Vertical Design

11% of exam

Equal-tangent parabolic curves: elevations and offsets, high/low points, PVI/PVC/PVT stationing, and crest/sag curve lengths from stopping and passing sight distance and the rate of vertical curvature K.

5 concepts

Exam areas

A. Vertical alignment

Equal-Tangent Parabolic Curves

How the symmetrical parabola joins two grades — PVC/PVI/PVT stationing, the tangent-offset elevation model, and locating the high or low point.

A vertical curve is the smooth transition that carries a road from one grade onto another — from a

+3\%

upgrade onto a

-2\%

downgrade over a crest, or out of a sag. The PE Civil exam tests the geometry of that transition relentlessly, because every elevation on the road profile, every sight-distance check, and every drainage low point depends on it. Get the parabola model and the stationing arithmetic right and a whole family of problems falls open. The governing relations are compact enough to live on a single page of the NCEES PE Civil Reference Handbook (§5.3.1, Symmetrical Vertical Curve Formula), so the points are won not by memorizing more, but by applying that one model cleanly — tangent elevation plus offset, every time.

Why a parabola, not a circular arc

Highway vertical curves are second-degree parabolas, not circular arcs, for one practical reason: a parabola produces a constant rate of change of grade. If the grade changes uniformly with distance, the vertical acceleration a driver feels is constant and the curve is comfortable and easy to stake. The defining property is that the second derivative of elevation with respect to horizontal distance is a constant

r

, the rate of change of grade. Integrate that twice and you get the elevation profile — a quadratic in the distance

x

measured from the curve's start.

\frac{d^2 Y}{dx^2} = r = \frac{g_2 - g_1}{L} = \text{constant}

The three control points and their stations

Every equal-tangent curve is anchored by three points. The PVI (point of vertical intersection) is where the two tangent grades, if extended, would meet — the vertex. The PVC (point of vertical curvature) begins the curve on the back tangent, and the PVT (point of vertical tangency) ends it on the forward tangent. Because the curve is symmetrical, the PVC and PVT sit one half-length to each side of the PVI. Station arithmetic is where careless points are lost: remember a station is

100\ \text{ft}

, so subtract

L/2

in feet from the PVI station and convert back to station form.

\text{Sta}_{PVC} = \text{Sta}_{PVI} - \tfrac{L}{2}, \qquad \text{Sta}_{PVT} = \text{Sta}_{PVI} + \tfrac{L}{2}

Elevations: tangent grade plus tangent offset

The cleanest way to compute any elevation on the curve is the handbook's tangent-offset model: first ride the back-tangent grade

g_1

from the PVC, then add the parabolic offset

y

that pulls the curve off that tangent. With

x

measured horizontally from the PVC, the curve elevation is

Y = Y_{PVC} + g_1 x + y

, where the offset grows with

x^2

. With the grade change

g_2 - g_1

in percent (and

g_1

as a decimal in the

g_1x

term) and

L

and

x

in feet, the offset takes the form below; the handbook writes it

x^2(g_2-g_1)/(2L)

with decimal grades. The sign of

g_2 - g_1

carries everything: negative is a crest (curve below the back tangent), positive is a sag. The handbook's

A = |g_2 - g_1|

is only the magnitude.

y = \frac{g_2 - g_1}{200\,L}\,x^2, \qquad Y = Y_{PVC} + g_1 x + \frac{g_2 - g_1}{200\,L}\,x^2

The middle ordinate E and the offset symmetry

At the PVI itself (

x = L/2

) the tangent offset reaches its maximum magnitude, the external or middle ordinate

E

. Substituting

x = L/2

into the offset gives

E = |A|L/800

(with

A

in percent and

L

in feet);

E

is reported as a positive distance, and whether the curve lies below the tangent (crest) or above it (sag) is carried by the sign of

g_2 - g_1

, not by

E

. A handy property follows: because offsets scale with the square of the distance from the nearest curve end, the offset at any fraction of the curve is that fraction squared times

E

— measured from whichever tangent is nearer. This lets you sketch and check a profile fast.

E = \frac{|A|\,L}{800}, \qquad y_x = E\left(\frac{x}{L/2}\right)^2

Locating the high or low point

The crest's high point (or sag's low point) is where the curve grade momentarily reaches zero — the turning point that controls drainage inlets, overpass clearance, and sight distance. Differentiate the elevation equation and set it to zero: the grade on the curve at distance

x

g_1 + (g_2 - g_1)(x/L)

(both grades in the same form), which vanishes at

x_m

below. Note

x_m

is measured from the PVC; if it comes out negative or larger than

L

, the grade never turns inside the curve (both grades have the same sign) and there is no interior high or low point.

x_m = -\,\frac{g_1}{2a} = \frac{g_1\,L}{g_1 - g_2}

Exam strategy

Build a tiny table every time: list PVC station and elevation first (from the PVI by backing off

g_1

over

L/2

), then march

x

from the PVC. Compute

Y = Y_{PVC} + g_1 x + [(g_2-g_1)/200L]x^2

— never mix this with a forward-tangent reference mid-problem. Keep grades in percent when you use the

200L

and

800

constants, and keep

x

L

in feet. For the high/low point, find

x_m

first, confirm

0 \le x_m \le L

, then feed it back into the same elevation equation. A fast sanity check: the offset magnitude at the PVI must equal

E = |A|L/800

, and the high-point offset magnitude exceeds

E

only when the turning point lies past the PVI (otherwise it is less).

