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Horizontal Design

11% of exam

Circular-curve elements (T, L, M, E, LC, degree of curve) and PC/PT stationing, superelevation rate and side friction, superelevation runoff, minimum radius, sight-distance offset, and compound/reverse curves.

6 concepts

A. Basic circular curve elements

Circular Curve Geometry & Stationing

The radius, tangent, length, middle ordinate, external distance, chord, and degree of curve of a simple circular curve, and how they fix PC and PT stations.

A horizontal alignment is a string of tangents joined by circular curves, and almost every layout, right-of-way, and earthwork quantity downstream depends on getting one curve's geometry exactly right. The whole curve is locked by just two numbers — the radius

R

and the central (intersection) angle

\Delta

— and from those two everything else follows: where the curve begins (PC), where it ends (PT), how long it runs, and how far it bows away from the tangents. The NCEES PE Civil Reference Handbook §5.2.1 Basic Curve Elements gives this family of relations, and the exam rewards an engineer who can move between them and the stationing without fumbling a sign or a unit. Points are lost here not on the trigonometry but on the bookkeeping: subtracting the tangent at the wrong place, adding arc length to the PI instead of the PC, or mixing the arc and chord definitions of degree of curve.

The two inputs and the points that name the curve

The point of intersection (PI) is where the two tangents would meet if extended; the deflection between them is the central angle

\Delta

, which equals the angle subtended at the center of the curve. The curve leaves the back tangent at the point of curvature (PC, sometimes BC) and rejoins the forward tangent at the point of tangency (PT, sometimes EC). Because the curve is tangent to both lines, the distance back from the PI to the PC equals the distance forward from the PI to the PT — that common distance is the tangent length

T

. Hold onto that symmetry: it is what lets you find the PC station by walking backward from the PI, and the PT station by walking forward from the PC along the arc.

Tangent, length, chord, ordinate, external

Every basic element is a one-line function of

R

and

\Delta

. The tangent comes from the right triangle PI-PC-center:

T = R\tan(\Delta/2)

. The arc length is the radius times the central angle in radians,

L = R\,\Delta_{\text{rad}} = \frac{\pi}{180}R\Delta

. The long chord spans PC to PT,

C = 2R\sin(\Delta/2)

. The middle ordinate

M

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

E

is the gap from the PI to the midpoint of the arc — both measured along the bisector of

\Delta

. Notice

M

uses

1-\cos

while

E

uses

\sec-1

; they are close for flat curves but diverge as

\Delta

grows.

T = R\tan\frac{\Delta}{2}, \quad L = \frac{\pi}{180}R\Delta, \quad C = 2R\sin\frac{\Delta}{2}, \quad M = R\!\left(1-\cos\frac{\Delta}{2}\right), \quad E = R\!\left(\sec\frac{\Delta}{2}-1\right)

Degree of curve — arc vs. chord

Highway practice often labels a curve by its degree of curve

D

rather than its radius — the central angle subtended by a standard

100\,\text{ft}

length. The catch is what that

100\,\text{ft}

measures. In the arc definition (the highway standard) it is

100\,\text{ft}

of arc, giving

R = 5{,}729.58/D

. In the chord definition (older railroad practice) it is a

100\,\text{ft}

chord, giving

R = 50/\sin(D/2)

. For flat curves the two radii are nearly identical, but for sharp curves they part company, and using the wrong one is a classic exam trap. With the arc definition the curve length collapses to a clean proportion,

L = 100\,\Delta/D

, because each degree of central angle is exactly

100/D

feet of arc.

R_{\text{arc}} = \frac{5{,}729.58}{D}, \qquad R_{\text{chord}} = \frac{50}{\sin(D/2)}, \qquad L = \frac{100\,\Delta}{D}

Deflection-angle layout

Curves are staked from the PC by turning deflection angles with an instrument and chaining chords. The total deflection from the back tangent to the PT is exactly

\Delta/2

, and the deflection to any intermediate point is proportional to the arc swept to that point. A point at arc distance

\ell

from the PC has deflection angle

\delta = \tfrac{1}{2}\cdot\tfrac{180}{\pi}\cdot\tfrac{\ell}{R}

degrees, and the field rule of thumb is that the deflection per

100\,\text{ft}

of arc equals

D/2

. Chaining short chords approximates the arc; the chord to a point subtending central angle

2\delta

is

c = 2R\sin\delta

.

