Geometric Design for the FE Civil Exam: Horizontal Curves, Vertical Curves and Sight Distance

NCEES FE Civil spec: Transportation Engineering, A. Geometric design (e.g., streets, highways, intersections). FE Reference Handbook: Civil Engineering chapter, Transportation section (Horizontal Curves, Vertical Curves, Stopping Sight Distance).

What the FE Civil Spec Covers

Section A of FE Civil Transportation is geometric design of streets, highways and intersections. In practice that means three families of problems: horizontal circular curves (stationing, tangent length, radius from degree of curve, superelevation), parabolic vertical curves (elevations, high or low points, required length for sight distance) and sight distance itself (stopping and intersection sight distance). Most questions are a single formula plus careful bookkeeping of stations and units.

Key Concepts

Horizontal curve geometry. Two tangents meet at the PI with intersection angle $I$ . The curve starts at the PC and ends at the PT. The tangent length $T$ runs from PC to PI and from PI to PT.
Degree of curve (arc definition) is the central angle subtended by a 100 ft arc, so a sharper curve has a larger $D$ .
Vertical curves are parabolas. The grade changes at a constant rate along the curve, so elevations follow a quadratic in distance from the PVC.
Sight distance controls curve length. A crest curve must be long enough for a driver to see an object in the road; a sag curve must be long enough for headlights to light the road at night.

Core Formulas

Horizontal curves

T = R tan \frac{I}{2} L = \frac{π R I}{18 0 ^{\circ}} D = \frac{5729.58}{R}

E = R (\frac{1}{cos ( I /2 )} - 1) M = R (1 - cos \frac{I}{2}) L C = 2 R sin \frac{I}{2}

Minimum radius for a design speed $V$ in mph, superelevation $e$ and side friction $f$ (both as decimals):

R_{min} = \frac{V ^{2}}{15 ( e + f )}

Vertical curves

y (x) = y_{P V C} + g_{1} x + \frac{g _{2} - g _{1}}{2 L} x^{2} x_{hi g h / l o w} = \frac{g _{1} L}{g _{1} - g _{2}}

Here $g_{1}$ and $g_{2}$ are decimals and $x$ is measured from the PVC. The rate of vertical curvature is $K = L / A$ , with $A$ the algebraic grade difference in percent.

Stopping sight distance

S S D = 1.47 V t + \frac{V ^{2}}{30 ( \frac{a}{32.2} \pm G )}

$V$ in mph, $t$ in seconds, $a$ in ft/s², and $G$ the grade as a decimal (plus for uphill).

Curve length for sight distance

With the standard AASHTO criteria (driver eye 3.5 ft, object 2.0 ft) and the sight distance $S$ shorter than the curve, the minimum lengths are:

L_{cr es t} = \frac{A S ^{2}}{2158} L_{s a g} = \frac{A S ^{2}}{400 + 3.5 S} (S \leq L)

If the result comes out shorter than $S$ , the assumption failed and you must use the $S > L$ form instead. For example, a 60 mph road with $t = 2.5$ s and $a = 11.2$ ft/s² on level ground needs $S S D = 220.5 + 345.0 = 565.5$ ft. A crest curve with $A = 5$ then needs $L = 5 (565.5)^{2} /2158 = 741$ ft, which is longer than 565.5 ft, so the $S \leq L$ formula was the right one.

On a horizontal curve, a building or cut slope on the inside of the curve can block the view. The clear distance needed from the centerline of the inside lane is the horizontal sight line offset:

H S O = R [1 - cos (\frac{28.65 S}{R})]

The angle is in degrees. With $R = 1200$ ft and $S = 565.5$ ft, the offset is about 33 ft. Intersection sight distance for a stopped vehicle entering a major road is $I S D = 1.47 V_{maj or} t_{g}$ , where $t_{g}$ is the time gap the entering driver needs.

