Vertical Curves

Transportation · FE Reference Handbook section

Transportation

37 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 Vertical 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 vertical curves describes physically and when it applies.
State every one of the 37 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. Vertical 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: vertical curves.
Wikimedia Commons, CC BY 2.0

Transportation — Vertical 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 37 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

y	Quantity produced by "y = ax 2 x A" — read its definition and unit from the handbook line directly above the equation.
A	Quantity produced by "A = g2 − g1 PVT g" — read its definition and unit from the handbook line directly above the equation.
a	Quantity produced by "a = 22L 1 FOR" — read its definition and unit from the handbook line directly above the equation.
E	Quantity produced by "E = ac 2 m CK NT" — read its definition and unit from the handbook line directly above the equation.
r	Quantity produced by "r = 2L 1" — read its definition and unit from the handbook line directly above the equation.
K	Quantity produced by "K = A" — read its definition and unit from the handbook line directly above the equation.
xm	Quantity produced by "xm =− 2a1 = g −" — read its definition and unit from the handbook line directly above the equation.
Tangent elevation	Quantity produced by "Tangent elevation = YPVC + g1x = YPVI + g2 (x – L/2)" — read its definition and unit from the handbook line directly above the equation.
Curve elevation	Quantity produced by "Curve elevation = YPVC + g1x + ax2 = YPVC + g1x + [(g2 – g1)/(2L)]x2" — read its definition and unit from the handbook line directly above the equation.
PVC	Quantity produced by "PVC= point of vertical curvature, or beginning of curve" — read its definition and unit from the handbook line directly above the equation.
PVI	Quantity produced by "PVI = point of vertical intersection, or vertex" — read its definition and unit from the handbook line directly above the equation.
PVT	Quantity produced by "PVT = point of vertical tangency, or end of curve" — 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.

y E
g −g PVC
TAN WARD
g1 GEN
BA NGE
TA YPVC
g −g
L DATUM
g g1L VERTICAL CURVE FORMULAS
1 g2
NOT TO SCALE
Compiled from AASHTO, A Policy on Geometric Design of Highways and Streets, 6th ed., 2011.
where
Vertical Curves: Sight Distance Related to Curve Length
Crest Vertical Curve 2
( h1 + h2 )
Standard Criteria:
AS2 2,158
2,158 A
Sag Vertical Curve
(based on standard headlight
criteria)
400 + 3.5 S
A )

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

Crest vertical curve geometry — Vertical Curves

A crest curve joins a +2.0% grade to a -2.5% grade over L = 681 ft. Find A, K, the mid-curve offset E and the station of the high point from the BVC.

Given

g₁ = +2.0%
g₂ = -2.5%
L = 681 ft

Find

A, K, E and the high-point location

Start with the thinking

A is the algebraic difference of grades — signs matter.
The high point occurs where the grade is zero, at x = g₁L/A.

Figure for Crest vertical curve geometry — Vertical Curves

Step-by-step solution

Grade change
$A = ∣ g_{1} - g_{2} ∣ = ∣2.0 - (- 2.5) ∣ = 4.50$
Rate of curvature
$K = L / A = 681/4.50 = 151.3 f tp er$
External offset
$E = A L /800 = 4.50 (681) /800 = 3.83 f t$
High point
$x = g_{1} L / A = 2.0 (681) /4.50 = 302.7 f t f r o m B V C$
Check — x < L (302.7 < 681) so the high point lies on the curve ✓

Answer: A = 4.50%, K ≈ 151.3, E ≈ 3.83 ft, high point 302.7 ft past the BVC

Why the other options are there

A = -0.50% (signs mishandled)
E = 383.1 ft (wrong constant)

Reference: FE Reference Handbook — Transportation → Vertical Curves

Example 5

Crest vertical curve geometry — Vertical Curves (2)

A crest curve joins a +4.0% grade to a -2.0% grade over L = 727 ft. Find A, K, the mid-curve offset E and the station of the high point from the BVC.

Given

g₁ = +4.0%
g₂ = -2.0%
L = 727 ft

Find

A, K, E and the high-point location

Start with the thinking

A is the algebraic difference of grades — signs matter.
The high point occurs where the grade is zero, at x = g₁L/A.