Key equations

Rate of change of grade

r = \frac{g_2 - g_1}{L}

Constant for a parabola. $g_1,g_2$ = grades (decimal or %), $L$ = curve length (ft or stations); the handbook's $A=|g_2-g_1|$ is the magnitude of the grade change.

Curve elevation (tangent-offset form)

Y = Y_{PVC} + g_1\,x + \frac{g_2 - g_1}{200\,L}\,x^2

Elevation at horizontal distance $x$ (ft) from the PVC. $g_2 - g_1$ in percent, $g_1$ in decimal, $L$ , $x$ in ft. Use $g_1$ as decimal here (e.g., $+0.03$ ).

Tangent offset

y = \frac{g_2 - g_1}{200\,L}\,x^2

Vertical distance from the back tangent to the curve. $g_2 - g_1$ in percent (its sign sets the direction: negative = crest, offset below the tangent; positive = sag, offset above), $x,L$ in ft. The reported offset is the magnitude $|y|$ ; the sign only flags crest vs sag.

PVC and PVT stations

\text{Sta}_{PVC}=\text{Sta}_{PVI}-\tfrac{L}{2},\quad \text{Sta}_{PVT}=\text{Sta}_{PVI}+\tfrac{L}{2}

Symmetrical curve: half the length each side of the PVI. Convert $L/2$ (ft) to station form ( $1\ \text{sta}=100\ \text{ft}$ ).

PVC elevation from PVI

Y_{PVC}=Y_{PVI}-g_1\,\tfrac{L}{2}

Back off the back-tangent grade over the half-length. Likewise $Y_{PVT}=Y_{PVI}+g_2\,L/2$ .

Middle ordinate (external) at PVI

E=\frac{|A|\,L}{800}

Maximum tangent offset, occurring at the PVI, reported as a positive magnitude. $A$ in percent, $L$ in ft, $E$ in ft. The sign of $g_2 - g_1$ (crest vs sag) sets which side of the tangent the offset falls on, not the size of $E$ .

High / low point station

x_m=-\frac{g_1}{2a}=\frac{g_1\,L}{g_1-g_2}

Distance from PVC to the turning point. Valid only if $0\le x_m\le L$ ; otherwise no interior high/low point. $g_1$ and $g_2$ in the same units.

Rate of vertical curvature

K=\frac{L}{|A|}

Horizontal length per 1% change of grade (ft/%). The single parameter that links curve length to sight-distance design.

Grade on the curve

g(x)=g_1+\frac{g_2-g_1}{L}\,x

Instantaneous grade (%) at distance $x$ from PVC; equals zero at the high/low point.

Worked examples

Elevation and high point on a crest curve

Problem. A crest vertical curve joins a $g_1=+3.00\%$ back grade to a $g_2=-2.00\%$ forward grade. The curve is $L=600\ \text{ft}$ long and the PVI is at station $50{+}00$ with elevation $1240.00\ \text{ft}$ . Find the PVC and PVT stations, the curve elevation at station $48{+}50$ , and the station and elevation of the high point.

Solution. Stations:

\text{Sta}_{PVC}=50{+}00-3{+}00=47{+}00

and

\text{Sta}_{PVT}=50{+}00+3{+}00=53{+}00

. PVC elevation:

Y_{PVC}=1240.00-0.03(300)=1240.00-9.00=1231.00\ \text{ft}

. Grade change:

g_2-g_1=-2.00-3.00=-5.00\%

(the handbook's

A=|g_2-g_1|=5.00\%

; negative

g_2-g_1

marks a crest). At station

48{+}50

x=4850-4700=150\ \text{ft}

Y = Y_{PVC} + g_1x + \frac{g_2-g_1}{200L}x^2 = 1231.00 + 0.03(150) + \frac{-5.00}{200(600)}(150)^2 = 1231.00 + 4.50 - 0.94 = 1234.56\ \text{ft}.

High point:

x_m=\dfrac{g_1 L}{g_1-g_2}=\dfrac{(3.00)(600)}{3.00-(-2.00)}=360\ \text{ft}

(inside

0

–

600

✓), at station

47{+}00+360=50{+}60

Y_{hp}=1231.00+0.03(360)+\frac{-5.00}{200(600)}(360)^2=1231.00+10.80-5.40=1236.40\ \text{ft}.