\delta = \frac{90}{\pi}\cdot\frac{\ell}{R} \;=\; \frac{\Delta}{2}\cdot\frac{\ell}{L} \quad(\text{deg}), \qquad c = 2R\sin\delta

Stationing through the curve

Stations run continuously along the centerline, written as hundreds of feet plus the remainder, so

39+45.54

means

3{,}945.54\,\text{ft}

. The PI station is set by the tangent geometry, but the curve does NOT pass through the PI — it cuts the corner. So the PC station is the PI station minus the tangent, and the PT station is the PC station plus the ARC length, never the PI plus anything. This is the single most common stationing error:

\text{Sta}_{\text{PT}} = \text{Sta}_{\text{PI}} - T + L

, not

\text{Sta}_{\text{PI}} + T

. The difference

2T - L

is the distance saved by following the arc instead of the two tangent legs.

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

Exam strategy

Start by writing down

R

and

\Delta

; if you are handed

D

, convert with the arc definition unless the problem explicitly says chord. Compute

T

and

L

first — they drive the stationing — then

M

,

E

, and

C

as needed. Always carry

\Delta/2

in the trig (a frequent slip is using

\Delta

where

\Delta/2

belongs). For stationing, march PI

\to

PC by subtracting

T

, then PC

\to

PT by ADDING

L

; if your PT comes out beyond the PI station by more than

T

, you added the arc to the PI by mistake. Keep

5{,}729.58

memorized as the arc-definition constant and remember

L=100\Delta/D

for a quick length when

D

is known.

Key equations

Tangent length

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

Distance from PI to PC (equals PI to PT). $R$ = radius (ft), $\Delta$ = central angle, $T$ in ft.

Curve (arc) length

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

Long chord

C = 2R\sin\dfrac{\Delta}{2}

Straight-line distance PC to PT (ft).

Middle ordinate

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

Offset from midpoint of long chord to midpoint of arc (ft); the sightline-clearance ordinate.

External distance

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

Distance from PI to midpoint of arc along the bisector (ft).

Degree of curve (arc)

R = \dfrac{5{,}729.58}{D}

Arc definition: $D$ = central angle subtended by 100 ft of arc (deg); $R$ in ft.

Degree of curve (chord)

R = \dfrac{50}{\sin(D/2)}

Chord definition: $D$ = central angle subtended by a 100 ft chord (deg). Railroad practice.

Deflection angle to a point

\delta = \dfrac{90}{\pi}\cdot\dfrac{\ell}{R}

Deflection from back tangent to a point at arc distance

PC and PT stations

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

Worked examples

Full curve from D and Δ, with stationing

Problem. A simple circular curve has a central angle $\Delta = 24^\circ$ and a degree of curve $D = 4^\circ$ (arc definition). The PI is at station $42+50.00$ . Find $R$ , $T$ , $L$ , $M$ , $E$ , the long chord $C$ , and the PC and PT stations.

Solution. Radius (arc def):

R = 5{,}729.58/4 = 1{,}432.40\,\text{ft}

. Tangent:

T = R\tan(12^\circ) = 1{,}432.40(0.21256) = 304.46\,\text{ft}

L = \frac{100(24^\circ)}{4^\circ} = 600\ \text{ft}, \qquad \text{Sta}_{\text{PT}} = (42+50) - 304.46 + 600 = 45+45.54

Arc vs. chord definition radius

Problem. A railroad curve is described as $D = 8^\circ$ . Compare the radius under the arc definition with the radius under the chord definition, and comment on the difference.

Solution. Arc:

R_a = 5{,}729.58/8 = 716.20\,\text{ft}

Common pitfalls

•Computing the PT as $\text{Sta}_{\text{PI}} + T$ . The curve cuts the corner — the PT station is $\text{Sta}_{\text{PC}} + L$ (PC = PI $- T$ ). Adding $L$ to the PI overstates the PT by $T$ .
•Using $\Delta$ in place of $\Delta/2$ inside the trig for $T$ , $C$ , $M$ , or $E$ . Every one of those uses the HALF-angle; only $L$
•Mixing the arc and chord definitions of degree of curve. $R=5{,}729.58/D$ is arc; $R=50/\sin(D/2)$ is chord. The exam states which — read for it.
•Confusing $M$ (midpoint of chord to arc, uses $1-\cos$ ) with $E$ (PI to arc, uses $\sec-1$ ). $E$ is always larger.
•Forgetting to convert $\Delta$ to radians in $L=\tfrac{\pi}{180}R\Delta$ , or applying $L=100\Delta/D$
•Reading a station like $45+45.54$ as 4,545 ft 54 in instead of $4{,}545.54\,\text{ft}$ ; the part after the '+' is feet and decimal feet.

References

NCEES PE Civil Reference Handbook — §5.2.1 Basic Curve Elements
AASHTO, A Policy on Geometric Design of Highways and Streets (Green Book, GDHS-7) — horizontal alignment background
Hickerson, Route Location and Design, 5th ed. — classic curve-layout reference

B. Sight distance considerations

Horizontal Sight Distance & the Middle Ordinate

The horizontal sightline offset (HSO) needed to clear a roadside obstruction on a curve, its middle-ordinate form, and how to back out the available sight distance.