Transportation Engineering

9% of exam

Horizontal and vertical geometric design, pavement design, traffic capacity and flow theory, traffic control devices, and transportation planning.

4 concepts

Exam areas

A. Geometric design

Horizontal Curves and Superelevation

Simple-curve geometry (R, T, L, M, E), degree of curve, the e+f superelevation balance, minimum radius, and horizontal sight distance for the FE Civil exam.

A horizontal curve is just a circular arc stitched between two straight tangents, but it is where a surprising number of FE Civil points are won or lost — because the same handful of right triangles generate every quantity you will ever be asked for, and because superelevation ties the geometry to a force balance on a vehicle. The NCEES FE Reference Handbook gathers all of this on two pages of the Civil Engineering chapter ("Horizontal Curves"), and the exam rewards you for knowing which formula answers which question without re-deriving anything. Master the labeled points —

PC

(point of curve),

PI

(point of intersection),

PT

(point of tangent) — and the intersection angle

I

(also written

\Delta

), and the rest is bookkeeping.

The geometry: one angle drives everything

The two tangents meet at the

PI

with deflection angle

I

. The curve of radius

R

is tangent to both lines, so the radius to

PC

and the radius to

PT

each make a right angle with their tangent, and the central angle of the arc equals

I

. That single fact gives you the tangent distance

T

(from

PC

PT

to the

PI

) as the leg of a right triangle with half-angle

I/2

. Commit the half-angle pattern to memory:

T

, the long chord

LC

, the middle ordinate

M

, and the external distance

E

all use

I/2

, while the arc length

L

uses the full angle

I

T = R\tan\!\frac{I}{2}, \qquad LC = 2R\sin\!\frac{I}{2}

Arc length, middle ordinate, and external distance

The arc length

L

is the curve measured from

PC

PT

and is simply the radius times the central angle in radians. The middle ordinate

M

is the distance from the midpoint of the long chord to the midpoint of the arc, and the external distance

E

is the distance from the

PI

to the midpoint of the arc; both fall straight out of the radius and the half-angle. These are the offsets a surveyor uses to stake the curve, and the exam loves to ask for one of them given

R

and

I

L = \frac{R\,I\,\pi}{180}, \quad M = R\left(1-\cos\tfrac{I}{2}\right), \quad E = R\left(\sec\tfrac{I}{2}-1\right)

Degree of curve: the American shorthand

US practice describes sharpness not by radius but by degree of curve

D

— the central angle that subtends a standard length of arc. The FE handbook uses the arc definition, where

D

subtends a

100\,\text{ft}

arc, giving the clean relation below. A large

D

means a sharp (small-radius) curve. Once you have

D

, the arc length has an even simpler form,

L = 100\,I/D

, because each

D

degrees of central angle buys exactly

100\,\text{ft}

of curve.

D = \frac{5729.58}{R} \quad\Longleftrightarrow\quad R = \frac{5729.58}{D}

Superelevation and side friction

On a curve, the centripetal demand

v^2/R

is supplied partly by banking the roadway (superelevation

e

) and partly by tire side friction

f

. Writing

V

in mph and

R

in feet, the handbook prints the point-mass balance as

0.01e + f = V^2/(15R)

with

e

in PERCENT (so a

8\%

bank enters as

e=8

); the constant

15

already folds in

g

and the mph-to-fps conversion. The mathematically equivalent reduced form

e+f = V^2/(15R)

instead takes

e

as a decimal (so

8\%

0.08

) — pick one convention and hold it through the whole problem. Rearranged for radius, it yields the minimum radius for a chosen design speed and a chosen split between banking and friction. Where the full bank cannot be introduced abruptly, a spiral (clothoid) transition curve of length

L_s

eases the driver from the straight tangent into the fully-superelevated circular arc.

0.01e + f = \frac{V^2}{15\,R}\;\;(e\text{ in }\%) \qquad\Longleftrightarrow\qquad e + f = \frac{V^2}{15\,R}\;\;(e\text{ decimal})

Horizontal sight distance around an obstruction

Inside a curve, a wall, barrier, or tree line can block the driver's line of sight even when the pavement is clear. The horizontal sightline offset

HSO

(sometimes

M_s

) is the clearance from the centerline of the inside lane to the obstruction needed to provide a sight distance

S

along the arc. Because

S

is measured along the arc, the angle in the formula is

28.65\,S/R

in degrees (this is

90S/(\pi R)

, i.e. half the central angle subtended by

S

). If your computed

HSO

exceeds the available clearance, you must flatten the curve or move the obstruction.

