Horizontal Curves

Transportation · FE Reference Handbook section

Transportation

33 formulas

10 exam-style examples

~60 min

All Transportation lectures

Learning objectives

What you must be able to do before leaving this section.

This chapter section covers Horizontal Curves within Transportation. Read it the way you would read a textbook chapter: the theory first so the relations mean something, then every equation with its use and its trap, then 10 fully worked examples with the arithmetic shown line by line, and finally a self-check you should be able to answer without notes.

Explain, in your own words, what horizontal curves describes physically and when it applies.
State every one of the 33 relations the handbook lists here and name each symbol with its unit.
Select the correct relation from the wording of an exam stem within 20 seconds.
Carry a complete solution from givens to a "most nearly" answer with the correct unit.
Recognise the distractors generated by the unit trap: grades as decimals in curve formulas, percent in the stem.

Lecture

Why this section exists. Horizontal Curves is the part of Transportation that lets you connect a vertical or horizontal alignment, or a traffic stream to a number you can defend. Before any equation is useful you must be able to picture the physical situation it describes; the schematic below is that picture.

How the theory is built. The handbook prints results, not derivations. Each relation in this section comes from one governing principle applied to the idealised system: state the principle, impose the stated assumptions, and the printed equation follows. Knowing which assumption each relation rests on is what lets you reject a wrong answer choice in seconds.

How it is examined. Items from this page are written as a curve geometry element or a capacity/flow relationship. Roughly two thirds are direct substitution, one third require one intermediate quantity from a neighbouring relation, and a small number are conceptual — testing whether you know the assumption, not the arithmetic.

The habit that earns the points. Unit discipline. grades as decimals in curve formulas, percent in the stem. Every relation below is dimensionally consistent only when that rule is honoured, and the distractor set is deliberately built from candidates who ignored it. Write the unit next to every number you substitute, every time.

How to study this page. Read the theory, then cover the formula cards and try to reproduce each relation from its description. Then work the examples with the solution hidden, revealing one line at a time. Finish with the self-check questions; if you cannot answer one, return to the matching formula card.

Dense peak-hour traffic queued on an urban arterial at dusk. — Photo 1. Where this shows up in practice: horizontal curves.
Wikimedia Commons, CC BY 2.0

Transportation — Horizontal Curves: reference schematic for orienting the symbols used in this section.

Theory, developed

Read this before the equations — it is what makes them memorable.

The physical situation

Every item from this section describes a vertical or horizontal alignment, or a traffic stream. Sketch it before you compute — a labelled sketch with the givens on it converts a wordy stem into a solvable problem and exposes the quantity the examiner left out on purpose.

The governing principle

The 33 relations on this page are consequences of one principle applied to that idealised system. Identify which quantity is conserved, balanced, or defined, and the correct equation follows without memorisation.

Assumptions and limits of validity

Each printed relation carries silent assumptions — linearity, steady state, uniformity, small deformation, or standard conditions, depending on the subject. Conceptual exam items are written by violating exactly one of these, so read the sentence above the equation as carefully as the equation itself.

Solution procedure you should automate

1) Read the last sentence of the stem to identify the requested quantity. 2) Locate the relation on this page whose left-hand side is that quantity. 3) Tabulate the givens with units and mark the missing symbol. 4) If a symbol is missing, find the one relation that produces it. 5) Rearrange symbolically, substitute once, evaluate, and round only at the end.

Notation used in this section

R	Quantity produced by "R = 5729.58" — read its definition and unit from the handbook line directly above the equation.
T	Quantity produced by "T = R tan _ I 2i = LC" — read its definition and unit from the handbook line directly above the equation.
L	Quantity produced by "L = RI r = I 100" — read its definition and unit from the handbook line directly above the equation.
M	Quantity produced by "M = R 81 - cos _ I 2iB" — read its definition and unit from the handbook line directly above the equation.
R - M	Quantity produced by "R - M = cos _ I 2i" — read its definition and unit from the handbook line directly above the equation.
c	Quantity produced by "c = 2R sin _ d 2i" — read its definition and unit from the handbook line directly above the equation.
l	Quantity produced by "l = Rd b r l" — read its definition and unit from the handbook line directly above the equation.
E	Quantity produced by "E = R= 1 - 1G" — read its definition and unit from the handbook line directly above the equation.
d	Quantity produced by "d = angle of sub-chord" — read its definition and unit from the handbook line directly above the equation.
D	Quantity produced by "D = degree of curve, arc definition" — read its definition and unit from the handbook line directly above the equation.
e	Quantity produced by "e = superelevation (%)" — read its definition and unit from the handbook line directly above the equation.
f	Quantity produced by "f = side friction factor" — read its definition and unit from the handbook line directly above the equation.

Handbook notes for this section

Definitions and conditions exactly as the handbook states them.

