R \sin\left(\frac{\Delta}{2}\right)\] \[E = R\left[\sec\left(\frac{\Delta}{2}\right) - 1\right] \qquad M = R\left[1 - \cos\left(\frac{\Delta}{2}\right)\right] \qquad D = \frac{5729.58}{R}\]
| T | tangent length (PC to PI, or PI to PT), ft |
|---|---|
| L | curve length along the arc, ft — needs Δ in radians |
| C | long chord (PC to PT straight line), ft |
| E | external distance, PI to the curve midpoint, ft |
| M | middle ordinate, curve midpoint to the long chord, ft |
| D | degree of curve (arc definition), degrees per 100-ft arc |
Stations are distances along the alignment in 100-ft units (sta 38+71.6 = 3,871.6 ft). The PT station is found through the curve — PC station + L — never by adding T to the PI station.
Worked example Curve elements and stationing
Given: a curve with \(R = 1{,}200\) ft and \(\Delta = 35^{\circ}\), with the PI at station 42+50.00. Find the tangent length, curve length, and the PC and PT stations.
Solution:
- Tangent: \(T = 1200 \tan(17.5^{\circ}) = 1200 \times 0.31530 = 378.4\) ft.
- Length: convert first, \(\Delta = 35^{\circ} \times \pi/180 = 0.61087\) rad, so \(L = 1200 \times 0.61087 = 733.0\) ft.
- PC station \(= 4250.00 - 378.4 = 3871.6\) ft → sta 38+71.6.
- PT station \(= 3871.6 + 733.0 = 4604.7\) ft → sta 46+04.7.
Answer: \(T \approx 378.4\) ft; \(L \approx 733.0\) ft; PC at sta 38+71.6; PT at sta 46+04.7.
Sight distance on horizontal curves
On the inside of a horizontal curve the driver’s sightline is a chord cutting across the curve while the road follows the arc — so any obstruction on the inside (a wall, a building, a cut slope, vegetation) can steal the stopping sight distance even when the vertical profile is fine. The geometry is the middle-ordinate idea turned sideways: the closer the obstruction sits to the travelled path, the shorter the visible chord. Exam questions in this subtopic ask you to recognise the controlling case (inside of curve, sightline as chord, clearance measured from the lane path) and to connect it to the curve elements above rather than to reach for a new formula family.
Superelevation
Superelevation banks the pavement so that part of the cornering demand is carried by the tilt of the road itself and the rest by side friction:
\[e + f = \frac{V^{2}}{15R}\]
| e | superelevation rate, decimal (0.04 = 4%) |
|---|---|
| f | side-friction factor (given in exam problems) |
| V | design speed, mph |
| R | curve radius, ft |
The full rate is not developed instantly at the PC: the superelevation runoff is the length over which the pavement rotates from normal crown to full e, and the tangent runout is the preceding length over which the adverse crown is removed. Designs distribute e and f between the point of zero superelevation demand and the maximum rate so that most of the runoff sits on the tangent, where steering is easiest.
Worked example Required superelevation rate
Given: design speed \(V = 55\) mph, radius \(R = 1{,}200\) ft, and a side-friction factor \(f = 0.13\) given for the check. Find the required superelevation rate.
Solution:
- Total demand: \(e + f = \dfrac{55^{2}}{15 \times 1200} = \dfrac{3025}{18{,}000} = 0.1681\).
- Superelevation: \(e = 0.1681 - 0.13 = 0.0381 \approx 0.038\).
Answer: \(e \approx 0.038\), i.e. about 3.8%.
Special curves, widening and coordination
Compound curves join two arcs of different radius turning the same way — the flatter curve goes first, so the driver meets the sharper radius already slowed and settled. Reverse curves turn opposite ways back to back; they need a length of tangent (or a designed transition) between them so superelevation can roll from one side to the other, and they are avoided where speeds are high. Widening compensates for off-tracking: the rear wheels of a long vehicle track inside the front wheels on a curve, sweeping extra width, so lanes on sharp curves are widened by an amount that grows with vehicle length and shrinks with radius. Finally, coordination: horizontal and vertical geometry are designed together — a sharp horizontal curve should not be hidden just over a crest (see Vertical Design), and superelevation transitions should not be placed where the vertical profile makes them drain badly.