HSO = R\left[\,1-\cos\!\left(\frac{28.65\,S}{R}\right)\right]

Exam strategy

First identify what is given — almost every problem hands you two of

\{R, D, I\}

plus a station, and asks for a length or a station. Convert

D\leftrightarrow R

immediately with

5729.58

, then pick the half-angle formula (

T, LC, M, E

) or the full-angle formula (

L

). For stationing, remember the curve is continuous:

\text{Sta }PC = \text{Sta }PI - T

and

\text{Sta }PT = \text{Sta }PC + L

(NOT

PI + T

, because you travel along the arc, not the tangent). For superelevation, keep

e

as a decimal and

V

in mph. A units slip or a

PI+T

stationing error is the most common way to miss these otherwise-free points.

Key equations

Tangent distance

T = R\tan\dfrac{I}{2}

Distance from $PC$ (or $PT$ ) to the $PI$ . $R$ in ft, $I$ = intersection/deflection angle in degrees.

Curve (arc) length

L = \dfrac{R\,I\,\pi}{180} = \dfrac{100\,I}{D}

Length of arc from $PC$ to $PT$ (ft). Uses the FULL central angle $I$ , not $I/2$ .

Long chord

LC = 2R\sin\dfrac{I}{2}

Straight-line distance from $PC$ to $PT$ (ft).

Middle ordinate

M = R\left(1-\cos\dfrac{I}{2}\right)

Offset from midpoint of long chord to midpoint of arc (ft).

External distance

E = R\left(\sec\dfrac{I}{2}-1\right)

Distance from $PI$ to midpoint of arc (ft); $\sec=1/\cos$ .

Degree of curve (arc definition)

D = \dfrac{5729.58}{R}

$D$ (degrees) subtends a $100\,\text{ft}$ arc. $R$ in ft. Larger $D$ = sharper curve.

Superelevation balance

0.01e + f = \dfrac{V^2}{15\,R}

Handbook form with $e$ = superelevation in PERCENT, $f$ = side-friction factor, $V$ in mph, $R$ in ft. The equivalent reduced form $e+f=V^2/(15R)$ takes $e$ as a decimal. Constant $15$ embeds $g$ and unit conversion.

Minimum radius

R_{\min} = \dfrac{V^2}{15\,(e_{\max}+f_{\max})}

Sharpest allowable curve for a design speed $V$ (mph) given maximum $e$ and $f$ as DECIMALS (reduced form). With $e$ in percent use $15\,(0.01e_{\max}+f_{\max})$ in the denominator.

Spiral transition length

L_s = \dfrac{3.15\,V^3}{R\,C}

Length of spiral easement (ft); $V$ in mph, $R$ in ft, $C$ = rate of change of lateral acceleration (use $1\,\text{ft/s}^3$ unless stated).

Horizontal sightline offset

HSO = R\left[1-\cos\!\left(\dfrac{28.65\,S}{R}\right)\right]

Lateral clearance to a sight obstruction (ft); $S$ = sight distance along arc (ft), $R$ in ft. Argument is in degrees.

Curve stationing

\text{Sta }PT = \text{Sta }PC + L,\quad \text{Sta }PC = \text{Sta }PI - T

Stations advance along the alignment; $PT$ is reached via the arc $L$ , never by adding $T$ to the $PI$ .

Worked examples

Full curve layout and stationing

Problem. A simple horizontal curve has $R = 1{,}000\,\text{ft}$ and intersection angle $I = 24^\circ$ . The $PI$ is at station $50{+}00.00$ . Find $D$ , $T$ , $L$ , $LC$ , $M$ , $E$ , and the stations of the $PC$ and $PT$ .

Solution. Degree of curve:

D = 5729.58/1000 = 5.7296^\circ

. Tangent:

T = 1000\tan 12^\circ = 1000(0.21256) = 212.56\,\text{ft}

. Arc length:

L = 1000(24)(\pi/180) = 418.88\,\text{ft}

(check:

100(24)/5.7296 = 418.88\,\text{ft}

). Long chord:

LC = 2(1000)\sin 12^\circ = 415.82\,\text{ft}

. Middle ordinate:

M = 1000(1-\cos 12^\circ) = 21.85\,\text{ft}

. External:

E = 1000(\sec 12^\circ - 1) = 22.34\,\text{ft}

. Stationing:

\text{Sta }PC = 5000.00 - 212.56 = 4787.44 = 47{+}87.44

;

\text{Sta }PT = 4787.44 + 418.88 = 5206.32 = 52{+}06.32

. Sanity check:

LC < L

(chord shorter than arc) and

M < E

, both as expected for a shallow curve.