100.00
PC c LC PT
I/2 I/2
R d
NOT TO SCALE
2 sin _ I 2i
2 cos _ I 2i
180 D
R cos _ I 2i
cos _ I 2i
where
Horizontal Curves
Side friction factor (based on superelevation) 15 R
3.15V 3
Spiral Transition Length RC
[use 1 ft/sec3 unless otherwise stated]
Sight Distance (to see around obstruction)

Core formulas for this FE topic

Definitions, applicability, units, assumptions and worked examples for each relation.

Worked exam-style examples

The four ways this section is written on the real exam — thoughts first, then equations, then substitution.

Example 1

Minimum radius for a superelevated curve

A highway is designed for 100 km/h with e = 0.06 and f = 0.12. Find the minimum horizontal curve radius.

Given

V = 100 km/h
e = 0.06
f = 0.12

Find

R_min

Start with the thinking

The SI formula uses V in km/h with the 127 constant.
e and f are added, not multiplied.

Figure for Minimum radius for a superelevated curve

Step-by-step solution

Formula
$R = V^{2} / [127 (e + f)]$
Denominator
$127 (0.06 + 0.12) = 127 (0.18) = 22.86$
Numerator
$V^{2} = 10 0^{2} = 10, 000$
Substitute
$R = 10, 000/22.86$
Result
$R_{m} in = 437 m$

Answer: R_min ≈ 437 m

Why the other options are there

1,312 m (e + f taken as 0.06)
78.7 m (V in m/s with the 127 constant)

Reference: FE Reference Handbook — Transportation — Horizontal curves

Example 2

Stopping sight distance on a downgrade

A vehicle travels 90 km/h on a −3% grade. With t = 2.5 s and a = 3.4 m/s², compute the stopping sight distance.

Given

V = 90 km/h = 25 m/s
t = 2.5 s
a = 3.4 m/s²
G = −0.03

Find

SSD

Start with the thinking

Reaction distance plus braking distance.
A downgrade lengthens the braking distance.

Step-by-step solution

Reaction distance
$d_{1} = V t = 25 (2.5) = 62.5 m$
Braking denominator
$2 (a / g + G) g = 2 (3.4 - 0.03 (9.81)) = 2 (3.4 - 0.294) = 6.21$
Braking distance
$d_{2} = V^{2} /6.21 = 625/6.21 = 100.6 m$
Total
$S S D = 62.5 + 100.6$
Result
$S S D = 163 m$

Answer: SSD ≈ 163 m

Why the other options are there

154 m (grade ignored)
100 m (reaction distance omitted)

Reference: FE Reference Handbook — Transportation — Stopping sight distance

Example 3

Crest vertical curve length

A +3.0% grade meets a −2.0% grade. For SSD = 190 m with h₁ = 1.08 m and h₂ = 0.60 m, find the minimum curve length (assume L > S).

Given

g₁ = +3.0%, g₂ = −2.0%
S = 190 m
h₁ = 1.08 m, h₂ = 0.60 m

Find

L_min

Start with the thinking

A = |g₁ − g₂| in percent.
The standard SI constant for crest curves with those eye/object heights is 658.

Figure for Crest vertical curve length

Step-by-step solution

Algebraic difference
$A = ∣3.0 - (- 2.0) ∣ = 5.0$
Formula
$L = A S^{2} /658 w h e n L > S$
Numerator
$5.0 (19 0^{2}) = 5.0 (36, 100) = 180, 500$
Substitute
$L = 180, 500/658$
Result
$L = 274 m; L > S = 190 m, so t h e a ss u m pt i o nh o l d s ✓$

Answer: L_min ≈ 274 m

Why the other options are there

55 m (S not squared)
L = 190 m (SSD reported as the curve length)

Reference: FE Reference Handbook — Transportation — Vertical curves

Example 4

Simple horizontal curve elements — Horizontal Curves

A simple circular curve has R = 1575 ft and Δ = 54°. Find the tangent length, curve length, long chord and middle ordinate.

Given

R = 1575 ft
Δ = 54°

Find

T, L, LC and M

Start with the thinking

Every element except L uses half the deflection angle.
Curve length uses radians: L = RΔ(π/180).

Figure for Simple horizontal curve elements — Horizontal Curves

Step-by-step solution

Tangent
$T = R tan (Δ/2) = 1575 tan (27. 0^{\circ}) = 802.5 f t$
Curve length — L = πRΔ/180 = π(1575)(54)/180 = 1,484 ft
Long chord
$L C = 2 R sin (Δ/2) = 2 (1575) sin (27. 0^{\circ}) = 1, 430 f t$
Middle ordinate
$M = R [1 - cos (Δ/2)] = 1575 [1 - 0.8910] = 171.7 f t$
Degree of curve
$D = 5729.58/ R = 3.63 8^{\circ}$

Answer: T ≈ 802.5 ft, L ≈ 1,484 ft, LC ≈ 1,430 ft, M ≈ 171.7 ft

Why the other options are there

T = 2,168 ft (full Δ used in the tangent)
L = 85,050 ft (degrees not converted to radians)

Reference: FE Reference Handbook — Transportation → Horizontal Curves

Example 5

Simple horizontal curve elements — Horizontal Curves (2)

A simple circular curve has R = 862 ft and Δ = 37°. Find the tangent length, curve length, long chord and middle ordinate.