Where It Is in the FE Reference Handbook

Search the Transportation section of the Civil Engineering chapter. It has a horizontal curve figure with all the labeled elements, the vertical curve formulas and figure, a table relating crest and sag curve length to sight distance, and the stopping sight distance equation next to signal timing.

Worked Example: Horizontal Curve Stationing

A horizontal curve with radius 1,200 ft joins two tangents that intersect at an angle of 40°. The PI is at station 52+40.00. Find the tangent length, curve length, PC and PT stations, and degree of curve. Then check whether the radius works for a 50 mph design speed with $e = 0.06$ and $f = 0.14$ .

Step 1: tangent length.

T = 1200 tan 2 0^{\circ} = 1200 (0.36397) = 436.76 ft

Step 2: curve length.

L = \frac{π ( 1200 ) ( 4 0 ^{\circ} )}{18 0 ^{\circ}} = 837.76 ft

Step 3: PC station. Back up the tangent length from the PI.

P C = 5240.00 - 436.76 = 4803.24 \Rightarrow 48 + 03.24

Step 4: PT station. Add the curve length, not the tangent length, to the PC.

P T = 4803.24 + 837.76 = 5641.00 \Rightarrow 56 + 41.00

(Carrying full precision gives 56+40.99; either rounding is fine on a multiple-choice question.)

Step 5: degree of curve.

D = \frac{5729.58}{1200} = 4.7 7^{\circ}

Step 6: minimum radius check.

R_{min} = \frac{5 0 ^{2}}{15 ( 0.06 + 0.14 )} = \frac{2500}{3.0} = 833 ft < 1200 ft ✓

A quick vertical curve check on the same idea: with $g_{1} = + 3%$ , $g_{2} = - 2%$ , $L = 600$ ft and PVC elevation 500.00 ft, the high point is at $x = 0.03 (600) /0.05 = 360$ ft, with elevation $500 + 0.03 (360) - \frac{0.05}{1200} (360)^{2} = 505.40$ ft.

Common Mistakes

Computing the PT as PI + T. Stations follow the curve, so PT = PC + L.
Using grades in percent inside the parabola equation. Convert 3% to 0.03 first.
Leaving the calculator in radians when taking tan(I/2), or using I instead of I/2.
Entering superelevation as 6 instead of 0.06 in the minimum radius formula.
Forgetting the 1.47 factor that converts mph to ft/s in the reaction distance term.

Practice Problems

Try these from the free PolyCalc FE Civil problem bank. Each one has a full worked solution.

A vehicle travels at 20 m/s on a level road. Perception-reaction time is 2.5 s and constant braking deceleration is 3 m/s². What is the approximate stopping sight distance? See the worked solution or practice the Transportation deck
Using e + f = V²/(127R), with speed V = 80 km/h, curve radius R = 400 m, and side-friction factor f = 0.10, what superelevation rate e is required as a decimal? See the worked solution or practice the Transportation deck
All else equal, increasing vehicle speed generally causes stopping sight distance to: See the worked solution or practice the Transportation deck
A circular horizontal curve has a deflection angle of 32° between its tangents and a radius of 450 m. The length of the curve is most nearly: See the worked solution or practice the Transportation deck
A parabolic vertical curve is 400 m long with an entry grade of +3% and an exit grade of −2%. The elevation at the beginning of the curve is 240.0 m. The elevation at the end of the curve is most nearly: See the worked solution or practice the Transportation deck

Related Calculators and Examples

Frequently Asked Questions

Is the PT station equal to the PI station plus the tangent length?

No. Stations are measured along the alignment, so the PT station is the PC station plus the curve length L. Adding T to the PI station gives a point on the old tangent, not the PT.

What reaction time and deceleration should I use for stopping sight distance?

Use the values the problem gives. When none are given, the AASHTO design values are a 2.5 second perception-reaction time and a deceleration of 11.2 ft/s squared.

Where is the high point of a crest vertical curve?

At a distance x = g1 L / (g1 - g2) from the PVC, where g1 and g2 are the grades as decimals. It only falls inside the curve when the grades have opposite signs.

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