\text{Sta }PT = 47{+}87.44 + 418.88 = 52{+}06.32

Minimum radius from superelevation

Problem. A rural highway is designed for $V = 65\,\text{mph}$ with a maximum superelevation $e_{\max} = 8\%$ and a maximum side-friction factor $f_{\max} = 0.10$ . What is the minimum allowable curve radius?

Solution. Use the reduced form with

e

as a decimal:

e_{\max} = 0.08

R_{\min} = \dfrac{V^2}{15(e_{\max}+f_{\max})} = \dfrac{65^2}{15(0.08+0.10)} = \dfrac{4225}{15(0.18)} = \dfrac{4225}{2.70} = 1564.8\,\text{ft}

. Answer:

R_{\min} \approx 1{,}565\,\text{ft}

(round UP to be conservative — a larger radius is flatter and safer; rounding down to

1{,}560

would permit a sharper curve than the balance allows). Sanity check: at

V=65

V^2/(15R) = 4225/(15\cdot1565) = 0.18 = e+f

, so the demand is exactly met.

R_{\min} = \frac{65^2}{15(0.18)} = 1564.8 \approx 1{,}565\ \text{ft}

Sight obstruction clearance on a curve

Problem. A curve of radius $R = 1{,}000\,\text{ft}$ must provide a stopping sight distance of $S = 500\,\text{ft}$ . A noise wall runs parallel to the inside lane. What minimum horizontal sightline offset (lateral clearance) is required?

Solution.

HSO = R\left[1-\cos\!\left(\dfrac{28.65\,S}{R}\right)\right]

. Argument:

28.65(500)/1000 = 14.325^\circ

HSO = 1000\left[1-\cos 14.325^\circ\right] = 1000(1-0.96891) = 31.1\,\text{ft}

. Answer: about

31.1\,\text{ft}

of clearance from the lane centerline to the wall. Sanity check: the offset must be a small fraction of

R

for a flat curve —

31/1000 \approx 3\%

, reasonable.

HSO = 1000\,[\,1-\cos 14.325^\circ\,] = 31.1\ \text{ft}

Common pitfalls

•Mixing $I$ and $I/2$ : tangent, chord, middle ordinate, and external all use the HALF angle $I/2$ ; only the arc length $L$ uses the full $I$ .
•Stationing the $PT$ as $\text{Sta }PI + T$ . You travel along the arc, so $\text{Sta }PT = \text{Sta }PC + L$ . Adding $T$ to the $PI$ double-counts the tangent and overshoots.
•Mixing the two superelevation forms. The handbook PRINTS $0.01e+f=V^2/(15R)$ with $e$ in PERCENT; the reduced form $e+f=V^2/(15R)$ takes $e$ as a decimal ( $8\% \to 0.08$ ). Feeding a percent $e$ into the decimal form (or a decimal into the $0.01e$ form) throws the balance off by 100× — pick one convention and hold it.
•Using metric or fps speed in the superelevation/sight formulas. $V$ must be in mph and $R$ in ft for the $15$ and $28.65$ constants to be valid.
•Confusing the arc and chord definitions of degree of curve. The FE handbook uses the ARC definition, $D = 5729.58/R$ ; the chord definition gives a slightly different $R$ .
•Forgetting that the $HSO$ angle $28.65S/R$ is in DEGREES, not radians — taking the cosine in radian mode gives a nonsense clearance.

References

NCEES FE Reference Handbook — Civil Engineering (Horizontal Curves)
AASHTO, A Policy on Geometric Design of Highways and Streets (Green Book) — Source of the superelevation and sight-distance design policy.

Vertical Curves and Sight Distance

Symmetric parabolic crest and sag curves: the parabola constant, rate K=L/A, high/low point, elevations, and the sight-distance equations that set curve length.

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B. Pavement system design

Pavement Design and Transportation Planning

Flexible vs rigid pavement, the AASHTO structural number, ESALs and load equivalency, subgrade strength (CBR/Mr), and the trip-based transportation planning process.

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C. Traffic capacity and flow theory

Traffic Flow Theory and Control

The q=k·v identity, the Greenshields speed-density-flow model, capacity and jam density, level of service, signal timing, and traffic-control-device warrants.

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