Given

R = 862 ft
Δ = 37°

Find

T, L, LC and M

Start with the thinking

Every element except L uses half the deflection angle.
Curve length uses radians: L = RΔ(π/180).

Figure for Simple horizontal curve elements — Horizontal Curves (2)

Step-by-step solution

Tangent
$T = R tan (Δ/2) = 862 tan (18. 5^{\circ}) = 288.4 f t$
Curve length — L = πRΔ/180 = π(862)(37)/180 = 556.7 ft
Long chord
$L C = 2 R sin (Δ/2) = 2 (862) sin (18. 5^{\circ}) = 547.0 f t$
Middle ordinate
$M = R [1 - cos (Δ/2)] = 862 [1 - 0.9483] = 44.55 f t$
Degree of curve
$D = 5729.58/ R = 6.64 7^{\circ}$

Answer: T ≈ 288.4 ft, L ≈ 556.7 ft, LC ≈ 547.0 ft, M ≈ 44.55 ft

Why the other options are there

T = 649.6 ft (full Δ used in the tangent)
L = 31,894 ft (degrees not converted to radians)

Reference: FE Reference Handbook — Transportation → Horizontal Curves

Example 6

Simple horizontal curve elements — Horizontal Curves (3)

A simple circular curve has R = 1408 ft and Δ = 45°. Find the tangent length, curve length, long chord and middle ordinate.

Given

R = 1408 ft
Δ = 45°

Find

T, L, LC and M

Start with the thinking

Every element except L uses half the deflection angle.
Curve length uses radians: L = RΔ(π/180).

Figure for Simple horizontal curve elements — Horizontal Curves (3)

Step-by-step solution

Tangent
$T = R tan (Δ/2) = 1408 tan (22. 5^{\circ}) = 583.2 f t$
Curve length — L = πRΔ/180 = π(1408)(45)/180 = 1,106 ft
Long chord
$L C = 2 R sin (Δ/2) = 2 (1408) sin (22. 5^{\circ}) = 1, 078 f t$
Middle ordinate
$M = R [1 - cos (Δ/2)] = 1408 [1 - 0.9239] = 107.2 f t$
Degree of curve
$D = 5729.58/ R = 4.06 9^{\circ}$

Answer: T ≈ 583.2 ft, L ≈ 1,106 ft, LC ≈ 1,078 ft, M ≈ 107.2 ft

Why the other options are there

T = 1,408 ft (full Δ used in the tangent)
L = 63,360 ft (degrees not converted to radians)

Reference: FE Reference Handbook — Transportation → Horizontal Curves

Example 7

Simple horizontal curve elements — Horizontal Curves (4)

A simple circular curve has R = 1308 ft and Δ = 52°. Find the tangent length, curve length, long chord and middle ordinate.

Given

R = 1308 ft
Δ = 52°

Find

T, L, LC and M

Start with the thinking

Every element except L uses half the deflection angle.
Curve length uses radians: L = RΔ(π/180).

Figure for Simple horizontal curve elements — Horizontal Curves (4)

Step-by-step solution

Tangent
$T = R tan (Δ/2) = 1308 tan (26. 0^{\circ}) = 638.0 f t$
Curve length — L = πRΔ/180 = π(1308)(52)/180 = 1,187 ft
Long chord
$L C = 2 R sin (Δ/2) = 2 (1308) sin (26. 0^{\circ}) = 1, 147 f t$
Middle ordinate
$M = R [1 - cos (Δ/2)] = 1308 [1 - 0.8988] = 132.4 f t$
Degree of curve
$D = 5729.58/ R = 4.38 0^{\circ}$

Answer: T ≈ 638.0 ft, L ≈ 1,187 ft, LC ≈ 1,147 ft, M ≈ 132.4 ft

Why the other options are there

T = 1,674 ft (full Δ used in the tangent)
L = 68,016 ft (degrees not converted to radians)

Reference: FE Reference Handbook — Transportation → Horizontal Curves

Example 8

Simple horizontal curve elements — Horizontal Curves (5)

A simple circular curve has R = 954 ft and Δ = 51°. Find the tangent length, curve length, long chord and middle ordinate.

Given

R = 954 ft
Δ = 51°

Find

T, L, LC and M

Start with the thinking

Every element except L uses half the deflection angle.
Curve length uses radians: L = RΔ(π/180).