Final: PVC

47{+}00

, PVT

53{+}00

; elevation at

48{+}50

\mathbf{1234.56\ \text{ft}}

; high point at

\mathbf{50{+}60}

, elevation

\mathbf{1236.40\ \text{ft}}

. Sanity check (offsets compared as magnitudes):

E=AL/800=5.00(600)/800=3.75\ \text{ft}

is the curve-to-tangent offset at the PVI (

x=L/2=300\ \text{ft}

). The back-tangent offset magnitude

|y|=|(A/200L)x^2|

keeps growing past the PVI, so at the high point (

x=360\ \text{ft}>300\ \text{ft}

) the offset is

|{-}5.40|=5.40\ \text{ft}

, larger than

E=3.75\ \text{ft}

— exactly what you expect for a point between the PVI and the PVT. Grade check:

g(360)=3.00+(-5.00/600)(360)=3.00-3.00=0

✓, confirming the turning point.

Sizing a sag curve to hold a low-point elevation

Problem. A sag vertical curve joins $g_1=-2.50\%$ to $g_2=+1.50\%$ with the PVI at station $70{+}00$ , elevation $842.00\ \text{ft}$ . To keep minimum cover over a buried storm drain, the curve's low point must sit at exactly $845.00\ \text{ft}$ . What curve length $L$ achieves that?

Solution. Let

L

be unknown.

g_2-g_1=1.50-(-2.50)=+4.00\%

(sag; the handbook's

A=|g_2-g_1|=4.00\%

). PVC elevation:

Y_{PVC}=Y_{PVI}-g_1\tfrac{L}{2}=842.00-(-0.025)\tfrac{L}{2}=842.00+0.0125L

. Low point from PVC:

x_m=\dfrac{g_1 L}{g_1-g_2}=\dfrac{(-0.025)L}{-0.025-0.015}=0.625\,L

(inside

[0,L]

✓). Low-point elevation, in terms of

L

Y_{lp}=Y_{PVC}+g_1 x_m+\frac{g_2-g_1}{200L}x_m^2=(842.00+0.0125L)+(-0.025)(0.625L)+\frac{4.00}{200L}(0.625L)^2.

The three

L

-terms combine cleanly:

0.0125L-0.015625L+0.0078125L=0.0046875L

, so

Y_{lp}=842.00+0.0046875\,L

. Set

Y_{lp}=845.00

0.0046875\,L=3.00\ \Rightarrow\ L=\dfrac{3.00}{0.0046875}=640\ \text{ft}

. Final:

\mathbf{L = 640\ \text{ft}}

. Verify:

Y_{PVC}=842.00+0.0125(640)=850.00\ \text{ft}

;

x_m=0.625(640)=400\ \text{ft}

;

Y_{lp}=850.00-0.025(400)+\frac{4.00}{200(640)}(400)^2=850.00-10.00+5.00=845.00\ \text{ft}

✓. Sanity check:

Y_{lp}

rises monotonically with

L

(

0.0046875\,\text{ft/ft}

), so a longer sag dips less and lifts the low point — to raise the low point you lengthen the curve, the direction the algebra confirms.

Common pitfalls

•Stationing arithmetic: $L/2$ is in feet, but stations read in hundreds. Subtracting $300\ \text{ft}$ from station $50{+}00$ gives $47{+}00$ ; writing $300\ \text{ft}$ as $0{+}30$ instead of $3{+}00$ gives the wrong $49{+}70$ . Convert deliberately.
•Sign of the grade change: a crest has $g_2-g_1<0$ (offset pulls the curve down). The handbook's $A=|g_2-g_1|$ has no sign — use $(g_2-g_1)$ in the offset and in $x_m$ , or a crest flips into a sag and the high/low point moves.
•Mixing grade units with the constants: the $200L$ and $800$ forms expect $A$ in percent. If you carry $A$ as a decimal, the offset is off by 100.
•Referencing the wrong tangent mid-problem: compute every elevation from the PVC and the back tangent $g_1$ . Switching to the forward tangent halfway corrupts the offset.
•Assuming a high/low point always exists. If $g_1$ and $g_2$ have the same sign, $x_m$ falls outside $[0,L]$ and the road keeps climbing or falling through the curve.
•Confusing $E=AL/800$ (offset at the PVI) with the high-point offset. They coincide only for a symmetric curve with $|g_1|=|g_2|$ .

References

NCEES PE Civil Reference Handbook — §5.3.1 Symmetrical Vertical Curve Formula
AASHTO, A Policy on Geometric Design of Highways and Streets (Green Book, GDHS-7) — Vertical Alignment — Profile design context and grade-control practice.

Unequal-Tangent (Asymmetric) Vertical Curves

When the two tangent legs are different lengths — two parabolas meeting at the CVC, the offset there, and elevations and the turning point on each branch.

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B. Stopping and passing sight distance

Crest Vertical Curves & Sight Distance

Sizing a crest curve so the pavement itself does not hide the road ahead — the S<L and S>L length equations, the K-rate, and AASHTO eye and object heights.

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Sag Vertical Curves

Why a sag curve is sized by headlight reach at night — plus the comfort, drainage, and overhead-clearance criteria and the K-value that governs.

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Sight Distance & Design K-Values

The braking-plus-reaction model for stopping and passing sight distance, the grade correction, and how design K = L/A ties sight distance to curve length.

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