Figure for Crest vertical curve geometry — Vertical Curves (2)

Step-by-step solution

Grade change
$A = ∣ g_{1} - g_{2} ∣ = ∣4.0 - (- 2.0) ∣ = 6.00$
Rate of curvature
$K = L / A = 727/6.00 = 121.2 f tp er$
External offset
$E = A L /800 = 6.00 (727) /800 = 5.45 f t$
High point
$x = g_{1} L / A = 4.0 (727) /6.00 = 484.7 f t f r o m B V C$
Check — x < L (484.7 < 727) so the high point lies on the curve ✓

Answer: A = 6.00%, K ≈ 121.2, E ≈ 5.45 ft, high point 484.7 ft past the BVC

Why the other options are there

A = 2.00% (signs mishandled)
E = 545.3 ft (wrong constant)

Reference: FE Reference Handbook — Transportation → Vertical Curves

Example 6

Crest vertical curve geometry — Vertical Curves (3)

A crest curve joins a +4.0% grade to a -1.5% grade over L = 556 ft. Find A, K, the mid-curve offset E and the station of the high point from the BVC.

Given

g₁ = +4.0%
g₂ = -1.5%
L = 556 ft

Find

A, K, E and the high-point location

Start with the thinking

A is the algebraic difference of grades — signs matter.
The high point occurs where the grade is zero, at x = g₁L/A.

Figure for Crest vertical curve geometry — Vertical Curves (3)

Step-by-step solution

Grade change
$A = ∣ g_{1} - g_{2} ∣ = ∣4.0 - (- 1.5) ∣ = 5.50$
Rate of curvature
$K = L / A = 556/5.50 = 101.1 f tp er$
External offset
$E = A L /800 = 5.50 (556) /800 = 3.82 f t$
High point
$x = g_{1} L / A = 4.0 (556) /5.50 = 404.4 f t f r o m B V C$
Check — x < L (404.4 < 556) so the high point lies on the curve ✓

Answer: A = 5.50%, K ≈ 101.1, E ≈ 3.82 ft, high point 404.4 ft past the BVC

Why the other options are there

A = 2.50% (signs mishandled)
E = 382.3 ft (wrong constant)

Reference: FE Reference Handbook — Transportation → Vertical Curves

Example 7

Crest vertical curve geometry — Vertical Curves (4)

A crest curve joins a +2.0% grade to a -3.0% grade over L = 731 ft. Find A, K, the mid-curve offset E and the station of the high point from the BVC.

Given

g₁ = +2.0%
g₂ = -3.0%
L = 731 ft

Find

A, K, E and the high-point location

Start with the thinking

A is the algebraic difference of grades — signs matter.
The high point occurs where the grade is zero, at x = g₁L/A.

Figure for Crest vertical curve geometry — Vertical Curves (4)

Step-by-step solution

Grade change
$A = ∣ g_{1} - g_{2} ∣ = ∣2.0 - (- 3.0) ∣ = 5.00$
Rate of curvature
$K = L / A = 731/5.00 = 146.2 f tp er$
External offset
$E = A L /800 = 5.00 (731) /800 = 4.57 f t$
High point
$x = g_{1} L / A = 2.0 (731) /5.00 = 292.4 f t f r o m B V C$
Check — x < L (292.4 < 731) so the high point lies on the curve ✓

Answer: A = 5.00%, K ≈ 146.2, E ≈ 4.57 ft, high point 292.4 ft past the BVC

Why the other options are there

A = -1.00% (signs mishandled)
E = 456.9 ft (wrong constant)

Reference: FE Reference Handbook — Transportation → Vertical Curves

Example 8

Crest vertical curve geometry — Vertical Curves (5)

A crest curve joins a +2.5% grade to a -3.0% grade over L = 892 ft. Find A, K, the mid-curve offset E and the station of the high point from the BVC.

Given

g₁ = +2.5%
g₂ = -3.0%
L = 892 ft

Find

A, K, E and the high-point location

Start with the thinking

A is the algebraic difference of grades — signs matter.
The high point occurs where the grade is zero, at x = g₁L/A.

Figure for Crest vertical curve geometry — Vertical Curves (5)

Step-by-step solution

Grade change
$A = ∣ g_{1} - g_{2} ∣ = ∣2.5 - (- 3.0) ∣ = 5.50$
Rate of curvature
$K = L / A = 892/5.50 = 162.2 f tp er$
External offset
$E = A L /800 = 5.50 (892) /800 = 6.13 f t$
High point
$x = g_{1} L / A = 2.5 (892) /5.50 = 405.5 f t f r o m B V C$
Check — x < L (405.5 < 892) so the high point lies on the curve ✓

Answer: A = 5.50%, K ≈ 162.2, E ≈ 6.13 ft, high point 405.5 ft past the BVC

Why the other options are there

A = -0.50% (signs mishandled)
E = 613.3 ft (wrong constant)

Reference: FE Reference Handbook — Transportation → Vertical Curves

Example 9

Crest vertical curve geometry — Vertical Curves (6)

A crest curve joins a +3.0% grade to a -2.5% grade over L = 437 ft. Find A, K, the mid-curve offset E and the station of the high point from the BVC.