Figure for Simple horizontal curve elements — Horizontal Curves (5)

Step-by-step solution

Tangent
$T = R tan (Δ/2) = 954 tan (25. 5^{\circ}) = 455.0 f t$
Curve length — L = πRΔ/180 = π(954)(51)/180 = 849.2 ft
Long chord
$L C = 2 R sin (Δ/2) = 2 (954) sin (25. 5^{\circ}) = 821.4 f t$
Middle ordinate
$M = R [1 - cos (Δ/2)] = 954 [1 - 0.9026] = 92.93 f t$
Degree of curve
$D = 5729.58/ R = 6.00 6^{\circ}$

Answer: T ≈ 455.0 ft, L ≈ 849.2 ft, LC ≈ 821.4 ft, M ≈ 92.93 ft

Why the other options are there

T = 1,178 ft (full Δ used in the tangent)
L = 48,654 ft (degrees not converted to radians)

Reference: FE Reference Handbook — Transportation → Horizontal Curves

Example 9

Simple horizontal curve elements — Horizontal Curves (6)

A simple circular curve has R = 778 ft and Δ = 45°. Find the tangent length, curve length, long chord and middle ordinate.

Given

R = 778 ft
Δ = 45°

Find

T, L, LC and M

Start with the thinking

Every element except L uses half the deflection angle.
Curve length uses radians: L = RΔ(π/180).

Figure for Simple horizontal curve elements — Horizontal Curves (6)

Step-by-step solution

Tangent
$T = R tan (Δ/2) = 778 tan (22. 5^{\circ}) = 322.3 f t$
Curve length — L = πRΔ/180 = π(778)(45)/180 = 611.0 ft
Long chord
$L C = 2 R sin (Δ/2) = 2 (778) sin (22. 5^{\circ}) = 595.5 f t$
Middle ordinate
$M = R [1 - cos (Δ/2)] = 778 [1 - 0.9239] = 59.22 f t$
Degree of curve
$D = 5729.58/ R = 7.36 4^{\circ}$

Answer: T ≈ 322.3 ft, L ≈ 611.0 ft, LC ≈ 595.5 ft, M ≈ 59.22 ft

Why the other options are there

T = 778.0 ft (full Δ used in the tangent)
L = 35,010 ft (degrees not converted to radians)

Reference: FE Reference Handbook — Transportation → Horizontal Curves

Example 10

Simple horizontal curve elements — Horizontal Curves (7)

A simple circular curve has R = 656 ft and Δ = 36°. Find the tangent length, curve length, long chord and middle ordinate.

Given

R = 656 ft
Δ = 36°

Find

T, L, LC and M

Start with the thinking

Every element except L uses half the deflection angle.
Curve length uses radians: L = RΔ(π/180).

Figure for Simple horizontal curve elements — Horizontal Curves (7)

Step-by-step solution

Tangent
$T = R tan (Δ/2) = 656 tan (18. 0^{\circ}) = 213.1 f t$
Curve length — L = πRΔ/180 = π(656)(36)/180 = 412.2 ft
Long chord
$L C = 2 R sin (Δ/2) = 2 (656) sin (18. 0^{\circ}) = 405.4 f t$
Middle ordinate
$M = R [1 - cos (Δ/2)] = 656 [1 - 0.9511] = 32.11 f t$
Degree of curve
$D = 5729.58/ R = 8.73 4^{\circ}$

Answer: T ≈ 213.1 ft, L ≈ 412.2 ft, LC ≈ 405.4 ft, M ≈ 32.11 ft

Why the other options are there

T = 476.6 ft (full Δ used in the tangent)
L = 23,616 ft (degrees not converted to radians)

Reference: FE Reference Handbook — Transportation → Horizontal Curves

Self-check

Answer these without notes before moving on.

Without looking, state the relation on this page whose left-hand side is the quantity most often requested, and name every symbol in it.
Which assumption, if violated, makes the main relation of this section invalid?
Given a vertical or horizontal alignment, or a traffic stream, what is the first quantity you would compute, and why that one first?
Which unit conversion in this subject most often produces a wrong answer choice, and what is its numerical factor?
Rework Example 1 above from the givens alone, without reading the solution lines.

Chapter summary

Horizontal Curves contains 33 relations; you must be able to find this page in under 15 seconds.
Exam style: a curve geometry element or a capacity/flow relationship.
Unit rule: grades as decimals in curve formulas, percent in the stem.
Work the 10 examples until the solution path, not the answer, is automatic.

Common traps in this section

grades as decimals in curve formulas, percent in the stem
Answering the intermediate quantity instead of the quantity requested.
Rounding intermediate values before the final step.
Using a relation from an adjacent handbook section that shares a symbol.
Skipping the sketch — most lost points on this page start with a misread geometry.