Given

g₁ = +3.0%
g₂ = -2.5%
L = 437 ft

Find

A, K, E and the high-point location

Start with the thinking

A is the algebraic difference of grades — signs matter.
The high point occurs where the grade is zero, at x = g₁L/A.

Figure for Crest vertical curve geometry — Vertical Curves (6)

Step-by-step solution

Grade change
$A = ∣ g_{1} - g_{2} ∣ = ∣3.0 - (- 2.5) ∣ = 5.50$
Rate of curvature
$K = L / A = 437/5.50 = 79.45 f tp er$
External offset
$E = A L /800 = 5.50 (437) /800 = 3.00 f t$
High point
$x = g_{1} L / A = 3.0 (437) /5.50 = 238.4 f t f r o m B V C$
Check — x < L (238.4 < 437) so the high point lies on the curve ✓

Answer: A = 5.50%, K ≈ 79.5, E ≈ 3.00 ft, high point 238.4 ft past the BVC

Why the other options are there

A = 0.50% (signs mishandled)
E = 300.4 ft (wrong constant)

Reference: FE Reference Handbook — Transportation → Vertical Curves

Example 10

Crest vertical curve geometry — Vertical Curves (7)

A crest curve joins a +2.5% grade to a -3.0% grade over L = 530 ft. Find A, K, the mid-curve offset E and the station of the high point from the BVC.

Given

g₁ = +2.5%
g₂ = -3.0%
L = 530 ft

Find

A, K, E and the high-point location

Start with the thinking

A is the algebraic difference of grades — signs matter.
The high point occurs where the grade is zero, at x = g₁L/A.

Figure for Crest vertical curve geometry — Vertical Curves (7)

Step-by-step solution

Grade change
$A = ∣ g_{1} - g_{2} ∣ = ∣2.5 - (- 3.0) ∣ = 5.50$
Rate of curvature
$K = L / A = 530/5.50 = 96.36 f tp er$
External offset
$E = A L /800 = 5.50 (530) /800 = 3.64 f t$
High point
$x = g_{1} L / A = 2.5 (530) /5.50 = 240.9 f t f r o m B V C$
Check — x < L (240.9 < 530) so the high point lies on the curve ✓

Answer: A = 5.50%, K ≈ 96.4, E ≈ 3.64 ft, high point 240.9 ft past the BVC

Why the other options are there

A = -0.50% (signs mishandled)
E = 364.4 ft (wrong constant)

Reference: FE Reference Handbook — Transportation → Vertical 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

Vertical Curves contains 37 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.

Vertical Curves

Learning objectives

Lecture

Theory, developed

Notation used in this section

Handbook notes for this section

Core formulas for this FE topic

y=ax2xA

A=g2​−g1​PVTg

a=22L1FOR

E=ac2mCKNT

r=2L1

K=A

xm=−2a1​=g−

Tangentelevation=YPVC+g1​x=YPVI+g2​(x−L/2)

Curveelevation=YPVC+g1​x+ax2​=YPVC+g1​x+[(g2​−g1​)/(2L)]x2​

PVC=pointofverticalcurvature,orbeginningofcurve

PVI=pointofverticalintersection,orvertex

PVT=pointofverticaltangency,orendofcurve

A=algebraicdifferenceingrades

a=parabolaconstant

E=tangentoffsetatPVI

g1​=gradeofbacktangent

g2​=gradeofforwardtangent

h1​=heightofdriver′seyesabovetheroadwaysurface(ft)

h2​=heightofobjectabovetheroadwaysurface(ft)

K=rateofverticalcurvature

L=lengthofcurve

r=rateofchangeofgrade

S=sightdistance(ft)

x=horizontaldistancefromPVCtopointoncurve

xm=horizontaldistancetomin/maxelevationoncurve

y=tangentoffset

V=designspeed(mph)

S≤LS>L

L=200

Generalequation:100(2h1​+2h2​)2L=2S−

h1=3.50ftandh2​=2.0ft:L=L=2S−

L =Needs technical review

L=2S−(400+3.5S

L =Needs technical review

L =Needs technical review

L=2S−

C=verticalclearanceforoverheadstructure(overpass)locatedwithin200

Worked exam-style examples

Self-check

Chapter summary

Common traps in this section